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The Born-Haber cycle, devised by Born and Haber in 1919, is the canonical application of Hess's law to ionic compound formation. Lattice enthalpy — the energy change when gaseous ions assemble into a crystalline lattice — is impossible to measure directly: you cannot prepare a mole of isolated gas-phase ions in a calorimeter. Instead, we route around the problem by constructing a thermodynamic cycle that links the elements in their standard states, the gaseous atoms, the gaseous ions, and the solid lattice, then solve for the unknown step. This lesson develops the canonical five-or-six-step Born-Haber cycle for NaCl, extends it to compounds with 2+ cations (MgO, CaCl₂) and 2− anions (Li₂O, MgO), and shows how comparing experimental Born-Haber lattice enthalpies with theoretical values from a perfect-ionic model exposes the degree of covalent character — a quantitative window onto Fajans's rules and ion polarisation.
Spec mapping (AQA 7405): This lesson maps to §3.1.8 (Born-Haber cycles, lattice enthalpies and the perfect-ionic model). It builds on lesson 2 of this course (Hess's law — the Born-Haber cycle is Hess applied to lattices) and §3.1.3 (ionic bonding fundamentals, lesson 0 of the bonding course). Ionisation-energy input data are drawn from §3.1.1.3 (electron configuration and successive ionisation energies). Lattice enthalpies of Group 1 and Group 2 halides feed §3.2 (inorganic — Period 3 chemistry and Group 2 trends). Refer to the official AQA specification document for the exact wording of each clause.
Assessment objectives: AO1 — define lattice formation enthalpy (ΔlattH, AQA convention, exothermic, gas-phase ions → solid lattice) versus lattice dissociation enthalpy (IUPAC convention, endothermic, solid lattice → gas-phase ions); define each step in a Born-Haber cycle (atomisation, first/second ionisation energy, first/second electron affinity, enthalpy of formation). AO2 — compute lattice enthalpy by closing a Born-Haber cycle for compounds such as NaCl, KCl, MgO, CaCl₂, MgCl₂ and Li₂O. AO3 — compare experimental (Born-Haber) lattice enthalpies with theoretical (perfect-ionic) values to assess covalent character via Fajans's rules; evaluate trends in lattice enthalpy down Group 1 and across Group 2 halides.
There are two sign conventions for lattice enthalpy, and AQA examiners deliberately test that you read the question carefully:
Lattice formation enthalpy (ΔlattH, AQA convention):
The enthalpy change when one mole of an ionic solid forms from its constituent gaseous ions under standard conditions.
M⁺(g) + X⁻(g) → MX(s) ΔlattH = exothermic (negative)
Lattice dissociation enthalpy (IUPAC convention, used by some textbooks):
The enthalpy change when one mole of an ionic solid is broken up into its constituent gaseous ions under standard conditions.
MX(s) → M⁺(g) + X⁻(g) ΔlattH = endothermic (positive)
The two values are numerically equal but opposite in sign. For NaCl, ΔlattH(formation) = −787 kJ mol⁻¹ and ΔlattH(dissociation) = +787 kJ mol⁻¹. AQA's specification §3.1.8 uses the formation convention as default, but past papers occasionally use the dissociation convention to test discrimination. Always state which convention you are using and check the sign on supplied data.
In this lesson, all lattice enthalpy values are quoted using the AQA lattice formation convention unless explicitly stated otherwise.
A Born-Haber cycle is constructed from five (or six) standard enthalpy terms. Each has a precise definition that examiners expect verbatim.
The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.
Always endothermic — bonds must be broken. For metals, ΔatH equals the enthalpy of sublimation. For diatomic non-metals, ΔatH equals half the bond dissociation enthalpy of the parent molecule (because the definition is per mole of atoms, not per mole of bonds broken). The factor of ½ on Cl₂ is one of the most common slip-ups in exams.
The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
Na(g) → Na⁺(g) + e⁻ IE₁(Na) = +496 kJ mol⁻¹
Always endothermic — work must be done against the nuclear attraction.
The energy required to remove one electron from each ion in one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions.
Mg⁺(g) → Mg²⁺(g) + e⁻ IE₂(Mg) = +1451 kJ mol⁻¹
Always more endothermic than IE₁ for the same element — the electron is being pulled off a positively charged ion against a greater net attraction.
The enthalpy change when one mole of gaseous atoms each gains one electron to form one mole of gaseous 1− ions.
Cl(g) + e⁻ → Cl⁻(g) EA₁(Cl) = −349 kJ mol⁻¹
Almost always exothermic for non-metals — the incoming electron is attracted to the partially screened nucleus.
The enthalpy change when one mole of gaseous 1− ions each gains a further electron to form one mole of gaseous 2− ions.
O⁻(g) + e⁻ → O²⁻(g) EA₂(O) = +798 kJ mol⁻¹
Always endothermic — work must be done against the electrostatic repulsion of the existing negative charge on the ion. This is a counterintuitive but examinable point: even though forming O²⁻ ions is favourable overall in the context of a lattice (because the lattice enthalpy is enormous), the EA₂ step itself is uphill.
The enthalpy change when one mole of a compound is formed from its elements in their standard states.
Na(s) + ½Cl₂(g) → NaCl(s) ΔfH(NaCl) = −411 kJ mol⁻¹
This is the bottom edge of the Born-Haber cycle — directly measurable, by combustion calorimetry or otherwise.
The cycle has two routes from Na(s) + ½Cl₂(g) to NaCl(s):
Direct route (bottom edge):
Na(s) + ½Cl₂(g) → NaCl(s) ΔfH = −411 kJ mol⁻¹
Indirect route (via gaseous ions, six steps):
(The electron transferred in step 3 is consumed in step 4, so it cancels and is not a separate step.)
Na⁺(g) + Cl(g) + e⁻
|
EA₁(Cl) = −349
|
Na⁺(g) + Cl⁻(g)
|
Na(g) + Cl(g) |
| |
IE₁(Na) = +496 |
| |
Na(g) + Cl(g) ──────┤
| |
ΔatH(Cl) = +122 ΔlattH = ?
| |
Na(g) + ½Cl₂(g) |
| |
ΔatH(Na) = +108 |
| |
Na(s) + ½Cl₂(g) ────┴──── NaCl(s)
ΔfH = −411
By Hess's law, the two routes give the same overall enthalpy change:
ΔfH = ΔatH(Na) + ΔatH(Cl) + IE₁(Na) + EA₁(Cl) + ΔlattH
Rearranging for ΔlattH:
ΔlattH = ΔfH − ΔatH(Na) − ΔatH(Cl) − IE₁(Na) − EA₁(Cl)
ΔlattH = (−411) − (+108) − (+122) − (+496) − (−349)
ΔlattH = −411 − 108 − 122 − 496 + 349
ΔlattH(NaCl) = −788 kJ mol⁻¹
(Literature: −787 kJ mol⁻¹ — agreement is within rounding tolerance on the input data.)
Exam Tip: Work systematically. Write the cycle out before substituting numbers. Assign each enthalpy its correct sign on supply (don't change signs implicitly when rearranging). If you get a positive value for lattice formation enthalpy, you have made a sign error somewhere — the answer must be exothermic.
When the cation has a 2+ charge, an extra step is added: the second ionisation energy. The Born-Haber cycle for MgO uses both IE₁(Mg) and IE₂(Mg), and because the oxide ion is 2−, it also uses both EA₁(O) and EA₂(O).
Given data:
| Step | Enthalpy (kJ mol⁻¹) |
|---|---|
| ΔatH(Mg) | +148 |
| IE₁(Mg) | +738 |
| IE₂(Mg) | +1451 |
| ΔatH(O) = ½ × bond enthalpy of O₂ | +249 |
| EA₁(O) | −141 |
| EA₂(O) | +798 |
| ΔfH(MgO) | −602 |
Steps from elements to ionic lattice:
Applying Hess:
ΔlattH = ΔfH − ΔatH(Mg) − IE₁ − IE₂ − ΔatH(O) − EA₁ − EA₂
ΔlattH = (−602) − (+148) − (+738) − (+1451) − (+249) − (−141) − (+798)
ΔlattH = −602 − 148 − 738 − 1451 − 249 + 141 − 798
ΔlattH(MgO) = −3845 kJ mol⁻¹
The lattice enthalpy of MgO is roughly five times more exothermic than that of NaCl. Two factors drive this:
The Coulomb-law scaling — ΔlattH ∝ −(q⁺ × q⁻) / (r⁺ + r⁻) — captures both effects in a single proportionality and underpins the Born-Landé equation discussed later.
When the formula contains two anions, every term involving that anion must be doubled. For CaCl₂, the cycle uses two ΔatH(Cl) and two EA₁(Cl).
Given data:
| Step | Enthalpy (kJ mol⁻¹) |
|---|---|
| ΔatH(Ca) | +178 |
| IE₁(Ca) | +590 |
| IE₂(Ca) | +1145 |
| ΔatH(Cl) | +122 |
| EA₁(Cl) | −349 |
| ΔfH(CaCl₂) | −796 |
Steps:
Applying Hess:
ΔlattH = ΔfH − ΔatH(Ca) − IE₁ − IE₂ − 2ΔatH(Cl) − 2EA₁(Cl)
ΔlattH = (−796) − (+178) − (+590) − (+1145) − (+244) − (−698)
ΔlattH(CaCl₂) = −2255 kJ mol⁻¹
Forgetting to double the chloride terms is the single most common Born-Haber error in past papers. Whenever the empirical formula contains a stoichiometric coefficient greater than 1, double-check that the corresponding atomisation and electron-affinity terms carry that coefficient too.
| Compound | ΔlattH (kJ mol⁻¹, experimental) | Drivers |
|---|---|---|
| LiF | −1031 | Small cation, small anion, 1+/1− |
| NaF | −918 | Slightly larger cation |
| NaCl | −787 | Cl⁻ larger than F⁻ |
| NaBr | −742 | Br⁻ larger than Cl⁻ |
| NaI | −705 | I⁻ larger than Br⁻ |
| KCl | −711 | K⁺ larger than Na⁺ |
| MgO | −3791 | 2+/2−, small ions |
| CaO | −3401 | Ca²⁺ larger than Mg²⁺ |
| CaCl₂ | −2255 | 2+/1−, two chlorides |
| MgCl₂ | −2526 | 2+/1−, Mg²⁺ smaller than Ca²⁺ |
| Li₂O | −2814 | 1+/2−, two lithiums |
Down a group (e.g. NaF → NaCl → NaBr → NaI): lattice enthalpy becomes less exothermic. The anion radius grows, increasing the internuclear distance, weakening the Coulomb attraction.
Increasing charge (NaCl → MgO): lattice enthalpy becomes much more exothermic. Doubling each ion charge quadruples the Coulomb factor while smaller radii compound the effect.
The Born-Haber cycle gives an experimental lattice enthalpy: it is built from measured ΔfH, IE, EA, and ΔatH values, so the lattice term inherits all the experimental information about the real solid, including any covalent character. A theoretical lattice enthalpy can be calculated independently from a model in which the ions are treated as perfect point (or hard-sphere) charges arranged on a known crystal structure — the Born-Landé equation (named after Born and Landé). This gives a "perfect-ionic" value that depends only on the charges, the ionic radii, and the Madelung constant for the geometry.
Comparing the two reveals how well the perfect-ionic model describes reality:
| Compound | Experimental (kJ mol⁻¹) | Theoretical (kJ mol⁻¹) | Difference | Interpretation |
|---|---|---|---|---|
| NaCl | −787 | −769 | 2% | Predominantly ionic — model works well |
| NaBr | −742 | −732 | 1% | Predominantly ionic |
| KCl | −711 | −686 | 4% | Predominantly ionic |
| AgF | −958 | −920 | 4% | Mostly ionic |
| AgCl | −916 | −769 | 19% | Significant covalent character |
| AgBr | −903 | −759 | 19% | Significant covalent character |
| AgI | −889 | −736 | 21% | Substantial covalent character |
| MgI₂ | −2327 | −1944 | 20% | Substantial covalent character |
When the experimental value is more exothermic than the theoretical value, extra bonding energy comes from sources not captured by the point-charge model — namely, partial covalent character arising from the cation polarising the anion's electron cloud, sharing some electron density. This is Fajans's rules in quantitative form. Covalent character is greatest when:
The Ag⁺ halide series (AgF mostly ionic, AgI substantially covalent) is the textbook illustration: as the halide grows, polarisability increases and the experimental-theoretical gap widens.
Examiners flag the same handful of slips year after year. Drilling these explicitly converts B-grade calculations into A*:
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