# 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