Question

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
$K{\underset{–}{I}}_3$

$H_2{\underset{–}{S}}_4{O}_6$

${\underset{–}{Fe}}_3{O}_4$

$\underset{–}{C}{H}_{3}C{H}_{2}OH$

$\underset{–}{C}{H}_3\underset{–}{C}OOH$

Answer 1

$K{\underset{–}{I}}_3$
In $K{I}_3$, 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 $K{I}_3$ to find the oxidation states.
In a $K{I}_3$ molecule, and atom of iodine forms a coordinate covalent bond with an iodine molecule.
$\stackrel{+1}{K^{+}}[\stackrel{0}{I}-\stackrel{0}{I}\leftarrow \stackrel{-I}{I}]$
Hence, in a KI3 molecule, the O.N. of the two I atoms forming the $I_2$ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

$H_2{\underset{–}{S}}_4{O}_6$
$\stackrel{+1}{H_2}\stackrel{x}{S{O}_4}\stackrel{-2}{O_6}$
Now, $2(+1)+4\left(x\right)+6(-2)=0\Rightarrow 2+4x-12=0\Rightarrow 4x=10\Rightarrow x=+2\frac{1}{2}$
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.

${\underset{–}{Fe}}_3{O}_4$
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.
$\stackrel{+2}{FeO,}\stackrel{+3}{F{e}_2{O}_3}$

$\underset{–}{C}{H}_{3}C{H}_{2}OH$
$\stackrel{x}{C_2}\stackrel{+1}{H_6}\stackrel{-2}{O}2\left(x\right)+4(+1)+1(-2)=0\Rightarrow 2x+6-2=0\Rightarrow x=-2$
Hence, the O.N. of C is - 2.

$\stackrel{x}{C_2}\stackrel{+1}{H_4}\stackrel{2}{O_1}2\left(x\right)+4(+1)+2(-2)=0\Rightarrow 2x+4-4=0\Rightarrow x=0$
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 $C{H}_{3}COOH.$