Write the molecular orbital diagram of N2+ and calculate their bond order

Asked by sonkarshiva009 | 13th Mar, 2019, 05:47: PM

Expert Answer:

Electronic configuration of N-atom(Z=7) is  1 s to the power of 2 space end exponent 2 s squared 2 p subscript x superscript 1 2 p subscript y superscript 1 2 p subscript z superscript 1.
 
The total number of electrons present in the N2 molecule is 14.
 
N2+ ion is formed by the loss of one electron from the N2 molecule. This lost electron will be lost from σ(2pz) orbital. Hence, the electronic configuration of  N2+ ion will be
 
N2+ = KK[σ(2s)]2 [σ*(2s)]2 [π(2px)]2 [π(2py)]2 [σ(2pz)]1
 
Here, Nb =7, Na=2 so that
 
Bond space order space space equals space space 1 half left square bracket straight N subscript straight b minus straight N subscript straight a right square bracket space space space space space space space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space space space space space space space equals space 1 half left square bracket 7 minus 2 right square bracket space space space space space space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space space space space space space space equals space space 2.5
 
The  molecular  orbital diagram (MO  Diagram)   of  N2+ ion is as follows:
 

Answered by Ramandeep | 14th Mar, 2019, 05:09: PM