In which of the following combination hybridisation of central atom `(**)` does not change ?

Options

A. `H_(2)O+ overset(**)CO_(2)`
B. `H_(3)overset(**)BO_(3)+OH^(-)`
C. `BF_(3)+overset(**)NH_(3)`
D. none of these

Answer

To determine in which combination the hybridization of the central atom does not change, we will analyze each option provided in the question using the formula for hybridization.

### Step-by-Step Solution:

1. **Understanding Hybridization**:
The hybridization of a central atom can be determined using the formula:
\[ Z = \frac{1}{2} \left( \text{Total number of valence electrons on the central atom} + \text{Number of monovalent atoms attached} - \text{Cationic charge} + \text{Anionic charge} \right) \]
Based on the value of Z:
- Z = 2 → sp
- Z = 3 → sp²
- Z = 4 → sp³

2. **Analyzing the First Option (H₂O + CO₂)**:
- **Central Atom**: Carbon in CO₂
- **Valence Electrons**: 4 (for Carbon)
- **Monovalent Atoms**: 0 (no hydrogen in CO₂)
- **Cationic Charge**: 0
- **Anionic Charge**: 0
- Calculation:
\[ Z = \frac{1}{2} (4 + 0 - 0 + 0) = \frac{4}{2} = 2 \quad \text{(sp hybridization)} \]
- In H₂CO₃:
- **Valence Electrons**: 4 (for Carbon)
- **Monovalent Atoms**: 2 (from H)
- Calculation:
\[ Z = \frac{1}{2} (4 + 2 - 0 + 0) = \frac{6}{2} = 3 \quad \text{(sp² hybridization)} \]
- **Conclusion**: Hybridization changes from sp to sp².

3. **Analyzing the Second Option (H₃BO₃ + OH⁻)**:
- **Central Atom**: Boron in H₃BO₃
- **Valence Electrons**: 3 (for Boron)
- **Monovalent Atoms**: 3 (from H)
- Calculation:
\[ Z = \frac{1}{2} (3 + 3 - 0 + 0) = \frac{6}{2} = 3 \quad \text{(sp² hybridization)} \]
- In H₄BO₄⁻:
- **Valence Electrons**: 3 (for Boron)
- **Monovalent Atoms**: 4 (from H)
- **Anionic Charge**: 1
- Calculation:
\[ Z = \frac{1}{2} (3 + 4 - 0 + 1) = \frac{8}{2} = 4 \quad \text{(sp³ hybridization)} \]
- **Conclusion**: Hybridization changes from sp² to sp³.

4. **Analyzing the Third Option (BF₃ + NH₃)**:
- **Central Atom**: Nitrogen in NH₃
- **Valence Electrons**: 5 (for Nitrogen)
- **Monovalent Atoms**: 3 (from H)
- Calculation:
\[ Z = \frac{1}{2} (5 + 3 - 0 + 0) = \frac{8}{2} = 4 \quad \text{(sp³ hybridization)} \]
- When BF₃ (Lewis acid) combines with NH₃ (Lewis base), a coordinate bond forms but the hybridization of nitrogen remains unchanged.
- **Conclusion**: Hybridization remains sp³.

5. **Final Analysis**:
- From the analysis, the only combination where the hybridization of the central atom does not change is when BF₃ combines with NH₃.

### Final Answer:
The hybridization of the central atom does not change in the combination of **BF₃ + NH₃**.

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