In solid state `PCl_(5)` exist as `[PCl_(4)]^(+) [PCl_(6)]^(-)`. The hybridisation of P is /are

Options

A. `sp^(3)`
B. `sp^(3)d`
C. `sp^(3)d^(2)`
D. `sp^(3) d^(3)`

Answer

To determine the hybridization of phosphorus in the solid state of \( PCl_5 \) which exists as \( [PCl_4]^+ \) and \( [PCl_6]^- \), we can follow these steps:

### Step 1: Calculate Hybridization for \( PCl_4^+ \)

1. **Identify the central atom**: The central atom is phosphorus (P).
2. **Count the valence electrons**: Phosphorus has 5 valence electrons.
3. **Count the number of monovalent atoms**: There are 4 chlorine atoms (Cl), which are monovalent.
4. **Account for the charge**: Since \( PCl_4^+ \) has a +1 charge, we subtract 1 from our total.

Using the formula:
\[ x = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right) \]

Substituting the values:
\[ x = \frac{1}{2} \left( 5 + 4 - 1 + 0 \right) = \frac{1}{2} \left( 8 \right) = 4 \]

5. **Determine hybridization**: Since \( x = 4 \), the hybridization is \( sp^3 \).

### Step 2: Calculate Hybridization for \( PCl_6^- \)

1. **Identify the central atom**: The central atom is still phosphorus (P).
2. **Count the valence electrons**: Phosphorus has 5 valence electrons.
3. **Count the number of monovalent atoms**: There are 6 chlorine atoms (Cl), which are monovalent.
4. **Account for the charge**: Since \( PCl_6^- \) has a -1 charge, we add 1 to our total.

Using the formula:
\[ x = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right) \]

Substituting the values:
\[ x = \frac{1}{2} \left( 5 + 6 - 0 + 1 \right) = \frac{1}{2} \left( 12 \right) = 6 \]

5. **Determine hybridization**: Since \( x = 6 \), the hybridization is \( sp^3d^2 \).

### Final Answer:
- The hybridization of phosphorus in \( PCl_4^+ \) is \( sp^3 \).
- The hybridization of phosphorus in \( PCl_6^- \) is \( sp^3d^2 \).