23.1 Lattice energy and Born-Haber cycles
- Syllabus
- 9701–2028–2029
- Topic
- 23.1
- Level
- A2
Learning objectives
23.1.1Terms• Define/use:: (a) enthalpy change of atomisation, ΔHat; (b) lattice energy, ΔHlatt (the change from gas phase ions to solid lattice)23.1.2Terms: first electron affinity, EA• (a) define and use the term first electron affinity, EA: (b) explain the factors affecting the electron affinities of elements; (c) describe and explain the trends in the electron affinities of the Group 16 and Group 17 elements23.1.3Born–Haber cycles for ionic solids• Construct/use: Born–Haber cycles for ionic solids (limited to +1 and +2 cations, –1 and –2 anions)23.1.4Calculations: Born–Haber cycles• Calculate: Born–Haber cycles23.1.5Explain, in qualitative terms, the effect• Explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of a lattice energy
Define atomisation and lattice energy by their exact particles and direction
| Term | Defining process for one mole | Typical sign |
|---|---|---|
| enthalpy change of atomisation, ΔHₐₜ | element in its standard state → 1 mol gaseous atoms | positive |
| lattice energy, ΔHₗₐₜₜ | stoichiometric gaseous ions → 1 mol solid ionic lattice | negative |
Na(s)Na(g)
21ClX2(g)Cl(g)
NaX+(g)+ClX−(g)NaCl(s)
Write the species, coefficients and states before inserting a value. Atomising a molecular element may require a fraction of its standard-state molecule; reversing lattice formation to separate the solid into gaseous ions changes the sign.
The official ΔHₗₐₜₜ convention here is gas-phase ions to solid lattice. Do not mix it with a positive lattice-dissociation value or call any solid-to-gas change atomisation without forming gaseous atoms.
First electron affinity balances nuclear attraction and electron repulsion
X(g)+eX−XX−(g)EA1=ΔH for this process
| Factor | Effect on attraction of the incoming electron |
|---|---|
| greater nuclear charge with similar shielding | makes EA₁ more exothermic |
| larger atomic radius / greater electron distance | makes EA₁ less exothermic |
| more inner-shell shielding | makes EA₁ less exothermic |
| strong repulsion in a compact or already occupied orbital | makes EA₁ less exothermic |
Group 17 atoms gain an electron to complete the p subshell, so their first electron affinities are strongly exothermic. Chlorine is more exothermic than fluorine because the incoming electron experiences greater repulsion in fluorine's very compact 2p orbital; from Cl down to I, increasing radius and shielding make EA₁ less exothermic.
Group 16 shows the parallel anomaly: sulfur has a more exothermic EA₁ than oxygen because oxygen's compact 2p orbital gives greater electron repulsion. From S down the group, increasing distance and shielding make EA₁ less exothermic. Group 17 values are generally more exothermic than the corresponding Group 16 values because the stronger nuclear attraction and p-subshell completion favour electron gain.
Electron affinity is not ionisation energy in reverse. A second electron affinity adds an electron to X⁻(g), so repulsion from the negative ion makes that separate process endothermic.
Construct a Born–Haber cycle by making the same gaseous ions
A Born–Haber cycle applies Hess's law between the elements in their standard states, the ionic solid, and one common set of gaseous ions. Every alternative path must end at exactly the same stoichiometric gaseous ions before lattice formation.
| MgCl₂ cycle step | Process / coefficient | Enthalpy term |
|---|---|---|
| formation | Mg(s) + Cl₂(g) → MgCl₂(s) | ΔH°f |
| atomise Mg | Mg(s) → Mg(g) | ΔHₐₜ(Mg) |
| atomise chlorine | Cl₂(g) → 2Cl(g) | 2ΔHₐₜ(Cl) |
| form Mg²⁺ | Mg(g) → Mg²⁺(g) + 2e⁻ | IE₁ + IE₂ |
| form 2Cl⁻ | 2Cl(g) + 2e⁻ → 2Cl⁻(g) | 2EA₁(Cl) |
| form lattice | Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s) | ΔHₗₐₜₜ |
For a +2 cation include both successive ionisation energies. For a –2 anion include EA₁ and the endothermic EA₂. Multiply every atomisation or electron-affinity term by the number of atoms/ions in one formula unit; the allowed charge range is ±1 and ±2.
Do not use bond dissociation and atomisation for the same non-metal atoms twice. A labelled cycle is valid only when atoms, electrons, charges, states and stoichiometric coefficients are conserved on every route.
Calculate a missing Born–Haber term with one signed Hess equation
Write the formation path first, then sum the alternative gas-ion path in the same direction. Insert tabulated values with their given signs and coefficients; only then rearrange for the unknown.
ΔHf∘=ΔHat(Na)+ΔHat(Cl)+IE1(Na)+EA1(Cl)+ΔHlatt
| NaCl term | Value / kJ mol⁻¹ |
|---|---|
| ΔH°f[NaCl(s)] | −411 |
| ΔHₐₜ[Na(s) → Na(g)] | +108 |
| ΔHₐₜ[½Cl₂(g) → Cl(g)] | +121 |
| IE₁(Na) | +496 |
| EA₁(Cl) | −349 |
ΔHlatt=−411−[108+121+496−349]=−787 kJ mol−1
The negative result matches the official lattice-formation direction: attraction releases energy when gaseous Na⁺ and Cl⁻ form NaCl(s). Check that a second IE/EA or a factor of two has not been omitted for multivalent ions.
Higher ionic charge and smaller radius increase lattice-energy magnitude
∣ΔHlatt∣ increases roughly with r++r−∣q+q−∣
| Change while other factors are comparable | Electrostatic consequence | Formation ΔHₗₐₜₜ |
|---|---|---|
| larger charge magnitude | larger charge product and stronger attraction | more negative; larger magnitude |
| smaller ionic radius | charge centres are closer | more negative; larger magnitude |
| larger ionic radius | charge centres are farther apart | less negative; smaller magnitude |
LiF has a larger lattice-energy magnitude than LiI because F⁻ is smaller than I⁻ at the same charges. MgO has a much larger magnitude than NaCl because the 2+/2− charge product is four times the 1+/1− product, alongside radius differences.
Say larger magnitude or more negative for the formation convention; “larger” alone is ambiguous. Charge and radius both matter, so compare one factor at a time where possible and acknowledge structural differences when compounds are not otherwise similar.