Question

Which compounds is prepared by the following reaction:
$\underset{\text{1:5 Volume ratio}}{Xe+F_2}\stackrel{873K,7bar}{\to }$

• A. $Xe{F}_4$
• B. $Xe{F}_2$
• C. $Xe{F}_6$
• D. None of these

Answer 1

The correct option is A $Xe{F}_4$

Xenon forms three binary fluorides, $Xe{F}_2,Xe{F}_4$ and $Xe{F}_6$ by the direct reaction of elements under appropriate experimental conditions. Xe has a complete filled 5p configuration. As a result, when it undergoes bonding with an odd number (3 or 5) of F atoms it leaves behind one unpaired electron. This causes the molecule to become unstable. As a result $Xe{F}_3$, $Xe{F}_5$ are not possible.
$Xe\left(g\right)+F_2\left(g\right)\stackrel{673K,1bar}{\to }Xe{F}_2\left(s\right)$
(xenon in excess)
$Xe\left(g\right)+2F_2\left(g\right)\stackrel{873K,7bar}{\to }Xe{F}_4\left(s\right)$
(1:5 ratio)
$Xe\left(g\right)+3F_2\left(g\right)\stackrel{573K,60-70bar}{\to }Xe{F}_6\left(s\right)$
(1:20 ratio)