Animated Solution for Chemistry - s and p-Block Elements: Correct statement(s) about the compounds P, Q and R is(are) $\text{Xe (g)}+{\text{F}}_2\text{(g)}\stackrel{873\text{ K, 7 bar}}{\to }\text{P}$ (1 : 5 ratio) $\text{P}+{\text{O}}_2{\text{F}}_2\stackrel{143\text{ K}}{\to }\text{Q}+{\text{O}}_2$ $\text{Q}+{\text{H}}_2\text{O}\stackrel{\text{complete hydrolysis}}{\to }\text{R}+\text{HF}$

In the fascinating world of noble gas chemistry, Xenon stands out as a surprisingly reactive element under the right conditions. This problem takes us on a thrilling journey through a sequence of reactions, transforming Xenon gas into various fluorides and eventually an oxide. Let's break down each step and uncover the molecular secrets hidden within.

Synthesizing Compound P

Our journey begins with Xenon gas reacting with Fluorine gas in a specific $1:5$ ratio at a high temperature of $873\text{ K}$ and a pressure of $7\text{ bar}$. These precise conditions are the classic recipe for synthesizing Xenon tetrafluoride ($Xe{F}_4$).

Now, let's visualize the structure of this newly formed compound P. Xenon, being a noble gas, has $8$ valence electrons. It forms four single bonds with the four fluorine atoms, utilizing $4$ of its electrons. The remaining $4$ electrons pair up to form two lone pairs.

According to VSEPR theory, a central atom with $4$ bond pairs and $2$ lone pairs adopts a square planar geometry. The lone pairs position themselves above and below the plane to minimize repulsion. Because $Xe{F}_4$ indeed has two lone pairs on the central atom, Statement (A) is absolutely correct.

Forming Compound Q

Next, we introduce $Xe{F}_4$ to $O_2{F}_2$ at a chilly $143\text{ K}$. $O_2{F}_2$ is an incredibly powerful fluorinating agent. It aggressively transfers fluorine atoms to $Xe{F}_4$, upgrading it to Xenon hexafluoride ($Xe{F}_6$) and releasing oxygen gas in the process.

Let's examine our new compound Q, $Xe{F}_6$. Xenon now uses $6$ of its $8$ valence electrons to form six single bonds with fluorine. This leaves just $2$ electrons, which constitute one lone pair.

With $6$ bond pairs and $1$ lone pair, the geometry cannot be a perfect octahedron. The presence of that single lone pair distorts the symmetry, resulting in a distorted octahedral (or capped octahedral) shape. Therefore, Statement (B) is incorrect.

However, $Xe{F}_6$ is renowned in chemistry as a highly reactive and strong fluorinating agent, readily donating fluorine to other species. Thus, Statement (C) is correct.

Complete Hydrolysis to Compound R

In the final step, we subject $Xe{F}_6$ to complete hydrolysis. When $Xe{F}_6$ reacts with an excess of water, a complete substitution occurs. All six fluorine atoms are replaced by three oxygen atoms, yielding Xenon trioxide ($Xe{O}_3$) and hydrogen fluoride.

To determine the shape of compound R ($Xe{O}_3$), we look at Xenon's valence shell again. Xenon forms three double bonds with the three oxygen atoms. Each double bond requires $2$ electrons from Xenon, consuming $6$ electrons in total. The remaining $2$ electrons form one lone pair.

Treating the double bonds as single bonding domains for VSEPR purposes, we have $3$ bond pairs and $1$ lone pair. This configuration pushes the bonding pairs downward, creating a trigonal pyramidal molecular structure, very similar to that of ammonia ($N{H}_3$). Consequently, Statement (D) is correct.

By carefully tracing the reaction conditions and applying VSEPR theory to each intermediate, we have successfully decoded the entire sequence. The correct statements describing the properties and geometries of compounds P, Q, and R are (A), (C), and (D).