(a) KI${}_3$
In KI${}_3$, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is $\frac{1}{3}$. However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI${}_3$ to find the oxidation states. In a KI${}_3$ molecule, an atom of iodine forms a coordinate covalent bond with an iodine

molecule.

Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

(b) H${}_2$S${}_4$O${}_6$

Now, 2(+1)+4(x)+6(-2)=0

$\Rightarrow$2+4x-12=0

$\Rightarrow$4x=10

$\Rightarrow$x=+2$\frac{1}{2}$

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) $\underline{\mathrm{F}}{\mathrm{e}}_3{\mathrm{O}}_4$

On taking the O.N. of O as –2, the O.N. of Fe is found to be $+2\frac{2}{3}$. However, O.N. cannot be fractional.
Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.

$\stackrel{+2}{\mathrm{F}}\mathrm{eO},\stackrel{+3}{\mathrm{F}}{\mathrm{e}}_2{\mathrm{O}}_3$

(d) $\underline{\mathrm{C}}{\mathrm{H}}_{3 }\underline{\mathrm{C}}{\mathrm{H}}_2\mathrm{OH}$

2(x)+4(+1)+1(-2)=0

$\Rightarrow$ 2x+6-2=0

$\Rightarrow$x=-2

Hence, the O.N. of C is –2.

(e) ${\mathrm{CH}}_3\underline{C}\mathrm{OOH}$

2(x)+4(+1)+2(-2)=0

$\Rightarrow$2x+4-4=0

$\Rightarrow$x=0

However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number.
Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH.