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

Options

A. `NH_(3)` combines with `H^(+)`
B. `AlH_(3)` combines with `H^(-)`
C. `NH_(3)` forms `NH_(2)^(-)`
D. `SiF_(4)` forms `SiCl_(4)`

Explanation

To determine when the hybridization of a central atom changes, we can analyze specific reactions or interactions involving the central atom. Let's break down the question step by step:

### Step 1: Understand Hybridization
Hybridization is the mixing of atomic orbitals to form new hybrid orbitals that can accommodate the bonding and lone pairs around a central atom. The type of hybridization depends on the steric number, which is the sum of sigma bonds and lone pairs around the central atom.

### Step 2: Analyze the Given Options
We will analyze the provided options to see if there is a change in hybridization.

#### Option A: Ammonia (NH3) combines with H+
- **Initial State (NH3)**:
- Nitrogen has 3 hydrogen atoms and 1 lone pair.
- Steric Number = 3 (sigma bonds) + 1 (lone pair) = 4.
- Hybridization = sp3.

- **Final State (NH4+)**:
- Nitrogen donates its lone pair to H+, forming NH4+.
- Steric Number = 4 (4 sigma bonds) + 0 (lone pairs) = 4.
- Hybridization = sp3.

**Conclusion**: No change in hybridization.

#### Option B: AlH3 combines with H-
- **Initial State (AlH3)**:
- Aluminum has 3 hydrogen atoms.
- Steric Number = 3 (sigma bonds) + 0 (lone pairs) = 3.
- Hybridization = sp2.

- **Final State (AlH4-)**:
- Aluminum accepts a hydride ion (H-) into its empty orbital.
- Steric Number = 4 (4 sigma bonds) + 0 (lone pairs) = 4.
- Hybridization = sp3.

**Conclusion**: Change in hybridization from sp2 to sp3.

#### Option C: Ammonia (NH3) forms NH2-
- **Initial State (NH3)**:
- Hybridization = sp3 (as analyzed in Option A).

- **Final State (NH2-)**:
- Nitrogen has 2 hydrogen atoms and 2 lone pairs.
- Steric Number = 2 (sigma bonds) + 2 (lone pairs) = 4.
- Hybridization = sp3.

**Conclusion**: No change in hybridization.

#### Option D: SiF4 forms SiCl4
- **Initial State (SiF4)**:
- Silicon has 4 fluorine atoms.
- Steric Number = 4 (sigma bonds) + 0 (lone pairs) = 4.
- Hybridization = sp3.

- **Final State (SiCl4)**:
- Silicon has 4 chlorine atoms.
- Steric Number = 4 (sigma bonds) + 0 (lone pairs) = 4.
- Hybridization = sp3.

**Conclusion**: No change in hybridization.

### Final Conclusion
The only option where the hybridization changes is Option B, where AlH3 (sp2) changes to AlH4- (sp3).

### Answer
The hybridization of the central atom will change when AlH3 combines with H-.

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