The shape of the lesson
By the end of the lesson, learners can
- Splitting and absorption: Use the figure: an electron moves up across ΔE, and the frequency of light absorbed matches the gap exactly.
- What changes the colour: ΔE depends on the metal, its oxidation state, the ligands and the shape.
- Octahedral: 3 lower, 2 higher. Tetrahedral: 2 lower, 3 higher, smaller ΔE.
- ΔE = hf = hc/λ; per mole multiply by L.
- Colour seen = complementary to colour absorbed.
- d0 and d10 ions are colourless.
One picture every idea returns to
In a free ion the five 3d orbitals are degenerate (equal in energy). Ligand lone pairs repel the d electrons unequally, so the orbitals become non-degenerate: in an octahedral complex they split into three lower and two higher orbitals; in a tetrahedral complex into two lower and three higher, with a smaller gap. When white light passes through, an electron is promoted from a lower to a higher d orbital, absorbing a photon of energy ΔE = hf. The colour seen is the complementary colour of the light absorbed. Ions with an empty (d0) or full (d10) d subshell cannot make this transition, so they are colourless. A different ligand, oxidation state or coordination number changes ΔE and so changes the colour.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Absorb one colour, see its complement.. In an isolated ion the five 3d orbitals have the same energy. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Splitting and absorption | Use the figure: an electron moves up across ΔE, and the frequency of light absorbed matches the gap exactly. Project tab 4 of the lab (Transition elements lab, colour from d-orbital splitting) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: What changes the colour | ΔE depends on the metal, its oxidation state, the ligands and the shape. Project tab 4 of the lab (Transition elements lab, colour from d-orbital splitting) and let a student drive it while the class predicts. | Think-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in What changes the colour, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The colour you see is not the colour absorbed: a blue solution absorbs red-orange light. | Octahedral splitting is 3 lower, 2 upper; tetrahedral is the reverse, 2 lower, 3 upper, with a smaller ΔE. Electrons are promoted within the 3d subshell, not from 3d to 4s. Colour needs a partly filled d subshell, so Zn 2+ and Sc 3+ compounds are white. |
Colour of complexes: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Explain why aqueous [Cu(H2O)6]2+ is coloured.
[Ti(H2O)6]3+ absorbs light of wavelength 500 nm. Calculate ΔE in kJ mol−1. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, L = 6.02 × 1023 mol−1)
Draw energy level diagrams to show how the 3d orbitals split in (a) an octahedral and (b) a tetrahedral complex.
Explain why copper(I) compounds such as CuCl are usually white, but copper(II) compounds are coloured.
Adding concentrated HCl to pink [Co(H2O)6]2+ gives blue [CoCl4]2−. Give two reasons why ΔE changes, and say which colour of light each complex mainly absorbs.
Explain why [Fe(H2O)6]2+ and [Fe(H2O)6]3+ have different colours although they have the same ligands and shape.
Answer key and marking notes
- Cu2+ is 3d9, a partly filled d subshell.
- The water ligands split the 3d orbitals into two non-degenerate sets (three lower, two higher), separated by ΔE.
- An electron is promoted from a lower to a higher d orbital by absorbing visible light of energy ΔE (red-orange light).
- The remaining (transmitted) light, the complementary colour, is seen: blue.
- ΔE = hc/λ = 6.63 × 10−34 × 3.00 × 108 ÷ 5.00 × 10−7 = 3.98 × 10−19 J.
- Per mole: 3.98 × 10−19 × 6.02 × 1023 = 2.39 × 105 J mol−1.
- ΔE = 239 kJ mol−1.
- (a) Octahedral: three orbitals at lower energy, two at higher energy, gap ΔE.
- (b) Tetrahedral: two lower, three higher, with a smaller ΔE.
- Cu+ is 3d10: the upper d orbitals are full, so no d-d electron transition can occur and no visible light is absorbed.
- Cu2+ is 3d9, so an electron can be promoted across ΔE, absorbing visible light.
- The ligand changes from H2O to Cl−, which splits the d orbitals differently.
- The shape changes from octahedral to tetrahedral, which gives a smaller, reversed splitting.
- Pink [Co(H2O)6]2+ absorbs mainly green light; blue [CoCl4]2− absorbs mainly orange-red light.
- Fe3+ has a higher charge and different d electron configuration (3d5) from Fe2+ (3d6).
- The ligands are held closer and the d orbitals split by a different ΔE, so a different frequency of light is absorbed: pale green versus yellow-brown.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The diagram shows the 3d orbitals of [Ti(H2O)6]3+ (3d1). What happens when the complex absorbs visible light?
- Athe d electron is promoted from a lower-energy d orbital to a higher-energy d orbital; the photon energy equals ΔE
- Bthe d electron is removed completely from the ion, so the complex is ionised
- Can electron drops from a higher d orbital to a lower one and the light given out is the colour we see
- Da 4s electron is promoted into an empty 3d orbital
How do the d orbitals split in a tetrahedral complex compared with an octahedral one?
- Atetrahedral: 2 orbitals lower and 3 higher, with a smaller ΔE; octahedral: 3 lower and 2 higher
- Btetrahedral: 3 lower and 2 higher, exactly the same as octahedral
- Ctetrahedral: the five d orbitals stay degenerate, because there are only four ligands
- Dtetrahedral: 2 lower and 3 higher, with a larger ΔE than octahedral
Use the colour wheel. A complex absorbs mainly violet light from the visible spectrum. What colour does its solution appear?
- Ayellow
- Bviolet
- Cgreen
- Dblue
Why are zinc(II) compounds white and their solutions colourless?
- AZn2+ is 3d10; the upper d orbitals are full, so no d electron can be promoted by visible light
- BZn2+ has no d electrons, so there is nothing to promote
- Czinc does not form complex ions, so its d orbitals are never split
- Dzinc ions absorb all visible frequencies equally
Which aqueous ion is colourless?
- A[Sc(H2O)6]3+
- B[Ti(H2O)6]3+
- C[Cu(H2O)6]2+
- D[Fe(H2O)6]2+
Pale blue [Cu(H2O)6]2+ turns deep blue when excess NH3 is added. Which statement explains the change of colour?
- Achanging the ligand changes ΔE, so a different frequency of visible light is absorbed
- Bthe oxidation state of copper changes from +2 to +1
- Cammonia molecules are themselves blue, so they deepen the colour
- Dthe d orbitals become degenerate when ammonia is bonded
Which list gives factors that ALL change the size of ΔE, and therefore the colour, of a transition metal complex?
- Athe metal, its oxidation state, the type of ligand and the coordination number (shape)
- Bthe metal and the ligand only; the oxidation state and shape have no effect
- Conly the number of d electrons in the metal ion
- Dthe concentration of the solution and the temperature
A complex absorbs most strongly at a wavelength of 500 nm. What is ΔE for one mole of complexes? (h = 6.63 × 10−34 J s; c = 3.00 × 108 m s−1; L = 6.02 × 1023 mol−1)
- A239 kJ mol−1
- B2.39 × 105 kJ mol−1
- C3.98 × 10−22 kJ mol−1
- D7.98 × 10−7 kJ mol−1
Quiz key
Ligands split the five degenerate 3d orbitals into two sets. A photon whose energy equals ΔE (with ΔE = hf) lifts the electron into the upper set; that frequency is removed from white light. Removing the electron would need far more energy than visible light carries. The colour we see is the light that is transmitted or reflected, not light emitted by electrons falling back. [Ti(H2O)6]3+ has no 4s electrons: they were the first to be lost when Ti3+ formed.
In an octahedral complex the ligands point at two of the d orbitals, so 3 are lower and 2 higher. In a tetrahedral complex the pattern is inverted (2 lower, 3 higher) and the splitting is smaller, because there are only four ligands and none points straight at a d orbital. Saying the pattern is the same misses the inversion. The orbitals are still split in a tetrahedral complex; any ligand field removes the degeneracy. A larger ΔE for tetrahedral gets the size the wrong way round.
The colour seen is the complementary colour of the light absorbed, which sits opposite on the wheel: violet is opposite yellow. Violet is the absorbed colour, the classic error of naming what is absorbed rather than what is seen. Green is opposite red, so a green solution absorbs red light. Blue is opposite orange; [Cu(H2O)6]2+ absorbs in the red-orange region and so looks blue.
Zn2+ is [Ar] 3d10. Every d orbital, upper and lower, is full, so no d electron can be promoted, no visible light is absorbed and the ion is colourless. The same holds for Cu+ (3d10). Having no d electrons describes Sc3+ (3d0), which is also colourless but for the other reason. Zinc does form complexes, such as [Zn(NH3)4]2+, and their d orbitals are split, but with no vacancy that does not matter. A substance absorbing all visible light would look black, not white.
Sc3+ is [Ar] 3d0: there are no d electrons to promote, so it absorbs no visible light. Ti3+ is 3d1 (violet), Cu2+ is 3d9 (blue) and Fe2+ is 3d6 (pale green); each has a partly filled d subshell, so a d-d transition absorbs some visible light and the solution is coloured.
ΔE depends on the ligand. NH3 splits the copper d orbitals more strongly than H2O, so [Cu(NH3)4(H2O)2]2+ absorbs a different (higher-energy) part of the spectrum and the transmitted colour changes to a deep blue. Copper stays at +2: no electrons are transferred in ligand exchange. Ammonia is colourless. Degenerate d orbitals would mean no d-d absorption at all, so the complex would be colourless, not deeper blue.
ΔE depends on the metal, its oxidation state (Fe2+(aq) is pale green, Fe3+(aq) yellow-brown), the ligand ([Cu(H2O)6]2+ versus [Cu(NH3)4(H2O)2]2+) and the coordination number and geometry (octahedral pink [Co(H2O)6]2+ versus tetrahedral blue [CoCl4]2−). Leaving out oxidation state and shape ignores these examples. The number of d electrons matters for whether a colour is possible, but it is not the only factor. Concentration changes how intense the colour is, not which frequency is absorbed.
Per photon: E = hc/λ = (6.63 × 10−34 × 3.00 × 108) / 5.00 × 10−7 = 3.98 × 10−19 J. Per mole: × 6.02 × 1023 = 2.39 × 105 J mol−1 = 239 kJ mol−1. 2.39 × 105 is the value in J mol−1 mislabelled as kJ. 3.98 × 10−22 converts one photon's energy to kJ but never multiplies by L. 7.98 × 10−7 leaves out c, using E = h/λ.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.