Question

\[Xe{F_6}\] on partial hydrolysis with water produces a compound X. The same compound X is formed when \[Xe{F_6}\] reacts with silica. The compound X is:
A. \[Xe{F_2}\]
B. \[Xe{F_4}\]
C. \[Xe{O_3}\]
D. \[XeO{F_4}\]

Answer 1

Hint:We know that hydrolysis involves the addition of oxygen in the molecule and when the molecule reacts with silica, then there is displacement of oxygen from silica molecule by another electronegative atom.

Complete answer
\[Xe{F_6}\] is having xenon in the compound hence it is known as the noble gas compound. The chemical name of \[Xe{F_6}\] is Xenon hexafluoride (Hexa as it contains six Fluorine atoms). This noble gas compound has an octahedral shape. The molecule is stable. \[s{p^3}{d^3}\] hybridization is of \[Xe\] in \[Xe{F_6}\] molecule. Water molecules will break down during hydrolysis, the \[Xe\] will form bonds with oxygen of the water and hydrogen will form a bond with fluorine. The product formed will be Xenon oxytetrafluoride which will be having square planar symmetry and square pyramidal geometry.
a. \[Xe{F_2}\] is not formed as in question it is said that the hydrolysis product and reaction with silica should be the same. The xenon difluoride is formed when Xenon reacts with fluorine.
b. \[Xe{F_4}\] is known as Xenon tetrafluoride; it is formed when xenon reacts with fluorine. It is not formed during the hydrolysis of \[Xe{F_6}\] nor it is formed when it reacts with silica.
c. \[Xe{O_3}\] the displacement reaction will take place but all the fluorine will not be displaced with oxygen.
d. Due to the octahedral structure of \[Xe{F_6}\], the two molecules will be replaced by the one oxygen as oxygen has a tendency to form double bonds.Therefore \[Xe{F_6}\] reacts with silica \[Si{O_2}\] to produce \[XeO{F_4}\] the same product as obtained from hydrolysis. Silica is the monomer the huge polymer will be formed which is present in the sand particles.
Let us discuss the reactions as follows;
\[Xe{F_6}{\rm{ }} + {\rm{ }}{H_2}O \to {\rm{ }}XeO{F_4}{\rm{ }} + {\rm{ }}2HF\]
\[2Xe{F_6}{\rm{ }} + {\rm{ }}Si{O_2}{\rm{ }} \to 2XeO{F_4}{\rm{ }} + {\rm{ }}Si{F_4}\]

Hence, the answer is (D) \[XeO{F_4}\].

Note:
The structure of the compound and the type of reactions should be known which will help to decide the correct answer.The geometry of any of the compounds can be found by knowing the no. of bond pairs and lone pairs present on the central atom.