In which case, hybridisation of the central atom is affected when :

Options

A. `NH_3` changes to `NH_4^+`
B. `AlH_3` changes to `AlH_4^+`
C. In both cases
D. Is none case

Answer

To determine in which case the hybridization of the central atom is affected, we will analyze the hybridization of the central atom in the given compounds: NH3 and NH4+, and AlH3 and AlH4+.

### Step-by-Step Solution:

1. **Identify the Central Atom**:
- For NH3 (Ammonia), the central atom is Nitrogen (N).
- For NH4+ (Ammonium ion), the central atom is also Nitrogen (N).
- For AlH3 (Aluminum Hydride), the central atom is Aluminum (Al).
- For AlH4+ (Tetrahydroaluminate ion), the central atom is also Aluminum (Al).

2. **Determine the Valency of the Central Atom**:
- **For Nitrogen (N)**:
- Atomic number = 7
- Electronic configuration = 1s² 2s² 2p³
- Valency = 5 (since it can form 3 bonds with hydrogen and has 2 lone pairs).
- **For Aluminum (Al)**:
- Atomic number = 13
- Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p¹
- Valency = 3 (since it can form 3 bonds with hydrogen).

3. **Calculate Hybridization Using the Formula**:
- The formula for hybridization is:
\[ \text{Hybridization} = \frac{1}{2} (B + M - C + A) \]
where:
- B = Valency
- M = Monovalent atoms (number of hydrogen atoms)
- C = Cation charge
- A = Anion charge

4. **Calculate Hybridization for NH3**:
- B = 5 (valency of N)
- M = 3 (3 hydrogen atoms)
- C = 0 (no cation)
- A = 0 (no anion)
- Hybridization = \( \frac{1}{2} (5 + 3 - 0 + 0) = \frac{1}{2} (8) = 4 \) → sp³ hybridization.

5. **Calculate Hybridization for NH4+**:
- B = 5 (valency of N)
- M = 4 (4 hydrogen atoms)
- C = 1 (cation charge of +1)
- A = 0 (no anion)
- Hybridization = \( \frac{1}{2} (5 + 4 - 1 + 0) = \frac{1}{2} (8) = 4 \) → sp³ hybridization.

6. **Calculate Hybridization for AlH3**:
- B = 3 (valency of Al)
- M = 3 (3 hydrogen atoms)
- C = 0 (no cation)
- A = 0 (no anion)
- Hybridization = \( \frac{1}{2} (3 + 3 - 0 + 0) = \frac{1}{2} (6) = 3 \) → sp² hybridization.

7. **Calculate Hybridization for AlH4+**:
- B = 3 (valency of Al)
- M = 4 (4 hydrogen atoms)
- C = 1 (cation charge of +1)
- A = 0 (no anion)
- Hybridization = \( \frac{1}{2} (3 + 4 - 1 + 0) = \frac{1}{2} (6) = 3 \) → sp² hybridization.

8. **Conclusion**:
- For NH3 and NH4+, the hybridization remains sp³.
- For AlH3 and AlH4+, the hybridization remains sp².
- Therefore, the hybridization of the central atom is not affected in either case.

### Final Answer:
The correct option is **Option 4: None Case**.