Why is $\ce{Co(SCN)_{4}^{2-}}$ blue?

I simply can't predict the colors of such inorganic salts. I have learned that $\ce {Co^{2+}}$ salts are pink in color. Also, if the ligand is a strong field ligand ($\ce {SCN-}$ in this case), then the splitting is more and the color should shift to the red side...

Is my concept wrong?


1 Answer 1


Let us consider this qualitatively with crystal field theory (CFT). Co is a transition metal with $d^7$ configuration. CFT says that depending on the geometry of the metal complex, the d-orbital energies will split into different energy levels. If the ligands (in this case, the four $\ce{SCN-}$ ligands) align with certain orbitals, then the electrons in these orbitals will feel an increased repulsion due to the closeness of the ligands, and increase in energy. The orbitals in-between the aligned axes, will not feel this repulsion, and will not increase in energy. The splitting diagram might look something like this

Splitting diagram for a tetrahedral complex

There is an equation that relates the (splitting) energy to wavelength:

$$ \Delta E = \frac{hc}{\lambda} $$

From this equation, we see that if the splitting energy increases, wavelength will have to decrease, i.e. be shorter than before. This means a shift towards the violet region. As you say, $\ce{SCN-}$ is a strong ligand, meaning it leads to a rather high $\Delta E$. So we could say that $\ce{Co(SCN)_4^{2-}}$ is blue due to the high energy gap between the d-orbitals. Electrons are excited from one of the lower orbitals up to one of the higher orbitals. Upon de-exciting, light is emitted with an energy equal to $\frac{hc}{\lambda}$. This happens to be in the blue/violet region.

Weaker ligands would split the energy levels less, and the complex's color would therefore have a color closer to the red part of the visible spectrum.

Did this make it a little clearer?

  • $\begingroup$ This is really a great answer. Concise and clear. $\endgroup$
    – ParaH2
    May 27, 2018 at 20:06

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