When `SO_(2)` is passed inoto an acidified potassium dichromate soltion, the oxidation number of sulphur and chromium in the final products respectively are:

Options

A. `+6,+6`
B. `+6,+3`
C. `+0,+3`
D. `+2,+3`

Answer

To solve the problem of determining the oxidation numbers of sulfur and chromium when SO₂ is passed into an acidified potassium dichromate solution, we can follow these steps:

### Step 1: Identify the Reactants and Products
- The reactant is sulfur dioxide (SO₂) and the acidified potassium dichromate (K₂Cr₂O₇).
- The products formed in the reaction are potassium sulfate (K₂SO₄) and chromium(III) sulfate (Cr₂(SO₄)₃).

### Step 2: Write the Balanced Chemical Equation
- The balanced chemical equation for the reaction is:

\[ 3 \, \text{SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 + 4 \, \text{H}_2\text{SO}_4 \rightarrow 3 \, \text{K}_2\text{SO}_4 + 2 \, \text{Cr}_2(SO_4)_3 + 4 \, \text{H}_2\text{O} \]

### Step 3: Determine the Oxidation States
- In the reactants:
- In SO₂, sulfur has an oxidation state of +4.
- In potassium dichromate (K₂Cr₂O₇), chromium has an oxidation state of +6.

- In the products:
- In potassium sulfate (K₂SO₄), sulfur has an oxidation state of +6.
- In chromium(III) sulfate (Cr₂(SO₄)₃), chromium has an oxidation state of +3.

### Step 4: Conclusion
- Therefore, the final oxidation numbers of sulfur and chromium in the products are:
- Sulfur: +6
- Chromium: +3

### Final Answer
The oxidation numbers of sulfur and chromium in the final products are +6 and +3, respectively.

---