Question

The $Xe{F_6}$ on hydrolysis produces:
A.$XeO{F_4}$
B.$Xe{O_2}{F_2}$
C.$Xe{O_3}$
D.$Xe{O_2}$

Answer 1

To answer this question, you should recall the concept of hydrolysis of fluorides of xenon. On hydrolysis, xenon fluoride forms oxide and HF.

Complete step by step solution:
A hydrolysis is a common form of a chemical reaction where water is mostly used to break down the chemical bonds that exist between a particular substance. Hydrolysis is derived from a Greek word hydro meaning water and lysis which translates to the word break or to unbind. Usually, in hydrolysis, the water molecules get attached to two parts of a molecule.
One molecule of a substance will get ${H^ + }$ ion and the other molecule receives the $O{H^ - }$ group.
The reaction can be represented by the equation:
$Xe{F_6} + 3{H_2}O \to Xe{O_{3}} + 6HF$
$Xe{F_6}$ one of the three binary fluorides formed by the xenon. The complete hydrolysis of $Xe{F_6}$ produces $Xe{O_3}$ which is xenon trioxide. This xenon trioxide is highly explosive and acts as a powerful oxidising agent in solution.

Hence, we can conclude that the correct answer to this question is option C.

Note:
$Xe{F_6}$ has seven electron pairs. It consists of 6 bond pairs and one lone pair. Xenon has 8 electrons in its valence shell and it forms six bonds with the fluorine atoms. When the fluorides of xenon have formed the electrons in the valence shell of xenon get unpaired and are promoted to vacant 5d orbitals.
After hybridization $Xe{F_6}$ molecular geometry will be distorted octahedral or square bipyramidal. What happens here is that the fluorine atoms are placed in the vertices of the octahedron while the lone pairs move in the space to avoid or reduce the repulsion.