# Burning magnesium ribbon [closed]

I know that this might sound wrong but, in the reaction

$$\ce{2 Mg(s) + O2(g) → 2 MgO(s) + energy}$$

isn't oxygen supposed to be oxidised as it has excess electrons $$(\ce{O^2-})$$ and can donate a few to magnesium as it has less electrons $$(\ce{Mg^2+})?$$

I have searched for it in almost all books and it is written that magnesium gets oxidised. Please clear my misconception. It would be helpful if I can get a reaction mechanism for this, because it includes transfers of electrons.

## closed as unclear what you're asking by Mithoron, Jon Custer, user55119, DrMoishe Pippik, Todd MinehardtJun 18 at 0:56

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• In layman's terms both magnesium and oxygen within MgO are "happy" since both $\ce{O^2-}$ and $\ce{Mg^2+}$ have configuration of neon, a noble gas. For a more in-depth explanation probably have a look at the elements' electronegativities or dive into physical chemistry (review free Gibbs energy for the burning process). – andselisk Jun 17 at 16:21
• But MgO is formed after the reaction. I am talking about their reaction to achieve the stable state of MgO. – Shishir Maharana Jun 17 at 16:28
• But, oxidation means loss of electrons. So, if magnesium is oxidized it has to lose electrons, but it clearly has less electrons, then why will it lose them? – Shishir Maharana Jun 17 at 16:37
• Well, then why did magnesium lose 2 electrons and oxygen gain 2 electrons? I mean why did Oxygen take 2 electrons from magnesium? – Shishir Maharana Jun 17 at 16:51
• The reactants are neutral Mg and neutral O2. The magnesium doesn't turn into Mg++ until after reaction, and it gets that way by being oxidized. The O2 doesn't get extra electrons and go to 2 O-- until after reaction. The MgO is the product and doesn't do any oxidizing or reducing, so we imagine it to be happy as a stable lattice of Mg++ and O-- ions. – James Gaidis Jun 18 at 13:32