`SF_(2),SF_(4)` and `SF_(6)` have the hybridisation at sulphur atom respectively as .

Options

A. `sp^(2),sp^(3),sp^(2)d^(2)`
B. `sp^(3),sp^(3),sp^(3)d^(2)`
C. `sp^(3),sp^(3)d,sp^(3)d^(2)`
D. `sp^(3),spd^(2),sp^(3)`

Explanation

To determine the hybridization of the sulfur atom in the molecules SF₂, SF₄, and SF₆, we will follow a systematic approach by calculating the number of valence electrons, bond pairs, and lone pairs for each molecule.

### Step 1: Hybridization of SF₂

1. **Calculate Valence Electrons:**
- Sulfur (S) has 6 valence electrons.
- Each fluorine (F) has 7 valence electrons, and there are 2 fluorine atoms.
- Total valence electrons = 6 (S) + 2 × 7 (F) = 6 + 14 = 20.

2. **Determine Bond Pairs and Lone Pairs:**
- Divide the total valence electrons by 8:
\[ \frac{20}{8} = 2 \text{ (quotient)} \quad \text{and} \quad 20 \mod 8 = 4 \text{ (remainder)} \]
- The quotient (2) represents the number of bond pairs, and the remainder (4) indicates twice the number of lone pairs.
- Therefore, the number of lone pairs = \(\frac{4}{2} = 2\).

3. **Determine Hybridization:**
- Number of bond pairs = 2 and number of lone pairs = 2.
- Hybridization = \(sp^3\).
- The molecular shape is V-shaped (bent).

### Step 2: Hybridization of SF₄

1. **Calculate Valence Electrons:**
- Total valence electrons = 6 (S) + 4 × 7 (F) = 6 + 28 = 34.

2. **Determine Bond Pairs and Lone Pairs:**
- Divide the total valence electrons by 8:
\[ \frac{34}{8} = 4 \text{ (quotient)} \quad \text{and} \quad 34 \mod 8 = 2 \text{ (remainder)} \]
- The quotient (4) represents the number of bond pairs, and the remainder (2) indicates twice the number of lone pairs.
- Therefore, the number of lone pairs = \(\frac{2}{2} = 1\).

3. **Determine Hybridization:**
- Number of bond pairs = 4 and number of lone pairs = 1.
- Hybridization = \(sp^3d\).
- The molecular shape is seesaw.

### Step 3: Hybridization of SF₆

1. **Calculate Valence Electrons:**
- Total valence electrons = 6 (S) + 6 × 7 (F) = 6 + 42 = 48.

2. **Determine Bond Pairs and Lone Pairs:**
- Divide the total valence electrons by 8:
\[ \frac{48}{8} = 6 \text{ (quotient)} \quad \text{and} \quad 48 \mod 8 = 0 \text{ (remainder)} \]
- The quotient (6) represents the number of bond pairs, and the remainder (0) indicates twice the number of lone pairs.
- Therefore, the number of lone pairs = \(\frac{0}{2} = 0\).

3. **Determine Hybridization:**
- Number of bond pairs = 6 and number of lone pairs = 0.
- Hybridization = \(sp^3d^2\).
- The molecular shape is octahedral.

### Final Answer:
- The hybridization at the sulfur atom in SF₂, SF₄, and SF₆ is respectively:
- **SF₂: \(sp^3\)**
- **SF₄: \(sp^3d\)**
- **SF₆: \(sp^3d^2\)**