JEE Main 2021 Answer Key Chemistry - Evening Shift (26-02-2021) Notes | EduRev

Mock Test Series for JEE Main & Advanced 2022

JEE : JEE Main 2021 Answer Key Chemistry - Evening Shift (26-02-2021) Notes | EduRev

 Page 1


JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
PART–B : CHEMISTRY
SECTION - I
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. Which pair of oxides is acidic in nature?
(1) CaO, SiO
2
(2) B
2
O
3
, CaO
(3) B
2
O
3
, SiO
2
(4) N
2
O, BaO
Answer (3)
Sol. CaO – Basic
SiO
2
– Acidic
B
2
O
3
– Acidic
N
2
O – Neutral
BaO – Basic
2. Identify A in the given reaction,
OH
CH OH
2
HO
 
2
SOCl
 A (major product) ???? ?
(1)
OH
CH Cl
2
OH
(2)
Cl
CH OH
2
OH
(3)
Cl
CH Cl
2
Cl
(4)
OH
CH Cl
2
Cl
Answer (4)
Sol.
OH
OH
SOCl
2
CH OH
2
OH
Cl
CH Cl
2
3. Match List-I with List-II
List-I List-II
(a)
NCl
2 
+–
Cu Cl
22
Cl
+N
2
(i) Wurtz reaction
(b)
NCl
2 
+–
Cu/HCl
Cl
+N
2
(ii) Sandmeyer
reaction
(c) 2CH
3
CH
2
Cl+2Na (iii) Fittig reaction
Ether
???? ?
C
2
H
5
 – C
2
H
5 
+
 
2NaCl
(d) 2C
6
H
5
Cl + 2Na (iv) Gatterman
Ether
???? ?
C
6
H
5
 – C
6
H
5 
+
 
2NaCl reaction
Choose the correct answer from the options
given below
(1) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
(2) (a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
(3) (a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
(4) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
Answer (2)
Sol. (a) – Sandmeyer reaction
(b) – Gatterman reaction
(c) – Wurtz reaction
(d) – Fittig reaction
(a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
4. Match list-I with list-II
List-I List-II
 (Molecule) (Bond order)
(a) Ne
2
(i) 1
(b) N
2
(ii) 2
(c) F
2
(iii) 0
(d) O
2
(iv) 3
Choose the correct answer from the options
given below
(1) (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i)
(2) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
(3) (a)-(i); (b)-(ii); (c)-(iii); (d)-(iv)
(4) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
Answer (4)
Page 2


JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
PART–B : CHEMISTRY
SECTION - I
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. Which pair of oxides is acidic in nature?
(1) CaO, SiO
2
(2) B
2
O
3
, CaO
(3) B
2
O
3
, SiO
2
(4) N
2
O, BaO
Answer (3)
Sol. CaO – Basic
SiO
2
– Acidic
B
2
O
3
– Acidic
N
2
O – Neutral
BaO – Basic
2. Identify A in the given reaction,
OH
CH OH
2
HO
 
2
SOCl
 A (major product) ???? ?
(1)
OH
CH Cl
2
OH
(2)
Cl
CH OH
2
OH
(3)
Cl
CH Cl
2
Cl
(4)
OH
CH Cl
2
Cl
Answer (4)
Sol.
OH
OH
SOCl
2
CH OH
2
OH
Cl
CH Cl
2
3. Match List-I with List-II
List-I List-II
(a)
NCl
2 
+–
Cu Cl
22
Cl
+N
2
(i) Wurtz reaction
(b)
NCl
2 
+–
Cu/HCl
Cl
+N
2
(ii) Sandmeyer
reaction
(c) 2CH
3
CH
2
Cl+2Na (iii) Fittig reaction
Ether
???? ?
C
2
H
5
 – C
2
H
5 
+
 
2NaCl
(d) 2C
6
H
5
Cl + 2Na (iv) Gatterman
Ether
???? ?
C
6
H
5
 – C
6
H
5 
+
 
2NaCl reaction
Choose the correct answer from the options
given below
(1) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
(2) (a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
(3) (a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
(4) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
Answer (2)
Sol. (a) – Sandmeyer reaction
(b) – Gatterman reaction
(c) – Wurtz reaction
(d) – Fittig reaction
(a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
4. Match list-I with list-II
List-I List-II
 (Molecule) (Bond order)
(a) Ne
2
(i) 1
(b) N
2
(ii) 2
(c) F
2
(iii) 0
(d) O
2
(iv) 3
Choose the correct answer from the options
given below
(1) (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i)
(2) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
(3) (a)-(i); (b)-(ii); (c)-(iii); (d)-(iv)
(4) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
Answer (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
Sol. Molecule Bond order
Ne
2
0
N
2
3
F
2
1
O
2
2
(a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
5. 2,4-DNP test can be used to identify
(1) Aldehyde
(2) Amine
(3) Ether
(4) Halogens
Answer (1)
Sol. 2,4 DNP test is used to identify 
O
– C – group. It
gives addition reaction with carbonyl
compounds. So, it can be used to identify
aldehyde in the given option. It gives yellow/
orange PPt with carbonyl containing
compounds.
6. Seliwanoff test and Xanthoproteic test are
used for the identification of _______and _____
respectively.
(1) Ketoses, aldoses (2) Proteins, ketoses
(3) Ketoses, proteins (4) Aldoses, ketoses
Answer (3)
Sol. Seliwanoff test is used to distinguish ketoses
from aldoses. On treatment with a concentrated
acid, ketones are dehydrated more rapidly to
give furfural derivative and on condensation
with resorcinol give cherry red complex.
Positive Seliwanoff’s test – Ketoses present
Positive Xanthoproteic test – Presence of
aromatic amino acid
The Xanthoproteic reaction is a method that
can be used to detect presence of protein
soluble in a solution, using concentrated nitric
acid.
7. The correct order of electron gain enthalpy is:
(1) O > S > Se > Te
(2) Te > Se > S > O
(3) S > O > Se > Te
(4) S > Se > Te > O
Answer (4)
Sol. Correct order of electron gain enthalpy is
S > Se > Te > O
8. A. Phenyl methanamine
B. N,N-Dimethylaniline
C. N-Methyl aniline
D. Benzenamine
Choose the correct order of basic nature of
the above amines.
(1) A > C > B > D (2) D > B > C > A
(3) D > C > B > A (4) A > B > C > D
Answer (4)
Sol.
NH
2
CH
3
N
HC
3
NH
2
NH – CH
3
(A) Phenyl methanamine pK = 4.7
b
() D Benzenamine pK = 9.38
b
1
Basicity
pK
b
?
(B) N, N-Dimethylaniline pK = 8.92
b
N-Methyl aniline pK = 9.3
b C (   )
(A) > (B) > (C) > (D)
9. Match List-I with List-II.
List - I List - II
(a) Sodium Carbonate(i) Deacon
(b) Titanium (ii) Castner-Kellner
(c) Chlorine (iii) van-Arkel
(d) Sodium hydroxide (iv) Solvay
Chose the correct answer from the options
given below:
(1) (a) ? (i), (b) ? (iii), (c) ? (iv), (d) ? (ii)
(2) (a) ? (iii), (b) ? (ii), (c) ? (i), (d) ? (iv)
(3) (a) ? (iv), (b) ? (i), (c) ? (ii), (d) ? (iii)
(4) (a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
Answer (4)
Sol. Compound Method of preparation
Sodium Carbonate Solvey
Titanium van-Arkel
Chlorine Deacon
Sodium hydroxide Castner-Kellner
(a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
Page 3


JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
PART–B : CHEMISTRY
SECTION - I
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. Which pair of oxides is acidic in nature?
(1) CaO, SiO
2
(2) B
2
O
3
, CaO
(3) B
2
O
3
, SiO
2
(4) N
2
O, BaO
Answer (3)
Sol. CaO – Basic
SiO
2
– Acidic
B
2
O
3
– Acidic
N
2
O – Neutral
BaO – Basic
2. Identify A in the given reaction,
OH
CH OH
2
HO
 
2
SOCl
 A (major product) ???? ?
(1)
OH
CH Cl
2
OH
(2)
Cl
CH OH
2
OH
(3)
Cl
CH Cl
2
Cl
(4)
OH
CH Cl
2
Cl
Answer (4)
Sol.
OH
OH
SOCl
2
CH OH
2
OH
Cl
CH Cl
2
3. Match List-I with List-II
List-I List-II
(a)
NCl
2 
+–
Cu Cl
22
Cl
+N
2
(i) Wurtz reaction
(b)
NCl
2 
+–
Cu/HCl
Cl
+N
2
(ii) Sandmeyer
reaction
(c) 2CH
3
CH
2
Cl+2Na (iii) Fittig reaction
Ether
???? ?
C
2
H
5
 – C
2
H
5 
+
 
2NaCl
(d) 2C
6
H
5
Cl + 2Na (iv) Gatterman
Ether
???? ?
C
6
H
5
 – C
6
H
5 
+
 
2NaCl reaction
Choose the correct answer from the options
given below
(1) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
(2) (a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
(3) (a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
(4) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
Answer (2)
Sol. (a) – Sandmeyer reaction
(b) – Gatterman reaction
(c) – Wurtz reaction
(d) – Fittig reaction
(a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
4. Match list-I with list-II
List-I List-II
 (Molecule) (Bond order)
(a) Ne
2
(i) 1
(b) N
2
(ii) 2
(c) F
2
(iii) 0
(d) O
2
(iv) 3
Choose the correct answer from the options
given below
(1) (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i)
(2) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
(3) (a)-(i); (b)-(ii); (c)-(iii); (d)-(iv)
(4) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
Answer (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
Sol. Molecule Bond order
Ne
2
0
N
2
3
F
2
1
O
2
2
(a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
5. 2,4-DNP test can be used to identify
(1) Aldehyde
(2) Amine
(3) Ether
(4) Halogens
Answer (1)
Sol. 2,4 DNP test is used to identify 
O
– C – group. It
gives addition reaction with carbonyl
compounds. So, it can be used to identify
aldehyde in the given option. It gives yellow/
orange PPt with carbonyl containing
compounds.
6. Seliwanoff test and Xanthoproteic test are
used for the identification of _______and _____
respectively.
(1) Ketoses, aldoses (2) Proteins, ketoses
(3) Ketoses, proteins (4) Aldoses, ketoses
Answer (3)
Sol. Seliwanoff test is used to distinguish ketoses
from aldoses. On treatment with a concentrated
acid, ketones are dehydrated more rapidly to
give furfural derivative and on condensation
with resorcinol give cherry red complex.
Positive Seliwanoff’s test – Ketoses present
Positive Xanthoproteic test – Presence of
aromatic amino acid
The Xanthoproteic reaction is a method that
can be used to detect presence of protein
soluble in a solution, using concentrated nitric
acid.
7. The correct order of electron gain enthalpy is:
(1) O > S > Se > Te
(2) Te > Se > S > O
(3) S > O > Se > Te
(4) S > Se > Te > O
Answer (4)
Sol. Correct order of electron gain enthalpy is
S > Se > Te > O
8. A. Phenyl methanamine
B. N,N-Dimethylaniline
C. N-Methyl aniline
D. Benzenamine
Choose the correct order of basic nature of
the above amines.
(1) A > C > B > D (2) D > B > C > A
(3) D > C > B > A (4) A > B > C > D
Answer (4)
Sol.
NH
2
CH
3
N
HC
3
NH
2
NH – CH
3
(A) Phenyl methanamine pK = 4.7
b
() D Benzenamine pK = 9.38
b
1
Basicity
pK
b
?
(B) N, N-Dimethylaniline pK = 8.92
b
N-Methyl aniline pK = 9.3
b C (   )
(A) > (B) > (C) > (D)
9. Match List-I with List-II.
List - I List - II
(a) Sodium Carbonate(i) Deacon
(b) Titanium (ii) Castner-Kellner
(c) Chlorine (iii) van-Arkel
(d) Sodium hydroxide (iv) Solvay
Chose the correct answer from the options
given below:
(1) (a) ? (i), (b) ? (iii), (c) ? (iv), (d) ? (ii)
(2) (a) ? (iii), (b) ? (ii), (c) ? (i), (d) ? (iv)
(3) (a) ? (iv), (b) ? (i), (c) ? (ii), (d) ? (iii)
(4) (a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
Answer (4)
Sol. Compound Method of preparation
Sodium Carbonate Solvey
Titanium van-Arkel
Chlorine Deacon
Sodium hydroxide Castner-Kellner
(a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
10. The nature of charge on resulting colloidal
particles when FeCl
3
 is added to excess of hot
water is:
(1) Sometimes positive and sometimes
negative
(2) Negative
(3) Neutral
(4) Positive
Answer (4)
Sol. Some FeCl
3
/Fe
3+
 will get hydrolyzed and form
Fe(OH)
3
. Over which some Fe
3+
 will get
adsorbed. So the resulting charge on colloidal
particle will be positive.
11. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : In TlI
3
, isomorphous to CsI
3
, the
metal is present in +1 oxidation
state.
Reason R : Tl metal has fourteen f electrons
in its electronic configuration.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct but R is NOT the
correct explanation of A
(2) Both A and R are correct and R is the
correct explanation of A
(3) A is correct but R is not correct
(4) A is not correct but R is correct
Answer (1)
Sol.
A : Due to inert pair effect, Tl is more stable in
+1 oxidation state
Hence TlI
3
 and CSI
3
 are isomorphous
R : Electronic configuration of Tl (81) =
Xe 4f
14
 5d
10
 6s
2
 6p
1
Both A and R are correct but R is not the
correct explanation of A.
12. Identify A in the following chemical reaction.
(i) HCHO , NaOH
A
CHO
CHO
3
(ii) CH CH Br , NaH, DMF
32
(iii) HI, ?
(1)
CHI
2
HO
(2)
CHOH
2
HO
(3)
C — OCH CH
23
HO
O
(4)
CH OH
2
CHO
3
Answer (1)
Sol.
CHO
CH O
3
HCHO, OH
–
(cannizzaro
  reaction)
CH OH
2
OCH
3
+ HCOO
–
CH – O
2
–
OCH
3
CH Br
25
CH OC H
225
OCH
3
NaH
CH I
2
OH
+ C H OH + CH OH
25 3
HI
13. Calgon is used for water treatment. Which of
the following statement is NOT true about
Calgon?
(1) It is polymeric compound and is water
soluble
(2) Calgon contains the 2
nd
 most abundant
element by weight in the Earth’s crust
(3) It is also known as Graham’s salt
(4) It doesnot remove Ca
2+
 ion by precipitation
Answer (2)
Sol. Calgon is sodium hexametaphosphate, a
polymeric compound also called as Graham’s
salt.
Silicon is the 2
nd
 most abundant element which
is absent in calgon.
14. Match List-I with List-II.
List-I List-II
(a) Siderite (i) Cu
(b) Calamine (ii) Ca
(c) Malachite (iii) Fe
(d) Cryolite (iv) Al
(v) Zn
Page 4


JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
PART–B : CHEMISTRY
SECTION - I
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. Which pair of oxides is acidic in nature?
(1) CaO, SiO
2
(2) B
2
O
3
, CaO
(3) B
2
O
3
, SiO
2
(4) N
2
O, BaO
Answer (3)
Sol. CaO – Basic
SiO
2
– Acidic
B
2
O
3
– Acidic
N
2
O – Neutral
BaO – Basic
2. Identify A in the given reaction,
OH
CH OH
2
HO
 
2
SOCl
 A (major product) ???? ?
(1)
OH
CH Cl
2
OH
(2)
Cl
CH OH
2
OH
(3)
Cl
CH Cl
2
Cl
(4)
OH
CH Cl
2
Cl
Answer (4)
Sol.
OH
OH
SOCl
2
CH OH
2
OH
Cl
CH Cl
2
3. Match List-I with List-II
List-I List-II
(a)
NCl
2 
+–
Cu Cl
22
Cl
+N
2
(i) Wurtz reaction
(b)
NCl
2 
+–
Cu/HCl
Cl
+N
2
(ii) Sandmeyer
reaction
(c) 2CH
3
CH
2
Cl+2Na (iii) Fittig reaction
Ether
???? ?
C
2
H
5
 – C
2
H
5 
+
 
2NaCl
(d) 2C
6
H
5
Cl + 2Na (iv) Gatterman
Ether
???? ?
C
6
H
5
 – C
6
H
5 
+
 
2NaCl reaction
Choose the correct answer from the options
given below
(1) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
(2) (a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
(3) (a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
(4) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
Answer (2)
Sol. (a) – Sandmeyer reaction
(b) – Gatterman reaction
(c) – Wurtz reaction
(d) – Fittig reaction
(a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
4. Match list-I with list-II
List-I List-II
 (Molecule) (Bond order)
(a) Ne
2
(i) 1
(b) N
2
(ii) 2
(c) F
2
(iii) 0
(d) O
2
(iv) 3
Choose the correct answer from the options
given below
(1) (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i)
(2) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
(3) (a)-(i); (b)-(ii); (c)-(iii); (d)-(iv)
(4) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
Answer (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
Sol. Molecule Bond order
Ne
2
0
N
2
3
F
2
1
O
2
2
(a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
5. 2,4-DNP test can be used to identify
(1) Aldehyde
(2) Amine
(3) Ether
(4) Halogens
Answer (1)
Sol. 2,4 DNP test is used to identify 
O
– C – group. It
gives addition reaction with carbonyl
compounds. So, it can be used to identify
aldehyde in the given option. It gives yellow/
orange PPt with carbonyl containing
compounds.
6. Seliwanoff test and Xanthoproteic test are
used for the identification of _______and _____
respectively.
(1) Ketoses, aldoses (2) Proteins, ketoses
(3) Ketoses, proteins (4) Aldoses, ketoses
Answer (3)
Sol. Seliwanoff test is used to distinguish ketoses
from aldoses. On treatment with a concentrated
acid, ketones are dehydrated more rapidly to
give furfural derivative and on condensation
with resorcinol give cherry red complex.
Positive Seliwanoff’s test – Ketoses present
Positive Xanthoproteic test – Presence of
aromatic amino acid
The Xanthoproteic reaction is a method that
can be used to detect presence of protein
soluble in a solution, using concentrated nitric
acid.
7. The correct order of electron gain enthalpy is:
(1) O > S > Se > Te
(2) Te > Se > S > O
(3) S > O > Se > Te
(4) S > Se > Te > O
Answer (4)
Sol. Correct order of electron gain enthalpy is
S > Se > Te > O
8. A. Phenyl methanamine
B. N,N-Dimethylaniline
C. N-Methyl aniline
D. Benzenamine
Choose the correct order of basic nature of
the above amines.
(1) A > C > B > D (2) D > B > C > A
(3) D > C > B > A (4) A > B > C > D
Answer (4)
Sol.
NH
2
CH
3
N
HC
3
NH
2
NH – CH
3
(A) Phenyl methanamine pK = 4.7
b
() D Benzenamine pK = 9.38
b
1
Basicity
pK
b
?
(B) N, N-Dimethylaniline pK = 8.92
b
N-Methyl aniline pK = 9.3
b C (   )
(A) > (B) > (C) > (D)
9. Match List-I with List-II.
List - I List - II
(a) Sodium Carbonate(i) Deacon
(b) Titanium (ii) Castner-Kellner
(c) Chlorine (iii) van-Arkel
(d) Sodium hydroxide (iv) Solvay
Chose the correct answer from the options
given below:
(1) (a) ? (i), (b) ? (iii), (c) ? (iv), (d) ? (ii)
(2) (a) ? (iii), (b) ? (ii), (c) ? (i), (d) ? (iv)
(3) (a) ? (iv), (b) ? (i), (c) ? (ii), (d) ? (iii)
(4) (a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
Answer (4)
Sol. Compound Method of preparation
Sodium Carbonate Solvey
Titanium van-Arkel
Chlorine Deacon
Sodium hydroxide Castner-Kellner
(a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
10. The nature of charge on resulting colloidal
particles when FeCl
3
 is added to excess of hot
water is:
(1) Sometimes positive and sometimes
negative
(2) Negative
(3) Neutral
(4) Positive
Answer (4)
Sol. Some FeCl
3
/Fe
3+
 will get hydrolyzed and form
Fe(OH)
3
. Over which some Fe
3+
 will get
adsorbed. So the resulting charge on colloidal
particle will be positive.
11. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : In TlI
3
, isomorphous to CsI
3
, the
metal is present in +1 oxidation
state.
Reason R : Tl metal has fourteen f electrons
in its electronic configuration.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct but R is NOT the
correct explanation of A
(2) Both A and R are correct and R is the
correct explanation of A
(3) A is correct but R is not correct
(4) A is not correct but R is correct
Answer (1)
Sol.
A : Due to inert pair effect, Tl is more stable in
+1 oxidation state
Hence TlI
3
 and CSI
3
 are isomorphous
R : Electronic configuration of Tl (81) =
Xe 4f
14
 5d
10
 6s
2
 6p
1
Both A and R are correct but R is not the
correct explanation of A.
12. Identify A in the following chemical reaction.
(i) HCHO , NaOH
A
CHO
CHO
3
(ii) CH CH Br , NaH, DMF
32
(iii) HI, ?
(1)
CHI
2
HO
(2)
CHOH
2
HO
(3)
C — OCH CH
23
HO
O
(4)
CH OH
2
CHO
3
Answer (1)
Sol.
CHO
CH O
3
HCHO, OH
–
(cannizzaro
  reaction)
CH OH
2
OCH
3
+ HCOO
–
CH – O
2
–
OCH
3
CH Br
25
CH OC H
225
OCH
3
NaH
CH I
2
OH
+ C H OH + CH OH
25 3
HI
13. Calgon is used for water treatment. Which of
the following statement is NOT true about
Calgon?
(1) It is polymeric compound and is water
soluble
(2) Calgon contains the 2
nd
 most abundant
element by weight in the Earth’s crust
(3) It is also known as Graham’s salt
(4) It doesnot remove Ca
2+
 ion by precipitation
Answer (2)
Sol. Calgon is sodium hexametaphosphate, a
polymeric compound also called as Graham’s
salt.
Silicon is the 2
nd
 most abundant element which
is absent in calgon.
14. Match List-I with List-II.
List-I List-II
(a) Siderite (i) Cu
(b) Calamine (ii) Ca
(c) Malachite (iii) Fe
(d) Cryolite (iv) Al
(v) Zn
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
Choose the correct answer from the options
given below
(1) (a) (i), (b) (ii), (c) (iii), (d) (iv)
(2) (a) (i), (b) (ii), (c) (v), (d) (iii)
(3) (a) (iii), (b) (v), (c) (i), (d) (iv)
(4) (a) (iii), (b) (i), (c) (v), (d) (ii)
Answer (3)
Sol. Siderite FeCO
3
Calamine ZnCO
3
Malachite CuCO
3
?Cu(OH)
2
Cryolite Na
3
AlF
6
15. Identify A in the given chemical reaction.
CHCH CHO
22
CHCH CHO
22
NaOH
CH OH, H O
25 2
?
 A (Major Product)
(1)
CHO
(2)
CHCH COOH
22
CHCH CHOH
22 2
(3)
O
O
(4)
C — H
O
Answer (4)
Sol.
CH – CH – CH
22
CH – CH – CHO
2
NaOH
CH – CH – CH
22
CH – CH
2
OH
CHO
?
O
H– OH
CHO
16. Ceric ammonium nitrate and CHCl
3
/alc. KOH
are used for the identification of functional
groups present in _____ and _____ respectively.
(1) Alcohol, phenol
(2) Amine, phenol
(3) Amine, alcohol
(4) Alcohol, amine
Answer (4)
Sol. Ceric ammonium nitrate is used for the
identification of alcohol.
??
??
42 3 6 2 3 4 4 3
Alkoxy cerium ion IV
Compound
Pink or Red colour
2R—OH (NH ) Ce(NO)(ROH) Ce(NO ) 2NH NO ?????
CHCl
3
/KOH is used for the identification of
primary amines.
?? ? ???? ?
warm
23
RNH CHCl 3KOH(alc)
??
2
R—NC 3KCl 3H O
17. In ?? ?
123 4
23
CH C CH CH molecule, the
hybridization of carbon 1, 2, 3 and 4
respectively, are :
(1) sp
2
, sp
2
, sp
2
, sp
3
(2) sp
2
, sp, sp
2
, sp
3
(3) sp
3
, sp, sp
3
, sp
3
(4) sp
2
, sp
3
, sp
2
, sp
3
Answer (2)
Sol.
?? ?
2 2 3
23
sp
sp sp sp
CH C CH CH
Hybridization of carbon 1, 2, 3 and 4
respectively are sp
2
,
 
sp,
 
sp
2 
and
 
sp
3
18. Which of the following forms of hydrogen emits
low energy ?
–
 particles?
(1) Proton H
+
(2) Tritium 
3
1
H
(3) Protium 
1
1
H (4) Deuterium 
2
1
H
Answer (2)
Sol. Out of isotopes of hydrogen, only tritium is
radioactive and emits low energy ?
–
 particles.
19.
(1) Zn/HCl
(2) Cr O , 773 K
23
10 - 20 atm O
Considering the above reaction, the major
product among the following is :
(1)
COCHCH
23
(2)
CH
3
CH
3
(3)
CHCH CH
22 3
(4)
CHCH
23
Answer (4)
Sol.
(1) Zn/HCl
(2) Cr O , 773 K
23
10 - 20 atm O
CH CH
23
Page 5


JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
PART–B : CHEMISTRY
SECTION - I
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. Which pair of oxides is acidic in nature?
(1) CaO, SiO
2
(2) B
2
O
3
, CaO
(3) B
2
O
3
, SiO
2
(4) N
2
O, BaO
Answer (3)
Sol. CaO – Basic
SiO
2
– Acidic
B
2
O
3
– Acidic
N
2
O – Neutral
BaO – Basic
2. Identify A in the given reaction,
OH
CH OH
2
HO
 
2
SOCl
 A (major product) ???? ?
(1)
OH
CH Cl
2
OH
(2)
Cl
CH OH
2
OH
(3)
Cl
CH Cl
2
Cl
(4)
OH
CH Cl
2
Cl
Answer (4)
Sol.
OH
OH
SOCl
2
CH OH
2
OH
Cl
CH Cl
2
3. Match List-I with List-II
List-I List-II
(a)
NCl
2 
+–
Cu Cl
22
Cl
+N
2
(i) Wurtz reaction
(b)
NCl
2 
+–
Cu/HCl
Cl
+N
2
(ii) Sandmeyer
reaction
(c) 2CH
3
CH
2
Cl+2Na (iii) Fittig reaction
Ether
???? ?
C
2
H
5
 – C
2
H
5 
+
 
2NaCl
(d) 2C
6
H
5
Cl + 2Na (iv) Gatterman
Ether
???? ?
C
6
H
5
 – C
6
H
5 
+
 
2NaCl reaction
Choose the correct answer from the options
given below
(1) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
(2) (a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
(3) (a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
(4) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
Answer (2)
Sol. (a) – Sandmeyer reaction
(b) – Gatterman reaction
(c) – Wurtz reaction
(d) – Fittig reaction
(a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
4. Match list-I with list-II
List-I List-II
 (Molecule) (Bond order)
(a) Ne
2
(i) 1
(b) N
2
(ii) 2
(c) F
2
(iii) 0
(d) O
2
(iv) 3
Choose the correct answer from the options
given below
(1) (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i)
(2) (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
(3) (a)-(i); (b)-(ii); (c)-(iii); (d)-(iv)
(4) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
Answer (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
Sol. Molecule Bond order
Ne
2
0
N
2
3
F
2
1
O
2
2
(a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
5. 2,4-DNP test can be used to identify
(1) Aldehyde
(2) Amine
(3) Ether
(4) Halogens
Answer (1)
Sol. 2,4 DNP test is used to identify 
O
– C – group. It
gives addition reaction with carbonyl
compounds. So, it can be used to identify
aldehyde in the given option. It gives yellow/
orange PPt with carbonyl containing
compounds.
6. Seliwanoff test and Xanthoproteic test are
used for the identification of _______and _____
respectively.
(1) Ketoses, aldoses (2) Proteins, ketoses
(3) Ketoses, proteins (4) Aldoses, ketoses
Answer (3)
Sol. Seliwanoff test is used to distinguish ketoses
from aldoses. On treatment with a concentrated
acid, ketones are dehydrated more rapidly to
give furfural derivative and on condensation
with resorcinol give cherry red complex.
Positive Seliwanoff’s test – Ketoses present
Positive Xanthoproteic test – Presence of
aromatic amino acid
The Xanthoproteic reaction is a method that
can be used to detect presence of protein
soluble in a solution, using concentrated nitric
acid.
7. The correct order of electron gain enthalpy is:
(1) O > S > Se > Te
(2) Te > Se > S > O
(3) S > O > Se > Te
(4) S > Se > Te > O
Answer (4)
Sol. Correct order of electron gain enthalpy is
S > Se > Te > O
8. A. Phenyl methanamine
B. N,N-Dimethylaniline
C. N-Methyl aniline
D. Benzenamine
Choose the correct order of basic nature of
the above amines.
(1) A > C > B > D (2) D > B > C > A
(3) D > C > B > A (4) A > B > C > D
Answer (4)
Sol.
NH
2
CH
3
N
HC
3
NH
2
NH – CH
3
(A) Phenyl methanamine pK = 4.7
b
() D Benzenamine pK = 9.38
b
1
Basicity
pK
b
?
(B) N, N-Dimethylaniline pK = 8.92
b
N-Methyl aniline pK = 9.3
b C (   )
(A) > (B) > (C) > (D)
9. Match List-I with List-II.
List - I List - II
(a) Sodium Carbonate(i) Deacon
(b) Titanium (ii) Castner-Kellner
(c) Chlorine (iii) van-Arkel
(d) Sodium hydroxide (iv) Solvay
Chose the correct answer from the options
given below:
(1) (a) ? (i), (b) ? (iii), (c) ? (iv), (d) ? (ii)
(2) (a) ? (iii), (b) ? (ii), (c) ? (i), (d) ? (iv)
(3) (a) ? (iv), (b) ? (i), (c) ? (ii), (d) ? (iii)
(4) (a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
Answer (4)
Sol. Compound Method of preparation
Sodium Carbonate Solvey
Titanium van-Arkel
Chlorine Deacon
Sodium hydroxide Castner-Kellner
(a) ? (iv), (b) ? (iii), (c) ? (i), (d) ? (ii)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
10. The nature of charge on resulting colloidal
particles when FeCl
3
 is added to excess of hot
water is:
(1) Sometimes positive and sometimes
negative
(2) Negative
(3) Neutral
(4) Positive
Answer (4)
Sol. Some FeCl
3
/Fe
3+
 will get hydrolyzed and form
Fe(OH)
3
. Over which some Fe
3+
 will get
adsorbed. So the resulting charge on colloidal
particle will be positive.
11. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : In TlI
3
, isomorphous to CsI
3
, the
metal is present in +1 oxidation
state.
Reason R : Tl metal has fourteen f electrons
in its electronic configuration.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct but R is NOT the
correct explanation of A
(2) Both A and R are correct and R is the
correct explanation of A
(3) A is correct but R is not correct
(4) A is not correct but R is correct
Answer (1)
Sol.
A : Due to inert pair effect, Tl is more stable in
+1 oxidation state
Hence TlI
3
 and CSI
3
 are isomorphous
R : Electronic configuration of Tl (81) =
Xe 4f
14
 5d
10
 6s
2
 6p
1
Both A and R are correct but R is not the
correct explanation of A.
12. Identify A in the following chemical reaction.
(i) HCHO , NaOH
A
CHO
CHO
3
(ii) CH CH Br , NaH, DMF
32
(iii) HI, ?
(1)
CHI
2
HO
(2)
CHOH
2
HO
(3)
C — OCH CH
23
HO
O
(4)
CH OH
2
CHO
3
Answer (1)
Sol.
CHO
CH O
3
HCHO, OH
–
(cannizzaro
  reaction)
CH OH
2
OCH
3
+ HCOO
–
CH – O
2
–
OCH
3
CH Br
25
CH OC H
225
OCH
3
NaH
CH I
2
OH
+ C H OH + CH OH
25 3
HI
13. Calgon is used for water treatment. Which of
the following statement is NOT true about
Calgon?
(1) It is polymeric compound and is water
soluble
(2) Calgon contains the 2
nd
 most abundant
element by weight in the Earth’s crust
(3) It is also known as Graham’s salt
(4) It doesnot remove Ca
2+
 ion by precipitation
Answer (2)
Sol. Calgon is sodium hexametaphosphate, a
polymeric compound also called as Graham’s
salt.
Silicon is the 2
nd
 most abundant element which
is absent in calgon.
14. Match List-I with List-II.
List-I List-II
(a) Siderite (i) Cu
(b) Calamine (ii) Ca
(c) Malachite (iii) Fe
(d) Cryolite (iv) Al
(v) Zn
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
Choose the correct answer from the options
given below
(1) (a) (i), (b) (ii), (c) (iii), (d) (iv)
(2) (a) (i), (b) (ii), (c) (v), (d) (iii)
(3) (a) (iii), (b) (v), (c) (i), (d) (iv)
(4) (a) (iii), (b) (i), (c) (v), (d) (ii)
Answer (3)
Sol. Siderite FeCO
3
Calamine ZnCO
3
Malachite CuCO
3
?Cu(OH)
2
Cryolite Na
3
AlF
6
15. Identify A in the given chemical reaction.
CHCH CHO
22
CHCH CHO
22
NaOH
CH OH, H O
25 2
?
 A (Major Product)
(1)
CHO
(2)
CHCH COOH
22
CHCH CHOH
22 2
(3)
O
O
(4)
C — H
O
Answer (4)
Sol.
CH – CH – CH
22
CH – CH – CHO
2
NaOH
CH – CH – CH
22
CH – CH
2
OH
CHO
?
O
H– OH
CHO
16. Ceric ammonium nitrate and CHCl
3
/alc. KOH
are used for the identification of functional
groups present in _____ and _____ respectively.
(1) Alcohol, phenol
(2) Amine, phenol
(3) Amine, alcohol
(4) Alcohol, amine
Answer (4)
Sol. Ceric ammonium nitrate is used for the
identification of alcohol.
??
??
42 3 6 2 3 4 4 3
Alkoxy cerium ion IV
Compound
Pink or Red colour
2R—OH (NH ) Ce(NO)(ROH) Ce(NO ) 2NH NO ?????
CHCl
3
/KOH is used for the identification of
primary amines.
?? ? ???? ?
warm
23
RNH CHCl 3KOH(alc)
??
2
R—NC 3KCl 3H O
17. In ?? ?
123 4
23
CH C CH CH molecule, the
hybridization of carbon 1, 2, 3 and 4
respectively, are :
(1) sp
2
, sp
2
, sp
2
, sp
3
(2) sp
2
, sp, sp
2
, sp
3
(3) sp
3
, sp, sp
3
, sp
3
(4) sp
2
, sp
3
, sp
2
, sp
3
Answer (2)
Sol.
?? ?
2 2 3
23
sp
sp sp sp
CH C CH CH
Hybridization of carbon 1, 2, 3 and 4
respectively are sp
2
,
 
sp,
 
sp
2 
and
 
sp
3
18. Which of the following forms of hydrogen emits
low energy ?
–
 particles?
(1) Proton H
+
(2) Tritium 
3
1
H
(3) Protium 
1
1
H (4) Deuterium 
2
1
H
Answer (2)
Sol. Out of isotopes of hydrogen, only tritium is
radioactive and emits low energy ?
–
 particles.
19.
(1) Zn/HCl
(2) Cr O , 773 K
23
10 - 20 atm O
Considering the above reaction, the major
product among the following is :
(1)
COCHCH
23
(2)
CH
3
CH
3
(3)
CHCH CH
22 3
(4)
CHCH
23
Answer (4)
Sol.
(1) Zn/HCl
(2) Cr O , 773 K
23
10 - 20 atm O
CH CH
23
JEE (MAIN)-2021 : Phase-1(26-02-2021)-E
20. Match List-I with List-II.
List-I List-II
(a) Sucrose (i) ?-D-Galactose and
?-D-Glucose
(b) Lactose (ii) ?-D-Glucose and
?-D-Fructose
(c) Maltose (iii) ?-D-Glucose and
?-D-Glucose
Choose the correct answer from the options
given below :
(1) (a) ??(i), (b) ??(iii), (c) ??(ii)
(2) (a) ??(iii), (b) ??(ii), (c) ??(i)
(3) (a) ??(ii), (b) ??(i), (c) ??(iii)
(4) (a) ??(iii), (b) ??(i), (c) ??(ii)
Answer (3)
Sol. Disaccharides Monomer present
Sucrose ?-D-glucose and
?-D-fructose
Lactose ?-D-Galactose and
?-D-Glucose
Maltose ?-D-Glucose and
?-D-Glucose
(a) ??(ii), (b) ??(i), (c) ??(iii)
SECTION - II
Numerical Value Type Questions: This section
contains 10 questions. In Section II, 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.
1. If the activation energy of a reaction is 80.9 kJ
mol
–1
, the fraction of molecules at 700 K,
having enough energy to react to form
products is e
–x
. The value of x is _________.
(Rounded off to the nearest integer)
[Use R = 8.31 J K
–1
 mol
–1
]
Answer (14)
Energy of activation, E
a
 = 80.9 kJ mol
–1
Temperature of reaction, T = 700 K
Fraction of molecules having enough energy
to react a
E/RT x
ee
? ?
??
a
E 80900
x13.914
RT 8.31 700
?? ? ?
?

2. The average S-F bond energy in kJ mol
–1
 of
SF
6
 is ____________. (Rounded off to the
nearest integer)
[Given : The values of standard enthalpy of
formation of SF
6(g)
, S
(g)
 and F
(g)
 are - 1100, 275
and 80 kJ mol
–1
 respectively.]
Answer (309)
SF
6
(g) ?? S(g) + 6F(g)
f ff6
HH (S) 6H (F) H (SF )
?? ?
???? ? ? ??
= 275 + 6 × 80 – (–1100)
= 1855 kJ mol
–1
Also, 
SF
H6H
?
??? ?
1
SF
1855
H 309.17 309kJmol
6
?
?
?? ? ? 
3. The NaNO
3
 weighed out to make 50 mL of an
aqueous solution containing 70.0 mg Na
+
 per
mL is __________ g. (Rounded off to the
nearest integer)
[Given : Atomic weight in g mol
–1
 - Na : 23; N :
14; O : 16]
Answer (13)
Mass of Na
+
 in 50 mL = 70 × 50 mg
Millimoles of NaNO
3
 = 
70 50
23
?
Mass of NaNO
3
 = 
3
70 50 85 10
23
?
?? ?
12.9 13 g ? 
4. When 12.2 g of benzoic acid is dissolved in
100 g of water, the freezing point of solution
was found to be –0.93°C (K
f
 (H
2
O) = 1.86 K kg
mol
–1
). The number (n) of benzoic acid
molecules associated (assuming 100%
association) is __________.
Answer (02.00)
ff
TiKm ??
12.2 1000
0.93 i 1.86
122 100
?
?? ?
?
i = 0.5
n(Benzoic acid) ?? (Benzoic acid)
n
Totalnumberof particlesafterassociation
i
Numberof particlesbeforeassociation
?
1
0.5
n
?
n = 2
Read More
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