# How do you find oxidation numbers of molecules?

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, $H - O - H$, and if we break each $O - H$ bond we get, FORMALLY, $2 \times {H}^{+}$ and ${O}^{2 -} :$

${H}_{2} O \rightarrow 2 \times {H}^{+} + {O}^{2 -}$

And thus the oxidation state of hydrogen in water is $+ I$, and that of oxygen is $- I I$. 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 ${M}^{2 +}$, or ${M}^{3 +}$, or ${X}^{-}$, or ${E}^{2}$, the oxidation state of each ion conforms to the give definitons, and are $+ I I$, $+ I I I$, $- I$, and $- I I$ 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, $S {O}_{4}^{2 -}$, and if oxygen retains its usual oxidation state, i.e. $- I I$ above, we get a sulfur oxidation state of $V I +$: i.e. $6 + 4 \times \left(- 2\right) = - 2$ as required. $S {O}_{3}^{2 -}$, we get $S \left(+ I V\right)$. And for ${S}_{2} {O}_{3}^{2 -}$, we get $S \left(V I +\right)$, and $S \left(- I I\right)$ respectively, i.e. an average oxidation state of $S \left(+ I I\right)$. Hydrogen peroxide, ${H}_{2} {O}_{2}$, throws a bit of a googly, and here you must know that peroxo-oxygen has a $- I$ oxidation state.

For $C l {O}_{4}^{-}$ we get $C l \left(V I I +\right)$. The maximum oxidation number of an element (and for the halogens $+ V I I$ 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:

$C {\left(s\right)}^{0} + {O}_{2}^{0} \rightarrow {C}^{+ I V} {O}_{2}^{- I I}$

There should be many other answers on these boards that illustrate these problems. Here is a start.