The hybridization of the central atom will change when `:`

Options

A. `NH_(3)` combines with `H^(+)`
B. `H_(3)BO_(3)` combines with`OH^(-)`
C. `NH_(3)` forms `NH_(2)^(-)`
D. `H_(2)O` combines with `H^(+)`

Explanation

To determine when the hybridization of the central atom changes, we need to analyze the given options step by step. We will use the formula for calculating the hybridization based on the value of Z:

\[ Z = \frac{1}{2} \left( \text{Number of valence electrons in the central atom} + \text{Number of monovalent atoms attached} - \text{Cationic charge} + \text{Anionic charge} \right) \]

### Step 1: Analyze NH3 combining with H+
1. **Identify the central atom**: In NH3, the central atom is nitrogen (N).
2. **Calculate Z for NH3**:
- Valence electrons in N = 5
- Number of monovalent atoms (H) = 3
- Cationic charge = 0
- Anionic charge = 0
- \( Z = \frac{1}{2} (5 + 3 - 0 + 0) = \frac{1}{2} (8) = 4 \)
- Hybridization = sp³ (since Z = 4)

3. **When NH3 combines with H+**:
- It forms NH4+.
- **Calculate Z for NH4+**:
- Valence electrons in N = 5
- Number of monovalent atoms (H) = 4
- Cationic charge = 1
- Anionic charge = 0
- \( Z = \frac{1}{2} (5 + 4 - 1 + 0) = \frac{1}{2} (8) = 4 \)
- Hybridization = sp³ (no change)

### Step 2: Analyze H3BO3 combining with OH-
1. **Identify the central atom**: In H3BO3, the central atom is boron (B).
2. **Calculate Z for H3BO3**:
- Valence electrons in B = 3
- Number of monovalent atoms (H) = 3
- Cationic charge = 0
- Anionic charge = 0
- \( Z = \frac{1}{2} (3 + 3 - 0 + 0) = \frac{1}{2} (6) = 3 \)
- Hybridization = sp² (since Z = 3)

3. **When H3BO3 combines with OH-**:
- It forms H4BO4-.
- **Calculate Z for H4BO4-**:
- Valence electrons in B = 3
- Number of monovalent atoms (H) = 4
- Cationic charge = 0
- Anionic charge = 1
- \( Z = \frac{1}{2} (3 + 4 - 0 + 1) = \frac{1}{2} (8) = 4 \)
- Hybridization = sp³ (change from sp² to sp³)

### Step 3: Analyze NH3 forming NH2-
1. **Calculate Z for NH2-**:
- Valence electrons in N = 5
- Number of monovalent atoms (H) = 2
- Cationic charge = 0
- Anionic charge = 1
- \( Z = \frac{1}{2} (5 + 2 - 0 + 1) = \frac{1}{2} (8) = 4 \)
- Hybridization = sp³ (no change)

### Step 4: Analyze H2O combining with H+
1. **Identify the central atom**: In H2O, the central atom is oxygen (O).
2. **Calculate Z for H2O**:
- Valence electrons in O = 6
- Number of monovalent atoms (H) = 2
- Cationic charge = 0
- Anionic charge = 0
- \( Z = \frac{1}{2} (6 + 2 - 0 + 0) = \frac{1}{2} (8) = 4 \)
- Hybridization = sp³ (no change)

2. **When H2O combines with H+**:
- It forms H3O+.
- **Calculate Z for H3O+**:
- Valence electrons in O = 6
- Number of monovalent atoms (H) = 3
- Cationic charge = 1
- Anionic charge = 0
- \( Z = \frac{1}{2} (6 + 3 - 1 + 0) = \frac{1}{2} (8) = 4 \)
- Hybridization = sp³ (no change)

### Conclusion
The hybridization of the central atom changes only in the second option (H3BO3 combining with OH-), where the hybridization changes from sp² to sp³.

### Final Answer
The hybridization of the central atom will change when:
- **H3BO3 combines with OH-**.