Question

What is the oxidation number of oxygen in oxide, peroxide and superoxide.

Answer 1

Hint: To find the answer to this question, write the formulas of oxide, peroxide and superoxide ions correctly. Here, all three of them have atoms of the same element, so to find the oxidation number you can divide the charge on the ion by the number of atoms present in the atom.

Complete step by step answer:
Before finding the oxidation number of oxygen in the given oxides, we will discuss the few common rules that we should keep in mind while trying to find the oxidation numbers of different elements.
- We should know that the oxidation number of a free element is 0.
- If we have a monatomic ion then its oxidation number is equal to the charge on it. It can be positive, negative or neutral.
- Oxidation number of oxygen is generally -2 but in some cases it can be -1or $(-{\displaystyle \frac{1}{2}})$ too. The sum of oxidation numbers of all the atoms in a neutral compound is 0. If the compound has an overall charge, the sum of all the oxidation numbers will be equal to that. Oxidation number of hydrogen is +1 in most cases but with less electronegative elements it can be -1 too.

-There are some general oxidation numbers that are more or less accurate in all compounds like for group 1 it is +1, for group 2 it is +2 and for the halogens in a binary compound it is -1.
Now, let us try to find the oxidation number of oxygen in oxide.
We know that oxide is $O^{2-}$ . From the above rules we know that the oxidation number of a monatomic ion is equal to its charge. Here we have a single oxygen atom with a charge of -2. Therefore, the oxidation state of oxygen in oxide is -2.

Next, we have peroxide. We know peroxide is ${O_2}^{2-}$ . Here, we have 2 oxygen atoms and a charge of -2 which is equally distributed among all the atoms. Therefore, the charge on each atom will be ${\displaystyle \frac{-2}{2}}=-1$ .
Therefore, the oxidation number each oxygen atom in peroxide is -1.

And lastly, we have superoxide which is ${O_2}^{1-}$ .
Here, we have 2 oxygen atoms and a charge of -1 which is evenly distributed among the two atoms.
Therefore, the oxidation number of each atom will be $-{\displaystyle \frac{1}{2}}$ .
We can see from the above discussion that the oxygen number of oxygen in oxide is -2, in peroxide is -1 and in superoxide is $-{\displaystyle \frac{1}{2}}$ , thus this is the correct answer.

Note: Oxidation number is the degree of oxidation i.e. number of electrons the atom of the considered element has lost in a chemical compound. We also refer to the oxidation number as oxidation state. It can be positive, negative or even zero. According to IUPAC, oxidation state is defined as- “Oxidation state of an atom is the charge of this atom after ionic approximation of its hetero-nuclear bonds.”