Structural aspects of the metal-insulator transitions in V0.985Al0.015O2

The crystal structure of V0.985Al0.015O2 has been refined from single-crystal X-ray data at four temperatures. At 373°K it has the tetragonal rutile structure. At 323°K, which is below the first metal-insulator transition, it has the monoclinic M2 structure, where half of the vanadium atoms are paired with alternating short (2.540 Å) and long (3.261 Å) V-V separations. The other half of the vanadium atoms form equally spaced (2.935 Å) zigzag V chains. At 298°K, which is below the second electric and magnetic transition, V0.985Al0.015O2 has the triclinic T structure where both vanadium chains contain V-V bonds, V(1)-V(1) = 2.547 Å and V(2)-V(2) = 2.819 Å. At 173°K the pairing of the V(1) chain remains constant: V(1)-V(1) = 2.545 Å, whereas that of the V(2) chain decreases: V(2)-V(2) = 2.747 Å. From the variation of the lattice parameters as a function of temperature it seems that these two short V-V distances will not become equal at lower temperatures. The effective charges as calculated from the bond strengths at 298 and 173°K show that a cation disproportionation has taken place between these two temperatures. About 20% of the V4+ cations of the V(1) chains have become V3+ and correspondingly 20% of the V4+ cations of the V(2) chains have become V5+. This disproportionation process would explain the difference between the two short V-V distances. Also it would explain why the T → M1 transition does not take at lower temperatures.