Total Number of Gases Released in Xenon Hexafluoride Hydrolysis Scheme

To determine the total number of gaseous species evolved as products in the final step of the given reaction scheme, we analyze the sequence step-by-step:

Step 1: Complete Hydrolysis of $\text{XeF}_6$

When xenon hexafluoride ($\text{XeF}_6$) undergoes complete hydrolysis, it yields xenon trioxide ($\text{XeO}_3$) and hydrogen fluoride ($\text{HF}$): $\text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF}$

Thus, compound $\mathbf{P}$ is xenon trioxide ($\text{XeO}_3$).

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Step 2: Reaction of $\text{XeO}_3$ with Hydroxide Ions ($\text{OH}^-$)

In an alkaline solution, $\text{XeO}_3$ reacts with $\text{OH}^-$ to form the hydrogen xenate ion ($\text{HXeO}_4^-$): $\text{XeO}_3 + \text{OH}^- \rightarrow \text{HXeO}_4^-$

Thus, species $\mathbf{Q}$ is the hydrogen xenate ion ($\text{HXeO}_4^-$).

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Step 3: Slow Disproportionation of $\text{HXeO}_4^-$

In an alkaline medium, the hydrogen xenate ion ($\mathbf{Q}$) slowly undergoes disproportionation to form perxenate ion ($\text{XeO}_6^{4-}$), xenon gas ($\text{Xe}$), oxygen gas ($\text{O}_2$), and water ($\text{H}_2\text{O}$): $2\text{HXeO}_4^-(aq) + 2\text{OH}^-(aq) \rightarrow \text{XeO}_6^{4-}(aq) + \text{Xe}(g) + \text{O}_2(g) + 2\text{H}_2\text{O}(l)$

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Conclusion

In the final reaction step, the gaseous species evolved are:

1. Xenon gas ($\text{Xe}$)
2. Oxygen gas ($\text{O}_2$)

Thus, the total number of gaseous species released in the final step is $2$.

Correct Option: (C)