Which one of the following conversions involve change in both hybridisation and shape?

Options

A. `CH_(4)toC_(2)H_(6)`
B. `NH_(3)toNH_(4)^(+)`
C. `BF_(3)toBF_(4)^(-)`
D. `H_(2)OtoH_(3)O^(+)`

Explanation

To solve the question of which conversion involves a change in both hybridization and shape, we will analyze each option step by step.

### Step 1: Analyze CH₄ (Methane)
- **Valence Electrons**: Carbon has 4 valence electrons.
- **Monovalent Atoms**: There are 4 hydrogen atoms.
- **Charge**: There is no charge.

Using the hybridization formula:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence Electrons} + \text{Monovalent Atoms} - \text{Charge} \right) \]
\[ = \frac{1}{2} \left( 4 + 4 + 0 \right) = \frac{8}{2} = 4 \]
- **Hybridization**: SP³
- **Shape**: Tetrahedral

### Step 2: Analyze C₂H₆ (Ethane)
- **Valence Electrons**: Each carbon has 4 valence electrons, total 8 for 2 carbons.
- **Monovalent Atoms**: There are 6 hydrogen atoms.
- **Charge**: There is no charge.

Using the hybridization formula:
\[ \text{Hybridization} = \frac{1}{2} \left( 8 + 6 + 0 \right) = \frac{14}{2} = 7 \]
However, for each carbon:
\[ \text{Hybridization for each C} = \frac{1}{2} \left( 4 + 4 + 0 \right) = 4 \quad \text{(for each carbon)} \]
- **Hybridization**: SP³ for each carbon
- **Shape**: Tetrahedral for each carbon

### Step 3: Analyze NH₃ to NH₄⁺ (Ammonia to Ammonium Ion)
- **NH₃**:
- Valence Electrons: 5 (N) + 3 (H) = 8
- Charge: 0
\[ \text{Hybridization} = \frac{1}{2} (5 + 3 + 0) = 4 \quad \text{(SP³)} \]
- **Shape**: Pyramidal

- **NH₄⁺**:
- Valence Electrons: 5 (N) + 4 (H) - 1 (charge) = 8
\[ \text{Hybridization} = \frac{1}{2} (5 + 4 - 1) = 4 \quad \text{(SP³)} \]
- **Shape**: Tetrahedral

### Step 4: Analyze BF₃ to BF₄⁻ (Boron Trifluoride to Tetrafluoroborate Ion)
- **BF₃**:
- Valence Electrons: 3 (B) + 3 (F) = 6
- Charge: 0
\[ \text{Hybridization} = \frac{1}{2} (3 + 3 + 0) = 3 \quad \text{(SP²)} \]
- **Shape**: Trigonal planar

- **BF₄⁻**:
- Valence Electrons: 3 (B) + 4 (F) + 1 (charge) = 8
\[ \text{Hybridization} = \frac{1}{2} (3 + 4 + 1) = 4 \quad \text{(SP³)} \]
- **Shape**: Tetrahedral

### Step 5: Analyze H₂O to H₃O⁺ (Water to Hydronium Ion)
- **H₂O**:
- Valence Electrons: 6 (O) + 2 (H) = 8
- Charge: 0
\[ \text{Hybridization} = \frac{1}{2} (6 + 2 + 0) = 4 \quad \text{(SP³)} \]
- **Shape**: Bent

- **H₃O⁺**:
- Valence Electrons: 6 (O) + 3 (H) - 1 (charge) = 8
\[ \text{Hybridization} = \frac{1}{2} (6 + 3 - 1) = 4 \quad \text{(SP³)} \]
- **Shape**: Pyramidal

### Conclusion
From the analysis:
- **CH₄**: No change in hybridization or shape.
- **C₂H₆**: No change in hybridization or shape.
- **NH₃ to NH₄⁺**: No change in hybridization, shape changes.
- **BF₃ to BF₄⁻**: Change in both hybridization (SP² to SP³) and shape (Trigonal planar to Tetrahedral).
- **H₂O to H₃O⁺**: No change in hybridization, shape changes.

Thus, the correct answer is **BF₃ to BF₄⁻**, as it involves a change in both hybridization and shape.