Chemistry E.M Vol 1

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INTERMEDIATE
CHEMISTRY
1
(CORE MODULES)
A.P. OPEN SCHOOL SOCIETY, GUNTUR, AMARAVATI.

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v
HOW TO USE THE STUDY MATERIAL
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vi
Overview of the Learning Material
1
Module I 7. The Liquid State
Atoms, Molecules and Chemical 8. The Solid State
Arithmatics 9. Solutions
1. Mole Concept 10. Colloids
2. Chemical Arithmatics
Module II Module IV
Atomic Structure and Chemical Chemical Energetics
Bonding 11. Chemical Thermodynamics
3. Atomic structure 12. Spontaneity of Chemical
4. Periodic table and atomic properties Reactions
5. Chemical Bonding
Module III Module V

States of Matter Chemical Dynamics
6. The Gaseous State 13. Chemical Equilibrium
14. Ionic Equilibrium
15. Electro Chemistry
16. Chemical Kinetics
17. Adsorption and Catalysis
2
Module VI Module - VII
Chemistry of Elements Chemistry of Organic Compounds
18. Occurance and extraction of metals 25. Nomenclature and General Principles
19. Hydrogen and s-block Elements 26. Hydrocarbons
20. General Characteristics of the p-block 27. Compounds of Carbon Containing
Elements Halogens (Haloalkanes and Haloarenes)
21. p-block Elements and their 28. Alcohols, Phenols and Ethers
Compounds - I 29. Aldehydes, Ketones and Carboxylic
22. p-block elements and their Acids
Compounds - II 30. Compounds of Carbon Containing
23. d-Block and f-Block Elements Nitrogen
24. Coordination Compounds 31. Biomlecules
3
Module VIIIA Module - VIIIB
Environmental Chemistry Chemistry and Industry
32. Environmental Concerns 32. Petrochemicals
33. Air Pollution 33. Polymers
34. Water Pollution 34. Dyes, Paints and Pigments
35. Heavy Metal Contamination 35. Drugs and Medicines
36. Radioactive Pollution 36. Building Materials
vii
CONTENTS
Module Name of the Lesson Page No.
I. Atoms Molecules and 1. Mole Concept 1

Chemical Arithmatics 2. Chemical Arithmatics 17
3. Atomic structure 35
4. Periodic table and atomic properties 68
Student's Assignment-1 88
5. Chemical Bonding 90
II. State of matter 6. The Gaseous State 123

7. The Liquid State 146
8. Solid State 161
9. Solution 186
10. Colloids 214
III. Chemical energetics 11. Chemical Thermodynamics 228

12. Spontaneity of chemical Reactions 251
IV. Chemical Dynamics 13. Chemical Equilibrium 266

14. Ionic Equilibrium 295
15. Electrochemistry 328
16. Chemical Kinetics 364
17. Adsorption and Catalysis 382
viii
MODULE - I
ATOMS, MOLECULES AND
CHEMICAL ARITHMATICS
1. Mole Concept
2. Chemical Arithmatics
1
MOLE CONCEPT
As you are aware, atoms and molecules are so small that we cannot see them with our
naked eyes or even with the help of a microscope. Any sample of matter which can be
studied consists of extremely large number of atoms or molecules. In chemical reactions,
atoms or molecules combine with one another in a definite number ratio. Therefore, it would
be pertinent if we could specify the total number of atoms or molecules in a given sample of
a substance. We use may number units in our daily life. For example, we express the number
of bananas or eggs in terms of 'dozen'. In chemistry we use a number unit called mole
which is very large.
Objectives
After studying this lesson you will be able to :
• state the need of SI units;
• list base SI units;
• explain the relationship between mass and number of particles;
• define Avogadro's constant and state its significance;
• calculate the molar mass of different element and compounds and
• define molar volume of gases at STP.
1.1. SI Units (Revisited)

Measurement is needed in every walk of life. As you know that for every measurement
a 'unit' or a 'reference standard' is required. In different countries, different systems of units
gradually developed. This created difficulties whenever people of one country had to deal
with those of another country. Since scientists had to often use each other's data, they faced
a lot of difficulties. For a practical use, data had to be first converted into local units and
then only it could be used.
-1-
In 1960, the 'General Conference of Weights and Measures." the international authority on
units proposed a new system which was based upon the metric system. This system is called
the "international System of Units' which is abbreviated as SI units from its French name,
Le Systeme Internationale d'Unites. You have learned about SI units in your earlier classes
also and know that they are based upon seven base units corresponding to seven base physical
quantities. Units needed for various other physical quantities can be derived from these
base SI units. The seven base SI units are listed in Table 1.1
Table 1.1 : SI Base Units
Physical Quantity Name of SI Unit Symbol of SI unit
Length Metre m
Mass Kilogram kg
Time Second s
Electrical current Ampere A
Temperature Kelvin K
Amount of substance Mole mol
Luminous intensity Candela cd
For measuring very large or very small quantities, multiples or sub-multiples of these units are
used. Each one of them is denoted by a symbol which is prefixed to the symbol of the unit. For
example, to measure long distances we use the unit Kilometer which is a multiple of metre, the
base unit of length. Here kilo is the prefix used for the multiple 103. Its symbol is k which is
prefixed to the symbol of metre, m. Thus symbol of kilometer is km and
1 km = 1.0 x 103 m = 1000 m
Similarly, for measuring small lengths we use centimetre (cm) and millimetre (mm) where
1 km = 1.0 x 10-2 m = 0.01 m
1 km = 1.0 x 10-3 m = 0.001 m
Some prefixes used with SI units are listed in Table 1.2.
Table 1.2 : Some Prefixes used with SI Units
Prefix Symbol Meaning Example
Tera T 10 12 1 terametre (Tm) = 1.0 x 1012 m

Giga G 10 9 1 gigametre (Gm) = 1.0 x 109 m
Mega M 10 6 1 megametre (Mm) = 1.0 x 106 m
-2-
Kilo k 10 3 1 kilometre (km) = 1.0 x 103 m
Hecta h 10 2 1 hectametre(hm) = 1.0 x 102 m
Deca da 10 1 1 decametre (dam) = 1.0 x 101 m
_1
Deci d 10 1 decimetre (dm) = 1.0 x 10-1 m
_2
Centi c 10 1 centimetre (cm) = 1.0 x 10-2 m
_3
Milli m 10 1 millimetre (mm) = 1.0 x 10-3 m
_6
Micro μ 10 1 micrometre (mm) = 1.0 x 10-6 m
_9
Nano n 10 1 nanometre (nm) = 1.0 x 10-9 m
_ 12
Pico p 10 1 picometre (pm) = 1.0 x 10-12 m
Before proceeding further try to answer the following questions:
Intext Questions 1.1

1. Name the SI Unit of mass
..................................................................................................................................
_
2. What symbol will represent 1.0 x 10 6 g?
..................................................................................................................................
2 _9
3. Name the prefixes used for (i) 10 and (ii) 10
..................................................................................................................................
4. What do the following symbols represent ?
i) Ms ii) ms
i) ................................................................................................................................
ii) ................................................................................................................................
1.2 Relationship between Mass and Number of Particles

Suppose you want to purchase 500 screws. How, do you think, the shopkeeper would give you
the desired quantity? By counting the screws individually? No, he would give the screws by weight
because it will take a lot of time to count them. If each screw weighs 0.8g, he would weight 400g
screws because it is the mass of 500 screws (0.8 x 500 = 400g). You will be surprised to note that
the Reserve Bank of India gives the desired number of coins by weight and not by counting. This
process of counting by weighing becomes more and more labour saving as the number of items
to be counted becomes large. We can carry out the reverse process also. Suppose we take 5000
very tiny springs (used in watches) and weigh them. If the mass of the springs is found to be 1.5g,
-3-
_
we can conclude that mass of each spring is 1.5 5000 = 3 x 10 4 g.
Thus, we see that mass and number of identical objects or particles are inter-related. Since atoms
and molecules are extremely tiny particles it is impossible to weigh or count them individually.
Therefore we need a relationship between the mass and number of atoms and molecules (particles).
Such a relationship is provided by 'mole concept'.
1.3 Mole - A Number Unit

Mass of an atom or a molecule is an important property. However, while discussing the quantitative
aspects of a chemical reaction, the number of reacting atoms or molecules in more significant than
their masses. Let us understand this with the help of the following activity.
Activity 1.1
Aim : To study whether during a reaction, the reactants react with each other in a simple ratio by
mass.
What is required?
China dish, sulphur powder, iron powder, a magnet and a magnifying glass.
What to do?
Mix 1g each of iron and sulphur powders in china dish and heat them till the reaction is complete
and the mixture becomes a hard mass. Now break it into small pieces. Repeat the procedure with
a mixture of 2g of iron and 1g of sulphur powder.
What to observe?
• Pieces obtained from the ‘reaction mixture containing iron and sulphur in 1:1 ratio by
mass (1g each) when observed through a magnifying glass show some yellowish
particles of sulphur. When a magnet is brought near them, they are not attracted showing
that there is no unreacted iron.
• Pieces obtained from the reaction mixture containing iron and sulphur in 2:1 ratio by
mass (2g iron and 1g sulphur) do not show yellow particles of unreacted sulphur but
are attracted by the magnet. This shows the presence of some unreacted iron.
Conclusion
You can conclude that iron and sulphur do not react with each other in a simple mass ratio.
When taken in 1:1 ration by mass (Fe:S), some sulphur is left unreacted and when taken in
2:1 ration by mass (Fe:S) some iron is left unreacted.
Let us now write the chemical equation of this reaction
Fe + S → FeS
-4-
From the above chemical equation, it is clear that 1 atom of iron reacts with 1 atom of sulphur to
form 1 molecule of iron (II) sulphide (FeS). It means that if we had taken equal number of atoms
of iron and sulphur, both of them would have reacted completely. Thus we may conclude that
substances react in a simple ratio by number of atoms or molecules.
From the above discussion it is clear that the number of atoms or molecules of a substance is
more relevant than their masses. In order to express their number we need a number unit. One
commonly used number unit is ‘dozen’, which, as you know, means a collection of 12. Other
number units that we use are ‘score’ (20)) and ‘gross’ (144 or 12 dozens). These units are useful
in dealing with small numbers only. The atoms and molecules are so small that even in the minute
sample of any substance, their number is extremely large. For example, a tiny dust particle contains
about 1016 molecules. In chemistry such large numbers are commonly represented by a unit known
as mole. Its symbol is ‘mol’ and it is defined as.
A mole is the amount of a substance that contains as many elementary entities (atoms,
molecules or other particles) as there are atoms in exactly 0.012 kg or 12g of the
carbon-12 isotope.
The term mole has been derived from the Latin word ‘moles’ which means a ‘heap’ or a
‘pile’. It was first used by the famous chemist Wilhelm Ostwald more than a hundred
years ago.
Here you should remember that one mole always contains the same number of entities, no
matter what the substance is . Thus mole is a number unit for dealing with elementary
entities such as atoms, molecules, formula units, electrons etc., just as dozen is a number
unit for dealing with bananas or oranges. In the next section you will learn more about this
number.
1.4 Avogadro’s Constant

In the previous section we have learned that a mole of a substance is that amount which
contains as many elementary entities as there are atoms in exactly 0.012 kilogram or 12
gram of the carbon-12 isotope. This definition gives us a method by which we can find out
the amount of a substance (in moles) if we know the number of elementary entities present
in it or vice versa. Now the question arises how many atoms are there in exactly 12g of
carbon-12. This number is determined experimentally and its currently accepted value is
6.022045 x 1023. Thus 1 mol = 6.022045 x 1023 entities or particles, or atoms or molecules.
For all practical purposes this number is rounded off to 6.022 x 1023.
The basic idea of such a number was first conceived by an Italian scientist Amedeo
Avogadro. But, he never determined this number. It was determined later and is known
as Avogadro’s constant in his honour.
This number was earlier known as Avogadro’s number. This number alongwith the unit, that
is, 6.022 x 1023 mol-1 is known as Avogadro constant. It is represented by the symbol NA.
Here you should be clear that mathematically a number does not have a unit. Avogadro’s
-5-
number 6.022 x 1023 will not have any unit but Avogradro’s constant will have unit of mol-1. Thus
Avogradro’s constant, NA = 6.022 x 1023 mol-1.
Significance of Avogadro’s Constant

You know that 0.012kg or 12g of carbon-12 contains its one mole of carbon atoms. A mole may
be defined as the amount of a substance that contains 6.022 x 1023 elementary entities like atoms,
molecules or other particles. When we say one mole of carbon-12, we mean 6.022 x 1023 atoms
of carbon-12 whose mass is exactly 12g. This mass is called the molar mass of carbon-12. The
molar mass is defined as the mass (in grams) of 1 mole of a substance. Similarly, a mole of
any substance would contain 6.022 x 1023 particles or elementary entities. The nature of elementary
entity, however, depends upon the nature of the substance as given below:
S.No. Type of Substance Elementary Entity
1. Elements like Na, K, Cu which Atom

exist in atomic form
2. Elements like O, N, H, which Molecule
exist in molecular form (O2, N2, H2)
3. Molecular compounds like NH3, H2O, CH 4 Molecule
_ _
4. Ions like Na+, Cu2+, Ag+, Cl , O2 Ion
5. Ionic compounds like NaCl, NaNO3, K 2SO4 Formula unit
Formula unit of a compound contains as many atoms or ions of different types as is given
by its chemical formula. The concept is applicable to all types of compounds. The following
examples would clarify the concept.
Formula Atoms / ions present in one formula unit
H2O Two atoms of H and one atom of O

NH 3 One atom of N and three atoms of H
_
NaCl One Na+ ion and one Cl ion
_
NaNO 3 One Na+ ion and one NO3 ion
2_
K 2SO 4 Two K+ ions and one SO4 ion
3_
Ba3(PO 4)2 Three Ba2+ ions and two PO4 ions
Now, let us take the examples of different types of substances and correlate their amounts and the
number of elementary entities in them.
-6-
1 mole C = 6.022 x 1023 C atoms
1 mole O2 = 6.022 x 1023 O2 molecules
1 mole H2O = 6.022 x 1023 H2O molecules
1 mole NaCl = 6.022 x 1023 formula units of NaCl
1 mole Ba2+ ions = 6.022 x 1023 Ba2+ ions
We may choose to take amounts other than one mole and correlate them with number of particles
present with the help of relation :
Number of elementary entities = number of moles x Avogadro’s constant
1 mole O2 = 1 x (6.022 x 1023) = 6.022 x 1023 molecules of O2
0.5 mole O2 = 0.5 x (6.022 x 1023) = 3.011 x 1023 molecules of O2
0.1 mole O2 = 0.1 x (6.022 x 1023) = 6.022 x 1022 molecules of O2

1. A sample of nitrogen gas consists of 4.22 x 1023 molecules of nitrogen. How many moles of
nitrogen gas are there ?
...................................................................................................................................
2. In a metallic piece of magnesium, 8.46 x 1024 atoms are present. Calculate the amount
of magnesium in moles.
...................................................................................................................................
3. Calculate the number of Cl2 molecules and Cl atoms in 0.25 mol of Cl2 gas.
...................................................................................................................................
1.5 Mole, Mass and Number Relationships

You know that 1 mol = 6.022 x 1023 elementary entities
and Molar mass = Mass of 1 mole of substance
= mass of 6.022 x 1023 elementary entities.
As discussed earlier the elementary entity can be an atom, a molecule, an ion or a formula
unit. As far as mole - number relationship is concerned it is clear that one mole of any
substance would contain 6.022 x 1023 particles (elementary entities). For obtaining the molar
mass, i.e., mole-mass relationship we have to use atomic mass scale.
-7-
1.5.1 Atomic Mass Unit
By international agreement, a unit of mass to specify the atomic and molecular masses has been
defined. This unit is called atomic mass unit and its symbol is ‘amu’. The mass of one C-12 atom,
is taken as exactly 12 amu. Thus, C-12 atom serves as the standard. The Atomic mass unit is
defined as a mass exactly equal to the 1/12th of the mass of one carbon-12 atom.
Atomic mass unit also called unified atomic mass unit whose symbol is ‘u’. Another name of
atomic mass unit is dalton (symbol Da). The latter is mainly used in biological sciences.
1.5.2 Relative Atomic and Molecular Masses

You are aware that atomic mass scale is a relative scale with C-12 atom (also written as 12C)
chosen as the standard. Its mass is taken as exactly 12. Relative masses of atoms and molecules
are the number of times each atom or molecules is heavier than 1/12th of the mass of one C-12
atom. Often, we deal with elements and compounds containing isotopes of different elements.
Therefore, we prefer to use average masses of atoms and molecules. Thus
Relative atomic mass =
and Rleative molecular mass =
Experiemnts show that one O-16 atom is 1.333 times as heavy as one C-12 atom. Thus
Relative atomic mass of O-16 = 1.333 x 12 = 15.996 ∼ 16.0
The relative atomic masses of all elements have been determined in a similar manner. Relative
molecular masses can aslo be determinsed experimentallty in a similar manner. In case we
know the molecular formula of a molecule. we can calculate its relative molecular mass by
adding the relative atomic masses of all its constituent atoms. Let us calculate the relative
molecular mass of water, H2O.
Relative molecular mass of water, H2O = (2 x relative atomic mass of H) + (relative atomic
mass of O)
= (2 x 1) + (16) = 2 + 16 = 18.
The relative atomic and molecular masses are just numbers and dimensionless, unit-less
quantities.
-8-
1.5.3 Atomic, Molecular and Formula Masses
From the definition of atomic mass unit, we can calculate the atomic masses. Let us again take the
example of oxygen-16 whose relative atomic atomic mass is 16. By definition.
Relative atomic mass of O-16 = 16 =
Since 1 amu = 1/12th the mass of one C-12 atom
Mass of one O-16 atom = 16 amu

Or Atomic mass of O-16 = 16 amu
From this example we can see that numerical value of the relative atomic mass and atomic mass is
the same. Only, the former has no unit while the latter has the unit amu.
Molecular and formula masses can be obtained by adding the atomic or ionic masses of all the
constituent atoms or ions of the molecule of formula unit respectively. Let us understand these
calculations with the help of following examples.
Example 1.1 : Calculate the molecular mass of ammonia, NH3.
Solution : One molecule of NH3 consists of one N atom and three H atoms.
Molecular mass of NH3 = (Atomic mass of N) + 3 (Atomic mass of H)
= [14+ (3x1)] amu
= 17 amu
Example 1.2 : Calculate the formula mass of sodium chloride (NaCl).
_
Solution : One formula unit of sodium chloride consists of one Na+ ion and one Cl ion.
_
Formula mass of NaCl = (Ionic mass of Na+) + Ionic mass of (Cl )
= 25 amu + 35.5 amu
= 58.5 amu
You would have noticed in the above example that ionic mass of Na+ ion has been taken as 23
amu which is the same as the atomic mass of Na atom. Since loss or gain of few electrons does not
change the mass significantly,
_
therefore atomic masses are used as ionic masses. Similarly
_
we
havetaken ionic mass of Cl as 35.5 amu which is the same as the atomic mass of Cl .
-9-
1.5.4 Molar Masses
We know that molar mass is the mass of 1 mol of the substance. Also, 1 mol of any substance is
the collection of its 6.022 x 1023 elementary entities. Thus
Molar mass = Mass of 6.022 x 1023 elementary entities.
i) Molar mass of an element
You know that the relative atomic mass of carbon-12 is 12. A 12g sample of it would contain
6.022 x 1023 atoms. Hence the molar mass of C-12 is 12g mol-1. For getting the molar masses
of other elements we can see relative atomic masses.
Since the relative atomic mass of oxygen -16 is 16, a 16g sample of it would contain 6.022 x 1023
oxygen atoms and would constitute its one mole. Thus, the molar mass of O-16 is 16g mol-1.
Relative atomic masses of some common elements have been listed the Table 1.3.
Table 1.3 : Relative atomic masses of some elements (upto 1st place of decimal)
Element Relative Element Relative
Atomic Mass Atomic Mass
Hydrogen, H 1.0 Phosphorus, P 31.0

Carbon, C 12.0 Sulphur, S 32.1
Nitrogen, N 14.0 Chlorine, Cl 35.5
Oxygen, O 16.0 Potassium, K 39.1
Sodium, Na 23.0 Iron, Fe 55.9
ii) Molar mass of a molecular substance

The elementary entity in case of a molecular substance is the molecule. Hence, molar mass
of such a substance would be the mass of its 6.022 x 1023 molecules, whcih can be obtained
from its relative molecular mass or by multiplying the molar mass of each element by the
number of its moles present in one mole of the substance and then adding them.
Let us take the example of water, H2O. Its relative molecular mass is 18. Therefore, 18g of it
would contain 6.022 x 1023 molecules. Hence, its molar mass is 18 g mol-1. Alternately we
can calculate it as :
Molar mass of water, H 2O = (2 x molar mass of H) + (molar mass of O)
= (2x 1 g mol-1) + (16g mol-1)
= 18g mol-1
Table 1.4 lists molecular masses and molar masses of some substances.
-10-
Table 1.4 : Molecular masses and molar masses of some substances
Element or Compound Molecular mass / amu Molar mass / (g mol -1)
O2 32.0 32.0
P4 124.0 124.0
S8 256.8 256.8
H2 O 18.0 18.0
NH 3 17.0 17.0
HCl 36.5 36.5
CH2Cl2 85.0 85.0
iii) Molar masses of ionic compounds

Molar mass of an ionic compound is the mass of its 6.022 x 1023 formula units. It can be
obtained by adding the molar masses of ions present in the formula unit of the substance. In
case of NaCl it is calculated as.
Molar mass of NaCl = molar mass of Na+ + molar mass of Cl_
= (23g mol-1) + (35.5g mol-1)
= 58.5 g mol-1
Let us take some more examples of ionic compounds and calculate their molar masses.
Example 1.3 : Caclulate the molar mass of

i) K2SO4 ii) Ba3(PO4)2
Solution :
_
i) Molar mass of K2SO4 = (2 x molar mass of K+) + (molar mass of SO24 )
= (2 x molar mass of K+) +
(molar mass of of S + 4 x molar mass of O)
= [(2 x 39.1) + (32.1 + 4 x 16)] g mol-1]
= (78.2 + 32.1 + 64) g mol-1 = 174.3 g mol-1
_
ii) Molar mass of Ba3(PO4)2 = (3 x molar mass of Ba2+) + molar mass of PO34 )
= (3 x molar mass of Ba2+) +
2 (molar mass of P + 4 x molar mass of O)
= [(3 x 137.3) + 2 (31.0 + 4 x 16.0)] g mol-1]
= (411.9 + 190.0) g mol-1 = 601.9 g mol-1
-11-
Now you have learned about the mole, mass and number relationships for all types of substances.
The following examples would illustrate the usefulness of these relationships.
Example 1.4 : Find out the mass of carbon-12 that would contain 1.0 x 1019 carbon-12 atoms.
Solution : Mass of 6.022 x 1023 carbon-12 atoms = 12g
Mass of 1.0 x 1019 carbon-12 = g
= 1.99 x 10-4 g
Example 1.5 : How many molecules are present in 100g sample of NH3 ?
Solution : Molar mass of NH3 = (14 + 3) gm mol-1 = 17 g mol-1
∴ 17g Sample of NH3 contains 6.022 x 1023 molecules
Therefore, 100g sample of NH3 would contain x 100g
= 35.42 x 1023 molecules

Example 1.6 : Molar mass of O is 16g mol-1. What is the mass of one atom and one molecule of
oxygen?
Solution : Mass of 1 mol or 6.022 x 1023 atoms of O = 16g.
Mass of 1 atom of O =
= 2.66 x 10-23g.
Since a molecule of oxygen contains two atoms (O2).
its mass = 2 x 2.66 x 10-23g = 5.32 x 10-23g.

1. Calculate the molar mass of hydrogen chloride, HCl.
...................................................................................................................................
2. Calculate the molar mass of argon atoms, given that the mass of single atom is 6.634 x 10-26 kg.
...................................................................................................................................
-12-
3. Calculate the mass of 1.0 mol of potassium nitrate, KNO3 (atomic masses : K = 39 amu;
N = 14 amu, O = 16 amu).
...................................................................................................................................
4. The formula of sodium phosphate is Na3PO4. What is the mass of 0.146 mol of Na3PO4?
(atomic masses : Na = 23.0 amu, P=31.0 amu; O = 16.0 amu)
...................................................................................................................................
1.6 Mass, Molar Mass and Number of Moles

Mass, molar mass number of moles of a substance are inter-related quantities. We know
what :
Molar mass (M) = Mass of one mole of the substance.
Molar mass of water is 18g mol-1. If we have 18g of water, we have 1mol of it. Suppose we
have 36g water (18 x 2), we have 2 mol of it. In general in a sample of water of mass
(n x 18) g, the number of moles of water would be n. We may generalize the relation as
Number of moles (amount) of a substance =
n =
m = nxM
These relations are useful in calculations involving moles of substance.
Example 1.7 : In a reaction, 0.5 mol of aluminium is required. Calculate the amount of
aluminium required in grams? (atomic mass of Al = 27 amu)
Solution : Molar mass of Al = 27g mol-1
Required mass = no. of moles x molar mas.
= (0.5 mol) x (27g mol-1)
= 13.5 g
1.7 Molar Volume, Vm

Molar volume is the volume of one mole of a substance. It depends upon temperature and
pressure. It is related to the density, by the relation.
Molar volume =
-13-
In case of gases, we use thier volumes at standard temperature and pressure (STP). For the
purpose 00C or 273 K temperature is taken as the standard temperature and 1 bar pressure is
taken as the standard pressure. At STP, the molar volume of an ideal gas is 22.7 litre*. You will
study that gases do not behave ideally and therefore their molar volume is not exactly 22.7 L.
However, it is very close to 22.7L and for all practical purposes we take the molar volume of all
gases at STP as 22.7 L mol-1.

1. How many moles of Cu atoms are present in 3.05 g of copper (Relatine atomic mass of Cu
= 63.5)
...................................................................................................................................
2. A piece of gold has a mass of 12.6 g. How many moles of gold are present in it ? (Relative
atomic mass of Au = 197)
...................................................................................................................................
3. In a combustion reaction of an organic compound, 2.5 mol of CO2 were produced. What
volume would it occupy at STP (273K, 1 bar) ?
...................................................................................................................................
* Earlier 1 atmosphere pressure was taken as the standard pressure and at STP (273K, 1 atm) the molar
volume of an ideal gas was taken as 22.4 L mol-1. The difference in the value is due to the change in the
standard pressure (1 bar) which is slightly less than 1 atm.
What You Have Learnt

• Mole is the amount of a substance which contains as may elementary entities as there
are atoms present in 0.012 kg or 12g of C-12. Thus mole denotes a number.
• The number of elementary entities present in one mole of substance is 6.022 x 1023.
• The mass of one mole of a substance is called its molar mass. It is numerically equal
to relative atomic mass or relative molecular mass expressed in grams per mole
(g mol-1) or kilogram per mole (kg mol-1).
• Molar volume is the volume occupied by one mole of a substance. One mole of an
ideal gas at standard pressure and temperature, STP (273 K and 1 bar) occupies 22.7
litres.
• In ionic substances, molar mass is numerically equal to the formula mass of the
compound expressed in grams.
• If the molar mass of a substance is known, then the amount of a substance present in
a sample having a definite mass can be calculated. If M is the molar mass, then, the
amount of substance n, present in a sample of mass m is expressed as n = .
-14-
Terminal Exercise
1. How many atoms are present in a piece of iron that has a mass of 65.0 g / (atomic mass; Fe
= 55.9 amu).
2. A piece of phosphorus has a mass of 99.2 g. How many moles of phosphorus, P4 are
present in it ? (atomic mass, P = 31.0 amu)
3. Mass of 8.46 x 1024 atoms of fluorine is 266.95g. Calculate the atomic mass of fluorine.
4. A sample of magnesium consists of 1.92 x 1022 Mg atoms. What is the mass of the
sample in grams ? (atomic mass - 24.3 amu)
5. Calculate the molar mass in g mol-1 for each of the following.
i) Sodium hydroxide, NaOH
ii) Copper Sulphate CuSO4.5H 2O
iii) Sodium Carbonate, Na2CO3.10H2O
6. For 150 gram sample of phosphorus trichloride (PCl3), calculate each of the following:
i) Mass of one PCl3 molecule.
ii) The number of moles of PCl3 and Cl in the sample.
iii) The number of grams of C1 atoms in the sample.
iv) The number of molecules of PC13 in the sample
7. Find out the mass of carbon-12, that would contain 1 x 1019 atoms.
8. How many atoms are present in 100g sample of C-12 atom?
9. How many moles of CaCO3 would weigh 5g ?
10. If you require 1.0 x 1023 molecules of nitrogen for the reaction N2 + 3H2 → 2NH 3.
i) What is the mass (in grams) of N2 required ?
ii) How many moles of NH3 would be formed in the above reaction from 1.0 x 1023
molecules of N2 ?
iii) What volume would NH3 gas formed in ii) occupy at STP ?
Answers to Intext Questions

1.1
1. Kilogram
2. μg
3. i) h ii) n
4. i) Megasecond, 106 s
ii) millisecond, 10-3 s.
1.2
1. Moles of N2 gas = = 0.70 mol
-15-
2. Amount of magnesium (moles = = 14.05 mol
3. No. of Cl2 molecules in 0.25 mol Cl2 = 0.25 x 6.022 x 1023 molecules
Since each Cl2 moecule has 2 Cl atoms, the number of Cl atoms = 2 x 1.5055 x 1023 =
3.011 x 1023 atoms.
1.3
1. Molar mass of hydrogen chloride = molar mass of HCl
= 1 mol of H + 1 mol of Cl
= 1.0 g mol-1 + 35.5 g mol-1
= 36.5 g mol-1
2. Molar mass of argon atoms = mass of 1 mol of argon
= mass of 6.022 x 1023 atoms of argon.
= 6.634 x 10-26 kg x 6.022 x 1023 mol-1
= 39.95 x 10-3 kg mol-1
= 39.95 g mol-1
3. Molar mass of KNO 3 = mass of 1 mol of K + mass of 1 mol of N + mass of 3 mol of O.
Since molar mass of an element is numerically equal to its atomic mass but has the
units of g mol-1 in place of amu = 39.1 g + 14.0 g + 3 x 16.0 g
∴ Molar mass of KNO 3 = 39.1 g + 14.0 g + 48.0 g = 101.1 g mol-1
4. Mass of 1 mol of Na3PO4 = 3 x (mass of 1 mol of Na) + mass of 1 mol of P +

4 x (mass of 1 mol of oxygen)
= 3 (23.0 g) + 31.0 g + 4 (16.0) g
= 69.0 g + 31.0 g + 64.0 g = 164.0 g
∴ Mass of 0.146 mol of Na3PO4 = 0.146 x 164.0 g = 23.94 g
1.4
1. Moles of Cu atoms in 3.05 g copper = = 0.048 mol
2. Moles of gold, Au = = 0.064 mol
3. Molar volume of any gas at STP (298 K, 1 bar) = 22.7 L

∴ Volume occupied by 2.5 mol CO2 at STP = 2.5 x 22.7 L = 56.75 L
-16-
2
CHEMICAL ARITHMATICS
We know that atoms of different elements combine in simple whole- number ratios to form
molecules. For example, hydrogen and oxygen atoms combine in the mass ratio of 1:8 and
form water, H 2O. However, it is impossible to deal with individual atoms because they are
so tiny we can neither see nor weigh them. Therefore, we must increase the size of these
quantities to the point where we can see them and weigh them. With the help of mole
concept it is possible to take a desired number of atoms / molecules by weighing (please
refer to lesson-1). Now, in order to study chemical compounds and reactions in the laboratory,
it is necessary to have adequate knowledge of the quantitative relationship among the amounts
of the reacting substances that take part and products formed in the chemical reaction. This
relationship is know as stoichiometry. Stoichiometry (derived from the Greek stoicheion =
element and metron = measure) is the term we use to refer to all the quantitative aspects of
chemical compounds and reactions. In the present lesson, you will see how chemical
formulae* are determined and how chemical equations prove useful in predicting the proper
amounts of the reactants that must be mixed to carry out a complete reaction. In other
words we can take reactants for a reaction in such a way that none of the reacting substances
is in excess. This aspect is very vital in chemistry and has wide application in industries.
Objectives
After reading this lesson, you will be able to:
• Define empirical and molecular formulae;
• Differentiate between empirical and molecular formulae;
• Calculate percentage by mass of an element in a compound and also work out empirical
formula from the percentage composition;
• Establish relationship between mole, mass and volume;
• Calculate the amount of substances consumed or formed in a chemical reaction using
a balanced equation and mole concept, and
• Explain that the amount of limiting reagent present initially limits the amount of the
products formed.
-17-
2.1. Molecular and Empirical Formulae
In your previous classes, you have studied how to write chemical formula of a substance.
For example, water is represented by H2O, carbon dioxide is represented by CO2, methane
is represented by CH 4, dinitrogen penta oxide is represented by N2O5, and so on. You are
aware, formula for a molecule uses a symbol and subscript number to indicate the number
of each kind of atoms present in the molecule (subscript 1 is always omitted). Such a formula
is called molecular formula as it represents a molecule of a substance. A molecule of water
consists of two hydrogen atoms and one oxygen atom. So its molecular formula is written
as H2O. Thus a molecular formula shows the actual number of atoms of different elements
in a molecule of a compound.
There is another kind of formula, the empirical formula of a compound, which gives only
relative number of atoms of different elements. These numbers are expressed as the simplest
ratio. For example, empirical formula of glucose, which consists of carbon, hydrogen and
oxygen in the ratio of 1:2:1 is CH2O (empirical formulae are also called simplest formulae).
Molecular formula = Xn where X is empirical formula and n is an integer). For example
molecular formula of glucose is C6H12O6 which is 6 x its empirical formula. Thus, while
empirical formula gives only a ratio of atoms, the molecular formula gives the actual number
of atoms of each element in an individual molecule. In some cases the ratio of atoms shown
in a molecular formula cannot be reduced to smaller integers. In such cases molecular and
empirical formulae are the same, for example, sucrose C12H22O11 which is popularly known
as cane-sugar. In case of certain elements, a molecule consists of several atoms for example
P4, S 8, etc. In such cases, empirical formula will be symbol of the element only.
As you know, common salt, which is chemically called sodium chloride is represented as
NaCl. This salt is ionic in nature and does not exist in molecular form. Therefore, NaCl is
its empirical formula which shows that sodium and chlorine atoms are present in NaCl in
the ratio of 1:1. Similar is the case with all ionic substances. KCl, NaNO3, MgO are examples
of empirical formulae as these are all ionic compounds. Table 2.1 provides a few more
examples.
Table 2.1 : Molecular and Empirical Formulae
Substance Molecular formula Empirical formula
Ammonia NH 3 NH 3
Carbon dioxide CO 2 CO 2
Ethane C2 H 6 CH 3
Fructose C6H12O6 CH 2O
Sulphur S8 S
Benzene C6 H 6 CH
Sodium chloride - NaCl
Calcium oxide - CaO
-18-
2.2. Chemical Composition and Formulae
How much carbon is present in one kilogram of methane whose molecular formula is CH4?
How much nitrogen is present in one kilogram of ammonia, NH3 ? If we have prepared a
substance that is made of 58.8% carbon, 28.4% oxygen, 8.28% nitrogen and 6.56% hydrogen,
what is its empirical formula? You have studied atomic masses, formulae, and the mole
concept. Can we solve the problem using these basic concepts? The answer is ‘yes. Atomic
masses, formulae and the mole concept are the basic tools needed to solve such problems.
What is percentage composition? Let us take up this aspect in a little detail and try to
understand.
2.2.1 Percentage Composition

If we know the formula of a compound, we can find out how much of each of the elements
is present in a given quantity of the compound. Aluminium is obtained from its oxide.
Al2O3 (which is found as the ore, bauxite). From the formula we can calculate how much
aluminium can be obtained, at least in principle, frm a given amount of aluminium oxide.
Calculation is done by making use of the idea of percentage composition.
Percentage mass of an element in a compound
mass of element in one molecular formula or in one empirical formula
= x 100
molecular mass or empirical formula mass of compound
= x 100
Let us calculate percentage composition of aluminium oxide, Al2O3
Percentage of Aluminium = x 100
Molar mass of A12O3 = (2 x 27.0) g + (3 x 16.0) g = 102.0 g
Since 1 mol of A12O3 contains 2 mol of A1 atoms, the mass of Al is 2 x 27.0 g = 54.0 g Al
Percentage of Aluminium = x 100 = 52.9%
We can calculate percentage of oxygen in the same way. One mole of A12O3 contains 3
mole of O atoms, that is, 3 x 16.0g oxygen therefore
Percentage of oxygen = x 100 = 47.1%
-19-
Example 2.1 : Butanoic acid, has the formula C4H8O2. What is the elemental analysis of
butanoic acid ?
Solution : Molecular formula of the butanoic acid is C4H8O2.
In one mole of butanoic acid there are 4 mol of carbon atoms. 8 mol of hydrogen atoms and
2 mol of oxygen atoms. Thus, 1 molar mass of butanoic acid will be equal to the sum of 4 x
molar mass of carbon atoms, 8 x molar mass of hydrogen atoms, and 2 x molar mass of
oxygen atoms.
Molar mass of butanoic acid = 4 x 12.0 g + 8 x 1.0 g + 2 x 16.0 g = 88.0 g
Percentage of C by mass = x 100 = 54.5%
Percentage of H by mass = x 100 = 9.1%
Percentage of O by mass = x 100 = 36.4%
The percentage of O is butanoic acid can also be calculated as follows :

Percentage of O by mass = 100 - (Percentage of C by mass + Percentage of H by mass)
= 100 - (54.5 + 9.1) = 36.4%
2.2. Determination of Empirical Formulae -

Formula Stoichiometry
We have just seen that if we know the formula of a compound we can calculate the percentage
composition. Now the question arises, can we determine the formula of the compound if we
know the percentage composition of compound. The answer will be ‘yes’, but this formula
will not be molecular formula; instead it would be empirical formula as it would give
simplest ratio of different atoms present in a compound. Normally we determine the percentag
composition of different elements present in an unknown compound and determine its
formula. Let us take a simple example of water. Water consists of 11.11% hydrogen and
88.89% oxygen by mass. From the data, we can determine empirical formula of water. Now
if we assume that we have a 100.00 g sample of water, then the percentage composition tells
us that 100.0 g of water contains 11.11 g of hydrogen atoms and 88.89 g of oxygen atoms.
From the atomic mass table, we find that 1 mol of hydrogen atoms has a mass of 1.0 g, and
1 mol of oxygen atoms has a mass of 16.0 g. Now we can write unit conversion factors so
that the mass of hydrogen can be converted to moles of H atoms and the mass of oxygen can
be converted to moles of O atoms. Since 1 mol of H atoms has a mass of 1.0g we get the
conversion factor as
-20-
Therefore
11.11 g H = (11.11 g H) = 11.11 mol H atoms
Similarly conversion factor for oxygen will be
Therefore, 88.89 g O = (88.89 g O) = 5.55 mol O atoms
Thus in water, the ratio of moles of hydrogen atoms to moles of oxygen atoms is 11.11:5.55.
Since a mole of one element contains the same number of atoms as a mole of another
element, the ratio of moles of atoms in a compound is also the ratio of the number of
atoms. Therefore, the ratio of hydrogen atoms to oxygen atoms is 11.1:5.55. Now by dividing
each by the smaller of the two numbers we can convert both numbers to integers.
= 2 and = 1.
Thus ratio hydrogen and oxygen atoms in water is 2:1 and empirical formula of water is
H2O.

1. For the compound Fe3O4, calculate percentage of Fe and O.
...................................................................................................................................
2. State percent composition for each of the following :
(a) C in SrCO3 b) SO3 in H2SO4
...................................................................................................................................
3. What are the empirical formulae of substances having the following molecular
formulae ?
H2O2, C6H12, Li2CO3, C2H4O2, S8, H2O, B2H6, O3, S3O9, N2O3
...................................................................................................................................
4. A compound is composed of atoms of only two elements, carbon and oxygen. If the
compound contains 53.1% carbon, what is its empirical formula.
...................................................................................................................................
-21-
2.4 Chemical Equation and Reaction Stoichiometry
You have studied that a reaction can be represented in the form of a chemical equation. A
balanced chemical equation carries a wealth of information qualitative as well as quantitative.
Let us consider the following equation and learn what all information it carries.
4Fe(s) + 3O2(g) → 2Fe 2O3(s) ...... (2.1)
(1) Qualitative Information

Qualitatively the equation (2.1) tells that iron reacts with oxygen to form iron oxide.
(2) Quantitative Information
Quantitatively a balanced chemical equation specifies numerical relationship among
the quantities of its reactants and products. These relationships can be expressed in
terms of :
i) Microscopic quantities, namely, atoms, molecules and formula units.
ii) Macroscopic quantities, namely, moles, masses and volumes (in case of gaseous
substances) of reactants and products.
Now let us again take the reaction (2.1) given earlier and get the quantitative
information out of it.
2.4.1 Microscopic Quantitative Information

The reaction (2.1)
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...... (2.1)
tells that 4 atoms of iron react with 3 molecules of oxygen to form 2 formula units of iron
oxide. Often this information is written below each reactant and product for ready reference
as shown below :
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...... (2.1)

4 atoms of Fe 3 molecules of O2 2 formula units of Fe 2O3
2.4.2 Macroscopic Quantitative Information

The microscopic quantitative information discussed in the previous section can be converted
into macroscopic information with the help of mole concept which you have learnt in
unit 1.
a) Mole Relationships
We know that Avogadro number of elementary entities like atoms, molecules, ions or formula
units of a substance consitute one mole of it. Let us multiply the number of atoms, molecules
and formula masses obtained in the previous section (Eq. 2.1a) by Avogadro’s constant, NA
-22-
2
F →
4Fe(s) + 3O2(g) 2Fe2O3(s) ...... (2.1)
4 atoms of Fe 3 molecules of O 2 2 formula units of Fe2O3
4 x N A atoms of Fe 3 x NA molecules of O2 2 x N A formula units of Fe2O3(s)

4 mol of Fe 3 mol of O2 2 mol of Fe2O3
We may rewrite the above equation as
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...... (2.1b)

4 mol of Fe 3 mol of O2 2 mol of Fe2O3
The above equation (2.1b) gives us the mole relationship between reactants and products.
Here 4 mol of Fe react with 3 mol of O2 and produce 2 mol of Fe2O3.
b) Mass Relationships
The mole relationships which you have learnt in the previous section, can be converted into
mass relationship by using the fact that mass of one mole of any substance is equal to its
molar mass which can be calculated from its formula with the help of relative atomic masses
of its constituent elements.
In the reaction that we are discussing, the relative atomic masses of iron and oxygen are
55.8 and 16.0 respectively. Therefore.
i) molar mass of Fe = 55.8 g mol-1
ii) molar mass of O2 = 2 x 16.0 = 32 g mol-1
iii) molar mass of Fe2 O3 = (2 x 55.8 + 3 x 16.0) g mol-1
= 159.6 g mol-1
Using these molar masses we can convert the mole relationship given by equation 2.1b into
mass relationship as given below :
4Fe(s) + 3O2(g) → 2Fe2O3(s)

4 mol Fe 3 mol O2 2 mol Fe2O3
(4 x 55.8) g Fe (3 x 32) g O2 → (2 x 159.6) g Fe2O3

223.2 g Fe 96 g O2 319.2 g FeO 3
Thus 223.2 g iron would react with 96g oxygen and produce 319.2 g iron oxide, We may
rewrite the above equation as
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...... (2.1c)

223.2 g Fe 96 g O2 319.2 g Fe2O3
-23-
c) Volume Relationships
We know that one mole of any gas occupaies a volume of 22.7 L* at STP (standard
temperature and pressure 00C and 1 bar pressure). We can use this information to arrive at
volume relationships between gaseous substances. The reaction that we are considering
involves only one gaseous substance, O2. We may rewrite the equation (2.1b) as
4Fe(s) + 3O 2(g) → 2Fe 2O3(s) ...... (2.1)
4 mol 3 mol 2 mol
(3 x 22.7) L at STP
68.1 L at STP
Thus 4 mol of iron would react with 68.1 L of oxygen at STP to produce 2 mol of iron
oxide. (The volume relationship becomes more useful for reactions involving 2 or more
gaseous substances).
We can express microscopic as well macroscopic quantitative relationships involved in the
above reaction as shown below.
4Fe(s) + 3O 2(g) → 2Fe2O3(s)
4 atoms 3 molecules 2 formula units
4 mol 3 mol 2 mol
223.2 g 96 g 319.2 g
- 68.1 L at STP -
We may use even mixed relations. For example, we may say 4 mol of iron would react with
68.1 L (at STP) of oxygen to produce 319.2 g of iron oxide.
Let us understand these, relationships with two more examples.
a) Let us work out the mole, mass and volume relationships for the reaction involved in
manufacture of ammonia by Haber’s process.
Microscopic relationship N2 (g) + 3H 2 (g) → 2NH3 (g) ..(2.2
Microscopic relationships 1 Molecules 3 Molecules 2 Molecules
i) Moles 1 mol 3 mol 2 mol

ii) Mass 28 g (3x2.0) = 6.0g (2x17.0) = 34 g
iii) Wolume 1 x 22.7 L (3 x 22.7) (2 x 22.7)
= 22.7 L = 68.1 L = 45.4 L
or 1 vol 3 vol 2 vol
* Earlier, the standard pressure was taken as 1 atmosphere and the volume of one mole of gas at STP was
taken as 22.4 L.
-24-
b) Let us take one more reaction, the combustion reaction of butane and work out the different
types of relationships. The reaction is :
2C4H10 (g) + 13O2 (g) → 8CO 2 (g) + 10H2O (g)
2 molecues 13 molecules 8 molecules 10 molecules
2 mol 13 mol 8 mol 10 mol
2x(4x12+10x1g) (13x32) g 8x(12+2x16) g 10x(2x1+16)g
116g 416g 352 g 180g
2x22.7=45.4L 13x22.7=295.1 L 8x22.7=181.6 L 10x22.7=227 L
2 vol 13 vol 8 vol 10 vol
Now let us use the mole, mass and volume reationships to make some calculations.
Eaxmple 2.2 In the manufacture of ammonia by Haber process, nitrogen reacts with hydrogen
at high temperature and high pressure inthe presence of a catalyst and gives ammonia.
N2 (g) + 3H2 (g) → 2NH3(s)
How much hydrogen would be needed to produce one metric ton of ammonia ?
Solution : We should first find out the mass relationships for the reaction.
N2 (g) + 3H 2 (g) → 2NH3(s)
1 mol 3 mol 2 mol
1x28g = 28 g 3x2g = 6.0 g 2x17g = 34 g
We know that :
1 metric ton = 1000 kg = 103 kg = 106 g
From the mass relationship 34 g NH3 requires 6.0 g H2 for its manufacture.
∴ 106g NH3 would require g = 1.76 x 105g of H2.
Thus 1 metric ton of ammonia will be obtained by using 1.176 x 105g of Hydrogen.
Example 2.3 In a rocket motor fuelled by butane, C4H10, how many kg of O2 should be provided
with each kg of butane to provide for complete combustion ?
Solution : The combustion reaction of butane is
2C4H10 (g) + 13O2 (g) → 8CO 2 + 10H2O(s)
2 mol 13 mol
2 x 58 = 116 g 13 x 32 = 416 g
-25-
Thus, to completely burn 116g butane, exygen required is 416 g.
Therefore, to completely burn 1 kg (1000 g) butane, oxygen required will be
= g O2
= 3586 g O2
= 3.586 kg O2 = 3.59 kg O2
Example 2.4 When lead sulphide; PbS and lead oxide, PbO, are heated together the product are
lead metal and sulphur dioxide, SO2.
If 14.0 g of lead oxide reacts according to the above equation, how many (a) moles of lead (b)
grams of lead, (c) atoms of lald and (d) grams of sulphur dioxide are formed?
(Atomic mass : Pb = 207.0, S = 32.1 ; O = 16.0

Solution : For each part of the question we well use the balanced equation
Now formula mass of PbO = (207.0 + 16.0) = 223.0 amu

Thus, one mole of lead oxide formula units have a mass of 223.0 g. Therefore, 14.0 g of
_
PbO is = 6.28 x 10 2 mol PbO
a) The balanced equation show that 2 mol of PbO form 3 mol of Pb. Therefore, 6.28 x
_
10 2 mol of PbO form
_ _
6.28 x 10 2 mol PbO x = 9.42 x 10 2 mol Pb
b) The atomic mass of Pb is 207.0; this tells us that one mol of lead has a mass 207.0g.
_
Thus, 9.42 x 10 2 mol of Pb has a mass of
_
9.42 x 10 2 mol Pb x = 19.5 g Pb
-26-
_
c) 9.42 x 10 2 mol of Pb is
_
9.42 x 10 2 mol of Pb x 6.022 x 1023 atoms mol-1 = 5.67 x 10 22 Pb atoms
d) The balanced equation shows that 2 mol of PbO form 1 mol of SO2.
_
6.28 x 10 2 mol PbO x
_
= 3.14 x 10 2 mol SO2
Now the relative molecular mass of SO 2 = 32.1 + 2 (16.0) = 64.1
Molar mass of SO2 = 64.1 g mol-1
_ _
Therefore 3.41 x 10 2 mol of SO2 molecules have a mass of 3.14 x 10 2 mol x 64.1 mol-1
= 2.01 g

1. How many grams of NH3 can be made according to the reaction
N2 (g) + 3H2 (g) → 2NH 3(s)
from (a) 0.207 mol of N 2 (b) 22.6 g of H2
...................................................................................................................................
2. In reaction
C2H4 (g) + 3O 2 (g) → 2CO 2 (g) + 2H2O(1)
How many (a) moles of O 2 are consumed and (b) moles of H 2O are formed when
4.16 x 10 -2 mol of C 2H 4 react ?
...................................................................................................................................
2.5 Limiting Reagents

We generally find that substances which react with each other are not present in exactly the
same proportionion a reaction mixture as stated by a balanced chemical equation. For
example, if 2 mol each of hydrogen and oxygen are mixed and a spark is passed through the
mixture, water is formed according to the equation.
Here, 2 mol of hydrogen react with only 1 mol of oxygen, and 1 mol of oxygen therefore
remains unreacted. In this example hydrogen is said to be the limiting reagent or reactant
because its amount becomes zero and the reaction therefore stops before the other reactant;
that is, the oxygen is used up. The amount of hydrogen present initially limits the amount of
product that is formed.
-27-
Example 2.5 3 mol of sulphur dioxide SO2 is mixed with 2 mol of oxygen O2, and after reaction
is over sulphur trioxide, SO3 is obtained.
i) Which is the limiting reagent ?
ii) What is the maximum amount of SO3 that can be formed ?
Solution : i) We must first write the balanced equation
2SO2 + O2 → 2SO3
According to the above equation
a) 2 mol of SO3 that can be formed from 3 mol of SO2.
.
. . Amount of SO3 that can be formed from 3 mol of SO2
= (3 mol SO2) x = 3 mol SO3
b) 2 mol of SO3 can be formed from 1 mol of O2. Therefore, the amount of SO3 that can be
formed from 2 mol of O2.
= (2 mol O2) x = 4 mol SO3
According to the definition, the limiting reactant is that reactant which gives the smallest amount.
In this case SO2 is the limiting reactant.
(ii) The maximum amount of product that can be obtained is the amount formed by the limiting
reagent. Thus a the maximum amount of SO3 that ca be obtained is 3 mol.
Example 2.6 2.3 g of sodium metal is introduced into a 2L flask filled with chlorine gas at STP
(273 K, 1 bar). After reaction is over, find :
i) What is the limiting reagent in this reaction?
ii) How many moles of sodium chloride are formed ?
iii) Which substance is left unconsumed at the end of the reaction ? Find out its mass in
grams.
iv) What percentage of the substance present in excess is converted into sodium chloride ?
(Given : Na = 23, Cl = 35.5)
Solution :
2 Na (s) + Cl2 (g) → 2NaCl
2 mol 1 mol 2 mol
or 22.7 L at STP
i) Moles of sodium introduced = = 0.1 mol
-28-
From the above equation, it is clear that 2 mol NaCl is formed from 2 mol Na
Therefore 0.1 mol Na can produce = = 0.1 mol NaCl
Molar volume at STP = 22.7 L
Therefore moles of chlorine in 2 L volume at STP = = 0.088 mol
From equation : 1 mol C12 can produce 2 mol NaCl

Therefore 0.088 mol Cl2 can produce 2 x 0.088 = 0.176 mol NaCl.
ii) Sodium being the limiting reagent, as calculated in (i), the moles of NaCl produced = 0.1 mol
iii) From above equation, 2 mol NaCl is produced from 1 mol Cl2
Therefore 0.1 mol NaCl is produced from = 0.05 mol Cl2
Initial moles of Cl2 = 0.088 mol

Moles of Cl2 left unconsumed = (0.088 - 0.05) mol = 0.038 mol
Therefore, mass of Cl2 left unconsumed = 0.038 g x 71.0 g mol-1 = 2.698 g
(because molar mass of Cl2 = 2 x 35.5 = 71.0 g mol-1)
iv) Moles of Cl2 comsumed = 0.05 mol out of 0.088 mol
∴ Percent of Cl2 consumed and converted into NaCl = x 100 = 56.8%
Eaxmple 2.7 : 2.0 g mixture of MgCO3 and CaCO3 are heated till no further loss of weight
takes place. The residue weighs 1.04g. Find the percentage composition of the mixture.
(Mg = 24, Ca = 40, C = 12, O = 16)
Solution : Mixture of MgCO3 and CaCO3 taken = 2.0 g
Let the mass of MgCO3 be = x g
Therefore the mass of CaCO3 = (2.0 - x) g
The decomposition reactions are
MgCO 3 (g) → MgO (s) + CO2 (g) (i)
(24+12+48) g (24+16) g
84 g 40 g (Residue)
-29-
CaCO3 (g) → CaO (s) + CO2 (g) (ii)
(40+12+48) g (40+16) g
100 g 56 g (Residue)
Form the equation (i)

84 g MgCO3 leave a residue = 40 g
x g MgCO3 will leave residue = g
Form the equation (ii)

100 g CaCO3 leave a residue = 56 g
(2.0 - x ) g CaCO3 will leave residue = g
Total mass of the residue = + = 1.04 g (given)
40 x 100x + 84 x 56 x 2 - 84 x 56x = 84 x 100 x 1.04

4000x + 9408 - 4704x = 8736
9408 - 8736 = (4704 - 4000)x
672 = 704x
Therefore, mass of MgCO3 in the mixture = x = = 0.96 g
Therefore, percentage of MgCO3 = x 100 = 48%
and percentage of CaCO3 = 100 - 48 = 52%

• A chemical formula is used not only to represent the name of a compound but also to
indicate its composition in terms (i) relative number of atoms and (ii) relative number
of moles of atoms.
• A molecular formula of a substance shows (i) the number of atoms of different elements
in one molecule. (ii) the number of moles of atoms of different elements in one mole
of molecule.
• An empirical formula shows only a ratio of (i) number of atoms, and (ii) moles of
atoms in a compound.
-30-
• Molecular formula is always an integral multiple of the empirical formula
• The empirical formula of a compound can be determined from its chemical analysis.
• In order to determine a compound’s molecular formula, molecular mass also must be
known.
• Stoichiometry is the quantitative study of the composition of chemical compounds
(compound or formula stoichiometry) and of the substances consumed and formed in
chemical reactions (reaction or equation stoichiometry).
• Chemical equations specify not only the identities of substances consumed and formed
in a reaction, but also relative quantities of these substances in terms of (a) atoms,
molecules and formula units and (b) moles of these entities.
• A balanced chemical equation demonstrates that all the atoms present in the reactants
are accounted for in the product; atoms are neither created nor destroyed in a reaction.
• The stoichiometric rations among the moles of reactants shown in a balanced equation
are useful for determining which substance is entirely consumed and which substance(s)
is (are) left over.
Terminal Exercise
1. Write empirical formulae of the following compounds:
CO, Na2SO3, C4H10, H2O2, KCl
...................................................................................................................................
2. The empirical formula of glucose is CH2O which has a formula mass of 30 amu. If the
molecular mass of glucose is 180 amu. Determine the molecular formula of glucose.
...................................................................................................................................
3. What is ratio of masses of oxygen that are combined with 1.0 gram of nitrogen in the
compound NO and N2O3 ?
...................................................................................................................................
4. A compound containing sulphur and oxygen on analysis reveals that it contains 50.1%
sulphur and 49.9% oxygen by mass. What is simplest formula of the compound?
...................................................................................................................................
5. Hydrocarbons are organic compound composed of hydrogen and carbon. A, 0.1647 g
sample of a pure hydrocarbon on burning in a combustion tube produced 0.5694 g of
CO 2 and 0.0845 g of H 2O. Determine the percentage of these elements in the
hydrocarbon.
...................................................................................................................................
6. On combustion 2.4g of a compound of carbon, hydrogen and oxygen gave 3.52 of CO2
and 1.44 g of H2O. The molecular mass of the compound was found to be 60.0 amu.
a) What are the masses of carbon, hydrogen and oxygen in 2.4 g of the compound ?
...................................................................................................................................
-31-
b) What are the empirical and molecular formulae of the compound ?
...................................................................................................................................
7. i) What mass of oxygen is required to react completely with 24g of CH4 in the following
reaction?
C2H4 (g) + 2O 2 (g) → CO2 (g) + 2H2O(1)
...................................................................................................................................
(ii) How much mass of CH4 would react with 96 g of oxygen.
...................................................................................................................................
8. Inthe reaction H2 + Cl2 → 2HCl
How many grams of chlorine, Cl2 are needed to react completely with 0.245 g of
hydrogen, H2, to give hydrogen chloride, HCl ? How much HCl is formed ?
...................................................................................................................................
9. 3.65 g of H2 and 26.7 g of O2 are mixed and reacted. How many grams of H2O are
formed ?
...................................................................................................................................
10. Cuastic soda NaOH can be commercially prepared by the reaction of Na2CO3 with
slaked lime, Ca(OH)2. How many grams of NaOH can be obtained by treating 2.0 kg
of Na2CO3 with Ca(OH)2 ?
...................................................................................................................................
11. A potable hydrogen generator utilizes the reaction
CaH2 + H2O → Ca(OH)2 + 2H2
How many grams of H2 can be produced by a 100 g cartridge of CaH2 ?
...................................................................................................................................
12. The reaction 2Al + 3MnO → Al2O3 + 3Mn proceeds till the limiting substance is
consumed. A mixture of 220g Al and 400g MnO was heated to initiate the reaction.
Which initial substance remained in excess and by how much ? (Al = 27, Mn = 55).
...................................................................................................................................
13. KClO4 may be prepared by means of following series of reactions
Cl2 + 2KOH → KCl + KClO + H2O
3KClO → 2KCl + KClO3
4KClO3 → 3KClO4 + KCl
How much Cl2 is needed to prepare 400 g KClO4 by the above sequence ?
(K = 39, Cl = 35.5, O = 16, H = 1)
...................................................................................................................................
-32-
14. 2.0 g of mixture of Na2CO3 and NaHCO3 was heated when its weight reduced to 1.876
g. Determine the percentage composition of the mixture.
...................................................................................................................................
15. Calculate the weight of 60% sulphuric required to decompose 150g of chalk (calcium
carbonate). Given Ca = 40, C = 12, O = 16, S= 32)
...................................................................................................................................

2.1
1. Molar mass of Fe3O4 = 3 x 56.0 + 4 x 16.0
= (168.0 + 64.0) = 232.0 g mol-1
Percentage of Fe = x 100 = 72.41%
Percentage of O = x 100 = 27.59%
2. a) Molar mass of SrCO3 = 87.6 + 12.0 + 48.0 = 147.6 g mol-1
Percentage of carbon C in SrCO3 = x 100 = 8.13%
b) Molar mass of H2SO4 = 2.0 + 32.1 + 64.0 = 98.1 g mol-1
Molar mass of SO3 = 32.1 + 48.0 = 80.1 g mol-1
Percentage of SO3 in H2SO4 = x 81.65%
3. Substance Empirical formula

H 2O 2 HO
C 6H 12 CH 2
Li 2CO 3 Li 2CO 3
C 2H 4 O 2 CH 2 O
S8 S
H2O H2O
B 2H 6 BH 3
O3 O3
S 3O 9 SO 3
N 2O 3 N 2O 3
-33-
4. Percentage of carbon = 53.1%
Percentage of Oxygen = 46.9%
Suppose we take 100g of the substance then moles of carbon = g = 4.43 mol
mole of oxygen = = 2.93 mol
molar ration of C and O =
= 1.50 : 1 or 3 : 2
Empirical formula of the compound is C3O2.
2.2
1. In equation
N2 (g) + 3H 2 (g) → 2NH3(s)
1 mol 3 mol 2 mol
0.207 mol N2 gives 0.414 mol of NH3

0.414 mol of NH3 = 0.414 mol x 17.0 g mol-1 = 7.038 of NH3
22.6 g of Hydrogen = = 11.3 mol of hydrogen
11.3 mol of hydrogen will give 2/3 x 11.3 mol of NH3 = 7.53 mol
Therefore, Mass of NH3 = 7.53 mol x 17.0 g mol-1 = 128.01 g
2. C2H4 (g) + 3O 2 (g) → 2CO2 (g) + 2H2O(g)

1 mol 3 mol 2 mol 2 mol
a) 4.16 x 10-2 mol of C2H4 will consume 3 x 4.16 x 10-2 mol of oxygen
= 12.48 x 10-2 = 1.248 x 10-1 mol of O 2
b) moles of H2O formed = 2 x 4.16 x 10-2 mol

= 8.32 x 10-2 mol of H2O
-34-
3
ATOMIC STRUCTURE
Chemistry has been defined as the study of matter in terms of its structure, composition and
the properties. As you are aware, matter is made up of atoms, and therefore an understanding
of the structure of atom is very important. You have studied in your earlier classes that the
earliest concept of atom (smallest indivisible part of matter) was given by ancient (600-400
BC) Indian and Greek philosophers. At that time there was no experimental evidence. The
origin of the concept of atom was based on their thoughts on ‘What would happen if we
continuously keep dividing matter’. John Dalton revived the concept of atom in the beginning
of nineteenth century in terms of his atomic theory which successfully explained the laws
of chemical combination. Later experiments showed that the atom is not indivisible but has
an internal structure.
In this lesson you will learn about the internal structure of an atom which will help you to
understand the correlations between is structure and properties. You would learn about these
in the later lessons.
Objectives
After reading this lesson you will be able to:
• recognize the fundamental particles of atom;
• describe Rutherford’s experiment and explain its results;
• define electromagnetic radiation;
• list and define the characteristic parameters of electromagnetic radiation;
• discuss line spectrum of hydrogen;
• explain Bohr’s postulates and discuss his model;
• draw energy level diagram of hydrogen atom showing different series of lines in its
spectrum;
-35-
• explain wave particle duality of matter and radiation;
• formulate Heisenberg’s uncertainty principle;
• explain the need for quantum mechanical model;
• draw probability pictures of an electron in an atom;
• list quantum numbers and discuss their significance;
• draw the shapes of s,p and d orbitals;
• recognize nodal plane;
• explain Pauli’s exclusion principle;
• define Aufbau principle and
• explain Hund’s rule of maximum multiplicity.
3.1 Fundamental Particles of Atom

In 1897 J.J. Thomson discovered electron as a constituent of atom. He determined that an
electron had a negative charge and had very little mass as compared to that of the atom.
Since an atom was found to be electrically neutral it was inferred that some source of positive
charge must be present in the atom. This soon led to the experimental discovery of the
proton, which is a positively charged subatomic particle. Proton was found approximately
1840 times heavier than an electron. Further experiments revealed that the atomic masses
were more than that expected from the presence of just protons and electrons in the atom.
For example, the mass of helium atom was expected to be double that of hydrogen atom but
was actually found to be almost four times the mass of hydrogen atom. This suggested the
presence of neutral particles with mass comparable to that of protons in the atom. Sir James
Chadwick discovered this neutral particle and called it neutron subsequently in 1932. Thus
we may conclude that atoms are not indivisible but are made up of three fundamental particles
whose characteristics are given in Table 3.1.
Table 3.1 Fundamental particles and their characteristics
Particle Symbol Mass / kg Actual Charge / C Relative charge
Electron e 9.109 389 x 10-31 - 1.602 177 x 10-19 -1

Proton p 1.672 623 x 10-27 1.602 177 x 10-19 +1
Neutron n 1.674 928 x 10-27 0 0
Since atoms are made up of still smaller particles, they must have an internal structure. In
the next section we shall take up some of the earlier ideas about the internal structure of
atom.
-36-
1. Compare the mass of an electron with that of the proton.
..................................................................................................................................
2. What is a fundamental particle?
..................................................................................................................................
3. What is the name given to neutral particles in the atom?
.................................................................................................................................
3.2 Earlier Models

Once it was established that the atom is not indivisible, the scientists made attempts to
understand the structure of the atom. A number of models have been proposed for the internal
structure of the atom. The first attempt to describe the structure of atom in terms of a model
was made by J.J. Thomson.
3.2.1 Thomson’s Model
On the basis of his experiments on discharge tubes. Thomson proposed that atoms can be
considered as a large positively charged body with a number of small negatively charged
electrons scattered throughout it. This model (Fig. 3.1) was called as plum pudding model
of the atom.
Fig 3.1 : A pictorial representation of Thomson’s plum-pudding model

The electrons represent the plums in the pudding made of positive charge. It is sometimes
also called as watermelon model. In this, the juicy pulp of the watermelon represents the
positive charge and the seeds represent the electrons.
J.J. Thomson Ernest Rutherford

(1856-1940) (1871-1937)
Won Nobel Prize in Physics in 1906 Won Nobel Prize in Chemistry in 1908
-37-
3.2.2 Rutherford’s Experiment
Ernest Rutherford performed an experiment called ‘God Foil Experiment’ or ‘a- ray scattering
experiment’ to test the structure of an atom as proposed by Thomson. In this experiment a
beam of fast moving alpha particles (positively charged helium ions) was passed through a
very thin foil of gold. He expected that the alpha particles would just pass straight through
the gold foil and could be detected by a photographic plate. But, the actual results of the
experiment (Fig. 3.2) were quite surprising. It was observed that most of the α - particles
did pass straight through the foil but a number of particles were deflected from their path.
Some of these deflected slightly while a few deflected through large angles and about 1 in
10,000 α- particles suffered a rebound.
electron
nucleus
Source of lead plate photographic gold foil

alpha particles plate
Fig. 3.2 Schematic representation of Rutheford’s Fig. 3.3 : Schematic representation
α−ray scattering experiment. of Rutherford’s model
These results led Rutherford to conclude that:

• the atom contained some dense and positively charged region located at the center of
the atom that he called as nucleus.
• all the positive charge of the atom and most of its mass was contained in the nucleus.
• the rest of the atom must be empty space which contains the much smaller and
negatively charged electrons (Fig. 3.3).
The model proposed by Rutherford explained the observation in the α–ray scattering
experiments as shown below in Fig. 3.4.
undeflected
α-rays
Large angle
deflection
Small angle
Fig. 3.4 Explanation of the results of α–ray deflection Fig. 3.5 : Failure of Rutherford’s
scattering experiment. model
-38-
However, there was a problem with the Rutherford’s model. According to the Maxwell’s
theory of electromagnetic radiation, a charged particle undergoing acceleration would
continnously emit radiation and lose energy. Since the electron in the atom is also a charged
particle and is under acceleration, it is expected to continuosly lose energy. As a consequence,
the eletron moving around the nucleus would approach the nucleus by a spiral path (Fig.
3.5) and the atom would collapse. However, since it does not happen we can say that the
Rutherford’s model failed to explain the stability of the atom.
The next attempt to suggest a model for atom was made by Neils Bohr-a student of
Rutherford. This model used the concept of quantisation of energy of electrons in the atom.
Since this fact was suggested by line spectrum of hydrogen atom it is worthwhile to
understand the meaning of a spectrum. For this we begin with the understanding of the
nature of an electromagnetic radiation.

1. List the three consituent particles of an atom.
...................................................................................................................................
2. What was the aim of Rutherford’s a-rays scattering experiment ?
...................................................................................................................................
3. Briefly describe Rutherford’s model of an atom.
...................................................................................................................................
4. On what basis was the Rutherford’s model rejected ?
...................................................................................................................................
3.3 Electromagnetic Radiations

Electromagnetic radiation is a kind of energy, which is transmitted through space in the
form of electric and magnetic fields. These do not require any medium to propagate. Visible
light, radiant heat, radio waves, X-rays and gamma radiation are some of the examples of
electromagnetic radiations. According to the Maxwell’s theory, an electromagnetic radiation
can be visualised as oscillating electric and magnetic fields. These travel as waves in the
planes perpendicular to each other and also the direction of propagation. (Fig. 3.6 (a) ).
These radiations travel with the velocity of light (3.0 x 108ms-1).
tic field
Elec on
a gati
of prop
c tion
Dire
ld
c fie
eti
gn
Ma
-39-
Crest Crest
Amplitude
Trough Trough
(a)
Electric field
component
CREST CREST
Direction of
propagation
Magnetic field THROUGH THROUGH
component
(a) (b)
Fig. 3.6 : (a) An electromagnetic wave showing electric and magnetic fields travelling
in planes perpendicular to each other and also to the direction of propagation
(b) Characteristics of electromagnetic wave
3.3.1 Characteristic Parameters of Electromagnetic Radiations

The electromagnetic radiations are characterized by a number of parameters. These are
Amplitude : This is refers to the maximum height to which the wave oscillates. It equals
the height of the crests or depth of the troughts.
Wavelength : It is the linear distance between two consecutive wave-crests or wave-
troughts as shown in Fig. 3.6(b). It is represented by a Greek letter lambda (λ) and is
expressed in terms of m, cm, nm or Angstrom (1Ao = 10-10 m).
Frequency : It is defined as the number of wave crests or wave troughs that pass through
a given point per second. It is represented by a Greek letter nu (ν) and is expressed in
terms of s-1 (second inverse or per second). It is also called as Hz (Hertz).
Wave number : It equals the number of waves per unit length. It is denoted as ν (nu bar)
and is equal to the reciprocal of the wavelength. The SI of ν is m-1 (meter inverse). However,
sometimes it is also expressed as cm-1 (centimeter inverse).
.... (3.1)
-40-
Velocity : it is defined as the linear distance travelled by the wave in one second. The
velocity in meters per second can be obtained by multiplying frequency in Hertz (s-1) with
wavelength in meters.
c=νλ or .... (3.2)
The velocity of a radiation depends on the medium. In vaccum the velocity is equal to
3.00 x 108 m s-1.
The electromagnetic radiations also show the characteristics of particles. These are called
as quanta. These quanta are actually bundles of energy. A quantum of visible light is called
a photon. The energy of the quantum (or photon) is proportional to the frequency of the
radiation. The two are related as
E = hv .... (3.3)
The energy of the quantum can also be related to the wavelength or wave number as
.... (3.4)
the energy of photon can be readily calculated from these equations if we know the frequency,
wavelength or wave number.
Example 3.1 : A microwave radiation has a frequency of 12 gigahertz. Calculate the energy
of the photon corresponding to this radiation. (h = 6.626 x 10-34 J s and 1 gigahertz = 109 Hz.)
Solution : The energy is given by the expression, E = hv
Substituting the values we get,
E = 6.626 x 10-34 Js x 1.2 x 1010 s -1 = 7.95 x 10-24 J
Example 3.2 : The green light has a wavelength of 535 nm. Calculate the energy of a
photon of green light.
Solution : We know that
3.3.2 Electromagnetic Spectrum

Depending on their characteristics (wavelength, frequency and wave number)
electromagnetic radiations are of many types and constitute what is called as an
electromagnetic spectrum. (Fig. 3.7) The part of the spectrum that we can see is called
visible spectrum and is a very small part of the overall spectrum.
-41-
Wavelength (in m)
-10 -9 -8
10 10 10 10-7 10-6 10-5 10-4 10-3 10-2 10-1
Radio waves
Gamma rays
Visble
Frequency (in Hz)
Visible Spectrum
Wavelength (in nm)

Fig. 3.7 : The electromagnetic spectrum

1. What is an electromagnetic radiation ?
...................................................................................................................................
2. List any three characteristics of electromagnetic radiation.
...................................................................................................................................
3. What is wave number? How is it related to wave length ?
...................................................................................................................................
4. What is the difference between a ‘quantum’ and a ‘photon’ ?
...................................................................................................................................
3.4 Line Spectrum

You know that when we pass a beam of sunlight through a prism we get a range of colours
from violet to red (VIBGYOR) in the form of a spectrum (like rainbow). This is called a
continuous spectrum because the wavelengths of the light varies continuously that is without
any break. Let us take another example. You are aware of the flame tests for identifying
cations in the qualitative analysis. Compounds of sodium impart a bright yellow colour to the
flame, copper gives a green flame while strontium gives a crimson red coloured flame. If we
pass such a light through a prism it gets separated into a set of lines. This is called as line
spectrum. Fig. 3.8 differentiates between a continuous and a line spectrum.
Prism een
Scr Prism
een
Fig. 3.8 : (a) a Continuous spectrum b) a Line spectrum Scr
-42-
3.4.1 Line Spectrum of Hydrogen Atom
When an electric discharge is passed through a discharge tube containing hydrogen gas at
low pressure, it emits some light. When this light is passed through a prism it splits up into
a set of five lines. This spectrum is called the line spectrum of hydrogen (Fig. 3.9).
EM radiation
Discharge Prism
Tube Johann Balmer
(1825-1898
Fig. 3.9 : A schematic diagram showing line spectrum

of hydrogen in the visible range
On careful analysis of the hydrogen spectrum it was found to consist of a few sets of lines
in the ultraviolet, visible and infrared regions. These sets of lines were observed by different
scientists. These spectral emission lines could be expressed in the form of a general formula
as :
ν=
1
λ
= RH [ 1
n12
1
n22 ]
cm-1; RH=109677cm-1 ...(3.5)
Where n1 and n 2 are positive integers (n1 < n2) and RH is called Rydberg’s constant. The
different sets of lines observed in the hydrogen atom spectrum, their discovers and the
values of n1 and n2 are given in the Table 3.2.
Table 3.2 : Summary of the emission lines observed in hydrogen spectrum
Series n1 n2 Region of spectrum
Lyman 1 2,3,4.... Ultraviolet

Balmer 2 3,4,5.... Visible
Paschen 3 4,5,6.... Infrared
Bracket 4 5,6,7.... Infrared
Pfund 5 6,7,8.... Infrared
The line Spectrum of hydrogen atom was explained by Bohr’s model, which is discussed in
section 3.5
-43-
Example 3.3 : Calculate the wavelength of the Balmer line corresponding to n2 = 3.
⎡1 1⎤
Solution : According to Balmer series ¯ = R H ⎢ 2 − 2 ⎥
⎣2 n2 ⎦
Where RH = 109,677 cm-1
For n2 = 3 ; ν = 109, 677 ⎡ 1 − 1 ⎤ = 109, 677 ⎛⎜ 5 ⎞⎟

⎢⎣ 22 32 ⎥⎦ ⎝ 36 ⎠
1 36
Since λ= ;λ =
v 109, 677 x 5
= 6.56 x 10-5 cm
= 656 nm
3.4.1.a Planck’s Quantam Theory
The wave nature of light could explain the phenomenon of diffraction and interference. However,
following are some of the observations which could not be explained with the help of
electromagnetic theory. (called classical physics)
• The nature of emission of radiation from hot bodies.
• Reflection of electrons from metal surface when radiation strikes it.
• Variation of heat capacity of solids as a function of Temperature.
• Line spectra of atoms with special reference to hydrogen.
The First concrete explanation for the above observations was given by Max Planck in
1900.
According to his theory light is emitted or absorbed in small packets of

energy called photons. Each photon has a definite amount of energy
associated with it.
the photons travel with a velocity, equal to velocity of light. The Energy
of radiation (E) is proportional to its frequency (ν).
Ε α ν ; Ε = hν
h is called the plank's constant.
h = 6.6256 x 10-34 Js = 6.6256 x 10-27 erg.s
-44-
Planck’s theory successfully explains the emission
of radiation from a hot body (black body radiations). When
solids are heated they emit radiation over a wide range of
wavelengths.
Example : When iron rod is heated in a furnace, it first

turns to dull red and then becomes more and more red as
the temperature increases. As this is heated further, the
radiation emitted becomes white, then blue. In terms of
frequency, it means that the radiation emitted goes from a
lower frequency to higher frequency region of
electromagnetic spectrum. The ideal body, which emits and
Fig. 3.9.a Radiations emitted by a
absorbs all frequencies is called a black body and the
black body at different
radiation emitted by such a body is called black body
temperatures
radiation. A hallow sphere coated inside with platinum black
and having a small hole in its wall acts as a near black
body.
The exact frequency distribution of the emitted radiation i.e intensity of the radiation versus frequency
give curves. Such curves obtained at different temperatures of the black body is shown in Fig. 3.4.1(a).
The study of the curves shows that the nature of the radiation depends on the temperature of the black
body. If the energy emitted is continuous the curve should be as shown by the dotted lines, but it is not
so.
The following conclusions are drawn from the study of the shape of the curves.
• At a given temperature, the intensity of the radiant energy increases with the wave length,
reaches a maximum and then decreases.
• As the temperature increases, the peak of the curve shifts to lower wavelength.
The above experimental results could not be explained satisfactorily on the basis of the wave theory of
light. The salient features of Planck’s theory are
* The oscillating particle in the black body does not emit energy continuously.
* Radiation is emitted only in discrete quantities, Planck gave the name
QUANTUM to the smallest quantity that can be emitted or absorbed in the
form of electromagnetic radiation.
* The phenomenon of wave propagation of radiant energy in the form of quanta
is called quantization of energy.
* Energy is emitted or absorbed in simple integral multiples of a quantum. It
cannot be in fractional values.
* E = n.hv; where n = an integer.
-45-
3.4.1.b Explanation of Photo Electric Effect
When a clean surface of an alkali metal (Ex:

Potassium, Rubidium etc) is exposed to a beam
of light, ejection of electrons from the metal
surface takes place. Only if the frequency of
incident light is greater than a certain minimum
value, characteristic of the metal. This
phenomenon is known as photo electric effect.
It is observed that violet light is able to eject
electrons from potassium but red light which is
of lower frequency has no effect. The explanation
from the frequency dependence of the
Fig. 3.9.b Equipment for studying the
photoelectric effect is given by ALBERT
photoelectric effect. Light of a particular
EINSTEIN. He could explain the photoelectric frequency strikes a clean metal surface inside a
effect using Planck’s quantum theory of vaccum chamber. Electrons are ejected from
electromagnetic radiation as a starting point. the metal and are counted by a detector that
Einstein argued that the wave model of light measures their kinetic energy.
cannot explain the observed facts. However, if
light is regarded as consisting of particles (called
photons) such that the energy (E) of a photon is
related to the frequency (ν) by the relation.
E = hν; where h = Planck’s constant.
It is easy to understand photoelectric effect.
• When photon of sufficient energy strikes an electron in the atom of the

metal, it transfers its energy instantaneously to the electron during the
collision and the electron is ejected without any timelog or delay.
• Greater the energy possessed by the photon, greater will be the transfer
of energy to the electron and greater the kinetic energy of the ejected
electron.
• When a photon strikes the surface of the metal, the energy (hν) of the
photon is absorbed by the electron in the metal and a part of this energy
is used to set free the electron from the attractive forces in the metal.
The remaining energy of the photon appears in the form of the kinetic
energy of the released electron.
Thus we can write hν = W+K.E where, hν = Energy of the photon.
W = Energy required to overcome the attractive forces on the electron
in the metal. Also called work function.
K.E = Kinetic energy of emitted electron.
∴ hν = hνο + 1/2 meV2
Me = mass of electron; ν = velocity of ejected electron.
ν0 = Threshold frequency.
-46-
3.5 Bohr’s Model
In 1913, Niels Bohr (1885-1962) proposed another model of the atom where electrons move
around the nucleus in circular paths. Bohr’s atomic model is built upon a set of postulates,
which are as follows :
1. The electrons move in a definite circular paths around the nucleus (Fig. 3.10). He
called these circular paths as orbits and postulated that as long as the electron is in a
given orbit its energy does not change (or energy remains fixed). these orbits were
therefore refered to as stationary orbits or stationary states or non radiating orbits.
Fig. 3.10 : Bohr’s model Bohr won the Nobel Prize in

Physics in 1922 for his work.
2. The electron can change its orbit by absorbing or releasing evergy. An electron at a
lower (initial) state of energy, Ei can go to a (final) higher state of energy, Ef by
absorbing (Fig 3.11) a single photon of energy given by
E = hv = Ef - Ei ....(3.6)
Similar, when electron changes its orbit from a higher initial state of energy Ei to a
lower final state of energy Ef, a single photon of energy hl is released.
Fig. 3.11 : Absorption and emission of photon causes the electron to change its energy level.
3. The angular momentum of an electron of mass me moving in a circular orbit of

radius r and velocity ν is an integral multiple of h/2π.
nh
m e νr =
2π
where n is a positive integer, known as the principal quantum number.
-47-
Bohr obtained the following expressions for the energy of an electron in stationary states of
hydrogen atom by using his postulates.
⎛ 1⎞
Energy of the orbit, E n = − R H ⎜⎝ 2 ⎟⎠ .... (3.8)
n
mz 2 e 4
Bohr could correlate RH to other properties as R H = ; .... (3.9)
8h 2 ε 02
where,
m = mass of the electron h = Plank’s constant
z = nuclear charge ε0 = permitivitty of the medium
e = electronic charge
The negative sign in the energy expression means that the there is an attractive interaction
between the nucleus and the electron. This means that certain amount of energy (called
ionisation energy) would be required to remove the electron from the influence of the nucleus
in the atom. You may note here that the energies of the Bohr orbits are inversely proportional
to the square of the quantum number n. As n increase the value of the energy increases
(becomes lesser negative or more positive). It means that as we go farther from the nucleus
the enrgy of the orbit goes on increasing.
3.5.1 Bohr’s Theory of Hydrogen atom:

Hydrogen atom contains a single proton in its nucleus with +e. An electron with –e revolve
round the nucleus in a circular orbit of radius ‘r’. As per coulomb’s law, electrostatic force of
attraction between the nucleus and the electron is given by
− e2
Attractive force =
r2
This is columbic force of attraction of the electron towards the nucleus. For the atom to be stable an
equal centrifugal force must act away from the nucleus. This centrifugal force arises as a result of the
2
electron revolving around the nucleus in the circular orbit. It is equal tomv where m is mass of electron
r
revolving around the nucleus in the circular orbit and 'r' is the radius of the orbit.
Therefore altrative force of the electron revolving round the nucleus in a stationary orbit
-e2 -mv2 or e2 = mv2 ---- (1)

=
r2 r r
As per Bohr’s quantum condition
mvr = nh
2π
-48-
nh n2h2
∴V = 2πmr (or) V2 = 4π2m2r2 ---- (2)
Substitute the value of V2 in equation (1) above

-e2 mn2h2 n2h2
= (or) r =
r2 4π2m2r2 4π2me2
if r is represented as rn for nth orbit
n2h2
Radius of nth orbit (rn ) = ----- (3)
4π2me2
Substitute the values of h=6.256 x 10-27 erg.sec
π = 3.14
(mass of electron) m = 9.1 x 10-28 g
(Charge on electron) e = 4.802 x 10-10 e.s.u in equation (3)
we get Radius of nth orbit (rn) = 0.529 x 10-8 n2 cm
If n = 1 Bohr’s orbit (r) = 0.529 x 10-8 cm = 0.529 A0.

Since ‘n’ can have values 1,2,3…... a simple integer. It is obvious that orbits of only certain
finite radii are possible and electron could occupy possibly only these permitted orbits. These
simple whole numbers denoted by ‘n’ are called Principal Quantum numbers.
Energy of electron:
The total energy of electron revolving in an orbit is obtained by summing up its kinetic energy
and potential energy.
1
Kinetic energy due to motion of an electron = 2 mv2
(m is the mass of electron and v is its velocity)
Potential energy of electron due to position = -e2/r
1 2 -e2 e2 e2 e2
Total energy of electron = K.E + P.E = 2 mv - r = 2r - r = mv = r
2
[ ]
− e2
Therefore total energy of electron (E) =
2r
n2h2
Substituting the value of 'r' =
4π2me2
-49-
Energy of electron in nth orbit = En = -e24π2me4
2n2h2
2 4 1
En = -2π me x
h2 n2
Except of 'n' all the other terms in the above equation are constant we can write
-K where K = 2π2me4
En=
n2 n2
Substituting the values of m,e,h and π we have
-21.72x10-12 erg = -21.72x10-19 J or -13.6 eV.

En=
n2 n2 n2
Thus, we can conclude that the energy of an electron (E) can have only certain discrete restricted values
depending upon the value of ‘n’. This is termed as quantization of energy of the electron. An electron
revolves only in certain orbits of fixed radii and also it has only certain allowed energy levels. These
orbits are also termed as main energy levels or the principal Quantum states.
In the equation energy of an electron -K

En=
n2
the energy of an electron is inversely proportional to the square of the principle quantum number n. As
the value of n increases, radius of the orbit increase and the absolute value of energy of an electron also
increases effectively.
As the value of n decrease, the radius of the orbit as well as the absolute value of its energy decreases.
This increases the stability of the atom. When the electron of a hydrogen atom revolves in the first orbit,
energy is at its lowest possible value and hence confers maximum stability to the hydrogen atom. This is
termed as ground state of the hydrogen atom.
When the electron is free from the influence of nucleus the energy is taken as zero. The electron in this
situation is associated with the stationary state of principle Quantum number.
(n) = ∞ and is called as ionized hydrogen atom. When the electron is attracted by the nucleus and is
present in orbit n the energy is emitted and it’s energy is lowered. This is the reason for the presence of
negative sign in the equation for energy and depicts its stability relative to the reference state of zero
energy and n = ∞.
3.5.2 Explanation of Line Spectrum of Hydrogen Atom

As per the second postulate mentioned above, the energy emitted in the transition of a single
electron from an initial stationary state of energy Ei to a final stationary state of energy Ef is
given as ho = Ei - Ef . Substituting the expresisons for energy from Eq. 1.8 we can get the
formula given in eg. 3.5. Thus Bohr’s model providess an explanation for the observed line
-50-
spectrum of hydrogen as summarized in Table 3.2 Fig. 3.12 shows the energy level diagram
for hydrozen atom and the transitions responsible for the observed line spectrum.
n=
8
n=6
n=5
n=4
Paschen seried (infrared)
n=3
Red
Blue
Violet
Green
Energy
Lyman series (ultraviolet)
Fig. 3.12 : Energy level diagram for H-atom, showing various transitions responsible
for the observed line spectrum.
1. What is the difference between a line spectrum and a continous spectrum ?

...................................................................................................................................
2. What are the main postulates of Bohr’s model ?
...................................................................................................................................
3. How does the energy of a Bohr orbit vary with the principle quantum number ‘n’.
...................................................................................................................................
4. What is ‘Quantum’?
............................................................................................................................... ....
5. Give the value of Planck’s constant in S.I. system
............................................................................................................................... ....
-51-
6. State photoelectric effect.
............................................................................................................................... ....
7. Define threshold energy.
............................................................................................................................... ....
3.6 Wave - Particle Duality

In section 3.3 you have learnt about the wave nature of light. As you are aware that some of
the properties of light e.g. diffraction and interference can be explained on the basis of its
wave nature. On the other hand some other properties like photoelectric effect and scattering
of light can be explained only on the basis of particle nature of light. Thus light has a dual
nature possessing the properties of both a wave and a particle, i.e., light could under some
conditions behave like a particle and other conditions behave as a wave.
In 1923 a young French physicist, Louis de Broglie, argued that if light can show wave as
well as particle nature, why should particles of matter (e.g., electron) not posses wave like
characteristics ? He proposed that matter particles should indeed have a wave nature and
said that a particle of mass m moving with a velocity v has an associated wavelength, e
(some times called de Broglie wavelength) given by the formula;
.... (3.10)
Where p (=mν) is the momentum of the particle. The de Broglie wavelength of a body is
inversely proportional to its momentum. Since the magnitude of h is very small, the
wavelength of the objects of our everyday world would be too small to be observed. Let us
make a calculation to see this.
Example 3.4 : Calculate the de Broglie wavelength associated with a cricket ball weighing
380g thrown at a speed of 140 km per hour.
Solution : Mass of the cricket ball = 380g = 380 x 10-3 kg = 0.38 kg
Speed or Velocity = 140 km/hr = (140 x 1000) / 3600
= 38.89 m s-1
The wavelength associated with the cricket ball will be
dr-Broglie
−34 (1892-1987)
h 6.626 x10 JS
λ= = de-Broglie proposed the
mv (0.380kg )(38.89ms −1 ) theory of wave-particle
dualism as a part of his PhD
= 4.48 x 10-35 m (J = kg m2 s-2) thesis in 1924. He got the
physics Nobel Prize in 1929
If the electrons show wave nature then a beam of these electrons is expected to show
diffraction which is a property of waves. In 1927 G.P. Thomson and C.J. Davidson
demonstrated the diffraction of electron waves by the crystal lattice of nikel (Fig. 3.13).
Thus electrons also show a dual nature. That is, sometimes these show particle nature while
at some other times they show wave nature.
-52-
Fig. 3.13 : Electron diffraction Werner Heisenberg
pattern from nickel crystal (1901-1976)
Heisenberg got the phys.
Nobel prize in 1932
3.7 Heisenberg’s Uncertainty Principle

An important consequence of the wave-particle duality of matter and radiation was discovered
by Werner Heisenberg in 1927 and is called the uncertainty principle. According to this
principle it is not possible to simultaneously measure both the position and momentum (or
velocity) of an electron accurately. In simple words we may state that more accurately you
measure a particle’s position, the less accurately you’re able to measure its momentum, and
vice versa. Mathematically, the Heisenberg principle can be expressed in terms of an
inequality.
....(3.11)
.. ..
Where Ax and Ap are the uncertainities in the measurements ..of position and momentum
respectively. If the position of an object is known exactly (i.e., Ax = 0), then the uncertainly
in the momentum must be infinite, meaning that we cannot say anything about the velocity.
Similarly, if the velocity is known exactly, then the position would be entirely uncertain
and the particle could be anywhere. It means that we cannot say anything about the position
of the particle. In actual practice none of the two properties can be measured with certainly.
Due to the small value of the Planck’s constant, h (6.626 x 10-34 J s) this principle is not
relevant while making measurements of large objects like car, bus or aeroplane etc. It is
relevant, only when you are making measurements on very small objects such as electrons.
Heisenberg’s principle questioned the validity of Bohr’s model. It is so because according
to Bohr’s model we can precisely calculate the radius of the orbit (i.e., the position of the
electron) and the velocity of electron in it. But it is not possible according to Heisenberg’s
principle. It motivated many scientists to develop newer models of the atom using the dual
nature of the electron. This resulted into the development of a Quantum mechanical or
Wave Mechanical Model of the atom discussed in the next section.
-53-
1. What do you understand by wave-particle duality ?
...................................................................................................................................
2. Name the experiment that established the wave nature of electron.
...................................................................................................................................
3. Compute the de-Broglie wavelength associated with an electron moving which a
velocity of 100 km / second? (me = 9.1 x 10-31 kg)
...................................................................................................................................
4. State Heisenberg’s Uncertainty Principle ?
...................................................................................................................................
3.8 Wave Mechanical Model of atom

Wave Mechanical Model of atom was proposed by Erwin Schrodinger - an Austrian physicist
in 1926. This model is basically a formalism or a mathematical recipe, which is based on
some postulates that have no foundation in classical physics. The correctness of these
postulates can be justified in terms of the correctness of the results predicted by them.
According to this Model, the motion of electron inside the atom could be described in terms
of a mathematical function called, wave function, φ (Greek letter, psi). The wave functions
are assumed to contain all the information about the electron and are obtained by solving a
differential equation called Schrodinger wave equation (SWE). The square of the wave
function φ2 is a measure of the probability of finding an electron in a three dimensional
space around the nucleus.
On solving the SWE for hydrogen atom we get a number of wave functions, which are
characterized by three quantum numbers viz.,
• Principal quantum number, n
• Azimuthal quantum number, l
• Magnetic quantum number, ml
These quantum numbers arise in the process of logically solving the wave equation. Every
electron in an atom has a unique (different) set of quantum numbers which help to describe
the three dimensional region where there is maximum probability of finding the electron.
This region is called as atomic orbital or simply orbital.
3.8.1 Significance of Quantum Numbers

The three quantum numbers describe the size, shape, and orientation of the atomic orbitals
-54-
in space. There is an additional quantum number which does not follow from the Schrodinger
3 wave equation but is introduced to account for electron spin. The fourth quantum number
F thus help in designating the electrons present in the atom. Let us understand the significance
of each of these quantum numbers.
Principal quantum number, n
The principal quantum number, n describes the energy level (or principal shell) of the
electron within the atom. n can have only positive non zero integral values (i.e., n =
1,2,3,4.....). This means that in an atom, the electron can have only certain energies. Thus
we may say that n quantizes energy of the electron. The principal quantum number also
determines the mean distance of the electron from the nucleus, i.e., its size. Greater the
value of n farther is the electron from the nucleus.
Each principal shell can accommodate a maximum of 2n2 electrons, i.e.
n=1 number of electrons : 2
n=2 number of electrons : 8
n=3 number of electrons : 18 and so on...
Azimuthal quantum number, l

The azimuthal quantum number, l is related to the geometrical shape of the orbital.
The value of l may be zero or a positive integer less than or equal to n-1 (n is the principal
quantum number), i.e., l = 0,1,2,3....... (n-1). Different values of l corrrespond to different
types of subshells and each subshell contains orbitals of a given shape.
l = O, corresponds to s-subshell and contains the orbital with spherical shape called as s-
orbital.
l = 1, corresponds to d-subshell and contains the orbitals with a dumb-bell shape called as
as p-orbitals. Therer are three p-orbitals in each p-subshell.
l = 2, corresponds to p-subshell and contains the orbitals with a cloverleaf shape called as
as d-orbitals.
l = 3, corresponds to f-subshell and contain f orbitals. There are seven f-orbitals in each f-
shubshell.
The shapes of s, p and d orbitals will be discussed in the next subsection.
Magnetic quantum number, ml
The quantum number, ml describes the direction or orientation of the orbital in space.
The quantum number ml may have any intergral value from - l to + l. For example, for l = 1;
ml can have the values as -1,0 and 1.
Magnetic quantum number, ms
The quantum number, m s describes the spin of the electron i.e., whether it is clockwise
-55-
or anticlockwise. The quantum number, ms does not arise while solving SWE. The clockwise
and anticlockwise direction of electron spin has arbitrarily been assigned the value as + 1/2
and -1/2 respectively.
To sumup, let us take an example of an electron belonging to the third shell (n=3). This
electron can be in as s-subshell (l = 0) or a p-subshell (l = 1) or a d-subshell (l = 2). If it
happens to be in a p-shell it may be in any of the three possible p-orbitals (corresponding to
m1 = -1, 0 + 1 directed along x, y or z - axis. And within the orbital it may have clockwise
(ms = +1/2) or anti-clockwise (ms = -1/2) direction of electron spin. The possible values of
different quantum numbers for an electron belonging to the third shell are given in
Table 3.3.
Table 3.3 : The quantum numbers for an electron belonging to the third shell
Principal Azimuthal Magnetic Magnetic spin

quantum quantum quantum quantum
number, n number, l number, ml number, ms
3 0 0 +1/2
-1/2
1 -1 +1/2
-1/2
0 +1/2
-1/2
+1 +1/2
-1/2
2 -2 +1/2
-1/2
-1 +1/2
-1/2
0 +1/2
-1/2
+1 +1/2
-1/2
+2 +1/2
-1/2
You may note here that the third shell can contain a maximum of 18 electrons and each of
them, has a distinct set of four quantum numbers.
-56-
1. What do you understand by a Wave Function ?
...................................................................................................................................
2. What is the difference between an orbit and an orbital ?
...................................................................................................................................
3. What are quantum numbers? List different quantum numbers obtained from
Schrodinger Wave Equation ?
...................................................................................................................................
4. Give the significance of the Principal, azimuthal and magnetic quantum numbers ?
...................................................................................................................................
3.8.2 Shapes of Orbitals

We have defined an orbital as “the three dimensional region of space around the nucleus
where there is maximum probability of finding the electron”. Lets us try to understand the
meaning of an orbital by taking the example of 1s orbital (n = 1 ; l = 0). This can be understood
in terms of a radial probability curve. Such a curve gives the variation of the probability of
finding the electron as a function of distance from the nucleus. For 1s orbital the radial
probability curve (Fig. 3.14 (a)) shows that the probability of finding the electron in 1s
orbital increase as we move away from the nucleus and reaches a maximum at a certain
distance (=0.0529 nm or 52.9 pm for hydrogen atom) and then decreases as we go further
away from it and at a certain distance it becomes close to zero. The curve shows the radial
probability for a given direction. The
Distance from the nucleus
1s orbital
(a) (b)
Fig. 3.14 : (a) Radial probability curve for 1s orbital (b) Boundary surface diagram for 1s orbital
probability would be same for all possible directions. if we put all such curves together it
would give a spherical distribution of the electron probability. Since the radial probability
-57-
does not become zero at any distance, we cannot specify the size of the sphere. Therefore, the
orbital is represented as a boundary surface diagram, which may be thought as a region of
space, which contains 95% of the probability of finding the electron, as indicated in Fig. 3.14
(b). Thus the 1s orbital is represented as a sphere.
Distance from the nucleus (nm) 2s orbital

(a) (b)
Fig. 3.15 : (a) Radial probability curve for 2s orbital (b) Boundary surface diagram for 2s
orbital
Similarly, the Fig 3.15 (a) gives the radial probability curve for a 2s orbital while the Fig 3.15
(b) shows the boundary surface diagram for the same. You can note two things here. First
you may note that for a 2s orbital the boundary surface diagram is bigger as compared to a 1s
orbital. Secondly, the radial probability curve shows two maxima. The probability initially
increases, reaches a maxima then it decreases and comes close to zero. It increases again and
decreases as we move further away from the nucleus. The region where the probability comes
close to zero (before increasing again) is called a spherical node. There are n-1-1 spherical
nodes in an orbital.
A node is a region in space where the probability of finding the electron is close to zero.
p- orbital: Now when we draw the shape of a p orbital (n = 1; l = 1) we get a shape as shown
in the Fig. 3.16. This picture shows the shape of one of the three possible p orbitals which is
directed towards the z-axis; Pz. You may note that the probability picture for a Pz orbital
consists of two lobes; one along the positive z-axis and the other along the negative z-axis.
Another important feature of a p-orbital is the absence of the electron probability in the XY-
plane. Such a plane is called a nodal plane. The shapes of the three p-orbitals are given in
Fig. 3.17
Fig. 3.16 : A p-orbital Fig. 3.17 : The boundary showing a nodal

surface diagrams (shapes) plane of the p-orbitals
-58-
The Fig. 3.18 gives shapes of five possible d-orbitals. The d-orbitals also contain nodal planes.
The five d-orbitals have different shapes but they all have same energies i.e., these are
degenerate.
Fig. 3.18 : The boundary surface diagrams (shapes) of the five d-orbitals

1. What are the shapes of s, p and d orbitals ?
...................................................................................................................................
2. Describe the shape of a 2s orbital. How is it different from that of a 1s orbital ?
...................................................................................................................................
3. What do you understand by
i) a spherical node ?
ii) a nodal plane ?
...................................................................................................................................
4. How many spherical nodes will be there in 3s orbtial ?
...................................................................................................................................
3.8 Electronic Configuration of Elements

You have so far learn that an atom consists of a positively charged nucleus surrounded by
electrons present in orbitals of different shapes and sizes. These orbitals are part of different
shells and sub-shells and are characterized by the three quantum numbers viz. n, l and ml.
Let us now take up the distribution of electrons in these shells and sub-shells. Such a
distribution of electrons is called Electronic Configuration and is governed by three basic
rules or principles.
-59-
3.9.1 Aufbau (or building up) Principle
This principle is concerned with the energy of the atom and states that the electrons should
occupy. The electrons occupy the orbitals in such a way that the energy of atom is minimum.
In other words the electrons in an atom are filled in the increasing order of their energies.
Now, how does one know the increasing order of the orbital energies? You have learnt
above that the principal quantum number determines the energy of the orbitals. Higher the
value of n higher the energy. This is true only for hydrogen atom. For other atoms, we need
to consider both n and l. This means that different sub-shells in a given shell have different
energies. The order of orbital energies can be determined by the following (n + l) rules.
Rule 1 : An orbital with a lower value for (n + l) has lower energy. For example, the 4s
orbital (n + l = 4 + 0 = 4) will fill lbefore a 3d orbital (n + l = 3 + 2 = 5).
Rule 2 : If the value of (n + l) is same for two orbitals then the orbital with lower value of
n will be filled first. For example, the 3d orbital (n + l = 3 + 2 = 5) will fill before a 4p
orbital (n + l = 4 + 1 = 5).
Following these rules the increasing order of the orbital energies comes out to be
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
3.9.2 Pauli’s Exclusion Principle

This principle concerns the spin of electrons present in an orbital. According to the Pauli’s
principle, no two electrons can have all the four quantum numbers to be same. For
example, if a given electron in an atom has the set of four quantum numbers as n = 2, l = 1,
ml = 1 and ms = +1/2 then no other electron in the atom can have the same set of quantum
numbers.
As you know that an orbital is characterized by three quantum numbers so the electrons
occupying a given orbital would have same values of these three quantum numbers. These
electrons are distinguished in terms of their spin quantum number, ms. Since the spin quantum
number can have only two values so only two electrons can occupy a given orbital. In fact
this fourth quantum number was introduced through Pauli’s principle only.
3.9.3 Hund’ Rule
This rule concerns the distribution of electrons in a set of orbitals of the same energy, i.e.
constituents of a subshell. According to this rule if a number of orbitals of the same sub-
shell are available then the electrons distribute in such a way that each orbital is first singly
occupied with same spin. For example, the six electrons in carbon distribute as
1s2 2s2 2p1x 2p1y 2p0z and not as 1s2 2s2 2p2x 2p0y 2p0z
Since electrons repel each other, they remain as far as possible from one another by occupying
different orbitals.
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The rules discussed above can be used to write the electronic configuration of different elements.
There are two common ways of representing the electronic configurations. These are
a) Orbital notation method : In this method the filled orbitals are written in the order of
increasing energies. The respective electrons in them are indicated as superscripts as shown
in the example given below. For example, the electronic configuration of nitrogen atom
(atomic number 7) is written as 1s2 2s2 2p1x 2p1y 2p1z.
b) Orbital diagram method : In this method the filled orbitals are represented by circles or
boxes and are written in the order of increasing energies. The respective electrons are
indicated as arrows whose direction represents their spin. For example, the electronic
configuration of nitrogen in the orbital diagram notation can be written as
1s 2s 2p
↑↓ ↓↑ ↑↑↑
Electonic configuration can also be written in a short hand form. In this method the last
completed orbital shell is represented in terms of a noble gas. For example, the electronic
configuration of lithium and sodium can be written as
Li [He]2s1
Na [Ne]2s1
The electrons in the noble gas configuration are termed as core electrons while the ones in
the outer shell are called valence electrons.

1. What do you understand bythe electronic configuration of an atom ?
...................................................................................................................................
2. What is Pauli’s exclusion principle ?
...................................................................................................................................
3. What is Aufbau principle ? What are (n + l) rules ?
...................................................................................................................................
4. Which of the following orbitals wil be filled first ?
i) 2p or 3s
ii) 3d or 4s
...................................................................................................................................
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• Atoms are made up of three fundamental particles namely, electrons, protons and
neutrons.
• J.J. Thomson made the first attempt to describe the structure of an atom in terms of a
model called plum pudding model. According to this atoms can be considered as a
large positively charged body (pudding) in which a number of small negatively charged
electrons (plums) are scattered..
• According to the Rutherford’s model, the positive charge of the atom and most of its
mass is contained in the nucleus and the rest of the atom is empty space which contains
the much smaller and negatively charged electrons.
• Electromagnetic radiation is a kind of energy, which is transmitted through space in
the form of electric and magnetic fields. It travels with the speed of light and does not
need any medium to travel.
• The electromagnetic radiations are characterized by a number of parameters like,
amplitude, wavelength, frequency, wave number and velocity.
• In 1913, Niels Bohr proposed ‘Planetary Model for atom. According to the model the
electrons move in definite circular paths of fixed energy around a central stationary
nucleus. The electrons can change their orbits by absorbing or emitting a photon of
energy (= hl) equal to the difference of the energies of the orbits.
• Bohr’s model did explain for the stability of atom and the line spectrum of hydrogen.
The model however was unable to explain the spectra of atoms other than hydrogen.
• Louis de Broglie, argued for the dual nature of electron and proposed that matter
particles should have a wave nature. The associated wave length, is given by the
formula;
• This was experimentally verified by Thomson and Davisson by diffraction of electron

waves by the crystal lattice of nickel.
• The wave-particle duality of matter led Werner Heisenberg to propose the uncertainty
principle. According to which it is not possible to measure simultaneously both the
position and memontum of a particle with a infinite precision.
• The dual nature of electron and Heisenberg’s uncertainy principle led to the
development of wave mechanical model.
• According tot he Wave Mechanical Model, the motion of electron inside the atom can
be described in terms of a mathematical function called, wave function, φ. This wave
function contains all the information about the system and can be found by solving a
wave equation called Schrodinger wave equation.
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• The square of the wave function, ψ2 is a measure of the probability of finding the electron
in a certain three dimensional space around the nucleus. This region is called as atomic
orbital or simply orbital.
• These wavefunctions are characterized by three quantum numbers. These quantum
numbers describe the size, shape and orientation of the atomic orbitals in space. Every
electron in an atom has a unique set of quantum numbers.
• The principal quantum number n concerns the quantisation of the energy of the electron
while the Azimuthal quantum number, l is related to the shape of the orbital. The
magnetic quantum number m1 describes the direction or orientation of the orbital in
space.
• An additional quantum number, ms is introduced to account for electron spin. This
quantum number does not follow from the wave mechanical model and describes the
spin of the electron.
• Different orbitals have different shapes. An s orbital is spherical; p orbitals are dumb-
bell shaped; d orbitals have cloverleaf shape while f orbitals have a eight lobed shape.
• The distribution of elections in the shells and subshells is called Electronic
Configuration. If is governed by three rules which are Aufbau principle; Pauli’s
exclusion principle and Hund’s Rule of maximum multiplicity.
• According to Aufbau principle the electrons in an atom are filled in the increasing
order of their energies which is determined by (n + 1) rules.
• According to the Pauli’s exclusion principle, no two electrons can have all the four
quantum numbers to be same.
• While filling electrons in the orbitals of same subshell, according to Hund’s rule,
each orbital is first singly occupied with same spin then the pairing up takes place.
• The photons travel with a velocity equal to the velocity of light. The energy of radiation
(E) is proportional to its frequency (ν)
E α ν ; E = hν
h is a Planck’s constant
h = 6.6256 x 10-34 JS
• The phenomenon of ejection of electrons from the metal surface when light is exposed on
it is called photo electric effect.
Terminal Exercise
1. a) What are the three fundamental particles that constitute an atom ?
b) Compare the charge and mass of an electron and of a proton.
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2. What do you think is the most significant contribution of Rutherford to the development
of atomic structure ?
3. What experimental evidence shows the dual nature of light?
a) Compute the energy of FM radio signal transmitted at a frequency of 100 MHz.
b) What is the energy of a wave of red light with λ = 670 nm?
4. In what way was the Bohr’s model better than the Rutherford’s model ?
5. What are the drawbacks of Bohr’s Model ?
6. What led to the development of Wave Mechancial Model of the atom ?
7. What do you understand by an orbital ? Draw the shapes of s and p-orbitals.
8. Explain the Hund’s rule of maximum multiplicity with the help of an example.
9. Explain Planck’s Quantum theory.
10. The threshold energy (work function) of a metal is 4.2 ev. If radiation of 2000 A0 falls on
the metal, what is the kinetic energy of the fastest photoelectron?

3.1
1. Proton is heavier than electron. The ratio of their masses is
= 1836
2. Main constituent particles like proton, neutron and electron etc. present in the atom
come in the category of fundamental particles.
3. Neutron
3.2
1. Electron, proton and neutron
2. The aim of Rutherford’s experiment was to test the Thomson’s plum-pudding model.
3. According to Rutherford’s model for atom, the positive charge of the atom and most
of its mass is contained in the nucleus. The rest of the atom is empty space which
contains the much smaller and negatively charged electrons.
4. Rutherford’s model was rejected because it could not explain the stability of the atom.
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3.3
1. Electromagnetic radiation is a kind of energy which is transmitted through space in the
form of electric and magnetic fields. It travels with the speed of light and does not need
any medium to travel.
2. The different characteristics of electromagnetic radiation are
i) Ampliture
ii) Wavelength,
iii) Frequency,
iv) Wave number and
v) Velocity
3. The wave number is defined as the number of waves per centimeter. It is equal to the
reciprocal of the wavelength.
4. A quantum of visible light is called photon. The energy of the quantum (or photon) is
proportional to the frequency of radiation.
3.4
1. A line spectrum consists of a series of discrete lines of characteristic wavelengths
while a continuous spectrum contains a broad band of radiations containing all possible
wavelengths in the range i.e., wavelengths of the radiation varies continuously.
2. The main postulates of Bohr’s model are
i) The electrons move in a definite circular paths called as stationary orbits or
stationary state around a central stationary nucleus.
ii) The electrons can change their orbits by absorbing or emitting a photon of energy
(= hl) equal to the difference of the energies of the orbits.
iii) The angular momentum of the electron is quantised.
3. The energy of a Bohr’s orbit increases with an increase in the value of the principal
quantum number, n. In fact it becomes lesser and lesser negative.
4. The smallest packet of energy is called quantum.
5. 6.625 x 10-34 Joules – second
6. When light is exposed to clean metallic surface, electrons are ejected from the surface.
This effect is called as photo electric effect.
7. The minimum energy required for emission of photo electrons is called threshold energy.
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3.5
1. The wave-particle duality refers to the fact that light and the material particles like
electrons could sometimes behave as a particle and as a wave at other times.
2. The wave nature of electron was established by the diffraction of electron waves by
the crystal lattice of nickel.
3. Mass of the electron = 9.1 x 10-31 kg
Speed or Velocity = 100 km s-1 = 105 m s-1
Using equation, the wavelength associated with the electron will be
h 6.626 x10 −34 Js

λ= = −31 −1
= 7.28x10 −9 m
mv (9.1x10 kg )(105ms )
4. According to Heisenberg’s Uncertainty Principle it is not possible to measure both

the position and momentum of a particle with any degree of certainity. More accurately
we measure a particle’s position, the less accurately we are able to measure it’s
momentum and vice versa.
3.6
1. It is a mathematical fucntion that describes the motion of an electron inside the atom.
It contains all the information about the system and can be found by solving a wave
equation called Schrodinger wave equation.
2. An orbit refers to definite cirular paths of fixed energy around a central stationary
nucleus while an orbital refers to the three dimensional region of space around the
nucleus where there is a probability of finding the electron.
3. The quantum numbers are integers that characterize the wavefunctions. These are
obtained in the process of solving Schrodinger wave equation and every electron in an
atom has a different set of quantum numbers. The three quantum numbers obtained
from Schrodinger Wave Equation are
i. The principal quantum number, n
ii. Azimuthal quantum number, l and
iii. The magnetic quantum number m1
4. The principal quantum number, n is concerned with the energy of the electron in a
shell. The quantum number l is related to the geometrical shape of the orbital and the
quantum, number, ml describes the orientation of the orbital in space.
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3.7
1. s-orbital : spherical;
p-orbitals : dumb-bell shaped ;
d-orbitals : cloverleaf shaped.
2. The 2s orbital is spherical in shape similar to the 1s orbital. However there are two
differences. Firstly, the size of a 2s orbital is bigger as compared to a 1s orbital and
secondly, it contains a spherical node.
3. i) It is a spherical region of zero probability in an s orbital (other than 1s).
ii) It is a planar region in an orbital (other than s orbitals) where the probability or
finding the electron is zero.
4. The 3s orbtial will have two spherical nodes.
3.8
1. The distribution of electrons in the shells and subshells of an atom is called Electronic
Configuration.
2. Pauli’s principle states that an atom no two electrons can have same set of the four
quantum numbers.
3. Aufbau principle states that the electrons in an atom are filled in the increasing order
of their energies which is determined by (n + l) rules.
There are two (n + l) rules. These are
An orbital with a lower value for (n + l) is filled first.
If the value of (n + l) is the same for two orbitals then the orbital with lower value of
n will be filled first.
4. i) 2p : (n + l) for 2p = 2 + 1 = 3; for 3s (n + l) = 3 + 0 = 3 ; Rule 2
ii) 4s : (n + l) for 4s = 4 + 0 = 4; for 3d (n + l) = 3 + 2 = 5 ; Rule 1
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4
PERIODIC TABLE AND
ATOMIC PROPERTIES
We have seen different heaps of onions & potatoes at vegetable shop. Imagine, they are
lying mixed and you want to buy 1kg of onion. What will happen? You will have to wait for
long to sort that and then weight them. When you possess a variety of material substances,
you have to keep them classified for an easy access and quick use. You cannot afford to mix
clothes with eatables, cosmetics or books. Classification assures you that your eatables are
in the kitchen, books on the study table or rack and your cosmetics are on the dressing table.
Shopeekers, business houses, storekeepers, administrators, managers, information technology
experts and scientists etc. have to keep their materials duly classified.
Chemists faced a similar problem when they were to handle a large number of elements.
The study of their physical and chemical properties and keeping a systematic record of
them had been a great challenge to chemists. Classification of elements finally could be
possible due to pioneering work of a few chemists. In the present lesson we shall discuss
the need, genesis of classification and periodic trends in physical and chemical properties
of elements.
Objectives
• recognise the need for classification of elements;
• recall the earlier attempts on classification of elements;
• define modern periodic law;
• name the elements with atomic number greater than 100 according to IUPAC
nomenclature;
• co-relate the sequence of arrangement of elements in the periodic table with the
electronic configuration of the elements.
• recall the designations of the groups (1-18) in the periodic table;
• locate the classification of elements into s-, p-, d- and f- blocks of the periodic table and
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• explain the basis of periodic variations of
a) atomic size b) ionic size
c) ionization enthalpy d) electron gain enthalpy within a group or a period.
4.1 Early Attempts

Attempts were made to classify elements ever since the discovery of metals or may be even
earlier. J.W. Dobereiner in 1817 discovered that when closely related elements are grouped
in a set of three, the atomic weight of the middle element was almost the arithmetical mean
of the two elements in the group e.g.,
Element Lithium Sodium Potassium
Atomic weight 6.94 22.99 39.10
mean atomic weight -------- 23.02 --------
He called such a group of three elements a triad. He could group only a few elements due to
lack of knowledge of correct atomic weights of the elements at that time.
In 1863, J.A.R. Newlands, developed a system of classification of elements and entitled it
as Law of Octaves. He arranged the elements is such a way that every eighth element had
similar properties, like the notes of music. The law could not apply to a large number of
known elements. However, the law indicated very clearly the recurrence of similar properties
among the arranged elements. Thus the periodicity was visualised for the first time in a
meaningful way.
Periodicity : Re-occurrence of properties after regular intervals.
More signifcant results were obtained when Lother Meyer’s work reflecting the periodicity
was found to be based on physical properties of the elements. He clearly showed that certain
properties showed a periodic trend.
4.2 Mendeleev’s Periodic Table

In 1869, Mendeleev’s a Russain Chemist made a thorough study of the relation between
the atomic weights of the elements and their physical and chemical properties. He then
constructed a table in which elements were arranged in order of their increasing atomic
weights. It was also found that every eighth elements had properties similar to that of the
first element. Thus, there was a periodic occurrence of elements with similer properties.
One of the most striking application of Mendeleev’s classification of elements was that in
his periodic table (Table 4.1) he left gaps for elements which were yet to be discovered. He
also predicted the properties of these elements. However, Mendeleev’s periodic table did
not provide any place for isotopes and noble gases which were discovered later on.
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Table 4.1 Mendeleev’s Table of 1871
Oxide
Hydride
Periods
First
series:
Second
series:
First
series:
Second
series:
First
series:
Second
series:
The extent of knowledge regarding the chemical properties of the elements and his insight
into the system of periodicty possessed by the elements under certain arrangement have no
parallel in the history of chemistry. This work laid strong foundation of the fundamental
principles of the periodic law. One of his most important conclusions was that the elements if
arranged according to their atomic weights, exhibit an evident systematic reoccurence of
properties (periodicity of properties) and even the properties of some elements were listed
much before their discovery. Mendeleev’s periodic Table (Table 4.1) was quite useful till the
discovery of atomic number there existed certain inherent defects which opposed the system.
4.3 Modern Approach

Atomic number was discovered in 1913 by a team lead by Moseley. The periodic table based
on atomic number is termed as Modern Periodic Table. Moseley arranged all the elements
according to increasing atomic number and showed that the properties of elements are periodic
function of their atomic numbers.
Modern periodic law : The properties of the elements are periodic function of their atomic
numbers.
4.4 Long Form of Periodic Table

The arrangement of elements in the long form of periodic table is a perfect matching of
electronic configuration of the elements on one hand and physical and chemical properties of
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the elements on the other. Some important considerations of the modern atomic structure
applied to the clasffication of elements are discussed below :
i) An atom loses electrons from or gains electrons in the outermost shell of an atom during
a chemical reaction.
ii) The sharing of an electron or electrons by an atom with other atom or atoms is largely
through the outer most shell. Thus the electrons in the outermost shell of an atom largely
determine the chemical properties of the elements.
We may therefore conclude that the elements possessing identical outer electronic configuration
should possess similar physical and chemical properties and therefore they should be placed
together for an easy and systematic study.
Keeping in mind the reasoning given above, when all the known elements are arranged in a
table according to their increasing atomic number, the properties of the elements show
periodicity (reappear at definite intervals). The periodicity is shown in Table in 4.2
4.5 Structural Features of the Long Form of Periodic Table

i) In this table there are 18 vertical columns called GROUPS. They are numbered from 1
to 18. Every group has a unique configuration.
ii) There are seven horizontal rows. These rows are called PERIODS. Thus the periodic
table has seven periods, numbered from 1 to 7.
iii) There are a total of 114 elements known to us till today. Of all the known elements 90
are naturally occurring and others are made through nuclear transformations or are
synthesised artificially. Either way they are Manmade Elements, but you will find the
term specifically applied to transuranic elements (elements listed after uranium) only.
iv) First period consists of only two elements (very short period). Second and third periods
consists of only eight elements each (short periods). Fourth and fifth periods consist of
18 elements each (long periods). Sixth period consists of 32 elements (long period).
Seventh period is yet incomplete and more and more elements are likely to be added as
the scientific research advances.
v) There are also nick names given to the groups or a cluster of groups on the basis of the
similarly of their properties, as given below :
Group 1 elements except hydrogen, are called Alkali Metals
Group 2 elements are called Alkaline Earth Metals
Group 3 to 12 elements are called Transition Metals.
Group 16 elements are called Chalcogens
Group 17 elements are called Halogens
Group 18 elements are called Noble Gases.
Apart from what has been said above elements with atomic numbers 58 to 71 are called
Lanthanoids - or Inner Transition elements (First series). Elements from atomic numbers 90
to 103 are called actinoids - Inner Transition elements (Second series). All elements except
transition and inner transition elements are also collectively called Main Group Elements.
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Atomic number (Z)
Name of element
Symbol of element
Atomic weight or mass number (A)
TRANSITION ELEMENTS
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4.6 Position of Metals, Non-Metals and Metalloids
In order to locate the position of metals, non-metals and metalloids in the periodic table, you
may draw a diagonal line joining the elements boron (At. no. 5) with that of tellurium (At. no.
52) and passing through silicon and arsenic. Now we are in a position to make the following
observations.
i) The elements above the diagonal line and to the far right are non-metals (except selenium
which shows slightly metallic character also). The non-metallic character is more marked
the farther an element is from the diagonal line and up.
ii) The elements below the diagonal line and to the left are metals. (Hydrogen is a non-
metal and is an exception) The metallic character is more marked the farther an element
is from the diagonal line and down. All lathanoids and actinoids are metals.
iii) The elements along the diagonal line are metalloids and possess the characteristics of
metals as well as of non-metals. In addition germanium, antimony and selenium also
show the characteristics of metalloids.

1. Classify the elements of group 14, 15 and 16 into metlas, non-metals and metalloids.
...................................................................................................................................
2. Compare the metallic character of aluminium and potassium.
...................................................................................................................................
3. Name the group number for the following types of elements.
i) Alkaline earth metals
ii) Alkali metals
iii) Transition metals
iv) Halogens
v) Noble gases.
...................................................................................................................................
4. Name five man made elements.
...................................................................................................................................
4.7 Catagorisation of Elements into ‘s’, ‘p’, ‘d’, and ‘f’ Blocks
Grouping of elements in the periodic table can be done in another way also, which is more
related to their electronic configuration. Under this categorisation, the location of the
differentiating electron (the last electron) is most important. If, for example, the electron
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has gone to ‘s-subshell’, the elements will fall in ‘s-block’ and if the last electron goes to ‘p-
subshell’, then the element will belong to p-block. Similarly if the defferentiating electron
enters the d-subshells of an atom, then the elements comprising all such atoms will belong to
d-block.
p-Block
d-Block
S-Block
f-Block
Fig. 4.1 : Blockwise categorization of elements
There are minor exceptions in Mn and Zn configurations. You will study more about the
reasons for such exceptions in Lesson 23.
The grouping of elements explained above can be related to the type of elements disussed
earlier.
i) s-block elements : All alkali metals and, alkaline earth metals.
ii) p-block elements : All elements of group no. 13 to group no. 18.
iii) d-block elements : All elements from group no. 3 to group no. 12 except Lanthanoids
and Actinoides.
iv) f-block elements : Lanthanoids (at no 58 to 71) and Actinoids (at no. 90 to 103)
This is shown in Fig. 4.1
Nomenclature of Elements with Atomic Numbers greater than 100
The naming of the new elements was earlier left entirely to its discoverer. The suggested
names were then later ratified by IUPAC. But due to certain disputes that arose over the
original discoverer of some of the elements of atomic numbers greater than 104, the IUPAC
in 1994 appointed a Commission on Nomenclature of Inorganic Chemistry (CNIC). After
consultation with the Commission and the chemists around the world, the IUPAC in 1997
recommended a nomenclature to be followed for naming the new elements with atomic
numbers greater than 103 until their names are fully recognised.
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• The names are derived directly from the atomic number of the element using the following
numerical roots for 0 and number 1-9.
0 = nil 3 = tri 6 = hex 9 = enn
1 = un 4 = quad 7 = sept
2 = bi 5 = pent 8 = oct
• The roots are put together in the order of the digits which make up the atomic number
and ‘ium’ is added at the end.
• Names, thus derives, and the IUPAC approved names of some elements with atomic
numbers greater than 103 are listed in Table 4.3
Table 4.3 Nomenclature of elements with atomic numbers greater than 103
Atomic Name Symbol IUPAC approved IUPAC

Number Name Symbol
104 Unnilquadium Unq Rutherfordium Rf

105 Unnilpentium Unp Dubnium Db
106 Unnilhexium Unh Seaborgium Sg
107 Unnilseptium Uns Bohrium Bh
108 Unniloctium Uno Hassium Hs
109 Unnilennium Une Meitnerium Mt
110 Ununnillium Uun
111 Unununnium Uuu
112 Ununbium Uub
113 Ununtrium Uul
114 Ununquadium Uuq
115 Ununpentium Uup
4.8 Periodicity in Atomic Properties

The term periodicity is used to indicate that some characteristic propeties occur in the periodic
table after definite intervals, however with a varying megnitude. Thus after starting from a
certain point on the periodic table, we are almost certain that the movement in particular
direction will show steady increase or decrease of a said property.
4.9 Atomic Size

In homonuclear diatomic molecules the distance from the centre of one nuclus to the centre
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of another nucleus gives the bond length and half of this bond length is atomic radius. (Fig.
4.2). The first number of each period is the largest in size. Thus we can say that the group 1
atom are the largest in their respective horizontal rows. Similarly, atoms of group 2 elements
are large but are definitely smaller than the corresponding atoms of group 1. This is due to the
reason that the extra charge on the nucleus draws the electrons inward resulting in smaller
size for the atoms under reference. This trend of decrease in size of atoms, continues from left
to right. An example is shown in Fig. 4.3. However there may be some exceptions and there
will be other reasons to explain them.
Fig. 4.2 : Atomic radium = 1/2 d AA

=r
Fig. 4.3 : From left to right, size of atoms decrease in the periodic table
In going down the group of elements (in any particular column) the atomic size increases at
each step.
This increase may be explained in terms of a new electron shell being added, when we pass
from one element to another in a group.
4.9.a Types of atomic radius
The atomic radius cannot be determined exactly. But the inter-nuclear distance of the bonded
atoms can be measured using x-ray difraction or other methods.
The atomic radius depends on many factors like the number of bonds formed by the atom,
nature of bonding, oxidation state etc.
The types of atomic radius are considered based on nature of bonding. They are
1. Crystal radius or Atomic radius
2. Vander waal's radius.
3. Covalent radius .
Crystalline Radius :
This type of radius is applicable for metal atoms.
Half of the distance between the centres of the nuclei of two adjacent metal atoms in the metallic
crystal is called crystal radius or atomic radius.
It is measured in 1A0 = 10-8 cm = 10-10 m = 0.1 m
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1 nm = 10-9 m
Example : The distance between two Na Atoms = 3.72 A0

Crystal radius of Na = 3.72/2 = 1.86 A0
Vander Waal's radius :

This type of radius is used only for molecular substances present in the solid state.
‘Vander Waal's radius is distance between the centres of two atoms of different molecules which are
closest to each other.
These atoms which approach closest to one another do not form chemical bond. But remain together
due to Vander waal's forces of attractions.
For example the distance between two adjacent Chlorine atoms of different molecules is 3.6A0 – Then
the Vander Waal's radius of Chlorine is 3.6/2 = 1.80 A0
Vander Waal forces of attraction are weak so it is found that Vander Waal's radius is 40% more than
half of the distance between two nuclei of chemically bonded atoms.
Covalent radius :
This radius is generally used while referring to non-metals.
It is half of the distance between the nuclei of two atoms held together by a covalent bond.
For Example the inter nuclear distance between Chlorine atoms in a molecule is 1.98A. The covalent
radius of chlorine is one half of the inter nuclear distance.
ie. 1.98/2 = 0.99A0
A C A1
Cl Cl Cl Cl
A B
Vander waal and Covalent radii in chlorine.
Vander waal's radius = AC = A C = r = 1.8A0

Covalent radius = AB = r = 0.99 A0
The inter nuclear distance = AA distance = 3.6A0
vander waal's radius = AA 1/2 = 3.6/2 = 1.80 A0
4.10Intext
IonicQuestions
Size 4.2
1. Give the order of penetration of orbitals toward, nucleus for a given principal quantum number.
..............................................................................................................................................
2. Define shielding effect.
..............................................................................................................................................
3. Define covalent radius of elements.
..............................................................................................................................................
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4.10 Ionic Size
An ion is formed when an atom undergoes a loss or gain of electron.
_
M (g) → M+ (g) + e (cation formation)
_ _
M (g) + e → M (g) (anion formation)
A cation is formed when an atom loses the most loosely bound electron from its outermost
shell. The atom acquires a positive charge and becomes an ion (a cation). A cation is smaller
than its atom. On the removal of an electron, the positive charge of the nucleus acts on
lesser number of electrons than in the neutral atom and thus greater pull is exerted by the
nucleus, resulting in a smaller size of the cation.
An anion is bigger than its atom because on receipt of an electron in the outermost orbit the
number of negative charges increase and it out weights the positive charges. Thus the hold
of the nucleus on the shells decrease resulting in an increase in the size of the anion.
A cation is always smaller than its atom and an anion is always bigger than its atom
_
e.g. Na+ is smaller than Na, Cl is bigger than Cl.
• In the main groups, the ionic radii increase on descending the group e.g., Li+ = 0.76 Ao,
Na+ = 1.02 Ao, K+ = 1.38 Ao, etc. it is due to the addition of extra shell at each step.
• There is a decrease in the ionic radii of the positive ions on moving from left to right
across a period in the periodic table. Na+ = 1.02 Ao, Mg2+ = 0.72 Ao, Al3+ = 0.535 Ao,
etc. It is due to the increase in the number of charges on the nucles and also due to the
increase in the charge on the ion.
• The ionic radii of the negative ions, also decrease on moving from left to right across
_ _
a period. e.g. O2 = 1.40 Ao, F = 1.33 Ao, etc. This is partly due to increase in the
number of charges on the nucleus and also due to the decreasing change on the ion.

1. Write the names of the elements with atomic numbers 105, 109, 112, 115 according to
IUPAC nomenclature.
...................................................................................................................................
2. Arrange the following in the order of increasing size.
_ _
Na+, Al3+, O2 , F
...................................................................................................................................
3. How does the size of atoms vary from left to right in a period and on descending a
group in the periodic table ?
...................................................................................................................................
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4.11 Ionization Enthalpy
Ionization Enthalpy is the energy required to remove the most loosely bound electron from an
isolated atom inthe gaseous state for one mole of an element. It is expressed in kJ mol-1 (kilo
joules per mole)
_
M(g) + IE → M+(g) + e
As we move from left to right in the periodic table, there is a nearly regular increase in the
magnitude of the ionization enthalpy of elements.
Similarly, on moving down a group the magnitude of the ionization enthalpy indicates a
regular decline. The ionization enthalpy of the first member of any group is the highest within
that group and the ionization enthalpy of the last member in the same group, is the least. This
is shown an table 4.4
Table 4.4 : First ionization enthalpies of the elements (in kJ mol-1)
The variation in the magnitude of ionization enthalpy of elements in the periodic table is
mainly dependent on the following factors :
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a) The size of the atom
b) The magnitude of the nuclear charge on the atom,
c) The extent of screening
d) The type of orbital involved (s, p, d or f)
• In small atoms, the electrons are tightly held whereas in large atoms the electron are less
strongly held. Thus, the ionization enthalpy decreases as the size of the atom increases.
• When an electron is removed from an atom, the effective nuclear charge, i.e., the ratio
of the number of charges onthe nucleus to the number of electrons, increases. As a
result the remaining electrons come closer to the nucleus and are held more tightly. The
removal of a second electron, therefore, requires more energy. e.g., Mg+ is smaller than
the Mg atom. The remaining elctrons in Mg+ are more tightly held. The second ionisation
enthalpy is, therefore, more than the first ionisation enthalpy.
• Since the orbitals (s, p, d and f) have different shapes, the ionization enthalpy depends
on the types of electrons removed. e.g. an electron in an s orbital is more tighly held as
compared to an electron in a p orbital. It is because an s electron is nearer to the nucleus
as compared to a p electron. Similarly a p-electron is more tightly held than a d-electron,
and a d-electron is more tightly held than a f-electron. If all other factors are equal, the
ionization enthalpies are in the order s > p > d > f.
These factors taken together contribute largely to decid the extent of the force of attraction
between the nucleus and the electrons around it. The resultant of these factos thus determine
the magnitude of ionization enthalpy of any element. You can see the variation in the magnitude
of the ionization enthalpy of elements with atomic number in the Fig. 4.4.
Ionization energy (kJ mol-1
Atomic number
Fig. 4.4 : Variation of ionization enthalpy of elemetns.
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It is clear from Fig. 4.4 that
i) The metals of group 1 (Li, Na, K, Rb, etc.) have the lowest ionization enthalpies in their
respective periods.
ii) The noble gases (He, Ne, Ar, Kr, Ze and Rn) have the highest ionization enthalpies in
their respective periods. It is because the energy required to remove an electron from
a stable fully filled shell is very large.
iii) The values of ionization energies do not increase smoothly. e.g. the first ionization
enthalpy of B (boron) is lower than that of Be (beryllium); the ionization enthalpy of
Al (aluminium) is lower than that of Mg (magnesium); the first ionization enthalpy of
O (oxygen) is lower than that of N (nitrogen). it can be explained as follows.
• The first ionization enthalpies of Be and Mg are higher than those of their preceding
elements because the electrons are removed from the fully filled s-orbitals.
• The first ionization enthalpy of N is higher than that of O because from N, the electron
is to be removed from a half-filled p-orbitals.
Ionization enthalpy is the energy required to remove the most loosely bound electron
form an atom (in the gaseous state) for one mole of an element. It is an absolute value
and can be determined experimentally.
Factors effecting Ionization energy

The magnitude of Ionization potential of an atom depends on the following factors :
1. Atomic radius
2. Nuclear Charge
3. Screening or shielding effect on the outer electrons
4. Extent of penetration of orbitals of valence electrons.
5. The nature of sub shells; Whether half filled or completely filled with electrons
or not.
1. Atomic radius: As the atomic size increases valence electrons move farther from the
nucleus and so are held by weak nuclear attraction. So less energy is enough for the
removal of the outer most electron of an atom. Hence, as the atomic radius increases, the
Ionization energy decreases.
2. Nuclear Charge : The Number of electronic shells remain same, as the nuclear charge
increase, the outer electrons are more strongly held by the nucleus due to attractive
forces As a result, the energy required to remove the most loosely held electron from the
atom is more. Hence IP increases, as the nuclear charge increases.
Element Li Be C N F
Nuclear Charge 3 4 6 7 9
Ionization Potential ev- 5.39 9.32 11.25 14.52 17.42
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3. Screening effect: In an atom having more than one electron, the valance electrons are attracted
by the nucleus and at the same time repelled by the inner core of electrons. The electrons
present in the inner orbits screen the electrons present in the outer most orbit from being attracted
by the nucleus. Hence, the nuclear attraction on the outer most electron decreases. This influence
of the inner core of electrons on the attraction of the nucleus towards outer electrons is referred
screening effect.
1
Screening Effect α
1p
The screening efficiency of orbitals falls off in the order s>p>d>f
4. Extent of penetration of Orbitals of valence electrons: For a given quantum number the
s-orbital penetrates more towards nucleus because of its spherical and symmetrical shape. So
an electron of an s-orbital requires large amount of energy to be removed. In the same shell P-
electrons require less energy to remove. Hence the Ionization energy follows the sequence of
the extent of penetration of the orbitals s>p>d>f.
5. Completely filled or half filled sub shells: Completely filled or half filled electronic sub
energy levels impart greater stability to the atom ie. the element. Such atoms required more
energy for Ionization. Hence, they have more Ionization energy values.
4.12 Electron Gain Enthalpy

Every atom, in general, has a tendency to gain or loose electrons in order to acquire a noble
gas configuration. The atom which have five, six or seven electrons in their outermost shell
show tendency to accept electrons and attain the nearest noble gas configuration. Halogens,
for example, have seven electrons in their outermost orbit. Thus they show a tendency to
accept, one more electron and attain the nearest noble gas configuration. The energy change
(ΔE) for this process is called electron gain enthalpy of that atom.
Electron gain enthalpy is the energy released or adsorbed for one mole of neutral atoms
in a gaseous state when electron is accepted by each atom.
_ _
X(g) + e → X (g)
Where X represents an atom.

→ Cl (g): ΔE = _349 kJ mol-1
_ _
Cl (g) + e
The negative value shows release of energy and hence tendency to greater stabilisation.
The eletron gain enthalpy becomes more in negative from left to right in a period. This is
because it is easier to add an electron to a smaller atom since the added electron would be
closer to the positvely charged nucleus. Halogens release maximum energy when they accept
an electron. On the otehr hand, metals do not accept electrons and show a high positive
value for ΔE. Thus electron enthalpy can be positive or negative.
Electron gain enthalpies becomes less in negative as we go down the group showing that
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the electropositive character of the atoms increases. This is because the size of the atom
increases down the group and the electron added goes to the higher shells. Electron affinity
values for some elements are shown in table 4.5, along with their position in the periodic
table. The electron gain enthalpy of chlorine is more in negative value as compared to that of
fluroine. This is due to the small size of the F atom. As the electron approaches the small F
atom, it expriences a repulsion from other electrons.
Table 4.5 : Electron gain enthalpy in kJ mol-1
4.13 Electronegativity
It is an indicator of the extent of attraction by which electrons of the bond pair are attracte by
an atom linked by this bond. The value of electronegativity is assigned arbitrarily to one atom
such as hydrogen. Then the value of electronegativity is assigned to all other atoms with
respect to hydrogen. One such scale is the Pasuling Scale of electronegativity (Table 4.6).
Electronegativity is defined as a measure of the ability of an atom to attract the electron
pair in a covalent bond to itself.
In a homonuclear diatomic molecule such as hydrogen (H2) or fluorine (F2), the electron pair
of the covalent bond in each molecule experiences equal attraction by each atom. Thus non of
the two atoms is able to shift the bond pair of electrons to itself. However in a heteronuclear
diatomic molecule, the bond pair electrons get shifted towards the atom which is more
electronegative than the other. For example, in HF or HCl the bond pair of electrons are not
shared equally but the more electronegative atom F or Cl is able to shift the bond pair towards
itself, resulting in the polarization of the molecule.
A large difference between electronegativities of the two atoms indicates highly ionic character
_
of the bond between them. For example Cs+F . On the other hand, zero difference in the
electronegativities between the two atoms indicates that the percentage ionic character is zero.
Therefore the molecule is purly covalent e.g. H2, Cl2, N2 etc.
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Table 4.6 : Electronegativities of elements on Pauling Scale.
The most electronegative elements have been placed on the farthest right hand upper corner
(noble gases are not included). The value of electronegativity decreases as we go down in
any group and increases from left to right in the period. Thus fluorine is the most
electronegative and caesium is the least electronegative element. (We have not considered
Francium being radioactive).

1. What is the correlation between atomic size and ionization enthalpy.
...................................................................................................................................
2. Which species, in each pair is expected to hav higher ionization enthalpy.
i) 3
Li, 11Na ii) 7
N, 15P
iii) 20
Ca, 12Mg iv) 13
Al, 14Si
v) 17
Cl, 18Ar vi) 18
Ar and 9K
vii) 13
Al, 14C
...................................................................................................................................
3. Account for the fact that there is a decrease in first ionization enthalpy from Be to B
and Mg to Al.
..................................................................................................................................
4. Why is the ionization enthalpy of the noble gases highest in their respective periods ?
...................................................................................................................................
5. Name the most electronegative element.
...................................................................................................................................
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• The classification of elements makes their systematic.
• The arrangement of elements in the long form of the periodic table depends on their
electronic configuration.
• The properties of the elements are periodic function of their atomic number.
• All the known elements are arranged in 18 groups in the long form of periodic table
• There are seven horizontal rows (periods) in the long form of the periodic table.
• Elements of groups 1 and 2 are known as alkali metals and alkaline earth metals
respectively.
• s, p, d and f are the four blocks in the periodic table classified on the basis of their
outer most electrons residing in s, p, d and f sub-shell.
• The elements can be classififed into metals, non-metals and metalloids on the basis of
their properties and their position in the periodic table.
• The atomic size, ionic size, ionization enthalpy, electron gain enthalpy and
electronegativity show regular trends along a group and a period.
• Various type of atoms radius possible for atoms
• Factors which effect the Ionization potential of different atoms.
Terminal Exercise
1. Define modern periodic law.
2. Refer the periodic table in Table 4.2 and answer the following questions.
i) The elements placed in group number 18 are called ................
ii) Alkali and alkaline earth metals are collectively called ................ block metals.
iii) The general configuration for halogens is ................
iv) Name a p-block elements which is a gas other than a noble gas or a halogen.
v) Name the groups that comprise the ‘s’ block of elements.
vi) Element number 118 has not yet been established, to which block, will it belong?
vii) How many elements should be there in total if all the 7s, 7p, 6d and 5f blocks are
to be full ?
3. Describe the variation of Electron Affinity and Ionization Enthalpy in the periodic
table.
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4. Define the following :
a) Electron gain enthalpy b) Ionization enthalpy
c) Ionic radius d) Electronegativity
5. What is electronegativity ? How is it related to the type of bond formed ?
6. Why is the electron gain enthalpy of Cl more in negative value as compared to that of F ?
7. Name the different types of atomic radius
8. What is shielding effect?
9. Explain 3 factors which influence the Ionization energy.

4.1
1. Metals Non metals Metalliods
Sn, Pb C Si, Ge
Sb, Bi N, P As
Te, Po O, S Se
2. Potassium is mroe mettalic than aluminum.
3. i) 2 ii) 2 iii) 3-12 iv) 17 v) 18
4. Np, Lw, No, Rf, Hs.
4.2
1) s>p>d>f
2) Protection of outer electrons from nuclear attraction by electrons present in inner shells of
an atom.
3) Covalent radius is half of the distance between the nuclei of two atoms held together by a
covalent bond.
4.3
1. i) Unnilpentium
ii) Unnilennium
iii) Ununbium
iv) Unnunpentium
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_ _
2. Al3+, Na+, F , O2
3. The atomic size decreases from left to right across a period and increases on moving
4
F down the group.
i) s>p>d>f
ii) Protection of outer electrons from nuclear attraction by electrons present in inner shells of
an atom.
iii) Covalent radius is half of the distance between the nuclei of two atoms held together by a
covalent bond.
4.4
1. Ionization enthalpy decreases with increase in atomic size and vice-versa.
2. i) 3Li ii) 7N iii) 12Mg iv) 14Si
v) 12Ar vi) 18Ar vii) 6C
3. The electronic configuration of Be is 1s2 2s2 whereas that of B is 1s2 2s2 2p1. In case of
Be, the electron is to be removed from completely filled s orbital whereas in case of B
it is to be removed from a singly occupied p orbital. Fully-filled orbitals are more
stable. Hence, ionization enthalpy decreases from be to B. Similarily it decreases
from Mg to Al.
4. The nole gases have fully filled and are stable. Hence, they have the highest ionization
enthalpies in their respective periods.
5. Fluorine.
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CHEMISTRY
Student's Assignment - 1
Maximum Marks : 50 Time : 2.30 Hrs.
INSTRUCTIONS :
• Answer all the questions on a separate sheet of paper.
• Give the following information on your answer sheet.
- Name
- Enrolment Number
- Subject
- Assignmnet Number
- Address
• Get your assignment checked by the subject teachmer at your study centre so that you
get positive feedback about your performance.
Do not send your assignment to APOSS
1. a) Write down symbols for prefixes micro and mega.

b) Write the empirical formula of C2H2 and C3H8.
c) State charles’ law and given mathematical expression for it.
d) Express 4.6 x 10-10 m in terms of pico metre.
e) What is the SI unit for measuring electric potential.
f) Atomic mass of C-12 is 12. What is the mass of one C-12 atom ?
g) Write down the names of four quantum numbers.
h) State Heisenberg’s uncertainly quantum numbers.
i) How many moles of oxygen atoms are present in 2 moles of CuSO4.5H2O ?
j) Define Orbital. (1x10=10)
2. a) Calculate the mass of one molecule of benzoic acid, C6H5COOH (Atomic masses;
C=12, H = 1, O = 16).
b) Write the empirical formulae of the following C6H6,N2O4,C6H12O6,NH3.
c) What are anode rays ? Give their two important properties.
d) Given values of n, l and m quantum numbers of an electron which is present in
3px orbital
e) Write electronic configuration of chromium (atomic number = 24)
f) Which of the following orbitals is not possible 3p, 4s, 2d, 5f.
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g) Write two differences between atom and ion.
h) What are electromagnetic radiations.
i) A solution of ferric sulphate containing 0.280 g of ferric ions is treated with an
excess of a solution of Ba (OH)2. Calculate the weight of the precipitate formed
if they react as
Fe2 (SO4)3 + 3Ba (OH)2 (aq) → 2Fe (OH)3 (s) + 3BaSO4 (s)
(Atomic Masses : Fe = 56, Ba = 137, O = 16, H = 1, S = 32)
j) Calculate and compare the energies of two radiations having wavelengths
λ1 = 4000A and λ2 = 8000A. (2x10=20)
3. a) Using de Broglie expression, calculate the momentum of a moving particle whose

wave length (λ) is 200 pm. (pm = picometers)
b) Determine the molecular formula of a compound which contains 2.19%H; 12.8%
C and 85.1% Br. 1g of the compound in gaseous state occupies 119 mL volume
at STP. (Atomic masses : C = 12, H = 1, Br = 80)
c) Calculate the wave length and frequency of the spectral line in Lyman series of
hydrogen spectra which has minium frequency.
d) 1.84 g of a mixture of CaCO3 and MgCO3 are heated strongly to decompose (as
given below) till no forther loss of weight occurs.
CaCO3 → CaO + CO2
and MgCO3 → MgO + CO2
The residue weighs 0.96g. Find the percentage composition of mixture (Atomic
masses : Mg = 24, Ca = 40, C = 12, O = 16) (3x4=12)
4. a) What is the uncertainty in the velocity of a moving cricket ball having mass
150g if its uncertainty in position is 1A.
b) A naturally occuring mineral was found to contain 42% MgCO3 and 55% CaCO3
and the rest was in purity. What volume of CO2 measured at 150C and 745 mm
Hg pressure will be evolved by heating 10g of the mineral hydrochloric acid if
the reation occurs as follows:
MCO3 + 2HC → MCl2 + CO 2 + H2O
Where M = mg or Ca (4x2=8)
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5
CHEMICAL BONDING
In lesson 1 you have learnt about the structue of atom while in the lesson 2 you studied
about the classfication of elements and the variation in atomic properites. You know that
molecules are obtained by the combination of two or more than two atoms of the same or
different elements. In this lesson you will study
• Why do atoms combine ?
• What are the different ways in which the atoms can combine ? and
• What are the shapes of different molecules ?
The answers to these questions are of fundamental importance to the study of chemistry, as
you would discover while studying the later parts of this course.
Objectives
• explain the formation of bond in terms of potential energy diagram and octet rule;
• list different types of bonds;
• define ionic bond and cite some examples;
• write Lewis structures of some simple molecules;
• list the charateristics of ionic compounds;
• define covalent bond and cite some examples;
• list the characteristics of covalent compounds;
• state valence shell electron pair repulson (VSEPR) theory;
• predict the geometry of molecules with the help of VSEPR thoery;
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• explain the hybridisation of atomic orbitals involving s, p and d orbitals and illustrate with
examples;
• tabulate the geometry of some molecules showing sp, sp2, sp3, dsp2 and dsp3 hybridisation;
• explain the formation of σ and π bonds in CH4, CH2H4 and C2H2;
• explain molecular orbital theory;
• write the molecular orbital configuration of H2, N2, O2 and F2 molecules;
• define bond length and bond order and relate them and
• explain hydrogen bonding with the help of examples.
5.1 What is a Chemical Bond

When two atoms of same or different elements approach each other, the energy of the
combination of the atoms becomes less than the sum of the energies of the two separate
atoms at a large distance. We say that the two atoms have combined or a bond is formed
between the two. The bond is called a chemcial bond. Thus a chemical bond may be visualised
as an effect that leads to the decrease in the energy. The combination of atoms leads to the
formation of a molecule that has distinct properties different from that of the consituent
atoms.
A question arises, “How do atoms achieve the decrease in energy to form the bond.” The
answer lies in the electronic configuration. As you are aware, the noble gases do not react
with other elements to form compounds. This is due to their stable electronic configuration
with eight electrons (two in case of helium) in their outermost shells. The formation of a
bond between two atoms may be visualised in terms of their acquiring stable electronic
configurations. That is when two atoms (other than that of noble gases) combine they will
do so in such a way that they attain an electronic configuration of the nearest noble gas.
The stable electronic configuration of the noble gases can be achieved in a number of ways;
by losing, gaining or sharing of electrons. Accordingly, there are different types of chemical
bonds, like,
• Ionic or electrovalent bond
• Covalent bond
• Co-ordinate covalent bond
In addition to these we have a special kind of bond called hydrogen bond. Let us discuss
about different types of bonds, their formation and the properties of the compounds so
formed.
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5.2 Ionic or Electrovalent Bond
Accordingt to these we have a special kind of bond called hydrogen bond. Let us discuss
about different types of bonds, their formation and the properties of the compounds so
formed.
The electronic configuration of sodium atom (atomic number 11) is 2, 8, 1. Since it is
highly electropositive, it readily loses an electron to attain the stable configuation of the
nearest noble gas (neon) atom. It becomes a positively charged sodium cation (Na+) in the
process
_
Na → Na+ + e ; ΔH = 493.8 k J mol-1
2, 8,1 2,8 (ΔH is enthalpy change)
On the other hand, a chlorine atom (electronic configuration: 2, 8, 7) requries one electron
to acquire the stable electronic arrangement of an argon atom. It becomes a negatively
_
charged chloride anion (Cl ) in the process.
_ _
Cl + e → Cl ; ΔH = 397.5 k J mol-1
2, 8,7 2,8,8
According to Kossel’s theory, there is a transfer of one electron from sodium atom to chlorine
atom and both the atoms attain noble gas configuration.
_
Na + Cl → Na+ + Cl
2, 8,1 2,8,7 2,8 2,8,8
The positively charged sodium ion and the negatively charged chloride ion are held together
by electrostatic atrractions. The bond so formed is called an electrovalent or an ionic bond.
Thus the ionic bond can be visualised as the electrostatic force of attraction that holds the
cation and anion together. The compounds so formed are termed as ionic or electrovalent
compounds.
Factors favour the Ionic bond formation :

a) Cation formation
1) Lower ionization energy : Lower the ionization energy of an atom, greater is the
case of formation of cation.
The IP of sodium is 519.82 kJ. mol-1 and that of Potassium is 495.57 kj. mol-1 so
K+ ion can readily form than Na+ ion.
2) Large size of the atom : Large atom can easily lose the valence electrons. If the
size is large, the distance between the nucleus and the valence electrons is more
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and therefore nuclear attraction is less. Therefore the electron can be removed
easily from the atom forming cation.
3) Ion with lower change : Small magnitude of the change favours the formation of
ions easily. eg. : the case of ion formation increases in the order Na+ > Mg+2 >
Al+3.
4) Cations with inert gas configuration : Ions pocessing eletronic configuration
similar to zero group elements are more stable than those ions which do not
have such configuration.
e.g. Ca+2 (2,8,8) in more stable than Zn+2 (2,8,18) because Ca+2 has inert gas
configuration.
b) Anion formation :
1) High Electron affinity : If the electron affinity of an element is high its anion can
be easily formed.
_ _ _
e.g. Cl > O2 > N3
2) Smaller size of atom : Smaller the atom lesser is the distance between the nucleus
and the valance orbit. Hence the nuclear attraction on incoming electron is more
So the anion is readily formed.
3) Lower change : Ions with lower change are more readily formed than those with
higher charges. (like O-2 and N-3).
Cl + e
_
→ _ _
Cl > O2 > N3
_
c) Electro Negativity Values :

If the two bonded atoms differ by more than 1.70 in their EN values, the bond between
them is ionic in nature.
5.2.1 Energetics of Ionic Compounds Formation

We have just described the formation of an ionic compound (NaCl) as a result of transfer of
electrons as proposed by Kossel. You may raise a question here that when more energy is
required (ionisation energy) to form a sodium ion from sodium atom, than that released
(electron affinity) in the formation of chloride ion from chlorine atom then how do we say
that the formation of NaCl is accompanied by a decrease in energy ? Your questions is quite
justified but let us assure you that there is no anomaly. Let us look at the whole process
somewhat closely to clarify your doubts.
The formation of NaCl from sodium and chlorine can be broken down into a number of steps
as :
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a) Sublimation of solid sodium to gaseous sodium atoms.
Na (s) → Na(g) ; ΔH = 108.7 kJ mol-1
b) Ionization of gaseous sodium atom to give sodium ion.
_
Na (s) → Na+ (g) + e ; ΔH = 493.8 kJ mol-1
c) Dissociation of gaseous chlorine molecule into chlorine atoms
1
/2 Cl2 (g) → Cl (g) ; ΔH = -120.9 kJ mol-1
d) Conversion of gaseous chlorine atom to chloride ion (addition of electron)
_ _
Cl (g) + e → Cl (g) ; ΔH = -379.5 kJ mol-1
e) Formation of NaCl from sodium and chloride ions. (Crystal or lattice formation).
_ _
Na+ (g) + Cl (g) → Na+ Cl (s) ; ΔH = -754.8 kJ mol-1
The energy released in this step is lattice energy.
The net reaction would be
_
Na(s) + 1/2Cl2 (g) → Na+ Cl (s) ; ΔH = -4109. kJ mol-1
The overall energy change can be computed by taking the sum of all the energy changes;
ΔH = (180.7 + 493.8 + 120.9 - 379.5 - 754.8) = -410.9 kJ mol-1
Thus we see that the net process of formation of NaCl from sodium and chlorine is accompanied
by a large decrease in the energy. The approach we have just followed is based on the law of
conservation of energy and is known as Born-Haber cycle.
Of the five different types of energies involved, two (sublimation and dissociation energies)
are generally have low values than the rest. Therefore, the three energy terms i.e., ionization
energy, electron affinity and lattice energy are important in determining the formation of an
ionic configuration the basis of the above discussion we can say that the formation of an nion
compound formed by
i. Low ionisation energy of the metal,
ii. High electron affinity of the other element (non-metal), and
iii. High lattice energy
5.2.2 Characteristic Properties of Ionic Compounds

• These exist as crystalline solids in which the ions are arranged in regular three dimensional
structure. The ionic compounds are generally hard and brittle in nature.
• These compounds have high melting and boiling points due to strong electrostatic
interactions between the ions.
• These are generally soluble in water and less soluble in non-polar solvents like ether,
alcohol, etc.
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• These conduct electricity when in molten state or in aqueous solutions.
Kossel’s theory explain bonding quite well but only for a small class of solids composed of
electropositive elements of Group 1 and 2 with highly eletronegative elements. Secondly, this
theory is incapable of explaining the formation of compounds like, SO2 or O2, etc. For example
in case of O2, there is no reason to expect that one atom of oxygen would lose two electrons
while the other accepts them. The problem was solved by Lewis theory of covalnet bonding.
Properties of ionic compounds :
1) Physical State : Due to close packing of ions, ionic compounds are crystalline solids.
2) Melting and Boiling points : In ionic crystals the oppositely charged ions are bound
by strong electrostatic force of attraction. To overcome these attractive force between
ions, more thermal energy is required. Hence the melting and boiling points of ionic
compounds are high.
3) Solubility : Ionic compounds are soluble in polar solvents like water, liquid ammonia
etc. But are insoluble in non-polar solvents like benzene, carbondisulphide etc.
4) Reactivity : Reaction between ionic compounds in aqueous solution are very fast due
to strong attraction among ions.
e.g. When AgNO3 solution is added to NaCl solution, a white precipitate of AgCl is
formed.
+ _ + _ + _
Ag+NO3+NaCl AgCl + NaNO3
5) Isomerism : Ionic bond is non directional so ionic compounds cannot exhibit isomersim.
6) Electrical, conductivity : Ionic substances conduct electricity in molten state and in
aqueous solution. The ionic compounds are, therefore, electrolytes.
5.3 Covalent Bond

Like Kossel, Lewis also assumed that atoms attain noble gas electronic configuration in the
process of bond formation. However, the way the noble gas electronic configuration is
achieved, is different. Lewis proposed that this is achieved by “sharing of pair of electrons”
between the two atoms. Both the atoms contribute an electron each to this pair. For example,
two hydrogen atoms form a molecule by sharing a pair of electrons. If electrons are indicated
as dots, formation of hydrogen molecule can be shows as
H. + .H → H : H → H _ H
This shared pair of electrons contributes towards the stability of both the atoms and is said to
be responsible for ‘bonding’ between the two atoms. Such a bond is called covalent bond
and the compounds so obtained are called covalent compounds. In the process of suggesting
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the process of chemical bonding Lewis provided a very convenient way of representing
bonding in simple molecues. This is called Lewis electron-dot structures or simply
Lewis structures.
In Lewis structure each element is represented by a Lewis symbol. This symbol consists of
the normal chemical symbol of the element surrounded by number of dots representing the
eletrons in the valence shell. Since the electrons are represented by dots, these are called
electron-dot structures. The Lewis symbols of some elements are as :
. . . . . . . .
. Li ; . Be . ; . B . ; . C. ; N. . ; O. . .
. .
. .
. .
; ;
.F. Ne
. .
. .
. . . .
You may note here that while writing the Lewis symbols, single dots are placed first on
each side of the chemical symbol then they are paired up. The Lewis structure of a molecule
is written in terms of these symbols.
In terms of Lewis symbols the ionic bond formation in NaCl can be represented as
_
: :
: :
Na. + . Cl : [Na]+ [ : Cl : ]
and the covalent bond formation in HCl is represented as

: :
: :
H. + . F : H :F:
Sometimes the electrons contributed by different atoms are represented by different symbols.
For example, formation of HF may also be shown as
:
Hx + . F : Hx. F :
:
In this case the hydrogen electron is shown as a cross while the electrons of fluorine are
represented by dots. There is no differece between electrons; it is just a presentation for the
sake of convenience.
In terms of Lewis structures the formation of a chlorine molecule from two chlorine atoms
may be represented as
: Cl _ Cl:
: :
: :
: :
: :
: :
: :
:Cl • + • Cl : :Cl : Cl:
Here each chlorine atom with seven valence electrons, contributes one electron to the shared
pair. In the process of bond formation both the chlorine atoms acquire the electronic
configuration of argon. In the same way, the formation of oxygen molecule involves sharing
of two pairs of electrons between the two oxygen atoms. In this case both the atoms contribute
two electrons each and acquire eight electrons or an octet in their valence shell.
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:
: :
: :
: :
: :
k
k
: O • + • O: O :: O O= O
.
You may have noticed that in the process of bond formation the elements of second period
acquire eight electrons in their valence shell.. This is called ‘Octet rule’. You may also
note that in case of H2 and Cl2 the atoms are linked by a single line while in case of O2 the
atoms are linked by two lines. These lines represent bonds. When two atoms are bound by
sharing a single pair of electron, they are said be joined by a single bond. And when, two
pairs of electrons are shared (as in case of O2), the two atoms are said to be bound by a
double bond. In nitrogen (N2) the two atoms are joined by a triple bond as they share three
pairs of electrons.
In a Lewis representation the electrons shown to be involved in the bond formation are
called bonding Electrons the pair of electrons is called ‘bond pair’ and the pairs of electrons
not involved in the bonding process are called ‘lone pairs’. The nature of the electron pair
plays an important role in determining the shapes of the molecules. This aspect is discussed
later in Section 5.4.
Valency Bond Theory

VSEPR theory gives the geometry of simple molecules but it does not explain them. Also it
has limited applications. To overcome these limitations, the Valence Bond Theory (VBT)
and the Molecular Orbital Theory (MOT) are introduced. Both the theories are based on
quantum mechanical principles.
Many attempts were made to apply this theory for the formation of covalent bonds. One such
theory, valence bond theory (VBT) was postulated by Heitler and London. This was extended
later by pauling and slater to explain the shapes of the molecules as well as the directions of
the bonds in them. The important postulates of this theory are as follows :
1) A covalent bond is formed by the overlap of two atomic orbitals.
2) The overlaping orbitals contain unpaired electrons of opposite spins.
3) Each of the bonded atoms retains its own atomic orbitals. But the electron pair in the
overlapping orbitals is shared by both atoms.
4) Greater the extent of over of orbitals, stronger is the bond formed.
5) As the atomic orbitals are directional in nature (except ‘s’ orbitals) the bonds that result
due to the overlap of orbitals are also directional. This gives the definite geometry to the
covalent molecule.
6) The increased electron density due to the overlap of atomic orbitals of the two combining
atoms, is along the internuclear axis and keeps two atoms attracted to each other. This
gives stability to the molecule.
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Examples :
i) Formation of H2 molecules :
Hydrogen molecule is formed due to overlapping of s-s orbitals. When two hydrogen
atoms come together, 1s orbitals of the Hydrogen atoms overlap to form a strong “σ”
bond this is σ s-s.
Formation of H2
Formation of Cl2 molecule :
The electronic configuration of chlorine atom is 1s2 2s2 2p6 3s2 3p2x 3p2y 3p1z. It has one
half filled 3pz orbital. The Pz orbital of one chlorine atom overlaps the Pz orbital of the
chlorine atom and the two electrons of opposite spins pair up to form covalent bond. As
the overlap along the internuclear axis is maximum a strong bond is formed. The bond
is formed due to σ P-P overlap.
Formation of Cl2
Formation of O2 molecule :
Th electronic configuration of oxygen atom is 1s2 2s2 2p2x 2p1y 2p1z. It has two half
filled 2p orbitals i.e., 2py and 2pz .
The Py orbital of one atom overlaps the Py orbital of the second atom to form a ‘σ’ bond
σ P y - P y.
The Pz orbital in the two atoms will be at right angles to the internuclear axis. These
two have lateral overlap. The electron density of the bonded pair is distributed in two
banana like regions lying on either side of the internuclear axis. Thus the oxygen
molecule has a double bond. The molecule has one σ P-P and one π P-P between the
two atoms.
Py
x
π
x Px + y Py y
σ
y z z
z
Formation of Oxygen O2
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Formation of HCl molecule :
The electronic configuration of chlorine atom is 1s2 2s2 2p 6 3s2 3px2 3py2 3pz1. It has one
half filled 3pz orbital. The 3Pz orbital of chlorine overlaps 1s orbital of hydrogen atom
and two electrons of opposite spins pair up to form covalant bond. As overlap along
the inter nuclear axis is maximum a strong bond is formed. The bond is formed due to
σ s-p overlap.
HCl formation
5.3.1 Polar Covalent Bond

In a chemical bond the shared electron pair is attracted by the nuclei of both the atoms. When
we write the electron dot formula for a given molecule the shared electron pair is generally
shown in the middle of the two atoms indicating that the two atoms attract it equally. However,
actually different kinds of atoms exert different degrees of attraction on the shared pair of
electrons. A more electronegative atom has greater attraction for the shared pair of electrons
in a molecule. As a consequence in most cases the sharing is not equal and the shared electron
pair lies more towards the atom with a higher electronegativity. For example, in HCl, the
shared pair of electron is attracted more toward more electronegative chlorine atom. As a
result of this unequal sharing of the electron pair, the bond acquires polarity or partial ionic
character.
_
: :
H : Cl : Hδ+ Clδ
In an extreme case, the difference in the electronegativity may be so high that the electron
pair is practically under the influence of a single atom. In other words the polarization of the
bond is complete i.e., we have a case of ionic bonding. Thus, though the Lewis theory talks
about covalent bonding it can account for the formation of ionic compounds also.
5.3.2 Coordinate Covalent Bond

You have learnt that in the formation of a covalent bond between the atoms, each atom
contributes one electron to the shared electron pair, However, in some cases both the electrons
of the shared pair are contributed by only one species (atom, molecule or ion) A common
example is the formation of a bond between boron trifluoride (BF3) and ammonia (NH3). BF 3
is an electron deficient molecule and can accept a pair of electrons. The molecule of ammonia
on the other hand is electron rich. It has a lone pair of electron on the nitrogen atom and that
can be donated. Electron rich ammonia donates a pair of electron to electron deficient BF3.
Such electron donor-acceptor bonds are called coordinate covalent or dative bonds.
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OR
A coordinate bond is normally represented by an arrow pointing from a donor atom to the
acceptor atom. A coordinate bond is idential to a covalent bond in terms of its polairty and
strength. The two are different only in the way they are formed. We cannot distinguish between
covalent and coordinate covalent bond, once these are formed. HNO3 and NH4+ ion are some
more common examples of formation of a coordinate bond.
1. Define, electrovalent bond.
...................................................................................................................................
2. Show the formation of a nitrogen molecule from two nitrogen atoms in terms of Lewis
theory.
...................................................................................................................................
3. What do you understand by a polar covalent bond? Give two example.
...................................................................................................................................
4. What is a coordinate covalent bond ? How is it different from a covalent bond ?
...................................................................................................................................
5. What are the factors which favour the formation of cation.
...................................................................................................................................
6. Ionic compounds do not exhibit isomerism why ?
...................................................................................................................................
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5.3.3 Characteristic Properties of Covalent Compounds
• The covalent compounds have low melting and boiling points due to weak forces of
interaction between the molecules.
• The covalent compounds are poor conductors of electricity as these lack ionic species.
• The covalent compounds are generally insoluble in water and dissolve in nonpolar
solvents like bebzene, carbon tetrachloride etc.
5.3.4 Hydrogen Bonding

It is a special type of attraction between a hydrogen atom bonded to a strongly
electronegative atom (like nitrogen, oxygen or fluorine) and the unshared pair of electrons
on another electronegative atom. Hydrogen bond is a weak bond, the strength being
just about 4-25 kJ mol -1 . It is quite small as compared to the covalent bond, which needs
a few hundreds of kJ mol -1 of energy to break. However, it is strong enough to be
responsible for the high boiling points of H 2O and HF etc. In fact it is due to hydrogen
bonding only that water exists as a liquid. The low density of ice also can be explained
in terms of hydrogen bonding.
Due to the difference in the electronegativity between hydrogen and the other electronegative
atom, the bond connecting them becomes polar. The hydrogen atom acquires a positive
charge while the electronegative atom bears the negative charge. Hyrogen bonding results
from the electrostatic interaction between the positively charged hydrogen atom and the
negatively charged electronegative atom. The second electronegative atom may be a part of
the same molecule or it may belong to a different molecule. Accordingly, there are two
types of hydrogen bonds. If the hydrogen bond is formed between two different molecules
it is called intermolecular hydrogen bond. When the hydrogen bond exists within the same
molecule, it is called intramolecular hydrogen bonding. Salicyldehyde and o-nitrophenol
are two common examples of the moelcules showing intramolecular hydrogen bonding
whereas in water, intermolecular hydrogen bonding exists.
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Hydrogen bonding plays an important role in the structure and function of many biomolecule
like proteins and nucleic acids.
Effects of Hydrogen bond on some properties with example :

Ring formation or chelaton through intramolecular H-bonding results in greater volatility lower
boiling points and lower solubility in water. These properties can be utilized in the separation
of the compounds with intramolecular hydrogen bonding.
The solubility of lower alcohols (ex. CH3OH ; C2H5OH etc.) is due to H-bonding between
molecules of alcohols and the molecules of water.
Intermolecular H-bonding has striking effect on the physical properties like metling points,
boiling points, enthalpies of vapourization and sublimation.
Comparision of boiling points of NH 3 and HCl : NH3 has higher boiling points has HCl
even though nitrogen and chlorine have nearly same electronegatie values. NH3 formsH-
bonds while HCl does not. This is because of smaller size of Nitrogen atom compared to the
chlorine atom.
H H
δ+ δ _ +δ δ _ δ+ δ _ δ+
............. H N H ............. N H ............. N H .............
H H H
NH3 as associated molecule
Comparision of boiling points of H 2O and HF :

The boiling point of H2O is higher than that of HF even though both the compounds form
associated molecules through intermolecular hydrogen bonding. They differ in the number
of hydrogen atoms. The number of hydrogen bonds are more in case of water due to the
presence of two hydrogen atoms. But HF exists as (HF)6 even in vapour state while water
molecules exist as simple H2O molecules in vapour state.
It means energy is not utilized to break the H-bonding in HF. But H-bonds in H 2O consume
energy for their cleave. Hence, the boiling point of water is higher than that of hydrogen
flouride.
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δ+ δ_ δ+ δ_ δ+ δ_
............. H F ............. H F ............. H F .............
HF as associated molecule
δ+ δ_ δ+ δ_ δ+ δ_
............. H O ............. H O ............. H O .............
H H H
δ+ δ+ δ+
H2O as associated molecule
5.4 Valence Shell Electron Pair Repulsion (VSEPR) Theory

In a molecule the constituent atoms have definite positions relative to one another ie. the
molecules have a definite shape. The theories of bonding that we have discussed so far do
not say anything about the shape of the molecules. A simple theory called VSEPR theory
was put forth by Sidgwick and Powell in 1940 to explain the shapes of molecules. It was
later refined and extended by Nyholm and Gillespie in 1957. This theory focuses on the
electron pairs present in the valence shell of the central atom of the molecule and can be
started in terms of two postulates.
POSTULATE 1
The electron pairs (both bonding and non-bonding) around the central atom in a molecule
arrange themselves in space in such a way that they minimize their mutual repulsion. In
other words, the chemical bonds in the molecule will be energetically most stable when they
are as far apart from each other as possible. Let us take up some examples.
BeCl2 is one of the simple triatomic molecules. In this molecules, the central atom, beryllium
has an electronic configuration of 1s2 2s2. That is it has two electrons in its valence shell. In
the process of covalent bond formation with two chlorine atoms two more electrons are
contributed (one by each chlorine atom) to the valence shell. Thus there are a total of 4
valence electrons or two pairs of valence electrons. According to the postulate given above,
these electron pairs would try to keep as far away as possible. It makes the two electron pairs
to be at an angle of 1800 which gives the molecule a linear shape.
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Other molecules of this type would also have a similar shape.
BF3 : In boron trifluoride, the central atom, boron has an electronic configuration of 1s2 2s2
2p1. That is, it has three electrons in its valence shell. In the process of covalent bond
formation with three fluorine atoms three more electrons are contributed (one by each fluorine
atom) to the valence shell. Thus there are a total of 6 valence electrons or three pairs of
valence electrons. According to the VSEPR postulate, these electron pairs would try to keep
as far apart as possible. It makes the three electron pairs to be located at an angle of 120 0
which gives the molecule a planar trigonal shape.
Thus different molecules would have different shapes depending on the number of valence
shell electrons involved. The geometric shapes associated with various numbers of electron
pairs surrounding the central atom are given in Table 5.1.
Table 5.1 : Geometric arrangements of electron pairs around central atom.
Molecule Number of Predicted Representative Example

Type electron pairs geometry Structure
Linear
Planer
trigonal
Tertahedral
Trigonal
bipyramidal
Octahedral
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POSTULATE 2
The repulsion of a lone pair of electrons for another lone pair is greater than that between
a bond pair and a lone pair which in turn is greater than that between two bond pairs.
The order of repulsive force between different possibilities is as under.
Lone pair – lone pair > lone pair – bond pair > bond pair – bond pair
The shapes of the molecules given in Table 5.1. correspond to the molecules containing only
bond pair electrons. The shapes of molecules containing a combination of lone pairs and
bond pairs would be distorted from above mentioned shapes.
Let us take an example of three molecules namely, methane, ammonia and water. All the
three contain a total of 4 electron pairs around their central atom. But the nature of these is
different in the three cases. In methane molecule the central carbon atom has 4 valence
electrons and it shares 4 electrons with four hydrogen atoms. So there are a total of 4 bond
pairs and according to Table 5.1 it should have a tetrahedral shape. In case of ammonia also
there are four pairs of electrons but their nature is different. Three of these are bond pairs
while one is a lone pair. Similarly, in case of water again there are four pairs of electrons;
two are bond pairs while two are lone pairs. Due to the differences in the mutual repulsion
between bond pair – bond pair and lone pair – bond pair the molecular shape would be
slightly distorted from the expected tetrahedral shape. The number and nature of electron
pairs and the geometries of these three molecules are given in Table 5.2.
Table 5.2 : Molecular geometries of molecules with 4 electron pairs
with different combinations of Ione pairs and bond pairs.
Bond angle
We have so far learnt that a chemical bond formation between two atoms can occur by transfer
(ionic bonding) or sharing (covalent bonding) of electrons. The processes of bond formation
and the bonding in simple molecules can be conveniently represented in terms of electron –
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dot structures. Further, the VSEPR theory provides a good idea of the shapes of the molecules.
But! Have you noticed that we have been representing electrons as well defined dots i.e.,
localized particles. This is in contradiction with the probabilistic (orbital) representation of
the electron that you have learnt in lesson 3. Let us learn how do we explain the process of
bond formation in terms of modern theories that incorporate the wave mechanical representation
of atom.

1. What are the basic postulates of VSEPR theory?
...................................................................................................................................
2. Predict the shape of methane (CH4) on the basis of VSEPR theory?
...................................................................................................................................
3. What is the effect of Hydrogen bond on boiling points?
...................................................................................................................................
5.5 Modern Theories of Chemical Bonding

The theories of chemical bonding proposed (in 1916) by Kossel and Lewis are called as
classical theories of bonding. These do not take into account the wave mechanical or quantum
mechanical principles. After the development of quantum mechanical description of atomic
structure two more theories were proposed to explain the bonding between atoms. These are
called modern theories of chemical bonding. These are Valence Bond Theory (VBT) and
Molecular Orbital Theory (MOT). Let us discuss these theories in brief.
5.5.1 Valence Bond Theory

Valence bond theory was proposed by Heitler and London in 1927, to describe the formation
of hydrogen molecule from its atoms. Linus Pauling and others further developed it. In this
approach the process of chemical bond formation can be visualised as the overlapping of
atomic orbitals of the two atoms as they approach each other. The strength of the bond
depends on the effectiveness or extent of the overlapping. Greater the overlapping of the
orbitals, stronger is the bond formed. Let us take the example of bonding in hydrogen
molecule to understand the VB approach.
Suppose that the two hydrogen atoms are at infinite distance from each other. Their electrons
are in their respective 1s orbitals and are under the influence of the corresponding nuclei.
As the two atoms approach each other their 1s orbitals begin to overlap which lead to decrease
in energy, Fig.5.1. At a distance equal to the bond length the overlapping is maximum and
the energy is minimum. The overlapping can be equated to the sharing of electrons between
the atoms. The electrons occupying the shared region of orbitals are under the influence of
both the nuclei.
-106-
Fig. 5.1: Formation of hydrogen molecule from overlapping of two hydrogen atoms
This simple approach can be used to explain the bonding in simple diatomic molecules like
HF, F 2 etc.. However, to explain bonding in molecules containing more than two atoms
some additional concepts like excitation and hybridization need to be used.
5.5.1.1 Hybridisation
Let us take up the example of bonding in a diatomic molecule; say beryllium hydride (BeH2)
to understand the concept of hybridization of orbitals and the need for the same. The atomic
number of beryllium is 4. Its electronic configuration is 1s2 2s2. In order to form bonds with
the 1s electrons of the two hydrogen atoms the valence electrons (2s2) of beryllium atom
must overlap with the 1s electrons of the two hydrogen atoms. Since the valence shell of
beryllium atom contains both the electrons in the same orbital (i.e.,2s) it cannot overlap
with the 1s orbital of hydrogen atoms containing one electron.[You know that an orbital can
contain a maximum of two electrons with opposite spin]. Pauling got over this problem by
suggesting that in the process of bond formation an electron from the 2s orbital of beryllium
atom gets momentarily excited to the empty 2p orbital as shown below.
Excitation
Beryllium Atom Beryllium Atom
↑↓ ↑↓ ↑↓ ↑ ↑
(Ground state) (Excited state)
1s 2s 2p 1s 2s 2p
Now the two valence electrons are in two singly occupied orbitals which can overlap with
the 1s orbitals of the two hydrogen atoms and form two bonds. The problem is still not over.
The two bonds formed by these overlaps would be of different nature. One of these would
involve overlapping of 2s orbital of beryllium with 1s orbital of hydrogen while the other
would involve overlapping of 2p orbital of beryllium with 1s orbital of hydrogen. However,
experimentally the two bonds are found to be equivalent.
This problem is solved with the help of a concept called hybridisation of orbitals. According
to this two or more than two non equivalent orbitals (having different energies and shapes)
of comparable energies mix or hybridize and give rise to an equal number of equivalent
(same energies and shapes) hybrid orbitals.
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In case of BeCl2 the two singly occupied orbitals (2s and 2p) hybridize to give two sp- hybrid
orbitals. This is called sp-hybridisation. These hybrid orbitals lie along the z-direction and
point in opposite directions.
Hybridization Beryllium Atom

Beryllium Atom ↑↓ ↑ ↑ ↑↓ ↑ ↑
(Excited state) (Hybridized)
1s 2s 2p 1s sp 2p
hybridized
orbitals
These hybrid orbitals can now overlap with the 1s orbitals of hydrogen atoms to give the
linear molecule of BeCl2 as shown below, Fig.5.2.
The concept of hybridisation as illustrated above can be used to describe the bonding and
shapes of different molecules by considering hybridization of suitable orbitals. Let us take
up some more cases involving hybridization of s and p orbitals.
Linear
Fig. 5.2 : Formation of BeCl2; sp hybridisation
Boron trichloride (Sp2 hybridisation) : In boron there are five electrons and the electronic
configuration is 1s2, 2s2, 2p1. There are three electrons in the valence shell of boron atom. In
order to form bonds with three chlorine atoms one of the electrons from the 2s orbital of
boron atom is excited to its 2p orbital.
Excitation
Boron Atom Boron Atom
↑↓ ↑↓ ↑ ↑↓ ↑ ↑ ↑
(Ground State) (Excited State)
1s 2s 2p 1s 2s 2p
One 2s orbital and two 2p orbitals hybridise to give three sp2 hybridized orbitals. This is
called sp2 - hybridisation.
Boron Atom Hybridization

(Exited State) ↑↓ ↑ ↑ ↑ ↑↓ ↑ ↑ ↑
2
1s 2s 2p 1s sp 2p
hybridized
orbitals
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The three hyrbidized orbitals are coplanar and directed towards the corners of an equilateral
triangle. These hybrid orbitals then form bonds with the p-orbitals of chlorine atoms as
shown below. fig. 5.3
Fig. 5.3 : Formation of BeCl3; sp2 hybridisation
Bonding in Methane (Sp 3 hybridisation) : In case of methane the central atom, carbon,
has an electronic configuration of 1s2, 2s2, 2p2. In order to form bonds with four hydrogen
atoms one of the electrons from the 2s orbital of carbon atom is excited to the 2p orbital
Excitation Carbon Atom

Carbon Atom
↑↓ ↑↓ ↑ ↑ (Exited State) ↑↓ ↑ ↑ ↑ ↑
(Ground State)
1s 2s 2p 1s 2s 2p
One 2s orbital and three 2p orbitals of the carbon atom then hybridize to give four sp3 hybridised
orbitals. This is called sp3-hybridisation.
Carbon Atom Excitation Carbon Atom

(Ground State) ↑↓ ↑↓ ↑ ↑ (Exited State) ↑↓ ↑ ↑ ↑ ↑
1s 2s 2p 1s sp3
Hybridised
These four sp3 hybrid orbitals are directed towards the corners of a regular tetrahedron. These
hybrid orbitals then form bonds with the 1s orbitals of hydrogen atoms to give a methane
molecule as shown below, Fig. 5.4.
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Fig. 5.4 : Formation of CH4 ; sp3 hybridisation
Phosphorus pentachloride (sp3d hybridisation):
P (ground state)
P (excited state)
sp3d hybridisation
Five sp3d hybrid orbitals are formed which are directed towards the corners of a trigonal
bipyramidal (Fig. 5.5a). These orbitals overlap with singly filled p-orbitals of five chlorine
atoms and five σ bons are formed. Thus PCl 5 molecule has a trigonal bipyramidal geometry.
Three P-Cl bonds (equatorial) make an angle of 1200 with each other and lie in one plane.
The other two P-Cl bonds (axial) are at 900 to the equatorial plane, one lying above and the
other lying below the plane.
SF6 (sp3d2 hybridisation) :
P (ground state)
P (excited state)
sp3d2 hybridisation
-110-
Six sp3d2 hybrid orbitals are formed which are directed towards the corners of a regular
octahedron. These orbitals overlap with singly filled orbitals of six F atoms and form σ
bonds giving a regular octahedral geometry (Fig. 5.5b).
Fig. 5.5(a) : Trigonal bipyramidal Fig. 5.5(b) : Octahedral geometry of

geometry of PCl 5 molecule SF6 molecule
5.5.1.2 Hybridisation and Multiple Bonds

So far we have discussed the bonding in those molecules in which the orbitals on a single
central atom are hybridized. Let us see how does the concept of hybridisation help us in
understanding bonding between pairs of such atoms. In the case of bonding in ethane (C2H6),
two carbon atoms are bonded to each other and each carbon atom is bonded to three hydrogen
atoms. You would recall that in the case of methane the valence orbitals of carbon atom
undergo sp3 hybridisation. In ethane each carbon atom undergoes sp3 hybridisation to give
four sp3 hybridized orbitals. The two carbon atoms form a carbon - carbon bond by sp3 - sp3
overlapping. The remaining six sp3 hybridized orbitals overlap with 1s orbitals of hydrogen
atoms to give a molecule of ethane, C2H6 as shown in fig. 5.6 The C-C bond so formed is
along the internuclear axis. Such a bond is called a 6 bond.
Fig. 5.6 : Formation of ethane molecule
-111-
Bonding in ethane : In case of ethene, the relevant orbitals of the carbon atoms undergo sp2
hybridisation. Here, only two of the three p orbitals of the carbon atoms hybridize with the 2s
orbital of from three sp2 hybrid orbitals each. The remaining p-orbitals (one on each carbon
atom) do not take part in hybridization. A carbon - carbon bond is formed by overlapping of
sp2 orbital on the two carbona atoms (Fig. 5.7(a)]. The remaining four sp2 hybridized orbitals
overlap with the 1s orbitals of hydrogen atoms to give the basic skeleton of the molecule.
This leaves an un-hybridized p orbital each on both the carbon atoms (Fig. 5.7(b) ]. These
are perpendicular to the molecular plane and undergo sideways overlap to give an electron
cloud in the plance above and below the molecule [Fig. 5.7 (b and c)]. This is called a π-
bond. In ethene there are two bonds between the carbon atoms (one sigma and one pi bond).
Fig. 5.7 : Formation of ethylene molecule : a) formation of the basic skeleton of the
molecule b) sideways overlapping of the un-hybridized p orbitals and c) a π−bond
d) and (e) complete picture of ethylene molecule.
Bonding in ethyne (acetylene) : In case of acetylene the bonding can be explained in terms
of sp-hybridisation in carbon atoms. One 2s and one 2p orbitals hybridize to give two sp-
hybridized orbitals. This leaves two mutually perpendicular unhybridized p orbitals each on
both the carbon atoms. The carbon - carbon bond is formed by sp - sp overlapping with each
other. The remaining sp orbital on each carbon overlaps with the 1s orbital of hydrogen to
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give C-H bonds (Fig. 5.8). The unhybridised p orbitals each on both the carbon atoms overlap
sideway to give two π-bonds.
Fig. 5.8 : Formation of acetylene molecule : a) formation of the basic skeleton of the
molecule b) sideways overlapping of the un-hybridized p orbitals and c) two mutually
perpendicular π−bond
2 Intext Questions 5.3

1. What do you understand by the term. “hybridisation”?
...................................................................................................................................
2. How would you explain the shape of ammonia molecule on the basis of hybridisation ?
...................................................................................................................................
5.5.2 Molecular Orbitals Theory

You have just learnt about valence bond theory. It describes bond formation as result of
overlapping of the atomic orbitals belonging to the constituent atoms. The overlapping region
responsible for bonding is situated between the two atoms i.e., it is localised. Molecular
orbital theory (MOT) developed by F. Hund and R.S. Mulliken in 1932, is based on the
wave mechanical model of atom. In contrast to the localized bonding in VBT, the molecular
orbital theory visualises the bonding to be delocalised in nature i.e., spread over the whole
molecule. According to MOT, in the process of bond formation
• The atomic orbitals of the constituent atoms combine to generate new types of
orbitals (called molecular orbitals). These are spread over the whole molecule i.e.,
they are delocalised. In other words these new orbitals, do not “belong” to any one
atom but extend over the entire region of the bonded atoms.
• These molecular orbitals are created by Linear Combination of Atomic Orbitals (LCAO)
approach in which, the atomic orbitals of comparable energies and of suitable
symmetry combine to give rise to an equal number of molecular orbitals.
• The available electrons then fill lthese orbitals in the order of increasing energy as
in the Aufbau princile used in the electron configurations of atoms.
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Let us take the example of hydrogen molecule to understand the molecular orbital approach
to chemical bonding. The two hydrogen atoms have an electron each in their respective 1s
orbitals. In the process of bond formation the atomic orbitals of two hydrogen atoms can
combine in two possible ways. In one, the MO wavefunction is obtained by addition of the
two atomic wave functions whereas in the other MO is obtained by substration of the atomic
orbitals. The combination of the 1s orbitals on the two hydrogen atoms are shown in fig. 5.9
Fig. 5.9 : Formation of bonding (σ) and anti bonding (σ*) molecular orbitals
The molecular orbital obained by the addition of atomic orbitals is of lower energy than that
of the atomic orbitals and is called a bonding orbital. On the other hand, the orbital obtained
by substraction of atomic orbitals is of higher energy and is called an anti-bonding orbital.
You can note here that the molecular orbitals obtained here are symmetric around the bond
axis (the line joining the two nuclei). Such molecular orbitals are called sigma (σ) molecular
orbitals. The bonding orbital obtained above is denoted as σ 1s while the anti-bonding orbital
is denoted as σ* 1s. Here σ indicates the type of molecular orbital; 1s tell about the atomic
orbital involved and * is indicative of the anti-bonding nature of the MO. There are a total of
2 electrons in a hydrogen molecule, according to Aufbau principle these are filled into σ1s
orbital. Since the σ1s orbital is a bonding orbital, its getting filled leads to stability or the bond
formation.
Like electronic configuration of atoms we write MO electronic configuration for molecules.
The MO configuration of hydrogen molecule is given as (σ 1s) 2. The molecular orbital
energy level diagram are given in Fig. 5.10 (a and b).
Fig. 5.10 : Molecular orbital energy level diagram for a)H2 and b)He2 molecules
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Bond Order : we may define a new parameter called bond order as
Bond order = (b.o.) = 1/2 (nb - na)
Where, nb and na refer to the number of electrons present in bonding and antibonding molecular
orbitals respectively. For hydrogen molecule the bond order will be 1/2(2-0) = 1, i.e., there is
a single bond between two hydrgen atoms.
Helium (He2) molecule
In case of He2, also there will be linear combination of 1s orbitals leading to the formation of
σ 1s and σ* 1s orbitals. The four electrons would be distributed as per the MO electronic
configuration : (σ 1s)2 (σ* 1s)2. The molecular orbital energy level diagram is given in fig.
3.10 (b). This is gives a bond order of 1/2 (2-2) = 0, that is there is no bond between two
helium atoms. In other words He2 molecule does not exist.
Li2 and Be2 molecules
The bonding in Li2 and Be2 can be explained by combining the 1s and 2s orbitals to give
appropriate MO’s. the molecular orbital diagrams for Li 2 and Be2 are given in Fig. 5.11
(a) (b)
Fig. 5.11 : Molecular orbital energy level diagram for a) Li2 and b) Be2 molecules
-115-
5.5.2.1 Moelcular Orbital Bonding in Diatomic Molecules of Second
Period
So far we have talked about bonding in the elements in which the MO’s were obtained from
the linear combination of s orbitals. Incase of the atoms of second period (beyond Be)
elements both s and p orbitals are involved in the formation of molecular orbitals. In such a
case a number of different molecular orbitals are obtained depending on the type and
symmetry of the atomic orbitals involved in the process. Let us try to understand the nature
of MO’s obtained in this case.
Here also the 1s and 2s orbitals of the two would combine to give corresponding bonding
and anti-bonding molecular orbitals as shown in Fig. 5.11 (b). Let us learn about the formation
of MO’s from the combination of p orbitals.
As mentioned above, in LCAO, the atomic orbitals of comparable energies and of suitable
symmetry combine to give molecular orbitals. A suitable symmetry means that the
combining orbitals should have same symmetry about the molecular axis. It is normally
assumed that the bond formation takes place along the z-direction. You have learnt in the
first unit that the three p orbitals are directed towards three mutually perpendicular direction
namely the x, y and z directions. Therefore the Pz orbitals of the two atoms would combine
along the bond axis to give two molecular orbitals as shown below fig. 5.12 Since these
molecular orbitals are symmetric around the molecular axis these are called σ orbitals. The
designation of the orbitals would be σ2pz and σ 2pz *.
Fig. 5.12 : Overlapping of two 2Pz orbitals to give molecular orbitals
Combination of a Pz - orbital with either a Px or Py orbital would not lead to any

bonding. On ther other hand a Px orbital will combine with a Px and the Py with a Py as
shown in Fig. 5.13
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Fig. 5.13 : Formation of molecular orbitals from two 2Px atomic orbitals
You may note here that these orbitals combine in a lateral fashion and the resulting molecular
orbitals are not symmetric around the bond axis. These MO’s are called π-molecular orbitals.
These have large electron density above and below the internuclear axis. The anti-bonding
π orbital, π* 2Px (or π* 2Py) have a node (a region of zero electron density) between the
nuclei.
The molecular orbitlas obtained as a result of combination of respectie AO’s of two atoms
can be represented in the form of following energy level diagram, Fig 5.14 (a). The MO’s
obtained from the combination of 1s orbitals are not shown. (these beong to the inner core
and are completely filled). The electrons in these molecular orbitals are filled in accordance
with Aufbau principle and Hund’s rule.
Fig. 5.14 : Molecular orbital energy level diagams (a) for O2 and F2 and (b) for diatomic
orbitals of lighter elements Li, Be, B, Cand N
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However, this energy level diagram is valied for the diatomic molecules O2 and F2 only; For
the diatomic molecules of the lighter elements like, B, C and N this energy level diagram is
somewhat modified. It is so because in case of lighter elements the difference in the energy
of 2s and 2p orbitals is very low and s and p orbitals on the two atoms get mixed up. In place
of normal pure 2s-2s or 2p-2p combinations we may have s-p combinations; for example 2s
orbital of first atom can have a reasonable overlapping with 2Pz, orbital of the second atom
and vice versa. The modified level diagram is given in Fi.g 5.14 (b).
5.5.2.2 Electronic Configuration and Properties of the Molecule

The MO energy level diagram discussed above can be used to find out the MO electronic
configuration of the molecule. This in turn provides the information about some properties
of the molecule. Let us take the example of nitrogen molecule. An atom of nitrogen has five
valence electrons; since there are two atoms, we have a total of ten valence electrons that
need to be filled in the MO’s Using Fig. 5.14, the MO electronic configuration can be
writeen as σ2s2, σ*2s2, π2px2, π2p y2, σ2sz2,
Bond order : 1/2 [nb-na] = 1/2 [8-2] = 1/2 [6] = 3 ; this means that in nitrogen molecule, a triple
bond exists between the two nitrogen atoms.
Megnetic nature : molecules show magnetic behaviour depending on their MO electronic
configuration. If all the MO’s are doubly occupied the substance shows diamagnetic
behaviour. In case one or more MO’s are singly occupied, the substance shows
paramagnetic behaviour. The MO electronic configuration of O2 (with 12 valence electrons)
is σ2s2, σ*2s2, π2sz2, π2p2, π2py2, π2py2 , π*2px1, π*2py1 ; Since it contains unpaired
electrons, oxygen shows paramagnetic behaviour. This has been found to be so experimentally
also. In fact, the explanation of the paramagnetic behaviour of oxygen is an achievement of
MOT.
The bond order and the magnetic behaviour of the molecular cations and anions can also be
obtained in the same way. In such cases we add one electron for every negative charge and
for every +ve charge we subtract an electron. For example O22 -(oxygen molecule dianion)
would have a total of 14 valence electrons (12 + 2) while oxygen molecule cation O2+ would
have 12-1 = 11 valence electrons.

1. What is the basic difference between the valence bond and molecular orbital theories ?
.............................................................................................................................................
2. Calculate the bond orders for Li2 and Be2 molecules using the molecular orbital
diagrams given in Fig. 5.12
.............................................................................................................................................
3. Predict the magnetic behaviour of O2.
.............................................................................................................................................
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• A chemical bond may be visualised as an effect that leads to the decrease in the energy
of the combination of two atoms when they come closer.
5
F • The atoms combine in such a way so as to attain stable electronic configuration of noble
gases.
• According to Kossel, transfer of an electron from one atom to the other achieves the
stable configuration. This leads to formation of ions, which are held together by
electrostatic interactions called ionic bond.
• According to Lewis, the stable configuration is achieved by sharing of electron pairs
between the bonding atoms. This leads to the formation of a covalent bond.
• Bonding in simple molecules can be conveniently represented in terms of Lewis
electron-dot structures.
• In some covalently bound atoms the shared pair of electron is more towards the atom
with greater electronegativity and leads to partial ionic character inthe molecule.
• Valence shell electron pair repulsion (VSEPR) theory is very helpful in predicting the
shapes of simple molecules. It is based onthe interactions between the electron pairs
around the central atom in the molecule.
• Valence bond theory (VBT) and Molecular orbital theory (MOT) are two modern
theories of chemical bonding. These are based onthe wave mechanical model of atom.
• According to the valence bond theory the process of chemical bond formation can be
visualised as the overlapping of atomic orbitals of the two atoms as they approach
each other. The overlap increases the electron charge density in the inter-nuclear region.
• In order to explain bonding in molecules containing more than two atoms, Pauling
proposed the concept of hybridisation. In hybridisation, the atomic orbitals of the valence
shell of the central atom ‘hybridise’ or merge and given never orbitals with proper
orientations, which explain the shape of the molecule.
• According to the Molecular orbital theory the atomic orbitals of comparable energies
and of suitable symmetry combine to give rise to an equal number of molecular orbitals.
These molecular orbitals extend over the entire region of the molecule i.e. these are
delocalised over the whole molecule.
• When two atomic orbitals combine it gives a pair of molecular orbitals; one is called
bonding molecular orbital of lower energy and the other of higher energy is called
anti-bonding orbital.
• The electrons present in the molecule are filled in these orbitals in the order of increasing
energy (Aufbau principle) to give the MO electronic configuration.
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• The number of bonds between the two atoms is called bond order and is defined as
Bond order = b.o. = 1/2 (nb = na).
• The MO electronic configuration can be used to predict the magnetic nature of the
molecule. If all the MO’s are doubly occupied the substance shows diamagnetic
behaviour and if one or more MO’s are singly occupied the substance shows paramagnetic
behaviour.
• The favourable conditions for the formation of ionic bond and properties of ionic
compounds
• Valence bond theory is discussed.
• The consequences of hydrogen bond.
Terminal Exercise
1. What do you understand by a chemical bond ?
2. Explain the process of bond formation as a decrease in energy.
3. What do you understand by the term, ‘bond length’ ?
4. Describe the two possible ways in which the noble gas electronic configuration is achieved
inthe process of bond formation.
5. What are Lewis electron-dot symbols ? Show the formation of MgCl2 in terms of Lewis
symbols.
6. Define a coordinate bond and give some examples.
7. What is VSEPR theory ? predict the shape of SF6 molecule using this theory.
8. Why do we need the concept of hybridisation ? How does it help in explaining the
shape of methane ?
9. Give the salient features of molecular orbital theory.
10. Be2 molecule does not exist. Explain on the basis of molecular orbital theory.
11. Write down the molecular orbital electronic configuration of the following species and
compute their bond orders.
_ _
O2 ; O 2+ ; O 2 ; O 2 2

5.1
1. An electrovalent bond is formed when one or more electrons are transferred from one
atom to another atom or atoms.
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• •
: :
:N • + N: : N .. N: : N ≡ N:
k
2. •
• •
3. In a covalnet bond the shared pair of electrons is closer to the more electronegative
atom. This leads to charge separation in the molecule and the bond becomes polar
4. A bond in which both the bonding electrons are contributed by one atom only
5. i) Low iozation energy ii) Low charge on the ion
iii) Large atomic size
6. As ionic bond is non directional hence ionic compounds does not show isomerison.
5.2
1. The two postulates of VSEPR theory are
i) The electron pairs (both bonding and non-bonding) around the central atom in a
molecule arrange themselves in space in such a way that they minimize their
mutual repulsion.
ii) The repulsion of a lone pair of electrons for another lone pair is greater than that
between a bond pair and a lone pair which in turn is stronger than that between
two bond pairs. The order of repulsive force between different possibilities is as
under.
lone pair - lone pair > lone pair - bond pair > bond pair - bond pair
2. In methane the central carbon atom would have four pairs of electrons in its valence
shell. According to VSEPR theory these would be placed tetrahedrally around the
carbon atom. Hence the methane molecule would have a tetrahedral shape.
3. Boiling point increase due to H-bonding.
5.3
1. Hybridisation is a concept which is quite useful in explaining the shapes of molecules.
According to this two or more than two non equivalent orbitals with comparable energies
and different shapes mix and give rise to an equal number of equivalent hybrid orbitals.
The hybrid ortbitals have identical energies and shapes.
2. In ammonia the 2s and three 2p orbitals hybridize to give four sp3 hybridized orbitals.
Three of these overlap with the 1s orbitals of hydrogen and one remains nonbonding
containing a lone pair. The sp3 hybridized orbitals are directed towards the corners of a
regular tetrahedron. But due to the pair the ammonia molecule has a distorted tetrahedras
shape which is some what like a trigonal pyramid.
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1. Valance bond theory visualises the bond formation to be localized whereas according
to MOT it is delocalised.
2. Bond order = b.o. = 1/2 (nb - na)
for Li2 ; Bond order = 1/2 [4-2] = 1/2 [2] = 1
for be2 : Bond order = 1/2 [4-4] = 1/2 [0] = 0
3. MO configuration of O2 is σ2s2, σ*2s2, σ2pz2, π2px2 = π2py2.
π*2p1x = π*2p1y
Due to 2 unpaired electrons O2 molecule is paramagnetic.
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6
THE GASEOUS STATE
We know that matter exist in three *different states namely solid, liquid and gas. The most
familiar example to show these different states of matter is water. Water exists as a solid
(ice), a liquid (water) and a gas (steam) under different conditions of temperature and
pressure. The difference between these three states of matter is due to the difference in their
intermolecular distance and intermolecular forces. In addition to these, temperature and
pressure also play an important role in deciding the states of matter.
In this lesson we shall first discuss the differences in properties of the solid, liquid and gaseous
state and the factors due to which these differences arise. We shall also study the effect of
pressure and temperature on the volume of the given amount of gas. These are governed by
the gas laws namely Boyles’ law, Charles’ law and Avogadros’ law.
Objectives
• differentiate between the three states of matter - solid, liquid and gas;
• list the characteristic properties of gases;
• state the gas laws (Boyle’s law, Charle’s law and Avogadro’s law) and express them
mathematically.
• draw the p-V, p-1/V, p-pV and V-T graphs.
• Interpret the effect of temperatue and pressure on the volume of a gas from the graph;
• derive the ideal gas equation from the gas laws;
• State the Dalton’s law of partial pressure and explain its significance;
• state Graham’s law of diffusion;
• state the postulates of Kinetic Molecular Theory of gases;
• explain the Maxwell’s distribution of velocities.
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• differentiate between urms, ump and uav,
• explain the deviation of real gases from ideal behaviour in term of compressibility factor;
• state the van der Waals equation and explain the significance of van der Waals constants
and
• explain the liquifaction of gases with the help of Andrews curves
6.1 The Three States of Matter

At any given conditions of temperature and pressure matter exists in one of the three states
namely solid, liquid and gas. The characteristic properties of solid, liquid and gaseous state
are listed in Table 6.1.
Table 6.1 : Properties of different states of matter
Property Solid Liquid Gas
Shape Definte Indefinite; takes up the Indefinite

upthe shape of the vessel.
Volume Definite Definite Indefinite (fills the
container completely)
Density High Less than solids but Low
much higher than gases.
Compressibility Incompressible Largely incompressible Highly compressible
The different characteristics of the three states of matter as listed above depend upon the
relative closeness of particles that make up the substance. In solid state, the particles are
held close together in a regular pattern by strong intermolecular forces. In liquid state,
intermolecular forces are weak as compared to solid state hence the particles are less tightly
held and allow them to move away from each other. In the gaseous state, the molecules are
farthest apart as compared to solid and liquid states and the intermolecular forces are
negligible so the particles move randomly. A simplified picture of particles in solid, liquid
and gaseous state is represented in Fig. 6.1.
Solid (a) Liquid (b) Gas (c)

Fig. 6.1 : A simplified picture of particles in solid, liquid and gaseous state
-124-
6.2 General behaviour of Gases : The Gas Laws
The volume of a given mass of gas depends upon the temperature and pressure under which
the gas exists. It is, therefore, possible to describe the behaviour of gases in terms of the four
variables : temperature, T; pressure p; volume V and amount (number of moles, n). For a
given amount of gas the volume of gas changes with change in variables such as temperature
and pressure. The relationship between any two of the variables is studied, keeping the other
variable constant by various laws which are described below.
6.2.1 Effect of Pressure on the Volume of the Gas (Boyle’s law)

The effects of pressure on the volume of gas for a given amount of gas at constant temperatue
was studied by Robert Boyle in 1662 for different gases. He observed that if the volume of
gas is doubled the pressure is halved and vice versa. Boyle’s law states that at constant
temperature, the volume of a given amount of a gas is inversely proportional to its pressure.
Mathematically Boyle’s law is expressed as shown below :
V ∝ 1
/p (at constant T and n)
or p1 V1 = p2 V 2
when the pressure of the gas, p is plotted against volume of the gas, V the exponential curve
1
is obtained (Fig. 6.2). However when the pressure, p of the gas is plotted against V a straight
line is obtained (Fig. 6.2). If the product of pressure and volume (pV) is plotted against pressure
(p) a straight line parellel to x-axis (pressure is axis) is obtained (Fig. 6.4)
T, n = contant T1 n = contant
T1n = contant
pV
-125-
Example 6.1 : The volume occupied by a given mass of a gas at 298K is 24 mL at 1 atmospheric
pressure. Calculate the volume of the gas if the pressure is increased to 1.25 atmosphere
keeping temperature constant.
Solution : Given that
V = 25 mL p1 = 1 atm
V2 = ? p2 = 1.25 atm
According to Boyle’s Law, p1 V1 = p2 V2

substituting the value of p1, V 1 and p2 in the above expression we get
The volume occupied by the gas is 20 mL at 298 K and 1.25 atm pressure.
Example 6.2 : The volume of a certain amount of a gas is decreased to one fifth of its initial
volume at a constant temperature. What is the final pressure ?
Solution : Let
Initial volume = V1 Initial pressure = p1
Final volume V2 = V1/5 Final pressure = p2
By Boyle’s law, we know that at constant temperature
p1 V1 = p2 V2
p1 V 1 p 1 V1
Thus when volume is decreased to 1/5th of the initial volume, the pressure in increased bv 5
times of the initial volume.
6.2.2 Effect of Temperature on the Volume of Gas (Charle’s Law)

The effects of temperature on the volume of the gas was studied by Jacques Carles in 1787
and Gay Lussac in 1802 at constant pressure for different gases. Their conclusion can be
given as Charles’ law which states that at a constant pressure, the volume of a given amount
of gas is directly proportional to the absolute temperature.
So, according to Charles’ Law, the volume of a gas increases as its absolute temperature is
being raised, if its absolute temperature is lowered, its volume will consequently decrease.
Mathematically, Charle’s Law is expressed as shown below:
-126-
V ∝ t (at constant p and n)
or V = k.1 (k is a constant)
Therefore V1/t1 = V 2/t2
Graphical representation of Charle’s Law is a straight line pointing away from the origin of
the graph as shown in Fig. 6.5
Here graph of the volume of a gas (V) plotted against its temperature at constant pressure
and amount (in moles). Notice that the graph is a straight line with a positive gradient
(slope).
Mathematically volume of a gas at temperature t is given as
v00 ⎛ t ⎞ ⎛ 273 + t ⎞
v t = vo + xt = v0 ⎜1 + = v0 ⎜
273 ⎝ 273 ⎠⎟ ⎝ 273 ⎟⎠
Thus at t = -2730C
⎛ 273 − 273 ⎞
v t = vo ⎜ =0
⎝ 273 ⎟⎠
This means that at -2730C, the volume of the gas is reduced to zero i.e., the gas ceases to exist.
Thus this temperatue (-2730C) at which the gas hypothetically ceases to exist is called Aboslute
zero. It is represented by zero K.
This is the theoretically lowest possible temperature. In actual practice, we cannot reduce the
temperature of the gas to zero kelvin.
Kelvin Scale of Temperature
The scale of temperature which has -2730C as zero is called Kelvin Scale. Degree celcius is
converterd to Kelvin by adding 273. Thus
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t /0C + 273 = T/K
where T = temperature in Kelvin
t = temperature in celcius
For example 150C can be converted in K by adding 273 to 15.
6.2.3 Effect of Temperature on Pressure

(Pressure-Temperature Law)
This law states that.
Pressure of given amount of gas at constant volume is directly proportional to its
absolute temperature.
p ∝ T
p = kT
Example 6.3 : A given amount of a gas is maintained at constant pressure and occupies a
volume of 2 liters at 1000 0C. What would be volume if gas is cooled to 00C keeping pressure
constant.
Solution : Given that,
Initial volume V1 = 2L T1 = 1000 + 273 = 1273 K
Final volume V2 = ? T2 = 0 + 273 = 273 K
Now using Charle’s Law V1 / T1 = V2 / T2 or V2 = (V1 / T/1) x T2
On substituting the values we get
V2 = (V1 / T1) x T2 = (2L / 1273 K) x 273 = 0.4291 L
6.2.4 Avogadros’ Law

The Italian physicist Amadeo Avogadro was the first to propose, in 1811, a relationship
between the volume of a gas and the number of molecules present in it. This, relationship is
known as Avogadros’ Law. It states that :
Equal volumes of all gases at the same temperatue and pressure contain equal number of
molecules.
Mathematically, Avogadros’ law is expressed as :
V ∝ N (at constant temperature and pressure)
Where V and N are volume and number of molecules respectively.
-128-
At a given temperature and pressure, the number of molecules present in the gas is directly
proportional to the number of moles.
therefore, N ∝ n
Where n is the number of moles
∴ V ∝ n
V
or = constant
n
V1 V2
n1 = n2 =
Also
He also found that the number of molecules present in 1 mole of any substance (22.4 litre of
any gas at 273K temperature and 1 atmosphere pressure) is 6.022 x 1023 molecules. This
number is known as Avogadros’ number.
It is the number of molecules (6.022 x 1023) of any gas present in a volume of 22.4 L(at 273 K
and 1 atm) and it is the same for the lightest gas (hydrogen) as for a heavy gas such as carbon
dioxide or bromine.
Example 6.4 : 0.965 mol of a gas occupies a volume of 5.0L at 298 K / temperature and 1
atm pressure. What would be the volume of 1.80 mol of the gas at the same temperature and
pressure ?
Solution : V1 n 2 = V 2 n 1
V2 n2 (5.0L)(1.8mol)
=
n1 (0.965mol)
V2 = 9.33 L
Example 6.5 : Compare the volumes of 16g of oxygen and 14g nitrogen at the same
temperature and pressure.
Solution : Number of moles of O2 = 16 g/32 g mol-1 = 0.5 mol
Number of moles of N2 = 14 g/28 g mol-1 = 0.5 mol
Since the two gases are at the same temperature and pressure, and contain equal number of
mole, hence according to the Avogadro’s Law they should also occupy the same volume.

1. The density of a gas is usually less than that of the liquid. Explain.
...................................................................................................................................
-129-
2. Calculate the pressure (atm) required to compress 500 mL of gas at 0.20 atm into a
volume of 10mL.
...................................................................................................................................
3. Equal volumes of oxygen gas and an unknown gas weigh 2.00 and 1.75 g respectively
under the same experimental conditions. What is the molar mass of the unknown
gas ?
...................................................................................................................................
6.3 The Ideal Gas Equation

Boyle’s Law, Charle’s Law and Avogadro’s Law can be combined to give a single equation
which represents the relation between the pressure, volume and kelvin temperature of a
given amount of gas under different conditions. Thus
V ∝ 1/p at constant temperature (Boyle’s Law)
V∝T at constant pressure (Charle’s Law)
V∝n at constant temperature and pressure (Avogadros’ Law)
All the three expressions can be combined into a single expression.
V ∝ nT/P or pV ∝ nT
or pV = constant x nT
The constant in this equation is called “universal gas constant’ or ‘molar gas constant’,
represented by R. Thus we can write for 1 mole of a gas.
pV = RT
Correspondigly, for n moles of a gas we have
pV = n RT
This is known as the ideal gas equation because it holds only when gases are behaving as
‘ideal’ gases.
Since for a given mass of gas we can write
pV/T = a constat, we have
p1 V1/T1 = p2 V2/T2
Where p1, V1 and T1 refer to one set of conditions and p2, V2 and T2 refer to a different set of
conditions.
The numerical value of R can be found by substituting experimental quantities in the equation.
At STP, T = 273.15 K, p = 1 atm and for 1 mol of gas (n = 1), v = 22.414 L. Consequently,
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R = pV/nT = (1 atm) (22.414L) / (1 mol) (273.15K)
= 0.082057 L atm mol-1 K-1
The value of R depednds on the units adopted for the quantities in the equation pV = nRT.
The various values are :
R = 0.082057 L atm K-1 mol-1 (for caluculation purpose the value is
taken as 0.0821 L atm K-1 mol-1)
R = 8.314 x 107 erg K-1 mol-1
R = 8.314 J K-1 mol-1
R = 1.987 cal K-1 mol-1
Example 6.6 : At 273 K, 10 mol of gas is confined in container of volume 224 L. Calculate
the pressure of the gas. R = 0.821 L atm mol-1 K -1.
Solution : The ideal gas equation pV = nRT will be used here
n = 10 mol, R = 0.0821 atm L K-1 mol-1
V = 224 L T = 273 K p=?
On substituting these values in the above equation we get
p = nRT/V = (10 mol x 0.0821 atm L mol-1 K-1 x273 K) / 224L = 0.99998 atm = 1 atm.
6.4 Dalton’s Law of Partial Pressure

The behaviour observed when two or more non-reacting gases are placed in the same
container is given by Dalton’s Law of partial pressures. Dalton’s Law states that.
The total pressure exerted by a mixture of non-reacting gases is equal to the sum of the
partial pressures of the various gases present in the mixture.
The partial pressure is defined as the pressure the gas would exert if it was alone in the
container. Suppose a sample of hydrogen is pumped into a one litre box and its pressure is
found to be 0.065 atm. Suppose, further a sample or argon is pumped into a second one litre
box and its pressure is found to be 0.027 atm. If both samples are now tranferred to a third
one litre box, the pressure is observed to be 0.092 atm. For the general case, Dalton’s Law
can be written as
Ptotal = PA + PB + PC + ........
Where PA , PB , PC ........ are the partial pressure of gases A, B, C..... respectively. This gas
laws provide a simple way of calculating the partial pressure of each component, given the
composition of the mixture and the total pressure. First we introduce the mole fractions
Xa and X B. These are defined as
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Where nA and nB are the number of moles of gas A and B respectively and n = nA +nB.
Since PA = nA RT/V, PB = nB RT/V, and p = n RT/V,
it follows that PA = XA p and pB = XBp
This is an exceptionally useful (and simple) way of calculating at partial pressures when the
composition and total (measured) pressure of a mixture of gas is known.
Example 6.7 : Atomosphere is oten considered mainly as a mixture of nitrogen and oxygen:
76.8% by mass of nitrogen and 23.2% by mass of oxygen. Calculate the partial pressure of
each gas when the total pressure is 1 atm.
Solution : The number of moles of each component is
PN = 76.8 g / 28 g mol-1 = 2.74 mol
2
PO = 23.2 g / 32 g mol-1 = 0.725 mol

2
The mole fractions of the components are therefore
XN = = 0.791 ; XO = = 0.209
2 2
The partial pressures are therefore given by

PN = 0.791 x 1 atm = 0.791 atm
2
PO = 0.209 x 1 atm = 0.209 atm

2
6.5 Graham’s Law of Diffusion

If we open a bottle of perfume in one corner of a room or burn an incense stick we can feel the
smell of the perfume or the incense stick all over the room also. The smell of perfume or
incense stick spreads from one point of the room to the other by mixing with air. This free
intermingling of gases when placed in contact with each other is known as diffusion.
Diffusion occurs in liquids as well as in gases. Effusion is the escape of a gas through a
small hole, as in case of a puncture in a tyre.
The experimental observation of the rate of effusion of gases through a small hole in the
side of the led Graham (1829) to formulate the following law;
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional
to the square root of its density. Graham’s law is applicable to both diffusion and effusion.
If the time for a given volume of gas A to escape is tA, while the time of the same volume of
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gas B to escape is tB, it follows, that,
tA/tB = (rate)B / (rate)A = ρ A / ρB where ρA and ρB are the densities of gases A and B respectively.
The ratio of the densities of the molecules is the same as the ratio of the molecular masses of
the gases at the same temperature and pressure.
hence, tA/tB = (rate)B / (rate)A = ρA / ρB = M A / M B where ΜA and ΜB are the molecular

masses of gases A and B respectively.
1. What is the difference between diffusion and effusion.

...................................................................................................................................
2. Explain why Dalton’s law is not applicable to a system of ammonia and hydrogen
chloride gas.
...................................................................................................................................
3. The rates of diffusion of CO2 and O3 were found to be 0.29 and 0.271. What is the
molecular mass of O3 if the molecular mass of CO2 is 44.
...................................................................................................................................
4. Calculate the pressure exerted by 5.0 mol of carbon dioxide in a 1 litre flask at 470C
using ideal gas equation.
...................................................................................................................................
6.6 Kinetic Molecular Theory of Gases

(Accounting for the Gas Laws)
To explain the behaviour of the gases theoretically, Claussius, Maxwell and Boltzmann
made the following assumptions :
1) Gases consist of large number of tiny particles called molecules.
2) The gas molecules are so small and so far apart that the total volume of the molecules
is a negligible fraction of the total volume occupied by the gas.
3) The molecules are in state of constant, rapid and random motion colliding with one
another and with the walls of the container.
4) There are no attractive or repulsive forced between the molecules of the gas.
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5) The collisions of the molecules among themselves and with the walls of the containing
vessel are perfectly elastic, so that there is no loss of energy during collisions.
6) The pressure exerted by a gas is due to the bombardment of the molecules on the walls
of the containing vessel.
7) The kinetic energy of a gas is directly proportional to the absolute temperature of the
gas.
On the basis of this model, it is possible to derive the following expression for a gas:
1
pV = mNC 2
3
Where p is pressure, V denotes volume, m is the mass of a gas molecule. N is the total number
of molecules. and C is the root mean square velocity of the gas molecules.
6.6.1 Root Mean Square Velocity

Root mean square velocity is the square root of the average of the squares of all the molecular
velocities. Mathematically.
RMS Velocity = (C12 + C22 + ... + C N2 ) / N
where C1, C2 ..... CN the molecular velocities.
Kinetic gas equation - Deduction of gas law’s from the kinetic gas equation
All the gas laws can be deduced from the kinetic gas equation.
1
Pv = mnc 2
3
a) Boyle’s law : According to kinetic gas equation Pv = 1/3 mnc2

Here, P is pressure of the gas molecules we can write
2 1
Pv = ( mnc 2 )
3 2
The kinetic energy of ‘m’ mole’s of the gas is 1/2 mnc2. Change the line to next row
according to kinetic molecular theory, kinetic energy is directly proportional to the
temperature in Kelvin, or 1/2 mnc2 = KT.
Where ‘K’ is constant on substituting this result in the above equation expression we
have
Pv = 2/3 KT ................ (1)
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At constant temperature (T) = Pv = constant.
This is Boyle’s law equation.
b) Charles law : From the equation (1) Pv = 2/3 KT
Or v = 2 KT ................ (2)
3 p
At constant pressure (P), we have v = constant x T (or) v α T (p, n are constants)

This is Charles law.
c) Avogadro’s law : consider equal volumes of the two different gases at same temperature
and pressure. As per the kinetic gas equation.
Pv = 1/3 m1 n1 c12 ......... Kinetic gas equation for the first gas
Pv = 1/3 m2 n2 c22 ......... Kinetic gas equation for the second gas
Since 1/3 m1 n1 c12 ......... 1/3 m2 n2 c22 ................... (3)
(Since on left side both equation have same Pv)
Since temperature (T) is the same for the two gases their average Kinetic energy per
molecule will also be the same
1 1
i.e. m 1 c 12 = m c 2 .......................... (4)
2 2 2 2
Dividing equation (3) by equation (4)
1 1
m1n1c12 m 2 n 2c22
3 = 3 (or )n1 = n 2
∴ 1 1
m1c12 m 2c22
2 2
It means that equal volumes of any two gases at the same temperatures and pressures
will have equal number of molecules. This is Avogadro’s law.
d) Dalton’s law of partial pressures : Consider a gas in a vessel of volume V. If ‘n1’ is
the number of molecules ‘m1’ is the mass of a molecule and ‘c1’is the RMS velocity.
According to Kinetic gas equation, pressure of the gas.
1
m1n1c12
P1 = 3
V
If the gas is replaced by another gas in the same vessel we will have
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1
m 2 n 2c2 2
P2 = 3 Where P2 is the pressure of the 2nd gas, m2, the mass of its molecules.
V
n2 the number of its molecule and C2 the R.M.S. velocity of its molecule.
Suppose that the two gases are taken in the same vessel. Let the total pressure’s of the
mixture be ‘P’.
PV = 1/3 m1 n 1 c12 + 1/3 m2 n2 c22
1 1
m1n1c12 m 2 n 2c22
P= 3 +3 or
V V
P + P1 + P2. This is Dalton’s law of particle pressures

e) Graham’s law of diffusion : According to Kinetic gas equation
Pv = 1/3 mnc2
‘mn’ is the expression that represents mass of the gas. If the gas contains Avogadro’s
number of molecules, then ‘mn’ becomes equal to gram molecular mass ‘M’ of the gas
3pv
Pv = 1/3 mnc2 or C =
M
V
Since = = d (density)
M
3p
∴ C=
d
At Constant pressure,
This shows that the RMS velocity or the rate of diffusion of gas is inversely proportional
to the square of its density. This is Graham law of diffusion.
6.6.2 Average Velocities

This is defined as
u1 + u2 + ... + u N
uav =
N
and is given by the expression
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8RT
u av =
πM
6.7 Distribution of Molecular Speeds

According to Maxwell in a gas all the molecules are in a state of constant rapid random
motion and they collide with one another and with the walls of the vessel. During collision
the redistribution of the energy takes place. As a result their speed and the kinetic energy
changes. Therefore at any instant different molecules have different speed and hence different
kinetic energy. At the given temperature even though the speed of the individual molecule
continuously changes, the fraciton of the molecules having the same speed remains constant
and this is known as Maxwell-Bolttzmann Distribution Law.
dN
At the given temperature this fraction is denoted by where dN is number of molecules
N
having the same velocity and N is the total number of the molecules present in the gas. At the
given temperature this fraction of the molecule is plotted against the molecular speed as shown
in figure 6.6.
Most Probable
Velocity
Velocity
Fig. 6.6 : Maxwells’ distribution of velocities at constant temperature
In above Fig. 6.6 the maximum in the distribution curve corresponds to the speed possessed
by the highest fraction of the molecule, this is known as most probable speed. It may be
noted that if the temperature is increased the fraction of the molecule with higher speeds
increase thus the most probable speed increases with increase of temperature. The temperature
dependence of the distribution of the speed is as shown in Fig. 6.7.
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Fig. 6.7 : Effect of temperature on distribution of velocities.
At the given temperature the most probable speed is given by the following expression.
2RT
U mp =
M
the three speeds, root mean square speed, average speed and most probable speed are related
by the following expressions
urms : uav : ump :: 3 : 8/ π : 2
and also 1.224 : 1.128 : 1, so

urms > uav > ump
The most probable velocity ump increases with the use in temperature of gas.
6.8 Deviation from Ideal Gas Behaviour

The gas laws mentioned above are strictly valid for an ideal gas under all conditions of
temperature and pressure. Real gases show deviation from these laws at low temperature and
high pressure. These deviations can be shown clearly by plotting pV as a function of
pressure at constant temperature, nRT
pV Vobserved
= V = Z (compressibility factor)
nRT ideal
Gases deviate from ideal behaviour due to the following faulty assumption of kinetic theory:
1. Volume of the molecules of a gas is negligibly small in comparison to the space

occupied by the gas.
2. There is no force of attraction between the molecules of a gas.
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Contrary to assumption (1), the volume occupied by the molecules of a gas becomes significant
at high pressures. If nb is the volume occupied by the molecules, the actual volume of the gas
is (V-nb). Assumption (2) too doesn’t hold good as at high pressures molecular interactions
start operating. Molecules are dragged back by other molecules which affects the pressure
exerted by them on the walls of the container.
an 2
Pideal = Preal +
V2
an 2
(Preal is observed pressure and is correction term)
V2
In view of the corrections for pressure and volume, ideal gas equation can be rewritten as
⎛ an 2 ⎞
⎜⎝ p +
V 2 ⎟⎠ (V - nb) = nRT
This is known as van der Waals equation

Volume
Fig. 6.8 : The plot volume versus P for real gases
Behaviour of real gases, deviation from ideal behaviour, Compressibility

factor Vs pressures diagram’s of real gases.
Intermolecular forces :
We know about molecular polarities. They give rise to some of the forces that occur between
the molecules. Take H2O for example. H2O molecule consists of two hydrogen atoms and
one oxygen atom joined together in a specific way by the intermolecular forces known as
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“Covalent bonds”. But water exists either as solid ice or liquid water or gaseous steam, depending
on its temperature. For this, there must be some intermolecular forces between molecules that hold
them together at certain temperatures. These intermolecular forces are usually called Vander Walls
forces after J. Vander walls (1837-1923). These forces are of several different types like ion -
dipole, dipole - dipole, dipole - induced dipole and induced dipole - induced dipole (London
dispersion) forces - Hydrogen bond is specific intemolecular force limited to certain kinds of
molecules. All these intermolecular forces are electrical. They result from the mutual attractions of
unlike charges or repulsions of like charges. If the species are ions full charges are present and ion-
ion interactions are very strong (500-1000 k J. mol-1). They give rise to the so called ionic bond’s.
If the molecules are neutral, then partial charges are possible to give substantial attractive forces.
Ion - Dipole force :
Water molecule’s are polar and in them hydrogen atoms possess

partial positive charges and oxygen atoms posses partial negative
charges due to electronegativity different between hydrogen and
_
δ
+
δ
+
oxygen atoms. When ionic compounds like NaCl dissolve in water,
_
δ
δ
they dissociate into its components. Negative and positve charges _
are possible to give substantial attractive forces, they disccociate
_
_
into component ions like. Na+ and Cl . Now the water molecules
δ
+
δ
+
_
δ
δ
orient in the presence of ion’s in such away that the positive end of
the dipole is near an anion and the negative end of the dipole is
near a cation. The magnitude of interaction energy depends on its
+
δ
+
δ
_
charge ‘z’, molecule’s dipole moment ‘m’ and on the inverse square
_
δ
δ
of the distance ‘r’ between the ion and the dipole E = zμ/r2. Ion
+
dipole forces are mainly important in aqueous solutions of ionic
_
_
substance. Such as NaCl in which dipolar water molecules surround

δ
δ
+
+
δ
the ion’s. δ
Dipole - Dipole forces :
Neutral but polar molecules experience dipole - dipole forces.
These are due to the electrical interaction’s among dipoles on
neighbouring molecules. These forces are again attractive
between unlike poles and repulsive between like poles and
depend on the orientation of the molecules. The net forces in a
large collection of molecules results from many individual
interactions of both types. The forces are generally weak δ+ δ− δ+ δ-
(energies of the order of 3-4 kJ mol-1) and are significant only
when the molecules are in close contact. The strength of a given
dipole - dipole interaction depends on the sizes of the dipole
moments involved. The more polar the molecules or the higher
the dipole moment the greater is the strength of interactions and
higher is the boiling points of those substances. Dipole-Dipole
interaction energy between stationary polar molecules as in solid’s
is proportional to 1/r3 and that between rotating molecules is
proportional to 1/r6 ‘r’ is the distance between the polar molecules.
-140-
Dipole - induced Dipole forces :
These forces are between polar molecules with permanent dipole moments and the molecules
with no permanent dipole - moment. Permanent dipole of the polar molecule induces dipole
on the electrically neutral molecule by deforming into electronic cloud. Here also the interaction
range is proportional to 1/r2 where ‘r’ is the distance between the molecules. The magnitude
of induced of dipole moment also depends on the magnitude of dipole moment of permanent
- dipole and polarisability of neutral molecule. Large molecules are easily polarized. Here
also commulative effect of dispersion forces and dipole-induced dipole interactions exist.
There are repulsive forces between the particles (atoms, molecules or ion’s) due to electron-
electron repulsions or (Nucleus-Nucleus) repulsion’s.
Thermal energy :
It is due to the motion of the atoms or molecules of the substance. This energy is directly
proportional to the absolute temperature of the substances. It is a measure of average kinetic
energy of the molecule’s of the substance. The movement of particles is called thermal motion.
Intermolecular forces Vs thermal energy :

Intermolecular forces tend to keep the molecules together but thermal energy tends to keep
them apart. If thermal energy predominates over inter molecular forces the substances would
change from.
Solid → liquid → gas
And if intermolecular forces predominate, then the substances change from
Gas → liquid → Solid.
6.9 Liquifaction of Gas

Any gas can be liquified at atmospheric pressure if it is cooled sufficiently. Many gases (but
not all) can be liquified at ordinary temperature by compressing them.
The conditions of temeperature and pressure under which gases liquify were first investigated
by Andrews in 1869.
Andrews subjected CO2 to various pressures at different temperatures and plotted the effect
of pressure on volume (Fig. 6.9). The curve obtained at a given temperature is called an
isotherm. As can be seen in the figure, at 321K the volume of the gas decreased with the
increased pressure approximately in accordance with the Boyle’s Law. At 294 K, however,
the volume first decreases in accordance with Boyle’s Law until the pressure was increased
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Boyle’s Law until the pressure was increased
to about 60 atm. At this pressure there was a
sudden break in the curve and liquid carbon
dioxide appeared. The pressure been converted
into liquid. Subsequent increase of pressure
caused practically no change in volume. In
accordance with the general rule that extremely
high pressures are required to compress liquids
appreciably.
Similar changes took place when the isotherms

were constructed for temperatures below 294
K, except that the pressure required to liquify
the gas became smaller as the temperature Fig. 6.9 : Isotherms of carbon dioxide
decreased. Andrews found that liquifaction
could be brought about at all temperatures
below 304.1 K. But above this temperature no liquifaction occured no matter how much
pressure was increased. This temperature was therefore called the critical temperature for
CO2. The pressure required to liquify the gas at the critical temperature was called critical
pressure, and the volume of 1 mole of the substance at the critical temperature and pressure,
the critical volume.
The temperature above which a gas cannot be liquified, however large the pressure
may be is known as critical temperature.
Table 6.2 list values of the critical temperature and pressure critical volume for some common
substances.
Table : 6.2 Critical temperatures and critical pressures
Substance Critical Critical
Temperature (K) Pressure (atm)
Water, H2O 647 217.7

Sulphur dioxide, SO2 430 77.7
Ammonia, NH3 406 112.5
Hydrogen Chloride, HCl 324 81.6
Carbon dioxide, CO2 304 73.0
Oxygen, O2 154 49.7
Nitrogen, N2 126 33.5
Hydrogen, H2 33 12.8
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1. What are the conditions under which real gases behave as ideal gas.
.............................................................................................................................................
2. Which term in van der waals equation accounts for the molecular volume.
.............................................................................................................................................
3. Calculate the root mean square velocity of ozone kept in a closed vessel at 200 C and 1 atm
pressure.
.............................................................................................................................................
4. What is compressibility factor.
.............................................................................................................................................

• Matter exists in three states, namely, solid, liquid and gas.
• The three states of matter differ in the relative closeness of the molecules constituting
them.
• There exists a definite relationship between the pressure, volume, temperature and
number of moles of a gas and they are given by Boyle’s law, Charle’s law and
Avogadro’s law.
• The gases obeying gas laws are known as ideal gases.
• Dalton’s law give the relationship between partial pressures exerted by the non-reacting
gases to the total pressure.
• Most of the gases deviate from the ideal behaviour. The deviations of gases from
ideal behaviour is due to the wrong assumptions of kinetic molecular theory.
• Real gases can be liquified under appropriate conditions.
Terminal Exercise
1. Draw the graphs of the following :
a) p vs V at constant T and n.
b) 1/T vs p at constant T and n.
c) T vs V at constant p.
2. What is the volume occupied by one mole of a gas at STP ?
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3. The volume of a sample of gas is 500mL at a pressure of 1.5 atm. If the temperature is
kept constant, what will be the volume of that gas at
i) 1 atm ii) 5.0 atm.
4. List the wrong assumptions of kinetic theory of gases which led to van der Waals
equation.
5. What is the standard temperature and pressure ?
6. What is the lowest possible temperature ?
7. CO2 can not be liquefied at 350C, however large the pressure may be, why ?
8. A sample of nitrogen gas weighing 9.3 g at pressure 0.99 atm occupies a volume of
12.4 litres when its temperature is 55K. What will be its volume when the temperature
is 220 K? Assume pressure is kept constant.
9. Calculate the volume of one mole of oxygen at 270C and 2 atm pressure, given that the
volume of oxygen at STP is 22.4 litres.
10. What is the Maxwell-Boltzmann Law?

6.1
1. Due to more intermolecular distances in gaseous molecule compared to liquid.
2. Boyle Law equation is
p1 V1 = p 2 V2
(.20 atm) (500 mL0 = p2 (10 mL)
(0.20atm)(500ml)
p2 =
10ml
p 2 = 10 atm.
3. By Avogadro’s Law
moles of O2 = moles of unknown gas
2.00g
=
32gmol −1
1.75x 32
Molar mass of unknown gas = = 28 g mol-1
2.00
-144-
6.2
1. Movement of gas molecules through another gas is called diffusion.
When gas escapes from a container through a very small opening it is called effusion.
2. Ammonia and hydrogen chloride gases are reacting gases and Dalton’s Law is applicable
to mixture of non-reacting gases.
1
rO3 ⎛ M co2 ⎞ 2
3. = ⎜ ⎟
rco 2 ⎝ M o3 ⎠
1
0.271 ⎛ 44 ⎞ 2
= ⎜ ⎟
0.290 ⎝ M o3 ⎠
Squaring both sides
(0.271) 2 44
2 =
(0.290) Mo3
44 x 0.29 x 0.29
Mo3 = = 50.4
0.271x 0.271
Molecular mass of O3 = 50.4

4. By ideal gas equation
pV = nRT
p x 1.0 = (5.0 mol) (0.0821 L atm K-1 mol-1) 320 K
p =
p = 131.3 atm.
6.3
1. Low pressure and high temperature.
2. b
3RT
3. U rms =
M
3.(8.314Jk −1mol −1 )(293K ) (8.314Kgm 2s −2 K −1mol −1 )(293K )

= =
(0.048kgmol −1 ) 0.048kgmol −1
= 390.3 ms -1 -145-
7
THE LIQUID STATE
You are familiar with gases, liquids and solids in your daily life. You are aware that water can
exist as a liquid, a solid (ice) or as a gas (vapour). These are called three states of matter. In
lesson 6, you have learnt about the differences in properties of these three states of matter.
The properties of gaseous state can be explained in terms of large separation of molecules and
very weak intermolecular forces. In this lesson we shall study about the intermolecular forces
in liquids and see how their properties can be explained in terms of these forces.
Objectives
• explain the properties of liquids in terms of their structure (molecular arrangement and
intermolecular forces);
• differentiate between evaporation and boiling;
• define vapour pressure of a liquid and correlate it with its boiling point;
• define surface tension and explain the effect of various factors on it;
• explain the consequences of surface tension and
• define viscosity of a liquid and correlate it with intermolecular forces.
7.1 Nature of Liquids

Look at Figure 7.1 in which the molecular arrangement has been shown in the three states of
matter. What do you notice ?
In figure 7.1a, you would find that the molecules are far apart, A gaseous state can be
represented by this arrangement. In liquid state (figure 7.1b), molecules are closer as compared
to gaseous state. You would notice that they have very little spaces between them. However,
there is no order in arrangement of molecules. Further we say that, these molecules can move
-146-
about, but with lesser speeds than those in gases. They can
still collide with one another as in the gaseous state. You would
recall that the molecules in gases have very little attraction
berween them. But in liquid state the attraction between the
molecules is comparatively much stronger as compared to that
in the gaseous state. The attractions are strong enough to keep
the molecules in aggregation. Contrary to this, in solids (Fi.g
7.1a) you notice that the molecules are arranged at the closest
possible distance.
Solid state is a well ordered state and has very strong

intermolecular forces. You would learn more about solids in
lesson 8.
We would say, in a gas there is complete choose due to very

weak termolecular forces, whereas in solids there is a
complete order due to strong forces. Liquids falls between
gases and solid state. Liquid molecules have some freedom
of gases state and some order of solid state. Intermolecular
forces in liqudis are strong enough to keep the molecules
close to one another but not strong enough to keep them in
perfect order.
7.2 Nature of Liquids

In this section you would learn how the properties of liquids can be explained in terms of
molecular arrangement and intermolecular forces. Let us consider a few properties of liquids
as examples.
7.2.1 Volume and Shape

You would recall that the liquids (for example water) take the shape of the container in
which they are kept. However, they have a definite volume. How can you explain the
properties of definite volume and variable shape? In liquids, the attractive forces are strong
enough to keep the molecules moving within a definite boundary. Thus, they maintain a
definite volume. These intermolecular forces are not strong enough to keep them in definite
positions. The molecules can, therefore, move around and take the shape of the container in
which they are kept.
7.2.2 Compressibility
Compressibility of a substance is its ability to be squeezed when a force is applied on it. Let
us study the compressibility of liquids with the help of the following activity.
-147-
Activity 7.1
Aim : To study the compressibility of water.

What is required ?
A 5 mL syringe and water.
What to do ?
i) Take the syringe and fill it with water by pulling out the plunger.
ii) Note the volume of water.
iii) Press the plunger while blocking the nozzle of the syringe with a finger.
What to observe ?
Observe the volume of water in the syringe while pressing the plunger. Does the volume of
water change by pressing the plunger? You would observe that it does not change.
The above activity clearly shows that liquids arr largely incompressible. It is because there
is very little empty space between the molecules. In contrast, the gases are highly
compressible because of large empty spaces between their molecules.
The large difference in the free space in gaseous and liquid states becomes evident from the
fact that the volume occupied by a given amount of a substance in liquid state is 100-1000
times less than that in the gaseous state.
7.2.3 Diffusion
Diffusion is the process of spreading of a substance from a region of higher concentration
to a region of lower concentration. Let us study the phenomenon of diffusion in liquids with
the help of the following activity.
Activity 7.2
Aim : To study the phenomenon of diffusion through water.

What is required ?
A glass, water, blue ink and a dropper.
What to do ?
i) Take some water in the glass.
ii) Add a few drops of blue ink into water with the help of a dropper.
What to observe ?
Observe the water and ink in the beaker.
-148-
Initially the ink does not mix with water. After some time it starts spreading slowly. After a
few hours the whole of water in the glass becomes coloured due to diffusion of ink through
water.
The above activity demonstrates that diffusion occurs in liquids. Why does it happen? Because
the molecules of both the liquids are moving and help in the diffusion process.
7.2.4 Evaporation
You know that water left in an open pan evaporates slowly until the pan becomes dry.
Evaporation is the process by which a liquid changes into vapour. It occurs at all temperatures
from freezing point to boiling point of the liquid.
In a liquid, at any temperature, a small fraction of the molecules is moving with relatively
high velocity. Such molecules have high kinetic energy. These can overcome the
intermolecular attractive forces and escape through the surface of the liquid.
Rate of evaporation of aliquid depends on a number of factors. For example, more is the
surface area, faster will be the evaporation. For faster dying, we increase the surface area by
spreading the wet clothes. If we supply heat to the liquid, evaporation is faster. The wet
clothes dry faster in the sun. The increase in temperature increases the kinetic energy of the
molecules of the liquid and the liquid evaporates at a faster rate. We feel cool after the bath.
Why do we feel so ? It is because during evaporation water takes the heat from our body and
we feel cold.
Now let us compare the rate of evaporation of two liquids, for example, water and alcohol.
Which of these two liquids evaporates faster ? You mut have experience that alcohol
evaporates faster. Why does this happen? The number of molecules escaping from a liquid
depends upon the attractive forces. When these forces are stronger, fewer molecule escape. In
alcohol, these attractive forces are weaker than those in the water. Hence, alcohol evaporates
faster than water.
7.3 Vapour Pressure and Boiling Point

In the previous section you have learnt that liquids evaporate when kept in an open vessel.
Different liquids evaporate to different extent under similar conditions. The extent of
evaporation of a liquid is measured with the help of vapour pressure of a liquid. In this
section, you will study about it and also about the boiling point of a liquid.
7.3.1 Vapour Pressure of a liquid

You know that a liquid placed in an open vessel evaporates completely. If, however, the
liquid is allowed to evaporate in a closed vessel say in stoppered bottle or a bell jar,
evaporation occurs, but after sometime the level of the liquid does not change any further
and become constant. Let us understand how does it happen. In the closed vessel, the
molecules evaporating from the liquid surface are confined to a limited space. These
-149-
molecules may collide among themselves or with the molecules of air and some of them may
start moving towards the surface of the liquid and enter into it. This is known as condensation.
In the beginning, rate of evaporation is greater than the rate of condensation. But as more and
more molecules accumulate in the space above the liquid, rate of condensation gradually
increases. After some time, rate of evaporation becomes equal to the rate of condensation and
an equilibrium state is reached (Fig. 7.2).
After some time At equilibrium the rates of evaporation

Initially molecules evaporate and condensation become equal
condensation begins
Fig. 7.2 : Establishing (vapour liquid) equilibirum under a evacuated jar
The number of molecules in the vapour above the liquid becomes constant. These molecules
exert certain pressure over the surface of the liquid. This pressure in known as (equilibrium
vapour pressure, saturated vapour pressure) or simply as (vapour pressure). The vapour pressure
of a liquid has a characteristic value at a given temperature. For example, vapour pressure
of water is 17.5 torr and that of benzene is 75.00 torr at 200C. The vapour pressure of a liquid
increases with increase in temperature. It is so because at a higher temperature more
molecules have sufficiently high energy to overcome the forces of attraction and escape to
form vapour. A plot of vapour pressure as a function of temperature is called vapour pressure
curve. Figure 7.3 depicts the vapour pressure curves of some liquids.
Temperature/ 0C
Fig. 7.3 : Vapour pressure curves of some liquids
What would happen if we remove some of the vapour from the closed vessel. Would the
vapour pressure of the liquid increase, decrease or remain constant? Vapour pressure of the
liquid would remain constant at that temperature. In the beginning, the vapour pressure
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would decrease after the removal of the vapour, but soon more liquid would evaporate to
maintain the equilibrium and the original vapour pressure would be restored. So the vapour
pressure of a liquid has a definite value at a particular temperature.
7.3.2 Boiling
You must have seen the formation of bubbles at the base of a vessel, in which a liquid is
6 heated. The rate of formation of bubbles increases with increase in heat supplied. What are
F the bubbles made up of? The first bubbles that you see are of the air, which is driven out of
the liquid by increase in temperature. After some time, bubbles of the liquid are formed
throughout it. These bubbles rise to the surface and break. When this happens, we say that the
liquid is boiling. The bubbles of the liquid would form only if its vapour pressure is equal to
the atmospheric pressure.
The temperature at which boiling occurs is called the boilling point of the liquid. At this
temperature the vapour pressure of the liquid is equal to the atmospheric pressure. The
boiling point, therefore, depends upon the atmospheric pressure. For example, water boils
at 1000C at 760 torr and at 97.70C at 700 torr.
The normal boiling point of a liquid is defined as the temperature at which the vapour
pressure of a liquid is equal to one atmosphere or 760 torr.
The boiling point of a liquid depends upon its nature. A more valatile liquid would boil at a
lower temperature than a less volatile liquid. You can refer to figure 7.3 and note that diethyl
ether boils at a much lower temperature than water, because it is highly volatile liquid. The
boiling point of ethanol lies in between those of diethyl ether and water. Vapour pressures
or boilin points of liquids give us an idea of the strength of attractive forces between molecules
in liquids. Liquids having lower boiling points have weaker attractive forces in comparison
tot hose havinng higher boiling points.
You can make a liquid boil at temperature other than normal boiling point. How? simply
alter the pressure above the liquid. If you increase this pressure, you can increase the boiling
point and if you can decrease this pressure you decrease the boiling point. On the mountains,
the atmospheric pressured decrease and therefore boiling point of water also decreases.
People living on hills face problem in cooking their meals. They, therefore, use pressure
cooker. How food is cooked fater in it? The lid of pressure cooker does not allow water
vapour to escape out. On heating the water vapour accumulate and the inside pressure
increases. This makes the water boil at a higher temperature and the food is cooked faster.
7.3.3 Evaporation and Boiling

Evaporation and boiling, both involve conversaion of a liquid into vapour and appear to be
similar. However, they differ from each other in some aspects. Evaporation occurs at all
temperatures from freezing point of a liquid to its boiling point, while boiling occurs at a
definite temperature only i.e., it is boiling point. Evaporation occurs slowly while boiling is
a fast process. Evaporation of a liquid occurs at its surface alone while boiling occurs
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throughout the liquid. These differences between evaporation and boiling have been
summarized in Table 7.1.
Table 7.1 : Differences between evaporation and boiling
S.No. Evaporation Boiling
1. It takes place at all temperatures. It takes place at a definite temperature.

2. It is a slow process. It is a fast process
3. It occurs only at the surface of It occurs throughout the liquid.
the liquid.

1. Match the following.
Column I Column II
i) liquids have a definite volume. a) The molecules in a liquid can

move about
ii) Liquids acquire the shape of b) The molecules in liquids are close
their container. and have very little free space.
iii) Liquid are largely incompressible. c) The inter molecular forces liquid
strong enough to keep the molecules
moving with in a definite space.
2. When a liquid is heated till it starts boiling.

i) What are the small bubbles that appears initially at the bottom and sides of the
vessel made up of ?
...................................................................................................................................
ii) What are the large bubbles that from in the boiling liquid made up of ?
...................................................................................................................................
3. Liquids A, B and C boil at 650C, 120 0C and 900C respectively. Arrange them in the
decreasing order of the strength of intermolecular forces.
7..................................................................................................................................4
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7.4 Surface Tension
Liquids show the effects of inter molecular forces most dramatically in another property,
namely, surface tension. Any molecule in the interior of liquid is equally attracted by
neighbour molecules from all sides and it does not experience any ‘net’ force. On the other
hand, any molecule at the surface of a liquid is attracted by other molecules at the surface of
the liquid or below it. Due to the imbalance of forces, any molecule at the surface experiences
a net inward pull (Figure 7.4). As a result the surface is under tension as if the liquid were
covered with a tight skin (or stretched membrane). The phenomenon is called surface tension,
Quantatively, the surface tension is defined as the force acting on an imaginary line of unit
length drawn on the surface of the liquid and acting perpendicular to it towards the liquid
side as shown in Figure 7.5 It is represented bythe Greek letter gamma, r. Its SI unit is
newton per meter (N m-1) and CGS unit is dyne per centimeter (dyne cm-1). The two units are
related as : 1 Nm-1 = 103 dyne cm-1.
Fig. 7.4 : Forces acting one molecules, at the surface and in bulk of liquids
Surface molecules of a liquid experience a constant inward force. Therefore they have a
higher energy than the molecules in the bulk of the liquid. Due to this reason liquids tend to
have minimum number of molecules at their surface. This is achieved by minimising the
surface area. In order to increase the surface area more molecules must come to the surface.
This can happen only if some energy is supplied or work is done. The energy supplied (or
work done) for increasing the surface area of a liquid by a unit amount is known as its
surface energy. Its units are joule per square meter J m-2 or N m-1 (since 1J = 1Nm). Thus
dimensionally, the surface tension and surface energy are similar quantities and they have
the same numerical value.
Fig. 7.5 : Surface tension force acting on the surface of a liquid.
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Effect of Temperature
One raising the temperature surface tension of a liquid decreases. It completely vanishes at
the critical temperature. This happens due to the following two factors :
i) On heating, the liquids expand. This increases the intermolecular distances.
ii) On heating, the average kinetic energy of molecules and hence their chaotic motion
increases.
Due to both of these factors, the intemolcular forces become weak and the surface tension
decreases.
Effect of adding Surface Active Solutes
The solutes which get more concentrated on the surface of the liquid than in the bulk called
surface active solutes or surfactants. Alcohols are examples of such substances. Their
addition to a liquid lowers its surface tension. The cleaning action of soaps and detergents
is based on this fact.
Some Effects Surface Tension
Surface tension results in many interesting and important properties of liquids. Let us now
study some of them.
i) Spherical Shape of liquid drops
You have already learnt that liquids tend to have a minimum surface area. For a given volume,
the geometrical shape having minimum surface area is a sphere. Hence, liquids have a natural
tendency to form spherical drops, when no external force acts on them. Rain drops are
distroted and the distortion is due to the friction of air.
ii) Wetting and Non- wetting propeties
When a drop of liquid is placed on solid surface, the force of gravity should cause it to
spread out and form a thin layer (Fig. 76). Such a liquid is called a wetting liquid. This
happens in case of most of the liquids. For example, drops of water or alcohol spreadout on
the surface of glass. Some liquids behave differently. When a drop of mercury is placed on
the surface of glass, it does not spread out (Fig. 7.6). Solid liquids are called non-wetting
liquids. non-wetting
liquid
Fig. 7.6 : Establishing (vapour liquid) equilibirum under a evacuated jar

Wetting or non-wetting nature of a liquid depands upon two types of forces. The
intermolecular attractive forces between molecules of a liquid are called cohesive force
while those between the molecules of the liquid and the solid (whose surface is in contact
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with the liquid) are called adhesive forces. If adhesive forces are stronger than cohesive
forces, the liquid would be wetting in nature and when cohesive forces are stronger than
adhesive forces it would be non-wetting in nature on the surface of a particular soild
iii) Capillary Action
Let us carry out the following activity.
Activity 7.3
Aim : To study the capillary action.

What is required ?
Glass capillary tubes, water, mercury and two petri dishes.
What to do ?
i) Take some water in a petri dish.
ii) Dip one end of a 3-4 cm long capillary in it.
iii) Take some mercury in another petri dish.
iv) Dip one end of another 3-4 cm long capillary in it.
What to observe ?
Observe the levels of water and mercury in the capillaries. Is it below or above the levels of
the liquid in petri dishes.
Fig. 7.7 : Capillary Action
You would observe that when end of a capillary tube is dipped in water, it rises in the
capillary as shown in Fig. 7.7(a). On the other hand when one end of a capillary tube is
dipped in mercury, its level falls in the capicllary as in Fig .7.7 (b).
The phenomenon of rise or fall of a liquid in a capillary is known as capillary action. The
rise of water in the glass capillary is due to its wetting nature as the adhesive forces are
stronger then cohesive forces. Water tends to increase the area of contact with glass wall of
the capillary by rising in it. Mercury being non-wetting with respect of glass (its cohesive
forces are stronger than adhesive forces) tends to minimise the area of contact by depressing
inside the capillary.
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iv) Curved menisus
When a wetting liquid such as water is taken in a glass tube, the liquid tends to rise lightly
along the walls of the tube for increasing its area of contact with glass. The surface of the
liquid (meniscus) becomes curved. It is concave in shape (Fig. 7.8 (a)]. When a non-wetting
liquid like mercury is taken a glass tube, it tends to decrease its area of contact and depresses
along the walls of the glass tube. The meniscus is convex in shape in this case (Fig. 7.8)(b)].
Fig. 7.8 : Curved meniscus of liquids

(a) Concavemeniscus (b) Convex meniscus
7.5 Viscosity
Every liquid has the ability to flow. It is due to the fact that molecules in a liquid move freely,
although within a limited space. Water flows down a hill under gravitational force or through
pipes when forced by a pump. Some external force is always required for a liquid to flow.
Some liquids like glycerol or honey flow slowly while others like water and alcohol flow
rapidly. This difference is due to the internal resistance to flow which is called viscosity. The
liquids with higher viscosity flow slowly and are more viscous in nature like glycerol or -
honey. Water and alcohol have lower viscosity and are less viscous in nature. They flow
more rapidly.
Surface of a solid
Fig. 7.9 : Flow of different layers of a liquid
The viscosity is related to the intermolecular forces. Stronger the intermolecular forces
more viscous are the liquids. Let us understand this with the help of Figure 7.9. When a liquid
flows steadily, it flows in different layers with one layer sliding over the other. Such a flow is
known as laminar flow. Consider a liquid flowing steadily on a plane surface. The layer
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closest to it is almost stationary due to adhesive forces. As the distance of the layer from the
surface increases, is velocity increases. Thus different layers move with different velocities.
Due to intermolecular forces (cohesive forces) each layer experiences a force of friction from
its adjacent layers. This force of friction, f between two layers depends upon :
i) area of contact between them A.
ii) distance between the layers, dx.
iii) difference in velocity between the layers, du.
These quantities are related as
du
f = ηA
dx
du
Hence η (Greek letter ‘eeta’) is called the coefficient of viscosity and is the velocity
dx
gradient between the layers.
If A = 1 cm2, du = 1 cm s-1 and dx = 1 cm, then
f=η
Thus, coefficient of viscosity is the force of friction between two parallel layer of the liquid
which have 1 cm2 area of contact, are separated by 1 cm and have a velocity difference of
1cm s-1. It may be noted that f is also equal to the external force which is required to
overcome the force of friction and maintain the steady flow between two parellel layers
having A area of contact, and which are dx distance apart and moving with a velocity
difference of du.
Units
CGS unit of viscosity is dyne cm-2 s. This unit is also known as poise(P). The SI unit of
viscosity is N m-2s or Pas. The two units are related as:
1 pas = 10 P
The unit paise is found to be too large and its submultiples centipoise (1 cP = 10-2 P) and milli
poise (1 mP = 10-3 P) are used for liquids and micropoise (μP = 10-6 P) is used for gases.
Effect of Temperature
Viscosity of a liquid decreases on raising the temperature. It is due to decrease in intermolecular
forces on heating as discussed in previous section (Section 7.4).
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1. Intext
Fill Questions
in the blanks. 7.2
i) A molecule at the surface of a liquid has ...................... energy than the one within the
liquid.
ii) Surface tension of liquid ...................... on cooling.
iii) Meniscus of a non-wetting liquid is ...................... in shape while that of a wetting
liquid is ...................... in shape.
iv) When one end of a glass capillary tube was dipped in a liquid, the level of liquid
inside the capillary was observed to fall. The adhesive forces in this liquid are
...................... than the cohesive forces between the liquid and glass.
v) Liquid X is more viscous than liquid Y. The intermolecular forces in Y are
...................... than in X.
2. What are the SI units of
i) Surface tension.
.............................................................................................................................................
ii) Coefficient of viscosity
.............................................................................................................................................
3. Why do liquids have a tendency to acquire minimum surface area ?
.............................................................................................................................................

• In liquids the intermolecular force are quite strong as compared to gases but weak
enough to allow the molecules to move within a limited space and athe intermolecular
distance is short.
• Liquids have definite volume but no definite shape, are almost incompressible and
can diffuse.
• Liquids eveporate and exert a definite vapour pressure at speficied temperature.
• Boiling point is the temperature at which the vapour pressure of the liquid becomes
eqaul to the external pressure.
• Surface tension is the force acting on an imaginary line of unit length drawn on the
surface of the liquid and acting perpendicular to it towards the liquid side.
• Due to surface tension, liquids tend to have minimum surface area and show the
phenomena of capillary rise or fall and curved meniscus.
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• Viscosity is the internal force of friction to the flow of liquid.
Terminal Exercise
1. Explain the following properties of liquids on in basis of their structure.
i) Volume ii) Shape iii) Compressibility iv) Ability to flow
2. Why diffusion can occur in liquids. Explain.
3. Define i) vapour pressure and ii) boiling point.
4. Differentiate between evaporation and boiling.
5. Explain the effect of temperature on vapour pressure of a liquid.
6. Define surface tension and give its CGS and SI units.
7. What is surface energy ?
8. Why is energy required to increase the surface area of a liquid ?
9. What is the effect of addition of a surface active substance on the surface tension of a
liquid.
10. Why are liquid drops spherical in shape ?
11. What are wetting and non-wetting liquids ?
12. The cohesive forces acting in liquids A and B are C1 and C2 respectively C1 > C2 which
of them would have higher surface tension.
13. Liquid A rises in glass capillary tube. If one drop of its is put on a plane glass surface,
would it spread out or not. Explain.
14. A liquid forms a convex meniscus in glass tube. Commnet on its nature.
15. Define viscosity.
16. What is coefficient of viscosity ?
17. Give CGS and SI units of coefficient of viscosity.
18. What is the effect of temperature on (i) vapour pressure (ii) surface tension and (iii)
viscosity of a liquid ?

7.1
1. i) C; ii) A; iii) B
2. i) Air ii) Liquid
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3. B > C >A
7.2
1. i) more ii) increases iii) convex; concave
iv) Stronger v) weaker
2. i) N m-1; ii) Nm-2s
3. Molecules in the surface of a liquid have higher energy due to an inward force on them.
Therefore liquids tend to have minimum number of molecules in the surface or have
minimum surface area.
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8
SOLID STATE
You are aware that the matter exists in three different states viz., solid liquid and gas. In
these, the constituent particles (atoms, molecules or ions) are held together by different
forces of attraction between them. However, the nature and magnitude of the forces varies.
In the first two lessons of this module you have learnt about the gaseous and the liquid
states of matter. In this lesson you would learn about solid state- a compact state of matter.
The solids are distinguished from a liquid or gas in terms of their rigidity which makes
them occupy definite volume and have a well defined shape. In solid state, the constituent
particles are in close contact and have strong forces of attraction between them. Here, you
would learn about structure, classification and properties of solids.
Objectives
After reading this lesson, you should be able to:
• explain the nature of solid state;
• explain the properties of solids in terms of packing of particles and intermolecular
attractions;
• explain the melting point of a solid;
• differentiate between crystalline and amorphous solids;
• classify the crystalline solids according to the forces operating between the constituent
particles;
• explain different types of packing in the solids;
• define coordination number;
• explain different types of unit cells;
• calculate the number of particles in simple cubic, face-central cubic and body centered
cubic unit cells;
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• define radius ratio;
• correlate the radius ratio with the structure of solids;
• explain the structure of simple ionic compounds and
• explain Frenkel and Schottky defects.
8.1 Nature of Solid State

You have learnt in lesson 6 that according to Kinetic Molecular Theory, the gases consist of
a large number of molecules, which are in constant random motion in all directions in the
available space. These molecules have very weak or negligible forces of attraction between
them. A sample of gas can be compressed, as there is a lot of free space between the molecules
Fig.8.1(a). In liquids Fig.8.1(b) on the other hand the molecules are also in constant motion
but this motion is relatively restricted. Since there is very little free space available between
the molecules the liquids are relatively incompressible.
(a) (b) (c)

Fig. 8.1 : A pictorial representation of the three states of matter :
(a) gas (b) liquid and (c) solid state.
In solid state the constituent particles are arranged in a closely packed ordered arrangement
Fig.8.1(c) with almost no free space. They can just vibrate about their fixed positions. These
are in close contact and cannot move around like the molecules of a gas or a liquid. As a
consequence, the solids are incompressible, rigid and have a definite shape. Like liquids,
the volume of a solid is independent of the size or the shape of the container in which it is
kept.
8.2 Classification of Solids

On the basis of nature of arrangements of the constituent particles the solids are classified
into amorphous and crystalline solids.
8.2.1 Amorphous and Crystalline Solids

In crystalline solids the constituent particles are arranged in a regular and periodic pattern
and give a well defined shape to it. The term ‘crystal’ comes from the Greek word, krustallos
meaning ice. The regular pattern extends throughout the solid and such solids are said to
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have long range order. On the other hand, some solids have only a short range of order. This
means that the particles are arranged regularly in only some regions of the solid and are
relatively disordered in other regions. Such solids are called amorphous solids. In Greek, a
means without and morph means form. Thus the word amorphous means without form.
Sodium chloride and sucrose are common examples of crystalline solids while glass, fused
silica, rubber and high molecular mass polymers are some examples of amorphous solids.
An important difference between the amorphous and crystalline solids is that while amorphous
solids are isotropic in nature (i.e., these exhibit same value of some physical properties in all
directions) the crystalline solids are anisotropic (i.e., the values of some physical properties
are different in different directions). Refractive index and coefficient of thermal expansion
are typical physical properties, Which have different values between amorphous and crystalline
solids is that while crystalline solids have a sharp or definite melting point, whereas the
amorphous solids do not have definite melting point, these melt over a range of temperature.
The crystalline solids can be further classified on the basis of nature of interaction between
the constituent particles as discussed below.
8.2.2 Classification of Crystalline Solids

In crystalline solids the constituent particles are arranged in an ordered arrangement and are
held together by different types of attractive forces. These forces could be coulombic or
electrostatic, covalent, metallic bonding or weak intermolecular in nature. The differences in
the observed properties of the solids are due to the differences in the type of forces between
the constituting particles. The types of forces binding the constituent particles can be used as
a basis for classification of crystalline solids. On this basis, the crystalline solids can be
classified into four different types- ionic- molecular, covalent and metallic solids. The
characteristics and the properties of different types of solids are compiled in Table 8.1.
Table 8.1 : Characteristics and properties of different types of solids.
Type of Constituent Nature of Appearance Melting Examples

Solid Particles interaction Point
between the
particles
Ionic Ions Coulombic Hard and High Sodium chloride

brittle zinc sulphide, etc
Molecular Molecules Water, carbon
Non polar van der Walls Soft and low dioxide, iodine etc
Polar Dipole-dipole brittle
Covalent Atoms Covalent bond- Hard Very high Diamond, graphite
ing silica, etc.
Metallic Atoms Metallic bonding Hard and Variable Copper, silver, etc.
malleable
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Sodium chloride is an example of an ionic solid because in this case the sodium ions and
chloride ions are attracted to each other by electrostatic interactions. Iodine on the other hand
is an example of a molecular solid because in this the molecules are held together by weak
vander Waals forces. Diamond, with strong covalent between the constituent carbon atoms is
an example of covalent solids while in metals a large number of positive cores of the atoms
are held together by a sea of electrons.
8.3 Properties of Crystalline Solids

You are familiar with the following properties of solids on the basis of handling solids in day
to day work.
• Solids are rigid in nature and have well defined shapes
• Solids have a definite volume irrespective of the size and shape of the container in
which they are placed.
• Solids are almost incompressible.
You are familiar with a number of crystalline solids like sugar, rock salt, alum, gem stones,
etc. You must have noticed that such solids have smooth surfaces. These are called ‘faces’
of the crystal. These faces are developed in the process of crystal formation by ordered
arrangement of the constituent particles. It is generally observed that the faces of crystals
are developed unequally. The internal angle between a pair of faces is called interfacial
angle and is defined as the angle between the normals to the intersecting faces. An important
characteristic of crystalline solids is that irrespective of the size and shape of the crystal of
a given substance, the interfacial angle between a pair of faces is always the same. This fact
was stated by Steno as the law of constancy of interfacial angles (Fig. 8.2).
Fig. 8.2 : The constancy of interfacial angles
8.3.1 Melting Point of a Solid

What is the effect of heat on a solid? You would have observed that when a solid is heated
it becomes hot and eventually gets converted into a liquid. This process of conversion of a
solid to a liquid on heating is called melting. You would also have observed that different
solids need to be heated to different extents to convert them to liquids. The temperature at
which a solid melts to give a liquid is called its melting point. Every solid is characterized
by a definite melting point. This in fact is a test of the purity of the solid. The melting point of
a solid gives us an idea about the nature of binding forces between constituent particles of the
solid. Solids like sodium chloride (m.p = 1077 K) have very high melting points due to strong
coulombic forces between the ions constituting it. On the other hand molecular solids like
naphthalene (m.P. 353 K) have low metling points.
The effect of heat on a solid can be understood in terms of energy and motion of the constituent
particles. You are aware that in a solid the constituent particles just vibrate about their mean
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position. As the heat is supplied to the solid, the constituent particles gain energy and start
vibrating more vigorously about their equilibrium positions. As more and more heat is supplied,
the energy keeps on increasing and eventually it becomes greater than the binding forces
between them. As a consequence the solid is converted into a liquid.

1) Differentiate between solid, liquid and gaseous state.
...................................................................................................................................
2) How are solids classified on the basis of the intermolecular forces.
...................................................................................................................................
3) What is Steno’s law of constancy of interfacial angels ?
...................................................................................................................................
8.4 Close Packed Structures of Solids

In the process of the formation of crystal the constituent particles get packed quite closely.
The crystal structures of the solids can be described in terms of a close packing of identical
spheres as shown in Fig. 8.3 These are held together by forces of attraction. Let us learn
about the possible close packed structures of solids and their significance.
Fig. 8.3 : Arrangement of identical spheres in one dimension
A linear horizontal arrangement of identical spheres in one dimension forms a row (Fig.
8.3). A two dimensional close packed structure can be obtained by arranging a number of
such rows to form a layer. This can be done in two possible ways. In one of these, we can
place these rows in such a way that these are aligned as shown in (Fig. 8.4 (a)). In such an
arrangement each sphere is in contact with four other spheres. This arrangement in two
dimensions is called square close packing.
(a) (b)
Fig. 8.4 : (a) Square close packing and (b) hexagonal close packing
of identical spheres in two dimensions
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In the other way we can place the spheres of the second row in the depressions of the first row
and so on so forth (Fig. 8.4 (b)). You may notice that in such an arrangement each sphere is
in contact with six other spheres. Such an arrangement in two dimensions is called hexagonal
close packing. In such a packing, the spheres of the third row are aligned with the first row.
You may also have noticed that in the hexagonal close packed the spheres are more effectively
packed. In Fig. 8.4 an equal number of identical spheres are arranged in two different types of
packing.
A three dimensional structure can be generated by placing such two dimensional layers on
top of each other. Before we move on to the three dimensional packing let us look at the
hexagonal close packed layer some what more closely (Fig. 8.5).
IIIrd row
IInd row
Ist row
Fig. 8.5 : A hexagonal close packed layer showing two types of triangular voids.
You may note from Fig. 8.5 that in a hexagonal close packed layer there are some unoccupied
spaces or voids. These are triangular in shape and are called trigonal voids. You can further
note that there are two types of triangular voids, one with the apex pointing upwards and the
other with the apex pointing downwards. Let us call these as X type and Y type voids
respectivly as makred in the Fig. 8.5
Close Packed Structures in three dimensions
Let us take a hexagonal close packed layer and call it A layer and place another hexagonal
close-packed layer (called the B layer) on it. There are two possibilities.
1. In one, we can place the second layer in such a way that the spheres of the second
layer come exactly on top of the first layer.
2. In other, the spheres of the second layer are in such a way that these are on the
depressions of the first layer. The first possibility is similar to square close packing
discussed above and is accompanied by wastage of space. In the second possibility
when we place the second layer into the voids of the first layer, the spheres of the
second layer can occupy either the X or Y types trigonal voids but not both. You may
verify this by using coins of same denomination. You would observe that when you
place a coin on the trigonal void of a given type, the other type of void becomes
unavailable for placing the next coin (Fig. 8.6).
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Fig. 8.6 : Two layers of close packed spheres, the second layer occupies
only one type (either X or Y) of triangular voids in the first layer.
In this process, the sphere of second layer covers the trigonal voids of the first layer. It
results into voids with four spheres around it, as shown in Fig. 8.7 (a). Such a void is called a
tetrahedral void since the four spheres surrounding it are arranged on the corners of a regular
tetrahedron, Fig. 8.7 (b). Similarly, the trigonal voids of the second layers will be placed over
the spheres of the first layer and give rise to tetrahedral voids.
(a) (b)
Fig. 8.7 : A tetrahedral void
In a yet another possibility, the trigonal voids of the first layer have another trigonal void of
the opposite type (B type over C and C type over B type) from the second layer over it. This
generates a void which is surrounded by six spheres, Fig. 8.9 (a). Such a void is called an
octahedral void because the six spheres surrounding the void lie at the corners of regular
octahedron, Fig. 8.8 (b).
(a) (b)
Fig. 8.8 : An octahedral void
A closer look at the second layer reveals that it has a series of regularly placed tetrahedral
and octahedral voids marked as ‘t’ and ‘o’ respectively in Fig. 8.9
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Fig. 8.9 : The top view of the second layer showing the tetrahedral and octahedral voids
Now when we place the third layer over the second layer, again there are two possibilities
i.e., either the tetrahedral or the octahedral voids of the second layer are occupied. Let us
take these two possibilities. If the tetrahedral voids of the second layer are occupied then
the spheres in the third layer would be exactly on top (i.e., vetically aligned) of the first or
A layer the next layer (4th layer) which is then placed would align with the B layer. In other
words, every alternate layer will be vertically aligned. This is called AB AB..... pattern or
AB AB ... repeat. On the other hand if the octahedral voids of the second layer are occupied,
the third layer is different from both the first as well as the second layer. It is called the C
layer. In this case the next layer, i.e., the fourth layer, howsoever it is placed will be aligned
with the first layer. This is called ABC ABC ... pattern or ABC ABC... repeat. In three
dimensional set up the AB AB ... pattern or repeat is called hexagonal closed packing
(hcp) (Fig. 8.10 (c) while the ABC ABC .... pattern or repeat is called cubic closed packing
(ccp) (Fig. 8.10 (a)).
A Layer
A Layer
A Layer
C Layer C Layer
B Layer
A Layer B Layer
B Layer
A Layer
A Layer
(a) (b) (c)
Fig. 8.10 : a) Cubic closed packing (ccp) as a result of ABC pattern of close
packed spheres: b) the layers in a) tilted and brought closer to
shows fcc arrangement c) hexagonal closed packing (hcp) as a
result of ABAB pattern of close packed spheres.
This process continues to generate the overall three dimensional packed structure. These
three dimensional structures contain a large number of tetrahedral and octahedral voids. In
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general there is one octahedral and two tetrahedral voids per atom in the close packed structure.
These voids are also called as interstices. As mentioned earlier, the identical spheres represent
the positions of only one kind of atoms or ions in a crystal structure. Other atoms or ions
occupy these interstices or voids.
In the close packed structures (hcp and ccp) discussed above each sphere is in contact with
six spheres in its own layer (as shown in Fig. 8.5) and is in contact with three spheres each of
the layer immediately above and immediately below it. That is each sphere is in contact with
a total of twelve spheres. This number of nearest neighbor is called its coordination number.
The particles occupying the interstices or the voids will have a coordination number depending
on the nature of the void. For example an ion in a tetrahedral void wil be in contact with four
neighbours i.e., would have a coordination number of four. Similarly the atom or ion in an
octahedral void would have a coordination number of six.

1) What is the difference between the square close packed and hexagonal close packed
structures ?
...................................................................................................................................
2) Which of the above two, is more efficient way of packing ?
...................................................................................................................................
3) Clearly differentiate between, triogonal, tetrahedral and octahedral voids.
...................................................................................................................................
8.5 Crystal Lattices and Unit Cells

You know, the crystalline solids have long-range order and the closely packed constituent
particles are arranged in an ordered three dimensional pattern. The structure of the crystalline
solids can be represented as an ordered three dimensional arrangement of points. Here each
point represented the location of a constituent particle and is known as lattice point and such
an arrangement is called a crystal lattice or space lattice or simply a lattice.
To understand the meaning or the term lattice let us first take a repetitive pattern in_ two
dimension. In the crystal structure of sodium chloride in two dimensions the Na+ and Cl ions
are arranged in an ordered fashion as shown in Fig. 8.11 (a). If the position of each ion is
represented as a point then the same crystal can be represented as an array of such points in
two dimensions (Fig. 8.11 (b)). It is called a two dimensional lattice.
Fig. 8.11 : a) A two dimensional arrangement of ions in sodium chloride
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(b) the lattice corresponding to the arrangement of ions in (a)
Similarly, in three dimensions, the crystal structure of a solid is represented as a three
dimensional array of lattice points. Remember that the lattice points represent the positions
of the constituent particles of the solid (Fig. 8.12).
Fig. 8.12 : Schematic representation of a three dimensional crystal lattice.
In a crystal lattice we can select a group of points which can be use to generate the whole
lattice. Such a group is called repeat unit or the unit cell of the crystal lattice. The shadded
region in the Fig. 8.12 represent a unit cell of the crystal lattice. The unit cell is characterized
by three distances along the three edges of the lattice and the angles between them as shown
in the figure. We can generate the whole crystal lattice by repeating the unit cell in the three
directions.
On the basis of the external appearance the known crystals can be classified into seven types.
These are called crystal systems. In terms of the internal structure also the crystal lattices
contain only seven types of unit cells. The seven crystal systems and the definition of their
unit cells in terms of their unit distances and the angles are compiled in Table 8.2 The seven
simple unit cells are given in Fig. 8.13
Table 8.2 : The seven crystal systems and their possible lattice types.
Systems Axes Angles Possible lattice

types
Cubic a=b=c a = β = γ = 900 P, F, I

Tetragonal a=b≠c a = β = γ = 900 P, I
Orthorhombic a≠b=c a = β = γ = 900 P, F, I, C
Rhombohedral a=b=c a = β = γ ≠ 900 P
Hexagonal a=b≠c a = β = 900 ; γ = 1200 P
Monoclinic a≠b=c a = γ = 900 ; β ≠ 900 P, I
Triclinic a≠b≠c a ≠ β ≠ γ ≠ 900 P
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* P = primitive, I = body centered, F = face centered and C = side centered
The unit cell shown in Fig. 8.12 and the ones given in Fig. 8.13 have the lattice points at the
corners only.
Cubic Orthorhombic Hexagonal
Triclinic Monoclinic Tetragonal Rhombohedral

Fig. 8.12.a : The primitive unit cells; the relative dimensions of the three repeat
distances (a, b and c) and the angles between them (a, β and γ) are given in Table 8.2
Such unit cells are called primitive (P) unit cells. Sometimes, the unit cell of crystal contains
lattice point(s) in addition to the ones at the corners. A unit cell containing a lattice point
each at the centers of its faces in addition to the lattice points at the corners is called a face
centered (F) unit cell. On the other hand a unit cell with lattice points at the center of the
unit cell and at the corners is called a body centered unit cell (I). In some cases, in addition
to the lattice points at the corners there are two lattice points located at the centers of any
two opposite faces. These are called as end centered (C) unit cells. The possible lattice types
in different crystal systems are also indicated in Table 8.2 The seven crystal systems when
combined with these possibilities give rise to 14 lattice types. These are called Bravais lattices.
X-Ray diffraction of Crystals (X-rays and Crystal system)

The arrangement of particles in solid structures are determined by x-ray diffraction and to a
lesser extent by electron diffraction. These processes are based on the fact that both x-rays
and electrons have wave lengths which are of the same order as interatomic distances
(10-10 m) and so crystals may be used as diffraction grating for them. The fundamentals of
x-rays are discussed below.
X-ray diffraction :
Bragg Method : The Bragg’s or reflection method gives results which are easier to interpret.
X-rays are produced when cathode rays fall on metlas. The effect of atomic sized particles
on metal is similar to the effect in a pond, a series of ripples (or water waves) are produced
when the handful of pebbles thrown into a pond. In a pond, a series of ripples (or water
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waves) are produced and the crests of some of them will meet and reinforce each other. While
some crests will meet through and cancel each other and cause interference. In the same way,
two x-ray waves that are in same phase reinforce each other and produce a wave that is
stronger than either of the original waves. The resultant wave has a greater amplitude than
either of the primary waves, but the wave length, λ, remains the same. Two x-rays waves that
are completely out of phase cancel each other, the resultant has negligible intensity.
In crystals there are regular layers of atoms or ions and it is possible to calculate the conditions
under which reinforcement will occur if a beam of x-ray strikes them. When x-rays are
diffracted by the layers they behave as if they are being reflected.
Fig. 8.5a illustrates the determination of crystal spacing’s by the use of x-rays of a single
wave length. The rays inspinge upon parallel planes of the crystal at an agnle θ, an angle of
reflection equals the angle of incidence. Some of the rays are reflected from the upper plane,
some from the second plane, and some from the lower planes. A strong reflected beam will
result only if the rays are in phase. In the illustration, the ray DFH travels further the ray ABC
by an amount equal to EF + FG. These rays will be in phase at CH only if this difference
equals a whole number of wave length. Thus EF + FG = nλ.
When ‘n’ is a single integer.
Fig. 8.13 : Bragg’s method
The line BE is drawn perpendicular to DF. Angle BEF, therefore equals 900, since the sum
of the angles of any triangle equals 1800, the sum of other two angles (EBF and EFB) of the
triangle BEF must also equal 900.
That is :
∠ EBF + ∠ EFB = 900
Angle XFB is a right angle, and angle XFE is θ, therefore angle EFB is equal to 900, minus
θ, consequently.
∠ EBF + (900 - θ) = 900
and angle EBF equals θ.
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The sine of this angle, θ, is equal EF/BF, and since BF is equal to d (the distance between the
planes of the crystal).
Sin θ =
or EF = d Sin θ
Likewise, FG = dSin θ
EF + FG = 2d Sin θ
or nλ = 2d Sin θ
This equation was derived by William Henry Bragg and his son William Lawrence Bragg.
The Bragg equation be rearranged as Sin θ =
Thus with x-rays of a definite wave length, reflections of various angles will be observed for
a given set of planes with a spacing equal to d. These reflections correspond to n = 1,2,3 and
so as, and are spoken of as first, second, third order and so on. With each successive order,
the angle increases, and the intensity of reflected beam weakens.
Sample Problem :
The diffraction of a crystal of barium with x-ray of wave length 2.29 A0 gives first order
reflection at 27081. What is the distance between the diffracted planes ?
Solution :
nλ = 2d Sin θ Bragg’s relation
d =
= (Since sin 27081 = 0.456)
= 2.51 x 10 -8 cm
= 2.51 A0
8.5.1 Cubic Unit Cells

Of the seven crystal systems, let us discuss unit cells belonging to the cubic crystal system in
somewhat details. As you can see from Table 8.2 that in the cubic crystal system the three
repeat distances are equal and all three angles are right angles. The unit cells of three possible
lattice types viz., primitive or simple cubic, body centered cubic and the face centered cubic,
belonging to cubic crystal system are shown in Figure 8.14.
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(a) (b) (c)
Fig. 8.14 : Bragg’s method
Number of atoms per unit cell

As you know that in unit cells the atoms can be on the corners, in the body center and on face
centers. All the atoms do not belong to a single unit cell. These are shared amongst diffrent
unit cells. It is important to know the number of atoms per unit cell. Let us learn how to
compute these for different cubic unit cells.
Simple Cubic Unit Cell
The simple or primitive unit cell has the atoms at the corners of the cube (Fig. 8.14 (a)). A
lattice point at the corner of the unit cell is shared by eight unit cells as you can see from the
encircled atom inthe Fig. 8.15. Therefore, the contribution of an atom at the corner to the unit
cell will be 1/8. The number of atoms per unit cell can be calculated as follows :
Number of corner atoms = 8.
Number of unit cells sharing atoms of the corner = 8.
The number of atoms in a simple cubic unit cell = 8 x 1/8 = 1.
Fig. 8.15 : A corner lattice point is shared by eight unit cells
Body Centered Cubic Unit Cell

A body centered cubic (bcc) unit cell has lattice points not only at the corners but also at the
center of the cube (Fig. 8.14 (b)). The atom in the center of the cube belongs entirely to the
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unit cell, i.e., it is not shared by other unit cells. The corner atoms, on the other hand, as in the
case of simple cubic unit cell, are shared by eight unit cells. Thus the number of atoms per
unit cell can be calculated as
Number of unit cells sharing atoms of the corner = 8.
Contribution to the unit cell = 8 x 1/8 = 1
Number of atoms at the center of the cube = 1.
Contribution to the unit cell = 1 (as it is not shared)
The number of atoms in a body centered cubic unit cell = 1 + 1 = 2
Face Centered Cubic Unit Cell
A face centered cubic (fcc) unit cell has atoms not only at the corners but also at the center of
each face. Thus it has eight lattice points at the corners and six at the face centers (Fig. 8.14
(c)). A face centered lattice point is shared by two unit cells, Fig. 8.16. As before,
Fig. 8.16 : A face centered lattice point is shared by two unit cells

Number of unit cells sharing these = 8.
Contribution to the unit cell = 8 x 1/8 = 1
Number of atoms at the center = 1.
Number of unit cells sharing a face centered lattice point = 2
Contribution of the face centered atoms to the unit cell = 6 x 1/2 = 3
The number of atoms points in a face centered cubic unit cell = 1 + 3 = 4.
The number of atoms per unit cell in different types of cubic unit cells is given in Table 8.3
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Table 8.3 : Atoms per unit cell
Simple cubic 1
Body centered cubic 2
Face centered cubic 4
8.5.2 Structures of Ionic Solids

In case of ionic solids that consist of ions of different sizes, we need to specify the positions
of both the cations as well as the anions in the crystal lattice. Therefore, structure adopted
by an ionic solid depends on the relative sizes of the two ions. In fact it depends on the
ratios of their ratios of their radii called radius ratio. Here r + is the radius of the cation and
r - is that of the anion. The radius ratios and the corresponding structures are compiled in
Table 8.4.
Table 8.4 : The radius ratios (r + / r-) and the corresponding structures
Radius ratio (r + /r-) Coordination number Structure adopted
0.225-0.414 4 Tetrahedral
0.414-0.732 6 Octahedral
0.732-0.91 8 Body centered cubic
> = 1.00 12 Close Packed structure
The common ionic compounds have the general formulae as MX, MX2 and MX3 where M
represents the metal ion and X denotes the anion. We would discuss the structures of some
ionic compounds of MX and MX 2 types.
8.5.2.1 Structures of the Ionic Compounds of MX Type

For the MX type of ionic compounds three types of structures are commonly observed. These
are sodium chloride, zinc sulphide and caesium chloride structures. Let us discuss these in
some details.
Caesium Chloride Structure
In CsC1 the cation and the anions are of comparable
Calsium Ion
sizes (the radius ratio = 0.93) and has a bcc structure
in which each ion is surrounded by 8 ions of
opposite type. The Cs+ ions is in the body center
position and eight Cl-ions are located at the corners Chloride Ions
(fig. 8.17) of the cube. Thus it has a coordination
number of 8. Fig. 8.17 : Caesium chloride structure
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Sodium Chloride Sturcture
In case of NaCl the anion (Cl-) is much larger than the cation (Na+). It has a radius ratio of
0.52. According to Table 3.3 it shoud have an octahedral arrangement. In sodium chloride
the (Cl-) form a ccp (or fcc) structure and the sodium ion occupy the octahedral voids. You
may visualise the structure having chloride ions at the corners and the face centers and the
sodium ions at the edge centers and in the middle of the cube (Fig. 8.18).
Fig. 8.18 : Sodium Chloride structure
Zinc Sulphide Structure

In case of zinc sulphide the radius ratio is just = 0.40. According to Table 3.3 it should have
an tetrahedral arrangement. In Zinc sulphide structure, the sulphide ions are arranged in a
ccp structure. The zinc ions are located at the corners of a tetrahedron, which lies inside the
cube as shown in the Fi.g 8.19. These occupy alternate tetrahedral voids.
Fig. 8.19 : Zinc Sulphide structure
Calcium flouoride or fluorite structure

In this structure the Ca2+ ions form a fcc arrangement and the fluoride ions are located in the
tetrahedral voids (Fig. 8.20).
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Fig. 8.20 : Calcium fluoride or Fluorite structure; calcium ions occupy the corners
of the cube and face centers the F- ions are on the corners of the smaller cube.
Antifluorite Sturcture
Some of the ionic compounds like Na2O have antiflourite structure. In this structure the
poisitions of cations and the anions in fluorite structures are interchanged. That is why it is
called anti fluorite structure. In Na2O the oxide ions form the ccp and the sodium ions
occupy the tetrahedral voids (Fig. 8.21).
Fig. 8.21 : Antifluorite structure adopted by Na2O; The oxide ions occupy the corners
of the cube and face centers and the Na+ ions (shown in black) are on the corners
of the smaller cube.
8.6 Defects in Ionic Crystals

you have leartn that in a crystalline solid the constituent particles are arranged in a ordered
three dimensional network. However, in actual crystals such a perfect order is not there.
Every crystal has some deviations from the perfect order. These deviations are called
imperfections or defects. These defects can be broadly grouped into two types. These are
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stoichiometric and non-stoichiometric defects depending on whether or not these disturb
the stiochiometry of the crystalline material. Here, we would deal only with stoichiometric
defects. In such compounds the number of positive and negative ions are in stoichiometry
proportions. There are two kinds of stoichiometric defects, these are
• Schottky defects
• Frenkel defects
Schottky defects : this type of defect are due to the absence of some positive and negative
ions from their positions. These unoccupied lattice sites are called holes. Such defects are
found in ionic compounds in which the positive and negative ions are of similar size e.g.,
NaCl and CsCl. the number of missing positive and negative ions is equal. The presence of
Schottky defects decreases the density of the crystal (Fig. 8.22(a)).
Frenkel defects : this type of defect arise when some ions move from their lattice positions
and occupy interstitial sites. The interstitial sites refer to the positions in between the ions.
When the ions leaves its lattice site a hole is created there. ZnS and AgBr are examples of
ionic compounds showing Frenkel defects. In these ionic compounds the positive and negative
ions are of quite different sizes. Generally the positive ions leave their lattice positions, as
these are smaller and can accommodate themselves in the interstitial sites. The Frenkel defects
do not change the density of the soilds (Fig. 8.22(b)).
Fig. 8.22 : Stoichiometric defects a) Schottky and b) Frenkel defects

These defects cause the crystal to conduct electricity to some extent. The conduction is due to
the movement of ions into the holes. When an ion moves into a hole it creates a new hole,
which in turn is occupied by another ion, and the process continues.

a) What do you understand by crystal lattice ?
.............................................................................................................................................
b) What is a unit cell ?
.............................................................................................................................................
c) How many atoms are there in a fcc unit cell ?
.............................................................................................................................................
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8.7 Treatment of metallic bond (elementary ideas)
The bonding in metals and their alloys is not completely understood. And therefore structures
are also not fully, realized. Any theory that is proposed for the bonding in metals must be
capable of explaning the following.
i) Bonding between atoms of the same element (identical atoms) and also between atoms
of widely differing metlas as well as bonding in alloys.
ii) Should not involve directional bonding.
iii) The properties as metals in solutions and in liquid states etc.
iv) The mobility of electrons.
Various theories of bonding in metals have been suggested. The elementary ideas about
these are given in the section that follows.
8.7.a Electron sea model :
This thoery was initially proposed by Drude in 1900. This was further refined by Lorentz in
1923. This theory is also known as. Drude - Lorentz theory.
According to this theory.
i) A metal lattice comprises of rigid spheres of metal ions.
ii) Each metal atom contributes its valence electrons to the sea.
iii) These electrons move freely in the lattice spaces i.e., interstices.
iv) Cohesive forces result from electrostatic attractions betwen the positive metal ions and
the electron cloud.
v) “The force that binds a metal ion to the mobile electrons within its sphere of influence is
known as metallic bond.”
_ mobile Electrons
+ metal ion Electrons
Fig. 8.23 : Electron sea model of a metal

Simply a metal was regarded as an asembly of positive ions immersed in a sea or pool of
valence electrons. This theory explain qualitatively the model of a metal lattice but fails to
explain quantitative calculations for lattice energies of ionic compounds.
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8.7.b Valence bond theory of metals :
The theory was poposed by Linus Pauling 1937. This theory also referred as Resonance
theory. According to this theory the metalic bond, is essentially a polar or a non-polar covalent
bond. This is covalent bond involves resonance, between a number of structures, having
one electron and electron pair bonds. As there is a possibility of insufficient valence electrons
for the formation of electron pair bonds with each atom of the metal, it is assumed that
resonance takes place throughout the solid metal. This resonance not only involves covalent
bonds but also ionic linkages. The atoms undergo hybridization. For example, the resonance
structure of sodium metal is represented taking four Na atoms only.
Na _ Na Na Na Na Na Na Na
Na _ Na Na Na Na Na Na Na
I II III IV
Na _ Na Na _ Na
Na Na Na Na
V VI
Fig. 8.7 b
This theory does not explain the conduction of the heat in solids or their luster or the reflection
of the metallic, properties either in the liquid state or in the solution.
• In solid state the constituent particles are arranged in a closely packed ordered arrangement
with almost no free space. These are held together by strong forces of attraction and
vibrate about their fixed positions. Solids are incompressible and rigid and have definite
shapes.
• Solids are classified into amorphous and crystalline solids. The crystalline solids have
long range order while amorphous solids have only short range order.
• The crystalline solids can be classified into four different types-ionic, molecular, covalent
and metallic solids on the basis of nature of forces of attraction between the constituent
particles.
• The temperature at which a solid melts to give a liquid is called its melting point.
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• The crystal structure of the solids can be described in terms of a close-packing of identical
spheres.
• In three dimensions there are two ways of packing identical spheres. These are hexagonal
closed packing (hcp) and cubic closed packing (ccp). The hcp arrangement is obtained
by ABAB repeat of the two dimensional layers whereas the ccp arrangement is obtained
by ABCABC repreat.
• The three dimensional internal structure of a crystalline solid can be represented in
terms of a crystal lattice in which the location of each constituent particle is indicated
by a point.
• The whole crystal lattice can be generated by moving the unit cell in the three directions.
• On the basis of the external appearance the known crystals can be classified into seven
types called crystal systems.
• The unit cells of cubic crystal system has three possible lattice types. These are simple
cubic, body centered cubic and the face centered cubic.
• The atoms at the corner of a cubic unit cell is shared by eight unit cells while a face
centered atom is shared by two unit cells. The atom at the body center, on the other
hand is exclusive to the unit cell as it is not shared.
• The number of atoms per unit cell for the simple cubic, bcc and fcc unit cells are 1.2
and 4 respectively.
• The structure adopted by an ionic solid depends on the ratios of their radii (r+/r-) called
radius ratio.
• The structures of some simple ionic solids can be described in terms of ccp of one type
of ions and the other ions occupying the voids.
• Actual crystals have some kind of imperfections in their internal structure. These are
called defects.
• There are two types of defects called stoichiometric and non-stoichiometric defects
depending on whether or not these disturb the stoichiometry of the crystalline material.
• There are two kinds of stoichiometric defects, these are called Schottky defects and
Frenkel defects.
Terminal Exercise
1. Outline the differences between a crystalline and an amorphous solid.
2. How can you classify solids on the basis of the nature of the forces between the
constituent particles ?
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3. What do you understand by the melting point of a solid ? What information does it
provide about the nature of interaction between the constituent particles of the solids?
4. What do you understand by coordination number? What would be an ion occupying
a/n octahedral void...?
5. Explain the following with the help of suitable examples.
a) Schottky defect
7 b) Frankel defect
F

8.1
1. Solids have definite shape and definite volume.
Liquids have indefinite shape but define volume.
gases have indefinite shape and indefinite volume.
2. Coulombic forces, dipole-dipole attractions, covalent bonding and metallic bonding.
3. Irrespective of the size and shape of the crystal of a substance, the interfacial angle
between a pair of faces is always the same.
8.2
1. Refer to section 8.4
2. Hexagonal close packed
3. Refer to sections 8.4
8.3
1. Ordered three dimensional arrangement of points representing the location of constituent
particles.
2. A select group of points which can be used generate the whole lattice. Unit cell is
characterised by three edges of the lattice and angles between them.
3. Four.
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CHEMISTRY
Maximum Marks : 50 Time : 1 1/2 Hrs.
INSTRUCTIONS :
· Answer all the questions on a separate sheet of paper.
· Give the following information on your answer sheet.
· Name
· Enrolment Number
· Subject
· Assignment Number
· Address
· Get your assignment checked by the subject teacher at your study centre so that you get
positive feedback about your performance.
1. a) Define Lattice energy.

b) What an electrovalent bond.
c) State charles’ law and given mathematical expression for it.
d) Why do molecular crystals have low melting point?
e) Why do you feel cool after a both?
f) State valence shell electron pair repulsion theory.
g) State down van der Waal’s equation.
h) What is Kelvin scale of temperature?
i) State Avogadro law.
j) Liquids have a definite volume. Explain
2. a) Name the parameters of a chemical bond and define any one of them.
b) List four characteristics of a covalent compound.
c) Differenciate π- bond from σ - bond
d) Define critical temperature and critical pressure of a gas
e) List the four types of crystalling solids and give one example of each.
f) Vapour pressure of a liquid at 250C is 35 mm Hg and its normal boiling points is
1100C. What will be its vapour pressure at 1100C?
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g) Which assumptions of kinetic theory of gases are wrong?
h) Write two differences between amorphous and crystalline solids.
i) How will you calculate the enthalpy of formations of NaCl using Born-Haber
Cycle.
j) At a certain altitude in the atmosphere, density is 10-9th the density of earth’s
atmosphere at STP and the temperature is _1000C. Assuming a uniform
atmospheric composition, find the pressure at that attitude.
(2x10=20)
3. a) Assign the geometry to the following molecule using the VSEPR theory. State
reasons for your decisions.
(i) Phosphorus (V) Chloride (PCl5)

(ii) Sulphur (VI) Flouride (SF6)
(iii) Boron (III) Flouride (BF3)
b) Explain that HNH bond angle in ammonia is 1070 while HOH bond angle in
water is 104.50
c) The carbon-oxygen bond length in CO is shorter than in CO2 molecule
d) Draw a diagram to show the arrangement of particles in gases, liquids and solids.
(3x4=12)
4. a) Calculate the volume of one mole of oxygen at 270C and 2 atm pressure, given
that the volume of oxygen at STP is 22.4 litres.
b) An element has a bcc structure and a cell edge of 288 pm. The density of the
element is 7.2 g/cm3. How many atoms are present in 208 g of the element?
(4x2=8)
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9
SOLUTIONS
You know that when sugar or salt is added to water, if dissolves. The resulting mixture is
called a solution. Solutions play an important role in our life. In industry, solutions of
various substances are used to carry out a large number of chemical reactions.
Study of solutions of various substances is very interesting.
In this lesson, let us learn about the various components of a solution and the ways in which
concentration of solutions is expressed. We shall also learn about some properties of solutions
which are dependent only on the number of solute particles. (you will learn about solute in
this lesson).
Objectives
After reading this lesson, you will be able to:
• identify the components of different types of solution;
• express the concentration of solutions in different ways;
• list different types of solutions;
• state Henry’s law;
• define vapour pressure;
• state and explain Raoult’s law for solutions;
• define ideal solutions;
• give reasons for non-ideal behavior of solutions;
• state reasons of positive and negative deviations from ideal behavior;
• explain the significance of colligative properties;
• state reasons for the elevation of boiling point and depression in freezing point of
solutions;
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• explain the abnormal colligative properties;
• define osmosis and osmotic pressure;
• define Van’t Hoff factor;
• correlate the degree of dissociation of solute and
• solve numerical problems.
9.1 Components of a Solution

When we put sugar into water, it dissolves to form a solution. We do not see any more sugar
in it. Like sugar, a large number of other substances such as common salt, urea, potassium
chloride etc dissolve in water forming solution. In all such solutions, water is the solvent
and substances which dissolve are the solutes.
Thus, solute and solvent are the components of a solution. Whenever a solute mixes
homogeneously with a solvent, a solution is formed.
solution + solvent → solution
A solution is a homogeneous mixture of two more substances.
Solvent is that component of a solution that has the same physical state as the solution
itself.
Solute is substance that is dissolved in a solvent to form a solution.
9.1.1 The Concentration of a Solution

Some of the properties of solutions, e.g... The sweetness of a sugar solution or the colour of
a dye solution, depend on the amount of solute compared to that of the solvent in it. This is
called the solution concentration. There are several ways for describing concentration of
solution. They include molarity, molality, normality mole fraction and mass percentage.
Molarity: Molarity is defined as the number of moles of solute dissolved per litre of solution
and is usually denoted by M. It is expressed as:
Where n is the number of moles of solute and V is the volume of the solution in litres. A 2.0
molar solution of sulphuric acid would be labeled as 2.0 M H2SO4. It is prepared by adding
2.0 mol of H2SO4 to water to make a litre of solution. Molarity of a solution changes with
temperature because of expansion or contraction of the solution.
Molality: It is defined as the number of moles of solute dissolved per kilogram of solvent.
It is designated by the symbol m. The label 2.0m H2SO4 is read “2 mole l sulphuric acid” and
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is prepared by adding 2.0 mol of H2SO4 to 1 kg of solvent Molality is expressed as:
where nB is the number of moles of the solute and WA is the mass in grams of solvent. The
molality of a solution does not change with temperature.
Example 9.1 : Find out the molarity of the solution whcih contains 32.0 g of methyl alcohol
(CH3OH) in 200 mL solution.
Solution : Molar mass of CH3OH = 12 + 1 x 3 + 16 + 1 = 32 g mol-1
Number of moles of CH3OH = = 1 mol
Volume of the solution = 200 mL = 0.2 litre
∴ Molarity = = =5M
Example 9.2 : What is the molality of a sulphiric acid solution of density 1.20 g/cm3
containing 50% sulphuric acid by mass.
Solution : Mass of 1cm 3 of H2SO4 solution = 1.20 g
Mass of 1 litre (1000 cm3) of H2SO4 solution = 1.20 x 1000 = 1200 g
Mass of H2SO4 in 100 g solution of H2SO4 = 50 g
Mass of H 2SO4 in 1200 g solution H2SO4 = 50 x 1200 = 600 g
100
∴ Mass of water in the solution = 1200 - 600 = 600 g
Molar mass of H2SO4 = 98 g mol-1
No. of moles of H 2SO4 = =
Molarity = x 1000
= x x 1000 = 6.8 m
Normality: Normality is another concentration unit. It is defined as the number of gram

equivalent weights of solute dissolved per litre of the solution.
The number of parts by weight of a substance (element of compound) that will combine
with or displace, directly or indirectly 1.008 parts by weight of hydrogen, 8 parts by
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weight of oxygen and 35.5 parts by weight of chlorine is known as equivalent weight.
Like atomic weight and molecular weight, equivalent weight is also a number and
hence no units are used to express it. However, when equivalent weight is expressed in
grams, it is known as gram equivalent weight of the substance.
Equivalent weight =
Equivalent weight of an acid =
Equivalent weight of a base =
Equivalent weight of salt =
oxidising and reducing agents may have different equivalent weights if they react to give
different products under different conditions. Thus, the equivalent weight of such substances
can be calculated from the reactions in which they take part.
Normality is denoted by the symbol N.
∴ Normality (N) =
= x
The lable 0.5 N KMnO4 is read “0.5 normal” and represents a solution which contains 0.5
gram equivalent of KMnO 4 per litre of solution.
Mole Fraction : The mole fraction of a component in a solution is the ratio of its number of
moles to the total number of moles of all the components in the solution. If a solution
contains 2 mol of alcohol and 3 mol of water, the mole fraction of alcohol is , an that of
water . The sum of mole fractions of all the components of a solution is equal to one. The
mole fraction (XA) of a component A in solution with B is :
nA
nA + nB
Where nA and nB are the number of the moles of A and B respectively.
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Mass Percentage : Mass percentage is the mass of solute present in 100 g of solution. Thus
5% solution of KMnO 4 in water means that 5 g of KMnO4 is present in 100g of the aqueous
solution of KMnO4 .
Example 9.3 : A solution contains 36.0 g water and 46.0 g ethyl alcohol (C2H5OH). Determine
the mole fraction of each component in the solution.
Solution : Molar mass of water = 18 g mol-1
Number of moles of water = = 2.0 mol
No. of moles of C2H5OH = = 1.0 mol
Total number of moles in the solution = 2.0 + 1.0 = 30
Mole fraction of water = = = 0.67
Mole fraction of C2H5OH = = = 0.33
Example 9.4 : Calculate the normality of a solution of NaOH if 0.4 g of NaOH is dissolved
in 100 ml of the solution.
Solution : Mass of NaOH present in 100 mL of the solution = 0.4 g
∴ Mass of NaOH present in 1000 mL of the solution = x 1000 = 4.0g
Mol. wat. of NaOH = 23 + 16 + 1 = 40amu
40
Eq. wt. of NaOH = = = 40
1
4
∴ Normality = = = N
100
Hence, the normality of the solution = N or 0.1 N
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1. List the various methods of expressing the concentration of a solution?
...................................................................................................................................
2. Define the following?
i) Molarity ii) Molality ii) Normality
...................................................................................................................................
9.2 Types of Solution

Solutions can be solid, liquid or gaseous. Depending upon the physical state of the solute and
the solvent, there are nine possible types of solutions consisting of two components (binary
solutions). Different types of solutions are given in Table 9.1.
Table 9.1 Different types of Solutions
Solute Solvent Solution
Gas Gas Air

Gas Liquid Soda Water
Gas Solid Hydrogen in palladium
Liquid Gas Humidity in air
Liquid Liquid Alcohol in water
Liquid Solid Mercury in gold
Solid Gas Camphor in air
Solid Liquid Sugar in water
Solid Solid Alloys such as brass (zinc in copper
and bronze (tin in copper)
Generally, we come across only the following three types of solutions:

(a) Liquids in Liquids : In the solution of liquids in liquids such as alcohol in water, the
constituent present in smaller amounts is designated as solute and the constituent
present in larger amounts is called the solvent. When two liquids are mixed, three
different situations may arise:
(i) Both the liquids are completely miscible, i.e., when, two liquids are mixed, they
dissolve in each other in all proportions, e.g., alcohol and water, benzene and
toluene.
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(ii) The liquids are partially miscible, i.e., they dissolve in each other only to a certain
extent, e.g., water and phenol.
(iii) The liquids are immiscible, i.e., they do not dissolve in each other, e.g., water and
benzene, water and toluene.
The solubility of liquids in liquids generally increases with rise in temperature.
(b) Gases in Liquids : Gases are generally soluble in liquids. Oxygen is sufficiently soluble
in water, which allows the survival of aquatic life in ponds, rivers and oceans. Gases
like CO2 and NH3 are highly soluble in water. The solubility of a gas in a liquid depends
on the pressure, temperature and the nature of the gas and the solvent. These factors are
discussed below in detail:
(i) Effect of Pressure : The variation of solubility of a gas in a liquid with pressure
is governed by Henry’s law. Henry’s law states that
The mass or mole fraction, of a gas dissolved in a solvent is directly proportional to the
partial pressure of the gas.
Henry’s law is represented by
x = Kp
where K is a constant, p is the partial pressure of the gas and x is the mole fraction of the gas
in the solution. Let us now see what are the conditions for the validity of Henry’s law.
(i) Conditions for validity of Henry’s law : It is found that gases obey Henry’s
law under the following conditions.
i) the pressure is not too high.
ii) the temperature is not too low.
iii) the gas does not dissociate, associate or enter into any chemical reaction
with the solvent.
(ii) Effect of temperature : The solubility of gas in a liquid at constant pressure

decreases with rise in temperature. For example, the solubility of CO2 in water
at 200C is 0.88 cm3 per cm3 of water, where as it is 0.53 cm3 per cm3 of water at
400C. This happens because on heating a solution, containing a dissolved gas,
some gas is usually expelled from the solution.
(iii) Effect of the nature of the gas and the solvent : Gases like CO2, HCl and, NH3
are highly soluble in water where as H2, O 2 and N2, are sparingly soluble.
(c) Solids in liquids : When a solid is dissolved in a liquid, the solid is referred as the
solute and the liquid as the solvent. For example, in a solution of sodium chloride in
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water, the solute is sodium chloride and water is the solvent. Different substances dissolve
to different extent in the same solvent.
9.3 Vapour Pressure

If we keep an inverted beaker over a small beaker containing a pure liquid, it is found that the
molecules of the liquid start evaporating in the form of vapours and fill the empty space above
the beaker containing the liquid. A time comes when the number of molecules evaporating
per unit time is equal to the number of molecules condensing during that time (Fig 9.1). An
equilibirum is thus established between the vapour and the liquid phase. The pressure exerted
by the vapour of the liquid in such a case is called the vapour pressure of the liquid.
Fig. 9.1 : Vapour pressure of a liquid
9.4 Raoult’s Law for Solutions

Did you ever think that if you mix two miscible volatile liquids A and B, what would be the
vapour pressure of the resulting solution ? The relationship between vapour pressure of liquid
and its mole fraction is given by Raoult’s law.
Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each
liquid in the solution is directly proportional to its mole fraction.
Raoult’s law is applicable only if the liquids are miscible. The vapour phase now consists of
vapour of both the liquids A and B. The partial vapour pressure of each liquid will depend
upon its mole fraction in the solution. Let the mole fractions of the liquids A and B be XA and
XB respectively. Aslo, if PA and PB are the partial vapour pressures of A and B respectively,
then
PA ∝ XA or PA = POA XA
Similar ly, PB = P OB X B
where POA and POB represent the vapour pressures of pure liquids A and B respectively.
If the values of PA and PB are plotted against the values of XA and XB for a solution, two
straight lines are obtained as shown in Fig. 9.2 the total vapour pressure P of the solution is
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given by the sum of partial vapour pressures PA and PB.
Thus,
P = PA + PB
or P = P OA X A + P OB X B
The total vapour pressure (P) of a solution is represented by the line joining POA and POB. The
solutions which obey Raoult’s law are known as ideal solutions.
A solution which obeys Raoult’s law over the entire range of concentration at all
temperatures is called an ideal solution.
pAO
Vapour pressure
XA = 1 Mole fraction XA = 0
XB = 0 XB = 1
Fig. 9.2 : Relationship between vapour pressure and mole fraction in a solution

1. State Raoult’s law.
...................................................................................................................................
2. State Henry’s law and list the conditions necessary for the validity of Henry’s law.
...................................................................................................................................
9.5 Raoult’s Law for Solution Containing Non-Volatile Solute

If we have an aqueous solution containing a non-volatile solute, such as sugar or salt, what
do you think about the vapour pressure exerted by such a solution? The vapour phase of
such a solution consists of vapours of solvent (A) only because the solute is non-volatile.
Since the mole fraction of the solvent in solution is less than one, therefore according to
Raoult’s law, the vapour pressure of the solution will be less than the vapour pressure of the
pure solvent. If the total vapour pressure of the solution is p, then
PA = P A O X A ......... (9.1)
for a binary mixture
therefore, XA + X B = 1
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XA = 1 - X B
Substituting the value of XA in equation (9.1) we get
PA = PAO (1- XB)
= 1 - XB
therefore, = XB
In the above equation, (PAO - PA) reperesents the lowering of the vapour pressure and
is called the relative lowering of the vapour pressure of the solution.
An alternative statement of Raoult’s law for solutions of non-volatile solute is :

The relative lowering of vapour pressure for a solution is equal to the mole fraction of
the solute, when only the solvent is volatile.
9.6 Ideal and Non-Ideal Solutions

Ideal solutions obey Raoult’s Law and during their formation there is no change in
heat and volume.
Non-ideal solutions are those solutions which do not obey Raoult’s law and whose
formation is accompanied by changes of heat and volume.
Most of the real solutions are non-ideal. They show considerable deviation from the ideal
behaviour. Generally deviations are of two types :
i) Positive deviation : Positive deviations are shown by liquid pairs for which the A-B
molecular interactions are weaker than the A-A and or B-B molecular interactions. The total
vapour pressure for such solutions is greater than predicted by Raoults law. The total vapour
pressure for such a solution will be maximum for a particular intermediate composition (Fig.
9.3).
Examples of non-ideal solutions showing positive deviation from the ideal behaviour are
mixtures of liquids such as water-propanol, ethanol-chloroform, acetone-carbon disulfide,
ethanol-cyclohexane etc.
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Vapour pressure
XB = 0 XB = 1
Fig. 9.3 : Positive deviation for a liquid pair
ii) Negative Deviation : Negative deviations are shown by liquid pairs for which the A-B
molecular interactions - are stronger than A-A or B-B molecular interactions. The total
vapour pressure for such solutions is less than that predicted by Raoult’s law. For a particular
intermediate composition, the total vapour pressure of such a solution will be minimum
(Fig. 9.4). Examples of such liquid pairs are chloroform acetone, water-sulphuric acid,
phenol-aniline, water-HCI etc.
Vapour pressure
XB = 0 XB = 1
Fig. 9.4 : Negative deviation for a liquid pair
9.7 Colligative Properties

Do you know that there are certain properties of dilute solutions which depend only on the
number of particles of solute and not on the nature of the solvent and the solute? Such
properties are called colligative properties. There are four colligative properties : relative
lowering of vapour pressure, elevation in boiling point, depression in freezing point and
osmotic pressure.
Weh shall discuss these colligative properties in detail in the following sections.
9.7.1 Relative Lowering of Vapour Pressue

According to Raoult’s law for solutions containing non-volatile solute
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= X B (see section 9.5) .........................(i)
Also XB =
In a dilute solution nB << nA Therefore the term nB can be neglected in the denominator.
WB
nB M B WB .M A
= =
Hence, X B = n A WA WA .WB
MA
Therefore equation (i) can be written as
P 0A - P A WB . MA
= XB =
P0A WA . MB
The above expression can be used to determine the molecular mass of the solute B, provided
the relative lowering of vapour pressure of a solution of known concentration and molecular
mass of the solvent are known. However, the determination of molecular mass by this method
is often difficult because the accurate determination of lowering of vapour pressure is difficult.
Example 9.5 : The relative lowering of vapour pressure produced by dissolving 7.2 g of a
substance in 100g water is 0.00715. What is the molecular mass of the substance ?
Solution : We know that
Substituting the values we get
0.00715 = or MB =
∴ Molecular mass of the substance = 181.26 amu
9.7.2 Elevation of Boiling Point

Boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes
equal to the atmospheric pressure.
As you know, the vapour pressure of a pure solvent is always higher than that of its solution.
So, the boiling point of the solution is always higher than that of the pure solvent. If you see
the vapour pressure curves for the solvent and the solution (Fig. 9.5), you will find that there
is an elevation in the boiling point of the solution.
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1 Atmosphere
Vapour pressure
ent
Solv
tion
Solu
Temperature K B.P. of B.P. of

solvent solution
Fig. 9.5 : Vapour pressure curves for solvent and solution
Now let ΔTb be the elevation in boiling point and Δp be the lowering in vapour pressure.
Then,
ΔTb α Δp α XB or ΔTb = K XB
K is the proportionality constant
As you know X B =
In a dilute solution, nB << nA and thus the term nB is neglected in the denominator.
WB
n M WB W
M M
W
Thus, X B = B = B = W x A =n x A
n A WA MB WA B WA
MA
Substituting the value of XB in the equation (i) we get
MA
W
ΔTb = K x nB x W
A
If we take the mass of the solvent WA in kilograms the term is molality m. Thus
ΔTb = Κ ΜΑ . m = KB m
The constant Kb is called the molal elevation constant for the solvent. Kb may be defined as
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the elevation in boiling point when one mole of a solute is dissolved in one kilogram of
the solvent. Kb is expressed in degree per molality.
9.7.3 Depression in Freezing Point

Freezing point is the temperature at which the solid and the liquid forms have the same vapour
pressure
The freezing point of the solution is always less than that of the pure solvent. Thus, there is a
depression in the freezing point of the solution. This is because the vapour pressure of the
solution is always less than that of the pure solvent.
Vapour pressure
Solvent
e nt
S olv
lid Solution
So
FP of FP of Temperature K
solvent solvent
Fig. 9.6 : Vapour pressure curves for solid, solvent and solution
Let ΔTf be the depression in freezing point. Then :

ΔTf α ΔXB
or ΔTf = KXB .............................................(ii)
Where K the proportionality constant
You know that XB =
In dilute solution nB << nA

Therefore, the term nB can be neglected from the denominator. Thus,
nB WB / M B WB WMA MA
W
XB = n = W / M = M x
A A A WB WA = nB x WA
⎛ WB ⎞
⎜⎝ sin ce n B = M ⎟⎠
B
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Substituting the valoue of XB in equation (ii) we get
M
ΔTf K x nB x
WA
If the mass of the solvent WA is taken in kg, then the term becomes molality m.
Thus, ΔTf = K MA. m = Kf . m
The constant (Kf ) for a solution is known as molal depression constant or molal cryoscopic
constant for the solvent. Kf may be defined as the depression in freezing point of a
solution when one mole of a solute is dissolved in 1 kilogram of the solvent.
Example 9.6 : Find the (i) boiling point and (ii) freezing point of a solution containing
0.520 g glocse (C6H12O6) dissolved in 80.2g of water. (Kf = 1.86 K/m, Kb = 0.52 k/m).
Solution : Molality of glocose = x
= = = 0.036
ΔTb = Kbm = 0.52 x 0.036 = 0.018 K
∴ Boiling point = 373 + 0.018 = 373.018 = 373.02 K
(ii) Kf = 1.86 K/m
m= = = 0.036
∴ ΔTf = 1.86 x 0.036 = 0.66 K
∴ Freezing point = 273 - 0.66 = 272.34 K
9.7.4 Osmosis and Osmotic Pressure

You must have observed that if rasins are soaked in water for some time, they swell. This is
due to the flow of water into the rasins through its skin which acts as a semipermeable membrane
(permeable only to the solvent molecules). This phenomenon is also observed when two
solutions of different concentrations in the same solvent are separated by a semipermeable
membrane. In this case the solvent flows from a solution of lower concentration to a solution
of higher concentration. The process continues till the concentrations of the solutions on both
sides of the membrane become equal.
The spontaneous flow of the solvent from a solution of lower concentration (or pure solvent)
to a solution of higher concentration when the two are separated by a semipermeable membrane
is known as osmosis.
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The flow of solvent into the solutionof higher concentration from a solution of lower
concentration (on pure solvent) can be stopped if some extra pressure is applied to that side of
the solution which has a higher concentration. The pressure that just stops the flow of the
solvent is called osmotic pressure.
Thus, ostmic pressure may be defined as the excess pressure that must be applied to the
solution side to just prevent the passage of pure solvent into it when the two are separated
by a perfect semipermeable membrane. This is illustrated in Fig. 9.7
Solvent
Solution
Semipermeable
membrane
Fig. 9.7 : Osmosis
The pressure that must be applied to the solution side to prevent it from rising in the tube is the
osmotic pressure. It is also equal to the hydrostatic pressure of the liquid column of heighth.
If the two solutions have the same osmotic pressure, they are known as isotonic solutions.
The osmotic pressure is a colligative property. It depends on the number of particles of solute
present in the solution and not on their nature. At a given temperature T, the osmotic pressure
(π) of a dilute solution is experimentally found to be proportional to the concentration of the
solution in moles per litre.
Mathematically, π = CRT
where π is the osmotic pressure and R is the gas constant
nB
or π = RT
V
where nB is the number of moles of solute present in V litres of the solution
w
or π V= M RT
solute
where w in the mass of solute dissolved in V litres of the solution and Msolute is the molar
mass of the solute. Thus, knowing, π, v and w, the molar mass of the solute can be calculated.
Thus, the molar masses of the solutes can be determined by measuring the osmotic pressure
of their solutions. This method has been widely used to determine the molar masses of
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macromolecules, protiens, etc., which have large molar masses and limited solubility. Therefore
their solutions have very low concentrations and the magnitudes of their other colligative
properties are too small to measure. Yet their osmotic pressures are large enough for
measurements. As the osmotic pressure measurements are done at around room temperature,
this method is particularly useful for determining the molar masses of biomolecules as they
are generally not stable at higher temperature.
Example 9.7 : The osmotic presure of an aqueous solution of a protein containing 0.63 g of
a protein in 100 g of water at 300 K was found to be 2.60 x 10-3 atm. Calculate the molar
mass of the protein. R = 0.082 L atm K-1 mol-1.
we know that osmotic pressure of a solution is given by the expression
w
π V= RT
M solute
w
or Msolute = RT
πV
Substituting the values, we get
(0.63 g ) x (0.082 L − atm K −1 mol −1 ) x (300 K )
Msolute =
(2.60 x 10 −3 atm) x (0.100 L)
= 61022 g mol-1
Thus, molar mass of the protein is 61022 g mol-1
Reverse Osmosis and Water Purification
If a pressure higher than the osmotic pressure is applied to the solution side, the direction of
flow of the solvent can be reversed. As a result, the pure solvent flows out of the solution
through the semipermeable membrane. This process is called reverse osmosis. It is of great
practical application as it is used for desalination of sea water to obtain pure water.
9.7.4.a Determination of Relative Lowering of Vapour pressure by

Walker and Ostwald’s dynamic method.
This method is very simple in which a slow current of dry air is drawn in succession through,
i) A series of two bulbs (A) containing the aqueous soltion (vapour pressue P).
ii) A similar series of two bulbs (B) containing the solvent (vapour pressure PO).
iii) A ‘U’ tube containing anhydrous CaCl2 solid.
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Fig. 9.8 : Ostwalds dynamic method
Dry air is passed through the solution bulbs first. The air carries with it some solvent (water)
in the form of vapour. Hence there will be loss of weight in the solution bulbs. This loss is
proportional to the vapour pressure of solution (P).
Now the air containing the solvent (water) vapour is allowed to pass through the solvent
bulbs. The vapour pressure of solvent (PO) is higher than the vapour pressure of solution (P).
Therefore the air containing the vapour already, takes some more vapour from the solvent
bulbs. This loss is proportional to (P O - P).
Now the air saturated with solvent vapour is passed through the ‘U’ tube, ‘U’ tube absorbs
the solvent vapour. (Water vapour, since water is solvent). The weight of ‘U’ tube increase.
Calculations and Results :
Loss (x) in ‘A’ bulbs (solution bulbs) α P
Loss (y) in ‘B’ bulbs (solvent bulbs) α PO - P
Total Loss = gain (z) in weight of ‘U’ tube (C)
x + y = P + (PO - P) = PO
PO - P = y
PO = (x+y) = z
PO − P y n
∴ = = X s = n Os
PO x+y no
y a W
∴ = X
x+y M b
aW x+y
M = =
b y
a = weight of solute
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b = weight of solvent
M = mol. Wt. of solute
W = mol. Wt. of solvent.
9.7.4.b Cottrell’s method of determining molecular weight of solute

using elevation of Boiling point
This method require (1) accurate determination of elevaiton of boiling point. Hence
conventional thermometers are not useful.
So, beckmann designed a thermometer which measures only the elevation in temperature
ΔT but not absolute values of B.P. This thermometer contains a reservoir of Hg at one end
of the capillary and as usual a bulb at the other end. These are internally connected through
the capillary. At any level of temperature (-600C to 3000C) the elevation can be measured by
adjusting the amount of Hg in the bulb unlike conventional thermometer temperature. This
adjustment is done with the help of the reservoir.
(2) The second requirement is elimination of super heating during boiling. For this, Cottrell
used a special device (pumping device) which resembles an inverted funnel, whose stem is
split into three symmetrically placed tubes as shown in the figure.
Fig. 9.9 : Cottrell’s method
The apparatus consists of tube provided with a side tube. The tube contains another tube
through which the thermometer is inserted. At the bottom of the outer tube a platinum wire is
fused which is heated during boiling. The funnel like device is placed in the solution in the
tube. This pumps the bubbles of the solution along with vapour on to the thermometer bulb.
So, the liquid and the vapour will be in equilibrium on the surface of the bulb at the boiling
point. This reduces super heating.
A known weight (b) of solvent (molecular weight - W) is taken in the Cottrell tube and the
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boiling point is determined (TO). A known weight (‘a’ g) of the solute (molecular weight m)
is introduced in the tube through the side tube in the form of a tablet. This dissolves completely
and the boiling point of solution (T) is measured. Molar mass is calculated as follows.
Wt. of solvent = ‘b’ g
Wt. of solute = ‘a’ g
B.P. of solvent = TO
B.P. of solution =T
Elevation in B.P. = T - To = ΔT
a 1000
ΔT = K b = x
m b
where K b = molal elevation constant of solvent.
Problem : The boiling point of CHCl3 was raised 0.325K when 5.141 x 10-4 kg. of anthracene
is dissolved in 3.5 x 10 -4 kg of CHCl 3 . Calculate the molar mass of anthracene.
(Kb = 3.9 K.mol-1).
Solution :
ΔT = 0.325 K
Kb = 3.9 Kg. mol-1
A = 0.5141 g ; b = 3.5 g
K b x a x 1000
∴M =
ΔTb x b
3.9 x 5.141 x 10 −4 x 1000

M =
0.325 x 3.5 x 10 −3
= 0.1763 kg of 176.3 g
9.7.4.c Rast’s Method :

This method was developed by Rast for Camphor as a solvent. This method is thus generally
used for solid solutions i.e., solid solute and solid solvent.
A known weight of campgor (b) is taken and finely powdered in dry mortar. To this a known
weight (a) of solid solute is added. The mixture is mixed throughly. The mixture is melted
to form a homogeneous solution, cooled dried and powdered. The mixture is taken in a
capillary tube whose one end is sealed. The melting point is determined carefully by the
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conventional melting point determination method. The melting point of the pure sample of
camphor is determined separately using another capillary tube. The difference in the melting
point gives ΔTf .
From the values a, b & , ΔTf M is calculated (Kf for camphor = 40 K.kg mol-1)
Wt. of camphor - b.g.
Wt. of solute - a.g.
Mol. wf. of solute - m
a 1000
T f = Kf x x
m b
K x a x 1000
∴ M = fb
ΔTbf x b
9.7.4.d Berkeley and Hartley’s Method of determining osmotic

Pressure :
It is method of applying pressure to the solution which was just sufficient to stop osmosis. A
porous pot, open at both the ends and having a copper ferrocyanide membrance fused in its
walls, is sealed into the outer bronze cylinder. The bronze cylinder contains the solution and
is fitted with a piston upon which weights may be placed to exert external pressure on the
solution.
Fig. 9.10 : Berkely - Hartley Method

1. Solvent Reservoir 2. Semipermeable membrance
3. solution 4. Capillary tube 5. Piston 6. Pressure guage 7. Solvent
This porous pot is connected with a side tube (capillary tube) one end and a reservior containing
water at the other end. The flow of water through the pot into the solution is indicated by the
downward motion of the water meniscus in the side tube. This flow of osmosis of water into
the solution can be stopped the application of external pressure on the solution with the help
of the piston and this would be indicated when the water meniscus in the side tube becomes
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stationary. The pressure so applied on the piston is equal to the osmotic pressure of the solution.
Thus substituting the values of osmotic pressure (π) molarity of the solution (C) and the
temperature of the solution in Kelvin scale (T), in the equation.
n
π = CRT (C = )
v
R = Gas Constant
The molecular weight of the solute can be calculated.
Problem :
What is the osmotic pressure of a solution containing 0.1 mole of non-volatile solute in 100
cm3 of solution at 270C.
Solution :
n
π= RT R = 0.82l.atm.K-1 mol-1
v
0.1
π= x 0.0082 x 300 V = 100 cm3 = 0.1l
0.1
π = 24.63 atm V = 270C = 300 K

1. Define colligative property. List two colligative properties.
...................................................................................................................................
2. What type of liquid pairs show (i) positive deviations (ii) negative deviations.
...................................................................................................................................
3. Why is the determination of osmotic pressure a better method as compared to other
colligative properties of determining the molar masses of biomolecules.
...................................................................................................................................
9.8 Abnormal Colligative Properties

The colligative properties of the solutions depend only upon the number of solute particles
present in the solution and not on their nature. But sometimes while measuring colligative
properties abnormal results are obtained due to the following reasons :
i) If the solution is very concentrated, the particles of the solute start interacting with each
other. Therefore, the solution should not be concentrated.
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ii) In case of association two or more solute molecules associate to form a bigger molecule.
The number of effective molecules in the solution, therefore decreases. Consequently
the value of the collgative property (relative lowering of vapour pressure, elevation of
boiling point, depression of freezing point, osmotic pressure) is observed to be less than
that calculated on the basis of unassociated molecules. Since, the colligative property is
inversely proportional to the molar mass, the molar mass of such solutes calculated on
the basis of colligative property will be greater than the true molar mass of the solute.
iii) In case of dissociation of the solute in the solution, the number of effective solute
particles increases. In such cases the value of the observed colligative property will be
greater than that calculated on the basis of undissociated solute particles. The molar
mass of the solute calculated from the measurement of colligative property will be lower
than the true molar mass of the solute.
Van’t Hoff factor
In order to account for extent of association or dissociation Van’t Hoff introduced a factor ‘i’.
Observed colligative property

i=
Normal (calculated or exp ected) colligative property
Since the colligative property is proportional to the number of solute particles or the number
of moles of solute
Total number of moles of solute in the solution
i=
Expected (calculated) number of moles of solute
Also, since colligative properties vary inversely as the molar mass of the solute, it follows that
Normal (calculated or exp ected) molar mass
i=
Observed molar mass
Here the observed molar mass is the experimentally determined molar mass whereas the normal
molar mass is the molar mass calculated on the basis of chemical formula of the solute. In case
of association the value of van’t Hoff factor, i, is less than unity while for dissociation it is
greater than unity. For example, benzoic acid association in benzene to form a dimer. The
value of i is, therefore, close to 1/2. The value of i for aqueous NaCl is close to 2.0 because
_
NaCl dissociates in water to form Na+ and Cl ions.
The inclusion of van’t Hoff factor, i, modifies the equations for the colligative properties as
follows :
= i XB
ΔTb = i Kb m
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ΔTf = i Kf m
πV = i CRT
Degree of Association
Degree of association may be defined as the fraction of the total number of molecules which
associate to form a bigger molecule. Let us consider the association of benzioc acid in
benzene. In benzene two molecules of benzoic acid associate to form a dimer. It can be
represented as
2C6H5COOH (C6H5COOH)2
If x represents the degree of association of benzoic acid in benzene (i.e out of one molecule of
benzoic acid, x molecules associate to form a dimer), then at equilibrium.
No. of moles of unassociated benzoic acid = 1 - x
x
No. of moles of associated benzoic acid = 2
x x
Total number of effective moles of benzoic acid = 1 - x + =1-
2 2
According to definition, Van’t Hoff factor is given by
x
Total number of moles of solute in the solution 1−
i= = 2
Expected (calculated) number of moles of solute
1
Example 9.8 : Acetic acid (CH3COOH) associates in benzene to form double molecules
1.60g of acetic acid when dissolved in 100g of benzene (C6H6) raised the boiling point by
273.35 K. Calculated the van’t Hoff factor and the degree of association of benzoic acid. Kb
for C6H6 = 2.57 K kg mol-1.
1000i Kb WB
Solution : ΔTb = i Kb m =
WA MB
Normal molar mass (MB) of CH3COOH = 60 g mol-1
ΔTb x WA x WB
Van’t Hoff factor, i, is = 1000 x K x W
b B
0.35 x 100 x 60
=
1000 x 2.57 x 1.60
= 0.51
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Since, acetic acid associates in benzene to form double molecules, the following equilibrium
exists in the solution.
2CH3COOH (CH3COOH) 2
If x represents the degree of association of the solute, then we would have (1-x) mol of acetic
acid left unassociated and x/2 moles of acetic acid at equilibrium.
Therefore, total number of particles at equilibrium = 1 - x + x/2
x
=1- 2
The total number of particles at equilibirum equal van’t Hoff factor. But van’t Hoff factor (i)
is equal to 0.51.
x
∴ 1- = 0.51
2
x
or = 1 - 0.51 = 0.49
2
x = 0.49 x 2 = 0.98
Therefore, degree of associaiton of acetic acid in benzene is 98%.

Degree of dissociation
Degree of dissociation may be defined as the fraction of the total number of particles that
dissociate, i.e., break into simpler ions. Consider a solution of KCl in water. When KCl is
_
dissolved in water, it dissociates into K+ and Cl ions.
_
KCl K + + Cl
Let x be the degree of dissociation of KCl, then at equilibrium, number of moles of undissociated
KCl = 1 - x.
According to the dissociation of KCl shown above, when x mol of KCl dissociates, x moles
_
of K + ions and x mol of Cl ions are produces.
Thus, the total number of moles in the solution after dissociation = 1 - x + x + x = 1 + x
Total number of moles of solute in the solution 1+ x

Hence, i - =
Expected (calculated) number of moles of solute 1
Example 9.9 : A 0.5 percent aqueous solution of potassium chloride was found to freeze at
27.276K. Calculate the van’t Hoff factor and the degree of dissociation of the solute at this
concentration. (Kf for H2O = 1.86 K kg mol-1).
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Solution : Normal molecular weight of KCl = 39 + 35.5 = 74.5 g mol-1
1000 x WB x K f
Observed molecular weight, MB =
ΔTf x WA
= 38.75 g mol-1
van’t Hoff factor (i) =
= = 1.92
Potassium chloride in aqueous solution dissociates as follow.

_
KCl K + + Cl
Let x be the degree of dissociation of KCl. Thus at equilibrium.
No. of moles of KCl left undissociated = (1- x) mol
No. of moles of K+ = x mol
_
No. of moles of Cl = x mol
Total number of moles at equilibrium = 1 - x + x + x = 1 + x.
1+ x
∴ Van’t Hoff factor = = 1.92
1
or x = 1.92 - 1 = 0.92
∴ Degree of dissociation of KCl = 92%

• Solution is a homogeneous mixture of to or more substances.
• solvent is that component of a solution that has the same physical state as the solution
it self.
• solute is the substance that is dissolved in a solvent to form a solution
• Molarity is expressed as the number of moles of solute per litre of solution.
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• Molality is expressed as the number of moles of solute per kilogram of solution.
• Normality is a concentration unit which tells the number of gram equivalents of solute
per litre of solution.
• Mole fraction is the ratio of the number of moles of one component to the total number
of moles in the solution.
• solutions can be solid, liquid or gaseous.
• Henry’s law states that mass or mole fraction of a gas dissolved in a solvent is directly
proportional to the partial pressure of the gas.
• Raout’s law states that for a solution of volatile liquids, the partial pressure of each
liquid in the solution is directly proportional to its mole fraction.
• A solution which obeys Raoult’s law over the entire range of concentration at all
temperatures is called an ideal solution.
• The relative lowering of vapour pressure for a solution is equal to the mole fraction of
the solute, when only the solvent is volatile.
• Those properties of dilute solutions which depend only on the number of particles of
solute and not on their nature are known as colligative properties.
• Molal elevation constant is the elevation in boiling point when one mole of solute is
dissolved in one kilogram of the solvent.
• Boiling point of a liquid is the temperatures at which the vapour pressure of the liquid
becomes equal to the atmospheric pressure.
• Freezing point is the temperature at which the solid and the liquid forms of the substance
have the same vapour pressure.
• Abnormal result are obtained when the solute associates or dissociates in the solution.
• Van’t Hoff factor is defined as the ratio of normal molar mass to experimentally
determined molar mass.
Terminal Exercise
1. What do you understand by ideal and non-ideal solutions ?
2. Define freezing point and boiling point.
3. Derive the relationship ΔTb = Kb m
4. A solution containing 7 g of a non-volatile solute in 250g of water boils at 373.26 K.
Find the molecular mass of the solute.
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5. 2g of a substance dissolved in 40g of water produced a depression of 274.5 K in the
freezing point of water. Calculate the molecular mass of the substance. The molal
depression constant for water is 274.85 K per molal.
6. Calculate the mole fraction of the solute in a solution obtained by dissolving 10g of urea
(mol wt 60) in 100 g of water.
7. A solution containing 8.6g of urea (molar mass = 60 per dm3) was found to be isotonic
with a 5 per cent solution of an organic non-volatile solute. Calculate the molar mass of
the non-volatile solute.
8. 2 g of benzoic acid (C6H5COOH) dissolved in 25g of benzene shows a depression in
freezing point equal to 1.62 K. Molar depression constant for benzene in 4.9 K kg
mol-1. What is the percentage association of C6H5COOH if it forms double molecules
in solution.
9. The freezing point depression of 5.0 x 10-3 M solution of Na2SO4 in water was found
to be 0.0265OC. Calculate the degree of dissociation of the salt at this concentration.
(Kf for H 2O is 1.86K kg mol-1)

9.1
1. Molarity, Molality, Normality, Mole fraction, Mass percentage.
Molarity is the number of moles of solute dissolved per litre of the soltion.
2. Molality is the number of moles of solute dissolved per kg of the solvent.
Normality is the number of gram equivalents of solute dissolved per litre of solution
9.2
1. For a solution of volatile liquids the partial vapour pressure of each liquid is proportional
to its mole fraction.
2. The mass of a gas dissolved in a solvent is directly proportional to its partial pressure.
Pressure should not be too high temperature should not be too low. The gas should not
associate or dissociate.
9.3
1. Properties that depend upon the number of particles of solute and not on the nature of
solute. e.g. Elevation of boiling point, depression of freezing point.
2. For which A-B molecular interactions are :
i) weaker than A-A and B-B interactions.
ii) stronger than A-A and B-B interactions.
3. At low concentration the magnitude of osmotic pressure is large enough for measurement.
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10
COLLOIDS
You are familiar with solutions. They play an important role in our life. A large number of
substances such as milk, butter, cheese, cream, coloured gems, boot polish, rubber, ink also
play an important role in our daily life. They are also solutions of another kind. They are
colloidal solutions. The term colloid has been derived from two terms, namely colla and oids.
‘Kolla’ means glue and ‘Oids’ means like i.e. glue-like. The size of the particles in colloidal
solutions is bigger than the size of particles present in solutions of sugar or salt in water. In
this lesson you will learn about the methods of preparation, properties and applications of
colloidal solutions.
Objectives
• Explain the difference between true solution, colloidal solution and suspension;
• Identify phases of colloidal solution;
• Classify colloidal solution;
• Describe methods of preparation of colloids;
• Explain some properties of colloidal solutions;
• Recognize the difference between gel and emulsion and,
• Cite examples of the application of colloids in daily life.
10.1 Distinction between a True Solution, Colloidal Solution

and Suspension.
You may recall that solution of sugar in water is homogeneous but milk is not. When you
closely look at milk you can see oil droplets floating in it. Thus, although it appears to be
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homogenous it is actually heterogenous in nature. The nature of the solution formed depends
upon the size of the solute particles. If the size of the solute particles is less than 1 nm it will
form true solution but when the size is between 1 to 100 nm then it will form colloidal
solution. When the size of solute particles is greater than 100 nm it will form a suspension.
Therefore we may conclude that colloidal solution is an intermediate state between true
solution and suspension (Table 10.1).
Table 10.1 : Some important properties of true solutions, colloids and suspensions
S.No. Name of True Solution Colloids Solution Suspension

8 Property
F
1. Size Size of particles Size of particles is Size of particles
is less than 1 nm between 1 nm and greater than 100nm
2. Filterability Pass through Pass through Do not pass
ordinary filter ordinary filter paper through filter paper
paper and also but not through or
through animal animal membrane. animal membrane.
membrane.
3. Settling Particles do not Particles do not Particles settle
settle down on settle down on down on their own
keeping their own but can under gravity.
be made to settle
down by
centrifugation.
4. Visibility Particles are Particles are Particles are visible
invisible to the invisible to the to the naked eye.
naked eye as well naked eye but their
as under a scattering effect can
microscope be observed with
the help of a
microscope
5. Separation The solute and The solute and The solute and
solvent cannot be solvent cannot be solvent can be
separated by separated by separated by
ordinary ordinary filteration ordinary filteration.
filteration or by but can be separated
ultra filteration. by ultra-filteration.
6. Diffusion Diffuse quickly Diffuse slowly Do not diffuse
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10.2 Phases of Colloids Solution
Colloids solutions are heterogenous in nature and always consist of at least two phases: the
dispersed phase and the dispersion medium.
• Dispersed Phase: It is the substance present in small proportion and consists of particles
of colloids size (1 to 100nm).
• Dispersion Medium: It is the medium in which the colloids particles are dispersed.
For example in a colloidal solution of sulphur in water, sulphur particles constitute the ‘dispersed
phase’ and water is the ‘dispersion medium’.
Each of the two phases namely, dispersed phase and dispersion medium can be solid, liquid
or gas. Thus, different types of colloidal solutions are possible depending upon the physical
state of the two phases. Different types of colloidal solutions and their examples are shown
in Table 10.2. You should note that gases cannot form a colloidal solution between themselves,
because they form homogenous mixtures.
Table 10.2 : Types of Colloidal Solutions
S.No. Dispersed Dispersion Type of Examples

Phase Medium Colloidal Solution
1. Solid Solid Solid Solution Gemstones,

2. Solid Liquid Sol Paints, muddy water,
gold sol, starch sol,
arsenious sulphide sol.
3. Solid Gas Aerosol of solids Smoke, dust in air
4. Liquid Solid Gel Jellies, Cheese
5. Liquid Liquid Emulsion Milk, Cream
6. Liquid Gas Aerosol Mist, fog, cloud
7. Gas Solid Solid foam Foam rubber,
pumice stone
8. Gas Liquid Foam Froth, whipped cream
Out of the various types of colloidal solutions listed above, the most common are sols (solid
in liquid type), gels (liquid in solid type) and emulsions (liquid in liquid type). If the
dispersion medium is water then the ‘sol’ is called a hydrosol; and if the dispersion medium
is alcohol then the ‘sol’ is called an alcosol
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1. Classify the following into suspension, colloidal solution and true solution. Milk, sugar
in water, clay in water, blood, boot polish, sand in water, face cream, jelly, foam.
...................................................................................................................................
2. Give one example each of
(a) Sol (b) Gel (c) Aerosol (d) Emulsion
...................................................................................................................................
3. What is the difference between an alcosol and hydrosol?
...................................................................................................................................
4. How does colloidal solution differ from true solution?
...................................................................................................................................
10.3 Classification of Colloids

Colloidal solutions can be classified in different ways:
(a) On the basis of interaction between the phases.
(b) On the basis of molecular size.
10.3.1 Classification Based Upon Interaction

Depending upon the interaction between dispersed phase and the dispersion medium colloidal
solutions have been classified into two categories.
(a) Lyophilic Colloids : The word Lyophilic means solvent lover. Lyophilic colloidal
solutions are those in which the dispersed phase have a great affinity (or love) for the dispersion
medium. Substances like gum, gelatin, starch etc when mixed with suitable dispersion medium,
directly pass into colloidal state and form colloidal solution. Therefore, such solutions are
easily formed simply by bringing dispersed phase and dispersion medium in direct contact
with each other. However these colloidal solution have an important property i.e. they are
reversible in nature. This means that once lyophilic colloidal solutions has been formed then
dispersed phase and dispersion medium can be separated easily. Once separated these can
again be formed by remixing the two phases. These sols are quite stable.
If water is used as dispersion medium then it is termed as hydrophilic colloid.
(b) Lyophobic Colloids : The word Lyophobic means solvent hating. Lyophobic colloidal
solutions are those in which the dispersed phase has no affinity for the dispersion medium.
Metals like Au, Ag and their hydroxides or sulphides etc., when simply mixed with dispersion
medium do not pass directly into colloidal state. These sols have to be prepared by special
methods. These sols can be readily precipitated and once precipitated they have little tendency
to go back into the colloidal state. Thus these sols are irreversible in nature. Also they are not
very stable and require a stability agent to remain in the colloidal form. In case water is used
as dispersion medium it is called as hydrophobic sol.
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10.3.2 Classification Based on Molecular Size
Depending upon the molecular size the colloids have been classified as
a) Marcromolecular colloids - In this type of colloids the size of the particles of the dispersed
phase are big enough to fall in the colloidial dimension as discussed earlier (i.e.-100
nm)
Examples of naturally occuring macromolecular colloids are starch, cellulose, proteins
etc.
b) Multi molecular colloids - Here individually the atoms are not colloidal size but they
aggregate to join together forming a molecule of colloidal dimension. For example sulphur
sol contains aggregates of S8 molecules which fall in colloidal dimension.
c) Associated colloids - These are substances which behave as normal electrolyte at low
concentration but get associated at higher concentration to form miscelle and behave as
colloidal solution. Soap is an example. Soap is sodium salt of long chain fatty acid R
_ _
COONa. When put in water, soap forms RCCO and Na+. These RCCO ions associate
themselves around dirt particles as shown below forming a miscelle (Fig. 10.1)
+
Na
Water
+
Na
+ Na
Misoells
+
Na
+
Na
+
Na
+
Na
+
Na
_
Fig. 10.1 : Aggregating of RCCO ions to form a micelle.
10.4 Preparation of Colloidal Solutions

As discused earlier, the lyophilic sols can be prepared directly by mixing the dispersed phase
with the dispersion medium. For example, colloidal solutions of starch, gelatin, gum etc. are
prepared by simply dissolving these substances in hot water. Similarly, a colloidal sol of
cellulose nitrate is obtained by dissolving it in alcohol. The resulting solution is called
Collodion.
However, lyophobic colloids cannot be prepared by direct method.
Hence two types of methods are used for preparing lyophobic colloids. These are :
i) Physical methods
ii) Chemical methods
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i) Physical methods
These methods are employed for obtaining colloidal solutions of metals like gold, silver,
platinium etc. (Fig. 10.2)
Electodes
Dispersion medium
Ice-Bath
Fig. 10.2 : Preparation of colloidal solution by Bredig’s Arc Method
An electric arc is struck between the two metallic electrodes placed in a container of water.
The intense heat of the arc converts the metal into vapours, which are condensed immediately
in the cold water bath. This results in the formation of particles of colloidal size. We call it
as gold so.
Peptisation : Peptisation is the process of converting a freshly prepared precipitate into
colloidal form by the addition of a suitable electrolyte. The electrolyte is called peptising
agent. For example when ferric chloride is added to a precipitate of ferric hydroxide, ferric
hydroxide gets converted into reddish brown coloured colloidal solution. This is due to
preferential adsorption of cations of the electrolyte by the precipitate. When FeCl3 is added
to Fe(OH) 3, Fe3+ ions from FeCl3 are adsorbed by Fe(OH)3 particles. Thus the Fe(OH)3
particles acquire + ve charge and they start repelling each other forming a colloidal solution.
ii) Chemical Methods : By oxidation
Sulphur sol is obtained by bubbling H2S gas through the solution of oxidizing agent like
HNO 3 or Br2 water, etc. according to the following equation :
Br2 + H2S → S + 2 HBr
2 HNO3 + H2S → 2 H2O + 2 NO 2 + S
Fe(OH)3 SOl, As2 S 3 sol can also be prepared by chemical methods
10.5 Purification of Colloidal Solution

When a colloidal solution is prepared it contains certain impurities. These impurities are
mainly electrolytic in nature and they tend to destabilise the colloidal solutions. Therefore
colloidal solutions are purified by the following methods :
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i) Dialysis
ii) Electrodialysis
Dialysis : The process of dialysis is based on the fact that colloidal particles cannot pass
through parchment or celloplane membrance while the ions of the eletrolyte can. The colloidal
solution is taken in a bag of cellophane which is suspended in a tub full of fresh water. The
impurities diffuse out leaving pure coloidal solution in the bag (Fig. 10.3). This process of
separating the particles of colloids from impurties by means of diffusion through a suitable
membrane is called dialysis.
Fig. 10.3 : A dialyser
Electrodialysis : The dialysis process is slow and to speed up its rate, it is carried out in the
presence of an electrical field. When the electric field is applied through the electrodes, the
ions of the electrolyte present as impurity diffuse towards oppositely charged electrodes at
a fast rate. The dialysis carried out in the presence of electric field is known as electrodialysis
(Fig. 10.4).
Addition of Water
Impuresol
Funnel
Electrodes
l Colloid
l Crystalloid
Solution of Orystalloid
in water
Cell ophanebag or
Parchment paper bag
Fig. 10.4 : Electrodialysis
-220-
The most important use of dialysis is the purification of blood in the artificial kidney machine.
The dialysis membrane allows the small particles (ions etc.) to pass through, whereas large
size particles like haemoglobin do not pass through the membrane.

1. Name two colloids that can be prepared by Bredig’s Arc method.
...................................................................................................................................
2. Name two colloids that can be prepared by chemical methods.
...................................................................................................................................
3. Differentiate between (a) Lyophilic and Lyophobic sol. (b) macromolecular and
multimolecular colloids.
...................................................................................................................................
4. Explain the formation of miscelle.
...................................................................................................................................
10.6 Properties of Colloids

The propeties of colloids are discussed below :
a) Heterogeneous character : Colloidal particles remain within their own boundary surfaces
which separate them from the dispersion medium. So a colloidal system is a
heterogeneous mixture of two phases. The two phases are dispersed phase and
dispersion medium.
b) Brownian movement : It is also termed as Brownian motion and is named after its
discoverer Robert Brown (a Botanist).
Brownian Motion is the zig-zag movement of colloidal particles in continuous random
manner (Fi.g 10.5). Brownian motion arises because of the impact of the molecules of
the dispersion medium on the particles of dispersed phase. The forces are unequal in
different directions. Hence it causes the particles to move in a zig-zag way.
Fig. 10.5 : Brownian Movement
-221-
c) Tyndall Effect : Tyndall in 1869, observed that if a strong beam of light is passed
through a colloidal solution then the path of light is illuminated. This phenomenon is
called Tyndall Effect. This phenomenon is due to scattering of light by colloidal
particles (fig. 10.6). The same effect is noticed when a beam of light enters a dark
room through a slit and becomes visible. This happens due to the scattering of light by
particles of dust in the air.
Light Source
True Solution Colloidal solution
Fig. 10.6 : The Tyndall Effect (Path of light visible due to
scattering)
d) Electrical Properties : The particles of a colloidal solution are electrically charged and
carry the same type of charge, either negative or positive. The dispersion medium has
an equal and opposite charge. The colloidal particles therefore repel each other and do
not cluster together to settle down. For example, arsenious sulphide sol, gold sol, silver
sol, etc. contain negatively charged colloidal particles whereas ferric hydroxide,
aluminium hydroxide etc. contain positively charged colloidal particles. Origin of charge
on colloidal particles is due to :
a) Preferential adsorption of cations or anions by colloidal particles.
b) Miscelles carry a charge on them.
c) During the formation of colloids especially by Bredig arc method, colloidal particles
capture electrons and get charged. The existence of charge on a colloidal particles is
shown by a process called electrophoresis.
Electrophoresis is a process with involves the movement of colloidal particles either
towards cathode or anode under the influence of electrical field. The apparants used is
as shown in Fig. 10.7.
Electrode
Coagulated sol
particles
As2S3 sol
(negative charged)
Fig. 10.7 : A set up for Electrophoresis
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10.7 Applications of Colloidal Solutions
Colloids play a very important role in our daily life. Some of these applications are discussed
below :
i) Sewage disposal : Colloidal particles of dirt, etc. carry elctric charge. When sewage
is allowed to pass through metallic plates kept at a high potential, the colloidal particles
move to the oppositely charged electrode and get precipitated there. Hence sewage
water is purified.
ii) Purification of Water in Wells : When alum is added to muddy water, the negatively
charged particles of the colloid are neutralized by Al3+ ions. Hence the mud particles
settle down and the water can be filtered and used.
iii) Smoke Precipitation : Smoke particles are actually electrically charged colloidal
particles of carbon in air. Precipitation of this carbon is done is a Cottrell’s
Precipitator. Smoke from chimneys is allowed to pass through a chamber having a
number of metallic plates connected to a source of high potential as shown in Fig.
10.8 Charged particles of smoke get attracted to the oppositively charged electrode
and get precipitated and hot purified air passes out.
High voltage electrode

(30000 volts or more)
Gases free from

carbon particles
Smoke
Precipitated ash
Fig. 10.8 : Cottrell smoke precipitator
Other applications in day to day life are :

i) Photography : A colloidal solution of silver bromide in gelatin is applied on glass
plates or celluloids films to form photo-senstive plates in photography.
ii) Clotting of Blood : Blood is a colloidal solution and is negatively charged. On applying a
solution of FeCl3 bleeding stops and clotting of the colloidal particles of blood takes place.
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iii) Rubber Plating : Latex is a colloidal solution of negatively charged rubber particles.
The object to be rubber plated is made the anode in ther rubber plating bath. The
negatively charged rubber particles move towards the anode and get deposited on it.
iv) Blue Colour of Sky : Have you ever wondered why is the sky blue? It is because the
colloidal dust particles floating about in the sky scatter blue light, that makes the sky
appear blue. In the absence of these colloidal particles the sky would have appeared
dark throughout.
10.8 Emulsion and Gel

Emulsions are colloidal solutions in which both the dispersed phase and dispersion medium
are liquids. However, the two liquids are immiscibel, as miscible liquids will form true
solution.
Emulsion are of two kinds :
a) Oil-in-water emulsion : Here the dispersed phase is oil while the dispersion medium is
water. Milk is an example of this kind as in milk liquid fats are dispersed in water. Vanishing
cream is another example.
b) Water-in-oil emulsion : Here dispersed phase is water and dispersion medium is oil.
Butter, cod-liver oil, cold creams are examples of this types.
The liquids forming emulsion i.e. oil and water will separate out on keeping as they are
immiscible. Therefore an emulsifying agent or emulsifier is added to stabilise the emulsion.
Soap is a common emulsifier. The preparation of emulsion in the presence of an emulsifier
is called emulsification.
How does an emulsifier work? It is believed that an emulsifier gets concentrated at the
interface between oil and water i.e. the surface at which oil and water come in contact with
each other. It acts as a binder between oil and water.
Application of Emulsions - Emulsions play very important role in our daily life. Some of
the common applications are given below :
1. The cleansing action of soap and synthetic detergents for washing clothes, bathing etc
is based upon the formation of oil in water type emulsion.
2. Milk is an emulsion of fat in water. Milk cream and butter are also emulsions.
3. Various cold creams, vanishing creams, body lotions etc. are all emulsions.
4. Various oily drugs such as cold liver oil are administered in the form of emulsion for
their better and faster absorption. Some ointments are also in the form of emulsions.
5. The digestion of fats in the intestine occurs by the process of emulsification.
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6. Emulsions are used for concentrating the sulphide ores by forth flotation process. Finely
powedered or is treated with an oil emulsion and the mixture is vigorously agitated by
compressed air when the ore particles are carried to the surface and removed.
Gels - Gels are the types of colloids in which the dispersed phase is a liquid and the dispersion
medium is a solid. Cheese, jelly, boot polish are common examples of gel. Most of the commonly
used gels are hydrophilic colloidal solution in which a dilute solution, under suitable conditions
set as elastic semi solid masses. For example 5% aqueous solution of gelatin in water on
cooling forms the jelly block.
Gels may shrink on keeping by loosing some of the liquid held by them. This is known as
syneresis or resetting on standing.
Gels are divided in two caterogies elastic gels and non elastic gels. Elastic gels are reversible.
When partly dehydrated on loosing water, they change back into the original form on addition
of water. The non elastic gels are not reversible.
Gels are useful in many ways. Silica, cheese, jelly, boot polish, curd are commonly used
gels. Solidified alcohol fuel is a gel of alcohol in calcium acetate.

• Size of the particles in the colloidal state is intermediate between that of suspension
and ture solution.
• There are eight different types of colloidal systems.
• Sols are classified on the basis of (a) interaction between dispersed phase and dispersion
medium (b) molecular size of dispered phase.
• Colloidal solutions are prepared by physical and chemical methods.
• The zig zag motion of colloidal particles is called Brownian motion.
• Colloidal size particles scatter light and so the path of light becomes visible in a semi
darkened room due to dust particles.
• Colloidal particles may carry electric charge.
• A colloidal dispersion of a liquid in another liquid is called an emulsion.
• A colloidal solution of liquid dispersed in a solid medium is called a gel.
• Colloids are extremely useful to mankind both in daily life and in industry.
Terminal Exercise
1. List three differences between a true solution and colloidal solution.
2. Describe one method of preparation of
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a) a lyophilic colloid
b) a lyophilic colloid
3. What are associated colloids ?
4. What is Brownian motion ? How does it originate ?
5. Why bleeding from a fresh cut stops on applying alum ?
6. Two beakers A and B contain ferric hydroxide sol and NaCl solution respectively.
When a beam of light is allowed to converge on them, (in a darkened room), beam of
light is visible in beaker A but not in breaker B. Give the reason. What is this effect
called ?
7. Define the following terms and give two examples of each
i) Gel
ii) Sol
8. Describe two important applications of colloidal solutions.
9. Give two examples of emulsions used in daily life.
10. Explain the role of emulsifier in an emulsion ?

10.1
1. Suspension - Clay in water, Sand in water
Colloidal - Milk, Blood, Boot polish, Face Cream, Jelly, Foam.
True Solution - Sugar in water
2. Sol - Starch in water
Gel - Silica gel
Aerosol - Fog
Emulsion - Milk
3. Alcosol - When alcohol is the dispersion medium.
Hydrosol - When water is the dispersion medium.
4. True solution Colloidal solution
i) size of solute in less than 1 nm i) Particle size (1-100) nm.
ii) Form transparent solution and allows ii) Path of light becomes visible.
light to pass through them.
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10.2
1. Gol sol, Platinum sol
2. As2S3, Fe (OH)3 (Arsenious sulphide sol, ferric hydroxide sol)
3. a) Lyophilic sol :
1) easy to prepare
2) affinity between dispersed phase and dispersion medium
3) reversible
Lyophobic
1) special method used for preparation
2) No affinity between the two phases.
3) Not reversible
b) Macromolecular - The size of the colloidal particles large enough to fall in the
colloidal dimensions.
Multimolecular - Individually the particles are not of colloidal dimensions but they
aggregate to join together to form molecules of colloidal side.
4. Refer to 10.3.2 (c)
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11
CHEMICAL
THERMODYNAMICS
When a chemical reaction occurs, it is accompanied by an energy change which may take
any of several different forms. For example, the energy change involved in the combustion
of fuels like kerosene, coal, wood, natural gas, etc., takes the form of heat and light. Electrical
energy is obtained from chemical reactions in batteries. The formation of glucose, C6H12O6
by the process of photosynthesis requires the absorption of light energy from the sun. Thus,
we see that the energy change that accompanies a chemical reaction can take different forms.
In this lesson, you shall study the reactions in which heat is either evolved or absorbed.
Objectives
• define the commonly used terms in thermodynamics;
• differentiate between exothermic and endothermic reactions;
• explain the first law of thermodynamics;
• define enthalpy and enthalpy change;
• state the relationship between enthalpy change and internal energy change;
• define enthalpy of formation, enthalpy of neutralisation and enthalpy of combustion;
• state the relationship between enthalpy of reaction and enthalpies of formation of
reactants and products;
• solve numerical problems based on the enthalpy changes;
• state Hess’s law;
• calculate enthalpy of a reaction using Hess’s law;
• define bond enthalpy and bond dissociation enthalpy and
• calculate enthalpy of a reaction using bond enthalpy data.
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11.1 Some Commonly used Terms
In this lesson you would come across some frequently used terms. Let us understand the
meaning of these terms first.
11.1.1 System and Surrounding

If we are studying the reaction of two substances A and B kept in a beaker, the reaction
mixture of A and B is a system and the beaker and the room where it is kept are surrounding
as shown in figure 11.1
Surroundings System
Fig. 11.1 : System and surroundings
System is the part of the physical universe which is under study, while the rest of the
universe is surroundings.
You know that hot tea/milk (let us call it a system) kept in a stoppered thermos flask remains
hot for a couple of hours. If this flask is made of perfect insulating material, then there
would be no exchange of matter or energy between the system and the surroundings. We
call such a system as isolated system.
Isolated system is a system which can exchange neither matter nor energy with the
surroundings.
If we keep hot tea/milk in a stoppered stainless steel flask, it will not remain hot after some
time. Here energy is lot to the surroundings through the steel walls, but due to stopper, the
matter will not be lost. We call this system a closed system.
Closed system is a system which can exchange energy but not matter with the
surroundings.
If we keep stainless steel flask or thermos flask open, some matter will also be lost due to
evaporation along with energy. We call such a system an open system. Plants, animals,
human beings are all examples of open systems, because they continuously exchange matter
(food, etc) and energy with the surroundings.
Open system is a system which can exchange both energy and matter with surroundings.
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11.1.2 State of a System
We describe the state of a system by its measurable properties. For example, we can describe
the state of a gas by specifying its pressure, volume, temperature etc. These variable properties
are called state variables or state functions. Their values depend only on the initial and final
state of the system and not on the path taken by the system during the change. When the
state of a system changes, the change depends only on the initial and the final state of the
system (Fig. 11.2).
Final State
Initial State
Fig. 11.2 : Change of state from initial state to final state through three paths
I, II and III. The difference P2 - P1 and T2 - T1 are independent of the path since pressure
and temperature are state functions.
State functions are those functions which depend only on the state of the system.
Change in state of a system is defined by giving the initial and the final state of the system.
We can understand it by considering another example. We travel from one point to another.
The distance travelled depends on the path or the route we take. But the separation between
these two points on the earth is fixed. Thus, separation is a state function, but not the distance
travelled.
11.1.3 Properties of a System

As stated earlier, the measurable properties of a system are called state variable. They may
be further divided into two main types.
i) Extensive property (variable) is one whose value depends upon the size of the system.
For example, volume, weight, heat, etc.
ii) Intensive property (variable) is one whose value is independent of the size of the system.
For example, temperature, pressure, refractive index, viscosity, density, surface tension,
etc.
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You may note that an extensive property can become an intensive property by specifying a
unit amount of the substance concerned. For example, mass and volume are extensive
properties, but density (mass per unit volume) and specific volume (volume per unit mass)
are intensive properties.
11.1.4 Types of Processes

Let us first understand what do we mean by a process. Suppose we want to raise the temperature
of the system. We may do it by heating it. Here, heating is the process.
The method of bringing about a change in state is called process.
Process could be of different types. The different types of processes are explained below.
i) Isothermal Process : Ice melts at 273 K and 1 atm pressure. The temperature does not
change as long as the process of melting goes on. Such process are examples of isothermal
process. We can define isothernal process as follows.
When the temperature of the system remains constant during various operations, then
the process is said to be isothermal. This is attained either by removing heat from the
system or by supplying heat to the system.
ii) Adiabatic Process : If an acid is mixed with a base in a closed thermos flask, the heat
evolved is retained by the system. Such process are known as adiabatic processes because
the thermos flask does not allow exchange of heat between the system and the
surroundings. Adiabatic process can be defined as follows :
In an adiabatic process there is no exchange of heat between the sytem and the
surroundings. Thus, in adiabatic process there is always a change in temperature.
iii) Reversible Process : In a reversible process, the initial and the final states are connected
through a succession of equilibrium states. All changes occuring in any part of the
process are exactly reversed when it is carried out in the opposite direction. Thus both
the systems and its surroundings must be restored exactly to their original state, when
the process has been performed and then reversed.
Let us understand it by an example. Imagine a liquid in equilibrium with its vapour in a
cylinder closed by a frictionless piston, and placed in a constant temperature bath as shown
in figure. 11.3. If the external pressure on the piston is increased by an infinitesimally small
amount, the vapours will condense, but the condenstion will occur so slowly that the heat
evolved will be taken up by the temperature bath. The temperature of the system will not
rise, and the pressure above this liquid will remain constant. Although condensation of the
vapor is taking place, the system at every instant is in the state of equilibrium. If the external
pressure is made just smaller than the vapour pressure, the liquid will vaporize extremely
slowly, and again temperature and pressure will remain constant.
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Cylinder with piston
Vapour
Liquid
Constant temperature bath
Fig. 11.3 : Reversible Process

Reversible process are those processes in which the changes are carried out so slowly
that the system and surroundings are always in equilibrium.
iv) Irreversible Processes : In the above example rapid evaporation or condensation by
the sudden decrease or increase of the external pressure, will lead to non-uniformity
in temperature and pressure within the system and the equilibrium will be disturbed.
Such processes are called as irreversible processes.
11.1.5 Standard States

You have seen that a system is described by the state variables. In order to compare the
energies for different compounds, a standard set of conditions is chosen. This refers to the
conditions of 1 bar pressure at any specified temperature, with a substance in its most stable
form.
11.2 Exothermic and Endothermic Reactions

i) Add a few cm3 of dilute hydrocholoric acid in a test tube containing a few pieces of
granulated zinc and observe the evolution of a gas. Feel the test tube. It would be hot.
ii) You must have observed that when some water is added to quick lime to prepare white
wash, a lot of heat is evolved.
iii) When a fuel like cooking gas or coal is burnt in air, heat is evolved besides light.
Many chemical reactions lead to release of energy (heat) to the surrounding. We call
these type of reactions as exothermic reactions.
Exothermic reactions are those reactions which proceed with the evolution of heat.
Let us now consider the following reactions :
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i) Add a small amount of solid ammonium chloride in a test tube half-filled with water.
Shake and feel the test tube. It will feel cold.
ii) Similarly repeat this experiment with potassium nitrate and feel the test tube, it will feel
cold.
iii) Mix barium hydroxide with ammonium chloride in small quantities in water taken in a
test tube. Feel the test tube. It will be cold.
In all the above processes we see that heat is absorbed by the system from the surroundings.
Such reactions are called endothermic reactions.
Endothermic reactions are those reactions which proceed with the adsorption of heat
from the surroundings.
11.3 Thermochemical Equations

You are familiar with equations for chemical reactions. Now we shall write the chemical
equations which will specify heat energy changes and state of the reactants and products.
These are called the thermochemical equations. For writing these equations, we follow the
conventions listed below :
i) The heat evolved or absorbed in a reaction is affected by the physical state of the
reacting substances. Therefore, gaseous, liquid and solid states are represented by
putting symbols (g), (l), and (s) along side the chemical formulae respectively.
For example, to represent burning of methane in oxygen, we write
CH4(g) + 2O2(g) → CO2(g) + 2H 2O + heat
In writing thermochemical reactions, we denote the amount of heat evolved or absorbed

by a symbol ΔH. The amount of heat evolved or absorbed is written after the equation
followed by semicolon. ΔH is negative for exothermic reactions and it is positive for
endothermic reactions.
For example :
An exothermic reaction is written as
CH4(g) + 2O2(g) → CO2(g) + 2H2O (l) ; ΔH = -891 kJ
Whereas an endothermic reaction is written as
H2(g) + I2(g) → 2HI (g); ΔH = 52.2 kJ
ii) In case of elements which exhibit allotropy, the name of allotropic modification is
mentioned. For example,
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C (graphite), C (diamond), etc.
iii) The substances in aqueous solutions are specified using the symbol (aq). For example
NaCl (aq) stands for any aqueous solution of sodium chloride.
iv) Thermochemical equations may be balanced even by using fractional coefficients, if
so required. The coefficients of the substances of the chemical euqation indicate the
number of moles of each substance involved in the reaction and the ΔH values given
correspond to these quantities of substances.
v) In case the coefficients are multiplied or divided by a factor, ΔH value must also be
multiplied or divided by the same factor. In such cases, the ΔH value will depend
upon the coefficients. For example, in equation.
H2 (g) + 1
/2 O2(g) → H2O (g) ; ΔH = - 242 kJ
If coefficients are multiplied by 2, we would write the equation
2H 2(g) + O2(g) → 2H2O (g) ; ΔH = 2 (-242) = - 484 kJ
11.4 The First law of Thermodynamics

You have learnt that chemical reactions are accompanied by energy changes. How do we
determine these energy changes? You know that we cannot create or destroy energy. Energy
only changes from one form to another. This is the observation made by many scientists
over the years. This observation has taken the form of first law of thermodynamics. It has
been found valid for various situations. We state this law as follows :
Energy can neither be created nor destroyed. The total energy of the universe or an
isolated system is constant.
Mathematically the first law of thermodynamics is stated as :
ΔU = q + w (11.1)
where ΔU = change in internal energy, q = heat absorbed by the system, and w = work done
on the system. These terms are explained as :
11.4.1 Internal Energy (U)

Every system has a definite amount of energy. This amount is different for different
susbstances. It includes translational, vibrational and rotational energies of molecules, energy
of electrons and nuclei.
The internal energy may be defind as the sum of the energies of all the atoms, molecules
or ions contained in the system.
It is a state variable. It is not possible to measure the absolute values of internal energy.
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However, we can calculate the change in internal energy. If the internal energy of the system
in the initial state is U1 and that in the final state is U2, then change in internal energy ΔU is
independent of the path taken from the intial to the final state.
We can write this change as :
ΔU = U2 _ U1
The internal energy of the system can be changed in two ways :
i) either by allowing heat to flow into the system or out of the system; and
ii) by work done on the system or by the system
11.4.2 Heat (q) and Work (w)

Heat and work are not state functions. This is because the value of both q and w depend upon
the way in which the change is carried out.
Since the law deals with the transfer of heat and work, we assign some signs to these quantities.
Any thing which increases the internal energy of a system is given a positive sign.
Heat given to the system (q) and work done on the system (w) are given positive signs. Let us
illustrate this with an example.
If a certain change is accompanied by absorption of 50kJ of heat and expenditure of 30 kJ
of work.
q = + 50 kJ
w = - 30 kJ
Change in internal energy ΔU = (+ 50kJ) + (-30 kJ) = + 20 kJ
Thus the system has undergone a net increase the internal energy of -20 kJ.
Change in the internal energy of the surrounding will be 20 kJ.
11.4.3 Work of Expansion

Let us assume that pressure p is constant and the volume of the system changes from V1 to
V2. The work done by a system is given as
w = -p (V2 - V1) = -p Δ V (11.2)
(Here we have taken minus sign, because the work is done by the system). Let us substitute
the expression given for w in equation 11.1.
ΔU = q - p Δ V (11.3)
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If the process is carried of at constant volume, i.e. ΔV = 0, then
ΔU = qV (11.4)
The subscript v in qV denotes that volume is constant.
The equation 11.4 shows that we can determine internal energy change if we measure the
heat gained or lost by the system at constant volume. However, in chemistry, the chemical
reactions are generally carried out at constant pressure (atmospheric presure). What do we do
then? Let us define another state function, called, enthalpy.
11.4.4 Enthalpy (H)

For meaning heat lost or gained at constant pressure, we define a new state function called
enthalpy. It is denoted by the symbol H and is given by
H=U+pV (11.5)
Enthalpy changes, Δ H, is given by
Δ H = Δ U + Δ (pV) (11.6)
or ΔH=ΔU+pΔV+VΔp
If the change is carried out at constant pressure, then Δp = 0. The equation 11.6 will become
Δ H = Δ U + p Δ V (at constant pressure) (11.7)
Substituting the value of ΔU from equation 11.3 in equation 11.7 we get
ΔH=q-p Δ V +pΔ V
= q (at constant pressure)
We denote q at constant pressure by qp hence
Δ H = qp (11.8)
Equation 11.8 shows that by measuring heat lost or gained at constant pressure, we can measure
enthalpy change for any process.
11.4.5 Reaction between Δ H and Δ U

For liquids and solids, the difference between ΔrH and ΔrU is not significant but for gasses,
the difference is significant as we will see here.
Let VA be the total volume of the gaseous reactants,
VB the total volume of the gaseous products,
nA the number of moles of gaseous reactants,
and nB the number of moles of gaseous products, at constant pressure and temperature.
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Then, using ideal gas law, we can write
p VA = nA RT (11.9)
p VB = nB RT (11.10)
Subtracting equation 11.9 from equation 11.10, we get
p VB = pVA = nB RT - n B RT = (nA - nB ) RT
p (VB - VA ) = p Δ V = Δ ng RT
At constant pressure
ΔH = Δ U + p Δ V
Therefore ΔH = Δ U + Δ ng RT
Here Δ ng = (number of moles of gaseous products) - (number of moles of gaseous reactants)
Thus we can find the value of ΔH from Δ U or vice versa.
For solids and liquids Δ V is very small. We can neglect the term p Δ V, hence Δ H is nearly
the same as Δ U.

1. Which of the following is false ?
a) The reaction
H2(g) + Cl2 (g) → 2HCl (g) + 185 kJ
in endothermic.
(b) Enthalpy change is a state function.
(c) Standard state condition for a gaseous system is 1 bar pressure at a specified
temperature.
...................................................................................................................................
2. For the reaction at 298 K,
1
/2N2(g) + 3/2H2(g) → NH2 (g) ; Δ H = - 46 kJ
(a) What is the value of Δ ng ?
...................................................................................................................................
b) Calculate the value of Δ U at 298 K ?
...................................................................................................................................
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3. Which of the following will increase the internal energy of a system ?
a) Heat given to the system
b) Work done by the system
...................................................................................................................................
11.5 Standard Enthalpy of Reactions

Let as denote total enthalpy of reactants as Hreactants and total enthalpy of products as Hproducts.
The difference between these enthalpies, Δ H, is the enthalpy of the reaction
Δ rH = Hproducts _ Hreactants
When Hproducts is greater than Hreactants then Δ H is positive and heat is absorbed in the reaction,
and the reaction will be endothermic. For example,
H2(g) + I2(g) → 2HI (g) ; Δ rH = 52.5 kJ
When Hproducts is greater than Hreactants then Δ H is negative and heat is evolved in the reaction,
and the reaction will be exothermic, For example,
CH4(g) + 2O2(g) → CO2 (g) + 2H2O (l) ; Δ rH = - 890.4 kJ

Enthalpy of a reaction changes with pressure and temperature. It is convenient to report
enthalpies of the reactions in terms of standard state of substances as we have defined
earlier. When substances are in their standard states, we call the enthalpy of reaction as
standard enthalpy of reaction. It is defined as the enthalpy change for a reaction, when the
reactants and the products are in their standard states. It is donoted by Δ rH0.
11.5.1 Enthalpy of formation (Δf H0)

The enthalpy change when one mole of a pure compound is formed from its elements
in their most stable states is called the enthalpy of formation and is denoted by Δf H0.
When the reacting elements and the products formed are all in their standard states, the
enthalpy change accompanying the chemical reaction is called the standard enthalpy of
formation and is denoted by Δf H0 . By convention, we take the standard enthalpy of formation
of an element in its most stable state as zero.
For example :
C(Graphite) + O2(g) → CO2 (g) ; Δf H0 = - 393.5 kJ mol-1
This means that carbon dioxide is formed from its elements in their most stable states.
Carbon in the form of graphite and at room temperature and gasesous O2 and CO 2 being at
1 bar.
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11.5.2 Enthalpy of Combustion (Δcomb H0)
Enthalpy of combustion is the enthalpy change (heat evolved) accompanying the complete
combustion of 1 mole of a substance in oxygen at a given temperature and 1 bar pressure.
For example :
C2H5OH (1) + 3O2(g) → 2CO2 (g) + 3H2O (1) ; Δcomb H0 = - 1365.6 kJ
Enthalpy of combustion of C2H5OH (1) is - 1365.6 kJ mol-1
11.5.3 Enthalpy of Neutralization (Δneut H0)

Enthalpy of neutralization is the enthalpy change (heat evolved) when one mole of hydrogen
_
ions (H+) is neutralized by one mole of hydroxyl ions (OH ) in dilute aqueous medium to
form water.
For example :
_
H+ (aq) + OH (aq) → H2O (1) + ; Δneut H0 = - 57 kJ/mol
Enthalpy of neutralization of a strong acid with a strong base is always constant having a
value of -57 kJ. However, enthalpy of neutralization of strong acid with a weak base or
weak acid with a strong base will be different, because of varying degree of ionization of
weak acids and bases.
11.6 Laws of Thermochemistry

There are two laws of thermochemistry : The Lavoisiter - Laplace law and the Hess’s Law
of Constant Heat Summation.
Lavoisier - Laplace Law : When a chemical equation is reversed, the sign of Δ rH is changed.
For example.
N2 (g) + O2 (g) → 2NO (g) ; Δr H = 180.5 kJ
2NO (g) → N2 (g) + O2 (g) ; Δr H = 180.5 kJ
Hess’s Law of constant heat summation : Hess’s law states that the enthalpy of reaction is
independent of the number and the nature of the intermediate steps.
You have learnt that standard enthalpy change of the reaction
C (graphite) + O2 (g) → CO2 (g)
is equal to - 393.5 kJ mol-1. This value can be determined with the help of a calorimeter.
However, there are some reactions for which the direct measurement of enthalpy in the
laboratory is not possible. For example the standard enthalpy change for the reactions.
C (graphite) + 1/2 O2 (g) → CO (g)
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cannot be to determined with the help of calorimeter because the combustion of carbon is
incomplete unless an excess of oxygen is used. If excess of oxygen is used, some of the CO
is oxidized to CO2. How can then we determine the enthalpy change for such reactions
when direct measurement is not possible ?
Since Δ H is a state function, it is not dependent on the way the reactions are carried out. Let
us carry out the reactions as follows.
1) First carry out the following reaction and let Δr H0 be the enthalpy change of the reaction.
C (graphite) + O2 (g) → CO2 (g) ; Δr H10 = -393.5 kJ/mol.
2) Now let us write the reaction for which we have to determine the enthalpy change of
the reaction and let it be Δr H20 .
C (graphite) + 1/2 O2 (g) → CO (g) ; Δr H20 = ?
3) Let us carry out the following reactions and let Δr H30 be the enthalpy change of the
reaction.
CO (g) + 1/2 O2 (g) → CO2 (g) ; Δr H30 = -283.0 kJ
We have obtained the products CO2 (g) from carbon and oxygen through two routes, namely
first as in step (1) and second as in step (2) + (3). According to Hess’s Law
Δr H10 = Δr H20 + Δr H30
or
Δr H20 = Δr H10 + Δr H30
Fig. 11.4 shows alternate paths for the conversion of carbon and oxygen to carbon dioxide.
Reactants=C (graphite)+ O2(g)
o
r
H2
C (graphite)+O2 (g) C (graphite)+ 1O2(g) CO(g)
2
o
r
H1 o
CO + 1 O2(g)
H3 2
Path I r
Path II
Fig. 11.4 : Alternate paths for the conversion of carbon and oxygen to carbon dioxide
The result of Hess’s law is that thermochemical equations can be added and substrated just
like algebraic equations to get the desired reaction. A useful practical application of this
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law is that we can calculate enthalpy changes for the reactions which cannot be studied directly
as in the above case.
To obtain the enthalpy change for the reactions, we algebraically combine the known values
of Δr H0 in the same way as the reactions themselves,
Thus, C (graphite) + O2 (g) → CO2 (g) ; Δr H10 = -393.5 kJ/mol.
- [CO (graphite) + 1/2 O2 (g) → CO2 (g)] ; Δr H30 = -283.0 kJ
C (graphite) + 1/2 O2 (g) → CO (g) ; Δr H20 = [(-393.5) - (-283.0)]
= -110.5 kJ/mol
Example 11.1 : The heat evolved in the combustion of glucose is shown in the following
equation.
C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O(1); Δcomb H = -2840 kJ/mol
How much energy will be required for the production of 1.08 g of glucose ?
Soltuion : Glucose will be prepared by the reverse reaction.
6CO2 (g) + 6HO2 (1) → C6H12O6 (s) + 6CO2 (g); Δ H = 2840 kJ
This equation refers to 1 mol of glucose (180g of glucose).

Production of 180g of glucose requires 2840 kJ of energy. Therefore, production of 1.08 g
will require :
2840 kJ
x 1.08 g = 17.04 kJ
180 g
Example 11.2 : Calculate the standard enthalpy of formation of ethane, given that
C (graphite) + O2 (g) → CO2 (g) ; Δf H0 = -394 kJ/mol-1
H2 (g) + 1/2 O2 (g) → H2O (g) ; Δf H0 = -286 kJ mol-1
C2H6 (g) + 7/2 O2 (g) → 2CO2 (g) + 3H2O (l); Δr H0 = -1560 kJ mol-1
Soltuion : Given that

C (graphite) + O2 (g) → CO 2 (g) ; Δr H10 = -394 kJ/mol (1)
H2 (g) + 1/2 O2 (g) → H2O (1) ; Δr H20 = -286 kJ/mol (2)
C2H6 (g) + 7/2 O2 (g) → 2CO2 (g) + 3H2O (l); Δr H30 = -1560 kJ/mol (3)
The required equation is

2C (graphite) + 3H2 (g) → C2H6 (g) ; Δf H0 = ?
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To obtain the above equation, multiply equations (1) by 2 and equation (2) by 3 and then add
both the equations we get
2C(graphite) + 3H2 (g) + 7/2 O2 (g) → 2CO 2(g) + 3H2O(l); Δr H50 = -1656 kJ/mol (5)
(where Δr H50 = 2Δr H10 + 3Δr H20 = 2 x (-394) + 3 x (-286) = - 1656 kJ/mol).
Substract equation (3) from equation (5) to get the equation (4)
2C(graphite) + 3H2 (g) → C2H6 ;
where Δf H0 = - 1656 - (-1560) = -96 kJ/mol
Thus, standard enthalpy of formation of ethane is -96 kJ/mol

1. Which of the following is ture ?
a) Enthalpy of formation is the heat evolved or absorbed when one gram of a substance
is formed from its elements in their most stable states.
b) When one mole of H+ (aq) and 1 mole of OH- (aq) react, 57.1 kJ of energy is
absorbed.
c) In the thermochemical equation.
C (graphite) + O 2 (g) → CO2 (g) ; Δf H0 = -394 kJ/mol-1
Δf H0 is known as enthalpy of formation of CO2 (g)
2. Calculate the enthalpy change for complete combustion of 29.0 g of butane, if
C4H10 (g) + 3/2 O2 (g) → 4CO2 (g) + 5H2O (l); Δcomb H0 = -2658 kJ
3. Calculate the standard enthalpy of the reaction
2H2S(g) + SO2 (g) → 3S (s) + 2H2O (l)
given that
Δf H0 (H2S) = - 20.6 kJ mol-1
Δf H0 (SO2) = - 296.9 kJ mol-1
Δf H0 (H2O) = - 289.9 kJ mol-1
11.7 Bond Enthalpies

In a chemical reactions, you have seen that energy is either absorbed or evolved. Do you
know the origin of this change in energy? You know that bonds are broken and new bonds
are formed in chemical reactions. Energy changes take place in breaking some bonds of the
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reactants and in forming new bonds of the products. So the energy changes in a chemical
reaction are a result of energy changes in breaking and forming of bonds. Let us consider the
gaseous reactions, because in these, we encounter the energy changes due to breaking and
forming of bonds only.
At high temperature, hydrogen molecules dissociate into atoms as

H2 (g) → H (g) + H (g) ; Δr H = 435 kJ/mol
The heat absorbed in this reaction is used to break the chemical bonds holding the hydrogen
atoms together in the H2 molecules. For a diatomic molecule like H2(g), we define bond
dissociation energy as the enthalpy change of the reaction in which one mole of the
gaseous molecules are dissociated into gaseous atoms.
Now, let us consider a polyatomic molecule like H2O(g). The dissociation may involve
fragmenting the molecules into an atom and a group of atoms, called a radical, as in
H2O (g) → H (g) + OH (g) ; Δr H0 = 502 kJ/mol
OH (g) → O (g) + H (g) ; Δr H0 = 472 kJ/mol
In the first reaction, one of the two OH bonds in H2O (g) dissociates with an enthalpy change
of 502 KJ/mol and in the second reaction, second OH bond dissociates with an enthalpy
change of 427 KJ/mol. It is calear that the dissociation energy of the O - H bond is sensitive
to its environment. However, the difference is not very large. We take the average value
(464.5 kJ/mol in this case) in case of polyatomic molecules and call it bond enthalpy.
Bond enthalpy is defined as the average amount of enthalpy change involved in the
dissociation of one mole of bonds present in different gaseous compounds.
Now you know the distinction between bond dissociation enthalpy and bond enthalpy. Bond
dissociation enthalpy refers to breaking a particular bond in a particular molecule wheareas
bond enthalpy is the average value of bond dissociation energies for a given type of bond.
The bond enthalpies of some bonds are listed in table 11.1.
By using bond enthalpies (B.E) it is possible to estimate the energy released when a gaseous
molecule is formed from its gaseous atoms. For example, the energy released at constant
pressure for the reaction (Δr H).
3H (g) + C (g) + Cl (g) → CH3Cl (g)
is the sum of the energies of three C - H bond, and one C - Cl bond, all taken with a negative
sign. because energy is released. Using the values of bond enthalpies (B.E.) from table 11.1
we get,
Δr H = - 3 x B.E (C - H) - B.E. (C - Cl)
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= (- 3 x 145 - 335) kJ mol-1
= (- 1245 - 335) kJ mol -1
= - 1574 kJ mol-1
We will now show you how to use bond enthalpy data to estimate the enthalpy of a reaction,
when direct calorimetric data are not available. Note that in section 11.7 we used enthalpy
of formation data to estimate enthalpy of a reaction. In principle, bond enthalpy data can be
used to calculate Δr H for a chemical reaction occuring in gaseous state by making use of
difference in energy absorbed in breaking the bonds in reactants and energy released in
formation of bonds in products.
Δr H = Σ B.E. (reactants) - Σ B.E. (products) (11.10)
Table 11.1 : Average Bond enthalpies
BOND BOND ENTHALPY / (kJ mol-1)
H-H 435
C-H 415
C - Br 284
C-C 356
C=C 598
Br - Br 193
Cl - Cl 242
C - Cl 339
F-F 155
H -Cl 431
H-O 462
H-N 390
H-F 563
H - Br 366
H-I 296
C-O 355
C=O 723
C-N 391
C=N 619
C≡ N 878
C≡ C 832
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Example 11.3 : Use bond enthalpy data given in table 11.1 to calculate the enthalpy of the
reaction.
CH3 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)
Soltuion : 1. Write the equation using structural formula
H H
| |
H C_H
_ + _
Cl Cl → H C _ Cl + H _ Cl
_
| |
H H
2. List the bonds broken and bonds formed under the equation as shown
H H
| |
H_C_H + Cl _ Cl → H _ C _ Cl + H _ Cl
| |
H H
Number of bonds broken number of bonds formed

C_H=4 C _ Cl = 1
Cl _ Cl - 1 H _ Cl = 1
C_H=3
3. Look up the values of bond enthalpies for the bonds in the reactants and products and
list them as shown
Reactants Products
B.E. (C _ H) = 435 kJ mol-1 B.E. (Cl _ C) = 339 kJ mol-1
B.E. (C _ Cl) = 242 kJ mol-1 B.E. (H _ Cl) = 431 kJ mol-1
B.E. (C _ H) = 435 kJ/mol
4. Use equation 11.10,

Enthalpy of the reaction Δr H = Σ B.E. (reactants) - Σ B.E. (products)
= 4 (B.E. (C _ H) + (B.E. (C _ Cl) - [B.E. (C _ Cl) + B.E. (H _ Cl) + 3 B.E. (C _ H)]
= [4 x 435 + 242] - [339 + 431 + 3 x 435] = - 93 kJ
Now let us take one example in which we calculate enthalpy of a reaction using
i) Enthalpy of formation data
ii) Bond enthalpy data.
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Example 11.4 : Calculate Δr H0 for the reaction
Cl2 (g) + 2HF (g) → 2HCl (g) + F2 (g)
a) Using enthalpy of formation data

Δr H0 (HCl) = - 92.5 kJ
Δf H0 (HF) = - 269 kJ
b) Using bond enthalpy data
B.E. (H _ Cl) = 431 kJ mol-1

B.E. (F _ F) = 155 kJ mol-1
B.E. (H _ F) = 563 kJ mol-1
B.E. (Cl _ Cl) = 242 kJ mol -1
Soltuion : a) Using enthalpy of formation of compounds,

Δr H = [2Δf H0 (HCl) + Δf H (F2)] - [2Δf H0 (HF) + Δf H0 (Cl2)]
f = [2 x (-92.5) + 0] - [2x - (269)+ 0] kJ
= - 185 kJ + 538 kJ
= + 353 kJ
b) Using bond enthalpies
Δr H0 = Σ B.E. (reactants bonds) - Σ B.E. (products bonds)
= [B.E. (Cl _ Cl) + 2 B.E. (H _ F)] _ (2B.E. (H _ Cl) + B.E. (F _ F)]
= [242 + 2 (563)] kJ - [2 x 431 + 155] kJ
= 1368 kJ - 1017 kJ
Δr H0 = 351 kJ
Δr H0 calculated by these two different methods are nearly the same.
11.8 The Third Law of Thermodynamics

In 1910M. Planck suggested that “The entropy of a pure and perfectly crystalline substance is
zero at the absolute zero of temperature” (-2730C)
Slim T =0
k
This is known as the third law of thermodynamics.

Unlike the first and the second laws, the third law does not lead to any new concept of
thermodynamics. It only imposes a limitation on the value of entropy.
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T
Cp
ST = ∫ dT
O
T
Thus entropy (S) of a substance at any temperature is calculated if the temperature dependence
of Cp known in evaluating the absolute value of entropy of any substance.
Contribution to entropy due to phase transformations such as melting, vaporization,
sublimation, transition from one allotropic from to the other allotropic form must be taken
into account. Accurate determination of entropy (ST) requires that the heat capacity at constant
pressure (CP) must be determined accurately. But CP cannot be measured at absolute zero
(-2730C) or around absolute zero.
For this heat capacity at constant volume (CV) measurements at various temperatures are
made upto as low temperature as possible. C V Value at absolute zero is obtained by using
9 the extrapolating technique and the Debye equation.
F
CV = aT3 ; a = constant for a substance
At the temperature in the vicinity of absolute zero (Cp - CV) is negligible. Hence Cp = C V.
Hence absolute entropy S0 is calculated using Cv Value.

1. Write True or False
a) Enthalpy of a reaction is equal to the sum of the enthalpy of formation of products
minus the sum of the enthalpy of formation of the reactants.
b) Enthalpy of formation of any elementary substance is equal to zero.
c) If a reaction can be carried out in several steps, the enthalpy change for the
overall reaction is equal to enthalpy change in the last step.
d) Bond enthalpy and bond dissociation energy are same for polyatomic molecules.
...................................................................................................................................
2. Calculate the bond enthalpy N _ H in NH3 (g), given
1
/2 N2 (g) + 3/2 H2 (g) → NH3 (g) ; Δr H0 = - 46 kJ mol-1
1
/2 H2 (g) → H 2 (g) ; Δf H0 = 218 kJ mol-1
1
/2 N2 (g) + N2 (g) ; Δr H0 = 973 kJ mol-1
...................................................................................................................................
3. Calculate the enthalpy of the reaction.
H2 (g) + Cl2 (g) → 2HCl (g)
given,
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Bond enthalpy (H - H) = 435 kJ mol-1
Bond enthalpy (Cl - Cl) = 242 kJ mol-1
Bond enthalpy (H - Cl) = 431 kJ mol-1
...................................................................................................................................

• System is the part of the physical universe which is under study, while the rest of the
universe is surrounding.
• Isolated system is a system which can exchange neither matter nor energy with the
surrounding.
• Closed system is a system which can exchange energy but not the matter with the
surrounding.
• Open sytem is a system which can exchange both energy and matter with the
surroundings.
• State functions are those functions which depend only on the state of the system.
• Extensive properties depend upon the size of the system whereas intensive properties
do not depend upon the size of the system.
• When the temperature of the system is kept constant during various operations then
the process is said to be isothermal.
• In an adiabatic process there is no exchange of heat between the system and the
surroundings.
• Reversible process are those processes in which the changes are carried out so slowly
that the system and surrounding are always in equilibrium.
• Exothermic reactions are those reactions which proceed with the evolution of heat.
• Endothermic reactions are those which proceed with absorption of heat from the
surroundings.
• First law of Thermodynamics states that energy can neither be created nor destroyed.
• Internal energy is the sum of the energies of all the atoms, molecules or ions contained
in the system.
• The state function enthalpy (H) is given by the relation H = U + PV.
• When a chemical equation is reversed, the sign of Δ H is also reversed.
• The enthalpy of reaction is dependent on the number and nature of the intermediate
steps.
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• Bond enthalpy is the average amount of bond dissociation enthalpies for a given type of
bond present in different gaseous compounds, when one mole of bonds is broken in the
gaseous state.
Terminal Exercise
1. Enthalpy of combustion of ethyl alcoho, C2H5OH, is - 950 kJ mol-1. How much heat is
evolved when one gram of ethyl alcohol burns ?
2. Given :
C2H2 (g) + 5/2 O2 (g) → 2CO2 (g) + H 2O (l); Δcomb H = -1299 kJ mol-1
C (graphite) + O2 (g) → CO2 (g) ; Δf H = -393 kJ/mol
H2 (g) + 1/2 O2 (g) → H 2O (g) ; Δf H = -285.5 kJ mol-1
Calculate enthalpy of formation of C2H2 (g)

3. Calculate the enthalpy of combustion of propane
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
Given the following :

H2 (g) + 1/2 O2 (g) → H2O (l) ; Δr H = -285.5 kJ mol
C (s) + O2 (g) → CO2 (g) ; Δf H = - 393 kJ/mol
3C (g) + 4H2 (g) → C3H8 (g) ; Δf H = - 104 kJ mol -1
4. When two moles of H2 and one mole of O2 react to produce two moles of gaseous
water at 373 K and 1 bar pressure, a total of 484 kJ are evolved. What are (a) Δ H and
(b) Δ U for the production of a single mole of H2O (g).
5. Calculate enthalpy of the reaction:
2Na2O2 (s) + 2H 2O (1) → 4NaOH (s) + O2 (g)
Enthalpies of formation of NaOH (s), Na2O2 (s) and H 2O (1) are - 426.4 kJ mol-1. 504
kJ mol-1 and -285 kJ mol-1 respectively.
6. Calculate the heat of formation of gaseous ethly alochol.
2C (graphite) + 3H 2 (g) + 1/2 O2 → C2H5OH (g)
given that enthalpy of sublimation of graphite is 714 kJ / mol and bond enthalpies of
H _ H, O = O, C _ C, C _ H, C _ O and O _ H are respectively 435 kJ/mol-1, 498 kJ/
mol-1, 347 kJ/mol-1, 415 kJ/mol-1 , 355 kJ/mol-1 and 462 kJ/mol-1 respectively.
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11.1
1. (a)
1 3
2. (a) Δn = 1 − − = −1
2 2
(b) Δ U = Δ H - Δ n RT
= 46000 (J mol-1) - (-1) (8.314 Jk mol-1) x (298 K)
= - 46000 (J mol-1) + 2247.6 (J mol-1)
= - 43.5 kJ mol-1
3. (a)
11.2
1. (c)
2. - 1329 kJ
3. Δf H0 = 2 Δf H0 (H2O) - 2 Δf H0 (H2S) - Δf H0 (SO2)
= - 241.7 kJ
11.3
1. (a) T
(b) T
(c) F
(d) F
2. Δr H0 = Σ B.E. (reactant bonds) - Σ B.E. (products bonds)
or - 46 (kJ mol-1) = 3 x 218 (kJ mol-1), + 973 (kJ mol-1) - B.E. (NH3(g) Bonds)
B.E. (NH3 (g) Bonds) = 1673 kJ mol-1
B.E. (N - H) = 557.7 kJ mol-1
3. Δr H0 = -185 kJ mol-1
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12
SPONTANEITY OF
CHEMICAL REACTIONS
We have studied about the first law of thermodynamics in lesson 11. According to this law
the processes occur in such a way that the total energy of the universe remains constant. But
it does not tell us whether a specified change or a process including a chemical reaction can
occur spontaneously i.e., whether it is feasible or not. For example, the first law does not
deny the possibility that a metal bar having a uniform temperature can spontaneously become
warmer at one end cooler at the other. But it is known from experience that such a change
does not occur without expenditure of energy from an external source.
The first law also states that energy can be converted from one form into an equivalent amount
of energy of another form. But it does not tell that heat energy cannot be completely converted
into an equivalent amount of work without producing some changes elsewhere. In this lesson
you shall learn to predict whether a given process or chemical reaction can occur spontaneously
or not.
Objectives
• define entropy;
• recognize that entropy change in a system is given by
q rev
ΔS =
T
• state entropy criterion for a spontaneous process Δ Sunivrse > 0 and at equilibrium
Δ Sunivrse = 0
• state third law of thermodynamics ;
• state the relationship between G, H and S;
• derive the relation Δ Gsystem = T Δ Ssystem;
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• state Gibbs energy criterion for spontaneous process
Δ G < 0 for a spontaneous process
Δ G = 0 at equilibrium
Δ G > 0 for a non-spontaneous process
• define standard Gibbs energy of formation of a substance;
• relate the standard Gibbs energy change with the equilibrium constant and
• solve numerical problems.
12.1 Spontaneous and Non-spontaneous Processes

We know that hot water kept in a container cools down by losing heat to the surroundings.
On the other hand, water at room temperature cannot become hot by gaining heat from the
surroundings. It can be made hot by heating it over a gas burner. The cooling down of hot
water is an example of a spontaneous process. Heating of water (at room temperature) is an
example of a non-spontaneous process because an outside agency (gas burner) has been
used.
A spontaneous process is a process that occurs in a system by itself; once started, no action
from outside the system (outside agency) is necessary to make the process continue. A non-
spontaneous process will not take place unless some external action is continuously applied.
Let us consider another example, we know that when iron objects are exposed to moist
atmosphere, rusting of iron takes place. Although the rusting iron is a slow process but it
always takes place in the same direction. We say that the rusting of iron is a spontaneous
process. During rusting of iron, iron is oxidised to iron (III) oxide.
4Fe(s) + 3O2(g) → 2Fe2O3(s)
The reverse of the above reaction is also possible but it is non-spontaneous. An external
agency has to be used to reduce iron (III) oxide to iron.
From our discussion it can be concluded that
• if a process is spontaneous, the reverse process in non-spontaneous.
• both spontaneous and non-spontaneous processes are possible.
• spontaneous processes occur naturally whereas non-spontaneous processes require the
help of an outside agency to occur.
12.2 Entropy
In fig 12.1(a) the bulb ‘I’ contains 1 mol of an ideal gas ‘A’ at a pressure of 1 bar and the bulb
II contains 1 mol of another ideal gas ‘B’ at 1bar. The two bulbs are joined together through
a valve.
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gas gas
Gas
A
Fig 12.1
When the valve between the two bulbs is opened[Fig 12.1 (b)], the two gases mix
spontaneously. The mixing of gases continues until the partial pressure of each gas becomes
equal to 0.5 bar in each bulb i.e., the equilibrium is attained. We know from experience that
the process cannot be reversed spontaneously-the gases do not unmix. What is the driving
force behind this process?
We know that the internal energy (U) and enthalpy (H) of an ideal gas depend only upon the
temperature of the gas and not upon its pressure or volume. Since there are no intermolecular
forces in ideal gases, Δ U = Δ H = 0 when ideal gases mix at constant temperature. Thus
energy change is not the driving force behind the spontaneous mixing of ideal gases. The
driving force is simply the tendency of the molecules of the two gases to achieve maximum
state of mixing, i.e., disorder. The thermodynamic property related to the disorder of the
system is called entropy. It is denoted by the symbol S.
The entropy is the measure of disorder or randomness in a system. The greater the
disorder in a system, the greater is the entropy of the system.
For a given substance,
(i) the crystalline state is the most ordered state, hence its entropy is the lowest.
(ii) the gaseous state is the most disordered state, hence its entropy is the maximum, and
(iii) the disorder in the liquid state is intermediate between the solid and the gaseous state.
When a system changes from one state to another, the change of entropy Δ S is given by
q rev
ΔS = (12.1)
T
Where qrev is the heat supplied reversibly at temperature T.
12.2 Criteria for Spontaneous Change :

The Second Law of Thermodynamics
So far we have studied about internal energy, enthalpy and entropy. Can we define the
spontaneity of a process in terms of these properties? Let us see whether these changes in
properties can be used as a criterion for determining the spontaneity of a process or not.
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(i) We know that most of the processes which occur spontaneously are exothermic. Water
kept in a container at room temperature evaporates spontaneously. It is an endothermic
process. Thus enthalpy change cannot be used as a criteria for spontaneous change.
(ii) Can we use the increase of entropy as a criteria for the spontaneous change? Then how
do we explain the spontaneous freezing of water at -100C? We know that crystalline ice
is more ordered then the liquid water and therefore the entropy must decrease. The
answer to this question is that we must consider simultaneously two entropy changes:
(a) the entropy change of the system itself, and
(b) the entropy change of the surroundings.
Δ Stotal = Δ Suniv = Δ Ssystem + Δ Ssurrounding > 0 (12.2)
The equation is one of the many forms of the second law of thermodynamics.
According to the second law of thermodynamics all spontaneous or natural processes
produce an increase in entropy of the universe.
Thus, for a spontaneous process when a system is at equilibrium, the entropy is at maximum,
and the change in entropy is zero.
Δ S = 0 (at equilibrium) (12.3)
12.4 Entropy Change in Phase Transitions

When a solid melts an produces liquid, the process occurs at the melting point of the solid.
For example, ice melts at 273 K and produces water at the same temperature.
at 273
H 2O(s) ⎯⎯⎯→ H 2O(l )
The heat involved in the process of melting is called enthalpy of fusion (Δ fusH). Therefore,
the entropy of fusion (Δ fusS) is given by
Δ fusS
Δ fusS = (∴ qrev at const p = Δ fus
H)
T
Similarly, for the equilibirum
at 373K
H 2O(l ) ⎯⎯⎯⎯
→ H 2 O(g )
Δ vap H
Δ vapS = (T is the boiling point)
T
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Example 12.1 : The enthalpy change for the transition of liquid water to steam at 373 K is
40.8 kJ mol-1. Calculate the entropy change for the process
373K
⎯⎯⎯
Solution : H 2O(l ) ← ⎯⎯⎯
→ H 2 O(g )
Δ vap H
Δ vapS =
T
Δ vapH = 40.8 kJ mol-1 = 40.8 x 10 3 J mol-1
T = 373 K
40.8 x 103 J mol −1

Δ vapS = = 109 J K-1 mol-1
373K

1. The enthalpy change for the transition of ice to liquid water at 273 K is 6.02 kJ mol-1.
Calculate the entropy change for the process.
...................................................................................................................................
2. Arrange the following systems in the order of increasing randomness.
i) 1 mol of gas A
ii) 1 mol of solid A
iii) 1 mol of liquid A
...................................................................................................................................
3. Indicate whether you would expect the entropy of the system to increase or decrease
a) 2SO2 (g) + O2(g) → 2SO3 (g)
b) N2 (g) + 3H2(g) → 2NH3 (g)
c) O2 (g) → 2O (g)
...................................................................................................................................
12.5 Absolute Entropy

When the temperature of a substance is increased, the translational, vibrational and rotational
motions become more vigorous. It leads to greater disorder and as a result the entropy of the
substance increases. Thus, on increasing the temperature of a substance the entropy of a
substance increases. It decreases on decreasing the temperature of a substance.
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According to Walther Nernst, the entropy of a pure perfectly crystalline substance approaches
zero as the temperature approaches absolute zero. This is known as the Third law of
thermodynamics. This third law of thermodynamics helps us to calculate the absolute value
of molar entropies (Sm) of substances at different temperatures. The standard molar entropies
of some substances at 298 K are given in Table 12.1
Table 12.1 : Standard molar entropies (S m0/J K -1 mol -1 ) at 298 K
Solids Entropy Liquid Entropy Gases Entropy

C (graphite) 5.7 H2O 69.9 H2 130.7
C (diamond) 2.4 Hg 76.0 O2 205.1
Fe 27.3 C 2H5OH 160.7 N2 191.6
Pb 64.8 C6H6 173.3 CO 2 213.7
Cu 33.1 CH 3COOH 159.8 NO 2 240.1
Al 96.2 N 2O 4 304.3
C 12H22O 11 360.8 NH 3 192.3
CaCO 3 92.9 CH 4 186.2
Entropy Change for a Reaction
The absolute entropies can be used for calculating standard entropies changes accompanying
chemical reaction. It can be determined by substracting the standard entropies of reactants
from those of products. Thus, for a general reaction.
aA + bB + ............ → pP + qQ + ....
Δ Smo = [pSmo (P) + qSmo (Q) + .....] - [aSmo (A) + bSmo (B) + .....]
Δr Smo = Σ Smo (products) - Σ Smo (reactants)
Example 12.2 : Calculate the entropy change, Δr Smo for the following reaction at 298 K.
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Given that the standard molar entropies of Fe(s), O2(g) and Fe2O3(s) at 298 K are 27.3,
205.0 and 87.4 J K-1 mol-1 respectively.
Solution : 4Fe(s) + 3O2(g) → 2Fe2O3(s)
Δr So = Σ vpSmo (products) - Σ vRSmo (reactants)
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Δr So = 2Smo (Fe 2O3) - [4Smo (Fe) + 3Smo (O2)]
= [2 x 87.4 - (4 x 27.3 + 3 x 205.0)] J K-1 mol-1
= - 549.4 J K -1 mol-1
12.6 Gibbs Energy and Spontaneity

When can use the expression
Δ Suniv = Δ Stotal = Δ Ssystem + Δ Ssurrounding > 0 (12.4)
as our basic criterion for a spontaneous change. But it is very difficult to apply it because we
have to evaluate the total entropy change i.e. the entropy change of system plus that of
surroundings. This is tedious process as it is difficult to figure out all the interactions between
the system and the surroundings. Thus, for a system which is not isolated from its surroundings.
Δ Stotal = Δ Ssystem + Δ Ssurrounding (12.5)
At constant temperature and pressure if q p is the heat given out by the system to the
surroundings, we can write
−q p ΔH system
Δ Ssurrounding = = − (12.6)
T T
(Since qp = Δ H at constant pressure)

Substituting Eq. 12.6 in Eq. 12.5, we get
ΔH system
Δ Stotal = Δ Ssystem _
T
T Δ Stotal = T Δ Ssystem _ Δ Hsystem
or _ TΔS
total
= Δ Hsystem _ T Δ Hsystem (12.7)
Now, let us define another thermodynamic property, Gibbs energy. It is defined by the equation
G = H - TS (12.8)
For a change in Giibs energy, we write
Δ G = Δ H _ Δ (TS)
ΔG=ΔH_TΔS_SΔT
For a change at constant temperature, ΔT = 0,
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Therefore ΔG=ΔH _ TΔS (12.9)
Since H, T and S are state functions, it follows that G is also a state function. Comparing
equations 12.7 and 12.9 we find that
Δ G = _ T Δ STotal (12.10)
We have seen that if Δ STotal is positive, the change will be spontaneous. Equations 12.10 can
be used to predict the spontaneity of a process based on the value of Δ G.
The use of Gibbs energy has the advantages that it refers to system only. Thus for a process
occurring at constant temperature and pressure, if
Δ G < 0 (negative), the process is spontaneous
Δ G > 0 (positive), the process is non-spontaneous
Δ G = 0 (zero), the process is at equilibrium
In deciding the spontaneity of a chemical reaction, the equation Δ G = Δ H - T Δ S takes into

account two factors (i) the energy factor Δ H, and (ii) the entropy factor Δ S. Based on the
signs of Δ H and Δ S there are four possibilities for Δ G. These possibilities are outlined in
table 12.2
Table 12.2 : Criterion for spontaneous change : Δ G = Δ H - T Δ S
S.No. ΔH Δ S ΔG Result
1. _ + _ Spontaneous at all temperatures.
2. _ _ _ Spontaneous at low temperatures.
+ Non-spontaneous at high temperatures
3. + + + Non-spontaneous at low temperatures.
_ Spontaneous at high temperatures.
4. + _ + Non-spontaneous at all temperatures.
Example 12.3 : For the reaction

2NO (g) + O 2 (g) → 2NO2 (g)
Calculate Δr G at 700 K when enthalpy and entropy changes (Δr H and Δr S) are respectively
- 113.0 kJ mol- and -145 JK-1 mol-1

Δ H = _ 133.0 kJ mol-1
Δ S = _ 145 J K-1 mol-1 = _ 145 x 10-3 kJ K -1 mol-1
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T = 700 K
Substituting the values in

Δ G = Δ H _ TΔ S
Δ G = (_ 113.0 kJ mol-1 ) _ (700 K) (_ 145 x 10-3 kJ K-1 mol-1
= (_ 113.0 kJ mol-1 ) + (101.5 kJ mol-1)
= _ 11.5 kJ mol-1

1. Determine whether the following reaction
CCl4 (l) + H 2 (g) → HCl (g) + CHCl3 (l)
is spontaneous at 298 K if ΔrH = 91.35 kJ mol-1 and ΔrS = 41.5 JK-1 mol-1 for this
reaction.
...................................................................................................................................
2. Which of the following conditions would predict a process that is always spontaneous ?
i) Δ H > 0, = Δ S > 0
ii) Δ H > 0, = Δ S < 0
iii) Δ H < 0, = Δ S > 0
iv) Δ H < 0, = Δ S < 0
...................................................................................................................................
12.7 Standard Gibbs Energy Change (ΔGO) and

Equilibrium Constant (K)
The standard Gibbs energy change is defined as the change in Gibbs energy for the process in
which the reactants in their standard states are converted into the products in their standard
states. It is denoted by the symbol ΔrGo.
The value of ΔrGo can be found from the standard energy of formation of substances.
The standard Gibbs energy of formation of a compound is defined as the change in Gibbs
energy when 1 mole of the compound is formed from its constituent elements in their standard
states. Like the standard enthalpy of formation of an element, the standard Gibbs energy of
formation of an element in its standard state is taken as zero.
Thus for a reaction
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aA + bB + ........ → pP + qQ + .......
ΔrGo = [p ΔfGPo + q ΔfGQo + .....] - [a ΔfGAo + b ΔfGBo + .....]
ΔrGo = Σ Δ f Go (products) - Σ Δ f Go (reactants)
The standard Gibbs energy change (Δr Go) is related to the equilibrium constant (K) of the
reaction by the expression.
ΔrGo = - RT in K = - 2.303 RT log K
Example 12.4 : The equilibrium constant of the reaction

3
P (s) + Cl2 (g ) U PCl3 (g )
2
is 2.00 x 1024 at 500 K. Calculate the value of Δr Go.
K = 2.00 x 1024
T = 500 K
Δr Go = _ 2.303 RT log K
= _ 2.303 x (8.314 J mol-1 K -1) (500 K) log 2.0 x 1024
= _ 2.303 x (8.314 J mol-1 K-1) (500 K) 24.30
= _ 232.6 kJ mol-1
Example 12.5 : Calculate the standard Gibbs energy change for the reaction
CH4 (g) + O 2 (g) → 2CO2 (g) + 2H2O (l)
at 298K. The standard Gibbs energies of formation of CH4, CO2 and H2O at 298 K are _ 50.8
kJ mol-1, _ 394.4 kJ mol-1 and _ 237.2 kJ mol-1 respectively.
Solution : CH 4 (g) + 2O 2 (g) → CO2 (g) + 2H 2O (l)
ΔrGo = ΔfGo (CO2) + 2 ΔfGo (H2O) _ ΔfGo (CH4) _ 2 ΔfGo (O2)
= _ 394.4 + 2 x (_ 237.2) _ (_50.8) _ 2 x 0
= _ 394.4 _ 474.4 + 50.8
= _ 818 kJ mol-1
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1. What is the relationship between the standard Gibbs energy change and the equilibrium
constant of the reaction ?
...................................................................................................................................
2. The standard Gibbs energy change for the reaction
CO (g ) + 2H 2 (g ) U CH3OH (l)
at 298 K is - 24.8 kJ mol-1. What is the value of the equilibrium constant at 298 K ?
...................................................................................................................................

• All spontaneous process lead to an increase in disorder or randomness.
• The thermodynamic function related to disorder in a system is called entropy, S.
• For a spontaneous change the total entropy change of the system and the surroundings
must increase.
• Gibbs energy is defined as G = H - TS.
• At a constant temperature, the change in Gibbs free energy is related to enthalpy and
entropy changes by the expression.
ΔG=ΔH-TΔS
For a spontaneous change, there must be a decrease in Gibbs energy, i.e., Δ G < 0. At
equilibrium Δ G = 0.
The standard Gibbs energy change is related to the equilibrium constant of the reaction
by the expression
ΔrG0 = -2.303 RT log K
• The standard Gibb’s energy change is given by
ΔrGo = Σ Δf Go (products) - Σ Δ f Go (reactants)
Terminal Exercise
1. What do you call the measure of disorder or randmoness in a system ?
2. Predict the sign of Δ S for each of the following processes.
i) H2 (g) → 2H (g)
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ii) O2 (g, 300 K) + O2 (g, 500 K)
3. Define entropy.
4. Explain why etropy is not a good criteria for determining the spontaneity of a process ?
5. What is the relationship between the enthalpy and the entropy change for a system at
equilibrium ?
6. For the reaction
O3 (g) + O (g) → 2O2 (g)
ΔrH = - 391.9 kJ mol-1 and ΔrS = 10.3 J K-1 mol-1 at 298 K. Calculate ΔrG at the
temperature and state whether the reaction is spontaneous or not.
7. What happens to ΔrG during
a) a spontaneous process
b) a non-spontaneous process
c) a process at equilibrium
8. Calculate ΔrG0 at 298 K for the reaction
2NO2 (g) → N2O4 (g)
Given ΔrH = - 57.20 kJ mol-1 and ΔrS = - 175.8 JK-1 mol-1
Is this reaction spontaneous ?
9. The standard Gibbs energies of formation at 298 are - 202.85 kJ mol-1 for NH4Cl (s), -
16.45 kJ mol-1 for NH3 (g) and - 95.3 kJ mol-1 for HCl (g)
a) What is ΔrG0 for the reaction
NH 4Cl (S) U NH3 (g ) + HCl (g )

b) Calculate the equilibrium constant for this decomposition.
10. For the following reaction
CCl4 (l) + H 2 (g) U HCl (g) + CHCl3 (l)
ΔrG0 = - 103.7 kJ mol-1 at 298 K. Calculate the equilibrium constant for this reaction.
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12.1
Δ fus H 6.02 kJ mol −1 6.02 x 103 J mol −1
1. ΔfusS = = =
T 273 K 273 K
= 22.0 J mol-1 k-1

2. 1 mol of solid, 1 mol of liquid, 1 mol of gas.
3. (a) Decrease (b) Decrease (c) Increase
12.2
1. Δ G = - 103.7 kJ. Therefore the reaction is spontaneous.
2. (iii)
12.3
1. Δ G0 = - 2.303 RT log K
2. 2.2 x 104
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CHEMISTRY
Maximum Marks : 50 Time : 2.30 Hrs.
INSTRUCTIONS :
- Name
- Enrolment Number
- Subject
- Assignment Number
- Address
• Get your assignment checked by the subject teachmer at your study centre so that you
get positive feedback about your performance.
1. a) Give one example each of the following types of solutions :

i) Liquid in solid
ii) Solid in gas
b) What is Brownian motion ?
c) Give one example each of the following of colloids :
i) Liquid in solid
ii) Solid in gas
d) Define enthalpy of formation of a compound.
e) What are state functions? Give one example.
f) Is enthalpy of reactants less or more than that of products in an endothermic
reaction ?
g) Define entropy.
h) What happens to the free energy of a system during a non-spontaneous process?
i) What is the relationship between free energy change and equilibrium constant ?
j) Mention the relation between Kp and Kc. (1x10=10)
2. a) What is an ideal solution ? Give one example.
b) What is a colligative property ? Name two such properies.
c) Distinguish between lyophilic and lyophobic colloids. Give one example of each.
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d) What are these causes of charge on colloidal particles ?
e) What is a (i) closed system (ii) isolated system.
f) Give relation between ΔH and ΔE of a reaction.
g) Indicate in each case whether the entropy of the system would increase or decrease
i) H2O (l) → H2O (s)
ii) H2 (300 K) → H2 (150 K)
iii) 3O 2 (s) → 2O 3 (g)
iv) I2 (s) → I2 (s)
h) Calculate the entropy change for the reaction :
2H 2 (s) + O2(g) → 2H 2O (l)
at 300 K if the standard entropies of H2(g), O2(g) and H2O(l) the 128.6 J K-1 mol-1.
201.2 JK-1 mol-1 and 68.0 J K-11 mol-1 respectively.
i) What are the characteristics of equilibirum state ?
j) When sulphur in the form S8 is heated at 900K its initial pressure of 1 atm falls by
0.299 at equilibrium. This is because of conversion of some S8 to S2. (2x10=20)
3. a) Calculate the molar mass of a substance, 0.4 gram of which when dissolved in 29g
benzene lowers its freezing point by 0.75 K (Kf for benzene = 5.1 K kg mol-1)
The normal molar mass of the substance is 68g mol-1. What is its molecular state
in the solution ?
b) Explain briefly how :
i) smoke is precipitated
ii) rubber is electrolysed.
c) Calculate the enthalpy of the following reaction
C2H4 (g) + HCl (g) → C 2H5Cl (s)
Given that the bond energies are :
C=C 615.1 kJ mol-1 ; C=C 347.7 kJ mol -1
C-H 413.4 kJ mol-1 ; H - Cl 431.8 kJ mol-1 and
C - Cl 328.4 kJ mol-1 ;
d) Calculate the temperature up to which the following reaction would not be spontaneous
M2O (s) → 2M (s) + 1/2 O2 (s)
ΔH0 = kJ mol-1 and ΔS0 = 70 JK-1 mol-1 (3x4=12)
4. a) 500 cm3 aqueous solution contains 8.0 g a dibasic acid (molar mass - 80 g mol-1). The
density of the solution is 0.99 g cm-3. Calculate molarity, molality and the mole fraction
of acid in the solution
b) 250 cm3 of H2 and 18 cm3 of I2 vapour were heated in a sealed vessel at 4600C. At
equilibrium 30.8 cm3 of HI was formed. Calculate the degree of dissociation of HI.
(4x2=8)
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13
CHEMICAL
EQUILIBRIUM
When reactants are mixed in exact stoichiometric proportion to perform a chemical reaction,
it is believed that all the reactants would be converted into products with the release or
absorption of energy. This is not true in all cases. Many chemical reactions proceed only to a
certain extent and stop. When analysed, the resulting mixture contains both the reactants and
products. It is because when reactants combine to form products, the products also start
combining to give back the reactants.
When such opposing processes take place at equal rates, no reaction appears to take place and
it is said that a state of equilibrium has reached. In this lesson, we will examine many aspects
of chemical equilibrium. We shall also discuss how can we control the extent to which a
reaction can proceed by changing the various conditions of the equilibrium.
Objectives
• differentiate between static and dynamic equilibirum;
• identify and differentiate between reversible and irreversible reactions;
• explain the reversible reaction occuring at the equilibrium state;
• list and explain characteristics of equilibrium state;
• apply the law of equilibrium and write expression of equilibrium constant for different
types of equilibria, namely physical, chemical, homogenenous and heterogenous;
• state and derive the relation between Kc and K p and carry out some calculations
involving them and
• list the factors which affect the state of equilibrium and state and apply Le-Chatelier
principle.
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13.1 Static and Dynamic Equilibrium
The state equilibrium can be observed in physical and chemical systems. Also, equilibrium
can be static or dynamic in nature. A book lying on the table is an example of static
equilibrium. The force of action and reaction cancel each other and no change takes place.
Thus it is a case of static equilibrium. On the other hand, when an escalator is coming down
and a passenger is going up at the same speed it is a case of dynamic equilibrium. Here,
because both are moving in opposite directions and at the same speed, no net change takes
place. The equilibrium established in the above examples are in physical systems.
13.2 Reversible and Irreversible Reactions

Chemical reactions can be classified as : Reversible and Irreversible reactions.
13.2.1 Reversible reactions

Consider the reaction between ethanol and acetic acid. When mixed in the presence of
dilute sulphuric acid they react and form ethyl acetate and water.
H+
C2 H5OH(l ) + CH3COOH(l ) ⎯⎯→ CH3COO C2 H5 (l ) + H 2O(l )
On the other hand, when ethyl acetate and water are mixed in the presence of dilute sulphuric
acid the reverse reaction occurs.
H+
CH3COOC2 H5 (l ) + H 2O(l ) ⎯⎯→ CH3COOH (l ) + C2 H5OH (l )
It may be noted here that the second reaction is reverse of the first one and under the same
conditions, the two reactions occur simultaneously. Such reactions which occur
simultaneously in opposite directions are called reversible reactions.
A reaction is said to be reversible if under certain conditions of temperature and
pressure, the forward and reverse reactions occur simultaneously.
Reversible reactions are indicated by placing two half arrows pointing in opposite directions
( U ) between the reactants and products. Thus the above reaction is more appropriately
written as
CH3COOH(l ) + C2 H5OH(l ) U CH3COOHC2 H5 (l ) + H 2O(l )
When ethyl acetate and water are formed in the forward reaction the reverse reaction also
starts in which ethanol and acetic are formed. After some time the concentrations of all
reactants and products become constant. This happens when the rates of forward and reverse
reactions become equal; and all the properties of the system become constant. It is said that
the system has attained state of equilibrium. However it may be noted that the state of
equilibrium is reached only if the reaction is carried out in a closed system. At the time of
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equilibrium, forward and reverse reactions are taking place and it is in a state of dynamic
equilibrium because no change is taking place.
A reversible reaction is said to be in the equilibrium state when the forward and
backward reaction occur simultaneously at the same rate in a closed system and the
concentrations of reactants and products do not change with time
A common example of reversible reactions of the type A + B U C + B
CH3COOH + C2 H5OH U CH3COOH + H 2O
The following graphs Fig. 13.1 shows the equilibrium state in a reversible reactions.
C or D
k
Concentration
A or B
k
Time Equilibrium
Fig. 13.1 : Equilibrium in reversible reaction
The graph depicts that the rate of forward reaction gradually decreases while the rate of
backward reaction increase till they become constant and equal to each other.
13.2.2 Irreversible reactions

Most of the reactions occur only in one direction. They are called irreversible reactions.
For example when carbon is burnt in air to form carbon dioxide the reaction goes only in
one direction i.e. in the direction of formation of carbon dioxide.
C (s) + O2 (g) → CO2 (g)
Strictly speaking all reactions are considered to be reversible. But the rate of reaction in one
particular direction is extremely small compared to the other. Thus the reaction proceeds
practically in one direction to near completion, leaving a negligibly small amount of reactant
at the end.
When hydrochloric acid is mixed with sodium hydroxide, a base, in equimolar quantities, a
neutralisation reaction takes place; with the formation of salt and water.
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HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
This reaction proceeds to completion in the forward direction. Similarly when a solution of
silver nitrate is added to a solution of sodium chloride silver chloride is precipitated
immediately.
NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
13.3 Characteristics of Equilibrium State

1. The state of chemical equilibrium is reached in a reversible reaction when;
i) the temperature of the system attains a constant value.
ii) the pressure of the system attains a constant value.
iii) the concentrations of all the reactants and products attain constant value.
The state of equilibirum has following characteristic properties :
i) Chemical equilibrium is dynamic in nature.
The chemical equalibrium is the result of two equal but opposite processes occurring in the
forward and reverse directions and there is no “net” change occurring in the system.
ii) Equilibrium can be attained from either side
The same state of equilibrium characterized by its equalibrium constant which is discussed
later can be reached whether the reaction is started from the reactants or products side. For
example, the same equilibrium
N2O4 (g) U 2 NO 2 (g)
is established whether we start the reaction with N2O4 or NO2 .
iii) Equilibrium can be attained only in a closed system

Equilibrium can be attained only if no substance among, reactants or products, is allowed to
escape i.e. the system is a closed one. Any system consisting of gaseous phase or volatile
liquids must be kept in a closed container, e.g.
N2 (g) + 3H2 (g) U 2 NH3 (g)
A system consisting of only non-volatile liquid and solid phases can be kept even in an
open container because such substances have no tendency to escape, e.g.
FeCl3 (aq) + 3NH 4SCN (aq) U Fe (SCN)3 (s) + 3 NH4Cl (aq)
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iv) A catalyst can not change the equilibrium state
Addition of a catelyst speeds up the forward and reverse reactions by same extent and help in
attaining the equilibrium faster. However, the equilibrium concentrations of reactants and
products are not affected in any manner.
13.4 Equilibrium in Physical Process; Phase Equilibrium

State of equilibrium can also be reached in physical processes.
13.4.1 Liquid - Vapour Equilibrium

Let us take some quantity of a liquid in an empty container and close it. Initially the vapour
pressure above the liquid will be zero. The liquid will evaporate and its vapour will fill the
empty space above it.
Liquid → Vapour
The rate of evaporation is maximum in begining. As vapour build up, its pressure increases
and the rate of evaporation slows down. Also the reverse process of condensation begins
(Fig. 13.2).
Vapour → Liquid
Vapour
Liquid
Fig. 13.2 : Liquid Vapour equilibrium
and its rate gradually increases with the increase in the vapour pressure. After some time the
two rates (of evaporation and condensation) become equal and the following equilibrium is
established.
Liquid U Vapour
At equilibrium the vapour pressure reaches its maximum value and is known as the saturated
vapour pressure or simply the vapour pressure. At a fixed temperature, each liquid has its
own characteristic vapour pressure. The vapour pressure of a liquid increases with rise in
temperature.
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13.4.2 Solid - Vapour Equilibrium
Volatile solids sublime form vapour. The situation is just similar to the liquid vapour system.
When kept in a closed container at a constant temperature the following equilibrium is
established.
Solid U Vapour
Iodine
Vapour
Solid
Iodine
Fig. 13.3 : Solid vapour equilibrium
Such an equilibrium can be established by keeping some solid iodine in a gas jar covered
with a lid. (Fig. 13.3). Gradually the purple coloured iodine vapours fill the jar and the
following equilibrium is established.
I2 (s) U I2 (g)
13.4.3 Solid - Liquid Equilibrium

Below its freezing point a liquid freezes spontaneously
Liquid → Solid
When heated above its melting point the solid melts spontaneously
Solid → Liquid
At the melting point, the two phases are in equilibrium
Solid U Liquid
because the above two processes occur simultaneously and at the same rate. Such an
equilibrium is characterized by its temperature i.e. the melting point of the solid.
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13.4.4 Solute - Solution Equilibria
Saturated
Solution
Undissdued
Sugar
Fig. 13.4 : Solute - Solution equilibrium
When sugar crystals are put in a saturated solution of sugar in water; it will appear that no
change is taking place and sugar appears to remain undissolved. Actually, the undissolved
sugar does dissolve in the saturated sugar solution; and an equal amount of sugar seperates
out from the solution. The solid sugar and the sugar solution form an equilibrium system
which is dynamic in nature.
sugar (s) U sugar solution (saturated)
The equilibrium is established when the rate of dissolution of sugar becomes equal to the
rate of crystallisation. In general such equilibrium can be represented as
solute (s) U solution (saturated)
The equilibrium is known as Solubility Equilibrium.
13.4.5 Phase and Phase Equilibrium

You must have noticed in each of the above equilibria the system consists of two distinct
parts; solid, liquid, solution or vapour. Each of these parts is called a phase.
A phase is defined as a homogenous part of a system which has uniform composition
and properties throughout.
A phase is not the same as physical state. A mixture of two solids, even when powdered
finely is a two phase system. This is because particles of the two solids have different
chemical compositions and physcial properties. Completely miscible liquids, solutions and
all gaseous mixture constitute only one phase each.
All the cases of physical equilibrium are in fact the systems in which different phases are in
equilibrium; only if they contain, at least one common component. A dynamic exchange of
the common component between two phases takes place. When the rates of exchange becomes
equal the equilibrium is established. In solid solute and solution equilibrium the example
given earlier, sugar is the common component.
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13.5 Equilibrium in Homogeneous and Heterogeneous Systems
13.5.1 Homogeneous and Heterogeneous Systems

Homogeneous system is one which has one phase. It has the same chemical composition and
uniform properties throughout. It is formed by particles of molecular size only. Pure solids,
liquids, gases and solutions are the examples of homogeneous systems.
A system consisting of only one phase is called a homogeneous system
Hetergeneous system, on the other hand has at least two phases - a mixture of solids or
immiscible liquids constitutes a heterogeneous system.
Any system consisting of two or more phases is called heterogeneous system
13.5.2 Homogeneous and Heterogeneous Equilibrium Systems

Equilibrium can be established in either types of systems. Since all physical equilibria involve
at least two phases, therefore these are all examples of heterogeneous equilibrium. But
chemical equilibrium can be homogeneous or heterogeneous in nature. It is homogeneous
if both the reactants and products are present only in one phase gas or liquid and
heterogeneous if present in more than one phase. In the following sections we shall study
such systems.
13.5.3 Homogeneous Chemical Equilibrium Systems

(a) Gas - Phase homogeneous systems
Such systems contain only gaseous reactants and products. Since all gaseous mixtures are
homogeneous in nature they constitute only one phase. Following are examples of this type
of equilibrium.
i) N2 (g) + 3H2 (g) U 2 NH3 (g)
ii) 2N2O5 (g) U 4NO2 (g) + O 2 (g)
(b) Liquid - Phase homogeneous systems

These are the sytems in which both the reactants and products are present in only one liquid
phase (as a solution) for example :
+
H
i) CH3 COOH(l ) + C2 H5OH (l ) Y
ZZZ
ZZXZ CH3COOC2 H5 (l ) + H 2O(l)
OH (l )
ii) KCN (aq ) + H 2O (l) Y

ZZ
ZXZ HCN (aq ) + KOH (aq )
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13.5.4 Heterogeneous Chemical Equilibrium Systems
The systems in which reactants and products are present in more than one phase belong to
this type. For example :
i) Fe (s) + 4H 2O (g ) Y
ZZ
ZXZ Fe3O 4 (s) + 4H 2 (g )
ii) ZZ
CaCO3 (s) YZXZ CaO (s) + CO 2 (g )

1. What is a reversible reaction ? Give two examples.
...................................................................................................................................
2. When does a reaction reach equilibrium state ?
...................................................................................................................................
3. How would you know whether a system has reached the equibrium state or not ?
...................................................................................................................................
4. Give two examples of physical equilibrium.
...................................................................................................................................
5. Give two example each of chemical homogeneous and heterogeneous equilibria.
...................................................................................................................................
13.6 Quantitative Aspect of Equilibrium State
13.6.1 Law of Equilibrium and Concentration Equilibrium Constant

Consider the following equilibrium
H2 (g) + I 2 (g) U 2HI (g)
At equilibrium the concentrations of H2, I2 and HI become constant. Also, it has been found
experimentally that irrespective of the starting concentration of H2, I 2 the following ratio of
concentration terms always remains constant.
[HI]2
Kc =
[H 2 ][I 2 ]
Here [H2], [I2] and [HI] represent the equilibrium molar concentrations of H2, I2 and HI
respectively and KC is called the concentration equilibrium constant (some times it is written
as simply K). In general, for reversible reaction
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aA + bB U cC + dD
at equilibrium, the following ratio of concentration terms always remains constant at a given
temperature.
[C]c [D]d
Kc =
[A]a [B]b
The above relation is known as the law of equilibrium. It may be noted here that all the
concentrations values in the law of equilibrium are the equilibrium concentrations of reactants
and products. The numerator of the law of equilibrium is the products of equilibrium molar
concentrations of products, each term being raised to the power equal to its stoichiometric
coefficient in the chemical equation and the denominator contains products of similar
concentration terms of reactants.
13.6.2 Pressure Equilibrium Constant Kp

In case of gases their partial pressures can also be used in place of molar concentrations
(since the two are directly proportional to each other) in the law of equilibrium. The new
equilibrium constant, Kp, is called the pressure equilibrium constant. For the reaction between
H2 and I2, Kp is given by
P2 HI
Kc =
PH x P I
2 2
Here PH , PI and PH are the equilibrium partial pressures of H2, I2 and HI respectively. For
2 2 1
the general gas phase reaction :
a A (g) + b B (g) U c C (g) + d D (g)
it is given by :
PCc x PDd
Kp =
PAa x PBb
13.6.3 Relation between Kp and Kc

For a general gas phase reaction at equilibrium
a A (g) + b B (g) U c C (g) + d D (g)
The pressure and concentration equilibrium constants kp and Kc are
PCc x PDd [C]c [D]d

Kp = and Kc =
PAa x PBb [A]a [B]b
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For a gaseous substance i, the ideal gas equation is
P1 V = n1 RT
where p1 and n1 are its partial pressure and amount in a gaseous mixture and V and T are its
volume and temperature and R is the gas constant. The relation may be written as
n1
P1 = RT = c1 RT
V
Where ci is the molar concentration or molarity of ‘i’ expressed in moles per litre. This
relation can be used for replacing the partial pressure terms in the expression for Kp.
(CC RT)c (CD RT)d

Kp =
(CA RT)a (CB RT) b
CcC x CdD (RT) (c + d) - (a + b)

=
CaA x CbB
Using the square bracket notation for molar concentration the relation can be written as
[C]c [D]d
Kp = (RT) (np + nR)
[A]a [B]b
= K C (RT) Δng
where Δng is the change in the moles of gaseous substances in the reaction and is equal to the
difference in the moles of gaseous products np and the moles of gaseous reactants, nR. Δng
may be zero positive or negative.
i) In the reaction
H2 (g) + I2 (g) U 2HI (g)
Here np = moles of the gaseous product is equal to 2

nR = moles of gaseous reactant H 2 and I2 is equal to 2 (as 1 + 1).
Hence Δng = np - ng = 2 - 2 = 0.
Δng = 0.
ii) In the reaction
N2 (g) + 3H 2 (g) U 2NH 3 (g)

np = 2, nR = 1 + 3 = 4.
and Δng = 2 - 4 = - 2
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iii) In the reaction involving solids and gases
CaCO3 (s) U CaO (s) + CO2 (g)
Δng = 1
13.6.4 Expressions of Equilibrium Constant for Some Reactions

The law of equilibrium can be applied to write down expressions of KC and KP for some reactions.
13.7 Homogeneous Equilibria

i) Decomposition of N 2O4
N2O4 (g) U 2 NO2 (g)
2
[ NO 2 ]2 PNO 2
KC = ; K P = KP =
[ N 2O 4 ] PN 2O 4
ii) Oxidation of sulphur dioxide
2SO2 (g) + O2 (g) U 2SO3 (g)

22
[SO3 ]2 PPSSO 3
KC = ; KP = 2 2
[SO 2 ]2 [O 2 ] PPSSO .P
2 O2
iii) Esterification of acetic acid with ethanol
CH3COOH (I) + C2H5OH (I) U CH3COOC2H5(I) + H2O (I)
[CH3COOC2 H5 ] [H 2O]
KC = [CH COOH] [C H OH]
3 2 5
In this reaction no gas is involved, therefore expression for KP is meaningless.
13.7.1 Heterogeneous Equilibrium

Consider the following equilibrium
According to the law of equilibrium
[CaO] [CO2]
KC =
[CaCO 3]
Here CaCO 3 and CaO are pure solids. The concentration of any solid is constant at a fixed
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temperature therefore these are not written in expression for equilibrium constant for
heterogenous reactions. Equilibrium constant for the reaction can be written as
KC = [CO2] and KP = Pco2
Following are some more examples of heterogenous equilibrium
i) Reaction between iron and steam
3Fe (s) + 4H2O (g) U Fe3O4 (s) + 4H2 (g)
4
[H 2 ]4 PH4 23
KC = ; KP =
[H 2O]4 PH4 2O
ii) Liquid - Vapour Equilibrium
H2O (I) U H2O (g)
KC = [H2O; g] ; KP = PH
2O
13.8 Characteristics of Equilibrium Constant
13.8.1 Equilibrium Constant and Chemical Equation

The expression of equilibrium constant depends upon the manner in which the chemical equation
representing it is written. For the reaction
H2 (g) + I2 (g) U 2HI (g)
[HI]2
The equilibrium constant K is given by K =
[H 2 ][I 2 ]
When the same reaction is written as
a) 1/2 H2 (g) + 1/2 I2 (g) U HI (g)

the corresponding equilibrium constant Kl is given by
[HI]
1 1
K1 =
[H 2 ] 2 [I 2 ]2
It may be noted that equilibrium constant K and K1 are related as K1 = K

b) When the reaction is written as reverse
2HI (g) U H2 (g) + I2 (g)
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[H 2 ][I 2 ]
K2 =
[HI]2
Here in can be seen that
1
K2 =
k
Similar relationship is also observed in the pressure equilibrium constant Kp. Thus the
expression of equilibrium constant depends on how the reaction is expressed in the form of a
chemical equation.
13.8.2 Units of Equilibrium Constant

Units of equilibrium constant KC or KP depend upon the fact whether during the reactions there is
any change in the moles of substance or not.
10
F a) The reactions in which there is no change in moles of substance i.e. Δn = 0.
The equilibrium constant for such reaction has no units. For example in the reaction between
H2 and I2
H2 (g) + I2 (g) U 2HI (g)
2
[HI]2 PHI
KC = KP = P .P
[H 2 ][I 2 ] H 2 l2
(mol L−1 ) 2 bar 2

KC = KP =
(mol L−1 ) (mol L−1 ) (bar ) (bar )
∴ Hence KP and KC have no units in such cases.

b) The reaction where there is change in the moles of substance i.e. Δn ≠ 0.
The equilibrium constant for such reactions has units which depend upon the change in moles of
substances.
For example :
N2 (g) + 3H2 (g) U 2NH3 (g)

Δn = Δ _ Δ
nP nR
The units of KC for this reaction would be (mol L-1)-2 or L2 mol-2 and those of KP would be bar-
2
as shown below :
The equilibirum constant for such reactions are
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2
[ NH3 ]2 PNH 3
KC = KP =
[ N 2 ][H 2 ]3 PN 2 x PH3 2
(mol L−1 ) 2
KC =
(mol L−1 ) (mol L−1 )3
= (mol L-1)-2
= L2 mol-2
pressure 2
KP =
pressure. pressure3
= pressure -2
= bar -2
For the reaction PCl5 (g) U PCl3 (g) + Cl2 (g)
Δ n = 2 - 1 = 1. Therefore,
The units for KC and KP are
KC = mol L-1 and KP = bar
13.8.3 Significance of the Magnitude of K

The equilibrium constant of a reaction has a constant and characteristic value at a given temperature.
The changes in starting concentration, pressure and the presence of a catalyst do not change the
value of the equilibrium constant. However if the temperature is changed. The value of the
equilibrium constant also changes.
The magnitude of the equilibrium constant is a measure of the extent upto which a reaction proceeds
before the equilibrium is reached. The magnitude of K is large when the products are present in
larger amounts than the reactants in the equilibrium mixture. For the reaction
H2 (g) + I2 (g) U 2HI (g) KC = 90 at 298 K
and for 2CO (g) + O2 (g) U 2CO2 (g) KP = 2.2 x 1022 at 1000 K
A large value of KC for the second reaction indicates that amount of products is much more than
the reactants present at the time of equilibrium. Thus the magnitude of equilibrium constant tells us
about the position of the equilibrium.
13.8.4 Calculation of Equilibrium Constants

Equilibrium constants KC and KP can be calculated if the equilibrium concentrations or partial
pressures are known or can be obtained from the given data. The following examples illustrate the
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calculations.
Example 13.1 : Calculate the equilibrium constant for the reaction
A (g) + B (g) U C (g) + D (g)
If at equilibrium 1 mol of A, 0.5 mole of B, 3.0 mole of C and 10 mol of D are present in a one litre
vessel.
Solution : From the law of equilibrium
[ C] [ D ]
KC =
[A] [B]
Since the volume of the vessel is one litre, the number of moles of A, B, C and D are equal to their
concentrations. Thus
[A] = 1 mol L-1, [B] = 0.5 mol L-1, [C] = 3.0 mol L-1 and [D] = 10 mol L-1 and
(3.0 mol L−1 ) (10 mol L−1 )

KC =
(1 mol L−1 ) (0.5 mol L−1 )
3.0 x 10
= = 60
1 x 0.5
Example 13.2 : In an experiment carried out at 298K, 4.0 mol of NOCl were placed in a 2 litre
flask and after the equilibrium was reached 1.32 mol of NO were formed. Calculate KC at 298 K
for the reaction
2 NOCl (g) U 2 NO (g) + Cl 2 (g)
Solution : Calculation of equilibrium concentrations
No. of moles of NO 1.32 mol

i) [NO] = = = 0.66 mol L-1
Volume 2L
1
No. of moles of Cl2 ( No. of moles of NO) 1.32 mol
ii) [Cl2] = = 2 = = 0.33 mol L-1
Volume Volume 2 x 2L
No. of moles NOCl (Initial moles − moles decomposed )

iii) [NOCl] = =
Volume Volume
(4.0 − 1.32 mol) 2.68 mol

= = 1.34 mol L-1
2L 2L
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For the reaction
2 NOCl (g) U 2 NO (g) + Cl 2 (g)
[ NO]2 [Cl2 ] (0.66 mol L−1 ) 2 (0.33 mol L−1 ) (0.66) 2 x 0.33
KC = = =
[ NOCl]2 (1.34 mol L−1 ) 2 (1.34) 2
= 0.080 mol L-1

KC = 0.080 mol L-1
Example 13.3 : 2 mol of HI were heated in a vessel of one litre capacity at 713 K till the equilibrium
was reached. At equilibrium HI was found to be 25% dissociated. Calculated KC and KP for the
reaction
Solution : Initial moles of HI = 2
25 x 2
Moles of HI dissociated = = 0.5 mol
100
Moles of HI at equilibrium = 2.0 - 0.5 = 1.5 mol

The dissociation of HI occures as
2HI (g) U H2 (g) + I2 (g)
Initial moles 2 0 0
Equilibrium moles (2 - 0.5) 0.25 0.25
1.5 mol 0.25 mol 0.25 mol
Volume of reaction vessel 1L 1L 1L
Equilibrium concentration 1.5 mol L-1 0.25 mol L-1 0.25 mol L-1
For the reaction
[H 2 ][I 2 ] (0.25 mol L−1 ) (0.25 mol L−1 )

KC = =
[HI]2 (1.5 mol L−1 ) 2
(0.25) 2
= = 0.028
(1.5) 2
Also Kp = Kc (RT)Δ ng
For this reaction = Δ ng = ΔnP - Δ nR = 2 - 2 = 0
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∴ Kp = Kc = 0.028
Example 13.4 : Calculate KP for the reaction COCl2 U CO + Cl2 in atm and Nm-2.
The equilibrium partial pressures of COCl 2, CO and Cl 2 are 0.20, 0.16 and 0.26 atm
respectively.
(1 atm = 101300 Nm-2)
Solution : i) Kp in atmospheres
COCl2 (g) U CO (g) + Cl2 (g)
PCO x PCl2 (0.16 atm) (0.26 atm) 0.16 x 0.26

KP = = = atm
PCOCl2 (0.20 atm) 0.20
= 0.21 atm.
ii) KP in Nm-2
KP = 0.21 atm and 1 atm = 101300 Nm-2
∴ Kp = (0.21 atm) (101300 Nm-2 atm-1) = 21273 Nm-2
Example 13.5 : When equal number of moles of ethanol and acetic acid. Were mixed at 300 K,
two-third of each had reacted when the equilibrium was reached. What is the equilibrium constant
for the reaction ?
CH3COOH (l) + C2H5OH (l) U CH3COOC2H5 (l) + H2O (l)
Solution : Let n moles each of acetic acid and ethanol be mixed initially. Then the number of
moles each reacted = 2/3 n.
Let V be the volume of the reaction mixture in litres.
CH3COOH (l) + C2H5OH (l) U CH3COOC2H5 (l) + H2O (l)
Initial mole n n 0 0
Equilibrium concentration (n - 2/3 n) (n - 2/3 n) 2
/3 n 2
/3 n
in moles
1 1 2 2
/3 n /3 n /3 n /3 n
n n 2n 2n
Equilibrium concentration
3V 3V 3V 3V
[CH3COOC2 H5 ] [H 2O]
KC = [CH COOH] [C H OH]
3 2 5
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⎛ 2n ⎞ ⎛ 2n ⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠
3V 3V
= ⎛ n ⎞ ⎛ n ⎞ =2x2=4
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠
3V 3V
KC = 4

1. For a reversible reaction
2A + B U 3C + 3D
Write the expression for the equilibrium constant

.............................................................................................................................................
2. What is the relation between KP and KC.
.............................................................................................................................................
3. i) Apply the law of equilibrium to the following and write the expression for KP and KC.
a) CO2 (g) + H2 (g) U CO (g) + H2O (g)
b) I2 (s) U I2 (g)
ii) For the above reaction write equation for KP and KC.
.............................................................................................................................................
4. The equilibrium constant for the reactions
i) N2 (g) + 3 H2 (g) U 2NH3 (g)

1 2
ii) /3 N2 (g) + H2 (g) U /3 NH3
are K1 and K2 respectively. What is the relation between them.

.............................................................................................................................................
5. What is the significance of the magnitude of equilibrium constant ?
.............................................................................................................................................
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13.9 Factors Affecting Equilibrium State
Supposing a reaction has reached the equilibrium state and then some conditions like
concentrations, temperature, pressure etc. are changed, would it be affecting the equilibrium
state. If yes how ?
In this section, we shall discuss these questions.
The state of equilibrium is in a dynamic balance between forward and backward reaction.
This balance can be disturbed by changing concentration, temperature or pressure. If done
so a cetain net change occurs in the system. The direction of change can be predicted with
the help of Le-Chatelier principle.
13.9.1 Le-Chatelier Principles

It states that when a system in equilibrium is disturbed by a change in concentration,
pressure or temperature, a ‘net’ change occurs in it in a direction that tends to decrease the
disturbing factor.
The principle can be applied to various situations.
13.8.9 Change in Concentration

Consider the state of equilibrium for the formation of ammonia from nitrogen and hydrogen.
N2 (g) + 3H2 (g) U 2NH3 (g) , Δ H = - 92.4 kJ/mol
The concentration of nitrogen, hydrogen and ammonia become constant at the point of
equilibrium. Now if any amount of reactants or ammonia is added or removed their concentration
will change and the equilibrium will get disturbed.
i) Increase concentration of reactant : When the concentration of either nitrogen or hydrogen
is increased; a net forward reaction will take place which consumes the added reactant.
ii) Increase in the concentration of any product : If the concentration of product ammonia is
increased, a net backward reaction would take place to utilise the added ammonia.
13.9.3 Change in Pressure

Change in pressure affects equilibrium involving gaseous phase either in a homogeneous or
heterogeneous system.
Le Chatelier principles for systems involving gases can be studied as follows :
i) When the number of moles of products is more than the total number of moles of reactants as in
the following system.
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N2O4 (g) U 2NO2 (g)
Increase in total pressure keeping the temperature constant, will cause a decrease in volume. This
means that the number of moles per unit volume will increase. A net change will take place in the
equilibrium in the direction where the number of moles decrease i.e. backward direction.
ii) When the number of moles of products is less than reactants. As in the following case
N2 (g) + 3H2 (g) U 2NH3 (g)
According to Le Chatelier’s principle increase in total pressure will bring a net change to the
equilibrium in the direction where the total number of moles is decreasing i.e. to the product
side as Δng = 2. Decrease in total pressure will bring the net change to equilibrium in the direction
where the total number of moles is increasing i.e. backward direction.
iii) When there is no change in the total number of moles of reactant and product in the following
state of equilibrium.
H2 (g) + I2 (g) U 2HI
There is no net change in equilibrium state when pressure is changed.
13.9.4 Change of Temperature

According to Le Chatelier principle when the temperature is changed (increased or decreased) the
equilibrium system reacts to nullify the change in heat content. However, the net change in
equilibrium is directed by the exothermic or endothermic nature of reaction.
i) Exothermic equilibrium : For the following system of equilibrium of exothermic nature:
N2 (g) + 3H2 (g) U 2NH3 (g) ; Δ H = - 92.4 kJ/mol
according to Le Chatelier principle increase in temperature brings a net change in the equilibrium
state in that direction where this extra heat is consumed. The net change is in the backward
direction and some ammonia will decompose producing nitrogen and hydrogen. Similarly if the
temperature is decreased the equilibrium shifts to the forward direction.
ii) Endothermic equilibrium
N2 (g) + O2 (g) U 2NO (g) ; Δ H = + 180.7 kJ/mol-1
If the temperature is increased the added heat will be absorbed by the reactant and the net change
takes place to the equilibrium in the forward direction. If the temperature is decreased it will bring
a ‘net’ change to equilibrium in the backward direction i.e. direction in which it is exothermic.
Addition of a Catalyst : It does not affect the equilibrium. However it helps to achieve the
equilibrium faster.
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13.9.5 Applications of Le Chatelier’s Principle
It can be applied to physical as well as chemical equilibria
A) Physical Equilibria
1) Melting of Ice
Ice U Water ; Δ H = + 6 kJ/mol-1
The change of ice to water is endothermic process. According to Le Chatelier principle if the
temperature is increased the net change will take place in the forward direction some ice will
melt into water.
When the pressure is increased on the equilibrium system, then the volume should decrease;
according to Le Chatelier principle the net change in equilibrium takes place in the forward
direction and ice melts. Therefore, ice melts on increasing the pressure.
2) Vaporization of Water
Water (l) U Water vapour ; Δ H = + ve
This process occurs with a large increase in volume since Δng = 1 - 0 = +1, and it occurs with
absorption of heat.
Increasing the temperature results in more vapour formation (endothermic process). Since
Δng = + 1, increase in pressure results in a net change in equilibrium in the backward direction as
the volume of water vapours is more than that of liquid water for a given mass of water.
3) Solubility Equilibrium
The equilibrium is
Solute (s) U Solute (solution)
The process of dissolution can be endothermic or exothermic. In case of solutes like KCl, KNO3
and NH4Cl, Δ H is positive (endothermic) and more solute will dissolve on heating. Thus, the
solubility increases with rise in temperature. In case of solutes like KOH and NaOH the ΔH is
negative (exothermic) and their solubility decrease on heating.
B) Chemical Equilibria
1) Favourable Conditions for Synthesis of Ammonia : This reaction is of great industrial
importance. During the synthesis of ammonia such conditions are maintained which favour the
‘net’ forward reaction namely low temperature and high pressure. Addition of catalyst makes
the reaction occur fast. Besides, nitrogen and hydrogen gases are continuosly fed into the reaction
chamber and ammonia is continuously removed. all this keeps the system under stress and
equilibrium is never permitted to be attained, so that the synthesis of ammonia continues to
occur.
In industry the reaction is carried out at 4500C and 200 atm pressure in the presence of finely
divided iron (catalyst) and molybdenum (promotor)
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2) Formation of SO3
The reaction
2SO2 (g) + O2 (g) U 2SO3 (g) ; Δ H = - ve
is extothermic and Δng = 2 - 3 = - 1. Formation of SO3 will be favoured bye high presure and low
temperature in the presence of a catalyst.
3) Formation of NO
The reaction
N2 (g) + O2 (g) U 2NO (g) , Δ H = + ve
is endothermic and Δng = 2 - 2 = 0. The reaction is not affected by pressure changes and is
favoured at high temperature. Presence of a suitable catalyst would be helpful.

1. What is Le Chatelier’ principle ?
...................................................................................................................................
2. What are the factors that can affect a system at equilibrium ?
...................................................................................................................................
3. What will happen to solid-vapour equilibrium when the temperature and pressure are
decreased.
...................................................................................................................................
4. a) Which of the following will result in ‘net’ forward reaction in case of
A (g) + 2B (g) U C (s) + D (g) ; Δ H = + ve
i) addition of C
ii) addition of A
iii) decrease in pressure
iv) increase in temperature
...................................................................................................................................
b) What are the most favourable conditions for the formation of C and D ?
...................................................................................................................................
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• A chemical reaction is said to be reversible under certain conditions, if along with the
reactants forming the products, the products also react and form back the reactants
simultaneously.
• Reversible reactions do not reach completion stage and result in a state of equilibrium
which is reached when two opposite processes occur at the same rate.
• The macroscopic properties of the system do not change once the equilibrium has
been established.
• Irreversible reactions are in fact the reversible reactions in which the equilibrium is
reached only when a negligible amount of the reactants is left unreacted.
• Chemical equilibrium is dynamic in nature. It can be attained by starting the reaction
from any side and only in a closed sytem.
• When equilibrium is reached as a result of two opposite physical changes, it is called
physical equilibrium and when as a result of two opposite chemical changes it is called
chemical equilibrium.
• A phase is a homogeneous system or a part of a system which has same composition
and unifrom properties throughout. It is not same a physical state.
• A system with only one phase is called a homogeneous system and the one with more
than one phases is called heterogeneous system.
• Chemical equilibrium can be homogeneous or heterogeneous while physical
equilibrium is always heterogeneous.
• For a general reaction aA + bB U cC + dD according to the law of equilibrium, the

equilibrium constant K is given by the expression
[C]c [D]d
K =
[A]a [B]b
• Concentration equilibrium constant Kc is obtained when molar concentration are used

for calculating K. Concentrations of pure solids and liquids are constant and are not
included in the expression of KC.
• In case of gaseous systems, the concentration of gases are expressed in terms of their
partial pressures. The equilibrium constant thus obtained is called the pressure
equilibrium constant, KP.
• The relation between KP and KC is = KC (RT)Δng is where Δng is the change in the
number of moles of gaseous substances during the reaction.
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• Expression of equilibrium constant depends upon how the chemical equation is written
for the reaction.
• Magnitude of the equilibrium constant is a measure of how close the reaction is to the
completion stage.
• Units of K depends upon the change in the number of moles of the substances during
the reaction.
• Concentration, pressure and temperature can affect the equilibrium systems and the
affect can be qualitatively predicted by Le Chatelier’s principle which states that when
a system at equilibrium is disturbed by changing concentration, pressure or temperature,
a ‘net’ change occurs in the direction that tends to neutralize the effect of the disturbing
factor.
• Changes in concentration and pressure do result in some chemical reaction, but the
value of the equilibrium constant is not changed.
• A catalyst does not change the equilibrium constant. It only helps in reaching the
equilibrium state quicker.
• A change in temperature change the value of the equilibrium constant.
Terminal Exercise
1. What do you understand by reversible and irreversible reactions ? Give on example of
each.
...................................................................................................................................
2. What is physical equilibrium ? Give one example ?
...................................................................................................................................
3. Give characteristics of equilibrium state.
...................................................................................................................................
4. Is the phase same as physical state? Illustrate your answer with one example of each.
...................................................................................................................................
5. How do homogeneous and heterogeneous systems differ from each other ? Which of
the following are homogeneous systems ?
a) Liquid U Vapour
b) N2O4 (g) U 2NO2 (g)
c) NH4Cl (s) U NH3 (g) + HCl (g)
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d) CH3COOH (l) + C2H5OH (l) U CH3COOC2H5 (l) + H2O (l)
...................................................................................................................................
6. What are KP and KC ? Derive a relation between them.
...................................................................................................................................
7. Write down the expression of KC for the following. Also give units in each case.
a) N2O5 (g) U 2NO2 (g) + 1/2 O2 (g)
b) CH4 (g) + H2O (l) U CO (g) + 3H2 (g)
c) FeCl3 (aq) + 3NH4SCN (aq) U Fe (SCN)3 (aq) + 3NH4Cl (aq)
...................................................................................................................................
8. Write down the expression of KP for the following and give its units (in terms of
atmosphere) in each case
a) CO2 (g) + H2 (g) U CO (g) + H2O (l)
b) 3Fe (s) + 4H2O (l) U Fe3O4 (g) + 4H2 (g)
c) 2SO3 (g) + 2SO2 (g) U O2 (g)
...................................................................................................................................
9. Give the relation between KC and KP for the reaction.

...................................................................................................................................
10. Using the reaction between write KC and KP the expression of
i) KP for the reaction given in Q. No. 7
ii) Kc for the reaction given in Q. No. 8
...................................................................................................................................
11. List the factors that can affect
i) a system at equilibrium and
ii) equilibrium constant of a sytem
...................................................................................................................................
12. State the Le Chatelier’s Principle.
...................................................................................................................................
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13. What will be the effect of the following factors on the following systems at equilibrium ?
2 x (g) U 2Y (s) + Z (g) ; ΔH = + ve
i) Addition of X,
ii) removal of Z
iii) addition of a catalyst
iv) increasing the pressure and
v) increasing the temperature.
...................................................................................................................................
14. 5 moles, of HI were produced by the reaction between 7.5 moles of H2 and 2.6 moles of I2
vapours at 4440C. What is the equilibrium constant of the reaction.
H2 (g) + I2 (g) U 2HI (g)
...................................................................................................................................
15. The equilibrium constant Kp for the reaction
N2O4 (g) U 2NO2 (g)
at 333K is found to be 1.33 atm under a total pressure of 1 atm. Calculate Kp for the reaction
2NO2 (g) U N2O4 (g)

at 333K and under 1 atm pressure.
...................................................................................................................................
16. At 4440C, 0.30 mole of H2 and 0.30 mole of I2 were taken in a one litre flask. After
some time the equilibrium H2 (g) + I2 (g) U 2HI (g) was established and it was found
that the concentration of I2 decreased to 0.06 mol L-1. Calculate the value of KC for the
reaction at this temperature.
...................................................................................................................................
17. The equilibrium constant for the reaction.
CH3COOH (l) + C2H5OH (l) U CH3COOC2H5 (l) + H2O (l) is 4.0

What will be the composition of the equilibrium mixture if 1 mole of acetic acid is taken
with 8 moles of ethanol ?
...................................................................................................................................
-292-
18. KC for the reaction
N2 (g) + 3H2 (g) U 2NH3 (g)
at 4000C was found to be 0.5 L2 mol-2. Calculate KP of this reaction in atm.

...................................................................................................................................

13.1
1. A chemical reaction is said to be reversible, if under cetain conditions its products also
react and form back the reactants.
Examples :
H2 (g) + I2 (g) U 2HI (g)
2SO2 (g) + O2 (g) U 2SO3 (g)
2. A reaction reaches an equilibrium state when two opposing reactions occur at the same rate
and balance each other at a particular temperature.
3. When a system reaches the equilibrium state, its temperature, pressure and concentrations of
all the reactants and products do not change any further with time.
4. i) Water-vapour system in a closed container at a constant temperature.
ii) A saturated solution containing some undissolved solute at a constant temperature.
5. i) Homogeneous systems :
H2 (g) + I2 (g) U 2HI (g)
2SO2 (g) + O2 (g) U 2SO3 (g)
ii) Heterogeneous systems :
Zn (s) + CuSO4 (aq) U Cu (s) + ZnSO4 (aq)
13.2
[C]3[D]2
1. K =
[A]2 [B]
2. KP = KC (RT)Δng
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[CO][H 2O] PCO x PH 2O
3. i) a) KC = [CO ][H ] ; KP = P
2 2 CO 2 x PH 2
b) Kc = [I2] ; K p = PI
2
ii) For the first reaction Δng = (1 + 1) - (1 - 1) = 0, hence KC = KP while for the second
reaction
Δng = 1 - 0 = + 1.
Kp
∴ KP = KC (RT) or KC = or KC < KP.
RT
[ NH3 ]2 [ NH3 ]2 / 3
4. Kl = and K2 =
[ N 2 ][H 2 ]3 [ N 2 ]1 / 3[H 2 ]
∴ K1 = [K2]3.
5. It is a measure of the extent up to which a reaction proceeds before the equilibrium is

reached.
13.3
1. Le Chatelier’s principle states that when a system at equilibrium is disturbed by changing
concentration, pressure or temperature, a ‘net’ change occurs in a direction that tends to
neuralize the effect of the disturbing factor.
2. Changes in pressure, temperature and concentrations of reactants or products.
3. When the temperature is decreased some vapour will condense and when the pressure
is decreased some solid will sublime.
4. (a) (ii) and (iv)
(b) High temperature, increase in pressure, presence of a catalyst and continuous
removal of D.
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14
IONIC EQUILIBRIUM
In the first lesson of this module you learnt about chemical equilibrium, its types of and the
factors affecting the state of equilibrium. In this the lesson you will learn about the equilibria
involving ionic species. The equilibria involving acids and bases are critically important for
a wide variety of reactions. The use of buffer solutions for pH control is of significance in
living systems, agriculture and industrial processes. Similarly, the solubility equilibrium
established in the solutions of sparingly soluble salts is also very important. For example, our
bones and teeth are mostly calcium phosphate, Ca3(PO4)2, a slightly soluble salt. We would
focus our attention primarily on the acid-base and the solubility equilibria and some interrelated
aspects concerning these equilibria such as pH, buffer solutions and common ion effect. In
this process you would apply the concepts learnt in the first lesson.
Objectives
• define and explain various concepts of acids and bases;
• define conjugate acid-base pairs and identify in an acid-base equilibrium;
• derive the expressions for the ionisation constants of weak acids and bases;
• correlate the ionisation constants and the strength of acids and bases;
• explain self ionisation of water and derive an expression for its ionic product;
• define pH and correlate it with the nature of aqueous solutions - neutral, acidic or basic
;
• define and explain common ion effect in ionisation of weak acids and bases;
• define buffer solutions;
• apply Henderson equation to calculate the pH of acidic and basic buffers;
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• explain hydrolysis of salts with the help of chemical equations;
• express solubility equilibrium with the help of an equation;
• identify the relationship between solubility and solubility product for salts of AB, AB2
A2B and A2B3 types;
• illustrate and explain the common ion effect on solubility equilibrium and
• recall some applications of common ion effect
14.1 General Concepts of Acids and Bases

You are quite familiar with the terms acid, base, acidic and acidity etc. But how do we define
an acid or a base? There is no general definition of acids and bases. There are three different
concepts of acids and bases (proposed by Arrhenius, Bronsted and Lowry and Lewis
respectively) which are well known. Each of these emphasize a different aspect of acid - base
chemistry. Let us try to understand these concepts.
14.1.1 Arrhenius Concept

The most commonly used concept of acids and bases was developed by Svante Arrhenius
(1884). According to this concept an acid is a substance that is capable of producing hydrogen
ion (H+) by dissociating in aqueous solution. The reaction can be represented as
_
HA (aq) → Η+ (aq) + A (aq) (14.1)
Where HA represents the acid and A- refers to the acid molecule without the hydrogen ion.
Hydrochloric acid, HCl is an example of an Arrhenius acid whose ionisation can be
represented as
_
HCl (aq) → H+ (aq) + Cl (aq) (14.2)
The proton or hydrogen ion binds itself to a water molecule and form H2O+ ion which is called
hydronioum ion.
H+ + H 2O → H 3O+
The hydronium ion is also known as oxonium ion or the hydroxonium ion.
In the light of this fact the equation 14.1 can be rewritten as
_
HA (aq) + H2O → H3O+ (aq) + A (aq) (14.3)
_
A base on the other hand is defined as a substance capable of providing a hydroxyl ion (HO ) on
dissociation in aqueous solutions.
_
MOH (aq) → M+ (aq) + OH (aq) (14.4)
Where M+ refers to the base molecule without the hydroxyl ion. Sodium hydroxide is an example
of a Arrhenius base, dissociating as,
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_
NaOH (aq) → Na+ (aq) + OH (aq) (14.5)
Arrhenius concept is quite useful and explains the acid-base behaviour to a good extent.
However it has certain drawbacks like,
• It is limited to only aqueous solutions and require dissociation of the substance.
• It does not explain the acidic behaviour of some substances which do not contain
hydrogen, for example, AlCl3. Silimarly it does not exaplin the basic character of
substances like NH3 and Na2CO3 which do not have a hydroxide groups.
14.1.2 Bronsted and Lowry Concept

In 1923, Bronsted and Lowry pointed out independently that acid-base reactions can be
interpreted as proton-transfer reactions. According to them, an acid is defined as a proton
(H+) donor, and a base is defined as a proton acceptor. The definition is sufficiently broad
and removes the first limitation of Arrhenius concept. Any hydrogen-containing molecule
or ion capable of donating or transferring a proton is an acid, while any molecule or ion that
can accpet a proton is a base.
For example, in a reaction between ammonia (NH3) and hydrogen fluoride (HF); ammonia
acts as a base (accepts a proton) while HF behaves as an acid (donates a proton).
_
NH3 + HF → NH4+ + F (14.6)
According to Bronsted-theory an acid is a substance that can donate a proton whereas
a base is a substance that can accept a proton.
You may note in this example that there is no role of a solvent. Let us consider the reaction
of hydrochloride acid with ammonia in a solvent like water. We can write ionic equation as
_ _
H3O+ (aq) + C (aq) + NH3 (aq) → H2O(l) + NH4+ (aq) + Cl (aq)
The hydronium and chloride ions are obtained from the ionisatio of HCl. After concelling
_
Cl on both sides, we get the following ionic equation as
NH3 (g) + H3O+ (aq) → H2O(l) + NH4+ (aq) (14.7)
Here, in aqueous solution, a proton is transferred from the hydronium ion, H3O+, to the NH3
molecule, giving H2O and NH4+ . In this case H3O+ acts as proton donor or an acid and NH3
as proton acceptor. We may visualise water (the solvent) playing the role of mediator in the
transfer of proton from hydrochloric acid to ammonia. It should be noted that in the Bronsted
- Lowry Concept, acids and bases can be either ions or molecular substances.
In any acid-base equilibrium both forward and reverse reactions involve proton transfers.
Let us consider the reaction of NH3 with H2O.
_
H2O (l) + NH3 (aq) U NH4+ (aq) + OH (aq) (14.8)
acid1 base 2 acid 2 base1
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In the forward reaction, NH3 accepts a proton from H2O. Thus, NH3 is a base and H2O is an
acid. In the reverse reaction, NH4+ donates a proton to -OH. Thus NH4+ ion acts as acid -
OH as a base. Here you find that NH 3 and NH4+ differ by a proton. That is, NH3 becomes
NH 4+ ion by gaining a proton, whereas NH4+ ion becomes NH3 molecule by losing a proton.
The species NH4+ and NH3 are called conjugate acid-base pair. A conjugate acid-base pair
consists of two species in an acid-base reaction, one acid and one base, that differ by the
gain or loss of a proton. It can be represented as NH 4+ /NH3. The acid in such a pair is a
called conjugate acid of the base, where as the base is the conjugate base of acid. Thus,
NH 4+ is the conjugate acid of NH3, and NH3 is the conjugate base of NH4+ . The members of
each conjugate pair are designated by the same numerical subscript e.g., NH4+ /NH3 pair is
marked as 2 while the H 2O/OH- pair is marked as 1.
Let us take another example of ionization of hydrozen fluoride in water. It can be represented
as
_
HF (g) + H2O U H3O+ (aq) + F (aq) (14.9)
acid1 base 2 acid2 base1
The equilibrium sign indicates that the reaction does not go to completion. The H+ ions obtained
from the ionisation of HF molecule can be accepted by either of the two bases, F- and H 2O.
The fact that the equilibrium is only slightly towards right tells us that it is the F- that accepts
the proton and make the reaction go to the left i.e., F- is a stronger base than H2O.
Similarly when HCI is dissolved in water, the HCI molecules give up H+ ions that can be
accepted by either of two bases, Cl- and H2O.
HCl + H2O → H3O+ + Cl- (14.10)
The fact HCl is completely ionized in dilute aqueous solution (indicated by a single arrow)
tells us that Cl- is a weaker base than H2O. We can conclude from these two examples that,
a stronger acid (HCl) has a weaker conjugate base, Cl- and the weaker acid, (HF) has a
stronger conjugate base, F-. We can generalize that in a conjugate acid-base pair the weaker
an acid is, stronger is its conjugate base. Likewise, the weaker a base is, the stronger is
its conjugate acid.
Here you should understand that the terms ‘strong’ and ‘weak’ like many other adjectives are
used in a relative sense. Consider the following reaction
_
F- + H2O U HF + OH (14.11)
In this case F- is a weaker base as compared to OH-.

Let us write a few reactions involving conjugate acid-base pairs.
HF (aq) + HCO3-
_
U H2CO3 (aq) + F (aq)
acid1 base2 acid2 base1
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HCO3- (aq) + OH (aq) U CO 23 (aq) + H2O (l)
_ _
acid1 base2 base 1 acid2
H2CO3 (aq) + CN (aq) U HCO3- (aq) + HCN (aq)

_
(14.12)
acid1 base2 base 1 acid2
If you examine tha above reactions closely, you would find that some species can act both as
an acid as a base. Such species are called amphiprotic species. In the above reactions HCO3-
(aq) acts as a base in presence of HF but an acid in presence of CN-. Similarly H2O behaves
as an acid and a base.
• Thus, we have seen that the Bronsted - Lowry concept of acids and bases has greater
scope than the Arrhenius concept.
14.1.3 Lewis Concept

As discussed above the Bronsted - Lowry concept does not depend on the nature of the
solvent (a short coming of the Arrhenius concept removed). However, like Arrhenius concept
it does not explain the acidity of the substances which do not have a hydrogen atom (e.g.
AlCl3) and the basicity of the substances devoid of a OH group (e.g. Na2CO3). G.N. Lewis
proposed (1923) a yet another concept of acids and bases that includes such substances
also. According to him, an acid may be defined as, ‘any atom, molecule or ion that can
accept an electron pair from any other atom, molecule or ion’. A lewis base on the other hand
can be defined as, ‘any atom, molecule or ion that can donate a pair of electrons, Let us take
an example
AlCl3 + NH3 → Cl3Al ← NH3 (14.13)
In the above example AlCl3 is an electron deficient species. It accepts an electron pair from a
molecule of NH3 which has a lone pair of electrons on N atom. Thus, AlCl3 is a Lewis acid
and NH3 is a Lewis base.
Types of Lewis acids

The Lewis acids can be divided into following 5 types.
* All Cations: Ex : Ag +, Co3+, Cu2+, Fe3+ etc.
* Compounds whose central atom has an incomplete octet and possessing an empty orbital.
Ex : BF3, BCl3, AlCl3, FeCl3
* Compounds in which the central atom has available d-orbitals and may expand its octet.
Ex : SiF6, SnCl 4, SF4, TeF4, FeCl3
* Molecules having multiple bonds between atoms of dissimilar electro negativities.
Ex : CO2, SO 2, SO3, NO2,Cl2O7, P4O10
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* Electrons with an electron sextet. Ex : S, O
14.1.4 Types of Lewis bases

The Lewis bases are divided into 3 types.
_ _ _ _ _ _
1. All anions : Ex : Cl , OH , CN , NH 2, F , SCN (thio cyanate ion)
2. Molecules with one or two lone pairs on the central atom.
.. .. .. .. .. ..
Ex : H2O, NH3, ROH, RNH 2, ROR, C 5H5N (pyridine)
.. .. ..
3. Molecules with multiple bonds.
Ex : CO, NO, H C ≡ C H, H2C= CH2
14.1.5 Examples of acid – base reactions

i) Formation of NH3 . BF3 :
In a reaction between BF3 and NH3, Boron of BF3 accepts a lone pair from nitrogen of
NH 3 and forms a coordinate covalent bond.
[ [
F H F H
F_ B + : N _ H → F_ B → N _H
F H F H
Lewis Lewis Ammonia - boron trifluoride
Acid base
ii) Formation of hydronium ion :
H+ Combines with H2O. Oxygen of water donates a pair of electrons to H+ ion.
H+ OH → H ← OH2+
..
.. 2
Lewis Lewis Hydronium ion
Acid base
14.1.6 Limitations of Lewis theory

1. One of the serious defects in the theory is that it cannot explain the strengths of acids
and bases.
2. Acids like HCl, H2SO 4 react with bases such as NaOH or KOH but do not form
coordinate covalent bond.
3. Generally acid – base interactions i.e., neutralization reactions are instantaneous reactions.
But some Lewis acid – base reactions go on very slowly.
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4. All the acid – base reactions do not involve, coordinate bond formation.
5. H_+ ion, as catalyst, cannot be explained by this theory.

1. Define Arrhenius and give two examples.
...................................................................................................................................
2. What are the limitations of Arrhenius definition ?
...................................................................................................................................
3. How does a Bronsted - Lowry base differ from an Arrhenius base ?
...................................................................................................................................
4. Classify the following into Bronsted - Lowry acid or Bronsted - Lowry base.
HCl, NH3, H3O+, CN-
...................................................................................................................................
14.2 Relative Strength of Acids and Bases

Different acids and bases have different strength depending on their nature. Since there are
different ways of defining acids and bases, there are different ways of comparing their
relative strengths also.
14.2.1 Relative strength according to Arrhenius concept

According to Arrhenius concept strong electrolytes (like HCl) which dissociate completely
in aqueous solutions and produce H+ (or H3O+) ions are called strong acids.
HCl (g) + H2O → H3O+ (aq) + Cl- (aq) (14.14)
Other examples of strong acids are H2SO4, HBr, HI, HNO 3 and HClO4. On the other hand
weak electrolytes like CH3COOH whose ionisation is not complete, (because the process is
reversible) and produce H+ (or H3O+) ions are called weak acids.
CH3COOH (aq) + H2O U H+ (aq) + CH3COO - (aq) (14.15)
Similarly strong bases may be defined as the electrolytes with complete ionisation and
weak bases as the electrolytes with incomplete ionisation. NaOH and NH 3 are typical
examples of strong and weak bases respectively. Their ionisaion can be represented as
NaOH (aq) → Na+ (aq) + OH - (aq) (14.16)
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NH3 (aq) + H2O U NH4+ (aq) + OH - (aq) (14.17)
The principal strong bases are hydroxides of Groups 1 and 2 elements (excpt Be). Most of the
other acids and bases we come across are weak bases.
14.2.2 Relative strength according to Bronsted - Lowry concept

You have learnt that according to Bronsted - Lowry concept an acid is a species that donates
a protons while a base is a species that accepts a protons. The tendency of a substance to
donate a proton also depends on the species accepting the proton. A given acid may have
different strengths in different solvents (of varying basic strength or proton accepting tendencies)
e.g.,
+ (14.18)
(14.19)
Acetic acid loses a proton to a basic solvent aniline but in sulphuric acid it infact accepts a
proton and acts as a base. Therefore the relative strengths of different acids and bases are
compared in the same solvent which normally is water. Therefore, in Bronsted - Lowry
concept, we may define the relative strength of an acid as its relative tendency to lose (or
donate) a proton to water. According to the Bronsted - Lowry concept strong acid are those
which lose their protons almost entirely to water. The reaction,
HCl (aq) + H2O (aq) → H3O+ (aq) + Cl- (aq) (14.20)
acid base acid base
goes completely to the right indicating that HCl is a strong acid. Acetic acid donates its
proton to water to the extent of only about 3% and the following equilibrium exists.
CH3COOH + H2O U H 3O+ + CH3COO - (14.21)
Acetic acid is, therefore, termed as a weak acid.
14.3 Quantitative Aspects of Strengths of Acids and Bases

We have so far discussed the relative strengths of acids and bases only qualitatively i.e.,
which is strong and which one is weak. Many a times we need to know the quantitative
aspects also i.e., how much ? That is if a acid is stronger than the other we would like to
know how many times stronger. Let us learn how do we quantify it ?
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14.3.1 Ionisation of weak acids
The dissociation or ionisation of weak acid, HA, can be represented as
HA (aq) + H2O (l) U H3O+ (aq) + A - (aq) (14.22)
As you know that in case of strong acids the ionisation is almost complete or close to 100% or
we may say that the equilibrium lies far to the right. In such cases the sign of equilibrium may
be replaced by a single arrow (→)
HA (aq) → H+ (aq) + A- (aq)

or
HA (aq) + H2O (l) U H3O+ (aq) + A - (aq) (14.23)
The reaction given above (eq. 14.22) is referred to as ionisation equilirbium and is characterized
by an equilibrium constant
[ H 3O + ] [ A − ]
Keq = (14.24)
[H 2O] [HA]
Since the concentration of a pure liquid or a solid is taken as 1, we can rewrite the above
expression can as
[ H 3O + ] [ A − ]
Keq = = Ka (14.25)
[HA]
where Ka is a new constant called acid dissociation constant or ionisation constant of the
acid.
The magnitude of the equilibrium constant is a measure of the strength of the acid. Higher
the value of the equilibrium constant the stronger is the acid. For all strong acids the values
of the equilibrium constants is quite high and does not help much in suggesting their relative
strengths. However, for a weak acid, this constant is quite useful.
Example 14.1 : Write down the expression for the dissociation constant for acetic acid - a
weak acid that ionizes as
CH3COOH (aq) + H2O U H 3O+ (aq) + CH3COO - (aq)
Solution : Applying the law of chemical equilibrium, we can write equilibrium constant K
as
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[CH3COO − ] [H3O + ]
K=
[CH3COOH] [H 2O]
Rearranging, we can write.
[H3O + ] [CH3COO − ]
K [H2O] = Ka =
[CH3COOH]
Dissociation constant of other weak acids also can be written in the same way.
The values of Ka show the extent to which the acids are dissociated in water. Relative strength
of acids is judged on the basis of dissociation constant. Like other equilibrium constants the
dissociated constant, Ka also depends on the temperature. Therefore, the ionisation constants
or dissociation constants are compared at the same temperature. For example
CH3COOH (aq) U H+ (aq) + CH3COO - (aq) Ka = 1.8 x 10-5 (14.26)
HCN (aq) U H+ (aq) + CN - (aq) Ka = 4.9 x 10-10 (14.27)
On the basis of Ka values we can say that acetic acid is much more ionized than hydrocyanic
acid. In other words acetic acid is stronger acid than hydrocyanic acid although both are
weak; netither is completely ionized.
14.3.2 Ionisation of weak bases

The ionisation of weak bases (BOH) can be expressed as
BOH (aq) U B+ (aq) + OH- (aq) (14.28)
(The equilibrium sign may be replaced by → in case of a strong base).

The solution contains the base, B the protonated base, BH+, hydroxide ion OH-, and water
in equilibrium. The equilibrium constant expression for the reaction is
[B+ ] [OH − ]
Kb = (14.29)
[BOH]
For example, the dissociation of NH4OH is represented as
NH4OH (aq) U NH4+ (aq) + OH - (aq)
and is characterized by
[ NH +4 ] [OH − ]
Kb = (14.30)
[ NH 4OH]
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The constant Kb is called dissociation constant of the base. Similar to values of Ka, Kb values
also give us the idea about the relative strengths of weak bases. Higher the value of Kb the
stronger is the base.
14.3.3 Polyprotic Acids

Many acids have more than one ionizable protons. These are called polyprotic acids. The
acids are called diprotic if there are two ionizable protons per molecule. (e.g. H2SO3, H2CO3),
and triprotic if there are three ionizable protons (e.g. H3PO4, etc.) Such acids dissociate in
more than one steps or stages, each with its own ionization constant. In the case of sulphurous
acid, H2SO3, these steps are
H2SO3 + H2O U HSO3- + H3O+ (14.31)
[H3O + ] [HSO3− ]
K1 = = 1.3 x 10-2
[H 2SO3 ]
HSO3- + H 2O U SO32- + H3O+ (14.32)
[H3O + ] [SO32− ]
K2 = = 6.3 x 10-8
[HSO3− ]
The values of the two ionisation constants (K1 and K2) are quite different : K1 being twenty
million times K2. It suggests that the first ionisation of sulphurous acid is much more than
the second one. In other words the sulphurous acid behaves as a much stronger acid than the
bisulphite ion.
14.3.4 Degree of Ionisation or Degree of Dissociation

As you know that the weak acids / bases do not ionize completely and an equilibrium exists
between the ionized and unionzed species. The degree of ionisation may be defined as the
fraction of total amount of a weak acid or a base that exists in the ionized form. It is denoted
by a Greek letter ‘α’. The equilibrium constant can be used to calculate the degree of
ionisation of weak acid or a base. An expression relating α and Ka or Kb can be derived as
follows.
Consider a weak acid HA which partially dissociated in its aqueous solutions and the following
equilibrium is established.
HA (aq) + H2O (l) U H3O+ (aq) + A- (aq)
Initial concentrations c -55 0 0

(in moles)
Equilibrium concentrations c (1-α) -55 cα cα
The equilibrium constant expression can be written as
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[ H 3O + ] [ A − ] [cα ] [cα ]
Ka = =
[H 2O] [HA] c [1 − α ] 55
rearranging we get,
[cα ] [cα ] c2 α 2 c α2
= 55K = Ka = = = (14.33)
c [1− α ] c [1 − α ] [1 − α ]
Since the acid HA is very weak, α < < 1; we can neglect α in comparison to 1 in the denominator
to get.
Ka Ka
Ka = cα2 or α2 = or α = (14.34)
c c
So if we know the value of the dissociation constant of the aicd and the concentration of the
weak acids we can find its degree of dissociation or ionisation. Let us take up an example to
see the application of this relationship.
Example 14.2 : Compare the degree of dissociation and percent dissociation of acetic acid
in its 0.1 M solution. Give Ka = 1.8 x 10-5.
Ka
Solution : Using the formula α = and substituting the values of Ka and c, we get,
c
1.8 x 10 −5
α = = 1.8 x 10 −4 = 1.34 x 10-2 = 0.0134
0.1
The percent dissociation of a weak acid is defined as
Percent dissociation = x 100 (14.35)
= Degree of dissociation x 100%

= α x 100% = 0.0134 x 100 = 1.34%
Thus acetic acid is dissociated to the extent of only 1.34% in a 0.1 M aqueous solution.
A similar expression can be obtained for the degree of dissocation of a weak base. The
desired expression is
Kb
α = (14.36)
c
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14.4 The Auto-Ionisation or Self-Ionisation of Water
We have seen that water can act as a very weak acid and also as a very weak base. In a sample
of water a small number of water molecules undergo auto ionisation. Half of them acting as
an acid while the other half acting as a base. As a consequence small concentration of H3O+
and OH- are formed in water. The self ionisation of water can be represented as
H2O + H2O U H3O+ + OH- (14.37)
The corresponding equilibrium constant expression can be written as

[H3O + ] [OH − ]
Keq = (14.38)
[H 2O]2
Since the concentration of H2O is constant we can rearrange the expression and define a new
constant Kw, as
[H3O+] [OH-] = Keq x [H2O]2 = Kw (a new constant) (14.39)
This constant Kw, is called the dissociation constant or ionic product constant of water. The
value of K w at 298 K has been determined from the measurement of electrical conductivity
of carefully purified water and has been found to be 1.0 x 10-14 mol2 dm-6.
Since the concentration of H3O+ and OH- ions is equal we may write
Kw = [H3O+] [OH-] = 1.0 x 10-14 mol2 dm-6
Kw = [H3O+]2 = 1.0 x 10-14 mol2 dm-6
[H3O+] = 1.0 x 10-7 mol dm-3
and similarly, [OH-] = 1.0 x 10-7 mol dm-3
Thus in pure water and in neutral solutions
[H3O+] = [OH-] = 1.0 x 10 -7 mol dm-3 at 298 K (14.40)
14.4.1 Acidic, Basic and Neutral Solutions

An acidic solution in defined as one in which the hydrogen ion (or hydronium ion)
concentration is greater than the hydroxide ion concentration. A basic solution is one in
which the reverse is true, that is, one in which [OH-] exceeds [H3O+] and a netural solution
is one in which [OH-] equals [H3O+].
Neutral solution [H3O+] = [OH-]
Acidic solution [H3O+] > [OH-]
Basic solution [H3O+] < [OH-] (14.41)
Since the product [H3O+] [OH-] is constant, if the concentration of either of these increase,
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the other one would decrease. In other words the concentrations of [H3O+] and [OH-] are not
independent but are linked by the relationship.
[H3O+] [OH-] = Kw (14.42)
This provides an easy way to calculate concentration of one of these if we know that of the
other.
You must note that the self ionisation equilibrium discussed above applies not only to pure
water but also to the self-ionization of water in any aqueous solution. Hydronium ions and
hydroxide ions are present in every aqueous solution, and they are always in equilibrium
with water molecules. Let us try to calculate the concentration of these ions in some simple
solutions.
Example 14.3 : Calculate the concentrations of OH- and H3O+ ions in 0.01 M solution of
HCl.
Solution : In an aqueous solution of HCl, the following two processes would occur
simultaneously.
H2O + H2O U H3O+ + OH -
HCl + H2O U H3O+ + Cl -
The ionisation of HCl goes to completion and that of water to a very limited extent. In
addition according to Le-Chatlier principle, the H3O+ from the ionization of HCl will shift
the position of the equilibrium of the self-ionization reaction to the left. As a consequence
the concentration of the OH ions would reduce further. Suppose concentration of OH is ‘x’
mol dm-3, then concentration of H3O+ from the self-ionization of water must also be x mol
dm-3. The concentration of H3O+ from ionization of HCl is 0.010 mol dm -3. Therefore, total
concentration of H3O+ (aq) = (0.010 + x) mol dm-3.
Thus Equilibrium Concentrations of H3O+ and OH- ions would be (0.01 + x) and x mol dm-3
respectively.
Substituting these values into the equilibrium constant for the self-ionization of water, we get
Kw = [H3O+] [OH-] = (0.01 x x) (x) mol2 dm-6 = 1.0 x 10-14 mol2 dm-6
Since x must be very small, we can assume that x << 0.01 and therefore we may assume that
the equilibrium concentration of H3O+ is equal to 0.01 M
(0.01 + x) = 0.01, so
0.01x = 1.0 x 10-14
or x = 1.0 x 10-14 / 0.01
x = 1.0 x 10-12
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[OH-] = 1.0 x 10-12 mol dm-3 and
[H3O+] = 0.01 + 1.0 x 10-12 mol dm-3 = 0.1 mol dm-3
Since the value of x (1.0 x 10-12) the hydronium ions obtained from the self ionization of
water is far smaller than 0.01, our assumption is justified. Thus you may note that in case of
aqueous solutions of strong acids it is reasonable to assume that the concentration of H3O+
ions equals the concentration of the acid itself.
14.4.2 pH Scale
In aqueous solutions of acids and bases the concentration of H3O+ and OH- ions may vary
from about 10 M to 10 -14 M. It is quite inconvenient to express these concentrations by
using powers of 10. in 1909 a Danish botanist S.P.L. Sorensen proposed a logarithmic scale
(called pH scale) for expressing the concentrations of H+ ions. He defined pH as the negative
logarithm of the molar concentration of hydrogen ions. That is,
pH = - log 10[H+] (14.43)
We now a days it is represented as
pH = - log 10[H3O+] (14.44)
For neutral solution (or pure water)
[H3O+] = [OH-] = 1 x 10-7
pH = - log 1 x 10 -7 = 7.0
For acidic solutions
[H3O+] > [OH-]
⇒ [H3O+] > 1 x 10-7
⇒ pH = -log (>1 x 10-7) = < 7.0

For basic solutions
[H3O+] < [OH-]
⇒ [H3O+] > 1 x 10-7
⇒ pH = -log (>1 x 10-7)
⇒ > 7.0
A strongly acidic solution can have a pH of less than zero (i.e., negative) and a strongly
alkaline solution can have a pH value greater than 14. However the pH range normally
observed is between 0 to 14.
The notation p is used in a number of places. It carries the meaning. ‘the negative logarithm
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of’. It has been extended to OH- (aq) and equilibrium constant like, Ka, Kb and Kw, etc.
pOH = - log10[OH-]
pKa = - log10 Ka
pKb = - log10 Kb
pKw = - log10 Kw (14.45)
This gives us an important relation. You would recall (eq 14.42)
Kw = [H3O+] [OH-]
Taking logs on both sides, we get
log Kw = log [H3O+] + log [OH-]
Multiplying through out by - 1
- log Kw = - log [H 3O+] - log [OH-]
pKw = pH + pOH
Since the value of Kw = 1.0 x 10-14 pKw = 14
i.e., pH + pOH = 14 (14.46)
If we know the value of pH of a given solution we can find its pOH and vice verse.
Let us take up some examples to understand the application of these important formulae.
Example 14.4 : What is the pH of a 0.01 M aqueous solution of HCl ?

Solution : Since HCl is a strong acid it would ionize completely.
Therefore, [H3O+] in 0.01 M HCl = 0.01 M
pH = - log10[H3O+] = - log10 -2
= - (-2.0) = 2.0
Example 14.5 : Calculate the pH of 0.010 M aqueous solution of NaOH ?

Solution : Since NaOH is a strong base, it is fully ionized to give Na+ and OH-
[OH-] = 1.0 x 10-2 mol L-1
Kw = [H3O+] [OH -] = 1.00 x 10-14 mol2 L-2
Kw 1.00 x 10 −14 mol2 dm −6

+
so [H3O ] = =
[OH − ] 1.00 x 10 −2 mol dm −3
= 1.00 x 10-12 mol dm-3
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pH = - log10 (1.0 x 10 -12) = 12
Example 14.6 : The pH of a sample of rain water at 250C is 5. What is the concentration of
hydronium ions ?
Solution : pH = - log [H3O+]
or 5 = - log [H3O+] ⇒ log [H3O+] = -5.
Taking antilogs, we get
[H3O+] = 10-5 mol dm-3
Example 14.7 : Calculate the pH of a 0.1 M aqueous solution of acetic acid. The dissociation
constant, Ka = 1.85 x 10-5 , α = 0.0134.
Solution : The following equilibrium exists in this solution
CH3COOH (aq) + H2O U H3O+ (aq) + CH3COO- (aq)

11
F If α be the degree of dissociation of acetic acid in this solution, the equilibrium concentrations
of various species would be
CH3COOH (aq) + H2O U H3O+ (aq) + CH3COO- (aq)

c (1 - α) cα cα
Since c= 0.1 M
0.1 (1 - α) 0.1 α 0.1 α
[H3O+] = c α
⇒ [H3O+] = 0.1 x 0.134 = 0.00134
pH = - log[H3O+] = - log[0.00134] = - log [1.34 x 10-3] = - (-2.87) = 2.87
14.4.3 Effect of Common-Ions on the Dissociation of Weak Acids

and Bases
In the previous lesson you have learnt about Le Chatelier’s principle. According to this principle
the presence of common ions in a solution of a weak acid or a base will affect its dissociation.
This in fact would suppress the dissociation of the acid or base.
In a solution containing a weak acid HA and its salt NaA, the following equilibrium exists.
HA (aq) H + (aq) + A- (aq)
NaA (aq) Na+ (aq) + A- (aq)
He A - (aq) is the common-ion

and in case of a weak base BOH and its salt BX the equilibria are
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BOH (aq) B+ (aq) + OH- (aq)
BX (aq) B+ (aq) + X- (aq)
Here, B+ is the common-ion. According to Le-Chatlier principle, the presence of common

ions would make the equilibrium to go to the left. It is said that the common ions suppress
the equilibrium.
Let us take example to understand the effect of common ions on such equilibria.
Example 14.8 : Calculate the pH, degree of dissociation and concentration of various species
in a solution of 0.1 M acetic acid which also contains 0.1 M sodium acetate (for acetic acid)
= 1.85 x 10-5 mol d-3).
Solution : In the given solution the following two equilibria exist simultaneously.
CH3COOH (aq) + H2O H3O+ (aq) + CH3COO- (aq)
CH3CONa (aq) → Na+ (aq) + CH3COO - (aq)
Let α be the degree of dissociation of acetic acid in this solution, the equilibrium concentrations
of various species would be
CH3COOH (aq) → H3O+ (aq) + CH3COO- (aq)
c (1 - α) cα cα
Since c= 0.1 M
0.1 (1 - α) 0.1 α 0.1 α
0 0.1 0.1
CH 3 COOH = 0.1 (1-α)
CH 3COO - = 0.1 + 0.1α = 0.1 (1 + α)
H 3O + = 0.1 α
[H3O + ] [CH3COO − ]
Ka =
[CH3COO − ]
rearranging the expression we get,
[CH3COOH]
[H3O+] = Ka =
[CH3COO − ]
Substituting the values, we get
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0.1 (1 − α )
[H3O+] = 1.85 x 10-5 x
0.1 (1 + α )
Since acetic acid is a weak acid, its degree of dissociation further decreases in presence of
acetate (ions the common ion). Therefore it is reasonable to assume that
α < < 1; and (1- α ) - 1; also (1 + α ) - 1
This is gives
[H3O+] = 1.85 x 10-5 x 0.1 / 0.1 = 1.85 x 10-5
and pH = -log (1.85 x 10-5 ) = 4.73
Also since [H3O+] = 0.1 α
∴ α = 1.85 x 10 -5 / 0.1 = 1.85 x 10-4 = 0.000185
The concentration of different species at equilibrium will be
CH3COOH = 0.1 (1-0.000185) = 0.1
CH3COO- = 0.1 (1+0.000185) = 0.1
H3O+ = 0.1 x α = 0.1 x 0.000185 = 1.85 x 10-5
⇒ the concentration of acid = initial concentration of acid
⇒ the concentration of CH COO - ions = initial concentration of the salt
3

1. HF is a weak acid in water. Write down the expression for Ka for the dissociation of
HF.
...................................................................................................................................
2. Consider a weak base BOH which partially dissociates in its aqueous solutions as per
the following equilibrium
B + H 2O BH+ + OH-
...................................................................................................................................
3. A sample of lime juice has a hydronium ion concentration of 6.3 x 10-2 M. Calculate
its pH.
...................................................................................................................................
4. Calculate the pH of 1.0 M aqueous solution of amino acid glycine - a weak acid. The
Ka = 1.67 x 10-10.
...................................................................................................................................
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14.5 Buffer Solutions
The example discussed above leads us to a very important conclusion that the extent of
dissociation of a weak acid can be decreased by adding a salt containing a common ion.
Further, it can be shown that the change in the extent of dissociation can also be varied by
changing the concentration of the common ion. The mixture of a weak base and a salt of
common ion also behave in the same way. These aqueous solutions containing weak acids/
bases and a salt of common ion are important in a yet another way. These act as buffer
solutions.
Buffer solutions are the ones that resist a change in its pH on adding a small amount of
an acid or a base.
In laboratory reactions, in industrial processes and in the bodies of plants and animals, it is
often necessary to keep the pH nearly constant despite the addition of acids and bases. The
oxygen carrying capacity of haemglobin in our blood and activity of the enzymes in our
cells depends very strongly on the pH of our body fluids. pH of the blood is very close to
7.4 and pH of saliva is close to 6.8. Fortunately, animals and plants are protected against
sharp changes in pH by the presence of buffers.
There are two kinds of commonly used buffer - solutions
i. A weak acid and a soluble ionic salt of the weak acid e.g. acetic acid and sodium
acetate ; CH3COOH + CH3COONa and,
ii. A weak base and a soluble ionic salt of the weak base e.g. ammonium hydroxide and
ammonium chloride ; NH4OH + NH4Cl.
The buffers with pH less than 7 are called acidic buffers and those with pH above 7 are
called basic buffers. Acetic acid - sodium acetate buffer is an example of acidic buffer while
Ammonium hydroxide - ammonium chloride is a basic buffer.
14.5.1 Buffer Action

A buffer system contains a conjugate acid-base pair and the concentrations of these two are
quite high as compared to that of the hydronium ions. These are called as the acid reserve
and the base reserve respectively. The added acid or base reacts with these reserves and gets
consumed without significantly altering the hydronium ion concentration and therefore the
pH does not change significantly. Let us consider a buffer solution containing acetic acid,
CH3COOH and sodium acetate CH3COONa to understand the buffer action.
In acetic acid - sodium acetate buffer CH3COOH is the acid reserve while CH3COONa (or
CH3COO - ions) is the base reserve. In the solution mixture the added components dissociate
as follows. The weak acid dissociates partially while the salt undergoes complete dissociation.
CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)
CH3COONa (aq) → Na+ (aq) + CH3COO - (aq)
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If we add a strong acid such as HCl to this solution, it produces H3O+. These added H3O+
(acid) react with an equivalent amount of the base reserve [CH3COO-] to generate undissociated
acetic acid. The reaction being
H3O+ (aq) + CH 3COO- (aq) → CH3COOH (aq) + H2O (l)
The net effect of this reaction is that there is a slight increase in the concentration of the acid
reserve and an equivalent decrease in the concentration of the base reserve. The effective
reaction being
HCl (aq) + CH3COONa (aq) → CH3COOH (aq) + NaCl (aq)
Similarly, when small amounts of a strong base like NaOH is added, it generates OH- ions.
These additional OH- neutralize some of the H3O+ ions present in the solution,
H3O+ (aq) + OH- (aq) H3O+ (aq) + CH3COO- (aq)
Since one of the products of the acid dissociation equilibrium (eq) is used up, there is some
more ionisation of CH3COOH to re-establish the equilibrium.
CH3COOH (aq) + H2O H3O+ (aq) + CH 3COO- (aq)
The net result is the neutralization of OH- by CH3COOH. In other words we can say that the
added OH- ions (base) react with the acid reserve to produce CH3COO- ions
OH- (aq) + CH3COOH (aq) CH3COONa- (l) + H2O (l)
The efffective reaction being the reaction of the added base with acid reserve.
NaOH (aq) + CH3COOH (aq) → CH3COONa (aq) + H2O (l)
The net effect of this reaction is that there is a slight increase in the concentration of the base
reserve and an equivalent decrease in the concentration of the acid reserve.
You may note here that the added acid or the base only cause minor changes in the
concentrations of the weak acid and the salt. The concentration of the hydronium ions and
thereby the pH does not changes significantly. Let us derive a mathematical expression for
determining the pH of a buffer solution.
14.5.2 Henderson - Hasselbalch Equation

This equation relates the pH of a given buffer solution to the concentrations of its components
viz. weak acid/salt or weak base/salt. Let us derive the expression for an acidic buffer system
that we have discussed above. In acetic acid - sodium acetate buffer the central equilibrium
is
CH3COOH (aq) + H2O H3O+ (aq) + CH 3COO- (aq)
which is characterized by the acid dissociation constant,
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[H3O + ] [CH3COO − ]
Ka =
[CH COO − ]
CH33COOH
rearranging, we get
[CH3COOH]
[H3O+] = K a x
[CH3COO − ]
The concentration of undissociated acetic acid can be taken as total acid concentration [Acid]
and that of sodium acetate as the total salt concentration [salt]. In the light of this the above
equation may be re written as
[Acid ]
[H3O+] = Ka
[Salt ]
Taking logarithm and multiplying through out by (-1) we get
[Acid ]
-log [H+] = - logKa - log
[Salt ]
Recall that pH = -log [H3O+] and pKa = - logKa. This gives the desired equation.
[Acid ] [Salt ]
pH = pKa - log = pKa + log
[Salt ] [Acid ]
The equation is known as Henderson-Hasselbalch equation. A similar expression can be

derived for a basic buffer (e.g., ammonium hydroxide and ammonium chloride). The
expression is
[Salt ]
pOH = pKb + log
[Base]
Let us take up some examples to see the application of this equation.
Example 14.9 : Calculate the pH of acetic acid - sodium acetate buffer containing 0.1 M
acetic acid and 0.1 M sodium acetate (Ka = 1.85 x 10-5 mol dm-3).
Solution : Here, [Acid] = 0.1 M and [Salt] = 0.1 M
Since Ka = 1.85 x 10-5 mol dm-3
pKa -logKa = - log 1.85 x 10-5
⇒ pKa = 4.73
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[Salt ]
According to Henderson equation, pH = pKa + log
[Acid ]
Substituting the values in Handerson equation, we get

pH = 4.73 + log (0.1/0.1) = 4.73 + log 1 = 4.73.
Example 14.10 : Calculate the pH of ammonium hydroxide - ammonium chloride buffer

solution that is 0.1 M in ammonium hydroxide and 0.01 M in ammonium chloride. (pKb of
NH4OH = 9.25).
Solution : Here, [Base] = 0.1 M and [Salt] = 0.01 M
Since pKb = 9.25 ;
[Salt ]
According to Henderson equation, pOH = pKb + log
[Base]

pOH = 9.25 + log (0.1/0.1) = 9.25 + log 0.1 = 9.25 - 1.0 = 8.25
14.6 Salt Hydrolysis

The aqueous solutions of certain salts also behave as acids or bases. They do so because of
the hydrolysis of the cation or anion or both. As you know, hydrolysis is a reaction with
water. Depending on the behaviour towards hydrolysis there are four different types of salts.
Salt of strong acid + strong base (eg. HCl + NaOH) NaCl
Salt of strong acid + weak base (eg. HCl + NH4OH) NH 4Cl
Salt of weak acid + strong base (eg. CH3COOH + NaOH) CH 3COONa
Salt of weak acid + weak base (eg. CH3COOH + NH4OH) CH 3COONH 4
Let us learn about the acid - base behaviour of the different types of salts.
Salt of strong acid + strong base : The cations of the strong bases and the anions of the
strong acids do not get hydrolysed. Therefore the salts of this category do not show any acid-
base behaviour and are neutral.
Salt of strong acid + weak base : The salts of this type dissociate in aqueous solutions to
give a cation of a weak base and the anion belonging to strong acid. For example, NH4Cl
dissociates as :
NH4OH (aq) → NH4+ (aq) + Cl- (aq)
As mentioned above, the anion does not get hydrolysed but the cation does get hydrolysed as
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per the following equation.
NH4+ (aq) + H2O (l) → NH4OH (aq) + H+ (aq)
since it generates H+ (aq) ions, the solution is acidic in nature.
Salt of weak acid and strong base : The salts of this type dissociate in aqueous solutions to
give a anion of a weak acid and the cation belonging to strong base. for example, CH3COONa
dissociates as :
In this case the cation does not get hydrolysed but the anion does get hdyrolysed as per the
following equation.
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)
Since it generates hydroxyl ions the solution is basic in nature.
Salt of weak acid and strong base : The salts of this type dissociate in aqueous solutions to
give a anion of a weak acid and the cation belonging to strong base. for example, ammonium
acetate CH3COONa dissociates as :
CH3COONa (aq) → Na+ (aq) + CH3COO - (aq)
in this case the cation does not get hydrolysed but the anion does get hdyrolysed as per the
following equation.
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH - (aq)
Since it generates hydroxyl ions the solution is basic in nature.
Salt of weak acid and weak base : The salts of this type dissociate in aqueous solutions to
give a anion of a weak acid and the cation belonging to a weak base. for example, ammonium
acetate, CH3COONH4 dissociates as :
CH3COONH4 (aq) NH 4+ (aq) + CH3COO - (aq)
In this case both the cation as well as the anion would undergo hydrolysis and the nature of
the solution, whether acidic, basic or neutral would depend on the relative strength of the
weak acid and the weak base.
14.7 The Solubility Equilibrium

When we try to dissolve a solid into water, if it dissolves, there are three possibilities:
1. The solid is a non-electrolyte and it dissolves as neutral molecules.
2. The solid is a highly soluble electrolyte; it dissolves almost completely.
3. The solid is a sparingly soluble electrolyte; it dissolves to a limited extent.
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It is the third possibility that interests us here. Let us take the example of dissolution of AgCl
to understand the equilibria in such cases. When silver chloride is added to water, the following
equilibrium is established.
AgCl (s) Ag+ (aq) + Cl - (aq)
This is an example of a heterogeneous equilibirum because it involves both a solid and a
solution. This equilibrium is known as the solubility equilibrium for which the equilibrium
constant expression is
[Ag + ] [Cl − ]
K=
(AgCl(s)]
As a matter of convention the concentration of the undissolved solid is taken as one. We can
rewrite the equilibrium as
Ksp = [Ag+] [Cl-]
The equilibrium constant now is the product of the concentrations of the ions. It is called
solubility product constant or simply solubility product. A new symbol Ksp, has been assigned
to this constant. The mass expression on the right, is called, ion product or ionic product. The
solubility product constant of a given salt is constant at a given temperature.
14.7.1 Relationship between Solubility and Solubility Product

Constant
The solubility product constant for a substance is related to its solubility. The nature of
relationship depends on the nature of the salt.
Salt of AB type : (For example AgCl, CaSO4). In such cases the solubility equilibrium can
be represented as
AB(s) A+ (aq) + B- (aq)
and Ksp = [A +] [B-]
If the solubility of salt ‘s’ is mol dm-3 then the concentrations of the cations and the anions
would be ‘s’ mol dm-3 each. Substituting the values in the expression of Ksp we get,
Ksp = [s’ mol dm-3] x [s’ mol dm-3] = s 2 mol2 dm-6
Salt of AB 2 type : (For example CaF2). In such cases the solubility equilibrium can be
represented as
AB2 (s) A2+ (aq) + 2B- (aq)
-319-
and Ksp = [A2+] [B-]2
If the solubility of salt is ‘s’ mol dm-3 then the concentrations of the cations and the anions
would be ‘s’ mol dm-3 and ‘2s’ mol dm-3 respectively. Substituting the values in the expression
of Ksp we get,
Ksp = [‘s’ mol dm-3] x [‘2s’ mol dm-3]2 = 4s3 mol3 dm-9
Salt of A 2B type : (For example Ag2CrO4). In such cases the solubility equilibrium can be
represented as
A2B (s) 2A+ (aq) + B2- (aq)
and Ksp = [A +]2 [B2-]
If the solubility of salt is ‘s’ mol dm-3 then the concentrations of the cations and the anions
would be ‘2s’ mol dm-3 and ‘s’ mol dm-3 respectively. Substituting the values in the expression
of Ksp we get,
Ksp = [‘2s’ mol dm-3]2 x [‘s’ mol dm-3] = 4s3 mol3 dm-9
Salt of A2B2 type : (For example Ca3(PO4)2). In such cases the solubility equilibrium can be
represented as
A3B2 (s) 3A2+ (aq) + 2B3- (aq)
and K sp = [A2+]3 [B3-]2
If the solubility of salt ‘s’ mol dm-3 then the concentrations of the cations and the anions
would be ‘3s’ mol dm -3 and ‘2s’ mol dm-3 respectively. Substituting the values in the
expression of Ksp we get,
Ksp = [‘3s’ mol dm-3]3 x [‘2s’ mol dm-3]2 = 108s5 mol5 dm-15
In general for a salt with the formula AxBy and a solubility of s mol dm-3 the relationship
between the solubility and Ksp can be given as
Ksp = [Ay+]x [Bx-]y = (xs)x (ys)y = xx yy sx+y
Example 14.11 : The solubility of calcium sulphate in water is 4.9 x 10-3 mol dm-3 at 298K.
Calculate the value of Ksp for CaSO4 at this temperature.
Solution : The following equilibrium would exist in this case
CaSO4 (s) Ca2+ (aq) + SO24- (aq)
For this reaction Ka = [Ca2+] [SO24-]
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From the equation we see that when 4.9 x 10-3 mol of CaSO4 dissolves to make 1 dm-3 of a
saturated solution, the ionic concentration are
[Ca2+] = 4.9 x 10-3 mol dm-3 ; [SO 24-] = 4.9 x 10-3 mol dm-3
Ksp = [Ca2+] [SO24-] = [4.9 x 10-3 mol dm-3 x 4.9 x 10-3 mol dm-3 ]
= 2.4 x 10-5 mol2 dm-6.
Example 14.12 : Solubility product of silver iodide, AgI is 8.5 x 10-17 at 250C. What is the
molar solubility of AgI in water at this temperature ?
Solution : Silver iodide dissolves according to the equation
AgI (s) Ag + (aq) + I- (aq)
Let the solubility of AgI be is ‘s’ mol dm-3 the concentrations of silver and iodide ions would
be is ‘s’ mol dm-3 each.
At equilibrium Ksp = [Ag+] [I-]; Substituting the values, we get
[‘s’ mol dm-3] [‘s’ mol dm-3] = s2 mol2 dm-6 = 8.5 x 10-17 mol2 dm-6
This gives, solubility (s) = [8.5 x 10-17 mol2 dm-6]1/2
= 9.2 x 10 -9 mol dm-3.
The solubility of AgI in water is therefore 9.2 x 10-9 mol dm-3 at 298 K.
14.7.2 Effect of Common Ion on Solubility Equilibria

What will happen if we add a soluble salt containing a common-ion to a solution of a sparingly
soluble salt ? You may reason out that according to Le Chatelier’s principle, the common-
ion will shift the equilibrium in backward direction which would reduce its solubility still
further. This actually is the case. Let us understand it with the help of an example.
Example 14.13 : Calculate the molar solubility of AgI in a solution containing 0.1 mol dm-3
AgNO 3. The solubility product of silver iodide, AgI is 8.5 x 10-17 mol2 dm-6 at 298 2K.
Solution : Silver nitrate is a strong electrolyte ionsing as
AgNO3 (s) Ag + (aq) + NO 3 (aq)
and for AgI the solubility equilibrium is

Agl (s) Ag+ (aq) + I- (aq)
If we take the solubility of Agl to be ‘s’ mol dm-3 , then the total concentration of Ag+ ions in
the solution would be [0.1 + s] mol dm-3 ~ [0.1] mol dm-3 because the value of ‘s’ is very
small. And the concentration of I- ions would be ‘s’ mol dm-3
-321-
Substituting in the expression of Ksp = [Ag+] [I-] ; we get
[0.1] 1mol dm-3] [‘s’ mol dm-3] = 0.1 s mol2 dm-6 = 8.5 x 10
-17
mol2 dm-6
This gives, solubility (s) = [8.5 x 10 ] / [0.1] mol dm-3
-17
= 8.5 x 10-16 mol dm-3.

(The value of ‘s’ is clearly negligible in comparison with 0.10 and thus justifies our assumption).
The solubility of Agl in 0.1 M AgNO3 is therefore 8.5 x 10-16 mol dm-3 at 298K. Compare this
value with the solubility of Agl in water as determined in the previous example
Solvent Water 0.1 M AgNO3
Solubility 9.2 x 10-9 mol dm-6 8.5 x 10-16 mol dm-3
Thus we see that the solubility of a sparingly soluble salt is decreased in the presence of
another salt that has common ion. This decrease in solubility is an example of the Common
Ion Effect.

1. Calculate the pH of a solution containing 0.05 M benzoic acid and 0.025 M sodium
benzoate. Benzoic acid has a pKa of 4.2
...................................................................................................................................
2. Calculate the solubility product for Ag2SO4 if [SO42-] = 2.5 x 10-2 M.
...................................................................................................................................
14.7.3 Application of Solubility Product in Qualitative Analysis

The qualitative analysis of cation is carried out by dividing them into several groups. This
groups separation of cations is based upon selective precipitation of some cations out of
many that are present in a solution. This is achieved by adjusting the solution conditions in
such a way that the Ksp of specific salts of some cations is exceeded and they precipitate out.
The remaining cations remain in the solution. A typical example is the use of H2S. The
dissociation of H2S can be written as
H2S (aq) 2H+ (aq) + S2- (aq)
Since the equilibrium involves hydrogen ions, the acidity of the solution would play an
important role in controlling the concentration of sulphide ions.
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You are aware, that in the separation of group II sulphides, the medium of the solution is kept
acidic. In this medium the concentration of the S2- ions is very small and only the sulphides of
group II are precipitated. On the other hand in the alkaline medium the concentration of
sulphide ions is fairly high and the sulphides of group IV cations precipitate out.

• There are three different concepts of acids and bases proposed by Arrhenius, Bronsted
and Lowry and Lewis respectively.
• According to Arrhenius Concept an acid is a substance capable of producing hydrogen
ions by dissociating in aqueous solution while a base is a substance capable of providing
a hydroxyl ion. The netralization reaction is basically the reaction between a proton
and a hydroxyl ion to give a molecule of water.
• Since a hydrogen ion H+ is very small with high charge density it does not exist free in
polar solvent like water. It binds itself to a water molecule and form a hydronium ion
(H3O+).
• According to Bronsted and Lowry, an acid is defined as a proton (H+) donor, and a
base is defined as a proton acceptor. An acid-base reactions can be thought of as a
proton-transfer from an acid to a base. In this concept, acids and bases can be either
ions or molecular substances.
• According to Bronsted and Lowry definition the species in either side of the acid-base
equilirbium, differ by the gain or loss of a proton. These are called a conjugate acid-
base pair. In such a pair a strong acid has a weak conjugate base while a weak acid has
a strong a conjugate base.
• Lewis definition is quite broad, according to him, an acid is defined as, any atom,
molecule or ion that can accept an electron pair from any other atom, molecule or ion,
while a base is any atom, molecule or ion that can donate a pair of electron. The
product of a reaction between an acid and a base is called an adduct.
• Strong Arrhenius acids and bases dissociate completely in aqueous solutions where as
the weak acids and bases undergo partial ionisation. Higher the extent of ionisation
stronger the acid or the base.
• In Bronsted - Lowry concept, the relative strength of an acid is defined as its relative
tendency to lose/donate a proton to water.
• The ionisation equilibria of weak acids and bases are characterized by equilibrium
constant called ionisation constants. The values of these constants is a measure of their
relative strength.
• Water can act both as a weak acid as well a weak base. In a sample of water a small
number of water molecules undergo autoionisation, in which half the ions act as an acid
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while the other half acts as a base.
• In aqueous solutions the concentration of H3O+ can be expressed in terms of a logarithmic
scale called pH scale. The pH of a solution is defined as pH = -log10[H+] or pH = -
log10[H 3O+].
• A neutral solution has a pH of 7; any solution with a pH less than 7 is acidic while the
ones with a pH of greater than 7 are basic in nature.
• The presence of common ion in a solution of a weak acid or a weak base suppress its
dissociation. Such solutions act as buffer solutions which resist a change in their pH
on addition of small amount of an acid or a base. The pH of buffer solutions depend
on their composition and can be found by using a simple equation called Henderson
Hasselbalch equation.
• The aqueous solutions of cetain salts also behave as acids or bases due to the hydrolysis
of their cation or anion or both.
• In an aqueous solution of a sparingly soluble salt an equilibrium exists between the
undissolved salt and the ions obtained from the dissolved salt. This is called solubility
equilibrium.
• The product of the concentration of the ions in the solubility equilibrium is a constant
called solubility product (Ksp) and is proportional to the solubility of the sparingly
soluble salt.
• The presence common ions decreases the solubility of a sparingly soluble sat. This is
called common ion effect and has widespread applications in qualitative analysis.
Terminal Exercise
1. Explain why a hydrogen ion can not exist free in an aqueous solution ?
2. Write the equilibrium constant expression for the following reaction ?
H2CO3 (aq) + H2O (l) U H3O (aq) + HCO3- (aq)
3. Explain why does a strong Bronsted - Lowry acid has a weak conjugate base ?
4. What do you understand by the term ‘amphoteric’ ? Show with the help of equations
that water is amphoteric in nature.
5. Calculate the pH of 1 x 10-3 M solution of NH4OH. The dissociation constant of NH4OH
is 1.85 x 10-5 mol dm-3.
6. The pH of an aqueous solution of HCl is 2.301. Find out the concentration of hydrogen
ions in this solution.
7. What is a buffer solution ? What are its main constituents ?
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8. Solubility of lead iodide Pbl2 is 1.20 x 10-3 mol dm-3 at 298K. Calculate its solubility
product constant.
9. Calculate the solubility of Bi2S3 in water at 298 K if its Ksp = 1.0 x 10 -97 mol5 dm-15.
10. Calculate the solubility of AgL in 0.10 M Nal at 298 K Ksp for AgL is 8.5 x 10-7 at this
temperature.
14.1 Answers to Intext Questions

1. According to Arrhenius concept an acid is defined as a substance that is capable of
producing hydrogen ion (H+) by ionisation in aqueous solution. For example, HCl and
CH 3 COOH.
2. Arrhenius definition has the following drawbacks.
• It is limited to only aqueous solutions and requires ionisation of the substance.
• It does not explain the acidic or basic behaviour of some substances which lack a
hydrogen (or a hydroxide) ion. For example, AlCl3 and Na2CO3 which lack a
hydroxide.
3. In the Bronsted - Lowry concept, any molecule or ion that can accept a proton is a
base wheareas in Arrhenius concept a base is the one which provides hydroxide ions
in solution.
4. Acids HCl, H 3O+
Bases NH 3, CN-
14.2
1. The ionisation of weak acid, HF, can be represented as
HF (aq) + H2O H3O+ (aq) + F- (aq)
[ H 3O + ] [ F − ]
The expression for Ka would be, Ka =
[HF]
2. For a weak base BOH which partially dissociates in aqueous solution, and has a degree
of dissociation as α we can write
B + H 2O BH+ (aq) + OH-
Initial concentrations c -55 0 0
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Equilibrium concentrations c(1-α) -55 cα cα
The equilibrium constant expression or base dissociation constant can be written as
[BH + ] [OH − ] [cα ] [cα ]

K = =
[H 2O] [B] c [1 − α ] 55
rearranging we get,
[cα ] [cα ] c2α 2 cα 2

55 = Kb = = =
c [1 − α ] c [1 − α ] [1 − α ]
Since the acid B is very weak, α << 1; we can neglected in comparison to 1 in the
denominator to get
Kb Kb
~ cα2 or α2 =
Kb ~ or α =
c c
3. Given hydronium ion concentration, [H3O+] = 6.3 x 10-2 M

As per definition pH = - log [H3O+]
⇒ pH = - log 6.3 x 10-2
⇒ pH = - (0.7993 - 2.0000)
⇒ pH = - (-1.2007) = 1.2007
4. Given Concentration of glycine = 1.0 M
Ka = 1.67 x 10-10.
K ab
For a weak acid α = =α = 1.67 x 10 −10 = 1.29 x 10-5
c
⇒ [H3O+] = 1 x 1.29 x 10-5 = 1.29 x 10-5 M

pH = - log[H3O+] = - log [1.29 x 10-5] = - (-4.8894) = 4.8894
14.3
1. Here, [Acid] = 0.05 M and [Salt] = 0.025 M; and pKa = 4.2
pH = 4.2 + log (0.05/0.025) = 4.2 + log 2 = 4.2 + 0.3010 = 4.5010
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2. Let the solubility of Ag2SO4 be ‘s’ mol dm-3
-2
The concentrations of the Ag + and the SO4would be ‘2s’ mol dm-3 and ‘s’ mol dm-3
respectively and Ksp = [Ag+]2 [SP4-2]
⇒
- -2 -2 -2
Given [SO4] = 2.5 x 10 M [Ag+] = 2 x 2.5 x 10 M = 5 x 10 M
Substituting the values in the expression of Ksp we get,
Ksp = [5 x 10-2]2 x [2.5 x 10-2] = 6.25 x 10-5 mol3 dm-9
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15
ELECTRO CHEMISTRY
Electrochemicsty deals with the conversion of electrical energy into chemical energy and
vice versa. When electric current is passed through an aqueous solution of certain substance
or through molten salts, it causes a chemical reaction to occur. On the other hand, in dry
cells, button cells or lead acid batteries chemical reactions occur which produce electrical
energy. In this lesson you will study some aspects of these processes.
Objectives
• understand oxidation and reduction in terms of electron transfer concept;
• calculate oxidation number (ON) of an atom in a molecule or ion;
• balance the chemical equation for redox reaction;
• explain electrolytic conduction, conductance and molar conductivity;
• describe the effect of dilution on conductiviy and molar conductivity of an electrolyte;
• differentiate between electrolytic and Galvanic cell;
• state standard electrode potential and use it for calculation of standard electrode potential
of a cell;
• explain standard Hydrogen electrode;
• describe electrochemical series and its application;
• state effect of concentration on eletrode potential (Nernst equation);
• solve numericals based on Nernst equation and
• find relationship between emf and Gibbs energy change.
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15.1 Oxidation and Reduction as Electron Transfer Process
Oxidation and reduction reactions constitute a very important class of chemical reaction. The
electronic concept looks at oxidation and reduction in terms of electron transfer : process in
which an atom or ion loses one or more electron to the other is called oxidation and the
process in which an atom or ion gains one or more electron is termed as reduction. In the
formation of NaCl from Na and Cal.
Na → Na+ + e (loss of e by Na; oxidation)
Cl + e → Cl- (gain of e- by Cl; reduction)
Sodium undergoes oxidation and chlorine undergoes reduction. Here, sodium helps chlorine
to undergo reduction and therefore it is called a reducing agent or reductant.
A reductant is a species in a chemical reaction which looses its electron to another reactant.
Chlorine, on the other hand accepts electron, therefore it is an oxidising agent or oxidant.
An oxidant is a species which accepts electrons in a chemical reaction.
It may be noted that oxidation and reduction processes do not take place independently but
occur simultaneously and are thus called oxidation-reduction reaction or redox reactions.
A redox reaction is a sum of oxidation and reduction half reactions in a chemical reaction.
15.2 Oxidation Number

It is easy to identify species undergoing oxidation or reduction in simple molecules. However,
in polyatomic molecuels, it is difficult to do the same. In the example of NaCl taken earlier
it was easy to identify as sodium undergoing oxidation and chlorine undergoing reduction but
in the reaction involving ferrous sulphate with potassium permanganate (KMnO4) it is difficult.
Therefore, a new term called Oxidation number has been introduced, Oxidation number is
the apparent charge which an atom appears to have when each pair of electrons is counted
with more electronegative atom. Oxidation number is always assigned to an atom. it is number
written with +ve or - ve sign. The number indicates the number of electrons that has been
shifted from an atom towards a more electro-negative atom, in a heteronuclear covalent bond.
The +ve sign for the atom shifting its electron away from itself and -ve is given to more
electro -ve atom. The concept of Oxidation Number is based on the assumption that in a
polyatomic covalent bonding, shared pair of electrons belongs to more electro-ve atom.
Oxidation state (OS) is also used for Oxidation Number.
15.2.1 Rules for Assigning Oxidation Number

There are certain rules that are followed for computing the oxidation number of an atom in a
molecule or ion.
1. Oxidation number is taken as zero if atoms are present in elemental form. For example,
O2, Na, P4 are elemental forms. They have oxidation number zero.
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2. The oxidation number of a monatomic ion is the same as the charge present on it. For
example, Na+, Mg2+, Al3+, Cl-, S2- will have oxidation no +1, +2, +3, -1, -2 respectively.
3. The oxidation number of oxygen is -2 in almost all the compounds except (a) in peroxides
e.g. Na2O2, H2O2 where oxidation number is -1 and (b) super oxides (KO2) where it is
-1/2.
4. The oxidation number of H of +1 when combined with non-metal and is -1 when
combined with metal e.g. in HCl the O.N. of H is +1 but in Ca H2 it is -1.
5. The Oxidation Numbers of alkali metal is +1 in its compounds.
6. In a compound made up of different elements the more electro negative element will
have negative oxidation number and less electro negative atoms will have positive
oxidation number e.g. in NCl3, N has +3 oxidation number and Cl has -1 oxidation
number.
7. The sum of the oxidation numbers of all the atoms in a neutral compound is zero.
8. In a polyatomic ion, the sum of oxidation numbers of all the atoms is equal to the charge
2-
on the ion. e.g. in CO3 , the sum of oxidation Number of carbon and oxygen is -2.
Let us illustrate the above rules taking few examples. The oxidation number of S, N and Cl
atoms in : (a) H2SO4 (b) NO3 (c) ClO4 respectively will be calculated as
(a) 1. Let the oxidation number of sulphur be x.
2. Since the oxidation number of O is -2. Therefore the sum of four O atoms is equal to -8.
3. The oxidation number of each H is +1 as bonded to a non-metal so two H atoms have
total oxidation number of +2.
4. H2SO4 is a neutral molecule. Therefore the sum of all the oxidation numbers is equal to
zero. Thus
+ 2 + x - 8 = 0.
x = + 6.
Therefore oxidation number of sulphur in H2SO4 is + 6.
b) NO 3 - first assign -2 oxidation number to each O atom. Here the sum of the
oxidation number of all the atoms will equal to charge present on the
ion.
.
. . x - 6 = -1.
x = + 5.
oxidation number of N is +5.
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c) In ClO4 x - 8 = -1.
x = + 7.
15.3 Balancing Redox Reaction

The redox reaction can be balanced by any of the following methods.
a) Oxidation number method.
b) Ion electron method
15.3.1 Balancing by Oxidation Number method

The steps involved in balancing redox reactions by this method are as follows :
1) Write the skeletal equation of reaction i.e. chemical equation without the stoichiometric
coefficient
2) Write the oxidation number of each atom above its symbol in the equation.
3) Identify the atoms undergoing change in oxidation number.
4) Calculate the increase or decrease in oxidation number per atom for the atom
undergoing a change in oxidation number. If more than one atom is involved, multiply
the increase or decrease in number with the number of atoms undergoing the change
to determine the total change in oxidation number.
5) Equate the increase and decrease in oxidation number on the reactant side by multiplying
the formulae of the oxidising and reducing agents suitably.
6) Balance the equation with respect to all the atoms except hydrogen and oxygen.
7) Finally balance H and O also.
8) If the reaction is taking place in acidic medium balance the O atoms by adding required
number of H2O molecule on the side where O atoms are less in number. Balance the H
atoms by adding H+ to the side deficient in H atoms.
9) In the basic medium by add required number of negative charges by adding required
number of OH- ions to the side deficient in the magnitude of charges, then add H2O
molecules to balance OH- ions.
For example : When Phosphorus is treated with nitric acid, nitric oxide is formed.
1. The skeletal equation is
P + HNO 3 → HPO3 + NO + H2O
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2. Write the oxidation number of each atom on the given skeletal equation
0 +1+5-2 +1+5-2 +2-2 +1-2
P + HNO 3 → HPO3 + NO + H2O
3. P and N are undergoing change in Odixation Number
0 +5 increase in ON = 5 +5 +2
P + HNO3 HPO3 + NO + H2O
decrease in ON = 3
4. Equating the increase and decrease in Oxidation Number of P and N on the reactant
side
3P + 5HNO 3 → HPO 3 + NO + H2O
5. Balance the P and N atoms on both sides of the equation
3P + 5HNO3 → 3HPO 3 + 5NO + H2O
6. O and H are already balanced in the equation.
15.3.2 Balancing by Ion Electron Method

This method is based on the pricnciple that electrons lost during oxidation half reaction is
equal to the electron gained in the reduction half reaction. The steps involved are
1. Write the skeleton equation.
2. Write the oxidation number of all the atoms above their symbols in the skeletal equation.
3. Find the atoms undergoing change in Oxidation Number. Thus find out the species
getting oxidized and reduced respectively.
4. Split the whole (net) equation into reactions i.e. oxidation half reaction and reduction
half reaction.
5. Balance the atoms, undergoing change in oxidation number in each half reaction.
6. Calculate the total change in oxidation number in each half reaction which is equal to
total number of electron transfer.
7. Add total number of electron tranfer as calculated above on the reactant side in reduction
half and on the right hand side on the oxidation half reaction.
8. Balance the charges by adding H+ (for reactions in acidic medium) or OH- (reactions
basic medium) either on left or right of equation.
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9. Finally balance H and O by adding H2O on the required side of the reaction.
10. Add the two half reactions such that total number of electrons cancel out on both sides.
To do so half reactions may be required to multiplied by some numbers to make the
number of electrons equal on both sides.
15.3.2 Balancing by Ion Electron Method

Example 15.1 : Balance the following skeletal reaction by ion electron method
Cr2O72- + Fe2+ → Cr3+ + Fe3+ in acid medium
Reffering to the rules given :
Step I and II
Write the oxidation number of the atoms above their symbol in the skeletal equation
+6 +2 +3 +3
Cr2O72- + Fe2+ → Cr3+ + Fe3+
Step III : Oxidation number of Fe2+ is increasing; therefore it is undergoing oxidation and
oxidation number of Cr is decreasing so it is undergoing reduction.
Step IV : Split the reactions in two half reactions
a) Reduction half reaction
Cr2O72- → Cr3+
b) Oxidation half reaction

Fe2+ → Fe3+
Balance the first reduction half reaction
Cr2O72- → Cr3+
Step V : Balance the atoms undergoing change in Oxidation Number.
Cr2O72- → 2Cr 3+
Step VI and VII : Write the total number of electron transfer taking place. Here each atom
undergoes change in in ON by 3 therefore two Cr atoms undergoes change in Oxidation
Number by 6.
Cr2O72- + 6e → 2Cr3+
Step VIII : Balance the charge by adding H + on the left side

Cr2O72- + 6e + 14H + → 2Cr3+
Step IX : Balance the H and O by adding H2O on either side

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Cr2O72- + 6e + 14H+ → 2Cr3+ + 7H2O
Balancing the Oxidation half reaction

According the steps as followed for re1duction half reaction
Fe2+ → Fe3+
i) Atoms are balanced on both side so we go to next step, that is number of electron
transfer taking place
Fe2+ → Fe3+ + e
ii) Balance the charge and it is balanced.
Step X : Add the two half reaction

→ [Fe3+ + e] x 6
Fe2+
Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O
___________________________________________
Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O

1. Determine the Oxidation number of element in the bold letter in the following :
SiH 4 BH 3 BF 3 S 2O 4
2
BrO 4 HPO 4 SH 12 O HNO 3
2
................................................................................................................................................................................................................................
2. How does oxidation number change in oxidation and reduction ?
................................................................................................................................................................................................................................
3. Mention the oxidising agent and reducing agent in the following.

H2S + HNO3 → NO + S + H2O
................................................................................................................................................................................................................................
4. Write the half reaction for the following

2I- (aq) + 2Fe3+ (aq) → I2 (s) + 2Fe3+ (aq)
Mg (s) + Cl2 (aq) → MgCl2 (s)
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I2 + HNO3 → HIO3 + NO 2 + H2O
................................................................................................................................................................................................................................
5. Balance the equation by oxidation number method

CuO + NH3 → Cu + N2 + H2O
MnO4 + HCl → MnCl2 + C2 + H 2O
................................................................................................................................................................................................................................
6. Balance the following by ion electro half reaction method

NO3- + Bi → Bi3+ + NO2 acidic medium
MnO4 + Fe 2+ → Mn2+ + Fe3+ acidic medium
Cr2O72- + Fe2+ → Fe3 + Cr3+ acidic medium
Al + NO3- → Al (OH)4- + NH3 basic medium
................................................................................................................................................................................................................................
15.4 Electrolytic Conduction

When electricity is passed through an aqueous solution, it may or may not conduct current.
The chemical substances whose aqueous solutions conduct electricity are called electrolytes
and those which do not conduct current are called as non-eletrolytes. This phenomenon of
conduction of current through a solution is called elctrolytic conduction.
Electrolytic conduction takes place due to the movement of cations and anions in a solution.
The electrical conductance of a solution, depends upon (a) nature of solute (b) valency of
its ion, (c) the concentration in solution and (d) the temperature. In this section we will
learn about various ways of expressing the conductance of electrolytes and the factors
affecting them.
15.4.1 Conductance and Conductivity
Like solid conductors, electrolytic solutions also obey Ohm’s Law. When a current of I
amperes flows through a solution which offers a resistance of R ohms and a potential
difference of V volts is applied, then according to ohm’s law.
V=I.R
If the solution is taken in a conductivity cell which has two parallel electrodes / cm apart
and each having an area of cross section A cm2, the resistance R of the electrolyte is found
to be directly proportional to l and inversity propoertional to A i.e.
R ∝ l/A
or R = ρ . l/A .... (i)
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Where ρ “rho” is a constant of proportionality and is called specific resistance or resistivity. it
is characteristic of the nature of electrolyte, its concentration and temperature.
In case of solution, it is preferred to discuss their conductance and conductivity rather than
their resistance and specific resistance. The conductance is reciprocal of resistance and
the conductivity is reciprocal of specific resistance.
Conductance is denoted by L and is measured in the unit of ohm-1 which has now been
named as siemens, S. The conductivity is denoted by k “kappa”. Thus by definition
1 1
L= and k = .... (ii)
R ρ
The units of k can be worked out from relation (i) as under :

The inverse of (i) is,
1 1 A
= .
R ρ l
A
or L=k
l
l
and K=L
A
cm
=S
cm 2
= S cm-1
The conductivity (K) is expressed in S cm-1 or 100 S m -1.
15.4.2 Measurement of Conductance

The conductance of an electrolyte is measured with the help of a conductivity cell. Conductivity
cell is a device which has two parallel platinum electrodes coated with platinum black.
The SI unit of length is metre, hence SI unit of conductivity (K) is Sm-1, but the commonly
l
used unit is Scm-1. In the expression for conductivity, is a constant. Here l represents the
A
distance between the two parallel electrodes and A represents the area of cross section of the
l
electrodes. Thus for a given conductivity cell, is a constant called cell constant.
A
K (Conductivity) = conductance x cell constant
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The conductivities of some substances are given in the table.
Table 15.1 : The values of conductivity of some selected substances at 298 K
Substance k/S cm-1 Substance k/S cm-1
Pure water 6.0 x 10-8 Silver metal 6.1 x 105

0.1 M HCl 3.5 x 10-2 Mercury metal 1.0 x 104
0.1 M NaCl 9.2 x 10 -3 Glass 1.0 x 10-14
0.1 M CH3COOH 4.7 x 10-4
0.1 M NH4OH 3.4 x 10 -4
We find from the table that the conductivities of metals are very high and that of pure water
and glass very low.
15.4.3 Molar Conductivity

The electrolytic conductivity of a solution depends on the concentration of the electrolyte in
the solution. Therefore, the conductivity of an electrolyte is normally expressed as molar
conductivity.
Molar conductivity is the conducting power of all the ions furnished by one mole of an
electrolyte in a solution of specified concentration.
It is denoted by λm and is related to K by the relation.
1000 K
λm = .... (iv)
M
Where M is the molarity of the solution. Its units are S cm2 mol-1.
15.5 Factors Affecting Conductivity

As mentioned the conductivity of an electrolyte depends upon the following aspects of the
electrolyte.
a) Nature of electrolyte : Conductivity of an electrolyte depends upon the nature of
electrolyte on the following points :
i) Weak or strong electrolyte : A weak electrolyte furnishes fewer ions therefore it
has lower conductivity than a strong electrolyte of same concentration.
ii) Valency of the ions : The ions with higher valency carry more charge and therefore
they conduct more charge than the ion of lower valency. Thus higher the valency
of the ion greater is the conducting power.
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ii) Speed of the ion : The ion which can move faster will carry the charge also faster
and therefore has more conducting power.
b) Temperature : Conductivity of an electrolyte generally increases by 2-3 percent for
each degree rise in temperature. With increase in temperature the viscosity of the solvent
decreased and thus ion can move faster. In case of weak electrolyte, when the temperature
is increased its degree of dissociation increases, thus conductivity increases.
c) Concentration :
i) Variation of conductivity (k) with concentration. When the solution is diluted
its conductivity also decreases. It is because k is the conducting power of all the
ions present per cm3 of the solution. When the solution is diluted the number of
ions per cm3 also decreases, hence k decreases.
KOM
k
CH3COOH
C
Fig. 15.1 : Variation of conductivity with concentration
ii) Variation of Molar and Equivalent conductivity with concentration : As the
solution is diluted its molar conductivity increases. λm is given as
1000 K
λm =
M
where k is conductivity and M is molar concentration.
This increase in λm is a resultant of two factors. On decreasing the concentration both k and
M decreases. Of the two (k) tries to decrease λm while the other factor (M) tries to increase
it. Since the decrease in M is much more, the net result is that λm increases. However, strong
and weak electrolyte as show different type of behaviour on dilution (Fig. 15.3)
KCL (strong electrolyte)
λm CH3 (weak electrolyte)
C
Fig. 15.2 : Variation of molar conductivity with concentration
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From the Fig. 15.3 we find that the increase in molar conductivity for a strong electrolyte like
KCl is very gradual on dilution and also the value is high at all concentrations. Whereas for a
weak electrolyte like CH3COOH, there is a gradual increase in λm on dilution which is
followed by a sharper increase on further dilution. These observations can be explained as:
Since KCl is a strong electrolyte, it is fully dissociated at all concentrations. In concentrated
solution, attraction between opposite ions is large and their conducting ability is less. On
dilution inter-ionic forces decrease and ions can move faster leading to the increase in molar
conductivity.
On the other hand, in weak electrolytes, molar conductivity is low in concentrated solution.
This is due to partial dissociation (ionisation) of weak electrolytes. On diluting the solution
the degree of ionisation increases, which increases the number of ions. This leads to a sharp
increase in molar conductivity in weak electrolytes.
15.5.1 Kohlrausch’s Law

Kohlrausch determined the molar conductivity at infinite dilution for a large number of strong
electrolytes. On the basis of his observations he conclude that at infinite dilution, each ion
makes a define contribution to the total molar conductivity of an electrolyte. This individual
contribution is called molar ionic conductivity. He generalised his observastion as
“At infinite dilution each ion of the electrolyte makes a definite contribution towards
conductivity of the electrolyte and it is independent of the presence of other ions of the
electrolyte.” This is called Kohlrausch’s Law of independent migration of ions.
For a salt like KCl, molar conductivity at infinite dilution can be written as
λm∝ KCl = λm∝ K+ + λm∝ Cl-
In general for a salt of formula AxBy the molar conductivity at infinite is written as
λm∝ (AxBy) = x λm∝ (Ay+) + y λm∝ (Bx+)
where λm∝ indicates molar conductivity at infinite dilution. This law is used to calculate the
molar conductivity at infinite dilution for weak electrolytes whose λm∝ can not be obtained
graphically.
Example 15.2 : 0 for NaCl. HCl and CH3COONa are 126.0, 426.0 and 91.0 S cm2 mol-1
>
respectively. Calculate 0 for CH3COOH.

>
Solution : 0
CH 3COOH = λ0 (H+) + λ0 (CH3COO-)
>
= λ0 (H+) + λ0 (Cl-) + λ0 (Na+) + λ0 (CH3COO-) − λ0 (Na+) − λ0 (Cl-)

= 426.0 + 91.0 - 126.0 = 391.0 S cm2 mol-1
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1. How does the solution of electrolytes conduct electricity ?
...................................................................................................................................
2. Define conductivity and molar conductivity.
...................................................................................................................................
3. Give the units of conductance and conductivity.
...................................................................................................................................
4. List factors affecting the conductance of an electrolyte.
...................................................................................................................................
5. Draw a graph showing variation in molar conductivity of weak and strong electrolytes.
...................................................................................................................................
6. Write the expression for molar conductivity at infinite dilution for Al2(SO4)3.
...................................................................................................................................
15.6 Electro Chemical Cell

An electrochemical cell is a device used for the interconversion of electrical and chemical
energy. An electrochemical cell contains two electrodes (cathode and anode) and an
electrolyte.
They are of two types ; based on the nature of conversion of energies
a) Electrolytic cell (Faradaic Cell)
b) Galvanic cell (Voltaic Cell)
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15.7 Electrolytic Cell
An electrolytic cell consists of two electrodes connected to a battery as shown in Fig. 15.4.
Anode Cathode
fused NaCl
Fig. 15.3 : Electrolytic Cell
In an electrolytic cell electrical energy is converted into chemical energy. The process of
decomposition of an electrolyte into its ions when an electric current is passed through it, is
called electrolysis.
When electricity is passed through an electrolyte, a chemical change i.e. decomposition of the
electrolyte into ions takes place at the electrode. Oxidation and reduction reactions occur in
the cell.
In the electrical field Cl- ions migrate to the +ve electrode (anode) and undergo oxidation by
loosing electrons. Na+ ions will go to -ve electrode (cathode) and undergo reduction.
The process can be represented as :
oxidation at
anode Cl- → Cl + e
Cl + Cl → Cl2 (g)
and reduction at cathode, Na+ + e → Na
15.7.1 Faraday’s laws of electrolysis

As a result of extensive studies on electrolysis, Faraday in 1834 has given the relationship
between electric current and chemical changes in the form of Faraday’s laws.
First law :
“The mass of a substance liberated at the electrodes by the passage of electric current are
proportional to the quantity of electricity passed”.
Mathematically this is expressed as,
m α Q or m α ct
m = ect
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Where m is the mass of the substance liberated at the electrode,
Q is the quantities of electricity in coulombs,
c is the strength of the current in amperes and
t is the time in seconds for which the current is passed through electrolyte,
e is the constant of proportionality called electrochemical equivalent of the substance liberated.
Its value is equal to m when c and t are units. Thus,
Electrochemical equivalent (e) :
Electro chemical equivalent is the weight of the substance deposited or liberated when 1
ampere current passed for 1 second during the electrolysis of electrolytic solution containing
that ion.
Ex : A current of 3 ampere strength passing through silver nitrate solution for 20 minutes
deposits 4g. of silver. What is the electrochemical equivalent of silver?
Solution: C = 3 amperes
t = 20 x 60 = 1200 seconds
m = e.c.t
4 = e x 3 x 1200
e = 4 / 3600 = 0.00111 gram / coulomb
Faraday (F)
Faraday is the quantity of electric charge carried by one mole of electrons
(i.e., 6.022 x 1023 electrons). Mathematically,
F = N x e (N = Avagadro Number & e = Charge of electron)
F = 6.022 x 1023 x 1.602 x 10-19 coulombs = 96500 Coulomb (approx).
Now, Faraday’s First law is written mathematically as
ExCxt
m= , E = Chemical equivalent of the substance.
96500
E=
Chemical Equivalent (E)

The mass of the substance deposited or liberated at an electrode during the passage of one
Faraday of electricity in the electrolysis is called chemical equivalent.
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Example:
A current of 10 amperes is passed through molten AlCl3 for 96.5 seconds. Calculate the mass
of Al deposited at the cathode. (At.wt. of Al = 27)
Solution
As per the first law of Faraday,
m=
At.wt. of Al = 27
C = 10 amp. t = 96.5 seconds valency = 3
m= = 0.09 g.
The mass of Al liberated = 0.09 g.

12
F
Faraday’s Second law:
When the same quantity of electricity is passed through different electrolytes (connected in
series) the masses of different species liberated or deposited at the electrodes are directly
proportional to their chemical equivalents.
For example, if the same current is passed through a number of solutions contained in different
Volta meters connected in series, as shown in Fig. 15.4
The masses of hydrogen, copper and silver liberated are proportional respectively to their
chemical equivalents.
Thus,
= =
mCu E mH2 EH
i.e., = Cu similarly, = 2
m Ag E Ag mCu E Cu Fig. 15.4 : Illustration of
Faraday’s Second law
From the second law, we conclude that the quantity of current that would deposit 1.008 g. of
H2 (1 gram – equivalent) from H2SO4 solution, would also deposit one g.eq.wt. each of Cu,
and Ag from the solution of their salts.
Relation between e (electrochemical equivalent) and E (Chemical equivalent weight)
As you are aware that, 1 Faraday (96500 coulombs) is the quantity of electricity consists of
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6.023 x 10 23 electrons which will deposit one chemical equivalent weight of the substance.
Thus 1 F = 6.023 x 10 23 electrons = 1 g. equivalent since 1 F i.e., 96500 coulombs liberate 1
g. equivalent of a substance i.e., E grams.
E
Therefore 1 coulomb liberates grams
96500
E
By definition e =
96500
108
eg. eAg =
96500
Hence, mathematically, the Faraday’s first law becomes

m = ect
Ect
m =
96500
At.wt
m = since E =
valency
Applications of Electrolysis:
i) Electrolysis process is used in the extraction of metals like sodium, potassium, Aluminium
etc.,
ii) It is used in the preparation of non metals like H2, Cl2 etc.
iii) It is used in the preparation of compounds like NaOH, KMnO4 etc.
iv) It is used in the preparation of alloys.
v) It is used in Electro plating.
Example:
Find the number of electrons passing for second through a cross section of copper rise carrying
10-6 amperes of current per one second.
Solution:
Quantity of electricity passed through a copper wire (Q) = c x t = 10-6 coulomb
But 1 Faraday = 96500 coulombs = 6.023 x 1023 electrons.
6.023 x 1023
∴ 1 Coulomb = = 6.249 x 1018
96500
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Hence 10-6 Coulombs = 6.249 x 1018 x 10-6 = 6.249 x 1012 electrons.
15.7.2 Galvanic Cell

Such cells are called voltaic cells. In such a cell chemical energy is converted into electrical
energy. Dry cells, car batteries and button cells used in wrist watches are all examples of
this type of cell. They are energy producing devices.
15.7.3 Redox Reaction and Galvanic Cell

You have already learnt that when electricity is passed through a solution, redox reaction
takes place. Now we shall learn how redox reaction can be used to produce electricity.
When a Zinc rod is dipped in CuSO4 solution, a reaction starts in the solution.
Zn(s) + CuSO 4 (aq) → Zn SO4 (aq) + Cu (s)
Zinc rod
Cuso4 solution
Fig. 15.5 : Redox reaction
It is an example of redox reaction. The two half reactions are
Zn(s) → Zn2+ (aq) + 2e- oxidation

Cu2+ (aq) + 2e- → Cu(s) reduction
In this redox reaction the electrons given by zinc rod have been directly consumed by Cu2+
ion. But, if somehow we make electrons given by Zinc rod to flow through a wire to reach
Cu2+ ions, we shall be producing electric current. To do so, the reaction is carried out in the
electrochemical cell as shown Fig. 15.6 :
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Ammeter
Anode(-) cathode (+)
zinc rod copper rod
Salt Bridge
CuSO4
Fig. 15.6 : Daniell cell having zinc and copper electrodes
The redox reaction in the electro chemical cell has been modified zinc rod is dipped in zinc
sulphate solution in one beaker while copper rod is dipped in another beaker containing CuSO4
solution. The two solutions are connected through a salt bridge and the two metals are
connected to an ammeter with the help of wire. We find electrons move through the wire
from zinc to copper rod.
A metal dipped in tis own salt solution is called as half cell. Zinc rod dipped in-zinc sulphate
solution is oxidation half cell because oxidation takes place.
Zn(s) → Zn2+ (aq) + 2e- oxidation

The released electrons are taken up by zinc rod and it becomes negatively charged.
Copper in copper shulphate is reduction half cell. Copper acts as cathode and reduction
take place here. Copper rod becomes positively charged. Copper gains electrons and in this
process, becomes positively charged.
Cu2+ (aq) + 2e- → Cu(s)

Here the electrons will move from negatively charged electrode to positively charged copper
electrode.
Flow of electrons in the external circuit
The electrons released at the anode during oxidation flow through the external circuit and
reach the cathode where they are taken up for reduction. Thus in a galvanic cell the electrons
always flow from anode to cathode while the conventional positive current flows in the
opposite direction i.e. from cathode to anode. Since the electric current always move in a
closed circuit salt bridge is used to make electrical contact between the two half cells.
15.7.4 Salt Bridge

A salt bridge is a inverted U tube filled with a concentrated solution of an inert electrolyte like
KCl or NH 4NO3 which does not take part in the cell reaction. The electrolyte is taken in the
form of solution and mixed with agar-agar. The mixture is heated and filled in the U tube
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when hot. On cooling it sets into a jelly like mass and does not flow out, during its use. Salt
bridge has two function.
i) It completes the inner circuit. It acts as a contact between the two half cells without any
mixing of electrolytes.
ii) It prevents accumulation of charges in two half cells and maintains electrical neutrality.
Cations amd anions of the salt bridge move into two half cells and neutralise the excess
charge. The anions move into oxidation half cell and neutralise the excess charge. The
cations move into the reduction half cell and neutralise the charge.
In a Daniell cell a salt bridge is replaced by a porous pot, to make the cell more handy to use.
15.7.5 Symbolic Representation of Galvanic Cells

In the previous section the cell was a Zn-Cu cell. But any two suitable metals can be used to
make the cell and everytime we do not always draw the diagrams showing the cell. It is
represented in the symbolic form with the help of standard notation. The rules of notations
are as follows :
1. Anode is written on the left hand side and cathode on the right hand side.
2. The metal electrode in the anode half cell is written by its symbol and this is followed
by the cation (metal ion) along with its concentration in a small bracket. The metal
and cations are separated by vertical line or a semicolon (:)
Zn(s) | Zn 2+ (aq) (1 M)
3. In the reduction half cell the anion along with its concentration is written first, then
vertical line and then the metal
Cu2+ (aq) (1 m) | Cu (s)
4. A salt bridge is represented by two vertical lines
Thus the Galvanic cell described above is written as
Zn(s) | Zn2+ (aq) (1 M) || Cu2+ (aq) (1 M) | Cu (s)
or
Zn | Zn2+ (1M) || Cu2+ (1M) | Cu
15.8 Electrode Potential

Metal atoms have tendency to lose electrons and go into solution as metal ions. Electrode
potential is a measure of the tendency of metal atoms to gain or loose electrons when in
contact with a solution of its own ions.
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When a metal strip M is immersed in a solution of its salt containing ions (Mn+), one of the
processes as shown in Fig. 15.7 (a) or (b) can occur.
Fig. 15.7 : Metal placed in a solution of its ions
i) The dissolution process where atoms of metal electrode M may loose some electrons to
the electrode and enter the solution as Mn+
M → M n+ + ne (metal is oxidised)
The metal electrode gets negative charge and the solution gets extra positive charge.
ii) The deposition process where Metal cations Mn+ from the solution may come in contact
with the metal strip, gain some electrons and get converted into metal atoms M, which
get deposited on the surfance of metal strip. Seperation of charges take place and a
potential is developed called electrode potential.
Mn+ + ne- → M (the ion is reduced)

The electrode reaction reaches an equilibrium as represented below
oxidation
ZZZZZZ
M(S) YZZZZZ
reduction
Z Mn+ (aq) + ne-
X
Electrode potential is the potential developed at the interface between metal and its salt
solution; when a metal is dipped it its own salt solution.
15.8.1 Standard Electrode Potential

An electrode is said to be in the standard state if the concentration of the electrolyte is one
molar and the temperature is 298K. Then its electrode potential is called standard electrode
potential and denoted by EO. If any gas is used to make the electrode then the pressure of the
gas should be 1 bar.
15.9 Measurement of Electrode Potential

It is not possible to measure single electrode potential. It is because the reaction taking
place at the electrodes is oxidation or reduction and these reactions do not take place in
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isolaiton. It can be measured with respect to a reference electrode. The electrode used as
reference electrode is standard hydrogen electrode (SHE).
15.9.1 Standard Hydrogen Electrode

Standard Hydrogen Electrode (SHE) consists of a container, containing 1 M HCl solution
kept at 298 K. A wire containing platinum electrode coated with platinum black is immersed
in the solution. Pure hydrogen gas in bubbled in the solution at 1 bar pressure.
(At 1 bar)
Platinum coated
with Pt black
Fig. 15.8 : Standard Hydrogen electode
The potential of SHE (EO) is taken as zero at all temperatures.

Standard hydogen electrode may act as anode or cathode depending upon the nature of the
other electrode. If its acts an anode, the oxidation reaction talking place is
H2 (g) → 2H+ (aq) + 2e-
If it acts as cathode then the reduction half reaction occuring is
2H + (aq) + 2e- → H2 (g)
15.9.2 Measurement of Standard Electrode Potential

i) Determination of magnitude : The standard electrode potential of an electrode can be
measured by combining it with standard hydrogen electrode. To illustrate, let us take the
example for the measurement of standard eletrode potential of zinc electrode. A zinc strip
is dipped in 1M ZnSO4 solution and it is Standard Hydrogen electrode. The cell emf is
found to be 0.76 V.
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voltmeter
Zinc Rod (1 bar pressure)
Standard
Hydrogen
Electrode
One molar solution One molar solution

+
of Zn2+ ions at 298 K of H at 298 0 K
Fig. 15.9 : Measurement of standard electrode potential of Zn/Zn2+ electrode
When copper electrode i.e. copper dipped in 1 M CuSO4 solution is connected to standard
hydrogen electrode then the cell emf is 0.34.
ii) Sign of electrode potential
The galvanic cell formed by the combination of SHE and electrode under study, the polarity
of the electrode is determined with the help of a voltmeter. In case the given electrode is
found to be positive electrode, its electrode potential is given the positive sign and if it is
negative then it is given the negative sign. In the case of zinc connected to SHE the polarity
is negative but in case of copper it is positive.
15.10 Electrochemical Series and its Applications

15.10.1 Cell emf and Potential difference
The difference in potential of the two electrodes (or half cells) of a galvanic cell, when measured
in the open circuit is called the cell electromotive force or cell emf. When it is measured in a
closed circuit with some external load it is called potential difference.
Cell emf can be measured by using a potentiometer. It depends on the nature of electrodes,
concentration of electrolyte and the temperature.
15.10.2 Standard cell emf

The emf of a cell has a standard value if both its half cells are in their standard states. It is
denoted by EO cell.
15.10.3 Cell emf and electrode potential

The standard cell emf is related to the standard electrode potentials of its anode and cathode.
EO cell = EO cathode - EO anode.
= EO Right - EO left
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Cell emf is related to the electrode potentials of its anode and cathode
E cell
=E cathode
-E anode
=E Right
-E left
15.10.4 Electrochemical Series

Standard potential of a large number of electrodes have been measured and they have been
listed in the increasing order of electrode potential in a series called electro chemical series.
The table 15.2 gives the standard reduction potentials of some electrodes.
Table 15.2 : Standared Electrode Potentials and electrochemical Series
Element Electrode rection EO (V)
Li +e–
kk
Li + Li -3.045
K K+ + e– K -2.925
Cs+ + e–
k
Cs Cs -2.923
–
kk
Ba Ba2+ + 2e Ba -2.906
. Ca Ca2++ 2e– Ca -2.866
Na+ + e–
k
Na Na -2.714
Mg2+ + 2e–
kkk kkk
Mg Mg -2.363
Al Al3+ + 3e– Al -1.662
H2 H2O+2e– H2 + 2OH– -0.829
Zn Zn2+ + 2e– Zn -0.763
Fe Fe2++ 2e– Fe -0.440
Cd Cd2 + 2e– Cd -0.403
– 2–
k
Pb PbSO4 + 2e Pb + SO4 - 0.310

Co2+ + 2e–
kkkk k
Co Co - 0.280
Ni Ni2+ + 2e– Ni -0.250
Sn Sn2+ + 2e– Sn -0.136
Pb Pb2+ + 2e– Pb -0.126
–
Fe Fe3+ + 3e Fe -0.036
2H+ + 2e–
kk
H2 H2 (SHE) 0
Cu Cu2++ e– Cu+ +0.153
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S S4O62– + 2e– 2S2O32– +0.170
k
Cu2+ + 2e–
kkkkkk
Cu Cu +0.337
I2 I2 + 2e– 21– +0.534
– 2–
Fe Fe3+ + e Fe +0.77
Ag Ag+ + e– Ag +0.799
Hg Hg2+ + 2e– Hg +0.854
Br2 Br2 + 2e– 2Br +1.066
O2 O2 + 4H+ + 2e– 2H2O +1.230
k
2- –
+14H+ + 6e 2Cr+3 + 7H2O
k
Cr Cr2O 7
+1.330
Cl2 Cl2 + 2e- 2Cl– +1.359
kk
Au Au3++3e- Au +1.498
Mn MnO4– + 8H+ + 5e– k Mn2+4H2O +1.510
F2 + 2e– 2F–
k
F2 +2.870
The most active metal lithium is placed at the top and the most active non metal fluorine at the
bottom. Thus we find that lithium is the most powerful reducing agent and flourine is most
powerful oxidising agent.
15.10.5 Applications of Electrochemical Series

i) It helps to predict a redox reaction. A given ion will oxidise all the metals below it
and a given metal will reduce ions of any metal placed above it in the series.
Example : Predict the redox rection between zinc and iron. Given EO of Zn2+ / Zn is -
0.763 and EO for Fe2+ / Fe is - 0.44 V.
The EO value of Zn2+ / Zn is lower than Fe2+ / Fe. it means Zn has a greater reducing
power than Fe or Zinc can undergo oxidation more quickly than Fe. Zinc will reduce
Fe2+ ions and iteself undergoes oxidation. The given reaction between Zn and Fe will
take place as shown.
Zn + Fe2+ → Fe + Zn2+
ii) It helps to calculate the emf of a galvanic cell.

EO cell = EO cathode - EO anode.
EO cell should always be positive. If EO cell comes as -ve it means the cell cannot work
and electrodes should be interchanged.
Example : Predict the EO for the cell
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Mg / Mg2+ (1 M) // Ag+ (1M) / Ag
From the table
EO cell = EO cathode - EO anode
EO Mg2+ / Mg = - 2.365 V and
EO Ag+ / Ag = 0.80 V
EO cell = 0.80 - (-2.365) V
= 0.80 + 2.365
= 3.165 V
iii) It helps to predict the feasibility of a redox reaction
The feasibility of a redox reaction can be predicted by calculating EO cell for the redox
reaction. The redox reaction is broken in two half reaction : oxidation half reaction acts as
anode and reduction half acts as cathode. The positive EO cell indicates the redox reaction is
possible.
Example 15.3 : Predict whether the following reaction is feasible or not ?

Cu2+ (aq) + 2Ag (s) → Cu (s) + 2Ag+ (aq)
Given EO Ag + / Ag = 0.80 V and EO Cu2+ / Cu = 0.34V
The given redox reaction can be written as two half reactions
Anode (oxidation) 2 Ag (s) → 2Ag+ (aq) + 2e-
Cathode (Reduction) Cu2+ (aq) + 2e- → Cu (s)
EO cell = EO cathode - EO anode
= EO Cu2+ / Cu - EO Ag + / Ag
= 0.34 V - 0.80 V
= - 0.46V
The -ve EO value indicates that the above reaction will never take place and silver cannot
displace Copper from a solution of Cu2+ ion. Instead the reverse reaction would be feasible.
iv) It helps to predict whether a metal can liberate hydrogen from acids. Any metal
which is above hydrogen in the electro chemical series can liberate hydrogen from
acid since it is a better reducing agent than hydrogen. Thus metals like, Zinc,
Magnesium, Calcium etc can displace hydrogen from HCl or H2SO4 but metals like
Copper, silver etc cannot displace hydrogen from acid.
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15.11 Nernst Equation for Electrode Potential
Nernst equation relates electrode potential to the concentration of ions in the electrolyte. An
electrode reaction is always written as reduction reaction. Let us take a general example for a
metal M.
Mn+ (aq) + ne → M (s)
The Nernst equation for its electrode potentials is :
2.303 RT [M]
E = EO - log (i)
nF [M n+ ]
where E = Electrode protential

EO = standard electrode Potential (Reduction)
R = gas constant in JK-1 mol-1
T = Temperature in Kelvin
F = Faraday constant
n = number of electrons involved in the electrode reaction
[Mn+] = molar concentration of Mn+ ion
[M] = concentration of pure solid metal taken as unity
2.303 RT 1
Therefore, E = EO - log
nF [M n+ ]
If we put the values of R, T and F in equation (i)

R = 8.314 J K-1 mol-1 F = 96500 Coulomb
T = 298 K
0.0591 1
we have E = E0 - log (ii)
n [M n+ ]
For example : For copper electrode as half cell

Cu2+ (aq) + 2e → Cu (s)
here n = 2. EO = + 0.34 V
Cu2+/Cu
0.0591 1
EO = EO - log
Cu2+/Cu Cu2+/Cu 2 C2+2+]
[Cu
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Since EO is 0.34 V the equation becomes
Cu2+/Cu
1
= 0.34 - 0.295 log
C2+2+]
[Cu
Example 15.4 : Calculate the reduction potential for the following half cell at 298 K
Ag + (0.1 M) + e → Ag (s)
E0 = 0.80 V
0.0591 1
E = E0 - log
n [Ag + ]
0.0591 1
= 0.80 - log
1 0.1
= 0.80 - 0.0591 log 10

= 0.80 - 0.0591 = 0.741 V
15.11.1 Nernst Equation for Cell emf

For a general cell reaction :
aA + bB → xX + yY
_ 2.303 RT [X]x [Y]y

0
E cell = E cell log
nF [A]a [B]b
Thus, for the cell reaction :

Ni (s) + 2Ag+ (aq) → Ni2+ (ag) + 2Ag
_ 2.303 RT [ Ni 2+ ]
0
E cell = E cell log
nF [Ag + ]2
Note : Only the ions are written in the fraction as concentration of pure solid or liquid is taken
us unity.
To dertermine the value of n the reaction is written in two half reactions.
Anode reaction (oxidation)
Ni(s) → Ni2+(aq) + 2e
Cathode reaction (reduction)
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2Ag+(aq) + 2e- → 2Ag (s)
This can be represented in the form of a cell as
Ni | Ni 2+ || Ag + | Ag
The value of n = 2 as 2e- are exchange between anode and cathode
Example 15.5 : Calculate the EMF of the following cell at 298 K

Ni (s) | Ni 2+ (0.001M) || Ag + (0.1 M) | Ag (s)
EO = 0.25 V EO = 0.80v
Ni2+ | Ni Ag+ | Ag
From the given values first find

E0 cell = E0 cathode - E0 anode
= 0.80 - (-0.25) V
= 1.05 V
_ 0.0591 [ Ni 2+ ]
E cell = E cell 0 log
2 [Ag + ]2
0.0591 0.001
= 1.05 _ log
2 (0.1) 2
10 −3
= 1.05 - 0.0295 log
10 −2
= 1.05 + 0.0295 log 10 -1

= 1.0795 V
15.12 Cell EMF and Gibbs Energy

The maximum amount of work that a cell in its standard state can perform is given by
Wmax = -n FE0
The significance of -ve sign is that the work is done by the cell. Since the maximum amount
of useful work which a system can perform is equal to decrease in Gibbs energy
thus
Wmax = ΔGO = -nFE0
If ΔGO calculated is negative, then the cell reaction is spontaneous otherwise not.
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Example 15.6 : Calculate the standard Gibb’s energy ΔGO for the reaction occurring in Daniell
cell
Zn (s) + Cu 2+ (aq) → Zn2+ (aq) + Cu (s)
at 298 K. The E0 cell at this temperature is 1.1V. Is the reaction spontaneous.
For Daniell cell, n = 2.
Solution : ΔG O = - nFE0
ΔG O = - 2 x 96500 x 1.1
= - 212, 300 J
= - 212.3 kJ
Since ΔEO is -ve, the cell reaction is spontaneous

1. Differentiate between electrolytic cell and galvanic cell.
...................................................................................................................................
2. What is a salt bridge ? What is the role of the salt bridge ?
...................................................................................................................................
3. What is electrochemical series ? List any two applications of the series.
...................................................................................................................................
4. With reference to the electrochemical series arrange the following metals in the orderin
which they displace each other from their salt solution.
Al, Cu, Ge, Mg, Zn, Ag
...................................................................................................................................
5. What is the charge in coulombs on the N3- (Nitride ion)?
...................................................................................................................................
6. Calculate the number of coulombs required to deposit 40.5 g. of Al when the electrode
reaction is Al3+ + 3e → Al.
...................................................................................................................................
7. Differentiate between the electro chemical equivalent and chemical equivalent weight.
...................................................................................................................................
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• Oxidation is a process in which electrons are lost by an atom or ion.
• Reduction is a process in which electrons are gained by an atom or ion.
• Atoms or ions undergoing oxidation are reducing agents and atoms or ions undergoing
reduction are oxidising agents.
• Oxidation number is the state of oxidation of an element in a compound, which is
calculated by a set of rules. It is based on the concept that electrons in a covalent bond
belong to the more electro negative element.
• Oxidation number of an atom in elemental form is zero. Otherwise the number is
always written with positive or negative sign.
• A substance, if in its molten state or aqueous solution conducts current is called electrolyte
and if it doesnot conduct electric current then it is called non-electrolyte.
• Ions carry charge in an electrolyte from one point to other. The conduction of electricity
follows Ohm’s law.
• Reciprocal of resistance and resistivity are called conductance and conductivity
respectively.
• On dilution of a solution, its conductivity decreases while molar conductivity increases.
• Electrolysis is a process in which electrical energy is used to decompose an electrolyte
into its ions and it is done in an electrolytic cell.
• Electrochemical cell or Galvanic cell produce electricity due to oxidation and reduction
reactions occurring in their half cells. Oxidation occurs at the anode (negative electrode)
and reduction at the cathode (positive electrode).
• A galvanic cell can be written in symbolic form as Anode | Electrolyte | | Electrolyte |
Cathode.
• The emf of a cell is the potential difference between two electrodes in the open circuit.
• When a metal is dipped in its own salt solution then the potential of metal with respect
to solution is called electrode potential. This potential is measured with respect to a
reference electrode called Standard Hydrogen electrode.
• Electrochemical series is the arrangement of electrodes in the order of their increasing
electrode potential.
• The cell emf is related to the electrode potential (reduction)
Ecell = Ecathode - Eanode
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• The Nernst equation is
2.303 RT [Re d ]
E = EO _ log
nF [Oxi]
• The standard Gibbs energy of the cell reaction ΔGO is related to the standard cell emf as
ΔGO = - nFE0.
Terminal Exercise
1. Calculate the Oxidation number of the elements written in bold letters
[Cr(H2O)6]3+, [Fe(CN6]3-, HCO-3, Pb 3O4
2. Balance the following reactions by oxidation number method
a) Fe2O3 + C → Fe + CO
b) C6H8 + O 2 → CO2 + H2O
3. Balance the following reaction by ion - electron method.
i) Zn + HNO3 → Zn (NO3)2 + NO2 + H2O
ii) ClO3- + Mn 2+ → MnO2 + Cl- in acidic medium
iii) Fe(OH)2 + H2O2 → Fe(OH) 3 + H2O in basic medium
4. Define the following and give its units :

i) Conductivity
ii) Molar Conductivity
5. Draw a graph showing the variation in molar conductiviy in a weak and a strong
electrolytes with concentration.
6. Explain why the molar conductivity of an electrolyte increases with dilution while
conductivity decreases on dilution.
7. The measured resistance of a conductance cell containing 7.5 x 10-3 M solution of KCL
at 250C was 1005 ohms. Calculate (a) conductivity (b) Molar conductivity of the solution.
Give the cell constant 1.25 cm-1.
8. The conductivity of 0.5M solution of an electrolyte at 298K is 0.0025 cm-1. Calculate
the molar conductivity.
9. Explain the term standard electrode potential. How is it determined experimentally.
10. Draw the diagram of the cell whose cell reaction is
Zn (s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag (s)
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11. For the cell
Mg | Mg 2+ || Zn2+ | Zn
i) Draw the cell diagram
ii) Identify anode and cathode
iii) Write cell reaction
iv) Write Nerst equation for the cell
v) Calculate EO use the data given the table 15.2
12. What are the functions of a salt bridge ?
13. Using electro chemical series predict whether the following reaction is feasible or not.
Ni (s) + Cu2+ (aq) → Cu (s) + Ni2+ (aq)
14. Explain withthe help of electro-chemical series whether any reaction will take place
when
i) Steam is passed over hot Cu.
ii) Tin is dipped in hydrochloric acid
15. Calculate ΔGO for the reaction
2Al(s) + 3Sn2+ (aq) → 2Al3+ (aq) + 2Sn2+ (aq)
16. Calculate emf of the cell
Cr | Cr3+ (.1M) || Fe 2+ (0.1M) | Fe
17. Calculate emf of the given cell reaction at 298 K
Sn4+ (1.5 M) + Zn (s) → Sn2+ (0.5 M) + Zn2+ (2 M)
18. The blue colour of CuSO4 is discharged when a rod of zinc is dipped in it ? Expalin.
19. Why oxidation cannot occur without reduction.
20. Knowing that
Cu2+ + 2e → Cu ; EO = .34 V
2Ag+ + 2e → 2Ag ; EO = +.80 V
reason out whether 1M silver nitrate solution can be stored in copper vessel or 1 M
copper sulphate can be stored in silver vessel.
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15.1
1. -4, -3, +3, +2, +7, +5, 0, +5
2. In oxidation O.N. increases
In reduction O.N. decreases
3. Oxidising agent HNO3
reducing agen H2S
3. I-2 (aq) → I2 (s)
i) oxidation
Fe3++3 (aq) → Fe2++2 (aq) (reduction)
reduction
ii) Mg + C12 → Mg Cl2

+2 -1
oxidation
iii) I2 → HIO3 oxidation
HNO 3 → NO2 reduction
oxidation
5. 3CuO + 2NH3 → 3Cu + N2 + 3H2O

+2 -3 0 0
reduction
oxidation
ii) MnO2 + 4HCl-1 → MnClO2 + Cl2 + 2H2O

+4 +2
reduction
6. Bi → Bi3+ + 3e-
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i) NO3 + 2H+ + e- → NO2 + H2O ] x 3
Bi + 3NO3- + 6H + → Bi3+ + 3NO2 + 3H2O
ii) MnO4 + 8H+ + 5e- → Mn2+ + 4H2O
Fe2+ → Fe3+ + e- ] x 5
MnO 4 + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Cr2O7 + 14H+ + 6e → 2Cr3+ + 7H2O

2-
iii)
Fe2+ → Fe3+ + e- ] x 6
Cr2O2-7 + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
iv) Al + 4 OH- → NH3 + 9 OH- ] x 8
NO-3 + 6H2O + 8e- → NH3 + 9OH-] x 3
8Al + 3NO-3 + 18H2O + 5OH- → 8Al(OH) -4 + NH3
15.2
1. Electrolytes give ions in the solution and these ions conducted electricity by moving to
their respective electrodes i.e. cations towards cathode and anions towards anode.
2. Specific conductance is the condutance of a solution taken in a cell in which two electrodes
are 1 cm apart and surface area of each one of them is 1cm2.
Equivalent conductance is the conductance of all the ions furnished by an equivalent of
the electrolyte in a solution of given concentration.
3. Conductance S; specific conductance Scm-1.
4. Nature of electrolyte (strong or weak), valency of its ions, speed of ions, concentration
and temperature.
5. See Fig. 15.2
6. λ∝ Al2 (SO4)3 = 2 λ∝ Al3+ + 3 λ∝ SO42-
m m m
15.3
1. In electrolytic cell electrical energy is used for carrying out redox reaction while in an
electrochemical cell a redox reaction is used for producing electrical energy.
2. See sec. 15.6.2
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3. See sec. 15.8
4. Mg > Al > Zn > Fe > Cu > Ag.
5. Charge of 1 electron = 1.6 x 10-19 Coulomb.
Charge on the N3- ion in Coulombs = 3 x 1.6 x 10-19 = = 4.8 x 10 -19
6. Al3+ + 3e → Al Shows that 3 Faraday liberate 1 mole Al
40.5
40.5g. Al = mole Al
27
3 x 40.5 3 x 40.5
Therefore F needed = F = x 96500 coulombs.
27 27
The no. of coulombs required to deposit 40.5 g. of Al is 4.34 x 105 C.

7. The electrochemical equivalent (e) is the mass of the substance liberated or deposited
during the passage of 1 coulomb of electricity through the electrolyte.
The chemical equivalent weight (E) is the mass of the substance deposited or liberated
at an electrode during the passage of one faraday of electricity in the electrolysis.
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16
CHEMICAL KINETICS
You know that a knowledge of Gibbs energy change in a given process can tell you whether
the process is feasible or not. However, the Gibbs energy decrease does not indicate anything
about the speed of a reaction. For example, the reaction between hydrogen and oxygen to
give water is accompanied by a large decrease in Gibbs energy. However, you would not be
able to observe the formation of water, even if hydrogen and oxygen are kept in contact with
each other for even. On the other hand, there are reactions which take place instantaneously.
For example, when HCl is added to AgNO 3 , white precipitate of Agcl is formed
instantaneously. In this lesson we shall study the speed or rate of reactions which are neither
very slow nor very fast. We shall also study the factors that control the rate of a reaction.
Objectives
• explain the rate of a chemical reaction;
• differentiate between average rate and instantaneous rate;
• correlate the average and instantaneous rates with changes in concentrations of various
reactants and products;
• explain the factors that affect the rate of a reaction;
• define rate law and rate constant;
• define order and molecularity of a reaction;
• distinguish the order from molecularity;
• derive rate law for first order reaction and define half life period for first order reaction;
• derive a relationship between half life and rate constant for first order reaction;
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• solve numericals on the relationships;
• explain the effect of temperature on reaction rate, and
• explain Arrhenius equation and activation energy.
16.1 Rate of Chemical Reaction

When the reactants are mixed to perform a reaction, there are no products present initially. As
time passes, the concentration of the products increases and that of the reactants decreases.
The rate of any chemical reaction can be expressed as the rate of the change in concentration
of a reactant (or a product).
Rate of a chemical reactio = of a reactant or a product
Let us consider the following chemical reaction :

2NO (g) + Br2 (g) → 2NOBr (g)
The rate for this reaction can be determined by measuring the increase in the molar
concentration of NOBr at different time intervals.
Let us see how can express the rate of this reaction. You know that molar concentration of a
substance is expressed by enclosing the formula of the substance is square bracket.
For example [NOBr] represents the molar concentration of NOBr.
Let us suppose that [NOBr]1 is the molar concentration at time t1 and [NOBr]2 is the molar
concentration at time t2.
Then, change in molar concentration = [NOBr]2 - [NOBr] 1 = Δ[NOBr]
Time required for the change = t2 - t1 = Δt
here, Δ means change in the respective quantity.
Δ[ NOBr ]
Therefore, the rate of formation of NOBr =
Δt
This expression gives the rate of reaction in terms of NOBr.

If the decrease in the molar concentration of NO or Br2 is measured we can write the rate of
the reaction with respect to NO as
Δ[ NO]
=
Δt
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Δ[Br 2 ]
and w.r.t. Br2 as =
Δt
Thus, the rate of a reaction can be expressed either in terms of reactants or products. We find
in the reaction mentioned above that two moles of NO react with one mole of Br2. Therefore,
the change in concentration of NO in a given time Δt will be double than that for Br2. Thus,
in order to make the rates with respect to different reactants or products equal, the rate
expression is divided by the storichrometric coefficient in the balanced chemical equation.
For example, in the equation,
2NO (g) + Br2 (g) → 2NOBr (g)
The rate of reaction with respect to reactants and products is expressed as
1 Δ[ NOBr ] 1 Δ[ NOBr ] 1 Δ[Br 2 ]

rate of reaction = + = - ==
2 Δt 2 Δt 2 Δtt
16.2 Average Rate and Instantaneous Rate

The rate of a reaction depends on the concentration of reactants. As the reaction proceeds the
reactants get consumed and their concentration decreases with time. Therefore, the rate of
reaction does not remain constant during the entire reaction.
Δ [concentration ]
The rate of reaction given as gives an average rate.
Δt
Δ[ NOBr ]
For example, gives the average rate of reaction. Instantaneous rate of a reaction is
Δt
the rate of reaction at any particular instant of time, we express instantaneous rate by making
Δt very small φ
[ NOBr ] d[ NOBr ]
Δ lim
t →0 =
Δt dt
When concentration of any of the reactants or products is plotted against time, the graph
obtained is as given below :
Product
Concentration
Reactant
Time →
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For the reaction,
2N 2O5 (g) → 2NO2 (g) + O2 (g)
Average rate of reaction
1 Δ[ N 2O5 ] 1 Δ[ NO 2 ] Δ[O 2 ]
= − = =
2 Δt 2 Δt Δt
1 d[ N 2O5 ] 1 d[ NO 2 ] d[O 2 ]
and instantaneous rate = − = =
2 dt 2 dt dt
16.3 Factors Affecting rate of a Reaction

The rate of chemical reaction is affected by the following factors :
1. The concentration of reactants : Generally the rate of a reaction increases as the
concentration of the reactants is increased.
2. Temperature - A reaction is faster when the reaction temperature is increased.
3. Presence of a catalyst - A catalyst alters the reaction rate without being consumed by
the reaction.
For example :
i) Reaction between hydrogen and oxygen to form water is extremely slow. It occurs
very fast in the presence of platinum as catalyst.
You would study these effects in more details in the following sections of this lesson.
Example 16.1 : Express the average and instantaneous rate of the following reaction
N2 (g) + 3H2 (g) → 2NH3 (g)
in terms of i) rate of formation of NH3,

ii) rate of disappearance N2, and
iii) rate of disappearance of H2.
Write the different expressions for the rate of reactions.
Solution : The expression for the three rates are :
Δ[ NH3 ]
Average rate of formation of NH3 =
Δt
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Δ[ N 2 ]
Average rate of disappearance of N2 =
Δt
Δ[H 2 ]
Average rate of disappearance of H2 =
Δt
Δ[ NH3 ]
Instantaneous rate of formation of NH3 =
dt
Δ[ N 2 ]
Instantaneous rate of disappearance of N2 =
dt
Δ[H 2 ]
Instantaneous rate of disappearance of N2 =
dt
To equate the three rates, divide each rate by the coefficient of the corresponding substances
in the balanced equation.
1 Δ[ NH3 ] Δ[ N 2 ] 1 Δ[H 2 ]
Average rate = + = − =−
2 Δt Δt 3 Δt
1 d[ NH3 ] d[ N 2 ] 1 d[H 2 ]
Instantaneous rate = = − =−
2 dt dt 3 dt

1. Which of the following units could be used to express the rate of reaction between
magnesium and hydrochloric acid ?
a) cm-3 s
b) cm3 min-1
c) cm3 s-1
d) mol dm-3 min-1
.................................................................................................................................................................................................................................

2NO2 (g) + F2 (g) → 2NO2F (g)
Write the expression for the average, rate of reaction in terms of :

a) rate of formation of NO2F
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b) rate of disappearance of NO2
c) rate of disappearance of F2
d) equivalent rate of formation of product and disappearance of reactants.
.................................................................................................................................................................................................................................
3. Express the instantaneous rates of formation of product and disappearance of reactants

for the above reaction.
.................................................................................................................................................................................................................................
4. Explain why the reaction CO (g) + NO2 (g) → CO2 (g) + NO (g) occurs more
slowly when the volume of the system is increased.
.................................................................................................................................................................................................................................
16.4 Dependence of Reaction Rate upon Concentration

Rate law : If we follow a chemical reaction over a period of time, we find that its rate
slowly decreases as the reactants are consumed. We can say that the rate is related to the
concentration of the reactants. Rate is directly proportional to the concentration of the
reactants raised to some power.
For example, consider the reaction
xA (g) + yB (g) → Products ....(i)
where a and b are coefficients of A and B respectively in the balanced equation. We can
express the rate as
Rate α [A]x [B]y
or Rate = k [A]x [B]y
where k is the constant of proportionality.
The above equation is known as the rate law for the reaction.
Rate law is defined as the mathematical relationship between rate of a reaction and
the concentration of the reactants.
It is important to note that x and y are not necessarily equal to the stoichiometric coefficient of
A and B in the reaction. The constant k in the rate law is called the rate constant. It is
numerically equal to the rate of the reaction if all the concentrations were set equal to unity.
Rate = k [1]x [1]y
∴ Rate = k
It means that large values of k indicate fast reaction and small values of k indicate slow
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reactions. Each reaction is characterised by its own rate constant, whose value is dependent
on the temperature, but independent of concentration of the reactants.
16.5 Order of a Reaction

The powers to which the concentration terms are raised in the rate expression describes the
order of the reaction with respect to that particular reactant
In the rate law,
rate = k [A]x [B]y
the values of x and y are order of the reaction with respect to the reactants A and B respectively.
The sum of the powers x + y represents the overall order of the reaction.
For example in the reaction 2NO (g) + O2 (g) → 2NO2 (g)
The rate law as found by experiment is
rate = k [NO] 2 [O2]1
Here the order of reaction with respect to NO is 2 and with respect to O2 is 1. The sum of the
powers is termed as overall order of reaction. In the above example it is 2 + 1 = 3.
It must be remembered that oder of a reaction must be determined experimentally and cannot
be deduced from the coefficients in the balanced equation
2N2O5 (g) → 4NO2 (g) + O2 (g)
The overall order is 1 and it is termed as first order reaction.
The order of reaction can be 0, 1, 2, 3 called as zero order, first order, second order and third
order respectively. The order of a reaction can be a fraction as well; for example the
decomposition of ethanal to methane and carbon monoxide.
723K
CH3 CHO (g ) ⎯⎯⎯→ CH 4 (g ) + CO (g )
is experimentally found to the following rate law :
rate = k [CH3CHO]3/2
16.5.1 Difference between Rate of Reaction and Rate Constant

Rate of reaction Rate constant
1. It is the speed at which the reactant It is the proportionality constant in the
are converted into products. It is rate law and is defined as the rate of
measured as the rate of decrease of the reaction when the concentration
concentration of reactant or rate of of each of the reactant is unity.
increase in the concentration of the
products.
2. It depends upon the initial concentration It is independent of the initial
of the reactant. concentration of reactant.
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16.6 Units of Rate Constant
For zero order
Rate = k [A]O
Rate = k
The unit for rate of reaction is mol L-1 s-1. The unit of k for zero order is same as that for rate.
For a first order reaction
Rate = k [A]1
mol L−1
= k mol L-1
s
k = s-1
For a first order reaction, the unit for rate constant is time-1.
For second order reaction
Rate = k [A]2
mol L−1
= k (mol L-1)2
s
Therefore k = mol -1 L s-1

In general for any order n the unit for rate constant k is given as
k = (mol L-1 ) 1-n s-1
16.6.1 Zero order Reaction

Zero order reactions are those, in which n = 0. Here rate is independent of reactant
concentration. Such reactions are uncommon. One such example is the decomposition of
ammonia on a platinum or tungsten metal surface. Under high pressure of ammonia the rate at
which ammonia decomposes is always the same regardless of its concentration.
16.6.2 First Order Reaction

We will now discuss how to determine the rate constant for a first order reaction. For the first
order reaction, this equation would tell us how does the concentration vary with time. The
predicted variations can then be compared with the experimental data to obtain the order of
the reaction.
Let us consider for the reaction A → Product.
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For first order reaction.
−d[A]
rate = = k1[A]
dt
where k1, is the rate constant,

Rearranging the rate expression, we have :
−d[A]
= k1dt
[A]
Integrating the two sides we get :

-1n [A] = k1t + constant ....(i)
where the constant can be determined from the initial conditions.
Let [A]O be concentration of [A] when t = 0 i.e. at the beginning of the reaction, then
- ln [A] = - ln [A]O, when t = 0.
or constant = - 1n [A]O
Putting the value of the constant in equation
- 1n [A], k1t - 1n [A]O
or - 1n [A] + 1n [A]O = k1t
1n [A]O
or = k 1t
[A]
1 [A]O
Rate constant, k1 = . in
t t
We can convert it in to log to the base 10
2.303 [A]
k1 = log
t [A]
k1t
= log [A]O - log [A]
2.303
k1t
- = log [A] - log [A]O
2.303
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⎛ k1t ⎞
log [A] = - ⎜⎝ ⎟ t + log [A]O
2.303 ⎠
The unit of k1 = time-1

This equation has the same form as the equation of a straight line.
y = mx + c
where m is the slope and c is the intercept. If we plot log10 [A] vs t, it would give a straight
k1t
line with a slope equal to - . The rate constant, k1 can then be calculated from the
2.303
slope as shown in Fig. 16.1
log10 (concentration)
k1t
Slope = _
2.303
Slope
Time
Fig. 16.1 : A graph of log10 (concentration) against time for a first-order reaction
16.6.3 Half-Life Period

The time taken for a reaction to reach the half-way stage i.e. when half of the starting material
has reacted is called half-life period of the reaction. It is denoted by t1/2 or t 0.5. Let us now
see how the half-life period of a first order reaction is calculated.
You know that
1n [A] = 1n [A] O - k1t
when [A] = 1/2 [A]O (reactant reduces to half of the initial concentration)
then t = t1/2 (half life period)
Therefore, we have
⎧⎪ [A]0 ⎫⎪
1n ⎨ 2 ⎬ = 1n [A]O - K1t1/2
⎪⎩ ⎪⎭
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⎧⎪ [A]0 ⎫⎪
or 1n ⎨ 2 ⎬ - 1n [A]O = - k1t1/2
⎪⎩ ⎪⎭
or 1n 1/2 = - k1t1/2
1n 2 0.693
or t1/2 = k = k
1 1
You would notice that half-life period is independent of the initial concentration.
Example 16.2 : The decomposition of hydrogen peroxide to water and oxygen.

2H 2O2 (l) → 2H2O (l) + O2 (g)
Is a first oder reaction with a rate constant of 0.0410 min-1. If we start with 0.20 M solution of
H2O2, what I will be concentration after 10 minutes?
Solution : We have equation for first order reaction as
1 [A]O
k= 1n
t [A]
converting it into log10 base
2.303 [A]0
k= log
t [A]
If [A] = x, after 10 minutes then substituting the values of k and t, we have
2.303 0.20
0.0410 (min-1) = log
10(min) x
0.20 10(min) (0.0410 min −1 )

or log = = 0.178
x 2.303
Taking antilog
0.20
= antilog 0.178 = 1.51
x
0.20
Solving x= = 0.132 mol litre-1.
1.51
Example 16.3 : In example 16.2 if we start initial concentration of 0.50 how long will it take
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for this concentration to drop to 0.10 M.
Solution : We have to find the time taken by the reaction starting from concentration of H2O2
of 0.5 M to concentration of 0.1 M.
We have the equation,
2.303 [A]0
k= log
t [A]
Substituting the values of k, [A]O and [A] we have
2.303 0.50
0.0410 min-1 = log
t 0.10
2.303 0.699 x 2.303

t = log 5 x −1 = min
0.0410 min 0.041
t = 39 minutes
Example 16.4 : In example 16.2, how long wil it take for the sample to decompose to 50%
13
F Solution : When half the sample has decomposed, we have
0.693
t1/2 =
k
Putting the value of k = 0.014 min-1 we have
0.693
t= min = 16.9 minutes
0.0414

1. The rate of a certain reaction, A → Product is 1.0 x 10-3 mol / litre.
When conc. A = 0.020 M, what is rate constant K, if the reaction is :
a) zero order
...................................................................................................................................
b) first order
...................................................................................................................................
C2H4 (g) + I2 (g) → C2H4I2 (g)
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the rate equation is
rate = K [C2H4(g)] [I 2 (g}]3/2
a) What is the order of reaction with respect to each reactant ?
...................................................................................................................................
b) What is the overall order of reaction ?
...................................................................................................................................
c) What is the unit of k, if concentrations are measured in mol dm-3 ?
...................................................................................................................................
3. The first order rate constant for the decomposition of C2H5Cl at 700K is 2.5 x 10-3 min-
1
. If the initial concentration is 0.01 mol L-1, calculate the time required for the
concentration of C2H5Cl to drop to half of its original value.
...................................................................................................................................
16.7 Dependence of Reaction Rate on Temperature

You have learnt earlier that an increase in temperarure causes an increase in the rate of reaction.
The rate is about doubled, for many reactions by a 10 degree rise in temperature. How can we
explain this behaviour ?
In order for a chemical reaction to occur, the reacting molecules must collide with each other.
Only fast moving molecules i.e. the molecules having energy are able to react. There must be
some minimum energy possessed by the reacting molecules during the collision. This is caleld
the threshold energy. All molecules having energy higher than the threshold energy are
therefore, capable of reacting. What would happen if we increase the number of molecules
having higher energy? More molecules will react. In other words rate of reaction would increase.
Thus, if we increase the temperature, we increase the rate of reaction. Let us see if we can
express it quantitatively.
energy of the Activated complex
Ea
1
Ea
A+B
Energy
Reactants
A+B
ABC
AB Products
Progress of reaction
Fig. 16.2 : Energy diagram for a reaction
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The change in energy that takes during the course of a reaction is shown in Fig. 16.2. The
horizontal axis denotes the progress of the reaction and it indicates the extent to which the
reaction has progressed towards completion. The graph indicates that the reactant molecules
A and B must posses enough energy. This is known as Activation energy, to climb the potential
energy hill to form the activated complex. All the reacting molecules do not posses energy
equal to threshold energy in their normal energy state. Therefore a cetain extra energy needs
to be supplied so that their energy equals the threshold energy. The potential energy of activated
complex is maximum. The reaction thereafter proceeds to completion (i.e. formation AB). Ea
indicates the activation energy of forward reaction and Ea’ is the activation energy of the
backward reaction.
You know that rate of reaction = k [concentration]. If we have value of concentration as
unity, then rate of reaction is equal to the rate constant, k. The rate constant k depends on the
magnitude of the activation energy, Ea, and also on the absolute temperature (T), k is small
when Ea is very large or the temperature is low.
We can express this relationship by a mathematical equation known as Arrehenius equation
after its discoverer, S. Arrhenius as follows :
k = Ae-Ea/RT
Where A is a proportionality constant and is also known as frequency factor, R is the gas
constant. How can we utilise this relationship between k, Ea and T? We can calculate
activation energy, if we measure the rate constant at two different temperatures.
Taking natural logarithm of equation, we get
Ea
1n k = 1n A -
T RT
We can compare this equatio with that of a straight line y = mx + c as follow ;
_E
1
1n k = a
R T ( )+1nA
The plot of in k vs 1/T gives a straight line whose slope is equal to -Ea/R and whose intercept
is In A (in Fig. 16.3).
k
− Ea
Slope =
R
1n k
k
1
T
Fig. 16.3 : Graphical determination of Ea.
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We can also obtain Ea from k at two temperatures by direct computation.
At T1, the equation becomes
k1 = Ae-Ea/RT1
At T2, we can write
k2 = Ae-Ea/RT2
k1 Ae − Ea / RT1
Dividing k1 by k2 we get =
k2 Ae − Ea / RT2
k1 − E a ⎛ T2 − T1 ⎞
ln =
taking natural logarithm k2 R ⎜⎝ T1 − T2 ⎟⎠
k1 −Ea ⎛ T2 − T1 ⎞
log =
Converting into logarithm (base 10) k2 2.303R ⎜⎝ T1 − T2 ⎟⎠
Multiply by -1 on both sides and inverting the fraction
k2 Ea ⎛ T2 − T1 ⎞
log =
k1 2.303R ⎜⎝ T1 − T2 ⎟⎠
This equation can also be used to calculate the rate constant at some specific temperature if Ea
and k at some other temperature are known.
Example 16.5 : What must be the value of Ea if the rate constant for a reaction is doubled
when the temperature increases from 300K to 310K?
k2
Solution : Given k = 2, R = 8.31 JK-1
1
T2 = 310 K
T1 = 300 K
Ea = ?
k2 Ea ⎛ T2 − T1 ⎞
log =
We have equation : k1 2.303R ⎜⎝ T1 − T2 ⎟⎠
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Ea ⎛ T2 − T1 ⎞
substituting the values we have log 2 = −1 ⎜⎝ T − T ⎟⎠
2.303 x 8.31 JK 1 2
solving, we have Ea = 53.5 kJ

1. As a rough rule of thumb, the rates of many reactions double for every ? 100C rise in
temperature.
...................................................................................................................................
2. The rate constant of a reaction at 288 K is 1.3 x 10-5 litre/mol. While at 323K its rate
constant is 8.0 x 10-3 lit./mol. What is the Ea for the reaction.
...................................................................................................................................
3. The rate of the reaction triples when temperature changes from 293 K to 323 K.
Calculate the energy of activation for such a reaction.
...................................................................................................................................
4. H2 (g) and O2 (g) combine to form H2O (l) by an exothermic reaction. Why do they not
combine when kept together for any period of time.
...................................................................................................................................

• The rate of a chemical reaction can be expressed as the rate of change in concentration
of a reactant or product.
• The factors that affect the rate of a chemical reaction are : the concentration of the
reactants, temperature and catalyst.
• Rate law is defined as the mathematical relationship between rate of a reaction with
the concentration of the reactant.
• The constant k in the rate law is called the rate constant. It is numerically equal to the
rate of the reaction if all the concentrations were set equal to unity.
• The sum of the powers to which the concentration terms are raised in the rate expression
describe the order of the reaction.
• The rate constant of a first order reaction, k1, is given by
1 [A]0
k1 = In
t [A]
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• The time taken for a reaction to reach the half way stage, i.e., when half of the starting
material has reacted is called half-life period of a reaction.
• The dependence of the rate constant on the magnitude of the activation energy, Ea, and
the absolute temperature, T, is given by the Arrhenius equation, k = Ae-Ea/RT.
• The activation energy for the decomposition of a compound corresponds to the difference
between the energy of the reactants and the maximum on the energy curve.
Terminal Exercise
1. Sketch an energy vs progress of reaction diagram for a reaction. Indicate the activation
energy for both forward (Ea) and reverse reaction (E’a).
2. For the reaction 2N2O5 (g) → 4NO2 (g) + O2 (g), it is found that N2O5 is decomposing,
Δ[ NO 2 ]
at the rate of 0.02 mol/litre sec. Calculate to rate of reaction defined as with Δt
Δt
in seconds.
3. The rate constant for a certain first order decomposition reaction is 0.23 s-1 at 673 K.
Calculate the half-change period for this reaction.
4. The rate constant for a certain first order reaction is 1.00 x 10-3 s -1 at 298K. The rate
constant for the reaction is 1.4 x 10-2 s-1 at 323K. Calculate the activation energy for
this reaction.
5. The rate of a reaction was found to be 3.0 x 10-4 mol/litre second. What will be the rate
if it were expressed in the units mol/litre minute.
6. List the three factors that affect the rate of a chemical reaction.
7. For a certain first order reaction the concentration of the reactant decreases from 200
mol/litre to 1.50 mol/litre in 64 minutes. Calculate the rate constant for this reaciton.
8. The rate constant at for a certain first order reaction is 1.0 x 10-3 min-1 at 298K. If the
activation energy is 10.0 k cal, calculate the rate constant at 323K.

16.1
1. d
Δ[ NO 2 F] Δ[ NO 2 ]
2. i) ii)
Δt Δt
−Δ[F2 ] 1 Δ[ NO 2 F] 1 Δ[ NO 2 ] −Δ[F2 ]
iii) iv) =− =
Δt 2 Δt 2 Δt Δt
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1 d[ NO 2 F] 1 d[ NO 2 ] d[F2 ]
3. =− =
2 dt 2 dt dt
4. Increase in volume would decrease the pressure which would decrease the rate of reaction.
16.2
1. a) 1.0 x 10-3 mol litre-1 s-1
b) 5.0 x 10-2 s-1
2. a) First order with respect to C2H4 and 1.5 with respect to I2.
b) The over all order of reaction is 2.5
sec −1
c) k=
(mol dm −3 )3 / 2
= mol-3/2 dm9/2 s -1.
2.303 0.01 mol L−1

3. -3
a) 2.5 x 10 (min ) = -1 log 10
10 min x
b) When half of the sample is decomposed
0.693
t1/2 = min = 0.277 x 103 min = 2.77 x 102 min
2.5 x 10 −3
16.3
1. 10
2. 34.0 K cal mol-1
3. 28.82 kJ
4. Molecules do not have enough energy to be equal to thershold energy.
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17
ADSORPTION AND CATALYSIS
Surface of solids plays a crucial role in many physical and chemical phenomena. There are
two main reasons for this special role. Firstly, the surface of a substance interacts first with
its surroundings. Secondly, the surface molecules are in a different state as compared to the
molecules in the interior of the solid. The surface molecules interact more readily with
other substances which come close by and are responsible for many special properties. In
this lesson we shall study about two such properties – adsorption and catalysis.
Objectives
• define adsorption;
• distinguish between physical adsorption and chemisorption;
• list and explain the various factors that affect adsorption;
• state Freundlich adsorption isotherm mathematically and explain it;
• explain Langmuir isotherm;
• define catalysis;
• distinguish between homogeneous and heterogeneous catalysis, and
• explain the role of activation energy in catalysis.
17.1 Adsorption
The surface of a solid attracts and retains molecules of a gas or a dissolved substance which
comes in its contact. These molecules remain only at the surface and do not go deeper into
the bulk Fig. 17.2(a).
The phenomenon of attracting and retaining the molecules of a gas or a dissolved
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substance by the surface of a solid, resulting in their higher concentration on the surface
is called adsorption.
The substance which gets adsorbed is called the adsorbate and the solid substance which
adsorbs is called the adsorbent.
A molecule in the interior of a solid is surrounded by other molecules in all directions
(Fig. 17.1). However, a molecule at the surface is surrounded by other molecules within the
solid phase but not from the outside. Therefore, these surface molecules have some
unbalanced or residual forces.
Fig. 17.1 : Molecules in the interior and at the surface of a solid
Fig. 17.2 : (a) Adsorption (b) Absorption
17.1.1 Adsorption and Absorption

The phenomenon of adsorption is different from that of adsorption. The latter term implies
that a substance is uniformly distributed throughout the body of a solid, Fig. 17.2(b). If we
leave a small lump of calcium chloride in open, it absorbs water vapour (moisture) from air
and after some time even starts dissolving in it. On the other hand if we keep a sample of
silica gel in open, it adsorbs water vapour on its surface as shown in Fig. 17.2(a).
17.1.2 Factors Affecting Absorption

Adsorption occurs on the surface of almost all solids. However the extent of adsorption of a
gas on the surface of a solid depends upon the following factors:
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(i) Nature and surface area of the adsorbent
(ii) Nature of the adsorbed gas
(iii) Temperature
(iv) Pressure of the gas
Let us now discuss these factors briefly.
(i) Nature and Surface Area of the Adsorbent
Different solids would adsorb different amounts of the same gas even under similar conditions.
Substances like charcoal and silica gel are excellent adsorbents. The substances that are porous
in nature and have rough surfaces are better adsorbents.
The extent of adsorption also depends upon the surface area of the solid. Greater the surface
area, more is the surface available for adsorption and greater is the adsorption. The surface
area depends upon the particle size of the substance. A cube of each side equal to 1cm has
six faces. Each of them is a square with surface area of 1cm2. Thus, the total surface area of
this cube is 6 cm2 Fig. 17.3(a). If its each side is divided into two equal halves, 1/2 cm long,
and the cube is divided into two equal halves, 1/2cm long, and the cube is cut along the
lines indicated in the Fig(b), the cube would be divided into 8 smaller cubes with each side
0.5cm long [Fig. 17.3 (b)]. Surface area of each small cube would be (6x0.5x0.5) = 1.5cm2
and the total surface area of all the 8 smaller cubes would be 12cm2 which is double the
surface area of the original cube. If it is subdivided into smaller cubes, each of side equal to
1x10 -6cm the surface area will increase to 6x106cm2 or 600m2. The increase in surface area
would result in greater adsorption.
Fig. 17.3 : Subdivision of a cube

Now we can explain why the solids that are porous in nature and have rough surfaces are
better adsorbents. It is so because each of these features increases the surface area.
(ii) The Nature of the Adsorbed Gas
The extent of adsorption also depends upon the nature of the gas. The gases which are more
easily liquifiable or are more soluble in water are more readily adsorbed than others. For
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example, under similar conditions, the amount of SO2 or NH3 adsorbed by charcoal is much
more than that of H2 or O 2 gases. It is because the intermolecular forces are stronger in more
easily liquifiable gases, therefore they get adsorbed more strongly.
(iii) Temperature
The extent of adsorption decreases with rise in temperature. For example, under one atmosphere
pressure, one gram of charcoal adsorbs about 10cm3 of N 2 gas at 272K, 20cm3 at 248K and
45cm3 at 195K.
Adsorption is an exothermic process. The change in enthalpy when one mole of a substance
is adsorbed is called enthalpy of adsorption. The adsorption process is similar to the
condensation process. The reverse process is called desorption and is endothermic in nature.
It is similar to the evaporation process. When a gas is kept in contact with a solid adsorbent
in a closed container, a dynamic equilibrium is established in due course of time.
gas + solid gas absorbed on the solid + heat
adsorbate absorbent
Since the forward process (adsorption) is exothermic in nature, according to the Le Chatelier’s
principle, it would be favoured at low temperature. Therefore, the extent of adsorption would
increase on decreasing the temperature and would decrease on increasing the temperature.
(iv) Pressure of the gas
At a constant temperature the extent of adsorption increases with increase in the pressure of
the gas (adsorbate). We shall study the relation between the two in detail a little later.
17.1.3 Physical and Chemical Adsorption

Adsorption can be divided into two main categories- physical and chemical adsorption.
(i) Physical Adsorption
It is the common type of adsorption. The basic feature of physical adsorption is that the
adsorbate molecules are held at the surface of the adsorbent by weak van der Waals forces.
These are the forces that exist between particles of all matter. Because of their universal
nature, these forces would operate between any adsorbent and adsorbate pair. Therefore,
the physical adsorption is observed on surface of any solid. Only, the extent of adsorption
varies according to the nature of the adsorbent and adsorbate as discussed earlier.
Physical adsorption is characterized by low enthalpy of adsorption, that is about 10-40 KJ
mol-1.
Another feature of the physical adsorption of a gas by a solid is that it is reversible in nature
and an equilibrium is established between the adsorbent and the adsorbate as discussed
earlier. Increase of pressure increases the adsorption and the release of pressure desorbs the
gas. When temperature is increased, the physical adsorption decreases and when it is lowered,
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the adsorption increases. In physical adsorption, several layers of adsorbate are adsorbed one
over the other.
(ii) Chemisorptions or Chemical Adsorption

We have seen earlier that some unsaturated valancies exist on the surface of a solid. Whenever
a chemical combination takes place between the adsorbent and the adsorbate the adsorption
becomes very strong. This type of adsorption caused by forces similar to chemical bonds
between the adsorbent and the adsorbate is called chemisorption or chemical adsorption.
The enthalpy of chemisorption is as high as that of chemical bonds (bond enthalpies) and is in
the range of 40-400 kJ mol-1. Chemisorption is highly specific and is possible between a
specific adsorbent - absorbate pair. Like most of the chemical changes it is irreversible.
Attempts to release the absorbed gas gives the gas and some amount of a definite compound.
For example, oxygen gas in chemisorbed on tungsten. It is released from the surface of
tungsten as a mixture of oxygen and tungsten oxide. Unlike physical adsorption,
chemisorption first increases and then decreases with rise in temperature [Fig. 17.4 (b)].
This shows that chemisorption has an energy of activation*. During chemisorption, only
one layer of adsorbate molecules is adsorbed. The main distinctions between physical
adsorption and chemisorption are summarized in Table 17.1.
Chemisorption
Physical
adsorption
Extent
Extent of
of adsorption
adsorption
Fig. 17.4 : Effect of temperature on (a) physical adsorption and Chemisorption

Table 17.1 Physical Adsorption and Chemisorption
Physical Adsorption Chemisorption
1. The forces operating between 1. The forces operating between

adsorbate and adsorbent are the absorbate and adsorbent are
weak van der Waals forces. strong and similar to chemical
bonds.
2. The enthalpy of adsorption is low and 2. The enthalpy of adsorption is high
ranges from 10 to 40 kJ mol-1. and ranges from 40 to 400 kJ mol-1.
* You will learn more about energy of activation later in this lesson.
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3. No activation energy is involved. 3. Significant activation energy is
involved.
4. Adsorption occurs more readily at low 4. Chemisorption occurs at relatively

temperature and high pressure. high temperature and high pressure.
5. It is not specific in nature. All gases 5. It is highly specific in nature and
are absorbed on all solids and no occurs between those adsorbents
compounds are formed. and adsorbates which have a
possibility of compound formation
between them.
6. It is reversible in nature. The gas is 6. It is irreversible in nature.
desorbed on increasing the temperature Desorption also separates some
or decreasing the pressure. amount of the compound formed.
7. Multilayer formation is common. 7. Monolayer formation occurs.
17.1.4 Adsorption Isotherms

x
The extent of adsorption is measured in terms of the quantity where, x is the mass of the
m
x
gas (adsorbate) adsorbed at equilibrium on mass m of the adsorbent. is the mass of the
m
x
adsorbate adsorbed per unit mass of the adsorbent. The graph showing variation in with
m
pressure (p) at a constant temperature is called adsorption isotherm. let us see the variation
in extent of adsorption in case of gases and of solutes from their solutions.
T1 < T 2 < T 3
T1
k
T2
x
m
T3
p
k
Fig. 17.5 : Adsorption isotherm of a gas
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i) Adsorption of Gases
The adsorption isotherm of a gas which is adsorbed on a solid is shown in Fig. 17.5. It shows
that the extent of adsorption of a gas on a solid increases with the increase in the pressure of
the gas, p at three different constant temperature. The curves also show that the extent of
adsorption, decreases at a fixed pressure as the temperature is increased (see the dotted line).
Freundlich Adsorption Isotherm
⎛ x⎞
Freundlich gave an empirical mathematical relationship between the extent of adsorption ⎜⎝ ⎟⎠
m
and the equilibrium pressure (p) of the gas as :
x 1
= k p n where n > 1
m
In this relation k is a constant at a given temperature and depends upon the nature of the
adsorbate and adsorbent. The value of n changes with pressure. It is 1 at low pressures and
increases with pressure. The relationship is valid at a constant temperature. Therefore, it is
called Freundlich Adsorption Isotherm. On taking logarithm of the above equation, we get
x 1
log = log k + log p
m n
x
This is an equation of a straight line and a plot of log against log p should be a straight line
m
1
with slope as depicted in Fig. 17.6. In actual practice, a straight line is obtained provided
n
the data at very low and very high pressures is neglected.
k
x
log
m
Intercept = log k
k
log p
x
Fig. 17.6 : Plot of log against log p
m
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17.1.5 Langmuir Adsorption Isotherm
Adsorption Isotherm : One of the drawbacks of the Freundlich adsorption isotherm is that it
fails at high pressure of the gas. Langmuir derived an adsorption isotherm on theoretical
consideration based on kinetic theory of gases. This is named as the Langmuir adsorption
isotherm. This isotherm is based on the assumption that every adsorption site is equivalent
and that the ability of a particle to bind there is independent of whether or not nearby sites are
occupied. In his derivation, Langmuir considered adsorption to consist of the following two
opposing processes :
Asorption of the gas molecules on the surface of the solid.
Desorption of the adsorbed molecules from the surface of the solid.
Langmuir believed that eventually a dynamic equilibrium is established between the above
two opposing processes. He also assumed that the layer of the adsorbed gas is only one
molecule thick i.e., unimolecular. Since such type of adsorption is obtained in the case of
chemisorption. Langmuir adsorption isotherm works particularly well for chemisorption.
The Langmuir adsorption isotherm is represented by the relation.
x ap
= .....(17.1)
m 1 + bP
where a and b are two Langmuir parameters. At very high pressure, the above isotherm acquires
the limiting form.
x a
= (at very high pressure) .....(17.2)
m b
At very low pressure, Eq. (17.1) is reduced to x/m = ap (at very low pressure).
.....(17.3)
In order to determine the parameters a and b, Eq. (17.1) may be written in its inverse form:
m 1 + bp b 1
= = + .....(17.4)
x ap a ap
A plot of m/x against 1/p gives a straight line the slope and intercept equal to 1/a and b/a,
respectively. Thus, both parameters can be determined.
The Langmuir isotherm, in the form of Eq. (17.1) is generally more successful in interpreting
the data than the Freundlich isotherm when a monolayer is formed. A plot of x/m versus p is
shown in (Fig. 17.7). At low pressures, according to Eq. (17.3), pressure x/m increases linearly
with p. At high pressure according to Eq. (17.2), x/m becomes constant i.e. the surface is
fully covered and change in pressure has no effect and no further adsorption takes place, as is
evident from Fig. 17.7.
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(x/m)
Pressure (p)
Fig. 17.7 : Langmuir Adsorption isotherm.
ii) Adsorption from Solutions.

Adsorption occurs from solutions also. The solute gets adsorbed on the surface of a solid
adsorbent. Charcoal, a good adsorbent, is often used to adsorb acetic acid, oxalic acid and
organic dyestuffs from their aqueous solutions.
x
The extent of adsorption, depends upon the concentration c of the solute. Freundlich
m
isotherm is applicable to adsorption from solutions when concentration is used in place of
pressure as shown below.
x 11
= k nn
c
m C
and in the logarithmic form as
x 1
log = log k + log c
m n
x
The plot of log against c is also a straight line, provided very low and very high
m
concentrations are avoided.
17.1.6 Application of Adsorption

The phenomenon of adsorption finds many applications, some of which are given below.
1. Activated charcoal is used in gas masks in which toxic gases are adsorbed and air passes
through it.
2. Silica gel packed in small cloth bags is used for adsorbing moisture in bottles of medicine
and in small electronic instruments.
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3. Animal charcoal is used for decolourizing many compounds during their manufacture.
4. In chromatography, the selective adsorption of different solutes on the surface of solid
adsorbent helps in their separation.
5. Mordants are used during dyeing process with certain dyes. In such cases, the mordants
fix the dye on the fabric by adsorption.

1. Indicate which of the following statements are true or false. (T/F)
i) More easily liquifiable gases are adsorbed more strongly.
ii) Non-porous adsorbents would adsorb more quantity of a gas than porous
adsorbents under similar conditions.
iii) The extent of adsorption increases with rise in temperature.
iv) Chemisorption is highly specific in nature.
v) Adsorption can occur from solutions also.
...................................................................................................................................
17.2 Catalysis
When hydrogen and oxygen gases are kept in contact with each other, no observable reaction
occurs. If we add a small piece of platinum gauge in the mixture of these gases, the reaction
occurs readily. Here platinum gauge speeds up the reaction and is called a catalyst.
A catalyst is a substance which changes the rate of a reaction but remains chemically
unchanged at the end of the reaction.
The phenomenon of change of reaction rate by addition of a substance which itself remains
unchanged chemically is called catalysis. The following are some more examples of
catalysis:
i) Decomposition of potassium chlorate occurs at high temperature. If a small amount
of the manganese dioxide is added, the decomposition occurs at much lower
temperature. Here, manganese dioxide acts are catalyst.
MnO (s )
2 KCIO3 (s) ⎯⎯⎯⎯
2
→ 2KCl(s) + 3O 2 (g )
ii) The evolution of hydrogen by the reaction between zinc and hdyrochloric acid is catalysed
by Cu2+ (aq) ions.
Cu 2 + ( aq )
Zn (s) + 2HCl(aq ) ⎯⎯⎯⎯→ ZnCl2 (aq ) + H 2 (g)
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iii) The oxidation of hydrogen chloride gas by oxygen occurs more quickly if the gases are
passed over cupric chloride.
CuCl (s )
4HCl(g ) + O 2 (g ) ⎯⎯⎯⎯
2
→ 2H 2O(g ) + Cl2 (g )
Auto Catalysis
In certain reactions, one of the products of the reaction acts as the catalyst. For example, the
oxidation of oxalic acid by acidified potassium permanganate occurs as
2KMnO 4(aq)+3H2SO4(aq)+5(COOH)2(aq) →K2SO 4(aq)+2MnSO4(aq)+8H2O(l)+10CO2(g)
At room temperature, the reaction is quite slow in the beginning. Gradually it becomes fast
due to the catalyst action of Mn2+ ions which are one of the products as MnSO4 in the reaction.
The phenomenon in which one of the products of a reaction acts as a catalyst is known as
auto-catalysis.
Negative Catalysis
Some catalysis retard a reaction rather than speed it up. They are known as negative catalysts.
For example :
i) Glycerol retards the decomposition of hydrogen peroxide.
ii) Phenol retards the oxidation of sulphurous acid.
Promoters and Poisons

Certain substances increase or decreases the activity of the catalyst, although, by themselves
they do not show any catalytic activity.
The substances which increase the activity of a catalyst are called promoters and those
which decrease the activity of a catalyst are called poisons. For example :
i) In Haber’s process for the manufacture of ammonia, the catalytic activity of iron is
enhanced by molybdenum which acts as promoter.
Fe (s )
N 2 (g ) + 3H 2 (g ) Y
ZZZZZ
ZZZZ X
Z 2NH
2 NH 23 (g )
Mo (s )
ii) Copper promotes the catalytic activity of nickel during hydrogenation of oils.
iii) In Haber’s process the catalyst iron is poisoned by hydrogen sulphide H2S.
iv) In contact process for the manufacture of sulphuric acid, the catalyst platinum is poisoned
by even the traces of arsenious oxide AS2O3.
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17.2.1 General Characteristics of a Catalyst
The following are the general characteristics of a catalyst :
i) A catalyst remains unchanged at the end of the reaction.
The amount and the chemical composition of a catalyst remain unchanged in a catalytic
reaction. However, the catalyst may undergo a physical change. For example, manganese
dioxide, which is used as a catalyst in thermal decomposition of potassium chlorate becomes
powder during the course of the reaciton.
ii) A small quantity of the catalyst is generally enough
In most of the reactions, only a minute amount of catalyst is required. Only one gram of
Cu2+ ions is sufficient to catalyse the oxidation of 109 litres of sodium sulphite solution. In
some cases, the rate of reaction is directly proportional to the concentration of catalyst
present. Catalysis by acids or bases is usually of this type.
iii) A catalyst does not alter the position of equilibrium state of a reversible reaction.
A catalyst allows the equilibrium to be reached faster. However, it does not alter the equilibrium
composition of the reaction mixture. It is because, a catalyst increases the rates of forward
and backward reaction equally.
iv) Catalysts are generally specific in their action.
Generally, one catalyst will change the rate only one reaction. For example, manganese dioxide
catalyses the decomposition of potassium chlorate but not of potassium perchlorate.
v) A catalyst cannot initiate a reaction.
A catalyst can change the rate of a reaction which occurs even in the absence of catalyst. It
cannot start a reaction.
vi) The activity of a catalyst can be increased by the presence of promoters and decreased
by the presence of poisons.
Presence of a promotor increases the activity of a catalyst, while the presence of a poison
decreases it.
17.2.2 Homogeneous and Heterogeneous Catalysis

The phenomenon of catalysis can be divided into two main types - homogeneous and
heterogeneous catalysis, on the bases of the number of phases present in the reaction mixture
(A phase is a homogeneous part of a system).
a) Homogeneous Catalysis
When the catalyst is present in the same phase as the reactants, the phenomenon is called
homogeneous catalysis. For example :
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i) Nitric oxide catalyses the oxidation of sulphur dioxide to sulphur trioxide in the lead
chamber process.
NO ( g )
2SO 2 (g ) + O 2 (g ) ⎯⎯⎯→ 2SO3 (g )
ii) Hydrogen ions catalyse the inversion of cane sugar
H + ( aq )
C12 H 22O11 (aq ) + H 2O(g ) ⎯⎯⎯⎯
→ C6 H12O6 (aq ) + C6 H12O6 (aq )
Glucose Fructose
b) Heterogeneous Catalysis
When the catalyst is present in a phase other than that of reactants the phenomenon is called
heterogeneous catalysis. For example :
i) Iron (s) catalyses the formation of NH3 gas.
Fe (sg )
N 2 (gN) 2 + 3H 2 (g ) ⎯⎯⎯→ 2 NH3
ii) In contact process for the manufacture of sulphuric acid, platinized asbestos is used as
the catalyst.
Pt (sg )
2SO 2 (g ) + O 2 (g ) ⎯⎯⎯ → 2SO3 (g )
17.2.3 Catalysis and Activation Energy

We have seen that a catalyst increases the rate of a reaciton. We explain it by considering the
Fig. 17.8
In this figure Ea is the activation energy of uncatalysed reaction and E’a is the activation
energy of the catalysed reaction. A catalyst lowers the activation energy as you can see in
the figure (E’a < Ea). The reduction in activation energy is achieved by providing an
alternative pathway of lower energy for the reaction.
You can also see in this figure that the relative energies of reactants and products are not
changed. The enthalpy change is the same for the catalysed and uncatalysed reaction.
Fig. 17.8 : Graphical representation of the effect of catalyst on a reaction
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1. List any two characteristics of a catalyst.
...................................................................................................................................
2. A small amount of alcohol when added to a solution of sodium sulphite slows down its
oxidation to sodium sulphate. What type of catalyst is alcohol ?
...................................................................................................................................
3. How would the activation energy be affected in the above reaction (given in Q.No. 2)
on adding the alcohol ?
...................................................................................................................................
4. Addition of molybdenum enhances the catalytic activity of iron in the Haber’s process
for the manufacture of ammonia. What are the substances like molybdenum called ?
...................................................................................................................................

• The pehnomenon of attracting and retaining the molecules of a gas or of a dissolved
substance on the surface of a solid is called adsorption.
• The substance which gets adsorbed is called the adsorbate and the solid substance
which adsorbs is called the adsorbent.
• The substances that are porous in nature and have rough surfaces are better adsorbent.
• Easily liquifiable gases are more readily adsorbed.
• Extent of adsorption decreases with rise in temperature and increases with the increase
in pressure of the gas.
• Physical adsorption is due to van der Waal forces and chemisorption is due to forces
similar to chemical bonds.
• Pressure dependence of adsorption of a gas at a constant temperature is given by
Freundich Adsorption Isotherm
x 1
= n
m kp
• A catalyst is the substance which changes the rate of a reaction, but itself remains
chemically unchanged during the reaction.
• The catalysts which increase the rate of a reaction are called the positive catalysts
while those which decrease the rate are called the negative catalysts.
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• Auto catalysed reactions the activity of a catalyst while a poison hampers it.
• A catalyst can’t initiate a reaction, nor can it alter the position of equilibrium state of
reversible reaction.
• When the catalyst is present in the same phase as the reactants it is called a
homogeneous catalyst.
• When the catalyst is present in a phase other than that of reactants it is called a
heterogenous catalyst.
• A catalyst changes the rate of a reaction by changing its path and the activation energy.
Terminal Exercise
1. What is the difference between adsorption and absorption ?
2. Distinguish between physical and chemical adsorption.
3. List the factors that affect adsorption.
4. What type of solids make better adsorbents ?
5. Easily liquifiable gases are adsorbed more readily. Explain.
6. What is ‘extent of adsorption’ ?
7. How does extent of adsorption vary with temperature in case of (i) physical adsorption
and (ii) chemisorption ? Depict graphically.
8. What is enthalpy of adsorption ?
9. Exaplain the effect of temperature on extent of physical adsorption with the help of
Le Chatelier’s Principle.
10. What is an adsorption isotherm ?
11. State mathematically Freundlich Adsorption Isotherm and depict it graphically. Under
what conditions is it applicable.
12. Give the mathematical equation of Frundlich Isotherm for adsorption of solutes from
solutions.
13. Give any three applications of adsorption.
14. What is a (i) catalyst and ii) negetive catalyst
15. What are promoters and poisons ? Give one example of each.
16. What is auto catalysis. Give one example.
17. Give any five characteristics of catalysis.
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18. Distinguich between homogeneous and heterogeneous catalysis.
19. Give two examples each of homogeneous and heterogenous catalysis.
20. How does a catalyst change the rate of reaction. Explain with the help of appropriate
example.

17.1
i) T, ii) F, iii) F, iv) T, v) T
17.2
1. See text section 17.2.1
2. Negative catalyst
3. Increase
4. Promoters
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CHEMISTRY
Maximum Marks : 50 Time :1 1/2 Hours.
INSTRUCTIONS :
- Name
- Enrolment Number
- Subject
- Assignment Number
- Address
• Get your assignment checked by the subject teacher at your study centre so that you get
positive feedback about your performance.
1. a) Classify the following as electrolytes and non-electrolytes : sodium chloride, iodine,

oxalic acid, alcohol.
b) Give units of equivalent conductance.
c) Identify anode and cathode in the following cell :
Zn | Zn2+ | Cn2+ | Cu.
d) List the factors which affect the rate of reaction.
e) Give the relation between half-change period of a first order reaction and its rate
constnat.
f) What is a buffer solution ?
g) What is the requirement for precipitating a sparingly soluble salt ?
h) What is overall order of a reaction ?
i) Name two characteristics of a catalyst.
j) Define the term pH. 1 x 10=10
2. a) How many electrons are required for discharging 0.5 mol of Al3+ ions.
b Draw a graph and label it showing the variation of equivalent conductance of
sodium chloride and sodium bicarbonate.
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c) Predict the redox reaction between zinc and lead. Given
EO of Zn2+ | Zn is - 0.763 V and that of Pb2+ / Pb is - 0.125
d) The rate constant of a first order reaction is 0.024 hr-1. What will be its initial rate
if the initial concentration of the reactant is 0.3 mol L-1 ?
e) Draw and label a diagram showing energy changes during the course of a
endothermic reaction
f) Which of the following pairs is the weaker acid or weaker base :
i) H2O and H3O+ (i) CH3COOH and HCN
ii) F- and NO -
3
iii) NH3 and SO43-

g) What is the hydrogen ion concentration in 10-4 NaOH solution ?
h) Differentiate adsorption from absorption.
i) Define the laws of electrolysis.
j) Write an expression for the half life period of first order reaction. (2x10=20)
3. a) How many molecules of chlorine would be deposited from molten sodium chloride
in five minutes by a current of 60 milliamperes ?’
b) The decomposition of Cl2O7 in gaseous phase into O2 and Cl2 is a first order
reaction.
i) After 55 seconds the pressure of Cl2O7 falls from 0.62 to 0.44 atmosphere at
450 K. Calculate the rate constant.
ii) Calculate the half change period of the reaction at 450 K.

c) What is the pH of a solution prepared by dissolving 5g NaOH in 500 mL water ?
d) Calculate Gib’s free energy ΔGO for the reaction occuring in the cell
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu (s); EOcell = 1.10 V (3x4=12)
4. a) At 300K solubility of AgCl in water in 1.30 x 10-3 mol L -1.

i) Calculate its solubility product.
ii) What will be the solubility of AgCl in 0.1 M Kcl at 300 K ?
b) What would be the activation energy of a reaction whose rate becomes three times
when the temperature is raised by 100C from 270C to 370C ?
(4x2=8)
-399-
TERMS YOU SHOULD KNOW
Acceleration due to gravity : Acceleration with which a body would fall freely under the
action of gravity in vacuum.
Amorphus Solid : That solid in which particles have only short range order.
Ampere : SI unit of electric current. It is the current which, if maintained in two parallel
conductors of infinite length, of negligible cross-section, and placed 1 metre apart in vacuum,
would produce between the conductors a force equal 2 x 10-7 newtons per metre of length.
Atom : The smallest portion of an element that can take part in a chemical reaction.
Avogadro’s constant : Number of atoms or molecules in mole of a substance. It is equal to
6.022 x 10-23 mol-1.
Boiling Point : That temperature at which the vapour pressure of a liquid is equal to the
atmospheric pressure is known as the boiling point of the liquid. Boiling point at one
atmosphere; 760 torr, or 760 mm Hg is referred to as normal boiling point.
Candel : Fundamental SI unit of luminous intensity.
Condensation : The change of vapour into liquid.
Covalent Solids : Those solids which have ordered arrangement of atoms held by continous
covalent bonds.
Critical temperatue : The temperature above which a given gass cannot be liquified.
Crystalline solid : That solid which has repetitive well-ordered arrangement of particles.
Density : The mass of unit volume of a substance, usually expressed in kg./m3.
Diffusion : General transport of matter where by molecules or ions mix through normal thermal
agitation.
Effusion : The passage of gasses through small apertures under pressure.
Empirical Formula : Formula deduced from the results of analysis which is the simplest
expression of the ratio of the atoms in substance.
Energy : The capacity of body for doing work.
Evaporation : The conversion of a liquid into vapour, at temperarues below the boiling
point.
Frequency : Number of oscillation in unit time, usually one second. SI unit of frequency is
hertz.
Gas : A state of matter in which the molecules move freely, thereby causing the matter to
-400-
expand indefinitely, occupying the total volume of any vessel in which it is contained.
Ideal Gas : Gas with molecules of neglibgile size and exerting no intermolecular forces such
a gas is hypothetical gas which would obey the ideal gas law under all conditions.
Intermolecular forces : The forces binding one molecule to another.
Ion : Any atom or molecule which as a resultant electric charge due to loss or gain of valence
electrons.
Kelvin : Thermodynamic scale of temperature.
Liquid : A state of matter between a solid and a gas, in which the shape of a given mass
depends on the containing vessel, the volume being independent.
Mass : Quantity of matter in a body.
Mole: The amount of substance that contains as many entities ( atoms molecules, ions electrons
etc) as there are atoms in 12 g of 12C.
Metallic solid: Those solids which have ordered arrangement of positive ions embeded in the
valence electrons belong to the crystal as a whole instead of any single atom.
Molecular solid: Those solids which have ordered arrangement of molecule or atoms held be
weak van Waals forces.
Molecule: The smallest portion of substance capable of existing independently and retaining
the properties of the original substance.
Vapour Pressure: The pressure exerted by the quantity of vapour above the liquid, when
equilibrium is established is called the vapour pressure of the liquid.
Weight: gravitational force acting on a body at the earth’s surface, Weight = mass x
acceleration due to gravity.
Alkali Metals : The chemical family headed by Lithium group 1 in the periodic
table.
Alkaline Earth metlas : the chemical family headed by beryllium; group 2 in the
periodic table.
Bond angle : The angle between the two lines representing bonds joining
the nuclei of these atoms.
Bond energy : The energy required to break a chemical bond between two
atoms into two neutral fragments
Deliquesant : A term used to describe compounds that remove enough water
from their surroundings to dissolve themselves completely.
Covalent bond : A chemical bond formed by two atoms sharing a pair, or pairs,
of electrons.
-401-
Double bond : A covalent bond that shares four electrons.
Electron-dot formular : A representation of atom, ion or molecule using dots to
sysmbolise the electrons.
Electronegativity : The ability of an atom in a covalent bond to attract the shared
pair of electron to itself.
Electrovalent bond : A chemical bond formed by the attraction of a positive ion
for negative ion.
Group : A vertical column of elements in the periodic table.
Halgoens : The chemical family headed by flourine; group 17 in the
periodic table.
Hydride : Compounds of hydrogen where hydrogen has oxidation
number-1.
Hydrogen bond : A weak electrostatic attraction between a hydrogen nucleus
bound to a highly electronegative atom (either oxygen, flourine
of nitrogen) in one molecule and one of these highly
electronegative atoms in a second molecule.
Hygroscopic : Having a tendency to absorb moisture.
Ionization energy : The energy required to remove an electron from an atom or
ion.
Lewis acid : A species that can accept a pair of electrons.
Lewis base : An electron-rice species that has one unshared pair of
electrons.
Lewis structures : Electron-dot representations of the arrangement of atoms,
ions or molecules.
Metalloid : An element that lies on the dividing line between metals and
non-metals. It has some properties intermediate between each
type.
Moderator : A substance used in a nuclear reactor core to slow down
neutrons.
Noble Gases : A chemical family headed by helium. Krypton and xenon of
this family are reactive to some extent. but the rest of the
family appears to be chemically inert.
Octet rule : Atoms is chemical changes (with few exceptions) tend to
acquire noble gas configurations of s2p6.
-402-
Oxidation number : A positive or negative number that represents the charge that
an atom appears to have when the bonding electrons are
counted using some rather arbitrary rules.
Oxide : A compound of oxygen.
Period : A horizontal row of elements in the periodic table.
Periodic law : The properties of elements are periodic functions of their
atomic numbers.
Periodic Table : A systematic arrangement of the known elements according
to increasing atomic number.
Polar bond : A covalent bond that is shared unequally between two atoms.
It has positive and negative side.
Triple bond : Two atoms sharing six electrons.
Valence electrons : The electrons located in the outer most principal energy level.
-403-
*313E* 313 – E
Intermediate (APOSS) Reg. No. : ....................................

July 2018
CHEMISTRY
Time : 3 Hours Maximum : 80 Marks
SECTION – A
Note : Answer all the questions. (2×13=26 Marks)

1. Calculate the molar mass of CH4.
2. What is the volume occupied by one mole of a gas at STP ?
3. State the bond angles in CH4 and NH3 molecules.

4. What is the radius ratio in tetrahedral structure of ionic compounds ?
5. State Avogadro’s law.
6. Define internal energy.
7. What is ionic product of water ? State it’s value at 25°C.
8. Write the molecular formula of Borax.
9. Calculate the oxidation number of copper in [Cu(NH3)4]2+ ion.
10. State any two uses of ethylene.
11. What is ‘decay of organic matter’ ?
12. What is the threshold limit of radiation for the people working in nuclear reactors ?
13. Name two plants that reduce mercury pollution.
SECTION – B
Note : Answer any six of the following. (5×6=30 Marks)
14. Calculate the no. of kilograms of oxygen required to completely burn 1 kg of

butane in a rocket.
15. What is an emulsion ? Write a note on various types of emulsions.
P.T.O.
313 – E *313E*
16. Calculate the standard entropy change of the following reaction at 298 K. At
298 K the standard molar entropies of Fe, O2 and Fe2O3 are 27.3, 205.0 and
87.4 J.K. mol–1 respectively. 4 Fe(s) + 3O2(g) }}m 2Fe2O3(s).
17. Write a note on hydrides of ‘p’ block elements.
18. What is Lanthanide contraction ? Explain.
19. Explain various resources of water on earth.
20. What are the hazards of sound pollution to human beings ?
21. Write any five examples of nucleophilic substitution reactions of alkyl halides.
SECTION – C
Note : Answer any three of the following. (3×8=24 Marks)
22. State and explain various quantum numbers.
23. Explain the factors influencing ionisation potential.
24. Write the differences between physical adsorption and chemical adsorption.
25. Write any two methods to prepare phenols in the laboratory with the help of
equations.
—————————

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1

A.P. OPEN SCHOOL SOCIETY, GUNTUR, AMARAVATI.

Published: 2018, 2019, 2020

All Rights Reserved

The Study Materials developed are of Self Learning in nature and

Best wishes to all the learners and happy learning experience at

Dr. K.V. SRINIVASULU REDDY

Ms. Poonam Barat Dr. Sunita Bansal

Mr. C. Narsimha Reddy M.Sc., M.Ed

Module III Module V

I. Atoms Molecules and 1. Mole Concept 1

II. State of matter 6. The Gaseous State 123

III. Chemical energetics 11. Chemical Thermodynamics 228

IV. Chemical Dynamics 13. Chemical Equilibrium 266

1.1. SI Units (Revisited)

Table 1.1 : SI Base Units

Physical Quantity Name of SI Unit Symbol of SI unit

Prefix Symbol Meaning Example

Tera T 10 12 1 terametre (Tm) = 1.0 x 1012 m

Before proceeding further try to answer the following questions:

Intext Questions 1.1

1.2 Relationship between Mass and Number of Particles

1.3 Mole - A Number Unit

1.4 Avogadro’s Constant

Significance of Avogadro’s Constant

S.No. Type of Substance Elementary Entity

1. Elements like Na, K, Cu which Atom

Formula Atoms / ions present in one formula unit

H2O Two atoms of H and one atom of O

Intext Questions 1.2

1.5 Mole, Mass and Number Relationships

1.5.2 Relative Atomic and Molecular Masses

Relative atomic mass =

and Rleative molecular mass =

Relative atomic mass of O-16 = 16 =

Since 1 amu = 1/12th the mass of one C-12 atom

Mass of one O-16 atom = 16 amu

Hydrogen, H 1.0 Phosphorus, P 31.0

ii) Molar mass of a molecular substance

Element or Compound Molecular mass / amu Molar mass / (g mol -1)

iii) Molar masses of ionic compounds

Example 1.3 : Caclulate the molar mass of

Mass of 1.0 x 1019 carbon-12 = g

Therefore, 100g sample of NH3 would contain x 100g

= 35.42 x 1023 molecules

Intext Questions 1.3

1.6 Mass, Molar Mass and Number of Moles

Number of moles (amount) of a substance =

1.7 Molar Volume, Vm

Intext Questions 1.4

What You Have Learnt

Answers to Intext Questions

1. Moles of N2 gas = = 0.70 mol

4. Mass of 1 mol of Na3PO4 = 3 x (mass of 1 mol of Na) + mass of 1 mol of P +

1. Moles of Cu atoms in 3.05 g copper = = 0.048 mol

2. Moles of gold, Au = = 0.064 mol

3. Molar volume of any gas at STP (298 K, 1 bar) = 22.7 L

2.2.1 Percentage Composition

Let us calculate percentage composition of aluminium oxide, Al2O3

Percentage of Aluminium = x 100