Q.Calculate

[!TLDR] Average O.N. of Fe in $\text{Fe}_3\text{O}_4$ from the neutral-formula sum; Cl oxidation numbers in the oxoacid series rise by 2 with each added oxygen. [!ANSWER] (a) average $\text{Fe} = +\tfrac{8}{3} \approx +2.67$;

(b) $\text{Cl}$ is $+1$ in $\text{HOCl}$, $+3$ in $\text{HClO}_2$, $+5$ in $\text{HClO}_3$, $+7$ in $\text{HClO}_4$.

(a) In $\text{Fe}_3\text{O}_4$, four O atoms contribute $4(-2) = -8$. Neutral formula: $3x + (-8) = 0 \Rightarrow x = \tfrac{8}{3} \approx +2.67$. This is an average value — $\text{Fe}_3\text{O}_4$ is chemically a mixed oxide, $\text{FeO}\cdot\text{Fe}_2\text{O}_3$, containing one $\text{Fe}^{2+}$ and two $\text{Fe}^{3+}$ per formula unit, whose true average is $\tfrac{2+3+3}{3} = \tfrac{8}{3}$, consistent with the formula-sum calculation.\n\n(b) In each oxoacid, H is $+1$ and each O is $-2$; solve for Cl using the neutral-formula sum. $\text{HOCl}$: $1 + x - 2 = 0 \Rightarrow x=+1$. $\text{HClO}_2$: $1+x-4=0 \Rightarrow x=+3$. $\text{HClO}_3$: $1+x-6=0 \Rightarrow x=+5$. $\text{HClO}_4$: $1+x-8=0 \Rightarrow x=+7$. Each additional oxygen atom raises the oxidation number of chlorine by exactly 2, since each O contributes $-2$ that chlorine's oxidation number must compensate for. [!ANSWER] (a) average $\text{Fe} = +\tfrac{8}{3} \approx +2.67$; (b) $\text{Cl}$ is $+1$ in $\text{HOCl}$, $+3$ in $\text{HClO}_2$, $+5$ in $\text{HClO}_3$, $+7$ in $\text{HClO}_4$.

Apply the neutral-formula sum rule, treating the whole formula as one unit for a mixed-oxidation-state compound to get an average; repeat the same rule for each acid in the series.

Assuming every Fe atom in $\text{Fe}_3\text{O}_4$ individually carries the fractional value $+2.67$, rather than understanding it as an average of two true integer oxidation states.