Sulphur and rest of the elements of group `16` are less electronegative than oxygen, Therefore, their atoms cannot take electrons easily. They can acquire `ns^2 np^6` configuration by sharing two electrons with the atoms of other elements and thus, exhibit `+2` oxidation state in their compounds. In addition to this, their atoms have vacant d-orbitals in their valence shell to which electrons can be promoted from the `p` and s-orbitals of the same shell. As a result, they can show `+4` and `+ 6` oxidation states.
The oxidation state of of sulphur in `S_8, SO_3` and `H_2 S` respectively are.

Options

A. `0, +6, -2`
B. `+2, +6, -2`
C. `0, +4 +2`
D. `-2, +6,+2`

Explanation

To determine the oxidation states of sulfur in \( S_8 \), \( SO_3 \), and \( H_2S \), we will analyze each compound step by step.

### Step 1: Oxidation State of Sulfur in \( S_8 \)

- **Explanation**: In its elemental form, sulfur exists as \( S_8 \). The oxidation state of any element in its standard state is always zero.
- **Calculation**:
\[ \text{Oxidation state of sulfur in } S_8 = 0 \]

### Step 2: Oxidation State of Sulfur in \( SO_3 \)

- **Explanation**: In sulfur trioxide (\( SO_3 \)), we need to consider the oxidation states of oxygen. Oxygen typically has an oxidation state of -2.
- **Calculation**: Let the oxidation state of sulfur be \( x \).
\[ x + 3(-2) = 0 \]
\[ x - 6 = 0 \]
\[ x = +6 \]
- **Conclusion**: The oxidation state of sulfur in \( SO_3 \) is +6.

### Step 3: Oxidation State of Sulfur in \( H_2S \)

- **Explanation**: In hydrogen sulfide (\( H_2S \)), hydrogen typically has an oxidation state of +1.
- **Calculation**: Let the oxidation state of sulfur be \( x \).
\[ 2(+1) + x = 0 \]
\[ 2 + x = 0 \]
\[ x = -2 \]
- **Conclusion**: The oxidation state of sulfur in \( H_2S \) is -2.

### Final Answer

- The oxidation states of sulfur in \( S_8 \), \( SO_3 \), and \( H_2S \) are:
- \( S_8 \): 0
- \( SO_3 \): +6
- \( H_2S \): -2

Thus, the final answer is: **0, +6, -2**.

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