Why is #["MnI"_4]^(2-)# favored over #["MnI"_6]^(4-)#?

1 Answer
Truong-Son N.
Sep 21, 2017

We can see this by considering that #"Mn"# atom has an electron configuration of

#[Ar]4s^2 3d^5#,

so the #+2# oxidation state gives the complex a #d^6# configuration. Maintaining this oxidation state across two compared complexes, we consider #["MnI"_6]^(4-)# vs. #["MnI"_4]^(2-)#...

Even without continuing any further, you can see that a #4-# charge is very hard to maintain... Even despite that, #"I"^(-)# is quite large, and that plays a factor as well.

A four-coordinate complex is more stable partly because the metal's low oxidation state is only able to sustain a moderate negative overall charge on the complex while also being charge-stabilized by the solvent.

The steric bulk of the ligands also limits the coordination number.

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