Question

Consider the following species, ${CN}^{+},{CN}^{-},NO$ and $CN$

Which one of these will have the highest bond order?

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

Answer 1

The correct option is D $C{N}^{-}$

$C{N}^{+}$: Valence electrons=$4+5-1=8$

$[{\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^2$

$BO=\frac{1}{2}\times (4-0)=2$

$NO$: Valence electrons = $5+6=11$

${\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^1$

$BO=\frac{1}{2}\times (6-1)=2.5$

$CN$: Valence electrons = $4+5=9$

${\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^3$

$BO=\frac{1}{2}\times (5-0)=2.5$

$C{N}^{-}$: Valence electrons = $4+5+1=10$

${\left(\sigma 2s\right)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4$

$BO=\frac{1}{2}\times (6-0)=3$

Hence, highest bond order is for $C{N}^{-}$

Hence, option $D$ is correct