Question

What is the oxidation state of K in $K{O_2}?$
A)-1
B)-2
C)+1
D)0

Answer 1

For solving this question, we need to consider that K here is an alkali metal. And here $K{O_2}$ is a neutral molecule. We will assign it with the net charge of 0 for further finding out the oxidation state.

Complete step by step answer:
Firstly, we need to see that $K{O_2}$ has K i.e. potassium which is a member of group 1. So, it can readily react with oxygen. We know that, as we move down the group the tendency of forming peroxides and superoxides increases.
So, as we know $K{O_2}$ is a superoxide of potassium. In $K{O_2}$, let’s consider that the oxidation number of oxygen is x.
Therefore, 1 + 2x = 0
Since, we know that K= +1 and $K{O_2}$​ is a neutral molecule. It has only one electron in its valence shell and can only take the +1 oxidation state.
Here, K shows an oxidation state of K+ where as other half forms ${O_2}^ -$. Thus, $K{O_2}$is paramagnetic in nature.
To find the oxidation state we can easily do it by using the formula which is the sum of the oxidation states of all the elements present in the compound is equal to the net charge present on the compound. In the case of $K{O_2}$we have seen that K belongs to group 1 so it will always possess an oxidation state of +1.

$\therefore$The option C is correct answer.

Note:
In case we have to find the oxidation state of O in $K{O_2}$ then we will consider x. Since $K{O_2}$is a neutral molecule so it has a net charge equal to zero on the compound.
So,$1 + 2x = 0$
Thus, the value of x=-1/2
Therefore, the Oxidation state of O in $K{O_2}$ is -1/2.