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

Answer: (D)

$NO :(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\sigma 2 p _{ z }\right)^{2}$

$\left(\pi 2 p _{ x }\right)^{2}=\left(\pi 2 p _{ y }\right)^{2},\left(\pi^{*} 2 p _{ x }\right)^{1}=\left(\pi^{2} 2 p _{ y }\right)^{0}$

$BO =\frac{10-5}{2}=2.5$

$CN ^{-}:(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\pi 2 p _{ x }\right)^{2}$

$=\left(\pi 2 p _{ y }\right)^{2},\left(\sigma 2 z _{ z }\right)^{2}$

$BO =\frac{10-4}{2}=3$

$CN :(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\pi 2 p _{ x }\right)^{2}$

$=\left(\pi 2 p _{ y }\right)^{2},\left(\sigma 2 p _{ z }\right)^{1}$

$BO =\frac{9-4}{2}=2.5$

$CN ^{+}:(\sigma 1 s )^{2},\left(\sigma^{*} 1 s \right)^{2},(\sigma 2 s )^{2},\left(\sigma^{*} 2 s \right)^{2},\left(\pi 2 p _{ x }\right)^{2}$

$=\left(\pi 2 p _{ y }\right)^{2}$

$BO =\frac{8-4}{2}=2$