Question

Complete the following equation
Xe{{F}_{6}}+{{H}_{2}}O\to \\_\\_\\_\\_\\_+2HF

Answer 1

Hint: Reaction of compounds with water is known as hydrolysis. Hydrolysis products are usually oxides or hydroxides.

Complete step by step solution:
Xenon hexafluoride is a compound of the noble gas Xenon with formula $Xe{{F}_{6}}$. It is one of the three binary fluorides of xenon. Xenon hexafluoride is the strongest fluorinating agent and is a colourless solid. The solid readily sublimes into intensely yellow vapours.
Structure of Xenon hexafluoride is distorted octahedral and has a hybridisation $s{{p}^{3}}{{d}^{3}}$.

In the question, hydrolysis of $Xe{{F}_{6}}$ is carried out. On hydrolysis we get $XeO{{F}_{4}}$. Further hydrolysis can be carried out to give $Xe{{O}_{3}}$. The reaction is as follows,

$Xe{{F}_{6}}+{{H}_{2}}O\to XeO{{F}_{4}}+2HF$

On further hydrolysis,

$XeO{{F}_{4}}+{{H}_{2}}O\to Xe{{O}_{2}}{{F}_{2}}+2HF$
$Xe{{O}_{2}}{{F}_{2}}+{{H}_{2}}O\to Xe{{O}_{3}}+2HF$

The total hydrolysis reaction can be shown as
$3Xe{{F}_{6}}+6{{H}_{2}}O\to 2Xe{{O}_{3}}+Xe+12HF$
Hence, the complete reaction is $Xe{{F}_{6}}+{{H}_{2}}O\to XeO{{F}_{4}}+2HF$.

Additional Information:
- Xenon hexafluoride can act as lewis acids too by binding with one or two anions.
-It is prepared by heating xenon difluoride at 300 degree Celsius under 60 atm pressure of fluorine. $Ni{{F}_{2}}$ is the catalyst used. These are exergonic (positive flow of energy from system to surroundings) and are stable at normal temperatures.
-Xenon hexafluoride has distorted octahedral structure because of the lone pair of electrons of xenon. All the six fluorides occupy positions just like in normal octahedral structure and the lone pair occupies a position as shown in the figure above.

Note: Note that in the question only 2 moles of HF is produced, so the reaction is the first hydrolysed product. If it was a complete hydrolysis reaction, then the number moles of HF will be 12.