Virtually all transition metal compounds are coloured, a consequence of the way the 3d orbitals in the transition metal ion interact with ligands in a complex.
There are five 3d orbitals which are degenerate in the transition metal atom i.e. they all have the same energy. The 3d orbitals have specific shapes orientated along or between the x, y and z axes in three dimensional space.
When an octahedral complex is formed, the ligands (and their electron clouds) are orientated at the vertices of an octahedron. There is a greater interaction / repulsion between ligands and the dz2 and dx2-y2 orbitals …
than there is between the ligands and the dxy, dxz and dyz orbitals.
The five d orbitals no longer have the same energy and we say that they have been split, with two orbitals destabilised because they have been more influenced by the negative charge (electron cloud) of the ligands and three orbitals stabilised because they are less influenced.
When we shine white light on a transition metal complex, specific frequencies of visible light are absorbed and electrons are promoted from a lower 3d orbital to a higher 3d orbital. The frequency of light absorbed corresponds to ΔE, the difference in energy between the split orbitals because
ΔE = ℎ𝜈
where ℎ is Planck’s constant and 𝜈 is the frequency of light.
You can find a detailed explanation of absorption spectroscopy here 😊.
If we are looking at a solid complex then the colour we observe is a result of all the frequencies of visible light that were not absorbed being reflected back at us. If we are looking at a solution of the complex, then we are observing all the frequencies of visible light that were not absorbed being transmitted through the solution.
We are seeing the complementary colour to that which was absorbed.
This distinction is important in terms of your wording when answering exam questions!
Let’s pull all this together with an example:
It’s also important to point out that if there are no electrons in the d orbitals e.g. in Sc3+ ions, or if the d orbitals are completely filled e.g. in Cu+ ions then no electrons movements between the split d orbitals is possible and no frequencies of visible light are absorbed. The compound is a white solid / colourless in solution.
(However, you should be aware that the colours of crystalline compounds also depend on the arrangement of ions in the lattice and the consequential merging of atomic orbitals into molecular bonding and anti-bonding orbitals for valence electrons. Movement between these molecular orbitals may also be the result of the crystal lattice absorbing specific frequencies of visible or UV light which is the reason for solid copper(I) oxide being red).
What affects the colour of a transition metal complex?
- The oxidation state of the transition metal ion because different oxidation states have different numbers of electrons in the d orbitals e.g. [V(H2O)6]2+ is purple and [V(H2O)6]3+ is yellow.
- The nature of the ligands in the complex have an effect on the splitting of the d orbitals (as in their relative energies) and hence ΔE
The colour also depends on the number of each type of ligand present e.g. [Cr(H2O)6]3+ is violet, [Cr(H2O)5Cl]2+ is green and [Cr(H2O)4Cl2]+ is dark green.
3. The coordination number of the complex / arrangement of ligands around the ion because the splitting of the d orbitals is different in a tetrahedral or square planar complex compared with an octahedral complex.
Practice questions
- Titanium(IV) oxide is a white pigment commonly used in paint. Explain why it is not coloured.
- Explain why there is a change in colour from red to purple when the water ligands in [Cr(H2O)6]3+(aq) are exchanged for ammonia in [Cr(NH3)6]3+(aq).
- Compare and contrast the origin of the colour of a copper(II) ion in a flame test with the origin of the colour of copper(II) complex in solution.
Answers
- Titanium has an oxidation state of +4 in titanium(IV) oxide which means it has no 3d electrons in the 3d orbitals. All frequencies of visible light are reflected so the compound is white.
2. In transition metal ion complexes, interaction / repulsion between the electron orbitals of the ligand and the 3d orbitals of the transition metal ion causes the 3d orbitals to ‘split‘ so that they are no longer of equal energy (degenerate).
Since the energy levels in the chromium (III) ion are quantised, ∆E is fixed and when the corresponding specific frequency of visible light is absorbed (∆E = hν).
An electron is promoted / excited from a lower energy 3d orbital to a higher energy level.
The colour of the complex ion is complementary to the frequencies of visible light absorbed (the complementary colour is transmitted).
Changing the ligand affects the degree of interaction / repulsion, so if ∆E changes then the specific frequency of visible light absorbed will change, and hence the colour of the complex changes.
Exam tip: you should use this a model answer – learn it and then you can easily adapt it for any long answer question about colour in transition metal complexes
3. The energy levels in a copper ion are quantised. In a flame test heat energy is absorbed and an electron is promoted from a lower energy level to a higher energy level. The electron falls back to a lower energy level releasing a specific frequency of visible light ( in this case blue) which corresponds to the difference in energy between the tow energy levels (∆E = hν).
In a copper(II) complex ion, the 3d orbitals are split. A specific frequency of visible light is absorbed and an electron is promoted from a lower 3d orbital to a higher 3d orbital. The complimentary colour is seen as these frequencies of visible light are not absorbed and are transmitted through the solution.