Ionic bonding and the Born-Haber cycle

Result

Ionic bonding and the Born-Haber cycle

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Statement

The lattice energy of an ionic solid — the energy released when gaseous ions come together to form the crystal, \(M^{n+}(g)+X^{n-}(g)\to MX(s)\) — cannot be measured directly, but it can be extracted exactly from measurable quantities by writing an indirect thermodynamic path from elements to the solid and applying Hess's law: the enthalpy change for a reaction is the same regardless of the path taken, because enthalpy is a state function.

Why it matters

Lattice energy is the single largest energy term governing whether an ionic compound forms at all and how stable it is once formed — it is what pays back the substantial energy cost of first ionising a metal atom (always endothermic) and, for many nonmetals, forming the anion. Without a large, favourable lattice energy, no ionic compound would be thermodynamically stable relative to its separate elements. The Born-Haber cycle is also one of the clearest classroom demonstrations that Hess's law (this unit's CU-106 result) is not just a bookkeeping trick but a genuine consequence of enthalpy being a state function: a quantity that cannot be measured in one step (lattice energy, since ions never actually assemble from separately-observable gaseous ions under normal conditions) is measured instead by closing a loop of five steps that all individually can be measured.

The same lattice-energy values, once extracted this way, are what explain why compounds like MgO (formed from doubly-charged ions) are dramatically more stable and higher-melting than singly-charged NaCl — a direct, checkable link to the \(1/r\) and \(q_1q_2\) dependence of Coulomb's law.

Hypotheses

Enthalpy is a state function.This is the entire content of Hess's law: the total enthalpy change from reactants to products does not depend on the path taken, only on the initial and final states. Without this, breaking the overall formation reaction into an arbitrary five-step path and summing the steps would tell us nothing about the direct reaction. Each intermediate species (gaseous metal atoms, gaseous ions) is treated as an ideal, isolated species in the gas phase, with well-defined standard enthalpies of sublimation, ionisation, dissociation, and electron affinity.Real solids and gases deviate slightly from ideal behaviour, and the tabulated thermodynamic quantities used in the cycle are themselves experimentally measured averages (e.g. bond dissociation enthalpy is temperature- and state-dependent). The cycle's accuracy is therefore bounded by the accuracy of these input values, not by any additional approximation in the cycle logic itself, which is exact given the inputs.

Proof

M(s) \xrightarrow{\Delta H_{\text{sub}}} M(g)

Sublimation: convert the solid metal directly to gaseous atoms. This is the standard enthalpy of sublimation (or, if starting from a diatomic nonmetal element, the enthalpy of atomisation), always endothermic (energy must be supplied to separate atoms from a solid or break bonds). A

M(g) \xrightarrow{\text{IE}_1,\ \text{IE}_2,\ \dots} M^{n+}(g) + ne^-

Successive ionisation energies remove \(n\) electrons one at a time to reach the metal cation's actual charge; each successive ionisation energy is larger than the last (removing an electron from an increasingly positively-charged species costs more), and all are endothermic. A

\tfrac{1}{2}X_2(g) \xrightarrow{\frac{1}{2}D_{X_2}} X(g)

Atomise the nonmetal: break one mole of \(X_2\) bonds and take half (matching the stoichiometry of one \(X\) atom per formula unit), always endothermic. For a nonmetal that is not naturally diatomic (e.g. solid iodine, or oxygen requiring both atomisation and this step combined with sublimation), the corresponding atomisation enthalpy is used directly in its place. A

X(g) + ne^- \xrightarrow{\text{EA}_1,\ \text{EA}_2,\ \dots} X^{n-}(g)

Successive electron affinities add \(n\) electrons to reach the anion's actual charge. The first electron affinity is typically exothermic (an isolated neutral atom generally has a favourable pull for one additional electron); each subsequent electron affinity is strongly endothermic, since adding an electron to an already-negative ion means pushing it against net electrostatic repulsion from the existing extra charge. A

M^{n+}(g) + X^{n-}(g) \xrightarrow{U} MX(s)

Lattice formation: the gaseous ions come together into the ordered crystal lattice, releasing the lattice energy \(U\) — large and exothermic, since it is dominated by the strong Coulombic attraction between oppositely-charged ions now held at their equilibrium separation throughout an extended, highly ordered structure, not just a single ion pair. This is the unknown quantity the entire cycle is built to isolate. A

\Delta H_f = \Delta H_{\text{sub}} + \sum\text{IE} + \tfrac{1}{2}D_{X_2} + \sum\text{EA} + U

Hess's law (Hypotheses): the five-step indirect path (Steps 1–5) and the single-step direct formation reaction \(M(s)+\tfrac{1}{2}X_2(g)\to MX(s)\), \(\Delta H_f\), share the same start and end states, so their enthalpy changes must be equal. This equates the sum of all five measurable steps to the one directly measurable quantity, \(\Delta H_f\) (the standard enthalpy of formation, tabulated for essentially every stable ionic compound). A

U = \Delta H_f - \left(\Delta H_{\text{sub}} + \sum\text{IE} + \tfrac{1}{2}D_{X_2} + \sum\text{EA}\right)

Solve Step 6 for the one unknown, \(U\), since every other quantity on the right-hand side is independently measurable (sublimation and atomisation enthalpies calorimetrically, ionisation energies spectroscopically, electron affinities from related spectroscopic and thermodynamic measurements, and \(\Delta H_f\) calorimetrically). A

Result

U = \Delta H_f - \Delta H_{\text{sub}} - \sum\text{IE} - \tfrac{1}{2}D_{X_2} - \sum\text{EA}

Reading. The lattice energy of an ionic solid is obtained entirely from five independently measurable quantities, by demanding that a five-step indirect thermodynamic path agree with the single-step direct formation enthalpy — a direct application of Hess's law to a quantity with no direct experimental route.

Scope. Applies to any binary ionic solid \(MX_n\) (or, with the natural generalisation to matched stoichiometric coefficients, \(M_pX_q\)) formed from a solid metallic element and a nonmetal element in its standard state, given tabulated sublimation/atomisation, ionisation, bond dissociation, electron affinity, and formation data.

Corollaries & converses

Lattice energies extracted this way (the "experimental" or thermochemical lattice energy) can be compared against lattice energies computed independently from a purely electrostatic (Born-Landé type) ionic model; close agreement supports the ionic-bonding picture for a given compound, while a large discrepancy is itself evidence of significant covalent character in the bonding (as famously found for compounds like silver halides).
Since \(U\) scales (to leading order, via Coulomb's law) with the product of ionic charges and inversely with the sum of ionic radii, doubly-charged ion pairs (e.g. \(\text{Mg}^{2+}\)/\(\text{O}^{2-}\)) have dramatically larger-magnitude lattice energies than singly-charged pairs of comparable size (e.g. \(\text{Na}^+\)/\(\text{Cl}^-\)) — consistent with, and explaining, the much higher melting point of MgO (2852°C) compared with NaCl (801°C).
Converse: given \(U\) from an independent source (e.g. a Born-Landé calculation) together with any four of the five cycle steps, the fifth (most often an electron affinity, the hardest of the five quantities to measure directly) can be solved for instead — historically, this is exactly how many electron affinities were first determined.

Fails without

Drop Hess's law (enthalpy as a state function): without it, there is no reason the five-step path should sum to the same value as the direct formation reaction, and Step 6 — the entire basis for solving for \(U\) — has no justification. The cycle would be an arbitrary sequence of reactions with no guaranteed relationship to \(\Delta H_f\).
Omit a step, or double-count a stoichiometric factor (e.g. forgetting the \(\tfrac{1}{2}\) on a diatomic nonmetal, or forgetting a second ionisation energy for a \(2+\) cation): the cycle no longer closes — the indirect path's start and end states no longer exactly match the direct reaction's — and the extracted \(U\) is wrong by exactly the value of the omitted or miscounted term. This is the single most common source of error in applying the cycle (see Common errors).

Common errors

Forgetting the factor of \(\tfrac{1}{2}\) on the nonmetal bond-dissociation term when the nonmetal is diatomic (e.g. using the full \(\text{Cl}\)–\(\text{Cl}\) bond enthalpy of \(244\,\text{kJ/mol}\) instead of half of it, \(122\,\text{kJ/mol}\), for one mole of \(\text{Cl}\) atoms).
Using only the first ionisation energy for a metal that forms a \(2+\) or \(3+\) cation, omitting the second (or third) ionisation energy entirely.
Sign errors: every step in the cycle has a definite, physically-determined sign (sublimation, ionisation, and bond dissociation are always endothermic/positive; the first electron affinity is usually exothermic/negative but later ones are endothermic/positive; lattice energy is exothermic/negative). Reversing any one sign in Step 6 or Step 7 propagates directly into an incorrect \(U\).
Confusing lattice energy (defined here as the enthalpy of the exothermic association reaction \(M^{n+}(g)+X^{n-}(g)\to MX(s)\), a negative quantity) with lattice enthalpy of dissociation (the reverse process, a positive quantity of the same magnitude) — both conventions appear in different textbooks, so the sign convention in use must always be stated.

Discussion

The cycle is named for Max Born and Fritz Haber, who independently developed it around 1919 as part of the broader effort (alongside Born's and Alfred Landé's purely electrostatic lattice-energy calculations) to test whether the still-new ionic-bonding picture of solids like NaCl was quantitatively correct. The historical significance runs in both directions: the cycle gives a thermochemical route to \(U\) with no assumptions about the detailed structure of the ionic bond, while the separately-computed Born-Landé electrostatic value gives a route to \(U\) built entirely on the assumption of a purely ionic bond — agreement between the two, to within a few percent for compounds like the alkali halides, was strong early evidence that the ionic model itself was essentially correct for those compounds.

Where the two methods disagree substantially, the discrepancy is itself informative: silver halides (AgCl, AgBr, AgI) show thermochemical lattice energies noticeably larger in magnitude than the pure electrostatic prediction, a signature of additional covalent (partial charge-sharing) character in the bonding beyond the simple point-charge ionic model — a theme that reappears in this unit's electronegativity-polarity result, and is treated in more detail there.

Common misconception: that the Born-Haber cycle somehow calculates or predicts the lattice energy from first principles. It does not — it is purely an application of Hess's law to already-known, independently measured quantities, extracting the one remaining unknown from a closed loop. The cycle contains no new physical assumption beyond Hess's law itself; the physics of why lattice energies are large and negative (Coulomb attraction between ions in an ordered array) is a separate topic (the electrostatic, Born-Landé-type calculation), only cross-checked against, not derived from, the cycle presented here.

Worked examples

\text{NaCl: } \Delta H_{\text{sub}}(\text{Na})=+107,\ \text{IE}_1(\text{Na})=+496,\ \tfrac12D_{\text{Cl}_2}=+122,\ \text{EA}(\text{Cl})=-349,\ \Delta H_f(\text{NaCl})=-411\ \text{kJ/mol}

All five quantities in \(\text{kJ/mol}\), standard values for the classic textbook example. A

U = -411 - (107+496+122-349) = -411-376 = -787\,\text{kJ/mol}

Applying the Result directly. The large negative value confirms that lattice formation is strongly exothermic, easily paying back the combined endothermic cost of sublimation and ionisation (\(107+496=603\,\text{kJ/mol}\)) and then some, which is exactly why solid NaCl is thermodynamically stable relative to separated \(\text{Na}(s)\) and \(\tfrac12\text{Cl}_2(g)\). A

U(\text{NaCl}) \approx -787\,\text{kJ/mol}

Reading. A single application of the Result, using five independently tabulated quantities, reproduces the well-established experimental lattice energy of sodium chloride.

Scope. The identical five-term calculation applies to any \(1{:}1\) ionic solid with tabulated sublimation, ionisation, dissociation, electron-affinity, and formation data.

Problems

For KCl: \(\Delta H_{\text{sub}}(\text{K})=+89\), \(\text{IE}_1(\text{K})=+419\), \(\tfrac12D_{\text{Cl}_2}=+122\), \(\text{EA}(\text{Cl})=-349\), \(\Delta H_f(\text{KCl})=-436\) (all \(\text{kJ/mol}\)). Find the lattice energy of KCl.
Solution
\(U=\Delta H_f-(\Delta H_{\text{sub}}+\text{IE}_1+\tfrac12D_{\text{Cl}_2}+\text{EA})=-436-(89+419+122-349)=-436-281=-717\,\text{kJ/mol}\). Slightly less negative (smaller magnitude) than NaCl's \(-787\,\text{kJ/mol}\), consistent with potassium's larger ionic radius increasing the \(\text{K}^+\)–\(\text{Cl}^-\) separation and weakening the Coulombic attraction, exactly as this unit's periodic-trends result predicts for ionic radius increasing down a group.
For MgO: \(\Delta H_{\text{sub}}(\text{Mg})=+148\), \(\text{IE}_1(\text{Mg})=+738\), \(\text{IE}_2(\text{Mg})=+1451\), \(\tfrac12D_{\text{O}_2}=+249\), \(\text{EA}_1(\text{O})=-141\), \(\text{EA}_2(\text{O})=+744\), \(\Delta H_f(\text{MgO})=-602\) (all \(\text{kJ/mol}\)). Find the lattice energy of MgO, and comment on why it is so much larger in magnitude than NaCl's.
Solution
Sum of endothermic/exothermic non-lattice steps: \(148+738+1451+249-141+744=3189\,\text{kJ/mol}\). \(U=\Delta H_f-3189=-602-3189=-3791\,\text{kJ/mol}\) — nearly five times NaCl's magnitude. This follows directly from the Corollaries: Coulomb's law scales lattice energy with the product of ionic charges, and MgO's ions are each doubly charged (\(\text{Mg}^{2+}\), \(\text{O}^{2-}\)) versus NaCl's singly-charged ions, a factor-of-four effect on the charge product alone, further amplified by MgO's smaller ionic radii (shorter ion–ion separation).
Explain, without doing any arithmetic, why the second electron affinity of oxygen, \(\text{O}^-(g)+e^-\to\text{O}^{2-}(g)\), is strongly endothermic (unfavourable), even though the first electron affinity of oxygen is exothermic.
Solution
The first electron affinity adds an electron to a neutral oxygen atom, which has a genuine electrostatic pull for one more electron (favourable, exothermic). The second electron affinity adds an electron to \(\text{O}^-\), an ion that is already negatively charged; the incoming electron is repelled by the net negative charge already present, so energy must be supplied to force the second electron on — unfavourable, endothermic. This is a general pattern: every electron affinity beyond the first, for any element, is endothermic for exactly this reason.
A student sets up a Born-Haber cycle for CaCl₂ but uses only the first ionisation energy of calcium and only a single (non-doubled) bond-dissociation term for chlorine. Identify both errors and state, in words, how each would change the calculated lattice energy compared with the correct value.
Solution
Error 1: \(\text{Ca}\) forms \(\text{Ca}^{2+}\), so both \(\text{IE}_1\) and \(\text{IE}_2\) must be included; omitting \(\text{IE}_2\) makes the summed endothermic cost too small, which (by the Result, \(U=\Delta H_f-\text{sum}\)) makes the calculated \(U\) too negative (an artificially large-magnitude, incorrect lattice energy). Error 2: \(\text{CaCl}_2\) requires two moles of \(\text{Cl}\) atoms per mole of \(\text{CaCl}_2\), so the bond-dissociation term must be the full \(D_{\text{Cl}_2}\) (not halved, since one full \(\text{Cl}_2\) molecule supplies exactly the two \(\text{Cl}\) atoms needed) — using only a single non-doubled term as stated understates this endothermic cost as well, compounding the same direction of error as Error 1: the calculated \(U\) comes out more negative than the true lattice energy.