What are the oxidation states of sulfur in the following :
`SO_(4)^-2`

Answer

To determine the oxidation state of sulfur in the sulfate ion, \( SO_4^{2-} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ion and Its Charge**:
The sulfate ion is represented as \( SO_4^{2-} \). The overall charge of the ion is -2.

2. **Assign Oxidation States to Known Elements**:
In the sulfate ion, we know that oxygen typically has an oxidation state of -2.

3. **Count the Number of Oxygen Atoms**:
There are 4 oxygen atoms in the sulfate ion. Therefore, the total contribution of the oxygen atoms to the overall charge is:
\[ 4 \times (-2) = -8 \]

4. **Set Up the Equation**:
Let the oxidation state of sulfur be \( x \). The equation representing the total charge of the ion can be set up as follows:
\[ x + (-8) = -2 \]

5. **Solve for the Oxidation State of Sulfur**:
Rearranging the equation gives:
\[ x - 8 = -2 \]
Adding 8 to both sides results in:
\[ x = -2 + 8 \]
Therefore:
\[ x = +6 \]

6. **Conclusion**:
The oxidation state of sulfur in the sulfate ion \( SO_4^{2-} \) is +6.

### Final Answer:
The oxidation state of sulfur in \( SO_4^{2-} \) is +6.

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