How is an ion formed?

Mar 11, 2018

Well, by $\text{oxidation}$ or $\text{reduction...}$

Explanation:

And formally, we introduce electrons as virtual particles that are added in a $\text{reductive process}$, and are lost in an $\text{oxidative process...}$

And as with any chemical reaction, both charge and mass are conserved.... Typically, metals are ELECTRON-RICH materials, which are $\text{OXIDIZED}$, i.e. they lose electrons:

$N a \left(s\right) \rightarrow N {a}^{+} + {e}^{-}$

or

Mg(s) rarr Mg^(2+)+2e^-)

or

$F e \left(s\right) \rightarrow F {e}^{3 +} + 3 {e}^{-}$

And non-metals, typically with high nuclear charge, from the RIGHT of the Periodic Table as we face it, are ELECTRON-POOR materials...that formally tend to accept electrons....and so they are $\text{REDUCED}$. Difluorine, and dioxygen are the most potent oxidants on the Periodic Table...

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

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

The electrons, as written, are particles of convenience. We add the oxidation, and reduction processes together such that the electrons are ELIMINATED...viz. for the oxidation of lithium metal...

$L i \left(s\right) \rightarrow L {i}^{+} + {e}^{-}$

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

We add TWO of the former and ONE of the latter to get...

$2 L i \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) + 2 {e}^{-} \rightarrow {\underbrace{2 L {i}^{+} + {O}^{2 -}}}_{\text{i.e. lithium oxide}} + 2 {e}^{-}$..and when we eliminate the electrons as particles of convenience...

$2 L i \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) + \cancel{2 {e}^{-}} \rightarrow {\underbrace{2 L {i}^{+} + {O}^{2 -}}}_{\text{i.e. lithium oxide}} + \cancel{2 {e}^{-}}$

to give finally....

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

...and the metal has been oxidized, and the non-metal has been reduced.