The shape of the lesson
By the end of the lesson, learners can
- Copper and cobalt swaps: Follow each arrow on the figure: write the equation, then say what happens to the colour, the coordination number and the shape.
- Hydroxide precipitates: Add aqueous sodium hydroxide or aqueous ammonia dropwise and then in excess.
- Excess NH3: Cu2+ deep blue [Cu(NH3)4(H2O)2]2+.
- Conc. HCl: [CuCl4]2− yellow-green, [CoCl4]2− blue, CN 4.
- A little OH−: neutral hydroxide precipitates.
- Colours: Cu blue, Fe(II) green, Fe(III) red-brown, Mn off-white, Cr grey-green.
One picture every idea returns to
In ligand exchange one ligand in a complex is replaced by another. Pale blue [Cu(H2O)6]2+ reacts with excess aqueous ammonia to give deep blue [Cu(NH3)4(H2O)2]2+; NH3 and H2O are similar in size, so the coordination number stays 6. With concentrated HCl it gives yellow-green [CuCl4]2−: the large Cl− ions change the coordination number to 4 and the shape to tetrahedral. Pink [Co(H2O)6]2+ similarly gives blue [CoCl4]2−. Adding a little NaOH(aq) or NH3(aq) removes H+ from water ligands and precipitates the neutral hydroxide, e.g. blue Cu(OH)2(H2O)4; some precipitates redissolve in excess reagent.
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, One ligand swaps for another.. Add ammonia, hydroxide or concentrated hydrochloric acid to a solution of a transition metal ion and the ligands around the metal are replaced. 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: Copper and cobalt swaps | Follow each arrow on the figure: write the equation, then say what happens to the colour, the coordination number and the shape. Project tab 3 of the lab (Transition elements lab, ligand exchange bench) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Hydroxide precipitates | Add aqueous sodium hydroxide or aqueous ammonia dropwise and then in excess. Project tab 3 of the lab (Transition elements lab, ligand exchange bench) and let a student drive it while the class predicts. | Anticipation / Reaction Guide |
| 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, 15 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 Hydroxide precipitates, 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.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
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 |
|---|---|
| A little ammonia acts as a base and gives the hydroxide precipitate; only excess ammonia acts as a ligand and gives the deep blue solution. | Concentrated HCl changes the coordination number from 6 to 4 because Cl − is large. [CuCl 4 ] 2− has a 2− charge (+2 + 4 × −1). |
Ligand exchange: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Describe what is seen, and write an equation for each stage, when aqueous ammonia is added dropwise until in excess to aqueous copper(II) sulfate.
Concentrated hydrochloric acid is added to aqueous cobalt(II) chloride. Describe the colour change and explain the change in shape of the complex.
Solutions X, Y and Z contain Fe2+, Fe3+ and Cr3+ (not in that order). Describe how aqueous NaOH alone would identify each.
Explain why the coordination number of copper does not change when excess ammonia reacts with [Cu(H2O)6]2+.
Write an equation for the reaction of [Cr(OH)3(H2O)3] with excess hydroxide ions, and state the colour of the product.
The yellow-green solution formed from copper(II) ions and concentrated HCl turns blue when diluted with water. Explain.
Answer key and marking notes
- Pale blue precipitate: [Cu(H2O)6]2+ + 2OH− → [Cu(OH)2(H2O)4] + 2H2O.
- The precipitate dissolves in excess to give a deep blue solution.
- [Cu(OH)2(H2O)4] + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2OH− + 2H2O.
- Pink to blue.
- [Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O.
- Cl− ions are larger than water molecules, so only four fit: octahedral (6) becomes tetrahedral (4).
- Fe2+: green precipitate, insoluble in excess, turning brown at the surface on standing.
- Fe3+: red-brown precipitate, insoluble in excess.
- Cr3+: grey-green precipitate that dissolves in excess NaOH to give a dark green solution.
- NH3 and H2O are similar in size and both are neutral.
- Four NH3 replace four H2O, so there are still six ligands: [Cu(NH3)4(H2O)2]2+, octahedral.
- [Cr(OH)3(H2O)3] + 3OH− → [Cr(OH)6]3− + 3H2O.
- Dark green solution.
- [Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O is an equilibrium.
- Adding water (and lowering [Cl−]) shifts it to the left, re-forming blue [Cu(H2O)6]2+.
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.
Excess concentrated aqueous ammonia is added to a solution containing [Cu(H2O)6]2+. What forms at the end?
- Aa deep blue solution of [Cu(NH3)4(H2O)2]2+
- Ba deep blue solution of [Cu(NH3)6]2+
- Ca yellow solution of [CuCl4]2−
- Da pale blue precipitate that does not dissolve
Concentrated hydrochloric acid is added to pink aqueous cobalt(II) ions, [Co(H2O)6]2+. What is observed, and what forms?
- Athe solution turns blue as tetrahedral [CoCl4]2− forms; the coordination number falls from 6 to 4
- Bthe solution turns blue as octahedral [CoCl6]4− forms; the coordination number stays 6
- Ca blue precipitate of cobalt(II) chloride forms
- Dthe solution stays pink because Cl− cannot replace water ligands
Aqueous sodium hydroxide is added to a solution of iron(II) sulfate. What is observed?
- Aa green precipitate of iron(II) hydroxide, which slowly turns orange-brown at the surface in air
- Ban orange-brown (red-brown) precipitate forms immediately
- Ca pale blue precipitate that dissolves in excess
- Dan off-white precipitate that dissolves in excess to give a colourless solution
Aqueous sodium hydroxide is added dropwise, then in excess, to aqueous Cu2+ ions. What is observed?
- Aa pale blue precipitate that does not dissolve in excess NaOH(aq)
- Ba pale blue precipitate that dissolves in excess NaOH(aq) to give a deep blue solution
- Ca grey-green precipitate that dissolves in excess NaOH(aq)
- Dno precipitate; the solution turns yellow-green
Which equation correctly shows the ligand exchange when concentrated HCl is added to aqueous copper(II) ions?
- A[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O
- B[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4(H2O)2]2− + 4H2O
- C[Cu(H2O)6]2+ + 6Cl− ⇌ [CuCl6]4− + 6H2O
- D[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2+ + 6H2O
A solution of an unknown ion gives a grey-green precipitate with a few drops of NaOH(aq). The precipitate dissolves in excess NaOH(aq) to give a dark green solution. Which ion is present?
- ACr3+
- BFe2+
- CMn2+
- DCu2+
A pale pink solution gives an off-white precipitate with NaOH(aq). The precipitate is insoluble in excess and turns brown on standing in air. Which ion is present?
- AMn2+
- BFe3+
- CZn2+
- DCo2+
Copper(II) hydroxide dissolves in excess aqueous ammonia but not in excess aqueous sodium hydroxide. Which statement explains this?
- Aammonia molecules are ligands that replace the water and hydroxide ligands to form the soluble complex [Cu(NH3)4(H2O)2]2+; hydroxide ions only remove H+ from water ligands
- Baqueous ammonia is a stronger base than aqueous sodium hydroxide
- Ccopper(II) hydroxide is amphoteric and reacts with ammonia as an acid
- Dammonia oxidises the copper(II) hydroxide to a soluble copper(III) compound
Quiz key
At first a little ammonia acts as a base and a pale blue precipitate of copper(II) hydroxide forms. In excess, ligand exchange gives the deep blue [Cu(NH3)4(H2O)2]2+: [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O. Writing [Cu(NH3)6]2+ assumes all six waters are replaced, which is the classic error. [CuCl4]2− needs a high concentration of Cl− (concentrated HCl), not ammonia. The precipitate does dissolve in excess ammonia; it is NaOH(aq) in excess that leaves it undissolved.
[Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O. The Cl− ions are larger than water molecules, so only four fit around the cobalt ion, giving a blue tetrahedral complex. Adding water reverses the change, back to pink. Six large chloride ions do not fit, so [CoCl6]4− is wrong. Cobalt(II) chloride is soluble, so no precipitate forms; the blue colour is a solution. Chloride ions can and do replace water: the high Cl− concentration drives the equilibrium to the right.
Fe2+(aq) + 2OH−(aq) → Fe(OH)2(s), a green precipitate. Oxygen in the air oxidises it to iron(III) hydroxide, so it darkens to orange-brown (red-brown) on standing. An immediate orange-brown precipitate shows Fe3+, not Fe2+. A pale blue precipitate is Cu(OH)2, and it does not dissolve in excess NaOH. An off-white precipitate is typical of Mn(OH)2, which does not dissolve in excess either.
[Cu(H2O)6]2+ + 2OH− → [Cu(OH)2(H2O)4] + 2H2O: the hydroxide ions remove H+ from two water ligands, giving a neutral pale blue precipitate, Cu(OH)2. It is insoluble in excess NaOH(aq). Dissolving to a deep blue solution happens with excess aqueous AMMONIA, not NaOH, which is the classic mix-up. A grey-green precipitate soluble in excess NaOH is Cr3+. A yellow-green solution is the result of adding concentrated HCl, not NaOH.
Four Cl− replace all six waters, giving tetrahedral [CuCl4]2− (charge +2 + 4 × −1 = 2−); the yellow complex mixed with remaining blue [Cu(H2O)6]2+ makes the solution look yellow-green. Keeping two waters gives an octahedral species that is not the product formed. Six chloride ions are too large to fit round the Cu2+ ion. A 2+ charge ignores the negative charge of the ligands, and the equation would not balance for charge.
Chromium(III) hydroxide is grey-green and amphoteric: with excess OH− it forms the soluble dark green [Cr(OH)6]3−. (With excess NH3(aq) the precipitate does not dissolve.) Fe2+ gives a green precipitate, but it does not dissolve in excess and it turns brown in air. Mn2+ gives an off-white precipitate, insoluble in excess, that darkens in air. Cu2+ gives a pale blue precipitate, insoluble in excess NaOH.
Manganese(II) solutions are very pale pink, and Mn(OH)2 is an off-white precipitate that air rapidly oxidises to a brown manganese(IV) compound. It does not dissolve in excess NaOH. Fe3+ gives an orange-brown precipitate straight away. Zn2+ gives a white precipitate that DOES dissolve in excess NaOH. Co2+ gives a blue precipitate, Co(OH)2, not an off-white one.
In excess ammonia, ligand exchange occurs: [Cu(OH)2(H2O)4] + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2H2O + 2OH−. The charged complex is soluble. Hydroxide ions only act as a base, so with NaOH the neutral precipitate remains. Ammonia is a weak base, much weaker than NaOH, so basic strength cannot be the reason. Copper(II) hydroxide is not amphoteric (that is why it stays in excess NaOH), and ammonia acts as a ligand, not an acid. The copper stays at +2 in the deep blue complex; nothing is oxidised.
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.