JEE Main 2021 Chemistry February 24 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
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 of the following are isostructural pairs?
A.
?? 22
44
SO and CrO
B.
44
SiCl and TiCl
C.
?
33
NH and NO
D.
33
BCl and BrCl
(1) A and C only (2) B and C only
(3) A and B only (4) C and D only
Answer (3)
Sol.
O
S
O
O
–
O
–
Tetrahedral
Cl
Si
Cl
Cl Cl
Tetrahedral
O
Cr
O
O
–
O
–
Tetrahedral
Cl
Ti
Cl
Cl Cl
Tetrahedral
Option A
Option B
Isostructural means same structure
So option-3 is the correct answer
2. Out of the following, which type of interaction
is responsible for the stabilisation of ?-helix
structure of proteins?
(1) Covalent bonding
(2) Hydrogen bonding
(3) Ionic bonding
(4) vander Waals forces
Answer (2)
Sol. “ ?-Helix is one of the most common ways in
which a polypeptide chain forms all possible
hydrogen bonds by twisting into a right handed
screw (helix) with the –NH group of each amino
acid residue hydrogen bonded to the C = O  of
an adjacent turn of the helix”
3. The major components in “Gun Metal” are:
(1) Cu, Ni and Fe (2) Cu, Sn and Zn
(3) Al, Cu, Mg and Mn (4) Cu, Zn and Ni
Answer (2)
Sol. Gun metal has composition of Cu, Zn, Sn
Cu – 87%
Zn – 3 %
Sn – 10%
4. In Freundlich adsorption isotherm, slope of AB
line is:
A
B
log P
log
x
m
(1) n with (n, 0.1 to 0.5)
(2)
??
?
??
??
11
 with 0 to 1
nn
(3) log n with (n > 1)
(4) log 
1
n
 with (n < 1)
Answer (2)
Sol. According to Freundlich adsorption isotherm
??
1/n
x
k.P (n 1)
m
??
x1
log logk logP
mn
so in the plot of 
x
log
m
 vs log P the slope is 
1
n
,
where 
1
n
varies from 0 to 1
5. Which of the following ore is concentrated
using group 1 cyanide salt?
(1) Sphalerite (2) Malachite
(3) Siderite (4) Calamine
Answer (1)
Page 2


JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
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 of the following are isostructural pairs?
A.
?? 22
44
SO and CrO
B.
44
SiCl and TiCl
C.
?
33
NH and NO
D.
33
BCl and BrCl
(1) A and C only (2) B and C only
(3) A and B only (4) C and D only
Answer (3)
Sol.
O
S
O
O
–
O
–
Tetrahedral
Cl
Si
Cl
Cl Cl
Tetrahedral
O
Cr
O
O
–
O
–
Tetrahedral
Cl
Ti
Cl
Cl Cl
Tetrahedral
Option A
Option B
Isostructural means same structure
So option-3 is the correct answer
2. Out of the following, which type of interaction
is responsible for the stabilisation of ?-helix
structure of proteins?
(1) Covalent bonding
(2) Hydrogen bonding
(3) Ionic bonding
(4) vander Waals forces
Answer (2)
Sol. “ ?-Helix is one of the most common ways in
which a polypeptide chain forms all possible
hydrogen bonds by twisting into a right handed
screw (helix) with the –NH group of each amino
acid residue hydrogen bonded to the C = O  of
an adjacent turn of the helix”
3. The major components in “Gun Metal” are:
(1) Cu, Ni and Fe (2) Cu, Sn and Zn
(3) Al, Cu, Mg and Mn (4) Cu, Zn and Ni
Answer (2)
Sol. Gun metal has composition of Cu, Zn, Sn
Cu – 87%
Zn – 3 %
Sn – 10%
4. In Freundlich adsorption isotherm, slope of AB
line is:
A
B
log P
log
x
m
(1) n with (n, 0.1 to 0.5)
(2)
??
?
??
??
11
 with 0 to 1
nn
(3) log n with (n > 1)
(4) log 
1
n
 with (n < 1)
Answer (2)
Sol. According to Freundlich adsorption isotherm
??
1/n
x
k.P (n 1)
m
??
x1
log logk logP
mn
so in the plot of 
x
log
m
 vs log P the slope is 
1
n
,
where 
1
n
varies from 0 to 1
5. Which of the following ore is concentrated
using group 1 cyanide salt?
(1) Sphalerite (2) Malachite
(3) Siderite (4) Calamine
Answer (1)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol. Ore Formula
Sphalerite ZnS
Siderite FeCO
3
Malachite Cu(OH)
2
?CuCO
3
Calamine ZnCO
3
ZnS + 4 NaCN ? [Zn(CN)
4
]
2–
 + 4 Na
+ 
+ S
2–
the reagent NaCN/KCN is used to suppress the
floating characteristics of ZnS by forming a
soluble complex with KCN.
6. ‘A’ and ‘B’ in the following reactions are :
NH
2
SnCl /HCl/H O
23
+
NaNO/HCl
2
KCN
(A) () B
(1)
N
2
CI
+ –
(A) :
CI
(B) :
(2)
CN
(A) :
CHO
(B) :
(3)
N
2
CI
+ –
(A) :
CHO
(B) :
(4)
CI
(B) :
CN
(A) :
Answer (2)
Sol.
NH
2
CN
CuCN/KCN
HNO
2
(HCl + NANO
2
)
SnCl /HCl
2
H
3
O
+
(Stephen 
reaction)
A
N
2
Cl
?
CHO
B
7. Consider the elements Mg, AI, S, P and Si, the
correct increasing order of their first ionization
enthalpy is :
(1) Mg < Al < Si < P < S
(2) Mg < Al < Si < S < P
(3) Al < Mg < S < Si < P
(4) Al < Mg < Si < S < P
Answer (4)
Sol. Across the period, generally ionization enthalpy
increases but half filled and fully filled
configuration are stable and may change the
regular trend.
P has more IE
1
 than S because of half filled.
Al has lower IE
1
 than Mg because of effective
shielding of 3P electrons from the nucleus by
3s-electrons.
Finally order should be
P > S > Si > Mg > Al
8. The electrode potential of M
2+
 / M of 3d-series
elements shows positive value for :
(1) Cu (2) Zn
(3) Co (4) Fe
Answer (1)
Sol. Only Cu
2+
/Cu has positive SRP among 3d-series
metals.
9. (A) HOCl + H
2
O
2
 ? H
3
O
+
 + CI
–
 + O
2
(B) I
2
 + H
2
O
2
 + 2OH
–
 ? 2I
–
 + 2H
2
O + O
2
Choose the correct option.
(1) H
2
O
2
 acts as oxidising agent in equations
(A) and (B).
(2) H
2
O
2
 act as oxidizing and reducing agent
respectively in equations (A) and (B).
(3) H
2
O
2
 acts as reducing agent in equations
(A) and (B).
(4) H
2
O
2
 acts as reducing and oxidising agent
respectively in equations (A) and (B).
Answer (3)
Page 3


JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
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 of the following are isostructural pairs?
A.
?? 22
44
SO and CrO
B.
44
SiCl and TiCl
C.
?
33
NH and NO
D.
33
BCl and BrCl
(1) A and C only (2) B and C only
(3) A and B only (4) C and D only
Answer (3)
Sol.
O
S
O
O
–
O
–
Tetrahedral
Cl
Si
Cl
Cl Cl
Tetrahedral
O
Cr
O
O
–
O
–
Tetrahedral
Cl
Ti
Cl
Cl Cl
Tetrahedral
Option A
Option B
Isostructural means same structure
So option-3 is the correct answer
2. Out of the following, which type of interaction
is responsible for the stabilisation of ?-helix
structure of proteins?
(1) Covalent bonding
(2) Hydrogen bonding
(3) Ionic bonding
(4) vander Waals forces
Answer (2)
Sol. “ ?-Helix is one of the most common ways in
which a polypeptide chain forms all possible
hydrogen bonds by twisting into a right handed
screw (helix) with the –NH group of each amino
acid residue hydrogen bonded to the C = O  of
an adjacent turn of the helix”
3. The major components in “Gun Metal” are:
(1) Cu, Ni and Fe (2) Cu, Sn and Zn
(3) Al, Cu, Mg and Mn (4) Cu, Zn and Ni
Answer (2)
Sol. Gun metal has composition of Cu, Zn, Sn
Cu – 87%
Zn – 3 %
Sn – 10%
4. In Freundlich adsorption isotherm, slope of AB
line is:
A
B
log P
log
x
m
(1) n with (n, 0.1 to 0.5)
(2)
??
?
??
??
11
 with 0 to 1
nn
(3) log n with (n > 1)
(4) log 
1
n
 with (n < 1)
Answer (2)
Sol. According to Freundlich adsorption isotherm
??
1/n
x
k.P (n 1)
m
??
x1
log logk logP
mn
so in the plot of 
x
log
m
 vs log P the slope is 
1
n
,
where 
1
n
varies from 0 to 1
5. Which of the following ore is concentrated
using group 1 cyanide salt?
(1) Sphalerite (2) Malachite
(3) Siderite (4) Calamine
Answer (1)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol. Ore Formula
Sphalerite ZnS
Siderite FeCO
3
Malachite Cu(OH)
2
?CuCO
3
Calamine ZnCO
3
ZnS + 4 NaCN ? [Zn(CN)
4
]
2–
 + 4 Na
+ 
+ S
2–
the reagent NaCN/KCN is used to suppress the
floating characteristics of ZnS by forming a
soluble complex with KCN.
6. ‘A’ and ‘B’ in the following reactions are :
NH
2
SnCl /HCl/H O
23
+
NaNO/HCl
2
KCN
(A) () B
(1)
N
2
CI
+ –
(A) :
CI
(B) :
(2)
CN
(A) :
CHO
(B) :
(3)
N
2
CI
+ –
(A) :
CHO
(B) :
(4)
CI
(B) :
CN
(A) :
Answer (2)
Sol.
NH
2
CN
CuCN/KCN
HNO
2
(HCl + NANO
2
)
SnCl /HCl
2
H
3
O
+
(Stephen 
reaction)
A
N
2
Cl
?
CHO
B
7. Consider the elements Mg, AI, S, P and Si, the
correct increasing order of their first ionization
enthalpy is :
(1) Mg < Al < Si < P < S
(2) Mg < Al < Si < S < P
(3) Al < Mg < S < Si < P
(4) Al < Mg < Si < S < P
Answer (4)
Sol. Across the period, generally ionization enthalpy
increases but half filled and fully filled
configuration are stable and may change the
regular trend.
P has more IE
1
 than S because of half filled.
Al has lower IE
1
 than Mg because of effective
shielding of 3P electrons from the nucleus by
3s-electrons.
Finally order should be
P > S > Si > Mg > Al
8. The electrode potential of M
2+
 / M of 3d-series
elements shows positive value for :
(1) Cu (2) Zn
(3) Co (4) Fe
Answer (1)
Sol. Only Cu
2+
/Cu has positive SRP among 3d-series
metals.
9. (A) HOCl + H
2
O
2
 ? H
3
O
+
 + CI
–
 + O
2
(B) I
2
 + H
2
O
2
 + 2OH
–
 ? 2I
–
 + 2H
2
O + O
2
Choose the correct option.
(1) H
2
O
2
 acts as oxidising agent in equations
(A) and (B).
(2) H
2
O
2
 act as oxidizing and reducing agent
respectively in equations (A) and (B).
(3) H
2
O
2
 acts as reducing agent in equations
(A) and (B).
(4) H
2
O
2
 acts as reducing and oxidising agent
respectively in equations (A) and (B).
Answer (3)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol. HOCl + H O
2 2
H O + CI + O
32
+–
oxidation
–1
IOH
22
 + H O  + 2
2
–
2I + 2H O + 
–
2
O
2
oxidation
–1
°
°
So, H
2
O
2
 is acting as reducing agent in both
reactions.
10. In the following reaction the reason why meta-
nitro product also formed is :
NH
2
[A]
51%
[B]
47%
[C]
2%
Conc. HNO
3
++
Conc. H SO
24
, 288 K
NH
2
NO
2
NH
2
NO
2
NH
2
NO
2
(1) Formation of anilinium ion
(2) low temperature
(3) –NO
2
 substitution always takes place at
meta-position
(4) –NH
2
 group is highly meta-directive
Answer (1)
Sol. Aniline itself is strong ortho/para director but
on addition of acid it becomes anilinium ion
which is a meta director.
H
+
NH
3
?
NH
2
So the answer should be 1.
11. Al
2
O
3
 was leached with alkali to get X. The
solution of X on passing of gas Y, forms Z. X, Y
and Z respectively are:
(1) X = Na[Al(OH)
4
], Y = SO
2
, Z = Al
2
O
3
(2) X = Al(OH)
3
, Y = SO
2
, Z = Al
2
O
3
.xH
2
O
(3) X = Al(OH)
3
, Y = CO
2
, Z = Al
2
O
3
(4) X = Na[Al(OH)
4
], Y = CO
2
, Z = Al
2
O
3
.xH
2
O
Answer (4)
Sol. ??
23 4
(aq)
(s) (aq)
Al O NaOH Na[Al(OH) ]
?? ? ?
42 23 2 3
(g) (aq) (s) (aq)
Na[Al(OH) ] CO Al O xH O NaHCO
12. Given below are two statements:
Statement-I : Colourless cupric metaborate is
reduced to cuprous metaborate in a luminous
flame.
Statement-II : Cuprous metaborate is obtained
by heating boric anhydride and copper
sulphate in a non-luminous flame.
In the light of the above statements, choose the
most appropriate answer from the options
given below.
(1) Statement I is false but statement II is true
(2) Both statement I and statement II are true
(3) Both statement I and statement II are false
(4) Statement I is true but statement II is false
Answer (3)
Sol. Statement-I : Cupric metaborate is blue in
colour
Hence statement-I is false
Statement-II : CuSO
4
 + B
2
O
3
 ? Cu(BO
2
)
2
cupric metaborate is obtained instead of
cuprous metaborate.
Hence statement-II is false
13. Which of the following compound gives pink
colour on reaction with phthalic anhydride in
conc. H
2
SO
4
 followed by treatment with NaOH?
(1)
CH
3
HO
(2)
CH
3
HO
HO
CH
3
(3)
CH
3
HO
H C
3
OH
(4)
CH
3
OH
Answer (4)
Page 4


JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
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 of the following are isostructural pairs?
A.
?? 22
44
SO and CrO
B.
44
SiCl and TiCl
C.
?
33
NH and NO
D.
33
BCl and BrCl
(1) A and C only (2) B and C only
(3) A and B only (4) C and D only
Answer (3)
Sol.
O
S
O
O
–
O
–
Tetrahedral
Cl
Si
Cl
Cl Cl
Tetrahedral
O
Cr
O
O
–
O
–
Tetrahedral
Cl
Ti
Cl
Cl Cl
Tetrahedral
Option A
Option B
Isostructural means same structure
So option-3 is the correct answer
2. Out of the following, which type of interaction
is responsible for the stabilisation of ?-helix
structure of proteins?
(1) Covalent bonding
(2) Hydrogen bonding
(3) Ionic bonding
(4) vander Waals forces
Answer (2)
Sol. “ ?-Helix is one of the most common ways in
which a polypeptide chain forms all possible
hydrogen bonds by twisting into a right handed
screw (helix) with the –NH group of each amino
acid residue hydrogen bonded to the C = O  of
an adjacent turn of the helix”
3. The major components in “Gun Metal” are:
(1) Cu, Ni and Fe (2) Cu, Sn and Zn
(3) Al, Cu, Mg and Mn (4) Cu, Zn and Ni
Answer (2)
Sol. Gun metal has composition of Cu, Zn, Sn
Cu – 87%
Zn – 3 %
Sn – 10%
4. In Freundlich adsorption isotherm, slope of AB
line is:
A
B
log P
log
x
m
(1) n with (n, 0.1 to 0.5)
(2)
??
?
??
??
11
 with 0 to 1
nn
(3) log n with (n > 1)
(4) log 
1
n
 with (n < 1)
Answer (2)
Sol. According to Freundlich adsorption isotherm
??
1/n
x
k.P (n 1)
m
??
x1
log logk logP
mn
so in the plot of 
x
log
m
 vs log P the slope is 
1
n
,
where 
1
n
varies from 0 to 1
5. Which of the following ore is concentrated
using group 1 cyanide salt?
(1) Sphalerite (2) Malachite
(3) Siderite (4) Calamine
Answer (1)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol. Ore Formula
Sphalerite ZnS
Siderite FeCO
3
Malachite Cu(OH)
2
?CuCO
3
Calamine ZnCO
3
ZnS + 4 NaCN ? [Zn(CN)
4
]
2–
 + 4 Na
+ 
+ S
2–
the reagent NaCN/KCN is used to suppress the
floating characteristics of ZnS by forming a
soluble complex with KCN.
6. ‘A’ and ‘B’ in the following reactions are :
NH
2
SnCl /HCl/H O
23
+
NaNO/HCl
2
KCN
(A) () B
(1)
N
2
CI
+ –
(A) :
CI
(B) :
(2)
CN
(A) :
CHO
(B) :
(3)
N
2
CI
+ –
(A) :
CHO
(B) :
(4)
CI
(B) :
CN
(A) :
Answer (2)
Sol.
NH
2
CN
CuCN/KCN
HNO
2
(HCl + NANO
2
)
SnCl /HCl
2
H
3
O
+
(Stephen 
reaction)
A
N
2
Cl
?
CHO
B
7. Consider the elements Mg, AI, S, P and Si, the
correct increasing order of their first ionization
enthalpy is :
(1) Mg < Al < Si < P < S
(2) Mg < Al < Si < S < P
(3) Al < Mg < S < Si < P
(4) Al < Mg < Si < S < P
Answer (4)
Sol. Across the period, generally ionization enthalpy
increases but half filled and fully filled
configuration are stable and may change the
regular trend.
P has more IE
1
 than S because of half filled.
Al has lower IE
1
 than Mg because of effective
shielding of 3P electrons from the nucleus by
3s-electrons.
Finally order should be
P > S > Si > Mg > Al
8. The electrode potential of M
2+
 / M of 3d-series
elements shows positive value for :
(1) Cu (2) Zn
(3) Co (4) Fe
Answer (1)
Sol. Only Cu
2+
/Cu has positive SRP among 3d-series
metals.
9. (A) HOCl + H
2
O
2
 ? H
3
O
+
 + CI
–
 + O
2
(B) I
2
 + H
2
O
2
 + 2OH
–
 ? 2I
–
 + 2H
2
O + O
2
Choose the correct option.
(1) H
2
O
2
 acts as oxidising agent in equations
(A) and (B).
(2) H
2
O
2
 act as oxidizing and reducing agent
respectively in equations (A) and (B).
(3) H
2
O
2
 acts as reducing agent in equations
(A) and (B).
(4) H
2
O
2
 acts as reducing and oxidising agent
respectively in equations (A) and (B).
Answer (3)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol. HOCl + H O
2 2
H O + CI + O
32
+–
oxidation
–1
IOH
22
 + H O  + 2
2
–
2I + 2H O + 
–
2
O
2
oxidation
–1
°
°
So, H
2
O
2
 is acting as reducing agent in both
reactions.
10. In the following reaction the reason why meta-
nitro product also formed is :
NH
2
[A]
51%
[B]
47%
[C]
2%
Conc. HNO
3
++
Conc. H SO
24
, 288 K
NH
2
NO
2
NH
2
NO
2
NH
2
NO
2
(1) Formation of anilinium ion
(2) low temperature
(3) –NO
2
 substitution always takes place at
meta-position
(4) –NH
2
 group is highly meta-directive
Answer (1)
Sol. Aniline itself is strong ortho/para director but
on addition of acid it becomes anilinium ion
which is a meta director.
H
+
NH
3
?
NH
2
So the answer should be 1.
11. Al
2
O
3
 was leached with alkali to get X. The
solution of X on passing of gas Y, forms Z. X, Y
and Z respectively are:
(1) X = Na[Al(OH)
4
], Y = SO
2
, Z = Al
2
O
3
(2) X = Al(OH)
3
, Y = SO
2
, Z = Al
2
O
3
.xH
2
O
(3) X = Al(OH)
3
, Y = CO
2
, Z = Al
2
O
3
(4) X = Na[Al(OH)
4
], Y = CO
2
, Z = Al
2
O
3
.xH
2
O
Answer (4)
Sol. ??
23 4
(aq)
(s) (aq)
Al O NaOH Na[Al(OH) ]
?? ? ?
42 23 2 3
(g) (aq) (s) (aq)
Na[Al(OH) ] CO Al O xH O NaHCO
12. Given below are two statements:
Statement-I : Colourless cupric metaborate is
reduced to cuprous metaborate in a luminous
flame.
Statement-II : Cuprous metaborate is obtained
by heating boric anhydride and copper
sulphate in a non-luminous flame.
In the light of the above statements, choose the
most appropriate answer from the options
given below.
(1) Statement I is false but statement II is true
(2) Both statement I and statement II are true
(3) Both statement I and statement II are false
(4) Statement I is true but statement II is false
Answer (3)
Sol. Statement-I : Cupric metaborate is blue in
colour
Hence statement-I is false
Statement-II : CuSO
4
 + B
2
O
3
 ? Cu(BO
2
)
2
cupric metaborate is obtained instead of
cuprous metaborate.
Hence statement-II is false
13. Which of the following compound gives pink
colour on reaction with phthalic anhydride in
conc. H
2
SO
4
 followed by treatment with NaOH?
(1)
CH
3
HO
(2)
CH
3
HO
HO
CH
3
(3)
CH
3
HO
H C
3
OH
(4)
CH
3
OH
Answer (4)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol.
O
O
O
OH
+
(i)  H SO
24
(ii)  NaOH
O
O
HO OH
Derivative of phenolphthalein
14. What is the major product formed by HI on
reaction with C
CH
3
CH
3
CH
3
CH
CH
2
?
(1) CH – C – CH – CH
33
I
CH
3
CH
3
(2) CH – CH – CH  – CH – CH
323
I CH
3
(3) CH – C – CH – CH I
32
CH
3
H CH
3
(4) CH – C – CH – CH
33
CH
3
I CH
3
Answer (1)
Sol.
H
+
1, 2
methyl
shift
I
–
I
15. The product formed in the first step of the
reaction of CH – CH – CH – CH – CH – CH
3223
Br
Br
 with
excess Mg/Et
2
O(Et = C
2
H
5
) is:
(1) CH CH – CH – CH – CH – CH
3223
MgBr
MgBr
(2)
CH – CH – CH – CH – CH – CH
3 2 2 3
CH – CH – CH – CH – CH – CH
32 2 3
(3) CH – CH
3
CH
2
CH – CH
3
(4) CH – CH – CH – CH – CH – CH
32 2 3
CH – CH – CH – CH – CH – CH
3 2 2 3
Answer (1)
Sol.
Br Br
Mg/ether
CH – CH – CH – CH – CH – CH
3 2 2 3
MgBr MgBr
Intramolecular substitution reaction
16. What is the final product (major) ‘A’ in the given
reaction?
(1)
(2)
(3)
(4)
Answer (4)
Page 5


JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
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 of the following are isostructural pairs?
A.
?? 22
44
SO and CrO
B.
44
SiCl and TiCl
C.
?
33
NH and NO
D.
33
BCl and BrCl
(1) A and C only (2) B and C only
(3) A and B only (4) C and D only
Answer (3)
Sol.
O
S
O
O
–
O
–
Tetrahedral
Cl
Si
Cl
Cl Cl
Tetrahedral
O
Cr
O
O
–
O
–
Tetrahedral
Cl
Ti
Cl
Cl Cl
Tetrahedral
Option A
Option B
Isostructural means same structure
So option-3 is the correct answer
2. Out of the following, which type of interaction
is responsible for the stabilisation of ?-helix
structure of proteins?
(1) Covalent bonding
(2) Hydrogen bonding
(3) Ionic bonding
(4) vander Waals forces
Answer (2)
Sol. “ ?-Helix is one of the most common ways in
which a polypeptide chain forms all possible
hydrogen bonds by twisting into a right handed
screw (helix) with the –NH group of each amino
acid residue hydrogen bonded to the C = O  of
an adjacent turn of the helix”
3. The major components in “Gun Metal” are:
(1) Cu, Ni and Fe (2) Cu, Sn and Zn
(3) Al, Cu, Mg and Mn (4) Cu, Zn and Ni
Answer (2)
Sol. Gun metal has composition of Cu, Zn, Sn
Cu – 87%
Zn – 3 %
Sn – 10%
4. In Freundlich adsorption isotherm, slope of AB
line is:
A
B
log P
log
x
m
(1) n with (n, 0.1 to 0.5)
(2)
??
?
??
??
11
 with 0 to 1
nn
(3) log n with (n > 1)
(4) log 
1
n
 with (n < 1)
Answer (2)
Sol. According to Freundlich adsorption isotherm
??
1/n
x
k.P (n 1)
m
??
x1
log logk logP
mn
so in the plot of 
x
log
m
 vs log P the slope is 
1
n
,
where 
1
n
varies from 0 to 1
5. Which of the following ore is concentrated
using group 1 cyanide salt?
(1) Sphalerite (2) Malachite
(3) Siderite (4) Calamine
Answer (1)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol. Ore Formula
Sphalerite ZnS
Siderite FeCO
3
Malachite Cu(OH)
2
?CuCO
3
Calamine ZnCO
3
ZnS + 4 NaCN ? [Zn(CN)
4
]
2–
 + 4 Na
+ 
+ S
2–
the reagent NaCN/KCN is used to suppress the
floating characteristics of ZnS by forming a
soluble complex with KCN.
6. ‘A’ and ‘B’ in the following reactions are :
NH
2
SnCl /HCl/H O
23
+
NaNO/HCl
2
KCN
(A) () B
(1)
N
2
CI
+ –
(A) :
CI
(B) :
(2)
CN
(A) :
CHO
(B) :
(3)
N
2
CI
+ –
(A) :
CHO
(B) :
(4)
CI
(B) :
CN
(A) :
Answer (2)
Sol.
NH
2
CN
CuCN/KCN
HNO
2
(HCl + NANO
2
)
SnCl /HCl
2
H
3
O
+
(Stephen 
reaction)
A
N
2
Cl
?
CHO
B
7. Consider the elements Mg, AI, S, P and Si, the
correct increasing order of their first ionization
enthalpy is :
(1) Mg < Al < Si < P < S
(2) Mg < Al < Si < S < P
(3) Al < Mg < S < Si < P
(4) Al < Mg < Si < S < P
Answer (4)
Sol. Across the period, generally ionization enthalpy
increases but half filled and fully filled
configuration are stable and may change the
regular trend.
P has more IE
1
 than S because of half filled.
Al has lower IE
1
 than Mg because of effective
shielding of 3P electrons from the nucleus by
3s-electrons.
Finally order should be
P > S > Si > Mg > Al
8. The electrode potential of M
2+
 / M of 3d-series
elements shows positive value for :
(1) Cu (2) Zn
(3) Co (4) Fe
Answer (1)
Sol. Only Cu
2+
/Cu has positive SRP among 3d-series
metals.
9. (A) HOCl + H
2
O
2
 ? H
3
O
+
 + CI
–
 + O
2
(B) I
2
 + H
2
O
2
 + 2OH
–
 ? 2I
–
 + 2H
2
O + O
2
Choose the correct option.
(1) H
2
O
2
 acts as oxidising agent in equations
(A) and (B).
(2) H
2
O
2
 act as oxidizing and reducing agent
respectively in equations (A) and (B).
(3) H
2
O
2
 acts as reducing agent in equations
(A) and (B).
(4) H
2
O
2
 acts as reducing and oxidising agent
respectively in equations (A) and (B).
Answer (3)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol. HOCl + H O
2 2
H O + CI + O
32
+–
oxidation
–1
IOH
22
 + H O  + 2
2
–
2I + 2H O + 
–
2
O
2
oxidation
–1
°
°
So, H
2
O
2
 is acting as reducing agent in both
reactions.
10. In the following reaction the reason why meta-
nitro product also formed is :
NH
2
[A]
51%
[B]
47%
[C]
2%
Conc. HNO
3
++
Conc. H SO
24
, 288 K
NH
2
NO
2
NH
2
NO
2
NH
2
NO
2
(1) Formation of anilinium ion
(2) low temperature
(3) –NO
2
 substitution always takes place at
meta-position
(4) –NH
2
 group is highly meta-directive
Answer (1)
Sol. Aniline itself is strong ortho/para director but
on addition of acid it becomes anilinium ion
which is a meta director.
H
+
NH
3
?
NH
2
So the answer should be 1.
11. Al
2
O
3
 was leached with alkali to get X. The
solution of X on passing of gas Y, forms Z. X, Y
and Z respectively are:
(1) X = Na[Al(OH)
4
], Y = SO
2
, Z = Al
2
O
3
(2) X = Al(OH)
3
, Y = SO
2
, Z = Al
2
O
3
.xH
2
O
(3) X = Al(OH)
3
, Y = CO
2
, Z = Al
2
O
3
(4) X = Na[Al(OH)
4
], Y = CO
2
, Z = Al
2
O
3
.xH
2
O
Answer (4)
Sol. ??
23 4
(aq)
(s) (aq)
Al O NaOH Na[Al(OH) ]
?? ? ?
42 23 2 3
(g) (aq) (s) (aq)
Na[Al(OH) ] CO Al O xH O NaHCO
12. Given below are two statements:
Statement-I : Colourless cupric metaborate is
reduced to cuprous metaborate in a luminous
flame.
Statement-II : Cuprous metaborate is obtained
by heating boric anhydride and copper
sulphate in a non-luminous flame.
In the light of the above statements, choose the
most appropriate answer from the options
given below.
(1) Statement I is false but statement II is true
(2) Both statement I and statement II are true
(3) Both statement I and statement II are false
(4) Statement I is true but statement II is false
Answer (3)
Sol. Statement-I : Cupric metaborate is blue in
colour
Hence statement-I is false
Statement-II : CuSO
4
 + B
2
O
3
 ? Cu(BO
2
)
2
cupric metaborate is obtained instead of
cuprous metaborate.
Hence statement-II is false
13. Which of the following compound gives pink
colour on reaction with phthalic anhydride in
conc. H
2
SO
4
 followed by treatment with NaOH?
(1)
CH
3
HO
(2)
CH
3
HO
HO
CH
3
(3)
CH
3
HO
H C
3
OH
(4)
CH
3
OH
Answer (4)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol.
O
O
O
OH
+
(i)  H SO
24
(ii)  NaOH
O
O
HO OH
Derivative of phenolphthalein
14. What is the major product formed by HI on
reaction with C
CH
3
CH
3
CH
3
CH
CH
2
?
(1) CH – C – CH – CH
33
I
CH
3
CH
3
(2) CH – CH – CH  – CH – CH
323
I CH
3
(3) CH – C – CH – CH I
32
CH
3
H CH
3
(4) CH – C – CH – CH
33
CH
3
I CH
3
Answer (1)
Sol.
H
+
1, 2
methyl
shift
I
–
I
15. The product formed in the first step of the
reaction of CH – CH – CH – CH – CH – CH
3223
Br
Br
 with
excess Mg/Et
2
O(Et = C
2
H
5
) is:
(1) CH CH – CH – CH – CH – CH
3223
MgBr
MgBr
(2)
CH – CH – CH – CH – CH – CH
3 2 2 3
CH – CH – CH – CH – CH – CH
32 2 3
(3) CH – CH
3
CH
2
CH – CH
3
(4) CH – CH – CH – CH – CH – CH
32 2 3
CH – CH – CH – CH – CH – CH
3 2 2 3
Answer (1)
Sol.
Br Br
Mg/ether
CH – CH – CH – CH – CH – CH
3 2 2 3
MgBr MgBr
Intramolecular substitution reaction
16. What is the final product (major) ‘A’ in the given
reaction?
(1)
(2)
(3)
(4)
Answer (4)
JEE (MAIN)-2021 : Phase-1(24-02-2021)-M
Sol.
OH
..
..
H
+
+
+
1, 2-hydride 
shift
+
(5 hyperconjugation 
and 3°-carbocation)
H
17. Identify Products A and B.
dil. KMnO
4
273 K
A
CrO
3
B
(1)
OH
A :
O
B :
(2)
OH
A :
O
B :
OH
(3)
OH
A :
O
B :
OH
(4)
A : OHC — 
O
B : HOOC — 
O
Answer (3)
Sol.
dil. KMnO
4
OH , H O 
–
2
273 K
CrO
3 OH
OH
H (Jones 
+
reagent)
OH
O
3°–alcohols do not undergo oxidation reaction
easily.
18. The gas released during anaerobic degradation
of vegetation may lead to :
(1) Acid rain
(2) Corrosion of metals
(3) Ozone hole
(4) Global warming and cancer
Answer (4)
Sol. During anaerobic degradation of vegetation
CO
2
 and CH
4
 are released which may lead to
cause global warming and cancer.
19. Match List I with List II.
List I List II
(Monomer unit) (Polymer)
(a) Caprolactum (i) Natural rubber
(b) 2-Chloro-1, (ii) Buna-N
3-butadiene
(c) Isoprene (iii) Nylon 6
(d) Acrylonitrile (iv) Neoprene
Choose the correct answer from the options
given below:
(1) (a) ? (iv), (b) ? (iii), (c) ? (ii), (d) ? (i)
(2) (a) ? (iii), (b) ? (iv), (c) ? (i), (d) ? (ii)
(3) (a) ? (i), (b) ? (ii), (c) ? (iii), (d) ? (iv)
(4) (a) ? (ii), (b) ? (i), (c) ? (iv), (d) ? (iii)
Answer (2)
Sol. Monomer Polymer
Isoprene Natural rubber
2-Chloro-1, 3-butadiene Neoprene
Caprolactum Nylon 6
Acrylonitrile Buna-N
20. Which of the following reagent is used for the
following reaction?
?? ?
?
32 3 3 2
CH CH CH CH CH CHO
(1) Copper at high temperature and pressure
(2) Manganese acetate
(3) Molybdenum oxide
(4) Potassium permanganate
Answer (3)
Sol.
?
?? ? ?????? ?
23
Mo O
23 2 3 3 2
O CH CHCH CH CHCHO
Mn(OAc)
2
 – manganese acetate oxidizes
alkanes to carboxylic acids in presence of O
2
and heat.