Past exam of the materials science course of the University of Cambridge / 2022 / Paper 1 / 2 / c

Each missing oxide ion leaves two units of excess positive effective charge, compensated by reducing two Ce${}^{4+}$ ions to Ce${}^{3+}$. For $0.2$ oxygen vacancies per conventional unit cell, $0.4$ of its four cerium ions must be trivalent: 10% Ce${}^{3+}$. This assumes the same number of cation sites per unit volume as in the doped calculation and that oxygen loss is compensated entirely by this cerium-valence change. If reduction changes the cell volume appreciably, matching the absolute vacancy density requires correcting the percentage for that change.