Two pi and half sigma bonds are present in | vedclass

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(D) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.For $N_2^+$,the total electrons are $13$. The molecular

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$N_2^+$

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(D) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.For $N_2^+$,the total electrons are $13$. The molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^1$.Here,$N_b = 9$ and $N_a = 4$. So,$B.O. = \frac{1}{2} (9 - 4) = 2.5$.In $N_2^+$,the triple bond structure of $N_2$ is modified by removing one electron from the bonding $\sigma 2p_z$ orbital,resulting in $2 \ \pi$ bonds and $0.5 \ \sigma$ bond.

Two $\pi$ and half $\sigma$ bonds are present in

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