In which of the following hybridisation of underlined atom changes

Options

A. `ulAlH_(3)` changes to `AlH_(4)^(-)`
B. `H_(2)ulO` changes to `H_(3)O^(+)`
C. `ulNH_(3)` changes to `NH_(4)^(+)`
D. in all cases

Explanation

To determine in which of the following cases the hybridization of the underlined atom changes, we will analyze each option step by step.

### Step 1: Analyzing AlH3 to AlH4-
1. **Identify the initial compound (AlH3)**:
- Aluminum (Al) has a maximum covalency of 3.
- In AlH3, aluminum forms three sigma bonds with three hydrogen atoms.
- **Steric Number Calculation**:
- Sigma bonds = 3
- Lone pairs = 0
- **Steric Number = 3 + 0 = 3**
- **Hybridization**:
- With a steric number of 3, the hybridization is **sp²**.
- **Shape**: The shape is trigonal planar.

2. **Identify the final compound (AlH4-)**:
- In AlH4-, aluminum forms four sigma bonds (three with hydrogen and one with a negative charge).
- **Steric Number Calculation**:
- Sigma bonds = 4
- Lone pairs = 0
- **Steric Number = 4 + 0 = 4**
- **Hybridization**:
- With a steric number of 4, the hybridization is **sp³**.
- **Shape**: The shape is tetrahedral.

3. **Conclusion for AlH3 to AlH4-**:
- The hybridization changes from **sp² to sp³**.

### Step 2: Analyzing H2O to H3O+
1. **Identify the initial compound (H2O)**:
- Oxygen in H2O forms two sigma bonds with two hydrogen atoms.
- **Steric Number Calculation**:
- Sigma bonds = 2
- Lone pairs = 2
- **Steric Number = 2 + 2 = 4**
- **Hybridization**:
- With a steric number of 4, the hybridization is **sp³**.
- **Shape**: The shape is bent.

2. **Identify the final compound (H3O+)**:
- In H3O+, oxygen forms three sigma bonds with three hydrogen atoms and has one lone pair.
- **Steric Number Calculation**:
- Sigma bonds = 3
- Lone pairs = 1
- **Steric Number = 3 + 1 = 4**
- **Hybridization**:
- With a steric number of 4, the hybridization remains **sp³**.
- **Shape**: The shape is trigonal pyramidal.

3. **Conclusion for H2O to H3O+**:
- The hybridization does not change; it remains **sp³**.

### Step 3: Analyzing NH3 to NH4+
1. **Identify the initial compound (NH3)**:
- Nitrogen in NH3 forms three sigma bonds with three hydrogen atoms and has one lone pair.
- **Steric Number Calculation**:
- Sigma bonds = 3
- Lone pairs = 1
- **Steric Number = 3 + 1 = 4**
- **Hybridization**:
- With a steric number of 4, the hybridization is **sp³**.
- **Shape**: The shape is trigonal pyramidal.

2. **Identify the final compound (NH4+)**:
- In NH4+, nitrogen forms four sigma bonds with four hydrogen atoms and has no lone pairs.
- **Steric Number Calculation**:
- Sigma bonds = 4
- Lone pairs = 0
- **Steric Number = 4 + 0 = 4**
- **Hybridization**:
- With a steric number of 4, the hybridization remains **sp³**.
- **Shape**: The shape is tetrahedral.

3. **Conclusion for NH3 to NH4+**:
- The hybridization does not change; it remains **sp³**.

### Final Conclusion:
- The only case where the hybridization changes is from **AlH3 (sp²)** to **AlH4- (sp³)**. Therefore, the answer is **Option A**.

---