What is the oxidation state of sulphur?

2 answer(s)
Answer # 1 #

That's a great chemistry question! Sulphur (S) is a fascinating element because it's a nonmetal that exhibits a very wide range of oxidation states. It’s a group 16 element, similar to oxygen. The most common oxidation states of Sulphur are: * $-2$: This is its lowest state and occurs when sulphur gains two electrons to fill its outermost shell, becoming the sulfide ion ($\text{S}^{2-}$). You see this in compounds like Hydrogen Sulfide ($\text{H}_2\text{S}$) and Iron(II) Sulfide ($\text{FeS}$). * $+6$: This is its highest state and occurs when it loses all six of its valence electrons. This is found in compounds like Sulfuric Acid ($\text{H}_2\text{SO}_4$) and Sulphur Trioxide ($\text{SO}_3$). * $+4$: Found in compounds like Sulfur Dioxide ($\text{SO}_2$) and Sulfite ion ($\text{SO}_3^{2-}$). * $0$: In its elemental form, like in a ring of $\text{S}_8$ or just the atom $\text{S}$. * $+2$ and $+3$ also occur, but are less common. So, the short answer is $-2, 0, +4, \text{ and } +6$ are the ones you see most often!

[1 Year]
Answer # 2 #

Sulfur oxidation state depends on the compound: -2 in H₂S, 0 in elemental S, +4 in SO₂, +6 in H₂SO₄. Always check the specific compound. Example: in H₂SO₄, S + 2(H) + 4(O) = 0 → S = +6.

[1 Year]