The average oxidation state of sulphur atom in `S_4O_6^(2-)` ion is

Options

A. `+2`
B. `+5`
C. 0
D. `+2.5`

Answer

To find the average oxidation state of sulfur in the \( S_4O_6^{2-} \) ion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
We have the ion \( S_4O_6^{2-} \). This consists of 4 sulfur (S) atoms and 6 oxygen (O) atoms.

2. **Assign Oxidation States**:
- The oxidation state of oxygen is generally \(-2\). Therefore, for 6 oxygen atoms:
\[ \text{Total oxidation state from oxygen} = 6 \times (-2) = -12 \]

3. **Set Up the Equation**:
Let \( X \) be the oxidation state of sulfur. Since there are 4 sulfur atoms, the contribution from sulfur is:
\[ 4X \]
The overall charge of the ion is \(-2\). Therefore, we can set up the equation:
\[ 4X + (-12) = -2 \]

4. **Solve for \( X \)**:
Rearranging the equation gives:
\[ 4X - 12 = -2 \]
Adding 12 to both sides:
\[ 4X = 10 \]
Dividing both sides by 4:
\[ X = \frac{10}{4} = 2.5 \]

5. **Conclusion**:
The average oxidation state of sulfur in the \( S_4O_6^{2-} \) ion is \( +2.5 \).

### Final Answer:
The average oxidation state of sulfur in \( S_4O_6^{2-} \) is \( +2.5 \).

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