Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section E / 26 / d

A triangular lattice has area per particle $\sqrt{3}{\ell }_{\mathrm{hex}}^2/2$, so ${\ell }_{\mathrm{hex}}=\sqrt{2A/(\sqrt{3}N)}$. Inserting this into the supplied accessible area yieldsThereforeThe prefactor $C$ remains $2/\Lambda$. The lattice is triangular in terms of particle centres; “hexagonal” here describes its sixfold coordination and cell symmetry.

Since $B_{\mathrm{hex}}<B_{\square }$, the hexagonal lattice has larger accessible cell area at any common allowed total area $A>B_{\square }$. HenceThe kinetic entropies cancel at the same temperature. The triangular lattice is favoured and has higher entropy within this cell model, even though both structures look ordered. More efficient packing leaves more local displacement freedom at a fixed number density. This comparison does not by itself establish whether either crystal is stable against the fluid.