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Bond order of a species is 2.5 and the n...

Bond order of a species is `2.5` and the number of electons in its bonding molecular orbital is found to be 8 The no. of electons in its antibonding molecular orbital is

A

three

B

four equivalent orbitals at `109^(@)` 28' to each other will be formed.

C

zero

D

cannot be calculated form the given information.

Text Solution

Verified by Experts

The correct Answer is:
A
Bond order `(1)/(2) (N_(b)-n_(a))`
`2.5 = (1)/(2)(8 - n_(a)) rArr 5 = 8 - n_(a)rArr n_(a)= 8 - 5 = 3`
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