Animated Solution for Chemistry - s and p-Block Elements: Comprehension Passage

The beauty of inorganic chemistry often lies in the interconnectedness of its reactions. In this problem, we are presented with a fascinating sequence of three chemical transformations. Our mission is to act as chemical detectives, identifying the unknown compounds $\text{W}$, $\text{X}$, $\text{Y}$, and $\text{Z}$ step by step. Let's dive into the sequence!

The journey begins with the thermal decomposition of potassium chlorate (${\text{KClO}}_3$). When heated in the presence of a catalytic amount of manganese dioxide (${\text{MnO}}_2$), it undergoes a classic decomposition reaction.

This is a standard laboratory method for the preparation of oxygen gas. The ${\text{MnO}}_2$ acts as a catalyst, significantly lowering the temperature required for the decomposition. Therefore, we can confidently identify our first unknown: Gas $\text{W}$ is ${\text{O}}_2$.

Moving to the second step, we take the oxygen gas ($\text{W}$) we just produced and react it with white phosphorus (${\text{P}}_4$). The problem explicitly states that an excess amount of $\text{W}$ is used.

When white phosphorus burns in an abundant supply of oxygen, it undergoes complete oxidation to form phosphorus pentoxide, which exists as the dimer ${\text{P}}_4{\text{O}}_{10}$. It's crucial to note that if the oxygen supply were limited, the product would have been phosphorus trioxide (${\text{P}}_4{\text{O}}_6$). The keyword "excess" is our guiding light here. Thus, Compound $\text{X}$ is ${\text{P}}_4{\text{O}}_{10}$.

Now we reach the grand finale. We react our newly formed ${\text{P}}_4{\text{O}}_{10}$ with pure nitric acid (${\text{HNO}}_3$). To predict the outcome, we must recall a key property of ${\text{P}}_4{\text{O}}_{10}$: it is an exceptionally powerful dehydrating agent.

When ${\text{P}}_4{\text{O}}_{10}$ encounters pure ${\text{HNO}}_3$, it aggressively extracts the elements of water from the acid. The dehydration of nitric acid yields its anhydride, dinitrogen pentoxide (${\text{N}}_2{\text{O}}_5$). In the process, the ${\text{P}}_4{\text{O}}_{10}$ itself becomes hydrated, forming metaphosphoric acid (${\text{HPO}}_3$).

This elegant reaction reveals our final two unknowns: Compound $\text{Y}$ is ${\text{N}}_2{\text{O}}_5$ and Compound $\text{Z}$ is ${\text{HPO}}_3$.

By systematically analyzing each step of the reaction sequence, we have successfully unmasked all the unknown compounds.

- $\text{W}={\text{O}}_2$ - $\text{X}={\text{P}}_4{\text{O}}_{10}$ - $\text{Y}={\text{N}}_2{\text{O}}_5$ - $\text{Z}={\text{HPO}}_3$

Matching these findings with the given options, we find that for Question 13, the correct pair is ${\text{O}}_2$ and ${\text{P}}_4{\text{O}}_{10}$ (Option B). For Question 14, the correct pair is ${\text{N}}_2{\text{O}}_5$ and ${\text{HPO}}_3$ (Option C). This problem beautifully illustrates how fundamental inorganic reactions can be chained together to test a deep understanding of chemical properties and reactivity.