Question

Oxidation state of oxygen in ${\text{K}}{{\text{O}}_{\text{2}}}$is:
A. $0$
B. $ - 1$
C.$ - 1/2$
D. $ - 2$

Answer 1

Hint:Oxidation number is the number of electrons gained or lost by the atoms. The charge of an atom represents the oxidation number of that atom. Some atoms show fixed oxidation states so, by adding or subtracting the oxidation number of known and unknown we can calculate the oxidation number of the desired atom.

Complete answer
An atom loss or gain electrons to form a bond with other atoms. So, the atoms get charged known as ions. The superscript of the ions represents the oxidation number of that atom. In neutral molecules, the sum of the oxidation number is equal to zero. In charged molecules, the sum of the oxidation number is equal to the charge of the molecule.The atoms in elemental numbers have zero oxidation number.

The alkali metals have the oxidation number$ + 1$. The transition metals show a variable oxidation number.

The oxidation number of oxygen in ${\text{K}}{{\text{O}}_{\text{2}}}$ is as follows:
Potassium is an alkali metal so the oxidation number of potassium is $ + 1$.
$\Rightarrow \,\left( { + 1 \times 1} \right)\, + \,\left( {x \times 2} \right)\, = \,0$
$\Rightarrow \,2x = \, - 1$
$\Rightarrow x = \, - 1/2$

The oxidation number of oxygen is$ - 1/2$.So, the oxidation state of oxygen in potassium oxide${\text{K}}{{\text{O}}_{\text{2}}}$ is $ \dfrac{-1}{2}$.

Therefore,the correct answer is option (C) $ \dfrac{-1}{2}$.

Note:In potassium oxide, the oxidation state of oxygen is ${\text{O}}_2^ - $ so, oxygen is present in superoxide form. We know that the general oxidation state shown by oxygen is $ - 2$. Oxygen also forms peroxide and superoxide. The oxygen atom in peroxides always has $ - 1$oxidation number. In superoxide oxygen is found as diatomic and oxygen in superoxide form is shown as${\text{O}}_2^ - $ . The more electronegative atom has a negative oxidation number, and less electronegative has a positive oxidation number.