Amongst the following the number of oxide(s) which are paramagnetic in nature is ______
$$Na_2O, KO_2, NO_2, N_2O, ClO_2, NO, SO_2, Cl_2O$$

To determine whether an oxide is paramagnetic or diamagnetic, we check for the presence of unpaired electrons.

Species containing unpaired electrons are paramagnetic.

Species with all electrons paired are diamagnetic.

$$\mathrm{Na_2O}$$

Contains:

$$\mathrm{Na^+ \ and\ O^{2-}}$$

The oxide ion has completely filled orbitals:

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

All electrons are paired.

Hence:

$$\mathrm{Na_2O \ is\ diamagnetic}$$

$$\mathrm{KO_2}$$

Contains the superoxide ion:

$$\mathrm{O_2^-}$$

The superoxide ion possesses one unpaired electron in the antibonding $$\mathrm{\pi^*}$$ orbital.

Hence:

$$\mathrm{KO_2 \ is\ paramagnetic}$$

$$\mathrm{NO_2}$$

Total valence electrons:

$$\mathrm{= 5 + 2(6) = 17}$$

Odd number of electrons ⇒ one unpaired electron.

Hence:

$$\mathrm{NO_2 \ is\ paramagnetic}$$

$$\mathrm{ClO_2}$$

Total valence electrons:

$$\mathrm{= 7 + 2(6) = 19}$$

Odd number of electrons ⇒ one unpaired electron.

Hence:

$$\mathrm{ClO_2 \ is\ paramagnetic}$$

$$\mathrm{NO}$$

Total valence electrons:

$$\mathrm{= 5 + 6 = 11}$$

Odd number of electrons ⇒ one unpaired electron.

Hence:

$$\mathrm{NO \ is\ paramagnetic}$$

$$\mathrm{N_2O}$$

Total valence electrons:

$$\mathrm{= 2(5) + 6 = 16}$$

All electrons are paired.

Hence:

$$\mathrm{N_2O \ is\ diamagnetic}$$

$$\mathrm{SO_2}$$

Total valence electrons:

$$\mathrm{= 6 + 2(6) = 18}$$

All electrons are paired.

Hence:

$$\mathrm{SO_2 \ is\ diamagnetic}$$

$$\mathrm{Cl_2O}$$

Total valence electrons:

$$\mathrm{= 2(7) + 6 = 20}$$

All electrons are paired.

Hence:

$$\mathrm{Cl_2O \ is\ diamagnetic}$$

Paramagnetic species:

$$\mathrm{KO_2,\ NO_2,\ ClO_2,\ NO}$$

Total number of paramagnetic oxides:

$$\boxed{\mathrm{4}}$$