How do you find oxidation numbers of molecules?

1 Answer
Feb 7, 2017

By a process of simple arithmetic............

Explanation:

Oxidation number is conceived to be the charge on the central atom (i.e. the atom under consideration) when all the bonds to that element are BROKEN, with the charge assigned to the most electronegative atom.

We start with water, a neutral molecule, HOH, and if we break each OH bond we get, FORMALLY, 2×H+ and O2:

H2O2×H++O2

And thus the oxidation state of hydrogen in water is +I, and that of oxygen is II. Typically, we use Roman numerals. Please note that this is an entirely conceptual exercise, that does not have much basis in reality.

For single element ions, such as M2+, or M3+, or X, or E2, the oxidation state of each ion conforms to the give definitons, and are +II, +III, I, and II respectively. This becomes important when we write redox equations, where the electrons appear as virtual particles.

For larger, complex molecules, and radical ions, the sum of the oxidation numbers equals the charge on the ion or the molecule. Thus the sum of oxidation numbers must equal 0 for a neutral molecule; it was for water above.

We could take sulfate, SO24, and if oxygen retains its usual oxidation state, i.e. II above, we get a sulfur oxidation state of VI+: i.e. 6+4×(2)=2 as required. SO23, we get S(+IV). And for S2O23, we get S(VI+), and S(II) respectively, i.e. an average oxidation state of S(+II). Hydrogen peroxide, H2O2, throws a bit of a googly, and here you must know that peroxo-oxygen has a I oxidation state.

For ClO4 we get Cl(VII+). The maximum oxidation number of an element (and for the halogens +VII is the maximum oxidation state) corresponds with its Group number on the Periodic Table, for the reason that the Periodic Table reflects electronic structure.

Note that elements, and elemental molecules, which are not conceived to have donated or received electrons, thus have a formal oxidation number of 0; cf. the oxidation of carbon by oxygen where there is a formal 4 electron transfer:

C(s)0+O02C+IVOII2

There should be many other answers on these boards that illustrate these problems. Here is a [start.](https://socratic.org/questions/how-do-you-write-oxidation-reduction-half-reactions)