# Why do ligands have such a small effect on overall absorption of a complexed ion?

When a metal cation is complexed, there is strong UV-Vis absorption due to the splitting of its $d$ orbitals, thereby allowing electronic transitions.

My understanding is that ligands contribute very little to overall observed absorption. This is evident, for example, in copper solutions. Copper(II) chloride, copper(II) sulfate, and copper(II) nitrate – all solutions of complexed ions – are all very similar in color and have similar absorption spectra.

I think the simple answer is that transition metals have the capacity to allow for electronic transitions while most ligands do not, but I don't believe it is this simple. Also, it is curious that sulfate, chloride, and nitrate are all colorless when dissolved in solution.

My question is: why do ligands have such a small effect on overall absorption of a complexed metal cation?

• An excellent question, though I will point out that in a few cases, ligands actually do affect the color significantly. See, for example, the cobalt chloride humidity indicator – chipbuster Aug 10 '15 at 22:48
• @chipbuster I was not aware of this. Thanks for bringing it to my attention. – T. Kent Aug 10 '15 at 22:55
• @chipbuster Interestingly enough, copper II nitrate is trihydrated and copper II sulfate is pentahydrated. In copper, the waters of solvation seemingly have very little color on absorption spectra. Yet cobalt chloride's color varies greatly depending on hydration. – T. Kent Aug 10 '15 at 22:59
• Copper salts solutions aren't colorless and influence of ligands on color is comparable to that of central ion. – Mithoron Aug 13 '15 at 0:07
• @Mithoron You misunderstand. What I was saying is that the ligand has little effect on the overall color (and absorption) of the solution. Copper(II) solutions are primarily blue, regardless of the ligand. – T. Kent Aug 13 '15 at 3:59

This is absolutely not true. Many ligands can strongly dictate the color of transition metal solutions.

Your example picks three simple salts and then questions whether the color of the solutions (dictated largely by $\ce{[Cu(H2O)6]^{2+}}$) is very different. No, because the resulting majority complex in aqueous solution is likely identical.

Even taking a simple ammonia complex, e.g. $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ you can see a substantial color change.

Many copper acetonitrile compounds are weakly colored or colorless, with UV/Vis optical absorption occurring near the edge of the red to near-IR.

When you get into octahedral complexes, you can find substantial changes in color due to MLCT and LMCT absorptions, as well as significant modulation of the d-d transitions due to the ligand (i.e., high-field and low-field ligands).

• Oh, I forgot to comment on your discussion of ligand transitions. Ligands like water, ammonia, etc. don't have visible electronic transitions - that's definitely true. But $\pi-\pi^*$ transitions are important with aromatic or otherwise conjugated ligands. Plus, as I mentioned MCLT and LMCT transitions involve both the metal and the ligand and are often the dominant absorption feature in complexes. – Geoff Hutchison Aug 15 '15 at 3:52

A large part of the colour of transition metal complexes are indeed explained more on the metal’s side of things. To explain this, as I did in a number of previous questions, it is always helpful to recall the molecular orbital scheme of a random transition metal complex; figure 1 shows such a scheme for an octahedral $\ce{[ML6]^n+}$ complex with σ and π interactions.

Figure 1: Molecular orbital scheme of an octahedral complex including σ and π interactions from the ligands to the metal. Based on the schemes given by Professor Klüfers in his coordination chemistry course at the LMU Munich.

Typically, ligands arrive as octet-rule fulfilling compounds, i.e. they have all bonding orbitals (the ones on the right) fully populated and the antibonding orbitals are energetically removed. Equally typically, the metal side includes a nonzero number of d-electrons which is rarely equal to 10 (if it is, the chemistry is much more complex and less ordered) and rarely equates to fully populated or completely unpopulated orbitals.

Therefore, it is typically the metal’s side of a coordination complex where the valence electrons are located. These are typically placed in d-type orbitals which are labelled $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ in this scheme. They are both antibonding with respect to $\ce{L\bond{->}M}$ bonds. The colour one can observe typically originates from excitations $\mathrm{t_{2g}^*\rightarrow e_g^*}$, which is defined to a significant extent by the metal’s side of things. The greatest difference between transition metals and thus between their hexaaquacomplexes is their d-orbital energy; mainly where they are located with respect to the orbital energies of the ligand group orbitals.

However note that the ligands can also have a significant effect on colours. Most importantly, changing the coordinating atom from e.g. oxygen to chlorine will move the entire set of ligand group orbitals; their new position will result in different recombination energies and thus a different resulting field split. While Geoff mentioned the colour change from $\ce{[Cu(H2O)6]^2+}$ to $\ce{[Cu(NH3)4(H2O)2]^2+}$ I consider that one a mild change (blue to blue) and offer the difference between $\ce{[Fe(CN)6]^3-}$ (red) and $\ce{[FeCl6]^3-}$ (bright yellow).[1]

And finally, sometimes an extreme colour change is due to a structural reorganisation. For example, adding chloride ions to a pink-ish solution of $\ce{[Co(H2O)6]^2+}$ will give the blue $\ce{[CoCl4]^2-}$ solution; however, the former complex is octahedral while the latter is tetrahedral; thus the molecular orbital schemes are very dissimilar and a wildly different colour is expected.

Note:

[1]: I’ll admit that it’s cheating a little bit. The red colour of $\ce{[Fe(CN)6]^3-}$ is a $\mathrm{t_{2g}^*\rightarrow e_g^*}$ transition while the yellow colour of $\ce{[FeCl6]^3-}$ is actually a ligand-to-metal–charge-transfer excitation, corresponding to $\mathrm{t_{1u} \rightarrow t_{2g}^*}$.