When `SO_(2)` is passed in acidified potassium dichromate solution, the oxidation number of S is changed from

Options

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

Answer

To solve the problem of determining the change in oxidation number of sulfur when sulfur dioxide (SO₂) is passed through acidified potassium dichromate (K₂Cr₂O₇), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Oxidation State of Sulfur in SO₂:**
- In sulfur dioxide (SO₂), the oxidation state of sulfur (S) can be calculated as follows:
- Let the oxidation state of sulfur be \( x \).
- The oxidation state of oxygen (O) is -2.
- The equation for SO₂ is:
\[ x + 2(-2) = 0 \]
- Simplifying this gives:
\[ x - 4 = 0 \implies x = +4 \]
- Therefore, the oxidation state of sulfur in SO₂ is +4.

2. **Identify the Final Oxidation State of Sulfur in SO₄²⁻:**
- In sulfate ion (SO₄²⁻), we can similarly calculate the oxidation state of sulfur:
- Let the oxidation state of sulfur be \( y \).
- The equation for SO₄²⁻ is:
\[ y + 4(-2) = -2 \]
- Simplifying this gives:
\[ y - 8 = -2 \implies y = +6 \]
- Therefore, the oxidation state of sulfur in SO₄²⁻ is +6.

3. **Determine the Change in Oxidation State:**
- The change in oxidation state of sulfur from SO₂ to SO₄²⁻ is:
\[ \text{Change} = +6 - (+4) = +2 \]
- Thus, the oxidation number of sulfur increases by 2.

### Final Answer:
The oxidation number of sulfur changes from +4 in SO₂ to +6 in SO₄²⁻.

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