
What is the oxidation state of K in $K{{O}_{2}}$?
A. -1
B. -2
C. +1
D. 0
Answer
561.9k+ views
Hint: To solve this question we have to consider that K given here as an alkali metal that is potassium and the we will take $K{{O}_{2}}$ as the neutral molecule with net charge 0 for further finding the oxidation state of K. The oxidation state of all the alkali metals is only of one type.
Complete Solution :
- From your chemistry lessons you have learned about the oxidation state. Oxidation state is also called as the oxidation number and it is defined as the number of electrons gained or lost by an atom. The oxidation state of any atom can be negative, positive as well as zero. The substance which is present in its elemental state will have the oxidation state as zero.
- If the oxidation state of an atom is increasing then it refers to the oxidation of the atom whereas if the oxidation state is decreasing then it refers to reduction of that atom. And the oxidation state of any compound will be equal to the overall charge that is present at the atom.
- As in the question we have asked to find the oxidation state of K in $K{{O}_{2}}$ compound, so the oxidation state of the compound will be zero due to zero net charge on the compound.
Now to find the oxidation state of K we should know the Oxidation state of oxygen. There are variable oxidation states shown by the oxygen because the oxidation states of the p-block element vary. The oxidation state shown by oxygen in most of the cases are -2. But in this case the oxidation state of oxygen will be $-\dfrac{1}{2}$ because the compound form here is Potassium superoxide. And as we know that the net charge on the compound will be zero. Therefore the oxidation of K will be taken as 'X' and the oxidation state can be calculated as:
\[Oxidation\,state\,of\,K+Oxidation\,state\,of\,O\times 2=0\]
\[\therefore X+\left( -\dfrac{1}{2} \right)\times 2=0\]
\[\therefore X=+1\]
Therefore the oxidation state of K in $K{{O}_{2}}$ is +1.
So, the correct answer is “Option C”.
Note: The other oxidation states shown by oxygen is -2, -1, +1 and +2. In case oxygen is bonded to more electronegative atoms then the oxidation state shown by that oxygen will be positive because the atom with more electronegativity will tend to pull the electron towards themselves and thus acquires negative charge on them. The more electronegative atom other than oxygen is only fluorine.
Complete Solution :
- From your chemistry lessons you have learned about the oxidation state. Oxidation state is also called as the oxidation number and it is defined as the number of electrons gained or lost by an atom. The oxidation state of any atom can be negative, positive as well as zero. The substance which is present in its elemental state will have the oxidation state as zero.
- If the oxidation state of an atom is increasing then it refers to the oxidation of the atom whereas if the oxidation state is decreasing then it refers to reduction of that atom. And the oxidation state of any compound will be equal to the overall charge that is present at the atom.
- As in the question we have asked to find the oxidation state of K in $K{{O}_{2}}$ compound, so the oxidation state of the compound will be zero due to zero net charge on the compound.
Now to find the oxidation state of K we should know the Oxidation state of oxygen. There are variable oxidation states shown by the oxygen because the oxidation states of the p-block element vary. The oxidation state shown by oxygen in most of the cases are -2. But in this case the oxidation state of oxygen will be $-\dfrac{1}{2}$ because the compound form here is Potassium superoxide. And as we know that the net charge on the compound will be zero. Therefore the oxidation of K will be taken as 'X' and the oxidation state can be calculated as:
\[Oxidation\,state\,of\,K+Oxidation\,state\,of\,O\times 2=0\]
\[\therefore X+\left( -\dfrac{1}{2} \right)\times 2=0\]
\[\therefore X=+1\]
Therefore the oxidation state of K in $K{{O}_{2}}$ is +1.
So, the correct answer is “Option C”.
Note: The other oxidation states shown by oxygen is -2, -1, +1 and +2. In case oxygen is bonded to more electronegative atoms then the oxidation state shown by that oxygen will be positive because the atom with more electronegativity will tend to pull the electron towards themselves and thus acquires negative charge on them. The more electronegative atom other than oxygen is only fluorine.
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