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.