Identification of Paramagnetic Compound Among Given Oxides

To determine which compound is paramagnetic, we analyze the electronic configuration and the presence of unpaired electrons in each species:

1. Sodium Peroxide ($\text{Na}_2\text{O}_2$):

   • It is an ionic compound composed of two $\text{Na}^+$ ions and a peroxide ion, $\text{O}_2^{2-}$.
   • The total number of electrons in $\text{O}_2^{2-}$ is $8 + 8 + 2 = 18$.
   • According to Molecular Orbital (MO) Theory, the electron configuration of the $\text{O}_2^{2-}$ ion is:
     $\sigma_{1s}^2\ \sigma_{1s}^{*2}\ \sigma_{2s}^2\ \sigma_{2s}^{*2}\ \sigma_{2p_z}^2\ (\pi_{2p_x}^2 = \pi_{2p_y}^2)\ (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2})$
   • All electrons are paired, so $\text{Na}_2\text{O}_2$ is diamagnetic.

2. Ozone ($\text{O}_3$):

   • Ozone has a total of $3 \times 8 = 24$ electrons ($18$ valence electrons).
   • In its ground electronic state, ozone is a closed-shell singlet molecule with a bent geometry, and all electron spins are paired.
   • Hence, $\text{O}_3$ is diamagnetic.

3. Nitrous Oxide ($\text{N}_2\text{O}$):

   • The total number of electrons is $2(7) + 8 = 22$ electrons ($16$ valence electrons).
   • Nitrous oxide is a linear molecule with closed-shell electronic structures in its resonance forms ($\text{N}\equiv\text{N}^+-\text{O}^- \longleftrightarrow \text{N}^-=\text{N}^+=\text{O}$).
   • All electrons are paired, making $\text{N}_2\text{O}$ diamagnetic.

4. Potassium Superoxide ($\text{KO}_2$):

   • It consists of $\text{K}^+$ and the superoxide anion, $\text{O}_2^-$.
   • The total number of electrons in $\text{O}_2^-$ is $8 + 8 + 1 = 17$ electrons (an odd-electron species).
   • Its molecular orbital configuration is:
     $\sigma_{1s}^2\ \sigma_{1s}^{*2}\ \sigma_{2s}^2\ \sigma_{2s}^{*2}\ \sigma_{2p_z}^2\ (\pi_{2p_x}^2 = \pi_{2p_y}^2)\ (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*1})$
   • Since there is one unpaired electron in the antibonding $\pi^*$ orbital, the superoxide ion $\text{O}_2^-$ possesses a net magnetic moment.
   • Consequently, $\text{KO}_2$ is paramagnetic.

Therefore, the paramagnetic compound is $\text{KO}_2$.

Correct Option: (D)