In the reaction `2PCI_(5) hArr PCI_(4)^(o+) + PCI_(6)^(Θ)` change in hybridisation is from .

Explanation

To determine the change in hybridization during the reaction \(2 \text{PCl}_5 \rightleftharpoons \text{PCl}_4^{+} + \text{PCl}_6^{-}\), we need to analyze the hybridization states of phosphorus in each species involved in the reaction.

### Step 1: Determine the hybridization of PCl5
1. **Identify the valence electrons of phosphorus**: Phosphorus (P) is in group 15 and has 5 valence electrons.
2. **Count the bond pairs**: In PCl5, phosphorus forms 5 bonds with 5 chlorine atoms, meaning there are 5 bond pairs and 0 lone pairs.
3. **Calculate the hybridization**: The hybridization can be determined using the formula:
\[ \text{Hybridization} = \text{Number of bond pairs} + \frac{\text{Number of lone pairs}}{2} \]
For PCl5:
\[ \text{Hybridization} = 5 + 0 = 5 \quad \Rightarrow \quad \text{sp}^3\text{d} \]

### Step 2: Determine the hybridization of PCl4+
1. **Identify the change in electrons**: The PCl4+ ion indicates that phosphorus has lost one electron.
2. **Count the bond pairs**: In PCl4+, phosphorus forms 4 bonds with chlorine, resulting in 4 bond pairs and 0 lone pairs.
3. **Calculate the hybridization**:
\[ \text{Hybridization} = 4 + 0 = 4 \quad \Rightarrow \quad \text{sp}^3 \]

### Step 3: Determine the hybridization of PCl6-
1. **Identify the change in electrons**: The PCl6- ion indicates that phosphorus has gained one electron.
2. **Count the bond pairs**: In PCl6-, phosphorus forms 6 bonds with chlorine, resulting in 6 bond pairs and 0 lone pairs.
3. **Calculate the hybridization**:
\[ \text{Hybridization} = 6 + 0 = 6 \quad \Rightarrow \quad \text{sp}^3\text{d}^2 \]

### Step 4: Summarize the change in hybridization
- The hybridization changes from:
- **PCl5**: sp³d
- **PCl4+**: sp³
- **PCl6-**: sp³d²

### Conclusion
The change in hybridization during the reaction is from **sp³d** (in PCl5) to **sp³** (in PCl4+) and to **sp³d²** (in PCl6-).