What is the oxidation number of S in SO42-?
The correct answer and explanation is:
The oxidation number of sulfur (S) in the sulfate ion (SO₄²⁻) is +6.
To determine this, we need to use the rules for assigning oxidation states:
- Oxidation state of oxygen: Oxygen in most compounds, including sulfates, has an oxidation state of -2.
- Total charge of the ion: The sulfate ion, SO₄²⁻, has a charge of -2. The sum of the oxidation numbers in this ion must equal -2.
- Assigning the oxidation state of sulfur: Let the oxidation number of sulfur be xx. Since there are four oxygen atoms, each with an oxidation state of -2, the total contribution from oxygen is 4×(−2)=−84 \times (-2) = -8. The sum of the oxidation numbers must equal the overall charge of the ion, which is -2. Therefore, the equation is: x+(−8)=−2x + (-8) = -2 Simplifying this equation: x−8=−2x – 8 = -2 Solving for xx: x=+6x = +6
Thus, the oxidation number of sulfur in SO₄²⁻ is +6.
This is consistent with sulfur’s behavior in various compounds, where it often has oxidation states ranging from +2 to +6. In the case of the sulfate ion, sulfur achieves its highest common oxidation state, +6, as it forms bonds with the oxygen atoms that contribute to the overall negative charge of the ion.