When sulphur dioxide is passed in an acidified `K_2Cr_2O_7` solution,the oxidation state of sulphur is changed from

Options

A. `+4" to "0`
B. `+4" to "+2`
C. `+4" to "+6`
D. `+6" to "+4`

Explanation

To determine the change in the oxidation state of sulfur when sulfur dioxide (SO₂) is passed through an acidified potassium dichromate (K₂Cr₂O₇) solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- The reactant is sulfur dioxide (SO₂).
- The oxidizing agent is acidified potassium dichromate (K₂Cr₂O₇).
- The products formed in the reaction are sulfate ions (SO₄²⁻) and chromium ions (Cr³⁺).

2. **Write the Balanced Chemical Equation:**
- The balanced equation for the reaction can be written as:
\[ 3 \text{SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 + 4 \text{H}_2\text{SO}_4 \rightarrow 3 \text{SO}_4^{2-} + 2 \text{Cr}^{3+} + 4 \text{H}_2\text{O} \]

3. **Determine the Oxidation States:**
- In sulfur dioxide (SO₂), the oxidation state of sulfur (S) is +4.
- In sulfate ion (SO₄²⁻), the oxidation state of sulfur is +6.

4. **Calculate the Change in Oxidation State:**
- The change in oxidation state of sulfur can be calculated as follows:
\[ \text{Change in oxidation state} = +6 - (+4) = +2 \]
- This indicates that sulfur is oxidized from an oxidation state of +4 to +6.

5. **Conclusion:**
- Therefore, when sulfur dioxide is passed in an acidified K₂Cr₂O₇ solution, the oxidation state of sulfur changes from +4 to +6.

### Final Answer:
The oxidation state of sulfur changes from +4 to +6.

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