Ch 3 · Colour of complexes · AS & A Level Chemistry 9701
Topic 3 of 5 in Chemistry of transition elements — AS & A Level Chemistry 9701 — 57 lessons.
Introduction
Most transition metal complexes are coloured, while the ions of scandium and zinc are not. This lesson explains that difference in terms of the splitting of d orbitals by surrounding ligands, and shows why changing the ligand — as in the familiar copper(II) and cobalt(II) ligand exchange reactions — changes the colour observed.
Key ideas — Degenerate and non-degenerate d orbitals
Orbitals of exactly the same energy are described as degenerate; orbitals split into different energy levels are non-degenerate. In an isolated, gaseous transition metal ion with no ligands attached, all five 3d orbitals are degenerate. Once ligands bond to the metal ion, this symmetry is broken and the d orbitals become non-degenerate.
🧑🏫 Teacher's explanation
Imagine you are in a room with five identical, empty chairs arranged in a circle. Because they are all the same, they have the same 'energy' or value to a person sitting in them. In chemistry, these are your five d orbitals. When the ion is alone in a vacuum, these orbitals are 'degenerate'—meaning they are identical in energy.
📖 Definitions
🔍 Worked example — Easy
A transition metal ion has five d orbitals that are all at an energy level of E_1. If the orbitals are degenerate, what is the energy difference (Δ E) between any two of these orbitals?
🔍 Worked example — Standard
An electron in a complex undergoes a transition from a lower energy d orbital to a higher energy d orbital. The energy gap Δ E is 2.5 × 10^-19 J. Calculate the frequency nu of the light absorbed. (Use h = 6.63 × 10^-34 J s).
🔍 Worked example — Exam-style
A complex ion has a Δ E of 3.0 × 10^-19 J. If the wavelength of light absorbed is λ, calculate λ in nm. (Use c = 3.00 \ 10^8 m/s and h = 6.63 \ 10^-34 J s).
⚠️ Common errors
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🎯 Exam question
(a) Explain why the presence of ligands causes the d orbitals to split into non-degenerate levels. [2] (b) Explain why [Cu(NH_3)_4(H_2O)_2]^2+ is a deeper blue than [Cu(H_2O)_6]^2+. [2] [4 marks]
Key ideas — Splitting of d orbitals in octahedral and tetrahedral complexes
In an octahedral complex, ligand lone pairs approach along the x, y and z axes. The two d orbitals that point directly along these axes (d_z^2 and d_x^2-y^2) experience greater electron–electron repulsion from the approaching ligands and are raised to a higher energy set, e_g; the three orbitals lying between the axes (d_xy, d_xz, d_yz) are raised less, forming a lower energy set, t_2g. The energy gap between the two sets is the splitting energy, Δ E. In a tetrahedral complex, with only four ligands approaching between the axes, the splitting pattern is reversed and the gap, Δ E_tet, is smaller than in an octahedral field of the same ligand.
🧑🏫 Teacher's explanation
Imagine you are in a crowded room where five people (the d-orbitals) are standing in a circle. Suddenly, six people (the ligands) rush in, but they only approach from specific directions (the axes). The people standing directly in the path of the newcomers feel much more crowded and 'uncomfortable' (higher energy), while those standing in the gaps between the newcomers feel less pressure (lower energy).
📖 Definitions
🔍 Worked example — Easy
An octahedral complex has a splitting energy Δ E of 2.0 × 10^-19 J. Calculate the wavelength of light absorbed by an electron making this transition.
🔍 Worked example — Standard
A transition metal ion in an octahedral field has its d-orbitals split into t_2g and e_g levels. If the energy gap Δ E increases due to a stronger ligand, what happens to the wavelength of light absorbed?
🔍 Worked example — Exam-style
Compare the splitting in octahedral vs tetrahedral complexes. If Δ E_oct for a specific metal is 20,000 cm^-1, qualitatively estimate Δ E_tet for the same metal and ligand set.
⚠️ Common errors
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🎯 Exam question
Key ideas — Why transition metal complexes are coloured
When white light falls on a complex, an electron in the lower energy set of d orbitals can absorb a photon of visible light whose energy exactly matches Δ E and be promoted to the higher energy set — an electron promotion between non-degenerate d orbitals. The wavelengths of light that are not absorbed pass through or are reflected, and the colour seen is complementary to the colour absorbed. This requires both an occupied lower orbital and a vacant higher orbital of different energy: ions with a completely empty d subshell (mathrmSc^3+, d^0) have no electron to promote, and ions with a completely full d subshell (mathrmZn^2+, d^10) have no vacant orbital to promote an electron into, so neither can absorb visible light in this way and both are colourless.
🧑🏫 Teacher's explanation
Imagine a ladder where the rungs are not evenly spaced. In a transition metal ion, the five d-orbitals usually have the same energy (they are 'degenerate'). However, when ligands approach the metal, they create an electric field that pushes some d-orbitals to a higher energy level and leaves others at a lower level. This splitting creates a gap, Δ E.
📖 Definitions
🔍 Worked example — Easy
A transition metal complex absorbs light at a wavelength of 600 nm. What is the energy Δ E in J required for the electron promotion? (Use c = 3.0 × 10^8 m/s and h = 6.6 × 10^-34 J s)
🔍 Worked example — Standard
Explain why Sc^3+ (a d^0 ion) and Zn^2+ (a d^10 ion) are both colourless.
🔍 Worked example — Exam-style
A solution of CuSO_4 is blue. If the complex absorbs red light, what is the likely colour observed?
⚠️ Common errors
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🎯 Exam question
(a) Explain why the addition of NH_3 ligands causes a change in the color of the solution. [3] (b) Predict whether the energy gap Δ E for the ammonia complex is larger or smaller than that for the aqua complex. [1] [4 marks]
Key ideas — Effect of the ligand on splitting and colour
Different ligands produce different sizes of splitting energy for the same metal ion and geometry: some ligands cause a large splitting (a large Δ E), so a higher-energy (shorter wavelength) photon is absorbed, while others cause a small splitting (a small Δ E), so a lower-energy (longer wavelength) photon is absorbed. Since the geometry also affects Δ E (octahedral fields split d orbitals more than tetrahedral fields), changing either the ligand or the geometry of a complex changes the colour observed.
🧑🏫 Teacher's explanation
Imagine a set of identical stairs. In an atom, the five d-orbitals are 'degenerate,' meaning they are all at the exact same energy level—like five identical steps on a staircase. When ligands (molecules or ions with lone pairs) approach the metal ion, they create an electric field that pushes some d-orbitals away and pulls others closer. This 'splits' the stairs into two different levels: a lower set and a higher set. The gap between these levels is called the splitting energy, Δ E.\n\nThink of this like a musical instrument. If you tighten a guitar string, the pitch (energy) increases. If you loosen it, the pitch decreases. Similarly, different ligands 'tighten' or 'loosen' the energy gap. When light hits the complex, an electron jumps from the lower level to the higher level by absorbing a specific color of light. The color you actually see is not the color absorbed, but the 'complementary' color—the leftover light that passes through. If a complex absorbs high-energy violet light, it will appear yellow/orange.
📖 Definitions
🔍 Worked example — Standard
A Cu^2+ complex with four H_2O ligands is octahedral. If a ligand exchange reaction replaces H_2O with NH_3, the splitting energy Δ E increases. Predict the change in the wavelength of light absorbed.
🔍 Worked example — Exam-style
Explain why [CoCl_4]^2- (tetrahedral) is blue while [Co(H_2O)_6]^2+ (octahedral) is pink, even though the metal ion is the same.
⚠️ Common errors
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🎯 Exam question
Key ideas — Colour changes on ligand exchange
Ligand exchange in copper(II) and cobalt(II) complexes illustrates this directly. Pale blue, octahedral mathrm[Cu(H_2O)_6]^2+ becomes deep blue octahedral mathrm[Cu(NH_3)_4(H_2O)_2]^2+ with excess ammonia, and yellow, tetrahedral mathrm[CuCl_4]^2- with excess concentrated hydrochloric acid, the colour shifting as both the ligand and the geometry, and hence Δ E, change. Pink, octahedral mathrm[Co(H_2O)_6]^2+ becomes blue, tetrahedral mathrm[CoCl_4]^2- with excess concentrated hydrochloric acid, for the same reason.
🧑🏫 Teacher's explanation
Imagine a group of people standing in a room. If everyone is at the same height, they are 'degenerate'. But if you put some people on chairs and some on the floor, you have split them into different energy levels. In transition metal complexes, the five d-orbitals start at the same energy (degenerate). When ligands approach, they create an electric field that pushes some d-orbitals to higher energy and leaves others at lower energy. This gap is called Δ E.\n\nWhen light hits the complex, an electron can 'jump' from a lower energy d-orbital to a higher one. The color we see is not the color of the light absorbed, but the color that is left over (the complementary color). If the gap Δ E changes—because you swapped one ligand for another or changed the shape of the complex—the energy of light needed to make the jump changes, and thus the color changes.
📖 Definitions
🔍 Worked example — Easy
A Cu^2+ ion in water exists as [Cu(H_2 extO)_6]^2+ (pale blue). When excess NH_3 is added, it forms [Cu(NH_3)_4(H_2 extO)_2]^2+ (deep blue). Explain why the color changes using Δ E.
🔍 Worked example — Standard
Compare the geometry and color of Co^2+ complexes: [Co(H_2 extO)_6]^2+ is pink (octahedral) and [CoCl_4]^2- is blue (tetrahedral). Why does the color change?
🔍 Worked example — Exam-style
Calculate the energy change (Δ E) in kJ mol^-1 for a transition where an electron absorbs a photon of wavelength 600 nm. (h = 6.63 × 10^-34 J s, c = 3.00 × 10^8 m s^-1, N_A = 6.02 × 10^23 mol^-1)
⚠️ Common errors
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🎯 Exam question
Worked example
Question. (a) Explain why an aqueous solution of Sc(NO_3)_3 is colourless while an aqueous solution of Fe(NO_3)_3 is yellow-brown. [4] (b) Predict and explain the colour change observed when excess concentrated hydrochloric acid is added to a pink aqueous solution of mathrm[Co(H_2O)_6]^2+. [3]
🔍 Worked example — Easy
Explain why Ti^4+ ions in aqueous solution are colorless.
🔍 Worked example — Standard
A transition metal complex has a d-orbital splitting energy (Δ E) of 2.5 imes 10^-19 J. Calculate the wavelength of light absorbed. (Use h = 6.63 imes 10^-34 J s and c = 3.00 imes 10^8 m/s).
🔍 Worked example — Exam-style
Predict the color change when excess Cl^- is added to a solution of [Cu(H_2O)_6]^2+ (blue). Explain using ligand exchange and geometry.
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