Consider the following species, CN ^ + , CN ^ - 1 , NOandCN Which...

Consider the following species,

{CN}^{+}, {CN}^{- 1}, NOandCN

Which one of these will have the highest bond order?

Options

B. ${CN}^{-}$

Number of

{e}^{-}

NO = 7 + 8 = 15

NO = \sigma1{s}^{2} < \overset{*}{\sigma}1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma}2{s}^{2} < \sigma2{P}_{z}^{2} < \pi2{P}_{y}^{2} = \pi2{Px}^{2} < \overset{*}{\pi}2{P}_{y}^{1} = \overset{*}{\pi}2{P}_{x}^{0}

Bond order

= \dfrac{\text{(number of electron in bonding molecular orbital - number of electron}inantibondingmolecularorbital )}{2}

BO = \dfrac{6 - 1}{2} = 2 .5

.

Number of

{e}^{-}

CN = 6 + 7 = 13

NO = \sigma1{s}^{2} < \overset{*}{\sigma}1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma}2{s}^{2} < \sigma2{P}_{z}^{2} < \pi2{P}_{y}^{2} = \pi2{Px}^{2} < \overset{*}{\pi}2{P}_{y}^{1} = \overset{*}{\pi}2{P}_{x}^{0}

.

Bond order

= \dfrac{5}{2} = 2.5

Number of

{e}^{-}

{CN}^{+} = 6 + 7 - 1 = 12

NO = \sigma1{s}^{2} < \overset{*}{\sigma}1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma}2{s}^{2} < \sigma2{P}_{z}^{2} < \pi2{P}_{y}^{2} = \pi2{Px}^{2} < \overset{*}{\pi}2{P}_{y}^{1} = \overset{*}{\pi}2{P}_{x}^{0}

Bond order

= \dfrac{4 - 0}{2} = 2

{CN}^{-} =

number of

{e}^{-}

{CN}^{-} = 6 + 7 + 1 = 14

{CN}^{-} = \sigma1{s}^{2} < \overset{*}{\sigma}1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma}2{s}^{2} < \sigma2{P}_{z}^{2} < \pi2{P}_{y}^{2} = \pi2{Px}^{2} < \overset{*}{\pi}2{P}_{y}^{1} = \overset{*}{\pi}2{P}_{x}^{0}

Bondorder = \dfrac{6 - 0}{2} = 3

.<br /><br />

{CN}^{-} as highest bond order.