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The acidity order of xenon oxides,fluorides and oxofluorides is: $\ce{XeF6 > XeO2F4 > XeO3F2 > XeO4 > XeOF4 > XeF4 > XeO2F2 > XeF3 > XeO3}$.

How to justify this order?

My thoughts were that the acidity order of xenon fluorides, oxides and oxofluorides is based on its tendency to accept an oxide(as these are the most common reactions). As fluorine is more Electronegative than oxygen, we can safely say that $\ce{XeF6}$ would be the most(As $\ce{Xe}$ would be most electron deficient and can accept an oxide easily). $\ce{XeO2F4}$ would be the second most and the last would be $\ce{XeO3}$, based on number Fluorine+Oxygen atoms, with priority to fluorine. However evidently, "the more number of fluorine atoms" logic doesn't apply to all of them. Looking at the structure of each molecule I'm unable to deduce any logical explanation for the above mentioned order.

Also, I know for a fact that $\ce{XeF6}$ corrodes glass and $\ce{XeF4}$ doesn't. Upto which xenon compound would corrode with glass?

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You have to consider the oxidation state of the xenon. Higher oxidation state makes the central atom a stronger Lewis acid in general, xenon is just one case.

So we should consider each oxidation state separately. Look at the ordering when we do that:

+8: $\ce{XeO2F4}>\ce{XeO3F2}>\ce{XeO4}$

+6: $\ce{XeF6}>\ce{XeO2F2}>\ce{XeOF4}>\ce{XeO3}$

+4, +3: Only the fluorides are given, no oxygen vs fluorine comparison possible from the list.

The +8 and +6 orderings are consistent with the expected effect of oxygen versus fluorine within a given xenon oxidation state.

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