What Lattice Enthalpy Means
When sodium and chlorine combine, the product is not a collection of NaCl molecules but a continuous three-dimensional array of Na+ and Cl− ions packed in an ordered crystal lattice. NCERT defines the governing energy term precisely: the lattice enthalpy of an ionic solid is the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions. The reference value for sodium chloride is $788\ \text{kJ mol}^{-1}$, meaning that 788 kJ must be supplied to pull one mole of solid NaCl apart into one mole of $\ce{Na+}$ ions and one mole of $\ce{Cl-}$ ions, each carried off to infinite separation.
Written as a single step, the defining process is endothermic because we are tearing the crystal apart against the coulombic attraction holding it together:
$\ce{NaCl(s) -> Na+(g) + Cl-(g)}\qquad \Delta H_{\text{lattice}} = +788\ \text{kJ mol}^{-1}$
The reverse process — gaseous ions collapsing into a crystal — releases the same magnitude of energy and is therefore strongly exothermic. This released energy, often called the lattice formation enthalpy, is what makes ionic solids stable. The process involves both the attractive forces between ions of opposite charge and the repulsive forces between ions of like charge, and because the crystal is three-dimensional it cannot be calculated directly from a simple pairwise interaction; the geometry of the lattice must be folded in.
Sign and direction of "lattice enthalpy"
The definitional lattice enthalpy (separation of the solid into gaseous ions) is positive — energy is absorbed. The lattice formation enthalpy (ions assembling into the solid) is the negative of it and is exothermic. Questions exploit this by quoting NaCl as both $+788\ \text{kJ mol}^{-1}$ and $-788\ \text{kJ mol}^{-1}$ depending on the direction written.
Read the equation arrow before you assign a sign — the number is the same, only the direction (and hence sign) changes.
The Energetics Puzzle of NaCl
There is a genuine puzzle hidden in the formation of NaCl. Making the cation costs energy: the ionisation enthalpy for $\ce{Na(g) -> Na+(g) + e-}$ is $495.8\ \text{kJ mol}^{-1}$, and ionisation is always endothermic. Making the anion releases energy, but not enough — the electron gain enthalpy for $\ce{Cl(g) + e- -> Cl-(g)}$ is only $-348.7\ \text{kJ mol}^{-1}$. Their sum, $+147.1\ \text{kJ mol}^{-1}$, is positive. If electron transfer were the whole story, the compound should not form.
NCERT resolves this directly: the positive $147.1\ \text{kJ mol}^{-1}$ is more than compensated by the enthalpy of lattice formation of NaCl(s), about $-788\ \text{kJ mol}^{-1}$. The energy released when the gaseous ions arrange themselves into the rock-salt lattice overwhelms the cost of producing those ions, so the net process is exothermic and the crystal is stable.
A qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation, and not simply by achieving an octet of electrons around the ionic species in the gaseous state.
This single insight is the reason the Born-Haber cycle exists. We cannot measure lattice enthalpy in one experiment, but we can measure or look up every other step along a path from the elements to the crystal. By conservation of energy, the path through gaseous ions and the direct path of formation must give the same overall enthalpy change.
The Five Steps of the Born-Haber Cycle
The NIOS treatment breaks the formation of NaCl from solid sodium and gaseous chlorine into five elementary steps. Each step is written as an mhchem equation with its enthalpy change. Together they form a complete route from the standard-state elements to one mole of solid NaCl.
Step (a) — Sublimation of sodium
$\ce{Na(s) -> Na(g)}\qquad \Delta H_{\text{sub}} = +108.7\ \text{kJ mol}^{-1}$
Solid metallic sodium is converted to free gaseous atoms. This is endothermic, as energy is needed to overcome metallic bonding.
Step (b) — Ionisation of gaseous sodium
$\ce{Na(g) -> Na+(g) + e-}\qquad \Delta H_{\text{ion}} = +493.8\ \text{kJ mol}^{-1}$
An electron is removed from the gaseous atom. Ionisation is always endothermic and is the largest of the "atom-making" costs here.
Step (c) — Dissociation of chlorine
$\ce{1/2 Cl2(g) -> Cl(g)}\qquad \Delta H_{\text{diss}} = +120.9\ \text{kJ mol}^{-1}$
Half a mole of $\ce{Cl2}$ is split to give one mole of chlorine atoms; only half the bond dissociation enthalpy is required because we need only one Cl per NaCl unit.
Step (d) — Electron gain by chlorine
$\ce{Cl(g) + e- -> Cl-(g)}\qquad \Delta H_{\text{eg}} = -379.5\ \text{kJ mol}^{-1}$
A gaseous chlorine atom accepts the electron released in step (b). This electron-gain step is exothermic.
Step (e) — Lattice formation
$\ce{Na+(g) + Cl-(g) -> NaCl(s)}\qquad \Delta H_{\text{lattice form}} = -754.8\ \text{kJ mol}^{-1}$
The gaseous ions condense into the crystal. This is the lattice formation enthalpy and is the single largest energy release in the cycle; it is the term the Born-Haber cycle is usually solved for.
| Step | Process | Enthalpy term | Value / kJ mol⁻¹ | Sign |
|---|---|---|---|---|
| a | Na(s) → Na(g) | Sublimation, $\Delta H_{\text{sub}}$ | +108.7 | Endothermic |
| b | Na(g) → Na⁺(g) + e⁻ | Ionisation, $\Delta H_{\text{ion}}$ | +493.8 | Endothermic |
| c | ½Cl₂(g) → Cl(g) | Dissociation, $\Delta H_{\text{diss}}$ | +120.9 | Endothermic |
| d | Cl(g) + e⁻ → Cl⁻(g) | Electron gain, $\Delta H_{\text{eg}}$ | −379.5 | Exothermic |
| e | Na⁺(g) + Cl⁻(g) → NaCl(s) | Lattice formation, $\Delta H_{\text{lattice form}}$ | −754.8 | Exothermic |
The net reaction is the direct formation of NaCl from its elements in their standard states:
$\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}\qquad \Delta H_f = -410.9\ \text{kJ mol}^{-1}$
The cycle assumes complete electron transfer — the idealised picture explained in Ionic (Electrovalent) Bond.
Assembling the Cycle
The five steps are best visualised as a closed thermochemical cycle. Starting from the elements, one path goes directly down to NaCl(s) with enthalpy $\Delta H_f$. The alternative path climbs up through sublimation, dissociation, ionisation and electron gain to reach the gaseous ions, then drops down to the crystal via lattice formation. Because enthalpy is a state function, both paths must arrive at the same point with the same total energy change — this is Hess's law.
Hess's law lets us write the formation enthalpy as the algebraic sum of every step. With $L$ standing for the lattice formation enthalpy:
$$\Delta H_f = \Delta H_{\text{sub}} + \tfrac{1}{2}\Delta H_{\text{diss}} + \Delta H_{\text{ion}} + \Delta H_{\text{eg}} + \Delta H_{\text{lattice form}}$$Of the five terms, the NIOS source notes that sublimation and dissociation generally have small values, so the three terms that genuinely control whether an ionic compound forms are the ionisation energy, the electron affinity (electron gain enthalpy) and the lattice energy.
Worked Calculation for NaCl
With every step pinned to a number, the Born-Haber cycle becomes simple arithmetic. The most common NEET-style task is to confirm the formation enthalpy from the five steps, or to invert the cycle and solve for the lattice term when $\Delta H_f$ is given.
Using the five Born-Haber steps for NaCl, compute the standard enthalpy of formation $\Delta H_f$ of NaCl(s).
Add the enthalpy changes of all five steps (the dissociation value already corresponds to ½Cl₂ → Cl):
$\Delta H_f = (+108.7) + (+493.8) + (+120.9) + (-379.5) + (-754.8)$
$\Delta H_f = (108.7 + 493.8 + 120.9) - (379.5 + 754.8) = 723.4 - 1134.3$
$\Delta H_f \approx -410.9\ \text{kJ mol}^{-1}$
The large negative result confirms that NaCl forms spontaneously in enthalpy terms, driven overwhelmingly by the lattice formation step. The endothermic ionisation cost is paid back many times over once the lattice closes.
If only $\Delta H_f = -410.9\ \text{kJ mol}^{-1}$ and the four other terms were known, how would you obtain the lattice formation enthalpy?
Rearrange the Hess-law sum to isolate the lattice term:
$\Delta H_{\text{lattice form}} = \Delta H_f - \big(\Delta H_{\text{sub}} + \tfrac{1}{2}\Delta H_{\text{diss}} + \Delta H_{\text{ion}} + \Delta H_{\text{eg}}\big)$
$= -410.9 - (108.7 + 493.8 + 120.9 - 379.5) = -410.9 - 343.9 = -754.8\ \text{kJ mol}^{-1}$
This is exactly how lattice enthalpy is determined in practice — indirectly, because the direct separation of the crystal into gaseous ions cannot be carried out and measured in a single experiment.
The factor of ½ on chlorine
Examiners often quote the full bond dissociation enthalpy of $\ce{Cl2}$ and expect you to halve it before inserting it into the cycle, because each formula unit of NaCl needs only one Cl atom. Forgetting the $\tfrac{1}{2}$ doubles the dissociation contribution and corrupts the answer.
For a 1:1 ionic solid MX from a diatomic non-metal, always use ½ × (bond dissociation enthalpy of X₂).
Factors Affecting Lattice Enthalpy
Lattice enthalpy is fundamentally a coulombic quantity — it tracks the strength of electrostatic attraction between the ions in the crystal. Two ionic properties dominate its magnitude, and a third structural factor sets the geometric context.
| Factor | Effect on lattice enthalpy | Reasoning |
|---|---|---|
| Ionic charge | Increases sharply with higher charge | Coulombic attraction is proportional to the product of the ionic charges; doubly charged ions bind far more strongly than singly charged ones. |
| Ionic size | Increases as ions get smaller | Attraction grows as the inter-ionic distance shrinks; smaller cations and anions sit closer, raising the lattice enthalpy. |
| Crystal geometry | Sets the summation of attractions and repulsions | The three-dimensional packing fixes how many neighbours of each charge surround a given ion, so geometry must be included rather than a single pairwise term. |
These same factors decide whether an ionic compound forms at all. The NIOS source summarises the favourable conditions as a low ionisation energy of the metal, a high electron affinity of the non-metal, and a high lattice energy. A large lattice enthalpy is the engine that pays back the endothermic ionisation cost, which is why compounds of small, highly charged ions are typically the most thermally stable.
The interplay with ion size also governs covalent (polarising) character. A small, highly charged cation distorts the electron cloud of a large anion, introducing covalent character into a nominally ionic bond — the basis of the order of ionic character probed in NEET 2018 (BeH₂ < CaH₂ < BaH₂).
Lattice enthalpy and the Born-Haber cycle in one screen
- Lattice enthalpy = energy to separate one mole of an ionic solid into gaseous ions at infinite distance; for NaCl it is $788\ \text{kJ mol}^{-1}$.
- Ionic compounds form because the lattice formation enthalpy ($\approx -788\ \text{kJ mol}^{-1}$ for NaCl) outweighs the positive sum of ionisation and electron gain ($+147.1\ \text{kJ mol}^{-1}$).
- The Born-Haber cycle = sublimation + dissociation + ionisation + electron gain + lattice formation, summed by Hess's law.
- For NaCl the steps sum to $\Delta H_f \approx -410.9\ \text{kJ mol}^{-1}$; lattice enthalpy is found by inverting this sum.
- Lattice enthalpy rises with higher ionic charge and smaller ionic size; ionisation energy, electron affinity and lattice energy are the decisive terms.