Normally, it can be obtained by reducing KMnO4 in HF solution. Is there different way to do it?
Different way - yes, cheaper - I don't think so, but higher yield (99,7 % in opposite to 78,6 % reported in "Efficient Preparations of Fluorine Compounds - Roesky"), e.g. as described in Acta Chim. Slov. 1999, 46 (2), pp. 229-238
K2MnF6 was prepared from a mixture of 9.00 mmol KF (0.523 g) (Merck, 99.9%) and 4.50 mmol MnF2 (0.418 g) (Riedel - De Haën) which was loaded in a PFA reaction vessel in a drybox. After evacuation of the reaction vessel on a vacuum line, a HF (6 ml) and KrF2 2.800 g (23 mmol) were condensed onto a mixture at 77 K. After warming to the room temperature the reaction was completed and clear orange solution was obtained. After the removal of aHF and gaseous decomposition products Kr and F2, a yellow solid K2MnF6 was obtained (calcd. 1.112 g; found 1.109 g), what wasconfirmed by X-ray powder diffraction pattern.
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