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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

Both $O_{2}$ and Xe are gases

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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.