Carbon monoxide has ten bonding electrons and four antibonding electrons, then bond order will be:

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1.5

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Solution

The correct option is A 3
We know that
Bond order = $\frac{n u m b e r o f e l e c t r o n s i n B . M . O - n u m b e r o f e l e c t r o n s i n A . B . M . O .}{2}$

where
B.M.O = Bonding molecular orbital
A.B.M.O = Antibonding molecular orbital
Given,
electrons in B.M.O = 10
electrons in A.B.M.O = 4
Bond order $= \frac{10 - 4}{2} = \frac{6}{2} = 3$

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