What is the oxidation state of sulphur in sulphate ion ?

Explanation

To determine the oxidation state of sulfur in the sulfate ion (SO₄²⁻), we can follow these steps:

### Step 1: Write the formula for the sulfate ion
The formula for the sulfate ion is SO₄²⁻. This indicates that the ion has a total charge of -2.

### Step 2: Assign the oxidation state to sulfur
Let the oxidation state of sulfur be represented by the variable \( x \).

### Step 3: Assign the oxidation state to oxygen
In compounds, oxygen typically has an oxidation state of -2. Since there are four oxygen atoms in the sulfate ion, the total contribution from oxygen will be:
\[ 4 \times (-2) = -8 \]

### Step 4: Set up the equation
Now, we can set up the equation based on the total charge of the sulfate ion:
\[ x + (-8) = -2 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ x - 8 = -2 \]
Adding 8 to both sides results in:
\[ x = -2 + 8 \]
Thus,
\[ x = +6 \]

### Conclusion
The oxidation state of sulfur in the sulfate ion (SO₄²⁻) is +6.

---