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

 



Q 11 :

In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one?           [2023]

  • N2N2+

     

  • NONO+

     

  • O2O2+

     

  • O2O22-

     

(2)

NO is paramagnetic with B.O = 2.5, NO+ is diamagnetic with B.O = 3.

 



Q 12 :

Consider the following statements:

(A) NF3 molecule has a trigonal planar structure.

(B) Bond length of N2 is shorter than O2.

(C) Isoelectronic molecules or ions have identical bond order.

(D) Dipole moment of H2S is higher than that of water molecule.

Choose the correct answer from the options given below:                       [2023]

  • (A) and (B) are correct

     

  • (B) and (C) are correct

     

  • (C) and (D) are correct

     

  • (A) and (D) are correct

     

(2)

(A) NF3 has triangular pyramidal structure

(B) Molecule:N2O2B.O.32B.L.N2<O2   
        
(C) Isoelectronic molecules or ions have identical bond order.

(D) Dipole moment H2S<H2O due to less electronegativity difference between H and S as compared to H and O.