In mildly alkaline medium, thiosulphate ion is oxidized by `MnO_(4)^(-)` to "A". The oxidation state of sulphur in "A" is ______

Answer

To solve the problem, we need to determine the oxidation state of sulfur in the product formed when thiosulfate ion (S₂O₃²⁻) is oxidized by permanganate ion (MnO₄⁻) in a mildly alkaline medium.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactant is thiosulfate ion (S₂O₃²⁻).
- The oxidizing agent is permanganate ion (MnO₄⁻).
- In a mildly alkaline medium, thiosulfate will be oxidized to sulfate ion (SO₄²⁻).

2. **Determine the Oxidation State of Sulfur in the Product**:
- The product formed is sulfate ion (SO₄²⁻).
- We need to find the oxidation state of sulfur (S) in sulfate.

3. **Set Up the Oxidation State Equation**:
- Let the oxidation state of sulfur in sulfate be \( x \).
- The sulfate ion has four oxygen atoms, each with an oxidation state of -2.
- The overall charge of the sulfate ion is -2.

The equation can be set up as follows:
\[ x + 4(-2) = -2 \]

4. **Solve the Equation**:
- Simplifying the equation:
\[ x - 8 = -2 \]
\[ x = -2 + 8 \]
\[ x = +6 \]

5. **Conclusion**:
- The oxidation state of sulfur in the product sulfate (SO₄²⁻) is +6.

### Final Answer:
The oxidation state of sulfur in "A" (sulfate) is **+6**.

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