Write the hybridization and also draw their molecular structures ?
(a)`XeF_(2)` , (b)`XeF_(4)` , (c )`XeF_(6)` ,

Answer

To solve the question regarding the hybridization and molecular structures of \( \text{XeF}_2 \), \( \text{XeF}_4 \), and \( \text{XeF}_6 \), we will follow a systematic approach for each compound.

### Step 1: Determine the Hybridization of \( \text{XeF}_2 \)

1. **Count the Valence Electrons**:
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) contributes 1 electron, and there are 2 fluorines.
- Total = \( 8 + 2 = 10 \) valence electrons.

2. **Draw the Lewis Structure**:
- Place Xe in the center and bond it to 2 F atoms.
- This uses 2 electrons (1 for each bond), leaving us with \( 10 - 2 = 8 \) electrons.
- These remaining electrons will form 3 lone pairs on Xe.

3. **Determine Hybridization**:
- The number of bonding pairs (2) + lone pairs (3) = 5.
- The hybridization is \( \text{sp}^3\text{d} \).

4. **Molecular Structure**:
- The molecular geometry is linear due to the arrangement of the 2 bonding pairs and 3 lone pairs.

### Step 2: Determine the Hybridization of \( \text{XeF}_4 \)

1. **Count the Valence Electrons**:
- Total = \( 8 + 4 = 12 \) valence electrons.

2. **Draw the Lewis Structure**:
- Place Xe in the center and bond it to 4 F atoms.
- This uses 4 electrons (2 for each bond), leaving us with \( 12 - 4 = 8 \) electrons.
- These remaining electrons will form 2 lone pairs on Xe.

3. **Determine Hybridization**:
- The number of bonding pairs (4) + lone pairs (2) = 6.
- The hybridization is \( \text{sp}^3\text{d}^2 \).

4. **Molecular Structure**:
- The molecular geometry is square planar due to the arrangement of the 4 bonding pairs and 2 lone pairs.

### Step 3: Determine the Hybridization of \( \text{XeF}_6 \)

1. **Count the Valence Electrons**:
- Total = \( 8 + 6 = 14 \) valence electrons.

2. **Draw the Lewis Structure**:
- Place Xe in the center and bond it to 6 F atoms.
- This uses 6 electrons (2 for each bond), leaving us with \( 14 - 6 = 8 \) electrons.
- These remaining electrons will form 1 lone pair on Xe.

3. **Determine Hybridization**:
- The number of bonding pairs (6) + lone pairs (1) = 7.
- The hybridization is \( \text{sp}^3\text{d}^3 \).

4. **Molecular Structure**:
- The molecular geometry is square bipyramidal due to the arrangement of the 6 bonding pairs and 1 lone pair.

### Summary of Results

- **\( \text{XeF}_2 \)**: Hybridization = \( \text{sp}^3\text{d} \), Molecular Structure = Linear
- **\( \text{XeF}_4 \)**: Hybridization = \( \text{sp}^3\text{d}^2 \), Molecular Structure = Square Planar
- **\( \text{XeF}_6 \)**: Hybridization = \( \text{sp}^3\text{d}^3 \), Molecular Structure = Square Bipyramidal