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Q.
The formation of $O_2^+ [PtF_6]^-$ is the basis for the formation of xenon fluorides. This is because:
The p-Block Elements - Part2
Solution:
In March $1962$, Neil Bartlett, then at the University of British Columbia, observed the reaction of a noble gas. First, he prepared a red compound which is formulated as $O_2[PtF_6]^-$ and it is already know that the first ionisation enthalpy of molecular oxygen ($1175 \,kJ \,mol^{-1})$ is almost similar with that xenon $(1170 \,kJ \,mol^{-1})$. Then he made efforts to prepare same type of compound by mixing $Pt F_6$ and Xenon $Xe^+ [PtF_6]^-$. After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesised.