When `SO_(2)` is passed in acidified `K_(2)Cr_(2)O_(7)` solution oxidation state of S changed from

Options

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

Answer

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

### Step 1: Identify the 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 formula for SO₂ can be expressed as:
\[ x + 2(-2) = 0 \]
- Simplifying this gives:
\[ x - 4 = 0 \implies x = +4 \]
Thus, the oxidation state of sulfur in SO₂ is +4.

### Step 2: Identify the oxidation state of sulfur in sulfate ion (SO₄²⁻)
Next, we need to find the oxidation state of sulfur in the sulfate ion (SO₄²⁻):
- Again, let the oxidation state of sulfur be \( x \).
- The sulfate ion has four oxygen atoms, each with an oxidation state of -2.
- The formula for SO₄²⁻ can be expressed as:
\[ x + 4(-2) = -2 \]
- Simplifying this gives:
\[ x - 8 = -2 \implies x = +6 \]
Thus, the oxidation state of sulfur in the sulfate ion (SO₄²⁻) is +6.

### Step 3: Determine the change in oxidation state
Now we can determine the change in the oxidation state of sulfur:
- Initial oxidation state in SO₂: +4
- Final oxidation state in SO₄²⁻: +6

The change in oxidation state is:
\[ +6 - (+4) = +2 \]

### Conclusion
When sulfur dioxide (SO₂) is oxidized to sulfate (SO₄²⁻) in the presence of acidified potassium dichromate, the oxidation state of sulfur changes from +4 to +6.

### Summary
- The oxidation state of sulfur in SO₂ is +4.
- The oxidation state of sulfur in SO₄²⁻ is +6.
- Therefore, the oxidation state of sulfur changes from +4 to +6.

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