Which of the following does not show octahedral geometry ?

Options

A. `SF_(6)`
B. `IF_(5)`
C. `SiF_(6)^(2-)`
D. `SF_(4)`

Explanation

To determine which of the given molecules does not show octahedral geometry, we need to analyze the molecular geometry of each compound based on their valence electrons and bonding arrangements.

### Step-by-Step Solution:

1. **Identify the Molecules**: The given molecules are SF6, IF5, SiF6^2-, and SF4.

2. **Understanding Octahedral Geometry**:
- Octahedral geometry typically involves a central atom surrounded by six bonding pairs of electrons and no lone pairs.
- The hybridization associated with octahedral geometry is sp³d².

3. **Analyze SF6**:
- Sulfur (S) has 6 valence electrons.
- In SF6, sulfur forms bonds with 6 fluorine (F) atoms.
- This results in 6 bond pairs and 0 lone pairs.
- **Geometry**: Octahedral.

4. **Analyze IF5**:
- Iodine (I) has 7 valence electrons.
- In IF5, iodine forms bonds with 5 fluorine atoms, leaving 2 electrons as a lone pair.
- This results in 5 bond pairs and 1 lone pair.
- **Geometry**: Despite having a lone pair, the steric number is 6 (5 bond pairs + 1 lone pair), which leads to an octahedral arrangement.
- **Geometry**: Octahedral.

5. **Analyze SiF6^2-**:
- Silicon (Si) has 4 valence electrons.
- With a 2- charge, Si has a total of 6 electrons available for bonding.
- In SiF6^2-, silicon forms bonds with 6 fluorine atoms.
- This results in 6 bond pairs and 0 lone pairs.
- **Geometry**: Octahedral.

6. **Analyze SF4**:
- Sulfur (S) has 6 valence electrons.
- In SF4, sulfur forms bonds with 4 fluorine atoms, leaving 2 electrons as a lone pair.
- This results in 4 bond pairs and 1 lone pair.
- The presence of the lone pair distorts the geometry from octahedral to a seesaw shape.
- **Geometry**: Not octahedral.

### Conclusion:
The molecule that does not show octahedral geometry is **SF4**.