JEE Main 2021 Answer Key Chemistry - Evening Shift (17-03-2021) JEE Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2021 Answer Key Chemistry - Evening Shift (17-03-2021) JEE Notes | EduRev

 Page 1


 
 
17
th
 March. 2021 | Shift 2 
1. Match List-I with List-II. 
  List – I    List – II 
 Chemical Compound   Used as 
 (a) Sucralose    (i) Synthetic detergent 
 (b) Glyceryl ester of stearic acid (ii) Artificial sweetener 
 (c) Sodium benzoate   (iii) Antiseptic 
 (d) Bithionol     (iv) Food preservative 
Choose the correct match: 
 (1) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
 (2) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) 
 (3) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)  
 (4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) 
Ans. (2) 
Sol. (a) Sucralose     ? ? ? ? Artificial sweetener 
 (b) Glyceryl ester of stearic acid   ? ? ? ? ?Synthetic detergent 
 (c) Sodium benzoate   ? ? ? ? Food preservative  
 (d) Bithionol    ? ? ? Antiseptic 
 
2. 
12 22 11 2 6 12 6 6 12 6
cos
? ? ?
Enzyme A
Sucrose Glu e Fructose
C H O H O C H O C H O 
 
6 12 6 2 5 2
cos
2 2 ? ? ?
Enzyme B
Glu e
C H O C H OH CO 
 In the above reactions, the enzyme A and enzyme B respectively are: 
 (1) Invertase and Amylase 
 (2) Amylase and Invertase 
 (3) Invertase and Zymase 
 (4) Zymase and Invertase 
Ans. (3) 
Sol. C
12
H
22
O
11
 + H
2
O       C
6
H
12
O
6
 + C
6
H
12
O
6
 
        Glucose    Fructose 
 C
6
H
12
O
6 
       2C
2
H
5
OH + 2 CO
2 
 
 
 
 
 
Invertase
Zymase
Page 2


 
 
17
th
 March. 2021 | Shift 2 
1. Match List-I with List-II. 
  List – I    List – II 
 Chemical Compound   Used as 
 (a) Sucralose    (i) Synthetic detergent 
 (b) Glyceryl ester of stearic acid (ii) Artificial sweetener 
 (c) Sodium benzoate   (iii) Antiseptic 
 (d) Bithionol     (iv) Food preservative 
Choose the correct match: 
 (1) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
 (2) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) 
 (3) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)  
 (4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) 
Ans. (2) 
Sol. (a) Sucralose     ? ? ? ? Artificial sweetener 
 (b) Glyceryl ester of stearic acid   ? ? ? ? ?Synthetic detergent 
 (c) Sodium benzoate   ? ? ? ? Food preservative  
 (d) Bithionol    ? ? ? Antiseptic 
 
2. 
12 22 11 2 6 12 6 6 12 6
cos
? ? ?
Enzyme A
Sucrose Glu e Fructose
C H O H O C H O C H O 
 
6 12 6 2 5 2
cos
2 2 ? ? ?
Enzyme B
Glu e
C H O C H OH CO 
 In the above reactions, the enzyme A and enzyme B respectively are: 
 (1) Invertase and Amylase 
 (2) Amylase and Invertase 
 (3) Invertase and Zymase 
 (4) Zymase and Invertase 
Ans. (3) 
Sol. C
12
H
22
O
11
 + H
2
O       C
6
H
12
O
6
 + C
6
H
12
O
6
 
        Glucose    Fructose 
 C
6
H
12
O
6 
       2C
2
H
5
OH + 2 CO
2 
 
 
 
 
 
Invertase
Zymase
 
 
3. The correct pair(s) of the ambident nucleophiles is (are) : 
 (A) AgCN/KCN 
 (B) RCOOAg/RCOOK 
 (C) AgNO
2
/KNO
2
 
 (D) AgI/KI  
 (1) (A) and (C) only   (2) (B) only 
 (3) (B) and (C) only   (4) (A) only 
Ans. (1) 
Sol. 
 
  C ? ?N  
  O –N = O 
? ? ? ?
O  O
O  O
O 
O 
More than one    donating side 
4. During which of the following processes, does entropy decrease? 
 (A)  Freezing of water to ice at 0°C 
 (B) Freezing of water to ice at -10°C 
 (C) N
2
(g) + 3H
2
(g) ?2NH
3
(g) 
 (D) Adsorption of CO(g) on lead surface. 
 (E) Dissolution of NaCI in water 
 (1) (A), (B), (C) and (D) only (2) (A), (C) and (E) only 
 (3) (A) and (E) only   (4) (B) and (C) only 
Ans. (1) 
Sol. A, B ? Freezing of water will decrease entropy as particles will move closer and forces of 
attraction will increase. This leads to decrease in randomness. So entropy decrease.   
 C ? No. of molecules decreasing  
 D ? Adsorption will lead to decrease in randomness of gaseous particles. 
 E ? NaCl(s) ? Na
+
 (aq) + Cl
–
(aq)  ?S > 0 
 So, (A, B, C, D) decreases entropy. 
?
5. Match List-I with List-II : 
  List-I     List-II 
(a) [Co(NH
3
)
6
] [Cr(CN)
6
]  (i) Linkage isomerism 
 (b) [Co(NH
3
)
3
 (NO
2
)
3
]  (ii) Solvate isomerism 
 (c) [Cr(H
2
O)
6
]Cl
3
   (iii) Co-ordination isomerism 
 (d) cis-[CrCl
2
(ox)
2
]
3-
   (iv) Optical isomerism  
 Choose the correct answer from the options given below: 
 1. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 2. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) 
 3. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv) 
 4. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i) 
Ans. (1) 
Sol. Theory based 
 
e 
Page 3


 
 
17
th
 March. 2021 | Shift 2 
1. Match List-I with List-II. 
  List – I    List – II 
 Chemical Compound   Used as 
 (a) Sucralose    (i) Synthetic detergent 
 (b) Glyceryl ester of stearic acid (ii) Artificial sweetener 
 (c) Sodium benzoate   (iii) Antiseptic 
 (d) Bithionol     (iv) Food preservative 
Choose the correct match: 
 (1) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
 (2) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) 
 (3) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)  
 (4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) 
Ans. (2) 
Sol. (a) Sucralose     ? ? ? ? Artificial sweetener 
 (b) Glyceryl ester of stearic acid   ? ? ? ? ?Synthetic detergent 
 (c) Sodium benzoate   ? ? ? ? Food preservative  
 (d) Bithionol    ? ? ? Antiseptic 
 
2. 
12 22 11 2 6 12 6 6 12 6
cos
? ? ?
Enzyme A
Sucrose Glu e Fructose
C H O H O C H O C H O 
 
6 12 6 2 5 2
cos
2 2 ? ? ?
Enzyme B
Glu e
C H O C H OH CO 
 In the above reactions, the enzyme A and enzyme B respectively are: 
 (1) Invertase and Amylase 
 (2) Amylase and Invertase 
 (3) Invertase and Zymase 
 (4) Zymase and Invertase 
Ans. (3) 
Sol. C
12
H
22
O
11
 + H
2
O       C
6
H
12
O
6
 + C
6
H
12
O
6
 
        Glucose    Fructose 
 C
6
H
12
O
6 
       2C
2
H
5
OH + 2 CO
2 
 
 
 
 
 
Invertase
Zymase
 
 
3. The correct pair(s) of the ambident nucleophiles is (are) : 
 (A) AgCN/KCN 
 (B) RCOOAg/RCOOK 
 (C) AgNO
2
/KNO
2
 
 (D) AgI/KI  
 (1) (A) and (C) only   (2) (B) only 
 (3) (B) and (C) only   (4) (A) only 
Ans. (1) 
Sol. 
 
  C ? ?N  
  O –N = O 
? ? ? ?
O  O
O  O
O 
O 
More than one    donating side 
4. During which of the following processes, does entropy decrease? 
 (A)  Freezing of water to ice at 0°C 
 (B) Freezing of water to ice at -10°C 
 (C) N
2
(g) + 3H
2
(g) ?2NH
3
(g) 
 (D) Adsorption of CO(g) on lead surface. 
 (E) Dissolution of NaCI in water 
 (1) (A), (B), (C) and (D) only (2) (A), (C) and (E) only 
 (3) (A) and (E) only   (4) (B) and (C) only 
Ans. (1) 
Sol. A, B ? Freezing of water will decrease entropy as particles will move closer and forces of 
attraction will increase. This leads to decrease in randomness. So entropy decrease.   
 C ? No. of molecules decreasing  
 D ? Adsorption will lead to decrease in randomness of gaseous particles. 
 E ? NaCl(s) ? Na
+
 (aq) + Cl
–
(aq)  ?S > 0 
 So, (A, B, C, D) decreases entropy. 
?
5. Match List-I with List-II : 
  List-I     List-II 
(a) [Co(NH
3
)
6
] [Cr(CN)
6
]  (i) Linkage isomerism 
 (b) [Co(NH
3
)
3
 (NO
2
)
3
]  (ii) Solvate isomerism 
 (c) [Cr(H
2
O)
6
]Cl
3
   (iii) Co-ordination isomerism 
 (d) cis-[CrCl
2
(ox)
2
]
3-
   (iv) Optical isomerism  
 Choose the correct answer from the options given below: 
 1. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 2. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) 
 3. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv) 
 4. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i) 
Ans. (1) 
Sol. Theory based 
 
e 
 
 
17
th
 March. 2021 | Shift 2 
6. The common positive oxidation states for an element with atomic number 24, are:  
 (1) +1 and +3    (2) +1 to +6 
 (3) +1 and +3 to +6    (4) +2 to +6 
Ans. (4) 
Sol. Fact 
 
7. The set of elements that differ in mutual relationship from those of the other sets is : 
 (1) Be – Al     (2) Li – Na 
 (3) B – Si     (4) Li – Mg 
Ans. (2) 
Sol. Li and Na does not have diagonal relationship.  
 
8. Given below are two statements : 
 Statement I : 2-methylbutane on oxidation with KMnO
4
 gives 2-methylbutan-2-ol. 
 Statement II : N-alkanes can be easily oxidized to corresponding alcohols with KMnO
4
.  
 Choose the correct option : 
 (1) Both statement I and statement II are incorrect 
 (2) Statement I is correct but statement II is incorrect 
 (3) Both statement I and statement II are correct  
 (4) Statement I is incorrect but statement II is correct 
Ans. (2) 
Sol. 
  
 
CH
3
—CH—CH
2
—CH
3
        CH
3 
KMnO
4
CH
3
—CH
2 
—C—CH
3
                 CH
3 
OH
     
 
   
 
n-Alkanes 
KMnO
4
No reaction 
 
9. Amongst the following, the linear species is : 
 (1) 
3
?
N
 
 (2) Cl
2
O  (3) O
3  
 (4) NO
2 
Ans. (1) 
Sol. (1) N N N
?
? ?
  SP(linear)
 
Page 4


 
 
17
th
 March. 2021 | Shift 2 
1. Match List-I with List-II. 
  List – I    List – II 
 Chemical Compound   Used as 
 (a) Sucralose    (i) Synthetic detergent 
 (b) Glyceryl ester of stearic acid (ii) Artificial sweetener 
 (c) Sodium benzoate   (iii) Antiseptic 
 (d) Bithionol     (iv) Food preservative 
Choose the correct match: 
 (1) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
 (2) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) 
 (3) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)  
 (4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) 
Ans. (2) 
Sol. (a) Sucralose     ? ? ? ? Artificial sweetener 
 (b) Glyceryl ester of stearic acid   ? ? ? ? ?Synthetic detergent 
 (c) Sodium benzoate   ? ? ? ? Food preservative  
 (d) Bithionol    ? ? ? Antiseptic 
 
2. 
12 22 11 2 6 12 6 6 12 6
cos
? ? ?
Enzyme A
Sucrose Glu e Fructose
C H O H O C H O C H O 
 
6 12 6 2 5 2
cos
2 2 ? ? ?
Enzyme B
Glu e
C H O C H OH CO 
 In the above reactions, the enzyme A and enzyme B respectively are: 
 (1) Invertase and Amylase 
 (2) Amylase and Invertase 
 (3) Invertase and Zymase 
 (4) Zymase and Invertase 
Ans. (3) 
Sol. C
12
H
22
O
11
 + H
2
O       C
6
H
12
O
6
 + C
6
H
12
O
6
 
        Glucose    Fructose 
 C
6
H
12
O
6 
       2C
2
H
5
OH + 2 CO
2 
 
 
 
 
 
Invertase
Zymase
 
 
3. The correct pair(s) of the ambident nucleophiles is (are) : 
 (A) AgCN/KCN 
 (B) RCOOAg/RCOOK 
 (C) AgNO
2
/KNO
2
 
 (D) AgI/KI  
 (1) (A) and (C) only   (2) (B) only 
 (3) (B) and (C) only   (4) (A) only 
Ans. (1) 
Sol. 
 
  C ? ?N  
  O –N = O 
? ? ? ?
O  O
O  O
O 
O 
More than one    donating side 
4. During which of the following processes, does entropy decrease? 
 (A)  Freezing of water to ice at 0°C 
 (B) Freezing of water to ice at -10°C 
 (C) N
2
(g) + 3H
2
(g) ?2NH
3
(g) 
 (D) Adsorption of CO(g) on lead surface. 
 (E) Dissolution of NaCI in water 
 (1) (A), (B), (C) and (D) only (2) (A), (C) and (E) only 
 (3) (A) and (E) only   (4) (B) and (C) only 
Ans. (1) 
Sol. A, B ? Freezing of water will decrease entropy as particles will move closer and forces of 
attraction will increase. This leads to decrease in randomness. So entropy decrease.   
 C ? No. of molecules decreasing  
 D ? Adsorption will lead to decrease in randomness of gaseous particles. 
 E ? NaCl(s) ? Na
+
 (aq) + Cl
–
(aq)  ?S > 0 
 So, (A, B, C, D) decreases entropy. 
?
5. Match List-I with List-II : 
  List-I     List-II 
(a) [Co(NH
3
)
6
] [Cr(CN)
6
]  (i) Linkage isomerism 
 (b) [Co(NH
3
)
3
 (NO
2
)
3
]  (ii) Solvate isomerism 
 (c) [Cr(H
2
O)
6
]Cl
3
   (iii) Co-ordination isomerism 
 (d) cis-[CrCl
2
(ox)
2
]
3-
   (iv) Optical isomerism  
 Choose the correct answer from the options given below: 
 1. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 2. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) 
 3. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv) 
 4. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i) 
Ans. (1) 
Sol. Theory based 
 
e 
 
 
17
th
 March. 2021 | Shift 2 
6. The common positive oxidation states for an element with atomic number 24, are:  
 (1) +1 and +3    (2) +1 to +6 
 (3) +1 and +3 to +6    (4) +2 to +6 
Ans. (4) 
Sol. Fact 
 
7. The set of elements that differ in mutual relationship from those of the other sets is : 
 (1) Be – Al     (2) Li – Na 
 (3) B – Si     (4) Li – Mg 
Ans. (2) 
Sol. Li and Na does not have diagonal relationship.  
 
8. Given below are two statements : 
 Statement I : 2-methylbutane on oxidation with KMnO
4
 gives 2-methylbutan-2-ol. 
 Statement II : N-alkanes can be easily oxidized to corresponding alcohols with KMnO
4
.  
 Choose the correct option : 
 (1) Both statement I and statement II are incorrect 
 (2) Statement I is correct but statement II is incorrect 
 (3) Both statement I and statement II are correct  
 (4) Statement I is incorrect but statement II is correct 
Ans. (2) 
Sol. 
  
 
CH
3
—CH—CH
2
—CH
3
        CH
3 
KMnO
4
CH
3
—CH
2 
—C—CH
3
                 CH
3 
OH
     
 
   
 
n-Alkanes 
KMnO
4
No reaction 
 
9. Amongst the following, the linear species is : 
 (1) 
3
?
N
 
 (2) Cl
2
O  (3) O
3  
 (4) NO
2 
Ans. (1) 
Sol. (1) N N N
?
? ?
  SP(linear)
 
 
 
 (2) 
 
sp
3
(Bent)
O
Cl Cl
 
 (3) 
 
sp
2
(Bent)
O
O O
 
 (4) 
 
sp
2
(Bent)
N
O
O
 
 
10. For the coagulation of a negative sol, the species below, that has the highest flocculating power 
is : 
 (1) 
2
4
?
SO
 
 (2) Na
+  
 (3) Ba
2+ 
 (4) 
3
4
?
PO
 
Ans. (3) 
Sol. For a negative sol, positive ion is required for flocculation. 
 Greater the valence  of the flocculating ion added, the greater is its power to cause 
precipitation. This is called Hardy-Schulz law. 
 So, Ba
+2
 has highest flocculating power. 
 
11. The functional groups that are responsible for the ion-exchange property of cation and anion 
exchange resins, respectively, are: 
 (1) –SO
3
H and –COOH 
 (2) –SO
3
H and –NH
2
 
 (3) –NH
2
 and –SO
3
H 
 (4) –NH
2
 and –COOH  
Ans. (2) 
Sol. –SO
3
H and –COOH are cation exchanger and –NH
2
 is anion exchanger. 
 
12. Choose the correct statement regarding the formation of carbocations A and B given. 
 
—
+
3 2 2 2 CH –
  CH –
 CH –
  CH + Br
"A"
 
CH
3
 – CH
2
 – CH = CH
2
 + HBr 
+
—
3 2 3 CH –
  CH –
  CH -  CH + B r
"B"
  
 
 
Page 5


 
 
17
th
 March. 2021 | Shift 2 
1. Match List-I with List-II. 
  List – I    List – II 
 Chemical Compound   Used as 
 (a) Sucralose    (i) Synthetic detergent 
 (b) Glyceryl ester of stearic acid (ii) Artificial sweetener 
 (c) Sodium benzoate   (iii) Antiseptic 
 (d) Bithionol     (iv) Food preservative 
Choose the correct match: 
 (1) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
 (2) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) 
 (3) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)  
 (4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) 
Ans. (2) 
Sol. (a) Sucralose     ? ? ? ? Artificial sweetener 
 (b) Glyceryl ester of stearic acid   ? ? ? ? ?Synthetic detergent 
 (c) Sodium benzoate   ? ? ? ? Food preservative  
 (d) Bithionol    ? ? ? Antiseptic 
 
2. 
12 22 11 2 6 12 6 6 12 6
cos
? ? ?
Enzyme A
Sucrose Glu e Fructose
C H O H O C H O C H O 
 
6 12 6 2 5 2
cos
2 2 ? ? ?
Enzyme B
Glu e
C H O C H OH CO 
 In the above reactions, the enzyme A and enzyme B respectively are: 
 (1) Invertase and Amylase 
 (2) Amylase and Invertase 
 (3) Invertase and Zymase 
 (4) Zymase and Invertase 
Ans. (3) 
Sol. C
12
H
22
O
11
 + H
2
O       C
6
H
12
O
6
 + C
6
H
12
O
6
 
        Glucose    Fructose 
 C
6
H
12
O
6 
       2C
2
H
5
OH + 2 CO
2 
 
 
 
 
 
Invertase
Zymase
 
 
3. The correct pair(s) of the ambident nucleophiles is (are) : 
 (A) AgCN/KCN 
 (B) RCOOAg/RCOOK 
 (C) AgNO
2
/KNO
2
 
 (D) AgI/KI  
 (1) (A) and (C) only   (2) (B) only 
 (3) (B) and (C) only   (4) (A) only 
Ans. (1) 
Sol. 
 
  C ? ?N  
  O –N = O 
? ? ? ?
O  O
O  O
O 
O 
More than one    donating side 
4. During which of the following processes, does entropy decrease? 
 (A)  Freezing of water to ice at 0°C 
 (B) Freezing of water to ice at -10°C 
 (C) N
2
(g) + 3H
2
(g) ?2NH
3
(g) 
 (D) Adsorption of CO(g) on lead surface. 
 (E) Dissolution of NaCI in water 
 (1) (A), (B), (C) and (D) only (2) (A), (C) and (E) only 
 (3) (A) and (E) only   (4) (B) and (C) only 
Ans. (1) 
Sol. A, B ? Freezing of water will decrease entropy as particles will move closer and forces of 
attraction will increase. This leads to decrease in randomness. So entropy decrease.   
 C ? No. of molecules decreasing  
 D ? Adsorption will lead to decrease in randomness of gaseous particles. 
 E ? NaCl(s) ? Na
+
 (aq) + Cl
–
(aq)  ?S > 0 
 So, (A, B, C, D) decreases entropy. 
?
5. Match List-I with List-II : 
  List-I     List-II 
(a) [Co(NH
3
)
6
] [Cr(CN)
6
]  (i) Linkage isomerism 
 (b) [Co(NH
3
)
3
 (NO
2
)
3
]  (ii) Solvate isomerism 
 (c) [Cr(H
2
O)
6
]Cl
3
   (iii) Co-ordination isomerism 
 (d) cis-[CrCl
2
(ox)
2
]
3-
   (iv) Optical isomerism  
 Choose the correct answer from the options given below: 
 1. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 2. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) 
 3. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv) 
 4. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i) 
Ans. (1) 
Sol. Theory based 
 
e 
 
 
17
th
 March. 2021 | Shift 2 
6. The common positive oxidation states for an element with atomic number 24, are:  
 (1) +1 and +3    (2) +1 to +6 
 (3) +1 and +3 to +6    (4) +2 to +6 
Ans. (4) 
Sol. Fact 
 
7. The set of elements that differ in mutual relationship from those of the other sets is : 
 (1) Be – Al     (2) Li – Na 
 (3) B – Si     (4) Li – Mg 
Ans. (2) 
Sol. Li and Na does not have diagonal relationship.  
 
8. Given below are two statements : 
 Statement I : 2-methylbutane on oxidation with KMnO
4
 gives 2-methylbutan-2-ol. 
 Statement II : N-alkanes can be easily oxidized to corresponding alcohols with KMnO
4
.  
 Choose the correct option : 
 (1) Both statement I and statement II are incorrect 
 (2) Statement I is correct but statement II is incorrect 
 (3) Both statement I and statement II are correct  
 (4) Statement I is incorrect but statement II is correct 
Ans. (2) 
Sol. 
  
 
CH
3
—CH—CH
2
—CH
3
        CH
3 
KMnO
4
CH
3
—CH
2 
—C—CH
3
                 CH
3 
OH
     
 
   
 
n-Alkanes 
KMnO
4
No reaction 
 
9. Amongst the following, the linear species is : 
 (1) 
3
?
N
 
 (2) Cl
2
O  (3) O
3  
 (4) NO
2 
Ans. (1) 
Sol. (1) N N N
?
? ?
  SP(linear)
 
 
 
 (2) 
 
sp
3
(Bent)
O
Cl Cl
 
 (3) 
 
sp
2
(Bent)
O
O O
 
 (4) 
 
sp
2
(Bent)
N
O
O
 
 
10. For the coagulation of a negative sol, the species below, that has the highest flocculating power 
is : 
 (1) 
2
4
?
SO
 
 (2) Na
+  
 (3) Ba
2+ 
 (4) 
3
4
?
PO
 
Ans. (3) 
Sol. For a negative sol, positive ion is required for flocculation. 
 Greater the valence  of the flocculating ion added, the greater is its power to cause 
precipitation. This is called Hardy-Schulz law. 
 So, Ba
+2
 has highest flocculating power. 
 
11. The functional groups that are responsible for the ion-exchange property of cation and anion 
exchange resins, respectively, are: 
 (1) –SO
3
H and –COOH 
 (2) –SO
3
H and –NH
2
 
 (3) –NH
2
 and –SO
3
H 
 (4) –NH
2
 and –COOH  
Ans. (2) 
Sol. –SO
3
H and –COOH are cation exchanger and –NH
2
 is anion exchanger. 
 
12. Choose the correct statement regarding the formation of carbocations A and B given. 
 
—
+
3 2 2 2 CH –
  CH –
 CH –
  CH + Br
"A"
 
CH
3
 – CH
2
 – CH = CH
2
 + HBr 
+
—
3 2 3 CH –
  CH –
  CH -  CH + B r
"B"
  
 
 
 
 
17
th
 March. 2021 | Shift 2 
(1) Carbocation A is more stable and formed relatively at faster rate 
(2) Carbocation B is more stable and formed relatively at faster rate 
(3) Carbocation A is more stable and formed relatively at slow rate 
 (4) Carbocation B is more stable and formed relatively at slow rate 
Ans. (2) 
Sol. B carbocation is more stable due to more hyperconjugation & it form relatively faster rate compared to A. 
 
13. 
? ?
7 7 2
A
C H N OCl+ C
2
H
5
OH ?? ??
?
+ N
2
 + “X” + “Y” 
 In the above reaction, the structural formula of (A), “X” and “Y” respectively are: 
 (1)  
 (2)  
 (3)  
3
+
2
N OCH 
Cl 
H 
H 
H 
H 
O 
, , HCl 
+ -
2
N Cl
OCH
3
 
    O 
   || 
, CH
3
 – C – H , HCl 
+ -
2
N Cl
OCH
3
 
H 
H 
H 
H 
O 
, , H
2
O 
OCH
3
 
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Extra Questions

,

MCQs

,

past year papers

,

video lectures

,

Objective type Questions

,

practice quizzes

,

JEE Main 2021 Answer Key Chemistry - Evening Shift (17-03-2021) JEE Notes | EduRev

,

Exam

,

ppt

,

Free

,

Viva Questions

,

pdf

,

Important questions

,

JEE Main 2021 Answer Key Chemistry - Evening Shift (17-03-2021) JEE Notes | EduRev

,

Previous Year Questions with Solutions

,

JEE Main 2021 Answer Key Chemistry - Evening Shift (17-03-2021) JEE Notes | EduRev

,

shortcuts and tricks

,

study material

,

mock tests for examination

,

Summary

,

Semester Notes

,

Sample Paper

;