Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

How are xenon fluorides $Xe{{F}_{2}}$, $Xe{{F}_{4}}$ and $Xe{{F}_{6}}$ obtained?


Answer
VerifiedVerified
554.1k+ views
Hint: Xenon is a noble gas element having eight electrons in their outermost shell making them stable in nature. The reactions of xenon with fluorine can only be possible in some environmental conditions.

Complete step by step solution:
Let us move directly towards the formation of xenon fluorides;
Xenon fluorides- There are three binary xenon fluorides which are formed directly under appropriate experimental conditions.
The reactions for the formation of the given xenon fluorides are;
For $Xe{{F}_{2}}$-
$X{{e}_{\left( g,excess \right)}}+{{F}_{2}}_{\left( g \right)}\xrightarrow[1bar]{673K}Xe{{F}_{2\left( s \right)}}$
Xe in excess and its gaseous state reacts with fluorine in its gaseous form at 673K temperature and 1 bar pressure to produce $Xe{{F}_{2}}$ solution.
$Xe{{F}_{2}}$ can also be formed by irradiating a mixture of xenon and fluorine with sunlight.
For $Xe{{F}_{4}}$-
$X{{e}_{\left( g,1mole \right)}}+2{{F}_{2}}_{\left( g,5moles \right)}\xrightarrow[7bar]{873K}Xe{{F}_{4\left( s \right)}}$
Xe in its gaseous state reacts with fluorine in its gaseous form at 873K temperature and 7 bar pressure to produce $Xe{{F}_{4}}$ solution. The mole ratio of xenon and fluorine is 1:5 and the mixture is sealed in Ni vessels while the reaction takes place.
For $Xe{{F}_{6}}$-
$X{{e}_{\left( g,1mole \right)}}+3{{F}_{2\left( g,20moles \right)}}\xrightarrow[65bar]{573K}Xe{{F}_{6\left( s \right)}}$
Xe in its gaseous state reacts with fluorine in its gaseous form at 573K temperature and 65 bar pressure to produce $Xe{{F}_{6}}$ solution. The mole ratio of xenon and fluorine is 1:20 and the mixture is sealed in Ni vessels while the reaction takes place.
$Xe{{F}_{6}}$ can also be synthesized by interacting $Xe{{F}_{4}}$ with ${{O}_{2}}{{F}_{2}}$ at 143 K as,
$Xe{{F}_{4}}+{{O}_{2}}{{F}_{2}}\xrightarrow{143K}Xe{{F}_{6}}+{{O}_{2}}$

Note: Do note that synthesis of all the xenon fluorides demand for the different experimental conditions, none of the reactions are possible at similar conditions. Also, the alternatives as described above are taken into account with respective precautions and specific conditions.