Question

Which of the following is formed on complete hydrolysis of $Xe{F_6}$?
A. $Xe{O_2}{F_2}$
B. $XeO{F_4}$
C. $Xe{O_3}$
D. $Xe{O_2}$

Answer 1

Hint: As we know that Xenon hexafluoride that is $Xe{F_6}$ is a noble gas compound and very less reactive but undergoes two types of hydrolysis that are partial hydrolysis and complete hydrolysis. Complete hydrolysis of $Xe{F_6}$ occurs when $1$ mole of $Xe{F_6}$ reacts with $3$ moles of water.

Complete Step by step answer: We know very well that Xenon $(Xe)$ is a group $18$ element that has $8$ electrons in the outermost shell possessing the atomic number $54$ and an electronic configuration as $\left[ {Kr} \right]4{d^{10}}5{s^2}5{p^6}$ . In Xenon hexafluoride $\left( {Xe{F_6}} \right)$, Xenon forms six bonds with the fluorine atoms. Xenon hexafluoride is one of the binary fluorides of gaseous element Xenon and it is a colorless compound and the strongest fluorinating agent as compared to other members of the noble family.
Now it is important to note that Xenon hexafluoride undergoes partial hydrolysis first. On partial hydrolysis of $Xe{F_6}$ it forms Xenon oxytetrafluoride $(XeO{F_4})$ and Xenon dioxide difluoride $(Xe{O_2}{F_2})$. We can show this through the reactions for partial hydrolysis that are as follows:
\[Xe{F_{6{\text{ }}}} + {\text{ }}{H_2}O{\text{ }} \to {\text{ }}XeO{F_4}{\text{ }} + {\text{ }}2HF\]
\[XeO{F_{{\text{4 }}}} + {H_2}O \to Xe{O_2}{F_{2{\text{ }}}} + 2HF\]
Now when partial hydrolysis has taken place the product formed will get further hydrolyzed to give Xenon trioxide and we can show this through the following reaction:
$Xe{O_2}{F_2} + {H_2}O \to Xe{O_3} + 2HF$
So we can say that complete hydrolysis of Xenon hexafluoride ultimately results into the formation of Xenon trioxide along with six molecules of hydrogen fluoride. We can show the whole reaction as:
$Xe{F_6} + 3{H_2}O \to Xe{O_3} + 6HF$

Therefore, the correct answer is option (C).

Note: Remember that noble gases are chemically unreactive due to their fully filled valence shells and high ionization enthalpy and are sparingly soluble in water. Three binary fluorides of xenon are known including $Xe{F_2},Xe{F_4}$ and $Xe{F_6}$ and among them $Xe{F_6}$ is strongest fluorinating agent. It hydrolysis to form $Xe{O_3}$which itself is a strong oxidizing agent but it can be toxic and explosive when comes in contact with the organic materials.