# Why do oxidation numbers relate to valence electrons?

Oct 15, 2016

Because, formally, oxidation numbers relate to the loss or gain of valence electrons.

#### Explanation:

Let us take a simple oxidation reaction:

$C \left(s\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right)$

Now the oxidation numbers of the elements, dioxygen and carbon, are quite properly regarded as $\text{ZERO}$. They have neither donated nor accepted electrons. But in the reaction, the dioxygen oxidant is conceived to have accepted 4 electrons, and the carbon reductant is conceived to have donated 4 electrons. This is quite clear from the oxidation numbers of the elements in $C {O}_{2}$, which feature $C \left(+ I V\right)$ and $O \left(- I I\right)$ as formal oxidation states.

The same redox transfer can also be invoked for dihydrogen combustion:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(g\right)$

How do electrons transfer here?