⚡ Oxidation Number Calculator

Find oxidation states for elements in compounds and polyatomic ions.

Enter elements and their known oxidation states plus total charge, and find the unknown element's oxidation state.

Element

Count

Ox. State

Unknown Element

Element

Count

Total Charge of Ion/Molecule

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Oxidation State

How to Use This Calculator

This tool finds the oxidation state of an unknown element in a compound or polyatomic ion. You enter the elements whose oxidation states you already know (following the standard rules), then specify the unknown element and the total charge of the species. The calculator does the algebra for you.

In the "Known elements" section, enter each element symbol, how many atoms of it appear (count), and its oxidation state. For KMnO₄: enter K with count 1 and oxidation state +1; enter O with count 4 and oxidation state -2.

In the "Unknown element" section, enter the element symbol you need to find (Mn in this example) and its count (1 here).

Enter the total charge of the ion or molecule. For a neutral molecule like KMnO₄ enter 0. For MnO₄⁻ enter -1.

Click "Find Oxidation State." The result shows the oxidation state with sign, for example +7 for Mn in KMnO₄.

The Oxidation State Formula

Sum of (oxidation state × atom count) = total charge of species Unknown ox. state = (total charge − sum of known contributions) / count of unknown

The key insight is that charges must balance. For a neutral molecule the contributions from all atoms sum to zero. For an ion, they sum to the ion's charge. Apply the fixed rules for known elements first, then solve the equation for the unknown element.

Common Oxidation States Reference

Hydrogen (H)+1 with nonmetals (HCl), −1 with metals (NaH)

Oxygen (O)−2 in most compounds, −1 in peroxides (H₂O₂)

Fluorine (F)Always −1 (most electronegative element)

Alkali metals (Li, Na, K)Always +1 in compounds

Alkaline earth metals (Mg, Ca)Always +2 in compounds

Where This Calculation Comes Up

Oxidation state problems appear constantly in redox chemistry. Before you can balance a redox equation by the half-reaction method, you need to identify which atoms changed oxidation state and by how much. In the reaction between permanganate (Mn = +7) and iron(II) (Fe = +2), Mn is reduced from +7 to +2 (gains 5 electrons per Mn) and Fe is oxidised from +2 to +3 (loses 1 electron per Fe). To balance the electron transfer you need 5 Fe²⁺ per MnO₄⁻. None of that is possible without first knowing the oxidation states.

In transition metal chemistry, oxidation state determines the colour, magnetism, and reactivity of complexes. Iron in haemoglobin is Fe²⁺ (ferrous), and oxidation to Fe³⁺ (ferric) produces methaemoglobin, which cannot carry oxygen. Chromium(VI) in CrO₄²⁻ and Cr₂O₇²⁻ is a known carcinogen, while Cr³⁺ is much less toxic. In industrial catalysis, knowing whether a metal centre is in a +2 or +4 state determines which substrate it will activate. These distinctions come up in inorganic, biochemistry, and environmental chemistry courses alike.

Frequently Asked Questions

What is an oxidation number?

An oxidation number (or oxidation state) is the charge an atom would have if all bonds were ionic. It tracks electron transfer in redox reactions.

What are the key rules for oxidation numbers?

1) Pure elements = 0. 2) Monatomic ions = charge. 3) O = −2 (usually). 4) H = +1 (with nonmetals), −1 (with metals). 5) F always = −1. 6) Sum equals overall charge.

What is the oxidation number of Mn in KMnO₄?

K=+1, O=−2×4=−8, overall=0. So +1+Mn−8=0, Mn=+7.

How are oxidation numbers used?

To identify oxidizing and reducing agents, balance redox equations, and classify reactions.

What is the oxidation number of S in H₂SO₄?

H₂=+2, O₄=−8, net charge=0. So +2+S−8=0, S=+6.