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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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    $\begingroup$ Color is due to electron transitions. Whether or not the electrons were paired before the transition is less important. $\endgroup$ – Ivan Neretin May 1 '16 at 6:48
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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 (which means no unpaired spins in the vast majority of cases) 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. $\pu{250 nm}$ — 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 $\pu{700 nm}$.

  • 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 a relatively large energy difference.

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    $\begingroup$ 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
  • $\begingroup$ Just wondering: Laporte and spin selection rules should have prevented the transitions, right? $\endgroup$ – William R. Ebenezer Mar 23 at 11:56
  • $\begingroup$ @WilliamR.Ebenezer The Laporte rule indeed should prevent a $\mathrm{t_{2g}\rightarrow e_g^*}$ transition, that is correct. However, that transition is not spin-forbidden as all $\mathrm{e_g^*}$ orbitals are empty. Which transition is actually responsible for the colour is a good question because right now I am at a loss. $\endgroup$ – Jan Mar 25 at 2:11

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