Topic 12 · page 10 of 17

Three definitions for ionic solids

Atomisation, electron affinity and lattice energy are the three new enthalpy changes behind a Born-Haber cycle. Each has a precise definition, and each is easiest to learn as an equation with state symbols.

12 min read12.12(i)12.12(ii)12.12(iii)

Read first: Thermodynamic and kinetic stability

A lattice energy cannot be measured directly. Page 11 builds it from a cycle of steps that can be measured or looked up. Two of the steps are old friends from earlier topics: the first ionisation energy (the enthalpy change when each atom in one mole of gaseous atoms loses one electron to form one mole of gaseous 1+ ions) and the second ionisation energy (the same for 1+ ions losing a second electron). This page defines the other three.

Every definition is about gaseous species, apart from the starting point for atomisation. If you can write the equation with state symbols, the definition follows.

Enthalpy change of atomisation 12.12(i)

The equation is always written so that one mole of atoms forms:

Na(s)→Na(g)ΔatH⊖=+107 kJ mol−1 \ce{Na(s) -> Na(g)} \qquad \Delta_\text{at}H^\ominus = +107\ \text{kJ mol}^{-1}
12 ClX2(g)→Cl(g)ΔatH⊖=+122 kJ mol−1 \ce{1/2 Cl2(g) -> Cl(g)} \qquad \Delta_\text{at}H^\ominus = +122\ \text{kJ mol}^{-1}

For a diatomic gas you need the half: 12 ClX2(g)\ce{1/2 Cl2(g)} makes one mole of Cl atoms. For bromine, the standard state is the liquid, so the equation is 12 BrX2(l)→Br(g)\ce{1/2 Br2(l) -> Br(g)}. Atomisation is always endothermic, because bonds (metallic or covalent) must be broken and nothing is formed.

Electron affinity 12.12(ii)

Cl(g)+eX−→ClX−(g)ΔH=−349 kJ mol−1 \ce{Cl(g) + e- -> Cl-(g)} \qquad \Delta H = -349\ \text{kJ mol}^{-1}

The first electron affinity is usually exothermic: the incoming electron is attracted by the nucleus. For oxygen the first electron affinity is −141 kJ mol⁻¹.

The second electron affinity adds an electron to a 1− ion to make a 2− ion:

OX−(g)+eX−→OX2−(g)ΔH=+798 kJ mol−1 \ce{O-(g) + e- -> O^2-(g)} \qquad \Delta H = +798\ \text{kJ mol}^{-1}

It is endothermic, because the electron is pushed towards an ion that is already negative. Energy must be supplied to overcome the repulsion.

Lattice energy 12.12(iii)

NaX+(g)+ClX−(g)→NaCl(s)ΔH=−787 kJ mol−1 \ce{Na+(g) + Cl-(g) -> NaCl(s)} \qquad \Delta H = -787\ \text{kJ mol}^{-1}
MgX2+(g)+2 ClX−(g)→MgClX2(s)ΔH=−2524 kJ mol−1 \ce{Mg^2+(g) + 2Cl-(g) -> MgCl2(s)} \qquad \Delta H = -2524\ \text{kJ mol}^{-1}

All the ions on the left are gaseous, and the solid on the right is one mole of the compound. Strong electrostatic attraction between oppositely charged ions releases a lot of energy as the lattice forms.

Enthalpy change General equation Usual sign
Atomisation element(standard state) → 1 mol of atoms(g) positive
First ionisation energy M(g)→MX+(g)+eX−\ce{M(g) -> M+(g) + e-} positive
First electron affinity X(g)+eX−→XX−(g)\ce{X(g) + e- -> X-(g)} negative
Second electron affinity XX−(g)+eX−→XX2−(g)\ce{X-(g) + e- -> X^2-(g)} positive
Lattice energy gaseous ions → 1 mol of solid negative

Check yourself

  1. Write equations, with state symbols, for the standard enthalpy change of atomisation of (a) magnesium, (b) bromine, (c) oxygen.
  2. Write equations for the first and second electron affinities of oxygen, and give the sign of each.
  3. Write an equation that represents the lattice energy of (a) calcium chloride, (b) aluminium oxide.
  4. Explain why the second electron affinity of oxygen is endothermic.
  5. Explain why atomisation is always endothermic.
  6. A student writes the lattice energy of sodium chloride as NaCl(s)→NaX+(g)+ClX−(g)\ce{NaCl(s) -> Na+(g) + Cl-(g)}, ΔH = +787 kJ mol⁻¹. What is wrong with this for this specification?

Answers cover these until you have tried the questions

  1. (a) Mg(s)→Mg(g)\ce{Mg(s) -> Mg(g)}. (b) 12 BrX2(l)→Br(g)\ce{1/2 Br2(l) -> Br(g)} (bromine is liquid in its standard state). (c) 12 OX2(g)→O(g)\ce{1/2 O2(g) -> O(g)}.
  2. First: O(g)+eX−→OX−(g)\ce{O(g) + e- -> O-(g)}, negative (−141 kJ mol⁻¹). Second: OX−(g)+eX−→OX2−(g)\ce{O-(g) + e- -> O^2-(g)}, positive (+798 kJ mol⁻¹).
  3. (a) CaX2+(g)+2 ClX−(g)→CaClX2(s)\ce{Ca^2+(g) + 2Cl-(g) -> CaCl2(s)}. (b) 2 AlX3+(g)+3 OX2−(g)→AlX2OX3(s)\ce{2Al^3+(g) + 3O^2-(g) -> Al2O3(s)}. Each forms one mole of solid from gaseous ions.
  4. The second electron is added to O⁻, which is already negative. The electron is repelled, so energy must be put in to force it on, and the process is endothermic.
  5. Atomisation breaks the bonds that hold the element together (metallic bonding or covalent bonds). Breaking bonds takes in energy and nothing is released, so ΔH is positive.
  6. That equation is the reverse of the definition. Lattice energy is the formation of one mole of the solid from its gaseous ions, NaX+(g)+ClX−(g)→NaCl(s)\ce{Na+(g) + Cl-(g) -> NaCl(s)}, which is exothermic with ΔH = −787 kJ mol⁻¹.