Explain lattice trends and use the difference between a Born–Haber value and a theoretical value as evidence about bonding.
Connect charge and separation to attraction
Oppositely charged ions attract electrostatically. Increasing ionic charge strengthens the attraction. For otherwise comparable structures and charges, reducing ionic radii decreases the distance between charge centres and also strengthens attraction. A stronger lattice has a more negative formation enthalpy and a more positive dissociation enthalpy.
Compare like with like. A value of −2500 is a larger magnitude but numerically smaller than −900 kJ mol⁻¹. State “more negative lattice formation enthalpy” to avoid ambiguity. Crystal structure and stoichiometry can also influence values; radius alone is not a universal ranking rule.
| Comparison | Expected trend | Reason |
|---|---|---|
| NaF versus NaI | NaF has the more negative lattice formation enthalpy. | F⁻ is smaller than I⁻; the ions are closer. |
| MgO versus NaCl | MgO has a much larger lattice enthalpy magnitude. | Mg²⁺ and O²⁻ have greater charge magnitudes. |
| LiCl to KCl | Formation becomes less negative in this comparison. | The Group 1 cation gets larger. |
What the theoretical calculation assumes
A perfect ionic model treats the particles as spherical ions with full integer charges and no covalent electron sharing. A lattice enthalpy obtained through a Born–Haber cycle uses measured thermochemical information and does not impose that perfectly ionic description. Neither value is a direct measurement of an isolated lattice-forming event.
Close agreement supports a predominantly ionic model. A substantial discrepancy suggests that the purely ionic description misses contributions to the bonding. For comparable examples, a Born–Haber formation value appreciably more negative than the ionic prediction is evidence of additional covalent character. Compare the same sign convention, units and formula amount before drawing a conclusion.
How covalent character develops
A cation attracts the electrons in a neighbouring anion. A small, highly charged cation has high charge density and a strong polarising effect. A larger anion has a more easily distorted electron cloud. Distortion towards the cation increases electron density between the nuclei, giving the bond some covalent character.
Do not say that all ionic bonding disappears, or that the cation literally transfers a second electron back. Ionic and covalent character describe a continuum. Comparing AlCl₃ with a Group 1 chloride can illustrate polarisation, but detailed molecular/solid structures need separate evidence.
Diagram placeholder
Polarisation comparison to add
Labels to include:
- Cation and anion nuclei
- Spherical anion cloud for ideal model
- Distorted cloud towards small high-charge cation
- Increased electron density between ions
The ideal diagram shows symmetrical charge clouds. The polarised diagram moves anion electron density towards the cation without changing the number of electrons or the total charge.
Worked comparison
For two illustrative salts, the theoretical and Born–Haber lattice formation values are respectively P: −810 and −825; Q: −1900 and −2210 kJ mol⁻¹. The discrepancies are 15 and 310 kJ mol⁻¹. Q shows the larger departure from the ionic model. This is evidence of greater covalent character in the comparison, not a calculated percentage of covalent bonding.
Lattice stabilisation also competes with the cost of forming ions. Successively higher metal charges may strengthen lattice formation but demand very large ionisation energies after an inner shell is reached. Compare complete formation cycles; lattice enthalpy alone cannot establish the most favourable formula. For a full thermodynamic claim, entropy and the balanced competing reaction also matter.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. Why is lattice formation more negative for LiF than LiI?Show answer
F⁻ is smaller than I⁻, bringing opposite charges closer for the same ionic charges; attraction is stronger.
Q2. Which has stronger binding: dissociation +760 or +920 kJ mol⁻¹, for otherwise comparable salts?Show answer
+920 kJ mol⁻¹: more energy is required to separate the ions.
Q3. A theoretical formation value is −1700 and a Born–Haber value −2050 kJ mol⁻¹. Interpret.Show answer
The 350 kJ mol⁻¹ discrepancy suggests extra stabilisation not described by the perfect ionic model, consistent with covalent character. It is not 350% covalent.
Q4. Which combination produces strong polarisation?Show answer
A small cation of high charge with a large, readily distorted anion. Explain the cation’s polarising power and the anion’s polarisability separately.
Q5. Why does a very negative predicted lattice enthalpy not prove that MX₃, containing M³⁺ and three X⁻ ions, is the favoured formula?Show answer
The energetic cost of forming M³⁺, including all three ionisation energies, must also be included. The total formation cycle, and ultimately Gibbs energy for balanced alternatives, determines stability.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- AQA 7405 physical chemistry specification — 3.1.8 coverage and required skills.
- Chemrevise: Thermodynamics — Coverage checklist; explanations, data exercises and quick checks on this page are original Finesse material.
- AQA June 2023 Paper 1 mark scheme — Q05, pp18–19: solution and Born–Haber cycles; report Q05, p4. Definitions and examples are independently written.
- AQA June 2023 Paper 1 examiner report — Read alongside the question-specific marking guidance; not a universal wording checklist.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.