What is the oxidation state of sulfur in FeSO4?
The correct answer and explanation is:
Correct Answer: The oxidation state of sulfur in FeSO₄ is +6.
Explanation:
To find the oxidation state of sulfur in iron(II) sulfate (FeSO₄), we need to analyze the compound’s structure and use known oxidation states of the other elements.
FeSO₄ is composed of three ions:
- Iron (Fe)
- Sulfate ion (SO₄²⁻)
Step 1: Determine the oxidation state of iron (Fe)
In FeSO₄, iron is in the +2 oxidation state. This is because the name includes the Roman numeral II, indicating Fe²⁺.
Step 2: Understand the charge on the sulfate ion (SO₄²⁻)
The sulfate ion (SO₄²⁻) has a total charge of -2. It is a polyatomic ion composed of one sulfur atom and four oxygen atoms.
Step 3: Assign oxidation states within SO₄²⁻
Oxygen typically has an oxidation state of -2. Since there are four oxygen atoms:
4 × (-2) = -8
Let the oxidation state of sulfur be x. The total charge on the sulfate ion must equal -2. Therefore:
x + (-8) = -2
x – 8 = -2
x = +6
Conclusion:
Sulfur has an oxidation state of +6 in FeSO₄.
This oxidation state indicates that sulfur is in a highly oxidized form. The +6 state is common for sulfur in oxyanions like sulfate (SO₄²⁻) and is one of the highest oxidation numbers sulfur can have. This means sulfur has lost six electrons compared to its elemental state. Understanding oxidation states helps in identifying redox reactions and balancing chemical equations accurately.