Colour of complexes
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Define the term 'degenerate d orbitals'.
Degenerate d orbitals are d orbitals that possess the same energy level. In an isolated atom or ion, the five d orbitals are degenerate. This degeneracy is removed when ligands are introduced.
Describe how the d orbitals split in an octahedral complex.
In an octahedral complex, the five degenerate d orbitals split into two sets. Three d orbitals (dxy, dxz, dyz) are lower in energy, and two d orbitals (dz2, dx2-y2) are higher in energy. This energy difference is denoted as ΔE.
Describe how the d orbitals split in a tetrahedral complex.
In a tetrahedral complex, the five degenerate d orbitals split into two sets. Two d orbitals (dxy, dxz, dyz) are higher in energy, and three d orbitals (dz2, dx2-y2) are lower in energy. The splitting pattern is the inverse of the octahedral complex.
Explain why transition metal compounds are coloured.
Transition metal compounds are coloured because electrons absorb specific frequencies of visible light to get promoted from a lower energy d orbital to a higher energy d orbital (d-d transition). The observed colour is the complementary colour to the light absorbed.
How does the identity of a ligand affect the magnitude of ΔE?
Different ligands cause different degrees of d-orbital splitting, and thus different values of ΔE. Strong-field ligands cause a large splitting (large ΔE), and weak-field ligands cause a small splitting (small ΔE).
Explain how ligand exchange can affect the observed colour of a complex.
Ligand exchange alters the magnitude of ΔE. This, in turn, changes the frequency of light absorbed and the complementary colour observed.
What happens to the frequency of light absorbed as ΔE increases?
As ΔE increases, the frequency of light absorbed also increases (E = hf). This means the compound will absorb light towards the blue/violet end of the spectrum and appear yellow/orange, as opposed to absorbing red/orange and appearing blue/green.
Give an example of a ligand exchange reaction using copper(II) ions and ammonia molecules.
The reaction between [Cu(H₂O)₆]²⁺ (pale blue) and ammonia (NH₃) initially forms a pale blue precipitate of copper(II) hydroxide. With excess ammonia, the precipitate dissolves, and [Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue) is formed.
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