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采用化学共沉淀法制备了Sn1-xSbxO2(x=0~1)固溶体,用XRD、TG-DTA和XPS手段分析了该类固溶体的物相组成,并根据分析的结果绘制了固相线下相图,重点考察了Sn1-xSbxO2(x=0~1)固溶体的形成作用机制。结果表明:在600~1000℃退火温度下,Sn0.95Sb0.05O2固溶体只有四方金红石型SnO2物相的存在,锑主要以Sb5+的形式进入SnO2晶格内形成有限固溶体,而Sn0.6Sb0.4O2和Sn0.2Sb0.8O2固溶体分别在700和800℃出现了斜方白安矿型Sb2O4和四方金红石型SnO2分相;固相线下Sn1-xSbxO2(x=0~1)固溶体相图分为3个区域:四方SnO2单相区,SnO2+Sb2O4两相区和斜方Sb2O4单相区,四方SnO2单相区是(Sn,Sb)O2ss固溶体(Sb3+或Sb5+进入SnO2晶格内,并达到饱和)。(Sn,Sb)O2ss固溶体可作为导电性好和耐腐蚀的半导体材料。
The solid solution of Sn1-xSbxO2 (x = 0 ~ 1) was prepared by chemical coprecipitation method. The phase composition of the solid solution was analyzed by XRD, TG-DTA and XPS. According to the results of the analysis, The focus is on the formation mechanism of Sn1-xSbxO2 (x = 0 ~ 1) solid solution. The results show that there is only tetragonal rutile SnO2 phase in the solid solution of Sn0.95Sb0.05O2 at the annealing temperature of 600 ~ 1000 ℃. Antimony mainly enters into the lattice of SnO2 as Sb5 + to form a limited solid solution, while the Sn0.6Sb0.4O2 and In the Sn0.2Sb0.8O2 solid solution, the orthorhombic Ba2O and tetragonal rutile SnO2 phases appear at 700 and 800 ℃, respectively. The phase diagram of solid solution of Sn1-xSbxO2 solid solution (x = 0 ~ 1) is divided into three regions: The tetragonal SnO2 single phase region, the SnO2 + Sb2O4 two phase region and the orthorhombic Sb2O4 single phase region, and the tetragonal SnO2 single phase region is the (Sn, Sb) O2ss solid solution (Sb3 + or Sb5 + enters the SnO2 lattice and reaches saturation). (Sn, Sb) O2ss solid solution can be used as a conductive and corrosion-resistant semiconductor material.