Part 1: Lattice formation and Born–Haber cycles

Build an energy route one physical change at a time and use Hess’s law to find a lattice enthalpy without losing signs or coefficients.

Start with the direction and the physical states

OCR uses lattice enthalpy for forming one mole of an ionic solid from its gaseous ions. Attraction between oppositely charged ions releases energy, so this formation value is negative. A lattice-dissociation value describes the reverse process and is positive with the same magnitude for the same solid and conditions.

For NaCl, Na⁺(g) + Cl⁻(g) → NaCl(s) forms one mole of formula units. For MgCl₂, Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s) still forms one mole of lattice, even though three moles of gaseous ions participate. State the amount of solid formed, not “one mole of bonds”.

The more negative the lattice formation enthalpy, the greater the energy released on forming the lattice and the stronger the ionic attraction represented by this measure. Avoid saying “larger” without specifying magnitude or sign: −2500 is more negative than −800 but numerically smaller.

Build the gaseous ions from the elements

A Born–Haber cycle compares direct formation from elements in their standard states with an indirect route through gaseous atoms and ions. The direct formation enthalpy makes one mole of compound. Along the alternative route, atomise each element, remove electrons from the metal and add electrons to the non-metal, then form the lattice.

First ionisation energy removes one electron from each atom in one mole of gaseous atoms to form gaseous 1+ ions; it is endothermic. A second ionisation removes an electron from gaseous 1+ ions, not from neutral atoms again. Electron affinity adds an electron to a gaseous species; the first is often exothermic, while adding a second electron to an already negative ion requires energy to overcome repulsion.

Atomisation must produce gaseous atoms. For chlorine, ½Cl₂(g) → Cl(g) produces one mole of atoms, whereas the Cl–Cl bond dissociation value is for Cl₂(g) → 2Cl(g). Inspect the supplied definition before dividing by two. States distinguish the steps; replacing Na(g) with Na(s) would change the process.

Worked example: label the route before substituting

Use original illustrative NaCl-like data in kJ mol⁻¹: formation −410, metal atomisation +110, chlorine atomisation +122, metal first ionisation +500 and chlorine first electron affinity −350. Hess’s law gives −410 = 110 + 122 + 500 − 350 + ΔLEH.

The known indirect steps sum to +382. Therefore ΔLEH = −410 − 382 = −792 kJ mol⁻¹. Check by recombining: +382 − 792 = −410, matching the direct route. The negative sign agrees with attraction releasing energy on lattice formation. These values illustrate the method and are not a measured data table.

Transfer the method to a 2+ metal or a 2− ion

For MgCl₂, the gaseous-ion route needs first and second ionisation energies of Mg, plus two chlorine atomisation and two first-electron-affinity terms. There is no second electron affinity of chlorine in this route: each of two chlorine atoms gains one electron.

For MgO, one oxygen atom gains two electrons, so both first and second electron affinities occur. The positive second electron affinity does not stop MgO forming overall: other steps, especially the strongly exothermic lattice formation, contribute to the total cycle. A single unfavourable step does not determine the sign of the overall enthalpy.

Draw each level with its actual species and count atoms and charge along the route. Multiplying the final lattice value by the number of ions is a common conceptual error; the value is already defined per mole of solid formed.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources