# What is a positive or negative number assigned to to an element?

May 8, 2017

Do you refer to $\text{oxidation number}$?

#### Explanation:

The $\text{oxidation number}$ is conceived to be A CONCEPTUAL charge assigned to atoms in a compound, where the individual atoms are considered either (i) to have accepted electrons, i.e. to have been $\text{reduced}$; or (ii), to have donated electrons, i.e. to have been $\text{oxidized}$.

And, typically, we can solve complex chemical reactions by invoking electrons as conceptual particles in our representation of $\text{oxidation-reduction reactions}$, so-called $\text{redox reactions}$.

Typically, metals are ELECTRON-RICH species, and tend to be oxidized, which we could represent by the following example...

$\stackrel{0}{F} e \rightarrow F {e}^{3 +} + 3 {e}^{-}$ $\left(i\right)$

Elements, which have neither donated not accepted electrons are assigned an oxidation numer of ZERO. And the oxidation number of the elemental ion is simply the CHARGE on the ion.

On the other hand, non-metals typically come from the right of the Periodic Table as we face it, and their high nuclear charge typically means that they ACCEPT electrons, i.e. they are reduced.......

$\frac{1}{2} {\stackrel{0}{O}}_{2} + 2 {e}^{-} \rightarrow {O}^{2 -}$ $\left(i i\right)$

OR........

$\frac{1}{2} {\stackrel{0}{F}}_{2} + {e}^{-} \rightarrow {F}^{-}$ $\left(i i i\right)$

(And note that oxygen and fluorine are the most potent oxidants on the Periodic Table).

Because, electrons are conceptual particles, when we combine the equations the electrons are eliminated. And this operation certainly models the stoichiometry we observe in the actual experiment (which is of course why we use the method; theory and representation follows experiment, not vice versa):

For the oxidation of iron, we cross multiply the individual redox reactions so as to eliminate the electrons: i.e. $2 \times \left(i\right) + 3 \times \left(i i\right) :$

$2 F e + \frac{3}{2} {O}_{2} + \cancel{6 {e}^{-}} \rightarrow F {e}_{2} {O}_{3} + \cancel{6 {e}^{-}}$

And finally,

$2 F e + \frac{3}{2} {O}_{2} \rightarrow F {e}_{2} {O}_{3}$

There should be many other examples of such reactions on this site. See here and enclosed links.