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Correct order of dissociation energy of $N_2$ and $N_2^+$ is :
A
$N_2 \gt N_2^+$
B
$N_2 = N_2^+$
C
$N_2^+ \gt N_2$
D
None
Detailed Solution
$N_2$ (14 electrons): $\sigma 1s^2\,\sigma^* 1s^2\,\sigma 2s^2\,\sigma^* 2s^2\,\pi 2p_x^2\,\pi 2p_y^2\,\sigma 2p_z^2$. Bond order $= \dfrac{10 - 4}{2} = 3$.
$N_2^+$ (13 electrons): one electron is removed from the bonding $\sigma 2p_z$ orbital. Bond order $= \dfrac{9 - 4}{2} = 2.5$.
Higher bond order means a stronger bond and greater dissociation energy.
So the dissociation energy of $N_2 \gt N_2^+$.
$N_2^+$ (13 electrons): one electron is removed from the bonding $\sigma 2p_z$ orbital. Bond order $= \dfrac{9 - 4}{2} = 2.5$.
Higher bond order means a stronger bond and greater dissociation energy.
So the dissociation energy of $N_2 \gt N_2^+$.