Question

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

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

Answer 1

Answer: (D)

$NO:(\sigma 1s{)}^2,{\left({\sigma }^{\ast }1s\right)}^2,(\sigma 2s{)}^2,{\left({\sigma }^{\ast }2s\right)}^2,{\left(\sigma 2p_z\right)}^2$

${\left(\pi 2p_x\right)}^2={\left(\pi 2p_y\right)}^2,{\left({\pi }^{\ast }2p_x\right)}^1={\left({\pi }^{2}2p_y\right)}^0$

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

$C{N}^{-}:(\sigma 1s{)}^2,{\left({\sigma }^{\ast }1s\right)}^2,(\sigma 2s{)}^2,{\left({\sigma }^{\ast }2s\right)}^2,{\left(\pi 2p_x\right)}^2$

$={\left(\pi 2p_y\right)}^2,{\left(\sigma 2z_z\right)}^2$

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

$CN:(\sigma 1s{)}^2,{\left({\sigma }^{\ast }1s\right)}^2,(\sigma 2s{)}^2,{\left({\sigma }^{\ast }2s\right)}^2,{\left(\pi 2p_x\right)}^2$

$={\left(\pi 2p_y\right)}^2,{\left(\sigma 2p_z\right)}^1$

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

$C{N}^{+}:(\sigma 1s{)}^2,{\left({\sigma }^{\ast }1s\right)}^2,(\sigma 2s{)}^2,{\left({\sigma }^{\ast }2s\right)}^2,{\left(\pi 2p_x\right)}^2$

$={\left(\pi 2p_y\right)}^2$

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