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Density of Unit Cell
d 
Z  M
d = Density (g/cm3)
a3  N A
Z = number of atoms per unit cell
a = Edge length in cm
Value of z for
M = Molar mass in g/mol
simple cubic lattice = 1
NA = Avogadro constant in per mol
bcc lattice = 2
ccp or fcc lattice = 4
Radius Ratio : It is the ratio of radius of cation to the radius of anion.
radius ratio 
r
r
S.No.
Radius Ratio
C.N.
Type of void or hole
Example
1.
0.155 – 0.225
3
Trigonal planar
B2O3
2
3
0.225 – 0.414
4
Tetrahedral
ZnS
0.414 – 0.732
6
Octahedral
NaCl
3
0.732 – 1.000
8
Cubic
CsCl
Relationship between atomic radius and edge length.
Simple Cubic
r 
a
2
BCC
r 
FCC
3a
4
r 
2a
4
In ccp or hcp packing two types of voids namely (i) tetrahedral (ii) octahedral are generated.
No. of octahendral voids present in a lattice = Number of close packed particles
No. of tetrahedral voids present in a lattice = 2 × number of close packed particles
In ionic solids, the larger ions (usually anions) form close packed structure and the smaller
ions (usually cations) occupy voids. If the cation is small enough then tetrahedral voids are occupied,
if bigger, then octahedral voids. Not all octrahedral or teterahedral voids are occupied. The fraction
of octahedral or tetrahedral voids that are occupied, depends upon the chemical formula of the
compound.
1
Chemistry
Imperfections in Crystals
Schottky Defects
Cations and equal number of anions are missing from the lattice site of a crystal of the type
B– e.g., there are 106 schottky pairs per cm3 at room temperature of NaCl. Schottkey defect
lowers the density of crystal. Ionic substances showing schottky defects have almost similar sizes
of cations and anions, e.g.; NaCl, KCl, KF, etc.
A+
Frenkel Defects
Cation is missing from the lattice site but trapped within interstitial position, e.g., Ag Br,
AgCl, Agl, AnS, etc. where cations and anions have much difference in their sizes, show this type
of defect.
F-Centre
Anions are missing from lattice sites and these anionic sites are occupied by unpaired electrons.
Anionic sites occupied by unpaired electrons are called F-centres and are responsible for colour
imparted to crystals.
Metal Deficiency Defect
Some metal oxides contain less amount of metal as compared to the stoichiometric proportion.
In iron oxide of composition Fe0.95.O, some Fe2+ are replaced by definite number of Fe3+ resulting
in the metal deficiency.
Packing Efficiency
It is the percentage of total space filled by particles.
Packing efficiency
hcp and ccp
74%
BCC
68%
Simple cubic lattice
52.4%
2
Chemistry
1.
2.
3.
4.
5.
6.
7.
8.
9.
In a solid lattice, the cation has left a lattic site and is located in an interstitial position.
The lattice defect is known as
(a)
Interstitial defect
(b)
Vacancy defect
(c)
Frenkel defect
(d)
Schottky defect
An octahedral void is surrounded by how many spheres
(a)
6
(b)
4
(c)
8
(d)
12
In NaCl crystal, number of Cl– ions around each Na+ ion will be
(a)
3
(b)
4
(c)
6
(d)
8
For an ionic crystal of general formula AB and coordination number 6, the radius ratio will
be
(a)
greater than 0.732
(b)
between 0.414 to 0.732
(c)
between 0.225 to 0.414
(d)
between 0.155 to 0.225
If pentavalent impurity is mixed in a crystal lattice of germanium, the semiconductor
will be
(a)
p-type
(b)
n-type
(c)
pnp
(d)
npn
A metallic crystal containing a sequence of layers AB AB AB....Any packing of spheres
leaves out voids in the lattice. What percent of volume of thus lattice is empty space.
(a)
74%
(b)
26%
(c)
52%
(d)
68%
When electrons are trapped into anion vacancies, the defect is known as
(a)
Schottky defect
(b) Frenkel defect
(c)
Non-stoichiometric
(d) Metal excess defect and F-centres are formed
The empty space in CCP unit cell is
(a)
74%
(b)
32%
(c)
47.6%
(d)
26%
The packing efficiency of HCP unit is
(a)
26%
(b)
74%
(c)
32%
(d)
47.6%
3
Chemistry
10.
11.
12.
13.
The coordination number of metal crystallizing in hexagonal close packed structure is
(a)
4
(b)
6
(c)
12
(d)
8
A substance consisting of two elements, ‘P’ and ‘Q’, has atoms of ‘P’ occupying each corner
of the cube and atoms of ‘Q’ occupying the body centre. The composition of the substance
is
(a)
PQ3
(b)
P4Q3
(c)
PQ
(d)
composition cannot be specified
A compound is formed by two elements ‘X’ and Y. Atoms of element Y (as anions) make ccp
arrangement and those of element X (as cations) occupy 50% of tetrahedral and octahedral
voids. The formula of compound is
(a)
X3Y2
(b)
X3Y3
(c)
X2Y3
(d)
XY2
On the basis of their magnetic properties, substances can be classified into various categories
like paramagnetic, diamagnetic, ferromagnetic, antiferromagnetic and ferrimagnetic.
Schematic alignment of magnetic moments in above-mentioned solids are :
(a)

(c)

(b)

These representation respectively are of :
14.
15.
16.
(a)
Ferromagnetic, ferrimagnetic, and antiferromagnetic
(b)
Ferrimagnetic; ferromagnetic, and antiferromagnetic
(c)
Ferromagnetic; antiferromagnetic, and ferrimagnetic
(d)
Ferrimagnetic, antiferromagnetic, and ferromagnetic
Which one of the following substances produce impurity defects when added to molten
NaCl?
(a)
KCl
(b)
AgCl
(c)
SrCl2
(d)
AgBr
Germanium crystal doped with equal number of phosphorus and antimony atoms is
(a)
an intrinsic semiconductor
(b)
p-type semiconductor
(c)
an n-type semiconductor
(d)
a superconductor
If ‘a’ stands for the edge length of cubic systems sc, bcc and fcc, then ratio of radii of the
spheres in these systems will be respectively,
(a)
1a :
(c)
1
a :
2
3a :
3a :
2a
1
2
a
4
(b)
1
3
1
a :
a :
a
2
4
2 2
(d)
1
3
2
a :
a :
a
2
2
2
Chemistry
17.
18.
19.
20.
21.
22.
23.
24.
Which of the following statements is not correct?
(a)
The number of Bravais lattices in which a crystal can be categorised is 14.
(b)
The fraction of total volume occupied by the atoms in a primitive cell is 0.524.
(c)
Molecular solids are generally volatile.
(d)
Number of carbon atoms in an unit cell of diamond is ‘4’
Match box is an example of
(a)
Tetragonal unit cell
(b)
Orthorhombic unit cell
(c)
Monoclinic unit cell
(d)
Triclinic unit cell
CsBr crystallises in a body centered cubic lattice. The edge length of unit cell is 436.6 pm.
Given that the atomic mass of Cs = 133 u and Br = 80 u, the Avogadro number being
6.02 × 1023 mol–1, the density of CsBr is
(a)
8.25 g/cm3
(b)
4.25 g/cm3
(c)
42.5 g/cm3
(d)
0.425 g/cm3
The Ca2+ ions and F– ions are located in CaF2 crystal respectively at face centered cubic
lattice points and in
(a)
tetrahedral voids
(b)
half of tetrahedral voids
(c)
octahedral voids
(d)
half of octahedral voids
Which is not the correct statement for ionic solids in which positive and negative ions are
held by strong electrostatic attractive forces?
(a)
The radius ratio r+/r– increases as coordination number increases.
(b)
As difference in the size of ions increases coordination number increases.
(c)
When coordination number is eight, the r+/r– ratio lies between 0.225 to 0.414
(d)
In ionic solids of type AX (ZnS, Wurzite) the coordination number of Zn2+ and S2–
respectively are 4 and 4.
The radii of Na+ and Cl– ions are 95 pm and 181 pm respectively. The edge length of NaCl
unit cell is
(a)
276 pm
(b)
138 pm
(c)
552 pm
(d)
415 pm
Potassium has a bcc structure with nearest neighbour distance of 4.52 Å. If atomic mass of
potassium is 39 and then its density is
(a)
454 kg m–3
(b)
804 kg m–3
(c)
852 kg m–3
(d)
900 kg m–3
Which of the following fcc structures contains cations in alternate tetrahedral voids?
(a)
ZnS
(b)
NaCl
(c)
Na2O
(d)
CaF2
5
Chemistry
25.
26.
Metallic lusture is explained by
(a)
Diffusion of metal ions
(b)
Oscillation of positive ions
(c)
Excitation of free electrons
(d)
Existence of bcc crystal lattice
Copper crystallises in fcc lattice with a unit length of 361 pm. What is the radius of copper
atom in pm?
(a)
157
(b)
181
(c)
108
(d)
128
1.
(c)
2. (a)
3. (c)
4. (c)
5.
(b)
6. (a)
7. (d)
8. (d)
9.
(b)
10. (c)
11. (a)
12. (a)
13.
(c)
14. (c)
15. (c)
16. (b)
17.
(d)
18. (b)
19. (b)
20. (b)
21.
(c)
22. (c)
23. (d)
24. (b)
25.
(c)
26. (d)
6
Chemistry
Molarity (M) is the number of moles of solute dissolved per litre of solution. Its unit is mol/L.
M 
WB
Number of moles of solute
1000


Volume of solution in litre
MB
Volume of solution  in ml 
For liquids, where concentration is expressed in percentage and density of solution.
M 
Percentage  density  10
Molar mass of solute
It is also defined as number of millimoles of solute dissolved in 1 mL of solution.
Molality (m) is the number of moles of solute dissolved per kg of solvent. Its unit is mol/kg
m 
WB
Number of moles of solute
1000


Mass of solvent in kilograms
MB
WA  in grams
Normality (N) is the number of gram equivalents of solute dissolved per litre of solution. Its
unit is equivalents per litre.
N 
WB
Number of g -equivalents
1000


Volume of solution in litres
Equ. mass of solute
Volume of solution  in mL 
Mole Fraction
It is the ratio of number of moles of one component to the total no. of moles of solution.
 
Number of moles of a component
Number of moles of solution
For binary solutions A + B = 1
ppm (Parts Per Million)
Mass of solute dissolved per million parts of the system.
ppm 
Number of parts by mass or volume of a component
 106
Total parts by mass or volume of the solution
Henry’s Law
The partial pressure of the gas in vapour phase (p) is directly proportional to the mole fraction
of the gas () in solution and is expressed as :
7
Chemistry
p   or p = KH ; where KH is Henry’s constant.
Raoult’s Law
For solution of volatile liquids, the partial vapour pressure of each component in solution is
directly proportional to its mole fraction.
PA  A
P
PA  PA0  A
Total
= PA +
PB
Vapour pressure
P°A
PB  PB0  B
PTotal  PA0  A  PB0  B
PTotal  PA0 1   B   PB0  B


PTotal  PA0  PB0  PA0 X B
PA
PB
A = 1
Raoult’s Law for the solution of Non-Volatile Solute B = 0
Mole Fraction
A = 0
B = 1
Relative lowering of vapour pressure of a solution is equal to the mole fraction of solute when
solvent alone is volatile and the solute in non-volatile and non-electrolyte [when solution is dilute,
i.e., nA + nB  nA]
PA0  Psoln
PA0
 B
B
W · MB
 B
 A  B
M B · WA
IDEAL SOLUTIONS
The solution, where unlike interactions (A – B) are identical to like interactions (A – A) and
(B – B) type, are known as ideal solutions.
Characteristics : (1) Vapour pressure of such solutions is the same as given by Raoult’s law.
(2) Hmix = 0
(3) Vmix = 0
Chorobenzene and bromobenzene, benzene and toluene, n-hexane and n-heptane form nearly
ideal solutions.
NON-IDEAL SOLUTIONS
Non-ideal solutions have Hmix  0, Vmix  0 and do not obey Raoult’s law.
Non-Ideal Solution Showing + ve and – ve Deviations from Raoult’s Law
Showing Positive Deviation
Showing Negative Deviation
Acetone + Carbon disulphide
Acetone + Ethyl alcohol
Acetone + Benzene
Methyl alcohol + Water
Ethyl alcohol + eater
Chloroform + Benzene
Chloroform + Acetone
Chloroform + Diethyl ether
Acetone + Aniline
8
Chemistry
Carbon tetrachloride + Chloroform
HCl + Water
Carbon tetrachloride + Benzene
HNO3 + Water
Carbon tetrachloride + Toluene
Acetic and + pyridine, phenol and aniline
Characteristics of Solution Showing –ve Deviation from Faoult’s Law
(a)
Hmix < 0
(b)
Vmix < 0
(c)
Vapour pressure of solution is less than what is
given by Raoult’s law.
Vapour pressure
P°A
Psoln.  PA0  A  pB0  B
PA  PA0  A
P°B
A = 1
B = 0
PB  PB0  A
A = 0
B = 1
Characteristics of Solution Showing +ve Deviation
(a)
Hmix > 0
(b)
Vmix > 0
(c)
Vapour pressure of solution is more than what is
given by Raoult’s law.
P°A
P°B
Psoln  PA0  A  pB0  B
PA  PA0  A
PB  PB0  B
Note : Curved lines show non-ideal behaviour
and dotted line, ideal behaviour.
A = 1
B = 0
A = 0
B = 1
AZEOTROPES
(i)
Azeotropic mixtures with minimum point i.e. whose boiling points is less than either
of the two pure components. This is formed by that composition of non-ideal solution showing
positive deviation for which the vapour pressure is maximum. (95% ethanol + 5% H2O).
(ii)
Azeotropic mixtures with maximum boiling point i.e. whose boiling is more than
either of the two pure components. This is formed by that composition of a non-deal solution
showing negative for which the vapour pressure is minimum. (68% HNO3 + 32% H2O), (20%
HCl + 80% H2O).
COLLIGATIVE PROPERTIES
The properties of a solution which depends on number of moles of solute but are independent
of its nature, are known as colligative properties. These are :
9
Chemistry
1.
Relative lowering of vapour pressure.
2.
Elevation of Boiling point.
3.
Depression of Freezing point.
4.
Osmotic pressure.
Osmotic Pressure
Pressure applied on solution to stop osmosis is known as osmotic pressure
 
nBT
n RT
or   B
   CRT
V
V
Osmotic pressure of a solution is directly proportional to the number of moles of solute dissolved
per litre of solution at a given temperature.
Solutions having equal molar concentration and equal osmotic pressure at a given temperature
are called isotonic solutions, e.g., A 0.90% (mass/volume) solution of NaCl is isotonic with human
RBC. A solution of NaCl with concentration less than 0.90% (mass/volume) is hypotonic and RBC
will swells up and even burst in solution. A NaCl solution with concentration > 0.90% (mass/
volume) is called hypertonic, RBC will shrink in such solution.
Elevation of Boiling Point
Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to that
of 1 atm or 1.013 bar. The elevation of b.p. is related to the molatily (m) of the solution as below:
Tb  K bm  K b
WB  1000
M B  WA
Tb  Tb  Tb0  Elevation of boiling point
1 atm = 1.013 bar
Tb
T°b
Tb
Temperature
Kb = Molal boiling point elevation constant or Ebullioscopic constant.
Depression of Freezing Point
Freezing point of liquid is the temperature at which liquid and solid phases coexist and have
the same vapour pressure. Freezing point of the liquid solvent in depressed when a non-volatile
solute is added to it.
Tf  K f m  K f
WB  1000
M B  WA
Tf  Tfo  Tf  Depression of freezing point
10
Chemistry
Kf = Molal freezing point depression constant or cryoscopic constant.
Tf
0
Tf Tf
Temperature
Abnormal Molecular Mass : When the molecular mass of a substance as determined by
using colligative properties, does not come out to be the same as expected theoretically, it is said
to show abnormal molecular mass.
Abnormal molecular mass is
dissociation (e.g. NaCl in water)
Dissociation results in the increases
colligative property and decrease in
obtained when the substance in the solution undergoes
or association (e.g. organic acids or phenols in benzene).
in the number of particles and hence increase in the value of
the molecular mass. Association results in the reverse.
Van’t Hoff factor (i) is given by
i 
No. of particles after dissociation or association of solute
Number of solute particles initially dissolved
i 
Experimental value of the colligative property
Calculated value of the colligative property
As molecular mass 
hence i 
1
Colligative property
M normal
Calculated or normal molecular mass

Observed molecular mass
M observed
Modified formulas for substance undergoing dissociation or association in the solution are
(i)
(iii)
Tb = i Kb m
  i
n
RT
V
(ii)
(iv)
Tf = i Kf m
PA0  psoln.
PA0
 iB
When solute dissociates to give ‘n’ ions, the degree of dissociation () is related to Van’t Hoff
factor as given below
 
i 1
n 1
If ‘n’ molecules of a solute undergo association in solution then degree of association () is
given by
 n 
  1  i  
 n  1 
11
Chemistry
1.
2.
An azeotropic mixture of two liquids has boiling point higher than either of them when it.
(a)
Shows positive deviation from Raoult’s law.
(b)
Show no deviation from Raoult’s law.
(c)
Show negative deviation from Raoult’s law.
(d)
is saturated.
Consider 0.02 M aqueous solutions of
1. NaCl,
2. BaCl2
3. Urea
The relative lowering of vapour pressures in these solutions will be such that
3.
(a)
2 < 1 < 3
(b)
2 > 3 > 1
(c)
1 > 2 > 3
(d)
3 < 1 < 2
1
C H OH 2 If ‘x’ is the degree of association
2 6 5
of phenol, then the total number of moles of particles present at equilibrium is
Phenol associates in benzene as C6 H 5OH 
(a)
(c)
4.
5.
6.
7.
1 – x
x
1
2
(b)
(d)
1 + x
x
1
2
On mixing two pure liquids to form an ideal solution, the statement which is not correct
(a)
no change in volume
(b)
(c)
no evolution or absorption of heat (d)
no change in Gibbs energy
entropy changes.
The dissolution of NH4Cl in water is endothermic even though dissolves in water spontaneously. Which one of the following statement best explains this behaviour
(a)
The bonds in solid are weak.
(b)
The entropy driving force causes dissolution.
(c)
Endothermic processes are energetically favourable.
(d)
The dissolving process is unrelated to energy.
Which of following 0.1 M aqueous solutions will have the lowest freezing point
(a)
K2SO4
(b)
NaCl
(c)
NH2CONH2
(d)
C6H12O6
0.2 molal acid, HX is 20% ionized in solution (Kb = 0.52 K/m) The boiling point of solution
is (standard boiling of water is 373K)
(a)
373.12 K
(b)
373.10 K
(c)
373.24 K
(d)
373.30 K
12
Chemistry
8.
9.
10.
11.
12.
13.
14.
At 80°C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is
1000 mmHg. If the solution of ‘A’ and ‘B’ boils at 80°C and 1 atm pressure, the amount of
liquid ‘A’ in the mixture is (1 atm = 760 mm Hg)
(a)
48 percent mol
(b)
50 percent mol
(c)
52 percent mol
(d)
34 percent mol
A solution ‘X’ is prepared by mixing ethanol and water. The mole fraction of ethanol in the
mixture is 0.9, then more water is added to the solution ‘X’ such that the mole fraction of
water in solution becomes 0.9, the boiling point of this solution is
(a)
380.4 K
(b)
376.2 K
(c)
375.5 K
(d)
354.7 K
During depression of freezing point in a solution, the following are in equilibrium
(a)
Liquid solution, solid solvent
(b)
Liquid solvent, solid solute
(c)
Liquid solute, solid solute
(d)
Liquid solute, solid solvent
2g of C6H5 COOH dissolved in 25g benzene shows a depression in freezing point equal to
1.62K. (Kf = 4.9 K kg mol–1) The value of van’t Hoff factor is :
(a)
0.5
(b)
1
(c)
2
(d)
1.5
Pick out the incorrect statement :
(a)
The vapour pressure of solution containing non-volatile and non-dissociative solute
is less than that of pure solvent.
(b)
Reverse osmosis occurs if a pressure smaller than the osmotic pressure of solution
is applied on solution side.
(c)
Only solvent molecules solidify at the freezing point.
(d)
Molar mass of NaCl determined by any of the colligative properties is found to be
less than that of normal (expected) value.
Which one of the following colligative properties can provide molar mass of proteins with
greatest precision?
(a)
Relative lowering of vapour pressure.
(b)
Osmotic pressure
(c)
Elevation of boiling point.
(d)
Depression of freezing point.
Which of the following is not true for an ideal solution?
(a)
Raoult’s law is obeyed in the entire concentration range of a binary solution.
(b)
 Hmix = 0
(c)
 Vmix = 0
(d)
 Smix = 0
13
Chemistry
15.
16.
17.
18.
19.
20.
21.
22.
If various terms in the below given expressions have usual meanings, the Van’t Hoff factor
(i) cannot be calculated by which one of the following expressions :
(a)
V 
i nRT
(b)
(c)
Tb = i kb . m
(c)
Tf = i kf . m
0
Psolvent
 Psolution
0
Psolvent
 n 
 i
 N  n 
A 5% solution of cane sugar (Mol. wt. = 342) is isotonic with 1% solution of ‘X’ under similar
condition. The mol. wt. of ‘X’ is :
(a)
136.2 u
(b)
68.4 u
(c)
34.2 u
(d)
171.2 u
Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of solution containing
1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further
added to this solution, vapour pressure of solution increases by 10 mm Hg. Vapour pressure
(in mm Hg) of X and Y in their pure states will be respectively :
(a)
500 and 600
(b)
200 and 300
(c)
300 and 400
(d)
400 and 600
The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm.
The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10
moles of water at 298 K and 5 atm. pressure is
(a)
4 × 10–4
(b)
4 × 10–5
(c)
5.0 × 10–4
(d)
4.0 × 10–6
What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with
30.0 mL of 0.10 M Ba(OH)2
(a)
0.40 M
(b)
0.0050 M
(c)
0.12 M
(d)
0.10 M
A 0.04 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at the temperature
298 K. The apparent degree of dissociation of Na2SO4 is
(a)
25%
(b)
50%
(c)
75%
(d)
85%
To neutralize completely 20 ml of 0.1 M aq solution of phosphorous acid, H3PO3, the volume
of 0.1 M aqueous KOH solution required is
(a)
10 ml
(b)
20 ml
(c)
40 ml
(d)
60 ml
If two substances A and B have PA0 : PB0  1 : 2 and have mole fraction in solution 1 : 2,
then mole fraction of A in vapours is
(a)
0.33
(b)
0.25
(c)
0.52
(d)
0.2
14
Chemistry
23.
24.
25.
26.
During osmosis, flow of water through a semipermeable membrane is
(a)
from both sides of semipermeable membrane with unequal flow rates,
(b)
from solution having lower concentration only,
(c)
from solution having higher concentration only,
(d)
from both sides of semipermeable membrane with equal flow rate.
A solution of acetone in ethanol
(a)
behaves like a near ideal solution,
(b)
obeys Raoult’s law,
(c)
shows a negative deviation from Raoult’s law,
(d)
shows a positive deviation from Raoult’s law.
A 0.002 m aqueous solution of an ionic compound Co(NH3)5 (NO2) Cl freezes at – 0.00732°C.
Number of moles of ions which 1 mol of ionic compound produces on being dissolved in
water will be (Kf = – 1.86°K/m)
(a)
3
(b)
4
(c)
1
(d)
2
A binary solution is prepared by mixing n-heptane and ethanol. Which one of the following
statements are correct regarding the behaviour of solution?
(a)
The solution formed is an ideal solution.
(b)
The solution is non-ideal showing +ve deviation from Raoult’s law.
(c)
The solution is non-ideal showing –ve deviation from Raoult’s law.
(d)
None of these.
1.
(c)
2. (c)
3. (d)
4 (b)
5.
(b)
6. (a)
7. (a)
8. (b)
9.
(b)
10. (a)
11. (a)
12. (b)
13.
(b)
14. (d)
15. (a)
16. (b)
17.
(d)
18. (a)
19. (b)
20. (c)
21.
(c)
22. (d)
23. (a)
24. (d)
25.
(d)
26. (b)
15
Chemistry
Electrochemical Cells
Galvanic cell is an electrochemical cell that converts the chemical energy of the spontaneous
redox reaction into electrical energy. The half-cell in which oxidation takes place is called anode
and it has negative potential with respect to the solution. The other half-cell in which reduction
takes place is called cathode and it has positive potential w.r.t. the solution. When the two halfcells are joined, the electrons start flowing from anode having negative polarity to cathode with
positive polarity.
The potential difference between the two electrodes is called cell potential. To maintain the
flow of current in the external circuit, the two half-cells are joined by a salt bridge. The salt bridge
maintains the electroneutrality in both the half-cells.
Ecell = ECathode – EAnode
If the concentrations of the oxidised and reduced forms of the species are unity at 298 K, then
cell potential is called standard cell potential (E°cell)



Ecell
 Ecathode
 E anode
The potential of individual half-cell is determined with the help of a reference electrode, e.g.,
standard hydrogen electrode (SHE) represented by Pt(s)/H2(g, 1bar)/H+ (aq, 1M), which is assigned
a zero potential at all temperatures corresponding to the reaction : 2H+ (aq) + 2e–  H2(g)
Significance of Sign with Standard Electrode Potential

EZn
2

 0.76V and Ecu
2
Zn

–ve sign with EZn
2
Zn
cu
 0.34V
means that reduction of Zn2+ to Zn is non-spontaneous but oxidation
of Zn to Zn2+ occur spontaneously.

+ve sign with Ecu
2
cu
means that reduction of Cu2+ is spontaneous but oxidation of Cu to Cu2+
is non-spontaneous.
Nernst Equation for an electrode Reaction : Mn+ + ne–  M(s)
Reaction quotient
Qc 

1
M n 
The Nernst equation for this electrode reaction is :
EM n 
M

 EM
n
M

2.303RT
log Qc
nF
16
Chemistry

 EM
n
M

0.059
V log Qc
n
Similarly Nernst equation for a cell reaction :
ne 
aA  bB  cC  dD
Qc 
C c  D d
 A a  B b
c

Ecell  Ecell

d
C   D 
0.059
V log
n
 A a  B b
At equilibrium, Ecell = 0 and Qc = Kc
0
0  Ecell

0.059
0.059
0
V log K c  Ecell

log K c
n
n
Concentration cells have the same electrodes but the concentrations of the solutions of the
same electrolyte is different in the two half-cells. For example,
Cu(s)/Cu2+(C1)||Cu2+(C2)/Cu(s)
2
2

 CuLHC
or Cu2   C2  
 Cu2   C1 
Cell Reactions : CuRHC
Ecell 
Cu2  LHC
0.059
V log
2
Cu2  RHC
Here RHC means right half-cell and LHC, left half-cell.
Ecell  
C
0.059
V log 1
2
C2
Cell potential and Gibbs energy of reaction (rG) are related as given below :
rG = – n Ecell F
Here rG is an extensive thermodynamic property and Ecell is an intensive parameter. ‘n’ is
the number of electrons involved in redox reaction. If the concentration of all the reacting species

are taken as unity, then we have  rG   nEcell
F.
1. Primary Batteries
(a) Dry Cell
Anode :
Zn(s)  Zn2+ + 2e –
Cathode :
MnO2 + NH4+ + e–  MnO (OH) + NH3
Mn is reduced from +4 oxidation state to +3 state. Ammonia produced forms a stable complex,
[Zn (NH3)4]2+. This cell is not rechargeable because the products formed during discharging cannot
be converted into the reactants. The cell potential of nearly + 1.5 V decreases with time since the
conc. of ions involved in the cell reaction changes during discharging.
(b) Mercury Cell
Anode :
Zn(Hg) + 2OH–  ZnO(s) + H2O + 2e –
17
Chemistry
Cathode :
Overall reaction :
HgO(s) + H2O + 2e–  Hg(l) + 2OH–
Zn(Hg) + HgO(s)  ZnO(s) + Hg(l)
Cell potential of nearly + 1.35 V remains constant during its life as the overall reaction does
not involve any ion whose concentration can change during its life.
2. Lead Storage Battery
Lead Storage Battery is a secondary battery and is rechargeable. It consists of a lead anode
and a grid of lead packed with PbO2 as cathode. A 38% solution of H2SO4 is used as an electrolyte.
During discharge, following reaction occurs at electrodes.
Anode : Pb  s   SO42   aq   PbSO4  s   2e –
Cathode : PbO2  SO42   aq   4H   aq  2e   PbSO4  2H 2O l 
Overallreaction : Pb  s   PbO2  s   2 H2 SO4  2 PbSO4  s   2H2O
On charging the battery, the reaction is reversed.
Fuel cell is a galvanic cell which converts the energy of combustion of fuels like H2, CH4,
C2H6, C3H8, CH3OH etc. directly into electrical energy.
In a fuel cell using H2 as a fuel, the following reactions are occurring at electrodes.
Cathode : O2 + 2H2O + 4e–  4 OH–
Anode :
2H 2  4OH   4 H 2O  4e 
,  rG   x kJ mol
2H 2  O2  2H 2O
rG = – n Ecell F. Here n = 4
In a fuel cell using CH4 as a fuel, the reactions are :
Cathode : O2 + 4H+ + 4e–  2H2O] × 2
Anode :
CH 4  2H 2O  CO2  8H   8e 
, here n  8
CH 4  2O2  CO2  2H2O,  r G   y kJ mol
rG = – n EcellF
Corrosion : Corrosion is an electrochemical phenomenon.
Anode :
2 Fe  s   2 Fe2   4 e  ;
Cathode :
O2  g   4 H   4e   2H 2 o  l  ;
Overall reaction :
2 Fe  s   O2  g   4 H   aq   2 Fe2   2H 2O

E Fe
2
Fe
 0.44V

EH

O2
H 2O
 1.23V

Ecell
 1.67V
H+ ions are furnished by H2CO3 formed by the dissolution of CO2 in water. More the concn.
of H+, faster is the reaction at cathode. Corrosion occurs at faster rate in saline water because the
salts present in water perform the functions of salt bridge.
of
Atmospheric O2, further oxidises Fe2+ to Fe3+ to form hydrated Fe2O3 with further production
ions which facilitate corrosion.
H+
We can prevent the corrosion by preventing the surface of the metal to come in contact with
atmosphere by covering the metallic surface with paint or by the layer of other metals (Sn, Zn, etc.).
18
Chemistry
Conductivity : Conductivity (k) is related to resistance as given below :
k 
Cell constant
l A

Resistance
R
Unit : ohm–1 cm–1 or S cm–1
Molar Conductivity (m) is given by m 
m= S cm2 mol–1.
k
where C = molar concentration and unit of
c
Conductivity (k) decreases but molar conductivity
(m) increases with the decrease in concentrations. It
slowly increases with the decrease in concentration for
strong electrolytes while the increase is very steep for
weak electrolytes in very dilute solutions.
Molar conductivity at infinite dilution or zero concentration is called limiting molar conductivity denoted by
°m. Limiting molar conductivity (°m) of strong electrolyte is obtained by the extrapolation of curve at conc  0
but °m for weak electrolyte is determined by the use of
Kohlrausch law of independent migration of ions which
states that limiting molar conductivity for an electrolyte
is the sum of the contribution of molar conductivity of the
0
ions in which it dissociates. m
 v   0  v   0 where v+
–
and v are number of cations and anions furnished by an
electrolyte.
1/2
Molar conductivity versus c for
ocetic acid (weak electrolyte) and
potassium chloride (strong
electrolyte) in aqueous solutions.
(For Faraday’s laws of electrolysis please refer the NCERT Text Book for class XII Part I page 83).
Product of Electrolysis
Product of electrolysis depend on the different oxidising and reducing species present in the
electrolytic cell and their electrode potentials. Moreover some electrochemical process although
feasible but do not seem to take place because extra potential (overvoltage) has to be applied.
Electrolysis of Aqueous NaCl
Al cathode there is the competition between the following reduction reactions
Na+ + e–  Na
H 2O  e  
1
H 2  2OH 
2

E Na

Na
 2.71V

EH
2O
H2
 0.83V
The reduction reaction with higher E is preferred at cathode and therefore, reduction of water
is preferred at cathode.
1
H 2O  e  
H  2OH 
Cathode :
2 2
At the anode there is the competition between the following oxidation reactions :
1
Cl   aq   Cl2  e  ;
E = 1.36 V
2
19
Chemistry
H 2O  l  
1
O  2 H   2e  ;
2 2
E = 1.23 V
At anode the reaction with lower E is preferred and therefore water should get oxidised in
preference to Cl–. However, on account of high over-potential of O2, the oxidation of Cl– ion is
preferred at anode.
1
Cl  e 
2 2
Note : During the electrolysis of aqueous solution of nitrates and sulphates, the oxidation of water
occurs at anode in preference to the oxidation of NO3– or SO42–. For example, during the electrolysis
of aq. Na2SO4, Na+ is not reduced at cathode and SO42– is not oxidised at anode. In this case water
is oxidised at anode and water is also reduced at cathode.
Anode :
Cl  aq  
20
Chemistry
1.
2.
3.
4.
5.
Kohlrausch law states that at
(a)
Infinite dilution, each ion makes definite contribution to the limiting molar
conductivity of an electrolyte whatever be the nature of other ion of the electrolyte.
(b)
Finite dilution, each ion makes definite contribution to the limiting molar conductivity
of an electrolyte whatever be the nature of other ion of the electrolyte.
(c)
Infinite dilution, each ion makes definite contribution to the limiting molar
conductivity of an electrolyte depending upon the nature of other ion of the electrolyte.
(d)
Infinite dilution, each ion makes definite contribution to electrolytic conductance of
an electrolyte whatever be the nature of the other ion of the electrolyte.
Electrolysis of dilute aqueous NaCl solution was carried out by passing 10mA current. The
time required to liberate 0.01 mol. of H2 gas at cathode is (1F = 96500 C/mol)
(a)
9.6 × 104 s
(b)
19.3 × 104 s
(c)
28.95 × 104 s
(d)
38.6 × 104 s
A dilute solution of NaBr is electrolysed using platinum electrodes. The products at anode
and cathode are :
(a)
O2, H2
(b)
Br2, H2
(c)
Br2, Na
(d)
O2, Na
The standard emf of cell : Zn(s)/Zn2+ (aq, 0.01 M) || Fe2+ (aq, 0.001 M)/Fe(s) at 298K is
0.3200 V, then the value of equilibrium constant for the cell reaction is :
(a)
0.32
0.0295
e
(c)
0.32
0.0295
10
0.32
(b)
e 0.0591
(d)
0.32
0.0591
10
The half-cell reaction for corrosion :
2H  
1
O2  2e   H 2O  E   1.23V 
2

Fe2   2e   Fe  s  E    0.44V

Find the AG (in kJ) for the overall reaction :
6.
(a)
–76
(b)
–161
(c)
–152
(d)
–322
Standard electrode potential data are useful for understanding the suitability of oxidant in
a redox titration. Some half-cell reactions and their standard potentials are given below :
MnO4– (aq) + 8H+ (aq) + 5e–  Mn2+ (aq) + 2H2O;
E = 1.51V
Cr2O72– (aq) + 14 H+ (aq) + 6e–  Cr3+ (aq) + 7H2O;
E = 1.38V
21
Chemistry
Fe3+ (aq) + e–  Fe2+ (aq)
E = 0.77V
Cl2 (g) + 2e–  2Cl– (aq)
E = 1.40V
Identify incorrect statement regarding the quantitative estimation of Fe(NO3)2 (aq)
7.
(a)
MnO4– can be used in aqueous HCl
(b)
Cr2O72– can be used in aqueous HCl
(c)
MnO4– can be used in aqueous H2SO4
(d)
Cr2O72– can be used in aqueous H2SO4
The emf of concentration cell :
Zn (s)/Zn2+ (0.0 1M) || Zn2+ (0.10)/Zn is given by
8.
9.
10.
11.
12.
(a)
0
Ecell  Ecell
 2.303
RT
1.10
log
F
0.01
(b)
0
Ecell  Ecell
 2.303
RT
0.01
log
2F
0.10
(c)
0
Ecell  Ecell
 2.303
RT
0.01
log
F
0.10
(d)
Ecell  2.303
RT
0.10
log
2F
0.01
Standard Gibbs energy of formation fG in kJ mol–1 at 298 K for water is –237.2, The value
of E cell for the hydrogen-oxygen fuel cell is :
(a)
1.458 V
(b)
1.0229 V
(c)
1.229 V
(d)
2.229 V
The standard reduction potentials of the three metallic cations X, Y and Z are 0.52, –3.03
and –1.18V respectively. The order of reducing powers of the corresponding metal is
(a)
y > z > x
(b)
x > y > z
(c)
z > y > x
(d)
z > x > y
The standard reduction potential for Fe2+/Fe and Sn2+/Sn are – 0.44 and – 0.14V respectively.

For the cell reaction Fe2+ + Sn  Fe + Sn2+, the Ecell
is :
(a)
+ 0.30 V
(b)
– 0.58 V
(c)
+ 0.58 V
(d)
– 0.30 V
The correct order of molar conductivity at infinite dilution of LiCl, NaCl, and KCl is
(a)
LiCl > NaCl > KCl
(b)
KCl > NaCl > LiCl
(c)
NaCl > KCl > LiCl
(d)
LiCl > KCl > NaCl
Standard electrode potentials for Fe2+/Fe and Fe3+/Fe2+ are – 0.44 and + 0.77V respectively.
Fe3+, Fe2+ and Fe blocks are kept together, then
(a)
Fe3+ increases
(b)
Fe3+ decreases
(c)
Fe2+/Fe3+ remains unchanged
(d)
Fe2+ decreases
22
Chemistry
13.
On the basis of information available from reaction :
4
2
Al  O2 
Al O , rG = – 827
3
3 2 3
kJ/mol of O2, the minimum emf required to carry out the electrolysis of Al2O3 is
14.
(a)
2.14 V
(b)
4.28 V
(c)
6.42 V
(d)
8.56 V
Given
(i)
Cu2+ + 2e–  Cu; E = 0.337 V
(ii)
Cu2+ + e–  Cu+; E = 0.153 V
Electrode potential, E for the reaction :
Cu+ + e–  Cu; will be
15.
(a)
0.90V
(b)
0.30 V
(c)
0.38 V
(d)
0.52 V
Given :
E Fe3+/Fe = – 0.036, E Fe2+/Fe = – 0.439 V
The value of standard electrode potential for the change, Fe3+ + e–  Fe2+ will be :
16.
17.
(a)
– 0.072 V
(b)
0.385 V
(c)
0.770V
(d)
– 0.27 V
In which of the following pairs the constants/quantities are not mathematically related to
each other?
(a)
Gibb’s free energy and standard cell potential.
(b)
Equilibrium constant and standard cell potential.
(c)
Rate constant and activation energy.
(d)
Rate constant and standard cell potential.

 1.10V  was allowed to be completely
The Cell : Zn/Zn2+ (1M) || Cu2+ (1M)/Cu  Ecell
  Zn2   
discharged at 298 K. the relative concentration of Zn2+ to Cu2+ 
 is
2
  Cu  
18.
(a)
anti log (24.08)
(b)
37.3
(c)
1037.3
(d)
9.65 × 104
The molar conductivities 0NaOAC and 0HCl at infinite dilution in water at 25°C are 91.0 and
426.2S cm2/mol respectively. To calculate 0HOAc, the additional value required is
(a)
0H2O
(b)
0KCl
(c)
 0NaOH
(d)
0NaCl
23
Chemistry
19.
What will be the e.m.f. of the given cell?
Pt s  H2  g , p1 bar  H   aq,1M  H2  g , P2 bar  Pt s 
20.
(a)
P
RT
ln 1
F
P2
(b)
P
RT
ln 1
2F
P2
(c)
P
RT
ln 2
F
P1
(d)
None of these
The charge required for the reduction of 1 mol of MnO4– to MnO2 is
(a)
1F
(b)
3F
(c)
5F
(d)
6F
1.
(a)
2. (b)
3. (b)
4 (c)
5.
(d)
6. (a)
7. (d)
8. (c)
9.
(a)
10. (d)
11. (b)
12. (b)
13.
(a)
14. (d)
15. (c)
16. (d)
17.
(c)
18. (d)
19. (b)
20. (b)
24
Chemistry
Consider the Reaction : 2 N2O5(g)  4NO2(g) + O2 (g).
When a reaction proceeds, the concentration of a reactant decreases while that of a product
increases with time. The rate of change of concentration of a reactant is called rate of consumption
or disappearance of the reactant.
Average rate of consumption of N 2O5  
  N 2O5 
t
since [N2O5] is a negative quantity, it is multiplied by – 1 to make the rate a positive
quantity.
Similarly, rates of formation of NO2 and O2 are given by
Average rate of formation of NO2  
Average rate of formation of O2  
  NO2 
t
  O2 
t
These rates can be equated if we divide the rate of change of concentration of a reactant or
a product by its stoichiometric coefficient appearing in balanced chemical equation, that is,
Average rate (rav)/or rate  
  O2 
1   N 2O5 
1   NO2 
 
 
2
t
4
t
t
Rate of a reaction at time ‘t’ is known as the instantaneous rate of reaction.
As t  0, rinst or rate  
d  O2 
1 d  N 2O5 
1 d  NO2 
 
 
2
dt
4
dt
dt
For the Reaction
5 Br– (aq) + BrO3– (aq) + 6H+ (aq)  3Br2 (aq) + 3H2O(l)
We do not use the concentration term for water for expressing the rate of reaction since
change in the concentration of water is negligibly small because the reaction is occurring in aqueous
medium. Hence, the rate of reaction is given as below :
Rate  
  BrO3 
1   Br  
1  H 
1   Br2 
  
 
 
5
t
t
6
t
3
t
Order and Molecularity of Reaction
Rate law : Consider a general reaction : a A + b B  c C + d D
25
Chemistry
The rate law for this reaction is :
Rate = k [A]x [B]y where x and y may or may not be equal to the stoichiometric coefficients a
and b of the reactants.
Rate law for any reaction cannot be predicted by merely looking at the balanced chemical
equation. It is determined experimentally.
x = order of reaction w.r.t. the reactant A
y = order of reaction w.r.t. the reactant B
and
x + y = overall order of the reaction.
Units of rate constant depend upon the order or reaction.
Rate = k [A]x [B]y
k 
Rate
 A x  B y
If x + y = n = order of reaction, then we have
unit of k 
 Concn. 1  n
concentration
1
M1  n



time
time
s
 concentration n
For zero order reaction (n = 0), unit of k is mol L–1 s–1 or Ms–1
For first order reaction (n = 1), unit of k is s–1
For second order reaction (n = 2), unit of k is L mol–1 s–1 or M–1 s–1.
Molecularity is the number of reacting species taking part in an elementary reaction, i.e., the
number of species that collide simultaneously in order to bring about a chemical reaction.
For a bimolecular or trimolecular elementary reactions, the order of reaction is the same as
its molecularity and order w.r.t. each reactant is equal to its stoichiometric coefficient.
For unimolecular elementary reactions, the order of reaction is one at high concentration or
pressure. At low pressure the reaction becomes second order but then the reaction is no longer an
elementary reaction. Order of reaction can be zero, 1, 2, 3 or even a fraction but molecularity can
never be zero or a non-integer. Order is applicable to elementary as well as complex reactions. For
complex reaction, molecularity has no meaning. For complex reaction, order of reaction is given by
slowest step and generally molecularity of the slowest step is the same as the order of the overall
reaction.
Zero Order Reaction : The rate of zero order reaction is independent of the concentration
change.
Rate  
k 
d R
0
 k R  k
dt
 R 0   R 
t
where
 R 0  Initial concentration
 R   Concentration at time t
Half-life of a reaction is the time in which one-half of the initial concentration of the reactant
is consumed.
t1 2 
 R 0
2k
or t1 2   R 0
26
Chemistry
The decomposition of gaseous ammonia on hot platinum surface is a zero order reaction at
high pressure.
 Rate = k [NH3]0 = k
First Order Reaction : 2N2O5  4NO2 is an example of first order reaction. For the reaction:
R  P

d R
 k R
dt
t 
 R 0
 R 1
1
1
ln
or t2  t1 
ln
R
 R 2
k
k
These relationships can be expressed in exponential form as given below
[R] = [R]0e–kt
Equation (ii) can be rewritten as :
t 
 R 0
 R 1
2.303
2.303
log
or t2  t1 
log
R
 R 2
k
k
Expression for half-life for first order reaction :
t1 2 
0.693
k
[R]
log
[R]0
Half-life period of first order reaction is independent of initial concentration of the reactant.
27
Chemistry
Rseudo First Order Reaction is a reaction which is first order w.r.t. each of the two
reactants but becomes first order reaction under certain experimental conditions, i.e., if one of the
reactants is taken in excess.
Note : Half-life period of nth order reaction in inversely proportional to initial concentration
of reactant raised to power (n – 1)
t1 2 
1
 R n0 1
Dependence of rate of reaction on temperature is described by Arrhenius equation :
k = Ae–Ea/RT where
Ea = Activation energy and is given by energy difference between the activated complex and
the reactant molecules.
A = Arrhenius factor or pre-exponential factor or frequency factor that has the units of rate
constant.
Natural logarithm of both sides of Arrehenius equation gives :
ln k  
Ea
 ln A
RT
The plot of ln k and
1
E
gives a straight line of slope   a .
T
R
If k1 and k2 are rate constants at temperature T1K and T2K then we have
log
Ea
k2
 
k1
2.303 R
Ea
 T2  T1 
1 
 1
 T  T   2.303 R  T T 
1 
 2
 2 1 
(Plot between ln k and 1/T)
(Solid line denotes the reaction path without
catalyst and dotted line, with catalyst)
Effect of Catalyst : A catalyst provides an alternative reaction path or reaction mechanism
by reducing the activation energy between reactants and products by lowering the potential energy
barrier. A catalyst can catalyse those reactions for which rG < 0. It does not affect the equilibrium
state, increases the rates of forward as well as reverse reactions in the same proportion without
affecting the rH and the equilibrium is attained sooner.
28
Chemistry
Collision Theory of Chemical Reactions : The molecules are assumed to be hard spheres and
the reaction is postulated to occur when molecules collides with each other. The rate of reaction
for the following bimolecular elementry reaction : A + B  Products, is given by
Rate = ZAB e–Ea/RT
where
ZAB = Collision frequency of the reactants, A and B.
e–Ea/R = Fraction of molecules with energy equal to or greater than Ea.
The collisions in which molecules collide with sufficient kinetic energy (or threshold energy)
and proper orientation are called effective collisions.
To account for effective collisions, another factor P, called the probability factor or steric factor
is introduced, i.e.,
Rate = pZABe–Ea/RT
29
Chemistry
1.
2.
3.
4.
5.
6.
The specific rate constant of a first order reaction depends on the
(a)
initial concentration of the reactants (s)
(b)
time of the reaction
(c)
temperature
(d)
extent of the reaction
The rate of the reaction : N2 (g) + 3H2 (g)  2NH3 (g) can be expressed in terms of time
derivative of concentration of N2, H2, or NH3. Identify the correct relationship amongst the
rate expressions.
(a)
Rate  
d  N 2 
1 d  H 2 
1 d  NH3 
 

dt
3
dt
2
dt
(b)
Rate  
d  NH 3 
d  N 2 
d  H 2 
 3 
 2 
dt
dt
dt
(c)
Rate 
(d)
Rate  
d  N2 
dt
 
1 d  H2 
1 d  NH 3 

3 dt
2
dt
d  NH 3 
d  N 2 
1 d  H 2 
 
 2 
dt
3
dt
dt
In a first order reaction, the concentration of the reactant decreases from 400 mol dm–3 to
25 mol dm–3 in 2 × 104 s. The rate constant in s–1 for this reaction is :
(a)
4 × 10–4
(b)
1.386 × 10–4
(c)
4 × 10–4
(d)
3.45 × 10–5
The rate constant for the reaction : 2N2O5  4NO2 + O2 is 3.0 × 10–5 s–1. If the rate of
reaction is 2.40 × 10–5 mol L–1 s–1, then [N2O5] in mol L–1 is :
(a)
1.4
(b)
1.2
(c)
0.04
(d)
0.8
Which one of the following statement for the order of reaction is not correct?
(a)
order can be determined experimentally.
(b)
Order of reaction is equal to the sum of the powers of the concentration terms in the
differential rate law.
(c)
Powers of the concentration terms may or may not be equal to the stoichiometric
coefficients of the reactants.
(d)
Order cannot be fractional.
The reaction : R  P follow first order kinetics. In 40 minutes the concentration of P changes
from 0.1 to 0.025 M. The rate of reaction, when the concentration of P is 0.01 M is :
(a)
3.47 × 10–4 M min–1
(b)
30
3.47 × 10–5 M min–1
Chemistry
(c)
7.
1.73 × 10–4 M min–1
(d)
1.73 × 10–5 M min–1
The rate constants k1, and k2 for two different reactions are 10 16 e
respectively. The temperature at which k1 = k2 is :
(a)
(c)
8.
9.
2000
K
2.303
2000 K
(b)
1000
K
2.303
(d)
1000 K
1000
2000
and 1015 e
T
T
For the reaction : R  products, it is found that the rate of reaction increases by a factor
of 6.25, when the concentration of R is increased by a factor of 2.5. The order of reaction
with respect to R is
(a)
2.5
(b)
2
(c)
1
(d)
0.5
The following data pertains to the reaction between A and B
S.No.
[A]/mol L–1
[B]/mol L–1
Rate/mol L–1 s–1
(i)
1 × 10–2
2 × 10–2
2 × 10–4
(ii)
2 × 10–2
2 × 10–2
16 × 10–4
(iii)
2 × 10–2
4 × 10–2
15 × 10–4
The rate law for the reaction is :
10.
11.
12.
(a)
Rate = k [A]2 [B]
(b)
Rate = k [A] [B]2
(c)
Rate = k [A]3 [B]0
(d)
Rate = k [A]1/2 [B]2.5
The activation energy of a reaction is zero. The rate constant (k) of reaction of 280K is
1.6 × 10–6 s–1. The value of k for this at 300 K is :
(a)
zero
(b)
3.2 × 10–5 s–1
(c)
1.6 × 10–5 s–1
(d)
1.6 × 10–6 s–1
If a reaction A + B  C is exothermic to the extent of 30 kJ/mol and forward reaction has
an activation energy of 70 kJ/mol, the activation energy for the reverse reaction is :
(a)
30 kJ/mol
(b)
40 kJ/mol
(c)
70 kJ/mol
(d)
100 kJ/mol
Given the following diagram for the reaction A + B  C + D
31
Chemistry
The enthalpy change and activation energy for the reverse reaction : C + D  A + B
respectively are
13.
(a)
x, y
(b)
x, x + y
(c)
– y, x + y
(d)
y, y + z
A reaction proceeds by a two step mechanism
k1


A2 
 2A
k
(fast reaction)
1
k
2
A  B 
 products
14.
15.
(a)
Rate = k [A2] [B]
(b)
Rate = k [A2]2 [B]
(c)
Rate = k [A2]1/2 [B]
(d)
Rate = k [A2]1/2
If k1 = Rate constant at temperature T1 and k2 = Rate constant at temperature T2 for a first
order reaction then which of the following velations is correct? (Ea : activation energy)
(a)
log
k1
2.303Ea  T2  T1 

 T T 
k2
RT
1 2
(b)
log
k2
Ea
 T2  T1 

k1
2.303 RT  T1T2 
(c)
log
k2
Ea
 T1T2 

k1
2.303 RT  T2  T1 
(d)
log
k1
Ea
 T1T2 

k2
2.303 RT  T2  T1 
For a reaction between A and B, the initial rate of reaction is measured for various initial
concentration of A and B. The data provided are
S.No.
16.
(slow reaction)
[A]
[B]
Initial reaction rate
(i)
0.20 M
0.30 M
5.07 × 10–5
(ii)
0.20 M
0.10 M
5.07 × 10–5
(iii)
0.40 M
0.05 M
1.43 × 10–4
(a)
One
(b)
Two
(c)
One and a half
(d)
Three
For the reaction : N2 + 3H2  2 NH3, if
d  NH3 
dt
 2  10 4 mol L–1 s 1 , the value of
d  H 2 
dt
would be
17.
(a)
4 × 10–4 mol L–1 s–1
(b)
6 × 10–4 mol L–1 s–1
(c)
1 × 10–4 mol L–1 s–1
(d)
3 × 10–4 mol L–1 s–1
Half-life period of a first order reaction is 1386 sec. The specific rate constant of the reaction
is :
(a)
0.5 × 10–2 s–1
(b)
0.5 × 10–3 s–1
(c)
5.0 × 10–2 s–1
(d)
5.0 × 10–3 s–1
32
Chemistry
18.
For the reaction A + B  products, it is observed that :
(i)
On doubling the initial concentration of A only, the rate of reaction is also doubled.
(ii)
On doubling the initial concentrations of both A and B there is a change by a factor
of 8 in the rate of reaction.
The rate of this reaction is given by
19.
(a)
Rate = k [A] [B]2
(b)
Rate = k [A]2 [B]2
(c)
Rate = k [A] [B]
(d)
Rate = k [A]2 [B]
For a first order reaction A  P, the temperature (T) dependent rate constant (k) was found
1
to follow the equation log k  2000    6.0. The pre-exponential factor A and the actuaT 
tion energy Ea respectively are
20.
(a)
1.0 × 106 s–1 and 9.2 kJ mol–1
(b)
6.0 s–1 and 16.6 kJ mol–1
(c)
1.0 × 106 s–1 and 16.6 kJ mol–1
(d)
1.0 × 106 s–1 and 38.3 kJ mol–1
1
A  2B , rate of disappearance of ‘A’ is related to the rate of appearance
2
of ‘B’ by the expression
For the reaction
(a)
d  A 
d B

dt
dt
(b)
d  A 
d B
 4
dt
dt
d  A 
1 d B
d  A 
1 d B
(d)


dt
2 dt
dt
4 dt
Under the same reaction conditions, initial concentration of 1.386 mol dm–3 of a substance
becomes half in 40 sec and 20 sec through first order and zero order kinetics respectively.
 k1 
Ratio  k  of the rate constants for first order (k1) and zero order (k0) of the reaction is
 0
(c)
21.
22.
23.
(a)
0.5 mol–1 dm3
(b)
1.0 mol dm–3
(c)
1.5 mol dm–3
(d)
2.0 mol–1 dm3
The energies of activation for forward and reverse reactions for A2 + B2  2AB are 180
kJ mol–1 and 200 kJ mol–1 respectively. The presence of a catalyst lowers the activation
energy of both (forward and reverse) reactions by 100 kJ mol–1. The enthalpy change for the
reaction (A2 + B2  2AB) in the presence of catalyst in (kJ mol–1) will be
(a)
20
(b)
300
(c)
120
(d)
280
The reaction of hydrogen and iodine monochloride is given as H2(g) + 2 ICl(g)  2HCl(g)
+ I2(g). This reaction is of first order with respect to H2(g) and ICl(g), following mechanisms
were proposed.
Mechanism A : H2(g) + ICl(g)  HCl(g) + HI(g)
33
Chemistry
Mechanism B :
slow
H 2  g   ICl  g  
 HCl  g   HI  g 
fast
HI  g   ICl  g   HCl  g   I 2  g 
which of the above mechanisms can be consistent with the given information about the
reaction.
24.
25.
26.
27.
(a)
A only
(b)
B only
(c)
A and B both
(d)
Neither A nor B

The total number of - and - particles excited in the nuclear reaction 238
92U 
(a)
6, 2
(b)
6, 10
(c)
4, 10
(d)
4, 2
214
82 Pb
is
The half life for radioactive decay of C–14 is 5730 years. An archeological artifact containing
wood had only 25% of the C–14 found in living tree. The age of the sample is
(a)
5730 years
(b)
14460 years
(c)
2865 years
(d)
1432 years
The relative penetrating power of    and neutron follows the order
(a)
 >  >  > n
(b)
n >  >  > 
(c)
 >  > n > 
(d)
None of these
Given the hypothetical reaction mechanism
I
II
III
IV
A 
 B 
 C  D  E and the data as
Species formed
Rate of its formation
B
0.002 mol/h per mole of A
C
0.030 mol/h per mol of B
D
0.011 mol/h per mol of C
E
0.420 mol/h per mol of D
The rate determining step is :
(a)
Step I
(b)
Step II
(c)
Step III
(d)
Step IV
34
Chemistry
1.
(c)
2. (a)
3. (b)
4. (d)
5.
(d)
6. (a)
7. (b)
8. (b)
9.
(c)
10. (d)
11. (d)
12. (c)
13.
(c)
14. (b)
15. (c)
16. (b)
17.
(b)
18. (a)
19. (d)
20. (d)
21.
(a)
22. (a)
23. (b)
24. (a)
25.
(b)
26. (b)
27. (a)
35
Chemistry
Adsorption is essentially a surface phenomenon. The accumulation of molecular species at
the surface rather than in the bulk of a solid or liquid is termed adsorption. For example, silica
gel adsorbs moisture.
Adsorbate is the substance whose concentration accumulates at the surface of a the solid or
a liquid called adsorbent.
Desorption is the process of removing the adsorbate from the surface of the adsorbent.
If a substance is uniformly distributed throughout the bulk of the solid, the phenomenon is
known as absorption. In adsorption, the substance is concentrated on the surface and does not
penetrate through the surface to the bulk of adsorbent. For example, anhydrous calcium chloride
absorbs moisture.
Both the adsorption and absorption can be also take place simultaneously. Then we use the
term sorption to describe both the processes. When a chalk stick is dipped in ink, coloured
molecules are adsorbed while solvent is absorbed. Adsorption is accompanied by the decrease in
enthalpy as well as decrease in the entropy of the system. Since adsorption is a spontaneous
process, therefore there is always decrease in the Gibbs energy. As the adsorption proceeds, H
becomes less and less negative, ultimately |H| = |TS| and G = 0. At this state, equilibrium
is attained.
Note : In exceptional cases, chemisorption may be endothermic. For example, H2 adsorbs
endothermically on glass. The S in the process : H2(g)  2H (glass) is sufficiently positive making
G = H – TS negative. Similarly, highly hydrated solutes, when adsorbed on solids have positive
H but there is large positive S due to release of water molecules during adsorption.
[For comparation between physisorption (Van der Waals adsorption) and chemisorption, refer
the NCERT text book part I pages 124-125].
Adsorption of Nitrogen on Iron
At 83 K nitrogen is physisorbed on iron surface as N2 molecules. At room temperature,
practically there is no adsorption of nitrogen on iron. At 773 K and above, nitrogen is chemisorbed
on iron surface as nitrogen atoms.
Factors Affecting Adsorption of Gases
(a)
Nature of adsorbate : Higher the critical temperature of gas, the more easily it will be
physisorbed on the solid. A gas can be chemisorbed on the solid if it is capable of forming
chemical bonds with the solid.
(b)
Surface Area of the Adsorbent : Greater the surface area, the more the extent of
adsorption.
36
Chemistry
(c)
Temperature : Adsorption is an exothermic process involving the equilibrium :
Gas (adsorbate) + Solid (adsorbent)  Gas adsorbed on solid + Heat
Applying Le Chatelier principle, increase of temperature decreases the adsorption and vice
versa.
(d)
Pressure : Adsorption increases with pressure at constant temperature. The effect is large
if temperature is kept constant at low value.
(e)
Activation of the Solid Adsorbent : This means increasing the adsorbing power of the
solid adsorbent. This can be done by subdividing the solid adsorbent or by removing the
gases already adsorbed by passing superheated steam.
(x/
Freundlisch Adsorption Isotherm
These curves show that at a fixed pressure, there is the decrease in the extent of physical
adsorption with the increase in temperature. The relationship between the quantity of gas adsorbed
by unit mass of adsorbent and pressure of the gas at constant temperature is expressed by the
following relation : x/m = KP1/n; (n > 1)
For adsorption from solution, the equilibrium concentration of solution is taken into account
x/m = KC1/n; (n > 1)
Homogeneous and Heterogeneous Catalysis
When the reactants and the catalyst are in the same phase, the process is called homogeneous
catalysis and if they are in different phases, then the processes are the examples of heterogeneous
catalysis. A catalyst enhances the rate of reaction without itself getting used up in the reaction.
Promoters are the substances that enhance the activity of a catalyst while poisons decrease
the activity of the catalyst.
Fe  s  as Catalyst
N 2  3H2 
 2NH3
Mo  s  as promoter
In heterogeneous catalysis, the reactants in gaseous state or in solution are adsorbed on the
surface of solid catalyst. This results in the increase in concentration of reactants on the surface
of solid Catalyst and hence, in the rate of reaction. Adsorption being an exothermic process, the
enthalpy of adsorption is utilised in increasing the rate of reaction.
37
Chemistry
[For shape selective catalysis and enzyme catalysis please refer to the NCERT text book part 1
pages 130-133].
Colloidal Solution
In Colloid or Colloidal dispersion, diameters of particles of the dispersed phase range from
1nm to 1000 nm.
1. Classification Based on the Physical State of Dispersed Phase and Dispersion Medium
Types of Colloidal Systems
Dispersed Phase
Dispersion Medium
Name
Examples
Solid
Solid
Solid sol
Some coloured glasses
Solid
Liquid
Sol
Paints, muddy water
Solid
Gas
Aerosol
Smoke, dust
Liquid
Solid
Gel
Cheese, butter, jellies
Liquid
Liquid
Emulsion
Milk, hair cream
Liquid
Gas
Aerosol
Fog, mist, cloud
Gas
Solid
Solid foam
Pumice stone, foam rubber
Gas
Liquid
Foam
Froth, whipped cream
2. Classification Based on Nature of Interaction between Dispersed Phase and Dispersion
Medium
The colloidal sols are divided in two categories as lyophilic (solvent attracting) and lyophobic
(solvent repelling).
Lyophilic Colloids/sols
Lyophobic Colloids/sols
(i) These are the organic substances like gum,
starch, gelatin etc. which, when mixed with
the liquid, directly form the colloidal sol.
(i) These are inorganic substances like metals, their
sulphides etc. which do not form the colloidal
sol directly. These solutions are prepared
indirectly.
(ii) They are reversible.
(ii) They are irreversible.
(iii) Their viscosity is higher and surface tension
is lower than that of the dispersion medium.
(iii) Their viscosity and surface tension are nearly
same as that of the dispersion medium.
(iv) They are quite stable and are not easily precipitated or coagulated.
(iv) They are easily precipitated by adding a small
amount of a suitable electrolyte.
3. Classification Based on Type of Particles of the Dispersed Phase
(a)
Multimolecular Colloids : A large number of atoms (as in case of gold sol) and smaller
molecules (as in case of sulphur sol) having diameters less than 1nm aggregate via weak
van der walls forces to form particles in the colloidal range.
(b)
Macromolecular Colloids : Macromolecules having colloidal dimensions form dispersion
in a suitable solvent. Examples of macromolecules are starch, cellulose, proteins, enzymes
and synthetic polymers like polythene, nylon 66, polystyrene, synthetic rubber, etc.
38
Chemistry
(c)
Associated Colloids (Micelles) : Surface active agents such as soaps and synthetic
detergents behave as strong electrolytes at low concentrations, but at higher concentrations
exhibit colloidal properties due to the formation of aggregates called micelles or associated
colloids.
Formation of micelles takes place only above a particular temperature called Kraft
temperature (Tk) and above a particular concentration called critical micelle concentration
(CMC). For soaps the CMC is 10–4 to 10–3 mol/L. These colloids have hydrocarbon part which is
hydrophobic and anionic part which is hydrophilic.
Formation of Colloids
(a)
Chemical methods like double decomposition, oxidation, reduction and hydrolyses are
used to form the molecules which the aggregate forming the colloids.
Double decomposition :
As2O3 + 3H2S  As2S3(sol) + 3H2O
Oxidation :
SO2 + 2H2S  3S(sol) + 2H2O
Reduction :
2AuCl3 + 3HCHO + 3H2O  2Au(sol) + 3HCOOH + 6HCl
Hydrolysis :
FeCl3 + 3 H2O  Fe2O3.xH2O + 6HCl
(b)
Bredig’s Arc Method : Colloidal sols of metals such as Au, Ag, Pt, etc. are prepared by
this method.
(c)
Preptisation : It is the process converting a precipitate into colloidal sols by shaking it
with a dispersion medium in presence of a small amount of electrolyte. The precipitate
adsorbs on its surface the ions of the electrolyte common to the lattice of the precipitate and
then the precipitate breaks into charged colloidal particles.
Purification of Colloidal Solutions
While the traces of electrolyte is essential for the stability of the colloidal solution, larger
quantities coagulate it. Hence, to reduce the concentration to a requisite minimum, dialysis,
electrodialysis and ultrafiltration are used.
Properties Exhibited by Colloidal Solution
(a)
Colligative Properties : The values of colligative properties are of small order as compared
to values shown by true solutions of the same concentration.
(b)
Tyndall Effect : The Tyndall effect observed in colloidal solution is due to scattering of
light by colloidal particles in all directions provided (i) the diameters of dispersed particles
are not much smaller than the wavelength of light used and (ii) refractive indicates of
dispersed phase and dispersion medium differ greatly in magnitude.
(c)
Brownian Movement : The zig-zag motion of colloidal particles is due to the unbalanced
bombardment of the colloidal particles by the molecules of dispersion medium. This movement
is responsible for the stability of sols.
(d)
Charge on Colloidal Particles : Colloidal particles always carry an electric charge.
39
Chemistry
Positively Charged Sols
Negatively Charged Sols
Hydrated metal oxides, e.g.,
Al2O3.xH2O, Fe2O3.xH2O
Metal sols, e.g., Ag, Au
Haemoglobin (blood)
Metallic sulphides, e.g., As2S3, Sb2S3, CdS
Oxide sols, e.g., TiO2
Sols of starch, gum, gelatin, clay, charcoal, etc.
Basic dyes, e.g., methylene blue sol.
Acid dyes, e.g., eosin, congo red sols.
The charge on sol particles is due to one or more reasons, viz, due to (i) electron capture
by sol during electrodispersion of metal (ii) preferential adsorption of ions from the solution
and/or formulation of electrical double layer.
Having acquired a positive or negative charge by selective adsorption on the surface of a
colloidal particle, the fixed layer of ions attract the counter ions from the medium forming
a mobile or diffused layer. The potential difference between the fixed layer of ions and
diffused layer of counter ions is called electrokinetic or zeta potential.
(e)
Electrophoresis : The movement of charged sol particles towards the oppositely charged
electrode under the effect of electric field applied across the colloidal solution, is called
electrophoresis.
If the movement of colloidal particles is prevented by some means, it is observed that
dispersion medium begins to move in an electric field. This phenomenon is called
electroosmosis.
(f)
Coagulation or Precipitation of Sols : It is the process of changing the colloidal particles
in a sol into the insoluble precipitate by addition of some suitable electrolytes.
The coagulation of lyophobic sols can be carried out in the following ways :
(i)
(ii)
By electrophoresis
by boiling
(iii)
by prolonged dialysis
(iv)
by the addition of electrolytes.
Coagulation of Lyophilic Sols
This is done by adding (i) an electrolyte and (ii) a suitable solvent which can dehydrate the
dispersed phase.
Hardy-Schulze Rule
A negative ion causes the precipitation of positively charged sol and vice versa. The greater
the valence of the coagulating ion added, the greater its power to cause the precipitation. In the
coagulation of a negative sol, the flocculating power is in the order Al3+ > Ba2+ > Na+. Similarly,
in the coagulation of a positive sol, the flocculating power is in the order :
 Fe  CN 6 
4
 PO43   SO42   Cl  .
Coagulating Value
The minimum concentration of an electrolyte in mmol per litre required to cause the
40
Chemistry
precipitation of a sol in two hours, is called coagulating value. The smaller the amount required,
the higher will be the coagulating power of an ion.
Lyophilic colloids are used to protect the lyophobic colloids. When a lyophilic sol is added, its
particles from a layer around the lyophobic particles. This layer of lyophilic particles is extensively
solvated.
To compare the protective action of different lyophilic colloids, Zsigmondy (1901) introduced
a term called gold number. Gold number of a protective colloid is the minimum mass of it in
milligrams which must be added to 10 mL of a standard red gold sol so that no coagulation of the
gold sol (i.e. the change of colour from red to blue) takes place when 1 mL of 10% sodium chloride
solution is rapidly added to it.
Evidently, smaller the gold number of a protective colloid, the greater is its protective action.
The gold numbers of a few protective colloids are given below :
Sol
Gold Number
Reciprocal
Gelatin
0.005 – 0.01
200 – 100
Casein
0.01 – 0.02
100 – 50
Haemoglobin
0.03 – 0.07
33 – 14
Emulsions
Colloidal systems in which both dispersed phase and dispersion medium are liquids. These
can be of :
(i)
(ii)
Water in Oil Type (W/O Type) : Examples are milk and vanishing cream.
Oil in Water Type (O/W Type) : Examples are butter, cream and cold cream. Soaps and
detergents are most frequently used as emulsifiers for the stabilisation of the emulsion of
O/W type and long chain alcohols, heavy metal salts of fatly acids for the stabilisation of
the emulsion of W/O type.
41
Chemistry
1.
2.
3.
4.
5.
6.
Rate of physisorption increases with the
(a)
decrease in temperature
(b)
increase in temperature
(c)
decrease in pressure
(d)
decrease in surface area
Adsorption of gases on solid surface is generally exothermic when
(a)
entropy of gas decreases during adsorption,
(b)
entropy of gas increases during adsorption,
(c)
enthalpy of adsorption is positive,
(d)
Gibbs energy increases.
Lyophilic sols are
(a)
irreversible
(b)
prepared from inorganic compounds like metal oxides and sulphides,
(c)
coagulated by adding electrolytes,
(d)
self stabilizing.
In a chemical reaction, a catalyst
(a)
alters the amounts of products,
(b)
lowers the activation energy,
(c)
decreases rH for forward reaction,
(d)
increases rH for the reverse section.
Which one of the following statement is correct?
(a)
Lyophobic colloids do not easily coagulate on adding electrolytes.
(b)
Lyophobic colloids are reversible in character.
(c)
Lyophilic colloids are reversible in character.
(d)
Lyophilic colloids are easily coagulated by electrolytes.
According to langumir adsorption isotherm, when the pressure of a gas is very large, the
extent adsorption is
(a)
directly proportional to pressure.
(b)
Inversely proportional to pressure.
(c)
Directly proportional to the square of pressure.
(d)
Independent of pressure.
42
Chemistry
7.
8.
9.
10.
11.
12.
13.
In the coagulation of arsenic sulphide solution, the flocculating powers of given ions are
such that
(a)
PO43– > SO42– > Cl–
(b)
Na+ > Ba2+ > Al3+
(c)
Cl– > SO42– > PO43–
(d)
Al3+ > Ba2+ > Na+
Of the following which is not correct?
(a)
As the adsorption proceeds, H becomes less and less negative, ultimately H becomes
equal to TS and G becomes zero.
(b)
The formation of micelles taken place above Kraft temperature Tk and above critical
micelle concentration (CMC).
(c)
The potential difference between the fixed layer and the diffused layer of opposite
charges around the collidal particle is called electrokinetic potential or zeta potential.
(d)
Hydrated aluminium oxide, Al2O3 . xH2O sol consists of positively charged particles.
When electric field is applied across the sol, charged sol particles move towards the
oppositely charged electrode. This phenomenon is called electroosmosis.
Gold numbers of protective colloids a, b, c and d are 0.50, 0.01, 0.10 and 0.005 respectively.
The correct order of their protective powers is
(a)
d < a < c < b
(b)
c < b < d < a
(c)
a < c < b < d
(d)
b < d < a < c
Which one of the following statements is correct? Peptisation is a process of
(a)
precipitation of colloidal particles,
(b)
purification of colloids,
(c)
dispersing precipitate into colloidal solution,
(d)
protection of colloidal solution.
Which of the following statements is incorrect regarding physisorption?
(a)
Enthalpy of adsorption ( Hadsorptions) is low and positive.
(b)
It occurs because of van der walls forces.
(c)
More easily liquefiable gases are adsorbed readily.
(d)
Under high pressure it results into multimolecular layer on adsorbent surface.
Freundlisch equation for adsorption of gases (in amount of ‘X’ g) on a solid (in amount of
‘m’ g) at constant temperature can be expressed as :
x
1
 log p  log k
m
n
(a)
log
(c)
x
 pn
m
x
1
 log k  log p
m
n
(b)
log
(d)
x
1
 log p  log k
m
n
If a liquid is dispersed in solid medium, then dispersion is called as :
(a)
Sol
(b)
Emulsion
(c)
liquid aesosol
(d)
Gel
43
Chemistry
14.
15.
16.
Among the following, surfactant that will form mecelles in aqueous solution at the lowest
miscelle concentration at ambient conditions is
(a)
CH3(CH2)15N+(CH3)3Br–
(b)
CH3(CH2)11OSO3–Na+
(c)
CH3(CH2)6COO–Na+
(d)
CH3(CH2)11N+(CH3)3Br–
Which among the following statements are correct with respect to adsorption of gases on a
solid?
1.
The extent of adsorption is equal to k P according to Freundlech isotherm.
2.
The extent of adsorption is equal to k P1/n according to Freundlech isotherm.
3.
The extent of adsorption is equal to (1 + bP)/aP according to Langmur isotherm.
4.
The extent of adsorption is equal to aP/(1 + bP) according to Langmur isotherm.
(a)
1 and 3
(b)
1 and 4
(c)
2 and 3
(d)
2 and 4
Match list I (colloidal dispersion) with list II (nature of the dispersion) and select the correct
answer using the codes given below the lists.
List I
(Colloidal Dispersion)
17.
18.
19.
List II
(Nature of Dispersion)
A
Milk
1. Solid in liquid
B
Clouds
2. Liquid in gas
C
Paints
3. Solid in solid
D
Jellies
4. Liquid in liquid
(a)
A – 4, B – 2, C – 1, D – 5
(b)
A – 1, B – 5, C – 3, D – 2
(c)
A – 4, B – 5, C – 1, D – 2
(d)
A – 1, B – 2, C – 3, D – 5
Which of the following statements about the zeolites is false?
(a)
They are used as cation exchanges.
(b)
They have open structure which enables them to take up small molecules.
(c)
Zeolites are aluminosilicates having three dimensional network.
(d)
Some of the SiO44– units are replaced by AlO45– and AlO69– ions in zeolites.
Which of the following forms catonic micelles above certain concentration
(a)
sodium dodecyl sulphate
(b)
sodium acetate
(c)
urea
(d)
cetyl trimethyl ammonium bromide.
Identify the correct statement regarding enzymes.
(a)
Enzymes are specific biological catalysts that can normally function at very low
temperature.
(b)
Enzymes are normally heterogeneous catalysts that are very specific in action.
(c)
Enzymes are specific biological catalysts that can not be poisoned.
(d)
Enzymes are specific biological catalysts that possess well defined active sites.
44
Chemistry
20.
21.
22.
23.
24.
25.
Bredig’s arc are method can not be used to prepare colloidal solution of which of the
following?
(a)
Pt
(b)
Fe
(c)
Ag
(d)
Au
The volume of a colloidal particle Vc as compared to the volume of a solute particle in a true
solution Vs, could be
(a)
Vc
 1
Vs
(b)
Vc
 1023
Vs
(c)
Vc
 103
Vs
(d)
Vc
 103
Vs
The disperse phase in hydrated iron (III) oxide sol and colloidal gold are positively and
negatively charged respectively. Which of the following statements is not correct?
(a)
Magnesium chloride solution coagulates the gold sol more readily than the hydrated
iron (III) oxide sol.
(b)
Sodium sulphate solution causes coagulation in both sols.
(c)
Mixing the sols has no effect.
(d)
Coagulation of both sols can be brought about by electrophoresis.
An example of autocatalysis is
(a)
oxidation of NO to NO2
(b)
oxidation of SO2 to SO3
(c)
decomposition of KClO3 to KCl and O2
(d)
oxidation of oxalic acid by acidified KMnO4
Given below, catalyst and corresponding process/reaction are matched. The mismatch is
(a)
[RhCl (PPh3)2] : hydrogenation.
(b)
TiCl4 + Al(C2H5) : polymerization.
(c)
V2O5 : Haber-Bosch Process.
(d)
Nickel : hydrogenation.
Which of the following is true in respect to adsorption
(a)
G < 0, S > 0, H < 0
(b)
G < 0, S < 0, H < 0
(c)
G > 0, S > 0, H < 0
(d)
G > 0, S > 0, H > 0
1.
5.
9.
13.
17.
(a)
(c)
(c)
(d)
(d)
2.
6.
10.
14.
18.
(a)
(d)
(c)
(a)
(d)
21.
(d)
22. (c)
25.
(b)
3.
7.
11.
15.
19.
(d)
(d)
(a)
(d)
(d)
23. (d)
45
4.
8.
12.
16.
20.
(b)
(d)
(b)
(a)
(b)
24. (c)
Chemistry
Minerals are naturally occurring chemical substances in earth’s crust obtainable by mining.
Ores of a metal are minerals which are used as source of that metal profitably.
Aluminium is the most abundant metal and the oxygen is most abundant element
in the earth’s crust. Iron is the second most abundant metal in the earth’s crust. Silver,
gold, Platinum, sulphur, oxygen and nitrogen are the elements that occur in native or free state.
Many gem-stones are impure form of Al2O3, the impurities range from Cr (in ruby)
to Co (in sapphire). [For principal ores of Al, Fe, Cu and Zn, please refer Table 6.1 in the NCERT
Text Book Part I, Class XII, Page 148]. Concentration of ores depends upon the difference in
physical properties of the compound of metal present and that of the gangue.
Forth floation process is used for the concentration of sulphide ores. A suspension of
powdered ore is made with water. To it collectors (e.g. pine oils, fatty acids, xanthates etc.) and
froth stabilisers (e.g., cresols, aniline which enhance the non-wettability of the mineral particles)
and froth stabilisers (e.g., cresol and aniline) which stabilise the forth) are added. Sometimes
depressants are added to separate the sulphide ores. For example, NaCN is used as a depressant
to selectively prevent ZnS from coming to froth but only allows PbS to come with froth. NaCN
forms a layer of Na2[Zn(CN)4] on the surface of ZnS that prevent it from coming to froth.
Leaching is useful in case the ore is soluble in a suitable solvent.
NaOH  aq 
dilution
Bauxite Al2O3  s  
 Na  Al  OH 4   aq  

CO2  g 
heat
Al2O3 · xH 2O  s  
 Al2O3  s 
Alumina
Conversion of Concentrated Ore to An Oxide
(a)
Calcination : The hydrated or carbonate ores are heated in presence of limited supply of
air when the volatile matter escapes leaving behind the metal oxide.

Fe2O3 · xH 2O 
 Fe2O3  s   xH 2O  g 

ZnCO3  s  
 ZnO  s   CO2  g 
(b)
Roasting : The sulphide ore is heated in a regular supply of air below the melting point
of metal to convert the metal into its oxide or sulphate. Sometimes a part of the sulphide
may act as reducing agent in the subsequent step.
46
Chemistry
2ZnS + 3O2  2ZnO + 2SO2
2PbS + 3O2  2PbO + 2SO2
PbS + 2O2  PbSO4
2Cu2S + 3O2  2Cu2O + 2SO2
Note : In auto reduction : For example, the partly converted Cu2O reduces Cu2S
Cu2S + 2Cu2O  6Cu + SO2
PbS + PbSO4  2Pb + 2SO2
PbS + 2PbO  3Pb + SO2
HgS + 2HgO  3Hg + SO2
Reduction of Oxide to Metal
Reduction means electron gain or electronation. For the reduction of metal oxides, heating is
required. To understand the variation in temperature requirement for thermal reduction
(pyrometallurgy) and to predict which element will suit as reducing agent for a metal oxide (MxOy),
Gibbs energy interpretations are made at any specified temperature.
G = H – TS
For any reaction :
G0 = – 2.303 RT log K
when a reaction proceeds towards products, K will be positive which implies that G will be
negative.
When the value of G is negative, only then the reaction proceed. If S is positive, on increasing
the temperature (T), the value of TS would increase i.e.; (H < TS) and then, TS will be –ve.
If the reactants and products of two reactions are put together in a system and the net G
of the two reactions is –ve, the overall reaction will occur spontaneously.
Ellingham Diagram
Gibbs energy (G) for formation of oxides per mol of O2 are plotted against temperature T.
It is evident that elements for which Gibbs energy of formation of oxides per mol of oxygen is more
negative, can reduce the oxides of elements for which Gibbs energy of formation per mol of O2 is
less negative, that is, the reduction of oxide represented by upper line is feasible by the element
represented by lower line.
Reduction by Carbon
Smelting : It is the process of extraction of metal from its roasted or calcined ore by heating
with powdered coke in presence of a flux. In smelting, oxides are reduced to molten metal by carbon
or carbon monoxide.
PbO + C  Pb + CO and Fe2O3 + 3CO  2Fe + 3CO2
 SiO2  CaCO3  CaSiO3  CO2 
Flux  impurities  slag  FeO  SiO2  FeSiO3 slag 
acidic basic
basic Acidic
 CaO  SiO2  CaSiO3 slag 
47
Chemistry
Extraction of Non-Metals or Metals by Oxidation
In simple electrolysis of molten salt, Mn+ are discharged at negative electrodes. Sometimes a
flux is added for making molten mass more conducting as in the electrolysis of molten alumina,
CaF2 or Na3[AlF6] is added to lower the melting point of mix and bring the conductivity.
Some extractions are based on oxidation for non-metals. For example, extraction of Cl2 from
brine is the oxidation of Cl– in aqueous medium.
2Claq   2H 2O  l   2OHaq   H 2  g   Cl2  g  ; G    422 KJ
Leaching of Ag or Au with CN– involves the oxidation of Ag  Ag+ or Au  Au+

4 Au  s   8CN   2H2O  O2  g   4  Au  CN  2  aq   4OHaq 
2  Au  CN 2 
–
 aq  
Zn  s 
 Reducing
agent 
 2 Au  s    Zn  CN 4 
2
 aq 
REFINING OF METALS
Distillation
Useful for low boiling metals like zinc and mercury.
Liquation
Useful for low boiling metals like tin and lead.
Electrolytic Refining
Anode : Impure metal; cathode strip of pure metal. Soluble metal salt solution is used as an
electrolyte.
A large no. of metals such as Cu, Ag, Au, Pb, Ni, Cr, Zn, Al etc. are refined by this method.
Zone Refining
Zone refining is based on the principles that impurities are more soluble in the melt than in
the solid state of metal. Pure metals are crystallised out of the melt and impurities move into
molten zone. This is used for metals of very high purity, e.g., Ge, Si, B, Ga and In.
Vapour Phase Refining
Impure metal is converted into volatile compound which is then decomposed to get pure metal
e.g. Mond’s process for the purification of Ni metal involves formation of Nickel carbonyl which on
further decomposition give pure nickel metal.
330  350 K
450  470 K
Ni  4CO 
 Ni  CO 4 
 Ni  4CO  g 
 Impure 
metal
Vapours
48
 Pure 
Chemistry
Van Arkel method for Refining Zr or Ti
This method is used to remove all oxygen and nitrogen present in the form of impurity in
certain metals like Zr or Ti.
Evacuated
Tungston filament
Zr  2 I 2 
ArI4 

Zr
 2 I2
Vessel
1800 K
 impure 
 Deposited on Filament 
Chromatographic Methods
This method is based on the principle that different components of a mixture are differently
adsorbed on a adsorbent. The adsorbed components are removed (eluted) by using suitable solvent
(elutant).
49
Chemistry
1.
2.
3.
4.
5.
6.
7.
8.
Froth floation process may be used to increase concentration of the mineral in
(a)
Bauxite
(b)
Calamine
(c)
Haemetite
(d)
Copper pyrites
The slag obtained during the extraction of copper from copper pyrites is mainly of
(a)
CuSiO3
(b)
FeSiO3
(c)
Cu2O
(d)
Cu2S
Heating the ore with carbon with the simultaneous removal of slag is called
(a)
roasting
(b)
calcination
(c)
smelting
(d)
leaching
Cryolite is used in electrolysis of alumina
(a)
to increase the conductivity and decrease the melting point of mix,
(b)
to decrease the conductivity and increase the melting point of mix,
(c)
to increase the conductivity and melting point of mix,
(d)
do decrease the conductivity and melting point of mix.
In the extraction of copper from sulphide ore, the metal is formed by reduction of Cu2O with
(a)
FeS
(b)
CO
(c)
Cu2S
(d)
SO2
The method of zone refining of metals is based on the principle of
(a)
greater mobility of pure metal than that of impurity,
(b)
higher melting point of impurity than that of pure metal,
(c)
higher noble character of solid metal than that of impurity,
(d)
greater solubility of impurity in molten state than in the solid.
Pyrolusite is an
(a)
sulphide orde
(b)
oxide ore
(c)
carbonate ore
(d)
phosphate ore
Pick out the incorrect statement
(a)
Calamine and siderite are carbonates.
(b)
Argentite and cuprite are oxides.
(c)
Zinc blende and iron pyrites are sulphides.
(d)
Malachite and azurite are the ores of copper.
50
Chemistry
9.
10.
11.
12.
13.
14.
15.
16.
17.
‘German silver’ does not have
(a)
Cu
(b)
Zn
(c)
Ni
(d)
Ag
The metal purified by fractional distillation is
(a)
Zn
(b)
Cu
(c)
Al
(d)
Si
Identify the reaction that does not take place in Blast furnace
(a)
2F e2O3
+ 3C  4Fe + 3CO2
(c)
CaCO3  CaO + CO2
(b)
CO2 + C  2CO
(d)
FeO + SiO2  FeSiO3
Native silver forms a water soluble complex with dilute solution of NaCN in presence of
(a)
Nitrogen
(b)
Oxygen
(c)
Carbon dioxide
(d)
Argon
Extraction of Zinc from zinc blende is achieved by
(a)
electrolytic reduction,
(b)
roasting followed by reduction with coke,
(c)
roasting followed by reduction with other metal,
(d)
roasting followed by self-reduction.
Blister copper is
(a)
impure copper having 10% FeSiO3 (b)
copper alloy
(c)
pure copper
copper having about 1% impurity.
(d)
Which one of the following metals has greater tendency to form oxide?
(a)
Al
(b)
Mg
(c)
Cr
(d)
Fe
Bauxite ore is made up of Al2O3 + SiO2 + TiO2 + Fe2O3. This ore is treated with conc. NaOH
solution at 500K and 35 bar pressure for few hours and filtered hot. In the filtrate, the
species present is/are
(a)
Na[Al(OH)4] only
(b)
Na[Al(OH)4] and Na2SiO3 both
(c)
Na2[Ti(OH)6] only
(d)
Na2SiO3 only
When copper pyrites is roasted in excess of air, a mixture of Cu2O + FeO is formed. FeO
is present as impurities. This can be removed as slag during reduction of Cu2O. The flux
added to from slag is
(a)
SiO2 which is an acid flux.
(b)
Lime stone which is basic flux.
(c)
SiO2 which is basic flux.
(d)
CuO which is basic flux.
51
Chemistry
18.
In the process of extraction of gold
O2
Roasted gold ore  CN   H 2O 
  X   OH 
 X   Zn 
 Y   Au
Identify the complexes [X] and [Y]
19.
20.
21.
(a)
X = [Au(CN)2]–; Y = [Zn(CN)4]2–
(b)
X = [Au(CN)4]2–; Y = [Zn(CN)4]2–
(c)
X = [Au(CN)2]–; Y = [Zn(CN)6]4–
(d)
X = [Au(CN)4]–; Y = [Zn(CN)4]2–
During electrolytic refining of copper, some metals present as impurity settle as anode mud.
These are
(a)
Sn and Ag
(b)
Pb and Zn
(c)
Ag and Au
(d)
Fe and Au
Heating mixture of Cu2O and Cu2S will give
(a)
Cu + SO2
(b)
Cu + SO3
(c)
CuO + CuS
(d)
CuO + SO3
Consider the following reactions at 1000°C
1
O  g   ZnO  s 
2 2
A
Zn  s  
B
C  graphite  
rG = – 360 KJ mol–1
1
O  g   CO  g 
2 2
rG = – 460 KJ mol–1
Choose the correct statement at 1000°C
22.
23.
(a)
Zinc can be oxidised by carbon monoxide.
(b)
Zinc oxide can be reduced by graphite.
(c)
Carbon monoxide can be reduced by zinc.
(d)
All above statements are false.
Which of the following factors is of no significance for roasting sulphide ores to oxides and
not subjecting the sulphide ores to carbon reduction directly.
(a)
CO2 is more volatile than CS2
(b)
Metal sulphides are thermodynamically more stable than CS2.
(c)
CO2 is thermodynamically more stable than CS2
(d)
Metal sulphides are less stable than the corresponding oxides.
Which of the following statements about the advantage of roasting of sulphide ore before
reduction is not true.
(a)
Roasting of the sulphide to the oxide is thermodynamically feasible.
(b)
Carbon and hydrogen are suitable reducing agents for metal sulphides.
(c)
The G0 for sulphide is greater than those for CS2 and H2S.
(d)
The G0 is negative for roasting sulphide ore to oxide.
52
Chemistry
24.
25.
Matte is a mixture of
(a)
Cu2S + FeS (small amount)
(b)
FeS + Cu2S (small amount)
(c)
Cu2O + FeO (small amount)
(d)
FeO + Cu2O (small amount)
During roasting copper pyrites are ultimately convented into a mixture of
(a)
FeS + Cu2S
(b)
FeS + Cu2O
(c)
FeO + Cu2S
(d)
FeS + Cu2S + FeO + Cu2O
1.
(d)
2. (b)
3. (c)
4. (a)
5.
(c)
6. (d)
7. (b)
8. (b)
9.
(d)
10. (a)
11. (d)
12. (b)
13.
(b)
14. (d)
15. (b)
16. (c)
17.
(a)
18. (a)
19. (c)
20. (a)
21.
(b)
22. (b)
23. (b)
24. (a)
25.
(d)
53
Chemistry
General Characteristics of 15 Group Elements : [N, P, As, Sb, Bi]
Oxidation States
Negative Oxidation States : The elements of this group show an oxidation state of – 3.
However, the tendency of these elements to show – 3 oxidation states decreases as we move down
the group from N to Bi due to a gradual decreases in the electronegativity and ionization enthalpy.
Positive Oxidation States : The elements of this group also show positive oxidation states
such as + 3 and + 5. The stability of + 3 oxidation state increases while that of + 5 decreases as
we go done the group. This is due to inert pair effect. The + 5 oxidation state in Bi is less stable
than in Sb. The only well characterized Bi (V) compound is BiF5.
Nitrogen shows – 1, – 2, – 3, + 1, + 2, + 3, + 4 and + 5 oxidation states in NH2OH, N2H4, NH3,
N2, N2O, NO, N2O3, N2O4 and N2O5 respectively.
..
..
Catenation : Nitrogen has little tendency for catenation since – N  N – single bond is weak
|
|
(167 kJ mol–1) due to repulsion between non-bonded electron pairs owing to small N – N bond
length. As we move down the group, the element-element bond enthalpies decrease rapidly viz.
N – N (167 kJ mol–1), P – P (201 kJ mol–1), As – As (146 kJ mol–1) and Sb – Sb (121 kJ mol–1) and
therefore, tendency for catenation decreases in the order P > N > As > Sb > Bi.
Elemental State
Because of small size and high electronegativity, nitrogen has a strong tendency to form
multiple (p – p) bonds with itself and with other elements like C and O having small size and
high electronegativily but other elements do no form multiple bonds. Thus, nitrogen exists as a
diatomic gas in which two nitrogen atoms are linked by a triple bond : N  N : (one  and two bonds). Because of small bond length and high bond strength (946 kJ mol–1), nitrogen is inert at
ordinary temperature. Other elements of this group do not exist as diatomic molecules due to their
reluctance to form multiple bonds. Phosphorus, arsenic and antimony exist as discrete teratomic
tetrahedral molecules, i.e., P4. As4 and Sb4 in which the four atoms lie at the corners of a regular
tetrahedron.
Formation of Hydrides
All the elements of group 15 form volatile hydride of the type EH3. They have a pyramidal
structure.
54
Chemistry
(i)
Thermal stability of these hydrides decreases gradually form NH3 to BiH3 due to the decrease
in bond dissociation enthalpy of E – H bond
NH3 > PH3 > AsH3 > SbH3 > BiH3
(ii)
Reducing Character : Because of decrease in thermal stability, the tendency of these
hydrides to give hydrogen and thus act as reducing agents gradually increases in the order:
NH3 > PH3 > AsH3 > SbH3 > BiH3
(iii)
Basic Character : The presence of the lone pair of electrons on the central atom E in EH3
makes these hydrides as Lewis bases.
EH3  H  
Base
EH4
Conjugate acid
As the size of the central atom increases, the stability of the conjugate acid decreases and
hence the basic character decreases in the order :
NH3 > PH3 > AsH3 > SbH3 > BiH3
Thus PH3 is weakly basic but AsH3, SbH3 and BiH3 are not at all basic.
(iv)
Hydrogen Bonding : Due to small size and high electronegativity of nitrogen and presence
of a lone pair of electrons, NH3 forms H-bonds resulting in exceptionally high m.p. and b.p.
(v)
Melting Points and Boiling Points : Due to H-bonding, the m.p. of NH3 is the highest
amongst the hydrides of group 15 elements. As we move from NH3 to PH3, there is a sharp
decrease in the m.p. and b.p. of PH3 as compared to NH3 due to absence of H-bonding.
However, the m.p. and b.p. of the hydrides of rest of the elements increase gradually as we
move down the group form PH3 to BiH3. This is due to increase in molecular size resulting
in increase in Van der Wall forces of attraction holding the molecules together.
Thus PH3 has the lowest and NH3 and SbH3 has the highest m.p. and b.p. respectively.
(vi)
Bond Angles : The hydrides of group 15 have pyramidal shapes, i.e., the central atom
undergoes sp3-hybridization. The HNH bond angle in NH3 is 107°. However, as we move
down the group the bond angles gradually decrease due to decrease in bond pair-bond pair
repulsion. NH3 (107.8°), PH3 (93.5°), AsH3 (91.8°), SbH3 (91.3°) and BiH3 (90°). Thus, the
bond angles show the following trend :
NH3 > PH3 > AsH3 > SbH3 > BiH3.
Pentahalides
P, As and Sb also form pentachlorides. Nitrogen, however, does not form pentahalides due to
the absence of d-orbitals in its valence shell.
Phosphorus pentaiodide, PI5, does not exist probably due to steric factors. Bismuth forms only
pentafluoride. The non-existence of pentachloride, bromide and iodide of Bi is probably due to the
strongly oxidizing properties of Bi5+ due to inert pair effect.
These pentahalides have trigonal bipyramidal shapes in the vapour phase involving sp3dhybridization. PF5 is not hydrolysed because the P – F bond is stronger that P – O covalent bond.
PCl5 is not very stable due to its unsymmetrical (trigonal bipyramidal) shape where some bond
angles are of 90° and the others are of 120°. It decomposes to give PCl5  PCl3 + Cl2. It is due to
this reason that PCl5 behaves as a good chlorinating agent. Solid PCl5 is an ionic compound
consisting of [PCl4]+ and [PCl6]– ions. Solid PbBr5 exists as [PBr4]+ Br–.
55
Chemistry
Oxides
All the elements of this group form two types of oxides, i.e., M2O3 and M2O5 and are called
trioxides and pentoxides.
The trioxides of N, P and As are acidic. Their acidic strength decreases in the order N2O3 >
P2O3 > As2O3. The acidic strength of pentoxides also decreases in the order N2O5 > P2O5 > AS2O5
> Sb2O5 > Bi2O5.
Dinitrogen (N2)
It is obtained
(i)
by the reaction of aqueous solution of NH4Cl and NaNO2
NH4Cl + NaNO2  N2 + 2H2O + NaCl
(ii)
by the thermal decomposition of ammonium dichromate, (NH4)2Cr2O7
 NH4 2 Cr2O7
Heat
 orange 

 N 2  4 H 2O  Cr2O3
 green
Few Typical Reaction of N2 are :
Heat
6Li  N 2 
 2Li3 N
Heat
3 Mg  N 2 
 Mg3 N 2
Heat
N 2  O2 
 2NO
Heat
N 2  3H 2 
 2NH 3
Ammonia (NH3)
Ammonium salts, when treated with caustic soda or lime, decompose to form NH3.
NH4Cl + NaOH  NH3 + NaCl + H2O
Properties
Aqueous solution of ammonia is basic due to the formation of OH– ions :



NH3  g   H2O  l  
 NH4  OH
It precipitates the hydroxides (hydrated oxides in case some metals like Al, Fe, etc.) of many
metals from their salt solutions. For example,
ZnSO4  2NH 4OH  Zn OH 2   NH 4 2 SO4
white ppt
Ammonia acts as a Lewis base due to the presence of a lone pair of electrons. It also acts as
ligand forming complexes.
Cu2   aq   4 NH 3  aq    Cu  NH 3 4 
 blue 
 deep blue 
2
 aq 
AgCl  s   2 NH3  aq    Ag  NH3 2  Cl  aq 
white ppt. 
colourless

solution
56

Chemistry
Oxides of Nitrogen
Formula
Oxidation state
of nitrogen
Common methods of preparation
Heat
Physical appearance and
chemical nature
N2O
+ 1
NH 4 NO3 
 N 2O  2H 2O
Colourless gas, neutral
NO
+ 2
2NaNO2 + 2FeSO4 + 3H2SO4 
Fe2(SO4)3 + 2NaHSO4 + 2H2O + 2NO
Colourless gas, neutral
N2O3
+ 3
2NO  N 2O4 
 2N 2O3
NO2
+ 4
2Pb  NO3 2 
 2NO2  2PbO
brown gas, acidic
N2O4
+ 4
Cool


2NO2 
 N 2O4
Heal
Colourless solid/liquid, acidic
N2O4
+ 5
4HNO3 + P4O10  4HPO3 + 2N2O5
colourless solid, acidic
250 K
673 K
blue solid, acidic
Structure of Oxides of Nitrogen
Formula
Resonance Structures
57
Bond Parameteres
Chemistry
Nitric Acid (HNO3)
On large scale it is manufactured by Ostwald’s process, i.e., by the catalytic oxidation of NH3
by atmospheric oxygen.
Pt Rh gauge
4 NH3  5O2 
 4 NO  6H2O
as Catalyst
2NO + O2  2NO2
3NO2 + H2O  2HNO3 + NO
Oxidising Action of Nitric Acid
4 HNO3  2e   2NO3  2H2O  2 NO2
 Cone 
8HNO3  6e   6 NO3  4 H2O  2 NO
 dil 
Nitric oxide predominate if the acid is dilute and nitrogen dioxide when the concentration of
acid is increased. Most metals form nitrates.
3Cu + 8HNO3 (dilute)  3cu (NO3)2 + 2NO + 4H2O
Cu + 4HNO3 (conc.)  Cu (NO3)2 + 2NO2 + 2H2O
Zn + 4HNO3 (conc.)  Zn (NO3)2 + 2NO2 + 2H2O
Fairly electropositive metals like Zn, Mg etc. reduce very dilute nitric acid further to give N2O,
NH3 or H2, e.g.
4Zn  10HNO3  4Zn  NO3 2  NH4 NO3  3H2O
 dilute 
4Zn  10HNO3  4 Zn  NO3 2  5H 2O  N 2O
Very dilute 
Mg  2HNO3  Mg 2   2NO3  H 2
Very dilute 
Concentrated nitric acid oxidises the non-metals. For example, I2 is oxidised to HIO3, carbon
to CO2, Sulphur to H2SO4, and P4 to H3PO4.
Chemistry of Brown Ring Test is based on the ability of Fe2+ to reduce NO3– to NO which
reacts with the Fe2+(aq) to form a brown coloured complex, [Fe(H2O)5(NO)]2+ which decomposes on
heating.
3 Fe2   NO3  4 H   NO  3 Fe3   2H 2O
 Fe  H 2O 6 


2
 NO   Fe  H 2O 5  NO  
 Brown complex 
2
 H 2O
Phosphine (PH3)
It is prepared by the reaction of
(a)
Ca3P2 with water or dilute HCl
58
Chemistry
Ca3P2 + 6H2O  3Ca(OH)2 + 2PH3
Ca3P2 + 6HCl  3CaCl2 + 2PH3
(b)
White phosphorus with concentrated NaOH solution in the inert atmosphere of CO2.
 1 
 3 
 
P4 0  3NaOH  3H 2O  P H 3  3NaH 2 PO2
 hypophosphite 
Here P4 molecule undergoes disproportionation reaction.
Properties
(i)
Phosphine is weakly basic and gives phosphonium compounds with acids, e.g.,
PH3 + HCl  PH4Cl
(ii)
When absorbed in CuSO4 or HgCl2 solution, the corresponding phosphides are obtained.
3CuSO4 + 2PH3  Cu3P2 + 3H2SO4
Phosphorus trichloride (PCl3) : It is prepared from the following reactions :
P4 + 6Cl2  4PCl3
P4 + 8SOCl2  4PCl3 + 4SO2 + 2S2Cl2
Properties
(i)
It reacts with organic compounds like RCOOH and R – OH containing – OH group :
3ROH + PCl3  3RCl + H3PO3
3RCOOH + PCl3  3RCOCl + H3PO3
(ii)
It hydrolyses to form H3PO3 and HCl
PCl3 + 3H2O  H3PO3 + 3HCl
Phosphorus Pentachloride
PCl5 molecule has trigonal bipyramidal structure, two axial P – Cl bonds are elongated and
weaker than the other three equatorial P – Cl bonds.
Preparation
(i)
By the reaction of white phosphorus with excess chlorine.
P4  10 Cl2
 excess 
(ii)
 4 PCl5
By the reaction of P4 with sulfuryl chloride.
P4 + 10 SO2Cl2  4PCl5 + 10 SO2
(iii)
It hydrolyses to POCl3 and finally to H3PO4
PCl5 + H2O  POCl3 + 2HCl
POCl3 + 3H2O  H3PO4 + 3HCl
Oxides of Phosphorus
Phosphorus forms two types of oxides : P2O3 and P2O5. P2O3 is formed by heating phosphorus
59
Chemistry
in the limited supply of air and P2O5 is formed by heating phosphorus in the excess of air or
oxygen.
 limited supply of air or oxygen
P4  3O2 
 2P2O3
 excess of air or oxygen 
P4  5O2 
 2P2O5
The reluctance of P to enter into p – p multiple bonding leads to cage structures for their
oxides existing as the dimers. The phosphorus atoms are at the corners of the tetrahedron and six
oxygen atoms are along the edges forming six P – O – P single bonds. In P = O; P forms d – p
bond with O atom.
O
P
P
O
O
°
P
O P
P
100°
O
O 0pm P
16
O
P
O
O
102°
O
°
123
3
12
O
O
P
O
O
O
143 pm
O
(b)
The structures of (a) phosphorus (III) oxide, P4O6 and (b) phosphorus (V) oxide, P4O10.
Structure of Oxoacids of Phosphorus
O
O
P
HO
HO
O
P
P
HO
OH
O
HO
H3PO4
Orthophosphoric acid
OH
P
O
O
OH
O
OH
H
O
P
O
OH
OH
H3PO2
Hypophosphorus acid
O
P
O
OH
H
H3PO3
Orthophosphorus acid
P
O
P
OH
H
O
O
P
O
P
OH
H4P2O7
Pyrophosphoric acid
O
HO
O
P
O
OH
O
OH
(HPO3)3
Cyclotrimatephosphoric acid
(HPO3)n
Polymetaphosphoric acid
Hydrogen directly attached with phosphorus is reducing in nature while hydrogen of O – H
bonds is acidic.
Group - 16 Elements
Electron Gain Enthalpies
The elements of group 16 have relatively high electron gain enthalpy values which decreases
down the group from S to PO. Oxygen has less negative electron gain enthalpy than sulphur. This
is due to its small size of oxygen atom. As a result there is strong interelectronic repulsions in the
60
Chemistry
relative small 2p orbitals of oxygen and thus the incoming electron experience less attraction
compared to sulphur.
Element
Electron gain enthalpy (kJ mol–1)
O
S
Se
Po
–140.9
–200.7
–195.1
–180.0
Oxidation State
The stability of –2 oxidation state decreases down the group. Due to very high electronegativity
of oxygen it may shows –2 oxidation state (Exception O 2F2 (+1), OF2 (+2), H2O2 (–1). Other elements
of the group exhibit +2, +4 and +6 oxidation states. The stability +6 oxidation state decreases and
that of +4 increases down the group (Inert pair effect).
Catenation : Oxygen has some but sulphur has greater tendency for catenation. Oxygen
chains are limited to two atoms as in peroxides but sulphur chains contain upto four atoms as in
polysulphides.
H  O  O  H, H  S  S  H, H  S  S  H, H  S  S  S  S  H
or H 2O2
Hydrogen peroxide
or H 2S2
or H 2S3 Polysulphides
or H 2S4
Sulphur possesses the maximum tendency of catenation due to the highest bond strength of
S – S bond (O – O = 142, S – S = 226, Se – Se = 172 and Te – Te = 126 kJ mol–1). The tendency
for catenation decreases markedly as we go down the group from S to Te. Oxygen possesses this
property to a very less extent. The decreasing order of catenation amongst group 16 elements is
S > Se > O > Te
Chemical Properties
Formation of Hydrides
All these elements form volatile, stable, bivalent hydrides of the formula H2E i.e., H2O, H2S,
H2Se, H2Te and H2P0. The central atom (E) in these hydrides is sp3 hybridized. Due to the presence
of two lone pairs on the central atom, these have bent (V) shapes.
(i)
Melting Points and Boiling Points : H2O has the highest and H2S has the lowest values
of melting and boiling points. Their decreasing order is
H2O > H2Te > H2Se > H2S
(ii)
Volatility : As H2O has the highest and H2S has the lowest boiling point, the volatility
increases abruptly from H2O to H2S and then decreases from H2S to H2Te. Thus, H2O is
least volatile and H2S is most volatile hydrides of group 16 elements. Therefore, volatility
increases in the order
H2O < H2Te < H2Se < H2S
(iii)
Thermal Stability of the hydrides decreases from H2O to H2Te i.e. H2O > H2S > H2Se >
H2Te. This is due to the reason that as the size of the central atom in H2E increases, the
H – E bond becomes weaker and breaks easily on heating.
(iv)
Reducing Character : Hydrides of all these elements except that of oxygen, i.e. H2O,
behave as reducing agents. The reducing character of these hydrides increases as the thermal
stability decreases from H2S to H2Te i.e.
61
Chemistry
H2S < H2Se < H2Te
Bond Angles : The hydrides of group 16 are V-shaped in which the bond angles decrease
on moving down the group as the electronegativity of the central atom decreases from O to
Te. Consequently, the bond pairs of electrons move away from the central atom and the
bond pair- bond pair repulsion decreases. The bond angles, therefore, decrease in the order
(v)
H 2O  H 2S  H 2Se  H 2Te
104 
 92 
 91 
 90  
Formation of Halides
(a)
Hexahalides : All the elements except oxygen form hexafluorides, i.e. SF6, SeF6 and TeF6.
No other halogen forms stable hexahalides. The hexafluorides have octahedral (sp3d2) shapes.
The stability of these hexafluorides decreases from SF6 to TeF6, i.e., SF6 is practically inert,
SeF6 is slightly more reactive while TeF6 is hydrolysed by water. This is due to the reason
that as the size of the atom increases, the polarity of the M – X bond increases and hence
the hexafluoride molecule becomes more susceptible to nucleophilic attack by water. Thus,
the ease of hydrolysis varies as SF6 < SeF6 < TeF6.
(b)
Tetrahalides : All the elements except oxygen form tetrafluorides (SF4, SeF4 and TeF4)
and tetrachlorides (SCl4, SeCl4, TeCl4). These have trigonal bipyramid (sp3d) geometry. As
one equatorial position is occupied by a lone pair of electrons which repels the axial bond
pairs thereby decreasing the angle from 180° to 173°, these halides have see-saw shape.
See-saw shape of SF4 molecule.
Formation of Oxides
These elements form a variety of oxides in different oxidation states from + 2 to + 6.
The acidic character of oxides of these elements in the same oxidation state decreases as we
move down the group. Thus SO2 and SeO2 are acidic while TeO2 and PoO2 are amphoteric. The
order varies as SO2 > SeO2 > TeO2 > PoO2 and SO3 > SeO3 > TeO3.
Reducing property of EO2 decreases from SO2 to TeO2. SO2 is reducing while TeO2 is an
oxidising agent.
(a) Sulphur Dioxide, SO2 is gas and forms discrete molecules even in the solid state. It is
acidic in nature and is also called anhydride of sulphurous acid (H2SO3). It can act as a reducing
agent and also as an oxidising agent. It can also act as a bleaching agent in the presence of a
moisture. It bleaches due to reduction and its bleaching action is temporary. (SO2 + 2H2O  H2SO4
+ 2H). The colour is, however, restored when the bleached article is exposed to air since the oxygen
62
Chemistry
of the air oxidises the colourless compound back to the original coloured substance. SO2 acts as a
Lewis base due to the presence of a lone pair of electrons on S atom. It also acts as a ligand and
forms numerous coordination compounds.
SO2 molecule has a bent structure with a O – S – O bond angle of 119°. The -bonds between
S and O are formed by sp2 – p overlap while one of the -bonds arises from p – p overlap and
the other from p – d overlap but even the both of the S – O bonds are identical (143 pm) due
to resonance.
Ozone (O3)
Ozone is an allotrope of dioxygen (O2) and is prepared by passing a silent electric discharge
through dioxygen.
3O2  2O3 rH = + 142 kJ/mol
O3 is thermodynamically unstable with respect to oxygen (because rG > 0)
It liberates nascent oxygen (O3  O2 + O). Therefore, it oxidises PbS to PbSO4, and KI to I2.
PbS + 4O3  PbSO4 + 4O2
2I– + H2O + O3  2OH– + I2 + O2
The liberated Iodine is titrated against a standard solution of sodium thiosulphate to
estimate O3.
I2 + 2S2O32–  2I– + S4O62–
Nitrogen oxides emitted from the exhaust system of supersonic aeroplanes deplete the
concentration of ozone layer in atmosphere.
Sulphurdioxide, SO2
It is prepared by
(i)
(ii)
(iii)
treating a sulphite with dilute H2SO4 : SO32– + 2H+  SO2 + H2O
S + O2  SO2
4FeS2 + 11O2  2Fe2O3 + 8SO2
Properties
It aqueous solution is called sulphurous acid (H2SO3)
SO2 + H2O  H2SO3
It reacts with NaOH forming Na2SO3, which then reacts with more SO2 to form NaHSO3.
2NaOH + SO2  Na2SO3 + H2O
Na2SO3 + H2O + SO2  NaHSO3
Reducing action of SO2 : It reduces Fe3+ ions to Fe2+ ions and acidified MnO4– to Mn2+.
2Fe3+ + SO2 + 2H2O  2Fe3+ + SO42– + 4H+
5SO2 + 2MnO4– + 2H2O  5SO42– + 4H+ + 2Mn2+
63
Chemistry
Sulphuric Acid, H2SO4
It is manufactured by contact process which involves the following steps :
(i)
(ii)
Burning of sulphur to form SO2
Catalytic conversion of SO2 to SO3
V2O5


2SO2  O2 
 2SO3
(iii)
rH  196.6 kJ mol
Absorption of SO3 into H2SO4 (conc.) to form oleum, H2S2O7
Properties
(a)
Acidic nature :



H 2SO4  H 2O 
 H 3O  HSO4  Virtually complete 

2

H 2SO4  H 2O 
 H 3O  SO4  about 10% complete 
(b)
Dehydrating properties :
H SO
2
4
C6 H12O6 
6C
6 H O
2
H SO
2
4
HCOOH 
CO
H O
2
(c)
Oxidising properties : Hot conc. H2SO4 is less powerful oxidising agent that conc. HNO3
H2SO4  SO2 + H2O + O
Cu  O  H 2SO4  CuSO4  H 2O
Cu  2H 2SO4  2H 2O  CuSO4
Zn  2H 2SO4  SO2  2H 2O  ZnSO4
90% acid
Zn  5H 2SO4  H 2S  4H 2O  4ZnSO4
90% acid
Zn  H 2SO4  H 2  ZnSO
 dilute 
HBr I 
H2 SO4
 Concentrated 
(d)
 2H 2O  SO2  Br2 I 2
Preparation of more volatile acids from their corresponding salts : 2MX + H2SO4(conc.) 
2HX + M2SO4
(M = Metal and X = F, Cl, NO3)
64
Chemistry
Oxoacids of Sulphur
O
S
HO
O
HO
Sulphurous acid
(H2SO3)
S
HO
O
HO
Sulphurous acid
(H2SO4)
O
O
O
S
S
O O
OH
HO
Peroxodisulphuric acid
(H2S2O8)
O
O
S
S
O
O
O
S
S
O
O
O
OH
OH
Pyrosulphuric acid (oleum)
H2S2O7
HO
OH
Sn–2
O
O
(H2SnO6) and n = 3 to 6
65
Chemistry
1.
2.
3.
4.
5.
6.
7.
8.
9.
Which of the following are isoelectronic and isostructural? NO3 , CO32  , CIO3 , SO3
(a)
NO3 , CO32 
(b)
SO3 , NO3
(c)
ClO3 , CO32 
(d)
CO32  , SO3
The acid having O – O bond is
(a)
H2S2O3
(b)
H2S2O6
(c)
H2S2O8
(d)
H2S4O6
The shape of O2F2 is similar to that of
(a)
C2F2
(b)
H2O2
(c)
H2F2
(d)
C2 H 2
The ONO angle is maximum in
(a)
NO3
(b)
NO2
(c)
NO2
(d)
NO2
The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is
(a)
H2S < SiH4 < NH3 < BF3
(b)
NH3 < H2S < SiH4 < BF3
(c)
H2S < NH3 < SiH4 < BF3
(d)
H2S < NH3 < BF3 < SiH4
The number of P – O – P bonds in the structures of P4O10 and P4O6 are respectively
(a)
6, 6
(b)
5, 5
(c)
5, 6
(d)
6, 5
Which of the following statement is not correct?
(a)
Solid PCl5 exists as tetrahedral [PCl4]+ and octahedral [PCl6]– ions.
(b)
Solid N2O5 exists as [NO2]+ [NO3]–
(c)
Solid PBr5 exists as [PBr4]+ Br–
(d)
Oxides of phosphorus P2O3 and P2O5 exist as monomers.
There is no S – S bond in
(a)
S2O42 
(b)
S2O32 
(c)
S2O52 
(d)
S2O72 
White phosphorus on reaction with NaOH gives PH3 as one of the products. This is a
(a)
dimerization reaction
(b)
disproportionation reaction
(c)
condensation reaction
(d)
precipitation reaction
66
Chemistry
10.
11.
In pyrophosphoric acid, H4P2O7, number of – and d–p bonds respectively
(a)
8 and 2
(b)
6 and 2
(c)
12 and zero
(d)
12 and 2
The following two reactions of HNO3 with Zn are given as (equations are not balanced)
Zn  conc. HNO3  Zn  NO3 2  X  H2O   A 
Zn  dilute HNO3  Zn  NO3 2  Y  H2O   B 
In reactions A and B, the compounds X and Y respectively are
12.
13.
14.
15.
16.
17.
18.
(a)
NO2 and NO
(b)
NO2 and NO2
(c)
NO and NO2
(d)
NO2 and NH4NO3
The reaction of P4 with X leads selectively to P4O6. The X is
(a)
dny O2
(b)
a mixture of O2 and N2
(c)
Moist O2
(d)
O2 in the presence of aqueous NaOH
The number of -bonds in P4O10 is
(a)
6
(b)
16
(c)
20
(d)
7
Which of the following has p – d bonding?
(a)
NO3
(b)
SO32 
(c)
BO33 
(d)
CO32 
The percentage of p-character in the orbitals forming p – p bonds in p4 is
(a)
25
(b)
33
(c)
50
(d)
75
The correct order of increasing bond angles in the following triatomic species is
(a)
NO2  NO2  NO2
(b)
NO2  NO2  NO2
(c)
NO2  NO2  NO2
(d)
NO2  NO2  NO2
Reaction of HNO3 with I2, S8, P4 and C give respectively
(a)
HIO3, H2SO4, H3PO4 and CO2
(b)
HIO3, H2SO4, H3PO3 and CO2
(c)
I2O5, H2SO4, H3PO4 and CO
(d)
I2O5, SO2, P2O5 and CO2
Two types of F – E – F angles are present in which of the following molecules [E = S,
Xe and C]
(a)
SF4
(b)
XeF4
(c)
SF6
(d)
CF4
67
Chemistry
19.
20.
The number of hydrogen atoms attached to phosphorus atom in hypophosphorus acid is
(a)
0
(b)
2
(c)
1
(d)
3
Which of the following is not correct about structure of white phosphorus
(a)
It has six P – P single bonds.
(b)
It has four P – P single bonds.
(c)
Four lone pairs of electrons.
(d)
P – P – P angle of 60°
1.
(a)
2. (c)
3. (b)
4. (d)
5.
(c)
6. (a)
7. (d)
8. (d)
9.
(b)
10. (d)
11. (d)
12. (b)
13.
(b)
14. (b)
15. (d)
16. (d)
17.
(a)
18. (a)
19. (b)
20. (b)
68
Chemistry
GENERAL CHARACTERISTICS OF 15 GROUP ELEMENTS
Electronegativity
Fluorine is the most electronegative element in the periodic table. With increase in atomic
number down the group, the electronegativity decrease.
Electron gain enthalpy : Electron gain enthalpies of chlorine, bromine and iodine become
less negative as the size of the atom increases. The electron gain enthalpy of fluorine is, however,
less negative than that of Cl because of its small size as a result of which inter electronic repulsions
present in its 2p-subshell are comparatively large. Thus, chlorine has the highest negative electron
gain enthalpy.
Element
Electron gain enthalpy (kJ mol –1)
F
Cl
Br
I
– 333
– 349
– 325
– 296
Oxidation states. All the halogens show an oxidation state of – 1. Fluorine being the most
electronegative element always shows an oxidation state of – 1 while other halogens also show
positive oxidation states up to a maximum of + 7 (i.e. + 1, + 3, + 5, and + 7) due to the availability
of vacant d-orbitals in the valence shell of these atoms. Some halogens also show + 4 and + 6
oxidation states in oxides and oxo acids.
Colour
All the halogens have characteristic colours. F2 is light yellow, Cl2 is greenish yellow, Br2 is
reddish brown and I2 is deep violet. The colour of halogens is due to the reason that their molecules
absorb light in the visible region as a result of which electrons are excited to higher energy levels.
Bonds Dissociations Energy or Enthalpy of Dissociation
Bond dissociation energies of chlorine, bromine and iodine decrease down the group as the size
of the atom increases. The bond dissociation energy of F2, is however, lower than those of Cl2 and
Br2 because of large inter electronic repulsions among the lone pairs in F2 molecule.
X2
Bond dissociation enthalpy (kJ mol –1)
F2
Cl2
Br2
I2
158.8
242.6
192.8
151.1
69
Chemistry
Oxidising Power : All the halogens act as strong oxidising agents since they have a strong
tendency to attract electrons and have positive values of electrode potential (E°). The oxidising
power, however, decreases as we move down the group from F to I i.e. F2 > Cl2 > Br2 > I2.
Since F2 is the strongest oxidising agent, it will oxidise all other halide ions to halogens in
solution or even in the solid phase.
F2 + 2X–  2F– + X2 (X = Cl, Br to I)
Similarly, Cl2 will displace Br– and I– ions from their solutions while Br2 will displace I– ions
only.
Cl2 + 2X–  2Cl– + X2 (X = Br or I); Br2 + 2I–  2Br– + I2
Hence F2 is the strongest and I2 is the weakest oxidising agent. This is also indicated by the
decrease in the electrode potential (E°) for the reaction X2(aq) + 2e–  2X– (aq) on moving down
the group.
Formation of Halides
Halogens combine with all the elements except He, Ne and Ar forming a large number of
binary halides.
(a) Halides of Metals
(i)
There is a regular gradation from ionic to covalent bonding as the atomic number of the
halogen increases for the same metal atom (M)
The ionic character of M – X bond and m.p. and b.p. of halides decrease in the order
M – F > M – Cl > M – Br > M – I.
(ii)
Metals of low ionization enthalpies such as alkali metals form ionic halides whereas metals
with high ionization enthalpies such as transition metals form covalent halides. Molecular
halides show decrease in m.p. and b.p. as MI > MBr > MCl > MF.
(iii)
Halides of metals in their higher oxidation states are more covalent than those formed in
lower oxidation states. For example, SnCl4 is more covalent than SnCl2.
Similarly PbCl4, SbCl5 are more covalent than PbCl2 and SbCl3 respectively. Less ionic
halides such as AgX show the solubility trends as AgI < AgBr < AgCl < AgF in water.
(b) Halides of Non-Metals
(i)
(ii)
Halides of non-metals are essentially covalent in nature.
The strength of M – X bond for a particular non metal (M) decreases in the order M – F
> M – Cl > M – Br > M – I. Thus for hydrogen halides HX, the bond strength of H – X bond
decreases from HF to HI as the atomic size of the halogen increases from F to I i.e. H – F
> H – Cl > H – Br > H – I.
(c) Reducing Character of Halides
The halide ions (X–) behave as reducing agents and their reducing power decreases in the
order :
I– > Br– > Cl– > F–
Aqueous solutions of hydrogen halides are known as hydrohalic acids.
70
Chemistry
(i)
Thermal stability of the hydrogen halides decrease from HF to HI i.e.
HF > HCl > HBr > HI
(ii)
Acidic strength : The Acidic strength of hydrogen halides decreases from HI to HF i.e. HI
> HBr > HCl > HF.
Oxoacids of Halogens
(i)
Fluorine forms only one oxoacid HOF since it is the strongest oxidising agent Chlorine,
bromine and iodine mainly form four series of oxoacids – namely halic (I) acid or hypohalous
acid (HOX), halic (III) acid or halous acid (HOXO), halic (V) acid or halic acid (HOXO2)
and halic (VII) acid or perhalic acid (HOXO3).
(ii)
Acidic Character : All these acids are monoprotic containing an – OH group. The acidic
character of the oxoacids increases with increase in oxidation number of the halogen, i.e.,
HClO < HClO2 < HClO3 < HClO4
Oxoacids
pKa
HOCl
HOClO
COClO2
HOClO3
7.5
2.0
– 1.2
– 10
Oxidising Power and Thermal Stability of Oxo Acids : The oxidising power of these
acids decreases as the oxidation number of the halogen increases, i.e., HClO > HClO2 >
HClO3 > HClO4. This is due to the reason that as the oxidation number incrases, the
halogen-oxygen bond becomes more covalent. As a result, the thermal stability of anions of
oxo acids increases. Thus, hypohalites are stronger oxidising agents than perhalates.
(iii)
Chlorine (Cl2)
It can be prepared from the following reactions
(i)
By the oxidation of HCl with MnO2
MnO2  4HCl  MnCl2 + Cl2 + H2O
Mixture of NaCl and conc. H2SO4 may be used in place of HCl.
(ii)
By the oxidation of HCl with KMnO4
2KMnO4 + 16HCl  2KCl + 2MnCl2 + 8H2O + 5Cl2
(iii)
By the electrolysis of brine (concentrated solution of NaCl)
Properties
(a)
(b)
Reaction with metals and non-metals
2Al + 3ACl  2AlCl3,
2Fe + 3Cl2  2FeCl3
P4 + 6Cl3  4PCl3,
S8 + Cl2  4S2Cl2
Reaction with ammonia :
3Cl2  8 NH 3  N 2  6 NH 4Cl
 Excess 
3Cl2  NH3  NCl3  3HCl
 Excess 
71
Chemistry
(c)
Reaction with water and alkalies :
Cl2  g   water  Cl2
+
Cl2(aq) + 2H2O
–
H3O + Cl + HOCl
–
–
OH
2H2O
OH
–
H2O + OCl
Cl2 reacts with cold and dilute NaOH solution to form NaCl and NaClO :
Cl2  2OH   Cl   OCl   H 2O
 hypochlorite 
cold and dilute
Cl2 reacts with hot and concentrated NaOH solution to form NaCl and NaClO3
Cl2
(d)

6OH 

 hot and conc. 
5Cl   ClO3  3H 2O
Oxidising and bleaching action of Cl2 :
Cl2 + H2O  HCl + HClO
HClO  HCl 
O
 nascent oxygen 
It oxidises Fe2+ to Fe3+, SO32– to SO42–, SO2 to H2SO4, I2 to HIO3 and Br–/I– to Br2/I2.
Bleaching action of Cl2 is due to oxidation in presence of moisture.
Cl2 + H2O  2HCl + O
Coloured substance + (O)  oxidised from (colourless)
Interhalogen Compounds
The compounds of one halogen with the other are called interhalogen compounds. The stability
of interhalogens increase as the size of central halogen increases (or with the increase in the
electronegativity difference of halogens).


For types of interhalogens XX ´, XX 3´ , XX 5´ and XX 7´ are formed where X is halogen of
larger size and X´ of smaller size. X´ is more electronegative than X. They are more reactive than
halogens except F2 because X – X´ bond dissociation enthalpy is less than that of X – X bond except
F – F bond. Their molecular geometries or structures can be predicted with the help of VSEPR
theory.
Groups 18 elements show least reactivity because :
(i)
(ii)
Noble gases except helium (1s2) have stable closed shell ns2np6 electronic configuration.
They have high ionisation enthalpies and more positive electron gain enthalpies.
Neil Bartlett prepared first noble gas compound, Xe+ [PtF6]–. True chemical compounds of
helium, neon and argon are not yet known because of their high ionisation enthalpies.
Xenon-fluorine compounds, XeF2, XeF4 and XeF6 are powerful fluorinating agents and are
formed by the direct reaction of xenon with fluorine under appropriate conditions.
72
Chemistry
Xenon fluorides react with F– acceptors like PF5, SbF5, etc. to form catonic species and F–
donor like alkali metal fluoride to form fluoroanions.
XeF2 + PF5  [XeF]+ [PF6]–
XeF4 + SbF5  [XeF3]+ [SbF6]–
XeF6 + MF  M+ [XeF7]–
Hydrolysis
XeF2 is hydrolysed to give Xe and O2 :
XeF2 + 2H2O  2Xe + 4HF + O2
Hydrolysis of XeF4 and XeF6 with water gives XeO3.
6XeF4 + 12H2O  4Xe + 2XeO3 + 24HF + 3O2
XeF6 + 3H2O  XeO3 + 6HF
Partial hydrolysis is XeF6 gives oxyfluorides, XeOF4 and XeOF2.
XeF6 + H2O  XeOF4 + 2HF
XeF6 + 2H2O  XeO2F2 + 4HF
XeO3 is a colourless explosive solid and has a pyramidal molecular structure XeOF4 is a
colourless volatile liquid and has square pyramidal molecular structure.
73
Chemistry
1.
2.
3.
4.
5.
6.
7.
8.
Which of the following is not true?
(a)
Among halide ions, iodide ion is the most powerful reducing agent.
(b)
Fluorine is the only halogen which does not show a variable oxidation state.
(c)
HOCl is a stronger acid than HOBr
(d)
HF is a stronger acid than HCl
Total number of lone pair of electrons in XeOF4 is
(a)
0
(b)
1
(c)
2
(d)
3
Among the following, the pair in which two species are not isostructural, is
(a)
SiF4 and SF4
(b)
IO3– and XeO3
(c)
BH4– and NH4+
(d)
PF6– and SF6
Which one of the following arrangements represents the correct order of electron gain
enthalpy (with negative sign) of the given atomic species.
(a)
F < Cl < O < S
(b)
S < O < Cl < F
(c)
O < S < F < Cl
(d)
Cl < F < S < O
In the case of alkalimetals, the covalent character decreases in the order :
(a)
MF > MCl > MBr > MI
(b)
MF > MCl > MI > MBr
(c)
MI > MBr > MCl > MF
(d)
MCl > MI > MBr > MF
Which is not the correct order for stated properly?
(a)
Ba > Sr > Mg ; atomic radius
(b)
F > O > N ; first ionisation enthalpy
(c)
Cl > F > I ; negative electron gain enthalpy
(d)
O > Se > Te ; electronegativity.
In which case the order of acidic strength is not correct?
(a)
HI > HBr > HCl
(b)
HIO4 > HBrO4 > HClO4
(c)
HClO4 > HClO3 > HClO2
(d)
HF > H2O > NH3
In which of the following arrangements, the sequence is not strictly according to the property
written against it?
(a)
B < C < 0 < N increasing first ionisation enthalpy
(b)
HF < HCl < HBr < HI, increasing acid strength
(c)
CO2 < SiO2 < SnO2 < PbO2, increasing oxidising power
(d)
NH3 < PH3 < AsH3 < SbH3, increasing basic strength.
74
Chemistry
9.
10.
11.
12.
Which one of the following reactions of Xenon compounds is not feasible?
(a)
Xe F6 + RbF  Rb [XeF7]
(b)
XeO3 + 6HF  XeF6 + 3H2O
(c)
3XeF4 + 6H2O  2Xe + XeO3 + 12HF + 3/2 O2
(d)
2XeF2 + 2H2O  2Xe + 4HF + O2
Identify the incorrect statement among the following
(a)
Ozone reacts with SO2 to give SO3
(b)
Silicon reacts with NaOH (aq) in the presence of air to give Na2SiO3 and H2O
(c)
Cl2 reacts with excess of NH3 to give N2 and HCl
(d)
Br2 reacts with hot and strong NaOH solution to give NaBr, NaBrO 3 and H2O.
Which is the most easily liquifiable rare gas?
(a)
Ar
(b)
Ne
(c)
Xe
(d)
Kr
Among the following molecules,
(i) XeO3
(ii) XeOF4
(iii) XeF6
those having same number of lone pairs on Xe are
13.
14.
15.
16.
(a)
(i) and (ii) only
(b)
(ii) and (iii) only
(c)
(ii) and (iii) only
(d)
(i), (ii) and (iii)
Which can do glass etching?
(a)
HIO4
(b)
SiF4
(c)
HF
(d)
HNO3
High concentration of fluoride are poisonous and harmful to bones and teeth at levels over
(a)
1 ppm
(b)
3 ppm
(c)
5 ppm
(d)
10 ppm
In which of the following molecules, are all the bonds not equal
(a)
NF3
(b)
ClF3
(c)
BF3
(d)
AlF3
In BrF3 molecule, the lone pairs occupy equatorial positions to minimise
(a)
lone pair–bond pair repulsions only
(b)
bond pair–bond pair repulsions only
(c)
lone pair–lone pair and lone pair–bond pair repulsions
(d)
lone pair–lone pair repulsions only.
75
Chemistry
17.
18.
19.
20.
Which of the following contains maximum number of lone pairs of electron on the central
atom?
(a)
ClO3–
(b)
XeF4
(c)
SF4
(d)
I3–
Which among the following factors is/are most important in making fluorine the strongest
oxidising agent?
(a)
Electron affinity
(b)
ionization energy
(c)
hydration energy
(d)
low bond dissociation energy and
high hydration energy
Which products are expected from the disproportionation reaction of hypochlorous acid?
(a)
HClO3 and Cl2O
(b)
HClO2 and HClO4
(c)
HCl and Cl2O
(d)
HCl and HClO3
Which of the following hydrogen halides is most volatile?
(a)
HF
(b)
HCl
(c)
HBr
(d)
HI
1.
(d)
2. (b)
3. (a)
4 (d)
5.
(c)
6. (b)
7. (d)
8. (d)
9.
(b)
10. (b)
11. (c)
12. (d)
13.
(c)
14. (b)
15. (b)
16. (c)
17.
(d)
18. (c)
19. (d)
20. (b)
76
Chemistry
Melting and Boiling Points : These metals have very high melting and boiling points due
to stronger metallic bonding. The melting points of the transition elements first rise to a maximum
and then fall as the atomic number increases Manganese and technetium have abnormally low
melting points.
In a particular series, the metallic bond strength increases upto the middle with increasing
number of unpaired electrons. Thus Cr, Mo and W have maximum number of unpaired d-electrons
and have highest melting points in their respective series. Tungsten has the highest melting point
(3683 K) among the d-block elements.
As there are no unpaired electrons in Zn, Cd and Hg, they have low melting points. Hg is
liquid at ordinary temperature with lowest melting point (234 K) among the transition metals.
Ionization Enthalpies : The first ionization enthalpies of d-block elements are higher than
those of s-block elements and are lesser than those of p-block elements. The ionization enthalpies
of 3d and 4 d-series are irregular but increase across the series while those 5d-series are much
higher than 3d and 4 d-elements. This is because of the weak shielding of nucleus by 4 f electrons
in 5d-transition series which result in greater effective nuclear charge acting on the outer valence
electrons.
Electrode Potentials and Reducing Character : Quantitatively the stability of transition
metal ions in different oxidation states in solution can be determined on the basis of electrode
potential data. The lower the electrode potential (i.e., more negative the standard reduction potential)
of the electrode, more stable is the oxidation state of the transition metal ion in aqueous solution.
Electrode potential values depend upon energy of sublimation of the metal, the ionization enthalpy
and the hydration enthalpy.
Oxidation States : All transition elements except the first and the last member in each
series show variable oxidation states. This is because difference of energy in the (n–1) d and ns
orbitals is very little. Hence electrons from both the energy levels can be used for bond formation.
(i) The highest oxidation states of transition metals are found in fluorides and oxides since
fluorine and oxygen are the most electronegative elements. The highest oxidation state shown by
any transition elements is + 8. Osmium shows highest oxidation state of + 8 in OsO4. Oxygen is
superior to fluorine in showing higher oxidation state because it has the ability to form multiple
bonds with metals.
(iv) Lower oxidation states (zero or + 1) are stabilized by ligands which can accept electrons
from the metal through -bonding (Such as CO), i.e., with -acceptor ligands.
(v) In going down a group, the stability of higher oxidation states increases while that of lower
oxidation states decreases.
(vii) The relative stability of different oxidation states of transition metal atoms can be
determined with the help of standard electrode potential data. For example, E° Values for the
77
Chemistry
couples Cr3+/Cr2+ = 0.41 V, Mn3+/Mn2+ = + 1.57V suggests that Cr2+ is unstable and is oxidised to
Cr3+ (which is more stable) and acts as a reducing agent whereas Mn3+ is unstable and is reduced
to Mn2+ (which is more stable) and acts as an oxidising agent. It may be noted that both Cr2+ and
Mn3+ are d4 species.
Catalytic Properties : Many transition metals (like Co, Ni, Pt Fe, Mo etc.) and their compounds
are used as catalysts because of the following reasons.
(i)
Transition metal ions change their oxidation states e.g.
Fe3+ Catalyses the reaction between 1– and S2O2–4 ions
2Fe3+ + 2I–  2Fe2+ + I2
2Fe2+ + S2O82–  2Fe3+ + 2SO42–
(ii)
Because of variable oxidation states, they easily combine with one of the reactants to form
intermediate which reacts with the second reactant to form the final products.
(iii)
They have tendency to adsorb reactants on the surface. This has the effect of increasing the
concentration of reactants and weakening of bonds in reactants.
Coloured Ions: Most of the transition metal compounds are coloured both in the solid state
and in aqueous solution. This is because of the presence of incompletely filled d-orbitals.
When white light falls on these compounds, some wavelength is absorbed for promotion of
electrons from one set of lower energy orbitals to another set of slightly higher energy within the
same d-subshell. This is called d-d transition. The remainder light is reflected which has a particular
colour.
Colours of Cr2O72–, CrO42–, MnO–4, MnO42–, Cu2O, AgBr, AgI are due to charge transfer
transitions.
Compounds of s and p-block elements are generally white as high energy is required for
promotion of s and p-electrons of incomplete subshells.
Magnetic Properties : Due to the presence of unpaired electrons in the (n-–1) d-orbitals, the
most of the transition metal ions and their compounds are paramagnetic i.e., they are attracted by
the magnetic field. As the number of unpaired electrons increases from 1 to 5, the magnetic
moment and hence paramagnetic character also increases. Those transition elements which have
paired electrons are diamagnetic i.e., they are repelled by the magnetic field.
The magnetic moment of species depends upon the sum of orbital and spin contributions for
each unpaired electron present. In transition metal ions, the orbital magnetic moment is largely
suppressed or quenched by the electrostatic field of other atoms, ions or molecules surrounding the
metal ion. Thus the effective magnetic moment arises mainly from the spin of electrons.
A paramagnetic substance is characterised by its effective magnetic moment () which is
calculated by using spin-only formula µ 
n  n  2  B. M .
where n is the number of unpaired electrons and B.M. stands for Bohr magneton.
Complex Formation : Transition metal ions form a large number of complexes in which the
central metal ion is linked to a number of ligands. This is because of the following characteristic
properties of transition metal ions :
(i)
(ii)
They have high nuclear charge and small size i.e., charge/size ratio (charge density is large.
Thus in complexes, metal ions behave as Lewis acids and the ligands as Lewis bases.
78
Chemistry
Formation of Interstitial Compounds : Transition metals form a number of interstitial
compounds in which small non-metal atoms such as H, C, B, N and He occupy the empty spaces
(interstitial sites) in their lattices and also form bonds with them.
Alloy Formation : Due to similarity in atomic sizes, atoms of one transition metal can easily
take up positions in the crystal lattice of the other in the molten state and are miscible with each
other forming solid solutions and smooth alloys on cooling. Alloys are generally harder, have higher
melting points and more resistant to corrosion than the individual metals.
Some Important Compounds of Transition Elements
(1) Potassium Dichromate (K2Cr2O7)
Preparation : It is prepared from chromite ore (Fe2Cr2O4 or FeO.Cr2O3) through the following
reactions :
4FeCr2O4 + 16NaOH + 7O2  8Na2CrO4 + 2Fe2O3 + 8H2O
2Na2CrO4 + H2SO4  Na2Cr2O7 + Na2SO4 + H2O
2Na2Cr2O7 + 2KCl  K2Cr2O7 + 2NaCl
(ii)
In the solution, dichromate ions (Cr 2O 72–) exist in equilibrium with chromate ions
(CrO42–) as follows :
–
OH
2



Cr2O72   H2O 
 2CrO4  2H
H+
Yellow
Orange
In alkaline solution, equilibrium shifts in the forward direction and the solution is yellow. In
the acidic medium, equilibrium shifts in the backward direction and the solution is orange. This
conversion takes place with change in pH of the solution
2. Potassium Permanganate (KMnO4)
Preparation : It is prepared from pyrolusite ore (MnO 2) through the following reactions :
2MnO2 + 4KOH + O2  2K2MnO4 + 2H2O
(green)
MnO42– undergoes disproportionation in a neutral or acid medium to give KMnO 4.
3MnO42– + 4H+  2MnO4– + MnO2 + 2H2O.
(green)
(purple)
(b) Electrolytic oxidation : This is the most preferred method. The manganate solution is
electrolysed. The electrode reactions are as follows :
At anode : MnO24 –  MnO4–  e 
Green
At cathode : 2H+ + 2e–  H2.
Purple
The solution is filtered and evaporated to get deep purple black crystals of KMnO 4.
OXIDISING ACTION OF K2Cr2O7 AND KMnO4
Cr2O72– + 14H+ + 6e–  2Cr3+ + 7H2O (acidic medium),
MnO4– + e–  MnO42– (highly alkaline medium),
79
E = 1.33 V
E = 0.56 V
Chemistry
MnO4– + 2H2O + 3e–  MnO2 + 4OH– (alkaline medium),
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O (acidic medium),
E = 1.69 V
E = 1.52 V
Acidified K2Cr2O7 Oxidises
(i)
(ii)
Iodides to iodine : 2l  I2 + 2e–
Surphides to sulphur : H2S  S + 2H+ + 2e–.
(iii)
tin (II) to tin (IV) : Sn2+  Sn4+ + 2e–
(iv)
Iron (II) to iron (III) : Fe2+  Fe3+ + e–
Acidified KMnO4 Oxidises
(i)
(ii)
Iodides to iodine : 2I–  I2 + 2e–
Iron (II) to iron (III) : Fe2+  Fe3+ + e–
(iii)
Hydrogen sulphides to sulphur : H2S  2H+ + S + 2e–
(iv)
Oxalate iron or oxalic acid to CO2 : C2O42–  2CO2 + 2e–
(v)
Sulphites to sulphate : SO32– + H2O  SO42– + 2H+ + 2e–
(vi)
(vii)
Nitrite to nitrate : NO2– + H2O  NO3– + 2H+ + 2e–
hydrogen peroxide to O 2 : H2O2  2H+ + O2 + 2e–
Neutral or Faintly Alkaline KMnO4 Oxidises
(i)
(ii)
Iodide to iodate : I– + 6OH–  IO3– + 3H2O + 6e–
Thiosulphate to sulphate : S2O32– + 10.OH–  2SO42– + 5H2O + 8e–
Lanthanoids and Actinoids
(a)
Lanthanoids : The elements with atomic numbers 58 to 71 i.e., cerium to letetium (which
come immediately after lanthanum Z = 57) are called lanthanoids. These elements involve
the filling of 4f-orbitals. Their general electronic configuration is [Xe]4f1–14 5d0–16s2
Promethium (Pm), At. No. 61 is the only synthetic radioactive lanthanide.
Lanthanoid contraction is due to imperfect shielding of 4f electrons. This causes the radii
of the members of third transition series to be very similar to those of the corresponding
members of second transition series.
Oxides and hydroxides of lanthanoids are basic in nature. Trivalent lanthanoid ions are
coloured (except La 3+ and Lr3+) due f – f transition which occurs by the absorption of visible
light.
(b)
Actionoids : The elements with atomic numbers 90 to 103 i.e., thorium to lawrencium
(which come immediately after actinium, Z = 89) are called actinoids or actinides.
These elements involve the filling of 5 f-orbitals. Their general electronic configuration is
[Rn] 5f1–14 6d0–1 7s2.
5f orbitals in actinoids penetrate less into the inner core of electrons than 4f orbitals in
lanthanoids, i.e., 5f electrons have poorer shielding effect than 4f electrons. Hence actioid
contraction from element to element is more than lanthanoid contraction.
5f orbitals are not as buried as 4f orbitals. Hence 5f electrons participate in bonding to a
far greater extent in actinoids compared with lanthanoids resulting in the participation of
5f electrons of actinoids in bond formation.
80
Chemistry
UNIT – 16
1.
2.
3.
4.
5.
6.
7.
MnO42– (1 mole) in neutral aqueous medium disproportionates to
(a)
2/3 mole of MnO4– and 1/3 mole of MnO2
(b)
1/3 mole of MnO4– and 2/3 mole of MnO2
(c)
1/3 mole of Mn2O7 and 1/3 mole of MnO2
(d)
2/3 mole of Mn2O7 and 1/3 mole of MnO2
In which of the following pairs of ions the lower oxidation state in aqueous solution is more
stable than the other?
(a)
Tl+, Tl3+
(b)
Cu+, Cu2+
(c)
Cr2+, Cr3+
(d)
V2+, VO2+
The basis character of the transition metal monoxide follows the order
(a)
VO > CrO > TiO > FeO
(b)
CrO > VO > FeO > TiO
(c)
TiO > FeO > VO > CrO
(d)
TiO > VO > CrO > FeO
When MnO2 is fused with KOH, a coloured compound is formed. The product and its
colour is
(a)
K2MnO4, purple green
(b)
KMnO4, purple
(c)
K2MnO4 brown
(d)
Mn2O4, black.
Lanthanoids are
(a)
14 elements in the sixth period (atomic number 90 to 103) that are filling 4f sub
level.
(b)
14 elements in the seventh period (atomic number 90 to 103) that are filling 5f
subshell.
(c)
14 elements in the sixth period (atomic number 58 to 71) that are filling the 4f
subshell.
(d)
14 elements in the seventh period (atomic number 58 to 71) that are filling the 4f
subshell.
The product of oxidation of I– with MnO4– in alkaline medium is
(a)
IO3–
(b)
I2
(c)
IO–
(d)
IO4–
Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statements
about cerium is incorrect?
(a)
The common oxidation sates of cerium are +3 and +4.
81
Chemistry
8.
9.
10.
11.
12.
13.
14.
(b)
The +3 oxidation state of cerium is more stable than +4 oxidation state.
(c)
The +4 oxidation state of cerium is not known in solutions.
(d)
Cerium (IV) acts as an oxidizing agents.
Lanthanoids and actinoids resemble in
(a)
electronic configuration
(b)
Oxidation state
(c)
ionization energy
(d)
formation of complexes.
For decolorisation of 1 mole of KMnO 4, the moles of H2O2 required is
(a)
1/2
(b)
3/2
(c)
5/2
(d)
7/2
The value of ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM.
The correct one is
(a)
d4 (in strong ligand field)
(c)
d3 (in weak as well as strong fields)
(d)
d5 (in strong ligand field)
(b)
d4 (in week ligand field)
Which of the following factor may be regarded as the main causes of lanthanoide contraction?
(a)
Poor shielding of one 4f-electron by another in the subshell.
(b)
Effective shielding of one 4f-electrons by another in the subshell.
(c)
Poorer shielding of 5d-electrons by 4f-electrons.
(d)
Greater shielding of 5d-electrons by 4f-electrons.
The aqueous solution containing which one of the following ions will be coloured :
(a)
CuCl
(b)
K3[Cu(CN)4]
(c)
CuF2
(d)
[Cu(CH2CN)4]BF4
The spin only magnetic moment value in Bohr magnition units of Cr(CO)6 is
(a)
6
(b)
2.84
(c)
4.90
(d)
5.92
For the reduction of NO 3– ion in an aqueous solution E° is 0.96V. Values of E° for some
metal ions are given below.
V3+(aq) + 2e–  V(s)
Fe3+(aq) + 3e–  Fe(s)
Au3+ (aq) + 3e–  Au(s)
Mg2+ (aq) + 2e–  Hg(l)
The metal/metals that can be oxidised by NO3– in aqueous medium are
(a)
V and Au
(b)
Fe and Au
(c)
Hg and Au
(d)
V, Fe and Hg
82
Chemistry
15.
16.
17.
18.
19.
20.
The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is
(a)
4
(b)
6
(c)
2
(d)
7
The number of unpaired electrons in gaseous species of Mn 3+, Cr3+ and V3+ respectively are
__________ and most stable species is __________.
(a)
4, 3, 2 and V3+ is most stable
(b)
3, 3 and 2 and Cr3+ is most stable
(c)
4, 3 and 2 and Cr3+ is most stable
(d)
3, 3 and 3 and Mn3+ is most stable.
Which is not correct statement about the chemistry of 3d and 4f series elements?
(a)
3d elements show more oxidation states than 4f series elements
(b)
The energy difference between 3d and 4s orbitals is very little.
(c)
Europium (II) is more stable than Ce(II)
(d)
The paramagnetic character in 3d series elements increases from scandium to copper.
A mixture of salts (Na2SO3 + K2Cr2O7) in a test tube is treated with dil H2SO4 and resulting
gas is passed through lime water. Which of the following observations is correct about this
test?
(a)
Solution in test tube becomes green and water turns milky.
(b)
Solution in test tube is colourless and lime water turns milky.
(c)
Solution in test tube becomes green and lime water remains clear.
(d)
Solution in test tube remains clear and lime water also remains clear.
Which one of the elements with the outer orbital configuration may exhibit the largest
number of oxidation states?
(a)
3d54s1
(b)
3d54s2
(c)
3d24s2
(d)
3d34s2
Out of TiF62–, CoF63–, Cu2Cl2 and NiCl42– the colourless species are
(Z of Ti = 22, Co = 27, Cu = 29, Ni = 28)
21.
(a)
Cu2Cl2 and NiCl42–
(b)
TiF62– and Cu2Cl2
(c)
CoF63– and NiCl42–
(d)
TiF62– and CoF63–
Knowing that chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which
of the following statement is incorrect?
(a)
Because of the large size of the Ln(III) ions the bonding in its compounds is
predominantly ionic in character.
(b)
The ionic sizes of Ln(III) decrease in general with increasing atomic number.
(c)
Ln(III) compounds are generally colourless
(d)
Ln(III) hydroxides are mainly basic in character.
83
Chemistry
22.
23.
24.
25.
Large number of oxidation states are exhibited by the actinoids than those by the lanthanoids,
the main reason being
(a)
more energy difference between 5f and 6d than between 4f and 5d orbitals.
(b)
more reactive nature of the actionoids than the lanthanoids.
(c)
4f orbitals are more diffused than 4f orbitals
(d)
lower energy difference between 5f and 6d than between 4f and 5d orbitals.
Amount of oxalic acid present in a solution can be determined by its titration with KMnO4
solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out
in the presence of HCl because HCl
(a)
reduces permanganate to Mn(II)
(b)
Oxidises oxalic acid to carbon dioxide and water
(c)
gets oxidised by oxalic acid to chlorine
(d)
furnishes H+ ions in addition to these from oxalic acid.
The correct order of decreasing second ionization enthalpy of Ti(22), Cr(24) and Mn(25) is
(a)
Cr > Mn > V > Ti
(b)
V > Mn > Cr > Ti
(c)
Mn > Cr > Ti > V
(d)
Ti > V > Cr > Mn
Identify the incorrect statement among the following:
(a)
4f and 5f orbitals are equally shielded
(b)
d-block elements show irregular and erratic chemical properties among themselves
(c)
La and Lu have partially filled d orbitals and no other partially filled orbitals
(d)
The chemistry of various lanthanoids is very similar.
1.
(a)
2. (a)
3. (d)
4. (a)
5.
(c)
6. (a)
7. (c)
8. (b)
9.
(c)
10. (a)
11. (a)
12. (c)
13.
(a)
14. (d)
15. (b)
16. (c)
17.
(d)
18. (a)
19. (b)
20. (b)
21.
(c)
22. (d)
23. (a)
24. (a)
25.
(a)
84
Chemistry
Consider the following coordination compound
[Co(H2O)Cl(en)2]Cl2
Here,

Co (Cobalt) : is the central metal atom/ion

(en)2, (H2O), (Cl–) are neutral didentate, neutral unidentate and anionic unidentate ligands
respectively.

Ligands and central metal atom/ion are bonded through coordinate bonds to from coordination
sphere, which is enclosed within square brackets.

Coordination number of a metal atom is number of  bonds between metal and ligands. e.g.,
in [Co(en)2(H2O)Cl]Cl2, coordination number of Co is six.

Cl– is a counter ion which is outside the coordination sphere.
IUPAC Nomenclature of Coordination Compounds
(i)
(ii)
Name of the cation is written before the name of the anion.
Within a complex ion or coordination entity, names of ligands are cited before the metal
atom/ion.
(iii)
Identical ligands are represented by prefixing di, tri, tetra etc. to their name.
(iv)
Different ligands are reported in alphabetical order.
(v)
(vi)
When the name of ligands includes a numerical prefix like mono or di. Their total number
is represented by prefixing bis for di, tris for tri, tetrakis for tetra and so on and the ligand
to which it refers is enclosed in parenthesis.
Name of anionic ligand ends in o
Cl– = Chlorido
Br– = Bromido
C2O42– = Oxalato
H– = Hydrido
CN– = Cyanido
(vii)
EDTA4– = Ethylenediaminetetracetato
The names of neutral and cationic ligands are the same except NH3(ammine), CH3NH2
(methylamine/methanamine), H 2O (aqua), CO (carbonyl), NO (nitrosyl) and
CH2
NH2
CH2
NH2
(ethylenediamine/ethane-1,2-diamine).
85
Chemistry
(viii)
When the charge on coordination complex is negative name of central metal atom/ion ends
in ‘ate.’
Iron (ferrum) – Ferrate
Cobalt = Cobaltate
Silver (argentum) – argentate
Nickel = Nickelate
Gold (aurum) – aurate
(ix)
(x)
Oxidation state of metal atom is expressed in Roman numeral and is written following its
name and enclosed in parenthesis.
Name of coordinate entity is a single word.
For e.g. [Co(H2O) Cl(en)2] Cl2 will be named as aquachloridobis (ethane-1, 2-diamine) cobalt
(III) chloride.
Isomerism in Coordination Compounds
(i)
Ionization Isomerism : This isomerism results when compounds with same molecular
formula give different ions in the solution e.g., {Co (NH3)5Br]2+ SO42– and [Co(NH3)5(SO4)]+
Br–. Here counter ion itself is a potential ligand.
(ii)
Solvate Isomerism : It arises when H2O molecule acting as ligand becomes water of
hydration outside the coordination sphere, i.e., here water forms a part of the coordination
entity or is outside it. For example, there are three hydrate isomers of CrCl 3.6H2O as shown
below :
[Cr(H2O)6]Cl3, [Cr(H2O)5Cl] Cl2.H2O, and [Cr(H2O)4Cl2] Cl.2H2O.
(iii)
Linkage Isomerism : This isomerism results when a ligand attaches with central atom/
ion in two different ways. Some common examples of such ligands are : nitrito–N–(–NO2)
and nitrito–N–(–ONO), cyano (–CN) and isocyano (–NC), thiocyanato (–SCN), and
isothiocyanato (–NCS), cyanato (–OCN) and isocyanato (–NCO).
Some example of linkage isomers are : [Co(NO 2) (NH3)5]Cl2 and [Co(ONO)(NH3)5]Cl2;
[Cr(H2O)5(SCN)]2+ and [Cr(H2O)5 (NCS)]2+
(iv)
Coordination Isomerism : This type of isomerism is shown by compounds in which both
cation as well as anion are complexes (coordination entities) and ligands may interchange
their position between the two complex ions. For example,
[CO(NH3)6] [Cr(C2O4)3] and [Cr(NH3)6] [Co(C2O4)3]
(v)
Geometrical Isomerism
(a)
Isomerism in complexes with coordination number 4 : Tetrahedral complexes
do not show geometrical isomerism because the relative position of the ligands with
respect to each other will be the same. The square planar complexes of the type
MA2X2, MA2XY, MA2BX and MABXY show this kind of isomerism, e.g., [Pt(NH3)2Cl2].
(b)
Isomerism in complexes with coordination number 6 : Octahedral complexes
of the type : MA4X2, MA3X3, MA4XY exist as cis-and trans-isomers. Common examples
are : [Co(NH3)4Cl2], [Co(NH3)4Cl(NO2)]
86
Chemistry
Complexes of the type : [M(AA) 2X2] and [M(AA)2XY] also show cis-trans isomers
(Here (AA) refers to symmetrical bidentate ligands and X and Y refer to anionic
ligands. Common examples are : [Co(en)2 Cl2]+, [Ni(OX)2Cl2]+ etc.
(ii)
Optical Isomerism : Chiral molecules i.e., molecules which do not have plane of symmetry,
exhibit optical isomerism. The optically active isomers called enantiomers or enantiomorphs
are non superimposable on mirror images of each other and rotate the plane of polarised
light equally but in opposite directions.
If the substance rotates the light to the right, i.e., in clockwise direction, it is dextrorotatory
or the d- form or (+) form. If the light is rotated in the anti-clockwise direction, the substance
is said to be levorotatory or l- form or (–) form. The optical isomers of a compound have
identical physical and chemical properties.
(b)
Isomerism in complexes with coordination number 6 : Octahedral complexes of
the type [M(AA)X2Y2], [M(AA) 2X2] and [M(AA)2XY] where (AA) is symmetrical
bidentate ligand and X and Y are monodentate ligands, can be resolved into a pair
of enantiomers. As seen earlier, the complexes of these types show geometrical
isomerism i.e., exist as cis-and trans-isomers. The trans isomer does not show optical
isomerism since it is symmetrical while only cis-isomer show optical activity as it is
unsymmetrical. Thus total three isomers may be possible in such cases.
Some common examples of the above type of complexes are :
[CO(en)(NH3)2Cl2]+, [Co(en)2Cl2]+
(ii)
Valence bond theory (VBT) : This theory, put forward by Linus Pauling (1931), is used
to explain the shapes of complexes and relation between the observed magnetic behaviour
and the bond type. It involves the hybridization of empty non equivalent orbitals of the
metal atom/ion, each of which then can accept a lone pair of electrons donated by the
ligands. The shape is octahedral, tetrahedral, square, planar, trigonal bipyramidal (or square
pyramidal) or linear when hybridization is d2sp3, sp3, dsp2, dsp3 or sp respectively.
87
Chemistry
Complexes of some transition metal ions/atoms on the basis of V.B. Theory.
Complex
Metal
Ion
[V(H2O)63]+
V3+
Cr(NH3)63+
Mn(CN)63–
Fe(CN)63–
Fe(CN)64–
[FeF6]3–
[Co(NH3)6]3+
[CoF6]3–
[Ni(CN)4]2–
Cr3+
Mn3+
Fe3+
Fe2+
Fe3+
Co3+
Co3+
Ni2+
Electron distribution
3s
3p
3d
2
2
2
2
2
2
2
2
2
Type of
Hybridisation
No. of
unpaired
2
d2sp3
2
6
3
d2sp3
3
6
4(le–
d2sp3
2
pair up)
d2sp3
1
pair up)
d2sp3
0
sp3d2
5
d2sp3
0
sp3d2
4
dsp 2
0
sp3
0
sp
0
dsp2
1
6
6
6
6
6
6
6
4s
4p
4d
pairs up)
5
(2e–
6
(2e–
5
6
(2e–
[pair up)
6
8
(1e–
pairs up)
[Ni(CO)4]
Ni
2
6
8
[Cu(NH3)2]+
[Cu(NH3)4]2+
Cu+
2
6
10
Cu2+
2
6
8
0
1
(1e– promoted to 4p)
[Cu(CN)4]3–
Cu+
2
6
10
sp3
0
[Zn(NH3)4]2+
Zn2+
10
sp3
0
(iii)
2
6
2 (Jump to 3d)
0
Crystal Field Theory (CFT) : Crystal field theory is based on the assumption that the
metal ion and the ligands act as point charges and the interaction between them is purely
electrostatic, i.e., metal-ligands bonds are 100% ionic. The five d-orbitals is an isolated
gaseous metal atom/ion are degenerate. This degeneracy is maintained in a spherically
symmetrical field of negative charges.
However, when this negative field is due to real ligands in a complex, the degeneracy of the
d-orbitals is lifted due to asymmetrical field. This results in splitting of the d-orbital energies.
The pattern of splitting depends upon (a) nature of the crystal field such as octahedral,
tetrahedral of square planar and (b) basic strength of ligands.
Crystal field effects in octahedral coordination entities : In an octahedral complex,
the six ligands approach the central metal atom lying at the origin symmetrical along the
dx2 –y2 , dz2 (eg)
0.6 0
Energy
1.
0 or 10Dq
–0.4 0
d-orbitals in free ion
Average energy of
dxy, dyz, dxz (t2g )
d-orbitals in spherical Splitting of d-orbitals in
crystal field
an octahedral crystal field
Splitting of five d-orbitals in an octahedral crystal field.
88
Chemistry
cartesian axes. Initially, there is an increase in the energy of d-orbitals relative to that of
the free ion (just like that in a spherical field). Next, the orbitals lying along the axes (dxy,
dyz and dxz) which are lowered in energy relative to the average energy in the spherical
crystal field as shown below :
This splitting of five degenerate d-orbitals of the metal ion into two sets of d-orbitals with
different energies is called crystal field splitting. The two sets of d orbitals, i.e., dx 2 – y2 and dz 2
and dxy , dyz and dzx are commonly called eg and t2g orbitals respectively. The crystal field splitting
is the energy difference between t2g and eg orbitals and is frequently measured in terms of a
parameter 0 where the subscript (0) stands for octahedral. (This is also measured in terms of
another parameter called Dq. The magnitude of splitting is arbitrarily fixed at 10Dq so that
(0 = 10Dq). The above energy level diagram shows that an electron will prefer to go into more
stable t2g orbital is stabilized by –0.40 (–4Dq) energy while each electron occupying the eg orbital
shall be destabilized by an amount 0.60(6Dq) energy. This gain in energy achieved by preferential
filling of stable t2g orbitals over the energy of a randomly filled d-orbital is called the crystal field
stabilization energy (CFSE).
The net CFSE is equal to (6x – 4y) Dq where x and y are the number of electrons in eg and
t2g orbitals respectively. The actual configuration adopted is decided by the relative value of 0 and
P (where P is the pairing energy). The energy required to cause pairing of electrons in the same
orbital is called pairing energy. If 0 < P, we have a weak field and high spin or spin free complex
and the fourth electron will enter one of the eg orbitals giving the configuration t2g3eg1. If now a fifth
electron is added to a weak field coordination entitity, the configuration becomes t2g3e2g. The
pairing of electrons will take place only if the gain in stability in terms of 0 is large enough to
overcome the loss in stability due to electron pairing. When 0 > P, we have a strong field and low
spin or spin paired complex and pairing will occur in the t2g level with eg level remaining unoccupied
in d1 to d6 systems.
89
Chemistry
1.
2.
3.
4.
5.
6.
7.
8.
9.
Which of the following is an organometallic compounds?
(a)
Ti(C2H4)4
(b)
Ti(OC2H5)
(c)
Ti(O COCH3)4
(d)
Ti (OC6H5)4
The oxidation state of Fe in the brown ring complex [Fe(H 2O)5 (NO)] SO4 is
(a)
+1
(b)
+2
(c)
+3
(d)
+4
In the compound, lithiumtetrahydrido aluminate (III), the ligands is
(a)
H+
(b)
H
(c)
H–
(d)
None of these
IUPAC name of [Pt(NH3)3 Br(NO2) Cl] Cl is
(a)
Triamminechloroidobromidonitrito-N-platinum (IV) chloride)
(b)
Triamminechloridobromidonitrito-N-platinum (IV) chloride
(c)
Triamminebromidochloridonitrito-N-platinum (IV) chloride
(d)
Triamminenitrito-N-chloridobromidoplatinum (IV) chloride.
The geometry of [Ni(CO) 4] and [Ni(PPh3)2Cl2] are
(a)
both square planar
(b)
tetrahedral and square planar respectively
(c)
both tetrahedral
(d)
square planar and tetrahedral respectively.
Which one of the following complexes will have four isomers?
(a)
[Co(en) (NH3)2Cl2]Cl
(b)
[Co(PPh3)2 (NH3)2Cl2]Cl
(c)
[Co(en)3]Cl3
(d)
[Co(en)2Cl2] Br (en = ethylenediamine)
The correct structure of Fe(CO)5 is
(a)
octahedral
(b)
tetrahedral
(c)
square pyramidal
(d)
trigonal bipyramidal.
Which of the following has magnesium?
(a)
Chlorophyll
(b)
Haemocyanin
(c)
Carbonic anhydrase
(d)
Vitamin B12
Which of the following shall form an octahedral complex?
(a)
d4 (low spin)
(b)
d8 (high spin)
(c)
d6 (low spin)
(d)
All of these
90
Chemistry
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
Which of the following is expected to be a paramagnetic complex?
(a)
[Ni(H2O)6]2+
(b)
[Ni(CO)4]
(c)
[Zn(NH3)4]2+
(d)
[Co(NH3)6]3+
Which of the following will give maximum numbers of isomers?
(a)
[Co(NH3)4Cl2]+
(b)
[Ni(en)(NH3)4]2+
(c)
[Ni(C2O4) (en)2]2–
(d)
[Cr(SCN)2(NH3)4]+
Which of the following organometallic compound is  and  bonded?
(a)
[Fe(n5 – C5H5)2]
(b)
K[PtCl3(n2–C2H4)]
(c)
[Co(CO)5(NH)3]2+
(d)
Fe(CH3)3
The complex which has no ‘d’ electrons in the central metal atom is
(a)
[MnO4]–
(b)
[Co(NH3)6]3+
(c)
[Fe(CN)6]3–
(d)
[Cr(H2O)6]3+
Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1
complex of BF3 and NH3
(a)
N : tetrahedral, sp3; B : tetrahedral, sp3
(b)
N : pyramidal, sp3; B : pyramidal sp3
(c)
N : pyramidal, sp3; B : planar, sp2
(d)
N : pyramidal, sp3; B : tetrahedral, sp3.
Among the following the most stable complex is
(a)
[Fe(H2O)6]3+
(b)
[Fe(CN)6]3–
(c)
[Fe(C2O4)3]3–
(d)
[FeCl6]3–
Among [Ni(CO)4], [Ni(CN)4]2–, [NiCl4]2– species, the hybridisation states at Ni atom are
respectively
(a)
sp3, dsp2, dsp2
(b)
sp3, dsp2, sp3
(c)
sp3, sp3, dsp2
(d)
dsp2, sp3, sp3
Considering H2O as a weak field ligand, the number of unpaired electrons in [Mn(H2O)6]2+
will be (atomic number of Mn = 25]
(a)
3
(b)
5
(c)
2
(d)
4
In Fe(CO)5, the Fe – CO bond possesses
(a)
ionic character
(b)
 – character only
(c)
 – character only
(d)
both  and  characters
How many EDTA (ethylenediaminetetraacetate) ions as ligands are required to make an
octahedral complex with a Ca2+ ion.
(a)
one
(b)
91
two
Chemistry
(c)
20.
21.
three
[C o(N H 3)4(NO2)2Cl]
(d)
four
exhibits
(a)
ionisation isomerism, geometrical isomerism and optical isomerism
(b)
linkage isomerism, geometrical isomerism, and optical isomerism
(c)
linkage isomerism, ionisation isomerism and optical isomerism
(d)
linkage isomerism, ionisation isomerism and geometrical isomerism.
One mole of coordination compound, Co(NH 3)5Cl3 gives three moles of ions on dissolution
in water. One mol of the same complex reacts with two moles of AgNO 3 solution to yield two
moles of AgCl(s). The structure of the complex is :
(a)
[Co(NH3)3Cl3].2NH3
(b)
[Co(NH3)4Cl2] Cl.NH3
(c)
[Co(NH3)4Cl] Cl2.NH3
(d)
[Co(NH3)5Cl] Cl2
1.
(a)
2. (b)
3. (c)
4. (c)
5.
(c)
6. (d)
7. (d)
8. (a)
9.
(c)
10. (a)
11. (d)
12. (b)
13.
(a)
14. (a)
15. (c)
16. (b)
17.
(b)
18. (d)
19. (a)
20. (d)
21.
(d)
92
Chemistry
1.
2.
3.
The total volume of atoms present in face-centered cubic unit cell of a metal is (r is atomic
radius)
(a)
20r3
3
(b)
24r3
3
(c)
12r3
3
(d)
16r3
3
An ionic compound has unit cell consisting of A ions at the corners of a cube and B ions at
the centres of the faces of the cube. The empirical formula of the compound would be
(a)
A3B
(b)
AB3
(c)
A2B
(d)
AB
What type of crystal defect is indicated in the diagram below?
Na+ Cl– Na+ Cl– Na+ Cl–
Cl –
Cl – Na 
Na  Cl –
Na 
Cl – Na  Cl –
Cl – Na  Cl – Na 
4.
5.
6.
Na 
(a)
Frenkel and schottky defect
(b)
Schottky defect
(c)
Interstitial defect
(d)
Frenkel defect.
Which among the following will show maximum osmotic pressure?
(a)
1M NaCl
(b)
1M MgCl2
(c)
1M (NH4)3 PO4
(d)
1M Na2SO4
If  is the degree of dissociation of Na2SO4, the vant Hoff ’s factor (i) used for calculating
the molecular mass is
(a)
1 + 
(b)
1 – 
(c)
1 + 2
(d)
1 + 2
At certain hill station pure water boils at 99.725°C. If Kb for water is 0.513°C Kg mol–1, the
boiling point of 0.69 m solution of urea will be
(a)
103°C
(b)
100.079°C
(c)
100.359°C
(d)
unpredictable.
93
Chemistry
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
If all the following four compounds were sold at same price, which would be the cheapest
for preparing an antifreeze solution for a car radiator?
(a)
CH3OH
(b)
C2H5OH
(c)
C2H4(OH)2
(d)
C3H5(OH)3
The molar solution of sulphuric acid is equal to
(a)
1 N solution of H2SO4
(b)
2N solution of H2SO4
(c)
N/2 solution of H2SO4
(d)
3N solution of H2SO4
A tooth paste has 0.2 g L–1 fluoride. Its concentration in ppm will be :
(a)
250
(b)
200
(c)
400
(d)
1000
Solution boils at a temperature T1 and the solvent at a temperature T2, the elevation of
boiling point is given by
(a)
T1 + T2
(b)
T1 – T2
(c)
T2 – T1
(d)
T1 + T2
If salt bridge is removed from two half cells, the voltage
(a)
Drops to zero
(b)
Does not change
(c)
Increase gradually
(d)
Increases rapidly
A smuggler could not carry gold by depositing iron on the gold surface because
(a)
gold is denser
(b)
iron rusts
(c)
gold has higher reduction potential than iron
(d)
gold has lower reduction potential than iron.
The charge required for the reduction of 0.4 mol of K 2Cr2O7 to Cr3+ ions is
(a)
0.6 × 96500 C
(b)
2.4 × 96500 C
(c)
6 × 96500 C
(d)
12.4 × 96500 C
A cell is constituted as follows Pt, H2(1 atm.) |HA1||HA2|H2(1 atm), Pt. The pH of two
acids solutions HA1 and HA2 are 3 and 5 respectively. The emf of the cell is
(a)
0.059 V
(b)
0.0295 V
(c)
0.118 V
(d)
–0.118 V
o
The values of m° for KCl and KNO3 are 149.86 and 154.96 –1 cm2 mol–1. Also Cl
– is
–1
2
–1
–
71.44 ohm cm mol . The value of °(NO3 ) is
(a)
76.54 ohm– 1 cm2 mol–1
(b)
133.08 ohm–1 cm2 mol–1
(c)
37.7 ohm–1 cm2 mol–1
(d)
unpredictable.
Which of the following is a primary cell?
(a)
Lead storage battery
(b)
Nickel–Cadmium cell
(c)
Mercury cell
(d)
Both (a) and (b)
94
Chemistry
17.
18.
19.
The units of rate constant for first order reaction is
(a)
s–1
(b)
mol L–1 s–1
(c)
mol s–1
(d)
L mol–1 s–1
A first order reaction is 75% complete after 32 minutes. When was 50% of the reaction
completed?
(a)
4 minutes
(b)
8 minutes
(c)
16 minutes
(d)
32 minutes
Which of the following represents zero order reaction :
(a)
t1/2
(b)
t1/2
[R]0
[R]0
(c)
t1/2
(d)
t1/2
[R]0
20.
21.
22.
[R]0
The rate constant of the reaction at temperature 200K is one-tenth of the rate constant at
400K. What is the activation energy of the reaction?
(a)
1842.4R
(b)
921.2R
(c)
460.6R
(d)
230.3R
For a first order reaction, the plot of log K against 1/T is a straight line. The slope of the
line is equal to
(a)
–
Ea
R
(b)
–
2.303
Ea R
(c)
–
Ea
2.303
(d)
–
Ea
2.303 R
Select the rate law that corresponds to the data shown for the following reaction A + B  C.
Exp. No.
(A)/M
(B)/M
Initial rate/Ms–1
1.
0.012
0.035
0.1
2.
0.024
0.070
0.8
3.
0.024
0.035
0.1
4.
0.012
0.070
0.8
95
Chemistry
23.
24.
(a)
Rate = k [B]3
(b)
Rate = k [B]4
(c)
Rate = k [A] [B]3
(d)
Rate = k [A]2 [B]2
A redioactive isotope has a half life of 10 days. If today 125 mg is left over, what was its
original weight 40 days earlier?
(a)
2g
(b)
600 mg
(c)
1g
(d)
1.5g
The relationship between standard reduction potential of a cell and requilibrium constant
is given by
n
0.059
o
log k c
log k c
(a) Ecell 
(b) Eocell 
0.059
n
(c)
25.
26.
27.
28.
29.
30.
31.
Eocell  0.059 n log k c
(d)
Eocell 
log k c
n
The specific conductance of 0.1M HNO3 is 6.3 × 10–2 ohm–1 cm–1. The molar conductance of
the solution is
(a)
630 ohm–1 cm2 mol–1
(b)
315 ohm–1 cm2 mol–1
(c)
100 ohm–1 cm2 mol–1
(d)
63 ohm–1 cm2 mol–1
When mercuric iodide is added to an aqueous solution of KI, the
(a)
freezing point is decreases
(b)
freezing point in raised
(c)
boiling point does not change
(d)
freezing point does not change
Cloud or fog is colloidal system in which the dispersed phase and dispersion medium are
respectively
(a)
Gas, Liquid
(b)
Liquid, Gas
(c)
Liquid, Liquid
(d)
Solid, Liquid
When a river enters into the sea, a delta is formed. Formation of delta is due to
(a)
peptization
(b)
coagulation
(c)
Emulsification
(d)
Dialysis.
Gold number is the number of milligrams of the protective colloid which prevents the
coagulation of 10 mL of a gold hydrosol on adding 1 mL of a ____ solution of sodium chloride.
(a)
1%
(b)
5%
(c)
25%
(d)
10%
Which of the following enzymes catalyzes the conversion of proteins into amino acids?
(a)
Pepsin
(b)
Amylase
(c)
Nuclease
(d)
Carbonic anhydrase
During the process of electrolytic refining of copper some metals present as impurity settle
as anode mud? These are
(a)
Sn and Ag
(b)
Pb and Zn
(c)
Ag and Au
(d)
Fe and Ni
96
Chemistry
32.
Extraction of gold is done by the following process
Au + CN–  X
(X) + Zn  Y + A
The (X) and (Y) are
33.
(a)
[Au(CN)2]–, [Zn(CN)4]2–
(b)
[Au(CN)2]+, [Zn(CN)4]2–
(c)
[Au(CN)4]–, [Zn(CN)2]
(d)
[Au(CN)4]–, [Zn(CN)4]–
Which method of purification is represented by the following equations?
773K
1675K
Ti  2I2 
 TiI4  Ti  2I2
34.
35.
36.
37.
38.
39.
(a)
Cupellation
(b)
Poling
(c)
Van Arkel method
(d)
Zone refining
Ellingham diagram is used to explain
(a)
Kinetics of metallurgical operations
(b)
Thermodynamics of metallurgy
(c)
Transition temperature
(d)
None of these.
The correct order of the thermal stability of hydrogen halides (H–X)
(a)
HI > HBr > HCl > HF
(b)
HF > HCl > HBr > HI
(c)
HCl > HF > HBr > HI
(d)
HI > HCl > HF > HBr
The number of  bonds in P4O10 is
(a)
6
(b)
16
(c)
20
(d)
7
Nitrogen forms N2 but phosphorus is converted into P4. The reason for this is
(a)
Triple bond is present between phosphorus atoms.
(b)
p – p bonding is weak
(c)
p – p bonding is strong
(d)
Multiple bond is formed easily.
(NH4)2 Cr2O7 on heating libertes a gas. The same gas will be obtained by heating :
(a)
NH4 NO2
(b)
NH4NO3
(c)
H2O2 with NaNO2
(d)
Mg3N2 with H2O
The formation of O2+[PtF6]– is the basis for the formation of Xenon fluorides. This is because
(a)
O2 and Xe have comparable electronegativies
(b)
O2 and Xe have comparable ionization ethalpies
97
Chemistry
40.
(c)
Both O2 and Xe are gases
(d)
O2 and Xe have comparable sizes.
In the following reaction :
P4 + 3NaOH + 3H2O  PH3 + 3NaH2PO2
(a)
Phosphorus is oxidised
(b)
Phosphorus is oxidised and reduced
(c)
Phosphorus is reduced
(d)
Phosphorus is neutralised
1.
(d)
2. (b)
3. (b)
4. (c)
5.
(c)
6. (b)
7. (a)
8. (b)
9.
(b)
10. (b)
11. (a)
12. (c)
13.
(b)
14. (a)
15. (a)
16. (c)
17.
(a)
18. (c)
19. (a)
20. (b)
21.
(d)
22. (a)
23. (a)
24. (b)
25.
(a)
26. (b)
27. (b)
28. (b)
29.
(d)
30. (a)
31. (c)
32. (a)
33.
(c)
34. (b)
35. (b)
36. (b)
37.
(b)
38. (a)
39. (b)
40. (b)
98
Chemistry