XII STD - Chemistry Vol-1 English Medium PDF
XII STD - Chemistry Vol-1 English Medium PDF
XII STD - Chemistry Vol-1 English Medium PDF
CHEMISTRY
VOLUME - I
Content Creation
The wise
possess all
II
III
Chemical Science
National Chemical Chemical Engineering
Laboratory(NCL) Catalytic materials www.ncl-india.org
Pune, Maharashtra Nano materials
Chemical looping combustion(CLC)
Li Batteries
Central Electrochemical
Corrosion www.cecri.res.in
Research Institute
(CECRI) Bio-Sensors
Karaikudi, Tamil Nadu. Materials Electrochemistry
Electro catalysis and Fuel Cells
IV
Biopolymer
Laboratory of Advanced Fuel Cells
Research in Polymeric Polymer Nano composite
www.larpm.gov.in
Materials (LARPM) Carbon Nanotubes
Bhubaneswar, disha. Polymer Blends & Alloys
E Waste Recycling
NEET
MBBS.,
(National Eligibility Written test www.cbscneet.nic.in
BDS.,
cum Entrance test
AIIMS
(All India Institute of Written test MBBS., www.aiimsexams.org
Medical Sciences)
JIPMER
(Jawaharlal Institute Of
Computer based test MBBS., www.jipmer.edu.in
Postgraduate Medical
Education & Research)
AFMC M.B.B.S.,
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College Entrance Exam in Armed Forces)
Integrated M.Tech.,
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Integrated M.Sc.,
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B.Sc.,
(Indian Institute of Written test www.iisc.ernet.in
(4 years)
Science Bangalore)
VI
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(National Aptitude Test Computer Based Test B.Arch., www.nata.in
in Architecture)
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BFTech.,
fashion technology)
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fashion technology) apparel design)
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Exam for Design
B.Tech.,
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(Avionics/ Aerospace
(Indian Institute of Written test www.iist.ac.in
Engineering/ Physical
Space Technology)
Science)
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(Common Law Written test Integrated LLB ( 5 years) www.cbscneet.nic.in
Admission Test)
NCHMCT
B.Sc.,
(National Council for Hotel
Written test (Hospitality and Hotel www.nchm.nic.in
Management Catering
Administration)
Technology Joint Entrance Exam)
B.Tech., Marine
AIMNET Engineering
(All India Merchant Navy Written Exam B.Sc., Nautical Science www.aim.net.co.in
Entrance Test) B.Tech.,Navel Architecture
and Ship Building
VII
Department of Science
Kishore Vaigyanik Protsahan
and Technology (DST), June to August
Yojana (KVPY)
Government of India
Department of Science
Inspire
and Technology (DST), October to December
Scholarship
Government of India
MOMA scholarship
Ministry of Minority Affairs,
(applicable only for Minority July to September
Government of India
students)
Saksham Scholarship
AICTE September to November
(Applicable for Disable students)
VIII
CHEMISTRY
UNIT I
Metallurgy 01
UNIT 2
p-Block Elements-I 26
UNIT 3
p-Block Elements - II 56
UNIT 4
Transition and Inner Transition Elements 100
UNIT 5
Coordination Chemistry 130
UNIT 6
Solid State 176
UNIT 7
Chemical Kinetics 204
PRACTICAL 256
IX
Learning Objectives
After studying this unit, students will be
Harold Johann Thomas Ellingham
able to
(1897–1975)
Ellingham was a British describe various methods of
physical chemist, best known concentrating ores
for his Ellingham diagrams. explain various methods of extraction
Ellingham diagram summarizes a of crude metals
large amount of information about
apply thermodynamic principles to
extractive metallurgy, and are
metallurgical processes
useful in predicting the favourable
thermodynamic conditions under predict the favourable conditions for
which an ore will be reduced to the reduction process using Ellingham
its metal. Ellingham was able to diagram
compare the temperature stability describe the electrochemical principles
of many different oxides. The of metallurgy
phenomenon of reduction of
apply the electrochemical principles in
metal oxides into free metal by
the extraction of metals
carbon or carbon monoxide was
known before Ellingham's time, explain the electrode reactions in
but Ellingham demonstrated it in electrolytic refining.
a scientific manner.
list the uses of Al, Zn, Fe, Cu and Au
Metallurgy relate to the science and technology of metals. In nature, only a few metals
occur in their native state, all other metals occur in a combined state as their oxides,
sulphides, silicates etc... The extraction of pure metals from their natural sources, is linked
to the history of human civilisation. Ancient people used the available materials in their
environment which includes fire and metals, and they were limited to the metals available
on the earth's surface. In the modern world, we use a wide range of metals in our daily life,
which is the result of the development of metallurgical knowledge over thousands of years.
Our need for the materials with specific properties have led to production of many metal
alloys. It is essential to design an eco-friendly metallurgical process that would minimize
waste, maximize energy efficiency. Such advances in metallurgy is vital for the economic
and technical progress in the current era. In this unit we will study the various steps
involved in the extraction of metals and the chemical principles behind these processes.
Azurite 2CuCO3.Cu(OH)2
Ammonia leaching
When a crushed ore containing nickel, copper and cobalt is treated with
aqueous ammonia under suitable pressure, ammonia selectively leaches these metals by
forming their soluble complexes viz. [Ni(NH3)6]2+, [Cu(NH3)4]2+, and [Co(NH 3)5H2O]3+
respectively from the ore leaving behind the gangue, iron(III) oxides/hydroxides and
aluminosilicate.
Alkali leaching
In this method, the ore is treated with aqueous alkali to form a soluble complex.
For example, bauxite, an important ore of aluminum is heated with a solution of sodium
hydroxde or sodium carbonate in the temperature range 470 - 520 K at 35 atm to form
soluble sodium meta-aluminate leaving behind the impurities, iron oxide and titanium
oxide.
Al2O3 (s) + 2NaOH (aq) + 3H2O (l) 2Na[Al(OH)4] (aq)
The hot solution is decanted, cooled, and diluted. This solution is neutralised by passing
CO2 gas, to the form hydrated Al2O3 precipitate.
Acid leaching
Leaching of sulphide ores such as ZnS, PbS etc., can be done by treating them with hot
aqueous sulphuric acid.
In this process the insoluble sulphide is converted into soluble sulphate and elemental sulphur.
Evaluate yourself 1
1. Write the equation for the extraction of silver by leaching with sodium cyanide and
show that the leaching process is a redox reaction.
∆
2ZnS + 3O2 2ZnO + 2SO2
S8 + 8O2 8SO2
P4 + 5O2 P4O10
Calcination
Calcination is the process in which the concentrated ore is strongly heated in the absence
of air. During this process, the water of crystallisation present in the hydrated oxide escapes as
moisture. Any organic matter (if present) also get expelled leaving behind a porous ore. This
method can also be carried out with a limited supply of air.
For examples,
During calcination of carbonate ore, carbon dioxide is expelled
Evaluate yourself 2
2. Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to
make refractory bricks. Write the decomposition reaction.
In this extraction, a basic flux, limestone (CaO) is used. Since the silica gangue present in
the ore is acidic in nature, the limestone combines with it to form calcium silicate (slag).
In the extraction of copper from copper pyrites, the concentrated ore is heated in a
reverberatory furnace after mixing with silica, an acidic flux. The ferrous oxide formed due
to melting is basic in nature and it combines with silica to form ferrous silicate (slag). The
remaining metal sulphides Cu2S and FeS are mutually soluble and form a copper matte.
The metallic copper is solidified and it has blistered appearance due to evolution of SO2
gas formed in this process. This copper is called blistered copper.
Reduction by carbon:
In this method the oxide ore of the metal is mixed with coal (coke) and heated strongly in
a furnace (usually in a blast furnace). This process can be applied to the metals which do not
form carbides with carbon at the reduction temperature.
Examples:
Reduction by hydrogen:
This method can be applied to the oxides of the metals (Fe, Pb, Cu) having less electro-
positive character than hydrogen.
Ag2O (s)+ H2 (g) 2Ag (s) + H2O (l)
Fe3O4 (s) + 4H2 (g) 3Fe (s) + 4H2O (l)
Nickel oxide can be reduced to nickel by using a mixture of hydrogen and carbon monoxide
(water gas)
2NiO (s) + CO (g) + H2 (g) 2Ni (s) + CO2 (g) + H2O (l)
Reduction by metal:
Metallic oxides such as Cr2O3 can be reduced by an aluminothermite process. In this
process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible. To
initiate the reduction process, an ignition mixture (usually magneisium and barium peroxide)
is used.
Active metals such as sodium, potassium and calcium can also be used to reduce the
metal oxide
B2O3 + 6Na 2B + 3Na2O
Rb2O3 + 3Mg 2Rb + 3MgO
TiO2 + 2Mg Ti + 2MgO
1250 K
ThO2 + 2Ca Th + 3CaO
Auto-reduction:
Simple roasting of some of the ores give the crude metal. In such cases, the use of
reducing agents is not necessary. For example, mercury is obtained by roasting of its ore
cinnabar (HgS)
The above reduction may be carried out with carbon. In this case, the reducing agent
carbon may be oxidised to either CO or CO2.
C + O2 CO2 (g) ------ (2)
10
100
O
= 2Ag 2
4A g + O2 O
0 2 Hg 2
=
O2
g+
2H
–100
–200 NiO
2C u 2O =2
+ O2 = O 2
4Cu 2N
i+
–300
= 2FeO
2 Fe+ O 2
–400
G° = RT Inp O2 / kJmol O2
C + O2 = CO2
–1
3 CrO 3
ZnO = 2/
–500 =2 r + O2
+O 2
4/3C
2Zn 2M nO
–600 + O2 =
2Mn
2C +
O2 =
–700 2CO
A l 2O 3
–800 = 2/3
O2
Al +
4/3
–900 O
Ca
=2
+O
2
a
–1000 2C
mgO
=2
+O
2
–1100 g
2M
–1200
0
00
00
00
00
00
00
00
00
0
0
20
40
60
80
10
12
14
16
18
20
22
24
T / °C
where, ΔH is the enthalpy change , T the temperature in kelvin and ΔS the entropy
change. For an equilibrium process, ΔG⁰ can be calculated using the equilibrium constant by
the following expression
ΔG⁰ =-RT lnKp
Harold Ellingham used the above relationship to calculate the ΔG⁰ values at various
temperatures for the reduction of metal oxides by treating the reduction as an equilibrium
process.
He has drawn a plot by considering the temperature in the x-axis and the standard free
energy change for the formation of metal oxide in y-axis. The resultant plot is a straight line with
ΔS as slope and ΔH as y-intercept. The graphical representation of variation of the standard
Gibbs free energy of reaction for the formation of various metal oxides with temperature is
called Ellingham diagram
Observations from the Ellingham diagram.
1. For most of the metal oxide formation, the slope is positive. It can be explained as follows.
Oxygen gas is consumed during the formation of metal oxides which results in the decrease
in randomness. Hence, ΔS becomes negative and it makes the term, TΔS positive in the
straight line equation.
2. The graph for the formation of carbon monoxide is a straight line with negative slope. In
this case ΔS is positive as 2 moles of CO gas is formed by the consumption of one mole of
oxygen gas. It indicates that CO is more stable at higher temperature.
3. As the temperature increases, generally ΔG value for the formation of the metal oxide
become less negative and becomes zero at a particular temperature. Below this temperature,
ΔG is negative and the oxide is stable and above this temperature ΔG is positive. This
general trend suggests that metal oxides become less stable at higher temperature and their
decomposition becomes easier.
4. There is a sudden change in the slope at a particular temperature for some metal oxides like
MgO, HgO. This is due to the phase transition (melting or evaporation).
12
13
-
Reaction at cathode 2Al3+ (melt) + 6e 2Al (l)
Reaction at anode -
6O2- (melt) 3O2 + 12e
Since carbon acts as anode the following reaction also takes place on it.
14
Due to the above two reactions, anodes are slowly consumed during the electrolysis. The
pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction
can be written as follows.
4Al3+ (melt) + 6O2- (melt) + 3C (s) 4Al (l) + 3CO2 (g)
Evaluate yourself 4
4. Metallic sodium is extracted by the electrolysis of brine (aq. NaCl). After electrolysis
the electrolytic solution becomes basic in nature. Write the possible electrode
reactions.
1.6.1 Distillation
This method is employed for low boiling volatile metals like zinc (boiling point 1180 K)
and mercury (630 K). In this method, the impure metal is heated to evaporate and the vapours
are condensed to get pure metal.
1.6.2 Liquation
This method, is employed to remove the impurities with high melting points from metals
having relatively low melting points such as tin (Sb; mp= 904 K), lead (Pb; mp=600 K), mercury
(Hg; mp=234 K), and bismuth (Bi; mp=545 K). In this process, the crude metal is heated to
form fusible liquid and allowed to flow on a sloping surface. The impure metal is placed on
sloping hearth of a reverberatory furnace and it is heated just above the melting point of the
metal in the absence of air, the molten pure metal flows down and the impurities are left
behind. The molten metal is collected and solidified.
15
During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The
positively charged silver cations migrate towards the cathode and get discharged by gaining
electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be
refined by this process in a similar manner.
1.6.4 Zone Refining
This method is based on the principles of fractional crystallisation. When an impure
metal is melted and allowed to solidify, the impurities will prefer to be in the molten region. i.e.
impurities are more soluble in the melt than in the solid state metal. In this process the impure
metal is taken in the form of a rod. One end of the rod is heated using a mobile induction
heater which results in melting of the metal on that portion of the rod. When the heater is
slowly moved to the other end the pure metal crystallises while the impurities will move on
to the adjacent molten zone formed due to the movement of the heater. As the heater moves
further away, the molten zone containing impurities also moves along with it. The process is
repeated several times by moving the heater in the same direction again and again to achieve
the desired purity level. This process is carried out in an inert gas atmosphere to prevent the
oxidation of metals . Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are
used as semiconductor are refined using this process.
1.6.5 Vapour phase method
In this method, the metal is treated with a suitable reagent which can form a volatile
compound with the metal. Then the volatile compound is decomposed to give the pure metal.
We can understand this method by considering the following process.
Mond process for refining nickel:
The impure nickel is heated in a stream of carbon monoxide at around 350 K. The nickel
reacts with the CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left
behind.
Ni (s) + 4 CO (g) Ni(CO)4 (g)
On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure
metal.
Ni(CO)4 (g) Ni (s) + 4 CO (g)
16
The volatile titanium tetraiodide vapour is passed over a tungsten filament at a temperature
aroud 1800 K. The titanium tetraiodide is decomposed and pure titanium is deposited on the
filament. The iodine is reused.
TiI4 (vapour) Ti (s) + 2I2 (s)
ÂÂ Many heat exchangers/sinks and our day to day cooking vessels are made of aluminium.
ÂÂ It is used as wraps (aluminium foils) and is used in packing materials for food items,
ÂÂ Aluminium is not very strong, However , its alloys with copper, manganese, magnesium
and silicon are light weight and strong and they are used in design of aeroplanes and other
forms of transport.
ÂÂ As Aluminium shows high resistance to corrosion, it is used in the design of chemical
reactors, medical equipments,refrigeration units and gas pipelines.
ÂÂ Aluminium is a good electrical conductor and cheap, hence used in electrical overhead
electric cables with steel core for strength.
1.7.1 Application of Zn
ÂÂ Metallic zinc is used in galvanising metals such as iron and steel structures to protect them
from rusting and corrosion.
ÂÂ Zinc is also used to produce die-castings in the automobile, electrical and hardware
industries
ÂÂ Zinc oxide is used in the manufacture of many products such as paints, rubber, cosmetics,
17
1.7.1 Application of Fe
ÂÂ Iron is one of the most useful metals and its alloys are used everywhere including bridges,
electricity pylons, bicycle chains, cutting tools and rifle barrels.
ÂÂ Cast iron is used to make pipes, valves and pumps stoves etc...
ÂÂ Magnets can be made from iron and its alloys and compounds.
ÂÂ An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in
architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for
making cables, automobiles and aeroplane parts. Chrome steels are used for maufacturing
cutting tools and curshing machines
1.7.1 Application of Cu
Copper is the first metal used by the human and extended use of its alloy bronze resulted in a
new era,'Bronze age'
Copper is used for making coins and ornaments along with gold and other metals.
Copper and its alloys are used for making wires, water pipes and other electrical parts
1.7.1 Application of Au
ÂÂ Gold, one of the expensive and precious metals. It is used for coinage, and has been used as
standard for monetary systems in some countries.
ÂÂ It is used extensively in jewellery in its alloy form with copper. It is also used in electroplating
to cover other metals with a thin layer of gold which are used in watches, artificial limb
joints, cheap jewellery, dental fillings and electrical connectors.
ÂÂ Gold nanoparticles are also used for increasing the efficiency of solar cells and also used
an catalysts.
18
Summary
19
EVALUATION
a) Al b) Ni c) Cu d) Zn
6. Which of the following statements, about the advantage of roasting of sulphide ore before
reduction is not true?
d) Carbon and hydrogen are suitable reducing agents for metal sulphides.
20
Column-I Column-II
(v) Purification of Ni
A B C B
a) Magnetite b) Haematite
c) Galena d) Cassiterite
21
a) Silver b) Lead
c) Copper d) iron
16. Extraction of gold and silver involves leaching with cyanide ion. silver is later recovered
by (NEET-2017)
a) Fe b) Cu
c) Mg d) Zn
18. The following set of reactions are used in refining Zirconium
Zr (impure) + 2I2 523
K
→ ZrI4
This method is known as
ZrI4 1800K
→ Zr (pure) + 2I2
a) Liquation b) van Arkel process
c) Zone refining d) Mond’s process
19. Which of the following is used for concentrating ore in metallurgy?
a) Leaching b) Roasting
c) Froth floatation d) Both (a) and (c)
20. The incorrect statement among the following is
d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution
22
a) Δ S Vs T b) Δ G0 Vs T
1
c) Δ G 0 Vs d) Δ G Vs T
0 2
T
23. In the Ellingham diagram, for the formation of carbon monoxide
∆ S0 ∆ G0
a) is negative b) is positive
∆ T ∆ T
∆ G0 ∆T
c) is negative d) initially 0
is positive, after 7000C ,
∆ T ∆G
∆G 0
∆ T is negative
a) Cr2O3 + 2Al →
Al 2O3 + 2Cr
b) Al 2O3 + 2Cr →
Cr2O3 + 2Al
c) 3TiO2 + 4Al →
2 Al 2O3 + 3Ti
d) none of these
25. Which of the following is not true with respect to Ellingham diagram?
a) Free energy changes follow a straight line. Deviation occurs when there is a phase
change.
b) The graph for the formation of CO2 is a straight line almost parallel to free energy axis.
c) Negative slope of CO shows that it becomes more stable with increase in temperature.
d) Positive slope of metal oxides shows that their stabilities decrease with increase in
temperature.
Answer the following questions:
1. What are the differences between minerals and ores?
2. What are the various steps involved in extraction of pure metals from their ores?
3. What is the role of Limestone in the extraction of Iron from its oxide Fe2O3 ?
4. Which type of ores can be concentrated by froth floatation method? Give two examples for
such ores.
5. Out of coke and CO, which is better reducing agent for the reduction of ZnO? Why?
6. Describe a method for refining nickel.
7. Explain zone refining process with an example using the Ellingham diagram given below.
23
24
minerals
magnetic separation
cyanide
leaching acid
Roasting
conversion of ore into oxides
smelting calcination
reduction by C
reduction of metal oxides
or H or metal
Auto reduction
Distillation
Liquation
Refining process principles of metalurgy
zone refining
Ellingham diagram
25
26
We have already learnt the classification of elements into four blocks namely s, p, d and
f. We have also learnt the properties of s-block elements and their important compounds in
the XI standard. This year we learn the elements of remaining blocks, starting with p-block
elements.
The elements in which their last electron enters the 'p' orbital, constitute the p-block
elements. They are placed in 13th to 18th groups of the modern periodic table and the first
member of the groups are B, C, N, O, F and He respectively. These elements have quite
varied properties and this block contains nonmetals, metals and metalloids. Nonmetallic
elements of this group have more varied properties than metals. The elements of this block
and their compounds play an important role in our day to day life, for example, without
molecular oxygen we cannot imagine the survival of living system. The most abundant
metal aluminium and its alloys have plenty of applications ranging from household
utensils to parts of aircraft. The semi conducting nature of elements such as silicon and
germanium made a revolutionary change in the field of modern electronics. In this unit
we discuss the properties of first three groups (Group 13 - 15) of p-block elements namely
boron, carbon and nitrogen family elements and their important compounds.
Evaluate yourself :
Why group 18 elements are called inert gases? Write the general electronic configuraton
of group 18 elements
27
Group No 13 14 15 16 17 EN-
IE1-800.63 IE1-800.63 IE1-1402.33 IE1-1313.94 IE1-1681.04 IE1-2080.67
28
29
30
31
Cr + nB 1500 K CrBn
Reduction of borontrihalides:
Reduction of borontrichloride with a metal assisted by dihydrogen gives metal borides.
1500 K
BCl3 + W WB + Cl2+HCl
H2
Formation of hydrides:
Boron does not react directly with hydrogen. However, it forms a variety of hydrides
called boranes. The simplest borane is diborane - B2H6. Other larger boranes can be prepared
from diborane. Treatment of gaseous boron trifluoride with sodium hydride around 450 K
gives diborane. To prevent subsequent pyrolysis, the product diborane is trapped immediately.
450 K
2BF3 + 6NaH B2H6 + 6NaF
32
Formation of oxides:
When boron is heated with oxygen around 900 K, it forms its oxide.
Boron reacts with fused sodium hydroxide and forms sodium borate.
2B + 6NaOH 2Na3BO3 + 3H2
Uses of boron:
1. Boron has the capacity to absorb neutrons. Hence, its isotope 10B5 is used as moderator in
nuclear reactors.
2. Amorphous boron is used as a rocket fuel igniter.
3. Boron is essential for the cell walls of plants.
4. Compounds of boron have many applications. For example eye drops, antiseptics, washing
powders etc.. contains boric acid and borax. In the manufacture of Pyrex glass , boric oxide
is used.
33
Properties:
Boric acid is a colourless transparent crystal. It is a very weak monobasic acid and,
it accepts hydroxyl ion rather than donating proton.
B(OH)3 + 2H2O H3O+ + [B(OH)4]-
It reacts with sodium hydroxide to form sodium metaborate and sodium tetraborate.
NaOH + H3BO3 NaBO2 + 2H2O
Acton of ammonia
Fusion of urea with B(OH)3, in an atmosphere of ammonia at 800 - 1200 K gives boron nitride.
Δ
B(OH)3 + NH3 BN + 3H2O
When boric acid or borate salt is heated with ethyl alcohol in presence of conc. sulphuric
acid, an ester, trialkylborate is formed. The vapour of this ester burns with a green edged flame
and this reaction is used to identify the presence of borate.
Conc.
H3BO3 + 3C2H5OH B(OC2H5)3+ 3H2O
H2SO4
Note: The trialkyl borate on reaction with sodium hydride in tetrahydrofuron to form a
coordination compound Na[BH(OR)3], which acts as a powerful reducing agent.
On heating magnesium boride with HCl a mixture of volatile boranes are obtained.
2Mg3B2 + 12HCl 6MgCl2 + B4H10 + H2
B4H10 + H2 2B2H6
Properties:
Boranes are colourless diamagnetic compounds with low thermal stability. Diborane is a
gas at room temperature with sweet smell and it is extremely toxic. It is also highly reactive.
At high temperatures it forms higher boranes liberating hydrogen.
388 K
5B2H6 U - tube 2B5H11 + 4H2
198 - 373 K
2B2H6 B4H10 + H2
373 K
5B2H6 B10H14 + 8H2
sealed tube
473 - 523 K
5B2H6 2B5H9 + 6H2
523 K
10B2H6 2B5H9 + 2B 5H10 + 11H2
Red hot
B2H6 2B + 3H2
Diboranes reacts with water and alkali to give boric acid and metaborates respectively.
B2H6 + 6H2 O 2H3BO3 + 6H2
B2H6 + 2NaOH +2H2 O 2NaBO2 + 6H2
Action of air:
At room temperature pure diborane does not react with air or oxygen but in impure form
it gives B2O3 along with large amount of heat.
B2H6 + 3O2 B2O3 + 3H2O ΔH = -2165 KJ mol-1
36
Hydroboration:
Diborane adds on to alkenes and alkynes in ether solvent at room temperature. This
reaction is called hydroboration and is highly used in synthetic organic chemistry, especially
for anti Markovnikov addition.
B2H6 + 6RCH =CHR 2B(RCH-CH2R)3
H N H H N H
High temp B B B B
H B H H B H
H H
Structure of diborane:
H
In diborane two BH2 units
are linked by two bridged H
H
37
Uses of diborane:
1. Diborane is used as a high energy fuel for propellant
2. It is used as a reducing agent in organic chemistry
3. It is used in welding torches
Properties:
Boron trifluoride has a planar geometry. It is an electron deficient compound and accepts
-
electron pairs to form coordinate covalent bonds. They form complex of the type [BX4] .
BF3 + NH3 F3 B ←NH3
BF3 + H2O F3B ←OH2
On hydrolysis, boric acid is obtained. This then gets converted into fluoro boric acid.
4BF3 + 12H2O 4H3BO3 + 12HF
+ -
3H3BO3 + 12HF 3H + 3[BF4] + 9H2O
+ -
4BF3 + 3H2O H3BO3 + 3H + 3[BF4]
38
McAfee Process:
Aluminium chloride is obtained by heating a mixture of alumina and coke in a current
of chlorine.
Al2O3 +3C + 3Cl2 2AlCl3 + 3CO2
On industrial scale it is prepared by chlorinating aluminium around 1000 K
1000 K
2Al + 3Cl2 2AlCl3
Properties:
Anhydrous aluminium chloride is a colourless, hygroscopic substance.
An aqueous solution of aluminium chloride is acidic in nature. It also produces hydrogen
chloride fumes in moist air.
AlCl3 + 3H2O Al(OH)3 + 3HCl
It behaves like a Lewis acid and forms addition compounds with ammonia, phosphine
and carbonylchloride etc... Eg. AlCl3.6NH3.
2.2.9 Alums:
The name alum is given to the double salt of potassium aluminium sulphate [K2SO4.
Al2(SO4)3.24.H2O]. Now a days it is used for all the double salts with M'2SO4.M"2(SO4)3.24H2O,
+
where M' is univalent metal ion or [NH4] and M" is trivalent metal ion.
39
Properties
Potash alum is a white crystalline solid it is soluble in water and insoluble in alcohol. The
aqueous solution is acidic due to the hydrolysis of aluminium sulphate it melts at 365 K on
heating. At 475 K loses water of hydration and swells up. The swollen mass is known as burnt
alum. Heating to red hot it decomposes into potassium sulphate, alumina and sulphur trioxide.
500 K
K2SO4.Al2(SO4)3.24 H2O K2SO4.Al2(SO4)3 + 24 H2O
K2SO4.Al2(SO4)3 Red hot K2SO4 + Al2O3 + 3SO3
Potash alum forms aluminium hydroxide when treated with ammonium hydroxide.
Uses of Alum:
1. It is used for purification of water
2. It is also used for water proofing and textiles
3. It is used in dyeing, paper and leather tanning industries
4. It is employed as a styptic agent to arrest bleeding.
40
41
42
On industrial scale carbon monoxide is produced by the reaction of carbon with air.
The carbon monoxide formed will contain nitrogen gas also and the mixture of nitrogen and
carbon monoxide is called producer gas.
2C + O2/N2 (air) 2CO + N2
Producers Gas
When carbon monoxide is treated with chlorine in presence of light or charcoal, it forms
a poisonous gas carbonyl chloride, which is also known as phosgene. It is used in the synthesis
of isocyanates.
CO + Cl2 COCl2
Carbon monoxide acts as a strong reducing agent.
CO + Fe2O3 2Fe + 3CO2
Under high temperature and pressure a mixture of carbon monoxide and hydrogen
(synthetic gas or syn gas) gives methanol.
CO + 2H2 CH3OH
In oxo process, ethene is mixed with carbon monoxide and hydrogen gas to produce
propanal.
43
Carbon monoxide forms numerous complex compounds with transition metals in which
the transition meal is in zero oxidation state. These compounds are obtained by heating the
metal with carbon monoxide.
Eg. Nickel tetracarbonyl [Ni(CO)4], Iron pentacarbonyl [Fe(CO)5], Chromium
hexacarbonyl [Cr(CO)6].
Structure:
It has a linear structure. In carbon monoxide, three electron pairs are shared between
carbon and oxygen. The bonding can be explained using molecular orbital theory as discussed
in XI standard. The C-O bond distance is 1.128Å. The structure can be considered as the
resonance hybrid of the following two canonical forms.
C O C O C O
44
Properties
It is a colourless, nonflammable gas and is heavier than air. Its critical temperature is 31⁰
C and can be readily liquefied.
Carbon dioxide is a very stable compound. Even at 3100 K only 76 % decomposes to
form carbon monoxide and oxygen. At still higher temperature it decomposes into carbon
and oxygen.
3100 K
CO2 CO + ½O2
high temperature
CO2 C + O2
Reducing behaviour:
At elevated temperatures, it acts as a strong reducing agent. For example,
CO2 + Mg 2MgO + C
Acidic behaviour:
The aqueous solution of carbon dioxide is slightly acidic as it forms carbonic acid.
CO2 + H2O + -
H2CO3 H + HCO3
Structure of carbon dioxide
Carbon dioxide has a liner structure with equal bond distance for the both C-O bonds. In
this molecule there is one C-O sigma bond. In addition there is 3c-4e bond covering all the
three atoms.
O C O O C O O C O
45
Properties:
Silicon tetrachloride is a colourless fuming liquid and it freezes at -70 ⁰C
In moist air, silicon tetrachloride is hydrolysed with water to give silica and hydrochloric acid.
SiCl4 + 4H2O 4HCl + Si(OH)4
When silicon tetrachloride is hydrolysed with moist ether, linear perchloro siloxanes are
formed [Cl-(Si Cl2O)nSiCl3 where n=1-6.
Alcoholysis
The chloride ion in silicon tetrachloride can be substituted by nucleophile such as OH,
OR, etc.. using suitable reagents. For example, it forms silicic esters with alcohols.
SiCl4 + C2H5OH Si(OC2H5)4 + 4HCl
Tetraethoxy silane
Ammonialysis.
Similarly silicon tetrachloride undergoes ammonialysis to form chlorosilazanes.
330 K
2SiCl4 + NH3 Ether
Cl3Si-NH-SiCl3
Uses:
1. Silicon tetrachloride is used in the production of semiconducting silicon.
2. It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for
ceramic materials.
2.3.8 Silcones:
Silicones or poly siloxanes are organo silicon polymers with general empirical formula
(R2SiO). Since their empirical formula is similar to that of ketone (R2CO), they were named
“silicones”. These silicones may be linear or cross linked. Because of their very high thermal
stability they are called high –temperature polymers.
46
The hydrolysis of dialkylchloro silanes R2SiCl2 yields to a straight chain polymer which
grown from both the sides
R R
+2H2O
Cl Si Cl HO Si OH
-2HCl
R R
R R R R R
HO Si OH + HO Si OH HO Si O Si OH + HO Si OH
-H2O
R R R R R
-H2O
R R R
Etc HO Si O Si O Si OH
-H2O
R R R
The hydrolysis of monoalkylchloro silanes RSiCl3 yields to a very complex cross linked
polymer.. Linear silicones can be converted into cyclic or ring silicones when water molecules
is removed from the terminal –OH groups.
Me Me O R
Me O Me
Me Si O Si Me Si Si R Si O Si O
Me Me
O O O O O O R
Me Si O Si Me Si O Si O Si O Si O
Me Me Me Me
R R O
Types of silicones:
(i) Liner silicones:
They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl
silicon chlorides.
47
48
49
Silicates which
contain (Si2O5)n 2n-
are
called sheet or phyllo
silicates. In these, Each
4-
[SiO ] tetrahedron
4
unit shares three oxygen
50
Summary
The elements in which their last electron enters the 'p' orbital, constitute the p-block
elements.
The p-block elements have a general electronic configuration of ns2, np1-6. The
elements of each group have similar outer shell electronic configuration and differ
only in the value of n (principal quantum number).
Generally on descending a group the ionisation energy decreases and hence the
metallic character increases.
The ionisation enthalpy of elements in successive groups is higher than the
corresponding elements of the previous group as expected.
As we move down the 13th group, the electronegativity first decreases from boron
to aluminium and then marginally increases.
51
52
a) SiO2 b) Si O
R
c) R O Si O d) Si O O R
R R
53
b) Mg2SiO4 is an orthosilicate
Column-I Column-II A B C D
(a) 2 1 4 3
A Borazole 1 B(OH)3
(b) 1 2 4 3
B Boric acid 2 B3N3H6
(c) 1 2 4 3
C Quartz 3 Na2[B4O5(OH)4] 8H2O (d) None of these
D Borax 4 SiO2
54
55
56
We have already learnt the general characteristics of p-block elements and the first two
group namely icosagens (boron group) and tetragens (carbon group) in the previous unit. In
this unit we learn the remaining p-block groups, pnictogens, chalcogens, halogens and inert
gases.
57
2NH 4 Cl + CaO
→ CaCl 2 + 2NH 3 + H 2 O
58
The dielectric constant of ammonia is considerably high to make it a fairly good ionising
solvent like water.
2NH 3
NH 4 + NH 2
+ −
Chemical Properties
Action of heat: Above 500°C ammonia decomposes into its elements. The decomposition may
be accelerated by metallic catalysts like Nickel, Iron. Almost complete dissociation occurs on
continuous sparking.
500 C
2NH 3 > → N 2 + 3H 2
0
Reaction with air/oxygen: Ammonia does not burn in air but burns freely in free oxygen with
a yellowish flame to give nitrogen steam.
4NH 3 + 3O 2
N 2 + 6H 2 O
In presence of catalyst like platinum, it burns to produce nitric oxide. This process is used for
the manufacture of nitric acid and is known as ostwalds process.
4NH 3 + 5O 2
4NO + 6H 2 O
Reducing property: Ammonia acts as a reducing agent. It reduces the metal oxides to metal
when passed over heated metallic oxide.
59
With metallic salts: Ammonia reacts with metallic salts to give metal hydroxides (in case of
Fe) or forming complexes (in case Cu)
−
Fe3+ + 3NH 4 + 3
OH
→ Fe(OH)3 + 3NH 4 +
Cu 2+ + 4NH 3
→ [Cu(NH 3 ) 4 ]2+
Tetraamminecopper(II)ion
(a coordinattion complex)
Formation of amines: Ammonia forms ammonated compounds by ion dipole attraction. Eg.
[CaCl2.8NH3]. In this, the negative ends of ammonia dipole is attracted to Ca2+ ion.
It can also act as a ligand and form coordination compounds such as [Co(NH3)6]3+, [Ag(NH3)2]+.
For example when excess ammonia is added to
aqueous solution copper sulphate a deep blue colour
compound [Cu(NH3)4]2+ is formed.
20 Å sp 3
Structure of ammonia 1.10
60
61
The reactions of metals with nitric acid are explained in 3 steps as follows:
Primary reaction: Metal nitrate is formed with the release of nascent hydrogen
M + HNO3
→ MNO3 + (H)
Secondary reaction: Nascent hydrogen produces the reduction products of nitric acid.
HNO3 + 2H
→ HNO 2 + H 2 O
Nitrous acid
HNO3 + 6H
→ NH 2 OH + 2H 2 O
Hydroxylamine
HNO3 + 8H
→ NH 3 + 3H 2 O
Ammonia
2HNO3 + 8H
→ H 2 N 2 O 2 + 4H 2 O
Hypo nitrous acid
62
2 HNO 2
→ NO 2 3 + H 2O
Nitrous acid Dinitrogentrioxide
H N O
→
2 2 2 NO 2 + H 2O
Hypo nitrous acid Nitrous oxide
Evaluate yourself :
Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.
63
Dinitrogen
3.1.6 Oxides and oxoacids of nitrogen
64
sesquoxide
4/2/2019 10:59:47 AM
Structures of oxides of nitrogen:
Name Formula Structure
Nitrous oxide N2O
N N O N N O
Nitric oxide NO N O
115 pm
Dinitrogen O O− O O
trioxide (or)
Nitrogen
N2O3 N N+ N N+
sesquoxide
O O−
Nitrogen
dioxide
NO2 O N O
O O
Nitrogen
tetraoxide
N2O4 N N
O O
O O
Nitrogen
N2O5 N O N
pentoxide
O O
Structures of oxoacids of nitrogen:
Hydronitrous N OH
H4N2O4
acid
HO N
OH
65
Hyponitrous
H2N2O2 +1 Ag 2 N 2O2 + 2HCl →
2AgCl + H2 N 2O2
acid
Pernitrous
HOONO +5 H 2 O 2 + ON OH
ON OOH + H 2 O
acid
4NH3 + 5O2
→ 4NO + 6H2O
2NO+O2
→ NO2
Nitric acid HNO3 +5
2NO2
→ N 2O 4
2N 2O4 + 2H2O+ O2
→ 4HNO3
66
P P P
P P P P P P
P P P
P4 + 5O 2 ∆
→ PO
4 10
Phosphorouspentoxide
Reaction with chlorine: Phosphorus reacts with chlorine to form tri and penta chloride.
Yellow phosphorus reacts violently at room temperature, while red phosphorous reacts on
heating
P4 + 6Cl2
→ 4PCl 3
Phosphorous tri chloride
P4 + 10Cl2
→ 4PCl 5
Phosphorous penta chloride
Reaction with alkali: Yellow phosphorous reacts with alkali on boiling in an inert atmosphere
liberating phosphine. Here phosphorus act as reducing agent.
P4 + 3NaOH + 3H 2 O
→ 3NaH 2 PO 2 + PH 3 ↑
sodium hypo phosphite Phosphine
Reaction with nitric acid: When phosphorous is treated with conc. nitric acid it is oxidised to
phosphoric acid. This reaction is catalysed by iodine crystals.
P4 + 20HNO3
→ 4H PO3 4 + 20NO 2 + 4H 2 O
Ortho phosphoric acid
Reaction with metals: Phosphorous reacts with metals like Ca and Mg to give phosphides..
Metals like sodium and potassium react with phosphorus vigorously.
P4 + 6Mg
→ 2Mg P3 2
Magnesium phosphide
P4 + 6Ca
→ 2Ca 3 P2
Calcium phosphide
P4 + 12Na
→ 4Na 3 P .
Sodium phosphide
Uses of phosphorous:
1. The red phosphorus is used in the match boxes
2. It is also used for the production of certain alloys such as phosphor bronze
68
AlP + 3HCl
→ PH 3 ↑+ AlCl3
Phosphine
Physical properties:
It is colourless, poisonous gas with rotten fish smell. It is slightly soluble in water and is
neutral to litmus test. It condenses to a colourless liquid at 188 K and freezes to a solid at 139.5 K .
Chemical properties:
Thermal stability: Phosphine decomposes into its elements when heated in absence of air at
317 K or when electric current is passed through it.
4PH 3 317K
→ P4 + 6H 2
Combustion: When phosphine is heated with air or oxygen it burns to give meta phosphoric
acid.
4PH 3 + 8O 2 ∆
→ PO4 10 + 6H 2 O
Phosphorous pentoxide
Basic nature: Phosphine is weakly basic and forms phosphonium salts with halogen acids.
PH 3 + HI
→ PH 4 I
PH 4 I + H 2 O ∆→ PH 3 + H 3O + + I −
Phosphine
69
This reaction involves the coordination of a water molecule using a vacant 3d orbital on
the phosphorous atom following by elimination of HCl which is similar to hydrolysis of SiCl4.
PCl3 + H2O PCl3.H2O P(OH)Cl2 + HCl
Similar reactions occurs with other molecules that contains alcohols and carboxylic acids.
C2H5OH + PCl3 C2H5Cl + H3PO3
C2H5COOH + PCl3 C2H5COCl + H3PO3
70
Phosphorous pentachloride:
Preparation
When PCl3 is treated with excess chlorine,
phosphorous pentachloride is obtained.
PCl 3 + Cl 2 →
PCl 5 Figure 3.5 Structure of
phosphorus trichloride
Chemical properties
On heating phosphorous pentachloride, it
decomposes into phosphorus trichloride and chlorine.
PCl 5 (g) →
PCl 3 (g) + Cl 2 (g) .
(Excess)
71
P1020
In P4O10 each P atoms form three bonds to oxygen O O
atom and also an additional coordinate bond with an 143 p
m
O
P P
oxygen atom. O O 1230 O
P
Terminal coordinate P-O bond length is 143 pm, O O
which is less than the expected single bond distance. O
This may be due to lateral overlap of filled p orbitals of Figure 3.7 Structure of P4O10
an oxygen atom with empty d orbital on phosphorous.
Oxoacids of Phosphorous-Structure:
O
O
Orthophosphrous
H3PO3 HO P OH
acid
H
O O
Hypophosphoric
H4P2O6 HO P P OH
acid
HO OH
O
Orthophosphoric
H3PO4 HO P OH
acid
OH
O O
Pyrophosphoric
H4P2O7 HO P O P OH
acid
HO OH
72
Oxidation
Formula
state
Name Preparation
Hypophosphorous
H3PO2 +1 P4 + 6H2O →
3H3PO2 + PH3
acid
Orthophosphrous
H3PO3 +3 P4O6 + 6H2O →
4H3PO3
acid
Hypophosphoric
H4P2O6 +4 2P + 2O2 + 2H2O →
H 4 P2O6
acid
Orthophosphoric
H3PO4 +5 P4O10 + 6H2O →
4H3PO4
acid
Pyrophosphoric
H4P2O7 +5 2H3PO3 →
H 4 P2O7 + H2O
acid
Atomic Mass
15.99 32.06 78.97 127.60 209
(g.mol at 293 K)
-1
73
3.2 Oxygen:
Preparation: The atmosphere and water contain 23% and 83% by mass of oxygen respectively.
Most of the world’s rock contain combined oxygen. Industrially oxygen is obtained by
fractional distillation of liquefied air. In the laboratory, oxygen is prepared by one of the
following methods.
The decomposition of hydrogen peroxide in the presence of catalyst (MnO2) or by
oxidation with potassium permanganate.
2H 2 O 2
2H 2 O + O 2
5H 2 O 2 + 2MnO 4 − + 6H +
→ 5O 2 + 8H 2 O + 2Mn 2+
O 2 + (O)
O3
Ozone
74
Rhombic sulphur also known as α sulphur, is the only thermodynamically stable allotropic
form at ordinary temperature and pressure. The crystals have a characteristic yellow colour
and composed of S8 molecules. When heated slowly above 96 ⁰C, it converts into monoclinic
sulphur. Upon cooling below 96 ⁰C the β form converts back to α form. Monoclinic sulphur
also contains S8 molecules in addition to small amount of S6 molecules. It exists as a long
needle like prism and is also called as prismatic sulphur. It is stable between 96 ⁰ - 119 ⁰C and
slowly changes into rhombic sulphur.
When molten sulphur is poured into cold water a yellow rubbery ribbon of plastic sulphur
is produced. They are very soft and can be stretched easily. On standing (cooling slowly) it
slowly becomes hard and changes to stable rhombic sulphur.
Sulphur also exists in liquid and gaseous states. At around 140 ⁰C the monoclinic sulphur
melts to form mobile pale yellow liquid called λ sulphur. The vapour over the liquid sulphur
consists of 90 % of S8, S7 & S6 and small amount of mixture of S2, S3, S4, S5 molecules.
75
- +
SO3 + 2H H2O + SO2
Properties:
Sulphur dioxide gas is found in volcanic eruptions. A large amount of sulphur dioxide gas
is released into atmosphere from power plants using coal and oil and copper melting plants.
It is a colourless gas with a suffocating odour. It is highly soluble in water and it is 2.2 times
heavier than air. Sulphur dioxide can be liquefied (boiling point 263 K) at 2.5 atmospheric
pressure and 288 K.
Chemical properties
Sulphur dioxide is an acidic oxide. It dissolves in water to give sulphurous acid.
SO 2 + H 2 O
H SO
2 3
Sulphurous acid
H 2SO3
+
2H + SO3
2−
Reaction with sodium hydroxide and sodium carbonate: Sulphur dioxide reacts with sodium
hydroxide and sodium carbonate to form sodium bisulphite and sodium sulphite respectively.
SO 2 + NaOH
→ NaHSO3
Sodium bisulphite
2SO 2 + Na 2 CO3 + H 2 O
→ 2NaHSO3 + CO 2
2 NaHSO3
→ Na 2SO3 + H 2 O + SO 2
Sodium sulphite
Oxidising property: Sulphur dioxide, oxidises hydrogen sulphide to sulphur and magnesium
to magnesium oxide.
2H 2S + SO2
→ 3S + 2H 2 O
2Mg + SO 2
→ 2MgO + S
76
Bleaching action of sulphur dioxide: In presence of water, sulphur dioxide bleaches coloured
wool, silk, sponges and straw into colourless due to its reducing property.
SO 2 + 2H 2 O
→ 2 H 2SO 4 + 2(H)
X + 2(H)
→ XH 2
Coloured Colourless
77
ii. Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst
such as V2O5 or platinised asbestos.
iii. The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum
(H2S2O7). The oleum is converted into sulphuric acid by diluting it with water.
SO3 + H 2SO 4
→ H 2S2 O7 H
2O
→ 2H 2SO 4
To maximise the yield the plant is operated at 2 bar pressure and 720 K. The sulphuric
acid obtained in this process is over 96 % pure.
Physical properties:
Pure sulphuric acid is a colourless, viscous liquid (Density: 1.84 g/mL at 298 K). High
boiling point and viscosity of sulphuric acid is due to the association of molecules together
through hydrogen bonding.
The acid freezes at 283.4 K and boils at 590 K. It is highly soluble in water and has strong
affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water,
it forms mono (H2SO4.H2O) and dihydrates (H2SO4.2H2O) and the reaction is exothermic.
The dehydrating property can also be illustrated by its reaction with organic compounds
such as sugar, oxalic acid and formic acid.
C12 H 22 O11 + H 2SO 4
→ 12C + H 2SO 4 .11H 2 O
Sucrose
HCOOH + H 2SO 4
→ CO + H 2SO 4 .H 2 O
Formic acid
(COOH) 2 + H 2SO 4
→ CO + CO 2 + H 2SO 4 .H 2 O
Oxalic acid
Chemical Properties:
Sulphuric acid is highly reactive. It can act as strong acid and an oxidising agent.
Decomposition: Sulphuric acid is stable, however, it decomposes at high temperatures to
sulphur trioxide.
H 2SO 4
→ H 2 O + SO3
Acidic nature: It is a strong dibasic acid. Hence it forms two types of salts namely sulphates
and bisulphates.
H 2SO 4 + NaOH
→ NaHSO 4 + H 2 O
sodium bisulphate
H 2SO 4 + 2NaOH
→ Na 2SO 4 + 2H 2 O
sodium sulphate
H 2SO 4 + 2NH 3
→ (NH 4 ) 2SO 4
Ammonium sulphate
78
Sulphuric acid oxidises elements such as carbon, sulphur and phosphorus. It also oxides
bromide and iodide to bromine and iodine respectively.
C + 2H 2SO 4
→ 2SO 2 + 2H 2 O + CO 2
S + 2H 2SO 4
→ 3SO 2 + 2H 2 O
P4 + 10H 2SO 4
→ 4H 3 PO 4 + 10SO 2 + 4H 2 O
H 2S + H 2SO 4
→ SO 2 + 2H 2 O + S
H 2SO 4 + 2HI
→ 2SO 2 + 2H 2 O + I 2
H 2SO 4 + 2HBr
→ 2SO 2 + 2H 2 O + Br2
Reaction with metals: Sulphuric acid reacts with metals and gives different product depending
on the reactants and reacting condition.
Dilute sulphuric acid reacts with metals like tin, aluminium, zinc to give corresponding
sulphates.
Zn + H 2SO 4
→ ZnSO 4 + H 2 ↑
2Al + 3H 2SO 4
→ Al2 (SO 4 )3 + 3H 2 ↑
Hot concentrated sulphuric acid reacts with copper and lead to give the respective
sulphates as shown below.
Cu + 2H 2SO 4
→ CuSO4 + 2H 2 O + SO 2 ↑
Pb + 2H 2SO 4
→ PbSO4 + 2H 2 O + SO 2 ↑
Sulphuric acid doesn’t react with noble metals like gold, silver and platinum.
Reaction with salts: It reacts with different metal salts to give metal sulphates and bisulphates.
KCl + H 2SO 4
→ KHSO 4 + HCl
KNO3 + H 2SO 4
→ KHSO 4 + HNO3
Na 2 CO3 + H 2SO 4
→ Na 2SO 4 + H 2 O + CO 2
2NaBr + 3H 2SO 4
→ 2NaHSO4 + 2H 2 O + Br2 + SO 2
Reaction with organic compounds: It reacts organic compounds such as benzene to give
sulphonic acids.
C6 H 6 + H 2SO 4
→ C H SO H + H 2 O
6 5 3
Benzene Benzene sulphonic acid
79
O
Sulphurous acid H2SO3
S
HO OH
O
Sulphuric acid H2SO4 HO S OH
O
S
Thiosulphuric acid H2S2O3 HO S OH
O
O O
Dithionous acid H2S2O4
HO S S OH
O O
Dithionic acid H2S2O5 HO S S OH
O
80
O O
Disulphuric acid or
H2S2O7 HO S O S OH
pyrosulphuric acid
O O
O
Peroxymono sulphuric acid H2SO5 HO S O OH
O
O O
Peroxodisulphuric acid.
H2S2O8 HO S O O S OH
Marshall’s acid
O O
O O
Dithionic acid H2S2O6 HO S S OH
O O
O O
Polythionic acid H2Sn+2O6 HO S (S)n S OH
O O
81
Properties:
Chlorine is highly reactive hence it doesn’t occur free in nature. It is usually distributed as various
metal chlorides. The most important chloride is sodium chloride which occurs in sea water.
Preparation:
Chlorine is prepared by the action of conc. sulphuric acid on chlorides in presence of
manganese dioxide.
4NaCl + MnO 2 + 4H 2SO 4
→ Cl2 + MnCl2 + 4NaHSO4 + 2H 2 O
It can also be prepared by oxidising hydrochloric acid using various oxidising agents such
as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
PbO 2 + 4HCl
→ PbCl2 + 2H 2 O + Cl2
MnO 2 + 4HCl
→ MnCl2 + 2H 2 O + Cl2
2KMnO 4 + 16HCl
→ 2KCl + 2MnCl2 + 8H 2 O + 5Cl2
K 2 Cr2 O7 + 14HCl
→ 2KCl + 2CrCl3 + 7H 2 O + 3Cl2
When bleaching powder is treated with mineral acids chlorine is liberated
CaOCl2 + 2HCl
→ CaCl2 + H 2 O + Cl2
CaOCl2 + H 2SO 4
→ CaSO 4 + H 2 O + Cl2
82
Deacon’s process: In this process a mixture of air and hydrochloric acid is passed up
a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are
placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.
0
4HCl + O2 Cu
400 C
Cl
→ 2H 2O + Cl2 ↑
2 2
The chlorine obtained by this method is dilute and is employed for the manufacture of
bleaching powder. The catalysed reaction is given below,
2Cu 2 Cl2 + O 2
→ 2Cu 2 OCl2
Cuprous oxy chloride
Cu 2 OCl2 + 2HCl
→ 2CuCl2 + H 2 O
Cupric chloride
2CuCl2
→ Cu 2 Cl2 + Cl2
Cuprous chloride
Physical properties:
Chlorine is a greenish yellow gas with a pungent irritating odour. It produces headache
when inhaled even in small quantities whereas inhalation of large quantities could be fatal. It
is 2.5 times heavier than air.
Chlorine is soluble in water and its solution is referred as chlorine water. It deposits
greenish yellow crystals of chlorine hydrate (Cl2.8H2O). It can be converted into liquid (Boiling
point – 34.6° C) and yellow crystalline solid (Melting point -102° C)
Chemical properties:
Action with metals and non-metals: It reacts with metals and non metals to give the
corresponding chlorides.
2Na + Cl2
→ 2NaCl
2Fe + 3Cl2
→ 2FeCl3
2Al + 3Cl2
→ 2AlCl3
Cu + Cl2
→ CuCl2
H 2 + Cl2
→ 2HCl ; ∆H = − 44kCal
83
P4 + 6Cl2
→ 4PCl3
2As + 3Cl2
→ 2AsCl3
2Sb + 3Cl2
→ 2SbCl3
Affinity for hydrogen : When burnt with turpentine it forms carbon and hydrochloric acid.
C10 H16 + 8Cl2
→10C + 16HCl
It forms dioxygen when reacting with water in presence of sunlight. When chlorine in
water is exposed to sunlight it loses its colour and smell as the chlorine is converted into
hydrochloric acid.
2Cl2 + 2H 2 O
→ O 2 + 4HCl
Chlorine reacts with ammonia to give ammonium chloride and other products as shown
below:
With excess ammonia,
2NH 3 + 3Cl2
→ N 2 + 6HCl
6HCl + 6 NH3
→ 6 NH 4 Cl
overall reaction
8NH 3 + 3Cl2
→ N 2 + 6 NH 4 Cl
With excess chlorine, NH 3 + 3Cl2
→ NCl3 + 3HCl
3HCl + 3NH 3
→ 3NH 4 Cl
overall reaction
4NH 3 + 3Cl2
→ NCl3 + 3NH 4 Cl
Chlorine oxidises hydrogen sulphide to sulphur and liberates bromine and iodine from
iodides and bromides. However, it doesn't oxidise fluorides
H 2S + Cl2
→ 2HCl + S
Cl2 + 2KBr
→ 2KCl + Br2
Cl2 + 2KI
→ 2KCl + I 2
Reaction with alkali: Chlorine reacts with cold dilute alkali to give chloride and hypochlorite
while with hot concentrated alkali chlorides and chlorates are formed.
Cl2 + H 2 O
→ HCl + HOCl
HCl + NaOH
→ NaCl + H2 O
HOCl + NaOH
→ NaOCl + H 2 O
overall reaction
Cl2 + 2NaOH
→ NaOCl + NaCl + H 2 O
sodium hypo chlorite
84
3NaOCl
→ NaClO3 + 2NaCl
overall reaction
3Cl2 + 6NaOH
→ NaClO3 + 5NaCl + 3H 2 O
sodium chlorate
Oxidising and bleaching action: Chlorine is a strong oxidising and bleaching agent because
of the nascent oxygen.
H 2 O + Cl2
→ HCl + HOCl
Hypo chlorous acid
HOCl
→ HCl + (O)
Colouring matter + Nascent oxygen → Colourless oxidation product
The bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to
sulphates and hydrogen sulphide to sulphur.
2FeCl2 + Cl2
→ 2FeCl3
Cl2 + H 2 O
→ HCl + HOCl
2FeSO 4 + H 2SO 4 + HOCl
→ Fe 2 (SO 4 )3 + HCl + H 2 O
overall reaction
2FeSO 4 + H 2SO 4 + Cl2
→ Fe 2 (SO 4 )3 + 2HCl
Cl2 + H 2 O
→ HCl + HOCl
Na 2SO3 + HOCl
→ Na 2SO 4 + HCl
overall reaction
Na 2SO3 + H 2 O + Cl2
→ Na 2SO 4 + 2HCl
Cl2 + H 2S
→ 2HCl + S
Preparation of bleaching powder: Bleaching powder is produced by passing chlorine gas
through dry slaked lime (calcium hydroxide).
Ca(OH) 2 + Cl2
→ CaOCl2 + H 2 O
Displacement redox reactions: Chlorine displaces bromine from bromides and iodine from
iodide salts.
Cl2 + 2KBr
→ 2KCl + Br2
Cl2 + 2KI
→ 2KCl + I 2
Formation of addition compounds: Chlorine forms addition products with sulphur
dioxide, carbon monoixde and ethylene. It forms substituted products with alkanes/arenes.
SO 2 + Cl2
→ SO 2 Cl2
Sulphuryl chloride
CO + Cl2
→ COCl2
Carbonyl chloride
85
CH 4 + Cl2
→ CH 3Cl + HCl
C6 H 6 + Cl2 FeCl
3
→ C6 H 5Cl + HCl
Uses of chlorine:
It is used in
1. Purification of drinking water
2. Bleaching of cotton textiles, paper and rayon
3. Extraction of gold and platinum
3.3.2 Hydrochloric acid:
Laboratory preparation:
It is prepared by the action of sodium chloride and concentrated sulphuric acid.
NaCl + H2SO4 NaHSO4 + HCl
NaHSO4 + NaCl Na2SO4 + HCl
Dry hydrochloric acid is obtained by passing the gas through conc. sulphuric acid
Properties:
Hydrogen chloride is a colourless, pungent smelling gas, easily liquefied to a colourless
liquid (boiling point 189K) and frozen into a white crystalline solid (melting point 159K). It
is extremely soluble in water.
+ -
HCl (g) + H2O (l) H3O + Cl
Chemical properties:
Like all acids it liberates hydrogen gas from metals and carbon dioxide from carbonate
and bicarbonate salts.
Zn + 2HCl ZnCl2 + H2
Mg + 2HCl MgCl2 + H2
Na2CO3 + 2HCl 2NaCl + CO2 + H2
CaCO3 + 2HCl CaCl2 + CO2 + H2
NaHCO3 + 2HCl 2NaCl + CO2 + H2O
It liberates sulphur dioxide from sodium sulphate
Na2SO4 + 2HCl 2NaCl + H2O + SO2
When three parts of concentrated hydrochloric acid and one part of concentrated nitric
acid are mixed, Aquaregia (Royal water) is obtained. This is used for dissolving gold, platinum
etc...
86
87
HF + F−
HF2
−
At high concentration, the equilibrium involves the removal of fluoride ions is important.
Since it affects the dissociation of hydrogen fluoride and increases and hydrogen ion
concentration Several stable salts NaHF2, KHF2 and NH4HF2 are known. The other hydrogen
halides do not form hydrogen dihalides.
Hydrohalic acid shows typical acidic properties. They form salts with acids, bases and
reacts with metals to give hydrogen. Moist hydrofluoric acid (not dry) rapidly react with silica
and glass.
SiO 2 + 4HF
→ SiF4 + 2H 2 O
Na 2SiO3 + 6HF
→ Na 2SiF6 + 3H 2 O
Oxidation: Hydrogen iodide is readily oxidised to iodine hence it is a reducing agent.
2HI
+
2H + I 2 + 2e
−
88
IF IF3
BrCl ICl3
ICl
IBr
89
2 ICl3
+
ICl2 + ICl4
−
Oxidation state +1 +4 +5 +6 +7 -
OF2 (-1)
F - - - - - O2F2 (-1)
O4F2 (-1)
90
I4O9
I - - I2O5 - -
I2O4 (+4)
Table 3.9
Type HOX HXO2 HXO3 HXO4
Common Name Hypohalous acid Halous acid Halic acid Perhalic acid
Oxidation state +1 +3 +5 +7
F HOF - - -
Cl HOCl HClO2 HClO3 HClO4
Br HOBr HBrO3 HBrO4
I HOI HIO3 HIO4
Structure
91
Xe + 2F2 Ni
/acetone
4000 C
→ XeF4
Xe + 3F2 Ni
/200 atm
4000 C
→ XeF6
92
93
94
When three parts of concentrated hydrochloric acid and one part of concentrated
nitric acid are mixed, Aquaregia (Royal water) is obtained. This is used for
dissolving gold, platinum etc...
Hydrogen halides are extremely soluble in water due to the ionisation.
Each halogen combines with other halogens to form a series of compounds called
inter halogen compounds.
Fluorine reacts readily with oxygen and forms difluorine oxide (F2O) and difluorine
dioxide (F2O2) where it has a -1 oxidation state.
All the noble gases occur in the atmosphere.
They are extremely stable and have a small tendency to gain or lose electrons.
Sodium per xenate is very much known for its strong oxidizing property.
95
a) Nessler’s reagent
d) Tollen’s reagent
3. An element belongs to group 15 and 3 rd period of the periodic table, its electronic
configuration would be
a) H3PO3 b) PH3
c) H3PO4 d) POCl3
7. P4O6 reacts with cold water to give
a) H3PO3 b) H4P2O7
c) HPO3 d) H3PO4
96
a) 4 b) 2
c) 3 d) 5
9. The molarity of given orthophosphoric acid solution is 2M. its normality is
a) 6N b) 4N
c) 2N d) none of these
10. Assertion : bond dissociation energy of fluorine is greater than chlorine gas
Reason: chlorine has more electronic repulsion than fluorine
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true but reason is not the correct explanation of
assertion.
c) Assertion is true but reason is false.
d) Both assertion and reason are false.
11. Among the following, which is the strongest oxidizing agent?
a) Cl2 b) F2
c) Br2 d) l2
12. The correct order of the thermal stability of hydrogen halide is
a) HI > HBr > HCl > HF b) HF > HCl > HBr > HI
c) HCl > HF > HBr > HI d) HI > HCl > HF > HBr
13. Which one of the following compounds is not formed?
a) XeOF4 b) XeO3
c) XeF2 d) NeF2
14. Most easily liquefiable gas is
a) Ar b) Ne
c) He d) Kr
15. XeF6 on complete hydrolysis produces
a) XeOF4 b) XeO2F2
c) XeO3 d) XeO2
16. On oxidation with iodine, sulphite ion is transformed to
a) S 4O6 b) S2O6
2− 2−
97
a) HI b) HF
c) HBr d) HCl
18. Which one of the following orders is correct for the bond dissociation enthalpy of
halogen molecules? (NEET)
a) Br2 > I2 > F2 > Cl2 b) F2 > Cl2 > Br2 > l2
c) I2 > Br2 > Cl2 > F2 d) Cl2 > Br2 > F2 > I2
19. Among the following the correct order of acidity is (NEET)
a) HClO2 < HClO < HClO3 < HClO4 b) HClO4 < HClO2 < HClO < HClO3
c) HClO3 < HClO4 < HClO2 < HClO d) HClO < HClO2 < HClO3 < HClO4
20. When copper is heated with conc HNO3 it produces
9. Give the balanced equation for the reaction between chlorine with cold NaOH and hot
NaOH.
98
a) BrF5 b) BrF3
23. Complete the following reactions.
1. NaCl + MnO2 + H2SO4 →
2. NaNO2 + HCl →
3. IO3 + I + H →
− − +
4. I2 + S2O3 →
2−
5. P4 + NaOH + H2O →
6. AgNO3 + PH3 →
7. Mg + HNO3 →
8. KClO3 →
∆
Hot conc
9. Cu + H2SO4 →
10. Sb + Cl 2 →
13. XeO6 + Mn + H →
4− 2+ +
100
H
1.0079
He
4.0026
lithium beryllium boron carbon nitrogen oxygen fluorine neon
3 4 5 6 7 8 9 10
Li
6.941
Be
9.0122
B
10.811
C
12.011
N
14.007
O
15.999
F
18.998
Ne
20.180
sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
11 12 13 14 15 16 17 18
Na Mg
22.990 24.305
d-Block Al
26.982
Si
28.086
P
30.974
S
32.065
Cl Ar
35.453 39.948
potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K
39.098
Ca Sc
40.078 44.956
Ti
47.867
V
50.942
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se
51.996 54.938 55.845 58.933 58.693 63.546 65.38 69.723 72.64 74.922 78.96
Br Kr
79.904 83.798
rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Rb Sr
85.468 87.62
Y
88.906
Zr Nb Mo Tc Ru Rh Pd Ag Cd In
91.224 92.906 95.96 [98] 101.07 102.91 106.42 107.87 112.41 114.82
Sn Sb Te
118.71 121.76 127.60
I
126.90
Xe
131.29
caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
lanthanum
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
Cs Ba La Hf Ta
132.91 137.33 178.49 180.95
W Re Os
183.84 186.21 190.23
Ir
192.22
Pt Au Hg Tl Pb Bi Po At Rn
195.08 196.97 200.59 204.38 207.2 208.98 [209] [210] [222]
138.91
francium radium actinium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium Copernicium Nahonium Flerovium Mascovium Livermorium Tennessine Oganessom
87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Nh
[223] [226] [227] [261] [262] [266] [264] [277] [268] [271] [272] [285] [286]
Fl Mc Lv
[289] [289] [293]
Ts Og
[294] [294]
cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
58 59 60 61 62 63 64 65 66 67 68 69 70 71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.12 140.91 144.24 [145] 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.05 174.97
thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Th Pa
232.04 231.04
U
238.03
Np Pu Am Cm Bk Cf
[237] [244] [243] [247] [247] [251]
Es Fm Md No Lr
[252] [257] [258] [259] [262]
f-Block
Figure 4.1-Position of d- block elements in the periodic table
101
Note: The extra stability due to symmetrical distribution can also be visualized as follows.
When the d orbitals are considered together, they will constitute a sphere. So the half filled and
fully filled configuration leads to complete symmetrical distribution of electron density. On
the other hand, an unsymmetrical distribution of electron density as in the case of partially
filled configuration will result in building up of a potential difference. To decrease this and to
achieve a tension free state with lower energy, a symmetrical distribution is preferred.
With these two exceptions and minor variation in certain individual cases, the general
electronic configuration of d- block elements can be written as [Noble gas] ( n −1) d1−10 ns1−2,
Here, n = 4 to 7 . In periods 6 and 7, (except La and Ac) the configuration includes
((n −2 ) f orbital ; [Noble gas] ( n −2 ) f 14 ( n −1) d1−10ns1−2 .
4.3 General trend in properties:
4.3.1 Metallic behavior:
All the transition elements are metals. Similar to all metals the transition metals are good
conductors of heat and electricity. Unlike the metals of Group-1 and group-2, all the transition
metals except group 11 elements are hard. Of all the known elements, silver has the highest
electrical conductivity at room temperature.
102
39 40 41 42 43 44 45 46 47 48
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
HCP HCP BCC BCC HCP HCP FCC FCC FCC HCP
73
57* 72 74 75 76 77 78 79 80
Ta
La Hf W Re Os Ir Pt Au Hg
BCC/
DHCP HCP BCC HCP HCP FCC FCC FCC RHO
TETR
89** 104 105 106 107 108 109 110 111 112
Ac Rf Db Sg Bh Hs Mt Ds Rg Cn
FCC [HCP] [BCC] [BCC] [HCP] [HCP] [FCC] [BCC] [BCC] [BCC]
Cr
V
Fe
Ti Ni
Sc Co
Mn
Cu
(K)
Zn
103
Nb Mo Cd
Tc Ag
charge and hence, the 2.1
Ru Rh Pd
Generally as we move
1.5
down a group atomic Atmoic Number
W Re Os Au
the atomic radii of the 2.1
Ir Pt
104
The increase in first ionisation enthalpy with increase in atomic number along a particular
series is not regular. The added electron enters (n-1)d orbital and the inner electrons act as a
shield and decrease the effect of nuclear charge on valence ns electrons. Therefore, it leads to
variation in the ionization energy values.
The ionisation enthalpy values can be used to predict the thermodynamic stability of their
compounds. Let us compare the ionisation energy required to form Ni2+ and Pt2+ ions.
For Nickel, IE1 + IE2 = ( 737 + 1753 )
−1
= 2490 kJmol
105
106
+6 +6
+7
Why iron is more stable in +3 +5 +6
Cr 2+ + 2e − →
Cr –0.91 Cr
Zn
1
Mn2+ + 2e − →
Mn –1.18 V Mn
1.5
Fe + 2e →
2+
Fe −
–0.44 Ti
Co2+ + 2e − →
Co –0.28 2
Ni + 2e →
2+
Ni −
–0.23 Sc
3d -Series
2.5
Cu + 2e →
2+
Cu −
+0.34 0
Figure 4.7 (a) E M 2+ -3d series
Zn + 2e →
2+
Zn −
–0.76 M
In 3d series as we move from Ti to Zn, the standard reduction potential E 0 2+ value
M M
is approaching towards less negative value and copper has a positive reduction potential. i.e.,
107
Transition metals in their high oxidation states tend to be oxidizing . For example,
Fe3+ is moderately a strong oxidant, and it oxidises copper to Cu2+ ions. The feasibility of the
reaction is predicted from the following standard electrode potential values.
Fe 3+ (aq) + e − Fe2+ E0 = 0.77V
Mn
Ti 3+ + e − →
Ti 2+ –0.37
V 3+ + e − →
V 2+ –0.26
Fe
Cr + e →
3+
Cr − 2+
–0.41
Mn + e →
Mn
3+ − 2+
+1.51
V
Ti
Fe + e →
Fe
3+ − 2+ Cr
+0.77 3d-Series
0.5
Co3+ + e − →
Co2+ +1.81 Figure 4.7 (b) M3+
M 2+ -3d series
The negative values for titanium, vanadium and chromium indicate that the higher
oxidation state is preferred. If we want to reduce such a stable Cr3+ ion, strong reducing agent
which has high negative value for reduction potential like metallic zinc ( E0 = − 0.76 V ) is
required.
3+
The high reduction potential of Mn Mn2+ indicates Mn2+ is more stable than Mn3+. For
Fe3+ 2+ the reduction potential is 0.77V, and this low value indicates that both Fe3+ and Fe2+ can
Fe
exist under normal conditions. The drop from Mn to Fe is due to the electronic structure of the ions
concerned.Mn3+ has a 3d4 configuration while that of Mn2+ is 3d5. The extra stability associated with a
half filled d sub shell makes the reduction of Mn3+ very feasible (E0 = +1.51V).
Paramagnetic solids having unpaired electrons possess magnetic dipoles which are
isolated from one another. In the absence of external magnetic field, the dipoles are arranged
at random and hence the solid shows no net magnetism. But in the presence of magnetic field,
the dipoles are aligned parallel to the direction of the applied field and therefore, they are
attracted by an external magnetic field.
Ferromagnetic materials have domain structure and in each domain the magnetic dipoles
are arranged. But the spin dipoles of the adjacent domains are randomly oriented. Some
transition elements or ions with unpaired d electrons show ferromagnetism.
3d transition metal ions in paramagnetic solids often have a magnetic dipole moments
corresponding to the electron spin contribution only. The orbital moment L is said to be
quenched. So the magnetic moment of the ion is given by
µ= g S (S + 1) µB
n n
µ= 2 2 2 + 1 µB
n ( n + 2)
µ= 2 µB
4
µ= n ( n + 2 ) µB
The magnetic moment calculated using the above equation is compared with the
experimental values in the following table. In most of the cases, the agreement is good.
109
H H H H
Alkene Alkane
In certain catalytic processes the variable oxidation states of transition metals find
applications. For example, in the manufacture of sulphuric acid from SO3, vanadium pentoxide
110
Co2(CO)8 CHO
+ CO + H2 +
Propene Butan-1-al 2-methylpropan-1-al
(ii) Preparation acetic acid from acetaldehyde.
Rh / Ir complex CH3- COOH
CH3- CHO + CO
Acetaldehyde Acetic acid
n
Propylene poly propylene
The roasted mass is treated with water to separate soluble sodium chromate from
insoluble iron oxide. The yellow solution of sodium chromate is treated with concentrated
sulphuric acid which converts sodium chromate into sodium dichromate.
2 Na 2 CrO4 + H2 SO4 →
Na 2 Cr2 O7 + Na 2 SO4 + H2 O
sodium chromate sodium dichromate
(yellow) (orange red)
The above solution is concentrated to remove less soluble sodium sulphate. The resulting
solution is filtered and further concentrated. It is cooled to get the crystals of Na2SO4.2H2O.
The saturated solution of sodium dichromate in water is mixed with KCl and then
concentrated to get crystals of NaCl. It is filtered while hot and the filtrate is cooled to obtain
K2Cr2O7 crystals.
112
Physical properties:
Potassium dichromate is an orange red crystalline solid which melts at 671K and it is
moderately soluble in cold water, but very much soluble in hot water. On heating it decomposes
and forms Cr2O3 and molecular oxygen. As it emits toxic chromium fumes upon heating, it is
mainly replaced by sodium dichromate.
4 K 2Cr2O7 ∆
→ 4 K 2CrO4 + 2 Cr2O3 + 3O2 ↑
potassium potassium chromium(III)
dichromate chromate oxide
113
Cr2O7 2− + 6Fe2+ + 14H+ →
2Cr 3+ + 6Fe3+ + 7H2O
Cr2O7 2− + 6I + 14H+ →
2Cr 3+ + 3I2 + 7H2O
Cr2O7 2− + 3S + 14H+ →
2Cr 3+ + 3S + 7H2O
Cr2O7 2− + 3SO2 + 2H+ →
2Cr 3+ + 3SO4 2− + H2O
Cr2O7 2− + 3Sn + 14H+ →
2Cr 3+ + 3Sn + 7H2O
The chromyl chloride vapours are dissolved in sodium hydroxide solution and then
acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate
is obtained.
CrO2Cl 2 + 4NaOH →
Na 2CrO4 + 2NaCl + 2H2O
Na 2CrO4 + ( CH3COO )2 Pb →
PbCrO4 ↓ + 2CH3COONa
Leadchromate
(Yellowprecipitate )
114
Electrolytic oxidation
In this method aqueous solution of potassium manganate is electrolyzed in the presence
of little alkali.
K 2 MnO4
2K + MnO4
+ 2−
H 2O
H + OH
+ −
Green purple
Physical properties:
Potassium permanganate exists in the form of dark purple crystals which melts at 513 K.
It is sparingly soluble in cold water but, fairly soluble in hot water.
115
Chemical properties:
1. Action of heat:
When heated, potassium permanganate decomposes to form potassium manganate and
manganese dioxide.
2KMnO4 →
2K 2 MnO4 + MnO2 + O2
3. Oxidising property:
Potassium permanganate is a strong oxidising agent, its oxidising action differs in different
reaction medium.
a) In neutral medium:
In neutral medium, it is reduced to MnO2.
MnO4 − + 2H2O + 3e− →
MnO2 + 4OH−
(i) It oxidises H2S to sulphur
2MnO4 − + 3H2S →
2MnO2 + 3S + 2OH− + 2H2O
(ii) It oxidises thiosulphate into sulphate
8MnO4 − + 3S2O32− + H2O →
6SO4 2− + 8MnO2 + 2OH−
116
117
Note HCl cannot be used for making the medium acidic since it reacts with KMnO4 as follows.
2MnO4 − + 10 Cl − + 16H+ →
2Mn2+ + 5Cl 2 + 8H2O
HNO3 also cannot be used since it is good oxidising agent and reacts with reducing agents
in the reaction.
However,H2SO4 is found to be most suitable since it does not react with potassium
permanganate.
Note
Equivalent weight of KMnO4 in Molecular weight of KMnO4 158
= = = 31.6
acid medium no of mols of electrons transferred 5
Equivalent weight of KMnO4 in Molecular weight of KMnO4 158
= = = 158
basic medium no of mols of electrons transferred 1
Equivalent weight of KMnO4 in Molecular weight of KMnO4 158
= = = 52.67
neutral medium no of mols of electrons transferred 3
118
H
1.0079
He
4.0026
lithium beryllium boron carbon nitrogen oxygen fluorine neon
3 4 5 6 7 8 9 10
Li
6.941
Be
9.0122
B
10.811
C
12.011
N
14.007
O
15.999
F
18.998
Ne
20.180
sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
11 12 13 14 15 16 17 18
Na Mg
22.990 24.305
d-Block Al
26.982
Si
28.086
P
30.974
S
32.065
Cl Ar
35.453 39.948
potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K
39.098
Ca Sc
40.078 44.956
Ti
47.867
V
50.942
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se
51.996 54.938 55.845 58.933 58.693 63.546 65.38 69.723 72.64 74.922 78.96
Br Kr
79.904 83.798
rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Rb Sr
85.468 87.62
Y
88.906
Zr Nb Mo Tc Ru Rh Pd Ag Cd In
91.224 92.906 95.96 [98] 101.07 102.91 106.42 107.87 112.41 114.82
Sn Sb Te
118.71 121.76 127.60
I
126.90
Xe
131.29
caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
lanthanum
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
Cs Ba La Hf Ta
132.91 137.33 178.49 180.95
W Re Os
183.84 186.21 190.23
Ir
192.22
Pt Au Hg Tl Pb Bi Po At Rn
195.08 196.97 200.59 204.38 207.2 208.98 [209] [210] [222]
138.91
francium radium actinium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium Copernicium Nahonium Flerovium Mascovium Livermorium Tennessine Oganessom
87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Nh
[223] [226] [227] [261] [262] [266] [264] [277] [268] [271] [272] [285] [286]
Fl Mc Lv
[289] [289] [293]
Ts Og
[294] [294]
cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
58 59 60 61 62 63 64 65 66 67 68 69 70 71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.12 140.91 144.24 [145] 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.05 174.97
thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Th Pa
232.04 231.04
U
238.03
Np Pu Am Cm Bk Cf
[237] [244] [243] [247] [247] [251]
Es Fm Md No Lr
[252] [257] [258] [259] [262]
f-Block
119
In Gadolinium (Gd) and Lutetium (Lu) the 4f orbitals, are half-filled and completely filled,
and one electron enters 5d orbitals. Hence the general electronic configuration of 4f series of
elements can be written as [Xe] 4 f 2−14 5d 0−1 6s 2
120
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
+2 +2 +2 +2 +2
+3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3
+4 +4 +4 +4 +4
121
Thorium 90 Th [Rn] 5f 0 6d 2 7s 2
Plutonium 94 Pu [Rn] 5f 6 6d 0 7s 2
Americium 95 Am [Rn] 5f 7 6d 0 7s 2
Berkelium 97 Bk [Rn] 5f 9 6d 0 7s 2
Californium 98 Cf [Rn] 5f 10 6d 0 7s 2
122
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
+2 +2
+3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3 +3
+4 +4 +4 +4 +4 +4 +4 +4
+5 +5 +5 +5 +5 +5
+6 +6 +6 +6
+7 +7 +7
Differences between lanthanoids and actinoids:
2 Binding energy of 4f orbitals are higher Binding energy of 5f orbitals are lower
They show less tendency to form They show greater tendency to form
3
complexes complexes
123
Summary
IUPAC defines transition metal as an element whose atom has an incomplete d
sub shell or which can give rise to cations with an incomplete d sub shell. They
occupy the central position of the periodic table, between s and p block elements,
d- Block elements composed of 3d series (4th period) Scandium to Zinc ( 10
elements), 4d series ( 5th period) Yttrium to Cadmium ( 10 elements) and 5d
series ( 6th period) Lanthanum, Haffinium to mercury.
the general electronic configuration of d- block elements can be written as
[Noble gas] ( n −1) d
1−10 1−2
ns ,
Here, n = 4 to 7 . In periods 6 and 7, the configuration includes
((n −2 ) f orbital ; [Noble gas] ( n −2 ) f 14 ( n −1) d1−10ns1−2 .
All the transition elements are metals. Similar to all metals the transition metals
are good conductors of heat and electricity. Unlike the metals of Group-1 and
group-2, all the transition metals except group 11 elements are hard.
As we move from left to right along the transition metal series, melting point first
increases as the number of unpaired d electrons available for metallic bonding
increases, reach a maximum value and then decreases, as the d electrons pair up
and become less available for bonding.
Ionization energy of transition element is intermediate between those of s and
p block elements. As we move from left to right in a transition metal series, the
ionization enthalpy increases as expected.
The first transition metal Scandium exhibits only +3 oxidation state, but all other
transition elements exhibit variable oxidation states by loosing electrons from
(n-1)d orbital and ns orbital as the energy difference between them is very small.
In 3d series as we move from Ti to Zn, the standard reduction potential E 0
M M
2+
value is approaching towards less negative value and copper has a positive
reduction potential. i.e., elemental copper is more stable than Cu2+.
124
EVALUATION
a) both Sc3+ and Zn2+ ions are colourless and form white compounds.
b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
c) last electron as assumed to be added to 4s level in case of zinc
d) both Sc and Zn do not exhibit variable oxidation states
2. Which of the following d block element has half filled penultimate d sub shell as well as half
filled valence sub shell?
a) Cr b) Pd
c) Pt d) none of these
125
M )
a) Ti b) Cu c) Mn d) Zn
4. Which one of the following ions has the same number of unpaired electrons as present in
V3+?
a) Ti3+ b) Fe3+
c) Ni2+ d) Cr3+
5. The magnetic moment of Mn2+ ion is
a) 5.92BM b) 2.80BM
c) 8.95BM d) 3.90BM
6. Which of the following compounds is colourless?
a) Fe3+ b) Ti4+
c) Co2+ d) Ni2+
7. the catalytic behaviour of transition metals and their compounds is ascribed
mainly due to
a) their magnetic behaviour
b) their unfilled d orbitals
c) their ability to adopt variable oxidation states
d) their chemical reactivity
8. The correct order of increasing oxidizing power in the series
a) VO2 < Cr2O7 < MnO4 b) Cr2O7 < VO2 < MnO4
+ 2− − 2− + −
c) Cr2O7 < MnO4 < VO2 d) MnO4 < Cr2O7 < VO2
2− − + − 2− +
126
c) Mn3+ d) MnO2
14. A white crystalline salt (A) react with dilute HCl to liberate a suffocating gas (B) and also
forms a yellow precipitate . The gas (B) turns potassium dichromate acidified with dil
H2SO4 to a green coloured solution(C). A,B and C are respectively
a) Na 2SO3 , SO2 , Cr2 ( SO4 )3 b) Na 2S2O3 , SO2 , Cr2 ( SO4 )3
c) Na 2S , SO2 , Cr2 ( SO4 )3 d) Na 2SO4 , SO2 , Cr2 ( SO4 )3
15. MnO4- react with Br- in alkaline PH to give
−
a) BrO3 , MnO2 b) Br2 , MnO4
2−
− 2−
c) Br2 , MnO2 d) BrO , MnO4
16. How many moles of I2 are liberated when 1 mole of potassium dichromate react with
potassium iodide?
a) 1 b) 2
c) 3 d) 4
17. The number of moles of acidified KMnO4 required to oxidize 1 mole of ferrous
oxalate(FeC2O4) is
a) 5 b) 3 c) 0.6 d) 1.5
18. When a brown compound of Mn (A) ids treated with HCl , it gives a gas (B) . The gas (B)
taken in excess reacts with NH3 to give an explosive compound (C). The compound A, B
and C are
a) MnO2 , Cl 2 , NCl 3 b) MnO, Cl 2 , NH 4Cl
c) Mn3O4 , Cl 2 , NCl 3 d) MnO3 , Cl 2 , NCl 2
19. Which one of the following statements related to lanthanons is incorrect?
a) Europium shows +2 oxidation state.
b) The basicity decreases as the ionic radius decreases from Pr to Lu.
c) All the lanthanons are much more reactive than aluminium.
d) Ce4+ solutions are widely used as oxidising agents in volumetric analysis.
127
128
c. MnO4 + Fe →
? − 2+
d. KMnO4 Red
∆
hot
→?
e. Cr2O7 2− + I− + H+ →
?
f. Na 2Cr2O7 + KCl →
?
129
130
We have already learnt in the previous unit that the transition metals have a tendency
to form complexes (coordination compounds). The name is derived from the Latin
words 'complexus' and 'coordinate' which mean 'hold' and 'to arrange' respectively. The
complexes of transition metals have interesting properties and differ from simple ionic
and covalent compounds. For example, chromium(III)chloride hexahydrate, CrCl3.6H2O,
exists as purple, pale green or dark green compound. In addition to metals, certain non
metals also form coordination compounds but have less tendency than d block elements.
Coordination compounds play a vital role in the biological functions, and have wide range
of catalytic applications in chemical industries. For example, haemoglobin, the oxygen
transporter of human is a coordination compound of iron, and cobalamine, an essential
vitamin is a coordination compound of cobalt. Chlorophyll, a pigment present in plants
acting as a photo sensitiser in the photosynthesis is also a coordination compound. Various
coordination compounds such as Wilkinson's compound, Ziegler Natta compound are
used as catalysts in industrial processes. Hence, it is important to understand the chemistry
of coordination compounds. In this unit we study the nature, bonding, nomenclature,
isomerism and applications of the coordination compounds.
Let us consider the different coloured complexes of cobalt(III) chloride with ammonia
which exhibit different properties as shown below.
131
CoCl3.5NH3 Purple 2
CoCl3.4NH3 Green 1
CoCl3.4NH3 Violet 1
In this case, the valences of the elements present in both the reacting molecules, cobalt(III)
chloride and ammonia are completely satisfied. Yet these substances react to form the above
mentioned complexes.
To explain this behaviour Werner postulated his theory as follows
1. Most of the elements exhibit, two types of valence namely primary valence and secondary
valence and each element tend to satisfy both the valences.In modern terminology, the
primary valence is referred as the oxidation state of the metal atom and the secondary
valence as the coordination number. For example, according to Werner, the primary and
secondary valences of cobalt are 3 and 6 respectively.
2. The primary valence of a metal ion is positive in most of the cases and zero in certain cases.
They are always satisfied by negative ions. For example in the complex CoCl3.6NH3, The
-
primary valence of Co is +3 and is satisfied by 3Cl ions.
3. The secondary valence is satisfied by negative ions, neutral molecules, positive ions or the
combination of these. For example, in CoCl3.6NH3 the secondary valence of cobalt is 6
and is satisfied by six neutral ammonia molecules, whereas in CoCl3.5NH3 the secondary
-
valence of cobalt is satisfied by five neutral ammonia molecules and a Cl ion.
4. According to Werner, there
are two spheres of attraction
Inner sphere or coordination sphere
around a metal atom/ion in a NH3
complex. The inner sphere is NH3 NH3 Cl
known as coordination sphere
and the groups present in this
Cl Co
sphere are firmly attached to
M
the metal. The outer sphere is
NH3 NH3
called ionisation sphere. The
groups present in this sphere NH3
are loosely bound to the Cl
central metal ion and hence outer sphere or ionization sphere
can be separated into ions
upon dissolving the complex Figure 5.1 inner and outer spheres of attraction in
in a suitable solvent. coordination compounds
132
-
Groups satisfy the No. of ionisable Cl No. of moles of
secondary valence ions in the complex AgCl formed =
Complex
(non-ionaisable, inner (outer coordination no. of moles of
-
coordination sphere) sphere) ionisable Cl
Evaluate yourself 1:
When a coordination compound CrCl3.4H2O is mixed with silver nitrate solution, one
mole of silver chloride is precipitated per mole of the compound. There are no free solvent
molecules in that compound. Assign the secondary valence to the metal and write the
structural formula of the compound.
134
Example 1:
4-
In [Fe(CN)6] ,
let the oxidation number of iron is x :
The net charge: -4 = x + 6 (-1) => x = +2
Example 2:
2+
In [Co(NH3)5Cl] ,
let the oxidation number of cobalt is x :
The net charge: +2 = x + 5 (0) + 1 (-1) => x = +3
Evaluate yourself 2:
2. In the complex, [Pt(NO2)(H2O)(NH3)2]Br , identify the following
i. Central metal atom/ion
ii. Ligand(s) and their types
iii. Coordination entity
iv. Oxidation number of the central metal ion
v. Coordination number
Types of complexes:
The coordination compounds can be classified into the following types based on (i) the
net charge of the complex ion, (ii) kinds of ligands present in the coordination entity.
Classification based on the net charge on the complex:
A coordination compound in which the complex ion
+
i. carries a net positive charge is called a cationic complex. Examples: [Ag(NH3)2] ,
3+ 2+
[Co(NH3)6] , [Fe(H2O)6] , etc
-
ii. carries a net negative charge is called an anionic complex. Examples: [Ag(CN)2] ,
3- 4-
[Co(CN)6] , [Fe(CN)6] , etc
iii. bears no net charge, is called a neutral complex. Examples: [Ni(CO)4], [Fe(CO)5] ,
[Co(NH3)3(Cl)3],
135
2. The simple ions are named as in other ionic compounds. For example,
3. To name a complex ion, the ligands are named first followed by the central metal atom/ion.
When a complex ion contains more than one kind of ligands they are named in alphabetical
order.
a. Naming the ligands:
i. The name of anionic ligands ends with the letter 'o' and the cationic ligand ends with
'ium'. The neutral ligands are usually called with their molecular names with fewer
exceptions namely, H2O (aqua), CO (carbonyl), NH3 (ammine) and NO (nitrosyl).
ii. A κ-term is used to denote an ambidendate ligand in which more than one coordination
mode is possible. For example, the ligand thiocyanate can bind to the central atom/
ion, through either the sulfur or the nitrogen atom. In this ligand, if sulphur forms a
coordination bond with metal then the ligand is named thiocyanato-κS and if nitrogen
is involved, then it is named thiocyanato-κN.
136
2-
Sulphate SO4 sulphato
2-
Sulphide S sulphido
2-
Oxalate (ox) C2O4 oxalato
H 2N
Ethylenediamine (en) NH2 ethane-1,2-diamine
O
-O O
Ethylenediaminetetraacetate -O N 2,2',2'',2'''-(ethane-1,2-
N O-
(EDTA) O O-
diyldinitrilo)tetraacetato
O
iii. If the coordination entity contains more than one ligand of a particular type, the
multiples of ligand (2, 3, 4 etc...) is indicated by adding appropriate Greek prefixes
(di, tri, tetra, etc...) to the name of the ligand. If the name of a ligand itself contains a
Greek prefix (eg. ethylenediamine), use an alternate prefixes (bis, tris, tetrakis etc..)
to specify the multiples of such ligands. These numerical prefixes are not taken into
account for alphabetising the name of ligands.
b. Naming the central metal: In cationic/neutral complexes, the element name is used
as such for naming the central metal atom/ion, whereas, a suffix 'ate' is used along with
the element name in anionic complexes. The oxidation state of the metal is written
immediately after the metal name using roman numerals in parenthesis.
137
4-
Anion (complex) [Fe(CN)6]
-
Ligands CN
138
-
ligands NH3 and Cl
Name of the ligand 4 ligands - prefix: tetra
(NH3) with prefix
Neutral ligand: ammine tetraamminiedichlorido
139
Dichloridobis(ethane-1,2-diamine)cobalt(III)
ii. [Co(en)2Cl2]Cl
chloride
Sodium 2,2',2'',2'''-(ethane-1,2-diyldinitrilo)
vii. Na2[Ni(EDTA)]
tetraacetatonickelate(II)
x. [Co(NO2)3(NH3)3] Triamminetrinirito-κNcobalt(III)
Pentaamminecyanido-κCcobalt(III)
xi. [Co(NH3)5CN][Co(NH3)(CN)5]
amminepentacyanido-κCcobaltate(III)
Tetrapyridineplatinum(II)
xii. [Pt(py)4][PtCl4]
tetrachloridoplatinate(II)
Tetraamminedichloridocobalt(III) hexacyanido-
xiii. [Co(NH3)4Cl2]3 [Cr(CN)6]
κCchromate(III)
140
4−
(v) Fe ( CN )6
15. Give the structure for the following compounds.
1. (i) diamminesilver(I) dicyanidoargentate(I)
2. (ii)Pentaammine nitrito-κNcobalt (III) ion
3. (iii)hexafluorido cobaltate (III) ion
4. (iv)dichloridobis(ethylenediamine) Cobalt (III) sulphate
5. (v) Tetracarbonylnickel (0)
Structural Stereo
Isomerism Isomerism
141
O
2+ 2+
O O N
N O
H3N NH3 H3N NH3
Co Co
H 3N NH3 H 3N NH3
NH3 NH3
Ionisation isomers:
This type of isomers arises when an ionisable counter ion (simple ion) itself can act as
a ligand. The exchange of such counter ions with one or more ligands in the coordination
entity will result in ionisation isomers. These isomers will give different ions in solution. For
-
example, consider the coordination compound [Pt(en)2Cl2]Br2. In this compound, both Br
-
and Cl have the ability to act as a ligand and the exchange of these two ions result in a different
- -
isomer [Pt(en)2Br2]Cl2. In solution the first compound gives Br ions while the later gives Cl
ions and hence these compounds are called ionisaiton isomers.
Some more example for the isomers,
Solvate isomers.
The exchange of free solvent molecules such as water , ammonia, alcohol etc.. in the
crystal lattice with a ligand in the coordination entity will give different isomers. These type
of isomers are called solvate isomers. If the solvent molecule is water, then these isomers are
called hydrate isomers. For example, the complex with chemical formula CrCl3.6H2O has three
hydrate isomers as shown below.
4.4.2 Stereoisomers:
Similar to organic compounds, coordination compounds also exhibit stereoisomerism. The
stereoisomers of a coordination compound have the same chemical formula and connectivity
between the central metal atom and the ligands. But they differ in the spatial arrangement of
ligands in three dimensional space. They can be further classified as geometrical isomers and
optical isomers.
Geometrical isomers:
Geometrical isomerism exists in heteroleptic complexes due to different possible three
dimensional spatial arrangements of the ligands around the central metal atom. This type of
isomerism exists in square planer and octahedral complexes.
In square planar complexes of the form [MA2B2]n± and [MA2BC]n± (where A, B and C are
mono dentate ligands and M is the central metal ion/atom), Similar groups (A or B) present
either on same side or on the opposite side of the central metal atom (M) give rise to two
different geometrical isomers, and they are called, cis and trans isomers respectively.
The square planar complex of the type [M(xy)2]n± where xy is a bidentate ligand with two
different coordinating atoms also shows cis-trans isomerism. Square planar complex of the
form [MABCD]n± also shows geometrical isomerism. In this case, by considering any one
of the ligands (A, B, C or D) as a reference, the rest of the ligands can be arranged in three
different ways leading to three geometrical isomers.
143
Example
Type
Cis Isomer Trans isomer
2+ 2+ 2+ 2+
H3N H3NNH3 NH3 H3 N H3N Cl Cl
MA2B2 Pt Pt Pt Pt
Cl Cl Cl Cl Cl Cl NH3 NH3
2+ 2+ 2+ 2+
H3N H3N NH3 NH3 H3 N H3 N Br Br
MA2BC Pt Pt Pt Pt
Cl Cl Br Br Cl Cl NH3 NH3
O O
H2 H H2N H2N NH2 NH2 O O NH2 NH2
C 2C C C
CH2 CH2 CH2 CH2
M(xy)2 O O Pt Pt H2C H2C Pt Pt
C C
C C C C
O O O O O O H2 N H2 N O O O O
+ + +
H3N Br Br NH Br NO
MABCD Pt Pt Pt
144
Z′
5 L
X′ L Mn+ L X
4 M 2
L
3
Z
L
Figure 5.5 Position of ligands in
6 octahedral complex Y′
In the above scheme, the positions (1,2), (1,3), (1,4), (1,5), (2,3), (2,5), (2,6), (3,4), (3,6),
(4,5), (4,6), and (5,6) are identical and if two similar groups are present in any one of these
positions, the isomer is referred as a cis isomer. Similarly, positions (1,6), (2,4), and (3,5) are
identical and if similar ligands are present in these positions it is referred as a trans-isomer.
Octahedral complex of the type [MA3B3]n± also shows geometrical isomerism. If the three
similar ligands (A) are present in the corners of one triangular face of the octahedron and the
other three ligands (B) are present in the opposing triangular face, then the isomer is referred
as a facial isomer (fac isomer)- Figure 5.6 (a).
If the three similar ligands are present around the meridian which is an imaginary
semicircle from one apex of the octahedral to the opposite apex as shown in the figure 5.6(b),
the isomer is called as a meridional isomer (mer isomer). This is called meridional because
each set of ligands can be regarded as lying on a meridian of an octahedron.
Cl Cl
CN CN
Cl Co3+ CN CN Co3+ CN
Cl Cl
CN Cl
Figure 5.6 (a) Facial isomer Figure 5.6 (b) Meridional isomer
145
Evaluate yourself 1:
5. Three compounds A ,B and C have empirical formula CrCl3.6H2O. they are kept in a
container with a dehydrating agent and they lost water and attaining constant weight
as shown below.
Compound Initial weight of the Constant weight after
compound(in g) dehydration (in g)
A 4 3.46
B 0.5 0.466
C 3 3
6. Indicate the possible type of isomerism for the following complexes and draw their
isomers
(i) [Co(en)3][Cr(CN)6] (ii) [Co(NH3)5(NO2)]2+ (iii) [Pt(NH3)3(NO2)]Cl
images of each other are called Co 3+
146
Co3+
Co3+
H2N-CH2-CH2-NH2
[()]+
(en)
Evaluate yourself 1:
10. Draw all possible stereo isomers of a complex Ca[Co(NH3)Cl(Ox)2]
147
7. In the octahedral complexes, if the (n-1) d orbitals are involved in hybridisation, then they
are called inner orbital complexes or low spin complexes or spin paired complexes. If the
nd orbitals are involved in hybridisation, then such complexes are called outer orbital or
high spin or spin free complexes. Here n represents the principle quantum number of the
outermost shell.
8. The complexes containing a central metal atom with unpaired electron(s) are paramagnetic.
If all the electrons are paired, then the complexes will be diamagnetic.
148
Complex [Ni(CO)4]
CO
Nature of ligand Strong field ligand causes the pairing of 4s electron with 3d
electrons in the metal
Coordination number - 4
Hybridisation
Hybridsation - sp3
Hybridised orbitals of
the metal atom in the
3d10 sp3 Hybridised orbitals
complex
Geometry Tetrahedral
Magnetic moment
(Using spin only µs = n(n+ 2) = 0
formula)
149
Complex [Ni(CN)4]4-
Central metal
atom/ion and its
Ni2+: 3d8, 4s0
outer electronic
configuration
Outer orbitals of metal
atom/ion 3d8 4s2 4p
-
CN
Nature of ligand
Strong field ligand causes the pairing of 3d electrons in the metal
Outer orbitals of metal
atom/ion in presence
3d8 4s0 4p0
of ligands
Coordination number - 4
Hybridisation
Hybridsation - dsp2
Hybridised orbitals of
the metal atom in the
3d8 dsp2 Hybridised orbitals 4pz0
complex
Geometry Square planar
No. of unparied electrons = 0;
Magnetic property
Hence diamagnetic
Magnetic moment
(Using spin only µs = n(n+ 2) = 0
formula)
Illustration 3
3-
Complex [Fe(CN)6]
Central metal
atom/ion and its
Fe3+: 3d5, 4s0
outer electronic
configuration
150
Illustration 4
3-
Complex [CoF6]
Coordination number - 6
Hybridisation
Hybridsation - sp3d2
151
Hybridised orbitals of
the metal atom in the
3d6 sp3d2 Hybridised orbitals 4d0
complex
Octahedral
Geometry In this complex outer d orbitals are involved in the hybridisaion
and hence the complex is called outer orbital complex
Magnetic moment
(Using spin only µs = n(n+ 2) = 4(4 + 2) = 4.899 BM
formula)
Limitations of VBT
Eventhough VBT explains many of the observed properties of complexes, it still has following
limitations
1. It does not explain the colour of the complex
2. It considers only the spin only magnetic moments and does not consider the other
components of magnetic moments.
3. It does not provide a quantitative explanation as to why certain complexes are inner orbital
complexes and the others are outer orbital complexes for the same metal. For example,
4- 4-
[Fe(CN)6] is diamagnetic (low spin) whereas [FeF6] is paramagnetic (high spin).
Evaluate yourself 1:
8. Predict the number of unpaired electrons in [CoCl4]2- ion on the basis of VBT.
9. A metal complex having composition Co(en)2Cl2Br has been isolated in two forms
A and B. (B) reacted with silver nitrate to give a white precipitate readily soluble in
ammonium hydroxide. Whereas A gives a pale yellow precipitate. Write the formula of
A and B. state the hybridization of Co in each and calculate their spin only magnetic
moment.
152
1. Crystal Field Theory (CFT) assumes that the bond between the ligand and the central metal
atom is purely ionic. i.e. the bond is formed due to the electrostatic attraction between the
electron rich ligand and the electron deficient metal.
2. In the coordination compounds, the central metal atom/ion and the ligands are considered
as point charges (in case of charged metal ions or ligands) or electric dipoles (in case of
neutral metal atoms or ligands).
3. According to crystal field theory, the complex formation is considered as the following
series of hypothetical steps.
Step 1: In an isolated gaseous state, all the five d orbitals of the central metal ion are degenerate.
Initially, the ligands form a spherical field of negative charge around the metal. In this filed, the
energies of all the five d orbitals will increase due to the repulsion between the electrons of the
metal and the ligand. L 4
L3
which the central dxy
Y
metal ion is located at dx2 - y2
-Z
L1
Figure 5.9 octahedral ligand field
153
+ 3 ∆o
5
Energy
∆o
- 2 ∆o
Average energy of 5
the d orbitals in a t2g
sphercial crystal dxy , dyz , dxz
d orbitals field
in free ion Splitting of d-orbitals
(dxy , dyz , dxz, dx2-y2 and dz2) in an octahedral
crystal field
dxz
X'
As a result, the energy of t2 Y
d
orbitals increases by 2/5Δt and that
xy
d x2 - y2
of e orbitals decreases by 3/5Δt as
Y'
shown below. when compared to X
2 ∆ t2
5 t
∆t = 94 ∆o
Energy
3 ∆
5 t
Average energy of e
the d orbitals in a
sphercial crystal dx2 y2, dz2
-
d orbitals field Splitting of d-orbitals
in free ion in an tetrahedral
(dxy , dyz , dxz, dx2-y2 and dz2) crystal field
where h is the Plank' s constant; c is velocity of light, υ is the wave number of absorption
maximum which is equal to 1/λ
From the above calculations, it is clear that the crystal filed splitting energy of the Ti3+ in
- -
complexes,the three ligands is in the order; Br < F < H2O. Similarly, it has been found form
the spectral data that the crystal field splitting power of various ligands for a given metal ion,
are in the following order
- - - - 2- - 2- - 4- - -
I <Br <SCN <Cl <S <F-<OH ≈urea< ox < H2O< NCS <EDTA <NH3<en<NO2 <CN < CO
The above series is known as spectrochemcial series. The ligands present on the right side
of the series such as carbonyl causes relatively larger crystal field splitting and are called strong
ligands or strong field ligands, while the ligands on the left side are called weak field ligands
and causes relatively smaller crystal field splitting.
156
The filling of electrons in the d orbitals in the presence of ligand field also follows Hund's
rule. In the octahedral complexes with d2 and d3 configurations, the electrons occupy different
degenerate t2g orbitals and remains unpaired. In case of d4 configuration, there are two
possibilities. The fourth electron may either go to the higher energy eg orbitals or it may pair
with one of the t2g electrons. In this scenario, the preferred configuration will be the one with
lowest energy.
If the octahedral crystal field splitting energy (Δo) is greater than the pairing energy
(P), it is necessary to cause paring of electrons in an orbital, then the fourth electron will pair
up with an the electron in the t2g orbital. Conversely, if the Δo is lesser than P, then the fourth
electron will occupy one of the degenerate higher energy eg orbitals.
For example, let us consider two different iron(III) complexes [Fe(H2O)6]3+ (weak field
complex; Δo is 14000 cm-1) and [Fe(CN)6]3- (Strong field complex; Δo is 35000 cm-1). The pairing
energy of Fe3+ is 30000 cm-1. In both these complexes the Fe3+ has d5 configuration. In aqua
complex, the Δo < P hence, the fourth & fifth electrons enter eg orbitals and the configuration is
t2g3, eg2. In the cyanido complex Δo > P and hence the fourth & fifth electrons pair up with the
electrons in the t2g orbitals and the electronic configuration is t2g5, eg0.
The actual distribution of electrons can be ascertained by calculating the crystal field
stabilisation energy (CFSE). The crystal field stabilisation energy is defined as the energy
difference of electronic configurations in the ligand filed (ELF) and the isotropic field/barycentre
(Eiso).
Here, nt2g is the number of electrons in t2g orbitals; neg is number of electrons in eg orbitals;
np is number of electron pairs in the ligand field; & n'p is the number of electron pairs in the
isotropic field (barycentre).
Complex: [Fe(H2O)6]3+
157
Magnetic moment
µs = n(n+ 2) = 5(5 + 2) = 5.916 BM
(Using spin only formula)
Complex: [Fe(CN)6]3-
158
The observed colour of a coordination compound can be explained using crystal field
theory. We learnt that the ligand field causes the splitting of d orbitals of the central metal atom
into two sets (t2g and eg). When the white light falls on the complex ion, the central metal ion
absorbs visible light corresponding to the crystal filed splitting energy and transmits rest of the
light which is responsible for the colour of the complex.
159
Evaluate yourself 1:
11. The mean pairing energy and octahedral field splitting energy of [Mn(CN)6]3- are
28,800 cm-1 and 38500 cm-1 respectively. Whether this complex is stable in low spin or
high spin?
12. Draw energy level diagram and indicate the number of electrons in each level for the
complex [Cu(H2O)6]2+. Whether the complex is paramagnetic or diamagnetic?
13. For the [CoF6]3- ion the mean pairing energy is found to be 21000 cm-1 . The magnitude
of Δ0 is 13000cm-1. Calculate the crystal field stabilization energy for this complex ion
corresponding to low spin and high spin states.
Metallic carbonyls
Metal carbonyls are the transition metal complexes of carbon monoxide, containing Metal-
Carbon bond. In these complexes CO molecule acts as a neutral ligand. The first homoleptic
carbonyl Ni ( CO ) 4 nickel tetra carbonyl was reported by Mond in 1890.These metallic
carbonyls are widely studied because of their industrial importance, catalytic properties and
their ability to release carbon monoxide.
160
These compounds contain only one metal atom, and have comparatively simple structures.
For example, Ni ( CO ) 4 - nickel tetracarbonyl is tetrahedral, Fe ( CO )5 - Iron pentacarbonyl
is trigonalbipyramidal, and Cr ( CO )6 - Chromium hexacarbonyl is octahedral.
b. Poly nuclear carbonyls
Metallic carbonyls containing two or more metal atoms are called poly nuclear carbonyls. Poly
(
nuclear metal carbonyls may be Homonuclear Co2 ( CO )8 , Mn2 ( CO )10 , Fe3 ( CO )12 )
(
or hetero nuclear MnCo ( CO )9 , MnRe ( CO )10 etc. )
(ii) Classification based on the structure:
The structures of the binuclear metal carbonyls involve either metal–metal bonds or
bridging CO groups, or both. The carbonyl ligands that are attached to only one metal atom
are referred to as terminal carbonyl groups, whereas those attached to two metal atoms
simultaneously are called bridging carbonyls. Depending upon the structures, metal carbonyls
are classified as follows.
a. Non-bridged metal carbonyls:
These metal carbonyls do not contain any bridging carbonyl ligands. They may be of two
types.
(i) Non- bridged metal carbonyls which contain only terminal carbonyls. Examples:
Ni ( CO ) 4 , Fe ( CO )5 and Cr ( CO )6
CO
CO
O
C OC
CO
O
Ni C
Cr
CO Fe C O
OC
CO C OC CO
O
C
O CO
(ii) Non- bridged metal carbonyls which contain terminal carbonyls as well as Metal-Metal
bonds. For examples,The structure of Mn2(CO)10actually involve only a metal–metal
bond, so the formula is more correctly represented as (CO)5Mn−Mn(CO)5.
161
OC CO
OC Mn Mn CO
2.79A0
OC CO
CO CO
Other examples of this type are,Tc2(CO)10, and Re2(CO)10.
b. Bridged carbonyls:
These metal carbonyls contain one or more bridging carbonyl ligands along with terminal
carbonyl ligands and one or more Metal-Metal bonds. For example,
(i) The structure of Fe2(CO)9, di-iron nona carbonyl molecule consists of three bridging CO ligands,
six terminal CO groups
CO
CO
CO CO
OC Fe Fe CO
CO CO
CO
(ii) For dicobaltoctacarbonylCo2(CO)8two isomers are possible. The one has a metal–metal
bond between the cobalt atoms, and the other has two bridging CO ligands.
CO
CO CO CO CO
OC CO
Co OC Co Co CO
0
Co
2.52 A
OC CO
CO CO
CO
CO CO
162
Cu ( NH3 ) 4
2+
Cu 2+ + 4 NH3
Cu ( NH3 ) 4
2+
β = 2+
Cu [NH3 ]
4
---------( 1 )
So, as the concentration of Cu ( NH3 ) 4 increases the value of stability complexes also
2+
increases. Therefore the greater the value of stability constant greater is the stability of the
complex.
Generally coordination complexes are stable in their solutions; however, the complex ion
can undergo dissociation to a small extent. Extent of dissociation depends on the strength of
the metal ligand bond, thus Stronger the M ← L , lesser is the dissociation.
163
Cu ( NH3 ) 4
2+
Cu + 4 NH3
2+
Cu 2+ [NH3 ]
4
α =
Cu ( NH3 ) 4 ---------( 2 )
2+
From (1) and (2) we can say that, the reciprocal of dissociation equilibrium constant ( α)
is called as formation equilibrium constant or stability constant ( β) .
1
β=
α
Cu ( NH3 ) 4
2+
−12
1.0 × 10 1.0 × 1012
−
Ag ( CN )2 1.8 × 10
−19
5.4 × 1018
Co ( NH3 )6
3+
−36
6.2 × 10 1.6 × 1035
2−
Hg ( CN ) 4 4.0 × 10
− 42
2.5 × 10 41
By comparing stability constant values in the above table, we can say that among the five
complexes listed, Hg ( CN ) 4 is most stable complex ion and Fe ( SCN )
2− 2+
is least stable.
5.11.1. Stepwise formation constants and overall formation constants
When a free metal ion is in aqueous medium, it is surrounded by (coordinated with)
water molecules. It is represented as [MS6]. If ligands which are stronger than water are added
to this metal salt solution, coordinated water molecules are replaced by strong ligands.
164
[MS ] 6
+ 6L [ML 6 ] + 6 S
[ML ] [S ]
6
βoverall = 6
[MS ] [L ]
6
6
βoverall is called as overall stability constant. As solvent is present in large excess, its
concentration in the above equation can be ignored.
∴ βoverall =
[ML ] 6
[MS ] [L ]
6
6
If these six ligands are added to the metal ion one by one, then the formation of complex
[ML6] can be supposed to take place through six different steps as shown below. Generally step
wise stability constants are represented by the symbol k.
[MS L ]
[MS ]
+ L [MS5L ] + S
k1 = 5
6
[MS ][L ]
6
[MS L ]
[MS L ]
+L [MS 4 L 2 ] + S
k2 = 4 2
5
[MS L ][L ]
5
[MS L ]
[MS L ]
+ L [MS3L 3 ] + S
k3 = 3 3
4 2
[MS L ][L ]
4 2
[MS L ]
[MS L ]
+ L [MS2 L 4 ] + S
k4 = 2 4
3 3
[MS L ][L ]
3 3
[MSL ]
[MS L ] +
L [MSL5 ] + S
k5 = 5
2 4
[MS L ][L ]
2 4
[ML ]
[MSL ]
+ L [ML 6 ]
+ S k6 = 6
5
[MSL ][L ]
5
In the above equilibrium, the values k1 , k 2 , k 3 , k 4 , k 5 and k 6 are called step wise stability
constants. By carrying out small a mathematical manipulation, we can show that overall
stability constant β is the product of all step wise stability constants k1 , k 2 , k 3 , k 4 , k 5 and k 6 .
β = k1 × k 2 × k 3 × k 4 × k 5 × k 6
165
6. Many of the complexes are used as catalysts in organic and inorganic reactions. For example,
(i) Wilkinson’s catalyst - ( PPh 3 )3 RhCl is used for hydrogenation of alkenes.
7. In order to get a fine and uniform deposit of superior metals (Ag, Au, Pt etc.,) over
− −
base metals, Coordination complexes Ag ( CN ) and Au ( CN ) etc., are used in
2 2
electrolytic bath.
8. Many complexes are used as medicines for the treatment of various diseases. For example,
(1) Ca-EDTA chelate, is used in the treatment of lead and radioactive poisoning. That is for
removing lead and radioactive metal ions from the body.
(2) Cis-platin is used as an antitumor drug in cancer treatment.
9. In photography, when the developed film is washed with sodium thio sulphatesolution
(hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called
sodiumdithiosulphatoargentate(I) which can be easily removed by washing the film with
water.
Na 3 Ag ( S2O3 )2 + 2 NaBr
AgBr + 2 Na 2S2O3 →
166
Cisplatin:
Cisplatin is a square planar coordination
complex (cis- [Pt (NH3)2Cl2]), in which two
similar ligands are in adjacent positions.
It is a Platinum-based anticancer drugThis drug undergoes
hydrolysis and reacts with DNA to produce various
crosslinks. These crosslinks hinder the DNA replication
and transcription, which results in cell growth inhibition and ultimately cell death.
It also crosslinks with cellular proteins and inhibits mitosis.
Summary
When two or more stable compounds in solution are mixed together and allowed
to evaporate, in certain cases there is a possibility for the formation of double
salts or coordination compounds. The double salts loose their identity and
dissociates into their constituent simple ions in solutions , whereas the complex
ion in coordination compound, does not loose its identity and never dissociate to
give simple ions.
According to werner, most of the elements exhibit, two types of valence namely
primary valence and secondary valence and each element tend to satisfy both the
valences.In modern terminology, the primary valence is referred as the oxidation
state of the metal atom and the secondary valence as the coordination number.
Coordination entity is an ion or a neutral molecule, composed of a central atom,
usually a metal and the array of other atoms or groups of atoms (ligands) that
are attached to it.
167
168
3. A complex has a molecular formula MSO4Cl. 6H2O .The aqueous solution of it gives white
precipitate with Barium chloride solution and no precipitate is obtained when it is treated
with silver nitrate solution. If the secondary valence of the metal is six, which one of the
following correctly represents the complex?
4. Oxidation state of Iron and the charge on the ligand NO in Fe ( H2O )5 NO SO4 are
a) chlorobisethylenediaminenitritocobalt(III) chloride
a) potassiumtrioxalatoaluminium(III)
b) potassiumtrioxalatoaluminate(II)
c) potassiumtrisoxalatoaluminate(III)
d) potassiumtrioxalatoaluminate(III)
169
a) TiCl 4 b) [CoCl 6 ]
4−
2−
c) Cu ( NH3 ) 4 d) Ni ( CN ) 4
2+
a) −0.6∆ 0 b) 0
c) 2(P −∆ 0 ) d) 2(P + ∆ 0 )
9. In which of the following coordination entities the magnitude of Δ0 will be maximum?
3−
b) Co ( C 2O4 )3
3−
a) Co ( CN )6
a) 3 b) 4 c) 0 d) 15
13. Which one of the following pairs represents linkage isomers?
14. Which kind of isomerism is possible for a complex Co ( NH3 ) 4 Br2 Cl ?
3+
c) Co ( NH3 )5 SO4 Cl d) Fe ( en )3
170
c) Fe ( H2 N-CH2 -CH2 -NH2 )3 ( PO4 )2 d) Fe ( H2 N-CH2 -CH2 -NH2 )3 3 ( PO4 )2
18. Which of the following is paramagnetic in nature?
b) Co ( NH3 )6
3+
a) Zn ( NH3 ) 4
2+
2−
d) Ni ( CN ) 4
2+
c) Ni ( H2O )6
19. Fac-mer isomerism is shown by
3+
a) Co ( en )3 b) Co ( NH3 ) 4 (Cl )2
+
2+
d) crystal field stabilization energy of V ( H2O )6 is higher than the crystal field
2+
stabilization of Ti ( H2O )
6
5
171
4. Ni2+ is identified using alcoholic solution of dimethyl glyoxime. Write the structural
formula for the rosy red precipitate of a complex formed in the reaction.
5. [CuCl 4 ] exists while [CuI4 ] does not exist why?
2− 2−
Ag +
6. Calculate the ratio of + in 0.2 M solution of NH3. If the stability constant for
Ag ( NH3 )2
the complex Ag ( NH3 )2 is 1.7 × 107
+
11. Give an example for complex of the type [Ma 2 b2 c2 ] where a, b, c are monodentate ligands
and give the possible isomers.
12. Give one test to differentiate Co ( NH3 )5 Cl SO4 and Co ( NH3 )5 SO4 Cl .
13. In an octahedral crystal field, draw the figure to show splitting of d orbitals.
14. What is linkage isomerism? Explain with an example.
15. Write briefly about the applications of coordination compounds in volumetric analysis.
16. Classify the following ligand based on the number of donor atoms.
a) NH3 b) en c) ox2- d) triaminotriethylamine e) pyridine
172
theory.
20. Why tetrahedral complexes do not exhibit geometrical isomerism.
21. Explain optical isomerism in coordination compounds with an example.
22. What are hydrate isomers? Explain with an example.
23. What is crystal field splitting energy?
24. What is crystal field stabilization energy (CFSE) ?
2+
25. A solution of Ni ( H2O )6 is green, whereas a solution of Ni ( CN ) 4
2−
is colorless -
Explain
26. Discuss briefly the nature of bonding in metal carbonyls.
27. What is the coordination entity formed when excess of liquid ammonia is added to an
aqueous solution of copper sulphate?
3−
28. On the basis of VB theory explain the nature of bonding in Co ( C 2O4 )3 .
29. What are the limitations of VB theory?
30. Write the oxidation state, coordination number , nature of ligand, magnetic property and
electronic configuration in octahedral crystal field for the complex K 4 Mn ( CN )6 .
173
Coordination
sphere Terms
Stereo isomerim Optical
isomerism
Coordination
polyhedron Hybridisation
Werner’s Theory
174
Charge on a coordination theories of
complex compounds coordination Primary valency
compounds
VBT
Secondary valency
CFT
4/2/2019 11:04:50 AM
ICT Corner
Steps
• Open the browser and type the URL given (or) Scan the QR Code. You will see the webpage as
shown in the figure.
Note: One time sign up is needed to access this webpage. Login using your username and password.
Once logged in click the simulation tab.
• You can select a suitable ligand field splitting using the drop down menu (box 1). Select a metal of
interest and a ligand using the drop down menu (box 2). Now crystal field splitting for the selected
complex appears on the screen.
• Apply crystal field theory to the selected complex and fill the d-electrons in the t2g and eg orbitals by
clicking each orbital. Click on the orbitals thrice to remove electrons. After completion, click submit
button (box 4). Now you can check the correctness of the electron distribution. If wrong try again.
• Enter the number of electrons in the t2g & eg orbitlals in the work sheet at the bottom of the page
(box 6), The calculated Crystal Field Stabilisation Energy (CFSE) will be displayed.
175
Sir William Henry Bragg was a After studying this unit, the students will
British physicist, chemist, and a be able to
mathematician. Sir William Henry
Bragg and his son Lawrence Bragg describe general characteristics of solids
worked on X-rays with much distinguish amorphous and crystalline
success. They invented the X-ray solids
spectrometer and founded the new
define unit cell
science of X-ray crystallography,
the analysis of crystal structure describe different types of voids and
using X-ray diffraction. Bragg close packed structures
was joint winner (with his son, calculate the packing efficiency of
Lawrence Bragg) of the Nobel different types of cubic unit cell
Prize in Physics in 1915, for their solve numerical problems involving
services in the “analysis of crystal unit cell dimensions
structure by means of ray”. The
explain point defects in solids
mineral Braggite (a sulphide ore of
platinum, palladium and Nickel) is
named after him and his son.
176
and their properties is very much useful Atomic solids - ex: frozen elements of Group 18
177
B A
178
179
180
There are seven primitive crystal systems; cubic, tetragonal, orthorhombic, hexagonal,
monoclinic, triclinic and rhombohedral. They differ in the arrangement of their crystallographic
axes and angles. Corresponding to the above seven, Bravis defined 14 possible crystal systems
as shown in the figure.
b
a
181
182
E
G
r
F
2rr a
D
r
A
C
2a
a a
B
r
r
a
Nc 8 1
∴ no of atoms in a SC unit cell = = +
8 8 1
8 = (1 + 1)
= =1
8 =2
183
a
r a The above equation is known as
A
Bragg’s equation.
Where
In a face centered cubic unit cell,
λ is the wavelength of X-ray used for
identical atoms lie at each corner as well diffraction.
as in the centre of each face. Those atoms θ is the angle of diffraction
in the corners touch those in the faces By knowing the values of θ,λ and n
but not each other. The atoms in the face we can calculate the value of d.
nλ
centre is being shared by two unit cells, d=
1 2sinθ
each atom in the face centers makes
2 Using these values the edge of the unit
contribution to the unit cell. cell can be calculated.
∴ Number of atoms
N N 6.5.5 Calculation of density:
in a fcc unitcell = c + f
8 2 Using the edge length of a unit cell, we
8 6 can calculate the density ( ρ) of the crystal
= + by considering a cubic unit cell as follows.
8 2
= (1 + 3) Density of
ρ =
mass of the unit cell
the unit cell volume of the unit cell
=4
...(1)
Drawing the crystal lattice on paper
184
M
m=
NA ...(3)
Substitute (3) in (2)
M
mass of the unit cell= n ×
NA ...(4)
For a cubic unit cell, all the edge lengths are equal i.e , a=b=c
volume of the unit cell = a × a × a = a 3 ...(5)
nM
∴ Density of the unit cell ρ = a 3 N ...(6)
A
Equation (6) contains four variables namely ρ , n , M and a . If any three variables are
known, the fourth one can be calculated.
Example 2
Barium has a body centered cubic unit cell with a length of 508pm along an edge.
What is the density of barium in g cm-3?
Solution:
nM
ρ =
a 3NA
In this case,
n=2 ; M=137.3 gmol-1 ; a = 508pm= 5.08X10-8cm
2 atoms × 137.3 g mol −1
ρ =
(5.08 × 10 cm ) ( 6.023 × 1023 atoms mol −1 )
−8 3
2 × 137.3
ρ = g cm −3
(5.08 )
3
× 10 −24
× 6.023 × 10 23
ρ = 3.5 g cm −3
185
186
a
Let ‘r’ is the radius of the sphere. From the figure, a=2r ⇒ r =
2
187
πa 3 E
6 100 π
G
= 52.31%
2rr a
189
Formation of third layer:
The third layer of spheres can be formed in two ways to achieve
closest
packing
The spheres can be arranged so as to fit into the depression in such a way that the third
layer is directly over a first layer as shown in the figure. This “aba’’ arrangement is known
as the hexagonal close packed (hcp) arrangement. In this arrangement, the tetrahedral
voids of the second layer are covered by the spheres of the third layer.
Alternatively, the third layer may be placed over the second layer in such a way that all
the spheres of the third layer fit in octahedral voids. This arrangement of the third layer is
different from other two layers (a) and (b), and hence, the third layer is designated (c). If
the stacking of layers is continued in abcabcabc… pattern, then the arrangement is called
cubic close packed (ccp) structure.
In both hcp and ccp arrangements, the coordination number of each sphere is 12 – six
neighbouring spheres in its own layer, three spheres in the layer above and three sphere in
the layer below. This is the most efficient packing.
190
Layer b
Layer a
Layer a
Layer c
Layer b
Layer a
191
Coordination
Structure
rC+
Example
4 2 2a 3
number
= π
3 64 rA−
2 πa 3
= 0.155 – Trigonal
24 0.225
3
planar
B2O3
192
194
195
Summary
196
EVALUATION
197
1 3 2
a) a :
2 2
a: a b) 1a : 3a : 2a
2
( )
1 3 1 1 1
c) a : a: a d) a : 3a : a
2 4 2 2 2 2
19. If ‘a’ is the length of the side of the cube, the distance between the body centered atom
and one corner atom in the cube will be
2 4
a) a b) a
3 3
3 3
c) a d) a
4 2
20. Potassium has a bcc structure with nearest neighbor distance 4.52 A0 . its atomic weight
is 39. its density will be
a) 915 kg m-3 b) 2142 kg m-3 c) 452 kg m-3 d) 390 kg m-3
21. S chottky defect in a crystal is observed when
a) unequal number of anions and anions are missing from the lattice
199
a) XY8 b) X4Y9
c) XY2 d) XY4
Answer the following questions:
1. Define unit cell.
2. Give any three characteristics of ionic crystals.
3. Differentiate crystalline solids and amorphous solids.
4. Classify the following solids
a. P4 b. Brass c. diamond d. NaCl e. Iodine
200
22. An atom crystallizes in fcc crystal lattice and has a density of 10 gcm with unit cell
−3
201
Ionic crystals
Covalent crystals
Crystalline structure
Molecular crystals
Sc Schottky defect
Unit cell
Packing fraction
202
CRYSTAL SYSTEMS
Steps
•
Open the Browser and type the URL given (or) Scan the QR Code. In the webpage click physical
science tab and then click solid state virtual lab. Then go to crystal structure and then click
simulator.
Note: One time sign up is needed to access this webpage. Login using your username and password.
Once logged in click the simulator tab.
• Now the using the menu (box 1) select any one of the seven crystal systems and the lattice type.
Now the unit cell of the selected crystal system will appear on screen (box 2) and the unit cell
parameters will also be displayed in the measurement tab (box 3)
2 1
203
204
205
1.0
0.8
Concentraon(M)
0.6
0.4
0.2
0
0 20 40 60 80 100 120
Time (mins)
* Schematic representation- Not to scale -A -B
206
207
1.50
[Cyclopropane] (M)
1.00
0.50
0.00
0 10 20 30 40 50 60 70 80 90 100
Time (mins)
208
[cyclopropane]
For experiment 3
Rate Rate
mol L-1
k=
mol L-1min-1 [cyclopropane] Rate3 = k [NO]m [O 2 ]n
2=
1.1
1.73 × 10 –2 0.5 3.46 × 10 –2
2=2n i.e., n=1
Let us consider an another example, the
oxidation of nitric oxide (NO) Therefore the reaction is first order with
respect to O 2
2NO(g) + O2 ( g )
→ 2NO2 (g)
(3) 76.8 x 10-2 k [2.6]m [1.1]n
⇒ =
Series of experiments are conducted (1) 19.26 x 10-2 k [1.3]m [1.1]n
by keeping the concentration of one of the m
2. 6
reactants constant and the changing the 4=
1. 3
concentration of the others.
4=2m i.e., m=2
Experiment
reaction reaction
Rate = k [NO]m[O2 ]n
For experiment 1, the rate law is It represents the
speed at which
Rate1 = k [NO]m[O2 ]n It is a
the reactants are
1 proportionality
19.26 X10-2 = k [1.3]m[1.1]n ...(1) converted into
constant
Similarly for experiment 2 products at any
instant.
Rate 2 = k [NO]m [O 2 ]n
209
reaction reaction
reaction mechanism is called an elementary
reaction. 1 It is the sum of It is the total
An elementary step is characterized the powers of number of
by its molecularity. The total number of concentration reactant species
reactant species that are involved in an terms involved in that are involved
elementary step is called molecularity of that the experimentally in an elementary
particular step. Let us recall the hydrolysis determined rate step.
of t-butyl bromide studied in XI standard. law.
Since the rate determining elementary step
involves only t-butyl bromide, the reaction 2 It can be zero (or) It is always a
is called a Unimolecular Nucleophilic fractional (or) whole number,
substitution (SN ) reaction.
1 integer cannot be zero
or a fractional
Let us understand the elementary
number.
reactions by considering another reaction,
the decomposition of hydrogen peroxide
3 It is assigned for a It is assigned
catalysed by I− .
overall reaction. for each
2H2O2 (aq)
→ 2H2O(l ) + O2 (g) elementary step
It is experimentally found that the of mechanism.
reaction is first order with respect to both
210
1).
Write the rate expression for the , if [x]=[y]=0.2 M and rate constant
following reactions, assuming them as at 400K is 2 x 10-2 s -1 , What is the
elementary reactions. overall order of the reaction.
i) 3A + 5B2
→ 4CD
Solution :
ii) X 2 + Y2
→ 2XY
Rate = k [x]n [y]m
2). Consider the decomposition of N 2 O5 (g) 4 x 10-3 mol L-1s -1 = 2 x 10-2 s -1 (0.2mol L-1 ) n (0.2mol L-1 ) m
to form NO 2 (g) and O 2 (g) . At a
4 x 10−3 mol L−1s −1
particular instant N 2 O5 disappears at a = (0.2) n + m ( mol L−1 )
n+m
−2 −1
2 x 10 s
rate of 2.5x10-2 mol dm -3s-1 . At what rates
0.2 ( mol L-1 ) = (0.2) n + m ( mol L-1 )
n+m
are NO 2 and O 2 formed? What is the
rate of the reaction? Comparing the powers on both sides
Example 2 The overall order of the reaction n + m = 1
211
reaction,
NO NOCl mol L-1s -1
A
→ product
Rate law can be expressed as
1 0.1 0.1 7.8 x10 −5
Rate = k [A]1
2 0.2 0.1 3.12 x10−4 Where, k is the first order rate constant.
-d[A]
= k [A]1
3 0.2 0.3 9.36 x10 −4 dt
-d[A]
⇒ = k dt
[A] ...(1)
7.5 The integrated rate equation:
Integrate the above equation between
We have just learnt that the rate of change the limits of time t = 0 and time equal to t,
of concentration of the reactant is directly while the concentration varies from the initial
proportional to that of concentration of the concentration [A 0 ] to [A] at the later time.
reactant. For a general reaction, [A]-d[A] t
A
→ product
∫[A 0 ] [A]
= k ∫ dt
0
A0
− d[A]
Rate = = k [A]x
dt - ln [A]- ( - ln [A0 ]) = k (t-0)
212
-0.5
-1
ln [A]
-1.5
-2
-2.5
0 10 20 30 40 50 60
Time (in mins)
2.303 [A ]
k= log 0 ----- (3)
t [A]
Equation (2) can be written in the form y = mx + c as below
ln [A 0 ] −ln [A] = kt
ln [A] = ln [A 0 ] −kt
⇒ y = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time
interval‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope.From this, the
rate constant is calculated.
Examples for the first order reaction
(i) Decomposition of dinitrogen pentoxide
1
N 2 O5 (g)
→ 2NO 2 (g) + O 2 (g)
2
213
Rate = k [CH3COOCH 3 ] [H 2 O]
If the reaction is carried out with the large excess of water, there is no significant change
in the concentration of water during hydrolysis. i.e.,concentration of water remains almost a
constant.
Now, we can define k [H 2 O] = k' ; Therefore the above rate equation becomes
0.75
[A] in M
0.5
0.25
0
0 10 20 30 40 50 60
Time (in mins)
Fig 7.4 : A plot of [A] Vs time for a zero order reaction A → product with
initial concentration of [A] = 0.5M and k = 1.5x10 mol−1L−1min −1
-2
214
[A 0 ] −[A] = kt
7.6 Half life period of a reaction:
[A 0 ] −[A]
k= The half life of a reaction is defined
t
as the time required for the reactant
Equation (2) is in the form of a straight concentration to reach one half its initial
line y = mx + c value. For a first order reaction, the half life
is a constant i.e., it does not depend on the
Ie., [A] = −kt + [A 0 ]
initial concentration.
⇒ y = c + mx
The rate constant for a first order
A plot of [A] Vs time gives a straight reaction is given by
line with a slope of −k and y - intercept of k=
2.303 [A ]
log 0
[A 0 ] . t [A]
[A 0 ]
Examples for a zero order reaction: =at t t= ; [A]
1
2 2
1. Photochemical reaction between H2 and I2
2.303 [A 0 ]
H2 (g)+Cl 2 (g) hv
→ 2HCl(g) k= log
t1 [A 0 ]
2. Decomposition of N2O on hot platinum
2 2
surface 2.303
k= log 2
1 t1
N 2 O(g) N 2 (g) + O 2 (g) 2
2
2.303x0.3010 0.6932
=k =
3. Iodination of acetone in acid medium is t1 t1
zero order with respect to iodine. 2 2
+
CH 3COCH3 + I 2 H→ ICH 2 COCH 3 + HI 0.6932
t1 =
2 k
Rate = k [CH3COCH3 ] [H + ]
215
Solution: [A 0 ] −[A]
x 100 = ?
For a first order reaction, [A 0 ]
We know that
2.303 [A ]
k= log 0 ....(1) 0.6932
t [A] For a first order reaction, t 1 =
2 k
Let [A 0 ] = 100M 0.6932
k=
When 6.932 x 104
t = t 90% ; [A]=10M (given that t90 % =8hours) k = 10−5 s −1
t = t 80% ; [A]=20M
216
217
Ea
Reactants
Products
Reaction progress Æ
218
improper allignment
A A + B B A A B B A A + B B
219
(4) –(3)
Example 8
E E
ln k 2 - ln k1 = - a + a
RT2 RT1
Rate constant k of a reaction varies
k E 1 1 with temperature T according to the
ln 2 = a - following Arrhenius equation
k1 R T1 T2
Ea 1
k E T −T log k = log A −
2.303 log 2 = a 2 1 2.303R T
k1 R T1T2
Where E a is the activation energy.
k E a T2 −T1 1
log 2 = When a graph is plotted for log k Vs
k1 2.303R T1T2 T
a straight line with a slope of -4000K
E E
ln k 2 −ln k1 = − a + a is obtained. Calculate the activation
RT2 RT1
energy
This equation can be used to calculate
E a from rate constants k1 and k 2 at
temperatures T1 and T2 . Solution
Example 7 Ea 1
log k = log A-
2.303R T
The rate constant of a reaction y = c + mx
at 400 and 200K are 0.04 and 0.02 s-1
Ea
respectively. Calculate the value of m= -
2.303R
activation energy.
E a = − 2.303 R m
221
The rate of a reaction is affected by the Let us consider another example that
following factors. you carried out in inorganic qualitative
analysis of lead salts. If you mix the
1. Nature and state of the reactant aqueous solution of colorless potassium
2. Concentration of the reactant iodide with the colorless solution of lead
nitrate, precipitation of yellow lead iodide
3. Surface area of the reactant take place instantaneously, whereas if
4. Temperature of the reaction you mix the solid lead nitrate with solid
potassium iodide, yellow coloration will
5. Presence of a catalyst appear slowly.
7.9.1 Nature and state of the reactant:
We know that a chemical reaction
involves breaking of certain existing bonds
of the reactant and forming new bonds
which lead to the product. The net energy
involved in this process is dependent on the
nature of the reactant and hence the rates
are different for different reactants.
Let us compare the following two
reactions that you carried out in volumetric
analysis.
1).
Redox reaction between ferrous
ammonium sulphate (FAS) and KMnO 4
2). Redox reaction between oxalic acid and
KMnO 4
The oxidation of oxalate ion by
KMnO 4 is relatively slow compared to the 7.9.2 Concentration of the reactants:
reaction between KMnO 4 and Fe2+ . In fact The rate of a reaction increases with
heating is required for the reaction between the increase in the concentration of the
KMnO 4 and Oxalate ion and is carried out reactants. The effect of concentration is
at around 600 C . explained on the basis of collision theory
The physical state of the reactant also of reaction rates. According to this theory,
plays an important role to influence the rate the rate of a reaction depends upon the
of reactions. Gas phase reactions are faster number of collisions between the reacting
as compared to the reactions involving solid molecules. Higher the concentration,
or liquid reactants. For example, reaction of greater is the possibility for collision and
sodium metal with iodine vapours is faster hence the rate.
222
A 10 40 0.02
B 20 30 0.04
C 40 10 0.08
1 1
Draw a graph between Vs is a direct measure of rate of
t t
concentration of sodium thiosulphate. A reaction and therefore, the increase in the
graph like the following one is obtained. concentration of reactants i.e Na 2S2 O3 ,
increases the rate.
(s )
1 1
t
A Known mass of marble chips are taken in a flask and a known volume of dilute HCl is
added to the content, a stop clock is started when half the volume of HCl is added. The mass
is noted at regular intervals until the reaction is complete. Same experiment is repeated with
the same mass of powdered marble chips and the observations are recorded.
Reaction
→ CaCl2 (aq) + H 2 O (l ) + CO 2 ( g )
CaCO3 (s) + 2HCl (aq)
r
ps
ma
chi
ble
ed
ar
flask, we can follow the rate of the reaction.
der
m
Mass
Pow
A plot of loss in mass Vs time is drawn and
it looks like the one as shown below.
For the powdered marble chips, the 0 10 20 30 40 50 60
224
Take two test tubes and label them as A and B. Add 7 ml of 0.1N oxalic acid solution, 5
ml of 0.1N KMnO 4 solution and 5 ml of 2N dilute H 2SO 4 in both the test tubes. The colour
of the solution is pink in both the test tubes.
Now add few crystals of manganese sulphate to the content in test tube A. the pink
colour fades up and disappears. In this case, MnSO4 acts as a catalyst and increases the rate
−
of oxidation of C2O 4 by MnO 4
2-
Summary
Chemical kinetics is the study of the rate and the mechanism of chemical reactions,
proceeding under given conditions of temperature, pressure, concentration etc.
The change in the concentration of the species involved in a chemical reaction per
unit time gives the rate of a reaction.
The rate of the reaction, at a particular instant during the reaction is called the
instantaneous rate. The shorter the time period, we choose, the closer we approach
to the instantaneous rate,
The rate represents the speed at which the reactants are converted into products at
any instant.
The rate constant is a proportionality constant and It is equal to the rate of reaction,
when the concentration of each of the reactants in unity
Molecularity of a reaction is the total number of reactant species that are involved
in an elementary step.
The half life of a reaction is defined as the time required for the reactant concentration
to reach one half its initial value. For a first order reaction, the half life is a constant
225
a)
EVALUATION
ln k
226
L s
2
2
c) 2.5 hours
8. The addition of a catalyst during a
d) Without knowing the rate constant,
t1/2 cannot be determined from the chemical reaction alters which of the
following quantities? (NEET)
given data
a) Enthalpy b)Activation
5. For the reaction, 2NH3 →
N 2 + 3H2 energy
−d [NH3 ] c) Entropy d) Internal energy
, if = k1 [NH3 ] ,
dt 9. Consider the following statements :
d [N 2 ]
= k 2 [NH3 ], [ 2 ] = k 3 [NH3 ]
d H (i) increase in concentration of the
dt dt reactant increases the rate of a zero
order reaction.
then the relation between k1 ,k 2 and k 3 is
(ii) rate constant k is equal to collision
a) k1 = k 2 = k 3
frequency A if Ea = 0
b) k1 = 3 k 2 = 2 k 3
(iii) rate constant k is equal to collision
c) 1.5 k1 = 3 k 2 = k 3 frequency A if Ea = °
d) 2k1 = k 2 = 3 k 3 (iv) a plot of ln ( k ) vs T is a straight
6. The decomposition of phosphine (PH3) line.
on tungsten at low pressure is a first 1
order reaction. It is because the (NEET) (v) a plot of ln ( k ) vs is a straight
T
a)
rate is proportional to the surface line with a positive slope.
coverage Correct statements are
b) rate is inversely proportional to the a) (ii) only b) (ii) and (iv)
surface coverage
c) (ii) and (v) d) (i), (ii) and (v)
c) rate is independent of the surface
coverage 10. In a reversible reaction, the enthalpy
change and the activation energy in
d) rate of decomposition is slow the forward direction are respectively
3
7. For a reaction Rate = k [acetone ] 2 then −x kJ mol −1 and y kJ mol −1 . Therefore ,
unit of rate constant and rate of reaction the energy of activation in the backward
respectively is direction is
a) ( y −x ) kJ mol
) , ( mol )
−1
a) ( mol L s
−1 1
−1 −1 −1
2
L2 s
b) ( x + y ) J mol
−1
(
b) mol
−1
2
1
)
L 2 s −1 , ( mol L−1s −1 ) c) ( x − y ) kJ mol −1
d) ( x + y ) × 103 J mol −1
227
k is the rate constant and the initial a) First order b) zero order
concentration of the reactant x is 0.1M,
c) Second order d) Third order
then, the half life is
log 2 0.693
a) b) 18. For the reaction
k ( 0.1) k 1
N 2O5 (g) →
2NO2 (g) + O (g) , the
ln2 2 2
c) d) none of these
k value of rate of disappearance of N2O5
15. Predict the rate law of the following is given as 6.5 × 10−2 mol L−1s −1 . The
reaction based on the data given below rate of formation of NO2 and O2 is given
228
229
reactions
. How would the rate of reaction change
when (i) Rusting of Iron
(ii) Radioactive disintegration of 92 U
238
(i) Concentration of [L] is quadrupled
products ; rate = k [A ] 2 [B ]
1 2
(ii) Concentration of both [A] and [B] (iii) 2A + 3B →
are doubled 19. A gas phase reaction has energy of
(iii) Concentration of [A] is halved activation 200 kJ mol-1. If the frequency
factor of the reaction is 1.6 × 1013 s −1
230
231
Temperature
factors affecting
Average rate Reaction rate
reaction rate
Catalyst
Instantaneous rate
232
Zero order [A 0 ] −[A]
k=
reaction t
Order Molecularity Rate constant
First order
reaction
2.303 [A ]
Collision theory Temperature k= log 0
dependence of t [A]
rate ( Arrhenius
theory)
4/2/2019 11:17:15 AM
ICT Corner
CHEMICAL KINETICS
Steps
• Open the Browser and type the URL given (or) Scan the QR Code. You can see a webpage
which displays the java applet called reactions and rates. You can click it and you will get a
window as shown in the figure. This applet contains three modules which can be selected by
clicking the appropriate tab (box1).
• Select single collision tab (tab 1) to visualise collision between two molecules. You can visualise
the progress of the reaction (box 5) by varying the temperature using the slider (box 2). You
can visualise that the raise in temperature, will raise the energy of the system and allows the
reactants to cross the energy barrier to form products. You can repeat this simulation with
more molecules in the many collisions tab (box 1).
• You can also perform virtual kinetic experiment, using rate experiments mode. Choose the types
reacting molecules and their stoichiometry using the options provided (box 2). The number of
reactant and product molecules at a given time will be displayed in panel (box-3). You can see
the effect of temperature on reaction rate by varying the temperature (box 4).
1
3
5
4
233
UNIT-2
Choose the correct answer:
1. c) basic
Na 2 B4O7 + 7H2O
2NaOH + 4H3BO3
Strong base Weak acid
234
3. b) B3H6 UNIT-3
nido borane : BnH4+n Choose the correct answer:
aracno borane : BnH6+n
1. a) Nessler’s reagent
B3H6 is not a borane
ability to form pπ − pπ bonds with
2. d)
4. a) Aluminium
itself
5. c) four
3. d) 1s2 2s2 2p6 3s2 3p3
There are two 3c – 2e bonds i.e., the
-
4. b) P4(white) and PH3
bonding in the bridges account for 4
electrons. 5. b) Nitroso ferrous sulphate
6. c) Lead 6. a) H3PO3
7. c) sp2 hybridised 7. a) H3PO3
8. a) +4 8. b) 2
Example : CH4+ in which the oxidation 9. a) 6N
state of carbon is +4 10. d) Both assertion and reason are wrong.
9. d) ( SiO4 )
4−
The converse is true.
R
11. b) F2
10. b) Si O
12. b) HF > HCl > HBr > HI
R
11. a) Me3SiCl 13. d) NeF2
12. d) dry ice 14. c) He
dry ice – solid CO2 in which carbon is 15. c) XeO3
in sp hybridized state
16. c) SO4 2−
13. a) Tetrahedral
17. a) HI
14. d) Feldspar is a three dimensional silicate
18. d) Cl2 > Br2 > F2 > I2
15. b) K 3 [AlF6 ]
19. d) HClO < HClO2 < HClO3 < HClO4
K 3 [AlF6 ]
AlF3 + 3KF →
20. c) Cu(NO3)2 and NO2
16. a) A-b , B-1 , C-4 , D-3
17. d) Al,Cu,Mn,Mg
Al-95% , Cu-4% , Mn-0.5% , Mn-0.5%
18. b) graphite
19. a) Metal borides
20. a) Al < Ga < In < Tl
235
236
237
5 H2O are ligands, remaining 1 H2O molecular and SO4 are in the outer coordination
2−
sphere.
Answer : option (c)
2+
+ +
4. Fe ( H2O )5 NO SO4
2−
+1 and +1 respectively
Answer : option(d)
5. Answer : option(d)
6. Answer : option(d)
7. Answer : option(c)
Ti 4+ ( d 0 ⇒ 0BM )
(
Co2+ d 7spin free ⇒ t 2 g 5 , e g 2 ; n = 3 ; µ = 3.9 BM )
Cu 2+ (d Low spin ⇒ t
9
2g
6
, e g 3 ; n = 1 ; µ = 1.732 BM )
Ni 2+ (d Low spin ⇒ t
8
2g
6
, e g 2 ; n = 2 ; µ = 2.44 BM )
8. Answer : option(b)
The electronic configuration t 2 g 3 , e g 2
3 × ( −0.4 ) + 2 ( 0.6 ) ∆ 0
238
Cl N Cl N N Cl
Co Co Co
N Cl Cl N N Cl
N N N
en en en
inactive Active forms ( enantiomorphs)
Complexes given in other options (a), (c) and (d) have symmetry elements and hence they
are optically inactive.
11. Answer : option(d)
Cl NH3 H3N Cl
Pt Pt
Cl NH3 Cl NH3
12. Three isomers. If we consider any one of the ligands as reference ( say Py), the arrangement
of other three ligands ( NH3, Br- and Cl-) with respect to (Py) gives three geometrical
isomers.
13. Answer : option(c)
(a)coordination isomers
(b) no isomerism ( different molecular formula)
(c) ← NCS , ← SCN coordinating atom differs : linkage isomers
14. Answer : option(a)
For [MA 4 B2 ] complexes-geometrical isomerism is possible
n+
239
Fe ( en )3 ( PO4 )
2+
3−
(
V 2+ t 2 g 3 , e g 0 ; CFSE = 3 × (-0.4)∆ 0 = −1.2∆ 0 )
(
Ti 2+ t 2 g 2 , e g 0 ; CFSE = 2 × (-0.4)∆ 0 = −0.8∆ 0 )
Statements given in option (a) ,(b), and (c) are wrong.
The current statements are
(a) since, the crystal field stabilization is more in octahedral field , octahedral complexes
are more stable than square planar complexes.
(b) Fe2+ −d 6 High Spin −t 2 g 4 , e g 2 )
[FeF ]
4−
−
CFSE = 4 × (-0.4) + ( 0.6 ) × 2 + P
6
UNIT-6
Choose the correct answer:
1. c) both covalent crystals
2. b) AB3
N 8
number of A ions = c = = 1
8 8
Nf 6
number of B ions = = = 3
2 2
simplest formula AB3
3. b) 1:2
if number of close packed atoms =N; then,
The number of Tetrahedral holes formed = 2N
number of Octahedral holes formed = N
therefore N:2N = 1:2
240
occupies corners and face centres and it is in the range of 0.414 - 0.732 ,
also occupying half of the tetrahedral hence the coordination number of each
voids. ion is 6
Nc N f
8 + 2 + 4 C atoms inTd voids 3
12. d) 2 × 400pm
8 6
8 + 2 + 4 = 8
9. d) M3N2
if the total number of M atoms is n,
then the number of tetrahedral voids
241
(
2 a = rCs + rCl + − ) bcc ⇒ 4r = 3a ⇒ r =
3a
4
3 2a a
2 400 = inter ionic distance fcc ⇒ 4r = 2a ⇒ r = =
4 2 2
a 3a a
100 2 : 4 :
13. a) 0.414 2 2
rX
= 0.414
+
π aa =
= 3= = 3
=
= 55..21
21 ×
× 100−−10
1 10
16. b) 3 3
6 22 × 39
ρ × 39
ρ==
43 πr 3 3 π a2 π ((55..21 10 ) 3 × ( 6.023 × 1023 )
3 3
21 × 10−−10
) × ( 6.023 × 10 )
4 23
× 10
a 3 = a 3 = 6 ρ
ρ=
−3
915 Kg
= 915 Kg mm −3
17. a) excitation of electrons in F centers NA
1 3 1 21. b) equal number of anions and anions
18. c) a : a: a are missing from the lattice
2 4 2 2
22. c) Frenkel defect
23. d) Both assertion and reason are false
24. b) FeO
25. a) XY8
242
In this case
k = x min−1 and [A 0 ] = 0.01M = 1 × 10−2 M
t = 1 hour = 60 min
[A ] = 1 × 10−2 ( e−60 x )
2. option (c)
n−1
−11
22n−1 −
for n
for n≠≠ 11 =
tt 112 =
n-1)) kk [[A
(( n-1 A 0 ]]n−1
n−1
2
0
11
for n
for n== 00 tt 1 =
=
22 kk [[A
A 0 ]]
−1
12 −1
2
0
tt 1 =
[[AA0 ]]
12
= 0 22 kk
2
tt 1 α
12
α [[A
A 00 ]] −−−−−
−−−−−((11))
2
given
given
[[AA00 ]] == 00..02
02MM ;; tt 112 =
= 10
10 min
min
2
[[AA ]] == 00..04
0
0
04MM ;; tt 1
12
=
= ??
2
substitute
substitute in
in (1)
(1)
10 min α 0.02M −−−−−
10 min α 0.02M −−−−−((22))
tt 1 α
12
0.04M −−−−−
α 0.04M −−−−−((33))
2
((33))
((22 ))
tt 1
00..04
⇒ = 04 M
M
12
⇒ 10 min
2
= 0.02 M
10 min 0.02 M
tt 1 == 22 ×
× 10 min =
10 min = 20
20 min
min
12
2
243
4. option(d)
For a first order reaction
0.693
t1 =
2 k
t1
2 does not depend on the initial concentration and it remains constant (whatever may
5. option(c)
−1 d [NH3 ] d [N 2 ] 1 d [H2 ]
Rate = = =
2 dt dt 3 dt
1 1
2 k1 [NH3 ] = k 2 [NH3 ] = 3 k 3 [NH3 ]
3
2 k1 = 3k 2 = k 3
1.5 k1 = 3k 2 = k 3
6. option(c)
Given :
At low pressure the reaction follows first order, therefore
Rate α[reactant ]
1
At high pressure due to the complete coverage of surface area, the reaction follows zero
order.
Rate α[reactant ]
0
244
−d [A ]
rate =
dt
mol L−1
unit of rate = = mol L−1s −1
s
in this case
3
rate = k [Acetone ] 2
n= 3
2
( 2 ) ( 3 2 )−1
1− 3
mol L s −1
−( 1 ) (1 )
mol 2 L 2 s −1
8. option(b)
A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers
the activation energy.
9. option(a)
In zero order reactions, increase in the concentration of reactant does not alter the rate.
So statement (i) is wrong.
E
− a
RT
k= Ae
if Ea = 0 so, statement (ii) is correct, and statement (iii) is wrong
k= Ae 0
k= A
E 1
ln k = ln A − a
R T
this equation is in the form of a straight line equatoion
y=c+mx
1
a plot of lnk vs is a straight line with negative slope
T
so statements (iv) and (v) are wrong.
245
(Ea)f = y kJ mol-1
(Ea)b = (x + y) kJ mol-1
Reactant
x kJ mol-1
( x + y ) kJmol −1
( x + y ) × 103 Jmol −1
11. option(c)
T1 = 200K ; k = k1
T2 = 400K ; k = k 2 = 2k1
k 2.303 Ea T2 −T1
log 2 = T T
k1 R 1 2
12. option(b)
2.303 [A ]
k= log 0
t [A ]
2.303 0.25
2.303 × 10-2 hour −1 = log
1806 min A
[ ]
2.303 × 10-2 hour −1 × 1806 min 0.25
= log A
2.303 [ ]
1806 × 10-2 0.25
= log
60 [A ]
0.25
0.301 = log
[A ]
0.25
log 2 = log
[A ]
0.25
2=
[A ]
0.25
[A ] = 2
= 0.125M
246
2.303 100
6.909 = log
t 25
2.303
t= log ( 4 )
6.909
1
t = log 22
3
2
t = log 2
3
14. option(c)
1 [A ]
k = ln 0
t [A ]
[A ]=0.1 ; [A ]=0.05
0
1 0. 1
k = ln
t 1 0.05
2
1
k = ln ( 2 )
t 1
2
ln ( 2 )
t1 =
2 k
247
19. option(d)
rate4 = k [0.2 ] [0.2 ] −−−−−(4)
n m
H2O2 →
H2O + 12 O2
(4)
(2) −d [H2O2 ] d [H2O ] 2d [O2 ]
Rate = = =
dt dt dt
8 x k [0.2 ] [0.2 ]
n m
= 48
2 x k [0.2 ]n [0.1]m no of moles of oxygen = = 1.5 mol
32
8
= 2m ∴ m =2 ∴ rate of formation of oxygen = 2 × 1.5 = 3 mol min-1
2
∴ rate = k [A ] [B ] 20. option(a)
1 2
t
For a first order reaction 12 is independent of
16. option(c)
initial concentration .i.e., ∴ n ≠ 1; for such cases
For a first reaction, If the 1
t1 α
concentration of reactant is n−1 ------- (1)
doubled, then the rate of reaction
2
[A0 ]
also doubled. If [A0] = 2[A0] ; then t 12 = 2t 12
Rate constant is independent of
concentration and is a constant 1
2 t1 α
at a constant temperature, 2
[2 A0 ]n−1 ------ (2)
17. option(a) (2) ⇒
The unit of rate constant is s-1 (1)
and it indicates that the reaction
[A ]
n−1
1
is first order. 2= × 0
[2A ]
n−1
0
1
[A ]
n−1
2= 0
[2A ]
n−1
0
248
21 = ( 2−n+1 )
n= 0
21. Answer : option(a)
A →
B C D
Initial a 0 0 0
Reacted at time t x x x x
After time t (a −x ) x x x
a α P0
(a + 2x ) α P
a P
= 0
(a + 2x ) P
x=
( P −P ) a0
2P0
( P −P0 ) a
(a −x ) = a −
2P 0
3P0 −P
(a −x ) = a
2P0
2.303
k=
[A ]
log 0
t [A ]
2.303 a
k= log
t a −x
a
2.303
k= log
t 3P0 −P
a
2P0
2.303 2P0
k= log
t 3P −P
0
249
n= 2
22−1 −1
t1 =
(2-1) k [A0 ]
2 −1
2
1
t1 =
2 k [A 0 ]
t 1 = 30 min
2
250
( 3
Rate = 8 k [A ] [B ][L ] 2 -----(2)
2
)
Comparing (1) and (2) ; rate is increased by 8 times.
(ii) when [A ] = [2A ]and [B ] = [2B ]
3
Rate = k [2A ] [2B ][L ] 2
2
( 3
Rate = 8 k [A ] [B ][L ] 2 -----(3)
2
)
Comparing (1) and (3) ; rate is increased by 8 times.
A
(iii) when [A ] =
2
2
A 3
Rate = k [B ][L ] 2
2
1
4 ( 2 3
Rate = k [A ] [B ][L ] 2 -----(4) )
Comparing (1) and (4) ; rate is reduced to ¼ times.
A
(iv ) when [A ] = and [L ] = [4L ]
3
2
A 3
Rate = k [B ][4L ] 2
3
8
9 ( 2 3
Rate = k [A ] [B ][L ] 2 -----(5) )
Comparing (1) and (5) ; rate is reduced to 8/9 times.
11. solution
Let us consider the dimerisation of a monomer M
2M → ( M )2
Rate= k [M ]
n
251
7.5 × 10−3
k= = 3 mol −1Ls −1
(0.05)
2
12. Solution
3 1
rate = k [x ] 2
[y] 2
3 1
overall order = + = 2
2 2
i.e., second order reaction.
Since the rate expression does not contain the concentration of z , the reaction is zero order
with respect to z.
15. Solution:
k=
2.303 [A ]
log 0
t [A ]
k=
2.303
log
[0.08 ]
1 min [0.04 ]
k = 2.303 log 2
k = 2.303 × 0.3010
k = 0.6932 min−1
0.6932 −1
k= s
60
k = 1.153 × 10−2 s −1
19. Solution
E
− a
RT
k= Ae
200 ×103 J mol −1
− −1 −1
13 −1 8.314 JK mol × 600 K
k = 1.6 × 10 s e
k = 1.6 × 1013s −1 e −( 40.1)
k = 1.6 × 1013s −1 × 3.8 × 10−18
k = 6.21 × 10−5s −1
252
t=
2.303 [A ]
log 0
0.15 = k [0.2 ] [0.02 ] −−−−−(1)
n m k [A ]
0.30 = k [0.4 ] [0.02 ] −−−−−(2)
n m
2.303 100
t= log 1
1.20 = k [0.4 ] [0.08 ] −−−−−(3) 0.087 min −1
n m
t = 52.93 min
(3)
(2 )
1.2 k [0.4 ] [0.08 ]
n m
25 Solution:
=
0.3 k [0.4 ]n [0.02 ]m i) Order of a reaction = 1; t1/2 = 60 ;
seconds, k = ?
[0.08 ]
m
4=
[0.02 ]
2.303
We know that, k =
t1/2
4 = (4)
m
2.303
∴ m =1 k= = 0.01155 s −1
60
(2) ii) [A 0 ] = 100% t = 180 s , k = 0.01155
(1) seconds-1, [A ] = ?
0.30 k [0.4 ] [0.02 ]
n m
For the first order reaction
=
0.15 k [0.2 ]n [0.02 ]m 2.303 [A ]
k= log 0
[0.4 ]
n
t [A ]
2= 100
[0.2 ]
2.303
0.01155 = log
180 [A ]
2 = (2)
n
log [A ] = 2 −0.9207
k = 37.5 mol L s −1 −1
log [A ] = 1.0973
[A ] = antilog of (1.0973)
23Solution:
[A ] = 12.5%
We know that, t1/2 = 0.693/ k
26 Solution:
t1/2 = 0.693/1.54x 10-3 s-1 = 450 s
i) Let A = 100M, [A0]–[A] = 20M,
For the zero order reaction
24.Solution:
We know that, k = 0.693/ t1/2 [A ] −[A ]
k= 0
k = 0.693/ 8.0 minutes = 0.087 minutes-1 t
253
2.303 V°
t (min) Vt V∞ - Vt log
t V° - Vt
2.303 58.3
6 19.3 58.3-19.3 = 39.0 k= log = 0.0670 min-1
6 39
2.303 58.3
12 32.6 58.3-32.6 = 25.7 k= log = 0.0683 min-1
12 25.7
2.303 58.3
18 41.3 58.3-41.3 = 17.0 k= log = 0.0685 min-1
18 17
2.303 58.3
24 46.5 58.3-46.5 = 11.8 k= log = 0.0666 min-1
24 11.8
254
29.Solution:
k=
2.303 [A ]
log 0
t [A ]
2.303 V
k= log 0
t Vt
2.303 V0
t (min) Vt k=
t
log V
t
2.303 46.1
10 29.8 k= log = 0.0436 min-1
10 29.8
2.303 46.1
20 19.3 k= log = 0.0435 min-1
20 19.3
Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first
order.
30. Solution:
255
Preliminary tests
1 Odour: (i) Fish odour (i) May be an amine
Note the Odour of the organic (ii) Bitter almond odour (ii) May be benzaldehyde
compound.
(iii) Phenolic odour (iii) May be phenol
(iv) Pleasant fruity (iv) May be an ester
odour
2 Test with litmus paper: (i) Blue litmus turns red (i) M ay be a carboxylic acid
Touch the Moist litmus paper or phenol
with an organic compound. (ii) Red litmus turns (ii) May be an amine
blue
(iii)
Absence of carboxylic
(iii) No colour change is acid , phenol and amine
noted
3 Action with sodium (i) Brisk effervescence (i) P
resence of a carboxylic
bicarbonate: acid.
(ii) No brisk
Take 2 ml of saturated sodium effervescence (ii) A
bsence of a carboxylic
bi carbonate solution in a test acid.
tube. Add 2 or 3 drops (or a
pinch of solid) of an organic
compound to it.
4 Action with Borsche’s reagent: yellow or orange or red Presence of an aldehyde or
Take a small amount of an precipitate ketone
organic compound in a test
tube. Add 3 ml of Borsche’s
reagent, 1 ml of Conc HCl to it,
then warm the mixture gently
and cool it.
256
257
258
Report:
The given organic compound contains /is
(i) Aromatic / aliphatic
(ii) Saturated / unsaturated
(iii) __________ functional group
259
REASONING
O2N O2N
R R
H H
C O + H2N N NO2 C N N NO2
H H
R R
H H
C O + H2N N NO2 C N N NO2
R R
260
C X ( H2O ) y
H SO
∆
2
→ x C + yH2O
4
6. Ignition test
Aromatic compounds burn with a strong sooty yellow flame because of the high
carbon–hydrogen ratio. Aliphatic compounds burn with non-sooty flame.
7.Test with bromine water:
In this test, the orange-red colour of bromine solution disappears when it is added
to an unsaturated organic compound.
Br Br
C C + Br Br C C
OH OH
C C + H2O + (O) C C
Aqueous solution Naphthols do not give any characteristic colour with neutral
ferric chloride. But alcoholic solution of α and β naphtholsgiveblue-violet and green
colouration respectively due to the formation of binaphthols.
261
Aldehydes react with Tollen’s reagent to form elemental silver, accumulated onto the
β-naphthol
inner surface of the test tube. Thus silver mirror is produced on the inner walls of the
test tube. Green complex
R-CHO 2 Ag NH 3 2 OH
2Ag R COONH 4 + H 2 O + 3NH 3
Aldehyde Metallic silver
Tollen'sreagent
(Silver mirror)
The Fehling’s solution is obtained by mixing equal volumes of both Fehling’s solution
A and Fehling’s solution B that has a deep blue colour. In Fehling’s solution, copper (II)
ions form a complex with tartrate ions in alkali. Aldehydes reduces the Cu(II) ions in the
Fehling’s solution to red precipitate of cuprous oxide(copper (I) oxide).
Note: Benzaldehyde may not give this test as the reaction is very slow.
262
OH
N N Cl + 00C
N N
O H2N O
NH2
O NH O
NH
H2N NH2
O O O
1800C Cu2+ Cu2+
NH2 H2N O
O
[Cu(Biuret)4]2+ complex
(violet colour)
263
H OH
H HO O
O
H3O+
HO
O
HO H -3H2O
H OH 5-(hydroxymethyl)furfural
H OH
OH O
OH
HO O H3O+ HO
+2 HO [O] O
O
-H2O O H3O+
5-(hydroxymethyl) α-naphthol
furfural
OH
OH
A purple dye
17.Osazone test:
Phenyl hydrazine in acetic acid, when boiled with reducing sugars forms Osazone.
The first two carbon atoms are involved in this reaction. The sugars that differ in their
configuration on these carbon atoms give the same type of Osazone. Thus glucose,
fructose and mannose give the same needle type yellow crystals.
H O H N NH
C C
H C OH C N NH
HO C H H2N HO C H
H C OH + HN H C OH
H C OH phenyl hydrazine H C OH
CH2OH CH2OH
Glucosazone
Glucose (Yellow crystals)
264
Principle:
During these titrations, Fe2+ ions (from ferrous salts) are oxidised to MnO4- ions and
MnO4- ion (from Mn2+) is reduced to Mn2+ ion.
Oxidation : 5 Fe 2+
→ 5 Fe3+ + 5e −
Reduction : → Mn 2+ + 4H 2 O
MnO 4 − + 8H + + 5e −
Pink colourless
Overall reaction :
5Fe 2+ + MnO 4 − + 8H +
→ 5Fe3+ + Mn 2+ + 4H 2 O
Short procedure:
Procedure :
Titration–I
(Link KMnO4)Vs (Standard FAS)
Burette is washed with water, rinsed with KMnO4 solution and filled with same
KMnO4 solution up to the zero mark. Exactly 20 ml of standard FAS solution is pipetted
out into the clean, washed conical flask. To this FAS solution, approximately 20ml of 2N
sulphuric acid is added. This mixture is titrated against KMnO4 Link solution from the
burette. KMnO4 is added drop wise till the appearance of permanent pale pink colour.
Burette reading is noted, and the same procedure is repeated to get concordant values.
265
Calculation :
Volume of KMnO4 (link) solution (V1) = ----------ml
V2 × N 2
N1 =
V1
Normality of KMnO4 (link) solution X
(N1) = _______________ N
Titration–II
(Unknown FeSO 4 ) Vs (Link KMnO4)
Burette is washed with water, rinsed with KMnO 4 solution and filled with same KMnO 4
solution up to the zero mark. Exactly 20 ml of unknown FeSO 4 solution is pipetted out into
the clean, washed conical flask. To this FeSO4 solution approximately 20ml of 2N sulphuric
acid is added. This mixture is titrated against KMnO4 Link solution from the burette. KMnO 4
is added drop wise till the appearance of permanent pale pink colour. Burette reading is noted
and the same procedure is repeated to get concordant values.
266
1 20
2 20
3 20
Calculation :
Volume of Unknown FeSO 4 solution V1 = 20 ml
Normality of Unknown FeSO4 solution N1 = ?N
Volume of KMnO4 (link) solution V2 = ml
Normality KMnO 4 (link) solution N2 = X N
According to normality equation: V1× N1 = V2 × N2
V2 × N 2
N1 =
V1
Y
N1= _______________ N
The normality of unknown FeSO 4 solution = ________________ N
Weight calculation:
Y × 278 × 3
N1 =
4
= g
Report :
The amount of FeSO 4 dissolved in 750 ml of the solution = g
267
Procedure :
Titration–I
(Link KMnO4)Vs (Standard FeSO4 )
Burette is washed with water, rinsed with KMnO4 solution and filled with same FeSO 4
solution up to the zero mark. Exactly 20 ml of standard FeSO4 solution is pipetted out into
the clean, washed conical flask. To this solution, approximately 20ml of 2N sulphuric acid
is added. This mixture is titrated against KMnO 4 Link solution from the burette. KMnO 4 is
added drop wise till the appearance of permanent pale pink colour. Burette reading are noted,
the same procedure is repeated to get concordant values.
268
Calculation :
Volume of KMnO4 (link) solution V1 = ml
Normality KMnO4 (link) solution N1 = ?N
Volume of standard FeSO4 solution V2 = 20 ml
Normality of standard FeSO4 solution N2 = 0.1024 N
V2 × N 2
N1 =
V1
Normality of KMnO4 (link) solution X
(N1) = _______________ N
Titration–II
(Unknown FAS) Vs (Link KMnO 4 )
Burette is washed with water, rinsed with KMnO4 solution and filled with same KMnO 4
solution up to the zero mark. Exactly 20 ml of unknown FAS solution is pipetted out into the
clean, washed conical flask. To this FAS solution approximately 20ml of 2N sulphuric acid
is added. This mixture is titrated against KMnO 4 Link solution from the burette. KMnO4 is
added drop wise till the appearance of permanent pale pink colour. Burette reading is noted
and the same procedure is repeated to get concordant values.
269
Calculation :
Volume of Unknown FAS solution V1 = 20ml
Normality of Unknown FAS solution N1 = ?N
Volume of KMnO4 (link) solution V2 = ml
Normality KMnO4 (link) solution N2 = N
According to normality equation: V1× N1 = V2 × N2
V2 × N 2
N1 =
V1
Y
N1= _______________ N
Weight calculation:
Y × 392 × 1500
=
1000
= g
Report :
The amount of FAS dissolved in 1500 ml of the solution = g
270
Reduction : → Mn 2+ + 4H 2 O
MnO 4 − + 8H + + 5e −
Pink colourless
Since one mole oxalic acid releases 2 moles of electrons, the equivalent weight of oxalic
106
acid = 2 = 63 (oxalic acid is dihydrated)
Short procedure:
s.no Content Titration-I Titration-II
1 Burette solution KMnO 4 KMnO 4
271
Calculation :
Volume of KMnO 4 (link) solution V1 = ml
Normality KMnO 4 (link) solution N 1 = ?N
Volume of standard FAS solution V2 = 20 ml
Normality of standard FAS solution N2 = 0.1 N
According to normality equation:
V1 x N1 = V2 x N 2
V2 x N 2
N1 = =
V1
Titration–II
(Unknown oxalic acid ) Vs (Link KMnO 4 )
Burette is washed with water, rinsed with KMnO 4 solution and filled with same KMnO4
solution up to the zero mark. Exactly 20 ml of unknown oxalic acid solution is pipetted out
into the clean, washed conical flask. To this oxalic acid solution approximately 20ml of 2N
sulphuric acid is added. This mixture is heated to 60 – 700C using Bunsen burner and that
hot solution is titrated against KMnO 4 Link solution from the burette. KMnO 4 is added drop
wise till the appearance of permanent pale pink colour. Burette reading are noted, the same
procedure is repeated to get concordant values.
272
V1 x N1 = V2 x N 2
V2 x N 2
N1 =
V1
Y
Normality of Unknown oxalic acid solution N1 = _______________ N
Weight calculation:
The amount of oxalic acid dissolved in 1 lit
=(Normality) x (equivalent weight)
of the solution
The amount of oxalic acid dissolved in 500 Y × 63 × 500
ml of the solution = 1000
x 63 x 500
=
1000
= g
Report :
The amount of oxalic acid dissolved in 500 ml of given the solution = g
273
Na 2CO3 + 2HCl
→ 2NaCl + CO2 + H2O
Neutralization of Sodium hydroxide by HCl is given below. To indicate the end point,
phenolphthalein is used as an indicator.
NaOH + HCl
→ NaCl + H 2 O
Short procedure:
s.no Content Titration-I Titration-II
1 Burette solution HCl ( link solution) HCl ( link solution)
20 ml of standard Na2CO3 20 ml of unknown NaOH
2 Pipette solution
solution solution
4 Temperature Lab temperature Lab temperature
5 Indicator Methyl orange Phenolphthalein
Colour change from straw
6 End point Disappearance of pink colour
yellow to pale pink
7 Equivalent weight of NaOH = 40
Procedure :
Titration–I
(Link HCl )Vs (standard Na2CO3)
Burette is washed with water, rinsed with HCl solution and filled with same HCl solution up to
the zero mark. Exactly 20 ml of standard Na2CO3solution is pipetted out into the clean, washed
conical flask. To This solution 2 to 3 drops of methyl orange indicator is added and titrated
against HCl link solution from the burette. HCl is added drop wise till the colour change from
straw yellow to pale pink. Burette reading is noted and the same procedure is repeated to get
concordant values.
274
Calculation :
Volume of HCl (link) solution V1 = ml
Normality HCl (link) solution N1 = ? N
Volume of standard Na2CO3 solution V2 = 20 ml
Normality of standard Na2CO3 solution N2 = 0.0948 N
V2 × N 2
N1 =
V1
X
Normality of HCl (link) solution (N1) = _______________ N
Titration–II
(Unknown NaOH ) Vs (Link HCl)
Burette is washed with water, rinsed with HCl solution and filled with same HCl solution up
to the zero mark. Exactly 20 ml of unknown NaOH solution is pipetted out into the clean,
washed conical flask. To This solution 2 to 3 drops of phenolphthalein indicator is added and
titrated against HCl link solution from the burette. HCl is added drop wise till the pink colour
disappears completely. Burette reading is noted and the same procedure is repeated to get
concordant values.
275
Calculation :
Volume of Unknown NaOH solution V1 = 20 ml
Normality of Unknown NaOH solution N1 = ? N
Volume of HCl (link) solution V2 = ml
Normality HCl (link) solution N 2 = N
According to normality equation:
V1 x N1 = V2 x N 2
V2 x N 2
N1 =
V1
Y
Normality of Unknown HCl solution N1 = _______________ N
Weight calculation:
= Y × 40 × 250
1000
x 40 x 250
= g
1000
Report :
The amount of NaOH dissolved in 750 ml of the solution = g
276
2NaOH + ( COOH )2
→ ( COONa )2 + 2H 2 O
Oxalic acid Sodium oxalate
Short procedure:
s.no Content Titration-I Titration-II
Oxalic acid ( unknown
1 Burette solution HCl (standard solution)
solution)
20 ml of NaOH link
2 Pipette solution 20 ml of NaOH link solution
solution
4 Temperature Lab temperature Lab temperature
5 Indicator Phenolphthalein Phenolphthalein
Disappearance of pink
6 End point Disappearance of pink colour
colour
7 Equivalent weight of oxalic acid = 63
Procedure :
Titration–I
(standard HCl )Vs (link NaOH)
Burette is washed with water, rinsed with HCl solution and filled with same HCl solution up
to the zero mark. Exactly 20 ml of NaOH is pipetted out into the clean, washed conical flask.
To This solution 2 to 3 drops of phenolphthalein indicator is added and titrated against HCl
solution from the burette. HCl is added drop wise till the pink colour disappears completely.
Burette reading is noted and the same procedure is repeated to get concordant values.
277
Calculation :
Volume of NaOH(link) solution V1 = 20 ml
Normality NaOH(link) solution N1 = ? N
Volume of standard HCl solution V2 = ml
Normality of standard HCl solution N 2 = 0.1010 N
According to normality equation:
V1 x N1 = V2 x N 2
× 0.1010
N1 = =
20
X
Normality NaOH (link) solution N1 =_______________ N
Titration–II
(Unknown oxalic acid ) Vs (Link NaOH)
Burette is washed with water, rinsed with oxalic acid solution and filled with same
oxalic acid solution up to the zero mark. Exactly 20 ml of NaOH solution is pipetted out into
the clean, washed conical flask. To This solution 2 to 3 drops of phenolphthalein indicator is
added and titrated against oxalic acid solution from the burette. oxalic acid is added drop wise
till the pink colour disappears completely. Burette reading is noted and the same procedure is
repeated to get concordant values.
278
Calculation :
Volume of Unknown oxalic acid solution V1 = ml
Normality of Unknown oxalic acid solution N1 = ?N
Volume of NaOH solution V2 = 20 ml
Normality NaOH solution N 2 = N
According to normality equation:
V1 x N1 = V2 x N 2
V2 x N 2
N1 =
V1
Y
Normality of Unknown oxalic acid solution N1 = _______________N
Weight calculation:
The amount of oxalic acid dissolved in 1 lit of
= (Normality) x (equivalent weight)
the solution
The amount of oxalic acid dissolved in 1250
= Normality x equivalentweight x 1250
ml of the solution 1000
Y × 63 × 1250
=
1000
x 63 x 1250
=
1000
= g
Report :
The amount of oxalic acid dissolved in 1250 ml of the solution = g
279
280
281
282
283
284
285
Dr.N.Kanagachalam
Art and Design Team Assistant Professor
Chikkanma Govt Arts Collage,
Tiruppur.
Layout QR Code Management Team
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Pums, Ganesapuram- Polur Green Garden Girls Matriculation Higher
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Thiruvarur.
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Pakkirisamy
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This book has been printed on 80 G.S.M.
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286