CBSE Chemistry Past year paper (Solutions) - 2014, Class 12 Class 12 Notes | EduRev

 Page 1


  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
CBSE 
Class XII Chemistry 
Board Paper 2014 – Set 1 Solution 
Time: 3 hrs. Total Marks: 70 
 
1. High temperature is favourable for adsorption, which increases with an increase in 
temperature. 
2. In the extraction of silver, zinc functions as a reducing agent. 
 
          Zn + 2Na[Ag(CN)2] ? Na2[Zn(CN)4] + 2Ag 
3. Basicity of H3PO3 is two as there are two replaceable hydrogen atoms. The structure 
of H3PO3 is as follows: 
 
4.    The two structures given are as follows: 
 
                                      
                                   2-chlorobutane                                   1-chlorobutane 
Between the two given compounds, 2-chlorobutane contains one carbon atom 
which is attached to four entities such as methyl group, chlorine atom, hydrogen 
atom and ethyl group.  
 
2-chlorobutane with chiral centre 
 
But in 1-chlorobutane, no carbon atom is attached to four atoms or groups. 
Hence, the chiral compound is 2-chlorobutane. 
 
5. Proteins are natural polymers.  
Buna-S and PVC are synthetic polymers. 
Page 2


  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
CBSE 
Class XII Chemistry 
Board Paper 2014 – Set 1 Solution 
Time: 3 hrs. Total Marks: 70 
 
1. High temperature is favourable for adsorption, which increases with an increase in 
temperature. 
2. In the extraction of silver, zinc functions as a reducing agent. 
 
          Zn + 2Na[Ag(CN)2] ? Na2[Zn(CN)4] + 2Ag 
3. Basicity of H3PO3 is two as there are two replaceable hydrogen atoms. The structure 
of H3PO3 is as follows: 
 
4.    The two structures given are as follows: 
 
                                      
                                   2-chlorobutane                                   1-chlorobutane 
Between the two given compounds, 2-chlorobutane contains one carbon atom 
which is attached to four entities such as methyl group, chlorine atom, hydrogen 
atom and ethyl group.  
 
2-chlorobutane with chiral centre 
 
But in 1-chlorobutane, no carbon atom is attached to four atoms or groups. 
Hence, the chiral compound is 2-chlorobutane. 
 
5. Proteins are natural polymers.  
Buna-S and PVC are synthetic polymers. 
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
6. Conversion of primary aromatic amines to diazonium salts is known as 
diazotisation. 
 
7. Sucrose on hydrolysis gives one molecule each of glucose and fructose. 
 
8. Structure of p-methyl benzaldehyde: 
 
p-methyl benzaldehyde 
9. Given: 
Density of f.c.c. unit cell = 2.8 g cm
-3
 
Edge length of f.c.c unit cell = 4 × 10
-8
 cm 
NA = 6.022 × 10
23
 mol
-1
 
Molar mass of the element = ? 
 
For f.c.c. unit cell, Z = 4 
Substituting the values in the equation: 
3
0
Z×M
?=
a ×N
4×M
-3
2.8gcm =
-8 3 23 -1
(4×10 cm) ×6.022×10 mol
-3 -8 3 23 -1
2.8gcm ×(4×10 cm) ×6.022×10 mol
M=
4
-1
M =26.98gmol
  
Therefore, the molar mass of the element is 26.98 gmol
-1
. 
 
10.    
(i) The metal excess defect due to anionic vacancies makes LiCl crystals pink.  
 
(ii) NaCl exhibits Schottky defect. In this type of defect, in an ionic crystal such as NaCl, 
equal number of cations and anions are missing from their lattice sites so that the 
electrical neutrality is maintained. This is a vacancy defect. 
OR 
 
Page 3


  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
CBSE 
Class XII Chemistry 
Board Paper 2014 – Set 1 Solution 
Time: 3 hrs. Total Marks: 70 
 
1. High temperature is favourable for adsorption, which increases with an increase in 
temperature. 
2. In the extraction of silver, zinc functions as a reducing agent. 
 
          Zn + 2Na[Ag(CN)2] ? Na2[Zn(CN)4] + 2Ag 
3. Basicity of H3PO3 is two as there are two replaceable hydrogen atoms. The structure 
of H3PO3 is as follows: 
 
4.    The two structures given are as follows: 
 
                                      
                                   2-chlorobutane                                   1-chlorobutane 
Between the two given compounds, 2-chlorobutane contains one carbon atom 
which is attached to four entities such as methyl group, chlorine atom, hydrogen 
atom and ethyl group.  
 
2-chlorobutane with chiral centre 
 
But in 1-chlorobutane, no carbon atom is attached to four atoms or groups. 
Hence, the chiral compound is 2-chlorobutane. 
 
5. Proteins are natural polymers.  
Buna-S and PVC are synthetic polymers. 
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
6. Conversion of primary aromatic amines to diazonium salts is known as 
diazotisation. 
 
7. Sucrose on hydrolysis gives one molecule each of glucose and fructose. 
 
8. Structure of p-methyl benzaldehyde: 
 
p-methyl benzaldehyde 
9. Given: 
Density of f.c.c. unit cell = 2.8 g cm
-3
 
Edge length of f.c.c unit cell = 4 × 10
-8
 cm 
NA = 6.022 × 10
23
 mol
-1
 
Molar mass of the element = ? 
 
For f.c.c. unit cell, Z = 4 
Substituting the values in the equation: 
3
0
Z×M
?=
a ×N
4×M
-3
2.8gcm =
-8 3 23 -1
(4×10 cm) ×6.022×10 mol
-3 -8 3 23 -1
2.8gcm ×(4×10 cm) ×6.022×10 mol
M=
4
-1
M =26.98gmol
  
Therefore, the molar mass of the element is 26.98 gmol
-1
. 
 
10.    
(i) The metal excess defect due to anionic vacancies makes LiCl crystals pink.  
 
(ii) NaCl exhibits Schottky defect. In this type of defect, in an ionic crystal such as NaCl, 
equal number of cations and anions are missing from their lattice sites so that the 
electrical neutrality is maintained. This is a vacancy defect. 
OR 
 
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
(i) Difference between tetrahedral and octahedral voids: 
 
Tetrahedral void Octahedral void 
Tetrahedral void is a vacant space 
formed by touching of four spheres. 
 
Octahedral void is formed by a 
combination of voids of two 
triangular layers of atoms when one 
triangular layer is placed above 
another. 
 
 
 
(ii) Difference between crystal lattice and unit cell: 
 
Crystal lattice Unit cell 
The crystal lattice is a symmetrical 
three-dimensional arrangement of 
atoms inside a crystal. 
The unit cell is the smallest portion 
of crystal lattice which when 
repeated in different directions 
generates an entire lattice.  
  
11.  Kohlrausch law: The limiting molar conductivity of an electrolyte (i.e. molar 
conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the 
cation and the anion each multiplied with the number of ions present in one formula 
unit of the electrolyte. 
On dilution, the volume of the solution increases, but the number of ions which 
carry a charge in a solution remains the same. Hence, the number of ions per unit 
volume which carry a charge in a solution decreases. Conductivity through a 
solution is due to the presence of these ions. Therefore, the more salt or ions 
dissolved in solution, the higher the conductivity (within limits). In a dilute solution, 
the   concentration of ions goes down, and the ability to pass a current is diminished. 
 
  
Page 4


  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
CBSE 
Class XII Chemistry 
Board Paper 2014 – Set 1 Solution 
Time: 3 hrs. Total Marks: 70 
 
1. High temperature is favourable for adsorption, which increases with an increase in 
temperature. 
2. In the extraction of silver, zinc functions as a reducing agent. 
 
          Zn + 2Na[Ag(CN)2] ? Na2[Zn(CN)4] + 2Ag 
3. Basicity of H3PO3 is two as there are two replaceable hydrogen atoms. The structure 
of H3PO3 is as follows: 
 
4.    The two structures given are as follows: 
 
                                      
                                   2-chlorobutane                                   1-chlorobutane 
Between the two given compounds, 2-chlorobutane contains one carbon atom 
which is attached to four entities such as methyl group, chlorine atom, hydrogen 
atom and ethyl group.  
 
2-chlorobutane with chiral centre 
 
But in 1-chlorobutane, no carbon atom is attached to four atoms or groups. 
Hence, the chiral compound is 2-chlorobutane. 
 
5. Proteins are natural polymers.  
Buna-S and PVC are synthetic polymers. 
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
6. Conversion of primary aromatic amines to diazonium salts is known as 
diazotisation. 
 
7. Sucrose on hydrolysis gives one molecule each of glucose and fructose. 
 
8. Structure of p-methyl benzaldehyde: 
 
p-methyl benzaldehyde 
9. Given: 
Density of f.c.c. unit cell = 2.8 g cm
-3
 
Edge length of f.c.c unit cell = 4 × 10
-8
 cm 
NA = 6.022 × 10
23
 mol
-1
 
Molar mass of the element = ? 
 
For f.c.c. unit cell, Z = 4 
Substituting the values in the equation: 
3
0
Z×M
?=
a ×N
4×M
-3
2.8gcm =
-8 3 23 -1
(4×10 cm) ×6.022×10 mol
-3 -8 3 23 -1
2.8gcm ×(4×10 cm) ×6.022×10 mol
M=
4
-1
M =26.98gmol
  
Therefore, the molar mass of the element is 26.98 gmol
-1
. 
 
10.    
(i) The metal excess defect due to anionic vacancies makes LiCl crystals pink.  
 
(ii) NaCl exhibits Schottky defect. In this type of defect, in an ionic crystal such as NaCl, 
equal number of cations and anions are missing from their lattice sites so that the 
electrical neutrality is maintained. This is a vacancy defect. 
OR 
 
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
(i) Difference between tetrahedral and octahedral voids: 
 
Tetrahedral void Octahedral void 
Tetrahedral void is a vacant space 
formed by touching of four spheres. 
 
Octahedral void is formed by a 
combination of voids of two 
triangular layers of atoms when one 
triangular layer is placed above 
another. 
 
 
 
(ii) Difference between crystal lattice and unit cell: 
 
Crystal lattice Unit cell 
The crystal lattice is a symmetrical 
three-dimensional arrangement of 
atoms inside a crystal. 
The unit cell is the smallest portion 
of crystal lattice which when 
repeated in different directions 
generates an entire lattice.  
  
11.  Kohlrausch law: The limiting molar conductivity of an electrolyte (i.e. molar 
conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the 
cation and the anion each multiplied with the number of ions present in one formula 
unit of the electrolyte. 
On dilution, the volume of the solution increases, but the number of ions which 
carry a charge in a solution remains the same. Hence, the number of ions per unit 
volume which carry a charge in a solution decreases. Conductivity through a 
solution is due to the presence of these ions. Therefore, the more salt or ions 
dissolved in solution, the higher the conductivity (within limits). In a dilute solution, 
the   concentration of ions goes down, and the ability to pass a current is diminished. 
 
  
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
12. For a chemical reaction R?P, the variation in the concentration (R) vs. time (t) plot 
is given as follows: 
 
 
(i) It is a zero-order reaction. 
(ii) Slope = -k 
 
13. Electrolytic refining: 
Principle:  
When electric current is passed, metal ions from electrolyte are deposited at the 
cathode in the form of pure metal, while metal from the anode goes into the 
electrolyte solution as metal ions. 
? ? ? ?
? ? ? ?
??
??
??
??
n
s aq
n
aq s
Anode: M M ne
cathode:M ne M
 
Example: In case of electrolytic refining of copper, crude copper metal acts as the 
anode, while a thin sheet of pure copper acts as the cathode. The electrolyte is 
copper sulphate solution acidified with sulphuric acid. 
? ? ? ?
? ? ? ?
??
??
??
??
2
s aq
2
aq s
Anode: Cu Cu 2e
cathode:Cu 2e Cu
 
14.  
(i) P4 + H2O ?  
     P4 + 6H2O? PH3 + 3H3PO2 
 
(ii) XeF6 can be prepared by the interaction of XeF4 and O2F2 at 143 K. 
      XeF4 + O2F2 ? XeF6 + O2 
 
  
Page 5


  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
CBSE 
Class XII Chemistry 
Board Paper 2014 – Set 1 Solution 
Time: 3 hrs. Total Marks: 70 
 
1. High temperature is favourable for adsorption, which increases with an increase in 
temperature. 
2. In the extraction of silver, zinc functions as a reducing agent. 
 
          Zn + 2Na[Ag(CN)2] ? Na2[Zn(CN)4] + 2Ag 
3. Basicity of H3PO3 is two as there are two replaceable hydrogen atoms. The structure 
of H3PO3 is as follows: 
 
4.    The two structures given are as follows: 
 
                                      
                                   2-chlorobutane                                   1-chlorobutane 
Between the two given compounds, 2-chlorobutane contains one carbon atom 
which is attached to four entities such as methyl group, chlorine atom, hydrogen 
atom and ethyl group.  
 
2-chlorobutane with chiral centre 
 
But in 1-chlorobutane, no carbon atom is attached to four atoms or groups. 
Hence, the chiral compound is 2-chlorobutane. 
 
5. Proteins are natural polymers.  
Buna-S and PVC are synthetic polymers. 
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
6. Conversion of primary aromatic amines to diazonium salts is known as 
diazotisation. 
 
7. Sucrose on hydrolysis gives one molecule each of glucose and fructose. 
 
8. Structure of p-methyl benzaldehyde: 
 
p-methyl benzaldehyde 
9. Given: 
Density of f.c.c. unit cell = 2.8 g cm
-3
 
Edge length of f.c.c unit cell = 4 × 10
-8
 cm 
NA = 6.022 × 10
23
 mol
-1
 
Molar mass of the element = ? 
 
For f.c.c. unit cell, Z = 4 
Substituting the values in the equation: 
3
0
Z×M
?=
a ×N
4×M
-3
2.8gcm =
-8 3 23 -1
(4×10 cm) ×6.022×10 mol
-3 -8 3 23 -1
2.8gcm ×(4×10 cm) ×6.022×10 mol
M=
4
-1
M =26.98gmol
  
Therefore, the molar mass of the element is 26.98 gmol
-1
. 
 
10.    
(i) The metal excess defect due to anionic vacancies makes LiCl crystals pink.  
 
(ii) NaCl exhibits Schottky defect. In this type of defect, in an ionic crystal such as NaCl, 
equal number of cations and anions are missing from their lattice sites so that the 
electrical neutrality is maintained. This is a vacancy defect. 
OR 
 
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
(i) Difference between tetrahedral and octahedral voids: 
 
Tetrahedral void Octahedral void 
Tetrahedral void is a vacant space 
formed by touching of four spheres. 
 
Octahedral void is formed by a 
combination of voids of two 
triangular layers of atoms when one 
triangular layer is placed above 
another. 
 
 
 
(ii) Difference between crystal lattice and unit cell: 
 
Crystal lattice Unit cell 
The crystal lattice is a symmetrical 
three-dimensional arrangement of 
atoms inside a crystal. 
The unit cell is the smallest portion 
of crystal lattice which when 
repeated in different directions 
generates an entire lattice.  
  
11.  Kohlrausch law: The limiting molar conductivity of an electrolyte (i.e. molar 
conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the 
cation and the anion each multiplied with the number of ions present in one formula 
unit of the electrolyte. 
On dilution, the volume of the solution increases, but the number of ions which 
carry a charge in a solution remains the same. Hence, the number of ions per unit 
volume which carry a charge in a solution decreases. Conductivity through a 
solution is due to the presence of these ions. Therefore, the more salt or ions 
dissolved in solution, the higher the conductivity (within limits). In a dilute solution, 
the   concentration of ions goes down, and the ability to pass a current is diminished. 
 
  
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
12. For a chemical reaction R?P, the variation in the concentration (R) vs. time (t) plot 
is given as follows: 
 
 
(i) It is a zero-order reaction. 
(ii) Slope = -k 
 
13. Electrolytic refining: 
Principle:  
When electric current is passed, metal ions from electrolyte are deposited at the 
cathode in the form of pure metal, while metal from the anode goes into the 
electrolyte solution as metal ions. 
? ? ? ?
? ? ? ?
??
??
??
??
n
s aq
n
aq s
Anode: M M ne
cathode:M ne M
 
Example: In case of electrolytic refining of copper, crude copper metal acts as the 
anode, while a thin sheet of pure copper acts as the cathode. The electrolyte is 
copper sulphate solution acidified with sulphuric acid. 
? ? ? ?
? ? ? ?
??
??
??
??
2
s aq
2
aq s
Anode: Cu Cu 2e
cathode:Cu 2e Cu
 
14.  
(i) P4 + H2O ?  
     P4 + 6H2O? PH3 + 3H3PO2 
 
(ii) XeF6 can be prepared by the interaction of XeF4 and O2F2 at 143 K. 
      XeF4 + O2F2 ? XeF6 + O2 
 
  
  
 
CBSE  |  Chemistry 
Board Paper ? 2014 
 
     
15.     
(i) XeF2 (linear) 
 
 
(ii) BrF3 (bent T-shaped) 
 
16.    
(i) Reimer–Tiemann reaction:  Treatment of phenol with chloroform in the 
presence of aqueous sodium or potassium hydroxide at 340 K, followed by 
hydrolysis of the resulting product gives 2-hydroxybenzaldehyde or 
salicylaldehyde as the major product. This reaction is called the Reimer–
Tiemann formylation reaction. 
 
 
 
(ii) Williamson synthesis: It involves the treatment of an alkyl halide with a 
suitable sodium alkoxide. The sodium alkoxide needed for the purpose is 
prepared by the action of sodium on a suitable alcohol. The reaction involves 
the nucleophillic displacement (substitution) of the halide ion from the alkyl 
halide by the alkoxide ion by the SN2 mechanism. 
 
R’–OH + 2Na ? 2R’–O
-
Na
+
 + H2 
R’–O
-
Na
+
 + R–X ? R’–O–R   + Na
+ 
X
-
 
                                                   Ether 
The second reaction is the substitution nucleophillic bimolecular (SN2) 
reaction.