JEE Main 2021 Answer Key Chemistry - Evening Shift (16-03-2021) JEE Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2021 Answer Key Chemistry - Evening Shift (16-03-2021) JEE Notes | EduRev

 Page 1


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
Page 2


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
Page 3


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
5. Ammonolysis of Alkylhalides followed by the treatment with NaOH solution can be used to 
prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is: 
 (1) to remove basic impurities 
 (2) to activate NH
3
used in the reaction 
 (3) to increase the reactivity of alkyl halide 
 (4) to remove acidic impurities 
Ans. (4) 
Sol.  
 
 
 
 
R–X 
NH
3
 
–HX 
R-NH
2
 
R– X 
–HX 
R
2
NH 
–HX R–X 
R–X 
R
3
N 
R
4
NX 
 
 
During the reaction HX (acid) is form 
 Hence, we use NaOH to remove this acidic impurities 
 
6. Arrange the following metal complex/compounds in the increasing order of spin only magnetic 
moment. Presume all the three, high spin system. 
 (Atomic numbers Ce = 58, Gd = 64 and Eu = 63) 
 (a) (NH
4
)
2
[Ce(NO
3
)
6
]  (b)Gd(NO
3
)
3
 and (c)Eu(NO
3
)
3 
 Answer is: 
 (1)(a)<(c)<(b)    (2)(a)<(b)<(c) 
 (3)(c)<(a)<(b)    (4)(b)<(a)<(c) 
Ans. (1) 
Sol. (NH
4
)
2
 [Ce(NO
3
)
6
]  (n = 0)   ? ? ? ?= 0 B.M 
 Eu (NO
3
)
3
   (n = 6)   ?  ? = 6.93 B.M 
 Gd(NO
3
)
3
   (n = 7)   ?  ? = 7.94 B.M 
 
Page 4


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
5. Ammonolysis of Alkylhalides followed by the treatment with NaOH solution can be used to 
prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is: 
 (1) to remove basic impurities 
 (2) to activate NH
3
used in the reaction 
 (3) to increase the reactivity of alkyl halide 
 (4) to remove acidic impurities 
Ans. (4) 
Sol.  
 
 
 
 
R–X 
NH
3
 
–HX 
R-NH
2
 
R– X 
–HX 
R
2
NH 
–HX R–X 
R–X 
R
3
N 
R
4
NX 
 
 
During the reaction HX (acid) is form 
 Hence, we use NaOH to remove this acidic impurities 
 
6. Arrange the following metal complex/compounds in the increasing order of spin only magnetic 
moment. Presume all the three, high spin system. 
 (Atomic numbers Ce = 58, Gd = 64 and Eu = 63) 
 (a) (NH
4
)
2
[Ce(NO
3
)
6
]  (b)Gd(NO
3
)
3
 and (c)Eu(NO
3
)
3 
 Answer is: 
 (1)(a)<(c)<(b)    (2)(a)<(b)<(c) 
 (3)(c)<(a)<(b)    (4)(b)<(a)<(c) 
Ans. (1) 
Sol. (NH
4
)
2
 [Ce(NO
3
)
6
]  (n = 0)   ? ? ? ?= 0 B.M 
 Eu (NO
3
)
3
   (n = 6)   ?  ? = 6.93 B.M 
 Gd(NO
3
)
3
   (n = 7)   ?  ? = 7.94 B.M 
 
 
 
7. Identify the elements X and Y using the ionisation energy values given below: 
 Ionization energy (kJ/mol) 
  1
st
 2
nd
 
 X 495 4563 
 Y 731 1450   
 (1) X = F;  Y = Mg   (2) X = Mg; Y = F 
 (3) X = Na;  Y = Mg   (4) X = Mg;  Y = Na 
Ans. (3) 
Sol. 2
nd
 I. E of Alkali metals is higher than their respective period. 
 
 
8. The INCORRECT statements below regarding colloidal solutions is: 
 (1) A colloidal solution shows colligative properties. 
 (2) An ordinary filter paper can stop the flow of colloidal particles. 
 (3) A colloidal solution shows Brownian motion of colloidal particles. 
 (4) The flocculating power of Al
3+
 is more than that of Na
+
. 
Ans. (2) 
Sol. Colloidal solutions can pass through ordinary filter paper but cannot pass through special filter 
 collodial solution coated paper. 
 
9. The characteristics of elements X, Y and Z with atomic numbers, respectively, 33, 53 and 83 
 are: 
 (1) X and Z are non-metals and Y is a metalloid. 
 (2) X and Y are metalloids and Z is a metal 
 (3) X, Y and Z are metals. 
 (4) X is a metalloid, Y is a non-metal and Z is a metal. 
Ans. (4) 
Sol. Atomic No.  Element 
 (1) 33     ? ? ? ? As (Metalloid) 
 (2) 53    ? ? ? ? I (Non metal) 
 (3) 83    ? ? ? ? Bi (Metal) 
 
10. The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H
3
PO
3
 solution and 
100 mL of 2 M H
3
PO
2
 solution, respectively, are: 
 (1) 100 mL and 50 mL   (2) 50 mL and 50 mL 
 (3) 100 mL and 100 mL   (4) 100 mL and 200 mL 
Ans. (4) 
Sol. (1) 2NaOH  +  H
3
PO
3
  ??  Na
2
HPO
3  
+  2H
2
O
 
  
100m mole  50m mole 
  100m mole = M × V
ml
 
  100m mole = 1 × V
ml 
  
V
ml
 = 100 ml 
 (2) NaOH  + H
3
PO
2
  ??  NaH
2
PO
2
 + H
2
O 
  200m mole  200m mole 
  200m mole = M × V
ml
 
  V
ml
 = 200 ml 
 
 
Page 5


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
5. Ammonolysis of Alkylhalides followed by the treatment with NaOH solution can be used to 
prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is: 
 (1) to remove basic impurities 
 (2) to activate NH
3
used in the reaction 
 (3) to increase the reactivity of alkyl halide 
 (4) to remove acidic impurities 
Ans. (4) 
Sol.  
 
 
 
 
R–X 
NH
3
 
–HX 
R-NH
2
 
R– X 
–HX 
R
2
NH 
–HX R–X 
R–X 
R
3
N 
R
4
NX 
 
 
During the reaction HX (acid) is form 
 Hence, we use NaOH to remove this acidic impurities 
 
6. Arrange the following metal complex/compounds in the increasing order of spin only magnetic 
moment. Presume all the three, high spin system. 
 (Atomic numbers Ce = 58, Gd = 64 and Eu = 63) 
 (a) (NH
4
)
2
[Ce(NO
3
)
6
]  (b)Gd(NO
3
)
3
 and (c)Eu(NO
3
)
3 
 Answer is: 
 (1)(a)<(c)<(b)    (2)(a)<(b)<(c) 
 (3)(c)<(a)<(b)    (4)(b)<(a)<(c) 
Ans. (1) 
Sol. (NH
4
)
2
 [Ce(NO
3
)
6
]  (n = 0)   ? ? ? ?= 0 B.M 
 Eu (NO
3
)
3
   (n = 6)   ?  ? = 6.93 B.M 
 Gd(NO
3
)
3
   (n = 7)   ?  ? = 7.94 B.M 
 
 
 
7. Identify the elements X and Y using the ionisation energy values given below: 
 Ionization energy (kJ/mol) 
  1
st
 2
nd
 
 X 495 4563 
 Y 731 1450   
 (1) X = F;  Y = Mg   (2) X = Mg; Y = F 
 (3) X = Na;  Y = Mg   (4) X = Mg;  Y = Na 
Ans. (3) 
Sol. 2
nd
 I. E of Alkali metals is higher than their respective period. 
 
 
8. The INCORRECT statements below regarding colloidal solutions is: 
 (1) A colloidal solution shows colligative properties. 
 (2) An ordinary filter paper can stop the flow of colloidal particles. 
 (3) A colloidal solution shows Brownian motion of colloidal particles. 
 (4) The flocculating power of Al
3+
 is more than that of Na
+
. 
Ans. (2) 
Sol. Colloidal solutions can pass through ordinary filter paper but cannot pass through special filter 
 collodial solution coated paper. 
 
9. The characteristics of elements X, Y and Z with atomic numbers, respectively, 33, 53 and 83 
 are: 
 (1) X and Z are non-metals and Y is a metalloid. 
 (2) X and Y are metalloids and Z is a metal 
 (3) X, Y and Z are metals. 
 (4) X is a metalloid, Y is a non-metal and Z is a metal. 
Ans. (4) 
Sol. Atomic No.  Element 
 (1) 33     ? ? ? ? As (Metalloid) 
 (2) 53    ? ? ? ? I (Non metal) 
 (3) 83    ? ? ? ? Bi (Metal) 
 
10. The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H
3
PO
3
 solution and 
100 mL of 2 M H
3
PO
2
 solution, respectively, are: 
 (1) 100 mL and 50 mL   (2) 50 mL and 50 mL 
 (3) 100 mL and 100 mL   (4) 100 mL and 200 mL 
Ans. (4) 
Sol. (1) 2NaOH  +  H
3
PO
3
  ??  Na
2
HPO
3  
+  2H
2
O
 
  
100m mole  50m mole 
  100m mole = M × V
ml
 
  100m mole = 1 × V
ml 
  
V
ml
 = 100 ml 
 (2) NaOH  + H
3
PO
2
  ??  NaH
2
PO
2
 + H
2
O 
  200m mole  200m mole 
  200m mole = M × V
ml
 
  V
ml
 = 200 ml 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
11. Which of the following reduction reaction CANNOT be carried out with coke?  
 (1) Fe
2
O
3
? Fe     (2)ZnO ? Zn 
 (3) Al
2
O
3
? Al     (4) Cu
2
O ? Cu 
Ans. (3) 
Sol. Al is extracted by electrolytic reduction of Al
2
O
3
 
 
12. An unsaturated hydrocarbon X on ozonolysis gives A. Compound A when warmed with 
ammonical silver nitrate forms a bright silver mirror along the sides of the test tube. The 
unsaturated  hydrocarbon X is: 
 (1) CH
3
–C ?C–CH
3
    (2)
3 3
3 3
CH C C CH
| |
CH CH
? ? ? 
 (3) HC ?C–CH
2
–CH
3
    (4) 
 
CH 3—C= 
CH 3 
 
Ans. (3) 
Sol.  
 
 
 
 
CH
3
CH
2
?CH 
(i) O
3
 
(ii) H
2
O 
CH
3
CH
2
COOH + H–C–OH 
O 
[Ag(NH
3
)
2
]
+
 
Tollen's 
reagent 
 
HCOOH CO
2
 + H
2
O+2Ag 
 
 
13. Statement-I: Sodium hydride can be used as an oxidising agent. 
 Statement-II: The lone pair of electrons on nitrogen in pyridine makes it basic: 
 Choose the CORRECT answer from the options given below: 
 (1) Statement I is true but statement II is false 
 (2) Both statement I and statement II are false 
 (3) Both statement I and statement II are true 
 (4) Statement I is false but statement II is true 
Ans. (4) 
Sol. ? ?NaH is used as reducing agent. 
? ? ? ? ?The ?p on nitrogen in pyridine makes it basic 
 
 
N 
 
 
14. Which of the following polymer is used in the manufacture of wood laminates? 
 (1) Melamine formaldehyde resin  (2)cis-poly isoprene 
 (3) Phenol and formaldehyde resin  (4) Urea formaldehyde resin 
Ans. (1) 
Sol. Melamine formaldehyde resin is used in the manufacture of wood laminates. 
 
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

mock tests for examination

,

study material

,

JEE Main 2021 Answer Key Chemistry - Evening Shift (16-03-2021) JEE Notes | EduRev

,

Summary

,

ppt

,

Sample Paper

,

Previous Year Questions with Solutions

,

practice quizzes

,

Extra Questions

,

pdf

,

JEE Main 2021 Answer Key Chemistry - Evening Shift (16-03-2021) JEE Notes | EduRev

,

Objective type Questions

,

Free

,

video lectures

,

MCQs

,

Viva Questions

,

shortcuts and tricks

,

Important questions

,

JEE Main 2021 Answer Key Chemistry - Evening Shift (16-03-2021) JEE Notes | EduRev

,

past year papers

,

Exam

,

Semester Notes

;