First slide

Molecular Orbital Theory

Question

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

Moderate

A

$NO$
B

${CN}^{-}$
C

${CN}^{+}$
D

$CN$

Solution

$NO : (σ 1 s)^{2}, {(σ^{⋆} 1 s)}^{2}, (σ 2 s)^{2}, {(σ^{} 2 s)}^{2}, {(σ 2 p_{z})}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}, {(π^{⋆} 2 p_{x})}^{1} = {(π^{⋆} 2 p_{y})}^{0}$
$B. O . = \frac{10 - 5}{2} = 2.5$
${CN}^{-} : (σ 1 s)^{2}, {(σ^{⋆} 1 s)}^{2}, (σ 2 s)^{2}, {(σ^{} 2 s)}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}, {(σ 2 p_{z})}^{2}$
$B.O. = \frac{10 - 4}{2} = 3$
$CN : (σ 1 s)^{2}, {(σ^{⋆} 1 s)}^{2}, (σ 2 s)^{2}, {(σ^{⋆} 2 s)}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}, {(σ 2 p_{z})}^{1}$
$B . O \cdot = \frac{9 - 4}{2} = 2.5$
${CN}^{+} : (σ 1 s)^{2}, {(σ^{⋆} 1 s)}^{2}, (σ 2 s)^{2}, {(σ^{⋆} 2 s)}^{2}, {(π 2 p_{x})}^{2} = {(π 2 p_{y})}^{2}$
$B.O. = \frac{8 - 4}{2} = 2$
$Hence, {CN}^{-} has highest bond order .$

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