Question

$Xe$ reacts with $F_2$ in $1:5$ ratio to form $A$ which then reacts with a fluoride ion acceptor to form cationic species $B.$ $B$ can be:

• A. $\left[Xe{F}_3{]}^{+}\right[Sb{F}_6{]}^{–}$
• B. $C{s}^{+}[Xe{F}_7{]}^{–}$
• C. $\left[Xe{F}_5{]}^{+}\right[Sb{F}_6{]}^{–}$
• D. All of the above

Answer 1

The correct option is A $\left[Xe{F}_3{]}^{+}\right[Sb{F}_6{]}^{–}$

$Xe+2F_2\underset{1:5}{\overset{873K,7bar}{\to }}\underset{\left(A\right)}{Xe{F}_4}$

$Xe{F}_4+Sb{F}_5\to \underset{\left(B\right)}{\left[Xe{F}_3{]}^{+}\right[Sb{F}_6{]}^{-}}$