Among the following the paramagnetic compound is

Options

A. `Na_2O_2`
B. `O_3`
C. `N_2O`
D. `KO_2`

Answer

To determine which of the given compounds is paramagnetic, we need to analyze the number of electrons in each compound and check for unpaired electrons. A paramagnetic compound has at least one unpaired electron, while a diamagnetic compound has all paired electrons.

### Step-by-Step Solution:

1. **Identify the Compounds**: We have the following compounds to analyze:
- Na⁺
- O₂²⁻
- Ozone (O₃)
- N₂O
- KO₂

2. **Calculate the Total Number of Electrons**:
- **Na⁺**: Sodium (Na) has an atomic number of 11, meaning it has 11 electrons. When it loses one electron to become Na⁺, it has 10 electrons (11 - 1 = 10).
- **O₂²⁻**: Oxygen (O) has an atomic number of 8, so O₂ has 16 electrons (8 + 8). Gaining 2 electrons makes it O₂²⁻, which has 18 electrons (16 + 2 = 18).
- **Ozone (O₃)**: Three oxygen atoms contribute 24 electrons (8 + 8 + 8 = 24).
- **N₂O**: Nitrogen (N) has an atomic number of 7, and each nitrogen contributes 7 electrons. Therefore, N₂ contributes 14 electrons. Oxygen contributes 8 electrons, making a total of 22 electrons (14 + 8 = 22).
- **KO₂**: Potassium (K) has an atomic number of 19, and O₂ contributes 16 electrons. So, KO₂ has a total of 19 + 16 = 35 electrons.

3. **Determine the Parity of Electrons**:
- **Na⁺**: 10 electrons (even) - diamagnetic
- **O₂²⁻**: 18 electrons (even) - diamagnetic
- **Ozone (O₃)**: 24 electrons (even) - diamagnetic
- **N₂O**: 22 electrons (even) - diamagnetic
- **KO₂**: 35 electrons (odd) - paramagnetic

4. **Conclusion**: Among the compounds analyzed, KO₂ is the only compound that has an odd number of electrons, indicating that it has unpaired electrons and is therefore paramagnetic.

### Final Answer:
The paramagnetic compound is KO₂.