Lattice Energy, Atomisation and Electron Affinity

Ionic compounds do not form in a single step. To understand why sodium chloride or magnesium oxide are so stable, chemists break the overall process into a series of smaller enthalpy changes that can each be measured or calculated separately. Three of the most important are the enthalpy of atomisation, lattice energy (lattice enthalpy), and electron affinity. Together with ionisation energy, these values are combined in Born-Haber cycles to find quantities that are difficult to measure directly.

The standard enthalpy of atomisation, \(\Delta H_{at}^{\ominus}\), is the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state, under standard conditions.

For example:

\(Na(s) \)→\( Na(g)\), \(\Delta H_{at}^{\ominus} = +107 \) kJ mol\(^{-1}\)

\(\frac{1}{2}Cl_2(g) \)→\( Cl(g)\), \(\Delta H_{at}^{\ominus} = +122 \) kJ mol\(^{-1}\)

Atomisation is always endothermic because bonds (metallic, covalent, or intermolecular forces) must be broken to separate a substance into individual gaseous atoms. Notice that for diatomic elements like chlorine, the equation is written for one mole of atoms, so only half a mole of \(Cl_2\) is used.

Lattice energy describes the strength of the ionic bonding within a crystal lattice. It can be defined in two directions, and it is essential to keep the sign convention straight:

Lattice energy of formation: the enthalpy change when one mole of an ionic solid is formed from its gaseous ions. This is always exothermic (negative), because attraction between oppositely charged ions releases energy.

\(Na^+(g) + Cl^-(g) \)→\( NaCl(s)\), \(\Delta H_{LE}^{\ominus} = -787 \) kJ mol\(^{-1}\)

Lattice energy of dissociation: the reverse process, breaking one mole of ionic solid apart into its gaseous ions. This is always endothermic (positive), with the same magnitude but opposite sign.

The size of the lattice energy depends on the charges of the ions and the distance between their centres (the sum of their ionic radii), following a Coulombic relationship:

\(\Delta H_{LE} \propto \dfrac{q_+ \times q_-}{r_0}\)

Greater ionic charge or smaller ionic radius produces a much more exothermic (more negative) lattice energy of formation. This is why the lattice energy of magnesium oxide, built from \(Mg^{2+}\) and \(O^{2-}\) ions, is far more negative than that of sodium chloride, which involves singly charged ions. Real lattice energies calculated from experimental data are often slightly different from those predicted by a purely ionic model, which is a clue that some bonds have partial covalent character; this idea is explored further under polarisability.

The first electron affinity is the enthalpy change when one mole of gaseous atoms each gains one electron to form one mole of gaseous 1- ions.

\(Cl(g) + e^- \)→\( Cl^-(g)\), \(\Delta H_{EA1}^{\ominus} = -349 \) kJ mol\(^{-1}\)

First electron affinities are usually exothermic, since the incoming electron is attracted to the positive nucleus. The second electron affinity, which adds a further electron to a negative ion, is always endothermic, because the incoming electron is repelled by the negative charge already present.

\(O^-(g) + e^- \)→\( O^{2-}(g)\), \(\Delta H_{EA2}^{\ominus} = +798 \) kJ mol\(^{-1}\)

Across a period, first electron affinity generally becomes more exothermic (more negative) as you move left to right, since nuclear charge increases and atomic radius decreases, pulling the incoming electron in more strongly. This trend peaks at the halogens, which have the most exothermic electron affinities in their period, since adding one electron completes a stable outer shell. Noble gases have positive or near-zero electron affinities because the extra electron must enter a new, higher-energy shell.

Down a group, electron affinity generally becomes less exothermic, because atomic radius increases and extra shielding reduces the attraction felt by the incoming electron. Fluorine is a well-known exception: its electron affinity is less exothermic than chlorine's, because fluorine's very small atomic radius causes strong repulsion between the incoming electron and the existing electrons already crowded around the nucleus.

Using a Born-Haber cycle and Hess's law, the lattice energy of formation of \(NaCl\) can be found from these standard enthalpy changes:

Enthalpy of atomisation of sodium: \(+107 \) kJ mol\(^{-1}\)

First ionisation energy of sodium: \(+496 \) kJ mol\(^{-1}\)

Enthalpy of atomisation of chlorine: \(+122 \) kJ mol\(^{-1}\)

First electron affinity of chlorine: \(-349 \) kJ mol\(^{-1}\)

Standard enthalpy of formation of \(NaCl(s)\): \(-411 \) kJ mol\(^{-1}\)

By Hess's law, the sum of all steps around the cycle equals the enthalpy of formation:

\(107 + 496 + 122 + (-349) + \Delta H_{LE} = -411\)

\(\Delta H_{LE} = -411 - (107 + 496 + 122 - 349) = -787 \) kJ mol\(^{-1}\)

This matches the accepted lattice energy of formation for sodium chloride. The full method for setting up and solving these cycles, including compounds with multiple charges like magnesium oxide, is covered in calculations using Born-Haber cycles. Once you can calculate lattice energy this way, the same cycle framework also connects to dissolving ionic solids in water, which is the focus of enthalpy of hydration and solution.

Enthalpy of atomisation, electron affinity and lattice energy each capture a different physical process: breaking a substance into separate gaseous atoms, adding an electron to a gaseous atom, and forming (or breaking apart) an ionic lattice. None of these values can usually be measured on their own for lattice energy, which is why they are combined through Hess's law in a Born-Haber cycle. Recognising the sign of each step (endothermic or exothermic) and matching the correct direction of each arrow is the key skill for using these enthalpy changes correctly in calculations.

Related lessons

• Uk A Level Chemistry