Explain why the Born–Haber cycle for potassium fluoride gives a value for lattice enthalpy that differs from the value calculated using a purely ionic model, and state what this difference indicates about the bonding in potassium fluoride.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The experimental lattice enthalpy of potassium fluoride, derived from its Born–Haber cycle, is −826 kJ mol⁻¹, whereas the theoretical value calculated assuming a perfect ionic model is −808 kJ mol⁻¹.
Model answer (4 marks)
The Born–Haber cycle value is called the experimental lattice enthalpy because it is obtained from measurable thermodynamic data using Hess’s law, whereas the theoretical value is calculated by assuming perfectly spherical, non‑polarising ions.
The experimental value (−826 kJ mol⁻¹) is larger in magnitude than the theoretical value (−808 kJ mol⁻¹), so more energy is released when the lattice forms than the purely ionic model predicts.
The discrepancy arises because the real bonding has some covalent character: the small, highly charged F⁻ ion polarises the electron cloud of the K⁺ ion, giving a slight sharing of electron density.
Thus the fact that the experimental lattice enthalpy exceeds the theoretical value indicates that potassium fluoride is predominantly ionic but contains a small degree of covalent character.
The experimental value (−826 kJ mol⁻¹) is larger in magnitude than the theoretical value (−808 kJ mol⁻¹), so more energy is released when the lattice forms than the purely ionic model predicts.
The discrepancy arises because the real bonding has some covalent character: the small, highly charged F⁻ ion polarises the electron cloud of the K⁺ ion, giving a slight sharing of electron density.
Thus the fact that the experimental lattice enthalpy exceeds the theoretical value indicates that potassium fluoride is predominantly ionic but contains a small degree of covalent character.
Examiner tips
- Use the word ‘experimental’ to describe the Born–Haber value and ‘theoretical’ for the ionic model; this shows you understand the difference. Include the sign and magnitude of the difference. Mention polarisation of the F⁻ ion to explain the extra energy. Finish by stating the bonding character implied.
Common mistakes
- Confusing the experimental and theoretical values or giving the wrong sign. Forgetting to mention polarisation or the covalent contribution. Writing that the experimental value is smaller rather than larger in magnitude.
Mark scheme (4 marks)
- The Born–Haber cycle value is described as the experimental lattice enthalpy because it is derived from measurable thermodynamic data (Hess's law), whereas the theoretical value assumes perfectly spherical, non-polarising ions.
- The experimental value is larger in magnitude than the theoretical value, meaning more energy is released when the lattice forms than the purely ionic model predicts.
- The discrepancy arises because the real bonding has some covalent character: the F⁻ ion (small, high charge density) polarises the electron cloud of the K⁺ ion to a slight degree, causing partial sharing of electron density.
- The fact that the experimental lattice enthalpy exceeds the theoretical value indicates that potassium fluoride has predominantly ionic bonding with a small degree of covalent character.
Key terms in this question
Born–Haber cycle · lattice enthalpy · ionic model
Related
- All IB DP Chemistry Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →