Ba(XeF2)5](AF6)2, A = Ru, Nb - Synthesis, Crystal Structure and Raman Spectra

Abstract

The reaction of Ba(NbF6)2 and excess XeF2 in anhydrous HF at room temperature yields [Ba(XeF2)5](NbF6)2. This is the first example of the coordination compound with XeF2 molecule as a ligand and NbF6- anion. It crystallizes in the orthorhombic space group Fmmm, with a = 12.084(6) Å, b = 13.646(7) Å, c = 13.927(7) Å, V = 2297(2) Å3, and Z = 4. The reaction of XeF2·Xe2F3·RuF6 and BaF2 in anhydrous HF yields [Ba(XeF2)5](RuF6)2 - the first example of the coordination compound containing RuF6- anion and XeF2 molecules. It appears to be isostructural with [Ba(XeF2)5](NbF6)2, with a = 11.9510(14) Å, b = 13.5174(14) Å, c = 13.8488(12) Å, V = 2237.2(4) Å3, and Z = 4. The Raman spectra of both compounds prove that all the XeF2 molecules are symmetrical. Four XeF2 molecules per formula unit act as bridges between Ba centres, while one molecule is held in the structure by electrostatic forces. In the series of compounds [Ba(XeF2)5](AF6)2 with A = As, Sb, Ru, Nb, the influence of the anions AF6- was analyzed.