Animated Solution for Chemistry - s and p-Block Elements: The reaction of ${\text{HClO}}_3$ with $\text{HCl}$ gives a paramagnetic gas, which upon reaction with ${\text{O}}_3$ produces

A Deep Dive into Redox Reactions

Welcome to a fascinating journey through the chemistry of the p-block elements! Inorganic chemistry often feels like a vast ocean of reactions to memorize, but when we look closely at the underlying principles—like oxidation states, redox potential, and molecular structure—the logic begins to reveal itself beautifully.

In this problem, we are tasked with tracing a two-step chemical transformation starting from common laboratory acids and ending with a highly specific oxide of chlorine. Let's break down the thought process step-by-step.

Chloric Acid Meets Hydrochloric Acid

Our journey begins with the reaction between chloric acid (${\text{HClO}}_3$) and hydrochloric acid ($\text{HCl}$). At first glance, you might think of an acid-base neutralization, but remember, both of these are acids! Instead, we must look at their redox properties.

Chloric acid is a strong oxidizing agent, containing chlorine in the $+5$ oxidation state. Hydrochloric acid, on the other hand, contains chloride ions (${\text{Cl}}^{-}$) in the $-1$ oxidation state, which can be oxidized. When these two react, a classic redox disproportionation-like scenario unfolds (though technically it's a comproportionation since two different oxidation states of the same element react to form intermediate states).

The balanced chemical equation for this reaction is:

This reaction yields a mixture of gases: chlorine dioxide (${\text{ClO}}_2$) and chlorine gas (${\text{Cl}}_2$), along with water.

The problem specifically directs our attention to a paramagnetic gas produced in the first step. Paramagnetism in molecules arises from the presence of unpaired electrons. To find our culprit, we need to examine the valence electrons of our gaseous products.

Chlorine gas (${\text{Cl}}_2$) is a stable diatomic molecule where all electrons are perfectly paired up in molecular orbitals. It is diamagnetic.

Now, let's look at chlorine dioxide (${\text{ClO}}_2$). We can calculate its total valence electrons by adding the valence electrons of one chlorine atom and two oxygen atoms:

Because $19$ is an odd number, it is mathematically impossible for all electrons to be paired. There must be at least one unpaired electron residing in a molecular orbital. This unpaired electron makes ${\text{ClO}}_2$ highly reactive and, crucially, paramagnetic.

Enter Ozone

Now that we have identified ${\text{ClO}}_2$ as our paramagnetic intermediate, we move to the second phase of the problem. The ${\text{ClO}}_2$ gas is reacted with ozone (${\text{O}}_3$).

Ozone is an exceptionally powerful oxidizing agent. When it encounters chlorine dioxide (where chlorine is in the $+4$ oxidation state), it forces the chlorine atom to give up even more electron density, pushing it to a higher oxidation state.

The reaction proceeds as follows:

The ozone oxidizes the chlorine dioxide to form hexoxide of chlorine (${\text{Cl}}_2{\text{O}}_6$), while the ozone itself is reduced to diatomic oxygen (${\text{O}}_2$).

Therefore, the final product produced upon reaction with ${\text{O}}_3$ is ${\text{Cl}}_2{\text{O}}_6$, which corresponds perfectly to Option (C).

While finding the answer is satisfying, the true joy of chemistry lies in understanding the nature of the molecules we create. ${\text{Cl}}_2{\text{O}}_6$ is a remarkably interesting compound.

At room temperature, it exists as a dark red liquid. However, when it freezes into a solid, it undergoes a fascinating structural reorganization. It does not exist as discrete ${\text{Cl}}_2{\text{O}}_6$ molecules in the solid state; instead, it forms an ionic lattice composed of chloryl cations and perchlorate anions:

Because it contains chlorine in both the $+5$ (in the cation) and $+7$ (in the anion) oxidation states, its average oxidation state is $+6$. Furthermore, when ${\text{Cl}}_2{\text{O}}_6$ reacts with water, it yields a mixture of chloric acid (${\text{HClO}}_3$) and perchloric acid (${\text{HClO}}_4$). For this reason, it is known as a mixed anhydride of these two acids.

By mastering these specific reactions and understanding the structural quirks of p-block compounds, you not only solve the problem at hand but also build a robust intuition for tackling the most challenging questions in advanced chemistry examinations.