Chemical Bonding and Molecular Structure - Result Question 92
####94. Consider the following species :
$CN^{+}, CN^{-}, NO$ and $CN$
Which one of these will have the highest bond order?
[2018]
(a) $NO$
(b) $CN^{-}$
(c) $CN$
(d) $CN^{+}$
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Solution:
- (b) $NO:(\sigma 1 s)^{2},(\sigma^{} 1 s)^{2},(\sigma 2 s)^{2},(\sigma^{} 2 s)^{2},(\sigma 2 p_z)^{2}$, $(\pi 2 p_x)^{2}=(\pi 2 p_y)^{2},(\pi^{*} 2 p_x)^{1}=(\pi * 2 p_y)^{0}$
B.O. $=\frac{10-5}{2}=2.5$
$CN^{-}:(\sigma 1 s)^{2},(\sigma^{} 1 s)^{2},(\sigma 2 s)^{2},(\sigma^{} 2 s)^{2}$, $(\pi 2 p_x)^{2}=(\pi 2 p_y)^{2},(\sigma 2 p_z)^{2}$
B.O. $=\frac{10-4}{2}=3$
$CN:(\sigma 1 s)^{2},(\sigma^{} 1 s)^{2},(\sigma 2 s)^{2},(\sigma^{} 2 s)^{2}$, $(\pi 2 p_x)^{2}=(\pi 2 p_y)^{2},(\sigma 2 p_z)^{1}$
B.O. $=\frac{9-4}{2}=2.5$
$CN^{+}:(\sigma 1 s)^{2},(\sigma^{} 1 s)^{2},(\sigma 2 s)^{2},(\sigma^{} 2 s)^{2}$,
$ (\pi 2 p_x)^{2}=(\pi 2 p_y)^{2} $
B.O. $=\frac{8-4}{2}=2$
Hence, option (2) should be the right answer.
