Concept used: Molecular Orbital (MO) theory and magnetism

A species is:

• diamagnetic if all electrons are paired.
• paramagnetic if it has one or more unpaired electrons.

MO configuration of $O_2$

For $O_2$, the valence MO electronic configuration is: $\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2=\pi(2p_y)^2\,\pi^*(2p_x)^1\,\pi^*(2p_y)^1$

Thus, $O_2$ has two unpaired electrons, so it is paramagnetic.

Check each option

A. $O_2^{2-}$

Start from $O_2$ and add 2 electrons.

These two electrons enter the antibonding $\pi^*$ orbitals and pair up the two unpaired electrons: $\pi^*(2p_x)^2\,\pi^*(2p_y)^2$

Now all electrons are paired.

Therefore, $O_2^{2-}$ is diamagnetic.

B. NO

Total electrons in NO: $7 + 8 = 15$ An odd-electron molecule must have at least one unpaired electron.

Therefore, NO is paramagnetic.

C. $O_2^+$

Start from $O_2$ and remove one electron from a $\pi^*$ orbital.

Then one $\pi^*$ orbital still contains one unpaired electron.

Therefore, $O_2^+$ is paramagnetic.

D. $O_2$

As discussed above, it has two unpaired electrons in $\pi^*$ orbitals.

Therefore, $O_2$ is paramagnetic.

Conclusion

The only species showing diamagnetic behaviour is: $\boxed{O_2^{2-}}$

Comparison with stored answer

Stored correct answer: A

Derived answer: A

They match.