The oxidation state of 'S' in Kal`(SO_(4))_(2) .12H_(2)O` is

Options

A. `-2`
B. `-1`
C. 2
D. `+6`

Answer

To find the oxidation state of sulfur (S) in the compound KAl(SO₄)₂·12H₂O, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the components of the compound**: The compound is potassium aluminum sulfate dodecahydrate, which can be broken down into its constituent ions: K⁺, Al³⁺, SO₄²⁻, and water (H₂O).

2. **Assign known oxidation states**:
- Potassium (K) has an oxidation state of +1.
- Aluminum (Al) has an oxidation state of +3.
- Oxygen (O) typically has an oxidation state of -2.

3. **Set up the equation**:
- Let the oxidation state of sulfur (S) be represented as X.
- The sulfate ion (SO₄²⁻) contains one sulfur atom and four oxygen atoms. The overall charge of the sulfate ion is -2.
- The equation for the sulfate ion can be written as:
\[ X + 4(-2) = -2 \]

4. **Simplify the equation**:
- This simplifies to:
\[ X - 8 = -2 \]
- Rearranging gives:
\[ X = -2 + 8 \]
\[ X = +6 \]

5. **Conclusion**: The oxidation state of sulfur (S) in KAl(SO₄)₂·12H₂O is +6.

### Final Answer:
The oxidation state of sulfur in KAl(SO₄)₂·12H₂O is +6.

---