Question

How is an ion formed?

Answer 1

Well, by `"oxidation"` or `"reduction..."`

Explanation:

And formally, we introduce electrons as virtual particles that are added in a `"reductive process"`, and are lost in an `"oxidative process..."`

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

`Na(s) rarr Na^(+) + e^(-)`

or

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

or

`Fe(s) rarr Fe^(3+)+3e^(-)`

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 `"REDUCED"`. Difluorine, and dioxygen are the most potent oxidants on the Periodic Table...

`1/2F_2(g) + e^(-) rarr F^-`

`1/2O_2(g) + 2e^(-) rarr 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...

`Li(s) rarr Li^(+) + e^(-)`

`1/2O_2(g) + 2e^(-) rarr O^(2-)`

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

`2Li(s)+1/2O_2(g) + 2e^(-) rarr underbrace(2Li^(+) + O^(2-))_"i.e. lithium oxide"+ 2e^(-)`..and when we eliminate the electrons as particles of convenience...

`2Li(s)+1/2O_2(g) + cancel(2e^(-)) rarr underbrace(2Li^(+) + O^(2-))_"i.e. lithium oxide"+ cancel(2e^(-))`

to give finally....

`2Li(s)+1/2O_2(g) rarr Li_2O(s)`

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