Question

The oxidation number of Iodine in each of the following compounds, $I{F_7},I{F_5},KI,{I_2},ICl$ and $HI{O_4}$ respectively, is-
(A) +7, +5, -1, 0, +1, +7
(B) +7, +5, -1, 0, +2, +7
(C) +7, +5, -1, 0, +4, +7
(D) None of these

Answer 1

Hint: Iodine is present in several stages of oxidation, namely iodide ($I^−$), and iodate ($IO^{−3}$) and separate anions of the periodate. It is the most stable halogen in abundance, being the 61st most abundant element. It is the most essential element in minerals. We will find out the oxidation number of iodine in each compound one by one.

Complete step by step solution:
Let the oxidation number of iodine be x.

Case-1
The given compound is $I{F_7}$.
As we know, the oxidation state of Fluorine is -1. It is the most electronegative atom.
There is no charge present on the compound which means that its total oxidation state will be 0. From this we get-
$\Rightarrow x + 7 \times - 1 = 0 \\\ \\\ \Rightarrow x = 7 \\\$
Hence, the oxidation number of iodine in $I{F_7}$ is +7.

Case-2
The compound given is $I{F_5}$.
As we know, the oxidation state of Fluorine is -1. It is the most electronegative atom.
There is no charge present on the compound which means that its total oxidation state will be 0. From this we get-
$\Rightarrow x + 5 \times - 1 = 0 \\\ \\\ \Rightarrow x = + 5 \\\$
Hence, the oxidation number of iodine in $I{F_5}$ is +5.

Case-3
The given compound is $KI$.
As we know, the oxidation state of potassium is +1.
There is no charge present on the compound which means that its total oxidation state will be 0. From this we get-
\Rightarrow + 1 + x = 0 \\\ \\\ \Rightarrow x = - 1 \\\
Hence, the oxidation number of iodine in $KI$ is -1.

Case-4
The given compound is ${I_2}$.
There is no charge present on the compound which means that its total oxidation state will be 0. From this we get-
$\Rightarrow 2 \times x = 0 \\\ \\\ \Rightarrow x = 0 \\\$
Hence, the oxidation number of iodine in ${I_2}$ is 0.

Case-5
The given compound is $ICl$.
As we know, the oxidation state of $Cl$ is -1.
There is no charge present on the compound which means that its total oxidation state will be 0. From this we get-
$\Rightarrow x + \left( { - 1} \right) = 0 \\\ \\\ \Rightarrow x = + 1 \\\$
Hence, the oxidation number of iodine in $ICl$ is +1.

Case-6
The given compound is $HI{O_4}$.
As we know, the oxidation state of $H$ is +1 and the oxidation state of $O$ is -2.
There is no charge present on the compound which means that its total oxidation state will be 0. From this we get-
$\Rightarrow + 1 + x + 4 \times \left( { - 2} \right) = 0 \\\ \\\ \Rightarrow 1 + x - 8 = 0 \\\ \\\ \Rightarrow x - 7 = 0 \\\ \\\ \Rightarrow x = + 7 \\\$
Hence, the oxidation number of iodine in $HI{O_4}$ is +7.
Thus, the oxidation number of Iodine in each compound is +7, +5, -1, 0, +1, +7. Hence, we can say that option A is the correct option.

Note: The cumulative number of electrons that an atom either gains in or loses to form a chemical bond with another atom, this is known as oxidation number. An oxidation number that represents the capacity to gain, transfer or disperse electrons is given for each atom that participates in an oxidation reduction reaction.