1. Identify the ionic species in each oxide

We check whether the oxide/peroxide/superoxide ion present has any unpaired electrons.

• $Li_2O \rightarrow 2Li^+ + O^{2-}$
• $CaO \rightarrow Ca^{2+} + O^{2-}$
• $Na_2O_2 \rightarrow 2Na^+ + O_2^{2-}$
• $KO_2 \rightarrow K^+ + O_2^{-}$
• $MgO \rightarrow Mg^{2+} + O^{2-}$
• $K_2O \rightarrow 2K^+ + O^{2-}$

2. Check magnetic nature of each anion

(i) Oxide ion: $O^{2-}$

Oxygen has 8 electrons, so $O^{2-}$ has 10 electrons. Its configuration is like Ne: $1s^2 2s^2 2p^6$ All electrons are paired, so $O^{2-}$ is diamagnetic.

Hence:

• $Li_2O$ is diamagnetic
• $CaO$ is diamagnetic
• $MgO$ is diamagnetic
• $K_2O$ is diamagnetic

(ii) Peroxide ion: $O_2^{2-}$

Using MO theory, $O_2$ has 12 valence electrons; $O_2^{2-}$ has 14 valence electrons. The extra electrons fill the antibonding $\pi^*$ orbitals completely, so all electrons are paired. Thus $O_2^{2-}$ is diamagnetic.

Hence:

• $Na_2O_2$ is diamagnetic

(iii) Superoxide ion: $O_2^{-}$

$O_2^{-}$ has 13 valence electrons. In MO theory, one electron remains unpaired in the $\pi^*$ antibonding orbital. Thus $O_2^{-}$ is paramagnetic.

Hence:

• $KO_2$ is paramagnetic

3. Count paramagnetic oxides

Only $KO_2$ is paramagnetic.

Therefore, the number of paramagnetic oxides is: $1$

4. Evaluate options

• A: $1$ ✅
• B: $2$ ❌
• C: $3$ ❌
• D: $0$ ❌

So the correct option is A.