In which reaction hybridisation of underlined atom does not change.

Options

A. `underset(-)(B)F_(3)+F^(-)toBF_(4)^(-)`
B. `underset(-)(N)H_(3)+H^(+)toNH_(4)^(+)`
C. `underset(-)(B)F_(3)+NH_(3)toBF_(3).NH_(3)`
D. `underset(-)(S)iF_(4)+2F^(-)toSiF_(6)^(2-)`

Answer

To determine in which reaction the hybridization of the underlined atom does not change, we will analyze the hybridization of the central atom in each reaction.

### Step-by-Step Solution:

1. **Identify the reactions and the central atoms**:
- The reactions provided are:
1. \( \text{BF}_3 \) forming \( \text{BF}_4^- \)
2. \( \text{NH}_3 \) forming \( \text{NF}_4^+ \)
3. \( \text{BF}_3 \) reacting with \( \text{NH}_3 \)
4. \( \text{SiF}_4 \) forming \( \text{SiF}_6^{2-} \)

2. **Calculate hybridization for each reaction**:
- **For \( \text{BF}_3 \) to \( \text{BF}_4^- \)**:
- Boron in \( \text{BF}_3 \):
\[ \text{Hybridization} = \frac{1}{2} \left( \text{valence electrons} + \text{monovalent atoms} + \text{charge} \right) = \frac{1}{2} (3 + 3 + 0) = \frac{6}{2} = 3 \quad \text{(sp}^2\text{)} \]
- Boron in \( \text{BF}_4^- \):
\[ \text{Hybridization} = \frac{1}{2} (3 + 4 + 1) = \frac{8}{2} = 4 \quad \text{(sp}^3\text{)} \]
- **Change in hybridization**: Yes, it changes from sp² to sp³.

- **For \( \text{NH}_3 \) to \( \text{NF}_4^+ \)**:
- Nitrogen in \( \text{NH}_3 \):
\[ \text{Hybridization} = \frac{1}{2} (5 + 3 + 0) = \frac{8}{2} = 4 \quad \text{(sp}^3\text{)} \]
- Nitrogen in \( \text{NF}_4^+ \):
\[ \text{Hybridization} = \frac{1}{2} (5 + 4 - 1) = \frac{8}{2} = 4 \quad \text{(sp}^3\text{)} \]
- **Change in hybridization**: No, it remains sp³.

- **For \( \text{BF}_3 \) and \( \text{NH}_3 \)**:
- Boron in \( \text{BF}_3 \): sp² (as calculated above).
- Nitrogen in \( \text{NH}_3 \): sp³ (as calculated above).
- **Change in hybridization**: Yes, both change.

- **For \( \text{SiF}_4 \) to \( \text{SiF}_6^{2-} \)**:
- Silicon in \( \text{SiF}_4 \):
\[ \text{Hybridization} = \frac{1}{2} (4 + 4 + 0) = \frac{8}{2} = 4 \quad \text{(sp}^3\text{)} \]
- Silicon in \( \text{SiF}_6^{2-} \):
\[ \text{Hybridization} = \frac{1}{2} (4 + 6 + 2) = \frac{12}{2} = 6 \quad \text{(sp}^3d^2\text{)} \]
- **Change in hybridization**: Yes, it changes from sp³ to sp³d².

3. **Conclusion**:
- The only reaction where the hybridization of the underlined atom does not change is the conversion of \( \text{NH}_3 \) to \( \text{NF}_4^+ \), where nitrogen remains sp³ in both cases.

### Final Answer:
The reaction in which the hybridization of the underlined atom does not change is \( \text{NH}_3 \) forming \( \text{NF}_4^+ \).