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Question: How are xenon fluorides \(Xe{{F}_{2}}\), \(Xe{{F}_{4}}\) and \(Xe{{F}_{6}}\) obtained?...

How are xenon fluorides XeF2Xe{{F}_{2}}, XeF4Xe{{F}_{4}} and XeF6Xe{{F}_{6}} obtained?

Explanation

Solution

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 XeF2Xe{{F}_{2}}-
Xe(g,excess)+F2(g)1bar673KXeF2(s)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 XeF2Xe{{F}_{2}} solution.
XeF2Xe{{F}_{2}} can also be formed by irradiating a mixture of xenon and fluorine with sunlight.
For XeF4Xe{{F}_{4}}-
Xe(g,1mole)+2F2(g,5moles)7bar873KXeF4(s)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 XeF4Xe{{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 XeF6Xe{{F}_{6}}-
Xe(g,1mole)+3F2(g,20moles)65bar573KXeF6(s)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 XeF6Xe{{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.
XeF6Xe{{F}_{6}} can also be synthesized by interacting XeF4Xe{{F}_{4}} with O2F2{{O}_{2}}{{F}_{2}} at 143 K as,
XeF4+O2F2143KXeF6+O2Xe{{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.