What is the oxidation state of S in FeSO4
The correct answer and explanation is:
The correct oxidation state of sulfur (S) in FeSO₄ is +6.
To determine the oxidation state of sulfur in FeSO₄, start by understanding the oxidation states of the other elements. FeSO₄ is composed of iron (Fe²⁺) and the sulfate ion (SO₄²⁻). The overall compound is neutral, meaning the positive and negative charges must balance out.
In this compound:
- Iron (Fe) has an oxidation state of +2, because it forms the ferrous ion Fe²⁺.
- The sulfate ion (SO₄²⁻) has a total charge of -2.
Within the sulfate ion (SO₄²⁻), oxygen always has an oxidation state of -2. There are four oxygen atoms, contributing a total of 4 × -2 = -8.
Let the oxidation state of sulfur be x. Since the total charge of the sulfate ion is -2, the sum of the oxidation states of sulfur and oxygen must equal -2:
x + (-8) = -2
Solving for x:
x = -2 + 8 = +6
Therefore, sulfur has an oxidation state of +6 in FeSO₄.
This value makes sense based on known chemistry. Sulfur is a multivalent nonmetal and can exhibit oxidation states ranging from -2 to +6. In sulfate compounds, sulfur is consistently found in the +6 oxidation state because it forms stable tetrahedral SO₄²⁻ ions. These ions are commonly found in salts like FeSO₄, CuSO₄, and ZnSO₄.
Knowing oxidation states is important in redox chemistry, chemical reactions, and determining electron transfer. It also helps classify compounds and understand their reactivity. In this case, FeSO₄ is often used in chemistry labs and medicine, and understanding its composition is essential to predict how it behaves chemically.