In which of the following process, no change in hybridisation state occur but bond angle increases

Options

A. ` BF _ 3 + bar F to B F _ 4 ^ - `
B. ` H _ 2O + H^+ to H _ 3 O ^+ `
C. `BeF _ 2 + 2 bar F to BeF _ 4 ^ ( - 2 ) `
D. ` BeCl _ 2 ( g ) to BeCl _ 2 (s ) `

Answer

To solve the question, we need to analyze the given options for changes in hybridization and bond angles. The question specifically asks for a process where there is no change in hybridization state but an increase in bond angle.

### Step-by-Step Solution:

1. **Understanding Hybridization and Bond Angles**:
- Hybridization refers to the mixing of atomic orbitals to form new hybrid orbitals.
- The bond angle is the angle between two bonds that originate from the same atom.

2. **Analyzing Option A: BF3 and BF4-**:
- For BF3:
- Valence electrons of Boron = 3
- Bonding pairs = 3 (from three F atoms)
- Hybridization: \( n = \frac{(3 + 3)}{2} = 3 \) → SP2
- For BF4-:
- Valence electrons of Boron = 3
- Bonding pairs = 4 (from four F atoms) + 1 (due to -1 charge)
- Hybridization: \( n = \frac{(3 + 4 + 1)}{2} = 4 \) → SP3
- **Conclusion**: There is a change in hybridization from SP2 to SP3, so this option does not satisfy the condition.

3. **Analyzing Option B: H2O and H3O+**:
- For H2O:
- Valence electrons of Oxygen = 6
- Bonding pairs = 2 (from two H atoms) + 2 (lone pairs)
- Hybridization: \( n = \frac{(6 + 2)}{2} = 4 \) → SP3
- For H3O+:
- Valence electrons of Oxygen = 6
- Bonding pairs = 3 (from three H atoms) + 1 (lone pair) - 1 (due to +1 charge)
- Hybridization: \( n = \frac{(6 + 3 - 1)}{2} = 4 \) → SP3
- **Conclusion**: There is no change in hybridization (both are SP3), but the bond angle increases from approximately 104.5° in H2O to around 113° in H3O+. This option satisfies the condition.

4. **Analyzing Option C: BeF2 and BeF4 2-**:
- For BeF2:
- Valence electrons of Beryllium = 2
- Bonding pairs = 2 (from two F atoms)
- Hybridization: \( n = \frac{(2 + 2)}{2} = 2 \) → SP
- For BeF4 2-:
- Valence electrons of Beryllium = 2
- Bonding pairs = 4 (from four F atoms) + 2 (due to -2 charge)
- Hybridization: \( n = \frac{(2 + 4 + 2)}{2} = 4 \) → SP3
- **Conclusion**: There is a change in hybridization from SP to SP3, so this option does not satisfy the condition.

5. **Analyzing Option D: BeCl2 gas to BeCl2 solid**:
- This is a phase change (gas to solid) and does not involve any change in hybridization or bond angles.
- **Conclusion**: No change in hybridization or bond angles occurs here, so this option does not satisfy the condition.

### Final Conclusion:
The only option where there is **no change in hybridization** but an **increase in bond angle** is **Option B: H2O to H3O+**.

### Answer:
**Option B** is the correct answer.