The relationship between the dissociation energy of `N_2` and `N_2^+` is

Options

A. dissociation energy of `N_2^+` gt dissociation energy of `N_2`
B. dissociation energy of `N_2` = dissociation energy of `N_2^+`
C. dissociation energy of `N_2` gt dissociation energy of `N_2^+`
D. dissociation energy of `N_2`can either be lower or higher than the dissociation energy of `N_2^+`

Answer

C

Explanation

The dissociation energy will be more when the bond order will be greater and bond order `prop` dissociation energy
Molecular orbital configuration of `N_2(14)=sigma1s^2 , overset***sigma1s^2, sigma2s^2 , overset***sigma2s^2, sigma2p_y^2~~ pi2p_z^2,pi2p_x^2`
So, bond order of `N_2=(N_b-N_a)/2=(10-4)/2=3`
and bond order of `N_2^(+)=(9-4)/2=2.5`
As the bond order of `N_2` is greater than `N_2^+` so , the dissociation energy of `N_2` will be greater than `N_2^+`