Which of the following has maximum number of lone pairs associated with `Xe`?

Options

A. `XeF_(4)`
B. `XeF_(6)`
C. `XeF_(2)`
D. `XeO_(3)`

Explanation

To determine which compound has the maximum number of lone pairs associated with xenon (Xe), we will analyze the given compounds one by one by calculating their hybridization and the corresponding lone pairs.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we need to analyze are:
- Xenon Tetrafluoride (XeF4)
- Xenon Hexafluoride (XeF6)
- Xenon Difluoride (XeF2)
- Xenon Trioxide (XeO3)

2. **Calculate Hybridization for Each Compound**:
- **Formula for Hybridization**:
\[ \text{Hybridization} = \frac{\text{Valence Electrons of Central Atom} + \text{Number of Monovalent Atoms} + \text{(Negative Charge)} - \text{(Positive Charge)}}{2} \]

- **XeF4**:
- Valence Electrons of Xe = 8
- Monovalent Atoms (F) = 4
- Hybridization = \(\frac{8 + 4}{2} = 6\) → sp³d²
- Bond Pairs = 4 (from 4 F atoms)
- Lone Pairs = \(6 - 4 = 2\)

- **XeF6**:
- Valence Electrons of Xe = 8
- Monovalent Atoms (F) = 6
- Hybridization = \(\frac{8 + 6}{2} = 7\) → sp³d³
- Bond Pairs = 6 (from 6 F atoms)
- Lone Pairs = \(7 - 6 = 1\)

- **XeF2**:
- Valence Electrons of Xe = 8
- Monovalent Atoms (F) = 2
- Hybridization = \(\frac{8 + 2}{2} = 5\) → sp³d
- Bond Pairs = 2 (from 2 F atoms)
- Lone Pairs = \(5 - 2 = 3\)

- **XeO3**:
- Valence Electrons of Xe = 8
- Monovalent Atoms (O) = 3 (Oxygen is divalent, so we consider it as 3)
- Hybridization = \(\frac{8 + 0}{2} = 4\) → sp³
- Bond Pairs = 3 (from 3 O atoms)
- Lone Pairs = \(4 - 3 = 1\)

3. **Compare the Number of Lone Pairs**:
- XeF4: 2 lone pairs
- XeF6: 1 lone pair
- XeF2: 3 lone pairs
- XeO3: 1 lone pair

4. **Conclusion**: The compound with the maximum number of lone pairs associated with xenon is **Xenon Difluoride (XeF2)**, which has **3 lone pairs**.

### Final Answer:
**Xenon Difluoride (XeF2) has the maximum number of lone pairs associated with Xe, which is 3.**