What is the bond order in N_2^{+}? | vedclass

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(D) The total number of electrons in $N_2^{+}$ is $13$. The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^*

Vedclass · Accepted Answer

$2.5$

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(D) The total number of electrons in $N_2^{+}$ is $13$. The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$. Number of bonding electrons $(N_b)$ $= 9$. Number of antibonding electrons $(N_a)$ $= 4$. Bond order $= \frac{N_b - N_a}{2} = \frac{9 - 4}{2} = \frac{5}{2} = 2.5$.

What is the bond order in $N_2^{+}$?

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