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Recently I came across the fact: Metal oxides basic and non-metals' acidic or neutral.

Searching for it, partly my query is solved, as I found:

Metal oxides or $\ce{O2-}$ forms $\ce{OH-}$ after hydrolysis in the following reaction so metal oxides are basic in aqeous solutions. $$\ce{O2- + 2H2O->4OH-}$$

I think this might happen as metal form ionic compunds and non-metals form covalent and some of them may not even react tus neutral.

I've got only a little $5\%$ of the exact reason, therefore I wish if someone could explain the whole process/fact. I considered Arhenius concept.

Note My question is partly answered here, but is not satisfactory.

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We have learnt that metallic oxides are basic nature. This is because the oxides of metals like $\ce{MgO}$ or $\ce{CaO}$ or $\ce{Na2O}$ form hydroxides in their aqueous solutions. For example: $$\ce{K2O + H2O -> 2KOH}$$ $$\ce{MgO + H2O -> Mg(OH)2}$$

But when non-metallic oxides dissociate into their constituent ions in water, they give $\ce{H+}$ ions in their aqueous solutions forming acids For example: $$\ce{CO2 + H2O -> H2CO3}$$ $$\ce{SO3 + H2O -> H2SO4}$$

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    $\begingroup$ This just restates information in a previous answer and the equations are also unbalanced. Perhaps you could clean this up a little and add some more detail. $\endgroup$
    – bon
    Commented Sep 6, 2015 at 13:22
  • $\begingroup$ so they are acidic because they form acids? Sorry I dont know much about Arrehinius definition...only know the Brownsted Lowry one $\endgroup$
    – MartianCactus
    Commented Sep 14, 2016 at 16:42
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Metal oxides like $\ce{CaO}$ and $\ce{Na2O}$ react with water and give hydroxides:

$\ce{CaO + H2O <=> Ca(OH)2}$

$\ce{Na2O + H2O <=> 2 NaOH}$

While non-metal oxides like $\ce{CO2}$ and $\ce{SO2}$ react with water and give acids:

$\ce{CO2 + H2O <=> H2CO3 <=> H+ + HCO3-}$

$\ce{SO2 + H2O <=> H2SO3 <=> H+ + HSO3-}$

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    $\begingroup$ can you use technical terms, thermodynamics, kinetics, bonding theories or other theoretical explanations? $\endgroup$
    – RE60K
    Commented Nov 14, 2014 at 16:02
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    $\begingroup$ It still didn't explain why. $\endgroup$
    – Simon-Nail-It
    Commented Nov 16, 2014 at 13:52

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