Construct Born–Haber cycles
Build an enthalpy cycle from elements to an ionic solid through gaseous atoms and ions, using the correct signs and stoichiometry.
Section 23 · Chemical energetics
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 23. Use the syllabus sign convention for lattice enthalpy (gaseous ions forming the ionic solid); state phases, signs, units and temperature. Written practice is self-checked
Before you begin
- Use AS enthalpy changes and Hess's law, formulae for ions, balanced equations, and temperature in kelvins.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What law lets you equate direct formation with a multi-step Born–Haber route?
What you will learn
- Arrange formation, atomisation, bond dissociation, ionisation and electron-affinity steps
- Use the correct ion charges and formula-unit ratios
- Check that a Born–Haber cycle returns to the stated formation equation
Follow a route through gaseous atoms and ions
A Born–Haber cycle applies Hess's law to formation of an ionic solid. One route is the direct standard enthalpy of formation from the elements. The alternative route atomises the elements, ionises the metal atoms, adds electrons to the non-metal atoms, then forms the lattice from gaseous ions.
For a diatomic non-metal, atomising one mole of atoms requires half a mole of X₂. For an MCl₂ formula unit, include two chlorine atomisation steps, two first electron affinities and the correct number of metal ionisation steps to reach the stated ion charge.
Close the cycle and preserve signs
For NaCl, the gaseous-ion route contains Na atomisation, half Cl₂ atomisation, Na first ionisation, Cl first electron affinity, and lattice formation. Adding those steps must equal ΔHf° for NaCl(s). The lattice arrow is gas ions → solid, so use the formation lattice enthalpy sign.
Before calculating, check that both routes start and finish at the same states. The syllabus limits Born–Haber cycles to +1 and +2 cations and −1 and −2 anions; keep ions within those charge limits in this pathway.
Pause and explain
How much Cl₂ is atomised to form one mole of Cl atoms in the NaCl cycle?
Worked example
List the steps in a Born–Haber cycle for NaCl(s) and state which step is the lattice energy.
Show the worked solution
- Start with Na(s) and ½Cl₂(g) in their standard states.
- Form gaseous atoms, ionise Na, and add an electron to Cl.
- Join Na⁺(g) and Cl⁻(g) to make NaCl(s).
Answer The route is Na atomisation, ½Cl₂ atomisation, Na(g) → Na⁺(g) + e⁻, Cl(g) + e⁻ → Cl⁻(g), then Na⁺(g) + Cl⁻(g) → NaCl(s). The last step is lattice formation.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
List the gaseous-ion steps needed to form CaCl₂(s) from Ca(s) and Cl₂(g).
Give me a hint
Track one calcium atom and two chlorine atoms to the ions in one formula unit.
Compare my reasoning
- Atomise Ca and one mole of Cl₂ to make Ca(g) and 2Cl(g).
- Remove two electrons from Ca; add one electron to each Cl atom.
- Form CaCl₂(s) from Ca²⁺(g) and 2Cl⁻(g).
Atomise Ca(s) and Cl₂(g), ionise Ca(g) twice to Ca²⁺(g), add one electron to each of two Cl(g) atoms, then form CaCl₂(s) from Ca²⁺(g) + 2Cl⁻(g).
Look for these in your work
- Atom counts correct
- Two ionisation steps/electrons tracked
- Lattice ion ratio correct
Why does a Born–Haber cycle for MgO require both first and second electron-affinity steps for oxygen?
Give me a hint
Follow the charge needed to reach O²⁻(g).
Compare my reasoning
- O(g) first gains one electron to form O⁻(g).
- A second electron is added to O⁻(g).
- The ion in the MgO formula is O²⁻, so both steps are needed.
The cycle must form O²⁻(g). It therefore includes two sequential electron additions: O(g) → O⁻(g), then O⁻(g) → O²⁻(g). The second addition can be endothermic because the electron is added to a negative ion.
Look for these in your work
- Final ion charge checked
- Both electron additions included
- Second-step repulsion recognised
A proposed NaCl cycle contains one mole of Cl₂ atomisation and ends with Na⁺(g) + Cl⁻(g) → NaCl(s). Find the mismatch.
Give me a hint
Count chlorine atoms on the direct route and the gas-ion route.
Compare my reasoning
- The formula unit contains one chlorine atom.
- One mole of Cl₂ contains two moles of chlorine atoms.
- Use half a mole of Cl₂ to produce one mole of Cl atoms.
The cycle atomises too much chlorine: one mole of Cl₂ gives two moles of Cl atoms, but one mole of NaCl needs one mole. Use ½Cl₂(g) → Cl(g).
Look for these in your work
- Formula-unit chlorine count used
- Mismatch explained
- Correct coefficient stated
Common mistakes
- Using a full mole of Cl₂ when only one Cl atom is needed
- Omitting the second halogen atom in an MCl₂ formula unit
- Using the wrong ion charge or number of electrons
- Adding cycle steps without checking the final equation
What can you explain now?
What law lets you equate direct formation with a multi-step Born–Haber route?
Compare with the explanation
Hess's law
Enthalpy change depends on initial and final states, not the route.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Which atom, ion and electron steps connect a formation equation to a lattice-formation step?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.