Which of the following conversions involves change in both shape and hybridization?

Options

A. `B(OH)_(3) to [B(OH)_(4)]^(-)`
B. `NH_(3) to NH_(4)^(+)`
C. `H_(2)O to H_(3)O^(+)`
D. None of these

Explanation

To determine which conversion involves a change in both shape and hybridization, we will analyze the hybridization and geometry of the given compounds.

### Step 1: Analyze Boric Acid (B(OH)₃)
- **Boron (B)** has 3 valence electrons.
- The formula for steric number (SN) is:
\[ \text{SN} = \text{Valence electrons of central atom} + \text{Number of monovalent atoms} - \text{Charge} \]
- For B(OH)₃:
- Valence electrons = 3 (from B) + 3 (from 3 OH groups) = 6
- SN = 6 / 2 = 3
- **Hybridization**: sp²
- **Shape**: Trigonal planar

### Step 2: Analyze Tetrahydroxyborate (BOH₄⁻)
- When B(OH)₃ gains an OH⁻ ion, it forms BOH₄⁻.
- For BOH₄⁻:
- Valence electrons = 3 (from B) + 4 (from 4 OH groups) + 1 (from the negative charge) = 8
- SN = 8 / 2 = 4
- **Hybridization**: sp³
- **Shape**: Tetrahedral

### Step 3: Analyze Ammonia (NH₃)
- For NH₃:
- Valence electrons = 5 (from N) + 3 (from 3 H) = 8
- SN = 8 / 2 = 4
- **Hybridization**: sp³
- **Shape**: Tetrahedral

### Step 4: Analyze Ammonium Ion (NH₄⁺)
- For NH₄⁺:
- Valence electrons = 5 (from N) + 4 (from 4 H) - 1 (from positive charge) = 8
- SN = 8 / 2 = 4
- **Hybridization**: sp³
- **Shape**: Tetrahedral

### Step 5: Analyze Water (H₂O)
- For H₂O:
- Valence electrons = 6 (from O) + 2 (from 2 H) = 8
- SN = 8 / 2 = 4
- **Hybridization**: sp³
- **Shape**: Bent

### Step 6: Analyze Hydronium Ion (H₃O⁺)
- For H₃O⁺:
- Valence electrons = 6 (from O) + 3 (from 3 H) - 1 (from positive charge) = 8
- SN = 8 / 2 = 4
- **Hybridization**: sp³
- **Shape**: Trigonal pyramidal

### Conclusion
- The conversion from B(OH)₃ to BOH₄⁻ involves a change in both hybridization (from sp² to sp³) and shape (from trigonal planar to tetrahedral).
- The conversions for NH₃ to NH₄⁺ and H₂O to H₃O⁺ do not involve a change in hybridization or shape.

### Final Answer
**The conversion that involves a change in both shape and hybridization is the conversion of B(OH)₃ to BOH₄⁻.**