To determine which of the given compounds is paramagnetic, we need to identify the species that has one or more unpaired electrons. We will analyze each option.

Step 1: Analyze Na₂O₂ (Sodium peroxide)

1. Na₂O₂ is an ionic compound formed from sodium (Na) and oxygen (O).
2. Sodium is an alkali metal and always forms a +1 ion (Na⁺).
3. Since there are two sodium atoms, the total positive charge is +2.
4. To maintain electrical neutrality, the anion must be the peroxide ion, O₂²⁻.
5. Now, let's determine the magnetic nature of the O₂²⁻ ion using Molecular Orbital Theory (MOT).
6. A neutral oxygen atom (O) has 8 electrons. An O₂ molecule has 16 electrons. The O₂²⁻ ion has 16 + 2 = 18 electrons.
7. The molecular orbital configuration for O₂²⁻ is: $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^2 \pi^*(2p_y)^2$
8. All the molecular orbitals are completely filled with paired electrons. Therefore, the peroxide ion (O₂²⁻) is diamagnetic.
9. Since both Na⁺ and O₂²⁻ ions are diamagnetic, Na₂O₂ is a diamagnetic compound.

Step 2: Analyze O₃ (Ozone)

1. Ozone (O₃) is a covalent molecule. Each oxygen atom contributes 6 valence electrons, so the total number of valence electrons is 3 × 6 = 18.
2. The Lewis structure of ozone involves resonance: $\text{Ö=Ö}^+-\text{Ö:}^- \longleftrightarrow \text{}^-\text{:Ö-Ö}^+=\text{Ö}$
3. In both resonance structures, all 18 valence electrons are accounted for as bonding pairs or lone pairs. There are no unpaired electrons.
4. Therefore, O₃ is a diamagnetic molecule.

Step 3: Analyze N₂O (Nitrous oxide)

1. N₂O is a covalent molecule. Each nitrogen atom contributes 5 valence electrons, and the oxygen atom contributes 6. The total number of valence electrons is (2 × 5) + 6 = 16.
2. The most stable resonance structure for N₂O is: $:\text{N} \equiv \text{N}^+ - \text{Ö:}^-$
3. In this structure, all 16 valence electrons are paired (in bonds or as lone pairs).
4. Therefore, N₂O is a diamagnetic molecule.

Step 4: Analyze KO₂ (Potassium superoxide)

1. KO₂ is an ionic compound formed from potassium (K) and oxygen (O).
2. Potassium is an alkali metal and forms a +1 ion (K⁺).
3. To maintain electrical neutrality, the anion must be the superoxide ion, O₂⁻.
4. Let's determine the magnetic nature of the O₂⁻ ion using MOT.
5. A neutral O₂ molecule has 16 electrons. The O₂⁻ ion has 16 + 1 = 17 electrons.
6. The molecular orbital configuration for O₂⁻ is: $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^2 \pi^*(2p_y)^1$
7. The π* antibonding molecular orbitals (π₂pₓ and π₂pᵧ) contain three electrons. One of these orbitals is fully filled with a pair of electrons, while the other contains a single, unpaired electron.
8. The presence of this unpaired electron makes the superoxide ion (O₂⁻) paramagnetic.
9. Since KO₂ contains the paramagnetic O₂⁻ ion, KO₂ is a paramagnetic compound.

Conclusion

Based on the analysis of each compound:

• Na₂O₂ is diamagnetic.
• O₃ is diamagnetic.
• N₂O is diamagnetic.
• KO₂ is paramagnetic.

Thus, the paramagnetic compound among the given options is KO₂.