JEE Main 2020 Chemistry September 6 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The INCORRECT statement is  :
(1) Cast iron is used to manufacture wrought iron.
(2) Brass is an alloy of copper and nickel.
(3) German silver is an alloy of zinc, copper and nickel.
(4) Bronze is an alloy of copper and tin
Sol. 2
Brass - (copper Zinc)
Bronze - (copper tin)
2. The species that has a spin-only magnetic moment of 5.9 BM, is : (T
d
 = tetrahedral)
(1) [Ni(CN)
4
]
2-
 (square planar) (2) Ni(CO)
4
(T
d
)
(3) [MnBr
4
]
2-
(T
d
) (4) [NiCl
4
]
2-
(T
d
)
Sol. 3
[MnBr
4
]
2–
? = 5(5 2) ? = 5.9 BM
3. For the reaction
2 2
3
Fe N(s) H (g)
2
? 2Fe(s) + NH
3
(g)
(1) 
1/2
c p
K K (RT) ? (2) 
1/2
c p
K K (RT)
?
?
(3) 
3
2
c p
K K (RT) ?
(4) 
c p
K K (RT) ?
Sol. 1
Fe
2
N(s) + 
3
2
H
2
(g)  2Fe(s) + NH
3
(g)
?n
g
 = 
3
1
2
? = 
1
2
?
p
c
K
K
 = 
g
n
(RT)
?
 = 
1
2
(RT)
?
K
c
 = 
p
1
2
K
(RT)
? = K
p
.
1
2
(RT)
Page 2


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The INCORRECT statement is  :
(1) Cast iron is used to manufacture wrought iron.
(2) Brass is an alloy of copper and nickel.
(3) German silver is an alloy of zinc, copper and nickel.
(4) Bronze is an alloy of copper and tin
Sol. 2
Brass - (copper Zinc)
Bronze - (copper tin)
2. The species that has a spin-only magnetic moment of 5.9 BM, is : (T
d
 = tetrahedral)
(1) [Ni(CN)
4
]
2-
 (square planar) (2) Ni(CO)
4
(T
d
)
(3) [MnBr
4
]
2-
(T
d
) (4) [NiCl
4
]
2-
(T
d
)
Sol. 3
[MnBr
4
]
2–
? = 5(5 2) ? = 5.9 BM
3. For the reaction
2 2
3
Fe N(s) H (g)
2
? 2Fe(s) + NH
3
(g)
(1) 
1/2
c p
K K (RT) ? (2) 
1/2
c p
K K (RT)
?
?
(3) 
3
2
c p
K K (RT) ?
(4) 
c p
K K (RT) ?
Sol. 1
Fe
2
N(s) + 
3
2
H
2
(g)  2Fe(s) + NH
3
(g)
?n
g
 = 
3
1
2
? = 
1
2
?
p
c
K
K
 = 
g
n
(RT)
?
 = 
1
2
(RT)
?
K
c
 = 
p
1
2
K
(RT)
? = K
p
.
1
2
(RT)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 2
4. Consider the following reactions :
'A'
(C H )
7 14
ozonolysis
'B' + 'C'
'B' 
yellow ppt
silver mirror
Ag O
2
(I +NaOH)
2
'C' 
no yellow ppt
Anhydrous ZnCl
2 LiAlH
4
(I +NaOH)
2
& Conc. HCl
'D' 
gives white
turbidity
within 5
minutes
'A' is :
(1) (2) 
(3) (4) 
Sol. 1
Ozonolysis
CH —CHO
3
+
O
OH
Cl
(C)
LiAlH
4
ZnCl /HCl
2
(turbidity in 5 min)
(A) (B)
5. Arrange the following solutions in the decreasing order of pOH :
(A) 0.01 M HCl (B) 0.01 M NaOH
(C) 0.01 M CH
3
COONa (D) 0.01 M NaCl
(1) (A) > (C) > (D) > (B) (2) (B) > (D) > (C) > (A)
(3) (B) > (C) > (D) > (A) (4) (A) > (D) > (C) > (B)
Page 3


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The INCORRECT statement is  :
(1) Cast iron is used to manufacture wrought iron.
(2) Brass is an alloy of copper and nickel.
(3) German silver is an alloy of zinc, copper and nickel.
(4) Bronze is an alloy of copper and tin
Sol. 2
Brass - (copper Zinc)
Bronze - (copper tin)
2. The species that has a spin-only magnetic moment of 5.9 BM, is : (T
d
 = tetrahedral)
(1) [Ni(CN)
4
]
2-
 (square planar) (2) Ni(CO)
4
(T
d
)
(3) [MnBr
4
]
2-
(T
d
) (4) [NiCl
4
]
2-
(T
d
)
Sol. 3
[MnBr
4
]
2–
? = 5(5 2) ? = 5.9 BM
3. For the reaction
2 2
3
Fe N(s) H (g)
2
? 2Fe(s) + NH
3
(g)
(1) 
1/2
c p
K K (RT) ? (2) 
1/2
c p
K K (RT)
?
?
(3) 
3
2
c p
K K (RT) ?
(4) 
c p
K K (RT) ?
Sol. 1
Fe
2
N(s) + 
3
2
H
2
(g)  2Fe(s) + NH
3
(g)
?n
g
 = 
3
1
2
? = 
1
2
?
p
c
K
K
 = 
g
n
(RT)
?
 = 
1
2
(RT)
?
K
c
 = 
p
1
2
K
(RT)
? = K
p
.
1
2
(RT)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 2
4. Consider the following reactions :
'A'
(C H )
7 14
ozonolysis
'B' + 'C'
'B' 
yellow ppt
silver mirror
Ag O
2
(I +NaOH)
2
'C' 
no yellow ppt
Anhydrous ZnCl
2 LiAlH
4
(I +NaOH)
2
& Conc. HCl
'D' 
gives white
turbidity
within 5
minutes
'A' is :
(1) (2) 
(3) (4) 
Sol. 1
Ozonolysis
CH —CHO
3
+
O
OH
Cl
(C)
LiAlH
4
ZnCl /HCl
2
(turbidity in 5 min)
(A) (B)
5. Arrange the following solutions in the decreasing order of pOH :
(A) 0.01 M HCl (B) 0.01 M NaOH
(C) 0.01 M CH
3
COONa (D) 0.01 M NaCl
(1) (A) > (C) > (D) > (B) (2) (B) > (D) > (C) > (A)
(3) (B) > (C) > (D) > (A) (4) (A) > (D) > (C) > (B)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 3
Sol. 4
(i) 10
–2
 M HCl ? [H
+
]  = 10
–2
 M ? pH
 
= 2
(ii) 10
–2
 M NaOH ? [OH
–
] = 10
–2
 M ? pOH = 2
(iii) 10
–2
 M CH
3
COO
–
Na
+
 ? [OH
+
] > 10
–7
 ? pOH < 7
(iv) 10
–2
 M NaCl ? Neutral pOH = 7
(i) > (iv) > (iii) > (ii)
6. The variation of equilibrium constant with temperature is given below :
Temperature Equilibrium Constant
T
1
 = 25
o
C K
1
 = 10
T
2
 = 100
o
C K
2
 = 100
The value of ?H
0 
, ?G
0
 
 
at T
1 
and ?G
0
 at T
2 
(in Kj mol
-1
) respectively, are close to
[use R = 8.314JK
-1
 mol
-1
]
(1) 28.4, -7.14 and -5.71 (2) 0.64, - 7.14 and -5.71
(3) 28.4, - 5.71 and -14.29 (4) 0.64, - 5.71 and -14.29
Sol. 3
In 
2
1
k
k
? ?
? ?
? ?
 = 
1 2
H 1 1
R T T
? ? ? ?
?
? ?
? ?
ln(10) = 
H 1 1
R 298 373
? ? ? ?
?
? ?
? ?
373 298 8.314 2.303
75
? ? ?
 = ?H° = 28.37 kJ mol
–1
1
T
G ? ?
 = –RT T
1
ln(K
1
) = –298R ln(10) = –5.71 kJ mol
–1
2
T
G ? ?
 = –RT T
2
 ln(K
2
) = –373R ln(100)
= –14.283 kJ/mol
7. Consider the following reactions
A P1; B  P2; C  P3; D  P4, ? ? ? ?
The order of the above reactions are a,b,c and d, respectively. The following graph is
obtained when log[rate] vs. log[conc.] are plotted :
[D]
[B]
log [Conc.]
log [rate]
[A]
[C]
Among the following the correct sequence for the order of the reactions is :
(1) c > a > b > d (2) d > a > b > c
(3) d > b > a > c (4) a > b > c > d
Page 4


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The INCORRECT statement is  :
(1) Cast iron is used to manufacture wrought iron.
(2) Brass is an alloy of copper and nickel.
(3) German silver is an alloy of zinc, copper and nickel.
(4) Bronze is an alloy of copper and tin
Sol. 2
Brass - (copper Zinc)
Bronze - (copper tin)
2. The species that has a spin-only magnetic moment of 5.9 BM, is : (T
d
 = tetrahedral)
(1) [Ni(CN)
4
]
2-
 (square planar) (2) Ni(CO)
4
(T
d
)
(3) [MnBr
4
]
2-
(T
d
) (4) [NiCl
4
]
2-
(T
d
)
Sol. 3
[MnBr
4
]
2–
? = 5(5 2) ? = 5.9 BM
3. For the reaction
2 2
3
Fe N(s) H (g)
2
? 2Fe(s) + NH
3
(g)
(1) 
1/2
c p
K K (RT) ? (2) 
1/2
c p
K K (RT)
?
?
(3) 
3
2
c p
K K (RT) ?
(4) 
c p
K K (RT) ?
Sol. 1
Fe
2
N(s) + 
3
2
H
2
(g)  2Fe(s) + NH
3
(g)
?n
g
 = 
3
1
2
? = 
1
2
?
p
c
K
K
 = 
g
n
(RT)
?
 = 
1
2
(RT)
?
K
c
 = 
p
1
2
K
(RT)
? = K
p
.
1
2
(RT)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 2
4. Consider the following reactions :
'A'
(C H )
7 14
ozonolysis
'B' + 'C'
'B' 
yellow ppt
silver mirror
Ag O
2
(I +NaOH)
2
'C' 
no yellow ppt
Anhydrous ZnCl
2 LiAlH
4
(I +NaOH)
2
& Conc. HCl
'D' 
gives white
turbidity
within 5
minutes
'A' is :
(1) (2) 
(3) (4) 
Sol. 1
Ozonolysis
CH —CHO
3
+
O
OH
Cl
(C)
LiAlH
4
ZnCl /HCl
2
(turbidity in 5 min)
(A) (B)
5. Arrange the following solutions in the decreasing order of pOH :
(A) 0.01 M HCl (B) 0.01 M NaOH
(C) 0.01 M CH
3
COONa (D) 0.01 M NaCl
(1) (A) > (C) > (D) > (B) (2) (B) > (D) > (C) > (A)
(3) (B) > (C) > (D) > (A) (4) (A) > (D) > (C) > (B)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 3
Sol. 4
(i) 10
–2
 M HCl ? [H
+
]  = 10
–2
 M ? pH
 
= 2
(ii) 10
–2
 M NaOH ? [OH
–
] = 10
–2
 M ? pOH = 2
(iii) 10
–2
 M CH
3
COO
–
Na
+
 ? [OH
+
] > 10
–7
 ? pOH < 7
(iv) 10
–2
 M NaCl ? Neutral pOH = 7
(i) > (iv) > (iii) > (ii)
6. The variation of equilibrium constant with temperature is given below :
Temperature Equilibrium Constant
T
1
 = 25
o
C K
1
 = 10
T
2
 = 100
o
C K
2
 = 100
The value of ?H
0 
, ?G
0
 
 
at T
1 
and ?G
0
 at T
2 
(in Kj mol
-1
) respectively, are close to
[use R = 8.314JK
-1
 mol
-1
]
(1) 28.4, -7.14 and -5.71 (2) 0.64, - 7.14 and -5.71
(3) 28.4, - 5.71 and -14.29 (4) 0.64, - 5.71 and -14.29
Sol. 3
In 
2
1
k
k
? ?
? ?
? ?
 = 
1 2
H 1 1
R T T
? ? ? ?
?
? ?
? ?
ln(10) = 
H 1 1
R 298 373
? ? ? ?
?
? ?
? ?
373 298 8.314 2.303
75
? ? ?
 = ?H° = 28.37 kJ mol
–1
1
T
G ? ?
 = –RT T
1
ln(K
1
) = –298R ln(10) = –5.71 kJ mol
–1
2
T
G ? ?
 = –RT T
2
 ln(K
2
) = –373R ln(100)
= –14.283 kJ/mol
7. Consider the following reactions
A P1; B  P2; C  P3; D  P4, ? ? ? ?
The order of the above reactions are a,b,c and d, respectively. The following graph is
obtained when log[rate] vs. log[conc.] are plotted :
[D]
[B]
log [Conc.]
log [rate]
[A]
[C]
Among the following the correct sequence for the order of the reactions is :
(1) c > a > b > d (2) d > a > b > c
(3) d > b > a > c (4) a > b > c > d
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 4
Sol. 3
A ? P1 B ? P2 C ? P3 D ? P4
Rate = K (conc.)
order
Slope = order
According graph
d > b > a > c order of slope
8. The major product obtained from the following reactions is :
O N
2
C C OCH
3 
2+
2
Hg /H+
H O
? ? ? ? ?
(1) 
O N
2
O
OCH
3
(2) 
O N
2
O
OH
(3) 
O N
2
O
OCH
3
(4) 
O N
2
O
OH
Sol. 3
O N
2
OCH
3
Hg /H
2+ +
H O
2
O N
2
OCH
3
CH —C
2
O
9. Which of the following compounds shows geometrical isomerism ?
(1) 2-methylpent-1-ene (2) 4-methylpent-2-ene
(3) 2-methylpent-2-ene (4) 4-methylpent-1-ene
Page 5


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 6
th
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The INCORRECT statement is  :
(1) Cast iron is used to manufacture wrought iron.
(2) Brass is an alloy of copper and nickel.
(3) German silver is an alloy of zinc, copper and nickel.
(4) Bronze is an alloy of copper and tin
Sol. 2
Brass - (copper Zinc)
Bronze - (copper tin)
2. The species that has a spin-only magnetic moment of 5.9 BM, is : (T
d
 = tetrahedral)
(1) [Ni(CN)
4
]
2-
 (square planar) (2) Ni(CO)
4
(T
d
)
(3) [MnBr
4
]
2-
(T
d
) (4) [NiCl
4
]
2-
(T
d
)
Sol. 3
[MnBr
4
]
2–
? = 5(5 2) ? = 5.9 BM
3. For the reaction
2 2
3
Fe N(s) H (g)
2
? 2Fe(s) + NH
3
(g)
(1) 
1/2
c p
K K (RT) ? (2) 
1/2
c p
K K (RT)
?
?
(3) 
3
2
c p
K K (RT) ?
(4) 
c p
K K (RT) ?
Sol. 1
Fe
2
N(s) + 
3
2
H
2
(g)  2Fe(s) + NH
3
(g)
?n
g
 = 
3
1
2
? = 
1
2
?
p
c
K
K
 = 
g
n
(RT)
?
 = 
1
2
(RT)
?
K
c
 = 
p
1
2
K
(RT)
? = K
p
.
1
2
(RT)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 2
4. Consider the following reactions :
'A'
(C H )
7 14
ozonolysis
'B' + 'C'
'B' 
yellow ppt
silver mirror
Ag O
2
(I +NaOH)
2
'C' 
no yellow ppt
Anhydrous ZnCl
2 LiAlH
4
(I +NaOH)
2
& Conc. HCl
'D' 
gives white
turbidity
within 5
minutes
'A' is :
(1) (2) 
(3) (4) 
Sol. 1
Ozonolysis
CH —CHO
3
+
O
OH
Cl
(C)
LiAlH
4
ZnCl /HCl
2
(turbidity in 5 min)
(A) (B)
5. Arrange the following solutions in the decreasing order of pOH :
(A) 0.01 M HCl (B) 0.01 M NaOH
(C) 0.01 M CH
3
COONa (D) 0.01 M NaCl
(1) (A) > (C) > (D) > (B) (2) (B) > (D) > (C) > (A)
(3) (B) > (C) > (D) > (A) (4) (A) > (D) > (C) > (B)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 3
Sol. 4
(i) 10
–2
 M HCl ? [H
+
]  = 10
–2
 M ? pH
 
= 2
(ii) 10
–2
 M NaOH ? [OH
–
] = 10
–2
 M ? pOH = 2
(iii) 10
–2
 M CH
3
COO
–
Na
+
 ? [OH
+
] > 10
–7
 ? pOH < 7
(iv) 10
–2
 M NaCl ? Neutral pOH = 7
(i) > (iv) > (iii) > (ii)
6. The variation of equilibrium constant with temperature is given below :
Temperature Equilibrium Constant
T
1
 = 25
o
C K
1
 = 10
T
2
 = 100
o
C K
2
 = 100
The value of ?H
0 
, ?G
0
 
 
at T
1 
and ?G
0
 at T
2 
(in Kj mol
-1
) respectively, are close to
[use R = 8.314JK
-1
 mol
-1
]
(1) 28.4, -7.14 and -5.71 (2) 0.64, - 7.14 and -5.71
(3) 28.4, - 5.71 and -14.29 (4) 0.64, - 5.71 and -14.29
Sol. 3
In 
2
1
k
k
? ?
? ?
? ?
 = 
1 2
H 1 1
R T T
? ? ? ?
?
? ?
? ?
ln(10) = 
H 1 1
R 298 373
? ? ? ?
?
? ?
? ?
373 298 8.314 2.303
75
? ? ?
 = ?H° = 28.37 kJ mol
–1
1
T
G ? ?
 = –RT T
1
ln(K
1
) = –298R ln(10) = –5.71 kJ mol
–1
2
T
G ? ?
 = –RT T
2
 ln(K
2
) = –373R ln(100)
= –14.283 kJ/mol
7. Consider the following reactions
A P1; B  P2; C  P3; D  P4, ? ? ? ?
The order of the above reactions are a,b,c and d, respectively. The following graph is
obtained when log[rate] vs. log[conc.] are plotted :
[D]
[B]
log [Conc.]
log [rate]
[A]
[C]
Among the following the correct sequence for the order of the reactions is :
(1) c > a > b > d (2) d > a > b > c
(3) d > b > a > c (4) a > b > c > d
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 4
Sol. 3
A ? P1 B ? P2 C ? P3 D ? P4
Rate = K (conc.)
order
Slope = order
According graph
d > b > a > c order of slope
8. The major product obtained from the following reactions is :
O N
2
C C OCH
3 
2+
2
Hg /H+
H O
? ? ? ? ?
(1) 
O N
2
O
OCH
3
(2) 
O N
2
O
OH
(3) 
O N
2
O
OCH
3
(4) 
O N
2
O
OH
Sol. 3
O N
2
OCH
3
Hg /H
2+ +
H O
2
O N
2
OCH
3
CH —C
2
O
9. Which of the following compounds shows geometrical isomerism ?
(1) 2-methylpent-1-ene (2) 4-methylpent-2-ene
(3) 2-methylpent-2-ene (4) 4-methylpent-1-ene
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-1), Chemistry     Page | 5
Sol. 2
4-Methylpent-2-ene
Can show G.I.
10. The lanthanoid that does NOT shows +4 oxidation state is :
(1) Dy (2) Ce
(3) Tb (4) Eu
Sol. 4
Fact
11. The major products of the following reactions are :
(1) 
CH
3
CH
3
COOH
+ HCOOH
(2) 
CH
3
CH
3 O
+ CH CHO
3
(3) 
CH
3
CH
3 O
+ CH COOH
3
(4) 
CH
3
CH
3
CHO
+ HCHO
Sol. 1