OxfordAQA A-Level Chemistry: Transition metals (9620) – Revision Notes

Section 3.2.5, Transition metals, is condensed here from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3 (exams from May/June 2020). This page covers 3.2.5.1 to 3.2.5.6. The whole section sits in International A2, so it is International A-level only. Full teaching and worked examples are in the study guide; start there.

After that, try the practice questions. Cell potentials are in electrode potentials, and ΔG = ΔH − TΔS is in thermodynamics. To plan the rest of your revision, use the course hub, a checklist to print and our free diagnostics.

3.2.5.1 Key terms

Term	Meaning
Transition metal properties (Ti to Cu)	Complexes, coloured ions, variable oxidation state, catalytic activity
Their cause	An incomplete d sub-level in the atom or ion
Ligand	Molecule or ion giving an electron pair to a transition metal ion to form a co-ordinate bond
Complex	Central metal atom or ion with ligands around it
Co-ordination number	Number of co-ordinate bonds formed by the central metal atom or ion

Ions lose 4s electrons before 3d. Cu²⁺ is [Ar] 3d⁹; V³⁺ is [Ar] 3d².

3.2.5.2 Ligand substitution

Swap	Co-ordination number	Why
H₂O ⇄ NH₃ (Co²⁺, Cu²⁺)	stays 6	same size, both uncharged
H₂O → Cl⁻ (Co²⁺, Cu²⁺, Fe³⁺)	6 → 4	Cl⁻ is larger

• Complete swap: [Co(NH₃)₆]²⁺. Incomplete swap: [Cu(NH₃)₄(H₂O)₂]²⁺ (blue to deep blue).
• With Cl⁻: [CoCl₄]²⁻ (pink to blue), [CuCl₄]²⁻, [FeCl₄]⁻.
• Bidentate: H₂NCH₂CH₂NH₂ (two N donors), C₂O₄²⁻ (two O donors). Multidentate: EDTA⁴⁻ (six donors).
• Haem: iron(II) with a multidentate ligand. O₂ bonds co-ordinately to Fe(II) for transport in blood. CO replaces the O₂, so it is toxic.

Method in steps: explain the chelate effect

1. Write the equation and count particles on each side.
2. More particles on the right, so ΔS is positive.
3. Similar bonds broken and formed (same donor atom type, same number), so ΔH is small.
4. ΔG = ΔH − TΔS is therefore negative; the chelating ligand replaces the monodentate ones.
5. If asked to calculate, change ΔS from J to kJ first.

3.2.5.3 Shapes and isomers

Shape	Bond angle	Forms with	Stereoisomerism
Octahedral	90°	small ligands such as H₂O and NH₃	cis–trans with monodentate ligands; optical with bidentate ones
Tetrahedral	109.5°	larger ligands such as Cl⁻	none listed
Square planar	90°	four ligands in a plane, as in [Pt(NH₃)₂Cl₂]	cis–trans; the cis form is cisplatin
Linear	180°	Ag⁺ with two NH₃ (Tollens’)	none

Cis–trans isomerism in complexes is a special case of E–Z isomerism.

Worked reminder. [Ni(H₂NCH₂CH₂NH₂)₂(H₂O)₂]²⁺: two bidentate ligands give four bonds and two waters give two more, so the co-ordination number is 6 and the ion is octahedral. Both diamines are neutral, so nickel is +2. The two water ligands can sit at 90° or at 180°, so cis and trans isomers are possible.

3.2.5.4 Colour

• The ion takes in certain visible wavelengths (absorbed); you see the ones it transmits or reflects.
• That energy lifts a d electron out of its ground state into an excited state.
• ΔE = hν = hc/λ. Changing the oxidation state, co-ordination number or ligand changes ΔE, so the colour changes.
• Spectroscopy relies on absorption of visible light; a simple colorimeter measures concentration.

Method in steps: ΔE calculations

1. Convert λ from nm to m (× 10⁻⁹).
2. ΔE (J, one ion) = hc/λ, or hν if you have frequency.
3. For kJ mol⁻¹: multiply by L, then divide by 1000.
4. Going the other way: divide kJ mol⁻¹ by L (after × 1000) before using λ = hc/ΔE.

Method in steps: colorimetry

Filter of the colour absorbed most → zero with solvent → standards → calibration graph → read unknown → multiply by any dilution factor.

Worked reminder. A calibration line has gradient 4.0 dm³ mol⁻¹. A solution diluted 1 in 20 reads 0.36. Diluted concentration = 0.36 ÷ 4.0 = 0.090 mol dm⁻³, so the original was 0.090 × 20 = 1.80 mol dm⁻³.

3.2.5.5 Variable oxidation states

Colour	V species	Oxidation state
yellow	VO₂⁺	+5
blue	VO²⁺	+4
green	V³⁺	+3
violet	V²⁺	+2

Zinc in acid reduces vanadate(V) through all four. Green also appears briefly as a yellow–blue mixture.

• pH effect: half-equations with H⁺ (such as MnO₄⁻/Mn²⁺) shift with acidity; more H⁺ makes the higher oxidation state easier to reduce.
• Ligand effect: E⦵(Ag⁺/Ag) = +0.80 V but E⦵([Ag(NH₃)₂]⁺/Ag) = +0.37 V.
• Tollens’ reagent: [Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃. Silver mirror with an aldehyde; no change with a ketone.

Titration ratios

Reaction	MnO₄⁻ : other
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O	1 : 5
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O	2 : 5

Method in steps: MnO₄⁻ titration

1. Moles of MnO₄⁻ = concentration × titre (in dm³).
2. Use the ratio to get moles of analyte in the portion titrated.
3. Scale up to the whole solution (for example × 10 for 25.0 cm³ from 250 cm³).
4. Convert to mass, percentage or Mr as asked.
5. For an unfamiliar reductant, build the equation from half-equations first, then repeat steps 1 to 4.

No indicator is needed: stop at the first pink that stays. Acidify with dilute H₂SO₄, never HCl.

3.2.5.6 Catalysts

Catalyst	Type	Reaction	How it works
V₂O₅	heterogeneous	Contact process	V +5 → +4 oxidising SO₂, then back to +5 with O₂
Fe	heterogeneous	Haber process	active sites on the surface
Fe²⁺	homogeneous	S₂O₈²⁻ + I⁻	Fe²⁺ → Fe³⁺ → Fe²⁺; each step between oppositely charged ions
Mn²⁺	homogeneous, autocatalyst	MnO₄⁻ + C₂O₄²⁻	Mn²⁺ → Mn³⁺ → Mn²⁺; a product speeds up the reaction

• Heterogeneous: different phase; support medium for a large surface area at low cost; poisoning blocks active sites and costs money.
• Homogeneous: same phase; goes through an intermediate species.
• Variable oxidation states let the catalyst pass electrons on and be regenerated.

Key equations:

V₂O₅ + SO₂ → V₂O₄ + SO₃        V₂O₄ + ½O₂ → V₂O₅
2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻        2I⁻ + 2Fe³⁺ → I₂ + 2Fe²⁺
MnO₄⁻ + 8H⁺ + 4Mn²⁺ → 5Mn³⁺ + 4H₂O        C₂O₄²⁻ + 2Mn³⁺ → 2CO₂ + 2Mn²⁺

Must-know distinctions

• Mono-, bi- and multidentate: ligands with one, two or many donor atoms.
• Number of ligands vs co-ordination number: [Co(C₂O₄)₃]⁴⁻ has three ligands but six bonds.
• Cis vs trans: identical ligands at 90° vs 180°.
• Absorbed vs seen: the colour you see is what is not absorbed.
• Heterogeneous vs homogeneous: different phase vs same phase.
• Catalyst vs autocatalyst: an autocatalyst is formed in the very reaction it catalyses.

Quick self-test

1. What are the oxidation state and co-ordination number of iron in [Fe(EDTA)]⁻?
2. Why does replacing H₂O by Cl⁻ around Cu²⁺ change the co-ordination number?
3. Count the particles on each side: [Co(H₂O)₆]²⁺ + 3C₂O₄²⁻ → [Co(C₂O₄)₃]⁴⁻ + 6H₂O.
4. A ligand exchange has ΔH = −2.0 kJ mol⁻¹ and ΔS = +96 J K⁻¹ mol⁻¹. Find ΔG at 310 K.
5. A d-electron transition needs 3.70 × 10⁻¹⁹ J. Find λ in nm (c = 3.00 × 10⁸ m s⁻¹; h = 6.63 × 10⁻³⁴ J s).
6. Give the formula and colour of vanadium(III) in solution.
7. What do you see when Tollens’ reagent is warmed with a ketone?
8. How many moles of Fe²⁺ react with 13.20 cm³ of 0.0150 mol dm⁻³ MnO₄⁻?
9. How many moles of C₂O₄²⁻ react with 21.60 cm³ of 0.0120 mol dm⁻³ MnO₄⁻?
10. Why does the MnO₄⁻ and C₂O₄²⁻ reaction start slowly?
11. What shape is the silver complex in Tollens’ reagent, and what is its bond angle?

Answers

1. Fe − 4 = −1, so +3; co-ordination number 6 (EDTA⁴⁻ has six donor atoms).
2. Cl⁻ is larger than H₂O, so only four fit round the ion: 6 → 4.
3. 4 on the left, 7 on the right; entropy increases.
4. ΔG = −2.0 − (310 × 0.096) = −31.8 kJ mol⁻¹.
5. λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 3.70 × 10⁻¹⁹ = 5.38 × 10⁻⁷ m = 538 nm.
6. V³⁺, green.
7. No change; ketones are not oxidised by it.
8. 0.0150 × 0.01320 = 1.98 × 10⁻⁴ mol MnO₄⁻, × 5 = 9.90 × 10⁻⁴ mol.
9. 0.0120 × 0.02160 = 2.592 × 10⁻⁴ mol MnO₄⁻, × 5/2 = 6.48 × 10⁻⁴ mol.
10. Two negative ions repel, and the catalyst, Mn²⁺, has not formed yet.
11. Linear, 180°.

Where marks are usually lost

• Writing [Ar] 4s² 3d⁷ for Co²⁺: the 4s electrons are removed first, so it is [Ar] 3d⁷.
• Calling [Cu(NH₃)₄(H₂O)₂]²⁺ four-co-ordinate; the two waters still count.
• Leaving out the entropy argument, or the particle count, in a chelate explanation.
• Forgetting to convert ΔS into kJ before using ΔG = ΔH − TΔS.
• Leaving λ in nm inside hc/λ.
• Stating that the colour seen is the colour absorbed.
• Writing the V₂O₅ steps without showing that the catalyst is regenerated.
• Using the Fe²⁺ ratio (1 : 5) for ethanedioate (2 : 5).
• Forgetting to scale up from the titrated portion to the whole solution.

Official syllabus

Source: OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for exams from May/June 2020. Section 3.2.5 Transition metals (International A2), 3.2.5.1 to 3.2.5.6.