An oxidation state is a number that tells you how many electrons an atom has effectively gained or lost in a compound. Transition metals can have more than one oxidation state, which is why iron compounds appear in two families — iron(II) and iron(III) — and why Roman numerals appear in chemical names at GCSE and beyond.

What is an oxidation state?

Oxidation state (also called oxidation number) is a bookkeeping device that tracks the effective charge on an atom within a compound, assuming all bonds are fully ionic. It is written as a signed integer: +2, −1, and so on.

Oxidation states are most useful for:

  • Naming compounds containing metals with variable valency
  • Deciding whether a reaction is redox (involves electron transfer)
  • Balancing half-equations in electrolysis and electrochemistry

What are the rules for assigning oxidation states?

Apply these rules in order of priority:

Rule Detail
1. Pure elements Oxidation state = 0 (e.g. Fe, O₂, Cl₂ all have oxidation state 0)
2. Simple (monatomic) ions Oxidation state = charge (e.g. Na⁺ = +1; Cl⁻ = −1; Fe³⁺ = +3)
3. Fluorine in compounds Always −1
4. Oxygen in compounds Almost always −2 (except in peroxides: −1, and OF₂: +2)
5. Hydrogen in compounds +1 with non-metals; −1 with metals (metal hydrides)
6. Sum rule In a neutral compound, all oxidation states sum to 0; in a polyatomic ion, they sum to the ion's charge

Worked examples: finding an unknown oxidation state

Example 1 — Iron in FeCl₃:

Chlorine = −1 (standard rule for non-metal halides except fluorides). Three Cl atoms contribute 3 × (−1) = −3. For the compound to be neutral: Fe + (−3) = 0, so Fe = +3. This compound is named iron(III) chloride.

Example 2 — Manganese in KMnO₄:

K = +1; O = −2; 4 O atoms = −8. Sum must equal 0: (+1) + Mn + (−8) = 0 → Mn = +7. This compound is named potassium manganate(VII).

Example 3 — Sulfur in SO₄²⁻ (sulfate ion):

4 O atoms = 4 × (−2) = −8. Sum must equal the ion charge (−2): S + (−8) = −2 → S = +6.

Why do transition metals have variable oxidation states?

Most main-group metals have only one common oxidation state (Na is always +1; Mg is always +2) because they lose all their outer-shell electrons to achieve a noble gas configuration. Transition metals have electrons in d sub-shells that are close in energy to the outer s electrons. Depending on conditions, different numbers of electrons can be removed, giving several stable oxidation states.

Iron can lose 2 or 3 electrons: Fe²⁺ (iron(II), oxidation state +2) and Fe³⁺ (iron(III), oxidation state +3). Copper can form Cu⁺ (copper(I)) or Cu²⁺ (copper(II)). Manganese can range from +2 up to +7.

This is why compounds of the same transition metal can have very different colours, reactivities, and uses.

How are Roman numerals used in compound names?

When naming a compound containing a transition metal with a variable oxidation state, the Roman numeral in brackets immediately after the metal name gives the oxidation state:

Formula Metal oxidation state Name
FeCl₂ Fe = +2 Iron(II) chloride
FeCl₃ Fe = +3 Iron(III) chloride
CuO Cu = +2 Copper(II) oxide
Cu₂O Cu = +1 Copper(I) oxide
MnO₂ Mn = +4 Manganese(IV) oxide
Cr₂O₃ Cr = +3 Chromium(III) oxide

Metals with only one common oxidation state do not need a Roman numeral (e.g. sodium chloride, not sodium(I) chloride), though you may see it in more formal contexts.

How do oxidation states track electron transfer in redox?

A change in oxidation state signals electron transfer. The rules are:

  • Oxidation = increase in oxidation state (loss of electrons: OIL — Oxidation Is Loss)
  • Reduction = decrease in oxidation state (gain of electrons: RIG — Reduction Is Gain)

Example — iron reacting with copper sulfate:

Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

  • Iron goes from 0 to +2 → oxidised (loses 2 electrons)
  • Copper goes from +2 to 0 → reduced (gains 2 electrons)

This is the displacement reaction used at GCSE. Oxidation state analysis confirms that both half-reactions happen simultaneously — you can never have oxidation without reduction.

Frequently asked questions

Why is the oxidation state of oxygen −1 in hydrogen peroxide?

In hydrogen peroxide (H₂O₂), the hydrogen atoms each contribute +1 (total +2). For the molecule to be neutral, both oxygen atoms together must contribute −2. With two oxygen atoms, each has an oxidation state of −1. This is the one common exception to the "oxygen = −2" rule. Peroxides contain an O–O single bond, which weakens the normal full assignment of −2 per oxygen.

Do I need to know oxidation states for KS3?

Oxidation states are not required at KS3 — the curriculum covers oxidation and reduction informally through the language of gaining or losing oxygen. At GCSE, oxidation states are needed mainly for naming transition metal compounds, writing half-equations for electrolysis, and discussing redox reactions. At A-level, oxidation states are used far more extensively for balancing complex equations and classifying reaction types.

How do I remember OIL RIG?

OIL RIG stands for: Oxidation Is Loss (of electrons); Reduction Is Gain (of electrons). A complementary mnemonic is LEORA: Loss of Electrons = Oxidation; Reduction is the Alternative (gain). The key point is that oxidation and reduction always occur together — if one atom loses electrons, another must gain them.

What happens to the colour of transition metal compounds when the oxidation state changes?

Transition metal ions absorb light in the visible spectrum because d electrons can be promoted between energy levels. The energy difference — and therefore the wavelength absorbed (and the complementary colour seen) — changes with oxidation state. Iron(II) compounds are typically pale green or white; iron(III) compounds are yellow-brown to orange. Copper(II) solutions are blue; copper(I) compounds are colourless or white. This colour change is used analytically: a colour change during a reaction can indicate a change in oxidation state.


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