Lattice enthalpy: the Born–Mayer equation and its trends

Lattice enthalpy is the price of pulling a crystal into ions

Intuition

An ionic crystal is held together by plus and minus charges pulling on each other. Lattice enthalpy is how much energy you must pay to pull every ion apart until they are far from each other. A big number means the crystal is hard to break.

A purely ionic crystal is held together by electrostatic forces. The lattice enthalpy $\Delta H^0_L$ is the energy that overcomes them: NaCl(s) → Na$^+$(g) + Cl$^-$(g), with $\Delta H = \Delta H^0_L$. It uses Coulomb’s law, $V(r) = -q_1q_2/(4\pi\varepsilon_0 r)$.

For ions $A^{z+}$ and $B^{z-}$ separated by $d = r_A + r_B$, the Born–Mayer equation gives

$\Delta H^0_L = \frac{N_A\,|z_A z_B|\,e^2}{4\pi\varepsilon_0 d}\left(1 - \frac{d^*}{d}\right)A$

where $N_A$ is Avogadro’s number, $e$ the fundamental electronic charge, $z_A$ and $z_B$ the ionic charges, $\varepsilon_0$ the vacuum permittivity, $d^*$ a short-range repulsion constant (typically 0.345 Å) and $A$ the Madelung constant.

Exam detail

Test 1 covers Born–Mayer and lattice enthalpy trends. Know every symbol in the equation and which ones change from one compound to another in the same structure.

The Madelung constant counts every neighbour, near and far

Intuition

Stand on one ion in the lattice. The six nearest neighbours of opposite charge pull you in. The twelve next ones are the same charge and push you away a bit. Then eight more pull, and so on. The Madelung constant is the net tally.

The Madelung constant $A$ depends on the crystal structure. It is the sum of the Coulomb attraction between each ion and successive coordination spheres. For NaCl, the first three spheres give $V = -6e^2|z_Az_B|/d$, then $V = +12e^2|z_Az_B|/(\sqrt{2}\,d)$, then $V = -8e^2|z_Az_B|/(\sqrt{3}\,d)$. Summing the whole series gives

$V = -\frac{e^2|z_Az_B|}{d}\left(6 - \frac{12}{\sqrt{2}} + \frac{8}{\sqrt{3}} - \cdots\right) = -\frac{e^2|z_Az_B|}{d}\cdot A$

Structure	$A$	Structure	$A$
NaCl	1.748	Sphalerite ZnS	1.638
CsCl	1.763	Fluorite CaF$_2$	2.520
Wurtzite (Zn,Fe)S	1.641	Rutile TiO$_2$	2.408

Bigger charge raises lattice enthalpy, bigger ions lower it

For a given structure, every term in the Born–Mayer equation is constant except $d$, $z_A$ and $z_B$. As ion size increases, $d$ increases, so lattice enthalpy decreases. As ion charge increases, lattice enthalpy increases. The lecture backs this with melting points. Varying the charges: m.p. CaO is 2572 °C and m.p. NaCl is 800 °C. Varying $d$: m.p. CaO is 2572 °C and m.p. BaO is 1923 °C.

Compound	$\Delta H_L$ (kJ mol$^{-1}$)	Compound	$\Delta H_L$ (kJ mol$^{-1}$)	Compound	$\Delta H_L$ (kJ mol$^{-1}$)
MgO	3938	LiF	1024	NaF	911
CaO	3566	LiCl	861	KF	815
SrO	3369	LiBr	803	RbF	777
BaO	3202	LiI	744	CsF	748

The lecture’s values are for rock-salt compounds, from Ladd and Lee. Down each column the cation or anion gets larger and the value falls.

Try it: worked example

Which has the larger lattice enthalpy, CaO or NaCl, and why? Use the equation only.

Show answer and steps

AnswerCaO. Its ions carry charges 2+ and 2-, so |zA zB| = 4 against 1 for NaCl.

1. Both have the rock-salt structure, so the Madelung constant $A$ is the same.
2. The prefactor contains $|z_Az_B|$. For CaO it is $2 \times 2 = 4$. For NaCl it is $1 \times 1 = 1$.
3. The lecture says higher ion charge gives higher lattice enthalpy.
4. The lecture’s melting points agree: CaO 2572 °C against NaCl 800 °C.

Pitfall

Larger ions mean a lower lattice enthalpy, not a higher one, because $d$ sits in the denominator. Charge and size push in opposite directions, so check which one the question varies.

Optional aside

The lecture also covers ReO$_3$ and CsCl as structure examples. ReO$_3$ is shown as an almost close-packed array of O$^{2-}$ and the slide gives no further text. CsCl is covered under ionic structures.