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, III, IV & V
D. I, II, III & IV only

Explanation

To determine which of the given molecules or ions involve sp³ hybridization, we will calculate the steric number for each molecule or ion. The steric number (S) can be calculated using the formula:

\[ S = \frac{1}{2} \left( V + M - C + A \right) \]

Where:
- \( V \) = number of valence electrons of the central atom
- \( M \) = number of monovalent atoms attached to the central atom
- \( C \) = positive charge (cationic charge)
- \( A \) = negative charge (anionic charge)

Let's analyze each molecule or ion step by step.

### Step 1: CH₂Cl₂
- Central atom: Carbon (C)
- Valence electrons (V) = 4 (from C)
- Monovalent atoms (M) = 4 (2 H and 2 Cl)
- Cationic charge (C) = 0
- Anionic charge (A) = 0

Calculating the steric number:
\[ S = \frac{1}{2} (4 + 4 - 0 + 0) = \frac{1}{2} (8) = 4 \]

Since the steric number is 4, CH₂Cl₂ has sp³ hybridization.

### Step 2: NH₄⁺
- Central atom: Nitrogen (N)
- Valence electrons (V) = 5 (from N)
- Monovalent atoms (M) = 4 (4 H)
- Cationic charge (C) = 1
- Anionic charge (A) = 0

Calculating the steric number:
\[ S = \frac{1}{2} (5 + 4 - 1 + 0) = \frac{1}{2} (8) = 4 \]

Since the steric number is 4, NH₄⁺ has sp³ hybridization.

### Step 3: SO₄²⁻
- Central atom: Sulfur (S)
- Valence electrons (V) = 6 (from S)
- Monovalent atoms (M) = 0
- Cationic charge (C) = 0
- Anionic charge (A) = 2

Calculating the steric number:
\[ S = \frac{1}{2} (6 + 0 - 0 + 2) = \frac{1}{2} (8) = 4 \]

Since the steric number is 4, SO₄²⁻ has sp³ hybridization.

### Step 4: ClO₄⁻
- Central atom: Chlorine (Cl)
- Valence electrons (V) = 7 (from Cl)
- Monovalent atoms (M) = 0
- Cationic charge (C) = 0
- Anionic charge (A) = 1

Calculating the steric number:
\[ S = \frac{1}{2} (7 + 0 - 0 + 1) = \frac{1}{2} (8) = 4 \]

Since the steric number is 4, ClO₄⁻ has sp³ hybridization.

### Step 5: NH₃
- Central atom: Nitrogen (N)
- Valence electrons (V) = 5 (from N)
- Monovalent atoms (M) = 3 (3 H)
- Cationic charge (C) = 0
- Anionic charge (A) = 0

Calculating the steric number:
\[ S = \frac{1}{2} (5 + 3 - 0 + 0) = \frac{1}{2} (8) = 4 \]

Since the steric number is 4, NH₃ has sp³ hybridization.

### Conclusion
All of the given molecules and ions (CH₂Cl₂, NH₄⁺, SO₄²⁻, ClO₄⁻, and NH₃) exhibit sp³ hybridization.

### Final Answer
The answer is that sp³ hybridization is involved in the formation of all the given molecules and ions.

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