Materials Data on Sr2Nb2Bi3BrO11 by Materials Project

Abstract

Bi3Sr2Nb2O11Br crystallizes in the tetragonal P4/mmm space group. The structure is three-dimensional. Sr2+ is bonded in a 4-coordinate geometry to eight O2- atoms. There are four shorter (2.40 Å) and four longer (3.18 Å) Sr–O bond lengths. Nb5+ is bonded to six O2- atoms to form distorted NbO6 octahedra that share corners with five equivalent NbO6 octahedra and faces with four equivalent BiO12 cuboctahedra. The corner-sharing octahedra tilt angles range from 0–22°. There are a spread of Nb–O bond distances ranging from 1.81–2.24 Å. There are two inequivalent Bi3+ sites. In the first Bi3+ site, Bi3+ is bonded to twelve O2- atoms to form BiO12 cuboctahedra that share corners with four equivalent BiO12 cuboctahedra, faces with four equivalent BiO12 cuboctahedra, and faces with eight equivalent NbO6 octahedra. There are eight shorter (2.71 Å) and four longer (2.81 Å) Bi–O bond lengths. In the second Bi3+ site, Bi3+ is bonded in a 4-coordinate geometry to four equivalent O2- and four equivalent Br1- atoms. All Bi–O bond lengths are 2.26 Å. All Bi–Br bond lengths are 3.57 Å. There are four inequivalent O2- sites. In the first O2- site, O2- is bonded to two equivalent Sr2+ and two equivalent Bi3+ atoms to form a mixture of edge and corner-sharing OSr2Bi2 tetrahedra. In the second O2- site, O2- is bonded in a distorted bent 150 degrees geometry to two equivalent Nb5+ and two equivalent Bi3+ atoms. In the third O2- site, O2- is bonded in a single-bond geometry to four equivalent Sr2+ and one Nb5+ atom. In the fourth O2- site, O2- is bonded in a distorted linear geometry to two equivalent Nb5+ and four equivalent Bi3+ atoms. Br1- is bonded in a body-centered cubic geometry to eight equivalent Bi3+ atoms.