Animated Solution for Chemistry - s and p-Block Elements: Which one of the following reaction of xenon compounds is not feasible?

For a long time, noble gases were considered completely inert, the "loners" of the periodic table. But the discovery of Xenon compounds shattered that myth. Xenon, being a larger noble gas, has outer electrons that are less tightly bound to the nucleus, allowing highly electronegative elements like Fluorine and Oxygen to coax it into forming bonds.

However, these Xenon fluorides are not exactly stable in all environments. They have a massive Achilles' heel: Water.

Xenon fluorides (${\text{XeF}}_2$, ${\text{XeF}}_4$, and ${\text{XeF}}_6$) are incredibly reactive towards water. They undergo a process called hydrolysis.

Let's look at ${\text{XeF}}_6$. When Xenon hexafluoride meets water, it doesn't just react; it reacts violently and spontaneously. The reaction is:

This forward reaction is highly favored thermodynamically. The formation of the strong $\text{H-F}$ bonds and the relatively stable $\text{Xe-O}$ bonds drives this reaction to completion.

Now, let's examine the options provided in the question. We are looking for the reaction that is not feasible.

Look closely at option (a):

Do you see it? This is the exact reverse of the hydrolysis of ${\text{XeF}}_6$! Because the forward hydrolysis reaction is so spontaneous and releases a lot of energy, trying to force the products (${\text{XeO}}_3$ and $\text{HF}$) back into the highly reactive ${\text{XeF}}_6$ and water is thermodynamically uphill. It simply will not happen under normal conditions. It's like trying to un-burn a piece of wood.

To be absolutely certain, let's quickly review the other options:

(b) $3{\text{XeF}}_4+6{\text{H}}_2\text{O}⟶2\text{Xe}+{\text{XeO}}_3+12\text{HF}+1.5{\text{O}}_2$: This is the standard, well-documented disproportionation/hydrolysis reaction of ${\text{XeF}}_4$. It is perfectly feasible. (c) $2{\text{XeF}}_2+2{\text{H}}_2\text{O}⟶2\text{Xe}+4\text{HF}+{\text{O}}_2$: This is the standard hydrolysis of ${\text{XeF}}_2$, which slowly yields Xenon gas, hydrogen fluoride, and oxygen. Highly feasible. (d)* ${\text{XeF}}_6+\text{RbF}⟶\text{Rb}[{\text{XeF}}_7]$: Here, ${\text{XeF}}_6$ is acting as a Lewis acid (fluoride acceptor) reacting with an alkali metal fluoride to form a complex ion. This is a known and feasible chemical property of ${\text{XeF}}_6$.

Therefore, the only reaction that defies thermodynamic logic is the reverse hydrolysis presented in option (a).