Question

The oxidation number of oxygen in $K{O_2}$ is:
A. -2
B. -1
C. $\dfrac{{ - 1}}{2}$
D. $\dfrac{{ - 1}}{3}$

Answer 1

Hint: Just try to recall that oxidation number is the number of electrons that an atom either gains or loses in order to form a chemical bond with another atom. Now by knowing the basic rules of oxidation state you can easily find the oxidation state of oxygen in $K{O_2}$.

Complete step by step solution:
There are some rules to assign an oxidation number to a given element or compound.
The rules are as follows:
Rule 1: The oxidation state of any free element is equal to 0. For example: the oxidation number of $Na$ is 0.
Rule 2: If any monatomic ion is given, then the oxidation number of that ion is its net charge.
Rule 3: In most compounds, oxidation state of oxygen atom is -2 except in case of peroxides where it is -1.
Rule 4: All alkali metals i.e. group 1 elements exhibit oxidation state of +1 in their compounds.
Rule 5: Oxidation number of all alkaline earth metals i.e. group 2 elements are taken +2 in their compounds.
Rule 6: Oxidation number of all halogens are taken as -1 except some cases. But the oxidation number of fluorine is always taken as -1 in their compounds
Rule 7: If the given species is neutral then, the sum of all the oxidation numbers of the constituent atoms equals to 0.
Coming to the question now you can easily calculate the oxidation state of oxygen in $K{O_2}$ by using above stated rules:
Calculation:
Since, Potassium is an alkali metal
So, from rule 4 oxidation number of potassium, K= +1
Let the oxidation number of oxygen be x in $K{O_2}$.
So, By Rule 7:
$
  ( + 1) + 2 \times x = 0 \\
   \Rightarrow 1 + 2x = 0 \\
   \Rightarrow 2x = ( - 1) \\
x = \dfrac{{ - 1}}{2} \\
$
Therefore, from the above calculation we can easily say that option C is the correct option to the given question.

Note: If a polyatomic ion is given then in that case, the sum of all oxidation numbers of all atoms that constitute them equals the net charge of the polyatomic ion.
Also, you should know that $K{O_2}$ is a very stable superoxide and is paramagnetic due to the presence of unpaired electrons.