Animated Solution for Chemistry - s and p-Block Elements: Ozonolysis of ${\text{ClO}}_2$ produces an oxide of chlorine. The average oxidation state of chlorine in this oxide is _______.

When we dive into the chemistry of the p-block, halogens stand out for their ability to form a wide variety of compounds, especially with oxygen. Chlorine, being highly electronegative but still less electronegative than oxygen, can exhibit a range of positive oxidation states from $+1$ all the way up to $+7$.

This versatility leads to the formation of several intriguing oxides, such as dichlorine monoxide ($C{l}_{2}O$), chlorine dioxide ($Cl{O}_2$), dichlorine hexoxide ($C{l}_2{O}_6$), and dichlorine heptoxide ($C{l}_2{O}_7$). Each of these oxides has unique structural properties and chemical behaviors. In this problem, we are specifically looking at the transformation of one such oxide into another through a powerful oxidation process.

Ozonolysis of Chlorine Dioxide

The problem states that chlorine dioxide ($Cl{O}_2$) undergoes ozonolysis. In organic chemistry, ozonolysis usually means breaking double bonds, but in inorganic chemistry, it simply refers to an oxidation reaction driven by ozone ($O_3$). Ozone is a incredibly potent oxidizing agent.

When ozone reacts with chlorine dioxide, it forces the chlorine atom to give up more electron density, elevating it to a higher oxidation state. The balanced chemical equation for this fascinating transformation is:

The product formed is dichlorine hexoxide ($C{l}_2{O}_6$). Knowing this specific reaction is a classic requirement for JEE Advanced, as it tests your memory of key p-block reactions.

Now that we have identified the mysterious oxide as $C{l}_2{O}_6$, the next step is to calculate the average oxidation state of the chlorine atoms within it. This is a straightforward application of algebraic rules for oxidation numbers.

We know that in almost all of its compounds (except when bonded to fluorine or in peroxides), oxygen exhibits an oxidation state of $-2$. Since the $C{l}_2{O}_6$ molecule is electrically neutral, the sum of the oxidation states of all its constituent atoms must equal zero.

Let the average oxidation state of chlorine be $x$. Since there are two chlorine atoms and six oxygen atoms, we can set up the following equation:

Now, we simply solve for $x$. Multiplying the terms gives us:

Moving the $12$ to the other side of the equation:

Finally, dividing by $2$, we arrive at our answer:

Thus, the average oxidation state of chlorine in dichlorine hexoxide is $6$.

Average vs. Actual Oxidation States

While the mathematical calculation gives us an average oxidation state of $+6$, the physical reality of the $C{l}_2{O}_6$ molecule is much more nuanced and beautiful.

Does a chlorine atom actually sit in a $+6$ state? The answer depends on the physical state of the compound! In its gaseous and liquid forms, $C{l}_2{O}_6$ exists as a covalent molecule, often represented as chloryl perchlorate ($O_{2}Cl-O-Cl{O}_3$).

However, when it freezes into a solid, it undergoes a remarkable transformation. It auto-ionizes to form an ionic crystal lattice consisting of chloryl cations and perchlorate anions:

If we calculate the oxidation state of chlorine in the chloryl cation ($[Cl{O}_2{]}^{+}$), we get $+5$. If we calculate it for the perchlorate anion ($[Cl{O}_4{]}^{-}$), we get $+7$.

This means that in the solid state, there is no single chlorine atom with a $+6$ oxidation state! Instead, we have a perfect mix of $+5$ and $+7$ states. When we take the mathematical average of these two distinct states:

We arrive right back at the $+6$ we calculated earlier. This is a brilliant example of why the question specifically asked for the average oxidation state. It is these deep, structural insights that make inorganic chemistry so rewarding to study.