I've been asking myself for a while now how counter-ions sometimes change the colors of coordination compounds. Like when I use d-metals as counter-ions to precipitate Cyanoferrates.

I looked at some crystal structures and saw that in the case of a cyanoferrate the terminal cyanide-N coordinate in an octahedral-prismatic geometry around the counter-ion. And I know that the cyano-ligand is a bridging ligand like in for example Prussian Blue, where the electron between Fe(II) and Fe(III) is moved across the cyano-ligand, probably through the π-systems. So I expect two ligand field splittings to happen, the one around the iron and the one around the counter-ion. These colors overlap to create a new color. Also depending on how lewis-acidic the counter-ion is it might pull electron density across the cyano-ligand lowering the gap and causing a red-shift.

I couldn't find much about that topic. It might be related to how colors derive in solid state compounds where I have no idea how this works. I know Lithium metal can be formulated as a huge molecule with many Li-Atoms, overlapping their atomic orbitals forming two big bands as the molecular orbitals create overlap so much. And usually salts crystallize in metal crystal structures so for example every second element is exchanged for the counter ion. If this was true then a colored solid state compound (salt) is a huge molecular orbital scheme, too creating two bands and depending on the size of their band gap the color is created (as I said I can't find anything on this topic). Sometimes the thermochromism in ZnO is explained by band gaps so I assume this concept of metals applies for salts as well. Then or course the counter-ion would have a much more complicated role here.

If anyone knows an answer to any of these questions I'd be really thankful.

  • 1
    $\begingroup$ The characteristic colour in Prussian blue arises due to intervalence charge transfer, which is actually what you described, up until the point where you said "two ligand field splittings" - yes, both Fe(II) and Fe(III) experience different LF splittings, but that does not really influence the colour. The LF splitting of Fe(II) would determine the energy gap between ground Fe(II) and excited Fe(II), for example. However the charge transfer is an electronic transition between Fe(II) and Fe(III). $\endgroup$ – orthocresol Apr 8 '17 at 11:57
  • $\begingroup$ in other words the LF splitting of Fe(II) determines this wavelength $$\ce{Fe^{II}-C#N-Fe^{III} ->[$h\nu$] ^*Fe^{II}-C#N-Fe^{III}}$$ (* denoting excited state) but the charge transfer is actually of the nature $$\ce{Fe^{II}-C#N-Fe^{III} ->[$h\nu$] Fe^{III}-C#N-Fe^{II}}$$ $\endgroup$ – orthocresol Apr 8 '17 at 12:02
  • $\begingroup$ Ah okay, thank you! So in my case where I have an M(II) at the right side it doesn't necessarily turn to M(III) I guess. So in the case of Prussian Blue I will have to descide between pushing the electron into the eg or across the bridge but if the redox reaction is not favored with the counter-ion it's back to ligand field splitting again, I guess. And if the M(II) pulls electrons stronger doesn't that descrease the electron denisty around the Iron causing a smaller gap? $\endgroup$ – Justanotherchemist Apr 8 '17 at 12:10
  • $\begingroup$ "if the redox reaction is not favored with the counter-ion it's back to ligand field splitting again" yep. I don't quite get your last sentence, though. Which iron and which gap? $\endgroup$ – orthocresol Apr 8 '17 at 12:16
  • $\begingroup$ Let's take copper(II)-ferricyanide. Potassium ferricyanide is yellow while the copper compound is dark brown. Well okay Fe(III) and Cu(I) might be likely here but I don't really know the colors of the other ferricyanides. So if there was another metal like Copper, maybe one which isn't reduced that easily as counterion for the Ferricyanide, does this element influence the ligand field splitting for the Fe(II) as it acts as a lewis acid for the same cyano-ligand as well? $\endgroup$ – Justanotherchemist Apr 8 '17 at 14:29

Your Answer


Browse other questions tagged or ask your own question.