Preparation and crystal structure of K5Te3

The structure of the first of the new alkali metal chalcogenides of the composition M5X3 has been solved (K5Te3: a=13.742(2) A, c=6.364(2) A, space group: I4/m, Z=4). It can be derived from the W5Si3 type of structure by substituting the ...Si...Si...Si... chains by those of Te22--dumbbells ...(Te-Te)...(Te-Te)...(Te-Te)..., both running along [001]. The Te22--dumbbells occur as a consequence of the necessity of charge compensation of the saltlike compound K5Te3, according to 2 K5Te3↔10 K+ + Te22- + 4 Te2-. The remaining tellurium atoms not involved in the dumbbells build up strings of tetrahedra 1∞[K Te4/2], with common edges along [001]. Therefore, the resulting K-K distances in these strings of tetrahedra are extremely short (3.18 A). (orig.)