Amongst hexahalides of sulphur, `SF_6` is exceptionally stable due to

Options

A. See- saw geometry
B. High polarization power
C. Least steric hindrance
D. Dimeric in nature

Answer

To determine why \( SF_6 \) is exceptionally stable compared to other hexahalides of sulfur such as \( SCl_6 \) or \( SBr_6 \), we can analyze the factors contributing to the stability of these compounds.

### Step-by-Step Solution:

1. **Understanding Hexahalides of Sulfur**:
- Sulfur can form hexahalides by bonding with halogens (F, Cl, Br, I). The most notable hexahalide is \( SF_6 \).

2. **Size of Halogens**:
- The halogens vary in size: Fluorine (F) is the smallest, followed by Chlorine (Cl), Bromine (Br), and Iodine (I), which is the largest.

3. **Steric Hindrance**:
- Steric hindrance refers to the repulsion between atoms that are too close to each other. Larger atoms create more steric hindrance due to their size.

4. **Analyzing \( SF_6 \)**:
- In \( SF_6 \), sulfur is surrounded by six fluorine atoms. The small size of fluorine minimizes steric hindrance, allowing for stable bonding.

5. **Comparing with \( SCl_6 \) and \( SBr_6 \)**:
- In \( SCl_6 \) and \( SBr_6 \), the larger size of chlorine and bromine increases steric hindrance. This leads to greater repulsion between the bonded atoms, making these compounds less stable.

6. **Conclusion**:
- The exceptional stability of \( SF_6 \) is primarily due to the small size of fluorine, which results in less steric hindrance compared to the larger halogens like chlorine and bromine.

### Final Answer:
\( SF_6 \) is exceptionally stable due to the least steric hindrance caused by the small size of fluorine compared to larger halogens like chlorine and bromine.