The oxidation state of `S` in `H_(2)S_(2)O_(8)` is

Options

A. `+6`
B. `+7`
C. `+8`
D. 0

Explanation

To determine the oxidation state of sulfur (S) in the compound \( H_2S_2O_8 \) (peroxy disulfuric acid), we can follow these steps:

### Step 1: Assign oxidation states to known elements
- Hydrogen (H) typically has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2, but in peroxy compounds, the oxygen in the peroxy linkage has an oxidation state of -1.

### Step 2: Set up the equation
Let the oxidation state of sulfur (S) be \( X \). The compound \( H_2S_2O_8 \) consists of:
- 2 hydrogen atoms
- 2 sulfur atoms
- 8 oxygen atoms

The equation to represent the sum of oxidation states in the compound is:
\[ 2(+1) + 2(X) + 6(-2) + 2(-1) = 0 \]
Here, we have:
- 2 hydrogen atoms contributing \( 2 \times +1 = +2 \)
- 2 sulfur atoms contributing \( 2X \)
- 6 oxygen atoms contributing \( 6 \times -2 = -12 \)
- 2 oxygen atoms in the peroxy linkage contributing \( 2 \times -1 = -2 \)

### Step 3: Simplify the equation
Now, we can simplify the equation:
\[ 2 + 2X - 12 - 2 = 0 \]
This simplifies to:
\[ 2X - 12 = 0 \]

### Step 4: Solve for \( X \)
Now, we can solve for \( X \):
\[ 2X = 12 \]
\[ X = 6 \]

### Conclusion
The oxidation state of sulfur in \( H_2S_2O_8 \) is +6.

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