CBSE Chemistry Past year paper (Solutions) - 2015, Class 12 Class 12 Notes | EduRev

Chemistry Class 12

Created by: Mohit Rajpoot

Class 12 : CBSE Chemistry Past year paper (Solutions) - 2015, Class 12 Class 12 Notes | EduRev

 Page 1


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2015 Solution 
 
1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal 
particles move towards the oppositely charged electrodes, get discharged and 
precipitate. 
 
2. Suppose the number of atoms Y in the hcp lattice = n 
 As the number of tetrahedral voids is double the number of atoms in close packing, 
the number of tetrahedral voids = 2n 
 As atoms X occupy 2/3
rd
 of the tetrahedral voids, the number of atoms X in the lattice  
 
2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?
  
 Hence, the formula of the compound is X4Y3. 
 
3. White phosphorus is less stable and therefore more reactive than the other solid 
phases under normal conditions because of the angular strain in the P4 molecule 
where the angles are only 60°. Red phosphorus is less reactive than white 
phosphorus. 
 
4.  
  
Page 2


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2015 Solution 
 
1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal 
particles move towards the oppositely charged electrodes, get discharged and 
precipitate. 
 
2. Suppose the number of atoms Y in the hcp lattice = n 
 As the number of tetrahedral voids is double the number of atoms in close packing, 
the number of tetrahedral voids = 2n 
 As atoms X occupy 2/3
rd
 of the tetrahedral voids, the number of atoms X in the lattice  
 
2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?
  
 Hence, the formula of the compound is X4Y3. 
 
3. White phosphorus is less stable and therefore more reactive than the other solid 
phases under normal conditions because of the angular strain in the P4 molecule 
where the angles are only 60°. Red phosphorus is less reactive than white 
phosphorus. 
 
4.  
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the 
carbocations, more easily will the product be formed and hence faster will be the rate 
of the reaction. Because the stability of the carbocations decreases in the order: 
3° carbocation > 2° carbocation > 1° carbocation > CH3
+
 
Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same 
order:  
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides 
The two structures are 
 
Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in 
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a 
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1 
reaction.  
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than 
bromoethane. 
 
6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant 
temperature is directly proportional to the pressure of the gas present in equilibrium 
with the liquid. 
Or 
The solubility of a gas in a liquid at a particular temperature is directly proportional 
to the pressure of the gas in equilibrium with the liquid at that temperature. 
 
The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied 
by the evolution of heat. Thus, 
 
 
Applying Le Chatelier’s principle, the increase of temperature would shift the 
equilibrium in the backward direction, that is, solubility would decrease. 
Therefore, gases always tend to be less soluble in liquids as the temperature is raised. 
 
 
OR 
Raoult’s law: In a solution, the vapour pressure of a component at a given 
temperature is equal to the mole fraction of that component in the solution 
multiplied by the vapour pressure of that component in the pure state. 
 
Page 3


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2015 Solution 
 
1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal 
particles move towards the oppositely charged electrodes, get discharged and 
precipitate. 
 
2. Suppose the number of atoms Y in the hcp lattice = n 
 As the number of tetrahedral voids is double the number of atoms in close packing, 
the number of tetrahedral voids = 2n 
 As atoms X occupy 2/3
rd
 of the tetrahedral voids, the number of atoms X in the lattice  
 
2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?
  
 Hence, the formula of the compound is X4Y3. 
 
3. White phosphorus is less stable and therefore more reactive than the other solid 
phases under normal conditions because of the angular strain in the P4 molecule 
where the angles are only 60°. Red phosphorus is less reactive than white 
phosphorus. 
 
4.  
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the 
carbocations, more easily will the product be formed and hence faster will be the rate 
of the reaction. Because the stability of the carbocations decreases in the order: 
3° carbocation > 2° carbocation > 1° carbocation > CH3
+
 
Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same 
order:  
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides 
The two structures are 
 
Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in 
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a 
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1 
reaction.  
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than 
bromoethane. 
 
6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant 
temperature is directly proportional to the pressure of the gas present in equilibrium 
with the liquid. 
Or 
The solubility of a gas in a liquid at a particular temperature is directly proportional 
to the pressure of the gas in equilibrium with the liquid at that temperature. 
 
The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied 
by the evolution of heat. Thus, 
 
 
Applying Le Chatelier’s principle, the increase of temperature would shift the 
equilibrium in the backward direction, that is, solubility would decrease. 
Therefore, gases always tend to be less soluble in liquids as the temperature is raised. 
 
 
OR 
Raoult’s law: In a solution, the vapour pressure of a component at a given 
temperature is equal to the mole fraction of that component in the solution 
multiplied by the vapour pressure of that component in the pure state. 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
Ideal Solution Non-ideal Solution 
1. An ideal solution of the 
components A and B is the 
solution in which the 
intermolecular interactions 
between the components (A–B 
attractions) are of the same 
magnitude as the 
intermolecular interactions 
found in the pure components 
(A–A attractions and B–B 
attractions). 
1. A non-ideal solution of the 
components A and B is the 
solution in which the 
intermolecular interactions 
between the components (A–B 
attractions) are of different 
magnitude as the intermolecular 
interactions found in the pure 
components (A–A attractions 
and B–B attractions). 
2. ?mixH = 0 and ?mixV = 0 2. ?mixH ? 0 and ?mixV ? 0 
 
7.  
(a) Reactions taking place during the electrolysis of aqueous sodium chloride solution: 
 
            
0.00
?
?
+ - o
+ - o
2
Na (aq)+e Na(s) E =-2.71 V (I)
1
H (aq)+ e H (g) E = V (II)
2
 
The reaction with more positive value of reduction potential would undergo 
reduction at the cathode. 
Here, in given reactions, reduction potential of H
+
 is more positive than reduction 
potential of Na
+
. The reduction potential value of Na
+
 is negative. 
So, the reduction of H
 +
 is feasible at the cathode. 
 
(b)  
? A mercury cell consists of a zinc container as the anode, a carbon rod as the 
cathode and a paste of mercuric oxide mixed with KOH as the electrolyte.  
? In this cell, the overall cell reaction does not involve any ion whose 
concentration may change.  
? Therefore, this cell gives a constant potential of 1.5 V throughout its life.  
 
8. Transition metals have high effective nuclear charge, greater number of valence 
electrons and some unpaired electrons. They thus have strong metal–metal bonding. 
Hence, transition metals have high enthalpies of atomisation.  
In the 3d series, from Sc to Zn, only zinc has filled valence shells. The valence shell 
electronic configuration of Zn is 3d
10
4s
2
. Because of the absence of unpaired electrons 
in ns and (n–1) d shells, the interatomic electronic bonding is the weakest in zinc. 
Consequently, zinc has the least enthalpy of atomisation in the 3d series of transition 
elements. 
Page 4


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2015 Solution 
 
1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal 
particles move towards the oppositely charged electrodes, get discharged and 
precipitate. 
 
2. Suppose the number of atoms Y in the hcp lattice = n 
 As the number of tetrahedral voids is double the number of atoms in close packing, 
the number of tetrahedral voids = 2n 
 As atoms X occupy 2/3
rd
 of the tetrahedral voids, the number of atoms X in the lattice  
 
2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?
  
 Hence, the formula of the compound is X4Y3. 
 
3. White phosphorus is less stable and therefore more reactive than the other solid 
phases under normal conditions because of the angular strain in the P4 molecule 
where the angles are only 60°. Red phosphorus is less reactive than white 
phosphorus. 
 
4.  
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the 
carbocations, more easily will the product be formed and hence faster will be the rate 
of the reaction. Because the stability of the carbocations decreases in the order: 
3° carbocation > 2° carbocation > 1° carbocation > CH3
+
 
Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same 
order:  
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides 
The two structures are 
 
Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in 
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a 
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1 
reaction.  
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than 
bromoethane. 
 
6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant 
temperature is directly proportional to the pressure of the gas present in equilibrium 
with the liquid. 
Or 
The solubility of a gas in a liquid at a particular temperature is directly proportional 
to the pressure of the gas in equilibrium with the liquid at that temperature. 
 
The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied 
by the evolution of heat. Thus, 
 
 
Applying Le Chatelier’s principle, the increase of temperature would shift the 
equilibrium in the backward direction, that is, solubility would decrease. 
Therefore, gases always tend to be less soluble in liquids as the temperature is raised. 
 
 
OR 
Raoult’s law: In a solution, the vapour pressure of a component at a given 
temperature is equal to the mole fraction of that component in the solution 
multiplied by the vapour pressure of that component in the pure state. 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
Ideal Solution Non-ideal Solution 
1. An ideal solution of the 
components A and B is the 
solution in which the 
intermolecular interactions 
between the components (A–B 
attractions) are of the same 
magnitude as the 
intermolecular interactions 
found in the pure components 
(A–A attractions and B–B 
attractions). 
1. A non-ideal solution of the 
components A and B is the 
solution in which the 
intermolecular interactions 
between the components (A–B 
attractions) are of different 
magnitude as the intermolecular 
interactions found in the pure 
components (A–A attractions 
and B–B attractions). 
2. ?mixH = 0 and ?mixV = 0 2. ?mixH ? 0 and ?mixV ? 0 
 
7.  
(a) Reactions taking place during the electrolysis of aqueous sodium chloride solution: 
 
            
0.00
?
?
+ - o
+ - o
2
Na (aq)+e Na(s) E =-2.71 V (I)
1
H (aq)+ e H (g) E = V (II)
2
 
The reaction with more positive value of reduction potential would undergo 
reduction at the cathode. 
Here, in given reactions, reduction potential of H
+
 is more positive than reduction 
potential of Na
+
. The reduction potential value of Na
+
 is negative. 
So, the reduction of H
 +
 is feasible at the cathode. 
 
(b)  
? A mercury cell consists of a zinc container as the anode, a carbon rod as the 
cathode and a paste of mercuric oxide mixed with KOH as the electrolyte.  
? In this cell, the overall cell reaction does not involve any ion whose 
concentration may change.  
? Therefore, this cell gives a constant potential of 1.5 V throughout its life.  
 
8. Transition metals have high effective nuclear charge, greater number of valence 
electrons and some unpaired electrons. They thus have strong metal–metal bonding. 
Hence, transition metals have high enthalpies of atomisation.  
In the 3d series, from Sc to Zn, only zinc has filled valence shells. The valence shell 
electronic configuration of Zn is 3d
10
4s
2
. Because of the absence of unpaired electrons 
in ns and (n–1) d shells, the interatomic electronic bonding is the weakest in zinc. 
Consequently, zinc has the least enthalpy of atomisation in the 3d series of transition 
elements. 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
9.   
(i) [Co(NH3)5(NO2)](NO3)2 
 IUPAC name: Pentaamminenitrocobalt (III) nitrate 
(ii) Potassium tetracyanidonickelate (II) 
 Formula of the complex: K2[Ni(CN)4] 
 
10.     
(i)  
  
 
(ii)   
  
 
11. Given: 
Molar mass of CaCl2 (MB)  = 111 g/mol 
Weight of water (wA)  = 500 g 
Kf for water   = 1.86 K kg/mol 
?Tf    = 2 K 
Formula: 
fB
f
AB
K × w × 1000
? T =
w × M
 
Solution: 
fB
f
AB
B
B
K × w × 1000
? T =
w × M
1.86 × w × 1000
2=
500 × 111
2 × 500 × 111
w = = 59.68
1.86 × 1000
 
Amount of CaCl2 required = 59.68 g 
 
  
Page 5


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
CBSE 
Class XII Chemistry 
Board Paper – 2015 Solution 
 
1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal 
particles move towards the oppositely charged electrodes, get discharged and 
precipitate. 
 
2. Suppose the number of atoms Y in the hcp lattice = n 
 As the number of tetrahedral voids is double the number of atoms in close packing, 
the number of tetrahedral voids = 2n 
 As atoms X occupy 2/3
rd
 of the tetrahedral voids, the number of atoms X in the lattice  
 
2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?
  
 Hence, the formula of the compound is X4Y3. 
 
3. White phosphorus is less stable and therefore more reactive than the other solid 
phases under normal conditions because of the angular strain in the P4 molecule 
where the angles are only 60°. Red phosphorus is less reactive than white 
phosphorus. 
 
4.  
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the 
carbocations, more easily will the product be formed and hence faster will be the rate 
of the reaction. Because the stability of the carbocations decreases in the order: 
3° carbocation > 2° carbocation > 1° carbocation > CH3
+
 
Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same 
order:  
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides 
The two structures are 
 
Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in 
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a 
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1 
reaction.  
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than 
bromoethane. 
 
6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant 
temperature is directly proportional to the pressure of the gas present in equilibrium 
with the liquid. 
Or 
The solubility of a gas in a liquid at a particular temperature is directly proportional 
to the pressure of the gas in equilibrium with the liquid at that temperature. 
 
The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied 
by the evolution of heat. Thus, 
 
 
Applying Le Chatelier’s principle, the increase of temperature would shift the 
equilibrium in the backward direction, that is, solubility would decrease. 
Therefore, gases always tend to be less soluble in liquids as the temperature is raised. 
 
 
OR 
Raoult’s law: In a solution, the vapour pressure of a component at a given 
temperature is equal to the mole fraction of that component in the solution 
multiplied by the vapour pressure of that component in the pure state. 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
Ideal Solution Non-ideal Solution 
1. An ideal solution of the 
components A and B is the 
solution in which the 
intermolecular interactions 
between the components (A–B 
attractions) are of the same 
magnitude as the 
intermolecular interactions 
found in the pure components 
(A–A attractions and B–B 
attractions). 
1. A non-ideal solution of the 
components A and B is the 
solution in which the 
intermolecular interactions 
between the components (A–B 
attractions) are of different 
magnitude as the intermolecular 
interactions found in the pure 
components (A–A attractions 
and B–B attractions). 
2. ?mixH = 0 and ?mixV = 0 2. ?mixH ? 0 and ?mixV ? 0 
 
7.  
(a) Reactions taking place during the electrolysis of aqueous sodium chloride solution: 
 
            
0.00
?
?
+ - o
+ - o
2
Na (aq)+e Na(s) E =-2.71 V (I)
1
H (aq)+ e H (g) E = V (II)
2
 
The reaction with more positive value of reduction potential would undergo 
reduction at the cathode. 
Here, in given reactions, reduction potential of H
+
 is more positive than reduction 
potential of Na
+
. The reduction potential value of Na
+
 is negative. 
So, the reduction of H
 +
 is feasible at the cathode. 
 
(b)  
? A mercury cell consists of a zinc container as the anode, a carbon rod as the 
cathode and a paste of mercuric oxide mixed with KOH as the electrolyte.  
? In this cell, the overall cell reaction does not involve any ion whose 
concentration may change.  
? Therefore, this cell gives a constant potential of 1.5 V throughout its life.  
 
8. Transition metals have high effective nuclear charge, greater number of valence 
electrons and some unpaired electrons. They thus have strong metal–metal bonding. 
Hence, transition metals have high enthalpies of atomisation.  
In the 3d series, from Sc to Zn, only zinc has filled valence shells. The valence shell 
electronic configuration of Zn is 3d
10
4s
2
. Because of the absence of unpaired electrons 
in ns and (n–1) d shells, the interatomic electronic bonding is the weakest in zinc. 
Consequently, zinc has the least enthalpy of atomisation in the 3d series of transition 
elements. 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
9.   
(i) [Co(NH3)5(NO2)](NO3)2 
 IUPAC name: Pentaamminenitrocobalt (III) nitrate 
(ii) Potassium tetracyanidonickelate (II) 
 Formula of the complex: K2[Ni(CN)4] 
 
10.     
(i)  
  
 
(ii)   
  
 
11. Given: 
Molar mass of CaCl2 (MB)  = 111 g/mol 
Weight of water (wA)  = 500 g 
Kf for water   = 1.86 K kg/mol 
?Tf    = 2 K 
Formula: 
fB
f
AB
K × w × 1000
? T =
w × M
 
Solution: 
fB
f
AB
B
B
K × w × 1000
? T =
w × M
1.86 × w × 1000
2=
500 × 111
2 × 500 × 111
w = = 59.68
1.86 × 1000
 
Amount of CaCl2 required = 59.68 g 
 
  
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2015 Solution 
 
     
12. Given: 
Density (D)   =  10 g/cm
3
  
Edge length (a)  =  3 × 10
-8
 cm 
Atomic mass   =  81 g/mol 
Formula: 3
0
Z × M
Density =
a × N
  
3
0
3 23
3 23
Z × M
Density =
a × N
Z × 81
10 =
(3 10) × 6.022 × 10
10 × (3 10) × 6.022 × 10
Z = = 2.0073
81
?
?
 
? Z = 2 
The unit cell contains 2 atoms, so it is a body-centred cubic unit cell. 
 
 
13. According to Nernst equation, 
 
0
cell cell
0.0592 [P]
E = E log
n [R]
?
 
 
2+ 2+
2
0 0 0
cell Oxi(anode) Oxi(cathode)
00
Sn|Sn H |2H
E E E
EE
0.14 0.0
0.14
??
??
? ? ?
??
 
 Reactions: 
 Anode (oxidation):   Sn(s) ? Sn2
+
(aq) + 2e
-
 
 Cathode (reduction):   2H
+
(aq) + 2e
- 
? H2(g) 
 Net reaction:    Sn(s) + 2H
+
(aq) ? Sn2
+
(aq) + H2(g) 
 ? n = 2  
 [P] = [Sn2
+
] = 0.001 M 
 [R] = [H
+
] = 0.01 M 
 So, putting the above values in the formula, 
 cell 2
0.0592 [0.001]
E = 0.14 log
2 [0.01]
??
 
 Ecell = –0.1696 V 
 
 
 
 
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