For group 13 elements M³⁺, why is the formation of MCl₄⁻ more favorable than for group 12 elements M²⁺?

I'm not a chemist and I'm trying to understand a chemistry point from a talk I went to recently. Any help would be much appreciated.


The speaker was talking about $+3$ (Group 13) elements in the periodic table. He found out that one of them forms $\ce{MCl4-}$ complexes, where $\ce{M}$ is a $+3$ element. 


He made a comment that I didn't understand, that the formation of $\ce{MCl4-}$ is electrostatically more reasonable than $\ce{MCl4^2-}$, where $\ce{M}$ would be a $+2$ metal. 


Why would this be true? I am hanging my head in shame that I don't know this.




Let's take an example. M can be Aluminum $Al$. In order for $Al$ to have an outer electronic layer equal to 0 or 8, it can either loose 3 electrons and become an ion similar to Neon. But it can also use its 3 electrons to make covalent bonds with three $Cl$ atoms, and still add a 4th $Cl^-$ ion, which brings one doublet. In this option, the Aluminum is surrounded by 3+3+2 = 8 electrons, and it looks like Argon. This is the most favorable choice, because it would cost much more energy to remove three electrons. Three ! 


It is quite different for metals having 2 electrons in their outer shell, like $M$g. If $Mg$ wants to look like a noble gas, it may loose 2 electrons, and become $Mg^{2+}$ which is similar to $Ne$. Of course it costs energy to remove 2 electrons, but not that much. And it would be much more difficult to make covalent bonds. With its 2 electrons, $Mg$ could make two covalent bonds with $Cl$ atoms. But 4 electrons are still missing to look like Argon. These 4 electrons may be brought by 2 $Cl^-$ ions, making $MgCl\_4^{2-}$. But this structure is not favorized, because the two approaching $Cl^-$ ions repell one another. 


All that to explain why, combined with Chlorine, $Al$ does not make $Al^{3+}$ and prefers covalent structures like $AlCl\_4^-$ions, and why $Mg$ prefers to make $Mg^{2+}$ ions


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