1. We need to count how many of the given oxides are paramagnetic.

2. A species is paramagnetic if it has one or more unpaired electrons.

3. Now check each oxide one by one:

(i) $\mathrm{Na_2O}$

• This is an ionic oxide: $\mathrm{Na^+}$ and $\mathrm{O^{2-}}$.
• $\mathrm{O^{2-}}$ has completely filled orbitals.
• Hence, diamagnetic.

(ii) $\mathrm{KO_2}$

• This is potassium superoxide: $\mathrm{K^+O_2^-}$.
• The superoxide ion $\mathrm{O_2^-}$ has one unpaired electron.
• Hence, paramagnetic.

(iii) $\mathrm{NO_2}$

• Total valence electrons: $5 + 2\times 6 = 17$
• Odd-electron molecule, so it has one unpaired electron.
• Hence, paramagnetic.

(iv) $\mathrm{N_2O}$

• Total valence electrons: $2\times 5 + 6 = 16$
• Even-electron molecule; usual Lewis structures show all electrons paired.
• Hence, diamagnetic.

(v) $\mathrm{ClO_2}$

• Total valence electrons: $7 + 2\times 6 = 19$
• Odd-electron molecule, so it has one unpaired electron.
• Hence, paramagnetic.

(vi) $\mathrm{NO}$

• Total valence electrons: $5 + 6 = 11$
• Odd-electron molecule, so it has one unpaired electron.
• Hence, paramagnetic.

(vii) $\mathrm{SO_2}$

• Total valence electrons: $6 + 2\times 6 = 18$
• All electrons are paired in the ground state.
• Hence, diamagnetic.

(viii) $\mathrm{Cl_2O}$

• Total valence electrons: $2\times 7 + 6 = 20$
• All electrons are paired.
• Hence, diamagnetic.

4. Therefore, the paramagnetic oxides are:

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

So, the total number of paramagnetic oxides is $4$

5. Comparison with stored answer:

• Stored correct answer = $4$
• Derived answer = $4$
• Hence, they agree.