In which of the following process, hybridisation of the central atom changes?

Options

A. `H_(2)O+H^(+)rarrH_(3)O^(+)`
B. `NF_(3)+F^()rarrNF_(4)^(+)`
C. `BF_(3)+F^(-)rarrBF_(4)^(-)`
D. `NH_(3)+H^(+)rarrNH_(4)^(+)`

Explanation

To determine in which of the given processes the hybridization of the central atom changes, we will analyze each reaction step by step.

### Step 1: Analyze the first reaction (H2O + H+ → H3O+)

1. **Identify the central atom**: In H2O, the central atom is oxygen (O).
2. **Determine the steric number**:
- H2O has 2 sigma bonds (to H) and 2 lone pairs.
- Steric number = Number of sigma bonds + Number of lone pairs = 2 + 2 = 4.
3. **Determine the hybridization**:
- With a steric number of 4, the hybridization is sp³.
4. **Product analysis (H3O+)**:
- In H3O+, there are 3 sigma bonds (to H) and 1 lone pair.
- Steric number = 3 + 1 = 4.
- Hybridization remains sp³.
5. **Conclusion**: No change in hybridization.

### Step 2: Analyze the second reaction (NF3 + F → NF4+)

1. **Identify the central atom**: In NF3, the central atom is nitrogen (N).
2. **Determine the steric number**:
- NF3 has 3 sigma bonds (to F) and 1 lone pair.
- Steric number = 3 + 1 = 4.
3. **Determine the hybridization**:
- With a steric number of 4, the hybridization is sp³.
4. **Product analysis (NF4+)**:
- In NF4+, there are 4 sigma bonds (to F) and no lone pairs.
- Steric number = 4 + 0 = 4.
- Hybridization remains sp³.
5. **Conclusion**: No change in hybridization.

### Step 3: Analyze the third reaction (BF3 + F- → BF4-)

1. **Identify the central atom**: In BF3, the central atom is boron (B).
2. **Determine the steric number**:
- BF3 has 3 sigma bonds (to F) and no lone pairs.
- Steric number = 3 + 0 = 3.
3. **Determine the hybridization**:
- With a steric number of 3, the hybridization is sp².
4. **Product analysis (BF4-)**:
- In BF4-, there are 4 sigma bonds (to F) and no lone pairs.
- Steric number = 4 + 0 = 4.
- Hybridization changes to sp³.
5. **Conclusion**: Change in hybridization from sp² to sp³.

### Step 4: Analyze the fourth reaction (NH3 + H+ → NH4+)

1. **Identify the central atom**: In NH3, the central atom is nitrogen (N).
2. **Determine the steric number**:
- NH3 has 3 sigma bonds (to H) and 1 lone pair.
- Steric number = 3 + 1 = 4.
3. **Determine the hybridization**:
- With a steric number of 4, the hybridization is sp³.
4. **Product analysis (NH4+)**:
- In NH4+, there are 4 sigma bonds (to H) and no lone pairs.
- Steric number = 4 + 0 = 4.
- Hybridization remains sp³.
5. **Conclusion**: No change in hybridization.

### Final Conclusion:
The only reaction where the hybridization of the central atom changes is **C: BF3 + F- → BF4-**, where the hybridization changes from sp² to sp³.

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