`SO_(2)` under atomspheric condition changes to `SO_(x)^(2-)`. If oxidation number of `S` in `SO_(x)^(2-)` is `+6`, what is the value of `x` in `SO_(x)^(2-)`?

Options

A. `2`
B. `1`
C. `3`
D. `4`

Explanation

To solve the problem, we need to determine the value of \( x \) in the compound \( SO_x^{2-} \) given that the oxidation number of sulfur (S) is +6. Let's break this down step by step.

### Step 1: Understand the oxidation states
In the compound \( SO_x^{2-} \):
- The oxidation number of sulfur (S) is given as +6.
- The oxidation number of oxygen (O) is -2.

### Step 2: Set up the equation
The overall charge of the compound is -2. We can express this using the oxidation numbers:
\[ \text{Total oxidation number} = (\text{Oxidation number of S}) + (\text{Oxidation number of O}) \times (\text{Number of O atoms}) \]
This can be written as:
\[ +6 + (-2) \times x = -2 \]

### Step 3: Simplify the equation
Now, let's simplify the equation:
\[ +6 - 2x = -2 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ -2x = -2 - 6 \]
\[ -2x = -8 \]
Now, dividing both sides by -2:
\[ x = 4 \]

### Conclusion
Thus, the value of \( x \) is 4. Therefore, the compound is \( SO_4^{2-} \).