Module#1: Theoretical Chemistry

Atoms, molecules and Stoichiometry

Relative Atomic Mass
In this topic, you will be able to learn the followings:

 Relative atomic mass

 Relative isotopic mass

 Mass spectrometry

 Relative molecular mass

 Relative formula mass

 Avogadro constant

Relative atomic mass (Ar)

The relative atomic mass of an element is the weighted average of the masses of its isotopes on a scale
on which a carbon-12 atom has a mass of exactly 12 units.

Chlorine has two isotopes. They are chlorine-35 and chlorine-37. The abundances are 75% and 25%
respectively. We get these data from an analytical technique, mass spectrometry.

So, the relative atomic mass of chlorine

= (35 × 0.75) + (37 × 0.25)

= 35.5

Relative isotopic mass

The relative isotopic mass of an isotope is the mass of the isotope on a scale on which a carbon-12 atom
has a mass of exactly 12 units.

The isotopes of chlorine-35 and chlorine-37 have relative isotopic masses are 35 and 37 respectively.
 Relative isotopic masses of
Chlorine. One is 35 and another 37

The mass spectrometer

The mass spectrometer is used to measure the mass of each isotope present in an element and its
relative abundances.

Using these two types of data we calculate the accurate atomic mass of an element.

The device, mass spectrometer has four main chambers:

 Ionization chamber

 Acceleration chamber

 Deflection chamber and

 Detection chamber.
 The complete diagram
of mass spectrometer with the spectrum.

The sample is changed to vapor and then the molecules produce +ve ions. The speed of the molecules is
increased by the ion-accelerating electric field.

The magnetic field deflects the ion beam into mass. After that Ions hit a detection plate. They take
electro from the metal wire in the detection plate. A tiny current is produced. This is captured and
amplified. Finally, a bar chart is drawn.

The X-axis of the chart has the mass-charge ratio and Y-axis has the percentage abundances. From the
bad chart, the number of peaks indicates the number of isotopes and their height indicates the
abundance, from these two data, we can calculate the accurate atomic mass. We calculate the relative
molecular mass from the data of relative molecular mass.

Calculation Practice#1:

A sample of magnesium has the following isotopic composition by mass.

 Isotope = 24 and % abundance by mass =78.60

 Isotope = 25 and % abundance by mass =10.11

  Isotope = 26 and % abundance by mass =11.29

Calculate the relative atomic mass, Ar, of magnesium to four significant figures.

Calculation Practice#2:

A sample of boron was found to have the following % composition by mass:

 Isotope = 10 and % abundance by mass =18.70

 Isotope = 11 and % abundance by mass =81.30

Calculate a value for the relative atomic mass of boron. Give your answer to 3 significant figures.

Calculation Practice#3:

A sample of sulfur has the following isotopic composition by mass.

 Isotope = 32 and % abundance by mass =95

 Isotope = 33 and % abundance by mass =0.77

 Isotope = 34 and % abundance by mass =4.23

Calculate the relative atomic mass, Ar, of sulfur to two decimal places.

Relative molecular mass (Mr):

The relative molecular mass of a substance is the weighted average of the masses of the molecules on a
scale on which a carbon-12 atom has a mass of exactly 12 units.

Mr of H2O = (2 x 1) + 16 = 18

Mr of CHCl3 = 12 + 1 + (3 x 35.5) = 119.5

For covalent molecules, we use the term relative molecular mass. But for ionic compounds, we use the
relative formula mass.

Relative formula mass

The relative formula mass of a substance is the weighted average of the masses of the formula units on
a scale on which a carbon-12 atom has a mass of exactly 12 units.

The relative formula mass of copper (II) sulfate crystals, CuSO4.5H2O: = 63.5 + 32 + (4 x 16) + 5 x [(2 x 1) +
16]

= 249.5

Avogadro Constant
The number of particles of 1.0 mole of a substance is called Avogadro constant. That number is called
the Avogadro Constant and is given the symbol L.

The Avogadro Constant is a huge number (approximately 6.02 x 1023). So, in 18 g of water, H2O, there
are 6.02 x 1023 water molecules.

Empirical Formula
A formula that gives the simplest whole-number ratio of atoms in a compound is called empirical
formula.

For example, the molecular formula of propene, C3H6, shows that each freely existing molecule of
propene contains three atoms of carbon and six atoms of hydrogen. After simplifying it we its empirical
formula is CH2.

Empirical Formula Calculation#1:

Empirical Formula Calculation#2:

Molecular formula

This is the formula, where the actual number of atoms of each element present in each molecule of a
compound.

It may be a multiple of the empirical formula.

For example, the molecular formula of propene, C3H6, shows that each freely existing molecule of
propene contains three atoms of carbon and six atoms of hydrogen.

Molecular formula calculation:

An organic compound has the empirical formula is CH3 and RMM is 30. What is the molecular formula?
Practice#1: Empirical and molecular formula

An organic compound, P has the following composition by mass:

C, 35.8%; H, 4.5%; O, 59.7%.

(i) Calculation the empirical formula of P.

(ii) The Mr of P is 134. Use this value to determine the molecular formula of P.

The answer is C4H6O5

Workout the problem and compare the answer with it.

Calculations of molecular formula using combustion data.

When 10 cm3 of Hydrocarbon, A was mixed at room temperature with 50 cm3 of oxygen (an excess) and
exploded, 40 cm3 of gas remained after cooling the apparatus to room temperature and pressure.

When this 40 cm3 of gas was shaken with an excess of aqueous potassium hydroxide, KOH, 30 cm3 of gas
still remained.

Calculate the molecular formula of A.

Answer:
Practice#1: Finding molecular formula form combustion data.

When 30 cm3 of hydrocarbon, A was mixed at room temperature with 150 cm3 of oxygen (an excess)
and exploded, 120 cm3 of gas remained after cooling the apparatus to room temperature and pressure.
When this 120 cm3 of gas was shaken with an excess of aqueous potassium hydroxide, KOH, 90 cm3 of
gas still remained. Calculate the molecular formula of A.

Workout the problem and compare the answer given below:

CH4

Practice#2:

Carbon disulphide vapour burns in oxygen according to the following equation.

CS2(g) + 3O2(g) → CO2(g) + 2SO2(g)

A sample of 20 cm3 of carbon disulphide was burned in 100 cm3 of oxygen. After measuring the volume
of gas remaining, the product was treated with an excess of aqueous sodium hydroxide and the volume
of gas measured again. All measurements were made at the same temperature and pressure; under
such conditions that carbon disulphide was gaseous. What were the measured volumes?

Answers are 100 cm3 and 40 cm3

Mole Concept and formulas

A mole is defined as the amount of substance that contains the Avogadro Number of particles. It is
indicated by L. The value of L = 6.0×1023

Formula of Moles:

Construct the balanced equation

1. Construct a balanced equation for the reaction of S2Cl2 with water

2. Ammonium sulfate reacts with NaOH in a 1: 2 ratio. Complete and balance the equation for this
reaction. (NH4)2SO4 + 2NaOH → …….NH3 + ……………. +…………….

3. Construct a balanced equation for the reaction between chlorine and potassium hydroxide at
high temperatures. The products: are potassium chlorate(v), KCl, and water

4. H2S(g) + SO2(g) → …….. + H2O(l)

5. Construct a balanced equation for the reaction between sodium thiosulphate (Na2S2O3) and
hydrochloric acid. The product are sulfur, sodium chloride, water and a colorless gas

Calculation#1:

When 0.42 g of a gaseous hydrocarbon is slowly passed over a large quantity of heated copper(II) oxide,
CuO. The hydrocarbon is completely oxidised.

The products are 1.32 g of CO2 and 0.54 g of H2O.

Copper is the only other product of the reaction. Find the mole ratio of Carbon and Hydrogen in that
compound.

Answer:

Calculation#2:

Using the information that one mole contains 6.02 × 1023 particles, calculate the number of electrons in
3 moles of NO2 gas molecules.

The answer is: 4.15 × 1025 electrons

Work out the problem and compare it with the answer given above.

Calculation#3:

A factory uses 100 tonnes of ammonia each day to produce 160 tonnes of nitrogen monoxide, NO.
Calculate the percentage yield of nitrogen monoxide.

Answer:

Calculation#4:

1.60 g of methanol was mixed with aqueous acidified potassium dichromate(VI). The desired product
(methanal) was collected by immediate distillation under gentle warming. The yield of the product was
70.0 %. What mass of the product was collected?
Answer: 1.05 g

Work out the problem and compare it with the answer given above.

Calculation#5:

When a sports medal with a total surface area of 100 cm2 was evenly coated with copper, using
electrolysis, its mass increased by 0.635 g. How many atoms of copper were deposited per cm2 on the
surface of the medal?

Answer: 6.02 1019

Work out the problem and compare it with the answer given above.

Calculation#6:

Sulphur and chlorine can be reacted together to form disulphur dichloride, S2Cl2. Disulphur dichloride,
S2Cl2, is decomposed by water forming sulphur and a mixture of hydrochloric acid and sulphurous acid.
When 2.7 g of S2Cl2 is reacted with an excess of water, 0.96 g of sulphur, S, is produced. What is the
amount, in moles, of S produced from 1.0 mol of S2Cl2?

Answer: 1.5 mol

Work out the problem and compare it with the answer given above.

Atomic Structure
Particles in the atoms

Atoms are the smallest particles. From the atoms, we get many sub-atomic particles. Protons, neutrons,
and electrons are the permanent sub-atomic particles. The other sub-atomic particles are not
permanent.
The atom: protons and neutrons in the nucleus and electrons in the shell

The relative mass of a proton is 1, the neutron is 1, and the is about two thousand times lighter than a
proton.

The charge of the proton is +1 electron is -1 and the neutron has no charge.

The relative and mass of proton, electron and neutron

If we perform the experiment of deflection of the beam of electrons, protons & neutrons, we get the
following results:

 The electrons are deflected towards the positive plate and more than a proton.

 The protons are deflected towards the negative plate.

 The neutrons are not deflected, because they have no charge.

The deflection of proton, electron and neutron

The nucleus consists of almost all the mass of an atom. So, the nucleolus is heavy. It contains proton
and neutrons. The number of protons plus neutrons is called the nucleon number or atomic mass. The
electrons are the lightest particle. They stay in the shells. If the mass of a proton is 1, the mass of an
electron is 1/1836.

An atom of the same proton can’t be different but a neutron can be different. The same element having
a different number of neutrons are called isotopes. Hydrogen element has three different types of
isotopes. They are:

 Protium: It has one 1 proton, 1 electron, and zero neutron

 Deuterium: It has one 1 proton, 1 electron, and 1 neutron

 Tritium: It has one 1 proton, 1 electron, and 2 neutrons

The three isotopes of hydrogen

We can express an atom using the atomic number (proton number) and atomic mass (proton +
neutron). In the periodic table, an atom is described with the Symbol of the element, atomic number,
and atomic mass. The general expression is shown here:

The atomic number and atomic mass

Electron configuration
Atomic Orbit

A stationary shell around the nucleus with a definite quantity of energy is called the atomic orbit. It is
denoted by n, where n = 1,2,3,4,5,6, and 7. It is the two-dimensional motion of electrons and the
average path is circular.
 The atomic
orbits or shells

Maximum Electrons in each shell = 2n2

n = 1, 1st orbit = 2

n = 2, 2nd orbit = 8

n = 3, 3rd orbit = 18 and so on

The energy of the different orbit is different.

Atomic Orbital

An atomic orbital is defined as the region in space around the nucleus in which the probability of finding
the electron is maximum. The orbitals are s, p, d, and f. They are three-dimensional motion of electrons.
Different shapes. S is spherical, p is dumb-bell.
 The shape of s and p
orbitals

Each orbital contains max 2 electrons

s2

px2, py2, pz2

dxy2, dyz2, dzx2 …………….

In an orbit, the orbitals have equal energy.

The energy of orbitals of different orbits is shown here:

The energy of orbit at the different orbit

Aufbau principle

According to the principle, electrons fill orbitals starting at the lowest available (possible) energy
levels before filling higher levels.

Aufbau principle of electronic configuration

The electron enter on the orbital of the lowest energy first. The order is as given below:

1s < 2s < 2p < 3s <3p < 4s < 3d < 4p

4s has lower energy than 3d.

So, electrons enter 4s first then 3d.

Electron configuration:

H (1): 1s1

He (2): 1s2

Li (3): 1s2 2s1

C (6): 1s2 2s2 2p2

N (7): 1s2 2s2 2p3

O (8): 1s2 2s2 2p4

Electronic structure showing unpaired electrons in the last orbital:

N (7): 1s2 2s2 2px1 2py1 2pz1

The unpair electrons

in the outershell of nitrogen atom

The nitrogen atom has 3 electrons on the 2p orbitals. The distribution of three electrons can be such a
way that the unpair will be maximum. So, the arrangement will be 1, 1, and 1. This Hund’s rule.

O (8): 1s2 2s2 2px22py1 2pz1 :

 The unpair electrons of
oxygen atom in the its outer shell

For oxygen atoms, 2p orbitals have 4 electrons. The distribution of four electrons can be such a way that
the unpair will be maximum. So, the arrangement will be 2,1, and 1. There is spin pair on the 2px
orbitals. The two electrons are in pairing. One is spinning in one direction and another is the opposite
direction. This is the exclusion principle of Pauli.

More electronic configurations:

P (15): 1s2 2s2 2p6 3s2 3p3

S (16): 1s2 2s2 2p6 3s2 3p4

Ca (20): 1s2 2s2 2p6 3s2 3p6 4s2

For Sc, the electrons enter at 4s he 3d

Sc (21): 1s2 2s2 2p6 3s2 3p6 4s2 3d1

Cr is an exception. Here the 1 electron comes at 4s and 3d will be 5. It will be half-filled. Half-filled is the
stable form. So, one electron comes from 4s.

Cr(24): 1s2 2s2 2p6 3s2 3p6 4s1 3d5

Another Cu is another exceptional. Here For this case, 4s will be 1 and 3d will be 10. 1o electron in 3d
orbital is stable. So, 1 electron will come from 4s.

Cu (29): 1s2 2s2 2p6 3s2 3p6 4s1 3d10

Electronic Configurations of ions

For making positive ions, electrons from the outer shell will leave first. For Ca ion, the electrons leave for
4s first.

So, the electronic configuration of calcium ion as given below:

Ca2+ : 1s2 2s2 2p6 3s2 3p6

Cu2+: 1s2 2s2 2p6 3s2 3p6 3d9

Zn2+: 1s2 2s2 2p6 3s2 3p6 3d10

Ionisation Energy
The explanation of ionisation energy:

The ionisation energy is the energy required to remove the most loosely held electron from one mole of
gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.

Things to notice about the equation

 The state symbols – (g) – are essential.

 Unit is kJ mol-1

For the element X

X(g) → X+(g) + e–

For the element C

C(g) → C+(g) + e– ;
First ionisation energy, ∆Hi1

C+(g) → C2+(g) + e– ; Second ionisation energy, ∆Hi2

C2+(g) → C3+(g) + e– ; Third ionisation energy, ∆Hi3

∆Hi3 > ∆Hi2 > ∆Hi1

Because more energy is needed to remove electrons from the positive ions.

The magnitude of ionisation energy depends on 4 factors:

The magnitude of ionization energy

 The distance of the electron from the nucleus

 The number of positive charges in the nucleus

 The degree of shielding outer electrons by the electrons of inner shells.

 Spin-pair repulsion

The distance of the electron from the nucleus

The distance between the nucleus and the outer electron

  Attraction falls off very rapidly with distance.

 An electron close to the nucleus will be much more strongly attracted than one further away.

 In a group up to down ionisation energy decreases.

The number of positive charges in the nucleus

The positive charge in the nucleus

The more protons there are in the nucleus, the more positively charged the nucleus is, and the more
strongly electrons are attracted to it.

In the periodic table from left to right nuclear charge increase and ionization energy increases.

The degree of shielding outer electrons by the electrons of inner shells.

The shielding or screening effect

 The repulsion by the electrons of the inner shell to the electrons of the outer shell is called
shielding.

 The more the inner shells, the greater the shielding.

 In a group up to down ionisation energy increases because of greater shielding.

 K is more shielded than Na.

Spin-pair repulsion

The spin-pair repulsion

  The electron removed from the oxygen is from the orbital which contains a pair of electrons.

 The extra repulsion between the pair of electrons in this orbital in less energy being needed to
remove an electron.

 So, ∆Hi1 for oxygen is lower, because of spin-pair repulsion.

The first ionisation energy of the third period

The first ionization energy of the element of period3

Overall ionisation energy increases because nuclear charge increases.

Two exceptions: Mg to Al decreases and P to S decreases.

For Mg, the last electron is on the 3s orbitals. And for Al the last electron is on the 3px orbital. This is in
higher energy or more shielded from the nucleus. So, less amount of energy is needed for Al.

For Phosphorus, 3p sub-shell is singly filled. For Sulphur, one 3p orbital has paired electrons. Paired
electrons repel.

Practice:

This question about the first ionisation of period 2 elements.

1. Why the Ne has the highest first ionisation energy?

 2. Compare the first ionization energy between Be and B.

3. Compare the first ionization energy between N and O.

The first ionisation energy of Group2 Elements

Down the group, atomic radius, and shielding increases. So, less amount of energy is to lose electrons.

Using ionisation energies to work out which group an element is in

Finding group in the periodic table using the successive ionization energy

There is a big jump, after the second ionisation energy. It means that there are 2 electrons that are
relatively easy to remove, while the third one is much more difficult.

Chemical Bonding
Sigma, Pi , Dative Bond, and Hybridisation
Many elements are unstable when they are alone. They make bonds with others to become stable.
Metal and non-metal atoms combine by ionic bonding. Non-metal and non-metal elements combine by
covalent bonding. Covalent bonding is the major part of this chapter.

In this part of chemical bonding, I’ll discuss the following points, which are the part of covalent bonding:

 Sigma bond

 Pi bond

 Dative bond

 Hybridization

Sigma bond

Sigma bond is one type of covalent bond, where a molecular orbital is formed by the end-to-end (direct)
overlap of atomic orbitals. It is denoted by the symbol σ.

Sigma bond is a strong covalent bond. Where there is a single covalent bond, a sigma bond is there.

In hydrogen molecule(H2), there is a single covalent bond of H-H. This bond is made by the direct
overlapping of s-s orbitals.

In the chlorine molecule (Cl2), there is the single covalent bond of Cl-Cl. This bond is made by the direct
overlapping of p-p orbitals.

In hydrogen chloride molecule (HCl), there is a single covalent bond of H-Cl. This bond is made by the
direct overlapping of s-p orbitals.

A clear description with diagrams is given in the following video:

Sigma bond

Pi bond

A pi bond is one type of covalent bond, where a molecular orbital is formed by the side to side (indirect)
overlap of atomic orbitals along a plane perpendicular to a line connecting the nuclei of the atoms. It is
denoted by the symbol π.
Pi bond is a less strong covalent bond than a sigma bond. One sigma bond and one pi bond are present
is double bonds. One sigma bond and two pi bonds are present in triple bond.

In the oxygen molecule (O2), there are the double covalent bonds of O=O. This bond is made by one
sigma and one pi bond.
One exercise for you about this topic.

The number of sigma and pi bonds

Dative bond

A coordinate bond is a covalent bond in which both electrons come from the same atom. It is also called
a coordination bond.
In ammonia, the nitrogen atom has one pair of lone pair electrons. On the other hand, hydrogen ion has
no electron. So, the nitrogen atom can make the dative bond with the hydrogen ion.

Dative Bond

The direction of the dative is from nitrogen atom to hydrogen atom.

Exercise: Show the dative bond in the following compounds:

Dative bond

Hybridization

Hybridization is the idea of mixing atomic orbitals into new hybrid orbitals. Hydrocarbons are formed
from Carbon and Hydrogen by hybridization.

Two things are involved here.

 Rearrangement of electrons

 Leveling energy of the electrons

Carbon has three types of hybridization:

1. SP3
 2. SP2

3. SP

SP3 hybridization:

One S orbital and three P orbitals are mixing together. All the alkanes have this type of hybridization.
The shape is tetrahedral and the bond angle is 109.50.

SP2 hybridization:

One S orbital and two P orbitals are mixing together. All the alkenes have this type of hybridization. The
shape is trigonal planar and the bond angle is 1200.
SP hybridization:

One S orbital and one P orbital are mixing together. All the alkynes have this type of hybridization. The
shape is linear and the bond angle is 1800.

Shapes of Covalent Molecules and Ions

Lone pair and bond pair electrons:

The lone pair and bond pair electrons

 You have to know the number of electron pairs there are at the bonding level. From the total electrons I
the bonding level you have to find both bonding pairs and lone pairs. In a covalent molecule, the
repulsion will be minimum. This is called valence shell electron repulsion theory.

The repulsion of the electrons as follows:

Lone pair -lone pair repulsion > Lone pair -bond pair repulsion> Bond pair -bond pair repulsion

Two electron pairs around the central atom:

Two electron pairs

It is forming 2 bonds so there are no lone pairs. The two bonding pairs arrange themselves at 180° to
each other, because that’s as far apart as they can get. The shape of the molecule is linear. An example
is barium chloride.

Three electron pairs around the central atom:

Three electron pairs

The 3 pairs arrange themselves as far apart as possible. They all lie in one plane at 120° to each other.
The arrangement is called trigonal planar. An example is boron trichloride.

Four electron pairs around the central atom:

Four electron pairs

Four electron pairs arrange themselves in space. A tetrahedron is a regular triangularly-based pyramid.
All the bond angles are 109.5°. An example is methane.

The bond angle and shape of Ammonia, NH3 :

The shape of ammonia molecule

Total pair of electrons around the nitrogen atom = 4. Amount electrons in the bonding level, the three
pair bond pair and one pair lone pair. The repulsion between lone pair and bond pair is greater. So, the
tetrahedral shape is deformed. The bond angle is from 109.5° to 107° and the shape is pyramidal.

The bond angle and shape of water, H2O:

The shape of water molecule

Total pair of electrons around the oxygen atom = 4. Amount electrons in the bonding level, the two pair
bond pair and two pair lone pair. The repulsion between lone pair and lone pair is greater. So, the
tetrahedral shape is deformed. The bond angle is from 109.5° to 104.5° ° and he shape is bent shape or
V shape.

The bond angle and shape of ammonium ion, NH4+:

The shape of ammonium ion

Methane and the ammonium ion are said to be isoelectronic. That means the same number of
electrons in the bonding level. All the bond angles are 109.5°. Four electron pairs arrange themselves in
space in what is called a tetrahedral arrangement.

The bond angle and shape of hydroxonium ion, H3O+ :

The shape of hydroxonium ion

This gives 4 pairs, 3 of which are bond pairs and 1 pair is lone pair. This is isoelectronic with ammonia. It
has an identical shape – pyramidal and bond angle 107°.

Five electron pairs around the central atom:

Five electron pairs

5 pairs around the central atom, all are bond pairs. These 5 electron pairs take up a shape described as
a trigonal bipyramid. The bond angles are 120° and 90°.

Six electron pairs around the central atom:

Six electron pairs

6 bond pairs and all are bond pairs. They arrange themselves entirely at 90°. The shape is octahedral.

The Shape of XeF4:

The shape xenon hexafluoride

4 bonding pairs and 2 lone pairs. There are two possible structures, but in one of them the lone pairs
would be at 90°.

Instead, they go opposite each other. XeF4 is described as square planar.

A critical example, ClF3:

This interhalogen compound. The chlorine is forming three bonds and 2 lone pairs. Bond angle is
900 and the shape is T-shaped.

The practice of bond angle and shapes:

Complete the blank places from the following table. For each case the formula for bond angle and shape
id given, you have to give examples.
 Electronegativity and Bond Polarity

Bond energy

Bond energy is the amount of energy needed to break a particular covalent bond. It is also called “bond
strength “.

The bond-breaking energy is positive and making energy is negative.

Cl-Cl bond energy is greater than Br-Br bond energy. Because the distance between the two nuclei of Br
atoms is greater and less energy is needed to break.

The bond energy is measured by gas molecules to gas atoms. For example, iodine solid at RTP. To
calculate the bond energy, first, you have to convert iodine solid to gas, then you can find the bond
energy.

I2 (g) → 2I-I (g)

I have given the more information about bond energy in the enthalpy change chapter.

Bond Length
Bond Length

Bond length is the distance between the two nuclei in a covalent bond. It is inversely proportional to the
bond energy. The greater the bond length, the less the bond energy.

Cl-Cl bond length is smaller than Br-Br bond length. It depends on the radius of the atom

Electronegativity

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It is a

periodic property.

Fluorine is having the highest electronegativity. In its period, it has the highest electronegativity. Also, F
has the highest electronegativity in its group.

Electronegativity depends on the following:

 the number of protons in the nucleus;

 the distance from the nucleus;

 the amount of screening by inner electrons

In H-Cl the molecule, Cl is more electronegative. So, Cl’s side is partially negative, and H’s side is partially
positive.

Bond Polarity

Bond Polarity

A bond of covalent the molecule has the polarity, where two atoms of the molecule have the unequal
electronegativity.

The electrons are pulled towards the atom with greater electronegativity. So, this side is negative and
the opposite side is positive.

H–Cl bond has the bond polarity. Cl’s side is partially negative due to the Cl atom is more
electronegative. So, the shared pair of electrons are near to the Cl atom. That is why, H-Cl bond is
polar.

N2 molecule has the triple bond between two nitrogen atoms. But these two atoms have the same
electronegativity. So, it has no bond polarity.

Dipole

HCl Dipole
A covalent molecule is a dipole, where the atoms of the molecule have substantially
different electronegativity. One side of the molecule is positive and another side is negative. This is also
called a permanent dipole.

Water is dipole because the difference of electronegativity of hydrogen atom and oxygen atom is
substantial.

Dipole-dipole force

Dipole-dipole force

The attraction force between two permanent dipoles is called dipole-dipole force. HF is a dipole. One
molecule of HF attracts another molecule of HF. The attraction force between these dipoles is called
dipole-dipole force. This dipole-dipole force is called the intermolecular force.

Intermolecular forces: H bond and vdW Forces

The general concept of intermolecular forces
Intermolecular force

Intermolecular force is the force between two molecules. There is an attractive force between one
molecule of water to another.

H2O —–>—-<—– H2O

The melting and boiling point of a substance depends on the strength of intermolecular forces. The
higher the strength of intermolecular force, the greater the boiling point. The strength of intermolecular
force depends on the dipole of the molecule. The details about dipole have been described in
the proceeding topic of this chapter.

The three major types of intermolecular interactions are:

 dipole–dipole interactions

 van der Waals forces

 hydrogen bonds

Hydrogen bond
Hydrogen bond

A hydrogen bond is an intermolecular force between polar molecules in which hydrogen (H) is bound to
a highly electronegative atom, such as

nitrogen (N),

oxygen (O) or

fluorine (F).

Some features of H bonding:

 The strongest intermolecular force.

 Permanent dipole-dipole force

 Permanent dipole-dipole force, where H is the common

 Permanent dipole-dipole force, were

N with H or

O with H or

F with H present

For drawing H Bonding 3 things must have to mention:

 Lone pair electrons

 Partially positive and partially negative

 Labeling the H bond.

Drawing Hydrogen bond mentioning lp electrons, partially positive and negative sign and labeling the H
bond

Some important information of H bonds:

 Water has the greater boiling point than H2S, because water has H bond but H2S has not.

 NH3 has greater boiling point than PH3 because NH3 has H bond but PH3 has not.

 In HCl molecules there is no H Bond. Only dipole – dipole force is there.

 Alcohols dissolve in water, because they can make hydrogen bond with water.

 Ethanoic acids make dimers due to hydrogen bond.

The hydrogen bond between normal and branched organic molecule:

Butanol-1 (CH3CH2CH2CH2OH) has a boiling point of 1170C. This is a normal organic molecule.

Another isomer of butanol-1 is 2-methyl propan-1-ol (CH3CH(CH3)CH2OH). This branched organic

molecule. It has a boiling point of 1080 C.
The unbranched alcohol has the greater boiling point because the intermolecular distance is greater due
to branch.

Practices about drawing H bond:

1. Water and water

2. Water and ammonia

3. Water and ethanol

4. Ammonia and ammonia

5. Ethanol and ammonia

6. Hydrofluoric acid and water

7. Hydrofluoric acid and Hydrofluoric acid

van der Waals’ Force

Van der Waal’s force between two temporary dipoles

Van der Waals’ force is the weak intermolecular force working between temporary fluctuating diploe
and temporary induced dipole.

The temporary dipole is produced, where there is very little difference of electronegativity or there is no
difference in electronegativity.

For the different arrangements of electrons, the temporary fluctuating and induced dipoles are created.
Temporary fluctuating and induced dipoles

At room temperature and pressure, chlorine is gas, bromine is liquid and iodine is solid due to the
greater van der Waal’s force.

The boiling point order from low to high:

Cl2(g) < Br2(l) < l2(s)

For inert gases of the periodic table from up to down (He to Rn), boiling points increase due to greater
van der Waal’s force.

The boiling point order from low to high:

He(g) < Ne(g) < Ar(g) < Kr(g) < Xe (g)

The boiling point of n-butane (CH3CH2CH2CH3) is greater than 2-methylpropane (CH3CH (CH3) CH3). The
branched alkane has a greater distance between the molecules, so der Waal’s force is lower.

The difference in boiling points of pentane (CH3CH2 CH2CH2CH3) and 2,2-dimethylpropane

(CH3C(CH3)2 CH3) can be explained by the strength of the van der Waals’ forces.
The vdW forces of Pentane and 2,2-dimethylpropane

HDPE (High-density polyethylene) has more crystalline regions where the molecules are closer together
than LDPE (Low-density polyethylene). The total van der Waals’ forces are greater, so HDPE is the
stronger of the two.

Q related to vdW force:

Why HI has a higher boiling point than HBr?