Fig. 6: Thermodynamic site preference of Ge in SrSnO3.
From: Hybrid molecular beam epitaxy of germanium-based oxides
Calculated region in the tin chemical potential (μSn) vs oxygen chemical potential (μO) plane showing where B-site alloys have lower formation enthalpy than the A-site alloys. The lines in the upper right corner (x = 0.0625, x = 0.125, and x = 0.25) separate the regions below which B-site alloys are preferred. This result indicates that for all allowed values of μSn and μO for which SrSnO3 is stable (orange region at the center), Ge will prefer to occupy the Sn site. The stability of SrSnO3 is limited by the formation of SrO on the left (dark-green line, corresponding to) and the formation of SnO2 on the right (light-green line), i.e., μSn +2μO \( > {\varDelta H}^{f}\left({{{{{\rm{SrSn}}}}}}{{{{{{\rm{O}}}}}}}_{3}\right)-{\varDelta H}^{f}\left({{{{{\rm{SrO}}}}}}\right)\), and obtain \({\mu }_{{{\mbox{Sn}}}}+2{\mu }_{{{\mbox{O}}}} < {\varDelta H}^{f}\left({{{{{\rm{Sn}}}}}}{{{{{{\rm{O}}}}}}}_{2}\right)\). The formation of GeO2 poses a lower limit to the oxygen potential, as indicated in the bottom region, i.e., \({\mu }_{{{\mbox{Ge}}}}+{2\mu }_{{{\mbox{O}}}} < {\varDelta H}^{f}\left({{{{{\rm{Ge}}}}}}{{{{{{\rm{O}}}}}}}_{2}\right)\). The formation enthalpy of A-site alloy will be lower than that of B-site alloy only in the upper right corner of the μSn vs. μO diagram, a region where SrSnO3 itself is unstable and SnO2 is favorable to form.
