What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?
$(1)$ $K\underline{I_3}$
$(2)$ $H_2\underline{S_4}O_6$
$(3)$ $\underline{Fe_3}O_4$
$(4)$ $\underline{C}H_3\underline{C}H_2OH$
$(5)$ $\underline{C}H_3\underline{C}OOH$

(N/A) $(1)$ $KI_3$: $K$ has an oxidation number of $+1$. The $I_3^-$ ion consists of an $I_2$ molecule coordinated to an $I^-$ ion. The oxidation numbers of the two iodine atoms in $I_2$ are $0$,and the oxidation number of the $I^-$ ion is $-1$. Thus,the oxidation numbers are $0, 0, -1$.
$(2)$ $H_2S_4O_6$ (Tetrathionic acid): Based on the structure $HO_3S-S-S-SO_3H$,the two central sulfur atoms have an oxidation number of $0$,while the two terminal sulfur atoms bonded to three oxygen atoms and one sulfur atom have an oxidation number of $+5$.
$(3)$ $Fe_3O_4$: This is a mixed oxide $FeO \cdot Fe_2O_3$. In $FeO$,$Fe$ is $+2$. In $Fe_2O_3$,$Fe$ is $+3$. The average oxidation number is $8/3$,but the actual oxidation states are $+2$ and $+3$.
$(4)$ $CH_3CH_2OH$: For $CH_3$,$C$ is bonded to $3H$ ($+1$ each) and $1C$ $(0)$. Thus,$C + 3(+1) = 0 \Rightarrow C = -3$. For $CH_2OH$,$C$ is bonded to $2H$ ($+1$ each),$1O$ $(-2)$,$1H$ $(+1)$,and $1C$ $(0)$. Thus,$C + 2(+1) + 1(-2) + 1(+1) = 0 \Rightarrow C = -1$.
$(5)$ $CH_3COOH$: For $CH_3$,$C$ is $-3$. For $COOH$,$C$ is bonded to $2O$ ($-2$ each) and $1OH$ $(-1)$. Thus,$C + 2(-2) + (-1) = 0 \Rightarrow C = +3$.