1. Identify the relevant molecular orbitals for oxygen species

For $\mathrm{O}_2$, $\mathrm{O}_2^+$, and $\mathrm{O}_2^-$, we only need the electrons in the antibonding pi orbitals, i.e. $\pi^*$ molecular orbitals.

For oxygen and heavier second-period homonuclear diatomic molecules, the MO order is:

2. Write the valence electron count

• $\mathrm{O}_2$: $6+6=12$ valence electrons
• $\mathrm{O}_2^+$: $11$ valence electrons
• $\mathrm{O}_2^-$: $13$ valence electrons

3. Fill the molecular orbitals

(a) $\mathrm{O}_2$

Configuration:

So, electrons in $\pi^*$ orbitals of $\mathrm{O}_2$:

(b) $\mathrm{O}_2^+$

One electron is removed from the highest occupied MO of $\mathrm{O}_2$, which is a $\pi^*$ orbital.

So configuration has:

(c) $\mathrm{O}_2^-$

One electron is added to the next available MO of $\mathrm{O}_2$, which is again a $\pi^*$ orbital.

So:

4. Total number of electrons in $\pi^*$ orbitals

5. Final Answer

The total number of electrons present in $\left(\pi^*\right)$ molecular orbitals of $\mathrm{O}_2, \mathrm{O}_2^+$ and $\mathrm{O}_2^-$ is

6. Comparison with stored correct answer

Stored correct answer = $6$

Our derived answer also = $6$, so they agree.