A LevelChemistryCambridge (CIE)Revision Notes6. ElectrochemistryRedox Processes: Electron Transfer & Changes in Oxidation Number (Oxidation State)Oxidation Number Rules

Oxidation Number Rules (Cambridge (CIE) A Level Chemistry): Revision Note

Exam code: 9701

Written by: Caroline Carroll

Reviewed by: Lucy Kirkham

Updated on 2 July 2026

Oxidation Numbers

The oxidation number (also known as oxidation state) is a number given to each atom or ion in a compound to keep track of how many electrons they have
In a single ion or molecular ion, the oxidation number tells us how many electrons have been lost or gained
Positive oxidation number = loss of electrons
Negative oxidation number = gain of electrons

Oxidation number rules

The oxidation number refers to a single atom in a compound

	Rule	Example
1	The oxidation number of any uncombined element is zero	H₂ Zn O₂
2	Many atoms or ions have fixed oxidation numbers in compounds	Group 1 elements are always +1 Group 2 elements are always +2 Fluorine is always –1 Hydrogen is +1, except in hydrides like NaH where it is –1 Oxygen is –2, except in peroxides where it is in –1 and in F₂O where it is +2
3	The oxidation number of an element in a monoatomic ion is always the same as the charge	Zn²⁺ = +2 Fe³⁺ = +3 Cl^– = –1
4	The sum of the oxidation numbers in a compound is zero	NaCl Na = +1 Cl = –1 Sum of oxidation numbers = 1 – 1 = 0
5	The sum of the oxidation numbers in an ion is equal to the charge on the ion	SO₄^2– S = +6 Four O atoms = 4 x (–2) = –8 Sum of oxidation numbers = 6 – 8 = –2
6	In either a compound or an ion, the more electronegative element is given the negative oxidation number	F₂O Two F atoms = 2 x (–1) = –2 O = +2

Worked Example

Deducing oxidation numbers

State the oxidation number of the bold atoms in these compounds or ions.

P₂O₅
SO₄^2–
H₂S
Al₂Cl₆
NH₃
ClO₂^–
CaCO₃

Answer

P₂O₅
- 5 O atoms = 5 x (–2) = –10
- The overall charge of the compound = 0
- 2 P atoms = +10
- Oxidation number of 1 P atom = (+10) / 2 = +5
SO₄^2–
- 4 O atoms = 4 x (–2) = –8
- The overall charge of the compound = –2
- The oxidation number of 1 S atom = +6
H₂S
- 2 H atoms = 2 x (+1) = +2
- The overall charge of the compound = 0
- The oxidation number of 1 S atom = –2
Al₂Cl₆
- 6 Cl atoms = 6 x (–1) = –6
- The overall charge of the compound = 0
- 2 Al atoms = +6
- The oxidation number of 1 Al atom = (+6) / 2 = +3
NH₃
- 3 H atoms = 3 x (+1) = +3
- The overall charge of the compound = 0
- The oxidation number of 1 N atom = –3
ClO₂^–
- 2 O atoms = 2 x (–2) = –4
- The overall charge of the compound = –1
- The oxidation number of 1 Cl atom = +3
CaCO₃
- 3 O atoms = 3 x (–2) = –6
- 1 Ca atom = +2
- The overall charge of the compound = 0
- The oxidation number of 1 C atom = +4

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