When sulphur is treated with excess of Fluorine, the compound formed is

Options

A. `SF_4`
B. `S_2F_2`
C. `SF_6`
D. `S_2F_6`

Explanation

To determine the compound formed when sulfur is treated with excess fluorine, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this case are sulfur (S) and fluorine (F).

2. **Determine the Oxidation States**:
- Sulfur has an atomic number of 16 and can exhibit various oxidation states. In the presence of a strong oxidizing agent like fluorine, sulfur can reach its highest oxidation state of +6.
- Fluorine, being highly electronegative, typically exists in the -1 oxidation state when it forms compounds.

3. **Reaction Analysis**:
- When sulfur reacts with fluorine, the reaction can be represented as:
\[ S + 6F \rightarrow SF6 \]
- Here, sulfur is oxidized to +6, and each fluorine atom is reduced to -1.

4. **Excess Fluorine**:
- The question specifies that there is an excess of fluorine. This means that all available sulfur can react with fluorine to form the highest stable compound, which is sulfur hexafluoride (SF6).

5. **Final Compound**:
- The compound formed when sulfur is treated with excess fluorine is sulfur hexafluoride (SF6).

### Conclusion:
The correct answer is **C - SF6**.

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