A specimen having a stoichiometric composition of KSbO3·(KSb) calcined at 800°C has an R (Formula presented.) rhombohedral structure (RS), and changes to a Pn (Formula presented.) cubic structure (CS) when calcined at 1100°C. Finally, a <111>-oriented rhombohedral phase is formed in the specimen calcined at 1230°C. K/Sb ratio decreases from 1.0 in RS, 0.93 in CS, and finally to 0.85 in <111>-oriented rhombohedral phases. On the other hand, a specimen having a K-excess composition of K1.1SbO3 calcined at 800°C shows a RS that is maintained in the K-excess specimen calcined at 1230°C. The composition of these specimens is very close to KSb. Therefore, the RS with a space group of R (Formula presented.) is a stable form of KSbO3. The formation of Pn (Formula presented.) cubic and <111>-oriented R (Formula presented.) phases can be explained by the evaporation of K2O during the calcination process at temperatures above 1100°C.
|Number of pages||4|
|Journal||Journal of the American Ceramic Society|
|Publication status||Published - 2016 Jul 1|
- potassium/potassium compounds
ASJC Scopus subject areas
- Ceramics and Composites
- Materials Chemistry