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 1: Determine the total charge on the ion
The \( S_4O_6^{2-} \) ion has a total charge of -2.

### Step 2: Assign oxidation states to oxygen
Oxygen typically has an oxidation state of -2. In the \( S_4O_6^{2-} \) ion, there are 6 oxygen atoms:
\[ \text{Total oxidation state from oxygen} = 6 \times (-2) = -12 \]

### Step 3: Set up the equation for the total oxidation state
Let the oxidation state of sulfur be \( x \). Since there are 4 sulfur atoms, the total oxidation state from sulfur will be \( 4x \). The overall charge of the ion can be expressed as:
\[ 4x + (-12) = -2 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ 4x - 12 = -2 \]
\[ 4x = -2 + 12 \]
\[ 4x = 10 \]
\[ x = \frac{10}{4} = 2.5 \]

### Step 5: Calculate the average oxidation state
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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