Oxidation state of S in compound 'SO2' is

Options

A. `+1`
B. `+4`
C. `+6`
D. `+2`

Answer

To find the oxidation state of sulfur (S) in the compound sulfur dioxide (SO2), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation state of oxygen**:
- Oxygen is more electronegative than sulfur. The common oxidation state of oxygen is -2.

2. **Set up the equation**:
- Let the oxidation state of sulfur be represented as \( x \).
- In the compound SO2, there are two oxygen atoms. Therefore, the total contribution of oxygen to the oxidation state is \( 2 \times (-2) = -4 \).

3. **Write the equation for the overall charge**:
- Since SO2 is a neutral molecule, the sum of the oxidation states must equal 0.
- We can write the equation as:
\[ x + 2 \times (-2) = 0 \]
- This simplifies to:
\[ x - 4 = 0 \]

4. **Solve for \( x \)**:
- Rearranging the equation gives:
\[ x = 4 \]
- Therefore, the oxidation state of sulfur in SO2 is +4.

5. **Conclusion**:
- The oxidation state of sulfur in SO2 is +4.

### Final Answer:
The oxidation state of sulfur (S) in sulfur dioxide (SO2) is +4.

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