Lattice Energy and Born–Haber Cycles

· Lattice energy, ΔH⦵latt = enthalpy change when 1 mol of an ionic solid is formed from its gaseous ions under standard conditions.
· CIE definition uses gas phase ions → solid lattice, so ΔH⦵latt is exothermic and usually negative.
· General equation: M⁺(g) + X⁻(g) → MX(s).
· For 2+ / 2– ions: M²⁺(g) + X²⁻(g) → MX(s).
· More negative lattice energy = stronger ionic lattice = greater electrostatic attraction.
· Lattice energy cannot usually be measured directly, so it is found using a Born–Haber cycle and Hess’s law.

This diagram shows how a Born–Haber cycle breaks ionic compound formation into smaller enthalpy changes. It is useful for seeing how lattice energy fits into the overall enthalpy change of formation. Source

Enthalpy Change of Atomisation

· Enthalpy change of atomisation, ΔH⦵at = enthalpy change when 1 mol of gaseous atoms is formed from an element in its standard state.
· Atomisation is always endothermic, so ΔH⦵at is positive.
· Examples:
· Na(s) → Na(g) = atomisation of sodium.
· ½Cl₂(g) → Cl(g) = atomisation of chlorine.
· Use fractions when needed because the definition is based on 1 mol of gaseous atoms, not 1 mol of molecules.
· Common exam mistake: writing Cl₂(g) → 2Cl(g) when only ½Cl₂(g) → Cl(g) is needed for 1 mol of NaCl.

Electron Affinity

· First electron affinity, EA₁ = enthalpy change when 1 mol of electrons is added to 1 mol of gaseous atoms to form 1 mol of gaseous 1– ions.
· General equation: X(g) + e⁻ → X⁻(g).
· EA₁ is usually exothermic, so it is usually negative, because the incoming electron is attracted to the positive nucleus.
· Second electron affinity, EA₂ = enthalpy change when 1 mol of electrons is added to 1 mol of gaseous 1– ions.
· General equation: X⁻(g) + e⁻ → X²⁻(g).
· EA₂ is always endothermic, so it is positive, because an electron is being added to an already negative ion.

Factors Affecting Electron Affinity

· Electron affinity depends on nuclear charge, atomic radius, and shielding.
· Greater nuclear charge → stronger attraction for the incoming electron → EA₁ more exothermic.
· Smaller atomic radius → incoming electron closer to nucleus → EA₁ more exothermic, unless electron–electron repulsion is very strong.
· Greater shielding → weaker attraction between nucleus and incoming electron → EA₁ less exothermic.
· Electron–electron repulsion in small atoms can make EA₁ less exothermic than expected.

Electron Affinity Trends in Group 16 and Group 17

· Group 17 halogens have very exothermic EA₁ because they gain one electron to form stable X⁻ ions with a full outer shell.
· Down Group 17, EA₁ generally becomes less exothermic because atomic radius and shielding increase.
· Exception: chlorine has a more exothermic EA₁ than fluorine because fluorine is very small, so the incoming electron experiences greater repulsion in the compact 2p orbital.
· Group 16 elements have less exothermic EA₁ values than Group 17 because adding one electron does not complete the outer shell.
· In Group 16, oxygen’s EA₁ is less exothermic than sulfur’s because oxygen is very small and has greater electron–electron repulsion.
· EA₂ for Group 16 is positive because forming O²⁻, S²⁻, etc. requires adding an electron to an already negative ion.

Born–Haber Cycles

· A Born–Haber cycle is an enthalpy cycle used to calculate lattice energy using Hess’s law.
· It links enthalpy change of formation, ΔH⦵f, to atomisation, ionisation energy, electron affinity, and lattice energy.
· Born–Haber cycles are only required for ionic solids with +1 and +2 cations and –1 and –2 anions.
· The cycle must start from elements in their standard states and end with 1 mol of ionic solid.
· The gaseous ion stage must contain the correct ions needed for the ionic formula.
· Use state symbols carefully: atomisation and ionisation/electron affinity steps involve gaseous atoms or ions.

Born–Haber Cycle Steps for MX

· For an ionic solid MX, typical steps are:
· Atomise the metal: M(s) → M(g).
· Atomise the non-metal: ½X₂(g) → X(g), if the non-metal is diatomic.
· Ionise the metal: M(g) → M⁺(g) + e⁻.
· Add electron to non-metal: X(g) + e⁻ → X⁻(g).
· Form the lattice: M⁺(g) + X⁻(g) → MX(s).
· Overall formation equation: M(s) + ½X₂(g) → MX(s).
· Equation template: ΔH⦵f = ΔH⦵at(M) + ΔH⦵at(X) + IE₁(M) + EA₁(X) + ΔH⦵latt.

Born–Haber Cycle Steps for 2+ or 2– Ions

· For M²⁺, include IE₁ + IE₂.
· For X²⁻, include EA₁ + EA₂.
· Example for MgO:
· Mg(s) → Mg(g) = atomisation of Mg.
· ½O₂(g) → O(g) = atomisation of O.
· Mg(g) → Mg⁺(g) + e⁻ = first ionisation energy.
· Mg⁺(g) → Mg²⁺(g) + e⁻ = second ionisation energy.
· O(g) + e⁻ → O⁻(g) = first electron affinity.
· O⁻(g) + e⁻ → O²⁻(g) = second electron affinity.
· Mg²⁺(g) + O²⁻(g) → MgO(s) = lattice energy.
· Equation template: ΔH⦵f = ΔH⦵at(Mg) + ΔH⦵at(O) + IE₁ + IE₂ + EA₁ + EA₂ + ΔH⦵latt.

Born–Haber Calculations

· Use Hess’s law: total enthalpy change is the same whichever route is taken.
· For lattice energy: rearrange the Born–Haber equation.
· For MX: ΔH⦵latt = ΔH⦵f – ΔH⦵at(M) – ΔH⦵at(X) – IE₁ – EA₁.
· Remember signs: ionisation energies are positive, atomisation is positive, EA₁ often negative, EA₂ positive, lattice formation is negative.
· Always check whether data is for lattice formation or lattice dissociation.
· Lattice dissociation enthalpy is the reverse process and has the opposite sign: MX(s) → M⁺(g) + X⁻(g).

Ionic Charge, Ionic Radius and Lattice Energy

· Lattice energy depends on the strength of electrostatic attraction between oppositely charged ions.
· Higher ionic charge → stronger attraction → lattice energy becomes more exothermic / more negative.
· Smaller ionic radius → ions closer together → stronger attraction → lattice energy becomes more exothermic / more negative.
· Larger ions have lower charge density, so attraction is weaker and lattice energy is less exothermic.
· Comparison examples:
· MgO has a more exothermic lattice energy than NaCl because Mg²⁺ and O²⁻ have higher charges than Na⁺ and Cl⁻.
· LiF has a more exothermic lattice energy than CsF because Li⁺ is smaller than Cs⁺.
· NaF has a more exothermic lattice energy than NaI because F⁻ is smaller than I⁻.

This graph shows how lattice energy changes with ionic size for alkali metal halides. It demonstrates that smaller ions produce larger lattice energies because the ions are closer together and attract more strongly. Source

Common Exam Traps

· Do not confuse lattice formation enthalpy with lattice dissociation enthalpy.
· Do not forget ½X₂(g) → X(g) for diatomic non-metals such as Cl₂, Br₂, I₂, F₂ and O₂.
· Do not omit IE₂ when forming M²⁺.
· Do not omit EA₂ when forming X²⁻.
· Do not use aqueous ions in Born–Haber cycles; use gaseous ions only.
· Do not ignore coefficients: the cycle must form 1 mol of ionic solid.
· Always include correct signs in calculations.