Abstract

It is well established that solid K2MnF6 reacts with excess SbF5 forming elemental F2. However, if the reaction is carried out in anhydrous HF (aHF) as a solvent, this is surprisingly not the case. Instead, the green Mn(IV) compound K3[(MnIVF)(SbF6)5]F is obtained. The reductive elimination of F2 was not observed under the applied conditions. The compound was characterized by its crystal structure, by Raman spectroscopy, and by quantum-chemical solid-state calculations. It crystallizes in the monoclinic space group P21/c, mP164, with the lattice parameters a = 12.2393(13), b = 12.167(2), c = 20.115(5) Å, β = 110.805(8)°, V = 2800.1(9) Å3, Z = 4 at T = 200 K. As the use of strictly anhydrous HF is crucial in this and other similar reactions, methods for drying moist HF are discussed.

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