This process occurs when a stronger ligand displaces a weaker ligand from the metal complex.
Let’s look at a sample worked example below:
Worked Example 2
Describe and explain what is seen when concentrated HCl is added to a solution of Cu²⁺ ions in water with the aid of a balanced equation.
[Cu(H₂O)₆]²⁺ (aq) + 4Cl⁻ (aq) ⇌ [CuCl₄]²⁻ (aq) + 6H₂O (l)
The blue solution turns green (mixture of blue [Cu(H₂O)₆]²⁺ and yellow [CuCl₄]²⁻) then yellow.
Ligand exchange has occurred where Cl⁻ displaces H₂O ligands.
Coloured compounds and ions
The fun thing about transition metals is their tendency to form coloured compounds, both in the solid and aqueous states. The not-so-fun thing however, is that memorising the different compounds and their respective colours can be overwhelming (and sadly, eventually necessary).
The table below summarises some of the colours students are required to know for common ions (with H₂O as the ligand)
The property of forming coloured compounds can be attributed to a phenomenon known as d-orbital splitting.
The five d-orbitals of a gaseous transition metal, Mⁿ⁺, are usually at the same energy level. But in octahedral complexes, upon the approach of ligands, the 3d orbitals are split into two different sets of energy levels.
The d𝑧² and dx² − y² orbitals point directly at the x,y and z axis. Due to stronger inter-electronic repulsion from direct interaction with the ligand’s lone pair of electrons, the energy level of the dz² and dx² - y² orbitals are raised by a greater extent.
Hence, the dyz, dxz , and dxy orbitals are at a lower energy level.
But in tetrahedral complexes, the four ligands approach the central metal atom or ion in between the axes. Hence, the dyz, dxz, and dxy orbitals now experience stronger inter-electronic repulsion from direct interaction with the ligand’s lone pair of electrons, the energy level of dyz, dxz, and dxy orbitals are raised by a greater extent. The dz² and dx²- y² orbitals are at a lower energy level.
However, It is unnecessary to state in your answers which is of the higher energy level. Writing that the d-orbitals are split into two different energy levels is sufficient.
Since the d-orbitals are split into two different energy levels, there is an energy gap, ΔE, which corresponds to the energy of the visible region on the electromagnetic spectrum. Transition metal complexes will absorb the light of wavelength with that energy, causing an electron to be promoted from the d-orbital of a lower energy to another d-orbital of a higher energy level. This is called the d-d transition.
Since light of that energy is absorbed, the complementary colour ( the colour opposite on the colour wheel) would be reflected into our eyes. For example, Cu²⁺ (aq) appears blue since light of orange wavelength is absorbed.
Hence, we can conclude that E is inversely proportional to λ.
Hence we can conclude that a bigger ΔE leads to light of a smaller wavelength being absorbed. Thus the colour of the compound reflected into our eyes is that of a larger wavelength, such as red.
The opposite can be said for a smaller ΔE as well. Light of a larger wavelength would be absorbed. The colour reflected into our eyes would be that of a smaller wavelength, such as blue.
ΔE is affected by two factors,
- 1. Strength of ligand
A stronger ligand would result in a larger ΔE. Hence compounds with stronger ligands are more likely to appear yellow.
- 2. Oxidation state of the cation
Different numbers of electrons in the d-orbitals of the transition metal ion would result in varied interactions with the electrons as the ligands approach the central ion. Hence, ΔE will vary.
Next, here is a table listing the common reactions undergone by transition metal ions