Lattice Enthalpy and the Born–Haber Cycle
Lattice enthalpy is the number that explains why sodium chloride is a hard, high-melting solid while sodium and chlorine are a soft metal and a gas. It cannot be measured directly — you cannot take a mole of gaseous Na⁺ and Cl⁻ ions and watch them assemble — so it is obtained indirectly, from a thermochemical cycle known as the Born–Haber cycle. This page builds that cycle step by step and works it in both directions.
First, settle the sign convention
More marks are lost here than anywhere else in the topic, because "lattice enthalpy" is defined two different ways in two different sets of textbooks, and they differ by a sign:
| Name | Process | Sign | Example, NaCl |
|---|---|---|---|
| Lattice formation enthalpy | Na⁺(g) + Cl⁻(g) → NaCl(s) | negative (exothermic) | about −786 kJ mol⁻¹ |
| Lattice dissociation enthalpy | NaCl(s) → Na⁺(g) + Cl⁻(g) | positive (endothermic) | about +786 kJ mol⁻¹ |
Neither is wrong; they are the same physical quantity read in opposite directions. Many Indian textbooks define lattice enthalpy as the energy required to separate one mole of the solid into gaseous ions, so they quote it as positive. This article uses the lattice formation enthalpy, written U, so that every term in the cycle points the same way as the enthalpy of formation. Whichever convention a question uses, state it in your answer — an examiner cannot penalise a stated convention, but can penalise an ambiguous sign.
The cycle for NaCl
The Born–Haber cycle is Hess's law applied to ionic compound formation. There are two routes from the elements in their standard states to the solid: the direct one, whose enthalpy change is ΔfH, and a long way round through gaseous ions.
| Step | Process | Quantity | Sign |
|---|---|---|---|
| 1 | Na(s) → Na(g) | enthalpy of atomisation / sublimation | + |
| 2 | Na(g) → Na⁺(g) + e⁻ | first ionisation enthalpy | + |
| 3 | ½Cl₂(g) → Cl(g) | half the bond dissociation enthalpy | + |
| 4 | Cl(g) + e⁻ → Cl⁻(g) | electron gain enthalpy | − (for a halogen) |
| 5 | Na⁺(g) + Cl⁻(g) → NaCl(s) | lattice formation enthalpy U | − |
Only one unknown appears in that equation — U — and every other term can be measured. That is the entire trick.
Electron affinity or electron gain enthalpy?
A second convention clash lives inside step 4. Electron gain enthalpy ΔegH is an enthalpy change and is negative when energy is released, so for chlorine it is about −349 kJ mol⁻¹. Electron affinity is traditionally quoted as the energy released, and so appears as +349 kJ mol⁻¹ for the same process. If your data table gives a positive number for a halogen, it is an electron affinity and you must subtract it where this cycle adds ΔegH.
Worked example 1 — lattice enthalpy of NaCl
Given, all in kJ mol⁻¹:
ΔfH(NaCl, s) = −411.2 · ΔsubH(Na) = +107.3 ·
IE₁(Na) = +495.8 · D(Cl–Cl) = +242 ·
ΔegH(Cl) = −349
Step 1 — halve the bond enthalpy. The equation forms one mole of NaCl
and so needs one mole of Cl atoms, from half a mole of Cl₂:
½D = 242 ÷ 2 = +121 kJ mol⁻¹
Step 2 — add the four measurable steps.
107.3 + 495.8 = 603.1
603.1 + 121 = 724.1
724.1 + (−349) = +375.1 kJ mol⁻¹
Step 3 — rearrange for U.
U = ΔfH − 375.1 = −411.2 − 375.1 = −786.3 kJ mol⁻¹
Reported the other way round, the lattice dissociation enthalpy of NaCl is +786.3 kJ mol⁻¹. Tables generally quote a value near 787 kJ mol⁻¹; small differences come from the bond enthalpy and electron gain enthalpy used.
Sanity check on the sign. Forming a lattice from separated ions must release energy — oppositely charged ions attract. A positive U here would mean an arithmetic error.
Worked example 2 — MgCl₂, where the counting gets harder
Given, all in kJ mol⁻¹:
ΔfH(MgCl₂, s) = −641.3 · ΔsubH(Mg) = +147.7 ·
IE₁(Mg) = +737.7 · IE₂(Mg) = +1450.7 · D(Cl–Cl) = +242 ·
ΔegH(Cl) = −349
Three things change compared with NaCl. Magnesium loses two electrons, so both ionisation enthalpies are needed. Two chloride ions are formed, so the electron gain enthalpy is counted twice. And two chlorine atoms require a whole Cl₂ molecule, so the full bond enthalpy is used, not half of it.
Ionisation: 737.7 + 1450.7 = 2188.4
Electron gain: 2 × (−349) = −698
Sum of the measurable steps:
147.7 + 2188.4 = 2336.1
2336.1 + 242 = 2578.1
2578.1 + (−698) = +1880.1 kJ mol⁻¹
Lattice formation enthalpy:
U = −641.3 − 1880.1 = −2521.4 kJ mol⁻¹
Notice how much larger this is than NaCl's −786 kJ mol⁻¹, even though the ionisation of magnesium is enormously expensive. The doubly charged Mg²⁺ ion binds the lattice far more strongly, and that is what pays for the second ionisation enthalpy and makes MgCl₂ stable at all.
Worked example 3 — running the cycle backwards
The cycle has one unknown, and it does not have to be U. Any single missing term can be found the same way — electron gain enthalpies of the halogens were historically obtained exactly like this.
Given the lattice formation enthalpy of NaCl as −786 kJ mol⁻¹, together with ΔfH = −411.2, ΔsubH(Na) = +107.3, IE₁(Na) = +495.8 and ½D(Cl₂) = +121, find ΔegH(Cl).
Rearranging the cycle:
ΔegH = ΔfH − [ΔsubH + IE₁ + ½D + U]
Inside the bracket: 107.3 + 495.8 = 603.1; 603.1 + 121 = 724.1; 724.1 + (−786) = −61.9
ΔegH = −411.2 − (−61.9) = −411.2 + 61.9 = −349.3 kJ mol⁻¹
Which is the accepted value for chlorine — the cycle is self-consistent ✔
What makes a lattice enthalpy large
The electrostatic models of ionic bonding, such as the Born–Landé and Kapustinskii expressions, all reduce to the same proportionality:
So lattice enthalpy grows with the product of the ionic charges and falls as the ions get bigger. Charge dominates, because it enters as a product while size enters as a sum in the denominator.
| Compound | Charges | Approximate lattice enthalpy | Reason |
|---|---|---|---|
| NaCl | 1 × 1 | about −787 kJ mol⁻¹ | singly charged, medium ions |
| MgCl₂ | 2 × 1 | about −2520 kJ mol⁻¹ | doubly charged cation, and two anions per formula unit |
| MgO | 2 × 2 | roughly −3800 kJ mol⁻¹ | charge product of 4 and two small ions |
This is why MgO melts far above 2000 °C while NaCl melts near 800 °C, and why lattice enthalpy decreases steadily down group 1 as the cation grows: LiCl to CsCl, same charges, bigger ion, weaker lattice.
A calculated lattice enthalpy from a purely ionic model can also be compared with the Born–Haber value. Where the two agree, as for NaCl, the bonding really is close to ionic. Where the Born–Haber value is much larger — silver iodide is the standard example — the extra stability signals covalent character, which is the experimental basis for Fajans' rules.
Common mistakes that cost marks
- Sign of the lattice enthalpy. Decide at the start whether you are quoting the formation or the dissociation value, write it down, and stay consistent.
- Forgetting to halve the bond enthalpy. NaCl needs one Cl atom, so ½D. MgCl₂ needs two, so the full D. Count the atoms in the formula.
- Using only the first ionisation enthalpy for a group 2 metal. Mg²⁺ requires IE₁ + IE₂, and Al³⁺ requires all three.
- Counting the electron gain enthalpy once for MgCl₂. Two chloride ions are made, so it is counted twice.
- Mixing electron affinity and electron gain enthalpy. They differ in sign; check whether your table's halogen value is positive or negative.
- Putting a solid or liquid inside the cycle. Every intermediate in a Born–Haber cycle is a gaseous atom or ion. That is what makes the steps well-defined.
- Explaining a lattice enthalpy trend by size alone. Charge is the stronger factor. Compare NaCl with MgO before you reach for ionic radii.
Where this appears in exams
| Level | Typical question |
|---|---|
| CBSE/ICSE Class 11 | Draw and label a Born–Haber cycle for NaCl; calculate the lattice enthalpy |
| JEE/NEET | MgCl₂ and CaO cycles, trends in lattice enthalpy down a group, solubility arguments |
| IIT-JAM / CUET-PG | Enthalpy of solution as lattice plus hydration; why some ionic solids do not dissolve |
| GATE / CSIR-NET | Kapustinskii-type estimates, covalent character from Born–Haber discrepancies, Fajans' rules |
A Born–Haber cycle is five signed numbers added carefully, so the suite has no dedicated tool for it — and pointing you at a screen that does not fit would waste your time. Open the suite for the Scientific Calculator and work down the cycle one step at a time, writing each sign before each value.
Open the ABC Chemistry Calculator Suite →Ionisation enthalpies, electron gain enthalpies and ionic radii all sit together in the Interactive Periodic Table, which is the fastest way to look up the terms this cycle needs.
Born–Haber questions are predictable once the cycle is drawn correctly. ABC Chemistry teaches Class 11–12 chemistry at its Gurugram centre and in online batches across India, with home tuition available in Delhi-NCR — abcchemistry.in.