Question

The relationship between the dissociation energy of $N_ 2$ and $N_2^+$ is

• A. dissociation energy of $N_2^+$ > dissociation energy of $N_2$
• B. dissociation energy of $N_2$ = dissociation energy of $N_2^+$
• C. dissociation energy of $N_2$ > 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 1

Answer: (C)

dissociation energy of $N_2$ > dissociation energy of $N_2^+$

Answer 2

The dissociation energy will be more when the bond order will be greater or bond order cc dissociation energy
Molecular orbital configuration of $N_2 (14) = \sigma 1s^2, _{\sigma}^{*}1s^2 , \sigma 2s^2 ,_{\sigma}^{*}2s^2 , \sigma 2 P_y^2 \approx \pi 2 p_z^2 , \sigma 2p_x^2$
So, bond order of $N_2 = \frac{ N_b -N_a }{ 2} = \frac{ 10 - 4}{2} =3$ and bond order of $N_2^+ = \frac{ 9 - 4 }{2} = 2.5$
As the bond order of $N_2$ is greater than $N_2^+$ so, the dissociation energy of $N_2$ will be greater than $N_2^+$.