For example, take $\ce{O2}$.

The oxidation state of oxygen is -2, yet once its in a molecule its oxidation state becomes zero?

How is this so?

  • 1
    $\begingroup$ Very strictly speaking, I don't think this must necessarily be true. It is possible to imagine allotropes of a pure element which would consist of ions and thus would require non-zero oxidation numbers. For example, the hypothetical octanitrogen $\ce{N8}$ allotrope could be composed of the azide anion $\ce{N3^{-}}$ and the pentazenium cation $\ce{N5^{+}}$. Granted, I do not believe this to be the case for any element, at least not in ambient conditions. $\endgroup$ Jul 16, 2017 at 12:13
  • $\begingroup$ By definition oxidation state or oxidation number is related to the number of electrons that an atom, molecule or any substance either gains or loses while forming a new substance in a chemical reaction. It is positive when electron leaves and negative when electron enters an atom. In some cases it can be zero because of sharing the electron pair to form a coordinate covalent bonds between identical atoms having same electro negativity values. $\endgroup$ Jul 8, 2022 at 7:56

3 Answers 3


Oxidation states are something of an "accounting fiction", used to describe where electrons are spending most of their time in a molecule. This is turn is dictated by electronegativity: the more electronegative, the more the element attracts electrons.

So when you say oxygen's oxidation state is -2, this is only with respect to other elements of lower electronegativity. Now in the case of oxygen, the only other element with higher electonegativity is fluorine. For example, the molecule Oxygen Difluoride ($\ce{OF2}$) has oxygen in the +2 oxidation state because fluorine is more electronegative! For any other element in combination with oxygen, however, we assume the electrons spend their time with the oxygen filling its outer valence shell, thus the -2 oxidation state.

But two oxygens in a molecule of gas have the same electronegativity, and there is no way to say that the electrons want to spend more time with one or the other oxygen. Therefore, the "average" electron state for an oxygen molecule is as for the individual atoms, and so the oxidation state is 0.


We do have at least one single-element species where the atoms have nonzero oxidation states: ozone. Unlike most single-element substances, ozone has oxygen atoms in nonequivalent positions (from https://commons.wikimedia.org/wiki/File:Ozone-resonance-Lewis-2D.png#mw-jump-to-license; created by Ben Mills; license information in the link):

enter image description here

The central oxygen atom shares more of its electrons and retains fewer unshared pairs than its neighbors and thus, according to the oxidation state formalism of nonpolar oxygen-oxygen bonds, takes on a positive charge. This atom thus has an oxidation state of +1. The other atoms, which take turns sharing fewer than normal electrons, are each considered to have an oxidation state of -1/2.

The charge separation implied above shows up experimentally as a nonzero dipole moment. Even with nonpolar bonds, the unequal distribution of nonbonding electrons would create a dipole.

Sulfur forms a similar triatomic molecule, $\ce{S3}$, with nonzero oxidation states of the individual atoms, which has been identified on Jupiter's moon Io.


Also while being 0 to the outside we know some compounds that are so called inner alloys. Take manganese for example. In its elemental form you have a very complicated structure consisting of various Frank-Kasper polyhedra. And there are some effects, which can be explained more easily if we assume different oxidation states for the Mn atoms in the Mn metal. To the outside it still is 0 but the amount of electrons each Mn gives into the conduction band is not the same for each Mn atom. At least this is how my Prof once explained the strange structure of Mn to me. You can even manipulate the amount of some of these oxidation states and make alloys with increased Mn(II)-high spin character for example, which show stronger magnetic effects.

  • $\begingroup$ Sounds interesting, but did your professor find it interesting enough to publish citeable results? Having such references is a good way to attract reads, upvotes, and even citations on answers elsewhere in SE. $\endgroup$ May 20, 2021 at 19:25

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