JEE Main 2020 Answer Key Chemistry - Morning Shift (09-01-2020) JEE Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2020 Answer Key Chemistry - Morning Shift (09-01-2020) JEE Notes | EduRev

 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. Determine wavelength of electron in 4
th
 Bohr's orbit ? 
4
th
 cksgj d{kk esa bysDVªkWu dh rjax)Sè;Z dk fuèkkZj.k dhft, ? 
 (1) 4 ?a0  (2) 2 ?a0  (3) 8 ?a0  (4) 6 ?a0 
Ans. (3) 
Sol.  2 ?r = n ? 
 2 ? × 
2
n
Z
a0 = n ? 
 2 ? × 
2
4
1
a0 = n ? 
 ? = 8 ?a0 
 
2. Which of the following species have one unpaired electron each? 
fuEu esa ls dkSulh Lih'kht ¼izR;sd esa½ ,d v;qfXer bysDVªkWu j[krh gSa\ 
 (1) O2, O2
–
  (2) O2, O2
+
   (3) O2
+
, O2
–
   (4) O2, O2
2 –
 
Ans. (3) 
Sol. O2 = ?1s
2
 ?*1s
2
??2s
2
??*2s
2
??2pz
2
??2px
2
 = ?2py
2
 ?*2px
1 
= ?2py
1 
 
3. For Br2( ?) 
Enthalpy of atomisation = x kJ/mol 
 Bond dissociation enthalpy of bromine = y kJ/mole 
 then 
 (1) x > y  (2) x < y  (3) x = y   (4)  Relation does not exist 
 Br2( ?) ds fy, 
ijek.kfodj.k dh ,sUFksYih = x kJ/mol 
 czksehu dh caèk fo;kstu ,sUFksYih = y kJ/mole 
 rc 
 (1) x > y  (2) x < y  (3) x = y   (4)  dksbZ lEcU„èk ugah gksrk gS 
 
Ans. (1)  
Sol. 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?H atomisation = x kJ/mole 
2Br(g) 
Bond energy = y kJ/mole 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?ijek.kfodj.k = x kJ/mole 
2Br(g) 
caèk ÅtkZ = y kJ/mole 
 
?Hatomisation = ?Hvap + Bond energy  
 Hence x > y 
?Hijek.kfodj.k = ?Hvap + caèk ÅtkZ 
 bl izdkj x > y 
 
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. Determine wavelength of electron in 4
th
 Bohr's orbit ? 
4
th
 cksgj d{kk esa bysDVªkWu dh rjax)Sè;Z dk fuèkkZj.k dhft, ? 
 (1) 4 ?a0  (2) 2 ?a0  (3) 8 ?a0  (4) 6 ?a0 
Ans. (3) 
Sol.  2 ?r = n ? 
 2 ? × 
2
n
Z
a0 = n ? 
 2 ? × 
2
4
1
a0 = n ? 
 ? = 8 ?a0 
 
2. Which of the following species have one unpaired electron each? 
fuEu esa ls dkSulh Lih'kht ¼izR;sd esa½ ,d v;qfXer bysDVªkWu j[krh gSa\ 
 (1) O2, O2
–
  (2) O2, O2
+
   (3) O2
+
, O2
–
   (4) O2, O2
2 –
 
Ans. (3) 
Sol. O2 = ?1s
2
 ?*1s
2
??2s
2
??*2s
2
??2pz
2
??2px
2
 = ?2py
2
 ?*2px
1 
= ?2py
1 
 
3. For Br2( ?) 
Enthalpy of atomisation = x kJ/mol 
 Bond dissociation enthalpy of bromine = y kJ/mole 
 then 
 (1) x > y  (2) x < y  (3) x = y   (4)  Relation does not exist 
 Br2( ?) ds fy, 
ijek.kfodj.k dh ,sUFksYih = x kJ/mol 
 czksehu dh caèk fo;kstu ,sUFksYih = y kJ/mole 
 rc 
 (1) x > y  (2) x < y  (3) x = y   (4)  dksbZ lEcU„èk ugah gksrk gS 
 
Ans. (1)  
Sol. 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?H atomisation = x kJ/mole 
2Br(g) 
Bond energy = y kJ/mole 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?ijek.kfodj.k = x kJ/mole 
2Br(g) 
caèk ÅtkZ = y kJ/mole 
 
?Hatomisation = ?Hvap + Bond energy  
 Hence x > y 
?Hijek.kfodj.k = ?Hvap + caèk ÅtkZ 
 bl izdkj x > y 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
4. Which of the following oxides are acidic, Basic Amphoteric Respectively. 
 fuEu esa ls dkSuls vkWDlkbM Øe'k% vEyh;] {kkjh;] mHk;èkehZ gS&  
 (1) MgO, P4O10, Al2O3 (2) N2O3, Li2O, Al2O3  (3) SO3, Al2O3, Na2O (4) P4O10, Al2O3, MgO 
Ans. (2) 
Sol. Non-metal oxides are acidic in nature 
 alkali metal oxides are basic in nature 
 Al2O3 is amphoteric. 
 vèkkfRod vkWDlkbM vEyh; izd`fr ds gksrs gSA 
 {kkjh; èkkrq vkWDlkbM {kkjh; izd`fr ds gksrs gSA 
 Al2O3 mHk;èkehZ gSA 
 
5. Complex Cr(H2O)6Cln shows geometrical isomerism and also reacts with AgNO3 solution. 
 Given : Spin only magnetic moment = 3.8 B.M.  
 What is the IUPAC name of the complex. 
 (1) Hexaaquachromium(III) chloride    
 (2) Tetraaquadichloridochromium(III) chloride dihydrate   
 (3) Hexaaquachromium(IV) chloride    
 (4)  Tetraaquadichloridochromium(IV) chloride dihydrate   
ladqy Cr(H2O)6Cln T;kferh; leko;ork n'kkZrk gS rFkk ;g AgNO3 foy;u ds lkFk Hkh vfHkd`r gksrk gSA 
 fn;k gS: izpØ.k dsoy pqEcdh; vk?kq.kZ = 3.8 B.M.  
 ladqy dk IUPAC uke D;k gS \ 
 (1) gsDlk,DokØksfe;e (III) DyksjkbM 
 (2) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV 
 (3) gsDlk,DokØksfe;e (IV) DyksjkbM 
 (4) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (IV) DyksjkbM MkbZgkbMªsV 
Ans. (2) 
Sol. Cr(H2O)6Cln ( ?complex)spin = 3.8 B.M. 
 From data of magnetic moment oxidation number of Cr should be +3 
 Hence complex is Cr(H2O)6Cl3. 
 Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl ?2H2O. 
 It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate   
 Cr(H2O)6Cln ( ?ladqy)pØ.k = 3.8 B.M. 
 pqEcdh; vk?kq.kZ ds eku ls Cr dk vkWDlhdj.k vad +3 gksuk pkfg,A  
 bl izdkj ladqy Cr(H2O)6Cl3 gSA 
 ladqy T;kferh; leko;ork n'kkZrk gSA blfy, ladqy dk lw=k [Cr(H2O)4Cl2]Cl ?2H2O gSA 
 bldk IUPAC uke VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV gSA ?
 
6. The electronic configuration of bivalent Europium and trivalent cerium respectively is:   
 (Atomic Number : Xe = 54, Ce = 58, Eu = 63) 
f}la;ksth ;wjksfi;e rFkk f=kla;ksth flfj;e ds bysDVªkWfud foU;kl Øe'k% gS%  
 (ijek.kq Øekd : Xe = 54, Ce = 58, Eu = 63) 
 (1) [Xe]4f
7
, [Xe]4f
1
    (2) [Xe]4f
7
 6s
2
, [Xe]4f
1
  
 (3) [Xe]4f
7
6s
2
, [Xe]4f
1
5d
1
6s
2
   (4) [Xe]4f
7
, [Xe]4f
1
5d
1
6s
2
 
Ans. (1) 
Sol. Eu
2+
 : [Xe]4f
7 
 
Ce
3+
 : [Xe]4f
1 
 
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. Determine wavelength of electron in 4
th
 Bohr's orbit ? 
4
th
 cksgj d{kk esa bysDVªkWu dh rjax)Sè;Z dk fuèkkZj.k dhft, ? 
 (1) 4 ?a0  (2) 2 ?a0  (3) 8 ?a0  (4) 6 ?a0 
Ans. (3) 
Sol.  2 ?r = n ? 
 2 ? × 
2
n
Z
a0 = n ? 
 2 ? × 
2
4
1
a0 = n ? 
 ? = 8 ?a0 
 
2. Which of the following species have one unpaired electron each? 
fuEu esa ls dkSulh Lih'kht ¼izR;sd esa½ ,d v;qfXer bysDVªkWu j[krh gSa\ 
 (1) O2, O2
–
  (2) O2, O2
+
   (3) O2
+
, O2
–
   (4) O2, O2
2 –
 
Ans. (3) 
Sol. O2 = ?1s
2
 ?*1s
2
??2s
2
??*2s
2
??2pz
2
??2px
2
 = ?2py
2
 ?*2px
1 
= ?2py
1 
 
3. For Br2( ?) 
Enthalpy of atomisation = x kJ/mol 
 Bond dissociation enthalpy of bromine = y kJ/mole 
 then 
 (1) x > y  (2) x < y  (3) x = y   (4)  Relation does not exist 
 Br2( ?) ds fy, 
ijek.kfodj.k dh ,sUFksYih = x kJ/mol 
 czksehu dh caèk fo;kstu ,sUFksYih = y kJ/mole 
 rc 
 (1) x > y  (2) x < y  (3) x = y   (4)  dksbZ lEcU„èk ugah gksrk gS 
 
Ans. (1)  
Sol. 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?H atomisation = x kJ/mole 
2Br(g) 
Bond energy = y kJ/mole 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?ijek.kfodj.k = x kJ/mole 
2Br(g) 
caèk ÅtkZ = y kJ/mole 
 
?Hatomisation = ?Hvap + Bond energy  
 Hence x > y 
?Hijek.kfodj.k = ?Hvap + caèk ÅtkZ 
 bl izdkj x > y 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
4. Which of the following oxides are acidic, Basic Amphoteric Respectively. 
 fuEu esa ls dkSuls vkWDlkbM Øe'k% vEyh;] {kkjh;] mHk;èkehZ gS&  
 (1) MgO, P4O10, Al2O3 (2) N2O3, Li2O, Al2O3  (3) SO3, Al2O3, Na2O (4) P4O10, Al2O3, MgO 
Ans. (2) 
Sol. Non-metal oxides are acidic in nature 
 alkali metal oxides are basic in nature 
 Al2O3 is amphoteric. 
 vèkkfRod vkWDlkbM vEyh; izd`fr ds gksrs gSA 
 {kkjh; èkkrq vkWDlkbM {kkjh; izd`fr ds gksrs gSA 
 Al2O3 mHk;èkehZ gSA 
 
5. Complex Cr(H2O)6Cln shows geometrical isomerism and also reacts with AgNO3 solution. 
 Given : Spin only magnetic moment = 3.8 B.M.  
 What is the IUPAC name of the complex. 
 (1) Hexaaquachromium(III) chloride    
 (2) Tetraaquadichloridochromium(III) chloride dihydrate   
 (3) Hexaaquachromium(IV) chloride    
 (4)  Tetraaquadichloridochromium(IV) chloride dihydrate   
ladqy Cr(H2O)6Cln T;kferh; leko;ork n'kkZrk gS rFkk ;g AgNO3 foy;u ds lkFk Hkh vfHkd`r gksrk gSA 
 fn;k gS: izpØ.k dsoy pqEcdh; vk?kq.kZ = 3.8 B.M.  
 ladqy dk IUPAC uke D;k gS \ 
 (1) gsDlk,DokØksfe;e (III) DyksjkbM 
 (2) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV 
 (3) gsDlk,DokØksfe;e (IV) DyksjkbM 
 (4) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (IV) DyksjkbM MkbZgkbMªsV 
Ans. (2) 
Sol. Cr(H2O)6Cln ( ?complex)spin = 3.8 B.M. 
 From data of magnetic moment oxidation number of Cr should be +3 
 Hence complex is Cr(H2O)6Cl3. 
 Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl ?2H2O. 
 It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate   
 Cr(H2O)6Cln ( ?ladqy)pØ.k = 3.8 B.M. 
 pqEcdh; vk?kq.kZ ds eku ls Cr dk vkWDlhdj.k vad +3 gksuk pkfg,A  
 bl izdkj ladqy Cr(H2O)6Cl3 gSA 
 ladqy T;kferh; leko;ork n'kkZrk gSA blfy, ladqy dk lw=k [Cr(H2O)4Cl2]Cl ?2H2O gSA 
 bldk IUPAC uke VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV gSA ?
 
6. The electronic configuration of bivalent Europium and trivalent cerium respectively is:   
 (Atomic Number : Xe = 54, Ce = 58, Eu = 63) 
f}la;ksth ;wjksfi;e rFkk f=kla;ksth flfj;e ds bysDVªkWfud foU;kl Øe'k% gS%  
 (ijek.kq Øekd : Xe = 54, Ce = 58, Eu = 63) 
 (1) [Xe]4f
7
, [Xe]4f
1
    (2) [Xe]4f
7
 6s
2
, [Xe]4f
1
  
 (3) [Xe]4f
7
6s
2
, [Xe]4f
1
5d
1
6s
2
   (4) [Xe]4f
7
, [Xe]4f
1
5d
1
6s
2
 
Ans. (1) 
Sol. Eu
2+
 : [Xe]4f
7 
 
Ce
3+
 : [Xe]4f
1 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
7. Ksp of PbCl
2
 = 1.6 × 10
–5 
 
On mixing  
 300 mL, 0.134M Pb(NO3)2(aq.) + 100 mL, 0.4 M NaCl(aq.) 
 (1) Q > Ksp  (2) Q < Ksp  (3) Q = Ksp   (4) Relation does not exit  
PbCl
2
 dk Ksp = 1.6 × 10
–5 
 300 mL, 0.134M Pb(NO3)2(aq.) + 100 mL, 0.4 M NaCl(aq.) feykus ij& 
(1) Q > Ksp     (2) Q < Ksp   
(3) Q = Ksp      (4) fdlh izdkj ds lEcUèk dk vfLrRo ugha gSA  
Ans. (1) 
Sol.  Q = [Pb
2+
][Cl
–
]
2
 
 
2
3 0 0 0 .1 3 4 1 0 0 0 .4
4 0 0 4 0 0
?? ??
??
??
??
 
 ? ?
2 3 0 .1 3 4
0 .1
4
?
?? 
 = 0.105 × 10
–2
 
 =1.005 ×10
–3 
 
Q > K sp 
 
8. Which of the following can not act as both oxidising and reducing agent ? 
fuEu esa ls dkSu vkWDlhdkjd rFkk vipk;d nksuks ds leku O;ogkj ugha dj ldrk gS\ 
 (1) H2SO3   (2) HNO2   (3) H3PO4   (4) H2O2 
Ans. (3) 
Sol. As in H3PO4 Phosphorous is present it's maximum oxidation number state hence it cannot act as 
reducing agent. 
 pwafd H3PO4 esa QkWLQksjl bldh mPpre vkWDlhdj.k voLFkk esa mifLFkr gSA blfy, ;g vipk;d ds leku dk;Z 
ugha dj ldrk gSA 
 
9. First Ionisation energy of Be is higher than that of Boron. 
 Select the correct statements regarding this 
 (i) It is easier to extract electron from 2p orbital than 2s orbital  
 (ii) Penetration power of 2s orbital is greater than 2p orbital 
 (iii) Shielding of 2p electron by 2s electron  
 (iv) Radius of Boron atom is larger than that of Be 
 (1) (i), (ii), (iii), (iv)  (2) (i), (iii), (iv)   (3) (ii), (iii), (iv)  (4)  (i), (ii), (iii) 
 Be dh izFke vk;uu ÅtkZ cksjksu dh rqyuk esa vfèkd gksrh gSA 
 mDr dFku ds lanHkZ esa lgh dFku@dFkuksa dk p;u dhft,sA 
 (i) 2p d{kd ls bysDVªkWu dk i`Fkddj.k 2s d{kd dh rqyuk esa ljy gksrk gSA 
 (ii) 2s d{kd dh Hksnu {kerk 2p d{kd dh rqyuk esa vfèkd gksrh gSA 
 (iii) 2s bysDVªkWu }kjk 2p bysDVªkWu dk ifjj{k.k gksrk gSA 
 (iv) cksjksu ijek.kq dh f=kT;k Be dh rqyuk esa vfèkd gksrh gSA 
 (1) (i), (ii), (iii), (iv)  (2) (i), (iii), (iv)   (3) (ii), (iii), (iv)  (4)  (i), (ii), (iii) 
Ans. (4) 
Sol. Theory Based. 
 
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. Determine wavelength of electron in 4
th
 Bohr's orbit ? 
4
th
 cksgj d{kk esa bysDVªkWu dh rjax)Sè;Z dk fuèkkZj.k dhft, ? 
 (1) 4 ?a0  (2) 2 ?a0  (3) 8 ?a0  (4) 6 ?a0 
Ans. (3) 
Sol.  2 ?r = n ? 
 2 ? × 
2
n
Z
a0 = n ? 
 2 ? × 
2
4
1
a0 = n ? 
 ? = 8 ?a0 
 
2. Which of the following species have one unpaired electron each? 
fuEu esa ls dkSulh Lih'kht ¼izR;sd esa½ ,d v;qfXer bysDVªkWu j[krh gSa\ 
 (1) O2, O2
–
  (2) O2, O2
+
   (3) O2
+
, O2
–
   (4) O2, O2
2 –
 
Ans. (3) 
Sol. O2 = ?1s
2
 ?*1s
2
??2s
2
??*2s
2
??2pz
2
??2px
2
 = ?2py
2
 ?*2px
1 
= ?2py
1 
 
3. For Br2( ?) 
Enthalpy of atomisation = x kJ/mol 
 Bond dissociation enthalpy of bromine = y kJ/mole 
 then 
 (1) x > y  (2) x < y  (3) x = y   (4)  Relation does not exist 
 Br2( ?) ds fy, 
ijek.kfodj.k dh ,sUFksYih = x kJ/mol 
 czksehu dh caèk fo;kstu ,sUFksYih = y kJ/mole 
 rc 
 (1) x > y  (2) x < y  (3) x = y   (4)  dksbZ lEcU„èk ugah gksrk gS 
 
Ans. (1)  
Sol. 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?H atomisation = x kJ/mole 
2Br(g) 
Bond energy = y kJ/mole 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?ijek.kfodj.k = x kJ/mole 
2Br(g) 
caèk ÅtkZ = y kJ/mole 
 
?Hatomisation = ?Hvap + Bond energy  
 Hence x > y 
?Hijek.kfodj.k = ?Hvap + caèk ÅtkZ 
 bl izdkj x > y 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
4. Which of the following oxides are acidic, Basic Amphoteric Respectively. 
 fuEu esa ls dkSuls vkWDlkbM Øe'k% vEyh;] {kkjh;] mHk;èkehZ gS&  
 (1) MgO, P4O10, Al2O3 (2) N2O3, Li2O, Al2O3  (3) SO3, Al2O3, Na2O (4) P4O10, Al2O3, MgO 
Ans. (2) 
Sol. Non-metal oxides are acidic in nature 
 alkali metal oxides are basic in nature 
 Al2O3 is amphoteric. 
 vèkkfRod vkWDlkbM vEyh; izd`fr ds gksrs gSA 
 {kkjh; èkkrq vkWDlkbM {kkjh; izd`fr ds gksrs gSA 
 Al2O3 mHk;èkehZ gSA 
 
5. Complex Cr(H2O)6Cln shows geometrical isomerism and also reacts with AgNO3 solution. 
 Given : Spin only magnetic moment = 3.8 B.M.  
 What is the IUPAC name of the complex. 
 (1) Hexaaquachromium(III) chloride    
 (2) Tetraaquadichloridochromium(III) chloride dihydrate   
 (3) Hexaaquachromium(IV) chloride    
 (4)  Tetraaquadichloridochromium(IV) chloride dihydrate   
ladqy Cr(H2O)6Cln T;kferh; leko;ork n'kkZrk gS rFkk ;g AgNO3 foy;u ds lkFk Hkh vfHkd`r gksrk gSA 
 fn;k gS: izpØ.k dsoy pqEcdh; vk?kq.kZ = 3.8 B.M.  
 ladqy dk IUPAC uke D;k gS \ 
 (1) gsDlk,DokØksfe;e (III) DyksjkbM 
 (2) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV 
 (3) gsDlk,DokØksfe;e (IV) DyksjkbM 
 (4) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (IV) DyksjkbM MkbZgkbMªsV 
Ans. (2) 
Sol. Cr(H2O)6Cln ( ?complex)spin = 3.8 B.M. 
 From data of magnetic moment oxidation number of Cr should be +3 
 Hence complex is Cr(H2O)6Cl3. 
 Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl ?2H2O. 
 It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate   
 Cr(H2O)6Cln ( ?ladqy)pØ.k = 3.8 B.M. 
 pqEcdh; vk?kq.kZ ds eku ls Cr dk vkWDlhdj.k vad +3 gksuk pkfg,A  
 bl izdkj ladqy Cr(H2O)6Cl3 gSA 
 ladqy T;kferh; leko;ork n'kkZrk gSA blfy, ladqy dk lw=k [Cr(H2O)4Cl2]Cl ?2H2O gSA 
 bldk IUPAC uke VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV gSA ?
 
6. The electronic configuration of bivalent Europium and trivalent cerium respectively is:   
 (Atomic Number : Xe = 54, Ce = 58, Eu = 63) 
f}la;ksth ;wjksfi;e rFkk f=kla;ksth flfj;e ds bysDVªkWfud foU;kl Øe'k% gS%  
 (ijek.kq Øekd : Xe = 54, Ce = 58, Eu = 63) 
 (1) [Xe]4f
7
, [Xe]4f
1
    (2) [Xe]4f
7
 6s
2
, [Xe]4f
1
  
 (3) [Xe]4f
7
6s
2
, [Xe]4f
1
5d
1
6s
2
   (4) [Xe]4f
7
, [Xe]4f
1
5d
1
6s
2
 
Ans. (1) 
Sol. Eu
2+
 : [Xe]4f
7 
 
Ce
3+
 : [Xe]4f
1 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
7. Ksp of PbCl
2
 = 1.6 × 10
–5 
 
On mixing  
 300 mL, 0.134M Pb(NO3)2(aq.) + 100 mL, 0.4 M NaCl(aq.) 
 (1) Q > Ksp  (2) Q < Ksp  (3) Q = Ksp   (4) Relation does not exit  
PbCl
2
 dk Ksp = 1.6 × 10
–5 
 300 mL, 0.134M Pb(NO3)2(aq.) + 100 mL, 0.4 M NaCl(aq.) feykus ij& 
(1) Q > Ksp     (2) Q < Ksp   
(3) Q = Ksp      (4) fdlh izdkj ds lEcUèk dk vfLrRo ugha gSA  
Ans. (1) 
Sol.  Q = [Pb
2+
][Cl
–
]
2
 
 
2
3 0 0 0 .1 3 4 1 0 0 0 .4
4 0 0 4 0 0
?? ??
??
??
??
 
 ? ?
2 3 0 .1 3 4
0 .1
4
?
?? 
 = 0.105 × 10
–2
 
 =1.005 ×10
–3 
 
Q > K sp 
 
8. Which of the following can not act as both oxidising and reducing agent ? 
fuEu esa ls dkSu vkWDlhdkjd rFkk vipk;d nksuks ds leku O;ogkj ugha dj ldrk gS\ 
 (1) H2SO3   (2) HNO2   (3) H3PO4   (4) H2O2 
Ans. (3) 
Sol. As in H3PO4 Phosphorous is present it's maximum oxidation number state hence it cannot act as 
reducing agent. 
 pwafd H3PO4 esa QkWLQksjl bldh mPpre vkWDlhdj.k voLFkk esa mifLFkr gSA blfy, ;g vipk;d ds leku dk;Z 
ugha dj ldrk gSA 
 
9. First Ionisation energy of Be is higher than that of Boron. 
 Select the correct statements regarding this 
 (i) It is easier to extract electron from 2p orbital than 2s orbital  
 (ii) Penetration power of 2s orbital is greater than 2p orbital 
 (iii) Shielding of 2p electron by 2s electron  
 (iv) Radius of Boron atom is larger than that of Be 
 (1) (i), (ii), (iii), (iv)  (2) (i), (iii), (iv)   (3) (ii), (iii), (iv)  (4)  (i), (ii), (iii) 
 Be dh izFke vk;uu ÅtkZ cksjksu dh rqyuk esa vfèkd gksrh gSA 
 mDr dFku ds lanHkZ esa lgh dFku@dFkuksa dk p;u dhft,sA 
 (i) 2p d{kd ls bysDVªkWu dk i`Fkddj.k 2s d{kd dh rqyuk esa ljy gksrk gSA 
 (ii) 2s d{kd dh Hksnu {kerk 2p d{kd dh rqyuk esa vfèkd gksrh gSA 
 (iii) 2s bysDVªkWu }kjk 2p bysDVªkWu dk ifjj{k.k gksrk gSA 
 (iv) cksjksu ijek.kq dh f=kT;k Be dh rqyuk esa vfèkd gksrh gSA 
 (1) (i), (ii), (iii), (iv)  (2) (i), (iii), (iv)   (3) (ii), (iii), (iv)  (4)  (i), (ii), (iii) 
Ans. (4) 
Sol. Theory Based. 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
10. [PdFClBrI]
2 – 
Number of Geometrical Isomers = n. For [Fe(CN)6]
n –6
, Determine the spin only magnetic 
moment and CFSE (Ignore the pairing energy) 
 (1) 1.73 B.M., –2 ?0  (2) 2.84 B.M., –1.6 ?0  (3) 0, –1.6 ?0   (4) 5.92 B.M., –2.4 ?0 
 [PdFClBrI]
2 – 
T;kfefr; leko;oh;ksa dh la[;k = n. [Fe(CN)6]
n –6 
ds fy;s, izpØ.k dsoy pqEcdh; vk?kw.kZ rFkk 
CFSE dk eku fu/kkZfjr dhft;s (;qXeu ÅtkZ dks ux.; ekurs gq;s) 
Ans. (1)  
Sol. Number of Geometrical Isomers in square planar [PdFClBrI]
2 –
 are = 3 
 Hence, n = 3  
 [Fe(CN)6]
3 –
 
 Fe
3+
 = 3d
5
, According to CFT configuration is 
2 2 1 0 0
2gg
te 
 
? ? 2 nn ??? = 1.73 B.M. 
 
0 2 0
– 0 . 4 0 . 6
g eg
CFSE nt n ? ? ? ? ? ? 
            = –0.4 ?0 × 5 = –2.0 ? 0 
Sol. oxZ leryh; [PdFClBrI]
2 – 
esa T;kfefr; leko;oh;ksa dh la[;k = 3 gSA 
 blfy;s, n = 3  
 [Fe(CN)6]
3 –
 
 Fe
3+
 = 3d
5
, CFT ds vuqlkj foU;kl 
2 2 1 0 0
2gg
te 
 ? ? 2 nn ??? = 1.73 B.M. 
 
0 2 0
– 0 . 4 0 . 6
g eg
CFSE nt n ? ? ? ? ? ? 
            = –0.4 ?0 × 5 = –2.0 ? 0 
 
11. A can reduce BO2 under which conditions.  
 fdl ifjfLFkfr esa A,  BO2 dks vipf;r dj ldrk gS   
1400°C 
?Gf 
A + O2 ? AO2 
B + O2 ??BO2 
T(0°C) 
 
 
 (1) > 1400°C     (2) < 1400°C 
 (3) > 1200°C and (rFkk) < 1400°C   (4) < 1200°C 
Ans. (1) 
Sol. A + BO2 ? ? ? B + AO2 
 ?G = –ve 
 Only above 1400°C 
 dsoy 1400°C ds Åij 
 
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. Determine wavelength of electron in 4
th
 Bohr's orbit ? 
4
th
 cksgj d{kk esa bysDVªkWu dh rjax)Sè;Z dk fuèkkZj.k dhft, ? 
 (1) 4 ?a0  (2) 2 ?a0  (3) 8 ?a0  (4) 6 ?a0 
Ans. (3) 
Sol.  2 ?r = n ? 
 2 ? × 
2
n
Z
a0 = n ? 
 2 ? × 
2
4
1
a0 = n ? 
 ? = 8 ?a0 
 
2. Which of the following species have one unpaired electron each? 
fuEu esa ls dkSulh Lih'kht ¼izR;sd esa½ ,d v;qfXer bysDVªkWu j[krh gSa\ 
 (1) O2, O2
–
  (2) O2, O2
+
   (3) O2
+
, O2
–
   (4) O2, O2
2 –
 
Ans. (3) 
Sol. O2 = ?1s
2
 ?*1s
2
??2s
2
??*2s
2
??2pz
2
??2px
2
 = ?2py
2
 ?*2px
1 
= ?2py
1 
 
3. For Br2( ?) 
Enthalpy of atomisation = x kJ/mol 
 Bond dissociation enthalpy of bromine = y kJ/mole 
 then 
 (1) x > y  (2) x < y  (3) x = y   (4)  Relation does not exist 
 Br2( ?) ds fy, 
ijek.kfodj.k dh ,sUFksYih = x kJ/mol 
 czksehu dh caèk fo;kstu ,sUFksYih = y kJ/mole 
 rc 
 (1) x > y  (2) x < y  (3) x = y   (4)  dksbZ lEcU„èk ugah gksrk gS 
 
Ans. (1)  
Sol. 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?H atomisation = x kJ/mole 
2Br(g) 
Bond energy = y kJ/mole 
 
Br2(
?
) 
Br2(g) 
?HVap. 
?ijek.kfodj.k = x kJ/mole 
2Br(g) 
caèk ÅtkZ = y kJ/mole 
 
?Hatomisation = ?Hvap + Bond energy  
 Hence x > y 
?Hijek.kfodj.k = ?Hvap + caèk ÅtkZ 
 bl izdkj x > y 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
4. Which of the following oxides are acidic, Basic Amphoteric Respectively. 
 fuEu esa ls dkSuls vkWDlkbM Øe'k% vEyh;] {kkjh;] mHk;èkehZ gS&  
 (1) MgO, P4O10, Al2O3 (2) N2O3, Li2O, Al2O3  (3) SO3, Al2O3, Na2O (4) P4O10, Al2O3, MgO 
Ans. (2) 
Sol. Non-metal oxides are acidic in nature 
 alkali metal oxides are basic in nature 
 Al2O3 is amphoteric. 
 vèkkfRod vkWDlkbM vEyh; izd`fr ds gksrs gSA 
 {kkjh; èkkrq vkWDlkbM {kkjh; izd`fr ds gksrs gSA 
 Al2O3 mHk;èkehZ gSA 
 
5. Complex Cr(H2O)6Cln shows geometrical isomerism and also reacts with AgNO3 solution. 
 Given : Spin only magnetic moment = 3.8 B.M.  
 What is the IUPAC name of the complex. 
 (1) Hexaaquachromium(III) chloride    
 (2) Tetraaquadichloridochromium(III) chloride dihydrate   
 (3) Hexaaquachromium(IV) chloride    
 (4)  Tetraaquadichloridochromium(IV) chloride dihydrate   
ladqy Cr(H2O)6Cln T;kferh; leko;ork n'kkZrk gS rFkk ;g AgNO3 foy;u ds lkFk Hkh vfHkd`r gksrk gSA 
 fn;k gS: izpØ.k dsoy pqEcdh; vk?kq.kZ = 3.8 B.M.  
 ladqy dk IUPAC uke D;k gS \ 
 (1) gsDlk,DokØksfe;e (III) DyksjkbM 
 (2) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV 
 (3) gsDlk,DokØksfe;e (IV) DyksjkbM 
 (4) VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (IV) DyksjkbM MkbZgkbMªsV 
Ans. (2) 
Sol. Cr(H2O)6Cln ( ?complex)spin = 3.8 B.M. 
 From data of magnetic moment oxidation number of Cr should be +3 
 Hence complex is Cr(H2O)6Cl3. 
 Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl ?2H2O. 
 It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate   
 Cr(H2O)6Cln ( ?ladqy)pØ.k = 3.8 B.M. 
 pqEcdh; vk?kq.kZ ds eku ls Cr dk vkWDlhdj.k vad +3 gksuk pkfg,A  
 bl izdkj ladqy Cr(H2O)6Cl3 gSA 
 ladqy T;kferh; leko;ork n'kkZrk gSA blfy, ladqy dk lw=k [Cr(H2O)4Cl2]Cl ?2H2O gSA 
 bldk IUPAC uke VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV gSA ?
 
6. The electronic configuration of bivalent Europium and trivalent cerium respectively is:   
 (Atomic Number : Xe = 54, Ce = 58, Eu = 63) 
f}la;ksth ;wjksfi;e rFkk f=kla;ksth flfj;e ds bysDVªkWfud foU;kl Øe'k% gS%  
 (ijek.kq Øekd : Xe = 54, Ce = 58, Eu = 63) 
 (1) [Xe]4f
7
, [Xe]4f
1
    (2) [Xe]4f
7
 6s
2
, [Xe]4f
1
  
 (3) [Xe]4f
7
6s
2
, [Xe]4f
1
5d
1
6s
2
   (4) [Xe]4f
7
, [Xe]4f
1
5d
1
6s
2
 
Ans. (1) 
Sol. Eu
2+
 : [Xe]4f
7 
 
Ce
3+
 : [Xe]4f
1 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
7. Ksp of PbCl
2
 = 1.6 × 10
–5 
 
On mixing  
 300 mL, 0.134M Pb(NO3)2(aq.) + 100 mL, 0.4 M NaCl(aq.) 
 (1) Q > Ksp  (2) Q < Ksp  (3) Q = Ksp   (4) Relation does not exit  
PbCl
2
 dk Ksp = 1.6 × 10
–5 
 300 mL, 0.134M Pb(NO3)2(aq.) + 100 mL, 0.4 M NaCl(aq.) feykus ij& 
(1) Q > Ksp     (2) Q < Ksp   
(3) Q = Ksp      (4) fdlh izdkj ds lEcUèk dk vfLrRo ugha gSA  
Ans. (1) 
Sol.  Q = [Pb
2+
][Cl
–
]
2
 
 
2
3 0 0 0 .1 3 4 1 0 0 0 .4
4 0 0 4 0 0
?? ??
??
??
??
 
 ? ?
2 3 0 .1 3 4
0 .1
4
?
?? 
 = 0.105 × 10
–2
 
 =1.005 ×10
–3 
 
Q > K sp 
 
8. Which of the following can not act as both oxidising and reducing agent ? 
fuEu esa ls dkSu vkWDlhdkjd rFkk vipk;d nksuks ds leku O;ogkj ugha dj ldrk gS\ 
 (1) H2SO3   (2) HNO2   (3) H3PO4   (4) H2O2 
Ans. (3) 
Sol. As in H3PO4 Phosphorous is present it's maximum oxidation number state hence it cannot act as 
reducing agent. 
 pwafd H3PO4 esa QkWLQksjl bldh mPpre vkWDlhdj.k voLFkk esa mifLFkr gSA blfy, ;g vipk;d ds leku dk;Z 
ugha dj ldrk gSA 
 
9. First Ionisation energy of Be is higher than that of Boron. 
 Select the correct statements regarding this 
 (i) It is easier to extract electron from 2p orbital than 2s orbital  
 (ii) Penetration power of 2s orbital is greater than 2p orbital 
 (iii) Shielding of 2p electron by 2s electron  
 (iv) Radius of Boron atom is larger than that of Be 
 (1) (i), (ii), (iii), (iv)  (2) (i), (iii), (iv)   (3) (ii), (iii), (iv)  (4)  (i), (ii), (iii) 
 Be dh izFke vk;uu ÅtkZ cksjksu dh rqyuk esa vfèkd gksrh gSA 
 mDr dFku ds lanHkZ esa lgh dFku@dFkuksa dk p;u dhft,sA 
 (i) 2p d{kd ls bysDVªkWu dk i`Fkddj.k 2s d{kd dh rqyuk esa ljy gksrk gSA 
 (ii) 2s d{kd dh Hksnu {kerk 2p d{kd dh rqyuk esa vfèkd gksrh gSA 
 (iii) 2s bysDVªkWu }kjk 2p bysDVªkWu dk ifjj{k.k gksrk gSA 
 (iv) cksjksu ijek.kq dh f=kT;k Be dh rqyuk esa vfèkd gksrh gSA 
 (1) (i), (ii), (iii), (iv)  (2) (i), (iii), (iv)   (3) (ii), (iii), (iv)  (4)  (i), (ii), (iii) 
Ans. (4) 
Sol. Theory Based. 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
10. [PdFClBrI]
2 – 
Number of Geometrical Isomers = n. For [Fe(CN)6]
n –6
, Determine the spin only magnetic 
moment and CFSE (Ignore the pairing energy) 
 (1) 1.73 B.M., –2 ?0  (2) 2.84 B.M., –1.6 ?0  (3) 0, –1.6 ?0   (4) 5.92 B.M., –2.4 ?0 
 [PdFClBrI]
2 – 
T;kfefr; leko;oh;ksa dh la[;k = n. [Fe(CN)6]
n –6 
ds fy;s, izpØ.k dsoy pqEcdh; vk?kw.kZ rFkk 
CFSE dk eku fu/kkZfjr dhft;s (;qXeu ÅtkZ dks ux.; ekurs gq;s) 
Ans. (1)  
Sol. Number of Geometrical Isomers in square planar [PdFClBrI]
2 –
 are = 3 
 Hence, n = 3  
 [Fe(CN)6]
3 –
 
 Fe
3+
 = 3d
5
, According to CFT configuration is 
2 2 1 0 0
2gg
te 
 
? ? 2 nn ??? = 1.73 B.M. 
 
0 2 0
– 0 . 4 0 . 6
g eg
CFSE nt n ? ? ? ? ? ? 
            = –0.4 ?0 × 5 = –2.0 ? 0 
Sol. oxZ leryh; [PdFClBrI]
2 – 
esa T;kfefr; leko;oh;ksa dh la[;k = 3 gSA 
 blfy;s, n = 3  
 [Fe(CN)6]
3 –
 
 Fe
3+
 = 3d
5
, CFT ds vuqlkj foU;kl 
2 2 1 0 0
2gg
te 
 ? ? 2 nn ??? = 1.73 B.M. 
 
0 2 0
– 0 . 4 0 . 6
g eg
CFSE nt n ? ? ? ? ? ? 
            = –0.4 ?0 × 5 = –2.0 ? 0 
 
11. A can reduce BO2 under which conditions.  
 fdl ifjfLFkfr esa A,  BO2 dks vipf;r dj ldrk gS   
1400°C 
?Gf 
A + O2 ? AO2 
B + O2 ??BO2 
T(0°C) 
 
 
 (1) > 1400°C     (2) < 1400°C 
 (3) > 1200°C and (rFkk) < 1400°C   (4) < 1200°C 
Ans. (1) 
Sol. A + BO2 ? ? ? B + AO2 
 ?G = –ve 
 Only above 1400°C 
 dsoy 1400°C ds Åij 
 
      
 
      
    
    
     
 
                       
 
 
 
 
 
 
12. A ? ? ? B 700 K 
 A 
C
? ? ? B 500 K 
 Rate of reaction in absence of catalyst at 700 K is same as in presence of catalyst at 500 K. If catalyst 
decreases activation energy barrier by 30 kJ/mole, determine activation energy in presence of catalyst. 
(Assume 'A' factor to be same in both cases) 
 700 K ij mRizsjd dh vuqifLFkfr esa vfHkfØ;k dk osx 500 K ij mRizsjd dh mifLFkfr esa vfHkfØ;k ds osx ds leku 
gS ;fn mRizssjd lfØ;.k ÅtkZ vojksèk esa 30 kJ/mole ls deh djrk gSA rc mRizsjd dh mifLFkfr esa lfØ;.k ÅtkZ 
dk fuèkkZj.k dhft,sA (ekuk fd 'A' dkjd nksuks ifjfLFkfr;ksa esa leku gSA) 
 (1) 75 kJ   (2) 135 kJ   (3) 105 kJ   (4) 125 kJ   
Ans.  (1) 
Sol.  Kcat = K 
 
1
1
Ea
RT
Ae
?
 = 
2
2
Ea
RT
Ae
?
 
 
12
12
Ea Ea
TT
? Ea1 = Ea2 – 30 
 
2
Ea 30
500
?
 = 
2
Ea
700
 
 5Ea2 = 7Ea2 – 210 
 Ea2 = 
210
2
 = 105 kJ/mole  
 Activation energy of the catalysed reaction = 105 – 30 = 75 kJ/mole 
 
13. A substance 'X' having low melting point, does not conduct electricity in both solid and liquid state. 'X' 
can be : 
 ,d inkFkZ 'X' de xyukad j[krk gS] Bksl rFkk æo izkoLFkk nksuks esa fo|qr dk dqpkyd gSA 'X' gks ldrk gSA 
 (1) Hg    (2) ZnS   (3) SiC    (4) CCl4   
Ans.  (4) 
Sol. CCl4 ? Non-conductor in solid  and liquid phase. 
CCl4 ? Bksl rFkk æo izkoLFkk esa dqpkyd gSA 
 
14. 
 NH2 
 
Br 
 
(1) NaNO2 + HCl 
 
(2) Cu2Br2 
 (3) HNO3, Conc. H2SO4 
 
 
The major product for above sequence of reaction is :   
mijksDr vfHkfØ;k Øe es fy;s izeq[k mRikn gS& 
 (1) 
 Br 
 
Br 
 
O2N 
   (2) 
 Br 
 
Br 
 
NO2 
 
  (3) 
 Br 
 
Br 
 
O2N 
 
NO2 
 
  (4)   
 Br 
 
Br 
 
NO2 
 
 
Ans.  (2) 
Sol.  
 NH2 
 
Br 
 
(1) NaNO2 + HCl 
 
(2) Cu2Br2 
 
N2
+
Cl
–
 
 
Br 
 
Br 
 
Br 
 
(3)HNO3 
 
Conc. H2SO4 
 
 Br 
 
Br 
 
NO2 
 
 
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

past year papers

,

Sample Paper

,

Objective type Questions

,

Free

,

ppt

,

pdf

,

Important questions

,

Summary

,

JEE Main 2020 Answer Key Chemistry - Morning Shift (09-01-2020) JEE Notes | EduRev

,

MCQs

,

Extra Questions

,

JEE Main 2020 Answer Key Chemistry - Morning Shift (09-01-2020) JEE Notes | EduRev

,

Exam

,

Previous Year Questions with Solutions

,

mock tests for examination

,

shortcuts and tricks

,

video lectures

,

JEE Main 2020 Answer Key Chemistry - Morning Shift (09-01-2020) JEE Notes | EduRev

,

Semester Notes

,

practice quizzes

,

Viva Questions

,

study material

;