Bonding in which of the following diatonic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?

(A) NO

(B) N2

(C) O2

(D) C2

(E) B2

Choose the most appropriate answer from the options given below

Option 1 -

(A), (B), (C) only            

Option 2 -

(B), (C), (E) only  

Option 3 -

(A), (C) only

Option 4 -

(D) only

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    Bond strength  Bond order

    N2σ1s2σ1s*2σ2s2σ2s*2π2px2π2py2σ2pz2

    O2σ1s2σ1s*2σ2s2σ2s*2σ2pz2π2px2π2py2π2px*1π2py*1

    C2σ1s2σ1s*2σ2s2σ2s*2π2px2π2py2

    B2σ1s2σ1s*2σ2s2σ2s*2π2px1π2py1

    NO Number of electron = 7 + 8 = 15

    B.O. Similar to N2

    N2σ1s2σ1s*2σ2s2σ2s*2π2px2π2py2σ2pz2π2px*1π2py*

    B.O. of N2 = 3B.O of C2842=2

    Removal of e form antibonding molecular orbital increases bond order.

    In NO & O2 has valance e in orbital.

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