Q. The relationship between the dissociation energy and $N_{2}$ and $N_{2}^{+}$ is
A) dissociation energy of $N_{2} =$ dissociation energy of$\ N_{2}^{+}$
B) dissociation energy of $N_{2}$ can either be lower or higher than the dissociation energy of $N_{2}^{+}$
C) dissociation energy of $N_{2} >$ dissociation energy of$\ N_{2}^{+}$
D) dissociation energy of${\ N}_{2}^{+}$ $>$ dissociation energy of$\ N_{2}$
Explanation
(c)
(i)$N_{2}$ (14 electrons)
= $KK^{*},\ \sigma 2s^{2},\ \sigma 2s^{2},\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2},\ \sigma 2p_{z}^{2}$
Bond Order $= \frac{1}{2}\left( N_{b} - N_{a} \right)$
$= \frac{1}{2}(8 - 2) = 3$
(ii)$N_{2}^{+}$ (13 electrons)
= $KK^{*},\ {\sigma 2s^{2},\ }_{\sigma}^{*}2s^{2},\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2},\ \sigma 2p_{z}^{1}$
Bond Order $= \frac{1}{2}(7 - 2)\$= 2.5
Since, bond dissociation energy $\propto$ bond order. Hence, bond dissociation energy of $N_{2}$ is greater than that of the bond dissociation energy of ${\ N}_{2}^{+}\$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.