A bottle labelled iron(II) chloride and a bottle labelled iron(III) chloride do not contain the same iron ion. The Roman numeral is a compact way of saying which oxidation number the named element has in that compound or ion.

Roman numerals used here are positive oxidation numbers written without a plus sign:

The numeral belongs to the element just named. In iron(III) oxide, iron has oxidation number +3. Oxygen is not being called "III"; oxygen is usually -2 in oxides. In sulfate(VI), sulfur has oxidation number +6 inside the sulfate group. In manganate(VII), manganese has oxidation number +7.

Worked example: naming from a formula

Name FeCl2 using a Roman numeral.

Chlorine in a simple chloride ion has oxidation number -1. There are two chloride ions, so the total contribution from chlorine is -2. The compound is neutral, so iron must be +2.

FeCl2 is iron(II) chloride.

The chemical meaning is that the formula contains Fe^2+ with two Cl^- ions. The Roman numeral shows the oxidation number of iron, not the number of chlorine atoms.

Active check

Feynman diagnostic

Explain to a younger student why copper(II) sulfate does not mean there are two sulfate ions. The trap is treating the Roman numeral as a subscript. A good explanation says that II tells you copper is +2; the sulfate ion is SO4^2-, so one sulfate ion balances one Cu^2+.

Formula writing is charge balance. Oxidation numbers tell you the charge needed on each ion or group, then the formula uses the smallest whole-number ratio that gives a neutral compound.

Use this route:

1. Write the positive ion or named element with its oxidation number as a charge.
2. Write the negative ion or group with its charge.
3. Find the smallest ratio that makes total positive charge equal total negative charge.
4. Use brackets around a compound ion when more than one of that ion is needed.

Worked example: iron(III) sulfate

Write the formula for iron(III) sulfate.

Iron(III) means Fe^3+. Sulfate is SO4^2-.

The lowest common total charge is 6:

So the formula is Fe2(SO4)3.

The brackets matter because there are three whole sulfate ions, not three oxygen atoms added separately.

Common-error contrast

FeSO4 is not iron(III) sulfate. In FeSO4, sulfate is -2 and the compound is neutral, so iron is +2. FeSO4 is iron(II) sulfate. For iron(III), the formula needs enough sulfate ions to balance two Fe^3+ ions: Fe2(SO4)3.

Guided practice

Write the formula for copper(II) nitrate.

Route:

• copper(II) gives Cu^2+
• nitrate is NO3^-
• two nitrate ions are needed for each copper ion
• formula: Cu(NO3)2

Active check

Rows 3.11 and 3.12 connect formulae to particles. Metals generally form positive ions by losing electrons; non-metals generally form negative ions by gaining electrons. Oxidation number tracks that electron transfer.

For a metal atom forming a positive ion:

Mg -> Mg^2+ + 2e-

Magnesium loses two electrons. Its oxidation number increases from 0 in the element to +2 in the ion, so this is oxidation.

For a non-metal atom forming a negative ion:

Cl2 + 2e- -> 2Cl^-

Each chlorine atom gains one electron. Chlorine's oxidation number decreases from 0 in Cl2 to -1 in Cl^-, so this is reduction.

The wording "metals, in general" and "non-metals, in general" matters. It is a broad pattern for ion formation, not a claim that every compound is purely ionic or that oxidation number is always the same as a real isolated ion charge. Oxidation number is a bookkeeping model that lets us track electron transfer consistently.

Worked example: deciding whether electron loss or gain happened

Aluminium forms Al^3+ from aluminium atoms.

The half-equation is:

Al -> Al^3+ + 3e-

Electrons are on the product side, so aluminium has lost electrons. Its oxidation number has increased from 0 to +3, so aluminium has been oxidised.

Active check

Feynman diagnostic

Explain why "metals form positive ions because protons move away" is wrong. The nucleus does not lose protons in ordinary chemical reactions; the atom loses electrons. Losing negative charge leaves the species positive and increases its oxidation number.

A half-equation shows one oxidation process or one reduction process. It is not complete by itself because electrons cannot be created or destroyed overall; it is one side of the electron-transfer story.

For simple ion changes, balance atoms first, then balance charge with electrons.

Worked example: iron(II) to iron(III)

Write the half-equation for Fe^2+ forming Fe^3+.

Atoms are already balanced: one iron atom on each side.

Charges:

• left: +2
• right before electrons: +3

Add one electron to the right to make the total right-hand charge +2:

Fe^2+ -> Fe^3+ + e-

Electrons are produced, so this is oxidation. The oxidation number of iron increases from +2 to +3.

Worked example: manganate(VII) to manganese(II) in acidic solution

Write the half-equation for MnO4^- forming Mn^2+ in acidic solution.

Start with the species containing manganese:

MnO4^- -> Mn^2+

Balance oxygen using water:

MnO4^- -> Mn^2+ + 4H2O

Balance hydrogen using H+:

MnO4^- + 8H+ -> Mn^2+ + 4H2O

Now balance charge with electrons. The left-hand side is -1 + 8 = +7. The right-hand side is +2. Add five electrons to the left so the left-hand charge becomes +2:

MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O

Electrons are used up, so this is reduction. Manganese decreases from +7 in MnO4^- to +2 in Mn^2+.

Half-equation checklist

• Atoms are balanced.
• Charge is balanced.
• Electrons are on the product side for oxidation.
• Electrons are on the reactant side for reduction.
• In acidic aqueous equations, use H+ and H2O to balance hydrogen and oxygen.

Active check

A full ionic redox equation is made by adding one oxidation half-equation to one reduction half-equation. The electrons must cancel because electrons transferred by one species are gained by another species.

Use this route:

1. Write the oxidation half-equation.
2. Write the reduction half-equation.
3. Multiply one or both half-equations so the number of electrons is the same.
4. Add the two equations.
5. Cancel electrons and any species that appear unchanged on both sides.
6. Check atoms and total charge.

Worked example: chlorine oxidises iodide ions

Half-equation for chlorine being reduced:

Cl2 + 2e- -> 2Cl^-

Half-equation for iodide being oxidised:

2I^- -> I2 + 2e-

The electron numbers already match. Add and cancel the electrons:

Cl2 + 2I^- -> 2Cl^- + I2

Check the charge: left side is -2; right side is -2. Atoms are balanced. This full ionic equation shows the electron transfer without spectator ions.

Worked example: acidified manganate(VII) oxidises iron(II) ions

Reduction half-equation:

MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O

Oxidation half-equation:

Fe^2+ -> Fe^3+ + e-

The manganate(VII) half-equation needs five electrons, but each iron(II) ion supplies one electron. Multiply the iron half-equation by 5:

5Fe^2+ -> 5Fe^3+ + 5e-

Add the equations and cancel 5e-:

MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+

Charge check:

• left: -1 + 8 + 10 = +17
• right: +2 + 15 = +17

That charge check is not optional decoration. In Pearson-style equation questions, balanced atoms without balanced charge is still an incomplete ionic equation.

Guided practice

Use the half-equations below to write the full ionic equation:

Mg -> Mg^2+ + 2e-

Cu^2+ + 2e- -> Cu

Because both half-equations contain two electrons, add them directly and cancel electrons:

Mg + Cu^2+ -> Mg^2+ + Cu

Magnesium is oxidised; copper(II) ions are reduced.

Feynman diagnostic

Explain why you must not leave electrons in the final full ionic equation. A good answer says a full ionic equation describes the overall chemical change; electrons are transferred internally from the reducing agent to the oxidising agent, so the number produced by oxidation must equal the number consumed by reduction and cancel out.

Pearson-aware recap

For write questions, the answer is the formula, half-equation or ionic equation itself, with charges and balancing correct. For explain questions, link the symbol work to electron transfer: metals usually lose electrons and increase oxidation number, non-metals usually gain electrons and decrease oxidation number, and a redox equation is only complete when both atoms and charge balance.