The bond order of NO molecule is | vedclass

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(C) The total number of electrons in $NO$ is $7 + 8 = 15$. According to Molecular Orbital Theory,the electronic configuration of $NO$ is: $\sigma 1s^2

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$2.5$

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(C) The total number of electrons in $NO$ is $7 + 8 = 15$. According to Molecular Orbital Theory,the electronic configuration of $NO$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order is calculated as: $B.O. = \frac{N_b - N_a}{2}$. Here,$N_b$ (bonding electrons) $= 10$ and $N_a$ (antibonding electrons) $= 5$. $B.O. = \frac{10 - 5}{2} = \frac{5}{2} = 2.5$.

The bond order of $NO$ molecule is

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