Why is aluminum a less active metal than magnesium despite having a lower first ionization energy?

Question

I was thinking about the general trend in metal activity (reactivity increases down a group and decreases across a period) when I started thinking about aluminum and how it has a lower first ionization energy than magnesium because its highest-energy electron is in a high-energy 3p orbital. Despite this, magnesium is the more active metal. Is this because activity measures the potential of the oxidation of Al t0 Al³⁺, not Al to Al⁺? Although it may not take much energy to remove the first electron from Al, oxidizing Al⁺ to Al³⁺ requires removing electrons from the lower energy 3s orbital, making the overall oxidation less favorable than the oxidation of Mg to Mg²⁺. Is this thinking correct?

Answer 1

Because neither aluminum nor magnesium forms many compounds by contributing one electron to a bond, whether ionic or covalent. So first ionization energy really does not have much to tell.

Neither do second or third ionization energies. The reactivities of aluminum and magnesium are hard to compare on the basis of any ionization energies because the elements have different common valences and aluminum does not form very many predominantly ionic compounds. No matter what you try to do with ionization energies, you're unavoidably comparing apples and olives.

Answer 2

Another reason is the fact that metallic aluminum is always covered by a thin film of alumina (aluminum oxide $\ce{Al2O3}$). This film is transparent, continuous and waterproof. It is formed spontaneously in the contact of air, and prevents the contact with reagents. If this film is removed by scratching, it will be immediately redone by atmospheric air. This is why, apparently the metallic aluminum objects do not react with many usual ionic solutions. For example, the reaction with dilute $\ce{HCl}$ solutions is relatively slow to start, because alumina film is rather resistant to acidic corrosion.

The production of this film can be prevented by mercury chloride $\ce{HgCl2}$. If a piece of aluminum is dipped into a solution of $\ce{HgCl2}$, some mercury chloride is able to cross this film and react with aluminum to produce mercury $\ce{Hg}$ (which soon makes an aluminum amalgam $\ce{Al(Hg)}$) according to the reaction : $$\ce{2 Al + 3 HgCl2 -> 2 AlCl3 + 3 Hg}$$ Of course, the aluminum amalgam $\ce{Al(Hg)}$ is soon oxidized in air, but the alumina film does not adhere on the outer surface of the amalgamated aluminum. As a consequence, the oxidation proceeds in a couple of minutes and produces long white filaments of alumina which are "growing" on the surface of the aluminum sample. This phenomena lasts as long as there is still some mercury on the outer surface of the aluminum sample.