Question

Which of the following pairs of species has the same bond order?

• A. $N{O}^{+}\text{and}C{N}^{+}$
• B. $N{O}^{+}\text{and}C{N}^{-}$
• C. $C{N}^{+}\text{and}C{N}^{-}$
• D. $O_2^{-}\text{and}C{N}^{-}$

Answer 1

The correct option is B $N{O}^{+}\text{and}C{N}^{-}$

Molecular electronic configuration of $N{O}^{+}$ is
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2=\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^0=\left({\pi }_{2p_y}^{\ast }{)}^0\right({\sigma }_{2p_z}^{\ast }{)}^0$
Therefore, bond order $=\frac{6}{2}=3$

Molecular electronic configuration of $C{N}^{+}$ is
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^1=\left({\pi }_{2p_y}{)}^1\right({\pi }_{2p_x}^{\ast }{)}^0=\left({\pi }_{2p_y}^{\ast }{)}^0\right({\sigma }_{2p_z}^{\ast }{)}^0$
Therefore, bond order $=\frac{4}{2}=2$

Molecular electronic configuration of $C{N}^{-}$ is
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2=\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^0=\left({\pi }_{2p_y}^{\ast }{)}^0\right({\sigma }_{2p_z}^{\ast }{)}^0$
Therefore, bond order $=\frac{6}{2}=3$

Molecular electronic configuration of $O_2^{-}$ is
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2=\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^2=\left({\pi }_{2p_y}^{\ast }{)}^1\right({\sigma }_{2p_z}^{\ast }{)}^0$
Therefore, bond order $=\frac{3}{2}=1.5$

$N{O}^{+}\text{and}C{N}^{-}$ has same bond order.