The bond order is not 3 for | vedclass

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Question

(A) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2}[N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$

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$N_2^+$

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(A) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2}[N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.$(A)$ For $N_2^+$ ($13$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. $B.O. = \frac{1}{2}[9 - 4] = 2.5$.$(B)$ For $O_2^{2+}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{1}{2}[10 - 4] = 3$.$(C)$ For $N_2$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{1}{2}[10 - 4] = 3$.$(D)$ For $NO^+$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{1}{2}[10 - 4] = 3$.Thus,the bond order is not $3$ for $N_2^+$.

The bond order is not $3$ for

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