Lattice Energy and Born–Haber Cycles

Detailed Explanations

A. Trend explanation (because → therefore)

Because the ionic lattice is held together by electrostatic attraction, stronger attraction releases more energy when the lattice forms.

Therefore higher ionic charge and smaller ionic radius make the lattice enthalpy more exothermic (more negative).

B. Reading a Born–Haber cycle: sodium chloride

A Born–Haber cycle is a Hess’s Law cycle drawn as enthalpy levels. Every level is a complete set of species with state symbols, and every arrow is one defined enthalpy change pointing from its starting state to its final state. Up arrows are endothermic (+); down arrows are exothermic (-).

The cycle always has two routes between the same start and end:

• the direct route: standard formation, Na(s) + 1/2 Cl₂(g) → NaCl(s);
• the alternative route: elements → gaseous atoms → gaseous ions → solid.

Read the alternative route one arrow at a time and check what each step does to the species:

The electron released by ionisation is the electron that chlorine accepts, so no free electron is left at the ionic level.

Because both routes start and finish at the same states, their enthalpy changes are equal. Write that as one signed equation before substituting numbers:

If you travel against an arrow, reverse its sign. That is how the same cycle answers for any single unknown step.

C. A cycle with doubly charged ions: magnesium oxide

The target ions set the electronic steps. MgO contains Mg²⁺ and O²⁻, so the route needs two ionisation energies and two electron affinities.

• Oxygen is atomised from 1/2O₂(g) to one mole of O(g): + 249, half the O=O bond energy.
• IE₁ and IE₂ take Mg(g) to Mg²⁺(g) + 2e⁻. Both are endothermic.
• EA₁(O), O(g) + e⁻ → O⁻(g), is exothermic (-141): the incoming electron is attracted by the nucleus.
• EA₂(O), O⁻(g) + e⁻ → O²⁻(g), is endothermic (+ 844): the electron is added to an ion that is already negative, so it must be pushed in against repulsion.

Count charge at every level: after IE₂ the level holds Mg²⁺ and 2e⁻ (total charge zero); after EA₁ it holds Mg²⁺, one e⁻ and O⁻; after EA₂ only the ions remain. Omitting either second step leaves the route unable to reach both Mg²⁺ and O²⁻; omitting both leaves Mg⁺ and O⁻. Neither ion pair matches the ions in MgO(s). If you simply leave a positive term out of the Hess sum, the calculated lattice formation enthalpy is too positive by 1451 kJ mol⁻¹ for IE₂ or 844 kJ mol⁻¹ for EA₂.

D. Formation or dissociation: one arrow, two directions

This lesson draws the lattice step as formation, gaseous ions → solid, so the arrow points down and Δ H_(latt(form)) is negative. A data table that quotes lattice dissociation enthalpy describes the same arrow reversed, solid → gaseous ions:

Before using a quoted “lattice energy”, check which direction it describes, convert it to the direction of the arrow in your cycle, and name the convention in your answer.

E. What the drawing does and does not show

The cycles above are schematic: each level sits above or below its neighbours in the order the data require, but the vertical gaps are not proportional to the enthalpy values. A quantitative enthalpy-level diagram must be drawn to scale from one stated reference level.

A Born–Haber cycle is not a reaction profile. It has no reaction-coordinate axis, no transition state and no activation energy; it compares the enthalpies of states, and says nothing about how fast any step happens.