The hybrid orbital having only 20% d character

Options

A. `"CCl"_(4)`
B. `SF_(6)`
C. `SF_(4)`
D. `Cl_(2)`O

Answer

To determine which hybrid orbital has only 20% d character, we will analyze the given compounds: CCl4, SF6, SF4, and Cl2O.

### Step-by-Step Solution:

1. **Identify the Central Atom and Valence Electrons:**
- For each compound, identify the central atom and count its valence electrons.
- **CCl4:** Carbon (C) is the central atom with 4 valence electrons.
- **SF6:** Sulfur (S) is the central atom with 6 valence electrons.
- **SF4:** Sulfur (S) is again the central atom with 6 valence electrons.
- **Cl2O:** Oxygen (O) is the central atom with 6 valence electrons.

2. **Count the Monovalent Atoms Attached:**
- Count the number of monovalent atoms (single-bonded atoms) attached to the central atom.
- **CCl4:** 4 Cl atoms attached.
- **SF6:** 6 F atoms attached.
- **SF4:** 4 F atoms attached.
- **Cl2O:** 2 Cl atoms attached.

3. **Use the Hybridization Formula:**
- The formula to determine hybridization is:
\[ \text{Hybridization} = \frac{1}{2} \left( V + M + C - A \right) \]
where:
- \( V \) = number of valence electrons of the central atom,
- \( M \) = number of monovalent atoms attached,
- \( C \) = charge (add for anions, subtract for cations),
- \( A \) = number of anions (not applicable here).

4. **Calculate Hybridization for Each Compound:**
- **CCl4:**
\[ \text{Hybridization} = \frac{1}{2} (4 + 4) = 4 \quad \text{(sp}^3\text{)} \]
- **SF6:**
\[ \text{Hybridization} = \frac{1}{2} (6 + 6) = 6 \quad \text{(sp}^3\text{d}^2\text{)} \]
- **SF4:**
\[ \text{Hybridization} = \frac{1}{2} (6 + 4) = 5 \quad \text{(sp}^3\text{d)} \]
- **Cl2O:**
\[ \text{Hybridization} = \frac{1}{2} (6 + 2) = 4 \quad \text{(sp}^3\text{)} \]

5. **Determine the d Character:**
- The percentage of d character can be calculated using:
\[ \text{Percentage d character} = \frac{\text{Number of d orbitals used}}{\text{Total number of orbitals}} \times 100 \]
- **CCl4:** No d orbitals used, 0% d character.
- **SF6:** 2 d orbitals used (sp³d²), total 6 orbitals:
\[ \text{Percentage d character} = \frac{2}{6} \times 100 = 33.33\% \]
- **SF4:** 1 d orbital used (sp³d), total 5 orbitals:
\[ \text{Percentage d character} = \frac{1}{5} \times 100 = 20\% \]
- **Cl2O:** No d orbitals used, 0% d character.

6. **Conclusion:**
- The hybrid orbital that has only 20% d character is found in **SF4**.

### Final Answer:
The hybrid orbital having only 20% d character is **SF4**.