What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species: N$$_2$$, N$$_2^+$$, O$$_2$$, O$$_2^+$$?

Explanation

For molecules up to $$N_2$$, the molecular orbital energy order is

$$\sigma_{2s}<\sigma_{2s}^{}<\pi_{2p_x}=\pi_{2p_y}<\sigma_{2p_z}<\pi_{2p}^{}$$

For $$O_2$$ and beyond, the order is

$$\sigma_{2s}<\sigma_{2s}^{}<\sigma_{2p_z}<\pi_{2p_x}=\pi_{2p_y}<\pi_{2p}^{}$$

For $$N_2$$,

$$N_2=(\sigma_{1s})^2(\sigma_{1s}^{})^2(\sigma_{2s})^2(\sigma_{2s}^{})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\sigma_{2p_z})^2$$

The HOMO is $$\sigma_{2p_z}$$ and it contains two paired electrons.

Therefore, the number of unpaired electrons in the HOMO is

$$0$$

For $$N_2^+$$, one electron is removed from the HOMO.

$$N_2^+=(\sigma_{2p_z})^1$$

The HOMO contains one unpaired electron.

Therefore,

$$1$$

unpaired electron is present in the HOMO.

For $$O_2$$,

$$(\pi_{2p_x}^{})^1(\pi_{2p_y}^{})^1$$

The HOMO consists of the degenerate $$\pi^{*}$$ orbitals. According to Hund's rule, the two electrons occupy separate orbitals.

Therefore, the number of unpaired electrons in the HOMO is

$$2$$

For $$O_2^+$$, one electron is removed from the $$\pi^{*}$$ level.

Only one unpaired electron remains in the HOMO.

Therefore, the number of unpaired electrons is

$$1$$

Hence, for

$$N_2,\ N_2^+,\ O_2,\ O_2^+$$

the numbers of unpaired electrons in the HOMO are

$$0,\ 1,\ 2,\ 1$$

Hence, the correct answer is

$$0,\ 1,\ 2,\ 1$$