The magnetic behaviour of `Li_(2)O,Na_(2)O_(2)` and `KO_(2)`,respectively,are

Options

A. diamagnetic,paramagnetic and diamagnetic
B. diamagnetic,diamagnetic and paramagnetic
C. paramagnetic,paramagnetic and diamagnetic
D. paramagnetic,diamagnetic and paramagnetic

Answer

To determine the magnetic behavior of the compounds \( \text{Li}_2\text{O} \), \( \text{Na}_2\text{O}_2 \), and \( \text{KO}_2 \), we need to analyze the oxidation state of oxygen in each compound and the total number of electrons present.

### Step-by-Step Solution:

1. **Identify the Oxidation State of Oxygen:**
- In all three compounds, oxygen is present in the \( O^{2-} \) (oxide) state.

2. **Count the Total Number of Electrons:**
- For \( \text{Li}_2\text{O} \):
- Each lithium ion \( \text{Li}^+ \) contributes 1 electron.
- Total electrons from 2 lithium ions = \( 2 \times 1 = 2 \)
- Oxygen in \( O^{2-} \) contributes 8 electrons (as \( O \) has 8 electrons in its neutral state).
- Total electrons in \( \text{Li}_2\text{O} = 2 + 8 = 10 \) electrons.

- For \( \text{Na}_2\text{O}_2 \):
- Each sodium ion \( \text{Na}^+ \) contributes 1 electron.
- Total electrons from 2 sodium ions = \( 2 \times 1 = 2 \)
- Oxygen in \( O^{2-} \) contributes 8 electrons.
- Total electrons in \( \text{Na}_2\text{O}_2 = 2 + 8 + 8 = 18 \) electrons (since there are two \( O^{2-} \) ions).

- For \( \text{KO}_2 \):
- Each potassium ion \( \text{K}^+ \) contributes 1 electron.
- Total electrons from 2 potassium ions = \( 2 \times 1 = 2 \)
- Oxygen in \( O^{2-} \) contributes 8 electrons.
- Total electrons in \( \text{KO}_2 = 2 + 8 + 8 + 1 = 19 \) electrons (since there is one \( O_2^{2-} \) ion, contributing 8 electrons, and one \( O^{-} \) ion contributing 1 additional electron).

3. **Determine Magnetic Behavior:**
- **Diamagnetic:** Species with an even number of total electrons (no unpaired electrons).
- **Paramagnetic:** Species with an odd number of total electrons (at least one unpaired electron).

- For \( \text{Li}_2\text{O} \): 10 electrons (even) → **Diamagnetic**
- For \( \text{Na}_2\text{O}_2 \): 18 electrons (even) → **Diamagnetic**
- For \( \text{KO}_2 \): 19 electrons (odd) → **Paramagnetic**

4. **Conclusion:**
- The magnetic behavior of \( \text{Li}_2\text{O} \), \( \text{Na}_2\text{O}_2 \), and \( \text{KO}_2 \) respectively is: **Diamagnetic, Diamagnetic, Paramagnetic**.

### Final Answer:
- The magnetic behavior of \( \text{Li}_2\text{O} \), \( \text{Na}_2\text{O}_2 \), and \( \text{KO}_2 \) is **Diamagnetic, Diamagnetic, Paramagnetic**.