Superoxides of alkali metals act as oxidising agents while normal oxids are basic in nature. The oxide which is paramagnetic in nature due to presence of unparied electron is

Options

A. `Na_(2)O_(2)`
B. `KO_(2)`
C. `Na_(2)O`
D. `K_(2)O_(2)`

Explanation

To solve the question regarding which oxide of alkali metals is paramagnetic due to the presence of unpaired electrons, we will analyze the superoxides and normal oxides of alkali metals step by step.

### Step-by-Step Solution:

1. **Identify the Oxides**:
- The question mentions superoxides and normal oxides of alkali metals. The superoxides include \( \text{Na}_2\text{O}_2 \) (sodium superoxide) and \( \text{K}_2\text{O}_2 \) (potassium superoxide), while normal oxides include \( \text{Na}_2\text{O} \) (sodium oxide) and \( \text{K}_2\text{O} \) (potassium oxide).

2. **Determine the Oxidation States**:
- In superoxides like \( \text{Na}_2\text{O}_2 \) and \( \text{K}_2\text{O}_2 \), the oxide ion is \( \text{O}_2^{-} \) (superoxide ion).
- In normal oxides like \( \text{Na}_2\text{O} \) and \( \text{K}_2\text{O} \), the oxide ion is \( \text{O}^{2-} \).

3. **Check for Unpaired Electrons**:
- To find out which oxide is paramagnetic, we need to check the electronic configuration of the superoxide ion \( \text{O}_2^{-} \) and the normal oxide ion \( \text{O}^{2-} \).

4. **Molecular Orbital Configuration**:
- For \( \text{O}_2^{-} \):
- Total electrons = 8 (for 2 O atoms) + 1 (for the negative charge) = 17 electrons.
- The molecular orbital filling will be:
- \( \sigma 1s^2 \)
- \( \sigma^* 1s^2 \)
- \( \sigma 2s^2 \)
- \( \sigma^* 2s^2 \)
- \( \sigma 2p_z^2 \)
- \( \sigma 2p_x^2 \)
- \( \sigma 2p_y^2 \)
- \( \sigma^* 2p_y^1 \) (this is where the unpaired electron is)
- Thus, \( \text{O}_2^{-} \) has one unpaired electron, making it paramagnetic.

- For \( \text{O}^{2-} \):
- Total electrons = 8 (for 2 O atoms) + 2 (for the negative charge) = 18 electrons.
- The molecular orbital filling will be:
- \( \sigma 1s^2 \)
- \( \sigma^* 1s^2 \)
- \( \sigma 2s^2 \)
- \( \sigma^* 2s^2 \)
- \( \sigma 2p_z^2 \)
- \( \sigma 2p_x^2 \)
- \( \sigma 2p_y^2 \)
- \( \sigma^* 2p_y^2 \)
- \( \sigma^* 2p_z^2 \)
- Thus, \( \text{O}^{2-} \) has no unpaired electrons, making it diamagnetic.

5. **Conclusion**:
- Since \( \text{O}_2^{-} \) has an unpaired electron and is paramagnetic, the oxide that is paramagnetic in nature due to the presence of unpaired electrons is \( \text{KO}_2 \) (potassium superoxide).

### Final Answer:
The oxide which is paramagnetic in nature due to the presence of unpaired electron is \( \text{KO}_2 \).