The colour of a complex compound is due to unpaired electrons. As per crystal field theory, $\ce{K4[Fe(CN)6]}$ has no unpaired electrons so it has to be colourless. But then why is it coloured?

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There is no direct connection between having unpaired electrons and having a colour. In fact, many colourful compounds have no unpaired electrons and yet display a bright colour: potassium permanganate, potassium dichromate, azobenzene, sudan red, phenolphthalein, iodine and many, many more. Arguably, there are more coloured compounds known in the singlet state (no unpaired spins) than in the dublet, triplet or any other state.

So what causes the colour? It is the possibility of electron transitions from a lower to a higher orbital. If that transition is equivalent to a photon with a visible wavelength, we observe a colour. Examples:

  • Most phenyl groups absorb light at c. $250~\mathrm{nm}$ — to short a wavelength to be visible; they are ‘ultravioletly coloured’.

  • The electronic transitions of most transition metal complexes (except $\mathrm{d^0}$, $\mathrm{d^5}$ and $\mathrm{d^10}$ complexes) are well within the visible range of $400$ to $700~\mathrm{nm}$.

  • In rare cases, extensive electronic systems may even allow the absorption maximum to be shifted outside of the visible range into the infrared range. Chlorophyll comes rather close to this, absorbing red light (the longest visible wavelengths).

For potassium hexacyanidoferrate(II), the colour we observe is yellow so the absorption must be the complementary blue/violet, indicating the large field split.

  • 1
    $\begingroup$ Some technicalities: "no unpaired electrons" is not synonymous with "singlet state" (usually it is, but not always). For example: the $^1\Sigma_\mathrm{g}^+$ excited state of dioxygen is a singlet state but has two unpaired electrons. Nor is "one or more unpaired electrons" synonymous with the triplet state. Free sodium atoms have one unpaired electron, but the ground state is a doublet ($^2S_{1/2}$) state. $\endgroup$ – orthocresol May 1 '16 at 13:09
  • $\begingroup$ @orthocresol Thanks, I was aware that unpaired $\ne$ triplet, but the organic chemist in me pays more attention to triplet than any other unpaired state, so I just … forgot that ;) Is writing unpaired spins technically correct? $\endgroup$ – Jan May 1 '16 at 13:18
  • $\begingroup$ Hmm, I guess so. $\endgroup$ – orthocresol May 1 '16 at 13:22

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