JEE Main 2020 Answer Key Chemistry - Morning Shift (08-01-2020) JEE Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2020 Answer Key Chemistry - Morning Shift (08-01-2020) JEE Notes | EduRev

 Page 1


     
 
      
    
    
      
 
                       
 
 
 
PAGE # 1 
 
 
PART : CHEMISTRY 
 
SECTION – 1 : (Maximum Marks : 80) 
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Number of S –O bond in S2O8
2 –
 and number of S –S bond in Rhombic sulphur are respectively: 
 S2O8
2 –
 esa S –O ca/kksa dh la[;k rFkk jksgEchd lYQj esa S –S ca/kksa dh la[;k Øe’’’'’k% gS &   
 
 (1) 8, 8   (2) 6, 8   (3) 2, 4  (4) 4, 2 
Ans. (1) 
Sol. S2O8
2 – 
 
–
O –S –O –O –S –O
–
 
O O 
O O 
 
 S8 
 
S 
S S 
S 
S 
S 
S 
S 
 
 
2. Following vanderwaal forces are present in ethyl acetate liquid   
 (1)  H-bond, london forces. 
 (2) dipole-dipole interation, H-bond  
 (3) dipole –dipole interation, London forces  
 (4) H-bond, dipole-dipole interation, London forces  
 nzOk ,fFky ,flVsV esa mifLFkr ok.MjokWy cy gSa&  
 (1)  H-cU/k, yanu cy 
 (2) f}/kzqo&f}/kzqo vUr%fØ;k] H-cU/k 
 (3) f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
 (4) H-cU/k, f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
Ans. (3) 
Sol. Ethyl acetate is polar molecule so dipole-dipole interaction will be present there. 
 ,fFky ,flVsV /kzqoh; v.kq gS blfy, ;gk¡ f}/kzqo&f}/kzqo vUr%fØ;k ik;h tkrh gSA 
 
3. Given, for H-atom        
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 Select the correct options regarding this formula for Balmer series.  
 (A) n1 = 2   
 (B) Ionization energy of H atom can be calculated from above formula.  
 (C) ?maximum is for n2 = 3.  
 (D) If ? decreases then spectrum lines will converse.  
Page 2


     
 
      
    
    
      
 
                       
 
 
 
PAGE # 1 
 
 
PART : CHEMISTRY 
 
SECTION – 1 : (Maximum Marks : 80) 
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Number of S –O bond in S2O8
2 –
 and number of S –S bond in Rhombic sulphur are respectively: 
 S2O8
2 –
 esa S –O ca/kksa dh la[;k rFkk jksgEchd lYQj esa S –S ca/kksa dh la[;k Øe’’’'’k% gS &   
 
 (1) 8, 8   (2) 6, 8   (3) 2, 4  (4) 4, 2 
Ans. (1) 
Sol. S2O8
2 – 
 
–
O –S –O –O –S –O
–
 
O O 
O O 
 
 S8 
 
S 
S S 
S 
S 
S 
S 
S 
 
 
2. Following vanderwaal forces are present in ethyl acetate liquid   
 (1)  H-bond, london forces. 
 (2) dipole-dipole interation, H-bond  
 (3) dipole –dipole interation, London forces  
 (4) H-bond, dipole-dipole interation, London forces  
 nzOk ,fFky ,flVsV esa mifLFkr ok.MjokWy cy gSa&  
 (1)  H-cU/k, yanu cy 
 (2) f}/kzqo&f}/kzqo vUr%fØ;k] H-cU/k 
 (3) f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
 (4) H-cU/k, f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
Ans. (3) 
Sol. Ethyl acetate is polar molecule so dipole-dipole interaction will be present there. 
 ,fFky ,flVsV /kzqoh; v.kq gS blfy, ;gk¡ f}/kzqo&f}/kzqo vUr%fØ;k ik;h tkrh gSA 
 
3. Given, for H-atom        
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 Select the correct options regarding this formula for Balmer series.  
 (A) n1 = 2   
 (B) Ionization energy of H atom can be calculated from above formula.  
 (C) ?maximum is for n2 = 3.  
 (D) If ? decreases then spectrum lines will converse.  
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 2 
 
  
H-ijek.kq ds fy, fn;k x;k gS&  
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 ckej Js.kh ds fy, fn;s x;s lw+=k ds lUnHkZ esa lgh fodYi dk p;u dhft;sA 
 (A) n1 = 2   
 (B) H ijek.kq dh vk;uu ÅtkZ dh x.kuk mijksDr lw+=k }kjk fd tk ldrh gSA 
 (C) n2 = 3 ds fy, ?vf/kdre gS 
 (D) ;fn ? ?esa deh gksrh gS rc LisDVªeh js[kkvksa ds e/; vUrj de gks tkrk gS ¼ladqfpr gksrh gS½A 
 (1) A, B   (2) C, D   (3) A & C   (4) A, B, C & D  
Ans. (3)  
Sol. Theory based.  
  
4. Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol
–1
 respectively 
are:  
 fuEu /kkrqvksa Na, Mg, Al, Si dh izFke vk;uu ÅtkZvksa (KJ mol
–1 
esa) dk lgh Øe Øe'k% gS& 
 (1) 497, 737, 577, 786     (2) 497, 577, 737, 786  
 (3) 786, 739, 577, 497     (4) 739, 577, 786, 487  
Ans. (1)  
Sol. Correct order of ionisation energy will be : Na < Al < Mg < Si  
 vk;uu ÅtkZvksa dk lgh Øe gksxk % Na < Al < Mg < Si  
 
5. Select the correct stoichiometry and its ksp value according to given graphs.  
 fn;s x;s vkjs[k ds vuqlkj lgh jllehdj.kerh rFkk blds ksp dk lgh eku gS& 
 
 
[X] 
2×10
–3 
M 
10
–3 
M [Y] 
   
 (1) XY, Ksp = 2×10
–6     
(2) XY2, Ksp = 4×10
–9 
 
(3) X2Y, Ksp = 9×10
–9    
(4) XY2, Ksp = 1×10
–9  
Ans. (1)  
Sol. XY(s) ?? 
3 3
10
–
10 2
) aq ( Y ) aq ( X
? ?
?
?
?
  
 Ksp = [X
+
] [Y
–
]  
 or ;k, Ksp = 2 × 10
–3
 × 10
–3
  
 or ;k, Ksp = 2 × 10
–6
 
 
 
 
 
 
 
 
Page 3


     
 
      
    
    
      
 
                       
 
 
 
PAGE # 1 
 
 
PART : CHEMISTRY 
 
SECTION – 1 : (Maximum Marks : 80) 
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Number of S –O bond in S2O8
2 –
 and number of S –S bond in Rhombic sulphur are respectively: 
 S2O8
2 –
 esa S –O ca/kksa dh la[;k rFkk jksgEchd lYQj esa S –S ca/kksa dh la[;k Øe’’’'’k% gS &   
 
 (1) 8, 8   (2) 6, 8   (3) 2, 4  (4) 4, 2 
Ans. (1) 
Sol. S2O8
2 – 
 
–
O –S –O –O –S –O
–
 
O O 
O O 
 
 S8 
 
S 
S S 
S 
S 
S 
S 
S 
 
 
2. Following vanderwaal forces are present in ethyl acetate liquid   
 (1)  H-bond, london forces. 
 (2) dipole-dipole interation, H-bond  
 (3) dipole –dipole interation, London forces  
 (4) H-bond, dipole-dipole interation, London forces  
 nzOk ,fFky ,flVsV esa mifLFkr ok.MjokWy cy gSa&  
 (1)  H-cU/k, yanu cy 
 (2) f}/kzqo&f}/kzqo vUr%fØ;k] H-cU/k 
 (3) f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
 (4) H-cU/k, f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
Ans. (3) 
Sol. Ethyl acetate is polar molecule so dipole-dipole interaction will be present there. 
 ,fFky ,flVsV /kzqoh; v.kq gS blfy, ;gk¡ f}/kzqo&f}/kzqo vUr%fØ;k ik;h tkrh gSA 
 
3. Given, for H-atom        
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 Select the correct options regarding this formula for Balmer series.  
 (A) n1 = 2   
 (B) Ionization energy of H atom can be calculated from above formula.  
 (C) ?maximum is for n2 = 3.  
 (D) If ? decreases then spectrum lines will converse.  
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 2 
 
  
H-ijek.kq ds fy, fn;k x;k gS&  
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 ckej Js.kh ds fy, fn;s x;s lw+=k ds lUnHkZ esa lgh fodYi dk p;u dhft;sA 
 (A) n1 = 2   
 (B) H ijek.kq dh vk;uu ÅtkZ dh x.kuk mijksDr lw+=k }kjk fd tk ldrh gSA 
 (C) n2 = 3 ds fy, ?vf/kdre gS 
 (D) ;fn ? ?esa deh gksrh gS rc LisDVªeh js[kkvksa ds e/; vUrj de gks tkrk gS ¼ladqfpr gksrh gS½A 
 (1) A, B   (2) C, D   (3) A & C   (4) A, B, C & D  
Ans. (3)  
Sol. Theory based.  
  
4. Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol
–1
 respectively 
are:  
 fuEu /kkrqvksa Na, Mg, Al, Si dh izFke vk;uu ÅtkZvksa (KJ mol
–1 
esa) dk lgh Øe Øe'k% gS& 
 (1) 497, 737, 577, 786     (2) 497, 577, 737, 786  
 (3) 786, 739, 577, 497     (4) 739, 577, 786, 487  
Ans. (1)  
Sol. Correct order of ionisation energy will be : Na < Al < Mg < Si  
 vk;uu ÅtkZvksa dk lgh Øe gksxk % Na < Al < Mg < Si  
 
5. Select the correct stoichiometry and its ksp value according to given graphs.  
 fn;s x;s vkjs[k ds vuqlkj lgh jllehdj.kerh rFkk blds ksp dk lgh eku gS& 
 
 
[X] 
2×10
–3 
M 
10
–3 
M [Y] 
   
 (1) XY, Ksp = 2×10
–6     
(2) XY2, Ksp = 4×10
–9 
 
(3) X2Y, Ksp = 9×10
–9    
(4) XY2, Ksp = 1×10
–9  
Ans. (1)  
Sol. XY(s) ?? 
3 3
10
–
10 2
) aq ( Y ) aq ( X
? ?
?
?
?
  
 Ksp = [X
+
] [Y
–
]  
 or ;k, Ksp = 2 × 10
–3
 × 10
–3
  
 or ;k, Ksp = 2 × 10
–6
 
 
 
 
 
 
 
 
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 3 
 
 
6. According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is :  
 Fe(OH)3 sol ds fy, gkMhZ&'kqYt fu;e ds vuqlkj Å.kZu eku dk lgh Øe gS& 
 (1) K2CrO4 > K3[Fe(CN)6] > KNO3 > KBr = AlCl3  
 (2) K3[Fe(CN)6] > K2CrO4 > KNO3 = KBr = AlCl3  
 (3) K3[Fe(CN)6] < K2CrO4 < KNO3 = KBr = AlCl3  
 (4) KNO3 > KBr = K2CrO4 > AlCl3 = K3[Fe(CN)6]  
Ans. (3)  
Sol. According to hardy-schultz rule,  
 Coagulation value or flocculation value ? 
pow er n Coagulatio
1
  
 gkMhZ&'kqYt fu;e ds vuqlkj  
 laxq.ku eku vFkok Å.kZu eku ? 
{kerk Å.kZu
1
 
7. Which of the following complex exhibit facial meridional geometrical isomerism. 
 fuEu esa ls dkSulk ladqy Qydh;&js[kkaf'kr T;kfefr; leko;ork n'kkZrk gS& 
 (1) [Pt(NH3)Cl3]
–
    (2) [PtCl2(NH3)2]   
(3)  [Ni(CO)4]    (4) [Co(NO2)3 (NH3)3] 
Ans. (4) 
Sol. [Ma3b3] type complex shows facial and meridional isomerism  
 [Ma3b3] izdkj dk ladqy Qydh;&js[kkaf'kr T;kfefr; leko;ork n'kkZrk gSA 
 
8.  
 
 
V.P. 
Temperature  
X Y 
Z 
 
(A) Intermolecular force of attraction of X > Y. 
(B) Intermolecular force of attraction of X < Y. 
(C) Intermolecular force of attraction of Z < X. 
Select the correct option(s). 
(1) A and C  (2) A and B  (3) B only   (4) B and C 
(A) vkd"kZ.k dk vUrj vkf.od cy X > Y. 
(B) vkd"kZ.k dk vUrj vkf.od cy X < Y. 
(C) vkd"kZ.k dk vUrj vkf.od cy Z < X. 
lgh fodYi dk p;u dhft,& 
(1) A rFkk C  (2) A rFkk B  (3) dsoy B   (4) B rFkk C 
Ans. (3) 
Sol. At a particular temperature as intermolecular force of attraction increases vapour pressure decreases. 
 ,d fuf'pr rki ij vkUrj vkf.od vkd"kZ.k cy esa o`f) ds lkFk ok"i nkc esa deh vkrh gSA 
 
 
 
Page 4


     
 
      
    
    
      
 
                       
 
 
 
PAGE # 1 
 
 
PART : CHEMISTRY 
 
SECTION – 1 : (Maximum Marks : 80) 
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Number of S –O bond in S2O8
2 –
 and number of S –S bond in Rhombic sulphur are respectively: 
 S2O8
2 –
 esa S –O ca/kksa dh la[;k rFkk jksgEchd lYQj esa S –S ca/kksa dh la[;k Øe’’’'’k% gS &   
 
 (1) 8, 8   (2) 6, 8   (3) 2, 4  (4) 4, 2 
Ans. (1) 
Sol. S2O8
2 – 
 
–
O –S –O –O –S –O
–
 
O O 
O O 
 
 S8 
 
S 
S S 
S 
S 
S 
S 
S 
 
 
2. Following vanderwaal forces are present in ethyl acetate liquid   
 (1)  H-bond, london forces. 
 (2) dipole-dipole interation, H-bond  
 (3) dipole –dipole interation, London forces  
 (4) H-bond, dipole-dipole interation, London forces  
 nzOk ,fFky ,flVsV esa mifLFkr ok.MjokWy cy gSa&  
 (1)  H-cU/k, yanu cy 
 (2) f}/kzqo&f}/kzqo vUr%fØ;k] H-cU/k 
 (3) f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
 (4) H-cU/k, f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
Ans. (3) 
Sol. Ethyl acetate is polar molecule so dipole-dipole interaction will be present there. 
 ,fFky ,flVsV /kzqoh; v.kq gS blfy, ;gk¡ f}/kzqo&f}/kzqo vUr%fØ;k ik;h tkrh gSA 
 
3. Given, for H-atom        
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 Select the correct options regarding this formula for Balmer series.  
 (A) n1 = 2   
 (B) Ionization energy of H atom can be calculated from above formula.  
 (C) ?maximum is for n2 = 3.  
 (D) If ? decreases then spectrum lines will converse.  
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 2 
 
  
H-ijek.kq ds fy, fn;k x;k gS&  
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 ckej Js.kh ds fy, fn;s x;s lw+=k ds lUnHkZ esa lgh fodYi dk p;u dhft;sA 
 (A) n1 = 2   
 (B) H ijek.kq dh vk;uu ÅtkZ dh x.kuk mijksDr lw+=k }kjk fd tk ldrh gSA 
 (C) n2 = 3 ds fy, ?vf/kdre gS 
 (D) ;fn ? ?esa deh gksrh gS rc LisDVªeh js[kkvksa ds e/; vUrj de gks tkrk gS ¼ladqfpr gksrh gS½A 
 (1) A, B   (2) C, D   (3) A & C   (4) A, B, C & D  
Ans. (3)  
Sol. Theory based.  
  
4. Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol
–1
 respectively 
are:  
 fuEu /kkrqvksa Na, Mg, Al, Si dh izFke vk;uu ÅtkZvksa (KJ mol
–1 
esa) dk lgh Øe Øe'k% gS& 
 (1) 497, 737, 577, 786     (2) 497, 577, 737, 786  
 (3) 786, 739, 577, 497     (4) 739, 577, 786, 487  
Ans. (1)  
Sol. Correct order of ionisation energy will be : Na < Al < Mg < Si  
 vk;uu ÅtkZvksa dk lgh Øe gksxk % Na < Al < Mg < Si  
 
5. Select the correct stoichiometry and its ksp value according to given graphs.  
 fn;s x;s vkjs[k ds vuqlkj lgh jllehdj.kerh rFkk blds ksp dk lgh eku gS& 
 
 
[X] 
2×10
–3 
M 
10
–3 
M [Y] 
   
 (1) XY, Ksp = 2×10
–6     
(2) XY2, Ksp = 4×10
–9 
 
(3) X2Y, Ksp = 9×10
–9    
(4) XY2, Ksp = 1×10
–9  
Ans. (1)  
Sol. XY(s) ?? 
3 3
10
–
10 2
) aq ( Y ) aq ( X
? ?
?
?
?
  
 Ksp = [X
+
] [Y
–
]  
 or ;k, Ksp = 2 × 10
–3
 × 10
–3
  
 or ;k, Ksp = 2 × 10
–6
 
 
 
 
 
 
 
 
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 3 
 
 
6. According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is :  
 Fe(OH)3 sol ds fy, gkMhZ&'kqYt fu;e ds vuqlkj Å.kZu eku dk lgh Øe gS& 
 (1) K2CrO4 > K3[Fe(CN)6] > KNO3 > KBr = AlCl3  
 (2) K3[Fe(CN)6] > K2CrO4 > KNO3 = KBr = AlCl3  
 (3) K3[Fe(CN)6] < K2CrO4 < KNO3 = KBr = AlCl3  
 (4) KNO3 > KBr = K2CrO4 > AlCl3 = K3[Fe(CN)6]  
Ans. (3)  
Sol. According to hardy-schultz rule,  
 Coagulation value or flocculation value ? 
pow er n Coagulatio
1
  
 gkMhZ&'kqYt fu;e ds vuqlkj  
 laxq.ku eku vFkok Å.kZu eku ? 
{kerk Å.kZu
1
 
7. Which of the following complex exhibit facial meridional geometrical isomerism. 
 fuEu esa ls dkSulk ladqy Qydh;&js[kkaf'kr T;kfefr; leko;ork n'kkZrk gS& 
 (1) [Pt(NH3)Cl3]
–
    (2) [PtCl2(NH3)2]   
(3)  [Ni(CO)4]    (4) [Co(NO2)3 (NH3)3] 
Ans. (4) 
Sol. [Ma3b3] type complex shows facial and meridional isomerism  
 [Ma3b3] izdkj dk ladqy Qydh;&js[kkaf'kr T;kfefr; leko;ork n'kkZrk gSA 
 
8.  
 
 
V.P. 
Temperature  
X Y 
Z 
 
(A) Intermolecular force of attraction of X > Y. 
(B) Intermolecular force of attraction of X < Y. 
(C) Intermolecular force of attraction of Z < X. 
Select the correct option(s). 
(1) A and C  (2) A and B  (3) B only   (4) B and C 
(A) vkd"kZ.k dk vUrj vkf.od cy X > Y. 
(B) vkd"kZ.k dk vUrj vkf.od cy X < Y. 
(C) vkd"kZ.k dk vUrj vkf.od cy Z < X. 
lgh fodYi dk p;u dhft,& 
(1) A rFkk C  (2) A rFkk B  (3) dsoy B   (4) B rFkk C 
Ans. (3) 
Sol. At a particular temperature as intermolecular force of attraction increases vapour pressure decreases. 
 ,d fuf'pr rki ij vkUrj vkf.od vkd"kZ.k cy esa o`f) ds lkFk ok"i nkc esa deh vkrh gSA 
 
 
 
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 4 
 
 
9. Rate of a reaction increases by 10
6
 times when a reaction is carried out in presence of enzyme catalyst 
at same temperature. Determine change in activation energy.  
 vfHkfØ;k dks leku rki ij ,Utkbe dh mifLFkr esa lEiUu djkus ij vfHkfØ;k ds osx esa 10
6 
ls o`f) gks tkrh gS 
rc lfØ;.k ÅtkZ esa ifjorZu dk fu/kkZj.k dhft;sA 
(1) –6 × 2.303 RT (2) +6×2.303RT  (3) + 6RT  (4) –6RT 
Ans. (1) 
Sol. K = Ae
–E/RT
 ……… ……….( 1) 
 10
6
k = 
RT / E –
C
Ae ……… …… ….( 2) 
 
1 equation
2 equation
 ? 10
6
 = 
? ? RT / E E
C
e
?
 
 or ;k 
 6 ln 10 = ? ? RT / E E
C
? 
 or ;k 
 
? ?
6 303 . 2
RT
E E
C
? ?
?
 
 or ;k, E –EC = 2.303 × 6RT 
 or ;k, RT 6 303 . 2 – E E E
C a
? ? ? ? ? 
 
10. Gypsum on heating at 393K produces  
(1) dead burnt plaster    (2) Anhydrous CaSO4 
(3) O H
2
1
CaSO
2 4
?     (4) CaSO4 ? 5H2O 
393K ij ftIle dks xeZ djus ij izkIr gksrk gS&  
(1) e`r tyk gqvk IykLVj    (2) vUkknz CaSO4 
(3) O H
2
1
CaSO
2 4
?     (4) CaSO4 ? 5H2O ?
Ans. (3) 
Sol. Theory based. 
 
11. Among the following least 3
rd
 ionization energy is for  
 fuEu esa ls fdl ds fy, 3
rd 
vk;uu ÅtkZ U;wure gS& 
(1) Mn    (2) Co   (3) Fe   (4) Ni ?
Ans. (3) 
Sol. ? ?
2 6
26
s 4 d 3 Ar Fe ? 
 
12. Accurate measurement of concentration of NaOH can be performed by following titration: 
(1) NaOH in burette and oxalic acid in conical flask 
(2) NaOH in burette and concentrated H2SO4 in conical flask  
(3) NaOH in volumetric flask and concentrated H2SO4 in conical flask  
(4) Oxalic acid in burette and NaOH in conical flask 
NaOH dh lkUnzrk dk ;FkkZr~ ekiu fuEu esa ls fdl vuqekiu }kjk fd;k tk ldrk gSA 
(1) C;wjsV esa NaOH rFkk dksfudy ¶ykLd esa vkWDtsfyd vEy 
(2) C;wjsV esa NaOH rFkk dksfudy ¶ykLd esa lkUnz H2SO4  
(3) vk;rufefr; ¶ykLd esa NaOH rFkk dksfudy ¶ykLd esa H2SO4  
(4) C;wjsV esa vkWDtsfyd vEy rFkk dksfudy ¶ykLd esa NaOH ?
Ans. (4) 
Sol. Oxalic acid is a primary standard solution while H2SO4 is a secondary standard solution.  
 vkWDtsfyd vEy izkFkfed ekud foy;u gS tcfd H2SO4 f}rh;d ekud foy;u gSA 
 
 
 
Page 5


     
 
      
    
    
      
 
                       
 
 
 
PAGE # 1 
 
 
PART : CHEMISTRY 
 
SECTION – 1 : (Maximum Marks : 80) 
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Number of S –O bond in S2O8
2 –
 and number of S –S bond in Rhombic sulphur are respectively: 
 S2O8
2 –
 esa S –O ca/kksa dh la[;k rFkk jksgEchd lYQj esa S –S ca/kksa dh la[;k Øe’’’'’k% gS &   
 
 (1) 8, 8   (2) 6, 8   (3) 2, 4  (4) 4, 2 
Ans. (1) 
Sol. S2O8
2 – 
 
–
O –S –O –O –S –O
–
 
O O 
O O 
 
 S8 
 
S 
S S 
S 
S 
S 
S 
S 
 
 
2. Following vanderwaal forces are present in ethyl acetate liquid   
 (1)  H-bond, london forces. 
 (2) dipole-dipole interation, H-bond  
 (3) dipole –dipole interation, London forces  
 (4) H-bond, dipole-dipole interation, London forces  
 nzOk ,fFky ,flVsV esa mifLFkr ok.MjokWy cy gSa&  
 (1)  H-cU/k, yanu cy 
 (2) f}/kzqo&f}/kzqo vUr%fØ;k] H-cU/k 
 (3) f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
 (4) H-cU/k, f}/kzqo&f}/kzqo vUr%fØ;k] yanu cy  
Ans. (3) 
Sol. Ethyl acetate is polar molecule so dipole-dipole interaction will be present there. 
 ,fFky ,flVsV /kzqoh; v.kq gS blfy, ;gk¡ f}/kzqo&f}/kzqo vUr%fØ;k ik;h tkrh gSA 
 
3. Given, for H-atom        
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 Select the correct options regarding this formula for Balmer series.  
 (A) n1 = 2   
 (B) Ionization energy of H atom can be calculated from above formula.  
 (C) ?maximum is for n2 = 3.  
 (D) If ? decreases then spectrum lines will converse.  
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 2 
 
  
H-ijek.kq ds fy, fn;k x;k gS&  
 
?
?
?
?
?
?
?
?
? ? ?
2
2
2
1
H
n
1
n
1
R  
 ckej Js.kh ds fy, fn;s x;s lw+=k ds lUnHkZ esa lgh fodYi dk p;u dhft;sA 
 (A) n1 = 2   
 (B) H ijek.kq dh vk;uu ÅtkZ dh x.kuk mijksDr lw+=k }kjk fd tk ldrh gSA 
 (C) n2 = 3 ds fy, ?vf/kdre gS 
 (D) ;fn ? ?esa deh gksrh gS rc LisDVªeh js[kkvksa ds e/; vUrj de gks tkrk gS ¼ladqfpr gksrh gS½A 
 (1) A, B   (2) C, D   (3) A & C   (4) A, B, C & D  
Ans. (3)  
Sol. Theory based.  
  
4. Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol
–1
 respectively 
are:  
 fuEu /kkrqvksa Na, Mg, Al, Si dh izFke vk;uu ÅtkZvksa (KJ mol
–1 
esa) dk lgh Øe Øe'k% gS& 
 (1) 497, 737, 577, 786     (2) 497, 577, 737, 786  
 (3) 786, 739, 577, 497     (4) 739, 577, 786, 487  
Ans. (1)  
Sol. Correct order of ionisation energy will be : Na < Al < Mg < Si  
 vk;uu ÅtkZvksa dk lgh Øe gksxk % Na < Al < Mg < Si  
 
5. Select the correct stoichiometry and its ksp value according to given graphs.  
 fn;s x;s vkjs[k ds vuqlkj lgh jllehdj.kerh rFkk blds ksp dk lgh eku gS& 
 
 
[X] 
2×10
–3 
M 
10
–3 
M [Y] 
   
 (1) XY, Ksp = 2×10
–6     
(2) XY2, Ksp = 4×10
–9 
 
(3) X2Y, Ksp = 9×10
–9    
(4) XY2, Ksp = 1×10
–9  
Ans. (1)  
Sol. XY(s) ?? 
3 3
10
–
10 2
) aq ( Y ) aq ( X
? ?
?
?
?
  
 Ksp = [X
+
] [Y
–
]  
 or ;k, Ksp = 2 × 10
–3
 × 10
–3
  
 or ;k, Ksp = 2 × 10
–6
 
 
 
 
 
 
 
 
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 3 
 
 
6. According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is :  
 Fe(OH)3 sol ds fy, gkMhZ&'kqYt fu;e ds vuqlkj Å.kZu eku dk lgh Øe gS& 
 (1) K2CrO4 > K3[Fe(CN)6] > KNO3 > KBr = AlCl3  
 (2) K3[Fe(CN)6] > K2CrO4 > KNO3 = KBr = AlCl3  
 (3) K3[Fe(CN)6] < K2CrO4 < KNO3 = KBr = AlCl3  
 (4) KNO3 > KBr = K2CrO4 > AlCl3 = K3[Fe(CN)6]  
Ans. (3)  
Sol. According to hardy-schultz rule,  
 Coagulation value or flocculation value ? 
pow er n Coagulatio
1
  
 gkMhZ&'kqYt fu;e ds vuqlkj  
 laxq.ku eku vFkok Å.kZu eku ? 
{kerk Å.kZu
1
 
7. Which of the following complex exhibit facial meridional geometrical isomerism. 
 fuEu esa ls dkSulk ladqy Qydh;&js[kkaf'kr T;kfefr; leko;ork n'kkZrk gS& 
 (1) [Pt(NH3)Cl3]
–
    (2) [PtCl2(NH3)2]   
(3)  [Ni(CO)4]    (4) [Co(NO2)3 (NH3)3] 
Ans. (4) 
Sol. [Ma3b3] type complex shows facial and meridional isomerism  
 [Ma3b3] izdkj dk ladqy Qydh;&js[kkaf'kr T;kfefr; leko;ork n'kkZrk gSA 
 
8.  
 
 
V.P. 
Temperature  
X Y 
Z 
 
(A) Intermolecular force of attraction of X > Y. 
(B) Intermolecular force of attraction of X < Y. 
(C) Intermolecular force of attraction of Z < X. 
Select the correct option(s). 
(1) A and C  (2) A and B  (3) B only   (4) B and C 
(A) vkd"kZ.k dk vUrj vkf.od cy X > Y. 
(B) vkd"kZ.k dk vUrj vkf.od cy X < Y. 
(C) vkd"kZ.k dk vUrj vkf.od cy Z < X. 
lgh fodYi dk p;u dhft,& 
(1) A rFkk C  (2) A rFkk B  (3) dsoy B   (4) B rFkk C 
Ans. (3) 
Sol. At a particular temperature as intermolecular force of attraction increases vapour pressure decreases. 
 ,d fuf'pr rki ij vkUrj vkf.od vkd"kZ.k cy esa o`f) ds lkFk ok"i nkc esa deh vkrh gSA 
 
 
 
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 4 
 
 
9. Rate of a reaction increases by 10
6
 times when a reaction is carried out in presence of enzyme catalyst 
at same temperature. Determine change in activation energy.  
 vfHkfØ;k dks leku rki ij ,Utkbe dh mifLFkr esa lEiUu djkus ij vfHkfØ;k ds osx esa 10
6 
ls o`f) gks tkrh gS 
rc lfØ;.k ÅtkZ esa ifjorZu dk fu/kkZj.k dhft;sA 
(1) –6 × 2.303 RT (2) +6×2.303RT  (3) + 6RT  (4) –6RT 
Ans. (1) 
Sol. K = Ae
–E/RT
 ……… ……….( 1) 
 10
6
k = 
RT / E –
C
Ae ……… …… ….( 2) 
 
1 equation
2 equation
 ? 10
6
 = 
? ? RT / E E
C
e
?
 
 or ;k 
 6 ln 10 = ? ? RT / E E
C
? 
 or ;k 
 
? ?
6 303 . 2
RT
E E
C
? ?
?
 
 or ;k, E –EC = 2.303 × 6RT 
 or ;k, RT 6 303 . 2 – E E E
C a
? ? ? ? ? 
 
10. Gypsum on heating at 393K produces  
(1) dead burnt plaster    (2) Anhydrous CaSO4 
(3) O H
2
1
CaSO
2 4
?     (4) CaSO4 ? 5H2O 
393K ij ftIle dks xeZ djus ij izkIr gksrk gS&  
(1) e`r tyk gqvk IykLVj    (2) vUkknz CaSO4 
(3) O H
2
1
CaSO
2 4
?     (4) CaSO4 ? 5H2O ?
Ans. (3) 
Sol. Theory based. 
 
11. Among the following least 3
rd
 ionization energy is for  
 fuEu esa ls fdl ds fy, 3
rd 
vk;uu ÅtkZ U;wure gS& 
(1) Mn    (2) Co   (3) Fe   (4) Ni ?
Ans. (3) 
Sol. ? ?
2 6
26
s 4 d 3 Ar Fe ? 
 
12. Accurate measurement of concentration of NaOH can be performed by following titration: 
(1) NaOH in burette and oxalic acid in conical flask 
(2) NaOH in burette and concentrated H2SO4 in conical flask  
(3) NaOH in volumetric flask and concentrated H2SO4 in conical flask  
(4) Oxalic acid in burette and NaOH in conical flask 
NaOH dh lkUnzrk dk ;FkkZr~ ekiu fuEu esa ls fdl vuqekiu }kjk fd;k tk ldrk gSA 
(1) C;wjsV esa NaOH rFkk dksfudy ¶ykLd esa vkWDtsfyd vEy 
(2) C;wjsV esa NaOH rFkk dksfudy ¶ykLd esa lkUnz H2SO4  
(3) vk;rufefr; ¶ykLd esa NaOH rFkk dksfudy ¶ykLd esa H2SO4  
(4) C;wjsV esa vkWDtsfyd vEy rFkk dksfudy ¶ykLd esa NaOH ?
Ans. (4) 
Sol. Oxalic acid is a primary standard solution while H2SO4 is a secondary standard solution.  
 vkWDtsfyd vEy izkFkfed ekud foy;u gS tcfd H2SO4 f}rh;d ekud foy;u gSA 
 
 
 
     
 
      
    
    
      
 
                       
 
 
 
PAGE # 5 
 
 
13. Arrange the following compounds in order of dehydrohalogenation (E1) reaction. 
 fuEu ;kSfxdks dks fMgkbMªksgSyksftfudj.k (E1) vfHkfØ;k ds Øe esa O;ofLFkr dhft,A 
 (A) 
 
Cl 
    
(B) 
 
Cl 
 
 
 
(C) 
 
Cl 
 
    
(D) 
 
Cl 
 
 
(1) C > B > D > A (2) C > D > B > A (3) B > C > D > A (4) A > B > C > D 
Ans. (2) 
Sol. E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster the 
E1 reaction. 
 E1 vfHkfØ;k dkcZ/kuk;u ds fuekZ.k ls lEiUu gksrh gS bl izdkj dkcZu/kuk;u dk vf/kd LFkkf;RkRo] E1 vfHkfØ;k dks 
rhoz djrk gSA 
 
14. 
 
CN 
 
  
peroxide
?
? ? ? ? ? ?
 [A]      
 [A]  +  
 
? ? ? [B] 
 Product A and B are respectively : (mRikn A rFkk B Øe'k% gS :) 
 (1) 
 
CN 
 
 
and (rFkk) 
 
CN 
 
  
(2)  
 
CN 
 
 and (rFkk) 
 
CN 
 
 
 (3) 
 
CN 
 
and
  
(rFkk) 
 
CN 
 
 
  
(4) 
 
CN 
 
and (rFkk) 
 
CN 
 
 
Ans. (3) 
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