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?

  • $\begingroup$ 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. $\endgroup$ – Nicolau Saker Neto Jul 16 '17 at 12:13

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.


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.


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