IB Chemistry HL Topic 2 — Ionic Bonding Paper 1 & 2 Core skill ~11 min read

Binary Ionic Compounds

Two elements, one metal and one non-metal, and a formula you are expected to produce from nothing but the periodic table. There is no memorising involved here — just charge balance, applied carefully and written down properly.

📘 What you need to know

What ionic bonding is

Students often describe ionic bonding as “the transfer of electrons”. That describes how the ions are made, not what the bond is. The bond is what happens afterwards.

Definition Ionic bonding is the electrostatic attraction between oppositely charged ions.
Transfer first, then attraction one electron moves across Na Cl Na⁺ Cl⁻ sodium atom chlorine atom 1 outer electron 7 outer electrons held by attraction this is the ionic bondThe bond is the attraction between the ions, not the transfer. Sodium ends up matching neon, and chloride matching argon.
Note the size change on the right: the sodium ion has shrunk and the chloride ion has grown, which is exactly what you would expect from the previous page.
A real ionic bond is not a private arrangement between one Na+ and one Cl. Every ion attracts every oppositely charged ion around it, in all directions. This diagram shows one pair only because it is easier to draw, not because that is how the solid works.

Naming binary ionic compounds

The rule is short. Metal first, exactly as it appears on the periodic table. Non-metal second, with its ending replaced by -ide.

Non-metalIon nameIon formulaCharge
chlorinechlorideCl1−
brominebromideBr1−
oxygenoxideO2−2−
sulfursulfideS2−2−
nitrogennitrideN3−3−
phosphorusphosphideP3−3−
hydrogenhydrideH1−
Hydrogen is the odd one. With a reactive metal it behaves as a non-metal and forms H, the hydride ion — so sodium and hydrogen give sodium hydride, NaH. With non-metals it does the opposite and forms H+.

Where the metal is a transition element, the name has to say which ion it is. Iron(II) sulfide is FeS; iron(III) sulfide is Fe2S3. Without the numeral the name is ambiguous, and an ambiguous name will not earn the mark.

Polyatomic ions

Some ions are not single atoms but small covalently bonded groups carrying an overall charge. They travel through reactions as a unit, so treat each one as a single indivisible ion with a single charge.

NameFormulaChargeSeen in
ammoniumNH4+1+ammonium sulfate fertiliser
hydroxideOH1−sodium hydroxide
nitrateNO31−potassium nitrate
hydrogencarbonateHCO31−baking soda
carbonateCO32−2−limestone, chalk
sulfateSO42−2−copper(II) sulfate
phosphatePO43−3−bone mineral, fertiliser
Notice these end in -ate, not -ide. That ending is a signal: an -ate ion almost always contains oxygen. So if a compound is called sodium sulfate rather than sodium sulfide, you know immediately that oxygen is in there and the ion is SO42−.

Working out the formula

Ionic compounds have no overall charge. That single fact gives you the formula every time.

The rule total positive charge = total negative charge
Balancing the charges the total positive charge must cancel the total negative Al³⁺ O²⁻lowest common multiple = 62 × 3+ = 6+ charges cancel 3 × 2− = 6− Al₂O₃The subscripts are how many of each ion you need, not the charges.
Some teachers show a criss-cross shortcut where the charges swap over as subscripts. It works, but it can leave you with Mg2O2 instead of MgO, so always simplify at the end.

🧩 Building an ionic formula

  1. Write the two ions with their charges. Get these from the group number, the Roman numeral, or the polyatomic ion table.
  2. Find the lowest common multiple of the two charge sizes. For 3+ and 2− that is 6.
  3. Work out how many of each ion you need to reach it: 2 × 3+ and 3 × 2−.
  4. Write the cation first, then the anion, with those numbers as subscripts. Leave out any subscript of 1.
  5. Simplify if both subscripts share a factor. Mg2O2 must be written MgO.
  6. Bracket any polyatomic ion that appears more than once, then put the subscript outside the bracket.

Worked examples

WORKED EXAMPLE

Naming the products

Give the name of the binary ionic compound formed in each reaction: potassium with bromine, magnesium with phosphorus, and barium with oxygen.

Rule: metal unchanged, non-metal ends in -ide Potassium + bromine Bromine becomes bromide. potassium bromide Magnesium + phosphorus Phosphorus becomes phosphide. magnesium phosphide Barium + oxygen Oxygen becomes oxide. barium oxide no Roman numerals needed — none of these metals is a transition element
WORKED EXAMPLE

A formula that needs both subscripts

Deduce the formula of magnesium nitride.

Step 1: write the ions Magnesium is Group 2, so Mg²⁺. Nitrogen is Group 15, so N³⁻. Step 2: lowest common multiple of 2 and 3 LCM = 6 Step 3: how many of each? 3 × 2+ = 6+   so 3 Mg²⁺ 2 × 3− = 6−   so 2 N³⁻ Mg₃N₂ the 3 goes on the magnesium even though its charge is 2 — that is the whole point of the method
WORKED EXAMPLE

A formula with a polyatomic ion

Deduce the formula of aluminium sulfate.

Step 1: write the ions Aluminium is Group 13, so Al³⁺. Sulfate is SO₄²⁻. Step 2: balance to the LCM of 3 and 2 LCM = 6 2 Al³⁺ gives 6+   3 SO₄²⁻ gives 6− Step 3: bracket the polyatomic ion Three sulfates are needed, so SO₄ goes inside brackets with the 3 outside. Al₂(SO₄)₃ writing Al₂SO₄₃ would mean something completely different
WORKED EXAMPLE

Working backwards to a name

Deduce the names of Cu2O and CuO.

Step 1: the oxide charge is fixed Oxygen is Group 16, so every oxide ion is O²⁻. Step 2: Cu₂O — one oxide means 2− to balance 2 copper ions share 2+, so each is 1+ copper(I) oxide Step 3: CuO — one copper balances one O²⁻ that single copper must carry 2+ copper(II) oxide the numeral and the subscript are different numbers here — a classic trap

💡 Exam tip

⚠ Common mix-up

Up next: Ionic Lattice Structures — why there is no such thing as a molecule of sodium chloride, and how the giant lattice explains every physical property ionic compounds have.

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