Chemistry: Sample Solution Paper- 4, Class 12 Notes | EduRev

 Page 1


  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
CBSE 
Class XII Chemistry   
Sample Paper 4 - Solution 
 
Time: 3 Hrs                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
1. Interstitial defect 
2. Brownian movement  
OR 
 
       The scattering of light by colloidal particles is known as the Tyndall effect. 
 
3. The maximum oxidation state shown by actinoids is +7. 
OR 
       The transition element which does not exhibit variation in the oxidation state in its 
compounds is zinc (Zn). 
 
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing 
or –I effect of three  groups. Thus, it gives a salt with Na2CO3 which is a weak base. 
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3. 
 
5. Elastomers are polymers in which polymer chains are held together by weak 
intermolecular forces and hence are stretchable. Fibres are polymers with strong 
intermolecular forces like hydrogen bonding and hence possess high tensile strength.  
 
Section B 
 
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly 
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a 
measure of its solubility, then it can be said that the mole fraction of a gas in the 
solution is proportional to the partial pressure of the gas over the solution. 
           p = KH . x                                                                                                                 
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these 
beverages, carbon dioxide is passed at high pressure to increase its solubility. 
 
OR 
Using the relation, 
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ? 
Solving for m, we get m= 
b.p.(solution) b.p.(l)
b.p.
TT
k
?
      
 
     
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
 
-NO
2
Page 2


  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
CBSE 
Class XII Chemistry   
Sample Paper 4 - Solution 
 
Time: 3 Hrs                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
1. Interstitial defect 
2. Brownian movement  
OR 
 
       The scattering of light by colloidal particles is known as the Tyndall effect. 
 
3. The maximum oxidation state shown by actinoids is +7. 
OR 
       The transition element which does not exhibit variation in the oxidation state in its 
compounds is zinc (Zn). 
 
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing 
or –I effect of three  groups. Thus, it gives a salt with Na2CO3 which is a weak base. 
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3. 
 
5. Elastomers are polymers in which polymer chains are held together by weak 
intermolecular forces and hence are stretchable. Fibres are polymers with strong 
intermolecular forces like hydrogen bonding and hence possess high tensile strength.  
 
Section B 
 
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly 
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a 
measure of its solubility, then it can be said that the mole fraction of a gas in the 
solution is proportional to the partial pressure of the gas over the solution. 
           p = KH . x                                                                                                                 
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these 
beverages, carbon dioxide is passed at high pressure to increase its solubility. 
 
OR 
Using the relation, 
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ? 
Solving for m, we get m= 
b.p.(solution) b.p.(l)
b.p.
TT
k
?
      
 
     
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
 
-NO
2
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
7. Fe + 2H
+
 ? Fe
2+
 + H2 
                                                                                                                                                
                                                                                                
        
 
 
8. Hexagonal    
       Example: Graphite      
                                                                                         
9.  
(a) 
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ? 
(b) 
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ? 
 
 
10. Compound (a) reacts faster than compound (b).                                               
This is due to the formation of a more stable carbocation in compound (a) in the 
rate-determining step than  carbocation in compound (b).                                                                                                    
 
 
11.  The increasing order of the acidic character of the given species is as follows: 
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH 
OR 
(a) pent-2-enal 
(b) 3-phenyl prop-2-enol 
 
 
12.  
(a) Monomers of Dacron:  
            ethylene glycol  and terephthalic acid                                                                                                           
                                                                                                                                                   
  
(b) Monomers of Buna-N: 
               1, 3-butadiene and acrylonitrile                                                                     
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
Page 3


  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
CBSE 
Class XII Chemistry   
Sample Paper 4 - Solution 
 
Time: 3 Hrs                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
1. Interstitial defect 
2. Brownian movement  
OR 
 
       The scattering of light by colloidal particles is known as the Tyndall effect. 
 
3. The maximum oxidation state shown by actinoids is +7. 
OR 
       The transition element which does not exhibit variation in the oxidation state in its 
compounds is zinc (Zn). 
 
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing 
or –I effect of three  groups. Thus, it gives a salt with Na2CO3 which is a weak base. 
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3. 
 
5. Elastomers are polymers in which polymer chains are held together by weak 
intermolecular forces and hence are stretchable. Fibres are polymers with strong 
intermolecular forces like hydrogen bonding and hence possess high tensile strength.  
 
Section B 
 
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly 
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a 
measure of its solubility, then it can be said that the mole fraction of a gas in the 
solution is proportional to the partial pressure of the gas over the solution. 
           p = KH . x                                                                                                                 
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these 
beverages, carbon dioxide is passed at high pressure to increase its solubility. 
 
OR 
Using the relation, 
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ? 
Solving for m, we get m= 
b.p.(solution) b.p.(l)
b.p.
TT
k
?
      
 
     
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
 
-NO
2
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
7. Fe + 2H
+
 ? Fe
2+
 + H2 
                                                                                                                                                
                                                                                                
        
 
 
8. Hexagonal    
       Example: Graphite      
                                                                                         
9.  
(a) 
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ? 
(b) 
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ? 
 
 
10. Compound (a) reacts faster than compound (b).                                               
This is due to the formation of a more stable carbocation in compound (a) in the 
rate-determining step than  carbocation in compound (b).                                                                                                    
 
 
11.  The increasing order of the acidic character of the given species is as follows: 
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH 
OR 
(a) pent-2-enal 
(b) 3-phenyl prop-2-enol 
 
 
12.  
(a) Monomers of Dacron:  
            ethylene glycol  and terephthalic acid                                                                                                           
                                                                                                                                                   
  
(b) Monomers of Buna-N: 
               1, 3-butadiene and acrylonitrile                                                                     
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
Section C 
13. NA = 6.023 × 10
23
 
Atomic mass of mercury = 200 
?Number of atoms present in 200 g of Hg = 6.023 × 10
23
 
So, the number of atoms of Hg present in 1 g of Hg = 
23
21
6.023 10
3.0115 10
200
?
?? 
The given density of mercury is 13.6 g/cc. 
? Volume of 1 atom of mercury (Hg) = 
23
21
1
2.44 10 cc
3.0115 10 13.6
?
??
??
 
Each mercury atom occupies a cube of edge length equal to its diameter.  
So, 
Diameter of an atom of mercury = (2.44 ×10
-23
)
1/3
cm = 2.905 ×10
-8
cm = 2.91A
° 
 
14.  
   
  T = 300 K 
                     c = 0.1 M 
                                             
                 i                                                                                                                                                                                                
  
 
 
OR 
(a) Vapour pressure is the pressure of the vapour at the equilibrium state when the rate 
of evaporation becomes equal to the rate of condensation. Equilibrium constant 
does not change at a particular temperature, and therefore, the vapour pressure 
remains constant. 
(b) As the solution becomes cool, heat gets absorbed; hence, enthalpy change is 
positive. Thus, the solution shows positive deviation.  
(c) B will show greater lowering of vapour pressure because 
0s
21
0
12
wM PP
P w M
?
? 
p = 4 . 6 a t m n = 2
= 1.87
? ? ? ?
+-
NaCl Na + Cl
Initial 1 0 0
After
dissociation 1 - a a a
1- a + a + a
i=
1
1+ a
= = 1 + a
1
1.87 = 1 + a
a = 1 .8 7 - 1
= 0.87
Page 4


  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
CBSE 
Class XII Chemistry   
Sample Paper 4 - Solution 
 
Time: 3 Hrs                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
1. Interstitial defect 
2. Brownian movement  
OR 
 
       The scattering of light by colloidal particles is known as the Tyndall effect. 
 
3. The maximum oxidation state shown by actinoids is +7. 
OR 
       The transition element which does not exhibit variation in the oxidation state in its 
compounds is zinc (Zn). 
 
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing 
or –I effect of three  groups. Thus, it gives a salt with Na2CO3 which is a weak base. 
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3. 
 
5. Elastomers are polymers in which polymer chains are held together by weak 
intermolecular forces and hence are stretchable. Fibres are polymers with strong 
intermolecular forces like hydrogen bonding and hence possess high tensile strength.  
 
Section B 
 
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly 
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a 
measure of its solubility, then it can be said that the mole fraction of a gas in the 
solution is proportional to the partial pressure of the gas over the solution. 
           p = KH . x                                                                                                                 
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these 
beverages, carbon dioxide is passed at high pressure to increase its solubility. 
 
OR 
Using the relation, 
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ? 
Solving for m, we get m= 
b.p.(solution) b.p.(l)
b.p.
TT
k
?
      
 
     
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
 
-NO
2
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
7. Fe + 2H
+
 ? Fe
2+
 + H2 
                                                                                                                                                
                                                                                                
        
 
 
8. Hexagonal    
       Example: Graphite      
                                                                                         
9.  
(a) 
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ? 
(b) 
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ? 
 
 
10. Compound (a) reacts faster than compound (b).                                               
This is due to the formation of a more stable carbocation in compound (a) in the 
rate-determining step than  carbocation in compound (b).                                                                                                    
 
 
11.  The increasing order of the acidic character of the given species is as follows: 
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH 
OR 
(a) pent-2-enal 
(b) 3-phenyl prop-2-enol 
 
 
12.  
(a) Monomers of Dacron:  
            ethylene glycol  and terephthalic acid                                                                                                           
                                                                                                                                                   
  
(b) Monomers of Buna-N: 
               1, 3-butadiene and acrylonitrile                                                                     
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
Section C 
13. NA = 6.023 × 10
23
 
Atomic mass of mercury = 200 
?Number of atoms present in 200 g of Hg = 6.023 × 10
23
 
So, the number of atoms of Hg present in 1 g of Hg = 
23
21
6.023 10
3.0115 10
200
?
?? 
The given density of mercury is 13.6 g/cc. 
? Volume of 1 atom of mercury (Hg) = 
23
21
1
2.44 10 cc
3.0115 10 13.6
?
??
??
 
Each mercury atom occupies a cube of edge length equal to its diameter.  
So, 
Diameter of an atom of mercury = (2.44 ×10
-23
)
1/3
cm = 2.905 ×10
-8
cm = 2.91A
° 
 
14.  
   
  T = 300 K 
                     c = 0.1 M 
                                             
                 i                                                                                                                                                                                                
  
 
 
OR 
(a) Vapour pressure is the pressure of the vapour at the equilibrium state when the rate 
of evaporation becomes equal to the rate of condensation. Equilibrium constant 
does not change at a particular temperature, and therefore, the vapour pressure 
remains constant. 
(b) As the solution becomes cool, heat gets absorbed; hence, enthalpy change is 
positive. Thus, the solution shows positive deviation.  
(c) B will show greater lowering of vapour pressure because 
0s
21
0
12
wM PP
P w M
?
? 
p = 4 . 6 a t m n = 2
= 1.87
? ? ? ?
+-
NaCl Na + Cl
Initial 1 0 0
After
dissociation 1 - a a a
1- a + a + a
i=
1
1+ a
= = 1 + a
1
1.87 = 1 + a
a = 1 .8 7 - 1
= 0.87
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
15.  
(a) Observing the graph, there is a decrease in the concentration of A in the time period 
of four hours from 1 hr to 5 hr, i.e. 0.5 – 0.3 = 0.2 M. 
In this same period of time, an increase in the concentration of B is twice the 
decrease in concentration of A. Thus, n = 2. 
(b) 
2
2
eq
eq
[B ]
(0.6)
k 1.2
[A ] (0.3)
? ? ? 
(c) Initial rate of conversion of A = change in conc. of A during 1 hour 
11
0.6 0.5
0.1mollitre hour
1
??
?
?? 
 
 
16.  
(a) It causes peptisation leading to the formation of a reddish brown colloidal solution 
of Fe(OH)3.                                                                                                              
(b) Scattering of light by colloidal particles takes place and the path of light becomes 
illuminated. This is called Tyndall effect.                                                                           
(c) Under the influence of current, the colloidal particles start moving towards the 
oppositely charged electrode. This is called electrophoresis.                                                                                                                                                
 
 
17.  
(a) Reduction reaction is feasible only at a high temperature; thus, it is not economically 
and practically viable.  
(b) This is done to remove basic impurities by the formation of slag. 
Example: 
    
(c) In electrometallurgy of Al, graphite rods act as the anode and get burnt away as CO 
and CO2 during electrolysis.  
 
18.  
(a) Mischmetal                                                                                                       
(b) Composition: 95% lanthanoid metal, 5% Fe and traces of S, C, Ca and Al. 
(c)  is more basic than .                                                                                            
Due to lanthanoid contraction, La is bigger in size than Lu. Larger the M-OH bond, 
weaker is the M-OH bond. This makes La(OH)3 more basic.                                                                                                                                                      
This is due to lanthanoid contraction because of the poor screening effect of f 
electrons. As a consequence of this, elements of the second and third transition 
series have similar size. Therefore, Zr and Hf have similar atomic radii.                                                                           
 
OR 
 
?
23
FeO + SiO FeSiO
(Slag)
? ? La OH
3
? ? Lu OH
3
Page 5


  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
CBSE 
Class XII Chemistry   
Sample Paper 4 - Solution 
 
Time: 3 Hrs                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
1. Interstitial defect 
2. Brownian movement  
OR 
 
       The scattering of light by colloidal particles is known as the Tyndall effect. 
 
3. The maximum oxidation state shown by actinoids is +7. 
OR 
       The transition element which does not exhibit variation in the oxidation state in its 
compounds is zinc (Zn). 
 
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing 
or –I effect of three  groups. Thus, it gives a salt with Na2CO3 which is a weak base. 
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3. 
 
5. Elastomers are polymers in which polymer chains are held together by weak 
intermolecular forces and hence are stretchable. Fibres are polymers with strong 
intermolecular forces like hydrogen bonding and hence possess high tensile strength.  
 
Section B 
 
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly 
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a 
measure of its solubility, then it can be said that the mole fraction of a gas in the 
solution is proportional to the partial pressure of the gas over the solution. 
           p = KH . x                                                                                                                 
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these 
beverages, carbon dioxide is passed at high pressure to increase its solubility. 
 
OR 
Using the relation, 
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ? 
Solving for m, we get m= 
b.p.(solution) b.p.(l)
b.p.
TT
k
?
      
 
     
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
 
-NO
2
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
7. Fe + 2H
+
 ? Fe
2+
 + H2 
                                                                                                                                                
                                                                                                
        
 
 
8. Hexagonal    
       Example: Graphite      
                                                                                         
9.  
(a) 
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ? 
(b) 
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ? 
 
 
10. Compound (a) reacts faster than compound (b).                                               
This is due to the formation of a more stable carbocation in compound (a) in the 
rate-determining step than  carbocation in compound (b).                                                                                                    
 
 
11.  The increasing order of the acidic character of the given species is as follows: 
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH 
OR 
(a) pent-2-enal 
(b) 3-phenyl prop-2-enol 
 
 
12.  
(a) Monomers of Dacron:  
            ethylene glycol  and terephthalic acid                                                                                                           
                                                                                                                                                   
  
(b) Monomers of Buna-N: 
               1, 3-butadiene and acrylonitrile                                                                     
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
Section C 
13. NA = 6.023 × 10
23
 
Atomic mass of mercury = 200 
?Number of atoms present in 200 g of Hg = 6.023 × 10
23
 
So, the number of atoms of Hg present in 1 g of Hg = 
23
21
6.023 10
3.0115 10
200
?
?? 
The given density of mercury is 13.6 g/cc. 
? Volume of 1 atom of mercury (Hg) = 
23
21
1
2.44 10 cc
3.0115 10 13.6
?
??
??
 
Each mercury atom occupies a cube of edge length equal to its diameter.  
So, 
Diameter of an atom of mercury = (2.44 ×10
-23
)
1/3
cm = 2.905 ×10
-8
cm = 2.91A
° 
 
14.  
   
  T = 300 K 
                     c = 0.1 M 
                                             
                 i                                                                                                                                                                                                
  
 
 
OR 
(a) Vapour pressure is the pressure of the vapour at the equilibrium state when the rate 
of evaporation becomes equal to the rate of condensation. Equilibrium constant 
does not change at a particular temperature, and therefore, the vapour pressure 
remains constant. 
(b) As the solution becomes cool, heat gets absorbed; hence, enthalpy change is 
positive. Thus, the solution shows positive deviation.  
(c) B will show greater lowering of vapour pressure because 
0s
21
0
12
wM PP
P w M
?
? 
p = 4 . 6 a t m n = 2
= 1.87
? ? ? ?
+-
NaCl Na + Cl
Initial 1 0 0
After
dissociation 1 - a a a
1- a + a + a
i=
1
1+ a
= = 1 + a
1
1.87 = 1 + a
a = 1 .8 7 - 1
= 0.87
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
15.  
(a) Observing the graph, there is a decrease in the concentration of A in the time period 
of four hours from 1 hr to 5 hr, i.e. 0.5 – 0.3 = 0.2 M. 
In this same period of time, an increase in the concentration of B is twice the 
decrease in concentration of A. Thus, n = 2. 
(b) 
2
2
eq
eq
[B ]
(0.6)
k 1.2
[A ] (0.3)
? ? ? 
(c) Initial rate of conversion of A = change in conc. of A during 1 hour 
11
0.6 0.5
0.1mollitre hour
1
??
?
?? 
 
 
16.  
(a) It causes peptisation leading to the formation of a reddish brown colloidal solution 
of Fe(OH)3.                                                                                                              
(b) Scattering of light by colloidal particles takes place and the path of light becomes 
illuminated. This is called Tyndall effect.                                                                           
(c) Under the influence of current, the colloidal particles start moving towards the 
oppositely charged electrode. This is called electrophoresis.                                                                                                                                                
 
 
17.  
(a) Reduction reaction is feasible only at a high temperature; thus, it is not economically 
and practically viable.  
(b) This is done to remove basic impurities by the formation of slag. 
Example: 
    
(c) In electrometallurgy of Al, graphite rods act as the anode and get burnt away as CO 
and CO2 during electrolysis.  
 
18.  
(a) Mischmetal                                                                                                       
(b) Composition: 95% lanthanoid metal, 5% Fe and traces of S, C, Ca and Al. 
(c)  is more basic than .                                                                                            
Due to lanthanoid contraction, La is bigger in size than Lu. Larger the M-OH bond, 
weaker is the M-OH bond. This makes La(OH)3 more basic.                                                                                                                                                      
This is due to lanthanoid contraction because of the poor screening effect of f 
electrons. As a consequence of this, elements of the second and third transition 
series have similar size. Therefore, Zr and Hf have similar atomic radii.                                                                           
 
OR 
 
?
23
FeO + SiO FeSiO
(Slag)
? ? La OH
3
? ? Lu OH
3
  
 
CBSE XII  |  Chemistry  
Sample Paper  4 - Solution 
 
     
(a) Cr
2+
 is strongly reducing because it changes to Cr
3+ 
which is more stable as Cr
3+
 has 
higher hydration energy. Manganese (III) is strongly oxidising because it changes 
from Mn(III) to Mn(II) which is more stable due to half-filled d-orbitals. 
 
(b) Co(III) has greater tendency to form coordination complexes than Co (II). Thus, in 
the presence of ligands, Co (II) changes to Co (III), i.e. it is easily oxidised.  
 
(c) The ions in d
1
 configuration have great tendency to lose one electron and acquire 
d
0
 configuration and hence act as reducing agents. All elements with d1 
configuration are either reducing or undergo disproportionation. 
                  Example: 
10
6 7 4
2
4 4 2 2
3d 3d
3MnO 4H 2MnO MnO H O
? ? ?
? ? ?
? ? ? ? ? ? 
 
19.  
(a) Ambidentate ligand: A unidentate ligand which can bind to the central metal atom 
through any of the two donor atoms present in it is called ambidentate ligand. 
Example:
2
NO ,SCN
??
 
M SCN M NCS
Thiocynato Isothiocynato
??
 
 
(b)  
(i) Potassium trioxalatochromate (III) 
(ii) Diamminedichloridoplatinum (II) 
 
(c)  
            
 
20.  
(a)   
(i) Dow’s process                                                                                         
(ii) Finkelstein reaction                                                                                                                                                                                              
(b) Iodobenzene has a higher boiling point than chlorobenzene. Due to the larger size of 
the halogen iodine, van der Waals forces are stronger in iodobenzene than 
chlorobenzene, making the boiling point of iodobenzene higher than chlorobenzene.