CBSE Chemistry Past year paper (Solutions) - 2018, Class 12 | Chemistry for Grade 12 PDF Download

 Page 1


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2018 Solution 
 
  
 
1. In the given complex [Pt(en)2Cl2],  
Ethylene diamine (en) is a bidentate ligand. 
Chloride is unidentate ligand.  
Therefore, 
                   [(2 × 2) + (2 × 1)] = 6 
 
Coordination number of Pt is 6. 
The oxidation state of Pt is +2. 
 
2. FeO has a non-stoichiometric composition with the formula Fe0.95O, because 
Fe is present in both +2 and +3 oxidation states. 
 
3. Benzyl chloride is more readily hydrolysed in aqueous NaOH. 
In chlorobenzene, the lone pair of chlorine is in conjugation with the ?-bond 
of benzene. 
This develops the partial double bond character which results in the reduced 
reactivity of chlorobenzene. 
 
4. The IUPAC name of the given compound is 3,3-dimethylpentan-2-ol. 
 
5.  
CO and H2 react to form different products in the presence of different 
catalysts. This shows the selectivity of a catalyst. 
Selectivity of a catalyst is the ability of a catalyst to direct the reaction to 
form a particular product excluding others. 
 
6.  
a. PH3 will have the lowest boiling point.  
b. Ammonia has the maximum basic character. 
c. NH3 will have the highest bond angle. 
d. BiH3 will have the maximum reducing character. 
 
  
Page 2


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2018 Solution 
 
  
 
1. In the given complex [Pt(en)2Cl2],  
Ethylene diamine (en) is a bidentate ligand. 
Chloride is unidentate ligand.  
Therefore, 
                   [(2 × 2) + (2 × 1)] = 6 
 
Coordination number of Pt is 6. 
The oxidation state of Pt is +2. 
 
2. FeO has a non-stoichiometric composition with the formula Fe0.95O, because 
Fe is present in both +2 and +3 oxidation states. 
 
3. Benzyl chloride is more readily hydrolysed in aqueous NaOH. 
In chlorobenzene, the lone pair of chlorine is in conjugation with the ?-bond 
of benzene. 
This develops the partial double bond character which results in the reduced 
reactivity of chlorobenzene. 
 
4. The IUPAC name of the given compound is 3,3-dimethylpentan-2-ol. 
 
5.  
CO and H2 react to form different products in the presence of different 
catalysts. This shows the selectivity of a catalyst. 
Selectivity of a catalyst is the ability of a catalyst to direct the reaction to 
form a particular product excluding others. 
 
6.  
a. PH3 will have the lowest boiling point.  
b. Ammonia has the maximum basic character. 
c. NH3 will have the highest bond angle. 
d. BiH3 will have the maximum reducing character. 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
7. Given, 
W2 = 60g 
M2 = 180g mol
-1
 
W1 = 250g 
Kf = 1.86 K kg mol
-1
 
?Tf = kf × m 
 
0 2 f
ff
1 2
f
K  w  1000
T - T
M  w 
1.86  60  1000
273 - T
180  250
??
?
?
??
?
?
 
 
273.15 – Tf = 2.48 
Tf = 273.15 – 2.48 = 270.67 K 
 
Hence, the freezing point is 270.67 K or -2.48°C. 
 
8.  
(a)  
 
 
 
(b)  
 
 
 
OR 
 
 
Page 3


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2018 Solution 
 
  
 
1. In the given complex [Pt(en)2Cl2],  
Ethylene diamine (en) is a bidentate ligand. 
Chloride is unidentate ligand.  
Therefore, 
                   [(2 × 2) + (2 × 1)] = 6 
 
Coordination number of Pt is 6. 
The oxidation state of Pt is +2. 
 
2. FeO has a non-stoichiometric composition with the formula Fe0.95O, because 
Fe is present in both +2 and +3 oxidation states. 
 
3. Benzyl chloride is more readily hydrolysed in aqueous NaOH. 
In chlorobenzene, the lone pair of chlorine is in conjugation with the ?-bond 
of benzene. 
This develops the partial double bond character which results in the reduced 
reactivity of chlorobenzene. 
 
4. The IUPAC name of the given compound is 3,3-dimethylpentan-2-ol. 
 
5.  
CO and H2 react to form different products in the presence of different 
catalysts. This shows the selectivity of a catalyst. 
Selectivity of a catalyst is the ability of a catalyst to direct the reaction to 
form a particular product excluding others. 
 
6.  
a. PH3 will have the lowest boiling point.  
b. Ammonia has the maximum basic character. 
c. NH3 will have the highest bond angle. 
d. BiH3 will have the maximum reducing character. 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
7. Given, 
W2 = 60g 
M2 = 180g mol
-1
 
W1 = 250g 
Kf = 1.86 K kg mol
-1
 
?Tf = kf × m 
 
0 2 f
ff
1 2
f
K  w  1000
T - T
M  w 
1.86  60  1000
273 - T
180  250
??
?
?
??
?
?
 
 
273.15 – Tf = 2.48 
Tf = 273.15 – 2.48 = 270.67 K 
 
Hence, the freezing point is 270.67 K or -2.48°C. 
 
8.  
(a)  
 
 
 
(b)  
 
 
 
OR 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
(a) Aromatic carboxylic acids undergo electrophilic substitution reactions in 
which the carboxyl group acts as a deactivating and meta-directing 
group. 
(b) They however do not undergo Friedel–Crafts reaction because the 
carboxyl group is deactivating and the catalyst aluminium chloride 
(Lewis acid) gets bonded to the carboxyl group. 
Para-nitro benzoic acid is more acidic than benzoic acid because of the 
electron-withdrawing nitro group at the para position to the acid group. 
The nitro group (-NO2) withdraws the electrons from the carboxylic acid 
group due to which the electron density on the hydrogen atom of the 
carboxylic acid becomes low, and it can be easily removed. Hence, it is 
more acidic than benzoic acid. 
 
 
 
9.  
(a) 5Fe
2+
 + MnO4
-
 + 8H
+
 ? 5Fe
3+
 + Mn
2+
 + 4H2O 
 
(b) 2MnO4
-
 + H2O + I
-
 ? 2MnO2 + 2OH
-
 + IO3
-
 
 
10.  
Rate of formation of NO2 = 2.8 × 10
-3
 M s
-1
 
 
Rate of disappearance of N2O5 =  
 
          
25 2
25 3
31
d N O d NO
11
2 dt 4 dt
d N O
1
2.8 10 2
dt 4
1.4 10 M s
?
??
?? ??
? ? ? ?
??
??
??
? ? ? ?
??
 
 
  
Page 4


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2018 Solution 
 
  
 
1. In the given complex [Pt(en)2Cl2],  
Ethylene diamine (en) is a bidentate ligand. 
Chloride is unidentate ligand.  
Therefore, 
                   [(2 × 2) + (2 × 1)] = 6 
 
Coordination number of Pt is 6. 
The oxidation state of Pt is +2. 
 
2. FeO has a non-stoichiometric composition with the formula Fe0.95O, because 
Fe is present in both +2 and +3 oxidation states. 
 
3. Benzyl chloride is more readily hydrolysed in aqueous NaOH. 
In chlorobenzene, the lone pair of chlorine is in conjugation with the ?-bond 
of benzene. 
This develops the partial double bond character which results in the reduced 
reactivity of chlorobenzene. 
 
4. The IUPAC name of the given compound is 3,3-dimethylpentan-2-ol. 
 
5.  
CO and H2 react to form different products in the presence of different 
catalysts. This shows the selectivity of a catalyst. 
Selectivity of a catalyst is the ability of a catalyst to direct the reaction to 
form a particular product excluding others. 
 
6.  
a. PH3 will have the lowest boiling point.  
b. Ammonia has the maximum basic character. 
c. NH3 will have the highest bond angle. 
d. BiH3 will have the maximum reducing character. 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
7. Given, 
W2 = 60g 
M2 = 180g mol
-1
 
W1 = 250g 
Kf = 1.86 K kg mol
-1
 
?Tf = kf × m 
 
0 2 f
ff
1 2
f
K  w  1000
T - T
M  w 
1.86  60  1000
273 - T
180  250
??
?
?
??
?
?
 
 
273.15 – Tf = 2.48 
Tf = 273.15 – 2.48 = 270.67 K 
 
Hence, the freezing point is 270.67 K or -2.48°C. 
 
8.  
(a)  
 
 
 
(b)  
 
 
 
OR 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
(a) Aromatic carboxylic acids undergo electrophilic substitution reactions in 
which the carboxyl group acts as a deactivating and meta-directing 
group. 
(b) They however do not undergo Friedel–Crafts reaction because the 
carboxyl group is deactivating and the catalyst aluminium chloride 
(Lewis acid) gets bonded to the carboxyl group. 
Para-nitro benzoic acid is more acidic than benzoic acid because of the 
electron-withdrawing nitro group at the para position to the acid group. 
The nitro group (-NO2) withdraws the electrons from the carboxylic acid 
group due to which the electron density on the hydrogen atom of the 
carboxylic acid becomes low, and it can be easily removed. Hence, it is 
more acidic than benzoic acid. 
 
 
 
9.  
(a) 5Fe
2+
 + MnO4
-
 + 8H
+
 ? 5Fe
3+
 + Mn
2+
 + 4H2O 
 
(b) 2MnO4
-
 + H2O + I
-
 ? 2MnO2 + 2OH
-
 + IO3
-
 
 
10.  
Rate of formation of NO2 = 2.8 × 10
-3
 M s
-1
 
 
Rate of disappearance of N2O5 =  
 
          
25 2
25 3
31
d N O d NO
11
2 dt 4 dt
d N O
1
2.8 10 2
dt 4
1.4 10 M s
?
??
?? ??
? ? ? ?
??
??
??
? ? ? ?
??
 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
11.  
(a) From the given molecules, (i) is a chiral molecule, as it has all different 
groups attached. Also, it does not have a centre of symmetry. 
                                           
 
(b) When a mixture of chlorobenzene and methyl chloride is treated with 
sodium metal in the presence of dry ether, toluene is obtained. 
 
 
 
 
(c)  
The elimination reaction can be represented as 
 
 
 
  
Page 5


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2018 Solution 
 
  
 
1. In the given complex [Pt(en)2Cl2],  
Ethylene diamine (en) is a bidentate ligand. 
Chloride is unidentate ligand.  
Therefore, 
                   [(2 × 2) + (2 × 1)] = 6 
 
Coordination number of Pt is 6. 
The oxidation state of Pt is +2. 
 
2. FeO has a non-stoichiometric composition with the formula Fe0.95O, because 
Fe is present in both +2 and +3 oxidation states. 
 
3. Benzyl chloride is more readily hydrolysed in aqueous NaOH. 
In chlorobenzene, the lone pair of chlorine is in conjugation with the ?-bond 
of benzene. 
This develops the partial double bond character which results in the reduced 
reactivity of chlorobenzene. 
 
4. The IUPAC name of the given compound is 3,3-dimethylpentan-2-ol. 
 
5.  
CO and H2 react to form different products in the presence of different 
catalysts. This shows the selectivity of a catalyst. 
Selectivity of a catalyst is the ability of a catalyst to direct the reaction to 
form a particular product excluding others. 
 
6.  
a. PH3 will have the lowest boiling point.  
b. Ammonia has the maximum basic character. 
c. NH3 will have the highest bond angle. 
d. BiH3 will have the maximum reducing character. 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
7. Given, 
W2 = 60g 
M2 = 180g mol
-1
 
W1 = 250g 
Kf = 1.86 K kg mol
-1
 
?Tf = kf × m 
 
0 2 f
ff
1 2
f
K  w  1000
T - T
M  w 
1.86  60  1000
273 - T
180  250
??
?
?
??
?
?
 
 
273.15 – Tf = 2.48 
Tf = 273.15 – 2.48 = 270.67 K 
 
Hence, the freezing point is 270.67 K or -2.48°C. 
 
8.  
(a)  
 
 
 
(b)  
 
 
 
OR 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
(a) Aromatic carboxylic acids undergo electrophilic substitution reactions in 
which the carboxyl group acts as a deactivating and meta-directing 
group. 
(b) They however do not undergo Friedel–Crafts reaction because the 
carboxyl group is deactivating and the catalyst aluminium chloride 
(Lewis acid) gets bonded to the carboxyl group. 
Para-nitro benzoic acid is more acidic than benzoic acid because of the 
electron-withdrawing nitro group at the para position to the acid group. 
The nitro group (-NO2) withdraws the electrons from the carboxylic acid 
group due to which the electron density on the hydrogen atom of the 
carboxylic acid becomes low, and it can be easily removed. Hence, it is 
more acidic than benzoic acid. 
 
 
 
9.  
(a) 5Fe
2+
 + MnO4
-
 + 8H
+
 ? 5Fe
3+
 + Mn
2+
 + 4H2O 
 
(b) 2MnO4
-
 + H2O + I
-
 ? 2MnO2 + 2OH
-
 + IO3
-
 
 
10.  
Rate of formation of NO2 = 2.8 × 10
-3
 M s
-1
 
 
Rate of disappearance of N2O5 =  
 
          
25 2
25 3
31
d N O d NO
11
2 dt 4 dt
d N O
1
2.8 10 2
dt 4
1.4 10 M s
?
??
?? ??
? ? ? ?
??
??
??
? ? ? ?
??
 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
11.  
(a) From the given molecules, (i) is a chiral molecule, as it has all different 
groups attached. Also, it does not have a centre of symmetry. 
                                           
 
(b) When a mixture of chlorobenzene and methyl chloride is treated with 
sodium metal in the presence of dry ether, toluene is obtained. 
 
 
 
 
(c)  
The elimination reaction can be represented as 
 
 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2018 
 
     
12. Given, 
t1/2 = 40 min  
T1 = 300 K 
t1/2 = 20 min  
T2= 320 K 
Half-life t1/2 for the first-order reaction is 
1/2 1/2
1 2
1 2
2
1
a
a
a
0.693 0.693
t  and t 
kk
0.693 0.693
k and k
40 20
0.693
k
40
20
2
0.693 k 20
10
E 320 300
log 2 = 
2.303 8.314 300 320
2.303 8.314 300 320 log 2
E
20
2.303 8.314 300 320 0.3010
E
20
   27663.79 J m
??
??
? ? ?
? ??
??
??
??
? ? ?
?
? ? ? ?
?
?
1
1
a
ol
E 27.66379 kJ mol
?
?
?
 
 
     Activation energy of the reaction is 27.66 kJmol
-1
.