The bond order in N_2 molecule is | vedclass

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Question

(C) The electronic configuration of $N_2$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma

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$3$

Vedclass · Answer

(C) The electronic configuration of $N_2$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.Number of bonding electrons $(N_b)$ = $2 + 2 + 2 + 2 + 2 = 10$.Number of antibonding electrons $(N_a)$ = $2 + 2 = 4$.Bond order = $\frac{N_b - N_a}{2} = \frac{10 - 4}{2} = \frac{6}{2} = 3$.Thus,the bond order in $N_2$ is $3$.

The bond order in $N_2$ molecule is

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