JEE Main 2022 Chemistry June 29 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


 
   
   
CHEMISTRY 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. Using the rules for significant figures, the correct 
answer for the expression 
0.02858 0.112
0.5702
?
 will be 
 (A) 0.005613 
 (B) 0.00561 
 (C) 0.0056 
 (D) 0.006 
Answer (B) 
Sol. 
0.02858 0.112
.00561
0.5702
?
= 
 Answer expressed in 3 significant figures. 
2. Which of the following is the correct plot for the 
probability density ( )
2
r ? as a function of distance 
‘r’ of the electron from the nucleus for 2s orbital?  
 (A)
 
 
 (B)
  
 (C)
 
 
 (D)
 
 
Answer (B) 
Sol. 2s  
 radial node = n – l – 1 
    = 2 – 0 – 1  
    = 1 
 It will have one radial node 
3. Consider the species 
+
44
CH ,NH and 
–
4
BH . Choose 
the correct option with respect to these species. 
 (A) They are isoelectronic and only two have 
tetrahedral structures 
 (B) They are isoelectronic and all have tetrahedral 
structures. 
 (C) Only two are isoelectronic and all have 
tetrahedral structures. 
 (D) Only two are isoelectronic and only two have 
tetrahedral structures. 
Answer (B) 
Sol. CH4,  
+
4
NH and 
–
4
BH are isoelectronic as well as 
tetrahedral. 
Page 2


 
   
   
CHEMISTRY 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. Using the rules for significant figures, the correct 
answer for the expression 
0.02858 0.112
0.5702
?
 will be 
 (A) 0.005613 
 (B) 0.00561 
 (C) 0.0056 
 (D) 0.006 
Answer (B) 
Sol. 
0.02858 0.112
.00561
0.5702
?
= 
 Answer expressed in 3 significant figures. 
2. Which of the following is the correct plot for the 
probability density ( )
2
r ? as a function of distance 
‘r’ of the electron from the nucleus for 2s orbital?  
 (A)
 
 
 (B)
  
 (C)
 
 
 (D)
 
 
Answer (B) 
Sol. 2s  
 radial node = n – l – 1 
    = 2 – 0 – 1  
    = 1 
 It will have one radial node 
3. Consider the species 
+
44
CH ,NH and 
–
4
BH . Choose 
the correct option with respect to these species. 
 (A) They are isoelectronic and only two have 
tetrahedral structures 
 (B) They are isoelectronic and all have tetrahedral 
structures. 
 (C) Only two are isoelectronic and all have 
tetrahedral structures. 
 (D) Only two are isoelectronic and only two have 
tetrahedral structures. 
Answer (B) 
Sol. CH4,  
+
4
NH and 
–
4
BH are isoelectronic as well as 
tetrahedral. 
 
   
   
4. 4.0 moles of argon and 5.0 moles of PCl5 are 
introduced into an evacuated flask of 100 litre 
capacity at 610 K. The system is allowed to 
equilibrate. At equilibrium, the total pressure of 
mixture was found to be 6.0 atm. The Kp for the 
reaction is [Given : R = 0.082 L atm K
–1
 mol
–1
] 
 (A) 2.25 (B) 6.24 
 (C) 12.13 (D) 15.24 
Answer (A) 
Sol.   
5 3 2
PCl PCl Cl + 
 t  = 0 5 0 0 
 t = t  5 – n n n 
   Total moles = 5 – n + n + n 
    = 5 + n. 
 For Argon 
  nAr = 4 
 Total moles = nAr + nPCl5 + nPCl3 + nPCl2  
     = 4 + 5 + n 
     = 9 + n 
 Kp = 
3
5
PCl 2
PCl
P .PCl
P
 PV = nRT 
    6 × 100 = (9 + n) × 0.082 × 610 
    n = 3  
 = 
33
66
12 12
2
6
12
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
 
 = 
27 9
2.25
12 4
== atm 
5. A 42.12% (w, v) solution of NaCl causes 
precipitation of a certain sol in 10 hours. The 
coagulating value of NaCl for the sol is 
 [Given : Molar mass : Na = 23.0 g mol
–1
; Cl = 35.5 
g mol
–1
] 
 (A) 36 mmol L
–1
 (B) 36 mol L
–1
 
 (C) 1440 mol L
–1
 (D) 1440 mmol L
–1
 
Answer (Bonus) 
Sol. Data is insufficient.  
6. Given below are two statements. One is labelled as 
Assertion A and the other is labelled as Reason R. 
 Assertion A: The first ionization enthalpy for 
oxygen is lower than that of nitrogen. 
 Reason R: The four electrons in 2p orbitals of 
oxygen experience more electron-electron 
repulsion. 
 In the light of the above statements, choose the 
correct answer from the options given below. 
 (A) Both A and R are correct and Rj is the correct 
explanation of A  
 (B) Both A and R are correct but R is NOT the 
correct explanation of A 
 (C) A is correct but R is not correct 
 (D) A is not correct but R is correct 
Answer (B) 
Sol. Nitrogen has half filled p-orbitals which is stable. 
Due to this it’s 1
st
 ionization energy is more than 
oxygen. 
7. Match List-I with List-II 
List-I Ore List-II Composition 
A. Siderite I. FeCO3 
B. Malachite II. CuCO3. Cu(OH)2 
C. Sphalerite III. ZnS 
D. Calamine IV. ZnCO3 
 Choose the correct answer from the options given 
below: 
 (A) A-I, B-II, C-III, D-IV (B) A-III, B-IV, C-II, D-I 
 (C) A-IV, B-III, C-I, D-II (D) A-I, B-II, C-IV, D-III 
Answer (A) 
Sol. Siderite ? FeCO3 
 Malachite ? CuCO3. Cu(OH)2 
 Sphalerite ? ZnS 
 Calamine ? ZnCO3 
Page 3


 
   
   
CHEMISTRY 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. Using the rules for significant figures, the correct 
answer for the expression 
0.02858 0.112
0.5702
?
 will be 
 (A) 0.005613 
 (B) 0.00561 
 (C) 0.0056 
 (D) 0.006 
Answer (B) 
Sol. 
0.02858 0.112
.00561
0.5702
?
= 
 Answer expressed in 3 significant figures. 
2. Which of the following is the correct plot for the 
probability density ( )
2
r ? as a function of distance 
‘r’ of the electron from the nucleus for 2s orbital?  
 (A)
 
 
 (B)
  
 (C)
 
 
 (D)
 
 
Answer (B) 
Sol. 2s  
 radial node = n – l – 1 
    = 2 – 0 – 1  
    = 1 
 It will have one radial node 
3. Consider the species 
+
44
CH ,NH and 
–
4
BH . Choose 
the correct option with respect to these species. 
 (A) They are isoelectronic and only two have 
tetrahedral structures 
 (B) They are isoelectronic and all have tetrahedral 
structures. 
 (C) Only two are isoelectronic and all have 
tetrahedral structures. 
 (D) Only two are isoelectronic and only two have 
tetrahedral structures. 
Answer (B) 
Sol. CH4,  
+
4
NH and 
–
4
BH are isoelectronic as well as 
tetrahedral. 
 
   
   
4. 4.0 moles of argon and 5.0 moles of PCl5 are 
introduced into an evacuated flask of 100 litre 
capacity at 610 K. The system is allowed to 
equilibrate. At equilibrium, the total pressure of 
mixture was found to be 6.0 atm. The Kp for the 
reaction is [Given : R = 0.082 L atm K
–1
 mol
–1
] 
 (A) 2.25 (B) 6.24 
 (C) 12.13 (D) 15.24 
Answer (A) 
Sol.   
5 3 2
PCl PCl Cl + 
 t  = 0 5 0 0 
 t = t  5 – n n n 
   Total moles = 5 – n + n + n 
    = 5 + n. 
 For Argon 
  nAr = 4 
 Total moles = nAr + nPCl5 + nPCl3 + nPCl2  
     = 4 + 5 + n 
     = 9 + n 
 Kp = 
3
5
PCl 2
PCl
P .PCl
P
 PV = nRT 
    6 × 100 = (9 + n) × 0.082 × 610 
    n = 3  
 = 
33
66
12 12
2
6
12
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
 
 = 
27 9
2.25
12 4
== atm 
5. A 42.12% (w, v) solution of NaCl causes 
precipitation of a certain sol in 10 hours. The 
coagulating value of NaCl for the sol is 
 [Given : Molar mass : Na = 23.0 g mol
–1
; Cl = 35.5 
g mol
–1
] 
 (A) 36 mmol L
–1
 (B) 36 mol L
–1
 
 (C) 1440 mol L
–1
 (D) 1440 mmol L
–1
 
Answer (Bonus) 
Sol. Data is insufficient.  
6. Given below are two statements. One is labelled as 
Assertion A and the other is labelled as Reason R. 
 Assertion A: The first ionization enthalpy for 
oxygen is lower than that of nitrogen. 
 Reason R: The four electrons in 2p orbitals of 
oxygen experience more electron-electron 
repulsion. 
 In the light of the above statements, choose the 
correct answer from the options given below. 
 (A) Both A and R are correct and Rj is the correct 
explanation of A  
 (B) Both A and R are correct but R is NOT the 
correct explanation of A 
 (C) A is correct but R is not correct 
 (D) A is not correct but R is correct 
Answer (B) 
Sol. Nitrogen has half filled p-orbitals which is stable. 
Due to this it’s 1
st
 ionization energy is more than 
oxygen. 
7. Match List-I with List-II 
List-I Ore List-II Composition 
A. Siderite I. FeCO3 
B. Malachite II. CuCO3. Cu(OH)2 
C. Sphalerite III. ZnS 
D. Calamine IV. ZnCO3 
 Choose the correct answer from the options given 
below: 
 (A) A-I, B-II, C-III, D-IV (B) A-III, B-IV, C-II, D-I 
 (C) A-IV, B-III, C-I, D-II (D) A-I, B-II, C-IV, D-III 
Answer (A) 
Sol. Siderite ? FeCO3 
 Malachite ? CuCO3. Cu(OH)2 
 Sphalerite ? ZnS 
 Calamine ? ZnCO3 
 
   
   
8. Given below are two statements. 
 Statement-I: In CuSO4.5H2O, Cu-O bonds are 
present. 
 Statement-II: In CuSO4.5H2O, ligands coordinating 
with Cu(II) ion are O-and S-based ligands. 
 In the light of the above statements, choose the 
correct answer from the options given below: 
 (A) Both Statement-I and Statement-II are correct 
 (B) Both Statement-I and Statement-II are incorrect 
 (C) Statement-I is correct but Statement-II is 
incorrect 
 (D) Statement-I is incorrect but Statement-II is 
correct. 
Answer (C) 
Sol. Statement I is true but statement II is false. Only 
oxygen atom forms Co-ordinate bond with Cu
+2
 in 
CuSO4.5H2O 
9. Amongst baking soda, caustic soda and washing 
soda, carbonate anion is present in 
 (A) Washing soda only  
 (B) Washing soda and caustic soda only 
 (C) Washing soda and baking soda only  
 (D) Baking soda, caustic soda and washing soda 
Answer (A) 
Sol. 
–2
3
CO ion is present only in washing soda. 
10. Number of lone pair(s) of electrons on central atom 
and the shape of BrF3 molecule respectively, are 
 (A) 0, triangular planar 
 (B) 1, pyramidal 
 (C) 2, bent T-shape 
 (D) 1, bent T-shape  
Answer (C) 
Sol.
 
 
11. Aqueous solution of which of the following boron 
compounds will be strongly basic in nature? 
 (A) NaBH4 
 (B) LiBH4 
 (C) B2H6 
 (D) Na2B4O7 
Answer (D) 
Sol. 
2 4 7 2 3 3 4
Na B O 7H O 2H BO 2Na[B(OH) ] + ?? + 
 Aqueous solution of borax is buffer whose pH 9 
 Other compounds are less basic than this. 
12. Sulphur dioxide is one of the components of 
polluted air. SO2 is also a major contributor to acid 
rain. The correct and complete reaction to 
represent acid rain caused by SO2 is 
 (A) 2SO2 + O2 ? 2SO3 
 (B) SO2 + O3 ? SO3 + O2 
 (C) SO2 + H2O2 ? H2SO4 
 (D) 2SO2 + O2 + 2H2O ? 2H2SO4 
Answer (D) 
Sol. 
2 2 2 2 4
2SO O 2H O 2H SO + + ?? 
 Acid rain occurs due to increased concentration of 
oxides of sulphur and Nitrogen. 
13. Which of the following carbocations is most stable? 
 (A) 
 
 (B) 
 
 (C) 
 
 (D) 
 
Answer (D) 
Page 4


 
   
   
CHEMISTRY 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. Using the rules for significant figures, the correct 
answer for the expression 
0.02858 0.112
0.5702
?
 will be 
 (A) 0.005613 
 (B) 0.00561 
 (C) 0.0056 
 (D) 0.006 
Answer (B) 
Sol. 
0.02858 0.112
.00561
0.5702
?
= 
 Answer expressed in 3 significant figures. 
2. Which of the following is the correct plot for the 
probability density ( )
2
r ? as a function of distance 
‘r’ of the electron from the nucleus for 2s orbital?  
 (A)
 
 
 (B)
  
 (C)
 
 
 (D)
 
 
Answer (B) 
Sol. 2s  
 radial node = n – l – 1 
    = 2 – 0 – 1  
    = 1 
 It will have one radial node 
3. Consider the species 
+
44
CH ,NH and 
–
4
BH . Choose 
the correct option with respect to these species. 
 (A) They are isoelectronic and only two have 
tetrahedral structures 
 (B) They are isoelectronic and all have tetrahedral 
structures. 
 (C) Only two are isoelectronic and all have 
tetrahedral structures. 
 (D) Only two are isoelectronic and only two have 
tetrahedral structures. 
Answer (B) 
Sol. CH4,  
+
4
NH and 
–
4
BH are isoelectronic as well as 
tetrahedral. 
 
   
   
4. 4.0 moles of argon and 5.0 moles of PCl5 are 
introduced into an evacuated flask of 100 litre 
capacity at 610 K. The system is allowed to 
equilibrate. At equilibrium, the total pressure of 
mixture was found to be 6.0 atm. The Kp for the 
reaction is [Given : R = 0.082 L atm K
–1
 mol
–1
] 
 (A) 2.25 (B) 6.24 
 (C) 12.13 (D) 15.24 
Answer (A) 
Sol.   
5 3 2
PCl PCl Cl + 
 t  = 0 5 0 0 
 t = t  5 – n n n 
   Total moles = 5 – n + n + n 
    = 5 + n. 
 For Argon 
  nAr = 4 
 Total moles = nAr + nPCl5 + nPCl3 + nPCl2  
     = 4 + 5 + n 
     = 9 + n 
 Kp = 
3
5
PCl 2
PCl
P .PCl
P
 PV = nRT 
    6 × 100 = (9 + n) × 0.082 × 610 
    n = 3  
 = 
33
66
12 12
2
6
12
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
 
 = 
27 9
2.25
12 4
== atm 
5. A 42.12% (w, v) solution of NaCl causes 
precipitation of a certain sol in 10 hours. The 
coagulating value of NaCl for the sol is 
 [Given : Molar mass : Na = 23.0 g mol
–1
; Cl = 35.5 
g mol
–1
] 
 (A) 36 mmol L
–1
 (B) 36 mol L
–1
 
 (C) 1440 mol L
–1
 (D) 1440 mmol L
–1
 
Answer (Bonus) 
Sol. Data is insufficient.  
6. Given below are two statements. One is labelled as 
Assertion A and the other is labelled as Reason R. 
 Assertion A: The first ionization enthalpy for 
oxygen is lower than that of nitrogen. 
 Reason R: The four electrons in 2p orbitals of 
oxygen experience more electron-electron 
repulsion. 
 In the light of the above statements, choose the 
correct answer from the options given below. 
 (A) Both A and R are correct and Rj is the correct 
explanation of A  
 (B) Both A and R are correct but R is NOT the 
correct explanation of A 
 (C) A is correct but R is not correct 
 (D) A is not correct but R is correct 
Answer (B) 
Sol. Nitrogen has half filled p-orbitals which is stable. 
Due to this it’s 1
st
 ionization energy is more than 
oxygen. 
7. Match List-I with List-II 
List-I Ore List-II Composition 
A. Siderite I. FeCO3 
B. Malachite II. CuCO3. Cu(OH)2 
C. Sphalerite III. ZnS 
D. Calamine IV. ZnCO3 
 Choose the correct answer from the options given 
below: 
 (A) A-I, B-II, C-III, D-IV (B) A-III, B-IV, C-II, D-I 
 (C) A-IV, B-III, C-I, D-II (D) A-I, B-II, C-IV, D-III 
Answer (A) 
Sol. Siderite ? FeCO3 
 Malachite ? CuCO3. Cu(OH)2 
 Sphalerite ? ZnS 
 Calamine ? ZnCO3 
 
   
   
8. Given below are two statements. 
 Statement-I: In CuSO4.5H2O, Cu-O bonds are 
present. 
 Statement-II: In CuSO4.5H2O, ligands coordinating 
with Cu(II) ion are O-and S-based ligands. 
 In the light of the above statements, choose the 
correct answer from the options given below: 
 (A) Both Statement-I and Statement-II are correct 
 (B) Both Statement-I and Statement-II are incorrect 
 (C) Statement-I is correct but Statement-II is 
incorrect 
 (D) Statement-I is incorrect but Statement-II is 
correct. 
Answer (C) 
Sol. Statement I is true but statement II is false. Only 
oxygen atom forms Co-ordinate bond with Cu
+2
 in 
CuSO4.5H2O 
9. Amongst baking soda, caustic soda and washing 
soda, carbonate anion is present in 
 (A) Washing soda only  
 (B) Washing soda and caustic soda only 
 (C) Washing soda and baking soda only  
 (D) Baking soda, caustic soda and washing soda 
Answer (A) 
Sol. 
–2
3
CO ion is present only in washing soda. 
10. Number of lone pair(s) of electrons on central atom 
and the shape of BrF3 molecule respectively, are 
 (A) 0, triangular planar 
 (B) 1, pyramidal 
 (C) 2, bent T-shape 
 (D) 1, bent T-shape  
Answer (C) 
Sol.
 
 
11. Aqueous solution of which of the following boron 
compounds will be strongly basic in nature? 
 (A) NaBH4 
 (B) LiBH4 
 (C) B2H6 
 (D) Na2B4O7 
Answer (D) 
Sol. 
2 4 7 2 3 3 4
Na B O 7H O 2H BO 2Na[B(OH) ] + ?? + 
 Aqueous solution of borax is buffer whose pH 9 
 Other compounds are less basic than this. 
12. Sulphur dioxide is one of the components of 
polluted air. SO2 is also a major contributor to acid 
rain. The correct and complete reaction to 
represent acid rain caused by SO2 is 
 (A) 2SO2 + O2 ? 2SO3 
 (B) SO2 + O3 ? SO3 + O2 
 (C) SO2 + H2O2 ? H2SO4 
 (D) 2SO2 + O2 + 2H2O ? 2H2SO4 
Answer (D) 
Sol. 
2 2 2 2 4
2SO O 2H O 2H SO + + ?? 
 Acid rain occurs due to increased concentration of 
oxides of sulphur and Nitrogen. 
13. Which of the following carbocations is most stable? 
 (A) 
 
 (B) 
 
 (C) 
 
 (D) 
 
Answer (D) 
 
   
   
Sol. 
 
14. 
 
 The stable carbocation formed in the above 
reaction is 
 (A) 
3 2 2
CH CH CH
?
 
 (B) 
32
CH CH
?
 
 (C) 
33
CH – CH– CH
?
 
 (D) 
 
Answer (C) 
Sol. Initially 2
32
CH — CH — CH
+
 is formed. On 
rearrangement 
33
CH — CH— CH
+
 stable 
carbocation is formed. 
15. Two isomers (A) and (B) with Molar mass 184 g/mol 
and elemental composition C, 52.2%; H, 4.9 % and 
Br 42.9% gave benzoic acid and p-bromobenzoic 
acid, respectively on oxidation with KMnO4. Isomer 
‘A’ is optically active and gives a pale yellow 
precipitate when warmed with alcoholic AgNO3. 
Isomers ‘A’ and ‘B’ are, respectively 
 (A) 
 
 (B) 
 
 (C) 
 
 (D) 
 
Answer (C) 
Sol. moles relative ratio simplest ratio
C 52.2 52.2 / 12 4.35 8.7
H 4.9 4.9 / 1 4.9 9.8
Br 42.9 42.9 / 80 0.5 1
= = ?
= = ?
= = ?
 
 C8H9Br 
 A is optically active 
 
3 6 5
Br
CH — CH— C H
*
?
 
 B forms para bromo benzoic acid on reaction with 
KMnO4. 
  
16. In Friedel-Crafts alkylation of aniline, one gets  
 (A) Alkylated product with ortho and para 
substitution. 
 (B) Secondary amine after acidic treatment.  
 (C) An amide product. 
 (D) Positively charged nitrogen at benzene ring. 
Answer (D) 
Sol.  
 
NH
2
RCl
AlCl
3
NH AlCl
2 3
?
Page 5


 
   
   
CHEMISTRY 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. Using the rules for significant figures, the correct 
answer for the expression 
0.02858 0.112
0.5702
?
 will be 
 (A) 0.005613 
 (B) 0.00561 
 (C) 0.0056 
 (D) 0.006 
Answer (B) 
Sol. 
0.02858 0.112
.00561
0.5702
?
= 
 Answer expressed in 3 significant figures. 
2. Which of the following is the correct plot for the 
probability density ( )
2
r ? as a function of distance 
‘r’ of the electron from the nucleus for 2s orbital?  
 (A)
 
 
 (B)
  
 (C)
 
 
 (D)
 
 
Answer (B) 
Sol. 2s  
 radial node = n – l – 1 
    = 2 – 0 – 1  
    = 1 
 It will have one radial node 
3. Consider the species 
+
44
CH ,NH and 
–
4
BH . Choose 
the correct option with respect to these species. 
 (A) They are isoelectronic and only two have 
tetrahedral structures 
 (B) They are isoelectronic and all have tetrahedral 
structures. 
 (C) Only two are isoelectronic and all have 
tetrahedral structures. 
 (D) Only two are isoelectronic and only two have 
tetrahedral structures. 
Answer (B) 
Sol. CH4,  
+
4
NH and 
–
4
BH are isoelectronic as well as 
tetrahedral. 
 
   
   
4. 4.0 moles of argon and 5.0 moles of PCl5 are 
introduced into an evacuated flask of 100 litre 
capacity at 610 K. The system is allowed to 
equilibrate. At equilibrium, the total pressure of 
mixture was found to be 6.0 atm. The Kp for the 
reaction is [Given : R = 0.082 L atm K
–1
 mol
–1
] 
 (A) 2.25 (B) 6.24 
 (C) 12.13 (D) 15.24 
Answer (A) 
Sol.   
5 3 2
PCl PCl Cl + 
 t  = 0 5 0 0 
 t = t  5 – n n n 
   Total moles = 5 – n + n + n 
    = 5 + n. 
 For Argon 
  nAr = 4 
 Total moles = nAr + nPCl5 + nPCl3 + nPCl2  
     = 4 + 5 + n 
     = 9 + n 
 Kp = 
3
5
PCl 2
PCl
P .PCl
P
 PV = nRT 
    6 × 100 = (9 + n) × 0.082 × 610 
    n = 3  
 = 
33
66
12 12
2
6
12
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
 
 = 
27 9
2.25
12 4
== atm 
5. A 42.12% (w, v) solution of NaCl causes 
precipitation of a certain sol in 10 hours. The 
coagulating value of NaCl for the sol is 
 [Given : Molar mass : Na = 23.0 g mol
–1
; Cl = 35.5 
g mol
–1
] 
 (A) 36 mmol L
–1
 (B) 36 mol L
–1
 
 (C) 1440 mol L
–1
 (D) 1440 mmol L
–1
 
Answer (Bonus) 
Sol. Data is insufficient.  
6. Given below are two statements. One is labelled as 
Assertion A and the other is labelled as Reason R. 
 Assertion A: The first ionization enthalpy for 
oxygen is lower than that of nitrogen. 
 Reason R: The four electrons in 2p orbitals of 
oxygen experience more electron-electron 
repulsion. 
 In the light of the above statements, choose the 
correct answer from the options given below. 
 (A) Both A and R are correct and Rj is the correct 
explanation of A  
 (B) Both A and R are correct but R is NOT the 
correct explanation of A 
 (C) A is correct but R is not correct 
 (D) A is not correct but R is correct 
Answer (B) 
Sol. Nitrogen has half filled p-orbitals which is stable. 
Due to this it’s 1
st
 ionization energy is more than 
oxygen. 
7. Match List-I with List-II 
List-I Ore List-II Composition 
A. Siderite I. FeCO3 
B. Malachite II. CuCO3. Cu(OH)2 
C. Sphalerite III. ZnS 
D. Calamine IV. ZnCO3 
 Choose the correct answer from the options given 
below: 
 (A) A-I, B-II, C-III, D-IV (B) A-III, B-IV, C-II, D-I 
 (C) A-IV, B-III, C-I, D-II (D) A-I, B-II, C-IV, D-III 
Answer (A) 
Sol. Siderite ? FeCO3 
 Malachite ? CuCO3. Cu(OH)2 
 Sphalerite ? ZnS 
 Calamine ? ZnCO3 
 
   
   
8. Given below are two statements. 
 Statement-I: In CuSO4.5H2O, Cu-O bonds are 
present. 
 Statement-II: In CuSO4.5H2O, ligands coordinating 
with Cu(II) ion are O-and S-based ligands. 
 In the light of the above statements, choose the 
correct answer from the options given below: 
 (A) Both Statement-I and Statement-II are correct 
 (B) Both Statement-I and Statement-II are incorrect 
 (C) Statement-I is correct but Statement-II is 
incorrect 
 (D) Statement-I is incorrect but Statement-II is 
correct. 
Answer (C) 
Sol. Statement I is true but statement II is false. Only 
oxygen atom forms Co-ordinate bond with Cu
+2
 in 
CuSO4.5H2O 
9. Amongst baking soda, caustic soda and washing 
soda, carbonate anion is present in 
 (A) Washing soda only  
 (B) Washing soda and caustic soda only 
 (C) Washing soda and baking soda only  
 (D) Baking soda, caustic soda and washing soda 
Answer (A) 
Sol. 
–2
3
CO ion is present only in washing soda. 
10. Number of lone pair(s) of electrons on central atom 
and the shape of BrF3 molecule respectively, are 
 (A) 0, triangular planar 
 (B) 1, pyramidal 
 (C) 2, bent T-shape 
 (D) 1, bent T-shape  
Answer (C) 
Sol.
 
 
11. Aqueous solution of which of the following boron 
compounds will be strongly basic in nature? 
 (A) NaBH4 
 (B) LiBH4 
 (C) B2H6 
 (D) Na2B4O7 
Answer (D) 
Sol. 
2 4 7 2 3 3 4
Na B O 7H O 2H BO 2Na[B(OH) ] + ?? + 
 Aqueous solution of borax is buffer whose pH 9 
 Other compounds are less basic than this. 
12. Sulphur dioxide is one of the components of 
polluted air. SO2 is also a major contributor to acid 
rain. The correct and complete reaction to 
represent acid rain caused by SO2 is 
 (A) 2SO2 + O2 ? 2SO3 
 (B) SO2 + O3 ? SO3 + O2 
 (C) SO2 + H2O2 ? H2SO4 
 (D) 2SO2 + O2 + 2H2O ? 2H2SO4 
Answer (D) 
Sol. 
2 2 2 2 4
2SO O 2H O 2H SO + + ?? 
 Acid rain occurs due to increased concentration of 
oxides of sulphur and Nitrogen. 
13. Which of the following carbocations is most stable? 
 (A) 
 
 (B) 
 
 (C) 
 
 (D) 
 
Answer (D) 
 
   
   
Sol. 
 
14. 
 
 The stable carbocation formed in the above 
reaction is 
 (A) 
3 2 2
CH CH CH
?
 
 (B) 
32
CH CH
?
 
 (C) 
33
CH – CH– CH
?
 
 (D) 
 
Answer (C) 
Sol. Initially 2
32
CH — CH — CH
+
 is formed. On 
rearrangement 
33
CH — CH— CH
+
 stable 
carbocation is formed. 
15. Two isomers (A) and (B) with Molar mass 184 g/mol 
and elemental composition C, 52.2%; H, 4.9 % and 
Br 42.9% gave benzoic acid and p-bromobenzoic 
acid, respectively on oxidation with KMnO4. Isomer 
‘A’ is optically active and gives a pale yellow 
precipitate when warmed with alcoholic AgNO3. 
Isomers ‘A’ and ‘B’ are, respectively 
 (A) 
 
 (B) 
 
 (C) 
 
 (D) 
 
Answer (C) 
Sol. moles relative ratio simplest ratio
C 52.2 52.2 / 12 4.35 8.7
H 4.9 4.9 / 1 4.9 9.8
Br 42.9 42.9 / 80 0.5 1
= = ?
= = ?
= = ?
 
 C8H9Br 
 A is optically active 
 
3 6 5
Br
CH — CH— C H
*
?
 
 B forms para bromo benzoic acid on reaction with 
KMnO4. 
  
16. In Friedel-Crafts alkylation of aniline, one gets  
 (A) Alkylated product with ortho and para 
substitution. 
 (B) Secondary amine after acidic treatment.  
 (C) An amide product. 
 (D) Positively charged nitrogen at benzene ring. 
Answer (D) 
Sol.  
 
NH
2
RCl
AlCl
3
NH AlCl
2 3
?
 
   
   
17. Given below are two statements : one is labelled  
as Assertion A and the other is labelled as  
Reason R. 
 Assertion A : Dacron is an example of polyester 
polymer.  
 Reason R : Dacron is made up of ethylene glycol 
and terephthalic acid monomers.  
 In the light of the above statements, choose the 
most appropriate answer from the options given 
below.   
 (A) Both A and R are correct and R is the correct 
explanation of A. 
 (B) Both A and R are correct but R is NOT the 
correct explanation of A. 
 (C) A is correct but R is not correct.  
 (D) A is not correct but R is correct.  
Answer (A) 
Sol.  
  
18. The structure of protein that is unaffected by 
heating is  
 (A) Secondary Structure 
 (B) Tertiary Structure  
 (C) Primary Structure 
  (D) Quaternary Structure 
Answer (C) 
Sol.  Primary structure is unaffected by heating 
19. The mixture of chloroxylenol and terpineol is an 
example of  
 (A) Antiseptic 
 (B) Pesticide  
 (C) Disinfectant 
 (D) Narcotic analgesic  
Answer (A) 
Sol.  Mixture of chloroxylenol and terpineol is known as 
Dettol. It acts as Antiseptic 
20. A white precipitate was formed when BaCl2 was 
added to water extract of an inorganic salt. Further, 
a gas 'X' with characteristic odour was released 
when the formed white precipitate was dissolved in 
dilute HCl. The anion present in the inorganic salt 
is  
 (A) I
– 
 (B) 
2
3
SO
-
 
 (C) S
2– 
 (D) 
2
NO
-
 
Answer (B) 
Sol. Anion is 
–2
3
SO 
 
dil HCl
32
X(gas)
BaSO SO ???? ? ? 
 Gas is released with smell of burning sulphur. 
SECTION - B 
Numerical Value Type Questions: This section 
contains 10 questions. In Section B, attempt any five 
questions out of 10. The answer to each question is a 
NUMERICAL VALUE. For each question, enter the 
correct numerical value (in decimal notation, 
truncated/rounded-off to the second decimal place; e.g. 
06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the 
mouse and the on-screen virtual numeric keypad in the 
place designated to enter the answer. 
CH – OH
2
CH – OH
2
n + n HOOC COOH
Terephthalic acid
C OCH – CH – O – C
2 2
O
O
n
Dacron