Determine the species present in each compound

• $\mathrm{Li_2O}$ contains the oxide ion, $\mathrm{O^{2-}}$.
• $\mathrm{Na_2O_2}$ contains the peroxide ion, $\mathrm{O_2^{2-}}$.
• $\mathrm{KO_2}$ contains the superoxide ion, $\mathrm{O_2^-}$.

Check magnetic behaviour from electron pairing

Magnetic behaviour depends on whether unpaired electrons are present:

• All electrons paired $\Rightarrow$ diamagnetic
• One or more unpaired electrons $\Rightarrow$ paramagnetic

For $\mathrm{Li_2O}$

In $\mathrm{Li_2O}$, oxygen is present as $\mathrm{O^{2-}}$.

Oxygen atom: $1s^2 2s^2 2p^4$

So, $\mathrm{O^{2-}}: 1s^2 2s^2 2p^6$

All electrons are paired, so $\mathrm{O^{2-}}$ is diamagnetic.

Hence, $\mathrm{Li_2O}$ is diamagnetic.

For $\mathrm{Na_2O_2}$

In $\mathrm{Na_2O_2}$, the anion is peroxide, $\mathrm{O_2^{2-}}$.

Using MO theory for dioxygen species:

• $\mathrm{O_2}$ has 2 unpaired electrons.
• Adding 2 electrons gives $\mathrm{O_2^{2-}}$.
• These extra electrons fill the antibonding $\pi^*$ orbitals completely.

Thus, in $\mathrm{O_2^{2-}}$, all electrons are paired.

Therefore, $\mathrm{Na_2O_2}$ is diamagnetic.

For $\mathrm{KO_2}$

In $\mathrm{KO_2}$, the anion is superoxide, $\mathrm{O_2^-}$.

Starting from $\mathrm{O_2}$, adding 1 electron gives $\mathrm{O_2^-}$. This results in one unpaired electron in the antibonding molecular orbitals.

Therefore, $\mathrm{O_2^-}$ is paramagnetic.

Hence, $\mathrm{KO_2}$ is paramagnetic.

Final sequence

$\mathrm{Li_2O}: \text{diamagnetic}$ $\mathrm{Na_2O_2}: \text{diamagnetic}$ $\mathrm{KO_2}: \text{paramagnetic}$

So the correct order is:

diamagnetic, diamagnetic and paramagnetic

Match with options

This corresponds to Option D.