Consider the following species: , NO and CN. Which one of these will have the highest bond order? [2018]

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Solution

Q.
Consider the following species: ${CN}^{+}, {CN}^{-}$ , NO and CN. Which one of these will have the highest bond order [2018]

1 NO

2 ${CN}^{-}$

3 ${CN}^{+}$

4 CN

Ans.
(2)

$N O (15) :$ ${(σ 1 s)}^{2}, {(σ^{} 1 s)}^{2}, {(σ 2 s)}^{2}, {(σ^{} 2 s)}^{2}, {(σ 2 p_{z})}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}, {(π^{} 2 p_{x})}^{1} = {(π^{} 2 p_{y})}^{0}$

$B.O. = \frac{10 - 5}{2} = 2.5$

${CN}^{-} (14) :$ ${(σ 1 s)}^{2}, {(σ^{} 1 s)}^{2}, {(σ 2 s)}^{2}, {(σ^{} 2 s)}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}, {(σ 2 p_{z})}^{2}$

$B.O. = \frac{10 - 4}{2} = 3$

$CN (13) :$ ${(σ 1 s)}^{2}, {(σ^{} 1 s)}^{2}, {(σ 2 s)}^{2}, {(σ^{} 2 s)}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}, {(σ 2 p_{z})}^{1}$

$B.O. = \frac{9 - 4}{2} = 2.5$

${CN}^{+} (12) :$ ${(σ 1 s)}^{2}, {(σ^{} 1 s)}^{2}, {(σ 2 s)}^{2}, {(σ^{} 2 s)}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}$

$B.O. = \frac{8 - 4}{2} = 2$

Hence, $C N^{-}$ has highest bond order.

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