Question

Which one of the following pair of species have the same bond order?

• A. $CO,NO$
• B. $O_2,N{O}^{+}$
• C. $C{N}^{-},CO$
• D. $N_2,O_2^{-}$

Answer 1

The correct option is C $C{N}^{-},CO$

MO electronic configuration of $NO$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^0$
$\text{Bond order}=\frac{10-5}{2}=2.5$

MO electronic configuration of $CO$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2$
$\text{Bond order}=\frac{10-4}{2}=3$
The pair $CO$ and $N{O}^{+}$ do not have same bond order.

MO electronic configuration of $N{O}^{+}$ is
$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2$
$\text{Bond order}=\frac{10-4}{2}=3$

MO electronic configuration of $O_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$
Bond order = 2.
the pair $N{O}^{+}and{O}_2$ do not have same bond order.


MO electronic configuration of $C{N}^{-}$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2=\pi 2p_y^2,\sigma 2p_z^2$
$\text{Bond order}=\frac{10-4}{2}=3$
Bond order of CO is also 3 as $C{N}^{-}$


MO electronic configuration of $O_2^{-}$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,{\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^1$
Bond order = 1.5


MO electronic configuration of $N_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2=\pi 2p_y^2,\sigma 2p_z^2$
Bond order = 3
$N_{2}and{O}_2^{-}$ also do not have same bond order.

Only $C{N}^{-}andCO$ have same bond order of 3.
Hence, option (c) is correct answer.