Number of paramagnetic oxides among the following given oxides is _________.
Li$$_2$$O, CaO, Na$$_2$$O$$_2$$, KO$$_2$$, MgO and K$$_2$$O

First, recall the basic idea: a species shows paramagnetism when it possesses one or more unpaired electrons. Diamagnetic species have all electrons paired. For oxides of s-block metals, the magnetic behaviour is governed by the type of oxygen-containing ion present - oxide $$\left(\mathrm O^{2-}\right)$$, peroxide $$\left(\mathrm O_2^{2-}\right)$$ or superoxide $$\left(\mathrm O_2^{-}\right)$$.

We now look at each given oxide.

1. Li2O, CaO, MgO and K2O
All four are simple ionic oxides. In each of them the oxygen is present as the monatomic oxide ion $$\mathrm O^{2-}$$.

The oxide ion has

$$\text{electrons} = 8\;(\text{from O atom}) + 2\;(\text{extra}) = 10$$

This is the same electron count as neon. Every orbital (1s, 2s, 2p) is completely filled, so

$$\text{number of unpaired electrons} = 0$$

Hence $$\mathrm O^{2-}$$ is diamagnetic, and therefore Li2O, CaO, MgO and K2O are all diamagnetic.

2. Na2O2 (peroxide)
This compound contains the peroxide ion $$\mathrm O_2^{2-}$$. Let us use the molecular-orbital (MO) picture of the O2 molecule and then add electrons.

For O2 (16 electrons) the last two electrons occupy the degenerate antibonding $$\pi^\ast_{2p_x}$$ and $$\pi^\ast_{2p_y}$$ orbitals singly, giving two unpaired electrons and paramagnetism. The peroxide ion has two extra electrons:

$$16 + 2 = 18\;\text{electrons}$$

These two extra electrons pair up in the antibonding $$\pi^\ast$$ orbitals, so now

$$\text{number of unpaired electrons} = 0$$

Thus $$\mathrm O_2^{2-}$$ and hence Na2O2 are diamagnetic.

3. KO2 (superoxide)
KO2 contains the superoxide ion $$\mathrm O_2^{-}$$. Starting again from O2 (16 electrons) and adding one extra electron gives

$$16 + 1 = 17\;\text{electrons}$$

With 17 electrons, one of the antibonding $$\pi^\ast$$ orbitals now contains a single unpaired electron:

$$\text{number of unpaired electrons} = 1$$

Because of this unpaired electron, $$\mathrm O_2^{-}$$ is paramagnetic, and therefore KO2 is paramagnetic.

Counting the paramagnetic oxides
Only KO2 is paramagnetic. All the others are diamagnetic.

So, the required number of paramagnetic oxides is

$$1$$

Hence, the correct answer is Option A.