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 solve the question, we need to analyze the hybridization of the central atom in each of the given combinations and determine if it changes or remains the same. Let's break down the analysis step by step.

### Step 1: Analyze H2O + CO2
1. **Identify the central atom**: In H2O (water), the central atom is Oxygen (O). In CO2 (carbon dioxide), the central atom is Carbon (C).
2. **Determine hybridization**:
- For CO2:
- Valence electrons (V) = 4 (from C) + 0 (no lone pairs) = 4
- Bonds (B) = 2 (double bonds with O)
- Hybridization = \( \frac{1}{2}(V + B) = \frac{1}{2}(4 + 0) = 2 \) → sp hybridization.
- For H2CO3 (product):
- Structure: C is bonded to 2 O atoms and 1 OH group.
- Valence electrons (V) = 4 (from C) + 2 (from 2 O) + 1 (from OH) = 7
- Bonds (B) = 3 (single bonds with OH and double bonds with O)
- Hybridization = \( \frac{1}{2}(7 + 3) = 5 \) → sp³ hybridization.
3. **Conclusion**: Hybridization changes from sp (in CO2) to sp³ (in H2CO3).

### Step 2: Analyze H3BO3 + OH⁻
1. **Identify the central atom**: In H3BO3 (boric acid), the central atom is Boron (B).
2. **Determine hybridization**:
- For H3BO3:
- Valence electrons (V) = 3 (from B) + 3 (from 3 O) = 6
- Bonds (B) = 3 (single bonds with OH groups)
- Hybridization = \( \frac{1}{2}(6 + 3) = 4.5 \) → sp³ hybridization.
- For B(OH)4⁻ (product):
- Valence electrons (V) = 3 (from B) + 4 (from 4 O) + 1 (from the negative charge) = 8
- Bonds (B) = 4 (single bonds with OH groups)
- Hybridization = \( \frac{1}{2}(8 + 4) = 6 \) → sp³ hybridization.
3. **Conclusion**: Hybridization changes from sp³ to sp³ (remains the same).

### Step 3: Analyze BF3 + NH3
1. **Identify the central atom**: In BF3 (boron trifluoride), the central atom is Boron (B).
2. **Determine hybridization**:
- For BF3:
- Valence electrons (V) = 3 (from B) + 0 (no lone pairs) = 3
- Bonds (B) = 3 (single bonds with F)
- Hybridization = \( \frac{1}{2}(3 + 3) = 3 \) → sp² hybridization.
- For BNH3 (product):
- Valence electrons (V) = 3 (from B) + 5 (from NH3) = 8
- Bonds (B) = 4 (3 from NH3 and 1 from B)
- Hybridization = \( \frac{1}{2}(8 + 4) = 6 \) → sp³ hybridization.
3. **Conclusion**: Hybridization changes from sp² (in BF3) to sp³ (in BNH3).

### Final Conclusion
- The only combination where the hybridization of the central atom does not change is **H3BO3 + OH⁻**.

### Answer
The correct answer is **H3BO3 + OH⁻**.