The chalcogenide compounds Bi2Se3, Bi2Te3 and Sb2Te3 are crystallize in Rhombohedral structure (S.G: R -3m, Z=3) and all the three compounds are ambient topological insulator. However, Sb2Se3 in this family crystallize as orthorhombic structure (SG: Pnma, Z=4 and band gap Eg= 1 eV). Some studies showed that Sb2Se3 could be a pressure induced topological insulator and it is not yet been proved. Since the crystal structure of Sb2Se3 is different as compared to other chalcogenide compounds, can i assume that the difference in structure responsible for the absence of topologically non-trivial state in Sb2Se3?. Is it possible to trnasform Sb2Se3 band insulator to a topological insulator under hydrostatic pressure?
The most relevant physical parameter in TIs is the spin-orbit splitting in the cation, not the crystal structure. In Bi 6p Shell the SO splitting is of the order of 2.2 eV while in Sb 5p Shell it is much lower (about 0.8 eV). According to band structure calculations by Li et al (PRB89, 035101 (2014)) Sb2Se3 becomes a TI under high pressure after a band inversion occurring at 3 GPa. I do not know of any experimental confirmation of this prediction.