Question

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

• A. 3
• B. 7
• C. 1
• D. 1.5

Answer 1

The correct option is A 3

We know that
Bond order = $\frac{numberofelectronsinB.M.O-numberofelectronsinA.B.M.O.}{2}$

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 $=\frac{10-4}{2}=\frac{6}{2}=3$