Oxidation number of S in `H_(2)S_(2)O_(7)` is -

Options

A. `+4`
B. `-6`
C. `-5`
D. `+6`

Answer

To find the oxidation number of sulfur (S) in the compound \( H_2S_2O_7 \), we can follow these steps:

### Step 1: Identify the oxidation states of other elements
In \( H_2S_2O_7 \):
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) generally has an oxidation state of -2.

### Step 2: Set up the equation for the oxidation state of sulfur
Let the oxidation state of sulfur be \( x \). Since there are two sulfur atoms in the compound, we will have \( 2x \) for sulfur.

### Step 3: Write the equation based on the total oxidation states
The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can write the equation as follows:

\[ 2 \times (+1) + 2x + 7 \times (-2) = 0 \]

### Step 4: Simplify the equation
Substituting the values into the equation gives:

\[ 2 + 2x - 14 = 0 \]

### Step 5: Solve for \( x \)
Now, we can simplify the equation:

\[ 2x - 12 = 0 \]

Adding 12 to both sides:

\[ 2x = 12 \]

Dividing both sides by 2:

\[ x = 6 \]

### Conclusion
The oxidation number of sulfur (S) in \( H_2S_2O_7 \) is +6.

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