What is the state of hybridisation of $Xe$ in the cationic part of solid $XeF_6$?

Balance the following equation: $XeF_{6} + H_{2}O \rightarrow XeO_{2}F_{2} + HF$

The structures of $XeF_2$ and $XeF_4$ respectively are

Assertion $(A)$: Noble gases have very low boiling points.
Reason $(R)$: All noble gases have general electronic configuration of $ns^2 np^6$ (except $He$).

$XeF_4$ reacts with $SbF_5$ to form $[XeF_m]^{n+}[SbF_y]^{z-}$. Calculate $m + n + y + z = .......$.

Which of the following is the correct decreasing order of the ease of liquefaction of noble gases?