3.1-3.2 - Oxidation Numbers

3.1-3.2 - Oxidation Numbers

Oxidation numbers are a bookkeeping system for tracking electron ownership in compounds and ions. This lesson covers what an oxidation number means and how to calculate it, including the Edexcel-required exceptions for peroxides and metal hydrides. It deliberately stops before classifying reactions as oxidation, reduction or disproportionation; that next step depends on being able to assign the numbers reliably first.

The Bookkeeping Model

Suppose you meet these three formulae in the same topic:

  • NaCl
  • HCl
  • ClO3^-

Chlorine is not doing the same electronic job in all three. In sodium chloride, chloride is a real Cl^- ion. In hydrogen chloride, the bond is covalent but polar. In ClO3^-, chlorine is surrounded by oxygen atoms and the whole species has a -1 charge. Oxidation numbers let chemists compare these cases using one consistent accounting method.

An oxidation number is the charge assigned to an atom by a set of rules. For a simple ion, it is the actual ionic charge: sodium in Na+ has oxidation number +1; chloride in Cl^- has oxidation number -1. In a covalent molecule or polyatomic ion, the oxidation number is usually not a real charge sitting on the atom. It is the charge the atom is treated as having when bonding electrons are assigned for bookkeeping.

This is why water can be written as:

SpeciesAtom consideredOxidation numberWhat this does and does not mean
H2OH+1Hydrogen is treated as electron-poor compared with oxygen.
H2OO-2Oxygen is treated as electron-rich compared with hydrogen.

The water molecule does not contain two free H+ ions and one free O^2- ion. Oxidation numbers are a model for electron accounting, not a claim that every covalent substance is ionic.

Feynman diagnostic: Explain to a younger student why oxygen in H2O can be assigned oxidation number -2 even though water is a neutral covalent molecule. The trap to avoid is saying "oxygen is an oxide ion in water." A stronger explanation is that oxidation numbers are assigned charges used for accounting; the total still has to match the overall charge of the species.

Rules And Totals

The most important rule is the total rule:

Species typeSum of oxidation numbers
Neutral element or compound0
Polyatomic ionthe ion charge

So in H2SO4, all the assigned oxidation numbers must add to 0. In SO4^2-, they must add to -2.

Use the fixed rules first, then use algebra for the unknown:

SituationUsual oxidation numberExamples
Uncombined element0Na, O2, Cl2, S8
Simple monatomic ionionic chargeMg^2+ is +2; Br^- is -1
Group 1 metal in a compound+1Na2O, KCl
Group 2 metal in a compound+2MgCl2, CaO
Aluminium in a compound+3Al2O3
Fluorine in a compound-1HF, NaF
Oxygen in most compounds-2H2O, CO2, SO4^2-
Oxygen in a peroxide-1H2O2, Na2O2
Hydrogen with non-metals+1HCl, H2O, NH4+
Hydrogen in a metal hydride-1NaH, CaH2

The order matters when there is an exception. In H2O2, oxygen is not -2; it is a peroxide, so each oxygen is -1. In NaH, hydrogen is not +1; it is a metal hydride, so hydrogen is -1.

Worked mini-example: carbon dioxide

Find the oxidation number of carbon in CO2.

Oxygen is usually -2, and there are two oxygen atoms:

x+2(2)=0x + 2(-2) = 0 x4=0x - 4 = 0 x=+4x = +4

Carbon has oxidation number +4 in CO2. The chemical sense check is that oxygen is more electronegative than carbon, so carbon is assigned a positive oxidation number.

Compounds And Ions

Most Edexcel calculations are not about memorising a table; they are about setting up the total correctly. Write the unknown as x, insert the fixed oxidation numbers, and make the sum equal to the overall charge.

Worked example 1: sulfur in sulfuric acid

Calculate the oxidation number of sulfur in H2SO4.

Decision points:

  • H2SO4 is neutral, so the total is 0.
  • Hydrogen is +1 with a non-metal.
  • Oxygen is -2 in this compound.

Set up the equation:

2(+1)+x+4(2)=02(+1) + x + 4(-2) = 0

Simplify:

2+x8=02 + x - 8 = 0 x=+6x = +6

Sulfur has oxidation number +6 in H2SO4.

Worked example 2: nitrogen in nitrate

Calculate the oxidation number of nitrogen in NO3^-.

This time the species is an ion, so the sum must equal -1, not 0.

x+3(2)=1x + 3(-2) = -1 x6=1x - 6 = -1 x=+5x = +5

Nitrogen has oxidation number +5 in NO3^-. A common error is to make the total 0 out of habit; that would give +6, which does not match the ion charge.

Worked example 3: nitrogen in ammonium

Calculate the oxidation number of nitrogen in NH4+.

Hydrogen is +1 because it is bonded to a non-metal, and the total is +1 because this is an ion:

x+4(+1)=+1x + 4(+1) = +1 x+4=+1x + 4 = +1 x=3x = -3

Nitrogen has oxidation number -3 in NH4+.

Guided practice: Calculate the oxidation number of the named element.

SpeciesElementRoute cueAnswer
CO3^2-Ctotal is -2+4
PO4^3-Ptotal is -3+5
ClO3^-Cltotal is -1+5

For a Pearson calculate item, the mark-earning step is the algebraic route, not just the final number. Show the total charge and the fixed oxidation numbers before the answer.

Peroxides And Metal Hydrides

Edexcel names two exceptions explicitly for this row: peroxides and metal hydrides. These are small exceptions, but they are high-value because they reveal whether the rule is being used thoughtfully or mechanically.

Peroxides

A peroxide contains an O-O link and each oxygen has oxidation number -1. The common examples at this level are H2O2 and compounds such as Na2O2.

For H2O2:

2(+1)+2x=02(+1) + 2x = 0 2x=22x = -2 x=1x = -1

Each oxygen is -1. If you forced oxygen to be -2, the total would be 2(+1) + 2(-2) = -2, which is impossible for a neutral molecule.

For Na2O2:

2(+1)+2x=02(+1) + 2x = 0 x=1x = -1

Each oxygen is again -1.

Metal hydrides

In a metal hydride, hydrogen has oxidation number -1. This is different from hydrogen in acids, water, ammonia or ammonium ions, where hydrogen is usually +1.

For CaH2, calcium is Group 2, so it is +2:

+2+2x=0+2 + 2x = 0 2x=22x = -2 x=1x = -1

Hydrogen has oxidation number -1 in CaH2.

Common-error contrast

FormulaTempting wrong routeCorrect route
H2O2oxygen is always -2peroxide: oxygen is -1
NaHhydrogen is always +1metal hydride: hydrogen is -1
CaH2two hydrogens must total +2calcium is +2, so two hydrides total -2

Feynman diagnostic: Explain why hydrogen is +1 in HCl but -1 in NaH. The trap is "hydrogen changes because the formula looks different." The useful explanation is that hydrogen is assigned +1 when bonded to a more electronegative non-metal, but in a metal hydride the metal is assigned a positive oxidation number and hydrogen is treated as hydride, H^-.

Non-Equivalent Atoms

Sometimes the same element appears in more than one chemical environment. Do not average those atoms unless the formula actually tells you they are equivalent.

Ammonium nitrate, NH4NO3, is a good test. It is made from NH4+ and NO3^-, so there are two different nitrogen environments.

For nitrogen in NH4+:

x+4(+1)=+1x + 4(+1) = +1 x=3x = -3

For nitrogen in NO3^-:

x+3(2)=1x + 3(-2) = -1 x=+5x = +5

So the two nitrogen atoms in ammonium nitrate have different oxidation numbers: -3 in NH4+ and +5 in NO3^-.

This matters because oxidation numbers are attached to atoms in their chemical environments, not just to element symbols in a whole empirical formula. Later redox lessons use changes in oxidation number, so the starting assignment has to identify which atom is being tracked.

Feynman diagnostic: A student says, "In NH4NO3, nitrogen must have one oxidation number because it is the same element." Correct the statement in two sentences. A strong answer should say that oxidation number depends on the atom's environment, and that the nitrogen in NH4+ and the nitrogen in NO3^- must be calculated separately.

Pearson Transfer

For Edexcel 9CH0, oxidation-number calculation is a small algebra skill with large consequences. It prepares you to classify redox reactions, recognise disproportionation, write formulae from oxidation states, and build half-equations in later lessons.

Use this route whenever you are asked to calculate or deduce an oxidation number:

  1. Identify whether the species is neutral or an ion.
  2. Write the total: 0 for a neutral species, or the ion charge for an ion.
  3. Insert fixed oxidation numbers, checking for peroxides and metal hydrides.
  4. Use x for the unknown atom.
  5. Solve the equation.
  6. Sense-check the sign and total.

Worked Pearson-style response

Calculate the oxidation number of manganese in MnO4^-.

The ion has overall charge -1. Oxygen is -2:

x+4(2)=1x + 4(-2) = -1 x8=1x - 8 = -1 x=+7x = +7

Manganese has oxidation number +7.

Pearson-aware recap: Oxidation number is an assigned charge used for electron bookkeeping. The calculation skill is to make the sum of assigned oxidation numbers equal the charge on the species, while applying the named exceptions for peroxides and metal hydrides. The next lesson uses these numbers to decide whether oxidation, reduction or disproportionation has happened.