SYNTHESIS AND ELECTRICAL CONDUCTIVITY OF SOLID SOLUTIONS OF THE SYSTEM RbF-PbF2-SnF2

In the system RbF–PbF2–SnF2 are formed solid solutions of the heterovalent substitution RbxPb0,86‑xSn1,14F4-x (0  0,2 on the X-ray diffractograms, in addition to the basic structure, additional peaks are recorded that do not correspond to the reflexes of the individual fluorides and can indicate the formation of a mixture of solid solutions of different composition. For single-phase solid solutions, the calculated parameters of the crystal lattice are satisfactorily described by the Vegard rule. The introduction of ions of Rb+ into the initial structure leads to an increase in the parameter a of the elementary cell from 5.967 for x = 0 to 5.970 for x = 0.20. The replacement of a part of leads ions to rubium ions an increase in electrical conductivity compared with β–PbSnF4 and Pb0.86Sn1.14F4. Insignificant substitution (up to 3.0 mol%) of ions Pb2+ at Rb+ at T 450 K is higher than the conductivity of the initial sample Pb0,86Sn1,14F4 and at temperatures below 450 K by an order of magnitude smaller. With further increase in the content of RbF the electrical conductivity of the samples increases throughout the temperature range, reaching the maximum values at x≥0.15 (σ573 = 0.34–0.41 S/cm, Ea = 0.16 eV and σ373 = (5.34–8.16)•10-2 S/cm, Ea = 0.48–0.51 eV, respectively). In the general case, the replacement of a part of the ions of Pb2+ with Rb+ to an increase in the electrical conductivity of the samples throughout the temperature range. The activation energy of conductivity with an increase in the content of RbF in the low-temperature region in the general case increases, and at temperatures above 400 K is inversely proportional decreasing. The nature of the dependence of the activation energy on the concentration of the heterovalent substituent and its value indicate that the conductivity of the samples obtained increases with an increase in the vacancies of fluoride ions in the structure of the solid solutions.