Animated Solution for Chemistry - s and p-Block Elements: Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is

Welcome to a fascinating journey into the world of noble gas chemistry! For a long time, noble gases were thought to be completely inert. However, the discovery of Xenon compounds shattered this belief. In this problem, we are exploring a classic reaction sequence involving Xenon hexafluoride (${\text{XeF}}_6$). Our ultimate goal is to trace the transformations and count the number of gaseous products released at the very end.

The Complete Hydrolysis of ${\text{XeF}}_6$

Imagine dropping a crystal of ${\text{XeF}}_6$ into water. It doesn't just dissolve; it reacts vigorously. Hydrolysis is essentially the cleavage of chemical bonds by the addition of water. During the complete hydrolysis of ${\text{XeF}}_6$, all six fluorine atoms are replaced by oxygen atoms from the water molecules.

The balanced chemical equation for this violent reaction is:

Here, the Xenon atom maintains its $+6$ oxidation state. The product formed, ${\text{XeO}}_3$ (Xenon trioxide), is our intermediate P. It is crucial to note that ${\text{XeO}}_3$ is not a gas; it is a highly explosive, colorless solid that remains dissolved in the aqueous solution.

The Acidic Nature of ${\text{XeO}}_3$

Now, the reaction scheme moves intermediate P into a basic medium (${\text{OH}}^{-}/{\text{H}}_2\text{O}$). How does ${\text{XeO}}_3$ behave in the presence of a base?

Interestingly, ${\text{XeO}}_3$ acts as an acidic oxide. When it encounters hydroxide ions, it readily accepts them to form a complex anion known as the xenate ion.

This xenate ion, ${\text{HXeO}}_4^{-}$, is our intermediate Q. At this stage, Xenon is still sitting comfortably in the $+6$ oxidation state.

The Beautiful Chaos of Disproportionation

Here is where the chemistry gets truly elegant. The xenate ion (${\text{HXeO}}_4^{-}$) is thermodynamically unstable in a basic solution. It undergoes a process called slow disproportionation. Disproportionation is a special type of redox reaction where a single substance is simultaneously oxidized and reduced.

Let's look at the master equation for this transformation:

This equation is a masterpiece of electron transfer. - One part of the Xenon($+6$) is oxidized to the $+8$ state, forming the highly stable perxenate ion (${\text{XeO}}_6^{4-}$). - Another part of the Xenon($+6$) is reduced all the way down to its elemental state, releasing pure Xenon gas ($\text{Xe}$). - Simultaneously, oxygen atoms from the xenate/hydroxide are oxidized to release Oxygen gas (${\text{O}}_2$).

The question asks us a very specific question: What is the total number of gases released as products in the final step?

Let's inspect our final products: 1. ${\text{XeO}}_6^{4-}$: This is an aqueous ion, happily swimming in the solution. 2. ${\text{H}}_2\text{O}$: This is liquid water, the solvent. 3. $\text{Xe}$: Xenon is a noble gas. It bubbles out of the solution. (Gas #1) 4. ${\text{O}}_2$: Oxygen is a diatomic gas. It also bubbles out. (Gas #2)

Therefore, under ambient conditions, exactly two distinct gases are released. The correct answer is 2.