JEE Main 2020 Chemistry September 5 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Chemistry
1. The major product formed in the following reaction is :
? ?
3 3
2
HBr
CH CH CHCH CH ? ? ? ? ? ?
(1) CH
3
CH(Br)CH
2
CH(CH
3
)
2
(2) CH
3
CH
2
CH
2
C(Br)(CH
3
)
2
(3) CH
3
CH
2
CH(Br)CH(CH
3
)
2
(4) Br(CH
2
)
3
CH(CH
3
)
2
Sol. 1
CH —CH=CH—CH
3
CH
3
CH
3
H Br
+ —
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
—
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
2. Hydrogen peroxide, in the pure state, is :
(1) Linear and blue in color (2) Linear and almost colorless
(3) Non-planar and almost colorless (4) Planar and bluein color
Sol. 3
H
2
O
2
 has openbook structure it is non planar
3. Boron and silicon of very high purity can be obtained through :
(1) Liquation (2) Electrolytic refining
(3) Zone refining (4) Vapour phase refining
Sol. 3
Fact
4. The following molecule acts as an :
(CH )
2 2
N
N
Br
(Brompheniramine)
(1) Anti-histamine (2) Antiseptic (3) Anti-depressant (4) Anti-bacterial
Sol. 1
Anti-histamine
Page 2


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Chemistry
1. The major product formed in the following reaction is :
? ?
3 3
2
HBr
CH CH CHCH CH ? ? ? ? ? ?
(1) CH
3
CH(Br)CH
2
CH(CH
3
)
2
(2) CH
3
CH
2
CH
2
C(Br)(CH
3
)
2
(3) CH
3
CH
2
CH(Br)CH(CH
3
)
2
(4) Br(CH
2
)
3
CH(CH
3
)
2
Sol. 1
CH —CH=CH—CH
3
CH
3
CH
3
H Br
+ —
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
—
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
2. Hydrogen peroxide, in the pure state, is :
(1) Linear and blue in color (2) Linear and almost colorless
(3) Non-planar and almost colorless (4) Planar and bluein color
Sol. 3
H
2
O
2
 has openbook structure it is non planar
3. Boron and silicon of very high purity can be obtained through :
(1) Liquation (2) Electrolytic refining
(3) Zone refining (4) Vapour phase refining
Sol. 3
Fact
4. The following molecule acts as an :
(CH )
2 2
N
N
Br
(Brompheniramine)
(1) Anti-histamine (2) Antiseptic (3) Anti-depressant (4) Anti-bacterial
Sol. 1
Anti-histamine
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 2
5. Among the following compounds, geometrical isomerism is exhibited by :
(1) 
CH
3
CHCl
(2) 
CH
3
CHCl
H C
3
 (3) 
CHCl
(4) 
CH
2
Cl
Sol. 1 & 2
C
H Cl
CH
3
 and 
C
Cl H
CH
3
 are geometrical isomer
C
H Cl
CH
3 CH
3
 and 
C
H Cl
CH
3 CH
3
 are geometrical isomer
6. Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the
gas adsorbed on mass m of the adsorbent, the correct plot of 
x
m
 versus p is :
(1) 
P
x
m
270 K 
250 K 
200 K (2) 
P
x
m
200 K 
250 K 
270 K 
(3) 
P
x
m
200 K 
250 K 
270 K 
(4) 
P
x
m
270 K 
250 K 
200 K 
Sol. 2
As temp. increases extent of Adsorption decreases
Therefore correct option (2)
x
m
 = KP
1/n
x
m
 
v/s P ? non linear curve e
Page 3


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Chemistry
1. The major product formed in the following reaction is :
? ?
3 3
2
HBr
CH CH CHCH CH ? ? ? ? ? ?
(1) CH
3
CH(Br)CH
2
CH(CH
3
)
2
(2) CH
3
CH
2
CH
2
C(Br)(CH
3
)
2
(3) CH
3
CH
2
CH(Br)CH(CH
3
)
2
(4) Br(CH
2
)
3
CH(CH
3
)
2
Sol. 1
CH —CH=CH—CH
3
CH
3
CH
3
H Br
+ —
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
—
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
2. Hydrogen peroxide, in the pure state, is :
(1) Linear and blue in color (2) Linear and almost colorless
(3) Non-planar and almost colorless (4) Planar and bluein color
Sol. 3
H
2
O
2
 has openbook structure it is non planar
3. Boron and silicon of very high purity can be obtained through :
(1) Liquation (2) Electrolytic refining
(3) Zone refining (4) Vapour phase refining
Sol. 3
Fact
4. The following molecule acts as an :
(CH )
2 2
N
N
Br
(Brompheniramine)
(1) Anti-histamine (2) Antiseptic (3) Anti-depressant (4) Anti-bacterial
Sol. 1
Anti-histamine
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 2
5. Among the following compounds, geometrical isomerism is exhibited by :
(1) 
CH
3
CHCl
(2) 
CH
3
CHCl
H C
3
 (3) 
CHCl
(4) 
CH
2
Cl
Sol. 1 & 2
C
H Cl
CH
3
 and 
C
Cl H
CH
3
 are geometrical isomer
C
H Cl
CH
3 CH
3
 and 
C
H Cl
CH
3 CH
3
 are geometrical isomer
6. Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the
gas adsorbed on mass m of the adsorbent, the correct plot of 
x
m
 versus p is :
(1) 
P
x
m
270 K 
250 K 
200 K (2) 
P
x
m
200 K 
250 K 
270 K 
(3) 
P
x
m
200 K 
250 K 
270 K 
(4) 
P
x
m
270 K 
250 K 
200 K 
Sol. 2
As temp. increases extent of Adsorption decreases
Therefore correct option (2)
x
m
 = KP
1/n
x
m
 
v/s P ? non linear curve e
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 3
7. The compound that has the largest H–M–H bond angle (M=N, O, S, C) is :
(1) CH
4
(2) H
2
S (3) NH
3
(4) H
2
O
Sol. 1
CH
4
> NH
3
> H
2
O > H
2
S
Sp
3
( ? p = 0) Sp
3
( ? p = 1 Sp
3
( ? p = 2) Sp
3
 ( ? p = 2)
BA 107º28
1
BA = 107º BA = 104º5
1
BA =92º
8. The correct statement about probability density (except at infinite distance from
nucleus) is :
(1) It can be zero for 3p orbital (2) It can be zero for 1s orbital
(3) It can never be zero for 2s orbital (4) It can negative for 2p orbital
Sol. 1
2
R /S
?
> 0 always
2
R /S
?
 can be = 0; As ‘2s’ has 1 Radial Node
2
R
?
 can never be negative e
2
R
?
(3P) can be = 0 as 3P has Radial Nodes
Ans. Option (1)
9. The rate constant (k) of a reaction is measured at differenct temperatures (T), and
the data are plotted in the given figure. The activation energy of the reaction in kJ
mol
–1
 is : (R is gas constant)
0
2 3 4 5 1
5
10
ln k
3
10
T
(1) R (2) 2/R (3) 1/R (4) 2R
Sol. 4
ln(k) = ln(A) – 
Ea
R
1
T
? ?
? ?
? ?
ln(A) = 10
Slope = 
–Ea
R
×10
–3
 = –10/5
E
a
 = 2000R J/mol
E
a
 = 2R KJ/mol
Page 4


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Chemistry
1. The major product formed in the following reaction is :
? ?
3 3
2
HBr
CH CH CHCH CH ? ? ? ? ? ?
(1) CH
3
CH(Br)CH
2
CH(CH
3
)
2
(2) CH
3
CH
2
CH
2
C(Br)(CH
3
)
2
(3) CH
3
CH
2
CH(Br)CH(CH
3
)
2
(4) Br(CH
2
)
3
CH(CH
3
)
2
Sol. 1
CH —CH=CH—CH
3
CH
3
CH
3
H Br
+ —
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
—
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
2. Hydrogen peroxide, in the pure state, is :
(1) Linear and blue in color (2) Linear and almost colorless
(3) Non-planar and almost colorless (4) Planar and bluein color
Sol. 3
H
2
O
2
 has openbook structure it is non planar
3. Boron and silicon of very high purity can be obtained through :
(1) Liquation (2) Electrolytic refining
(3) Zone refining (4) Vapour phase refining
Sol. 3
Fact
4. The following molecule acts as an :
(CH )
2 2
N
N
Br
(Brompheniramine)
(1) Anti-histamine (2) Antiseptic (3) Anti-depressant (4) Anti-bacterial
Sol. 1
Anti-histamine
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 2
5. Among the following compounds, geometrical isomerism is exhibited by :
(1) 
CH
3
CHCl
(2) 
CH
3
CHCl
H C
3
 (3) 
CHCl
(4) 
CH
2
Cl
Sol. 1 & 2
C
H Cl
CH
3
 and 
C
Cl H
CH
3
 are geometrical isomer
C
H Cl
CH
3 CH
3
 and 
C
H Cl
CH
3 CH
3
 are geometrical isomer
6. Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the
gas adsorbed on mass m of the adsorbent, the correct plot of 
x
m
 versus p is :
(1) 
P
x
m
270 K 
250 K 
200 K (2) 
P
x
m
200 K 
250 K 
270 K 
(3) 
P
x
m
200 K 
250 K 
270 K 
(4) 
P
x
m
270 K 
250 K 
200 K 
Sol. 2
As temp. increases extent of Adsorption decreases
Therefore correct option (2)
x
m
 = KP
1/n
x
m
 
v/s P ? non linear curve e
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 3
7. The compound that has the largest H–M–H bond angle (M=N, O, S, C) is :
(1) CH
4
(2) H
2
S (3) NH
3
(4) H
2
O
Sol. 1
CH
4
> NH
3
> H
2
O > H
2
S
Sp
3
( ? p = 0) Sp
3
( ? p = 1 Sp
3
( ? p = 2) Sp
3
 ( ? p = 2)
BA 107º28
1
BA = 107º BA = 104º5
1
BA =92º
8. The correct statement about probability density (except at infinite distance from
nucleus) is :
(1) It can be zero for 3p orbital (2) It can be zero for 1s orbital
(3) It can never be zero for 2s orbital (4) It can negative for 2p orbital
Sol. 1
2
R /S
?
> 0 always
2
R /S
?
 can be = 0; As ‘2s’ has 1 Radial Node
2
R
?
 can never be negative e
2
R
?
(3P) can be = 0 as 3P has Radial Nodes
Ans. Option (1)
9. The rate constant (k) of a reaction is measured at differenct temperatures (T), and
the data are plotted in the given figure. The activation energy of the reaction in kJ
mol
–1
 is : (R is gas constant)
0
2 3 4 5 1
5
10
ln k
3
10
T
(1) R (2) 2/R (3) 1/R (4) 2R
Sol. 4
ln(k) = ln(A) – 
Ea
R
1
T
? ?
? ?
? ?
ln(A) = 10
Slope = 
–Ea
R
×10
–3
 = –10/5
E
a
 = 2000R J/mol
E
a
 = 2R KJ/mol
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 4
10. The variation of molar conductivity with concentration of an electrolyte (X) in aque-
ous solution is shown in the given figure.
Molar
Conductivity
c
The electrolyte X is :
(1) HCl (2) CH
3
COOH (3) NaCl (4) KNO
3
Sol. 2
C
Such type of variation is always for weak electrolyte
Hence Ans (2) CH
3
COOH
11. The final major product of the following reaction is :
Me
NH
2
(i) Ac O/Pyridine
2
(ii) Br /FeCl
2 3
(1) 
Me
NH
2
Br
(2) 
Me
NH
2
Br
(3) 
Me
NH
2
Br
(4) 
Me
NH
2
Br
Page 5


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 1
Date : 5
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Chemistry
1. The major product formed in the following reaction is :
? ?
3 3
2
HBr
CH CH CHCH CH ? ? ? ? ? ?
(1) CH
3
CH(Br)CH
2
CH(CH
3
)
2
(2) CH
3
CH
2
CH
2
C(Br)(CH
3
)
2
(3) CH
3
CH
2
CH(Br)CH(CH
3
)
2
(4) Br(CH
2
)
3
CH(CH
3
)
2
Sol. 1
CH —CH=CH—CH
3
CH
3
CH
3
H Br
+ —
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
—
CH —CH—CH —CH
3 2
CH
3
CH
3
Br
2. Hydrogen peroxide, in the pure state, is :
(1) Linear and blue in color (2) Linear and almost colorless
(3) Non-planar and almost colorless (4) Planar and bluein color
Sol. 3
H
2
O
2
 has openbook structure it is non planar
3. Boron and silicon of very high purity can be obtained through :
(1) Liquation (2) Electrolytic refining
(3) Zone refining (4) Vapour phase refining
Sol. 3
Fact
4. The following molecule acts as an :
(CH )
2 2
N
N
Br
(Brompheniramine)
(1) Anti-histamine (2) Antiseptic (3) Anti-depressant (4) Anti-bacterial
Sol. 1
Anti-histamine
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 2
5. Among the following compounds, geometrical isomerism is exhibited by :
(1) 
CH
3
CHCl
(2) 
CH
3
CHCl
H C
3
 (3) 
CHCl
(4) 
CH
2
Cl
Sol. 1 & 2
C
H Cl
CH
3
 and 
C
Cl H
CH
3
 are geometrical isomer
C
H Cl
CH
3 CH
3
 and 
C
H Cl
CH
3 CH
3
 are geometrical isomer
6. Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the
gas adsorbed on mass m of the adsorbent, the correct plot of 
x
m
 versus p is :
(1) 
P
x
m
270 K 
250 K 
200 K (2) 
P
x
m
200 K 
250 K 
270 K 
(3) 
P
x
m
200 K 
250 K 
270 K 
(4) 
P
x
m
270 K 
250 K 
200 K 
Sol. 2
As temp. increases extent of Adsorption decreases
Therefore correct option (2)
x
m
 = KP
1/n
x
m
 
v/s P ? non linear curve e
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 3
7. The compound that has the largest H–M–H bond angle (M=N, O, S, C) is :
(1) CH
4
(2) H
2
S (3) NH
3
(4) H
2
O
Sol. 1
CH
4
> NH
3
> H
2
O > H
2
S
Sp
3
( ? p = 0) Sp
3
( ? p = 1 Sp
3
( ? p = 2) Sp
3
 ( ? p = 2)
BA 107º28
1
BA = 107º BA = 104º5
1
BA =92º
8. The correct statement about probability density (except at infinite distance from
nucleus) is :
(1) It can be zero for 3p orbital (2) It can be zero for 1s orbital
(3) It can never be zero for 2s orbital (4) It can negative for 2p orbital
Sol. 1
2
R /S
?
> 0 always
2
R /S
?
 can be = 0; As ‘2s’ has 1 Radial Node
2
R
?
 can never be negative e
2
R
?
(3P) can be = 0 as 3P has Radial Nodes
Ans. Option (1)
9. The rate constant (k) of a reaction is measured at differenct temperatures (T), and
the data are plotted in the given figure. The activation energy of the reaction in kJ
mol
–1
 is : (R is gas constant)
0
2 3 4 5 1
5
10
ln k
3
10
T
(1) R (2) 2/R (3) 1/R (4) 2R
Sol. 4
ln(k) = ln(A) – 
Ea
R
1
T
? ?
? ?
? ?
ln(A) = 10
Slope = 
–Ea
R
×10
–3
 = –10/5
E
a
 = 2000R J/mol
E
a
 = 2R KJ/mol
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 4
10. The variation of molar conductivity with concentration of an electrolyte (X) in aque-
ous solution is shown in the given figure.
Molar
Conductivity
c
The electrolyte X is :
(1) HCl (2) CH
3
COOH (3) NaCl (4) KNO
3
Sol. 2
C
Such type of variation is always for weak electrolyte
Hence Ans (2) CH
3
COOH
11. The final major product of the following reaction is :
Me
NH
2
(i) Ac O/Pyridine
2
(ii) Br /FeCl
2 3
(1) 
Me
NH
2
Br
(2) 
Me
NH
2
Br
(3) 
Me
NH
2
Br
(4) 
Me
NH
2
Br
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Chemistry     Page | 5
Sol. 3
Me
NH
2
AC O/P
2 4
Me
NH—C—CH
3
O
Br /FeCl
2 3
Me
NH—C—CH
3
O
Br
OH/
Me
NH
2
Br
12. The major product of the following reaction is :
HO CH CH
2 3
O
H SO
2 4
(1) 
CH=CH
2
O
(2) 
CH CH
2 3
O
(3) 
CH CH
2 3
O
(4) 
CHCH
3
O
Sol. 3
CH —CH
2 3 OH
O
H
+
–H O
2
CH —CH
2 3
O
–H
+
CH —CH
2 3
O
13. Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol
–1
, and 4 kJ mol
–1
,
respectively. The hydration enthalpy of NaCl is :
(1) –780 kJ mol
–1
(2) 784 kJ mol
–1
(3) –784 kJ mol
–1
(4) 780 kJ mol
–1
Sol. 3
?H
sol
 = L.E. + ?H
hyd
4 = 788 + ?H
Hyd
?H
Hyd 
= – 784 KJ/mol Ans
14. Reaction of ammonia with excess Cl
2
 gives :
(1) NH
4
Cl and N
2
(2) NH
4
Cl and HCl
(3) NCl
3
 and HCl (4) NCl
3
 and NH
4
Cl
Sol. 3
(1) NH
3
 + 3Cl
2
 
? ? ? ?
NCl
3
 + 3HCl
     (excess)
(2) 8NH
3
 + 3Cl
2
 
? ? ? ?
6NH
4
Cl+N
2
      (excess)