Question

Which of the following options is/are true for $Xe{F}_6$?

• A. It is a colourless solid that readily sublimes into intensely white vapors
• B. Hybridization of $Xe{F}_6$ is $s{p}^3{d}^3$
• C. It reacts with $Ru{F}_5$ to form $\left[Ru{F}_4{]}^{+}\right[Xe{F}_7^{-}]$
• D. It gives $Xe{O}_3$ as final product on complete hydrolysis with water along with $HF$

Answer 1

Answer: (B), (D)

The correct options are
B Hybridization of $Xe{F}_6$ is $s{p}^3{d}^3$
D It gives $Xe{O}_3$ as final product on complete hydrolysis with water along with $HF$

(a) $Xe{F}_6$ is colourless solid which sublimes into intensely yellow vapours.

(b) According to VSEPR theory, $Xe{F}_6$ has 6 bond pairs and 1 lone pair so hybridises into $s{p}^3{d}^3$.

(c) $Xe{F}_6+Ru{F}_5⟶\left[Xe{F}_5{]}^{+}\right[Ru{F}_6{]}^{-}$

(d) $Xe{F}_6+3H_{2}O⟶Xe{O}_3+6HF$