What does Oxidation number actually signify? If it means the number of electrons gained or accepted, why is it a fraction in some cases like in super oxides?


1 Answer 1


Quoting from Wikipedia:

The oxidation state, often called the oxidation number, is an indicator of the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds.

Coming to fractional oxidation state:

It is not actually a fraction rather it is the average of various oxidation states of the element present in the compound.

For example:

  1. $\ce{Fe3O4}$

enter image description here

Here the two Iron atoms that are bonded to three oxygen atoms have an oxidation state of +3 each, and the Iron atom that is bonded to two oxygen atom has an oxidation state of +1.

Thus taking the average (3+3+2)/3 = 8/3;

You would have got the same if you used the fact that $\ce{Fe3O4}$ is neutral.If you assume the oxidation state of Iron to be x, then 3x+4(-2)=0. Solving you would get x as 8/3(However this oxidation state would be the average oxidation state of Iron).

  1. $\ce{KO2}$

enter image description here

In this case the oxygen on the left has an oxidation state of zero as it is only bonded to itself and the oxygen on the right hand side is in oxidation state of -1 (1 bond to another O and a charge of -1). In reality the -1 charge would be spread over the whole species - oxidation state is only a system of chemical " book-keeping". The oxidation state of -1/2 can be regarded as the mean of zero and -1.

There are many more such examples.

Image courtesy:ChemSpider

  • 1
    $\begingroup$ So the two oxygen atoms in $\ce{KO2}$ are different? $\endgroup$
    – DHMO
    Dec 31, 2016 at 16:49
  • $\begingroup$ No, they aren't and these pictures are quite bad... $\endgroup$
    – Mithoron
    Aug 31, 2020 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.