Question

What is the product formed when $Xe{F}_4$ reacts with $Sb{F}_5$?

• A. $\left[Xe{F}_3{]}^{+}\right[Sb{F}_6{]}^{-}$
• B. $\left[Xe{F}_5{]}^{-}\right[Sb{F}_4{]}^{+}$
• C. $\left[Xe{F}_5{]}^{+}\right[Sb{F}_3{]}^{-}$
• D. $\left[Xe{F}_6{]}^{+}\right[Sb{F}_4{]}^{-}$

Answer 1

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

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

$Sb{F}_5$ being a strong lewis acid act as a flouride acceptor.