Step-by-step Solution:

1. Identify the Reactants and Reaction: The problem describes the ozonolysis of chlorine dioxide ($ClO_2$). This means $ClO_2$ is reacting with ozone ($O_3$). Both are strong oxidizing agents, but ozone is generally stronger and will oxidize the chlorine in $ClO_2$.

2. Determine the Oxidation State of Chlorine in the Reactant: In chlorine dioxide ($ClO_2$), let the oxidation state of chlorine be $x$. The oxidation state of oxygen is typically -2. $x + 2(-2) = 0$ $x - 4 = 0$ $x = +4$ So, the initial oxidation state of chlorine is +4.

3. Write the Chemical Reaction: Ozone ($O_3$) will oxidize $ClO_2$ to a higher oxidation state oxide of chlorine. The balanced chemical reaction for the ozonolysis of $ClO_2$ is: $2ClO_2 + 2O_3 \rightarrow Cl_2O_6 + 2O_2$ This reaction produces dichlorine hexoxide ($Cl_2O_6$) and oxygen gas.

4. Identify the Product Oxide: The oxide of chlorine formed is dichlorine hexoxide, with the chemical formula $Cl_2O_6$.

5. Calculate the Average Oxidation State of Chlorine in the Product: To find the average oxidation state of chlorine in $Cl_2O_6$, we again use the rule that the sum of oxidation states in a neutral compound is zero. Let the average oxidation state of chlorine be $y$. The oxidation state of oxygen is -2. The equation for the sum of oxidation states in $Cl_2O_6$ is: $2(y) + 6(-2) = 0$ $2y - 12 = 0$ $2y = 12$ $y = +6$

   Alternatively, $Cl_2O_6$ exists in an ionic form as chloryl perchlorate, $[ClO_2]^+[ClO_4]^-$.

   • In the cation $[ClO_2]^+$, the oxidation state of Cl is +5 (since $x + 2(-2) = +1 \Rightarrow x = +5$).
   • In the anion $[ClO_4]^-$, the oxidation state of Cl is +7 (since $x + 4(-2) = -1 \Rightarrow x = +7$). The average oxidation state is the average of these two values: $\text{Average Oxidation State} = \frac{(+5) + (+7)}{2} = \frac{12}{2} = +6$

6. Final Answer: The average oxidation state of chlorine in the resulting oxide ($Cl_2O_6$) is 6.