`F_(2)` oxidises `KHSO_(4)` to `K_(2)S_(2)O_(8)`. The change in oxidation number of 'S' from reactnat to product is 2

Options

A. 2 units
B. 4 units
C. no change
D. one unit

Explanation

To solve the problem, we need to determine the change in the oxidation number of sulfur (S) when potassium hydrogen sulfate (KHSO₄) is oxidized to dipotassium peroxodisulfate (K₂S₂O₈) by fluorine (F₂).

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- Reactant: KHSO₄ (Potassium Hydrogen Sulfate)
- Product: K₂S₂O₈ (Dipotassium Peroxodisulfate)

2. **Determine the Oxidation State of Sulfur in KHSO₄:**
- The formula for KHSO₄ can be broken down into its components:
- K (Potassium) = +1
- H (Hydrogen) = +1
- O (Oxygen) = -2 (4 oxygen atoms contribute -8)
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation states is:
\[ +1 + 1 + x + 4(-2) = 0 \]
- Simplifying this gives:
\[ 1 + 1 + x - 8 = 0 \implies x - 6 = 0 \implies x = +6 \]
- Therefore, the oxidation state of sulfur in KHSO₄ is +6.

3. **Determine the Oxidation State of Sulfur in K₂S₂O₈:**
- The formula for K₂S₂O₈ can be analyzed similarly:
- K (Potassium) = +1 (2 potassium atoms contribute +2)
- O (Oxygen) = -2 (6 oxygen atoms contribute -12)
- The two oxygen atoms that are part of the peroxide linkage contribute -1 each, so they contribute -2.
- Let the oxidation state of sulfur be \( y \).
- The equation for the oxidation states is:
\[ 2(+1) + 2y + 6(-2) = 0 \]
- Simplifying gives:
\[ 2 + 2y - 12 = 0 \implies 2y - 10 = 0 \implies 2y = 10 \implies y = +5 \]
- Therefore, the oxidation state of sulfur in K₂S₂O₈ is +6.

4. **Calculate the Change in Oxidation Number:**
- The change in oxidation state of sulfur from KHSO₄ to K₂S₂O₈ is:
\[ +6 \text{ (in KHSO₄)} - +6 \text{ (in K₂S₂O₈)} = 0 \]
- Thus, there is no change in the oxidation number of sulfur.

### Conclusion:
The change in oxidation number of sulfur from KHSO₄ to K₂S₂O₈ is **0**.