How is an ion formed?

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.