What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
$(a)$ $K\underline {I_3}$
$(b)$ $H_2\underline {S_4}O_6$
$(c)$ $\underline {Fe_3}O_4$
$(d)$ $\underline {C}H_3\underline {C}H_2OH$
$(e)$ $\underline {C}H_3\underline {C}OOH$

(N/A) $(a)$ $K\underline {I_3}$: The average oxidation number of $I$ is $-1/3$. Since $O.N.$ cannot be fractional,we consider the structure: $K^+ [I-I \leftarrow I]^-$. The two terminal $I$ atoms have $O.N.$ of $0$,and the central $I$ atom has $O.N.$ of $-1$.
$(b)$ $H_2\underline {S_4}O_6$: The average $O.N.$ of $S$ is $+2.5$. In the structure $HO_3S-S-S-SO_3H$,the two terminal $S$ atoms have $O.N.$ of $+5$ and the two central $S$ atoms have $O.N.$ of $0$.
$(c)$ $\underline {Fe_3}O_4$: The average $O.N.$ of $Fe$ is $+8/3$. $Fe_3O_4$ is a mixed oxide $FeO \cdot Fe_2O_3$,where one $Fe$ is $+2$ and two $Fe$ atoms are $+3$.
$(d)$ $\underline {C}H_3\underline {C}H_2OH$: The average $O.N.$ of $C$ is $-2$. Specifically,the $CH_3$ carbon is $-3$ and the $CH_2OH$ carbon is $-1$.
$(e)$ $\underline {C}H_3\underline {C}OOH$: The average $O.N.$ of $C$ is $0$. Specifically,the $CH_3$ carbon is $-3$ and the $COOH$ carbon is $+3$.