Question

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your result?
a. $K{\underline I _3}$
b. ${H_2}{\underline S _4}{O_6}$
c. ${\underline {Fe} _3}{O_4}$
d. $\underline C {H_3}\underline C {H_2}OH$
e. $\underline C {H_3}\underline C OOH$

Answer 1

In simple words, the oxidation number is the number assigned to the components in a chemical combination. The oxidation number is the total number of electrons that atoms in a molecule can share, lose, or gain while forming chemical interactions with atoms of another element.

Complete answer:
(a) $K{\underline I _3}$
The oxidation number (O.N.) of $K$ in $K{I_3}$ is $+ 1$ . As a result, average oxidation number of $I$ is $- \dfrac{1}{3}$ . O.N., on the other hand, cannot be fractional. To determine the oxidation states, we must first consider the structure of $K{I_3}$ . An iodine atom forms a coordinate covalent bond with an iodine molecule in a $K{I_3}$ molecule.
$\mathop {{K^ + }}\limits^{ + 1} {\left[ {\mathop I\limits^0 - \mathop {I\,}\limits^0 \leftarrow \mathop I\limits^{ - 1} } \right]^ - }$
As a result, the O.N. of the two $I$ atoms that make up the ${I_2}$ molecule in a $K{I_3}$ molecule is $0$, whereas the O.N. of the $I$ atom that makes up the coordinate bond is $-1.$
(b) ${H_2}{\underline S _4}{O_6}$
${\mathop H\limits^{ + 1} _2}\mathop S\limits^x {O_4}{\mathop O\limits^{ - 2} _6}$
Now,
$2\left( { + 1} \right) + 4\left( x \right) + 6\left( { - 2} \right) = 0 \\\ \Rightarrow 2 + 4x - 12 = 0 \\\ \Rightarrow x = + 2\dfrac{1}{2} \\\$
O.N., on the other hand, cannot be fractional. As a result, $S$ in the molecule must exist in several oxidation states.

Two of the four S atoms have an O.N. of +5, whereas the other two have an O.N. of 0.
c. ${\underline {Fe} _3}{O_4}$
The O.N. of $Fe$ is determined to be when the O.N. of $O$ is set to $-2$ . ${}^{ + 2}\left( {\dfrac{2}{3}} \right)$ O.N., on the other hand, cannot be fractional.
One of the three $Fe$ atoms in this example has an O.N. of $+ 2$ , whereas the other two $Fe$ atoms have an O.N. of $+ 3$ .
$\mathop {Fe}\limits^{ + 2} O,\mathop {Fe}\limits^{ + 3} {O_3}$
d. $\underline C {H_3}\underline C {H_2}OH$
${\mathop C\limits^x _2}{\mathop H\limits^{ + 1} _6}{\mathop O\limits^{ - 2} _2}$

$$2\left( x \right) + 4\left( { + 1} \right) + 2\left( { - 2} \right) = 0 \\\ \Rightarrow 2x + 4 - 4 = 0 \\\ \Rightarrow x = 0 \\\$$

Hence, the O.N of $C$ is $-2$ .
e. $\underline C {H_3}\underline C OOH$
${\mathop C\limits^x _2}{\mathop H\limits^{ + 1} _4}{\mathop O\limits^{ - 2} _2}$
$2\left( x \right) + 4\left( { + 1} \right) + 2\left( { - 2} \right) = 0 \\\ \Rightarrow 2x + 4 - 4 = 0 \\\ \Rightarrow x = 0 \\\$
The average O.N. of $C$ , on the other hand, is $0$. This molecule's two carbon atoms are found in two separate settings. As a result, their oxidation numbers cannot be the same. As a result, $C$in $C{H_3}COOH$ has the oxidation states $+ 2$ and $- 2$

Note:
A negative oxidation state is attributed to the more electronegative element in a material. A positive oxidation state is attributed to the less electronegative element. Keep in mind that electronegativity is highest at the top-right corner of the periodic table and falls toward the bottom-left corner.