JEE Main 2021 Answer Key Chemistry - Evening Shift (25-02-2021) Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2021 Answer Key Chemistry - Evening Shift (25-02-2021) Notes | EduRev

 Page 1


JEE (MAIN)-2021 : Phase-1(25-02-2021)-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. Carbylamine test is used to detect the
presence of primary amino group in an organic
compound. Which of the following compound is
formed when this test is performed with
aniline?
(1)
CONH
2
(2)
CN
(3)
NHCH
3
(4)
NC
Answer (4)
Sol.
NH
2
aniline
+ CHCl + 3KOH
3
?
NC
+ 3KCl + 3H O
2
2. What is ‘X’ in the given reaction?
CH OH
2
CH OH
2
+ oxalic acid X
(major product)
210°C
(1)
CHO
CHO
(2)
CH
2
CH
2
(3)
CH – OH
CH
2
(4)
CHOH
2
CHO
Answer (2)
Sol.
CH – OH
2
CH – OH
2
HO – C
O
+
HO – C
O
CH – O – C
2
CH – O – C
2
210°C
CH
2
CH
2
(major)
O
O
3. The major components of German Silver are:
(1) Cu, Zn and Ni (2) Zn, Ni and Ag
(3) Ge, Cu and Ag (4) Cu, Zn and Ag
Answer (1)
Sol. German silver contains Cu (50%), Zn (30%),
Ni (20%) respectively.
4. Which among the following species has
unequal bond lengths?
(1) XeF
4
(2)
?
4
BF
(3) SF
4
(4) SiF
4
Answer (3)
Sol. F
S
F
F
F
:
axial bonds are longer than equatorial bonds.
Only SF
4
 has unequal bond length.
5. Which of the following compound is added to
the sodium extract before addition of silver
nitrate for testing of halogens?
(1) Hydrochloric acid (2) Sodium hydroxide
(3) Ammonia (4) Nitric acid
Answer (4)
Sol. The sodium fusion extract is acidified with nitric
acid and then treated with silver nitrate.
6. The major product of the following reaction is:
NO
2
HSO
24
(1)
NO
2
(2)
NO
2
(3)
NO
2
(4)
NO
2
Answer (2)
Page 2


JEE (MAIN)-2021 : Phase-1(25-02-2021)-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. Carbylamine test is used to detect the
presence of primary amino group in an organic
compound. Which of the following compound is
formed when this test is performed with
aniline?
(1)
CONH
2
(2)
CN
(3)
NHCH
3
(4)
NC
Answer (4)
Sol.
NH
2
aniline
+ CHCl + 3KOH
3
?
NC
+ 3KCl + 3H O
2
2. What is ‘X’ in the given reaction?
CH OH
2
CH OH
2
+ oxalic acid X
(major product)
210°C
(1)
CHO
CHO
(2)
CH
2
CH
2
(3)
CH – OH
CH
2
(4)
CHOH
2
CHO
Answer (2)
Sol.
CH – OH
2
CH – OH
2
HO – C
O
+
HO – C
O
CH – O – C
2
CH – O – C
2
210°C
CH
2
CH
2
(major)
O
O
3. The major components of German Silver are:
(1) Cu, Zn and Ni (2) Zn, Ni and Ag
(3) Ge, Cu and Ag (4) Cu, Zn and Ag
Answer (1)
Sol. German silver contains Cu (50%), Zn (30%),
Ni (20%) respectively.
4. Which among the following species has
unequal bond lengths?
(1) XeF
4
(2)
?
4
BF
(3) SF
4
(4) SiF
4
Answer (3)
Sol. F
S
F
F
F
:
axial bonds are longer than equatorial bonds.
Only SF
4
 has unequal bond length.
5. Which of the following compound is added to
the sodium extract before addition of silver
nitrate for testing of halogens?
(1) Hydrochloric acid (2) Sodium hydroxide
(3) Ammonia (4) Nitric acid
Answer (4)
Sol. The sodium fusion extract is acidified with nitric
acid and then treated with silver nitrate.
6. The major product of the following reaction is:
NO
2
HSO
24
(1)
NO
2
(2)
NO
2
(3)
NO
2
(4)
NO
2
Answer (2)
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
Sol. NO
2
HSO
24
NO
2
?
+
NO
2
?
NO
2
?
H
(Major)
NO
2
7. In which of the following order the given
complex ions are arranged correctly with
respect to their decreasing spin only magnetic
moment?
(i)
3–
6
[FeF ]
(ii)
? 3
36
[Co(NH ) ]
(iii)
2–
4
[NiCl ]
(iv)
? 2
34
[Cu(NH ) ]
(1) (iii) > (iv) > (ii) > (i)
(2) (ii) > (iii) > (i) > (iv)
(3) (i) > (iii) > (iv) > (ii)
(4) (ii) > (i) > (iii) > (iv)
Answer (3)
Sol.
Unpaired e (n)
–
(i)  FeF 
6
3–
(ii)  [Co(NH 
36
)]
3+
(iii)  [NiCl 
4
)
2–
(iv)  [Cu(N H) ]
34
2+
5
0
2
1
Fe 
3+ 
(W.F.L)
Co (S.F.L)
3+ 
 
Ni (W.F.L)
2+ 
 
Cu
2+ 
?=n(n+2)B.M
So, correct order of spin only magnetic
moment is
(ii) < (iv) < (iii) < (i)
8. Water does not produce CO on reacting with:
(1) C
(2) CH
4
(3) CO
2
(4) C
3
H
8
Answer (3)
Sol. H
2
O + CO
2
 ?? H
2
CO
3
all other will produce CO on reaction with
water .
CH + nH O 
n2n+2 2
nCO + (2n+1) H
2
1270 K
Ni
CH (g) + H O (g)
42
CO (g) + 3H (g) 
2
1270 K
Ni
C(s) + H O (g)
2
CO (g) + H (g) 
2
1270 K
9. The correct order of bond dissociation enthalpy
of halogens is:
(1) Cl
2 
> Br
2 
> F
2
> I
2
(2) F
2 
> Cl
2 
> Br
2 
> I
2
(3) Cl
2 
> F
2 
> Br
2
 > I
2
(4) I
2 
> Br
2 
> Cl
2 
> F
2
Answer (1)
Sol. Cl
2
 > Br
2
 > F
2
 > I
2
Bond dissociation enthalpy of F
2
 is lower than
Cl
2
 and Br
2
. It is done to presence of e
–
 on
fluorine atom, which create greater repulsion
due to small size of fluorine.
10. The correct sequence of reagents used in the
preparation of 4-bromo-2-nitroethyl benzene
from benzene is:
(1) CH
3
COCl/AlCl
3
, Zn-Hg/HCl, Br
2
/AlBr
3
,
HNO
3
/H
2
SO
4
(2) HNO
3
/H
2
SO
4
, Br
2
/AlCl
3
,CH
3
COCl/AlCl
3
,
Zn-Hg/HCl
(3) CH
3
COCl/AlCl
3
, Br
2
/AlBr
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
(4) Br
2
/AlBr
3
, CH
3
COCl/AlCl
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
Answer (1)
Sol.
CH COCl
3
AlCl
3
Zn – Hg
HCl
O
CH
25
Br /AlBr
23
CH
25
Br
HNO /H SO
32 4
CH
25
Br
NO
2
Page 3


JEE (MAIN)-2021 : Phase-1(25-02-2021)-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. Carbylamine test is used to detect the
presence of primary amino group in an organic
compound. Which of the following compound is
formed when this test is performed with
aniline?
(1)
CONH
2
(2)
CN
(3)
NHCH
3
(4)
NC
Answer (4)
Sol.
NH
2
aniline
+ CHCl + 3KOH
3
?
NC
+ 3KCl + 3H O
2
2. What is ‘X’ in the given reaction?
CH OH
2
CH OH
2
+ oxalic acid X
(major product)
210°C
(1)
CHO
CHO
(2)
CH
2
CH
2
(3)
CH – OH
CH
2
(4)
CHOH
2
CHO
Answer (2)
Sol.
CH – OH
2
CH – OH
2
HO – C
O
+
HO – C
O
CH – O – C
2
CH – O – C
2
210°C
CH
2
CH
2
(major)
O
O
3. The major components of German Silver are:
(1) Cu, Zn and Ni (2) Zn, Ni and Ag
(3) Ge, Cu and Ag (4) Cu, Zn and Ag
Answer (1)
Sol. German silver contains Cu (50%), Zn (30%),
Ni (20%) respectively.
4. Which among the following species has
unequal bond lengths?
(1) XeF
4
(2)
?
4
BF
(3) SF
4
(4) SiF
4
Answer (3)
Sol. F
S
F
F
F
:
axial bonds are longer than equatorial bonds.
Only SF
4
 has unequal bond length.
5. Which of the following compound is added to
the sodium extract before addition of silver
nitrate for testing of halogens?
(1) Hydrochloric acid (2) Sodium hydroxide
(3) Ammonia (4) Nitric acid
Answer (4)
Sol. The sodium fusion extract is acidified with nitric
acid and then treated with silver nitrate.
6. The major product of the following reaction is:
NO
2
HSO
24
(1)
NO
2
(2)
NO
2
(3)
NO
2
(4)
NO
2
Answer (2)
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
Sol. NO
2
HSO
24
NO
2
?
+
NO
2
?
NO
2
?
H
(Major)
NO
2
7. In which of the following order the given
complex ions are arranged correctly with
respect to their decreasing spin only magnetic
moment?
(i)
3–
6
[FeF ]
(ii)
? 3
36
[Co(NH ) ]
(iii)
2–
4
[NiCl ]
(iv)
? 2
34
[Cu(NH ) ]
(1) (iii) > (iv) > (ii) > (i)
(2) (ii) > (iii) > (i) > (iv)
(3) (i) > (iii) > (iv) > (ii)
(4) (ii) > (i) > (iii) > (iv)
Answer (3)
Sol.
Unpaired e (n)
–
(i)  FeF 
6
3–
(ii)  [Co(NH 
36
)]
3+
(iii)  [NiCl 
4
)
2–
(iv)  [Cu(N H) ]
34
2+
5
0
2
1
Fe 
3+ 
(W.F.L)
Co (S.F.L)
3+ 
 
Ni (W.F.L)
2+ 
 
Cu
2+ 
?=n(n+2)B.M
So, correct order of spin only magnetic
moment is
(ii) < (iv) < (iii) < (i)
8. Water does not produce CO on reacting with:
(1) C
(2) CH
4
(3) CO
2
(4) C
3
H
8
Answer (3)
Sol. H
2
O + CO
2
 ?? H
2
CO
3
all other will produce CO on reaction with
water .
CH + nH O 
n2n+2 2
nCO + (2n+1) H
2
1270 K
Ni
CH (g) + H O (g)
42
CO (g) + 3H (g) 
2
1270 K
Ni
C(s) + H O (g)
2
CO (g) + H (g) 
2
1270 K
9. The correct order of bond dissociation enthalpy
of halogens is:
(1) Cl
2 
> Br
2 
> F
2
> I
2
(2) F
2 
> Cl
2 
> Br
2 
> I
2
(3) Cl
2 
> F
2 
> Br
2
 > I
2
(4) I
2 
> Br
2 
> Cl
2 
> F
2
Answer (1)
Sol. Cl
2
 > Br
2
 > F
2
 > I
2
Bond dissociation enthalpy of F
2
 is lower than
Cl
2
 and Br
2
. It is done to presence of e
–
 on
fluorine atom, which create greater repulsion
due to small size of fluorine.
10. The correct sequence of reagents used in the
preparation of 4-bromo-2-nitroethyl benzene
from benzene is:
(1) CH
3
COCl/AlCl
3
, Zn-Hg/HCl, Br
2
/AlBr
3
,
HNO
3
/H
2
SO
4
(2) HNO
3
/H
2
SO
4
, Br
2
/AlCl
3
,CH
3
COCl/AlCl
3
,
Zn-Hg/HCl
(3) CH
3
COCl/AlCl
3
, Br
2
/AlBr
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
(4) Br
2
/AlBr
3
, CH
3
COCl/AlCl
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
Answer (1)
Sol.
CH COCl
3
AlCl
3
Zn – Hg
HCl
O
CH
25
Br /AlBr
23
CH
25
Br
HNO /H SO
32 4
CH
25
Br
NO
2
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
11. Which of the following is correct structure of
?-anomer of maltose ?
(1)
O
OH
H
CHOH
2
O
H
HO
OH
H
H
O
HH
H
CHOH
2
OH
H
H
H
OH
OH
(2)
O
OH
H
CHOH
2
O
H
HO
H
HO
H
O
HH
H
CHOH
2
H
HO
H
OH
OH
H
(3)
O
OH
H
CHOH
2
O
H
HO
H
OH
H
O
HH
H
CHOH
2
H
HO
H
H
OH
OH
(4)
O
OH
H
CHOH
2
O
H
HO
OH
H
H
O
HH
H
CHOH
2
OH
H
H
OH
OH
H
Answer (4)
Sol. Maltose is composed of two units of ?-D
glucose which are joined through C
1
 – C
4
glycosidic linkage
CH OH
2
H
OH
H
H
OH
H
HO
O
CH OH
2
H
OH
H
H
OH
OH
H
O
HH
O
1
2 3
4
5
6
1
2 3
4
5
6
12. The solubility of Ca(OH)
2
 in water is :
[Given : The solubility product of Ca(OH)
2
 in
water = 5.5 × 10
–6
]
(1) 1.77 × 10
–2
(2) 1.11 × 10
–2
(3) 1.77 × 10
–6
(4) 1.11 × 10
–6
Answer (2)
Sol. Let s be the solubility of Ca(OH)
2
 in water
??
?
2
2
s2s
Ca(OH) Ca 2OH 
K
sp
= [Ca
+2
] [OH
–
]
2
= s × (2s)
2
5.5 × 10
–6
 = 4s
3
??
?? ? ?
36 6
5.5
s 10 1.375 10
4
?
??
1
6
3
s (1.375 10 )
   = 1.11 × 10
–2
13. Which one of the following statements is FALSE
for hydrophilic sols ?
(1) Their viscosity is of the order of that of H
2
O
(2) They do not require electrolytes for stability
(3) These sols are reversible in nature
(4) The sols cannot be easily coagulated
Answer (1)
Sol. The viscosity of the hydrophilic sols are much
higher than that of the dispersion medium.
14. The major product of the following reaction is :
? ? ?????? ?
2
H/CO
32 2
Rh catalyst
CH CH CH CH
(1) CHCH C
32
CH
2
CHO
(2) CH
3
CH
2
CH
2
CHO
(3) CH
3
CH
2
CH
2
CH
2
CHO
(4) CH
3
CH
2
CH = = CH — CHO
Answer (3)
Sol. CH CH CH = CH
32 2
CH CH CH – CH – CHO
32 2 2
H /CO
2
Rh catalyst
15. Given below are two statements :
Statement I : The identification of Ni
2+
 is carried
out by Dimethylglyoxime in the presence of
NH
4
OH.
Statement II : The Dimethylglyoxime is a
bidentate neutral ligand.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Answer (2)
Page 4


JEE (MAIN)-2021 : Phase-1(25-02-2021)-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. Carbylamine test is used to detect the
presence of primary amino group in an organic
compound. Which of the following compound is
formed when this test is performed with
aniline?
(1)
CONH
2
(2)
CN
(3)
NHCH
3
(4)
NC
Answer (4)
Sol.
NH
2
aniline
+ CHCl + 3KOH
3
?
NC
+ 3KCl + 3H O
2
2. What is ‘X’ in the given reaction?
CH OH
2
CH OH
2
+ oxalic acid X
(major product)
210°C
(1)
CHO
CHO
(2)
CH
2
CH
2
(3)
CH – OH
CH
2
(4)
CHOH
2
CHO
Answer (2)
Sol.
CH – OH
2
CH – OH
2
HO – C
O
+
HO – C
O
CH – O – C
2
CH – O – C
2
210°C
CH
2
CH
2
(major)
O
O
3. The major components of German Silver are:
(1) Cu, Zn and Ni (2) Zn, Ni and Ag
(3) Ge, Cu and Ag (4) Cu, Zn and Ag
Answer (1)
Sol. German silver contains Cu (50%), Zn (30%),
Ni (20%) respectively.
4. Which among the following species has
unequal bond lengths?
(1) XeF
4
(2)
?
4
BF
(3) SF
4
(4) SiF
4
Answer (3)
Sol. F
S
F
F
F
:
axial bonds are longer than equatorial bonds.
Only SF
4
 has unequal bond length.
5. Which of the following compound is added to
the sodium extract before addition of silver
nitrate for testing of halogens?
(1) Hydrochloric acid (2) Sodium hydroxide
(3) Ammonia (4) Nitric acid
Answer (4)
Sol. The sodium fusion extract is acidified with nitric
acid and then treated with silver nitrate.
6. The major product of the following reaction is:
NO
2
HSO
24
(1)
NO
2
(2)
NO
2
(3)
NO
2
(4)
NO
2
Answer (2)
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
Sol. NO
2
HSO
24
NO
2
?
+
NO
2
?
NO
2
?
H
(Major)
NO
2
7. In which of the following order the given
complex ions are arranged correctly with
respect to their decreasing spin only magnetic
moment?
(i)
3–
6
[FeF ]
(ii)
? 3
36
[Co(NH ) ]
(iii)
2–
4
[NiCl ]
(iv)
? 2
34
[Cu(NH ) ]
(1) (iii) > (iv) > (ii) > (i)
(2) (ii) > (iii) > (i) > (iv)
(3) (i) > (iii) > (iv) > (ii)
(4) (ii) > (i) > (iii) > (iv)
Answer (3)
Sol.
Unpaired e (n)
–
(i)  FeF 
6
3–
(ii)  [Co(NH 
36
)]
3+
(iii)  [NiCl 
4
)
2–
(iv)  [Cu(N H) ]
34
2+
5
0
2
1
Fe 
3+ 
(W.F.L)
Co (S.F.L)
3+ 
 
Ni (W.F.L)
2+ 
 
Cu
2+ 
?=n(n+2)B.M
So, correct order of spin only magnetic
moment is
(ii) < (iv) < (iii) < (i)
8. Water does not produce CO on reacting with:
(1) C
(2) CH
4
(3) CO
2
(4) C
3
H
8
Answer (3)
Sol. H
2
O + CO
2
 ?? H
2
CO
3
all other will produce CO on reaction with
water .
CH + nH O 
n2n+2 2
nCO + (2n+1) H
2
1270 K
Ni
CH (g) + H O (g)
42
CO (g) + 3H (g) 
2
1270 K
Ni
C(s) + H O (g)
2
CO (g) + H (g) 
2
1270 K
9. The correct order of bond dissociation enthalpy
of halogens is:
(1) Cl
2 
> Br
2 
> F
2
> I
2
(2) F
2 
> Cl
2 
> Br
2 
> I
2
(3) Cl
2 
> F
2 
> Br
2
 > I
2
(4) I
2 
> Br
2 
> Cl
2 
> F
2
Answer (1)
Sol. Cl
2
 > Br
2
 > F
2
 > I
2
Bond dissociation enthalpy of F
2
 is lower than
Cl
2
 and Br
2
. It is done to presence of e
–
 on
fluorine atom, which create greater repulsion
due to small size of fluorine.
10. The correct sequence of reagents used in the
preparation of 4-bromo-2-nitroethyl benzene
from benzene is:
(1) CH
3
COCl/AlCl
3
, Zn-Hg/HCl, Br
2
/AlBr
3
,
HNO
3
/H
2
SO
4
(2) HNO
3
/H
2
SO
4
, Br
2
/AlCl
3
,CH
3
COCl/AlCl
3
,
Zn-Hg/HCl
(3) CH
3
COCl/AlCl
3
, Br
2
/AlBr
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
(4) Br
2
/AlBr
3
, CH
3
COCl/AlCl
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
Answer (1)
Sol.
CH COCl
3
AlCl
3
Zn – Hg
HCl
O
CH
25
Br /AlBr
23
CH
25
Br
HNO /H SO
32 4
CH
25
Br
NO
2
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
11. Which of the following is correct structure of
?-anomer of maltose ?
(1)
O
OH
H
CHOH
2
O
H
HO
OH
H
H
O
HH
H
CHOH
2
OH
H
H
H
OH
OH
(2)
O
OH
H
CHOH
2
O
H
HO
H
HO
H
O
HH
H
CHOH
2
H
HO
H
OH
OH
H
(3)
O
OH
H
CHOH
2
O
H
HO
H
OH
H
O
HH
H
CHOH
2
H
HO
H
H
OH
OH
(4)
O
OH
H
CHOH
2
O
H
HO
OH
H
H
O
HH
H
CHOH
2
OH
H
H
OH
OH
H
Answer (4)
Sol. Maltose is composed of two units of ?-D
glucose which are joined through C
1
 – C
4
glycosidic linkage
CH OH
2
H
OH
H
H
OH
H
HO
O
CH OH
2
H
OH
H
H
OH
OH
H
O
HH
O
1
2 3
4
5
6
1
2 3
4
5
6
12. The solubility of Ca(OH)
2
 in water is :
[Given : The solubility product of Ca(OH)
2
 in
water = 5.5 × 10
–6
]
(1) 1.77 × 10
–2
(2) 1.11 × 10
–2
(3) 1.77 × 10
–6
(4) 1.11 × 10
–6
Answer (2)
Sol. Let s be the solubility of Ca(OH)
2
 in water
??
?
2
2
s2s
Ca(OH) Ca 2OH 
K
sp
= [Ca
+2
] [OH
–
]
2
= s × (2s)
2
5.5 × 10
–6
 = 4s
3
??
?? ? ?
36 6
5.5
s 10 1.375 10
4
?
??
1
6
3
s (1.375 10 )
   = 1.11 × 10
–2
13. Which one of the following statements is FALSE
for hydrophilic sols ?
(1) Their viscosity is of the order of that of H
2
O
(2) They do not require electrolytes for stability
(3) These sols are reversible in nature
(4) The sols cannot be easily coagulated
Answer (1)
Sol. The viscosity of the hydrophilic sols are much
higher than that of the dispersion medium.
14. The major product of the following reaction is :
? ? ?????? ?
2
H/CO
32 2
Rh catalyst
CH CH CH CH
(1) CHCH C
32
CH
2
CHO
(2) CH
3
CH
2
CH
2
CHO
(3) CH
3
CH
2
CH
2
CH
2
CHO
(4) CH
3
CH
2
CH = = CH — CHO
Answer (3)
Sol. CH CH CH = CH
32 2
CH CH CH – CH – CHO
32 2 2
H /CO
2
Rh catalyst
15. Given below are two statements :
Statement I : The identification of Ni
2+
 is carried
out by Dimethylglyoxime in the presence of
NH
4
OH.
Statement II : The Dimethylglyoxime is a
bidentate neutral ligand.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Answer (2)
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
Sol.
CH – C = N – OH
3
H C – C = N – OH
3
Dimethylglyoxime
+ Ni
+2
NH OH
4
CH – C = N
3
CH – C = N
3
Ni
+2
N = C – CH
3
N = C – CH
3
O
–
H – O
O
–
O – H
+ 2H
+
Ni(DMG)
2
Identification of Ni
+2
 is carried out by dimethyl
glyoxime in presence of NH
4
OH
16. The correct order of acid character of the
following compounds is :
OH
I
COOH
NO
2
II
COOH
III
COOH
CH
3
IV
(1) IV > III > II > I
(2) II > III > IV > I
(3) I > II > III > IV
(4) III > II > I > IV
Answer (2)
Sol. Acidic strength
> >>
–I, –R +I, +R
Since carboxylic acids are more acidic than
phenols.
–I and – R  effect increase the acidic strength
where as +I and +R effect decrease the acidic
strength of carboxylic acids.
17. The method used for the purification of Indium
is :
(1) Vapour phase refining
(2) Zone refining
(3) Liquation
(4) van Arkel method
Answer (2)
Sol. Indium is purified by zone refining method.
18.
NH
2
HNO , H SO
34 2
288 K
NH
2
NO
2
(A) (B)
+
NH
2
NO
2
+
(C)
NH
2
NO
2
Correct statement about the given chemical
reaction is :
(1) The reaction will form sulphonated product
instead of nitration.
(2) Reaction is possible and compound (B) will
be the major product.
(3) Reaction is possible and compound (A) will
be major product.
(4)
NH
2
:
 group is ortho and para directive,
so product (B) is not possible.
Answer (3)
Sol. A will be the major product.
NH
2
HNO , H SO
34 2
288 K
NH
2
NO
2
(51)%
A
(47%)
B
+
NH
2
NO
2
+
(2%)
C
NH
2
NO
2
In strongly acidic medium aniline is protonated
to form anilinium ion which is a meta directing.
That is why besides the ortho para derivatives,
significant amount of meta derivatives is also
formed.
Page 5


JEE (MAIN)-2021 : Phase-1(25-02-2021)-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. Carbylamine test is used to detect the
presence of primary amino group in an organic
compound. Which of the following compound is
formed when this test is performed with
aniline?
(1)
CONH
2
(2)
CN
(3)
NHCH
3
(4)
NC
Answer (4)
Sol.
NH
2
aniline
+ CHCl + 3KOH
3
?
NC
+ 3KCl + 3H O
2
2. What is ‘X’ in the given reaction?
CH OH
2
CH OH
2
+ oxalic acid X
(major product)
210°C
(1)
CHO
CHO
(2)
CH
2
CH
2
(3)
CH – OH
CH
2
(4)
CHOH
2
CHO
Answer (2)
Sol.
CH – OH
2
CH – OH
2
HO – C
O
+
HO – C
O
CH – O – C
2
CH – O – C
2
210°C
CH
2
CH
2
(major)
O
O
3. The major components of German Silver are:
(1) Cu, Zn and Ni (2) Zn, Ni and Ag
(3) Ge, Cu and Ag (4) Cu, Zn and Ag
Answer (1)
Sol. German silver contains Cu (50%), Zn (30%),
Ni (20%) respectively.
4. Which among the following species has
unequal bond lengths?
(1) XeF
4
(2)
?
4
BF
(3) SF
4
(4) SiF
4
Answer (3)
Sol. F
S
F
F
F
:
axial bonds are longer than equatorial bonds.
Only SF
4
 has unequal bond length.
5. Which of the following compound is added to
the sodium extract before addition of silver
nitrate for testing of halogens?
(1) Hydrochloric acid (2) Sodium hydroxide
(3) Ammonia (4) Nitric acid
Answer (4)
Sol. The sodium fusion extract is acidified with nitric
acid and then treated with silver nitrate.
6. The major product of the following reaction is:
NO
2
HSO
24
(1)
NO
2
(2)
NO
2
(3)
NO
2
(4)
NO
2
Answer (2)
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
Sol. NO
2
HSO
24
NO
2
?
+
NO
2
?
NO
2
?
H
(Major)
NO
2
7. In which of the following order the given
complex ions are arranged correctly with
respect to their decreasing spin only magnetic
moment?
(i)
3–
6
[FeF ]
(ii)
? 3
36
[Co(NH ) ]
(iii)
2–
4
[NiCl ]
(iv)
? 2
34
[Cu(NH ) ]
(1) (iii) > (iv) > (ii) > (i)
(2) (ii) > (iii) > (i) > (iv)
(3) (i) > (iii) > (iv) > (ii)
(4) (ii) > (i) > (iii) > (iv)
Answer (3)
Sol.
Unpaired e (n)
–
(i)  FeF 
6
3–
(ii)  [Co(NH 
36
)]
3+
(iii)  [NiCl 
4
)
2–
(iv)  [Cu(N H) ]
34
2+
5
0
2
1
Fe 
3+ 
(W.F.L)
Co (S.F.L)
3+ 
 
Ni (W.F.L)
2+ 
 
Cu
2+ 
?=n(n+2)B.M
So, correct order of spin only magnetic
moment is
(ii) < (iv) < (iii) < (i)
8. Water does not produce CO on reacting with:
(1) C
(2) CH
4
(3) CO
2
(4) C
3
H
8
Answer (3)
Sol. H
2
O + CO
2
 ?? H
2
CO
3
all other will produce CO on reaction with
water .
CH + nH O 
n2n+2 2
nCO + (2n+1) H
2
1270 K
Ni
CH (g) + H O (g)
42
CO (g) + 3H (g) 
2
1270 K
Ni
C(s) + H O (g)
2
CO (g) + H (g) 
2
1270 K
9. The correct order of bond dissociation enthalpy
of halogens is:
(1) Cl
2 
> Br
2 
> F
2
> I
2
(2) F
2 
> Cl
2 
> Br
2 
> I
2
(3) Cl
2 
> F
2 
> Br
2
 > I
2
(4) I
2 
> Br
2 
> Cl
2 
> F
2
Answer (1)
Sol. Cl
2
 > Br
2
 > F
2
 > I
2
Bond dissociation enthalpy of F
2
 is lower than
Cl
2
 and Br
2
. It is done to presence of e
–
 on
fluorine atom, which create greater repulsion
due to small size of fluorine.
10. The correct sequence of reagents used in the
preparation of 4-bromo-2-nitroethyl benzene
from benzene is:
(1) CH
3
COCl/AlCl
3
, Zn-Hg/HCl, Br
2
/AlBr
3
,
HNO
3
/H
2
SO
4
(2) HNO
3
/H
2
SO
4
, Br
2
/AlCl
3
,CH
3
COCl/AlCl
3
,
Zn-Hg/HCl
(3) CH
3
COCl/AlCl
3
, Br
2
/AlBr
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
(4) Br
2
/AlBr
3
, CH
3
COCl/AlCl
3
, HNO
3
/H
2
SO
4
,
Zn/HCl
Answer (1)
Sol.
CH COCl
3
AlCl
3
Zn – Hg
HCl
O
CH
25
Br /AlBr
23
CH
25
Br
HNO /H SO
32 4
CH
25
Br
NO
2
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
11. Which of the following is correct structure of
?-anomer of maltose ?
(1)
O
OH
H
CHOH
2
O
H
HO
OH
H
H
O
HH
H
CHOH
2
OH
H
H
H
OH
OH
(2)
O
OH
H
CHOH
2
O
H
HO
H
HO
H
O
HH
H
CHOH
2
H
HO
H
OH
OH
H
(3)
O
OH
H
CHOH
2
O
H
HO
H
OH
H
O
HH
H
CHOH
2
H
HO
H
H
OH
OH
(4)
O
OH
H
CHOH
2
O
H
HO
OH
H
H
O
HH
H
CHOH
2
OH
H
H
OH
OH
H
Answer (4)
Sol. Maltose is composed of two units of ?-D
glucose which are joined through C
1
 – C
4
glycosidic linkage
CH OH
2
H
OH
H
H
OH
H
HO
O
CH OH
2
H
OH
H
H
OH
OH
H
O
HH
O
1
2 3
4
5
6
1
2 3
4
5
6
12. The solubility of Ca(OH)
2
 in water is :
[Given : The solubility product of Ca(OH)
2
 in
water = 5.5 × 10
–6
]
(1) 1.77 × 10
–2
(2) 1.11 × 10
–2
(3) 1.77 × 10
–6
(4) 1.11 × 10
–6
Answer (2)
Sol. Let s be the solubility of Ca(OH)
2
 in water
??
?
2
2
s2s
Ca(OH) Ca 2OH 
K
sp
= [Ca
+2
] [OH
–
]
2
= s × (2s)
2
5.5 × 10
–6
 = 4s
3
??
?? ? ?
36 6
5.5
s 10 1.375 10
4
?
??
1
6
3
s (1.375 10 )
   = 1.11 × 10
–2
13. Which one of the following statements is FALSE
for hydrophilic sols ?
(1) Their viscosity is of the order of that of H
2
O
(2) They do not require electrolytes for stability
(3) These sols are reversible in nature
(4) The sols cannot be easily coagulated
Answer (1)
Sol. The viscosity of the hydrophilic sols are much
higher than that of the dispersion medium.
14. The major product of the following reaction is :
? ? ?????? ?
2
H/CO
32 2
Rh catalyst
CH CH CH CH
(1) CHCH C
32
CH
2
CHO
(2) CH
3
CH
2
CH
2
CHO
(3) CH
3
CH
2
CH
2
CH
2
CHO
(4) CH
3
CH
2
CH = = CH — CHO
Answer (3)
Sol. CH CH CH = CH
32 2
CH CH CH – CH – CHO
32 2 2
H /CO
2
Rh catalyst
15. Given below are two statements :
Statement I : The identification of Ni
2+
 is carried
out by Dimethylglyoxime in the presence of
NH
4
OH.
Statement II : The Dimethylglyoxime is a
bidentate neutral ligand.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Answer (2)
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
Sol.
CH – C = N – OH
3
H C – C = N – OH
3
Dimethylglyoxime
+ Ni
+2
NH OH
4
CH – C = N
3
CH – C = N
3
Ni
+2
N = C – CH
3
N = C – CH
3
O
–
H – O
O
–
O – H
+ 2H
+
Ni(DMG)
2
Identification of Ni
+2
 is carried out by dimethyl
glyoxime in presence of NH
4
OH
16. The correct order of acid character of the
following compounds is :
OH
I
COOH
NO
2
II
COOH
III
COOH
CH
3
IV
(1) IV > III > II > I
(2) II > III > IV > I
(3) I > II > III > IV
(4) III > II > I > IV
Answer (2)
Sol. Acidic strength
> >>
–I, –R +I, +R
Since carboxylic acids are more acidic than
phenols.
–I and – R  effect increase the acidic strength
where as +I and +R effect decrease the acidic
strength of carboxylic acids.
17. The method used for the purification of Indium
is :
(1) Vapour phase refining
(2) Zone refining
(3) Liquation
(4) van Arkel method
Answer (2)
Sol. Indium is purified by zone refining method.
18.
NH
2
HNO , H SO
34 2
288 K
NH
2
NO
2
(A) (B)
+
NH
2
NO
2
+
(C)
NH
2
NO
2
Correct statement about the given chemical
reaction is :
(1) The reaction will form sulphonated product
instead of nitration.
(2) Reaction is possible and compound (B) will
be the major product.
(3) Reaction is possible and compound (A) will
be major product.
(4)
NH
2
:
 group is ortho and para directive,
so product (B) is not possible.
Answer (3)
Sol. A will be the major product.
NH
2
HNO , H SO
34 2
288 K
NH
2
NO
2
(51)%
A
(47%)
B
+
NH
2
NO
2
+
(2%)
C
NH
2
NO
2
In strongly acidic medium aniline is protonated
to form anilinium ion which is a meta directing.
That is why besides the ortho para derivatives,
significant amount of meta derivatives is also
formed.
JEE (MAIN)-2021 : Phase-1(25-02-2021)-E
19. Given below are two statements :
Statement I  :
The pH of rain water is normally ~5.6.
Statement II :
If the pH of rain water drops below 5.6, it is
called acid rain.
In the light of the above statements, choose the
correct answer from the options given below:
(1) Both Statement I and Statement II are false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Statement I is true but Statement II is false.
Answer (3)
Sol. pH of rain water is normally ? 5.6 due to
presence of H
+
 formed by the reaction of water
and CO
2
 present in atmosphere
?



22 23
CO (g) HO() HCO (aq)
??
??


23 3
HCO(aq) H(aq) HCO(aq)pH 5.6
The pH of acid rain drop below 5.6 due to
presence of other acidic gases like SO
2
 and
NO
2
 present in atmosphere
?? ?? ? 
22 2 24
2SO (g) O (g) 2H O( ) 2H SO (aq)
?? ?? ?
?

22 2 3
4NO (g) O (g) 2H O() 4HNO (aq)
pH 5.6
20. Given below are two statements :
Statement I  :
? and ? forms of sulphur can change reversibly
between themselves with slow heating or slow
cooling.
Statement II :
At room temperature the stable crystalline form
of sulphur is monoclinic sulphur.
In the light of the above statements, choose the
correct 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 true but Statement II is false.
(4) Statement I is false but Statement II is true.
Answer (3)
Sol. The stable form at room temperature is rhombic
sulphur, which transformed to monoclinic
sulphur on heating at 369 K.
? and ? form of sulphur can change reversibly
between themselves with slow heating or slow
cooling.
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. The unit cell of copper corresponds to a face
centered cube of edge length 3.596 Å with one
copper atom at each lattice point. The
calculated density of copper in kg/m
3
 is _____.
[Molar mass of Cu : 63.54 g; Avogadro Number
= 6.022 × 10
23
]
Answer (9077)
Sol. Copper crystallises is fcc unit cell with edge
length, a = 3.596Å
3
A
4M
density
N (a)
?
?
3
23 10 3
4 63.54 10
6.022 10 (3.596 10 )
?
?
??
?
??
3
9077kgm
?

2. The spin only magnetic moment of a divalent
ion in aqueous solution (atomic number 29) is
Answer (2)
Sol. The element having atomic no. 29 is copper
The electronic configuration of Cu
2+
 is
Cu
2+
 : 3d
9
It has 1 unpaired electron
31.73BM 2 ?? ? 
3. Copper reduces 
?
3
NO into NO and NO
2
depending upon the concentration of HNO
3
 in
solution. (Assuming fixed [Cu
2+
] and
?
NO NO
2
PP
), the HNO
3
 concentration at which
the thermodynamic tendency for reduction of
?
3
NO into NO and NO
2
 by copper is same is 10
x
M. The value of 2x is ______. (Rounded-off to
the nearest integer)
[Given, 
?? ?
?? ?
??
2
Cu /Cu NO /NO NO /NO
332
E0.34V,E 0.96V,E
= 0.79 V and at 298 K, 
?
RT
(2.303) 0.059]
F
Answer (*)
Incomplete data.
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