The formation of O2+ [PtF6]- is the basis for the formation of xenon fluorides. This is because:

Question

The formation of O2+ [PtF6]- is the basis for the formation of xenon fluorides. This is because: (A) O2 and Xe have comparable sizes (B) Both O2 and Xe are gases (C) O2 and Xe have comparable ionisation energies (D) O-2 and Xe have comparable electronegativities

Tardigrade Admin · Accepted Answer

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 O2[PtF6]- 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 F6 and Xenon Xe+ [PtF6]-. After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesised.

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

$O_{2}$ and $X e$ have comparable sizes

Both $O_{2}$ and $X e$ are gases

$O_{2}$ and $X e$ have comparable ionisation energies

$O - 2$ and $X e$ have comparable electronegativities

Q. The formation of O2+​[PtF6​]− is the basis for the formation of xenon fluorides. This is because:

O2​ and Xe have comparable sizes

Both O2​ and Xe are gases

O2​ and Xe have comparable ionisation energies

O−2 and Xe have comparable electronegativities

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

$O_{2}$ and $X e$ have comparable sizes

Both $O_{2}$ and $X e$ are gases

$O_{2}$ and $X e$ have comparable ionisation energies

$O - 2$ and $X e$ have comparable electronegativities