Lesson 1 · 30 min

Ionic bonding: opposite charges together

Moving an electron from sodium to chlorine costs energy — so why does salt form at all? The answer is the Coulomb attraction between ions, and above all the collective attraction of a whole crystal.

In this lesson

Do the energy bookkeeping for moving an electron from sodium to chlorine, and explain why the transfer itself costs energy
Explain the minimum of an ion-pair potential curve with Coulomb attraction and short-range repulsion
Build a Born–Haber cycle and compute a lattice enthalpy from measured data
Estimate lattice enthalpies from ionic charges and radii, and read covalent character from where the model fails

Drop a small piece of sodium into chlorine gas and it bursts into a brilliant yellow flame, leaving white sodium chloride and a great deal of heat. Yet the bookkeeping looks wrong: pulling an electron off a gaseous sodium atom costs 496 kJ/mol (the ionisation energy), and a gaseous chlorine atom releases only 349 kJ/mol when it takes the electron (the electron affinity). The electron transfer alone loses about 147 kJ per mole. So where does all the heat come from?

Predict first

A $\ce{Na+}$ ion and a $\ce{Cl-}$ ion approach each other slowly in the gas phase, starting far apart. How does their total energy change?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

It keeps falling until the two nuclei coincide It falls, then rises steeply when they get too close — with a minimum in between It stays the same, because the total charge is zero It rises, crosses a barrier, then falls

Attraction, repulsion and the minimum

The attraction between ions obeys Coulomb’s law: the energy is proportional to the product of the charges and inversely proportional to the distance. At 236 pm a $\ce{Na+}$ – $\ce{Cl-}$ pair has a Coulomb energy of about −589 kJ/mol. But ions are not point charges; they carry filled electron shells, and when two clouds overlap they repel strongly. Born and Mayer described this repulsion with an exponential, which dies away so fast that it matters only at close range.

Add the two terms and the minimum sits at 236 pm, about −503 kJ/mol below the separated ions. That settles the opening puzzle: the 147 kJ/mol lost in the electron transfer is repaid many times over by the Coulomb attraction. The explorer’s “electron-jump distance” shows that the attraction covers the cost once the ions are closer than about 943 pm — which is where the electron actually jumps when a sodium atom meets a chlorine atom.

E(r) = -\frac{M\,|z_+ z_-|\,e^2 N_\mathrm{A}}{4\pi\varepsilon_0\, r} + B\,e^{-r/\rho}

The Born–Mayer potential: Coulomb attraction plus the repulsion of overlapping electron clouds.

$z_+,\ z_-$: charge numbers of the cation and anion
$r$: distance between the ion centres
$M$: Madelung constant: 1 for an isolated pair, 1.748 for the rock-salt crystal
$e,\ \varepsilon_0,\ N_\mathrm{A}$: elementary charge, vacuum permittivity, Avogadro constant
$B,\ \rho$: strength and range of the repulsion (ρ ≈ 34.5 pm)

Coulomb attraction + Born–Mayer repulsion

Ion-pair potential energy

Ion separation (or click the chart) 331 pm

Ion pair, total
Crystal, per formula unit
Coulomb attraction
Repulsion

$\ce{Na+ Cl-}$

Separation: 331 pm
Total energy: −414 kJ/mol
of which Coulomb: −420
of which repulsion: +5
Net force: attractive: the ions are pulled together

Energy minimum

At 236.1 pm the energy is −503 kJ/mol (relative to separated ions).

The crystal releases more than the pair by 253 kJ/mol

In the crystal each ion is attracted by many oppositely charged neighbours at once (Madelung constant M = 1.7476), at a longer equilibrium distance.

The “Compare” view: the same ions release about 250 kJ/mol more per formula unit in the crystal than as an isolated pair. Switch to LiF and CsCl — the smaller the ions, the deeper the well.

From ion pairs to crystals: lattice enthalpy

Table salt is not made of NaCl pairs but of an enormous ionic lattice. In the rock-salt structure every $\ce{Na+}$ is surrounded by 6 $\ce{Cl-}$ at 282 pm, then 12 $\ce{Na+}$ of the same sign, then 8 more $\ce{Cl-}$ … Summing all these attractions and repulsions gives 1.748 times the energy of a single pair — the Madelung constant. That is why ionic compounds are solids at room temperature rather than gases of ion pairs.

The lattice enthalpy $U_\mathrm{L}$ is the enthalpy needed to break one mole of an ionic crystal into its gaseous ions; it is always positive. It cannot be measured directly — nobody can pull a crystal apart into gaseous ions in one step and catch the heat. Hess’s law offers a detour: turn the elements into gaseous ions, then let the ions assemble into the crystal. Both routes must add up to the same enthalpy change.

U_\mathrm{L} = \Delta_\mathrm{at}H(\mathrm{M}) + \Delta_\mathrm{at}H(\mathrm{X}) + \sum IE + \sum \Delta_\mathrm{eg}H - \Delta_\mathrm{f}H^\circ(\mathrm{MX,\,s})

The Born–Haber cycle: going from the elements to the crystal via gaseous ions has the same enthalpy change as forming the crystal directly.

$\Delta_\mathrm{at}H$: atomisation enthalpy: forming one mole of gaseous atoms from the element (subliming Na, half a Cl–Cl bond)
$IE$: ionisation energy: a gaseous atom loses an electron
$\Delta_\mathrm{eg}H$: electron-gain enthalpy, minus the electron affinity; the first one is usually exothermic
$\Delta_\mathrm{f}H^\circ$: standard enthalpy of formation of the solid (measured)

Hess’s law · enthalpy staircase

Born–Haber cycle builder

Step 0 of 5

The elements sit on the zero line at the left; the measured enthalpy of the solid is at the right. Press “Next step” to turn the elements into gaseous ions, one step at a time.

#	Step	Process	ΔH / (kJ/mol)
—	Enthalpy of formation ΔfH° (measured)	$\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}$	−411.2
①	Atomise Na (sublimation)	$\ce{Na(s) -> Na(g)}$	…
②	Split Cl₂ into atoms	$\ce{1/2 Cl2(g) -> Cl(g)}$	…
③	First ionisation energy of Na	$\ce{Na(g) -> Na+(g) + e-}$	…
④	Cl gains an electron (electron affinity)	$\ce{Cl(g) + e- -> Cl-(g)}$	…
−U	unknown: lattice enthalpy	$\ce{Na+(g) + Cl-(g) -> NaCl(s)}$	?

Press “Next step” to build NaCl’s enthalpy staircase; the last step is forced by Hess’s law: 787 kJ/mol. Then try MgO (note that the second electron gain is endothermic) and AgCl (where the ionic model is off by a fifth).

Worked example

Lattice enthalpy of KCl from a Born–Haber cycle

Given (kJ/mol): $\Delta_\mathrm{f}H^\circ(\ce{KCl, s}) = -436.5$ ; atomisation of K 89.0; first ionisation energy of K 418.8; half the $\ce{Cl2}$ bond enthalpy 121.3; electron affinity of Cl 348.6. Find the lattice enthalpy of KCl.

Charge and size: what sets the lattice enthalpy

Coulomb energy grows with the product of the charges and shrinks with distance, so lattice enthalpies follow two simple rules: higher charges and smaller ions give larger lattice enthalpies. Kapustinskii condensed this into a formula that needs only charges and ionic radii — not even the crystal structure. MgO has four times NaCl’s charge product and smaller ions, and a lattice enthalpy about five times larger; that is why MgO melts only near 2800 °C and lines furnaces, while NaCl melts at 801 °C.

U_\mathrm{L} \approx \frac{1213.8\ \nu\,|z_+ z_-|}{r_+ + r_-}\left(1 - \frac{0.345}{r_+ + r_-}\right)\ \mathrm{kJ\,mol^{-1}}

The Kapustinskii equation (radii in Å). It averages the Madelung constant per ion and works for most simple ionic crystals.

$\nu$: ions per formula unit (2 for NaCl, 3 for CaCl₂)
$r_+ + r_-$: sum of cation and anion radii (Å)

Concept check

Which of these compounds has the largest lattice enthalpy?

\ce{NaCl}

\ce{KCl}

\ce{MgO}

\ce{CaCl2}

Try the calculation

Find the lattice enthalpy of LiF from these data (kJ/mol): $\Delta_\mathrm{f}H^\circ(\ce{LiF, s}) = -616.0$ ; atomisation of Li 159.3; ionisation energy of Li 520.2; half the $\ce{F2}$ bond enthalpy 79.4; electron affinity of F 328.2.

Key takeaways

Moving an electron from a metal atom to a non-metal atom costs energy; ionic compounds owe their stability to the Coulomb attraction between the ions.
An ion pair’s potential curve combines Coulomb attraction with the repulsion of overlapping electron clouds; the minimum sets the equilibrium distance.
Lattice enthalpy cannot be measured directly; a Born–Haber cycle finds it through Hess’s law — about 787 kJ/mol for NaCl.
Higher charges and smaller ions give larger lattice enthalpies; where the ionic model misses (AgCl) the bonding is partly covalent.

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Data and references

W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (CRC Press, 2016): standard enthalpies of formation; CODATA Key Values for gaseous atoms (J. D. Cox, D. D. Wagman, V. A. Medvedev, 1989).
NIST Atomic Spectra Database (ionisation energies); J. C. Rienstra-Kiracofe et al., Chem. Rev. 102, 231 (2002) (electron affinities).
K. P. Huber, G. Herzberg, Constants of Diatomic Molecules (1979), via the NIST Chemistry WebBook (gas-phase NaCl, KCl, LiF … bond lengths).
R. D. Shannon, Acta Crystallogr. A32, 751 (1976) (ionic radii); L. Glasser, H. D. B. Jenkins, J. Am. Chem. Soc. 122, 632 (2000) (Kapustinskii constant).
C. E. Housecroft, A. G. Sharpe, Inorganic Chemistry, 5th ed., Ch. 6 (lattice energy, Born–Haber cycles, O⁻ → O²⁻ electron-gain enthalpy).