The oxidation states of S in `S_2O_8^(2-)` is

Options

A. `+2`
B. `+4`
C. `+6`
D. `+7`

Answer

To find the oxidation state of sulfur (S) in the ion \( S_2O_8^{2-} \), we can follow these steps:

### Step 1: Identify the total charge of the ion
The given ion is \( S_2O_8^{2-} \). The overall charge of this ion is -2.

### Step 2: Determine the oxidation state of oxygen
In most compounds, oxygen has an oxidation state of -2. Since there are 8 oxygen atoms in \( S_2O_8^{2-} \), the total contribution from oxygen is:
\[ 8 \times (-2) = -16 \]

### Step 3: Set up the equation for the oxidation states
Let the oxidation state of sulfur be \( x \). Since there are 2 sulfur atoms, the total contribution from sulfur will be \( 2x \). The equation can be set up as follows:
\[ 2x + (-16) = -2 \]

### Step 4: Solve for the oxidation state of sulfur
Now, we can simplify the equation:
\[ 2x - 16 = -2 \]
Adding 16 to both sides gives:
\[ 2x = 14 \]
Dividing both sides by 2 results in:
\[ x = 7 \]

### Step 5: Conclusion
The oxidation state of each sulfur atom in \( S_2O_8^{2-} \) is +7.

### Summary
Thus, the oxidation state of sulfur in \( S_2O_8^{2-} \) is +7.

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