Explain why the electron affinity of oxygen has two values — a first electron affinity that is exothermic and a second electron affinity that is endothermic — and discuss how both values are nevertheless incorporated into a Born–Haber cycle to give a valid lattice enthalpy for an ionic oxide such as magnesium oxide.

The first electron affinity of oxygen is exothermic because the incoming electron is attracted to the nuclear charge of the neutral O atom, releasing energy. The second electron affinity is endothermic because the incoming electron is added to an already negatively charged O⁻ ion, so electrostatic repulsion must be overcome, requiring energy input.

In a Born–Haber cycle the two electron affinities are simply added to the other steps. Hess’s law means that the sign of each step is irrelevant – the cycle is a closed loop of enthalpy changes between the same initial and final states. Thus both the exothermic first and the endothermic second electron affinities appear in the cycle.

The lattice enthalpy of MgO is very large and exothermic. When the two electron affinities are combined with the other steps (ionisation of Mg, sublimation of Mg, dissociation of O₂, formation of O²⁻), the large negative lattice enthalpy compensates for the positive second electron affinity, giving a negative overall enthalpy of formation for MgO.