Class 12 Chemistry: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical PDF Download

 Page 1


SAMPLE PAPER (2023 -24) 
CHEMISTRY THEORY (043) 
MARKING SCHEME 
SECTION A 
1. (b) 0.1 M HCl solution, conductivity is higher for strong electrolyte, conductivity decreases with 
dilution 
2. (c)  A= Benzaldehyde , B= Acetophenone. This is an example of crossed Aldol condensation. 
3. (d) Vitamin A,D, E and K These are fat soluble vitamins 
4. (a) 3-Methylpent-3-en-2-one 
5. (b) The carbon-magnesium bond is covalent and non-polar in nature . 
6. (a) (i) (r) , (ii) (q), (iii) (p) 
Zinc has no unpaired electrons in 3d or 4s orbitals, so enthalpy of atomization is low 
Mn = 3d
5
4s
2 
shows +2,+3,+4,+5,+6 and+7 oxidation state , maximum number in 3d series 
7. (b) molecularity has no meaning for complex reactions. 
8. (c) water 
9. (b) no turbidity will be observed, given compound is a primary alcohol 
10. (a) 0.00625 molL
-1
s
-1 
for zero order k = [Ro] –[R] / t = 1.5 -0.75 /120 
11. (a) Due to the activation of benzene ring by the methoxy group. 
12. (b) atomic radii 
for visually challenged learners 
12. (d) Zinc 
13. (c) A is true but R is false 
14. (b) Both A and R are true but R is not the correct explanation of A 
15. (b) Both A and R are true and R is not the correct explanation of A. 
16. (d) A is false but R is true. 
Cu will deposit at cathode 
SECTION B 
17. (a)  for first order reaction   
                   half life of X = 12 hours                                                            
2 days = 48 hours means 4 half lives , amount of X left = 1/16 of initial value 
half life of Y = 16 hours                                                                    ( ½) 
          2 days = 48 hours means 3 half lives, amount left = 1/8 of initial value 
Ratio of X:Y = 1:2                                                                              ( ½) 
     (b) b. mol
1/2
L
-1/2
s
-1         
as
 
Rate = k [P]
1/2    
                                                                 (1)
                                                                                                         
  
18. p1 = p2 (1/2) 
iC 1RT = C 2RT 
 
(1/2) 
Page 2


SAMPLE PAPER (2023 -24) 
CHEMISTRY THEORY (043) 
MARKING SCHEME 
SECTION A 
1. (b) 0.1 M HCl solution, conductivity is higher for strong electrolyte, conductivity decreases with 
dilution 
2. (c)  A= Benzaldehyde , B= Acetophenone. This is an example of crossed Aldol condensation. 
3. (d) Vitamin A,D, E and K These are fat soluble vitamins 
4. (a) 3-Methylpent-3-en-2-one 
5. (b) The carbon-magnesium bond is covalent and non-polar in nature . 
6. (a) (i) (r) , (ii) (q), (iii) (p) 
Zinc has no unpaired electrons in 3d or 4s orbitals, so enthalpy of atomization is low 
Mn = 3d
5
4s
2 
shows +2,+3,+4,+5,+6 and+7 oxidation state , maximum number in 3d series 
7. (b) molecularity has no meaning for complex reactions. 
8. (c) water 
9. (b) no turbidity will be observed, given compound is a primary alcohol 
10. (a) 0.00625 molL
-1
s
-1 
for zero order k = [Ro] –[R] / t = 1.5 -0.75 /120 
11. (a) Due to the activation of benzene ring by the methoxy group. 
12. (b) atomic radii 
for visually challenged learners 
12. (d) Zinc 
13. (c) A is true but R is false 
14. (b) Both A and R are true but R is not the correct explanation of A 
15. (b) Both A and R are true and R is not the correct explanation of A. 
16. (d) A is false but R is true. 
Cu will deposit at cathode 
SECTION B 
17. (a)  for first order reaction   
                   half life of X = 12 hours                                                            
2 days = 48 hours means 4 half lives , amount of X left = 1/16 of initial value 
half life of Y = 16 hours                                                                    ( ½) 
          2 days = 48 hours means 3 half lives, amount left = 1/8 of initial value 
Ratio of X:Y = 1:2                                                                              ( ½) 
     (b) b. mol
1/2
L
-1/2
s
-1         
as
 
Rate = k [P]
1/2    
                                                                 (1)
                                                                                                         
  
18. p1 = p2 (1/2) 
iC 1RT = C 2RT 
 
(1/2) 
3 ? 5 
?
 2  
322 M 
M ? 
2 ? 322 
3 ? 5 
M = 42.9 g 
(1/2) 
 
(1/2) 
19. (a)  m-dicholrobenzene < o-dicholrobenzene < p-dicholrobenze (1/2) 
symmetrical structure and close packing in para isomer 
ortho has a stronger dipole dipole interaction as compared to meta (1/2) 
(b) the halogen atom because of its –I effect has some tendency to withdraw electrons from the 
benzene ring. As a result, the ring gets somewhat deactivated as compared to benzene and hence 
the electrophilic substitution reactions in haloarenes occur slowly and require more drastic 
conditions as compared to those in benzene. (1) 
20. (a) p-nitrobenzaldehyde is more reactive towards the nucleophilic addition reaction than p- 
tolualdehyde as Nitro group is electron withdrawing in nature . Presence of nitro group decrease 
electron density, hence facilitates the attack of nucleophile . Presence of -CH 3 leads to +I effect as - 
CH 3 is electron releasing group. (1) 
 
(b) 
 
 
 
 
OR 
(a) 
 
 
 
 
 
 
(b) 
( (1) 
 
 
 
 
 
(1) 
 
 
 
 
 
(1) 
21. (a) Replication 
A sequence of bases on DNA is unique for a person and is the genetic material transferred to the 
individual from the parent which helps in the determination of paternity. (1) 
 
 
  
 
Page 3


SAMPLE PAPER (2023 -24) 
CHEMISTRY THEORY (043) 
MARKING SCHEME 
SECTION A 
1. (b) 0.1 M HCl solution, conductivity is higher for strong electrolyte, conductivity decreases with 
dilution 
2. (c)  A= Benzaldehyde , B= Acetophenone. This is an example of crossed Aldol condensation. 
3. (d) Vitamin A,D, E and K These are fat soluble vitamins 
4. (a) 3-Methylpent-3-en-2-one 
5. (b) The carbon-magnesium bond is covalent and non-polar in nature . 
6. (a) (i) (r) , (ii) (q), (iii) (p) 
Zinc has no unpaired electrons in 3d or 4s orbitals, so enthalpy of atomization is low 
Mn = 3d
5
4s
2 
shows +2,+3,+4,+5,+6 and+7 oxidation state , maximum number in 3d series 
7. (b) molecularity has no meaning for complex reactions. 
8. (c) water 
9. (b) no turbidity will be observed, given compound is a primary alcohol 
10. (a) 0.00625 molL
-1
s
-1 
for zero order k = [Ro] –[R] / t = 1.5 -0.75 /120 
11. (a) Due to the activation of benzene ring by the methoxy group. 
12. (b) atomic radii 
for visually challenged learners 
12. (d) Zinc 
13. (c) A is true but R is false 
14. (b) Both A and R are true but R is not the correct explanation of A 
15. (b) Both A and R are true and R is not the correct explanation of A. 
16. (d) A is false but R is true. 
Cu will deposit at cathode 
SECTION B 
17. (a)  for first order reaction   
                   half life of X = 12 hours                                                            
2 days = 48 hours means 4 half lives , amount of X left = 1/16 of initial value 
half life of Y = 16 hours                                                                    ( ½) 
          2 days = 48 hours means 3 half lives, amount left = 1/8 of initial value 
Ratio of X:Y = 1:2                                                                              ( ½) 
     (b) b. mol
1/2
L
-1/2
s
-1         
as
 
Rate = k [P]
1/2    
                                                                 (1)
                                                                                                         
  
18. p1 = p2 (1/2) 
iC 1RT = C 2RT 
 
(1/2) 
3 ? 5 
?
 2  
322 M 
M ? 
2 ? 322 
3 ? 5 
M = 42.9 g 
(1/2) 
 
(1/2) 
19. (a)  m-dicholrobenzene < o-dicholrobenzene < p-dicholrobenze (1/2) 
symmetrical structure and close packing in para isomer 
ortho has a stronger dipole dipole interaction as compared to meta (1/2) 
(b) the halogen atom because of its –I effect has some tendency to withdraw electrons from the 
benzene ring. As a result, the ring gets somewhat deactivated as compared to benzene and hence 
the electrophilic substitution reactions in haloarenes occur slowly and require more drastic 
conditions as compared to those in benzene. (1) 
20. (a) p-nitrobenzaldehyde is more reactive towards the nucleophilic addition reaction than p- 
tolualdehyde as Nitro group is electron withdrawing in nature . Presence of nitro group decrease 
electron density, hence facilitates the attack of nucleophile . Presence of -CH 3 leads to +I effect as - 
CH 3 is electron releasing group. (1) 
 
(b) 
 
 
 
 
OR 
(a) 
 
 
 
 
 
 
(b) 
( (1) 
 
 
 
 
 
(1) 
 
 
 
 
 
(1) 
21. (a) Replication 
A sequence of bases on DNA is unique for a person and is the genetic material transferred to the 
individual from the parent which helps in the determination of paternity. (1) 
 
 
  
 
(b) During denaturation secondary and tertiary structures are destroyed but the primary structure 
remains intact. (1) 
 
SECTION C 
22. (a) [Cr(en)2(OH)2]Cl or [Cr(H 2NCH 2CH 2NH 2)2(OH)2]Cl (1) 
(b) No, ionization isomers are possible by exchange of ligand with counter ion only and not by 
exchange of central metal ion. (1) 
(c) The central atom is electron pair acceptor so it is a Lewis acid. (1) 
23. (a) Yes, if the concentration of ZnSO 4 in the two half cell is different , the electrode potential will be 
different making the cell possible. (1) 
(b) ?
0
m (MgCl2) = ?
0
m ( Mg
2+
) + 2 ?
0
m ( Cl
-
) 
258.6 = 106 + 2 ?
0
m ( Cl
-
) 
?
0
m ( Cl
-
) = 76.3 Scm
2
mol
–1 
(1) 
(c) cell constant G* = k x R 
k= G*/R = 0.146/ 1000 = 1.46 x 10 
-4 
Scm
-1
. (1) 
24. (a) Reimer Tiemann , (1/2) 
OH 
CHO 
2-Hydroxybenzaldehyde (1/2+1/2) 
 
(b) Williamson synthesis, CH 3CH 2CH(CH 3)CH(CH 3)O C 2H 5 
2- Ethoxy-3 -methylpentane (1/2+1/2+1/2) 
         (c) Stephen reaction, CH 3CH 2CHO,  Propanal                                                         (1/2+1/2 +1/2) 
25. A, B and C contain carbonyl group as they give positive 2,4 DNP test 
A and B are aldehydes as aldehydes reduce Tollen’s reagent 
C is a ketone, as it contains carbonyl group but does not give positive Tollen’s test (1/2) 
C is a methyl ketone as it gives positive iodoform test 
B is an aldehyde that gives positive iodoform test (1/2) 
D is a carboxylic acid 
Since the number of carbons in the compounds A,B,C and D is three or two 
B is CH 3CHO as this is only aldehyde which gives a positive i odoform test (1/2) 
The remaining compounds A, C and D have three carbons 
A is CH 3CH 2CHO, C is CH 3COCH 3 and D is CH 3CH 2COOH (1/2 each) 
26. (a) The reactant Sucrose is dextrorotatory. On hydrolysis it give glucose dextrorotatory and 
fructose which is leavoroatatory. The specific rotation of fructose is higher than glucose 
Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory 
fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), 
the mixture is laevorotatory. (1) 
(b) Invert sugar, The hydrolysis of sucrose brings about a change in the sign of rotation, from 
dextro (+) to laevo (–) and the product is named as invert sugar. (1) 
(c) Glucose (1) 
Page 4


SAMPLE PAPER (2023 -24) 
CHEMISTRY THEORY (043) 
MARKING SCHEME 
SECTION A 
1. (b) 0.1 M HCl solution, conductivity is higher for strong electrolyte, conductivity decreases with 
dilution 
2. (c)  A= Benzaldehyde , B= Acetophenone. This is an example of crossed Aldol condensation. 
3. (d) Vitamin A,D, E and K These are fat soluble vitamins 
4. (a) 3-Methylpent-3-en-2-one 
5. (b) The carbon-magnesium bond is covalent and non-polar in nature . 
6. (a) (i) (r) , (ii) (q), (iii) (p) 
Zinc has no unpaired electrons in 3d or 4s orbitals, so enthalpy of atomization is low 
Mn = 3d
5
4s
2 
shows +2,+3,+4,+5,+6 and+7 oxidation state , maximum number in 3d series 
7. (b) molecularity has no meaning for complex reactions. 
8. (c) water 
9. (b) no turbidity will be observed, given compound is a primary alcohol 
10. (a) 0.00625 molL
-1
s
-1 
for zero order k = [Ro] –[R] / t = 1.5 -0.75 /120 
11. (a) Due to the activation of benzene ring by the methoxy group. 
12. (b) atomic radii 
for visually challenged learners 
12. (d) Zinc 
13. (c) A is true but R is false 
14. (b) Both A and R are true but R is not the correct explanation of A 
15. (b) Both A and R are true and R is not the correct explanation of A. 
16. (d) A is false but R is true. 
Cu will deposit at cathode 
SECTION B 
17. (a)  for first order reaction   
                   half life of X = 12 hours                                                            
2 days = 48 hours means 4 half lives , amount of X left = 1/16 of initial value 
half life of Y = 16 hours                                                                    ( ½) 
          2 days = 48 hours means 3 half lives, amount left = 1/8 of initial value 
Ratio of X:Y = 1:2                                                                              ( ½) 
     (b) b. mol
1/2
L
-1/2
s
-1         
as
 
Rate = k [P]
1/2    
                                                                 (1)
                                                                                                         
  
18. p1 = p2 (1/2) 
iC 1RT = C 2RT 
 
(1/2) 
3 ? 5 
?
 2  
322 M 
M ? 
2 ? 322 
3 ? 5 
M = 42.9 g 
(1/2) 
 
(1/2) 
19. (a)  m-dicholrobenzene < o-dicholrobenzene < p-dicholrobenze (1/2) 
symmetrical structure and close packing in para isomer 
ortho has a stronger dipole dipole interaction as compared to meta (1/2) 
(b) the halogen atom because of its –I effect has some tendency to withdraw electrons from the 
benzene ring. As a result, the ring gets somewhat deactivated as compared to benzene and hence 
the electrophilic substitution reactions in haloarenes occur slowly and require more drastic 
conditions as compared to those in benzene. (1) 
20. (a) p-nitrobenzaldehyde is more reactive towards the nucleophilic addition reaction than p- 
tolualdehyde as Nitro group is electron withdrawing in nature . Presence of nitro group decrease 
electron density, hence facilitates the attack of nucleophile . Presence of -CH 3 leads to +I effect as - 
CH 3 is electron releasing group. (1) 
 
(b) 
 
 
 
 
OR 
(a) 
 
 
 
 
 
 
(b) 
( (1) 
 
 
 
 
 
(1) 
 
 
 
 
 
(1) 
21. (a) Replication 
A sequence of bases on DNA is unique for a person and is the genetic material transferred to the 
individual from the parent which helps in the determination of paternity. (1) 
 
 
  
 
(b) During denaturation secondary and tertiary structures are destroyed but the primary structure 
remains intact. (1) 
 
SECTION C 
22. (a) [Cr(en)2(OH)2]Cl or [Cr(H 2NCH 2CH 2NH 2)2(OH)2]Cl (1) 
(b) No, ionization isomers are possible by exchange of ligand with counter ion only and not by 
exchange of central metal ion. (1) 
(c) The central atom is electron pair acceptor so it is a Lewis acid. (1) 
23. (a) Yes, if the concentration of ZnSO 4 in the two half cell is different , the electrode potential will be 
different making the cell possible. (1) 
(b) ?
0
m (MgCl2) = ?
0
m ( Mg
2+
) + 2 ?
0
m ( Cl
-
) 
258.6 = 106 + 2 ?
0
m ( Cl
-
) 
?
0
m ( Cl
-
) = 76.3 Scm
2
mol
–1 
(1) 
(c) cell constant G* = k x R 
k= G*/R = 0.146/ 1000 = 1.46 x 10 
-4 
Scm
-1
. (1) 
24. (a) Reimer Tiemann , (1/2) 
OH 
CHO 
2-Hydroxybenzaldehyde (1/2+1/2) 
 
(b) Williamson synthesis, CH 3CH 2CH(CH 3)CH(CH 3)O C 2H 5 
2- Ethoxy-3 -methylpentane (1/2+1/2+1/2) 
         (c) Stephen reaction, CH 3CH 2CHO,  Propanal                                                         (1/2+1/2 +1/2) 
25. A, B and C contain carbonyl group as they give positive 2,4 DNP test 
A and B are aldehydes as aldehydes reduce Tollen’s reagent 
C is a ketone, as it contains carbonyl group but does not give positive Tollen’s test (1/2) 
C is a methyl ketone as it gives positive iodoform test 
B is an aldehyde that gives positive iodoform test (1/2) 
D is a carboxylic acid 
Since the number of carbons in the compounds A,B,C and D is three or two 
B is CH 3CHO as this is only aldehyde which gives a positive i odoform test (1/2) 
The remaining compounds A, C and D have three carbons 
A is CH 3CH 2CHO, C is CH 3COCH 3 and D is CH 3CH 2COOH (1/2 each) 
26. (a) The reactant Sucrose is dextrorotatory. On hydrolysis it give glucose dextrorotatory and 
fructose which is leavoroatatory. The specific rotation of fructose is higher than glucose 
Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory 
fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), 
the mixture is laevorotatory. (1) 
(b) Invert sugar, The hydrolysis of sucrose brings about a change in the sign of rotation, from 
dextro (+) to laevo (–) and the product is named as invert sugar. (1) 
(c) Glucose (1) 
 
27. C
2
H
5   
CH CH
3
 
Br 
(1) 
       (1+1) 
 
28. 
 
 
 
 
 
 
 
 
 
 
                                                            (                                                                         ( 1+1+1) 
       
SECTION D 
29. (a) D. Energy is directly proportional to the wave number. Maximum energy of light is required 
for an electron to jump from t 2g to eg in case of [CrD 6]
3- 
(1/2+1/2) 
OR 
(a) A, The splitting caused in least in this case as the energy required for electron to jump from t 2g 
to eg., is minimum. (1/2 +1/2) 
(b) [CrB 6]
3+
, wavelength of light absorbed is 1/17830 = 560nm for the complex while 1/13640 = 733 
nm for [CrA 6]
3- 
complex. (1/2+1/2) 
(c) (i) [CrCl 6]
3- 
, Hexachloridochromate(III) ion (1 each) 
(ii) [Cr(NH 3) 6]
3+
, Hexaamminechromium(III) ion 
A = Cl
–
, B= H 2O, C= NH 3, D= CN
–
 
30. (a) 2mol e
– 
(or 2F) have been transferred from anode to cathode to consume 2 mol of H 2SO 4 
Page 5


SAMPLE PAPER (2023 -24) 
CHEMISTRY THEORY (043) 
MARKING SCHEME 
SECTION A 
1. (b) 0.1 M HCl solution, conductivity is higher for strong electrolyte, conductivity decreases with 
dilution 
2. (c)  A= Benzaldehyde , B= Acetophenone. This is an example of crossed Aldol condensation. 
3. (d) Vitamin A,D, E and K These are fat soluble vitamins 
4. (a) 3-Methylpent-3-en-2-one 
5. (b) The carbon-magnesium bond is covalent and non-polar in nature . 
6. (a) (i) (r) , (ii) (q), (iii) (p) 
Zinc has no unpaired electrons in 3d or 4s orbitals, so enthalpy of atomization is low 
Mn = 3d
5
4s
2 
shows +2,+3,+4,+5,+6 and+7 oxidation state , maximum number in 3d series 
7. (b) molecularity has no meaning for complex reactions. 
8. (c) water 
9. (b) no turbidity will be observed, given compound is a primary alcohol 
10. (a) 0.00625 molL
-1
s
-1 
for zero order k = [Ro] –[R] / t = 1.5 -0.75 /120 
11. (a) Due to the activation of benzene ring by the methoxy group. 
12. (b) atomic radii 
for visually challenged learners 
12. (d) Zinc 
13. (c) A is true but R is false 
14. (b) Both A and R are true but R is not the correct explanation of A 
15. (b) Both A and R are true and R is not the correct explanation of A. 
16. (d) A is false but R is true. 
Cu will deposit at cathode 
SECTION B 
17. (a)  for first order reaction   
                   half life of X = 12 hours                                                            
2 days = 48 hours means 4 half lives , amount of X left = 1/16 of initial value 
half life of Y = 16 hours                                                                    ( ½) 
          2 days = 48 hours means 3 half lives, amount left = 1/8 of initial value 
Ratio of X:Y = 1:2                                                                              ( ½) 
     (b) b. mol
1/2
L
-1/2
s
-1         
as
 
Rate = k [P]
1/2    
                                                                 (1)
                                                                                                         
  
18. p1 = p2 (1/2) 
iC 1RT = C 2RT 
 
(1/2) 
3 ? 5 
?
 2  
322 M 
M ? 
2 ? 322 
3 ? 5 
M = 42.9 g 
(1/2) 
 
(1/2) 
19. (a)  m-dicholrobenzene < o-dicholrobenzene < p-dicholrobenze (1/2) 
symmetrical structure and close packing in para isomer 
ortho has a stronger dipole dipole interaction as compared to meta (1/2) 
(b) the halogen atom because of its –I effect has some tendency to withdraw electrons from the 
benzene ring. As a result, the ring gets somewhat deactivated as compared to benzene and hence 
the electrophilic substitution reactions in haloarenes occur slowly and require more drastic 
conditions as compared to those in benzene. (1) 
20. (a) p-nitrobenzaldehyde is more reactive towards the nucleophilic addition reaction than p- 
tolualdehyde as Nitro group is electron withdrawing in nature . Presence of nitro group decrease 
electron density, hence facilitates the attack of nucleophile . Presence of -CH 3 leads to +I effect as - 
CH 3 is electron releasing group. (1) 
 
(b) 
 
 
 
 
OR 
(a) 
 
 
 
 
 
 
(b) 
( (1) 
 
 
 
 
 
(1) 
 
 
 
 
 
(1) 
21. (a) Replication 
A sequence of bases on DNA is unique for a person and is the genetic material transferred to the 
individual from the parent which helps in the determination of paternity. (1) 
 
 
  
 
(b) During denaturation secondary and tertiary structures are destroyed but the primary structure 
remains intact. (1) 
 
SECTION C 
22. (a) [Cr(en)2(OH)2]Cl or [Cr(H 2NCH 2CH 2NH 2)2(OH)2]Cl (1) 
(b) No, ionization isomers are possible by exchange of ligand with counter ion only and not by 
exchange of central metal ion. (1) 
(c) The central atom is electron pair acceptor so it is a Lewis acid. (1) 
23. (a) Yes, if the concentration of ZnSO 4 in the two half cell is different , the electrode potential will be 
different making the cell possible. (1) 
(b) ?
0
m (MgCl2) = ?
0
m ( Mg
2+
) + 2 ?
0
m ( Cl
-
) 
258.6 = 106 + 2 ?
0
m ( Cl
-
) 
?
0
m ( Cl
-
) = 76.3 Scm
2
mol
–1 
(1) 
(c) cell constant G* = k x R 
k= G*/R = 0.146/ 1000 = 1.46 x 10 
-4 
Scm
-1
. (1) 
24. (a) Reimer Tiemann , (1/2) 
OH 
CHO 
2-Hydroxybenzaldehyde (1/2+1/2) 
 
(b) Williamson synthesis, CH 3CH 2CH(CH 3)CH(CH 3)O C 2H 5 
2- Ethoxy-3 -methylpentane (1/2+1/2+1/2) 
         (c) Stephen reaction, CH 3CH 2CHO,  Propanal                                                         (1/2+1/2 +1/2) 
25. A, B and C contain carbonyl group as they give positive 2,4 DNP test 
A and B are aldehydes as aldehydes reduce Tollen’s reagent 
C is a ketone, as it contains carbonyl group but does not give positive Tollen’s test (1/2) 
C is a methyl ketone as it gives positive iodoform test 
B is an aldehyde that gives positive iodoform test (1/2) 
D is a carboxylic acid 
Since the number of carbons in the compounds A,B,C and D is three or two 
B is CH 3CHO as this is only aldehyde which gives a positive i odoform test (1/2) 
The remaining compounds A, C and D have three carbons 
A is CH 3CH 2CHO, C is CH 3COCH 3 and D is CH 3CH 2COOH (1/2 each) 
26. (a) The reactant Sucrose is dextrorotatory. On hydrolysis it give glucose dextrorotatory and 
fructose which is leavoroatatory. The specific rotation of fructose is higher than glucose 
Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory 
fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), 
the mixture is laevorotatory. (1) 
(b) Invert sugar, The hydrolysis of sucrose brings about a change in the sign of rotation, from 
dextro (+) to laevo (–) and the product is named as invert sugar. (1) 
(c) Glucose (1) 
 
27. C
2
H
5   
CH CH
3
 
Br 
(1) 
       (1+1) 
 
28. 
 
 
 
 
 
 
 
 
 
 
                                                            (                                                                         ( 1+1+1) 
       
SECTION D 
29. (a) D. Energy is directly proportional to the wave number. Maximum energy of light is required 
for an electron to jump from t 2g to eg in case of [CrD 6]
3- 
(1/2+1/2) 
OR 
(a) A, The splitting caused in least in this case as the energy required for electron to jump from t 2g 
to eg., is minimum. (1/2 +1/2) 
(b) [CrB 6]
3+
, wavelength of light absorbed is 1/17830 = 560nm for the complex while 1/13640 = 733 
nm for [CrA 6]
3- 
complex. (1/2+1/2) 
(c) (i) [CrCl 6]
3- 
, Hexachloridochromate(III) ion (1 each) 
(ii) [Cr(NH 3) 6]
3+
, Hexaamminechromium(III) ion 
A = Cl
–
, B= H 2O, C= NH 3, D= CN
–
 
30. (a) 2mol e
– 
(or 2F) have been transferred from anode to cathode to consume 2 mol of H 2SO 4 
therefore, one mole H 2SO4 requires one faraday of electricity or 96500 coulombs.                           (1) 
(b) wmax = – nFE° = – 2× 96500 × 2.0 = 386000 J of work can be extracted using lead storage cell when 
the cell is in use.                    (1) 
(c) Both yes and no should be accepted as correct answers depending upon what explanation is provided. 
Yes, Hydrogen is a fuel that on combustion gives water as a byproduct. There are no carbon emissions 
and no pollutions caused.  
However, at present the means to obtain hydrogen are electrolysis of water which use electricity obtained 
from fossil fuels and increase carbon emissions.  
Inspite of the problems faced today in the extraction of hydrogen, we cannot disagree on the fact that 
hydrogen is a clean source of energy. Further research can help in finding solutions and greens ways like 
using solar energy for extraction of hydrogen.                                                                               (2) 
No. It is true that Hydrogen is a fuel that on combustion gives water as a byproduct. There are no carbon 
emissions and no pollutions caused.  
However, at present the means to obtain hydrogen are electrolysis of water which use electricity obtained 
from fossil fuels and increase carbon emissions.  
Hydrogen is no doubt a green fuel, but the process of extraction is not green as of today. At present, 
looking at the process of extraction, hydrogen is not a green fuel.                                              (2) 
OR 
Both answers will be treated as correct 
(i) Lead batteries are currently the most important and widely used batteries. These are 
rechargeable. The problem is waste management which needs research and awareness. 
Currently, these are being thrown into landfills and there is no safe method of disposal or 
recycling. Research into safer method of disposal will reduce the pollution and health hazards 
caused to a great extent.  
( 1 mark for importance, 1 for need for the research) 
(ii) Fuel cell is a clen source of energy. Hydrogen undergoes combustion to produce water. The 
need of the hour is green fuel and hydrogen is a clean fuel. The current problem is obtaining 
hydrogen. Research that goes into this area will help solve the problem of pollution and will be 
a sustainable solution. 
( 1 mark for importance, 1 for need for the research) 
 
SECTION E 
31. (a) Both Ti
3+ 
and Cu
2+ 
have 1 unpaired electron, so the magnetic moment for both will be 1.73 BM 
(b) Zn, it has a more negative electrode potential so will corrode itself in place of iron. 
(c) Mn
+ 
has 3d
5
4s
1 
configuration and configuration of Cr
+ 
is 3d
5 
, therefore, ionisation enthalpy of 
Mn
+ 
is lower than Cr
+ 
. 
(d) Sc and Zn both form colourless compound and are diamagnetic. 
(e) The decrease in the atomic and ionic radii with increase in atomic number of actinoids due to 
poor shielding effect of 5f electron. 
(f) In both chromate and dichromate ion the oxidation state of Cr is +6 
(g) 10I
– 
+ 2MnO4 
– 
+ 16H
+ 
? 2Mn
2+ 
+ 8H2O + 5I2 (1 each, any 5)