Question

Oxidation number of potassium in${{\text{K}}_{\text{2}}}{\text{O}}$, ${{\text{K}}_{\text{2}}}{{\text{O}}_2}$ and ${\text{K}}{{\text{O}}_2}$, respectively, is:
(A) ${\text{ + 1, + 4 and + 2}}$
(B) ${\text{ + 1, + 1 and + 1}}$
(C) ${\text{ + 1, + 2 and + 4}}$
(D) ${\text{ + 2, + 1 and + }}\dfrac{{\text{1}}}{{\text{2}}}$

Answer 1

Hint: Potassium is the element which belongs to the alkali metal group. All the alkali metals show only one type of oxidation state. There is no trend in case of oxidation state as we move down the group.

Complete step by step answer:
Oxidation state is also referred to as oxidation number. It indicates the number of electrons lost or gained by an atom. Oxidation state may be positive, negative or zero. Any substance which is in its elemental state will have zero oxidation state. The increase in the oxidation state refers to oxidation of that atom while decrease in the oxidation state refers to reduction of that atom.

Oxidation state of any compound is equal to the overall charge that it carries. For example a compound like ${\text{CaC}}{{\text{O}}_{\text{3}}}$, the compound does not carry any charge which means the compound is neutral overall. In this case the sum of the oxidation state of all the atoms will be equal to zero. Now for an anion such as${\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$, it carries net charge of -2. Therefore the sum of oxidation states of Sulphur and oxygen atoms will be equal to -2.
As per the question there are three different types of potassium oxides. So before we find the oxidation of Potassium, we should know the oxidation state of oxygen.

Oxygen shows oxidation state of -2 in most of the compounds. As it has the valency of 2. But there is an exception regarding the oxidation state of oxygen as p-block elements show variable oxidation state. The different oxidation states of oxygen are as follows:

1) It shows -2 in most of the compounds like${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{, ZnO}}$.
2) It shows -1 in compound like peroxide ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$
3) It shows $ - \dfrac{1}{2}$ oxidation state in superoxide like ${{\text{K}}_{\text{2}}}{{\text{O}}_{\text{2}}}$
4) Oxygen shows +1 oxidation state in compounds like ${{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$
5) It shows +2 oxidation states in ${\text{O}}{{\text{F}}_{\text{2}}}$

Oxidation state of Potassium in ${{\text{K}}_{\text{2}}}{\text{O}}$: ${{\text{K}}_{\text{2}}}{\text{O}}$ is Potassium oxide. So the oxidation state of oxygen in oxides is -2. The compound has no net charge. Thus the oxidation state of potassium can be calculated as follows:
${\text{oxidation state of K }} \times {\text{ 2 + oxidation state of oxygen = 0}}$
$\therefore {\text{ x }} \times {\text{ 2 + ( - 2) = 0}}$
$\therefore {\text{ 2x + - 2 = 0}}$
$\therefore {\text{ 2x = + 2}}$
$\therefore {\text{ x = + 1}}$
So the oxidation state of potassium is +1 in ${{\text{K}}_{\text{2}}}{\text{O}}$.

Oxidation state of Potassium in ${{\text{K}}_{\text{2}}}{{\text{O}}_2}$: ${{\text{K}}_{\text{2}}}{{\text{O}}_2}$ is Potassium peroxide. So the oxidation state of oxygen in oxides is -1. The compound has no net charge. Thus the oxidation state of potassium can be calculated as follows:
${\text{oxidation state of K }} \times {\text{ 2 + oxidation state of oxygen }} \times {\text{ 2 = 0}}$
$\therefore {\text{ x }} \times {\text{ 2 + - 1 }} \times {\text{ 2 = 0}}$
$\therefore {\text{ 2x + - 2 = 0}}$
$\therefore {\text{ 2x = + 2}}$
$\therefore {\text{ x = + 1}}$
So the oxidation state of potassium is +1 in ${{\text{K}}_{\text{2}}}{{\text{O}}_2}$.

Oxidation state of Potassium in ${\text{K}}{{\text{O}}_2}$: ${\text{K}}{{\text{O}}_2}$ is Potassium superoxide. So the oxidation state of oxygen in oxides is $ - \dfrac{1}{2}$. The compound has no net charge. Thus the oxidation state of potassium can be calculated as follows:
\[{\text{oxidation state of K + oxidation state of oxygen}} \times {\text{ 2 = 0}}\]
\[\therefore {\text{ x + ( - }}\dfrac{1}{2}{\text{) }} \times {\text{ 2 = 0}}\]
$\therefore {\text{ x + - 1 = 0}}$
$\therefore {\text{ x = + 1}}$

So the oxidation state of potassium is +1 in ${\text{K}}{{\text{O}}_2}$.
Thus oxidation of potassium in all the three different oxides is +1
So, the correct answer is “Option B”.

Note: When oxygen is bonded to more electronegative atoms then in that case it shows positive oxidation state. In this case the atom which is more electronegative will pull the electrons towards it and acquire negative charge. The only atom which is more electronegative than oxygen is fluorine. The second important point is elements of s block do not show variable oxidation state while elements of p, d and f block show variable oxidation states.