The big idea: An ionic solid is a giant lattice of oppositely charged ions held together by strong electrostatic attraction. Pulling that lattice completely apart costs a lot of energy.
Lattice enthalpy, ΔHlat⊖, is the enthalpy change when one mole of a solid ionic compound is converted into its gaseous ions:
MX(s) → M⁺(g) + X⁻(g)
Because you are breaking attractions, lattice enthalpy is always endothermic (large and positive).
Why it can't be measured directly: You can't sit in a lab and tear a salt crystal into a gas of free ions and read the heat off a thermometer — there is no single experiment that does it.
Instead we find ΔHlat⊖ indirectly, using a Born–Haber cycle: a closed enthalpy cycle built from steps that can be measured, combined with Hess's law.
Why Hess's law works for a Born–Haber cycle: forming the solid directly from its elements (ΔHf⊖) must equal the long way round — make gaseous atoms, ionise/add electrons to get gaseous ions, then let those ions snap together (the reverse of ΔHlat⊖). Same start, same end → same total.
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A Born–Haber cycle is a step-by-step 'paper route' from the elements to the gaseous ions, then back down to the solid. Each step is its own measurable enthalpy term, and each has a characteristic sign.
| Step (for MX) | Enthalpy term | What happens | Sign |
|---|---|---|---|
| 1 | ΔHf⊖ (formation) | elements in standard states → 1 mol solid MX | usually − |
| 2 | ΔHatom⊖ (atomisation/sublimation) | metal solid → gaseous metal atoms | + |
| 3 | Bond dissociation (½ × per mole of bonds) | non-metal molecule → gaseous atoms | + |
| 4 | 1st ionisation energy | M(g) → M⁺(g) + e⁻ | + (always) |
| 5 | 1st electron affinity | X(g) + e⁻ → X⁻(g) | usually − |
| 6 | ΔHlat⊖ (lattice enthalpy) | MX(s) → M⁺(g) + X⁻(g) | + (large) |
Get the signs right: - Atomisation / sublimation ΔHatom⊖ — turning an element into gaseous atoms: endothermic (+). - Bond dissociation — splitting a diatomic molecule (e.g. ½Cl2 → Cl): endothermic (+). Use half the bond enthalpy if only one atom is needed. - Ionisation energy (IE) — M(g) → M⁺(g) + e⁻: always endothermic (+) (you remove an electron against attraction). - First electron affinity (EA) — X(g) + e⁻ → X⁻(g): usually exothermic (−) for the first electron added. - Enthalpy of formation ΔHf⊖ — usually exothermic (−). - Lattice enthalpy ΔHlat⊖ — solid → gaseous ions: endothermic (+).
The classic sign trap: A first electron affinity is exothermic, but adding a second electron (e.g. O⁻ → O²⁻) is endothermic (+) — you push an electron onto an already-negative ion. For Group-16 oxides you must add both EA values, with the second one positive.
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Set up the cycle so the direct route (elements → solid, ΔHf⊖) equals the indirect route (elements → gaseous atoms → gaseous ions → solid). Then rearrange to make the unknown term — almost always the lattice enthalpy — the subject.
- standard enthalpy of formation of the ionic solid (kJ mol⁻¹)
- sum of atomisation (sublimation) + bond-dissociation steps to make gaseous atoms
- sum of the ionisation energies (always endothermic, +)
- sum of the electron affinities (1st EA usually exothermic, −)
- lattice enthalpy — solid → gaseous ions (endothermic, +)
- lattice enthalpy (kJ mol⁻¹) — the unknown
- atomisation + bond-dissociation enthalpies
- ionisation energies (+)
- electron affinities (1st usually −)
- enthalpy of formation of the solid (−)
Worked example — lattice enthalpy of sodium fluoride
Determine the lattice enthalpy of NaF, NaF(s) → Na⁺(g) + F⁻(g), from: ΔHf⊖(NaF) = −574; ΔHatom⊖(Na) = +108; ½ bond dissociation of F2 = +79; 1st IE(Na) = +496; 1st EA(F) = −328 (all kJ mol⁻¹).
Solution
- Formula first — rearrange the Born–Haber relation for the lattice enthalpy:
- Add up the 'route up to gaseous ions' (atomise Na, atomise ½F2, ionise Na, add e⁻ to F):
- Now subtract the (negative) formation enthalpy — subtracting a negative adds:
- Work it out — keep the sign and unit:
Final answer
ΔHlat⊖(NaF) = +929 kJ mol⁻¹ (endothermic, as expected for breaking a lattice apart).
Born–Haber energy ladder for NaF, matching the worked example: climb up via atomisation of Na (+108), atomisation of ½F₂ (+79), 1st ionisation of Na (+496); the 1st electron affinity of F (−328) drops back down; then the gaseous ions snap into the solid lattice (−929, the reverse of ΔHlat⊖ = +929), ending at NaF(s). Endothermic steps rise (rose); exothermic steps fall (emerald).
Interactive diagram
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Sign discipline: The single biggest source of lost marks is a wrong sign, especially the − ΔH_{f}⊖ term: subtracting a negative number adds its magnitude. Write every value with its sign in brackets before you add them up.
How this is tested: HL Paper 2 loves two things here:
- a calculate ΔH_{lat}⊖ Born–Haber question (set up the cycle, watch the signs), and - an explain / compare question: 'why is the lattice enthalpy of MgO larger than that of NaCl?'
The magnitude of lattice enthalpy is governed by the electrostatic attraction between the ions, which depends on two factors.
The two factors: Attraction (and so ΔHlat⊖) is larger when:
- Ionic charge is bigger — higher charges attract more strongly (the dominant factor). Mg²⁺/O²⁻ attract far more strongly than Na⁺/Cl⁻. - Ionic radius is smaller — smaller ions sit closer together, so the attraction is stronger. F⁻ gives a larger lattice enthalpy than I⁻ with the same cation.
Increasing charge
- Na⁺Cl⁻ (1+, 1−): ΔHlat⊖ ≈ +790 kJ mol⁻¹
- Mg²⁺O²⁻ (2+, 2−): ΔHlat⊖ ≈ +3791 kJ mol⁻¹
- Doubling both charges has a huge effect — charge is the dominant factor.
Decreasing radius
- NaF (small F⁻): ΔHlat⊖ ≈ +930 kJ mol⁻¹
- NaI (large I⁻): ΔHlat⊖ ≈ +700 kJ mol⁻¹
- Smaller ion → ions closer → stronger attraction → larger ΔHlat⊖.
IB-style question — compare two lattice enthalpies
Explain why the lattice enthalpy of magnesium oxide, MgO, is much larger than that of sodium chloride, NaCl. [3]
How to score the marks
- State that lattice enthalpy depends on the electrostatic attraction between the ions, which increases with greater ionic charge and smaller ionic radius. [1]
- Identify the charges: Mg²⁺ and O²⁻ are both 2+/2−, whereas Na⁺ and Cl⁻ are only 1+/1− — the larger charges give a much stronger attraction. [1]
- Add the size point: Mg²⁺ and O²⁻ are also smaller than Na⁺ and Cl⁻, so the ions sit closer together, further increasing the attraction → MgO has the much larger lattice enthalpy. [1]
Final answer
MgO has the larger lattice enthalpy because its ions carry larger charges (2+/2− vs 1+/1−) and are smaller, giving a much stronger electrostatic attraction.