Question

Complete the following reactions:
(i) $Xe{{F}_{2}}+{{H}_{2}}\to$
(ii) $Xe{{F}_{6}}+Si{{O}_{2}}\to$
(iii) $Xe{{F}_{6}}+N{{H}_{3}}\to$
(iv) $Xe{{F}_{6}}+{{H}_{2}}O\to$
(v) $Xe{{F}_{6}}+Sb{{F}_{5}}\to$
(vi) $Xe{{F}_{4}}+KI\to$
(vi) $Xe{{F}_{4}}+BC{{l}_{3}}\to$

Answer 1

Hint : Xenon is a noble gas whose atomic number is 54. It only makes compounds with high electronegative elements, such as F. Some compounds of Xe are made with O also. Any kind of reaction with Xe or Xenon Fluoride requires very high energy and hence high temperature, pressure are prerequisites.

Complete Step By Step Answer:
$Xe{{F}_{2}}$ reacts with hydrogen to make hydro-fluoride and it gives Xe as a by-product because HF is more stable than $Xe{{F}_{2}}$ .
$Xe{{F}_{2}}+{{H}_{2}}\to HF+Xe$
$Xe{{F}_{6}}$ reacts with $Si{{O}_{2}}$ to make a compound of Xenon, Oxygen and Fluorine. The reaction is:
$Xe{{F}_{6}}+Si{{O}_{2}}\to XeO{{F}_{4}}+Si{{F}_{4}}$ . This happens because of the high electronegativity of oxygen, which makes a double bond with Xe in $XeO{{F}_{4}}$ .
When $Xe{{F}_{6}}$ reacts with $N{{H}_{3}}$ the reaction results in the separation of Xe and F, formation of
$N{{H}_{4}}F$ and releasing of ${{N}_{2}}$ .
$Xe{{F}_{6}}+8N{{H}_{3}}\to Xe+6N{{H}_{4}}F+{{N}_{2}}$
The Reaction of $Xe{{F}_{6}}$ with water makes hydro-fluoride and $Xe{{O}_{3}}$
$Xe{{F}_{6}}+3{{H}_{2}}O\to Xe{{O}_{3}}+6HF$
The Reaction of $Xe{{F}_{6}}$ and $Sb{{F}_{5}}$ makes an ionic compound in which $Xe{{F}_{5}}$ makes the cation part and $Sb{{F}_{6}}$ makes the anion part.
$Xe{{F}_{6}}+Sb{{F}_{5}}\to {{[Xe{{F}_{5}}]}^{+}}{{[Sb{{F}_{6}}]}^{-}}$
The reaction of $Xe{{F}_{4}}$ and $KI$ is a replacement reaction in a way that F replaces I in KI. The by-products are $Xe$ and ${{I}_{2}}$
$Xe{{F}_{4}}+4KI\to Xe+4KF+2{{I}_{2}}$
The reaction of $Xe{{F}_{4}}$ and $BC{{l}_{3}}$ is similar to the reaction of $Xe{{F}_{4}}$ and $KI$ . Here F replaces Cl and the by-products are Xe and $C{{l}_{2}}$ .
$Xe{{F}_{4}}+BC{{l}_{3}}\to 4B{{F}_{3}}+3Xe+6C{{l}_{2}}$

Note :
Xenon forms three fluorides $Xe{{F}_{2}},Xe{{F}_{4}}$ and $Xe{{F}_{6}}$ .
$Xe{{F}_{2}}$ is formed when the excess of Xe is reacted with ${{F}_{2}}$ under high temperature and pressure.
Similarly when instead of one mole of ${{F}_{2}}$ , two moles are taken, then the same reaction yields $Xe{{F}_{4}}$ . And again the same reaction forms $Xe{{F}_{6}}$ when 3 moles of ${{F}_{2}}$ are reacted with excess of Xe.