In which reaction, the hybridisation on the central atom changes from `sp^(2)` to `sp^(3)`.

Options

A. `NH_(3)+H^(+) to NH_(4)^(+)`
B. `BF_(3)+F^(-) to BF_(4)^(-)`
C. `H_(2)O+H^(+) to H_(3)O^(+)`
D. `C_(2)H_(2)+2H_(2) to C_(2)H_(6)`

Explanation

To determine in which reaction the hybridization of the central atom changes from sp² to sp³, we need to analyze the hybridization of the central atoms in the reactions provided.

### Step-by-Step Solution:

1. **Identify the Reactions**: We are given three reactions to analyze:
- Reaction 1: NH₃ (Ammonia)
- Reaction 2: BF₃ (Boron Trifluoride)
- Reaction 3: C₂H₂ (Ethylene) to C₂H₆ (Ethane)

2. **Calculate Hybridization for Each Reaction**:
- **For NH₃**:
- Valence electrons (V) of Nitrogen = 5
- Number of single bonds (M) = 3 (N-H bonds)
- Charge (C) = 0
- Charge on anion (A) = 0
- Steric Number = 1/2 * (V + M - C + A) = 1/2 * (5 + 3 - 0 + 0) = 4
- Hybridization = sp³

- **For BF₃**:
- Valence electrons (V) of Boron = 3
- Number of single bonds (M) = 3 (B-F bonds)
- Charge (C) = 0
- Charge on anion (A) = 0
- Steric Number = 1/2 * (V + M - C + A) = 1/2 * (3 + 3 - 0 + 0) = 3
- Hybridization = sp²

- **For C₂H₂ to C₂H₆**:
- **For C₂H₂**:
- Each Carbon has 4 valence electrons.
- Number of single bonds = 2 (C-H and C-C)
- Steric Number for each Carbon = 1/2 * (4 + 2 - 0 + 0) = 3
- Hybridization = sp.
- **For C₂H₆**:
- Each Carbon has 4 valence electrons.
- Number of single bonds = 4 (3 C-H and 1 C-C)
- Steric Number for each Carbon = 1/2 * (4 + 4 - 0 + 0) = 4
- Hybridization = sp³.

3. **Identify the Change in Hybridization**:
- In the reaction from BF₃ (sp²) to C₂H₆ (sp³), we see that the hybridization changes from sp² to sp³.
- Therefore, the reaction where the hybridization changes from sp² to sp³ is the conversion of BF₃ to a compound that contains sp³ hybridized carbon, such as C₂H₆.

### Final Answer:
The hybridization on the central atom changes from sp² to sp³ in the reaction involving BF₃.