Topic 17
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Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.
Definitions to learn word for word
- Transition metal: a d-block element that forms one or more stable ions with an incompletely filled d sub-shell.
- Ligand: a molecule or ion with a lone pair of electrons that forms a dative (coordinate) covalent bond to a metal ion.
- Complex ion: a central metal ion surrounded by ligands.
- Coordination number: the number of dative bonds from ligands to the central metal ion.
- Bidentate / hexadentate: a ligand that forms two / six dative bonds.
- Amphoteric: reacts with both acids and bases.
- Ligand exchange: one ligand in a complex is replaced by another.
- Heterogeneous catalyst: in a different phase from the reactants; the reaction occurs at its surface.
- Homogeneous catalyst: in the same phase as the reactants; the reaction goes via an intermediate species.
- Autocatalysis: a product of the reaction catalyses it.
See Transition metals and their electrons, Ligands and complex ions and Two kinds of catalyst.
Electrons and oxidation numbers
| Rule | Detail |
|---|---|
| Filling | 4s before 3d; exceptions Cr [Ar] 3d⁵ 4s¹ and Cu [Ar] 3d¹⁰ 4s¹ |
| Ions | lose 4s first, then 3d: Fe²⁺ [Ar] 3d⁶, Fe³⁺ [Ar] 3d⁵, Cu²⁺ [Ar] 3d⁹ |
| Not transition metals | Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) |
| Variable oxidation number | 4s and 3d close in energy, so ionisation energies rise gradually |
| Highest oxidation number (Sc to Mn) | number of 4s + 3d electrons |
Shapes
| Coordination number | Shape | Angle | Examples |
|---|---|---|---|
| 6 | octahedral | 90° | [Cu(H₂O)₆]²⁺, [Ni(NH₃)₆]²⁺, [Cr(OH)₆]³⁻ |
| 4 | tetrahedral (large Cl⁻) | 109.5° | [CuCl₄]²⁻, [CoCl₄]²⁻ |
| 4 | square planar | 90° | cis-platin, [Pt(NH₃)₂Cl₂] |
Ligands that count: H₂O, NH₃, OH⁻, Cl⁻ (one bond each); NH₂CH₂CH₂NH₂ (two); EDTA⁴⁻ (six).
Colour
- Ligands split the d-orbitals into two levels; a d electron absorbs light with ΔE = hf and is promoted; the remaining light is seen.
- Colourless: d⁰ (Sc³⁺, Ti⁴⁺) and d¹⁰ (Zn²⁺, Cu⁺): no promotion possible.
- Colour changes with the oxidation number, the ligand or the coordination number.
Vanadium and chromium
| Oxidation number | Vanadium | Chromium |
|---|---|---|
| +6 | Cr₂O₇²⁻ orange; CrO₄²⁻ yellow | |
| +5 | VO₂⁺ yellow | |
| +4 | VO²⁺ blue | |
| +3 | V³⁺ green | Cr³⁺ green; [Cr(OH)₆]³⁻ dark green |
| +2 | V²⁺ violet | Cr²⁺ blue |
- Zinc in acid reduces V(+5) to V(+2) and Cr₂O₇²⁻ to Cr²⁺. Every step has a positive E°cell. The first green in the vanadium series is a mixture of yellow and blue.
- A reagent stops at a state when its E° is less positive than that step but more positive than the next.
- Cr(III) → Cr(VI) with H₂O₂ needs alkali; then acid gives dichromate(VI).
- : alkali gives yellow, acid gives orange. Not redox: Cr is +6 in both.
Observations with NaOH and NH₃
| Ion | A few drops NaOH | Excess NaOH | A few drops NH₃ | Excess NH₃ |
|---|---|---|---|---|
| Cr³⁺ | green ppt | dark green solution | green ppt | purple solution |
| Mn²⁺ | off-white ppt, browns in air | no change | off-white ppt | no change |
| Fe²⁺ | green ppt, browns in air | no change | green ppt | no change |
| Fe³⁺ | orange-brown ppt | no change | orange-brown ppt | no change |
| Co²⁺ | blue ppt, pink on standing | no change | blue ppt | straw solution, darkens in air |
| Ni²⁺ | green ppt | no change | green ppt | blue solution |
| Cu²⁺ | blue ppt | no change | blue ppt | deep blue solution |
| Zn²⁺ | white ppt | colourless solution | white ppt | colourless solution |
Three kinds of reaction:
- Deprotonation (OH⁻ or a few drops of NH₃):
- Amphoteric (Cr, Zn with excess NaOH):
- Ligand exchange (excess NH₃):
Details are on pages 11, 12 and 13.
Ligand exchange, stability, haemoglobin
- : blue to yellow (green mixture); 6 to 4; octahedral to tetrahedral.
- : pink to blue.
- Bidentate or hexadentate replaces monodentate: more particles, ΔSsystem positive, ΔH ≈ 0, so ΔStotal positive and the complex is more stable.
- Haemoglobin is an iron(II) complex with a polydentate ligand. CO replaces O₂ by ligand exchange and binds more strongly.
Catalysts
| Catalyst | Type | How it works |
|---|---|---|
| V₂O₅(s), contact process | heterogeneous | V +5 → +4 (SO₂ → SO₃), then +4 → +5 (O₂) |
| Pt/Rh(s), catalytic converter | heterogeneous | CO and NO adsorbed, bonds weakened, react; CO₂ and N₂ desorbed |
| Fe²⁺(aq), I⁻ with S₂O₈²⁻ | homogeneous | Fe²⁺ → Fe³⁺ → Fe²⁺; intermediate Fe³⁺ |
| Mn²⁺(aq), MnO₄⁻ with C₂O₄²⁻ | homogeneous, autocatalysis | product catalyses; slow, fast, slowing |
Standard electrode potentials used in this topic
| Couple | E° / V |
|---|---|
| V²⁺/V | −1.18 |
| Zn²⁺/Zn | −0.76 |
| Cr³⁺/Cr²⁺ | −0.41 |
| V³⁺/V²⁺ | −0.26 |
| CrO₄²⁻/Cr(OH)₃ (alkali) | −0.13 |
| Sn⁴⁺/Sn²⁺ | +0.15 |
| VO²⁺/V³⁺ | +0.34 |
| I₂/I⁻ | +0.54 |
| Fe³⁺/Fe²⁺ | +0.77 |
| H₂O₂/OH⁻ (alkali) | +0.88 |
| VO₂⁺/VO²⁺ | +1.00 |
| O₂/H₂O | +1.23 |
| Cr₂O₇²⁻/Cr³⁺ | +1.33 |
| S₂O₈²⁻/SO₄²⁻ | +2.01 |
The ten traps
- Writing Fe²⁺ as [Ar] 3d⁴ 4s²: remove the 4s electrons first.
- Defining a transition metal by the atom: it is the ion with an incompletely filled d sub-shell.
- Counting ligands, not dative bonds, for the coordination number.
- Drawing the bond from H, not O, in an aqua complex; leaving out brackets or the charge.
- Saying the electron emits the colour. The colour seen is the light not absorbed.
- Saying the first green in the vanadium series is V³⁺. It is a mixture.
- Treating yellow to orange (chromate to dichromate) as redox. Cr is +6 in both.
- Writing NH₃ as a ligand in the first step. A few drops of ammonia are a base.
- Calling the dissolving in excess ammonia amphoteric. It is ligand exchange.
- Explaining chelate stability by "more bonds" instead of more particles and a positive ΔSsystem.
Mixed test
Try these with no notes, then check the answers.
- Write the configuration of Co²⁺ (Z = 27) and explain why cobalt is a transition metal.
- Give the oxidation number of manganese in MnO₄⁻ and the coordination number of nickel in [Ni(NH₂CH₂CH₂NH₂)₃]²⁺.
- Explain why [Ni(H₂O)₆]²⁺ is green but [Zn(H₂O)₆]²⁺ is colourless.
- Will acidified dichromate(VI) oxidise Fe²⁺? Use E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V and E°(Fe³⁺/Fe²⁺) = +0.77 V, and write the equation.
- Describe what you would see when sodium hydroxide is added to Zn²⁺(aq) dropwise and then in excess. Write both equations.
- Describe what happens when excess ammonia is added to Ni²⁺(aq), and write the equation starting from [Ni(H₂O)₆]²⁺.
- Pink [Co(H₂O)₆]²⁺ turns blue in concentrated hydrochloric acid. Give the equation and the changes in coordination number and shape.
- Explain why [Cu(en)₃]²⁺ is more stable than [Cu(H₂O)₆]²⁺, given .
- Why is carbon monoxide toxic?
- A student starts the preparation of chromium(II) ethanoate with 4.00 g of K₂Cr₂O₇ ( = 294.2) and obtains 2.60 g of [Cr₂(CH₃COO)₄(H₂O)₂] ( = 376.0). Calculate the percentage yield.
Answers cover these until you have tried the questions
- Co²⁺ has 25 electrons: [Ar] 3d⁷ 4s² loses the 4s electrons, giving [Ar] 3d⁷. Cobalt forms stable ions (Co²⁺, 3d⁷) with an incompletely filled d sub-shell, so it is a transition metal.
- MnO₄⁻: x − 8 = −1, so +7. Three bidentate ligands, each forming 2 dative bonds: coordination number 6.
- Ni²⁺ is 3d⁸: ligands split the d-orbitals, an electron absorbs visible light and is promoted, and the remaining light is seen (green). Zn²⁺ is 3d¹⁰: the d sub-shell is full, no electron can be promoted, so no visible light is absorbed.
- E°cell = +1.33 − (+0.77) = +0.56 V, positive, so yes.
- A white precipitate forms, which dissolves in excess to a colourless solution. ;
- A green precipitate forms with a few drops, and dissolves in excess ammonia to a blue solution.
- . Coordination number 6 to 4; octahedral to tetrahedral.
- Particles: 1 + 3 = 4 on the left, 1 + 6 = 7 on the right, so ΔSsystem is positive. Six dative bonds are broken and six made, so ΔH ≈ 0. ΔStotal is positive, the exchange is feasible and K is large.
- CO replaces O₂ on the Fe²⁺ of haemoglobin by ligand exchange. It binds more strongly, so it is not easily released and the blood cannot carry oxygen.
- n(K₂Cr₂O₇) = 4.00 ÷ 294.2 = 0.013596 mol, which gives 0.013596 mol of product. Theoretical mass = 0.013596 × 376.0 = 5.112 g. Yield = 2.60 ÷ 5.112 × 100 = 50.9%.