The bond order of $C O$ molecule on the basis of molecular orbital theory is:

Zero

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2

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3

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1

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Solution

The correct option is B $3$
Species $C O (14)$ , contain bonding electrons $= 10$ and antibonding electrons $= 4.$

Bond order $= \frac{N_{b} - N_{a}}{2} = \frac{10 - 4}{2} = 3.$

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