Question
Find the oxidation state of sulfur (S) in H₂SO₄ (sulphuric acid).
This is one of the most frequently tested problems in NCERT Class 11 redox chapter, and the same rule set applies to dozens of similar problems — CrO₄²⁻, MnO₄⁻, Cr₂O₇²⁻ — so getting the method right here pays dividends.
Solution — Step by Step
Hydrogen is +1 (Rule: H is +1 in all compounds except metal hydrides like NaH). Oxygen is −2 (Rule: O is −2 in all compounds except peroxides like H₂O₂ and OF₂).
Let oxidation state of S = .
The molecule H₂SO₄ is neutral — overall charge is zero.
So the sum of all oxidation numbers must equal zero:
Sulfur’s maximum possible oxidation state is +6 (it has 6 valence electrons). H₂SO₄ is a fully oxidised sulfur compound — the +6 makes complete chemical sense.
Oxidation state of S in H₂SO₄ = +6
Why This Works
Oxidation numbers are a bookkeeping tool. We assign them based on a priority hierarchy of rules, then use algebra to find the unknown.
The sum rule is the key: for a neutral molecule, all oxidation numbers add to zero. For a polyatomic ion, they add to the charge of the ion. This single rule, combined with the fixed values for H and O, lets us crack almost every problem.
The numbers you assign to H and O are not arbitrary — they come from electronegativity. Oxygen is more electronegative than nearly every element, so it “pulls” both electrons in each bond, giving it a −2 charge in our accounting system.
| Species | Oxidation State |
|---|---|
| Free element (Na, O₂, S₈) | 0 |
| Monatomic ion (Na⁺, Cl⁻) | = ionic charge |
| H in compounds | +1 (except metal hydrides: −1) |
| O in compounds | −2 (except peroxides: −1, OF₂: +2) |
| Halogens (F, Cl, Br, I) | −1 when bonded to metals |
| Sum rule — neutral molecule | = 0 |
| Sum rule — polyatomic ion | = charge of ion |
Alternative Method — Structural Approach
If you know the Lewis structure of H₂SO₄, you can verify: sulfur forms 4 bonds with oxygen (two S=O double bonds and two S–OH single bonds). Each S=O bond assigns −2 to O and +2 contribution to S. Each S–OH: oxygen gets −2, hydrogen gets +1. Summing the contributions to S gives +6. Same answer, different path — useful for building intuition on why the number is what it is.
For ions like SO₄²⁻, the method is identical — just set the sum equal to the charge (−2), not zero. So: , giving . Same sulfur, same oxidation state.
Common Mistake
Students count the number of H and O atoms correctly but then forget to multiply. Writing instead of — ignoring the subscripts. H₂SO₄ has two hydrogens and four oxygens. Always multiply oxidation state × number of atoms before summing.
This exact slip turns a +6 answer into a completely wrong value and costs marks on a question that is essentially free if the method is clean. Write out each term explicitly: 2(+1) + x + 4(−2) = 0 — never shortcut the algebra on paper.