Consider the following species :
` CN^(+) ,CN^(-) ,NO and CN`
Which one of these will have the highest bond order?

Options

A. `CN^(+)`
B. `CN^(-)`
C. `NO`
D. `CN`

Answer

B

Explanation

The formula of bond order is given as

Energy level pattern for molecular orbitals of different molecules depends upon their central atom.
No : Central atom is N
(Total number of electrons=15)
`therefore sigma 1s^(2),sigma^(* *)1s^(2), sigma2s^(2),sigma^(* *)2s^(2)`
`(pi2p_(x)^(2)=pi2p_(y)^(2)),sigma2p_(z)^(2),(pi2_(px)^(* *1)~~pi2_(py)^(* *0))`
`B.O.=(10-5)/(2)=2.5`
`CN^(-)` : Central atom is C
[Total number of electrons=14]
`sigma1s^(2),sigma^(* *)1s^(2),sigma2s^(2),sigma^(* *)2s^(2),(pi2px^(2)~~pi 2py^(2)sigma2p_(z)^(2))`
`B.O.=(10-4)/(2)=3`
CN : Central atom is C
[Total number of electrons=13]
`sigma1s^(2),sigma^(* *)1s^(2),sigma2s^(2),sigma^(* *)2s^(2),(pi2p_(x)^(2)~~pi2p_(y)^(2)),sigma2p_(z)^(1)`
`B.O. =(9-4)/(2)=2.5`
`CN^(+)` : Central atom is C
[Total number of electrons=12]
`sigma1s^(2),sigma^(* *)1s^(2),sigma2s^(2),sigma2s^(2),sigma^(* *)2s^(2),(pi2p_(x)^(2)~~pi2p_(y)^(2))`
`B.O. =(8-4)/(2)=2`
Therefore, option (b) is correct.