Question

How Are ${\mathrm{XeF}}_2$And ${\mathrm{XeF}}_4$ Prepared?

Answer 1

Part 1: Preparation of ${\mathrm{XeF}}_2$

• ${\mathrm{XeF}}_2$can be prepared by heating a mixture of Xenon and Fluorine in the ratio of 2:1 at 400 $°\mathrm{C}$ and 1 bar of pressure in a sealed nickel tube.
• The balanced chemical reaction can be represented as:

$\underset{\left(\mathrm{Xenon}\right)}{\mathrm{Xe}\left(\mathrm{g}\right)}+\underset{\left(\mathrm{Fluorine}\right)}{{\mathrm{F}}_2\left(\mathrm{g}\right)}\stackrel{400°\mathrm{C}(\mathrm{Ni})}{\to }\underset{(\mathrm{Xenon}\mathrm{fluoride})}{{\mathrm{XeF}}_2\left(\mathrm{s}\right)}$

• Also it can be prepared by reacting Dioxygen difluoride with xenon at 118$°\mathrm{C}$.

$\underset{\left(\mathrm{Xenon}\right)}{\mathrm{Xe}\left(\mathrm{g}\right)}+\underset{(\mathrm{Dioxygen}\mathrm{difluoride})}{{\mathrm{O}}_2{\mathrm{F}}_2\left(\mathrm{g}\right)}\stackrel{118°\mathrm{c}}{\to }\underset{(\mathrm{Xenon}\mathrm{fluoride})}{{\mathrm{XeF}}_2\left(\mathrm{s}\right)}+\underset{\left(\mathrm{Oxygen}\right)}{{\mathrm{O}}_2\left(\mathrm{g}\right)}$

Part 2: Preparation of ${\mathrm{XeF}}_4$

• Xenon fluoride can be prepared when a mixture of xenon and fluorine (1:5) is passed at 400$°\mathrm{C}$ and 5-6 atm.
  $\underset{\left(\mathrm{Xenon}\right)}{\mathrm{Xe}\left(\mathrm{g}\right)}+\underset{\left(\mathrm{Flourine}\right)}{2{\mathrm{F}}_2\left(\mathrm{g}\right)}\stackrel{400°\mathrm{C},5-6\mathrm{atm}}{\to }\underset{(\mathrm{Xenon}\mathrm{fluoride})}{{\mathrm{XeF}}_4\left(\mathrm{s}\right)}$

Hence, the above is the appropriate process by which ${\mathrm{XeF}}_2$and ${\mathrm{XeF}}_4$ can be prepared.