The oxidation number of sulphur in `S_(2)O_(7)^(-2)` is `"______________"`.

Options

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

Answer

To find the oxidation number of sulfur in the ion \( S_2O_7^{2-} \), we can follow these steps:

### Step 1: Assign oxidation states
We start by assigning oxidation states to the elements in the compound. We know that:
- The oxidation state of oxygen (O) is typically \(-2\).

### Step 2: Set up the equation
Let the oxidation state of sulfur (S) be \( x \). Since there are two sulfur atoms in the ion, the total contribution from sulfur will be \( 2x \).

### Step 3: Calculate the total oxidation state
The total oxidation state for the ion can be expressed as:
\[ 2x + 7(-2) = -2 \]
Here, \( 7(-2) \) accounts for the seven oxygen atoms contributing \(-2\) each.

### Step 4: Simplify the equation
Now, we can simplify the equation:
\[ 2x - 14 = -2 \]

### Step 5: Solve for \( x \)
To isolate \( x \), we add 14 to both sides:
\[ 2x = -2 + 14 \]
\[ 2x = 12 \]
Now, divide by 2:
\[ x = 6 \]

### Conclusion
The oxidation number of sulfur in \( S_2O_7^{2-} \) is \( +6 \).

### Final Answer
The oxidation number of sulfur in \( S_2O_7^{2-} \) is \( +6 \).

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