Question

Which of the following is not obtained by direct reaction of constituent elements
(a) \[Xe{F_2}\]
(b) \[Xe{F_4}\]
(c) \[Xe{O_3}\]
(d) \[Xe{F_6}\]

Answer 1

Hint: Among all the gases of the zero group only xenon can react with fluorine and form a variety of compounds. Whereas other gases of the zero group almost remain inert and do not form compounds with fluorine.

Complete Step by Step Solution:
Based on compound forming ability the noble gases can be divided into two parts.
(1) Reactive noble gases: Krypton (\[Kr\]), Xenon (\[Xe\]), and Radon (\[Rn\]).
(2) Non-reactive noble gases: Argon (\[Ar\]), Neon (\[Ne\]), and Helium (\[He\]).

Xenon can directly react with fluorine and can form three compounds such as., \[Xe{F_2}\], \[Xe{F_4}\], and \[Xe{F_6}\]. The \[Xe{F_2}\], \[Xe{F_4}\], and \[Xe{F_6}\]can be synthesised by the following methods. The \[Xe{F_2}\]can be prepared by using an excess of xenon gas with fluorine at 673K and 1 bar. The\[Xe{F_4}\]is isolated by using a 1:5 ratio of xenon and fluorine at 873K and 7 bar.
Whereas the \[Xe{F_6}\]is isolated by employing the 1:20 ratio of xenon and fluorine at 873K and 7 bar

The\[Xe{O_3}\]compound of xenon is prepared by the hydrolysis of \[Xe{F_4}\]and \[Xe{F_6}\]. The following schematic representation is showing the hydrolysis of xenon fluorides.
\[Xe{F_6} + 3{H_2}O \to Xe{O_3} + 6HF\]
\[6Xe{F_4} + 12{H_2}O \to 2Xe{O_3} + 24HF + 4Xe + 3{O_2}\]
Therefore from the above explanation we can conclude that the fluorides of xenon can be prepared by direct reaction. Whereas \[Xe{O_3}\]cannot be prepared by direct reaction. Hence, option (c) will be the correct option:

Note: The \[Xe{F_2}\]has linear geometry and \[s{p^3}d\] hybridisation. The \[Xe{F_4}\]has square planar geometry and \[s{p^3}{d^2}\] hybridisation. Whereas the \[Xe{F_6}\] has distorted octahedral geometry and\[s{p^3}{d^3}\] hybridisation. The \[Xe{O_3}\]possess tetrahedral geometry and \[s{p^3}\] hybridization.