JEE Main 2021 Chemistry February 24 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 : Phase-1(24-02-2021)-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 one of the following carbonyl
compounds cannot be prepared by addition of
water on an alkyne in the presence of HgSO
4
and H
2
SO
4
?
(1) C
O
CH
3
(2)
O
CH – C – CH CH
323
(3)
O
CH – C – H
3
(4)
O
CH – CH – C – H
32
Answer (4)
Sol. CH
3
 – CH
2
 – CHO (Propanaldehyde) cannot be
prepared by addition of water on alkyne in the
presence of HgSO
4
 and H
2
SO
4
.
C
O
CH
3
C  CH ?
HO
2
Hg , H
2+ +
CH – CHO
3
HC  CH ?
H O
2
Hg , H
2+ +
O
CH – C – CH
33
CH – C  CH
3
?
HO
2
Hg , H
2+ +
2. Match List-I with List-II.
List-I List-II
(Metal) (Ores)
(a) Aluminium (i) Siderite
(b) Iron (ii) Calamine
(c) Copper (iii) Kaolinite
(d) Zinc (iv) Malachite
Choose the correct answer from the options
given below:
(1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Answer (3)
Sol. Ores Formula Metal present
Siderite FeCO
3
Fe
Calamine ZnCO
3
Zn
Kaolinite Al
2
Si
2
O
5
(OH)
4
Al
Malachite Cu(OH)
2
?CuCO
3
Cu
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
3. Which one of the following compounds is non-
aromatic?
(1)
O
(2)
(3)
(4)
Answer (2)
Sol.
O
Aromatic
Non-aromatic
sp hybridized
3
Aromatic
Aromatic
Page 2


JEE (MAIN)-2021 : Phase-1(24-02-2021)-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 one of the following carbonyl
compounds cannot be prepared by addition of
water on an alkyne in the presence of HgSO
4
and H
2
SO
4
?
(1) C
O
CH
3
(2)
O
CH – C – CH CH
323
(3)
O
CH – C – H
3
(4)
O
CH – CH – C – H
32
Answer (4)
Sol. CH
3
 – CH
2
 – CHO (Propanaldehyde) cannot be
prepared by addition of water on alkyne in the
presence of HgSO
4
 and H
2
SO
4
.
C
O
CH
3
C  CH ?
HO
2
Hg , H
2+ +
CH – CHO
3
HC  CH ?
H O
2
Hg , H
2+ +
O
CH – C – CH
33
CH – C  CH
3
?
HO
2
Hg , H
2+ +
2. Match List-I with List-II.
List-I List-II
(Metal) (Ores)
(a) Aluminium (i) Siderite
(b) Iron (ii) Calamine
(c) Copper (iii) Kaolinite
(d) Zinc (iv) Malachite
Choose the correct answer from the options
given below:
(1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Answer (3)
Sol. Ores Formula Metal present
Siderite FeCO
3
Fe
Calamine ZnCO
3
Zn
Kaolinite Al
2
Si
2
O
5
(OH)
4
Al
Malachite Cu(OH)
2
?CuCO
3
Cu
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
3. Which one of the following compounds is non-
aromatic?
(1)
O
(2)
(3)
(4)
Answer (2)
Sol.
O
Aromatic
Non-aromatic
sp hybridized
3
Aromatic
Aromatic
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
4. The correct shape and I – I – I bond angles
respectively in 
?
3
I ion are:
(1) Distorted trigonal planar; 135° and 90°
(2) Trigonal planar; 120°
(3) T-shaped; 180° and 90°
(4) Linear; 180°
Answer (4)
Sol.
?
3
I
I
I
I
Shape = Linear
Angle ?I – I – I is 180°
5. Given below are two statements : one is
labelled as Assertion A and the other is
labelled as Reason R.
Assertion A : Hydrogen is the most abundant
element in the Universe, but it is not the most
abundant gas in the troposphere.
Reason R : Hydrogen is the lightest element.
In the light of the above statements, choose
the correct answer from the options given
below.
(1) Both A and R are true but R is NOT the
correct explanation of A
(2) A is true but R is false
(3) Both A and R are true and R is the correct
explanation of A
(4) A is false but R is true
Answer (3)
Sol. Both A and R are true and R is the correct
explanation of A.
Hydrogen is the lightest element. Because it is
lighter than air, and so it can easily escape the
earth’s gravity.
6. Given below are two statements :
Statement I : The value of the parameter
“Biochemical Oxygen Demand (BOD)” is
important for survival of aquatic life.
Statement II : The optimum value of BOD is 6.5
ppm.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) Both statement I and statement II are true
(2) Both statement I and statement II are false
(3) Statement I is false but statement II is true
(4) Statement I is true but statement II is false
Answer (4)
Sol. The amount of BOD in the water is a measure
of the amount of organic material in the water,
in terms of how much oxygen will be required
to break it down biologically. Clean water
would have BOD value of less than 5 ppm,
whereas highly polluted water would have a
BOD value of 17 ppm or more.
Statement I is true and statement II is false.
7. The calculated magnetic moments (spin only
value) for species [FeCl
4
]
2–
, [Co(C
2
O
4
)
3
]
3–
 and
2–
4
MnO
 respectively are :
(1) 5.82, 0 and 0 BM
(2) 4.90, 0 and 1.73 BM
(3) 5.92, 4.90 and 0 BM
(4) 4.90, 0 and 2.83 BM
Answer (2)
Sol. [FeCl
4
]
2–
Fe
2+
 + Weak ligand n = 4
(High spin complex) ? = 4.9 BM
[Co(C
2
O
4
)
3
]
3–
Co
3+
 + Strong field n = 0
(Low spin complex) ? = 0 BM
2–
4
MnO
Mn
+6
n = 1
(Low spin complex) ? = 1.73 BM
(n is number of unpaired electron and ? is spin
only magnetic moment)
8. Match List-I and List-II.
List-I List-II
(a) Valium (i) Antifertility drug
(b) Morphine (ii) Pernicious anaemia
(c) Norethindrone (iii) Analgesic
(d) Vitamin B
12
(iv) Tranquilizer
(1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(2) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
(3) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(4) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
Answer (1)
Page 3


JEE (MAIN)-2021 : Phase-1(24-02-2021)-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 one of the following carbonyl
compounds cannot be prepared by addition of
water on an alkyne in the presence of HgSO
4
and H
2
SO
4
?
(1) C
O
CH
3
(2)
O
CH – C – CH CH
323
(3)
O
CH – C – H
3
(4)
O
CH – CH – C – H
32
Answer (4)
Sol. CH
3
 – CH
2
 – CHO (Propanaldehyde) cannot be
prepared by addition of water on alkyne in the
presence of HgSO
4
 and H
2
SO
4
.
C
O
CH
3
C  CH ?
HO
2
Hg , H
2+ +
CH – CHO
3
HC  CH ?
H O
2
Hg , H
2+ +
O
CH – C – CH
33
CH – C  CH
3
?
HO
2
Hg , H
2+ +
2. Match List-I with List-II.
List-I List-II
(Metal) (Ores)
(a) Aluminium (i) Siderite
(b) Iron (ii) Calamine
(c) Copper (iii) Kaolinite
(d) Zinc (iv) Malachite
Choose the correct answer from the options
given below:
(1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Answer (3)
Sol. Ores Formula Metal present
Siderite FeCO
3
Fe
Calamine ZnCO
3
Zn
Kaolinite Al
2
Si
2
O
5
(OH)
4
Al
Malachite Cu(OH)
2
?CuCO
3
Cu
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
3. Which one of the following compounds is non-
aromatic?
(1)
O
(2)
(3)
(4)
Answer (2)
Sol.
O
Aromatic
Non-aromatic
sp hybridized
3
Aromatic
Aromatic
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
4. The correct shape and I – I – I bond angles
respectively in 
?
3
I ion are:
(1) Distorted trigonal planar; 135° and 90°
(2) Trigonal planar; 120°
(3) T-shaped; 180° and 90°
(4) Linear; 180°
Answer (4)
Sol.
?
3
I
I
I
I
Shape = Linear
Angle ?I – I – I is 180°
5. Given below are two statements : one is
labelled as Assertion A and the other is
labelled as Reason R.
Assertion A : Hydrogen is the most abundant
element in the Universe, but it is not the most
abundant gas in the troposphere.
Reason R : Hydrogen is the lightest element.
In the light of the above statements, choose
the correct answer from the options given
below.
(1) Both A and R are true but R is NOT the
correct explanation of A
(2) A is true but R is false
(3) Both A and R are true and R is the correct
explanation of A
(4) A is false but R is true
Answer (3)
Sol. Both A and R are true and R is the correct
explanation of A.
Hydrogen is the lightest element. Because it is
lighter than air, and so it can easily escape the
earth’s gravity.
6. Given below are two statements :
Statement I : The value of the parameter
“Biochemical Oxygen Demand (BOD)” is
important for survival of aquatic life.
Statement II : The optimum value of BOD is 6.5
ppm.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) Both statement I and statement II are true
(2) Both statement I and statement II are false
(3) Statement I is false but statement II is true
(4) Statement I is true but statement II is false
Answer (4)
Sol. The amount of BOD in the water is a measure
of the amount of organic material in the water,
in terms of how much oxygen will be required
to break it down biologically. Clean water
would have BOD value of less than 5 ppm,
whereas highly polluted water would have a
BOD value of 17 ppm or more.
Statement I is true and statement II is false.
7. The calculated magnetic moments (spin only
value) for species [FeCl
4
]
2–
, [Co(C
2
O
4
)
3
]
3–
 and
2–
4
MnO
 respectively are :
(1) 5.82, 0 and 0 BM
(2) 4.90, 0 and 1.73 BM
(3) 5.92, 4.90 and 0 BM
(4) 4.90, 0 and 2.83 BM
Answer (2)
Sol. [FeCl
4
]
2–
Fe
2+
 + Weak ligand n = 4
(High spin complex) ? = 4.9 BM
[Co(C
2
O
4
)
3
]
3–
Co
3+
 + Strong field n = 0
(Low spin complex) ? = 0 BM
2–
4
MnO
Mn
+6
n = 1
(Low spin complex) ? = 1.73 BM
(n is number of unpaired electron and ? is spin
only magnetic moment)
8. Match List-I and List-II.
List-I List-II
(a) Valium (i) Antifertility drug
(b) Morphine (ii) Pernicious anaemia
(c) Norethindrone (iii) Analgesic
(d) Vitamin B
12
(iv) Tranquilizer
(1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(2) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
(3) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(4) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
Answer (1)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
Sol. Valium Tranquilizer
Morphine Analgesic
Norethindrone Antifertility drug
Vitamin B
12
Pernicious anaemia
(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
9.
O
?
O
Which of the following reagent is suitable for
the preparation of the product in the above
reaction?
(1) Red P + Cl
2
(2) Ni/H
2
(3) NaBH
4
(4) NH –NH/CH ONa
2225
? –
Answer (4)
Sol.
O O
NH–NH
22
CHONa
25
–+
10. The correct set from the following in which
both pairs are in correct order of melting point
is
(1) LiCl > LiF; NaCl > MgO
(2) LiF > LiCl; MgO > NaCl
(3) LiCl > LiF; MgO > NaCl
(4) LiF > LiCl; NaCl > MgO
Answer (2)
Sol. Correct melting point order is
LiF > LiCl
22 1 1
MgONaCl
?? ? ?
?
m.p. ? q
1
 q
2
?
1
r
11. The incorrect statement among the following
is:
(1) VOSO
4
 is a reducing agent
(2) RuO
4
 is an oxidizing agent
(3) Red colour of ruby is due to the presence
of Co
3+
(4) Cr
2
O
3
 is an amphoteric oxide
Answer (3)
Sol. Red colour of the ruby is due to the presence
of Cr
3+
.
RuO
4
 is an oxidizing agent
VOSO
4
 ? VO
2+
 ? V
4+
 (it can oxidized) So it is
a reducing agent.
Cr
2
O
3
 is amphoteric oxide
12. What is the correct sequence of reagents
used for converting nitrobenzene into
m-dibromobenzene?
NO
2
Br
Br
(1)
NaNO 
2 HCl KBr H
+
(2)
Sn/HCl Br
2
NaNO
2 NaBr
(3)
Sn/HCl KBr Br
2 H
+
(4)
Br /Fe
2
Sn/HCl NaNO /HCl
2
CuBr/HBr
Answer (4)
Sol.
NO
2
Br /Fe
2
NO
2
Br
Sn/HCl
NH
2
Br
NaNO  HCl
2
NCl
2
Br
CuBr/HBr
Br
Br
– +
13. The diazonium salt of which of the following
compounds will form a coloured dye on
reaction with ?-Naphthol in NaOH?
(1)
N—CH
3
CH
3
(2)
CHNH
22
(3)
NH
2
(4)
NH—CH
  
3
Answer (3)
Page 4


JEE (MAIN)-2021 : Phase-1(24-02-2021)-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 one of the following carbonyl
compounds cannot be prepared by addition of
water on an alkyne in the presence of HgSO
4
and H
2
SO
4
?
(1) C
O
CH
3
(2)
O
CH – C – CH CH
323
(3)
O
CH – C – H
3
(4)
O
CH – CH – C – H
32
Answer (4)
Sol. CH
3
 – CH
2
 – CHO (Propanaldehyde) cannot be
prepared by addition of water on alkyne in the
presence of HgSO
4
 and H
2
SO
4
.
C
O
CH
3
C  CH ?
HO
2
Hg , H
2+ +
CH – CHO
3
HC  CH ?
H O
2
Hg , H
2+ +
O
CH – C – CH
33
CH – C  CH
3
?
HO
2
Hg , H
2+ +
2. Match List-I with List-II.
List-I List-II
(Metal) (Ores)
(a) Aluminium (i) Siderite
(b) Iron (ii) Calamine
(c) Copper (iii) Kaolinite
(d) Zinc (iv) Malachite
Choose the correct answer from the options
given below:
(1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Answer (3)
Sol. Ores Formula Metal present
Siderite FeCO
3
Fe
Calamine ZnCO
3
Zn
Kaolinite Al
2
Si
2
O
5
(OH)
4
Al
Malachite Cu(OH)
2
?CuCO
3
Cu
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
3. Which one of the following compounds is non-
aromatic?
(1)
O
(2)
(3)
(4)
Answer (2)
Sol.
O
Aromatic
Non-aromatic
sp hybridized
3
Aromatic
Aromatic
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
4. The correct shape and I – I – I bond angles
respectively in 
?
3
I ion are:
(1) Distorted trigonal planar; 135° and 90°
(2) Trigonal planar; 120°
(3) T-shaped; 180° and 90°
(4) Linear; 180°
Answer (4)
Sol.
?
3
I
I
I
I
Shape = Linear
Angle ?I – I – I is 180°
5. Given below are two statements : one is
labelled as Assertion A and the other is
labelled as Reason R.
Assertion A : Hydrogen is the most abundant
element in the Universe, but it is not the most
abundant gas in the troposphere.
Reason R : Hydrogen is the lightest element.
In the light of the above statements, choose
the correct answer from the options given
below.
(1) Both A and R are true but R is NOT the
correct explanation of A
(2) A is true but R is false
(3) Both A and R are true and R is the correct
explanation of A
(4) A is false but R is true
Answer (3)
Sol. Both A and R are true and R is the correct
explanation of A.
Hydrogen is the lightest element. Because it is
lighter than air, and so it can easily escape the
earth’s gravity.
6. Given below are two statements :
Statement I : The value of the parameter
“Biochemical Oxygen Demand (BOD)” is
important for survival of aquatic life.
Statement II : The optimum value of BOD is 6.5
ppm.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) Both statement I and statement II are true
(2) Both statement I and statement II are false
(3) Statement I is false but statement II is true
(4) Statement I is true but statement II is false
Answer (4)
Sol. The amount of BOD in the water is a measure
of the amount of organic material in the water,
in terms of how much oxygen will be required
to break it down biologically. Clean water
would have BOD value of less than 5 ppm,
whereas highly polluted water would have a
BOD value of 17 ppm or more.
Statement I is true and statement II is false.
7. The calculated magnetic moments (spin only
value) for species [FeCl
4
]
2–
, [Co(C
2
O
4
)
3
]
3–
 and
2–
4
MnO
 respectively are :
(1) 5.82, 0 and 0 BM
(2) 4.90, 0 and 1.73 BM
(3) 5.92, 4.90 and 0 BM
(4) 4.90, 0 and 2.83 BM
Answer (2)
Sol. [FeCl
4
]
2–
Fe
2+
 + Weak ligand n = 4
(High spin complex) ? = 4.9 BM
[Co(C
2
O
4
)
3
]
3–
Co
3+
 + Strong field n = 0
(Low spin complex) ? = 0 BM
2–
4
MnO
Mn
+6
n = 1
(Low spin complex) ? = 1.73 BM
(n is number of unpaired electron and ? is spin
only magnetic moment)
8. Match List-I and List-II.
List-I List-II
(a) Valium (i) Antifertility drug
(b) Morphine (ii) Pernicious anaemia
(c) Norethindrone (iii) Analgesic
(d) Vitamin B
12
(iv) Tranquilizer
(1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(2) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
(3) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(4) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
Answer (1)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
Sol. Valium Tranquilizer
Morphine Analgesic
Norethindrone Antifertility drug
Vitamin B
12
Pernicious anaemia
(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
9.
O
?
O
Which of the following reagent is suitable for
the preparation of the product in the above
reaction?
(1) Red P + Cl
2
(2) Ni/H
2
(3) NaBH
4
(4) NH –NH/CH ONa
2225
? –
Answer (4)
Sol.
O O
NH–NH
22
CHONa
25
–+
10. The correct set from the following in which
both pairs are in correct order of melting point
is
(1) LiCl > LiF; NaCl > MgO
(2) LiF > LiCl; MgO > NaCl
(3) LiCl > LiF; MgO > NaCl
(4) LiF > LiCl; NaCl > MgO
Answer (2)
Sol. Correct melting point order is
LiF > LiCl
22 1 1
MgONaCl
?? ? ?
?
m.p. ? q
1
 q
2
?
1
r
11. The incorrect statement among the following
is:
(1) VOSO
4
 is a reducing agent
(2) RuO
4
 is an oxidizing agent
(3) Red colour of ruby is due to the presence
of Co
3+
(4) Cr
2
O
3
 is an amphoteric oxide
Answer (3)
Sol. Red colour of the ruby is due to the presence
of Cr
3+
.
RuO
4
 is an oxidizing agent
VOSO
4
 ? VO
2+
 ? V
4+
 (it can oxidized) So it is
a reducing agent.
Cr
2
O
3
 is amphoteric oxide
12. What is the correct sequence of reagents
used for converting nitrobenzene into
m-dibromobenzene?
NO
2
Br
Br
(1)
NaNO 
2 HCl KBr H
+
(2)
Sn/HCl Br
2
NaNO
2 NaBr
(3)
Sn/HCl KBr Br
2 H
+
(4)
Br /Fe
2
Sn/HCl NaNO /HCl
2
CuBr/HBr
Answer (4)
Sol.
NO
2
Br /Fe
2
NO
2
Br
Sn/HCl
NH
2
Br
NaNO  HCl
2
NCl
2
Br
CuBr/HBr
Br
Br
– +
13. The diazonium salt of which of the following
compounds will form a coloured dye on
reaction with ?-Naphthol in NaOH?
(1)
N—CH
3
CH
3
(2)
CHNH
22
(3)
NH
2
(4)
NH—CH
  
3
Answer (3)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
Sol.
NH
2
NaNO/HCl
2
NCl
2
+ –
OH
N     N
OH
+ HCl
(Orange colour dye)
14. The correct order of the following compounds
showing increasing tendency towards
nucleophilic substitution reaction is :
Cl
(i) 
Cl
NO
2
(ii) 
Cl
NO
2
NO
2
(iii) 
Cl
NO
2
NO
2
ON
2
(iv)
(1) (iv) < (i) < (ii) < (iii)
(2) (iv) < (i) < (iii) < (ii)
(3) (iv) < (iii) < (ii) < (i)
(4) (i) < (ii) < (iii) < (iv)
Answer (4)
Sol. More the number of EWG attached to benzene
ring, more will be the tendency towards
nucleophilic substitution reaction
Cl Cl
NO
2
Cl
NO
2
NO
2
Cl
NO
2
NO
2
ON
2
>> >
iv > iii > ii > i
15. In polymer Buna-S : ‘S’ stands for:
(1) Strength
(2) Sulphonation
(3) Styrene
(4) Sulphur
Answer (3)
Sol. In polymer Buna-S : S stands for styrene
Buna-S is a polymer of buta 1,3 diene and
styrene.
16. According to Bohr’s atomic theory:
(A) Kinetic energy of electron is ?
2
2
Z
n
(B) The product of velocity (v) of electron and
principal quantum number (n), ‘vn’ ? Z
2
(C) Frequency of revolution of electron in an
orbit is ?
3
3
Z
n
(D) Coulombic force of attraction on the
electron is ?
3
4
Z
n
Choose the most appropriate answer from the
options given below:
(1) (A) only (2) (A) and (D) only
(3) (C) only (4) (A), (C) and (D) only
Answer (2)
Sol. In Bohr’s atomic theory:
2
2
Z
K.E.
n
?
Velocity (v) ? 
Z
n
? V .n ? Z
Frequency of revolution 
V
2r
?
?
Frequency of revolution 
V
r
?
  
2
ZZ
nn
?
?
?
  
2
3
Z
n
?
Force of attraction 
2
mV
r
?
Force of attraction 
2
V
r
?
2
22
Zz
nn
?
?
?
3
4
Z
n
?
Correct statements : (A) and (D).
Page 5


JEE (MAIN)-2021 : Phase-1(24-02-2021)-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 one of the following carbonyl
compounds cannot be prepared by addition of
water on an alkyne in the presence of HgSO
4
and H
2
SO
4
?
(1) C
O
CH
3
(2)
O
CH – C – CH CH
323
(3)
O
CH – C – H
3
(4)
O
CH – CH – C – H
32
Answer (4)
Sol. CH
3
 – CH
2
 – CHO (Propanaldehyde) cannot be
prepared by addition of water on alkyne in the
presence of HgSO
4
 and H
2
SO
4
.
C
O
CH
3
C  CH ?
HO
2
Hg , H
2+ +
CH – CHO
3
HC  CH ?
H O
2
Hg , H
2+ +
O
CH – C – CH
33
CH – C  CH
3
?
HO
2
Hg , H
2+ +
2. Match List-I with List-II.
List-I List-II
(Metal) (Ores)
(a) Aluminium (i) Siderite
(b) Iron (ii) Calamine
(c) Copper (iii) Kaolinite
(d) Zinc (iv) Malachite
Choose the correct answer from the options
given below:
(1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Answer (3)
Sol. Ores Formula Metal present
Siderite FeCO
3
Fe
Calamine ZnCO
3
Zn
Kaolinite Al
2
Si
2
O
5
(OH)
4
Al
Malachite Cu(OH)
2
?CuCO
3
Cu
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
3. Which one of the following compounds is non-
aromatic?
(1)
O
(2)
(3)
(4)
Answer (2)
Sol.
O
Aromatic
Non-aromatic
sp hybridized
3
Aromatic
Aromatic
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
4. The correct shape and I – I – I bond angles
respectively in 
?
3
I ion are:
(1) Distorted trigonal planar; 135° and 90°
(2) Trigonal planar; 120°
(3) T-shaped; 180° and 90°
(4) Linear; 180°
Answer (4)
Sol.
?
3
I
I
I
I
Shape = Linear
Angle ?I – I – I is 180°
5. Given below are two statements : one is
labelled as Assertion A and the other is
labelled as Reason R.
Assertion A : Hydrogen is the most abundant
element in the Universe, but it is not the most
abundant gas in the troposphere.
Reason R : Hydrogen is the lightest element.
In the light of the above statements, choose
the correct answer from the options given
below.
(1) Both A and R are true but R is NOT the
correct explanation of A
(2) A is true but R is false
(3) Both A and R are true and R is the correct
explanation of A
(4) A is false but R is true
Answer (3)
Sol. Both A and R are true and R is the correct
explanation of A.
Hydrogen is the lightest element. Because it is
lighter than air, and so it can easily escape the
earth’s gravity.
6. Given below are two statements :
Statement I : The value of the parameter
“Biochemical Oxygen Demand (BOD)” is
important for survival of aquatic life.
Statement II : The optimum value of BOD is 6.5
ppm.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) Both statement I and statement II are true
(2) Both statement I and statement II are false
(3) Statement I is false but statement II is true
(4) Statement I is true but statement II is false
Answer (4)
Sol. The amount of BOD in the water is a measure
of the amount of organic material in the water,
in terms of how much oxygen will be required
to break it down biologically. Clean water
would have BOD value of less than 5 ppm,
whereas highly polluted water would have a
BOD value of 17 ppm or more.
Statement I is true and statement II is false.
7. The calculated magnetic moments (spin only
value) for species [FeCl
4
]
2–
, [Co(C
2
O
4
)
3
]
3–
 and
2–
4
MnO
 respectively are :
(1) 5.82, 0 and 0 BM
(2) 4.90, 0 and 1.73 BM
(3) 5.92, 4.90 and 0 BM
(4) 4.90, 0 and 2.83 BM
Answer (2)
Sol. [FeCl
4
]
2–
Fe
2+
 + Weak ligand n = 4
(High spin complex) ? = 4.9 BM
[Co(C
2
O
4
)
3
]
3–
Co
3+
 + Strong field n = 0
(Low spin complex) ? = 0 BM
2–
4
MnO
Mn
+6
n = 1
(Low spin complex) ? = 1.73 BM
(n is number of unpaired electron and ? is spin
only magnetic moment)
8. Match List-I and List-II.
List-I List-II
(a) Valium (i) Antifertility drug
(b) Morphine (ii) Pernicious anaemia
(c) Norethindrone (iii) Analgesic
(d) Vitamin B
12
(iv) Tranquilizer
(1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(2) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
(3) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(4) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
Answer (1)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
Sol. Valium Tranquilizer
Morphine Analgesic
Norethindrone Antifertility drug
Vitamin B
12
Pernicious anaemia
(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
9.
O
?
O
Which of the following reagent is suitable for
the preparation of the product in the above
reaction?
(1) Red P + Cl
2
(2) Ni/H
2
(3) NaBH
4
(4) NH –NH/CH ONa
2225
? –
Answer (4)
Sol.
O O
NH–NH
22
CHONa
25
–+
10. The correct set from the following in which
both pairs are in correct order of melting point
is
(1) LiCl > LiF; NaCl > MgO
(2) LiF > LiCl; MgO > NaCl
(3) LiCl > LiF; MgO > NaCl
(4) LiF > LiCl; NaCl > MgO
Answer (2)
Sol. Correct melting point order is
LiF > LiCl
22 1 1
MgONaCl
?? ? ?
?
m.p. ? q
1
 q
2
?
1
r
11. The incorrect statement among the following
is:
(1) VOSO
4
 is a reducing agent
(2) RuO
4
 is an oxidizing agent
(3) Red colour of ruby is due to the presence
of Co
3+
(4) Cr
2
O
3
 is an amphoteric oxide
Answer (3)
Sol. Red colour of the ruby is due to the presence
of Cr
3+
.
RuO
4
 is an oxidizing agent
VOSO
4
 ? VO
2+
 ? V
4+
 (it can oxidized) So it is
a reducing agent.
Cr
2
O
3
 is amphoteric oxide
12. What is the correct sequence of reagents
used for converting nitrobenzene into
m-dibromobenzene?
NO
2
Br
Br
(1)
NaNO 
2 HCl KBr H
+
(2)
Sn/HCl Br
2
NaNO
2 NaBr
(3)
Sn/HCl KBr Br
2 H
+
(4)
Br /Fe
2
Sn/HCl NaNO /HCl
2
CuBr/HBr
Answer (4)
Sol.
NO
2
Br /Fe
2
NO
2
Br
Sn/HCl
NH
2
Br
NaNO  HCl
2
NCl
2
Br
CuBr/HBr
Br
Br
– +
13. The diazonium salt of which of the following
compounds will form a coloured dye on
reaction with ?-Naphthol in NaOH?
(1)
N—CH
3
CH
3
(2)
CHNH
22
(3)
NH
2
(4)
NH—CH
  
3
Answer (3)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
Sol.
NH
2
NaNO/HCl
2
NCl
2
+ –
OH
N     N
OH
+ HCl
(Orange colour dye)
14. The correct order of the following compounds
showing increasing tendency towards
nucleophilic substitution reaction is :
Cl
(i) 
Cl
NO
2
(ii) 
Cl
NO
2
NO
2
(iii) 
Cl
NO
2
NO
2
ON
2
(iv)
(1) (iv) < (i) < (ii) < (iii)
(2) (iv) < (i) < (iii) < (ii)
(3) (iv) < (iii) < (ii) < (i)
(4) (i) < (ii) < (iii) < (iv)
Answer (4)
Sol. More the number of EWG attached to benzene
ring, more will be the tendency towards
nucleophilic substitution reaction
Cl Cl
NO
2
Cl
NO
2
NO
2
Cl
NO
2
NO
2
ON
2
>> >
iv > iii > ii > i
15. In polymer Buna-S : ‘S’ stands for:
(1) Strength
(2) Sulphonation
(3) Styrene
(4) Sulphur
Answer (3)
Sol. In polymer Buna-S : S stands for styrene
Buna-S is a polymer of buta 1,3 diene and
styrene.
16. According to Bohr’s atomic theory:
(A) Kinetic energy of electron is ?
2
2
Z
n
(B) The product of velocity (v) of electron and
principal quantum number (n), ‘vn’ ? Z
2
(C) Frequency of revolution of electron in an
orbit is ?
3
3
Z
n
(D) Coulombic force of attraction on the
electron is ?
3
4
Z
n
Choose the most appropriate answer from the
options given below:
(1) (A) only (2) (A) and (D) only
(3) (C) only (4) (A), (C) and (D) only
Answer (2)
Sol. In Bohr’s atomic theory:
2
2
Z
K.E.
n
?
Velocity (v) ? 
Z
n
? V .n ? Z
Frequency of revolution 
V
2r
?
?
Frequency of revolution 
V
r
?
  
2
ZZ
nn
?
?
?
  
2
3
Z
n
?
Force of attraction 
2
mV
r
?
Force of attraction 
2
V
r
?
2
22
Zz
nn
?
?
?
3
4
Z
n
?
Correct statements : (A) and (D).
JEE (MAIN)-2021 : Phase-1(24-02-2021)-E
17. Most suitable salt which can be used for
efficient clotting of blood will be
(1) FeCl
3
(2) FeSO
4
(3) NAHCO
3
(4) Mg(HCO
3
)
2
Answer (1)
Sol. More the positive charge, more effective is the
coagulation.
Hardy-Schuldze rule - “Higher the valency or
charge greater its coagulating power”.
18. Match List - I with List - II.
List - I List - II
(Salt) (Flame colour
wavelength)
(a) LiCl (i) 455.5 nm
(b) NaCl (ii) 670.8 nm
(c) RbCl (iii) 780.0 nm
(d) CsCl (iv) 589.2 nm
Choose the correct answer from the options
given below:
(1) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(2) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(3) (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
(4) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
Answer (2)
Sol. Salt Flame colour wavelength
(a) LiCl 670.8 nm (Crimson red)
(b) NaCl 589.2 nm (Yellow)
(c) RbCl 780.0 nm (Red violet)
(d) CsCl 455.5 (Blue)
19. Match List - I and List - II.
List - I List - II
(a) ?? ? ?
O
||
RC Cl R CHO (i) Br
2
/NaOH
(b) R – CH
2
 – COOH ? (ii) H
2
/Pd – BaSO
4
?? R CH COOH
|
Cl
(c) R – C – NH R – NH
22
?
O
(iii) Zn(Hg)/Conc. HCl
(d) R – C – CH R – CH – CH
32 3
?
O
(iv) Cl
2
/Red P, H
2
O
Choose the correct answer from the options
given below:
(1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(2) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(3) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(4) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Answer (2)
Sol. (a)
2
4
H/Pd
BaSO
O
||
R C Cl R CHO(Rosenmund's
reduction)
?? ???? ??
(b)
22
Cl /RedP,H O
2
R CH COOH R CH COOH
|
Cl
(Hell-volhard-zelinsky
reaction)
? ?????? ? ? ?
(c)
2
Br /NaOH
22
O
||
RC NH R NH (Hoffmann
bromamide)
? ? ????? ??
(d)
Zn(Hg)
323
ConHCl
O
||
RC CH R CHCH
(Clemmensen reduction)
?? ??????
20. What is the correct order of the following
elements with respect to their density?
(1) Cr < Zn < Co < Cu < Fe
(2) Cr < Fe < Co < Cu < Zn
(3) Zn < Cu < Co < Fe < Cr
(4) Zn < Cr < Fe < Co < Cu
Answer (4)
Sol.
3
Cu Co Fe Cr Zn densitying/cm
8.9 8.7 7.8 7.19 7.1
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. 1.86 g of aniline completely reacts to form
acetanilide. 10% of the product is lost during
purification. Amount of acetanilide obtained
after purification (in g) is ______ × 10
–2
.
Answer (243)