What is the bond order in N_{2} molecule? | vedclass

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(D) 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, \si

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The electronic configuration of $N_{2}$ ($14$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.Bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.Here,$N_b = 10$ and $N_a = 4$.$	ext{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3$.Thus,the bond order in $N_{2}$ is $3$.

What is the bond order in $N_{2}$ molecule?

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