Consider the following molecules or ions
`CH_2Cl_2` (ii) `NH_4^+` (iii) `SO_4^(2-)` (iv) `ClO_4^(-)` (v) `NH_3`
`sp^3`-hybridization is involved in the formation of

Options

A. (i), (ii), (v) only
B. (i). (ii) only
C. (i),(ii),(iv) only
D. (i),(ii),(iii),(iv) and (v)

Answer

To determine which of the given molecules or ions involve `sp^3` hybridization, we will calculate the steric number for each molecule or ion. The steric number can be calculated using the formula:

**Steric Number = (Valence Electrons of Central Atom) + (Number of Monovalent Atoms Attached) + (Charge Contribution)**

Where:
- If the molecule has a cationic charge, subtract 1 for each positive charge.
- If the molecule has an anionic charge, add 1 for each negative charge.

If the steric number is 4, the hybridization is `sp^3`.

### Step-by-Step Solution:

1. **For CH₂Cl₂:**
- Central Atom: Carbon (C)
- Valence Electrons of C: 4
- Monovalent Atoms Attached: 2 (H) + 2 (Cl) = 4
- Charge Contribution: 0 (neutral molecule)
- Steric Number = 4 + 4 + 0 = 8
- Since we need the number of lone pairs, we divide by 2: 8/2 = 4
- Therefore, CH₂Cl₂ has an `sp^3` hybridization.

2. **For NH₄⁺:**
- Central Atom: Nitrogen (N)
- Valence Electrons of N: 5
- Monovalent Atoms Attached: 4 (H)
- Charge Contribution: -1 (due to +1 charge)
- Steric Number = 5 + 4 - 1 = 8
- Divide by 2: 8/2 = 4
- Therefore, NH₄⁺ has an `sp^3` hybridization.

3. **For SO₄²⁻:**
- Central Atom: Sulfur (S)
- Valence Electrons of S: 6
- Monovalent Atoms Attached: 0 (since it forms double bonds with O)
- Charge Contribution: +2 (due to -2 charge)
- Steric Number = 6 + 0 + 2 = 8
- Divide by 2: 8/2 = 4
- Therefore, SO₄²⁻ has an `sp^3` hybridization.

4. **For ClO₄⁻:**
- Central Atom: Chlorine (Cl)
- Valence Electrons of Cl: 7
- Monovalent Atoms Attached: 0 (since it forms double bonds with O)
- Charge Contribution: +1 (due to -1 charge)
- Steric Number = 7 + 0 + 1 = 8
- Divide by 2: 8/2 = 4
- Therefore, ClO₄⁻ has an `sp^3` hybridization.

5. **For NH₃:**
- Central Atom: Nitrogen (N)
- Valence Electrons of N: 5
- Monovalent Atoms Attached: 3 (H)
- Charge Contribution: 0 (neutral molecule)
- Steric Number = 5 + 3 + 0 = 8
- Divide by 2: 8/2 = 4
- Therefore, NH₃ has an `sp^3` hybridization.

### Conclusion:
All the given molecules or ions (CH₂Cl₂, NH₄⁺, SO₄²⁻, ClO₄⁻, and NH₃) involve `sp^3` hybridization.