Question

$Xe{{F}_{2}}$ reacts with $P{{F}_{5}}$ to give
a.) $Xe{{F}_{6}}$
b.) ${{\left[ XeF \right]}^{+}}{{\left[ P{{F}_{6}} \right]}^{-}}$
c.) $Xe{{F}_{4}}$
d.) ${{\left[ P{{F}_{4}} \right]}^{+}}{{\left[ Xe{{F}_{3}} \right]}^{-}}$

Answer 1

$Xe{{F}_{6}}$ or xenon difluoride is an example of a noble gas compound. Generally noble gases are inert in nature due to their stable electron configuration but in certain circumstances heavier noble gases react with highly reactive elements like fluorine to produce these compounds. $Xe{{F}_{6}}$ is used as a fluorinating agent and is considered to be a possible replacement for elemental fluorine as the latter is highly reactive and can become more difficult to control the reaction.
$P{{F}_{5}}$ or phosphorus pentafluoride is a corrosive gas that fumes in air. It is a Lewis acid and therefore can accept lone pairs of electrons. It can accept a fluoride ion to become $P{{F}_{6}}^{-}$.

Complete step by step answer:
$Xe{{F}_{2}}$ is a fluorinating agent and P{{F}_{5}}$$$$P{{F}_{5}} is a Lewis acid.$Xe{{F}_{2}}$ donates a fluoride ion to $P{{F}_{5}}$ which results in the formation of ${{\left[ XeF \right]}^{+}}{{\left[ P{{F}_{6}} \right]}^{-}}$. This is possible because of the availability of vacant d-orbitals in phosphorus .

Hence the answer is option B which is ${{\left[ XeF \right]}^{+}}{{\left[ P{{F}_{6}} \right]}^{-}}$.
So, the correct answer is “Option B”.

Note: $Xe{{F}_{2}}$ is not the only xenon compound that can be synthesised: there are several oxides, fluorides and oxofluorides of xenon, like $Xe{{F}_{4}}$,$Xe{{F}_{6}}$,$Xe{{O}_{3}}$ and $XeO{{F}_{4}}$ all of these compounds are highly reactive and exist either as a highly volatile liquid or as a gas.