In order to explain the characteristic geometrical shapes of polyatomic molecules, Pauling introduced the concept of hybridisation. The orbitals undergoing hybridisation should have nearly same energy. There are various types of hybridisations involving s, p and d - type of orbitals. The type of hybridisation gives the characteristic shape of the molecular or ion.
Which molecule does not have the same type of hybridisation as P has in `PF_(5)`?

Options

A. `ClF_(3)`
B. `SF_(4)`
C. `XeF_(4)`
D. `XeF_(2)`

Explanation

To determine which molecule does not have the same type of hybridization as phosphorus in PF5, we will follow these steps:

### Step 1: Determine the hybridization of PF5
- **Valence Electrons of Phosphorus (P)**: Phosphorus has 5 valence electrons.
- **Number of Monovalent Bonds**: PF5 has 5 fluorine atoms, which means there are 5 monovalent bonds.
- **Hybridization Formula**: The formula for hybridization is given by:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence Electrons} + \text{Number of Monovalent Bonds} - \text{Charge} \right) \]
For PF5:
\[ \text{Hybridization} = \frac{1}{2} \left( 5 + 5 - 0 \right) = \frac{1}{2} \times 10 = 5 \]
- **Type of Hybridization**: Since the hybridization number is 5, the type of hybridization is **sp³d**. This corresponds to a trigonal bipyramidal geometry.

### Step 2: Analyze the hybridization of the given options
1. **ClF3**:
- **Valence Electrons of Cl**: 7
- **Monovalent Bonds**: 3 (with F)
- **Hybridization**:
\[ \text{Hybridization} = \frac{1}{2} \left( 7 + 3 - 0 \right) = \frac{1}{2} \times 10 = 5 \]
- **Type**: sp³d (trigonal bipyramidal with 2 lone pairs)

2. **XeSF4**:
- **Valence Electrons of S**: 6
- **Monovalent Bonds**: 4 (with F)
- **Hybridization**:
\[ \text{Hybridization} = \frac{1}{2} \left( 6 + 4 - 0 \right) = \frac{1}{2} \times 10 = 5 \]
- **Type**: sp³d (seesaw geometry with 1 lone pair)

3. **XeF4**:
- **Valence Electrons of Xe**: 8
- **Monovalent Bonds**: 4 (with F)
- **Hybridization**:
\[ \text{Hybridization} = \frac{1}{2} \left( 8 + 4 - 0 \right) = \frac{1}{2} \times 12 = 6 \]
- **Type**: sp³d² (square planar geometry with 2 lone pairs)

4. **XeF2**:
- **Valence Electrons of Xe**: 8
- **Monovalent Bonds**: 2 (with F)
- **Hybridization**:
\[ \text{Hybridization} = \frac{1}{2} \left( 8 + 2 - 0 \right) = \frac{1}{2} \times 10 = 5 \]
- **Type**: sp³d (linear geometry with 3 lone pairs)

### Step 3: Compare the hybridizations
- PF5 has sp³d hybridization.
- ClF3 has sp³d hybridization.
- XeSF4 has sp³d hybridization.
- **XeF4 has sp³d² hybridization**, which is different from PF5.

### Conclusion
The molecule that does not have the same type of hybridization as phosphorus in PF5 is **XeF4**.