Q 1 :

Number of molecules/species from the following having one unpaired electron is __________.

O2,O2-1,NO,CN-1,O22-                            [2024]



(2)

Molecules Number of electrons Electronic configuration Number of
unpaired electrons
     O2     16 σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px1=π*2py1       2
     O2-     17 σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px2=π*2py1       1
    NO     15 σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px1=π*2py0       1 
    CN-     14 σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz2       0
    O22-     18 σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px2=π*2py2       0

 



Q 2 :

Number of molecules having bond order 2 from the following molecule is _________ .

C2,O2,Be2,Li2,Ne2,N2,He2                         [2024]



(2)

Molecule Electronic configuration Bond order =12[Nb-Na]
      C2 σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2                     12[8-4]=2
      O2   σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px1=π*2py1                     12[10-6]=2
     Be2 σ1s2<σ*1s2<σ2s2<σ*2s2                     12[4-4]=0
     Li2 σ1s2<σ*1s2<σ2s2                      12[4-2]=1
     Ne2 σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px2=π*2py2<σ*2pz2                      12[10-10]=0
     N2 σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz2                      12[10-4]=3
     He2 σ1s2<σ*1s2                      12[2-2]=0

 



Q 3 :

The total number of species from the following in which one unpaired electron is present, in ____________ .

N2,O2,C2-,O2-,O22-,H2+,CN-,He2+                              [2024]



(4)

Molecule Electronic configuration Number of unpaired electrons
       N2 σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz2         0           
       O2 σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px1=π*2py1         2
       C2- σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz1         1
       O2- σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px2=π*2py1         1
       O22- σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px2=π*2py2         0
      H2+ σ1s1         1
     CN- σ1s2<σ*1s2<σ2s2<σ*2s2<π2px2=π2py2<σ2pz2         0
      He2+ σ1s2<σ*1s1         1

 



Q 4 :

Total number of electrons present in (π*) molecular orbitals of O2,O2+ and O2- is _________ .               [2024]



(6)

Molecule Electronic configuration Electrons present in π*
       O2 σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px1=π*2py1                           2
       O2+ σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px1=π*2py0                           1
      O2- σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px2=π2py2<π*2px2=π*2py1                           3

Sum = 2 + 1 + 3 = 6



Q 5 :

When ψA and ψB are the wave function of atomic orbitals, then σ* is represented by:                   [2024]

  • ψA-2ψB

     

  • ψA+ψB

     

  • ψA+2ψB

     

  • ψA-ψB

     

(4)

Constructive overlap of wavefunctions (ψA+ψB) give bonding molecular orbital (σ or π) and destructive overlap of wavefunctions (ψA-ψB) give antibonding molecular orbitals (σ* or π*). Axial overlap of orbitals give σ (or σ*) bond and lateral overlap of orbitals give π (or π*) bond.

 



Q 6 :

Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z-direction]?

(A)  2pz and 2px

(B)  2s and 2px

(C)  3dxy and 3dx2-y2

(D)  2s and 2pz

(E)  2pz and 3dx2-y2

Choose the correct answer from the options given below:                                [2025]

  • (A) and (B) only

     

  • (E) only

     

  • (D) only

     

  • (C) and (D) only

     

(3)

 



Q 7 :

Which of the following molecules(s) show/s paramagnetic behavior?                       [2025]

(A)  O2
(B)  N2
(C)  F2
(D)  S2
(E)  Cl2

Choose the correct answer from the options given below:

  • B only

     

  • A and C only

     

  • A and E only

     

  • A and D only

     

(4)

Molecule Electronic Configuration Number of Unpaired Electrons Magnetic Nature
O2 σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2, π2px2, π2py2, π*2px1, π*2py1 2 Paramagnetic
N2 σ1s2, σ*1s2, σ2s2, σ*2s2, π2px2, π2py2, σ2pz2 0 Diamagnetic
F2 σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2, π2px2, π2py2, π*2px2, π*2py2 0 Diamagnetic
S2 KK, LL, σ3s2, σ*3s2, σ3pz2, π3px2, π3py2, π*3px1, π*3py1 2 Paramagnetic
Cl2 KK, LL, σ3s2, σ*3s2, σ3pz2, π3px2, π3py2, π*3px2, π*3py2 0 Diamagnetic

 



Q 8 :

What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species:

N2; N2+; O2; O2+ ?                           [2023]

  • 0, 1, 2, 1

     

  • 2, 1, 2, 1

     

  • 0, 1, 0, 1

     

  • 2, 1, 0, 1

     

(1)

Species Number of Unpaired Electrons
(N2):(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px2=π2py2)(σ2pz)2 0
(N2+):(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2px2=π2py2)(σ2pz)1 1
(O2): (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2=π2py2)(π*2px1=π*2py1) 2
(O2+): (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2=π2py2)(π*2px1=π*2py0) 1

 



Q 9 :

According to MO theory the bond orders for O22-, CO and NO+ respectively, are                     [2023]

  • 1, 3 and 3

     

  • 2, 3 and 3

     

  • 1, 2 and 3

     

  • 1, 3 and 2

     

(1)

Species Molecular Orbital configuration Bond order = 12(Nb-Na)
CO KK σ2s2*σ2s*2(π2px2π2py2)σ2pz2   12(10-4)=3
NO+ KK σ2s2*σ2s*2σ2pz2(π2px2π2py2)(π2px0*π2py0*) 12(10-4)=3
O2-2 KK σ2s2*σ2s*2σ2pz2(π2px2π2py2)(π2px*2π2py*2) 12(10-8)=1

 



Q 10 :

The bond order and magnetic property of acetylide ion are same as that of                [2023]

  • N2+

     

  • NO+

     

  • O2+

     

  • O2-

     

(2)

Species Total number of electrons Bond order Unpaired electrons (n) Magnetic momentum (n×(n+2)BM)
Acetylides ion (C2-2) 14 3 0 0
O2- 17 1.5 1 3
O2+ 15 2.5 1 3
N2+ 13 2.5 1 3
NO+ 14 3 0 0