Oxidation States

Oxidation states (also called oxidation numbers) provide a systematic way of tracking electron distribution in chemical species. They are essential for identifying redox reactions, naming compounds, and balancing redox equations. Mastering the rules for assigning oxidation states is a fundamental A-Level skill.

What Is an Oxidation State?

An oxidation state is a number assigned to an atom in a chemical species that represents the number of electrons it has lost, gained, or shared unequally compared to its elemental form. It is essentially a bookkeeping tool for tracking electrons.

A positive oxidation state means the atom has (effectively) lost electrons.
A negative oxidation state means the atom has (effectively) gained electrons.
An oxidation state of zero means the atom is in its elemental form or has not been oxidised/reduced.

Oxidation states are written with the sign before the number: +2, −1, +7 (not 2+, 1−, 7+, which are used for ion charges).

Rules for Assigning Oxidation States — Summary Table

Rule	Description	Priority
1	Uncombined elements = 0	Highest
2	Monatomic ions = charge on ion
3	Fluorine = −1 always
4	Oxygen = −2 (except peroxides: −1, and OF₂: +2)
5	Hydrogen = +1 (except metal hydrides: −1)
6	Sum in neutral compound = 0
7	Sum in polyatomic ion = charge on ion	Lowest

Applying the Rules in Detail

Rule 1: The oxidation state of an uncombined element is 0. Examples: Fe(s) = 0, O₂(g) = 0, S₈(s) = 0, Na(s) = 0, Cl₂(g) = 0

Rule 2: The oxidation state of a monatomic ion equals its charge. Examples: Na⁺ = +1, Cl⁻ = −1, Fe³⁺ = +3, O²⁻ = −2, Al³⁺ = +3

Rule 3: In compounds, fluorine is always −1 (it is the most electronegative element).

Rule 4: In compounds, oxygen is usually −2. Exception: In peroxides (e.g., H₂O₂, Na₂O₂), oxygen is −1. Exception: In OF₂, oxygen is +2 (fluorine takes priority). Exception: In superoxides (e.g., KO₂), oxygen is −½ (rarely tested at A-Level).

Rule 5: In compounds, hydrogen is usually +1. Exception: In metal hydrides (e.g., NaH, CaH₂), hydrogen is −1.

Rule 6: The sum of oxidation states in a neutral compound is 0.

Rule 7: The sum of oxidation states in a polyatomic ion equals the charge on the ion.

Worked Examples

Example 1: Find the oxidation state of sulfur in H₂SO₄

Using rules:

H = +1 each, so 2 hydrogens = +2
O = −2 each, so 4 oxygens = −8
Sum = 0 (neutral compound)

(+2) + S + (−8) = 0 S = +6

Example 2: Find the oxidation state of manganese in MnO₄⁻

O = −2 each, so 4 oxygens = −8
Sum = −1 (charge on the ion)

Mn + (−8) = −1 Mn = +7

Example 3: Find the oxidation state of chromium in Cr₂O₇²⁻

O = −2 each, so 7 oxygens = −14
Sum = −2 (charge on the ion)

2Cr + (−14) = −2 2Cr = +12 Cr = +6

Example 4: Find the oxidation state of nitrogen in NO₃⁻

O = −2 each, so 3 oxygens = −6
Sum = −1 (charge on the ion)

N + (−6) = −1 N = +5

Example 5: Find the oxidation state of nitrogen in NH₄⁺

H = +1 each, so 4 hydrogens = +4
Sum = +1 (charge on the ion)

N + (+4) = +1 N = −3

Example 6: Oxidation state of each element in Na₂Cr₂O₇

Na = +1 (Group 1 metal), so 2 × Na = +2
O = −2 each, so 7 × O = −14
Sum = 0 (neutral compound)

(+2) + 2Cr + (−14) = 0 2Cr = +12 Cr = +6, Na = +1, O = −2

Roman Numeral Naming Convention

When a metal can have more than one oxidation state, the oxidation state is shown using a Roman numeral in the name of the compound:

Formula	Name	Metal oxidation state
FeCl₂	Iron(II) chloride	+2
FeCl₃	Iron(III) chloride	+3
CuO	Copper(II) oxide	+2
Cu₂O	Copper(I) oxide	+1
MnO₂	Manganese(IV) oxide	+4
KMnO₄	Potassium manganate(VII)	Mn = +7
K₂Cr₂O₇	Potassium dichromate(VI)	Cr = +6
VO₂⁺	Vanadium(V)	V = +5
V₂O₃	Vanadium(III) oxide	V = +3

Identifying Redox Using Oxidation States

A reaction is a redox reaction if any element changes its oxidation state during the reaction.

If an element's oxidation state increases, it has been oxidised.
If an element's oxidation state decreases, it has been reduced.

Worked Example 7

Is the following a redox reaction?

2FeCl₂ + Cl₂ → 2FeCl₃

Oxidation states:

Fe in FeCl₂: Cl = −1, so Fe = +2
Fe in FeCl₃: Cl = −1, so Fe = +3
Cl in Cl₂: 0 (element)
Cl in FeCl₃: −1

Fe goes from +2 to +3 → oxidised (increase) Cl goes from 0 to −1 → reduced (decrease)

Yes, this is a redox reaction. Fe²⁺ is the reducing agent; Cl₂ is the oxidising agent.

Worked Example 8

Identify which species is oxidised and which is reduced.

Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Cu: 0 → +2 (oxidised — lost 2 electrons)
Ag: +1 → 0 (reduced — gained 1 electron)
N and O: unchanged (+5 and −2 throughout)

Cu is the reducing agent; Ag⁺ is the oxidising agent.

Disproportionation

A disproportionation reaction is one in which the same element is simultaneously oxidised and reduced.

Example

Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Chlorine in Cl₂ has oxidation state 0. In NaCl, Cl = −1 (reduced). In NaClO, Cl = +1 (oxidised).

The same element (Cl) has been both oxidised (0 → +1) and reduced (0 → −1). This is disproportionation.

Another important example is the reaction of copper(I) oxide in acid: Cu₂O + H₂SO₄ → Cu + CuSO₄ + H₂O

Cu in Cu₂O is +1. In Cu metal it is 0 (reduced). In CuSO₄ it is +2 (oxidised). This is disproportionation of copper.

Common Mistakes with Oxidation States

Mistake	Example	Fix
Forgetting sign convention	Writing "2" instead of "+2"	Always include + or −
Confusing ion charge and oxidation state	Writing Fe²⁺ as oxidation state 2⁺	Oxidation state is +2 (sign first)
Not recognising peroxides	Giving O as −2 in H₂O₂	O is −1 in peroxides
Forgetting that elements are zero	Giving O₂ as −2	Uncombined elements = 0
Not using the ion charge for polyatomic ions	Using sum = 0 for SO₄²⁻	Sum = −2 for SO₄²⁻
Ignoring Roman numerals in names	Not connecting "iron(III)" to +3	Roman numeral = oxidation state

Practice: Quick-Fire Oxidation States

Test yourself — find the oxidation state of the underlined element:

Species	Element	Answer
SO₂	S	+4
SO₃²⁻	S	+4
ClO⁻	Cl	+1
ClO₃⁻	Cl	+5
HNO₂	N	+3
PO₄³⁻	P	+5
Na₂O₂	O	−1
CaH₂	H	−1

Variable Oxidation States of Common Elements

Many elements, especially transition metals and non-metals, can exhibit multiple oxidation states. Being aware of the common ones helps you recognise what a compound contains.

Element	Common oxidation states	Example compounds
Iron (Fe)	+2, +3	FeCl₂, FeCl₃
Copper (Cu)	+1, +2	Cu₂O, CuO
Manganese (Mn)	+2, +4, +7	MnCl₂, MnO₂, KMnO₄
Chromium (Cr)	+3, +6	Cr₂O₃, K₂Cr₂O₇
Sulfur (S)	−2, +4, +6	H₂S, SO₂, H₂SO₄
Nitrogen (N)	−3, +3, +5	NH₃, HNO₂, HNO₃
Carbon (C)	−4, +2, +4	CH₄, CO, CO₂
Vanadium (V)	+2, +3, +4, +5	V₂O₅, VO₂⁺, V³⁺, V²⁺

The ability to move between these oxidation states is what makes transition metals such effective catalysts and gives them their characteristic coloured compounds.

Real-World Application: Oxidation States in Everyday Life

Oxidation states are not just an abstract exam concept — they explain real phenomena:

Rust is hydrated iron(III) oxide: iron has been oxidised from 0 to +3.
Bleach (sodium hypochlorite, NaClO) contains chlorine in the +1 oxidation state. It works by oxidising coloured compounds.
Breathalyser tests traditionally used potassium dichromate(VI), where Cr is reduced from +6 (orange) to +3 (green) by ethanol in the breath.
Photography relied on silver(I) halides being reduced to silver metal (Ag⁺ → Ag⁰) when exposed to light.

Being able to assign and track oxidation states lets you understand and predict these everyday chemical processes.

Oxidation states are the foundation for everything in redox chemistry. If you can assign them quickly and accurately, you will find redox equations, half-equations, and titration calculations much more straightforward.

A-Level Deep Dive: Oxidation States

Spec mapping

Edexcel 9CH0 specification Topic 3 — Redox I, sub-strand on oxidation numbers, requires students to assign oxidation states using the standard rules (uncombined element = 0; sum to overall charge; specific assignments for H, O, F, Group 1 and 2 metals); identify oxidising and reducing agents; and use changes in oxidation state to deduce electron transfer (refer to the official specification document for exact wording). Examined across Paper 1 (inorganic and physical contexts) and Paper 3 (synoptic with Topic 14 redox titrations and Topic 15 transition metals). The data sheet does not list oxidation-state rules; they must be memorised. The Periodic Table on the data sheet supplies group numbers, which inform Group 1/2 oxidation-state assignment.

Worked example with full mark scheme

Question (8 marks):

(a) Assign the oxidation state of:

(i) S in $\text{H}_2\text{SO}_4$ . (1)

(ii) Cr in $\text{K}_2\text{Cr}_2\text{O}_7$ . (1)

(iii) S in $\text{Na}_2\text{S}_2\text{O}_3$ . (1)

(iv) N in $\text{NH}_4\text{NO}_3$ (consider both N atoms separately). (2)

(b) In the reaction $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$ :

(i) State the change in oxidation state of Mn and Fe. (2)

(ii) Identify the oxidising and reducing agents, with reasoning. (1)

Solution with mark scheme:

(a) (i) Sum to 0 (neutral): $2(+1) + S + 4(-2) = 0$ , so $S = +6$ . B1.

(ii) Sum to 0: $2(+1) + 2\text{Cr} + 7(-2) = 0$ , so $2\text{Cr} = +12$ , $\text{Cr} = +6$ . B1.

(iii) Sum to 0: $2(+1) + 2\text{S} + 3(-2) = 0$ , so $2\text{S} = +4$ , $\text{S} = +2$ (average; structurally the two sulfurs are inequivalent, but Edexcel A-Level accepts +2 average). B1.