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Oxidation State Calculator

Enter a chemical formula to determine the oxidation state of each element. For polyatomic ions, enter the charge.

How oxidation states are assigned

Oxidation states distribute the electrons in a compound by pretending every bond is fully ionic, with the more electronegative atom keeping the shared pair. The result is an integer (sometimes a fraction in superoxides and a few odd cases) that may not correspond to any real charge but accounts for electron transfer in a way that makes redox chemistry tractable.

The assignment rules form a strict priority list — apply them top-down, and the unknown atom is whatever the sum demands:

  1. Free elements: 0.
  2. Fluorine in compounds: −1 (always).
  3. Group 1 metals in compounds: +1. Group 2 metals: +2.
  4. Hydrogen: +1, except −1 in metal hydrides like NaH.
  5. Oxygen: −2, except −1 in peroxides (H₂O₂, Na₂O₂), −½ in superoxides (KO₂), and +2 in OF₂.
  6. The sum equals the overall charge — 0 for a neutral compound, the ion charge for a polyatomic ion.

How this calculator works

Type a formula (KMnO4, H2SO4, Cr2O7) and the optional ionic charge if the species is an ion. The parser counts atoms, applies the rules in priority order, and solves the resulting linear equation for the unknown element. The output table shows each element, its assigned oxidation number, and which rule produced it. A verification line confirms the assignments sum to the expected charge.

Worked examples

KMnO₄. K = +1 (rule 3), O = −2 × 4 = −8 (rule 5). Neutral compound: +1 + Mn − 8 = 0 → Mn = +7.

H₂SO₄. H = +1 × 2 = +2 (rule 4), O = −2 × 4 = −8 (rule 5). Neutral: +2 + S − 8 = 0 → S = +6.

Na₂O₂ (sodium peroxide). Na = +1 × 2 = +2 (rule 3). The compound is a peroxide, so O = −1 each by the rule 5 exception. Verify: +2 + (−2) = 0. Each O is −1, not −2.

Cr₂O₇²⁻ (dichromate). O = −2 × 7 = −14. Sum equals ion charge: 2(Cr) − 14 = −2 → 2(Cr) = +12 → each Cr = +6.

The peroxide and dichromate cases are the most common places students slip — the calculator surfaces the rule-exception path so the reasoning is visible, not just the answer.

Frequently Asked Questions

What is an oxidation state?
An oxidation state is the charge an atom would carry if every bond in the compound were treated as fully ionic, with shared electrons assigned to the more electronegative partner. It is a bookkeeping device, not a measured charge — but it tracks electron transfer in redox reactions cleanly enough to balance equations and predict which species is the oxidizer.
What are the rules for assigning oxidation states?
Apply them in this priority order: free elements are 0; fluorine is always −1; alkali metals are +1 and alkaline earths +2 in compounds; hydrogen is +1 except in metal hydrides where it is −1; oxygen is −2 except in peroxides (−1), superoxides (−1/2), and OF2 (+2); the sum across all atoms equals the overall charge of the species. The unknown element falls out of the sum.
How do you find oxidation states in polyatomic ions?
Same rule set, but the sum of oxidation states equals the charge of the ion rather than zero. For MnO4⁻ the four oxygens contribute −8, the total must equal −1, so Mn is +7. For Cr2O7²⁻ the seven oxygens contribute −14, the total must equal −2, so the two chromiums share +12 and each Cr is +6.
What is the oxidation state of Mn in KMnO4?
K is +1 by the alkali metal rule. The four O atoms are −2 each for −8 total. The compound is neutral, so +1 + Mn + (−8) = 0, giving Mn = +7. That is manganese's highest accessible oxidation state, which is why permanganate is such an aggressive oxidizer.
Why are oxidation states important?
They identify which atoms are oxidized (oxidation state increases) and which are reduced (decreases) in a reaction. That tells you what the oxidizing and reducing agents are, lets you balance redox equations by half-reaction, and connects directly to electrochemistry — the change in oxidation state times the moles of substance equals the moles of electrons that moved.