The formation of $O_{2}^{+} {[P t F_{6}]}^{-}$ is the basis for the formation of Xenon fluorides. This is because:

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$O_{2}$ and $X e$ does not have comparable sizes

Both $O_{2}$ and Xe are gases

$O_{2}$ and Xe have comparable ionization energies

$O_{2}$ and Xe have comparable electro negativities.

answer is C.

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Detailed Solution

In march 1962, Neil Bartlett, then at the university of British Columbia, observed the reaction of noble gas. First, he prepared a red compound which is formulated as $O_{2}^{+} {[P t F_{6}]}^{-}$ and it is already know that the first ionisation enthalpy of molecular oxygen $(1175 k J m o l^{- 1})$ is almost similar with that xenon $(1170 k J m o l^{- 1})$ . Then he made efforts to prepare same type of compound by mixing $P t F_{6}$ and Xenon $X e^{+} {[P t F_{6}]}^{-}$ . After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygene, have been synthesised.