IB DP Chemistry Structure 2.1: The ionic model

This guide teaches Structure 2.1, The ionic model, from the Diploma Programme Chemistry guide that the IB issued in February 2023, with first assessment in 2025. It covers all three understandings, Structure 2.1.1 to 2.1.3. Every one of them is studied at both SL and HL; this sub-topic has no AHL extension. The guiding question is what makes a compound ionic, and what that does to its properties.

Shorter revision notes and a set of practice questions go with this page. The chemistry hub and the printable checklist place the unit within the whole course, and the free diagnostics help you find weak spots.

What this unit covers

Understanding	Skills to show	SL/HL
Structure 2.1.1	Explain how cations and anions form; predict an ion’s charge from the atom’s electron configuration; know that a transition element can form ions of more than one charge	SL and HL
Structure 2.1.2	Describe the ionic bond; deduce formulas and names from the ions present, including the seven listed polyatomic ions; switch between names and formulas of binary compounds	SL and HL
Structure 2.1.3	Describe the 3D lattice and why it is written as an empirical formula; explain volatility, electrical conductivity and solubility; use lattice enthalpy to compare bond strength through ion charge and radius	SL and HL

Forming ions (2.1.1)

An atom becomes an ion when it gains or loses electrons. The number of protons does not change, so the charge equals protons minus electrons.

• Metal atoms lose electrons and become positive ions, called cations.
• Non-metal atoms gain electrons and become negative ions, called anions.

Atoms of main-group elements usually lose or gain enough electrons to reach the configuration of the nearest noble gas. That is the link between the electron configuration and the charge.

Group	Outer electrons	Usual change	Ion charge
1	s¹	lose 1	1+
2	s²	lose 2	2+
13	s² p¹	lose 3	3+
15	s² p³	gain 3	3−
16	s² p⁴	gain 2	2−
17	s² p⁵	gain 1	1−

Group 18 atoms already have full outer levels and do not form ions, and carbon and silicon in group 14 do not form simple ions in their compounds.

Worked example 1: charge from a configuration

Predict the ion formed by each atom, and say which noble gas configuration the ion has.

(a) Z = 31: [Ar] 3d¹⁰ 4s² 4p¹
    Outer level (n = 4) holds 3 electrons: 4s² 4p¹
    Losing 3 is easier than gaining 5 -> gallium forms Ga³⁺
    Ga³⁺ has 31 − 3 = 28 electrons: [Ar] 3d¹⁰ (a full 3d sublevel)

(b) Z = 34: [Ar] 3d¹⁰ 4s² 4p⁴
    Outer level holds 6 electrons; it needs 2 more to fill 4p
    Selenium forms Se²⁻, with 34 + 2 = 36 electrons: [Kr]

(c) Z = 15: [Ne] 3s² 3p³
    Gains 3 electrons -> P³⁻, with 18 electrons: [Ar]

Notice that Ga³⁺ does not have a noble gas configuration. After losing its 4s and 4p electrons it keeps a full 3d sublevel, which is still a stable arrangement.

Transition elements: more than one charge

Transition elements lose their 4s electrons first. Many can then lose one or more 3d electrons as well, so a single element can form ions of different charge. Iron forms Fe²⁺ and Fe³⁺. Copper forms Cu⁺ and Cu²⁺. You cannot predict one fixed charge from the group number, so the name of the compound carries the charge as a Roman numeral: iron(II), iron(III), copper(I), copper(II).

Deducing the full electron configurations of transition element ions is part of HL Structure 3.1, so you will meet that again later.

The ionic bond, formulas and names (2.1.2)

An ionic bond is the electrostatic attraction between oppositely charged ions. It is not a bond between one particular pair of ions. Each ion attracts every oppositely charged ion around it, in all directions.

Writing a formula

A compound has no overall charge. The total positive charge must cancel the total negative charge.

1. Write each ion with its charge.
2. Take the smallest number that both charge sizes divide into exactly.
3. Work out how many of each ion give that total.
4. Write the cation first. Put brackets around a polyatomic ion if you need more than one of it.

The polyatomic ions you must know

Name	Formula
ammonium	NH₄⁺
hydroxide	OH⁻
nitrate	NO₃⁻
hydrogencarbonate	HCO₃⁻
carbonate	CO₃²⁻
sulfate	SO₄²⁻
phosphate	PO₄³⁻

Ammonium is the only cation in the list. A polyatomic ion stays as one unit in the formula, so the atoms inside it are never shared out or changed.

Naming binary compounds

A binary ionic compound contains two elements. Name the cation first, then the anion. The anion takes the ending -ide: nitride, oxide, fluoride, phosphide, sulfide, chloride, selenide, bromide, iodide. The cation keeps the name of the metal. With polyatomic ions, use the ion’s own name, so K₃PO₄ is potassium phosphate, not potassium phosphide.

Worked example 2: names to formulas and back

(a) barium phosphide
    Ba²⁺ and P³⁻; LCM of 2 and 3 is 6
    3 × (2+) = 6+ and 2 × (3−) = 6−  ->  Ba₃P₂

(b) chromium(III) sulfate
    Cr³⁺ and SO₄²⁻; LCM of 3 and 2 is 6
    2 Cr³⁺ and 3 SO₄²⁻  ->  Cr₂(SO₄)₃

(c) ammonium phosphate
    NH₄⁺ and PO₄³⁻; need 3 NH₄⁺ for each PO₄³⁻  ->  (NH₄)₃PO₄

(d) Name Cu₂O
    Oxide is O²⁻, so the two copper ions share 2+: each is Cu⁺
    ->  copper(I) oxide

(e) Name Sr(HCO₃)₂
    Strontium is group 2, so Sr²⁺; HCO₃⁻ is hydrogencarbonate
    ->  strontium hydrogencarbonate

In (d), the charge of a transition metal ion comes from working backwards from the anion. This is the usual way to find the Roman numeral.

Lattices and empirical formulas (2.1.3)

Solid ionic compounds are three-dimensional lattices: regular, repeating arrangements of cations and anions that extend in every direction. Each ion is surrounded by ions of the opposite charge. In the rock-salt arrangement found in potassium bromide, every K⁺ has six Br⁻ neighbours and every Br⁻ has six K⁺ neighbours.

There are no separate molecules in a lattice. A crystal of potassium bromide contains a huge, variable number of ions. The formula therefore shows only the simplest whole-number ratio of ions, so it is an empirical formula. KBr means one K⁺ for every Br⁻, and Ga₂Se₃ means two Ga³⁺ for every three Se²⁻. It is wrong to call KBr “a molecule”.

Physical properties (2.1.3)

All three properties follow from strong, non-directional attractions between ions that are fixed in place in the solid.

Volatility

Ionic compounds have low volatility. To melt or boil one, you must overcome many strong electrostatic attractions throughout the lattice. That takes a lot of energy, so melting and boiling points are high. Most ionic compounds are solids at 298 K.

Electrical conductivity

A current needs mobile charged particles.

• Solid: each ion is locked into its lattice site. None can travel towards an electrode, so the solid does not conduct.
• Molten (liquid): melting has freed the ions from their fixed sites, so they drift towards the electrodes. The liquid conducts.
• Aqueous solution: the ions separate and move freely among water molecules, so the solution conducts.

In every case the charge carriers are ions, not electrons.

Solubility

Many ionic compounds dissolve in water. Water molecules are polar. The partially negative oxygen end is attracted to cations, and the partially positive hydrogen ends are attracted to anions. These attractions pull ions out of the lattice and surround them. Non-polar solvents such as hexane cannot attract the ions in this way, so ionic compounds are generally insoluble in them.

Not every ionic compound dissolves in water. If the attractions inside the lattice are too strong compared with the attractions to water, the compound stays largely undissolved. Calcium carbonate is an example.

Lattice enthalpy (2.1.3)

Lattice enthalpy is a measure of the strength of the ionic bonding in a compound. Here it is taken as the enthalpy change, per mole, for pulling an ionic solid apart into separate gaseous ions, for example:

KBr(s) → K⁺(g) + Br⁻(g)

Separating ions takes energy, so lattice enthalpy written this way is endothermic, with a positive value. The larger the value, the stronger the ionic bonding.

Two factors control it:

• Ion charge: higher charges attract more strongly. Doubling both charges multiplies the attraction by about four. This is usually the bigger effect.
• Ion radius: smaller ions sit closer together, so their centres are nearer and the attraction is stronger. Larger ions give a smaller lattice enthalpy.

Worked example 3: ranking lattice enthalpies

Use these ionic radii (in pm): Na⁺ 102, K⁺ 138, Mg²⁺ 72, Ba²⁺ 135, Cl⁻ 181, I⁻ 220, S²⁻ 184. Rank NaCl, KI, MgS and BaS in order of decreasing lattice enthalpy, and explain.

Step 1: compare charges
  MgS and BaS: 2+ and 2−   (charge product 4)
  NaCl and KI: 1+ and 1−   (charge product 1)
  -> MgS and BaS come before NaCl and KI

Step 2: within each pair, compare distances between ion centres
  MgS:  72 + 184 = 256 pm      BaS:  135 + 184 = 319 pm
  NaCl: 102 + 181 = 283 pm     KI:   138 + 220 = 358 pm
  Shorter distance -> stronger attraction

Order: MgS > BaS > NaCl > KI

The charge effect dominates. NaCl has a shorter distance between ion centres than BaS, yet BaS still has the larger lattice enthalpy because its ions carry double charges.

A higher lattice enthalpy usually goes with a higher melting point, because the ions are harder to pull apart. This is why magnesium sulfide melts far above potassium iodide.

Common errors

• Writing the charge as a superscript but forgetting the sign order: write 2+ and 3−, as in Ca²⁺ and PO₄³⁻.
• Losing the brackets: Cr2SO43 or CrSO₄₃ instead of Cr₂(SO₄)₃.
• Rewriting a polyatomic ion to make the charges balance, for example merging two nitrate ions into N₂O₆ instead of writing (NO₃)₂.
• Using -ide for polyatomic anions (sulfide for SO₄²⁻) or using -ate for a simple anion.
• Leaving out the Roman numeral for a transition metal, or using it for a group 1 or 2 metal.
• Saying an ionic solid conducts because electrons move, or saying a solution conducts because it contains “free electrons”.
• Explaining a high melting point by “breaking covalent bonds” or “strong intermolecular forces”. It is the electrostatic attraction between ions throughout the lattice.
• Ranking lattice enthalpy by radius alone when the charges differ.

Where to go next

• Condensed recall: Structure 2.1 revision notes
• Exam-style work: Structure 2.1 practice questions
• The background to ion charges: Structure 1 overview and the course syllabus guide
• The IB DP chemistry hub, a topic checklist and the free diagnostics

Official syllabus

Source: the International Baccalaureate’s Diploma Programme Chemistry guide, issued in February 2023 for courses examined from 2025. This page follows its sub-topic Structure 2.1, The ionic model: understandings 2.1.1, 2.1.2 and 2.1.3, which are all common to SL and HL.