Q.Balance the following equation: XeF6 + H2O → XeO2F2 + HF

In 1962, Neil Bartlett had shown that platinum hexafluoride, $PtF_6$, is a powerful enough oxidiser to remove an electron from molecular oxygen, forming $O_2^+PtF_6^-$; noting that xenon's first ionisation enthalpy (1170 kJ/mol) is very close to that of $O_2$ (1175 kJ/mol), he reasoned that $PtF_6$ should be able to oxidise xenon as well, and on carrying out the reaction he obtained the first noble-gas compound, $Xe^+PtF_6^-$. Xenon, the most reactive noble gas because of its comparatively low ionisation enthalpy and large, polarisable electron cloud, combines directly with fluorine under different conditions of temperature, pressure and reactant ratio to give three fluorides: $Xe + F_2 \rightarrow XeF_2$ (excess xenon, about 673 K), $Xe + 2F_2 \rightarrow XeF_4$ (1:5 Xe:F2 ratio, about 673 K and 7 atm), and $Xe + 3F_2 \rightarrow XeF_6$ (1:20 ratio, high pressure, or alternatively from $XeF_4$ reacting with $O_2F_2$). Controlled hydrolysis of these fluorides gives xenon's oxygen compounds: complete hydrolysis of $XeF_6$ gives $XeO_3$, $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$, while partial hydrolysis gives the oxyfluorides, $XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$ and $XeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF$; despite involving water, these are simple ligand-exchange (substitution) reactions rather than redox reacti …