What is the bond order in CO molecule? | vedclass

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(B) According to Molecular Orbital Theory,the bond order is calculated as $\frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_

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$3$

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(B) According to Molecular Orbital Theory,the bond order is calculated as $\frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons. For $CO$ molecule,the total number of electrons is $6 + 8 = 14$. The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Here,$N_b = 10$ and $N_a = 4$. Bond order = $\frac{10 - 4}{2} = \frac{6}{2} = 3$.

What is the bond order in $CO$ molecule?

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