(A) 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, \sigm

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(A) 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: $\text{Bond order} = \frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3$.

Class 11 Chemistry Chemical Bonding and Molecular Structure Molecular orbital theory

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What is the bond order of $N_2$?

TS EAMCET 2015 All TS EAMCET

A
$3$
B
$4$
C
$2$
D
$1$

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What is the bond order of $N_2$?

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