What is writing formulae?

Ionic formulae balance the charges of the ions (for example Al^3+ and O^2- give Al_2O_3). Covalent formulae follow the combining numbers of the atoms. You should know common ions such as SO_4^2-, NO_3^-, CO_3^2- and NH_4^+.

State the oxidation number of sulfur in the sulfate ion SO_4^2-. [1 mark]

WalesChemistrySyllabus dot point

How do we write what happens in a chemical reaction?

Oxidation numbers, formulae of ionic and covalent compounds, writing and balancing full, ionic and half equations, and state symbols.

A focused answer to WJEC A-Level Chemistry Unit 1, covering oxidation numbers, deducing formulae of ionic and covalent compounds, writing and balancing full, ionic and redox half equations, and using state symbols correctly.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

WJEC wants you to assign oxidation numbers, deduce the formulae of ionic and covalent compounds, and write and balance full equations, ionic equations and redox half equations, including correct state symbols.

The answer

Oxidation numbers

Writing formulae

Ionic formulae balance the charges of the ions (for example $\text{Al}^{3+}$ and $\text{O}^{2-}$ give $\text{Al}_2\text{O}_3$ ). Covalent formulae follow the combining numbers of the atoms. You should know common ions such as $\text{SO}_4^{2-}$ , $\text{NO}_3^-$ , $\text{CO}_3^{2-}$ and $\text{NH}_4^+$ .

Balancing equations

Half equations and redox

The rules for assigning oxidation numbers

Apply the rules in order of priority. An uncombined element is always $0$ . A simple monatomic ion equals its charge. Fluorine is always $-1$ ; oxygen is $-2$ except in peroxides ( $-1$ ) and with fluorine; hydrogen is $+1$ except in metal hydrides ( $-1$ ). The sum of oxidation numbers equals zero in a neutral compound and equals the charge in an ion. Working through these in priority order lets you find the oxidation number of any element in a formula, even an unfamiliar one, which is essential for naming and for identifying redox changes.

Ionic equations and spectator ions

An ionic equation strips a reaction down to the species that actually change. To write one, split all soluble ionic compounds into their separate aqueous ions, then cancel any ion that appears unchanged on both sides (a spectator ion). For the neutralisation of any strong acid by any strong alkali, the ionic equation reduces to $\text{H}^+_{(aq)} + \text{OH}^-_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)}$ , which is why the enthalpy of neutralisation is almost constant for such reactions. Recognising spectator ions makes equations simpler and reveals the underlying chemistry.

Examples in context

Half equations in batteries and electrolysis. The oxidation and reduction half equations you balance here describe the electrode reactions in cells, central to the electrochemistry of Unit 3. Oxidation numbers and naming. Roman numerals in names like iron(III) chloride and manganate(VII) come directly from oxidation numbers, so assigning them correctly lets you name and identify unfamiliar compounds.

Try this

Q1. State the oxidation number of sulfur in the sulfate ion $\text{SO}_4^{2-}$ . [1 mark]

Cue. $+6$ (four oxygens at $-2$ give $-8$ ; $x - 8 = -2$ , so $x = +6$ ).

Q2. Write the ionic equation for the reaction of aqueous silver nitrate with aqueous sodium chloride. [1 mark]

Cue. $\text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)} \rightarrow \text{AgCl}_{(s)}$ .

Q3. Deduce the formula of the compound formed between calcium and nitrate ions. [1 mark]

Cue. $\text{Ca}(\text{NO}_3)_2$ .

Q4. State the oxidation number of oxygen in hydrogen peroxide, $\text{H}_2\text{O}_2$ . [1 mark]

Cue. $-1$ .

Q5. Write the ionic equation for the neutralisation of any strong acid by any strong alkali. [1 mark]

Cue. $\text{H}^+_{(aq)} + \text{OH}^-_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)}$ .

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20203 marksDetermine the oxidation number of manganese in the manganate(VII) ion,

\text{MnO}_4^-

, and in manganese(II),

\text{Mn}^{2+}

, then state whether manganese is oxidised or reduced when

\text{MnO}_4^-

is converted to

\text{Mn}^{2+}

Show worked answer →

In $\text{MnO}_4^-$ each oxygen is $-2$ , total $-8$ . The overall charge is $-1$ , so manganese is $+7$ (since $x + (-8) = -1$ gives $x = +7$ ).

In $\text{Mn}^{2+}$ the oxidation number equals the ionic charge, $+2$ .

The oxidation number falls from $+7$ to $+2$ , so manganese is reduced (gain of electrons).

Markers reward both oxidation numbers and identifying reduction from the decrease.

WJEC 20173 marksWrite a balanced half equation for the reduction of dichromate(VI) ions,

\text{Cr}_2\text{O}_7^{2-}

, to chromium(III) ions,

\text{Cr}^{3+}

, in acidic solution.

Show worked answer →

Balance chromium first: $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$ .

Balance oxygen with water: add $7\text{H}_2\text{O}$ to the right.

Balance hydrogen with $\text{H}^+$ : add $14\text{H}^+$ to the left.

Balance charge with electrons: left side charge is $(-2) + 14 = +12$ , right side is $+6$ , so add $6e^-$ to the left.

Final: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ .

Markers reward balancing atoms, then $\text{H}^+$ and water, then electrons for charge.

Related dot points

Sources & how we know this

WJEC A-level Chemistry specification — WJEC (2015)