The oxidation numbers of sulphur in `S_(8), SO_(2)` and `H_(2)S` respectively are (M-2010 & IIT 1999) 

Options

A. `0, +4 and -2`
B. `+2, +4 and -2`
C. `0, +4 and +2`
D. `-2, +4 an d-2`

Answer

To determine the oxidation numbers of sulfur in the compounds \( S_8 \), \( SO_2 \), and \( H_2S \), we will analyze each compound step by step.

### Step 1: Determine the oxidation number of sulfur in \( S_8 \)

- **Analysis**: \( S_8 \) is elemental sulfur, which means it is in its natural state.
- **Oxidation Number**: The oxidation number of any element in its elemental form is always 0.

**Conclusion**: The oxidation number of sulfur in \( S_8 \) is **0**.

### Step 2: Determine the oxidation number of sulfur in \( SO_2 \)

- **Analysis**: In sulfur dioxide (\( SO_2 \)), we know that oxygen is more electronegative than sulfur. The oxidation state of oxygen is typically \(-2\).
- **Setting up the equation**: Let the oxidation number of sulfur be \( x \). The compound is neutral, so we can write the equation:

\[ x + 2(-2) = 0 \]

- **Solving the equation**:

\[ x - 4 = 0 \implies x = +4 \]

**Conclusion**: The oxidation number of sulfur in \( SO_2 \) is **+4**.

### Step 3: Determine the oxidation number of sulfur in \( H_2S \)

- **Analysis**: In hydrogen sulfide (\( H_2S \)), hydrogen has an oxidation state of \( +1 \).
- **Setting up the equation**: Let the oxidation number of sulfur be \( y \). The compound is neutral, so we can write the equation:

\[ 2(+1) + y = 0 \]

- **Solving the equation**:

\[ 2 + y = 0 \implies y = -2 \]

**Conclusion**: The oxidation number of sulfur in \( H_2S \) is **-2**.

### Final Summary of Oxidation Numbers

- For \( S_8 \): **0**
- For \( SO_2 \): **+4**
- For \( H_2S \): **-2**

Thus, the oxidation numbers of sulfur in \( S_8 \), \( SO_2 \), and \( H_2S \) respectively are **0, +4, -2**.

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