0
$\begingroup$

I am facing problems in indentifying colour of complex compunds just by seeing the their molecular formula. Is there any method or concept to predict the colour of complex compunds?

$\endgroup$
  • 1
    $\begingroup$ The coordination geometry + central metal ion + spin state often gives a good guess what kind of color you will see. However, it is not exact science (if you don't do any calculations), especially with charge transfer bands. $\endgroup$ – Greg Mar 1 '17 at 17:04
  • $\begingroup$ Can you provide some examples using the three properties just mentioned ? $\endgroup$ – Aviral Agarwal Mar 1 '17 at 17:06
3
$\begingroup$

The basic principle of coloured compounds is that there is some kind of electronic transition whose energy difference corresponds to a photon whose wavelength is in the visible region ($\pu{400nm}<\lambda<\pu{700nm}$). Thus, to determine the colour of a compound we should always be looking at the molecular orbital scheme; for reference, I have attached the MO scheme of a typical octahedral $\ce{[ML6]^n+}$ compound in figure 1.

octahedral
Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

The most common electronic transition corresponding to visible wavelengths is the transition between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$, marked with $\pu{10Dq}$ in figure 1. In simple terms, these five orbitals correspond to the central metal’s $\mathrm{d}$ orbitals meaning that the metal’s electronic configuration plus the strength of the $\ce{L\bond{->}M}$ interaction will determine the colour.

The problem with attempting to predict colours ab initio is that you cannot really predict the difference in energy levels between the ligand orbitals (right-hand side) and metal orbitals (left hand side) — it is mostly the metal and its oxidation state that determines this, a higher oxidation state corresponding to lower-lying metal orbitals, therefore a greater value of $\pu{10Dq}$ and therefore a blue-shift in colour.

However, there are still a few shortcuts that you can use assuming you have memorised the ‘general colour’ of a certain metal in a certain oxidation state. For example, you may consider stronger and weaker field ligands. When going from a slightly weaker ligand to a slightly stronger ligand — e.g. going from $\ce{[Cu(H2O)6]^2+}$ to $\ce{[Cu(NH3)4(H2O)2]^2+}$ — $\pu{10Dq}$ is predicted to increase and thus the absorbed wavelength predicted to decrease.

Unfortunately, the generality of this scheme is strongly limited. It fails as soon as:

  • the complex transitions from a high-spin state to a low-spin state (e.g. $\ce{[Fe(H2O)6]^3+}$ to $\ce{[Fe(CN)6]^3-}$)
  • the geometry of the complex changes from (distorted) octahedral to something else like tetrahedral (e.g. $\ce{[Co(H2O)6]^3+}$ to $\ce{[CoCl4]-}$).

So you might just be better off memorising the colours of certain complexes rather than attempting to learn the pattern and remember the exceptions.

$\endgroup$
-1
$\begingroup$

See presence of unpaired electrons, complexes having unpaired elctrons absorb visible radiations for d d transitions.

$\endgroup$
  • $\begingroup$ I am asking if a compound is given to us and we need to determine whether it is yellow coloured or not so can we predict that? $\endgroup$ – Aviral Agarwal Mar 1 '17 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.