Oxidation State Calculator

Calculate oxidation states of elements in compounds and ions using charge balance rules.

Overall charge of formula unit

0=neutral, -1=anion, +2=cation

Sum of known oxidation states

e.g. 4 oxygen atoms x -2 = -8

Number of unknown element atoms

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How Oxidation States Are Assigned

Oxidation state asks the opposite question to formal charge: if every bond were broken and both electrons given to the more electronegative atom, what charge would remain? It is a deliberately extreme assumption — treating every bond as fully ionic — but it is exactly what makes redox bookkeeping work.

Assignment follows a priority order. Work down the list, and once an atom is assigned by a higher rule, that assignment holds.

Priority	Rule	Value	Common exceptions
1	Free element	0	None
2	Monatomic ion	Equal to its charge	None
3	Fluorine in compounds	−1	None — fluorine is always −1
4	Group 1 metals	+1	None in normal compounds
5	Group 2 metals	+2	None in normal compounds
6	Hydrogen	+1	−1 in metal hydrides such as NaH
7	Oxygen	−2	−1 in peroxides, −½ in superoxides, +2 in OF₂
8	Everything else	Whatever makes the total work	—

The final rule does the real work: oxidation states must sum to the overall charge on the species. For a neutral compound that is zero; for a polyatomic ion it is the ion charge. Assign everything you can from the priority list, then solve for the unknown.

Unlike formal charge, oxidation states can legitimately be fractional. In Fe₃O₄ the average iron oxidation state is +8/3, reflecting a mixture of Fe(II) and Fe(III) centres rather than a genuinely fractional charge on any single atom.

Worked Examples

Example 1: MnO4-: charge=-1, 4x O=-8, n=1 Mn

Mn=(-1-(-8))/1

Result: Mn = +7

Permanganate - strong oxidizer

Example 2: Cr2O72-: charge=-2, 7x O=-14, n=2 Cr

Cr=(-2-(-14))/2

Result: Cr = +6

Dichromate - Cr(VI) toxic

Example 3: Sulfur in sulfate SO4^2-

4 oxygens at −2 = −8; total must equal −2

Result: S = +6

Solve S + (−8) = −2. The sum equals the ion charge, not zero — the most common slip in this calculation.

Example 4: Oxygen in hydrogen peroxide H2O2

2 hydrogens at +1 = +2; total must be 0

Result: O = −1 each

The oxygen–oxygen bond is split evenly between identical atoms, so each oxygen holds −1 rather than the usual −2.

Example 5: Iron in magnetite Fe3O4

4 oxygens at −2 = −8; total 0 across 3 iron atoms

Result: Fe = +8/3 average

A fractional average is legitimate here — the crystal contains one Fe(II) and two Fe(III) centres.

Common Mistakes

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Assuming oxygen is always −2

Peroxides such as H₂O₂ have oxygen at −1, superoxides at −½, and in OF₂ oxygen is +2 because fluorine outranks it.

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Assuming hydrogen is always +1

In metal hydrides such as NaH and CaH₂, hydrogen is the more electronegative partner and takes −1.

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Forgetting the ion charge in the sum

For a polyatomic ion the oxidation states sum to the ion charge, not zero. In sulfate the total must be −2, giving sulfur +6.

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Confusing oxidation state with formal charge

Oxidation state assumes fully ionic bonds; formal charge assumes perfectly even sharing. Carbon in methane is −4 by oxidation state but 0 by formal charge.

Frequently Asked Questions

Oxidation state rules?

Priority: 1. Element=0. 2. Monatomic ion=charge. 3. F=-1 always. 4. O=-2 (except peroxides). 5. H=+1 (with nonmetals) or -1 (with metals). 6. Sum=overall charge.

Uses?

Identifying redox reactions, balancing half-reactions, naming compounds (iron(II) vs iron(III)), predicting reactivity.

How do I assign oxidation states?

Work down the priority rules — free elements 0, fluorine −1, group 1 +1, group 2 +2, hydrogen +1, oxygen −2 — then solve for the remaining atom so the total equals the species charge.

When is oxygen not −2?

In peroxides it is −1, in superoxides −½, and in OF₂ it is +2 because fluorine is more electronegative and always takes −1.

Can oxidation states be fractional?

Yes, as averages. Fe₃O₄ gives +8/3 because the structure contains a mixture of Fe(II) and Fe(III), not because any atom carries a fractional charge.

What is the difference from formal charge?

Oxidation state assumes fully ionic bonding; formal charge assumes perfectly covalent sharing. They are two extremes bracketing the real situation.

Why do oxidation states matter?

They identify what is oxidised and what is reduced in a reaction, which is the basis for balancing redox equations and understanding electron transfer.

Formula Explorer connections

Interpretation: This formula links electron transfer, charge, potential, current or ionic transport in an electrochemical system. Assumption: Balance electron count and half-reactions, preserve sign conventions, and use consistent concentration, temperature and electrical units. Real cells include losses and overpotential.

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