Colour of Transition metal Ions
A concise revision guide to colour in transition metal chemistry: how ligands split the d orbitals, why some light is absorbed and the complementary colour is seen, the equation ΔE = hν, why some ions are colourless, and what changes the colour, with colorimetry.
- 3.2.5.4i
- 3.2.5.4ii
- 3.2.5.4iii
- 3.2.5.4iv
- 3.2.5.4v-a
- 3.2.5.4v-b
- 3.2.5.4v-c
- 3.2.5.4vi
What these spec points say
- 3.2.5.4i know that transition metal ions can be identified by their colour
- 3.2.5.4ii understand that colour arises when some wavelengths of visible light are absorbed and the remaining wavelengths are transmitted or reflected
- 3.2.5.4iii know that d electrons move from the ground state to an excited state when light is absorbed
- 3.2.5.4iv know that the energy difference between ground and excited states is given by DeltaE = hv = hc/lambda
- 3.2.5.4v-a know that changes in oxidation state alter DeltaE and lead to a change in colour
- 3.2.5.4v-b know that changes in co-ordination number alter DeltaE and lead to a change in colour
- 3.2.5.4v-c know that changes in ligand alter DeltaE and lead to a change in colour
- 3.2.5.4vi know that a simple colorimeter can be used to determine the concentration of coloured ions in solution
Why Complexes Are Coloured
Choose a step to watch six ligands split the 3d orbitals, an electron absorb light to jump the gap, and a full or empty 3d set let all light through.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
Before You Start
Colour depends on which parts of white light a substance absorbs.
- White light is a mixture of all the colours of the visible spectrum, from red (long wavelength) to violet (short wavelength).
- A coloured substance absorbs some of these colours and transmits or reflects the rest.
- The colour seen is the complementary colour of the light absorbed.
- The 3d configurations of the ions come from Electron Configurations and Properties of Transition metals, and the complexes from Ligands, Complex Ions and Their Shapes.
Absorbed Colour and Observed Colour
Solutions of transition metal ions are usually coloured, and an ion can often be identified by its colour. The solution removes part of the visible spectrum from white light, and the eye sees what is left.
| Colour absorbed | Colour seen |
|---|---|
| red | green |
| orange | blue |
| yellow | violet |
| green | red |
| blue | orange |
| violet | yellow |
Complementary colours sit opposite each other on the colour wheel. A solution of [Cu(H₂O)₆]²⁺ absorbs orange and red light, so it looks blue.
- A solution transmits the colours it does not absorb. A solid reflects them.
- A blue solution is not absorbing blue light. It is absorbing light from the orange and red part of the spectrum.
- A substance that absorbs no visible light looks colourless in solution, or white as a solid.
Common mistake: Do not write that a blue solution “absorbs blue light” or “emits blue light”. It absorbs the complementary colour and transmits blue.
Splitting of the d Orbitals
In an isolated transition metal ion, the five 3d orbitals all have the same energy. Ligands change this.
- Ligands bond to the metal ion, and their lone pairs repel the electrons in the 3d orbitals.
- Some 3d orbitals point more directly at the ligands than others, so they are raised in energy by different amounts.
- The five 3d orbitals split into two sets with different energies.
- The energy gap between the two sets is called ΔE.
| Complex | Lower set | Higher set |
|---|---|---|
| octahedral | three 3d orbitals | two 3d orbitals |
Five 3d orbitals at one energy in the isolated ion split into three lower and two higher in an octahedral complex. The gap is ΔE.
Remember: The splitting is caused by the ligands. Without ligands there is no energy gap, so the explanation of colour always starts with the ligands splitting the 3d sub-level.
Check: Splitting and Complementary Colours
Link the colour absorbed to the colour seen, and describe what ligands do to the 3d orbitals.
How Absorbing Light Gives Colour
The gap ΔE between the two sets of 3d orbitals matches the energy of visible light. The full explanation has four steps.
- The ligands split the 3d sub-level into two sets of orbitals with different energies.
- An electron in the lower set absorbs visible light of energy ΔE and is promoted to the higher set: it moves from the ground state to an excited state.
- Light of one colour is therefore removed from the white light.
- The remaining light is transmitted or reflected, and the complementary colour is seen.
Example: [Cu(H₂O)₆]²⁺
- Cu²⁺ is [Ar]3d⁹, so there is one space in the higher set of orbitals.
- An electron from the lower set is promoted into that space.
- The energy needed corresponds to orange and red light.
- The rest of the spectrum is transmitted, so the solution looks blue.
Exam focus: A full-mark answer has three ideas: the 3d orbitals are split by the ligands, electrons absorb visible light to move to the higher level, and the light that is not absorbed is the colour seen.
Quick Check: What Really Happens to the Light?
Pick the one accurate statement in each set, using complexes and solids that are new to this page.
The Energy Gap: ΔE = hν
The energy gap is linked to the frequency and wavelength of the light absorbed.
Equation: ΔE = hν = (hc) / (λ), where h is the Planck constant (6.63 × 10⁻³⁴ J s), ν is the frequency in Hz (s⁻¹), c is the speed of light (3.00 × 10⁸ m s⁻¹) and λ is the wavelength in m.
- A larger ΔE means light of higher frequency and shorter wavelength is absorbed.
- Convert a wavelength in nm to m before using it: 1 nm = 1 × 10⁻⁹ m.
Worked example: ΔE from a wavelength
A complex absorbs light of wavelength 600 nm. Calculate ΔE in J, and in kJ mol⁻¹.
Step 1. λ = 600 nm = 6.00 × 10⁻⁷ m.
Step 2. ν = (c) / (λ) = (3.00 × 10⁸) / (6.00 × 10⁻⁷) = 5.00 × 10¹⁴ Hz.
Step 3. ΔE = hν = 6.63 × 10⁻³⁴ × 5.00 × 10¹⁴ = 3.32 × 10⁻¹⁹ J.
Worked example: a wavelength from ΔE
In another complex, ΔE = 3.98 × 10⁻¹⁹ J. Calculate the wavelength of the light absorbed.
Step 1. Rearrange ΔE = (hc) / (λ) to give λ = (hc) / (ΔE).
Step 2. λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (3.98 × 10⁻¹⁹) = 5.00 × 10⁻⁷ m.
Remember: ΔE from the equation is the energy for one ion. Multiply by the Avogadro constant, 6.02 × 10²³ mol⁻¹, and divide by 1000 to give kJ mol⁻¹.
Quick Check: Calculate the Energy Gap
Work each step on paper, then flip the card to check your working.
Why Some Ions Are Colourless
Colour needs an electron that can move from the lower set of 3d orbitals to the higher set. That requires a partly filled 3d sub-level.
| Ion | 3d electrons | Why it is colourless |
|---|---|---|
| Sc³⁺ | 3d⁰ | there are no 3d electrons to promote |
| Zn²⁺ | 3d¹⁰ | the higher set is full, so no electron can be promoted |
| Cu⁺ | 3d¹⁰ | the higher set is full, so no electron can be promoted |
- In each case no visible light is absorbed, so all of it is transmitted.
- Other 3d⁰ ions, such as Ti⁴⁺, are colourless for the same reason.
- Copper(II), 3d⁹, is coloured but copper(I), 3d¹⁰, is not: the same element, with a different number of 3d electrons.
Exam focus: Give the configuration and its consequence: Zn²⁺ is 3d¹⁰, a full sub-level, so no electron can move between the split 3d orbitals and no visible light is absorbed.
Check: Explaining Colour and Lack of Colour
Put the steps of the explanation in order and decide from their configurations which ions are colourless.
What Changes the Colour
The size of ΔE decides which colour is absorbed. Anything that alters ΔE alters the colour. Three changes do this.
| Change | Example | Colour change |
|---|---|---|
| Oxidation state of the metal | [Fe(H₂O)₆]²⁺ → [Fe(H₂O)₆]³⁺ | pale green → yellow |
| Ligand | [Cu(H₂O)₆]²⁺ → [Cu(NH₃)₄(H₂O)₂]²⁺ | blue → deep blue |
| Co-ordination number (and shape) | [Cu(H₂O)₆]²⁺ → [CuCl₄]²⁻ | blue → yellow |
| Co-ordination number (and shape) | [Co(H₂O)₆]²⁺ → [CoCl₄]²⁻ | pink → blue |
Test tubes showing the three changes: iron(II) and iron(III), copper(II) with water and with ammonia, and aqua ions against chloride complexes.
- A change in oxidation state changes the number of 3d electrons and the attraction between the ion and the ligands.
- Different ligands split the 3d orbitals by different amounts.
- A change in co-ordination number changes the shape, and so the pattern and size of the splitting.
- Pure [Fe(H₂O)₆]³⁺ is pale violet. Its solutions look yellow because some of the ions react with water.
- The reactions that swap the ligands are covered in Ligand substitution and the Stability of Complexes.
Remember: Colour depends on the metal, its oxidation state, the ligands and the co-ordination number. Change any one of these and ΔE changes, so a different colour of light is absorbed.
Using a Colorimeter
The more concentrated a coloured solution, the more light it absorbs. A simple colorimeter measures this absorbance and uses it to find an unknown concentration.
- Choose a filter of the complementary colour, so that the light reaching the sample is the colour it absorbs most strongly.
- Zero the colorimeter with a cuvette of water (the solvent).
- Measure the absorbance of several solutions of known concentration.
- Plot a calibration curve of absorbance against concentration.
- Measure the absorbance of the unknown and read its concentration from the curve.
The parts of a simple colorimeter, and a calibration curve used to read off an unknown concentration from its absorbance.
Worked example: reading a calibration curve
A calibration line passes through the origin and through an absorbance of 0.60 at 0.050 mol dm⁻³. An unknown solution has an absorbance of 0.42.
Answer. Concentration = (0.42) / (0.60) × 0.050 = 0.035 mol dm⁻³.
- If the colour is too pale to measure, a suitable ligand is added first to give a more intensely coloured complex.
- Use the same cuvette, or identical ones, so that the path length does not change.
Exam focus: A red filter is used for a blue solution, because the blue solution absorbs red light. The filter is the complementary colour of the solution.
Check: What Changes the Colour
Explain colour changes in terms of oxidation state, ligand and co-ordination number, using complexes whose colours are given in the question.
Common Exam Points
- Start every colour explanation with the ligands splitting the 3d orbitals into two energy levels.
- Electrons are promoted by absorbing visible light. Never write that light is emitted.
- The colour seen is the complementary colour: it is the light transmitted or reflected, not the light absorbed.
- Sc³⁺ (3d⁰), Zn²⁺ and Cu⁺ (both 3d¹⁰) are colourless because no electron can move between the split orbitals.
- Colour changes when the oxidation state, the ligand or the co-ordination number changes, because ΔE changes.
- Quote the formula and the colour together: [CuCl₄]²⁻ is yellow and [CoCl₄]²⁻ is blue.
- In ΔE = (hc) / (λ), convert nm to m, and remember that ΔE is for one ion, not one mole.
- For colorimetry, describe the filter, the calibration curve from known concentrations and the reading for the unknown.
FAQs
Use these quick answers to check why transition metal ions are coloured and what changes the colour.
Why are transition metal complexes coloured?
Ligands split the 3d orbitals into two sets with an energy gap, ΔE. An electron absorbs visible light of energy ΔE and moves to the higher set. The light that is not absorbed is transmitted, so the complementary colour is seen.
Why does [Cu(H₂O)₆]²⁺ look blue?
It absorbs orange and red light, which promotes a 3d electron to the higher set of orbitals. The remaining wavelengths pass through the solution, and together they look blue. Blue is the complementary colour of orange.
Why are Sc³⁺ and Zn²⁺ compounds colourless?
Colour needs a partly filled 3d sub-level. Sc³⁺ is 3d⁰, so it has no d electron to promote. Zn²⁺ and Cu⁺ are 3d¹⁰, so the higher orbitals are already full. No visible light is absorbed, so the ions are colourless.
What changes the colour of a complex?
Changing the oxidation state of the metal, the ligand or the co-ordination number changes ΔE, so light of a different frequency is absorbed. [Cu(H₂O)₆]²⁺ is blue, [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue and [CuCl₄]²⁻ is yellow.
How is a colorimeter used to find a concentration?
Choose a filter of the complementary colour. Measure the absorbance of several solutions of known concentration and plot a calibration curve. Then measure the absorbance of the unknown solution and read its concentration from the graph.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
Keep this note — free
Save your progress across every AQA topic. A free account remembers which topics you have covered, saves your question scores, and syncs across your phone and laptop.
- Track every topic you have finished
- Keep your practice-question scores
- No payment, no card, free forever
Already registered? Log in