Question

Which one of the following reactions of Xenon compounds is not feasible?
(A) $Xe{{O}_{3}}+6HF\to Xe{{F}_{6}}+3{{H}_{2}}O$
(B) $3Xe{{F}_{4}}+6{{H}_{2}}O\to 2Xe+Xe{{O}_{3}}+12HF+1.5{{O}_{2}}$
(C) $2Xe{{F}_{2}}+2{{H}_{2}}O\to 2Xe+4HF+{{O}_{2}}$
(D) $Xe{{F}_{6}}+RbF\to Rb\left[ Xe{{F}_{7}} \right]$

Answer 1

Hint: The irreversible reactions will be feasible than that of reversible ones. The formation of stable products in a reaction is what we call as feasible ones. One observation can be made that in the given equations, three of them are about reacting xenon fluorides and one is about xenon oxide.

Complete step by step solution:
Firstly, let us discuss xenon and its compounds. Xenon is an atom in the modern periodic table having atomic number as 54 with symbol Xe. It is a noble gas found in trace amounts in earth’s crust. It is generally unreactive, but can undergo some chemical reactions to form some compounds. The three commonly known fluorides of xenon are $Xe{{F}_{2}}$, $Xe{{F}_{4}}$ and $Xe{{F}_{6}}$. Whereas, the three oxides known are $Xe{{O}_{3}}$, $Xe{{O}_{4}}$ and $Xe{{O}_{2}}$. According to the illustration given, we need to see which reaction would not be possible. Thus, we need to see that the reactions are reversible or irreversible simply.
Xenon fluorides- These compounds of xenon can be obtained under appropriate conditions when we react with fluorine with xenon directly. For example,
$Xe+{{F}_{2}}\xrightarrow[673K]{Nickel-tube}Xe{{F}_{2}}$ and similarly, other fluorides are formed.
-All the Xenon fluorides are the powerful fluorinating agents and can be readily hydrolysed by the traces of water. Thus,
$2Xe{{F}_{2}}+2{{H}_{2}}O\to 2Xe+4HF+{{O}_{2}}$ is feasible.
 -Xenon hexafluoride acts as a Lewis acid when bound to a fluoride ion. Thus, it would form a salt when added to rubidium fluoride to give a complex as rubidium heptafluoroxenonate.
Hence, $Xe{{F}_{6}}+RbF\to Rb\left[ Xe{{F}_{7}} \right]$ is feasible.
-The oxides of xenon are obtained from the other compounds of xenon i.e. xenon fluorides. Hydrolysis of xenon fluorides ($Xe{{F}_{4}}$ or $Xe{{F}_{6}}$) results in the formation of xenon oxides. Thus,
$3Xe{{F}_{4}}+6{{H}_{2}}O\to 2Xe+Xe{{O}_{3}}+12HF+1.5{{O}_{2}}$ is feasible.
-Now, as we know that the xenon fluorides easily hydrolysed to give xenon oxides. But the vice versa reaction is impossible as xenon fluorides will again reverse the reaction to give xenon oxides. Thus, $Xe{{O}_{3}}+6HF\to Xe{{F}_{6}}+3{{H}_{2}}O$ is not feasible.

Therefore, option (A) is correct.

Note: Xenon does not react with oxygen directly to give xenon oxides. Always xenon oxides are formed from the hydrolysis of xenon fluorides. So, do note that the vice versa reaction would never be possible as the stability of reaction and reversibility matters.