In which of the following central atom makes use of `sp^3` hybrid orbitals

Options

A. `SO_2`
B. `SO_3`
C. `H_2S`
D. `H_2O`

Explanation

To determine which central atom makes use of `sp^3` hybrid orbitals, we need to analyze the hybridization of the given compounds. We will use the hybridization formula:

**Hybridization formula:**
\[ \text{Hybridization} = \frac{1}{2} \times (V + M + C - A) \]

Where:
- \( V \) = Valence electrons of the central atom
- \( M \) = Monovalent atoms attached to the central atom
- \( C \) = Cationic charge (positive charge)
- \( A \) = Anionic charge (negative charge)

Now, let's analyze each option step by step.

### Step 1: Analyze SO2
1. **Identify the central atom:** Sulfur (S)
2. **Determine the valence electrons of sulfur:** Sulfur has 6 valence electrons (group 16).
3. **Count the monovalent atoms:** There are 0 monovalent atoms (Oxygen is divalent).
4. **Count the charges:** There are no cationic or anionic charges (0).
5. **Substitute into the formula:**
\[ \text{Hybridization} = \frac{1}{2} \times (6 + 0 + 0 - 0) = \frac{1}{2} \times 6 = 3 \]
- This corresponds to `sp^2` hybridization.

### Step 2: Analyze SO3
1. **Identify the central atom:** Sulfur (S)
2. **Determine the valence electrons of sulfur:** 6 valence electrons.
3. **Count the monovalent atoms:** 0 (Oxygen is divalent).
4. **Count the charges:** 0.
5. **Substitute into the formula:**
\[ \text{Hybridization} = \frac{1}{2} \times (6 + 0 + 0 - 0) = \frac{1}{2} \times 6 = 3 \]
- This also corresponds to `sp^2` hybridization.

### Step 3: Analyze H2S
1. **Identify the central atom:** Sulfur (S)
2. **Determine the valence electrons of sulfur:** 6 valence electrons.
3. **Count the monovalent atoms:** 2 (Hydrogen).
4. **Count the charges:** 0.
5. **Substitute into the formula:**
\[ \text{Hybridization} = \frac{1}{2} \times (6 + 2 + 0 - 0) = \frac{1}{2} \times 8 = 4 \]
- This corresponds to `sp^3` hybridization.

### Step 4: Analyze H2O
1. **Identify the central atom:** Oxygen (O)
2. **Determine the valence electrons of oxygen:** 6 valence electrons.
3. **Count the monovalent atoms:** 2 (Hydrogen).
4. **Count the charges:** 0.
5. **Substitute into the formula:**
\[ \text{Hybridization} = \frac{1}{2} \times (6 + 2 + 0 - 0) = \frac{1}{2} \times 8 = 4 \]
- This also corresponds to `sp^3` hybridization.

### Conclusion
The central atoms that make use of `sp^3` hybrid orbitals are found in H2S and H2O. Therefore, the correct answer is:
- **H2O** (Option D).