A binary ionic compound is just a metal cation and a non-metal anion locked together by electrostatic attraction. Two skills come out of it — naming the compound and writing its formula — and both come down to one rule: the charges must cancel.
📚 What you need to know
A binary ionic compound contains ions of exactly two elements: a metal cation and a non-metal anion.
Ionic bonding is the electrostatic attraction between oppositely charged ions.
Naming: cation first, unchanged; anion second with the ending changed to –ide.
Ionic compounds are electrically neutral — total positive charge = total negative charge.
Use Roman numerals for transition metals: iron(II) sulfide, copper(II) chloride.
If a polyatomic ion is needed more than once, put it in brackets: Mg(NO3)2.
What holds the compound together
Once electrons have transferred, you have a positive ion sitting next to a negative ion — and opposite charges attract. That attraction is the bond:
Definition — ionic bonding
the electrostatic attraction between oppositely charged ions
Three things follow from that definition, and they explain almost every property later in the topic:
The attraction is strong, so a lot of energy is needed to break it.
It is non-directional — a cation attracts every anion around it, not just one, which is why ionic solids build lattices rather than molecules.
It gets stronger with higher charge and smaller ions.
Write the definition as one clean sentence and stick to it. “Attraction between a metal and a non-metal” won’t score — the marking point is electrostatic attraction between oppositely charged ions.
Naming binary ionic compounds
The rule is short: metal first, unchanged — non-metal second, ending in –ide.
Metal name unchanged, non-metal ending replaced with –ide. Polyatomic ions are the exception — they keep their given name.
WORKED EXAMPLE
Give the IUPAC name of the binary ionic compound formed from (a) potassium + bromine, (b) magnesium + nitrogen, (c) calcium + hydrogen.
(a) metal = potassium; bromine → bromidepotassium bromide(b) metal = magnesium; nitrogen → nitridemagnesium nitride(c) metal = calcium; hydrogen → hydridecalcium hydrideNotice no numbers appear in the name — none of these metals has a variable charge.
Writing the formula: balancing the charges
An ionic compound has no overall charge, so the positives must exactly cancel the negatives. That single condition fixes the ratio of ions, and the formula you write is the empirical formula — the simplest whole-number ratio.
Find the lowest common multiple of the two charges, then work out how many of each ion you need to reach it.
🧩 Formula in four steps
Write both ions with their charges, cation first.
Find the lowest common multiple of the two charge sizes.
Work out how many of each ion gets you there, and write those as subscripts.
Simplify to the smallest whole-number ratio, and drop any subscript of 1.
WORKED EXAMPLE
Determine the formulae of (a) calcium chloride, (b) sodium oxide, (c) iron(III) sulfide.
(a) Ca²⁺ and Cl⁻ — need 2 chlorides per calciumCaCl₂(b) Na⁺ and O²⁻ — need 2 sodiums per oxideNa₂O(c) Fe³⁺ and S²⁻ — LCM of 3 and 2 is 62 × 3+ = 6+ and 3 × 2– = 6–Fe₂S₃The Roman numeral (III) is what tells you the iron charge here.
Compounds containing polyatomic ions
Polyatomic ions balance in exactly the same way — the only extra rule is bracketing. If you need more than one of a polyatomic ion, put it in brackets with the number outside, so the subscript applies to the whole group:
Mg(NO3)2 — one Mg2+ with two NO3− ions. Writing MgNO32 would mean something completely different.
Al2(SO4)3 — two Al3+ (6+) with three SO42− (6−).
No brackets are needed when only one polyatomic ion is present: NaOH, K2SO4.
WORKED EXAMPLE
Determine the formulae of (a) ammonium sulfate, (b) calcium hydroxide, (c) aluminium sulfate.
(a) NH₄⁺ and SO₄²⁻ — two ammoniums needed(NH₄)₂SO₄(b) Ca²⁺ and OH⁻ — two hydroxides neededCa(OH)₂(c) Al³⁺ and SO₄²⁻ — LCM 6, so 2 and 3Al₂(SO₄)₃Every one of these needs brackets — the polyatomic ion appears more than once.
Naming when the metal has a variable charge
If the metal is a transition element, you must work its charge out backwards from the anion, then state it as a Roman numeral:
Formula
Anion charge
Metal charge
Name
FeCl2
2 × Cl− = 2−
Fe2+
iron(II) chloride
FeCl3
3 × Cl− = 3−
Fe3+
iron(III) chloride
Cu2O
1 × O2− = 2−
Cu+ (two of them)
copper(I) oxide
CuO
1 × O2− = 2−
Cu2+
copper(II) oxide
💡 Exam tip
The Roman numeral is the charge on one metal ion — not the number of metal ions in the formula.
Never write charges inside a compound formula. It is MgCl2, not Mg2+Cl2−.
Ionic formulae are empirical: always reduce to the simplest ratio.
Watch the ammonium exception — NH4Cl is ionic with no metal at all.
⚠️ Common mix-up
–ide vs –ate: –ide means a simple non-metal anion (chloride, oxide). –ate means a polyatomic ion containing oxygen (nitrate, sulfate, carbonate).
Sulfide (S2−) is not sulfate (SO42−) — same charge, completely different ion.
Forgetting brackets around a repeated polyatomic ion changes the formula’s meaning entirely.
Up next: Ionic Lattice Structures — why these ions don’t pair off into molecules, and how the giant lattice explains melting points, brittleness, conductivity and solubility.
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