Which of the following have distored octahedral structure ? .

Options

A. `SF_(6)`
B. `PF_(6)^(Theta)`
C. `SiF_(6)^(2-)`
D. `XeF_(6)`

Explanation

To determine which of the given compounds has a distorted octahedral structure, we will analyze each compound step by step, focusing on their hybridization and molecular geometry.

### Step 1: Analyze SF6
1. **Valence Electrons**: Sulfur (S) has 6 valence electrons.
2. **Bonding with Fluorine**: Each fluorine (F) atom requires 1 electron for bonding. Since there are 6 fluorine atoms, all 6 of sulfur's electrons are used for bonding.
3. **Bond Pairs and Lone Pairs**:
- Bond pairs = 6 (one for each F)
- Lone pairs = 0
4. **Hybridization**: Using the formula:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence electrons} + \text{Monatomic atoms} - \text{Cationic charge} + \text{Anionic charge} \right) \]
\[ = \frac{1}{2} (6 + 6 - 0 + 0) = 6 \]
This corresponds to \( sp^3d^2 \) hybridization, indicating an octahedral structure.

### Step 2: Analyze PF6^-
1. **Valence Electrons**: Phosphorus (P) has 5 valence electrons, and with the negative charge, it has a total of 6 valence electrons.
2. **Bonding with Fluorine**: Each fluorine atom requires 1 electron, and there are 6 fluorine atoms.
3. **Bond Pairs and Lone Pairs**:
- Bond pairs = 6
- Lone pairs = 0
4. **Hybridization**:
\[ = \frac{1}{2} (6 + 6 - 0 + 0) = 6 \]
This also corresponds to \( sp^3d^2 \) hybridization, indicating an octahedral structure.

### Step 3: Analyze SiF6^2-
1. **Valence Electrons**: Silicon (Si) has 4 valence electrons, and with the 2 negative charges, it has a total of 6 valence electrons.
2. **Bonding with Fluorine**: Each fluorine atom requires 1 electron, and there are 6 fluorine atoms.
3. **Bond Pairs and Lone Pairs**:
- Bond pairs = 6
- Lone pairs = 0
4. **Hybridization**:
\[ = \frac{1}{2} (6 + 6 - 0 + 0) = 6 \]
This corresponds to \( sp^3d^2 \) hybridization, indicating an octahedral structure.

### Step 4: Analyze XeF6
1. **Valence Electrons**: Xenon (Xe) has 8 valence electrons.
2. **Bonding with Fluorine**: Each fluorine atom requires 1 electron, and there are 6 fluorine atoms.
3. **Bond Pairs and Lone Pairs**:
- Bond pairs = 6 (for the 6 F atoms)
- Lone pairs = 2 (since 8 - 6 = 2)
4. **Hybridization**:
\[ = \frac{1}{2} (8 + 6 - 0 + 0) = 7 \]
This corresponds to \( sp^3d^3 \) hybridization, which typically indicates a pentagonal bipyramidal geometry. However, due to the presence of lone pairs, the structure becomes distorted octahedral.

### Conclusion
Among the compounds analyzed, only **XeF6** has a distorted octahedral structure due to the presence of lone pairs.

### Final Answer
The compound with a distorted octahedral structure is **XeF6**.

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