Question

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

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

Answer 1

The correct option is B $C{N}^{–}$

According to molecular orbital theory (MOT), bond order is given as

bond order = (no. of bonding electrons - no. of antibonding electrons )/2

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

bond order = (8-4)/2 = 2

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

bond order = (9-4)/2 = 2.5

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

bond order = (10-4)/2 = 3

$NO$ : $(\sigma 1s{)}^2$, $({\sigma }^{\ast }1s{)}^2$, $(\sigma 2s{)}^2$, $({\sigma }^{\ast }2s{)}^2$, $(\pi 2p_z{)}^2$, $(\pi 2p_x{)}^2$ = $(\pi 2p_y{)}^2$, $({\pi }^{\ast }2p_x{)}^1$ = $({\pi }^{\ast }2p_y{)}^0$

bond order = (10-5)/2 = 2.5

so, $C{N}^{-}$ has the highest bond order