HookA faulty boiler kills by ligand substitution

Around fifty people die in Britain each year from accidental carbon monoxide poisoning — the 'silent killer' of faulty boilers and blocked flues. The chemistry of those deaths is on this page. Haemoglobin carries oxygen on an iron(II) ion held in a multidentate porphyrin ligand; oxygen binds to the iron reversibly, loads in the lungs and unloads in the tissues. Carbon monoxide is the wrong passenger for the same seat: it binds to the iron roughly 200 times more strongly than oxygen and will not get off. Every CO molecule that boards takes a haem site out of service, and no amount of fresh air quickly reverses it — which is why treatment is high-pressure oxygen, a brute-force attempt to shove the equilibrium back.

That is a ligand substitution reaction, and it is the template for everything in 3.2.5. A transition-metal ion sits at the centre; molecules or ions with lone pairs — ligands — bond to it; swap the ligands and you change the complex's colour, shape, stability, sometimes whether a person survives the night. Add the d-block's other party tricks — variable oxidation states, catalysis, and ions coloured brilliantly enough to have founded the pigment industry — and you have the most visually distinctive section of the AQA course. It is also one of the most learnable: a handful of rules generates every equation.

ModelWhat makes a metal 'transition' — and why Sc and Zn are out

AQA's definition: a transition metal forms at least one stable ion with a partially filled d subshell. That last clause does real work. Scandium's only common ion, Sc\(^{3+}\), has an empty d subshell (3d\(^{0}\)); zinc's only ion, Zn\(^{2+}\), has a full one (3d\(^{10}\)). Both sit in the d block; neither is a transition metal — and 'because its ion has no partially filled d subshell' is the exact justification the mark scheme wants.

The partially filled d subshell buys four characteristic properties: complex formation (the ion accepts lone pairs from ligands into vacant orbitals), coloured ions (electrons shuffle between split d orbitals — block four of this page), variable oxidation states (the 4s and 3d electrons are close in energy, so iron can be +2 or +3, manganese anything from +2 to +7), and catalysis (those variable states let the metal lend and reclaim electrons mid-reaction).

Vocabulary to nail now, because every later block leans on it: a ligand is a molecule or ion with a lone pair that forms a coordinate bond to the metal; the coordination number is the number of coordinate bonds to the central ion. Water, ammonia and chloride are monodentate (one lone pair donated); 1,2-diaminoethane and the ethanedioate ion are bidentate (two); EDTA\(^{4-}\) is hexadentate, wrapping a metal ion in a single molecular hug.

MechanismLigand substitution — same seat, different passengers

Ammonia and water are similar in size and both uncharged, so swapping one for the other leaves the coordination number at six. With cobalt the exchange runs to completion: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 6\text{NH}_3 \rightarrow [\text{Co}(\text{NH}_3)_6]^{2+} + 6\text{H}_2\text{O}\). With copper, excess ammonia only replaces four of the six waters, giving the royal-blue \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).

Chloride is a different shape of passenger: larger than water and negatively charged. Only four fit, so substitution with concentrated hydrochloric acid changes the coordination number: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\) — blue octahedral to yellow-green tetrahedral, and reversible on dilution. Cobalt does the same trick: pink \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) to blue \([\text{CoCl}_4]^{2-}\).

Why do multidentate ligands win so decisively? The chelate effect, and it is an entropy argument, not a bond-strength one. Take \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + \text{EDTA}^{4-} \rightarrow [\text{Cu(EDTA)}]^{2-} + 6\text{H}_2\text{O}\): two particles become seven. The enthalpy change is close to zero — six coordinate bonds broken, six made — but the entropy change is strongly positive, so \(\Delta G = \Delta H - T\Delta S\) is firmly negative and the multidentate complex is the more stable. Haemoglobin's porphyrin holds iron the same way; carbon monoxide's fatal grip is the substitution the body cannot afford.

ModelShapes and stereoisomerism — four geometries, two kinds of twin

Four shapes cover every complex AQA asks about. Octahedral, coordination number 6, 90° angles: most aqua ions, \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}\). Tetrahedral, coordination number 4, 109.5°: the chloro complexes such as \([\text{CoCl}_4]^{2-}\), because chloride's bulk holds neighbours apart. Square planar, coordination number 4, 90°: platinum(II) complexes, most famously cisplatin. Linear, coordination number 2, 180°: \([\text{Ag}(\text{NH}_3)_2]^{+}\) — Tollens' reagent from organic analysis, moonlighting here.

Two kinds of stereoisomerism follow from these shapes. Cis-trans isomerism needs two identical ligands that can sit adjacent (90°, cis) or opposite (180°, trans): octahedral \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^{+}\) exists as both, and square-planar Pt(NH\(_3\))\(_2\)Cl\(_2\) gives cisplatin and transplatin — only the cis isomer is the anticancer drug, because only its geometry lets both chlorides be replaced by neighbouring guanines on one DNA strand.

Optical isomerism appears in octahedral complexes with three bidentate ligands, such as \([\text{Co(en)}_3]^{3+}\): the complex and its mirror image are non-superimposable, like a left and right propeller. State the shape, the bond angle and the coordination number as a trio — AQA's diagram questions award them separately.

MechanismWhy the colours? Split d orbitals and a formula from physics

In an isolated ion the five d orbitals are degenerate — equal in energy. Ligands change that: their lone pairs repel the d orbitals unequally, splitting them into a lower and a higher set separated by an energy gap \(\Delta E\). An electron in the lower set can absorb a photon of exactly the right energy and jump — and for transition-metal complexes that energy sits in the visible region. The light that reaches your eye is what is left over: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) absorbs at the red end of the spectrum, so it looks blue. The governing equation is \(\Delta E = h\nu = \dfrac{hc}{\lambda}\) — absorption at 700 nm, say, corresponds to \(\Delta E = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{7.00 \times 10^{-7}} = 2.84 \times 10^{-19}\ \text{J}\).

Anything that changes the splitting changes the colour, and AQA lists exactly four levers: the metal's oxidation state (pale green Fe\(^{2+}\) vs yellow Fe\(^{3+}\)), the ligand (blue aqua-copper vs royal-blue ammine-copper), the coordination number (six-coordinate blue to four-coordinate yellow-green in the chloro swap) — and, trivially, no d electrons or a full set means no transition and no colour, which is why Sc\(^{3+}\) and Zn\(^{2+}\) solutions are colourless.

Colour also makes concentration measurable. In colorimetry you choose a filter of the complementary colour (the wavelength the complex absorbs most), measure the absorbance of standard solutions of known concentration, plot a calibration curve, then read the unknown off it. Pale complexes are first intensified with a ligand swap — thiocyanate turns dilute Fe\(^{3+}\) blood-red precisely so there is something to measure.

DataVariable oxidation states — vanadium's rainbow and the manganate(VII) titration

Reduce ammonium vanadate(V) with zinc in acid and vanadium climbs down the oxidation-state ladder in full colour: +5 yellow (VO\(_2^{+}\)), +4 blue (VO\(^{2+}\)), +3 green (V\(^{3+}\)), +2 violet (V\(^{2+}\)). Each step is a one-electron redox reaction, and the sequence is the standard exam illustration that oxidation states in the d block are a staircase, not a single rung.

The workhorse quantitative chemistry is the manganate(VII) titration. Acidified potassium manganate(VII) oxidises iron(II): the half-equations are \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\) and \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\), combining (×5 on the iron) to \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\). The titration is self-indicating: manganate(VII) in the burette is intense purple, its product effectively colourless, so the end-point is the first permanent pale-pink tinge. Acidify with dilute sulfuric acid — hydrochloric is itself oxidised by manganate(VII) and inflates the titre.

Worked example

A 0.700 g sample of iron wire is dissolved in excess dilute sulfuric acid (forming Fe\(^{2+}\)) and made up to 250 cm³. A 25.0 cm³ portion is titrated with 0.0200 mol dm⁻³ KMnO\(_4\); the mean titre is 12.00 cm³. Find the percentage purity of the wire (A\(_r\) Fe = 55.8).

Step 1 — moles of the known: \(n(\text{MnO}_4^-) = 0.0200 \times \dfrac{12.00}{1000} = 2.40 \times 10^{-4}\ \text{mol}\).

Step 2 — ratio from the equation (1 : 5): \(n(\text{Fe}^{2+}) = 5 \times 2.40 \times 10^{-4} = 1.20 \times 10^{-3}\ \text{mol}\) in the 25.0 cm³ portion.

Step 3 — scale to the flask: the portion is one tenth of 250 cm³, so total \(n(\text{Fe}^{2+}) = 1.20 \times 10^{-2}\ \text{mol}\).

Step 4 — mass and purity: \(m = 1.20 \times 10^{-2} \times 55.8 = 0.670\ \text{g}\), so purity \(= \dfrac{0.670}{0.700} \times 100 = 95.7\%\).

The four-step spine — moles of the known, ratio, scale-up, convert — solves every redox titration on the paper. The two marks students drop: forgetting the ×10 scale-up to the volumetric flask, and inverting the 5 : 1 ratio.

CaseCatalysis — the d block earning its keep

A heterogeneous catalyst works in a different phase from the reactants, doing its chemistry at active sites on the surface: reactants adsorb, bonds weaken, products desorb. The Contact process runs on vanadium(V) oxide, and its mechanism is a two-step loan of oxygen that only works because vanadium has two accessible oxidation states: \(\text{SO}_2 + \text{V}_2\text{O}_5 \rightarrow \text{SO}_3 + \text{V}_2\text{O}_4\), then \(2\text{V}_2\text{O}_4 + \text{O}_2 \rightarrow 2\text{V}_2\text{O}_5\). The catalyst is consumed and remade — write both equations and show the regeneration. Surface area is money: catalysts are spread on inert supports, and they die by poisoning when impurities block the active sites — sulfur compounds poison the iron of the Haber process, lead wrecks the platinum-rhodium of a catalytic converter, which is why leaded petrol and 'cats' never coexisted.

A homogeneous catalyst shares a phase with the reactants and works through an intermediate oxidation state. The classic: peroxodisulfate oxidising iodide, \(\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2\), is slow because both ions are negative and repel. Fe\(^{2+}\) mediates in two mutually attractive steps: \(\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}\), then \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\).

Sometimes a reaction brews its own catalyst — autocatalysis. Manganate(VII) oxidising ethanedioate starts sluggishly, then accelerates sharply as the product Mn\(^{2+}\) begins catalysing the reaction, then slows as reactants run out. On a rate-time graph that signature — slow, surge, fade — is the fingerprint AQA asks you to explain, and 'the product is the catalyst' is the sentence that earns it.

VocabularyKey terms the mark scheme pays for

Transition metal

A d-block element that forms at least one stable ion with a partially filled d subshell — which excludes scandium (Sc³⁺ is d⁰) and zinc (Zn²⁺ is d¹⁰).

Ligand

A molecule or ion with a lone pair of electrons that forms a coordinate bond to a central metal ion — H₂O, NH₃, Cl⁻, CN⁻, en, EDTA⁴⁻.

Coordination number

The number of coordinate bonds from ligands to the central metal ion: 6 in octahedral aqua ions, 4 in tetrahedral chloro complexes, 2 in linear [Ag(NH₃)₂]⁺.

Bidentate ligand

A ligand donating two lone pairs from different atoms, forming two coordinate bonds — 1,2-diaminoethane (en) and the ethanedioate ion are AQA's standard pair.

Chelate effect

The extra stability of complexes with multidentate ligands: substitution releases more particles than it consumes, so ΔS is strongly positive while ΔH is roughly zero.

Degenerate orbitals

Orbitals of equal energy. Ligands split the five d orbitals into two sets; electrons absorbing visible light to jump the gap ΔE = hν is the origin of colour.

Heterogeneous catalyst

A catalyst in a different phase from the reactants, working by adsorption at active sites on its surface — V₂O₅ in the Contact process, Fe in the Haber process.

Autocatalysis

Catalysis of a reaction by one of its own products, giving a slow start then a sharp acceleration — Mn²⁺ in the manganate(VII)/ethanedioate reaction.

TrapsMisconceptions that cost marks

“Every d-block element is a transition metal.”

Actually: The definition turns on the ion, not the block. Sc³⁺ has an empty d subshell and Zn²⁺ a full one, so neither element forms an ion with a PARTIALLY filled d subshell — no colour, no characteristic catalysis, not transition metals.

“A copper solution looks blue because it absorbs blue light.”

Actually: Exactly backwards. The complex absorbs the red-orange end of the visible spectrum (an electron jumps the d-orbital gap) and transmits or reflects what is left — the blue you see is the light that was NOT absorbed.

“Multidentate complexes are more stable because their coordinate bonds are stronger.”

Actually: The bonds are similar — six broken, six made, ΔH ≈ 0. The stability is entropic: [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ turns two particles into seven, ΔS is strongly positive, so ΔG is negative. Examiners specifically reward the particle count.

“You add an indicator to a manganate(VII) titration.”

Actually: Never — it is self-indicating. The purple MnO₄⁻ decolourises as it reacts, so the end-point is the first permanent pale pink from one drop of excess. Adding an indicator, or acidifying with HCl (which MnO₄⁻ oxidises), both wreck the titre.

ExamWhat examiners want

Colours are pure AO1 and the paper asks for them verbatim: learn AQA's specific pairs — [Cu(H₂O)₆]²⁺ blue to [CuCl₄]²⁻ yellow-green, [Co(H₂O)₆]²⁺ pink to [CoCl₄]²⁻ blue, vanadium's +5/+4/+3/+2 as yellow/blue/green/violet, Fe²⁺ pale green vs Fe³⁺ yellow. Vague answers ('darker blue') score nothing where the scheme says 'deep/royal blue'.

In ligand-substitution equations the marks are in the bookkeeping: charges must balance, the correct number of water molecules must leave, and coordination-number changes must be stated with the reason (chloride ligands are larger than water, so only four fit). For the chelate effect, the creditable argument is quantitative in shape: count the particles on each side, state ΔS > 0 and ΔH ≈ 0, and conclude via ΔG = ΔH − TΔS. 'Stronger bonds' scores zero.

Redox-titration calculations are AO2's favourite: set out moles of the known, the 5 : 1 ratio, the scale-up to the volumetric flask, then the conversion to mass or percentage — and quote three significant figures with units. In catalysis six-markers, name the type (heterogeneous/homogeneous), give BOTH steps of the mechanism including catalyst regeneration, and for surface catalysts mention active sites, support medium and poisoning. The Contact-process equations with V₂O₅ regenerated is the single most-recycled catalysis question in the 7405 archive.

Transition metals — colour, catalysis and the chemistry of complexes.