Question

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

• A. $CN$
• B. $NO$
• C. $ CN^{+}$
• D. $CN^{-}$

Answer 1

The Correct Option is D

$NO :(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\sigma 2 p _{ z }\right)^{2}$
$\left(\pi 2 p _{ x }\right)^{2}=\left(\pi 2 p _{ y }\right)^{2},\left(\pi^{*} 2 p _{ x }\right)^{1}=\left(\pi^{2} 2 p _{ y }\right)^{0}$
$BO =\frac{10-5}{2}=2.5$
$CN ^{-}:(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\pi 2 p _{ x }\right)^{2}$
$=\left(\pi 2 p _{ y }\right)^{2},\left(\sigma 2 z _{ z }\right)^{2}$
$BO =\frac{10-4}{2}=3$
$CN :(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\pi 2 p _{ x }\right)^{2}$
$=\left(\pi 2 p _{ y }\right)^{2},\left(\sigma 2 p _{ z }\right)^{1}$
$BO =\frac{9-4}{2}=2.5$
$CN ^{+}:(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\pi 2 p _{ x }\right)^{2}$
$=\left(\pi 2 p _{ y }\right)^{2}$
$BO =\frac{8-4}{2}=2$