JEE Main 2020 Chemistry January 9 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 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.  5 g of Zn reacts with 
 (I) Excess of NaOH  (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
5 g Zn fuEu ds lkFk fØ;k djrk gSA 
 (I) NaOH dk vkf/kD; (II) ruq HCl, rc (I) rFkk (II) es mRlftZr H2 xSl dk vk;ru vuqikr gS& 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
Ans.  (3) 
Sol. Zn + 2NaOH ? ? Na2ZnO2 + H2 
 Zn + 2HCl ? ? ZnCl2 + H2 
 According to stoichiometry in both the reactions, equal number of moles of H2 are evolved. 
 nksuks vfHkfØ;kvksa es jllehdj.kfefr ds vuqlkj H2 ds eksyks dh leku la[;k mRlftZr gksrh gSA 
 
2. Given Ksp for Cr(OH)3 is 6 × 10
–31
 then determine [OH
–
]. 
 (Neglect the contribution of OH
–
 ions from H2O) 
 fn;k gS Cr(OH)3 ds fy;s Ksp, 6 × 10
–31 
gS rc [OH
–
] dk fu/kkZj.k dhft;sA 
 (H2O ls OH
–
 vk;uksa dk ;ksxnku ux.; ekusaA) 
 (1) (18 × 10
–31
)
1/4 
M  (2) (18 × 10
–31
)
1/2
 M  (3) (6 × 10
–31
)
1/4
 M  (4) 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? M 
Ans.  (1) 
Sol. Cr(OH)3 ? ? ?Cr
+3
 + 3OH
– 
   s 3s 
  Ksp = s . (3s)
3
  
 ? 6 × 10
–31
 = 27.s
4
 
? ? ? s = 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? 
[OH
–
] = 3s 
 = 3 × 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? = (18 × 10
–31
)
1/4
 M 
 
3. Select the correct statements among the followings 
 (A) LiCl does not dissolve in pyridine 
 (B) Li does not react ethyne to form ethynide. 
 (C) Li and Mg react slowly with water. 
(D) Among alkali metals Li has highest hydration tendency. 
fuEu es lgh dFkuksa dk p;u dhft;s& 
 (A) LiCl fifjMhu es ugha ?kqyrk gSA 
 (B) Li ,FkkbZu ds lkFk fØ;k djds ,FkkbZukbM ugha cukrk gSA 
 (C) Li rFkk Mg ty ds lkFk /khjs&/khjs fØ;k djrs gSA 
(D) {kkjh; e`nk /kkrqvks esa Li mPpre ty;kstu izo`fÙk j[krk gSA 
(1) B, C, D  (2) A, B, C, D  (3) A, B, C  (4) C, D 
Ans.  (1) 
Sol.  Theory based ¼lS}kfUrd½ 
Page 2


      
 
    
    
    
     
 
                       
 
  
 
 
 
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.  5 g of Zn reacts with 
 (I) Excess of NaOH  (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
5 g Zn fuEu ds lkFk fØ;k djrk gSA 
 (I) NaOH dk vkf/kD; (II) ruq HCl, rc (I) rFkk (II) es mRlftZr H2 xSl dk vk;ru vuqikr gS& 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
Ans.  (3) 
Sol. Zn + 2NaOH ? ? Na2ZnO2 + H2 
 Zn + 2HCl ? ? ZnCl2 + H2 
 According to stoichiometry in both the reactions, equal number of moles of H2 are evolved. 
 nksuks vfHkfØ;kvksa es jllehdj.kfefr ds vuqlkj H2 ds eksyks dh leku la[;k mRlftZr gksrh gSA 
 
2. Given Ksp for Cr(OH)3 is 6 × 10
–31
 then determine [OH
–
]. 
 (Neglect the contribution of OH
–
 ions from H2O) 
 fn;k gS Cr(OH)3 ds fy;s Ksp, 6 × 10
–31 
gS rc [OH
–
] dk fu/kkZj.k dhft;sA 
 (H2O ls OH
–
 vk;uksa dk ;ksxnku ux.; ekusaA) 
 (1) (18 × 10
–31
)
1/4 
M  (2) (18 × 10
–31
)
1/2
 M  (3) (6 × 10
–31
)
1/4
 M  (4) 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? M 
Ans.  (1) 
Sol. Cr(OH)3 ? ? ?Cr
+3
 + 3OH
– 
   s 3s 
  Ksp = s . (3s)
3
  
 ? 6 × 10
–31
 = 27.s
4
 
? ? ? s = 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? 
[OH
–
] = 3s 
 = 3 × 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? = (18 × 10
–31
)
1/4
 M 
 
3. Select the correct statements among the followings 
 (A) LiCl does not dissolve in pyridine 
 (B) Li does not react ethyne to form ethynide. 
 (C) Li and Mg react slowly with water. 
(D) Among alkali metals Li has highest hydration tendency. 
fuEu es lgh dFkuksa dk p;u dhft;s& 
 (A) LiCl fifjMhu es ugha ?kqyrk gSA 
 (B) Li ,FkkbZu ds lkFk fØ;k djds ,FkkbZukbM ugha cukrk gSA 
 (C) Li rFkk Mg ty ds lkFk /khjs&/khjs fØ;k djrs gSA 
(D) {kkjh; e`nk /kkrqvks esa Li mPpre ty;kstu izo`fÙk j[krk gSA 
(1) B, C, D  (2) A, B, C, D  (3) A, B, C  (4) C, D 
Ans.  (1) 
Sol.  Theory based ¼lS}kfUrd½ 
      
 
    
    
    
     
 
                       
 
  
 
 
4. Given an element having following ionisation enthalpies IE1 = 496 
mol
kJ
 and IE2 = 4562 
mol
kJ
 one mole 
hydroxide of this element is treated separated with HCl and H2SO4 respectively. Moles of HCl and 
H2SO4 reacted respectively is 
 fuEu vk;uu ,UFkSYih j[kus okyk ,d rRo fn;k x;k gS IE1 = 496 
mol
kJ
 rFkk IE2 = 4562 
mol
kJ
 bl rRo ds ,d 
eksy gkbMªkWDlkbM dks i`Fkd :i ls Øe'k% HCl rFkk H2SO4 ds lkFk mipkfjr fd;k tkrk gSA vfHkÑr HCl rFkk 
H2SO4 ds eksy Øe'k% gS& 
 (1) 1, 0.5   (2) 0.5, 1  (3) 2, 0.5  (4) 0.5, 2 
Ans.  (1) 
Sol.  According to the given data of I.E, This element must belong to group 1 and thus is monovalent & form 
hydroxide of the type M(OH). 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
Sol.  I.E. ds fn;s x;s vkadM+ks ds vuqlkj ;g rRo oxZ 1 ls lEcfU/kr gksuk pkfg;s blfy;s ;g ,dy la;ksth gSS rFkk 
M(OH) izdkj dk gkbMªkWDlkbM cukrk gSA 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
 
5. Reactant A represented by square is in equilibrium with product B represented by circles. Then value of 
equilibrium constant is 
 oxZ }kjk iznf'kZr vfHkdkjd A o`Ùkksa }kjk iznf'kZr mRikn B ds lkFk lkE; esa gSA rc lkE; fu;rkad dk eku gSA 
     
 
 
 (1) 1   (2) 2   (3) 3   (4) 4 
Ans.  (2) 
 
6. Given following complexes 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 Correct order of spin only magnetic moment for the above complexes  is. 
 fuEu ladqy fn;s x;s gS% 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 mijksDr ladqyksa ds fy, dsoy pØ.k pqEcdh; vk?kw.kZ dk lgh Øe gS& 
(1) (II) > (III) > (IV) > I     (2) (II) > (IV) > (III) > (I) 
(3) (I) > (IV) > (III) > (II)    (4) (II) > (I) > (IV) > (III) 
Ans. (1) 
Sol. As, ?s = ) 2 n ( n ? 
 Complex (I) ? Fe
+2
 ? S.F.L ? 
2 , 2 , 2
g 2
t eg
0,0
 ? ?s = 0 
Complex (II) ? Cr
+2
 ? W.F.L ?  
1 , 1 , 1
g 2
t eg
1,0
 ? ?s = 24 B.M. 
Complex (III) ? Co
+2
 ? W.F.L ? 
2 , 2
g
e 
1 , 1 , 1
g 2
t ? ?s = 15 B.M. 
Complex (IV) ? Fe
+3
 ? S.F.L ? 
1 , 2 , 2
g 2
t eg
0,0
 ? ?s = 3 B.M. 
Page 3


      
 
    
    
    
     
 
                       
 
  
 
 
 
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.  5 g of Zn reacts with 
 (I) Excess of NaOH  (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
5 g Zn fuEu ds lkFk fØ;k djrk gSA 
 (I) NaOH dk vkf/kD; (II) ruq HCl, rc (I) rFkk (II) es mRlftZr H2 xSl dk vk;ru vuqikr gS& 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
Ans.  (3) 
Sol. Zn + 2NaOH ? ? Na2ZnO2 + H2 
 Zn + 2HCl ? ? ZnCl2 + H2 
 According to stoichiometry in both the reactions, equal number of moles of H2 are evolved. 
 nksuks vfHkfØ;kvksa es jllehdj.kfefr ds vuqlkj H2 ds eksyks dh leku la[;k mRlftZr gksrh gSA 
 
2. Given Ksp for Cr(OH)3 is 6 × 10
–31
 then determine [OH
–
]. 
 (Neglect the contribution of OH
–
 ions from H2O) 
 fn;k gS Cr(OH)3 ds fy;s Ksp, 6 × 10
–31 
gS rc [OH
–
] dk fu/kkZj.k dhft;sA 
 (H2O ls OH
–
 vk;uksa dk ;ksxnku ux.; ekusaA) 
 (1) (18 × 10
–31
)
1/4 
M  (2) (18 × 10
–31
)
1/2
 M  (3) (6 × 10
–31
)
1/4
 M  (4) 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? M 
Ans.  (1) 
Sol. Cr(OH)3 ? ? ?Cr
+3
 + 3OH
– 
   s 3s 
  Ksp = s . (3s)
3
  
 ? 6 × 10
–31
 = 27.s
4
 
? ? ? s = 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? 
[OH
–
] = 3s 
 = 3 × 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? = (18 × 10
–31
)
1/4
 M 
 
3. Select the correct statements among the followings 
 (A) LiCl does not dissolve in pyridine 
 (B) Li does not react ethyne to form ethynide. 
 (C) Li and Mg react slowly with water. 
(D) Among alkali metals Li has highest hydration tendency. 
fuEu es lgh dFkuksa dk p;u dhft;s& 
 (A) LiCl fifjMhu es ugha ?kqyrk gSA 
 (B) Li ,FkkbZu ds lkFk fØ;k djds ,FkkbZukbM ugha cukrk gSA 
 (C) Li rFkk Mg ty ds lkFk /khjs&/khjs fØ;k djrs gSA 
(D) {kkjh; e`nk /kkrqvks esa Li mPpre ty;kstu izo`fÙk j[krk gSA 
(1) B, C, D  (2) A, B, C, D  (3) A, B, C  (4) C, D 
Ans.  (1) 
Sol.  Theory based ¼lS}kfUrd½ 
      
 
    
    
    
     
 
                       
 
  
 
 
4. Given an element having following ionisation enthalpies IE1 = 496 
mol
kJ
 and IE2 = 4562 
mol
kJ
 one mole 
hydroxide of this element is treated separated with HCl and H2SO4 respectively. Moles of HCl and 
H2SO4 reacted respectively is 
 fuEu vk;uu ,UFkSYih j[kus okyk ,d rRo fn;k x;k gS IE1 = 496 
mol
kJ
 rFkk IE2 = 4562 
mol
kJ
 bl rRo ds ,d 
eksy gkbMªkWDlkbM dks i`Fkd :i ls Øe'k% HCl rFkk H2SO4 ds lkFk mipkfjr fd;k tkrk gSA vfHkÑr HCl rFkk 
H2SO4 ds eksy Øe'k% gS& 
 (1) 1, 0.5   (2) 0.5, 1  (3) 2, 0.5  (4) 0.5, 2 
Ans.  (1) 
Sol.  According to the given data of I.E, This element must belong to group 1 and thus is monovalent & form 
hydroxide of the type M(OH). 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
Sol.  I.E. ds fn;s x;s vkadM+ks ds vuqlkj ;g rRo oxZ 1 ls lEcfU/kr gksuk pkfg;s blfy;s ;g ,dy la;ksth gSS rFkk 
M(OH) izdkj dk gkbMªkWDlkbM cukrk gSA 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
 
5. Reactant A represented by square is in equilibrium with product B represented by circles. Then value of 
equilibrium constant is 
 oxZ }kjk iznf'kZr vfHkdkjd A o`Ùkksa }kjk iznf'kZr mRikn B ds lkFk lkE; esa gSA rc lkE; fu;rkad dk eku gSA 
     
 
 
 (1) 1   (2) 2   (3) 3   (4) 4 
Ans.  (2) 
 
6. Given following complexes 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 Correct order of spin only magnetic moment for the above complexes  is. 
 fuEu ladqy fn;s x;s gS% 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 mijksDr ladqyksa ds fy, dsoy pØ.k pqEcdh; vk?kw.kZ dk lgh Øe gS& 
(1) (II) > (III) > (IV) > I     (2) (II) > (IV) > (III) > (I) 
(3) (I) > (IV) > (III) > (II)    (4) (II) > (I) > (IV) > (III) 
Ans. (1) 
Sol. As, ?s = ) 2 n ( n ? 
 Complex (I) ? Fe
+2
 ? S.F.L ? 
2 , 2 , 2
g 2
t eg
0,0
 ? ?s = 0 
Complex (II) ? Cr
+2
 ? W.F.L ?  
1 , 1 , 1
g 2
t eg
1,0
 ? ?s = 24 B.M. 
Complex (III) ? Co
+2
 ? W.F.L ? 
2 , 2
g
e 
1 , 1 , 1
g 2
t ? ?s = 15 B.M. 
Complex (IV) ? Fe
+3
 ? S.F.L ? 
1 , 2 , 2
g 2
t eg
0,0
 ? ?s = 3 B.M. 
      
 
    
    
    
     
 
                       
 
  
 
 
 
7. Select the correct option : 
 (1) Entropy is function of temperature and also entropy change is function of temperature. 
 (2) Entropy is a function of temperature & entropy change is not a function of temperature. 
 (3) Entropy is not a function of temperature & entropy change is a function of temperature. 
 (4) Both entropy & entropy change are not a function of temperature. 
 lgh fodYi dk p;u dhft,A 
 (1) ,UVªkWih rki dk Qyu gS rFkk ,UVªkWih ifjorZu Hkh rki dk Qyu gSA 
 (2) ,UVªkWih rki dk Qyu gS rFkk ,UVªkWih ifjorZu rki dk Qyu ugha gSA 
 (3) ,UVªkWih rki dk Qyu ugh gSa rFkk ,UVªkWih ifjorZu rki dk Qyu gSA 
 (4) ,UVªkWih rFkk ,UVªkWih ifjorZu nksuksa rki dk Qyu ugha gSA 
Ans. (1) 
Sol. ?S = 
?
T
dq
 
 ST = 
?
T
O
T
ncdT
 
 
8. A compound (A ; B3N3H3Cl3) reacts with LiBH4 to form inorganic benzene (B). (A) reacts with (C) to 
form B3N3H3(CH3)3. (B) and (C) are respectively. 
 (1) Boron nitride, MeMgBr   (2) Boron nitride, MeBr 
 (3) Borazine, MeBr    (4) Borazine, MeMgBr 
 ,d ;kSfxd (A ; B3N3H3Cl3) LiBH4 ds lkFk fØ;k djds vdkcZfud cSathu (B) cukrk gSA (A), (C) ds lkFk fØ;k 
djds B3N3H3(CH3)3 cukrk gSA (B) rFkk (C) Øe'k% gSA 
 (1) cksjksu ukbVªkbM, MeMgBr   (2) cksjksu ukbVªkbM, MeBr 
 (3) cksjksthu, MeBr    (4) cksjksthup, MeMgBr 
Ans. (4) 
Sol. B3N3H3Cl3 + LiBH4 ? ? ? B3N3H6 + LiCl + BCl3 
        (A)        (B) 
 B3N3H3Cl3 + 3MeMgBr ? ? ? B3N3H3 (CH3)3 + 3MgBrCl  
        (A)  (C) 
 
9. In a box a mixture containing H2, O2 and CO along with charcoal is present then variation of pressure 
with the time will be as follows : 
 ,d ckWDl es pkjdksy ds lkFk H2, O2 rFkk CO ;qDr ,d feJ.k mifLFkr gS rks le; ds lkFk nkc dk vUrj fuEu 
izdkj gksxk& 
 (1) 
 
Time 
P 
  
  (2) 
 
Time 
P 
 
 (3) 
 
Time 
P 
   
 (4)  
 
Time 
P 
 
Ans. (3) 
Sol. Theory based. ¼lS}kfUrd½ 
Page 4


      
 
    
    
    
     
 
                       
 
  
 
 
 
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.  5 g of Zn reacts with 
 (I) Excess of NaOH  (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
5 g Zn fuEu ds lkFk fØ;k djrk gSA 
 (I) NaOH dk vkf/kD; (II) ruq HCl, rc (I) rFkk (II) es mRlftZr H2 xSl dk vk;ru vuqikr gS& 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
Ans.  (3) 
Sol. Zn + 2NaOH ? ? Na2ZnO2 + H2 
 Zn + 2HCl ? ? ZnCl2 + H2 
 According to stoichiometry in both the reactions, equal number of moles of H2 are evolved. 
 nksuks vfHkfØ;kvksa es jllehdj.kfefr ds vuqlkj H2 ds eksyks dh leku la[;k mRlftZr gksrh gSA 
 
2. Given Ksp for Cr(OH)3 is 6 × 10
–31
 then determine [OH
–
]. 
 (Neglect the contribution of OH
–
 ions from H2O) 
 fn;k gS Cr(OH)3 ds fy;s Ksp, 6 × 10
–31 
gS rc [OH
–
] dk fu/kkZj.k dhft;sA 
 (H2O ls OH
–
 vk;uksa dk ;ksxnku ux.; ekusaA) 
 (1) (18 × 10
–31
)
1/4 
M  (2) (18 × 10
–31
)
1/2
 M  (3) (6 × 10
–31
)
1/4
 M  (4) 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? M 
Ans.  (1) 
Sol. Cr(OH)3 ? ? ?Cr
+3
 + 3OH
– 
   s 3s 
  Ksp = s . (3s)
3
  
 ? 6 × 10
–31
 = 27.s
4
 
? ? ? s = 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? 
[OH
–
] = 3s 
 = 3 × 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? = (18 × 10
–31
)
1/4
 M 
 
3. Select the correct statements among the followings 
 (A) LiCl does not dissolve in pyridine 
 (B) Li does not react ethyne to form ethynide. 
 (C) Li and Mg react slowly with water. 
(D) Among alkali metals Li has highest hydration tendency. 
fuEu es lgh dFkuksa dk p;u dhft;s& 
 (A) LiCl fifjMhu es ugha ?kqyrk gSA 
 (B) Li ,FkkbZu ds lkFk fØ;k djds ,FkkbZukbM ugha cukrk gSA 
 (C) Li rFkk Mg ty ds lkFk /khjs&/khjs fØ;k djrs gSA 
(D) {kkjh; e`nk /kkrqvks esa Li mPpre ty;kstu izo`fÙk j[krk gSA 
(1) B, C, D  (2) A, B, C, D  (3) A, B, C  (4) C, D 
Ans.  (1) 
Sol.  Theory based ¼lS}kfUrd½ 
      
 
    
    
    
     
 
                       
 
  
 
 
4. Given an element having following ionisation enthalpies IE1 = 496 
mol
kJ
 and IE2 = 4562 
mol
kJ
 one mole 
hydroxide of this element is treated separated with HCl and H2SO4 respectively. Moles of HCl and 
H2SO4 reacted respectively is 
 fuEu vk;uu ,UFkSYih j[kus okyk ,d rRo fn;k x;k gS IE1 = 496 
mol
kJ
 rFkk IE2 = 4562 
mol
kJ
 bl rRo ds ,d 
eksy gkbMªkWDlkbM dks i`Fkd :i ls Øe'k% HCl rFkk H2SO4 ds lkFk mipkfjr fd;k tkrk gSA vfHkÑr HCl rFkk 
H2SO4 ds eksy Øe'k% gS& 
 (1) 1, 0.5   (2) 0.5, 1  (3) 2, 0.5  (4) 0.5, 2 
Ans.  (1) 
Sol.  According to the given data of I.E, This element must belong to group 1 and thus is monovalent & form 
hydroxide of the type M(OH). 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
Sol.  I.E. ds fn;s x;s vkadM+ks ds vuqlkj ;g rRo oxZ 1 ls lEcfU/kr gksuk pkfg;s blfy;s ;g ,dy la;ksth gSS rFkk 
M(OH) izdkj dk gkbMªkWDlkbM cukrk gSA 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
 
5. Reactant A represented by square is in equilibrium with product B represented by circles. Then value of 
equilibrium constant is 
 oxZ }kjk iznf'kZr vfHkdkjd A o`Ùkksa }kjk iznf'kZr mRikn B ds lkFk lkE; esa gSA rc lkE; fu;rkad dk eku gSA 
     
 
 
 (1) 1   (2) 2   (3) 3   (4) 4 
Ans.  (2) 
 
6. Given following complexes 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 Correct order of spin only magnetic moment for the above complexes  is. 
 fuEu ladqy fn;s x;s gS% 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 mijksDr ladqyksa ds fy, dsoy pØ.k pqEcdh; vk?kw.kZ dk lgh Øe gS& 
(1) (II) > (III) > (IV) > I     (2) (II) > (IV) > (III) > (I) 
(3) (I) > (IV) > (III) > (II)    (4) (II) > (I) > (IV) > (III) 
Ans. (1) 
Sol. As, ?s = ) 2 n ( n ? 
 Complex (I) ? Fe
+2
 ? S.F.L ? 
2 , 2 , 2
g 2
t eg
0,0
 ? ?s = 0 
Complex (II) ? Cr
+2
 ? W.F.L ?  
1 , 1 , 1
g 2
t eg
1,0
 ? ?s = 24 B.M. 
Complex (III) ? Co
+2
 ? W.F.L ? 
2 , 2
g
e 
1 , 1 , 1
g 2
t ? ?s = 15 B.M. 
Complex (IV) ? Fe
+3
 ? S.F.L ? 
1 , 2 , 2
g 2
t eg
0,0
 ? ?s = 3 B.M. 
      
 
    
    
    
     
 
                       
 
  
 
 
 
7. Select the correct option : 
 (1) Entropy is function of temperature and also entropy change is function of temperature. 
 (2) Entropy is a function of temperature & entropy change is not a function of temperature. 
 (3) Entropy is not a function of temperature & entropy change is a function of temperature. 
 (4) Both entropy & entropy change are not a function of temperature. 
 lgh fodYi dk p;u dhft,A 
 (1) ,UVªkWih rki dk Qyu gS rFkk ,UVªkWih ifjorZu Hkh rki dk Qyu gSA 
 (2) ,UVªkWih rki dk Qyu gS rFkk ,UVªkWih ifjorZu rki dk Qyu ugha gSA 
 (3) ,UVªkWih rki dk Qyu ugh gSa rFkk ,UVªkWih ifjorZu rki dk Qyu gSA 
 (4) ,UVªkWih rFkk ,UVªkWih ifjorZu nksuksa rki dk Qyu ugha gSA 
Ans. (1) 
Sol. ?S = 
?
T
dq
 
 ST = 
?
T
O
T
ncdT
 
 
8. A compound (A ; B3N3H3Cl3) reacts with LiBH4 to form inorganic benzene (B). (A) reacts with (C) to 
form B3N3H3(CH3)3. (B) and (C) are respectively. 
 (1) Boron nitride, MeMgBr   (2) Boron nitride, MeBr 
 (3) Borazine, MeBr    (4) Borazine, MeMgBr 
 ,d ;kSfxd (A ; B3N3H3Cl3) LiBH4 ds lkFk fØ;k djds vdkcZfud cSathu (B) cukrk gSA (A), (C) ds lkFk fØ;k 
djds B3N3H3(CH3)3 cukrk gSA (B) rFkk (C) Øe'k% gSA 
 (1) cksjksu ukbVªkbM, MeMgBr   (2) cksjksu ukbVªkbM, MeBr 
 (3) cksjksthu, MeBr    (4) cksjksthup, MeMgBr 
Ans. (4) 
Sol. B3N3H3Cl3 + LiBH4 ? ? ? B3N3H6 + LiCl + BCl3 
        (A)        (B) 
 B3N3H3Cl3 + 3MeMgBr ? ? ? B3N3H3 (CH3)3 + 3MgBrCl  
        (A)  (C) 
 
9. In a box a mixture containing H2, O2 and CO along with charcoal is present then variation of pressure 
with the time will be as follows : 
 ,d ckWDl es pkjdksy ds lkFk H2, O2 rFkk CO ;qDr ,d feJ.k mifLFkr gS rks le; ds lkFk nkc dk vUrj fuEu 
izdkj gksxk& 
 (1) 
 
Time 
P 
  
  (2) 
 
Time 
P 
 
 (3) 
 
Time 
P 
   
 (4)  
 
Time 
P 
 
Ans. (3) 
Sol. Theory based. ¼lS}kfUrd½ 
      
 
    
    
    
     
 
                       
 
  
 
 
 
10. Given complex [Co(NH3)4Cl2]. In it if Cl – Co – Cl bond angle is 90º then it is :  
 (1) Cis-isomer     (2) Trans- isomers 
 (3) Meridional and trans    (4) Cis and trans both 
ladqy [Co(NH3)4Cl2] fn;k x;k gSA ;fn bles Cl – Co – Cl cU/k dks.k 90º gS rks ;g gS& 
 (1) lei{k&leko;oh    (2) foi{k&leko;oh 
 (3) esjhfM;ksuy rFkk foi{k    (4) lei{k rFkk foi{k nksuksa 
Ans. (1) 
Sol. 
 
H3N 
Cl 
NH3 
Co 
H3N 
Cl 
NH3 
90º 
Cis form 
 
 
11. Amongst the following which has minimum conductivity.  
 (1) Distilled water    (2) Sea water   
 (3) Saline water used for intra venous injection   (4) Well-water   
fuEu es ls dkSu U;wure pkydrk j[krk gS& 
 (1) vklqr ty     (2) leqnzh ty 
 (3) var%f'kjk batsD'ku ds fy;s iz;qDr yo.kh; ty (4) dq,sa dk ty 
Ans. (1) 
Sol. Theory based. ¼lS}kfUrd½ 
 
12. Number of sp
2
 hybrid orbitals in Benzene is :  
 csfUtu es sp
2 
ladfjr d{kdksa dh la[;k gS& 
 (1) 18    (2) 24    (3) 6    (4) 12   
Ans. (1)  
Sol. In benzene total six sp
2
 hybrid carbon atoms are present. Each carbon atom has 3 sp
2
 hybrid orbitals. 
 Therefore total sp
2
 hybrid orbitals are 18 in benzene. 
Sol. csfUtu es dqy N% sp
2
 ladfjr dkcZu ijek.kq mifLFkr gSA izR;sd dkcZu ijek.kq 3 sp
2
 ladfjr d{kd j[krk gSA 
blfy;s csfUtu es dqy sp
2
 ladfjr d{kd 18 gSA 
 
13. Which of the following reaction will not form racemic mixture as product? 
 fuEu esa ls fdl vfHkfØ;k esa jslsfed feJ.k mRikn ds :i esa izkIr ugha gksrk gS\ 
 (1) 
 
CH3CH2 –C –CH3 
O 
HCN 
     (2) (CH3)2CH –CH=CH2
HCl
? ? ? ?   
(3) 
 
HCl 
    (4) 
 
CH3CH2 –CH=CH2 
HBr 
  
Ans. (2)  
Sol. (CH3)2CH –CH=CH2 
HCl
? ? ? ?
3 2 3
(CH ) CH CH CH
?
??
Hydride shift
? ?? ?? ? ?
 
CH3 –C –CH2 –CH3 
CH3 
? ?
 
 
 
 
CH3 –C –CH2 –CH3 
CH3 
Cl 
 
 In this reaction, major product is not chiral. 
mijksDr vfHkfØ;k eas eq[; mRikn fdjSy ugh gSA 
Page 5


      
 
    
    
    
     
 
                       
 
  
 
 
 
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.  5 g of Zn reacts with 
 (I) Excess of NaOH  (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
5 g Zn fuEu ds lkFk fØ;k djrk gSA 
 (I) NaOH dk vkf/kD; (II) ruq HCl, rc (I) rFkk (II) es mRlftZr H2 xSl dk vk;ru vuqikr gS& 
 (1) 2 : 1   (2) 1 : 2   (3) 1 : 1   (4) 3 : 1 
Ans.  (3) 
Sol. Zn + 2NaOH ? ? Na2ZnO2 + H2 
 Zn + 2HCl ? ? ZnCl2 + H2 
 According to stoichiometry in both the reactions, equal number of moles of H2 are evolved. 
 nksuks vfHkfØ;kvksa es jllehdj.kfefr ds vuqlkj H2 ds eksyks dh leku la[;k mRlftZr gksrh gSA 
 
2. Given Ksp for Cr(OH)3 is 6 × 10
–31
 then determine [OH
–
]. 
 (Neglect the contribution of OH
–
 ions from H2O) 
 fn;k gS Cr(OH)3 ds fy;s Ksp, 6 × 10
–31 
gS rc [OH
–
] dk fu/kkZj.k dhft;sA 
 (H2O ls OH
–
 vk;uksa dk ;ksxnku ux.; ekusaA) 
 (1) (18 × 10
–31
)
1/4 
M  (2) (18 × 10
–31
)
1/2
 M  (3) (6 × 10
–31
)
1/4
 M  (4) 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? M 
Ans.  (1) 
Sol. Cr(OH)3 ? ? ?Cr
+3
 + 3OH
– 
   s 3s 
  Ksp = s . (3s)
3
  
 ? 6 × 10
–31
 = 27.s
4
 
? ? ? s = 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? 
[OH
–
] = 3s 
 = 3 × 
4 / 1
31 –
10
27
6
?
?
?
?
?
?
? = (18 × 10
–31
)
1/4
 M 
 
3. Select the correct statements among the followings 
 (A) LiCl does not dissolve in pyridine 
 (B) Li does not react ethyne to form ethynide. 
 (C) Li and Mg react slowly with water. 
(D) Among alkali metals Li has highest hydration tendency. 
fuEu es lgh dFkuksa dk p;u dhft;s& 
 (A) LiCl fifjMhu es ugha ?kqyrk gSA 
 (B) Li ,FkkbZu ds lkFk fØ;k djds ,FkkbZukbM ugha cukrk gSA 
 (C) Li rFkk Mg ty ds lkFk /khjs&/khjs fØ;k djrs gSA 
(D) {kkjh; e`nk /kkrqvks esa Li mPpre ty;kstu izo`fÙk j[krk gSA 
(1) B, C, D  (2) A, B, C, D  (3) A, B, C  (4) C, D 
Ans.  (1) 
Sol.  Theory based ¼lS}kfUrd½ 
      
 
    
    
    
     
 
                       
 
  
 
 
4. Given an element having following ionisation enthalpies IE1 = 496 
mol
kJ
 and IE2 = 4562 
mol
kJ
 one mole 
hydroxide of this element is treated separated with HCl and H2SO4 respectively. Moles of HCl and 
H2SO4 reacted respectively is 
 fuEu vk;uu ,UFkSYih j[kus okyk ,d rRo fn;k x;k gS IE1 = 496 
mol
kJ
 rFkk IE2 = 4562 
mol
kJ
 bl rRo ds ,d 
eksy gkbMªkWDlkbM dks i`Fkd :i ls Øe'k% HCl rFkk H2SO4 ds lkFk mipkfjr fd;k tkrk gSA vfHkÑr HCl rFkk 
H2SO4 ds eksy Øe'k% gS& 
 (1) 1, 0.5   (2) 0.5, 1  (3) 2, 0.5  (4) 0.5, 2 
Ans.  (1) 
Sol.  According to the given data of I.E, This element must belong to group 1 and thus is monovalent & form 
hydroxide of the type M(OH). 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
Sol.  I.E. ds fn;s x;s vkadM+ks ds vuqlkj ;g rRo oxZ 1 ls lEcfU/kr gksuk pkfg;s blfy;s ;g ,dy la;ksth gSS rFkk 
M(OH) izdkj dk gkbMªkWDlkbM cukrk gSA 
 MOH + HCl ? ? MCl + H2O 
 1 mole 1 mole 
 2MOH + H2SO4 ? ? M2SO4 + H2O 
 1 mole 1/2 mole 
 
5. Reactant A represented by square is in equilibrium with product B represented by circles. Then value of 
equilibrium constant is 
 oxZ }kjk iznf'kZr vfHkdkjd A o`Ùkksa }kjk iznf'kZr mRikn B ds lkFk lkE; esa gSA rc lkE; fu;rkad dk eku gSA 
     
 
 
 (1) 1   (2) 2   (3) 3   (4) 4 
Ans.  (2) 
 
6. Given following complexes 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 Correct order of spin only magnetic moment for the above complexes  is. 
 fuEu ladqy fn;s x;s gS% 
 (I) Na4[Fe(CN)6]    (II) [Cr(H2O)6] Cl2 
 (III) (NEt4)2 [CoCl4]   (IV) Na3[Fe(C2O4)3] ( ?0 > P) 
 mijksDr ladqyksa ds fy, dsoy pØ.k pqEcdh; vk?kw.kZ dk lgh Øe gS& 
(1) (II) > (III) > (IV) > I     (2) (II) > (IV) > (III) > (I) 
(3) (I) > (IV) > (III) > (II)    (4) (II) > (I) > (IV) > (III) 
Ans. (1) 
Sol. As, ?s = ) 2 n ( n ? 
 Complex (I) ? Fe
+2
 ? S.F.L ? 
2 , 2 , 2
g 2
t eg
0,0
 ? ?s = 0 
Complex (II) ? Cr
+2
 ? W.F.L ?  
1 , 1 , 1
g 2
t eg
1,0
 ? ?s = 24 B.M. 
Complex (III) ? Co
+2
 ? W.F.L ? 
2 , 2
g
e 
1 , 1 , 1
g 2
t ? ?s = 15 B.M. 
Complex (IV) ? Fe
+3
 ? S.F.L ? 
1 , 2 , 2
g 2
t eg
0,0
 ? ?s = 3 B.M. 
      
 
    
    
    
     
 
                       
 
  
 
 
 
7. Select the correct option : 
 (1) Entropy is function of temperature and also entropy change is function of temperature. 
 (2) Entropy is a function of temperature & entropy change is not a function of temperature. 
 (3) Entropy is not a function of temperature & entropy change is a function of temperature. 
 (4) Both entropy & entropy change are not a function of temperature. 
 lgh fodYi dk p;u dhft,A 
 (1) ,UVªkWih rki dk Qyu gS rFkk ,UVªkWih ifjorZu Hkh rki dk Qyu gSA 
 (2) ,UVªkWih rki dk Qyu gS rFkk ,UVªkWih ifjorZu rki dk Qyu ugha gSA 
 (3) ,UVªkWih rki dk Qyu ugh gSa rFkk ,UVªkWih ifjorZu rki dk Qyu gSA 
 (4) ,UVªkWih rFkk ,UVªkWih ifjorZu nksuksa rki dk Qyu ugha gSA 
Ans. (1) 
Sol. ?S = 
?
T
dq
 
 ST = 
?
T
O
T
ncdT
 
 
8. A compound (A ; B3N3H3Cl3) reacts with LiBH4 to form inorganic benzene (B). (A) reacts with (C) to 
form B3N3H3(CH3)3. (B) and (C) are respectively. 
 (1) Boron nitride, MeMgBr   (2) Boron nitride, MeBr 
 (3) Borazine, MeBr    (4) Borazine, MeMgBr 
 ,d ;kSfxd (A ; B3N3H3Cl3) LiBH4 ds lkFk fØ;k djds vdkcZfud cSathu (B) cukrk gSA (A), (C) ds lkFk fØ;k 
djds B3N3H3(CH3)3 cukrk gSA (B) rFkk (C) Øe'k% gSA 
 (1) cksjksu ukbVªkbM, MeMgBr   (2) cksjksu ukbVªkbM, MeBr 
 (3) cksjksthu, MeBr    (4) cksjksthup, MeMgBr 
Ans. (4) 
Sol. B3N3H3Cl3 + LiBH4 ? ? ? B3N3H6 + LiCl + BCl3 
        (A)        (B) 
 B3N3H3Cl3 + 3MeMgBr ? ? ? B3N3H3 (CH3)3 + 3MgBrCl  
        (A)  (C) 
 
9. In a box a mixture containing H2, O2 and CO along with charcoal is present then variation of pressure 
with the time will be as follows : 
 ,d ckWDl es pkjdksy ds lkFk H2, O2 rFkk CO ;qDr ,d feJ.k mifLFkr gS rks le; ds lkFk nkc dk vUrj fuEu 
izdkj gksxk& 
 (1) 
 
Time 
P 
  
  (2) 
 
Time 
P 
 
 (3) 
 
Time 
P 
   
 (4)  
 
Time 
P 
 
Ans. (3) 
Sol. Theory based. ¼lS}kfUrd½ 
      
 
    
    
    
     
 
                       
 
  
 
 
 
10. Given complex [Co(NH3)4Cl2]. In it if Cl – Co – Cl bond angle is 90º then it is :  
 (1) Cis-isomer     (2) Trans- isomers 
 (3) Meridional and trans    (4) Cis and trans both 
ladqy [Co(NH3)4Cl2] fn;k x;k gSA ;fn bles Cl – Co – Cl cU/k dks.k 90º gS rks ;g gS& 
 (1) lei{k&leko;oh    (2) foi{k&leko;oh 
 (3) esjhfM;ksuy rFkk foi{k    (4) lei{k rFkk foi{k nksuksa 
Ans. (1) 
Sol. 
 
H3N 
Cl 
NH3 
Co 
H3N 
Cl 
NH3 
90º 
Cis form 
 
 
11. Amongst the following which has minimum conductivity.  
 (1) Distilled water    (2) Sea water   
 (3) Saline water used for intra venous injection   (4) Well-water   
fuEu es ls dkSu U;wure pkydrk j[krk gS& 
 (1) vklqr ty     (2) leqnzh ty 
 (3) var%f'kjk batsD'ku ds fy;s iz;qDr yo.kh; ty (4) dq,sa dk ty 
Ans. (1) 
Sol. Theory based. ¼lS}kfUrd½ 
 
12. Number of sp
2
 hybrid orbitals in Benzene is :  
 csfUtu es sp
2 
ladfjr d{kdksa dh la[;k gS& 
 (1) 18    (2) 24    (3) 6    (4) 12   
Ans. (1)  
Sol. In benzene total six sp
2
 hybrid carbon atoms are present. Each carbon atom has 3 sp
2
 hybrid orbitals. 
 Therefore total sp
2
 hybrid orbitals are 18 in benzene. 
Sol. csfUtu es dqy N% sp
2
 ladfjr dkcZu ijek.kq mifLFkr gSA izR;sd dkcZu ijek.kq 3 sp
2
 ladfjr d{kd j[krk gSA 
blfy;s csfUtu es dqy sp
2
 ladfjr d{kd 18 gSA 
 
13. Which of the following reaction will not form racemic mixture as product? 
 fuEu esa ls fdl vfHkfØ;k esa jslsfed feJ.k mRikn ds :i esa izkIr ugha gksrk gS\ 
 (1) 
 
CH3CH2 –C –CH3 
O 
HCN 
     (2) (CH3)2CH –CH=CH2
HCl
? ? ? ?   
(3) 
 
HCl 
    (4) 
 
CH3CH2 –CH=CH2 
HBr 
  
Ans. (2)  
Sol. (CH3)2CH –CH=CH2 
HCl
? ? ? ?
3 2 3
(CH ) CH CH CH
?
??
Hydride shift
? ?? ?? ? ?
 
CH3 –C –CH2 –CH3 
CH3 
? ?
 
 
 
 
CH3 –C –CH2 –CH3 
CH3 
Cl 
 
 In this reaction, major product is not chiral. 
mijksDr vfHkfØ;k eas eq[; mRikn fdjSy ugh gSA 
      
 
    
    
    
     
 
                       
 
  
 
 
 
14. In which compound C –Cl bond length is shortest? 
 fuEu esa fdl ;kSfxd esa C –Cl caèk yEckbZ lcls de gS\ 
 (1) Cl –CH=CH2      (2) Cl –CH=CH –CH3 
 (3) Cl –CH=CH –OCH3     (4) Cl –CH=CH –NO2 
Ans. (4)  
Sol. Resonance form of Cl –CH=CH –NO2 is more stable than resonance form of any other given 
compounds. Hence, double bond character in carbon-chlorine bond is maximum and bond length is 
shortest. 
Cl –CH=CH –NO2 dh vuquknh lajpuk vU; lHkh ;kSfxdksa dh vuquknh lajpukvksa ls lokZfèkd LFkk;h gSA vr% 
dkcZu&Dyksjhu caèk esa f}caèk vfHky{k.k vfèkdre gS rFkk caèk yEckbZ lcls de gksxhA 
 
15. Biochemical oxygen demand (BOD) is defined as ............ in ppm of O2. 
 (1) Required to sustain life    
(2) The amount of oxygen required by bacteria to break down the organic matter present in a certain 
volume of a sample of water.  
(3) The amount of oxygen required by anaerobic bacteria to break down the inorganic matter present 
in a certain volume of a sample of water.  
(4) Required photochemical reaction to degrade waste. 
tSojklk;fud vkWDlhtu ek¡x* (BOD) ........ dks dgk tkrk gSA ¼O2 dh ppm esa ek=kk½ 
 (1) thou ds fy, vko';d  
(2) ty ds ,d uewus ds fuf'pr vk;ru esa mifLFkr dkcZfud inkFkZ dks fo[kafMr djus ds fy, thok.kq }kjk 
vko';d vkWDlhtu  
 (3) ty ds ,d uewus ds fuf'pr vk;ru esa mifLFkr vdkcZfud inkFkZ dks fo[kafMr djus ds fy, vok;oh; thok.kq 
}kjk vko';d vkWDlhtu 
(4) tSo vi?kVuh; vif'k"V gsrq izdk'k jklk;fud vfHkfØ;k 
Ans. (2) 
Sol. The amount of oxygen required by bacteria to break down the organic matter present in a certain 
volume of a sample of water, is called Biochemical Oxygen Demand (BOD).  
ty ds ,d uewus ds fuf'pr vk;ru esa mifLFkr dkcZfud inkFkZ dks fo[kafMr djus ds fy, thok.kq }kjk vko';d 
vkWDlhtu dks ^tSojklk;fud vkWDlhtu ek¡x* (BOD) dgk tkrk gSA  
 
16. Monomer(s) of which of the given polymer is chiral? 
 (1) Buna-S   (2) Neoprene  (3) Nylon-6,6   (4) PHBV 
fuEu esa ls dkSulk ,dyd (monomer) fdjSy gS\ 
 (1) C;quk-S   (2) fu;ksizhu  (3) uk;ykWu-6,6   (4) PHBV 
Ans.  (4) 
Sol. Polymers     Monomers  
(1) Buna-S  
Ph 
-[CH
2
-CH-CH
2
-CH=CH-CH
2
]- 
 
  CH2=CH –Ph & CH2=CH –CH=CH2 
 (2) Neoprene –[CH2-CH=CCl-CH2]n –   
 
CH2=CH –C=CH2 
Cl 
   
 (3) Nylon-6,6     HO 2C –(CH 2) 4 –CO 2H  & H 2N –(CH 2) 6 –NH 2 
~[CO(CH 2) 4CO –NH(CH 2) 6NH] n~ 
  (4) 
 
   
 
) acid oic tan Hydroxybu 3 ( 
COOH CH CH CH 
| 
OH 
2 3 
? 
? ? ? 
 &  
 
) acid oic tan Hydroxypen 3 ( 
COOH CH CH CH CH 
| 
OH 
2 2 3 
? 
? ? ? ? 
 
 In PHBV, both monomers have chiral centre.