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 $Xe{O}_3$
 $Xe{F}_6+3H_{2}O\to Xe{O}_3+6HF$
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 $I_2$
 $Xe{F}_4+4KI\to Xe+4KF+2I_2$
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.