Answer:
The oxidation state of sulfur, S, is +6, and the oxidation state of oxygen, O, is - 2:
Explanation:
Application of rules
1. <u>Ion SO₄²⁻</u>
In a neutral compound, the sum of the oxidation states of all the atoms that constitute the compound is equal to zero. In ions the sum of the oxidation states of the atoms is equal to the charge of the ion.
- In this case, the ion SO₄²⁻ has charge negative 2: - 2
<u>2. Oxygen</u>
With the exception of the peroxides, when oxygen is combined with other elements, it always has oxidation state - 2.
Hence, the total charge of the four atoms of oxygen in SO₄²⁻ is 4×(-2) = - 8.
<u>3. Sulfur</u>
Therefore, naming x the unknown oxidation state of S, you can set this equation:
- x + ( - 8) = - 2
- ⇒ x = - 2- ( - 8) = - 2 + 8 = + 6
Conclusion
- In SO₄²⁻, the oxidation state of sulfur, S, is +6, and the oxidation state of oxygen, O, is - 2.
