Flashcards · 9701 · A Level Topic 28

Chemistry of transition elements (A Level) flashcards

29 cards on A Level Topic 28, subtopics 28.1 to 28.5. Pick a subtopic or study them all.

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28.1 General physical and chemical properties of the first row of transition elements, titanium to copper

Transition element: A d-block element that forms one or more stable ions with a partially filled d subshell.
Why are scandium and zinc not transition elements?: Scandium forms only Sc³⁺, which is d⁰. Zinc forms only Zn²⁺, which is d¹⁰ (full). Neither has a stable ion with a partially filled d subshell.
Electron configurations of Cr, Cu and Fe²⁺: Cr: [Ar] 3d⁵4s¹ Cu: [Ar] 3d¹⁰4s¹ Fe²⁺: [Ar] 3d⁶ (4s electrons are lost first)
Why do transition elements have variable oxidation states?: The 4s and 3d sub-shells are very close in energy, so different numbers of electrons can be lost (or used in bonding).
Why are transition elements and their compounds good catalysts?: Variable oxidation states let them form intermediates, and d orbitals allow reactants to be adsorbed onto their surface.
Four characteristic properties of transition elements: Variable oxidation states; coloured compounds; catalytic activity; formation of complex ions.

28.2 General characteristic chemical properties of the first set of transition elements, titanium to copper

Ligand and complex ion: Ligand: a species with a lone pair that forms a dative covalent bond to a central metal ion. Complex: a central metal ion surrounded by ligands bonded by dative bonds.
Monodentate, bidentate and multidentate ligands (examples): Monodentate: H₂O, NH₃, Cl⁻, CN⁻, OH⁻. Bidentate: ethanedioate (C₂O₄²⁻), 1,2-diaminoethane (en). Multidentate: EDTA⁴⁻ (hexadentate).
Shapes of complexes by coordination number: 6: octahedral. 4: tetrahedral (e.g. [CuCl₄]²⁻) or square planar (e.g. [Pt(NH₃)₂Cl₂]). 2: linear.
[Cu(H₂O)₆]²⁺ with NaOH(aq), and with a little then excess NH₃(aq): NaOH: pale blue precipitate [Cu(OH)₂(H₂O)₄], insoluble in excess. NH₃: the same pale blue precipitate, which dissolves in excess to give deep blue [Cu(NH₃)₄(H₂O)₂]²⁺.
[Cu(H₂O)₆]²⁺ with excess concentrated HCl: [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O: yellow-green, tetrahedral. Coordination number falls from 6 to 4 because Cl⁻ is larger than H₂O; adding water reverses it.
Colours of MnO₄⁻, Cr₂O₇²⁻, [Fe(H₂O)₆]²⁺ and [Fe(H₂O)₆]³⁺: MnO₄⁻: purple. Cr₂O₇²⁻: orange. [Fe(H₂O)₆]²⁺: pale green. [Fe(H₂O)₆]³⁺: yellow-brown.
25.0 cm³ Fe²⁺(aq) reacts with 23.50 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find [Fe²⁺].: n(MnO₄⁻) = 0.02350 × 0.0200 = 4.70 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 4.70 × 10⁻⁴ = 2.35 × 10⁻³ mol; [Fe²⁺] = 2.35 × 10⁻³ ÷ 0.0250 = 0.0940 mol dm⁻³.
Reaction of Cu²⁺ with I⁻, and how it is used: 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂. White CuI precipitates; the I₂ released is titrated with thiosulfate to find the amount of Cu²⁺.

28.3 Colour of complexes

Degenerate and non-degenerate d orbitals: Degenerate: the same energy (all five 3d orbitals in an isolated ion). Non-degenerate: different energies; ligands split the d orbitals into two sets separated by ΔE.
How do the d orbitals split in octahedral and tetrahedral complexes?: Octahedral: two higher, three lower. Tetrahedral: three higher, two lower, with a smaller ΔE for comparable ligands.
Why are transition metal complexes coloured?: Ligands split the d orbitals. An electron absorbs visible light of frequency given by ΔE = hf and is promoted to the higher set. The colour seen is the complementary colour of the light absorbed.
Why are Sc³⁺ and Zn²⁺ compounds colourless?: Sc³⁺ is d⁰ (no electron to promote) and Zn²⁺ is d¹⁰ (no vacancy in the higher set), so no d–d transition is possible.
Why does ligand exchange change the colour of a complex?: Different ligands split the d orbitals by different amounts, so ΔE changes, a different frequency is absorbed and a different complementary colour is seen.
Cobalt(II): aqua ion, with excess NH₃(aq), and with concentrated HCl: [Co(H₂O)₆]²⁺: pink. Excess NH₃: [Co(NH₃)₆]²⁺, pale yellow-brown (straw), darkens in air. Conc. HCl: [CoCl₄]²⁻, blue, tetrahedral.

28.4 Stereoisomerism in transition element complexes

Cis and trans isomers of [Pt(NH₃)₂Cl₂]: Square planar. Cis: the two Cl⁻ at 90° (cisplatin, an anti-cancer drug). Trans: the two Cl⁻ at 180°.
Which octahedral complexes in 9701 show cis–trans isomerism?: [Co(NH₃)₄(H₂O)₂]²⁺ and [Ni(en)₂(H₂O)₂]²⁺: two H₂O at 90° (cis) or 180° (trans).
Why does [Ni(en)₃]²⁺ show optical isomerism?: The three bidentate en ligands wrap round Ni²⁺ like a propeller, so the complex and its mirror image are non-superimposable.
How many stereoisomers does [Ni(en)₂(H₂O)₂]²⁺ have?: Three: the trans isomer (has a plane of symmetry, not optically active) and the two optical isomers of the cis form.
Are cis and trans complexes polar?: Trans: the bond dipoles are opposite and cancel, so non-polar. Cis: the dipoles do not cancel, so polar.

28.5 Stability constants, Kstab

Stability constant, Kstab: The equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.
Write Kstab for [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O: Kstab = [Cu(NH₃)₄(H₂O)₂²⁺] ÷ ([Cu(H₂O)₆²⁺][NH₃]⁴), units mol⁻⁴ dm¹². [H₂O] is left out.
What does the size of Kstab tell you about ligand exchange?: A large Kstab means a stable complex. When ligands compete, exchange favours the complex with the larger Kstab, though a large excess of a ligand can still shift the equilibrium.
Kstab = 1.0 × 10¹³ mol⁻⁴ dm¹² (illustrative) and [NH₃] = 0.20 mol dm⁻³. Ratio of ammine to aqua complex?: Ratio = Kstab × [NH₃]⁴ = 1.0 × 10¹³ × 1.6 × 10⁻³ = 1.6 × 10¹⁰, so almost all the copper is in the ammine complex.