Question

The species having bond order different from that of CO is

• A. ${NO}^{-}$
• B. ${NO}^{+}$
• C. ${CN}^{-}$
• D. $N_2$

Answer 1

The correct option is A ${NO}^{-}$

CO has 14 electrons. If we write down the electronic configuration, we will end up with $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p^4\sigma 2p^2$

${NO}^{-}$ has 16 electrons. The difference here as compared with CO is that $\sigma 2p^2$ will be before $\pi 2p^4$ and there are 2 extra electrons. So, the final configuration will be $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p^2\pi 2p^4{\pi }^{\ast }2p^2$

In the other 3 cases, i.e. ${NO}^{+}$, ${CN}^{-}$ and $N_2$ we have 14 electrons. So, the electronic configurations of all these species will be the same as CO.

So, ${NO}^{-}$ is the answer.

Note: This question could have been directly solved just by looking at the number of electrons in each species. It is still a good practice to keep track of the molecular orbitals.