In which of the following change in hybridisation is taking place ?

Options

A. `BF_(3)+NH_(3) to [H_(3)N to BF_(3)]`
B. `NH_(3)+H^(+) to NH_(4)^(+)`
C. `SbF_(5)+F^(-) to [SbF_(6)]^(-)`
D. `2HClO_(2) to HClO+HClO_(3)`

Explanation

To determine where a change in hybridization occurs, we need to analyze the hybridization states of the molecules involved in the reactions. Let's break down the analysis step by step.

### Step 1: Analyze BF3 and NH3
1. **BF3 (Boron Trifluoride)**:
- Boron has 3 valence electrons.
- Each fluorine contributes 1 valence electron, leading to 3 bond pairs.
- Total steric number = 3 (bond pairs) + 0 (lone pairs) = 3.
- Hybridization = sp².

2. **NH3 (Ammonia)**:
- Nitrogen has 5 valence electrons.
- It forms 3 bond pairs with hydrogen and has 1 lone pair.
- Total steric number = 3 (bond pairs) + 1 (lone pair) = 4.
- Hybridization = sp³.

3. **Change in Hybridization**:
- When NH3 interacts with BF3, nitrogen donates its lone pair to boron.
- After donation, nitrogen retains its sp³ hybridization, but boron changes from sp² to sp³ because it now has 4 bond pairs and 0 lone pairs.

### Step 2: Analyze NH4⁺ Formation
1. **NH4⁺ (Ammonium Ion)**:
- The nitrogen donates its lone pair to a proton (H⁺).
- It forms 4 bond pairs and 0 lone pairs.
- Total steric number = 4 (bond pairs) + 0 (lone pairs) = 4.
- Hybridization = sp³.
- **No change in hybridization occurs here.**

### Step 3: Analyze SF5⁻ and SF6²⁻
1. **SF5⁻ (Sulfur Pentafluoride)**:
- Sulfur has 6 valence electrons.
- It forms 5 bond pairs with fluorine and has 0 lone pairs.
- Total steric number = 5 (bond pairs) + 0 (lone pairs) = 5.
- Hybridization = sp³d.

2. **SF6²⁻ (Sulfur Hexafluoride)**:
- When one of the fluorine atoms donates a lone pair, sulfur now has 6 bond pairs and 0 lone pairs.
- Total steric number = 6 (bond pairs) + 0 (lone pairs) = 6.
- Hybridization = sp³d².
- **Change in hybridization occurs here.**

### Step 4: Analyze HClO2 and HClO3
1. **HClO2 (Chlorous Acid)**:
- Chlorine has 7 valence electrons.
- It forms 1 double bond with oxygen and 1 single bond with OH, leaving 3 lone pairs.
- Total steric number = 2 (bond pairs) + 3 (lone pairs) = 5.
- Hybridization = sp³d.

2. **HClO3 (Chloric Acid)**:
- Chlorine forms 2 double bonds with oxygen and 1 single bond with OH, leaving 1 lone pair.
- Total steric number = 3 (bond pairs) + 1 (lone pair) = 4.
- Hybridization = sp³.
- **No change in hybridization occurs here.**

### Conclusion
The changes in hybridization occur in the following reactions:
- From NH3 to BF3 (Boron changes from sp² to sp³).
- From SF5⁻ to SF6²⁻ (Sulfur changes from sp³d to sp³d²).

### Summary of Changes
1. **BF3 to NH3**: Change in hybridization from sp² to sp³ for boron.
2. **SF5⁻ to SF6²⁻**: Change in hybridization from sp³d to sp³d² for sulfur.