Cambridge International AS & A Level Chemistry 9701 · A Level topic 28

Chemistry of transition elements

What this chapter covers28.1–28.5

Rust, the blue of copper sulfate, the purple of potassium manganate(VII) in a titration, the iron catalyst in the Haber process, the platinum drug cisplatin and the iron at the centre of haemoglobin all belong to the chemistry of the transition elements. These metals share a set of properties that the s-block metals do not have: they exist in several oxidation states, they catalyse reactions, they form complex ions with molecules and ions called ligands, and most of their compounds are coloured.

Every one of these properties can be traced to a single feature of electronic structure: a partly filled 3d sub-shell whose energy is close to that of the 4s sub-shell. This chapter begins with that structure, then studies complexes — how ligands bond, what shapes result, how one ligand replaces another and how stable the products are — before turning to redox chemistry and titrations, the origin of colour, and the isomerism that makes cisplatin an anticancer drug while its isomer is not.

Transition elements and their electronic configurations28.1.1

The d block occupies the ten columns between Group 2 and Group 13. In the fourth period it runs from scandium to zinc, and across it electrons are added to the 3d sub-shell. Because the 4s sub-shell fills before 3d, the atoms have configurations [Ar]3dⁿ4s², with two exceptions: chromium and copper take one 4s electron into the 3d sub-shell, giving the more stable half-filled 3d⁵ and full 3d¹⁰ arrangements.

When these atoms form positive ions, the 4s electrons are removed before the 3d electrons. Once the 3d sub-shell is occupied it lies slightly lower in energy than 4s, so 4s electrons are the outermost and the first to go. Fe is [Ar]3d⁶4s², so Fe²⁺ is [Ar]3d⁶ — not [Ar]3d⁴4s² — and Fe³⁺ is [Ar]3d⁵.

The definition excludes two d-block elements. Scandium forms only Sc³⁺, which has no d electrons (3d⁰); zinc forms only Zn²⁺, which has a complete 3d¹⁰ sub-shell. Neither has an ion with a partly filled d sub-shell, and — as later sections show — neither shows the typical colour and variable oxidation states. Copper qualifies because Cu²⁺ is 3d⁹, even though Cu⁺ is 3d¹⁰. The syllabus considers the first row from titanium to copper.

The shapes of the 3d orbitals28.1.2

There are five 3d orbitals. In an isolated atom or ion they have the same energy — they are degenerate — but they have different shapes and orientations, and that difference becomes important when ligands approach (section 11). The syllabus requires sketches of two of them (Figure 28.1).

• 3d_xy has four lobes lying in the xy plane, pointing between the x and y axes. 3d_xz and 3d_yz have the same shape in the xz and yz planes, and 3d_x²−y² has four lobes pointing along the x and y axes.
• 3d_z² has two lobes along the z axis, meeting at the nucleus, with a doughnut-shaped ring (torus) of electron density around the middle in the xy plane.

Characteristic properties and variable oxidation states28.1.3, 28.1.4

Transition elements are typical metals — dense, hard, high-melting and good conductors — but their chemical properties are what set them apart. The syllabus lists four:

1. they have variable oxidation states;
2. they behave as catalysts;
3. they form complex ions;
4. they form coloured compounds.

High density and high melting point are physical properties; a question that asks for "chemical properties" does not credit them.

Why the oxidation states vary

For a Group 2 metal such as calcium, the two 4s electrons are removed easily, but the third electron would come from the 3p sub-shell, far lower in energy and closer to the nucleus: the third ionisation energy is enormous, and calcium is only ever +2. For a transition element the 3d and 4s sub-shells are close in energy. After the 4s electrons have gone, the 3d electrons can be removed one after another without a sudden large jump in ionisation energy, and the extra ionisation energy can be repaid by stronger bonding or hydration in the compound. So a range of oxidation states is accessible.

Two patterns are worth noticing. The maximum oxidation state rises from Ti (+4) to Mn (+7) — it equals the total number of 4s and 3d electrons — and then falls, because beyond manganese the increasing nuclear charge holds the 3d electrons more tightly. And the highest oxidation states are found only in compounds with very electronegative oxygen or fluorine, usually as oxoanions such as MnO₄⁻ and Cr₂O₇²⁻, where the bonding is covalent: a free Mn⁷⁺ ion does not exist.

Why transition elements form complexes and act as catalysts28.1.5, 28.1.6

Complex formation

A complex forms when molecules or ions with lone pairs — ligands — form dative covalent bonds to a central metal ion (section 6). To accept those electron pairs, the metal ion needs empty orbitals of suitable energy. Transition element ions have vacant d orbitals that are energetically accessible — the 3d orbitals not fully occupied, together with the empty 4s and 4p orbitals close in energy to them — which can accept lone pairs from ligands and form dative bonds. The small size and high charge of the ions also attract the ligand lone pairs strongly.

Catalysis

Two features explain catalytic activity:

• More than one stable oxidation state. The catalyst can accept electrons from one reactant and pass them to another, changing oxidation state and then changing back — exactly how Fe²⁺/Fe³⁺ catalyse the I⁻/S₂O₈²⁻ reaction and MnO₂/Mn²⁺ the decomposition of hydrogen peroxide (topic 26).
• Vacant d orbitals that are energetically accessible and can form dative bonds with ligands. Reactant molecules can bond temporarily to a transition-metal surface or ion (adsorption onto iron in the Haber process), which weakens bonds within them and lowers the activation energy.

Past-paper practice · Set 28A · Transition elements: definition, orbitals and properties

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 28A.1[3]

Answer and marking guidance

(a) a d-block element that forms one or more stable ions with incomplete (partially filled) d sub-shell ✔. (b)(i) any two of: behave as catalysts; form complex ions; form coloured compounds ✔. (ii) the 3d and 4s sub-shells are close (similar) in energy ✔. Examiner insight: the definition was well known; the common error in (b)(i) was giving a physical property such as high density or high melting point.

Question 28A.2[2]

Parts (a) and (b); part (c) of the same question is 28D.1.

Answer and marking guidance

(a) the 3d and 4s sub-shells (orbitals) are close / similar in energy ✔. (b) four lobes lying between the x and y axes ✔. Examiner insight: (a) proved difficult for many, some confusing it with the explanation of why transition elements form complex ions; (b) was usually correct — the common error was drawing the lobes along two axes.

Question 28A.3[6]

Part (c) revises heterogeneous catalysis from topic 26.

Answer and marking guidance

(a) forms one or more stable ions with an incomplete (partially filled) 3d sub-shell ✔. (b) two lobes along the z axis with a ring (torus) around the middle ✔. (c)(i) the catalyst is in a different phase (state) from the reactants ✔. (ii) reactants adsorb onto the surface of the catalyst ✔; bonds in the reactants weaken ✔; reaction occurs and the products desorb ✔. Examiner insight: many recognised the importance of stable ions with partially filled d sub-shells; the 3d_z² hour-glass must narrow to zero at the origin with the ring round the narrowest part; weak answers in (c)(ii) used "catalyst", "reactants" and "products" in the wrong places or wrote only "adsorption occurs", "bonds weaken", "desorb".

Question 28A.4[2]

Answer and marking guidance

(a) +4 and any of +1, +2, +3 ✔ (removing up to the two 4s and two 3d electrons). (b) the 4s and 3d sub-shells are similar in energy, so 3d electrons can also be lost; in calcium the next electron after 4s would come from the much lower 3p sub-shell ✔. (No examiner report was published for this session.)

Ligands and complexes28.2.2, 28.2.3, 28.2.4

When copper(II) sulfate dissolves, each Cu²⁺ ion does not float free: six water molecules bond to it through lone pairs on their oxygen atoms, forming the ion [Cu(H₂O)₆]²⁺. This is a complex, and the water molecules are ligands.

Ligands are classified by the number of dative bonds each one can form to the same metal ion — its denticity (Figure 28.3):

• monodentate ligands form one dative bond: H₂O and NH₃ (neutral), Cl⁻ and CN⁻ (negative; cyanide bonds through carbon);
• bidentate ligands have two donor atoms, far enough apart to both bond to the same ion: 1,2-diaminoethane, H₂NCH₂CH₂NH₂ (abbreviated en), bonding through both N atoms, and the ethanedioate ion, C₂O₄²⁻, bonding through two O atoms;
• polydentate ligands have more than two donor atoms: EDTA⁴⁻ forms six dative bonds — through two N atoms and four O atoms — wrapping round a single metal ion.

Shapes, coordination number, formula and charge28.2.5, 28.2.6

The shape of a complex is decided mainly by the coordination number, which depends on the size of the metal ion and of the ligands. Four geometries are required (Figure 28.2):

Large ligands such as Cl⁻ crowd around a small metal ion, so four chloride ions (tetrahedral) often replace six water molecules (octahedral). Square planar geometry is typical of platinum(II) and some nickel(II) complexes.

Formula and charge

The overall charge of a complex is the charge of the metal ion plus the charges of all the ligands. The formula is written in square brackets with the charge outside:

Reactions of copper(II) and cobalt(II) complexes28.2.1, 28.2.7

The syllabus names two metal ions and four ligands. In aqueous solution both metals start as hexaaqua complexes: [Cu(H₂O)₆]²⁺ (pale blue) and [Co(H₂O)₆]²⁺ (pink). Two different kinds of reaction then occur, and it matters which is which.

Precipitation: hydroxide ions, or a little ammonia

Hydroxide ions remove H⁺ from two of the water ligands, leaving a neutral complex that is insoluble. This is an acid–base (deprotonation) reaction and a precipitation:

The products may also be written as Cu(OH)₂ and Co(OH)₂, with 6H₂O released. Because the precipitate is neutral, it carries no charge. Aqueous ammonia added a drop at a time does the same thing, because ammonia is a weak base: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

Ligand exchange: excess ammonia

With an excess of ammonia, the precipitate dissolves as NH₃ molecules replace ligands on the metal ion. This is ligand exchange (substitution): one ligand replaces another without any change in oxidation state.

Only four water molecules are replaced for copper: the product is [Cu(NH₃)₄(H₂O)₂]²⁺, still octahedral. Writing [Cu(NH₃)₆]²⁺ is the formula error examiners report most often. For cobalt all six waters are replaced; [Co(NH₃)₆]²⁺ is usually described as a pale brown (straw) solution that darkens on standing in air as cobalt(II) is oxidised to cobalt(III). Colour descriptions of this ion vary between sources, and one published mark scheme records it as a blue solution; check the wording expected in the paper you are working from.

Ligand exchange: concentrated hydrochloric acid

Chloride ions are larger than water molecules, so only four fit around the metal ion, and the octahedral complex becomes tetrahedral. The charge also changes, from 2+ to 2−:

Ligand exchange as equilibrium

Ligand exchange reactions are reversible, and Le Chatelier's principle predicts how they respond to changes in concentration. Adding water to yellow [CuCl₄]²⁻ shifts the equilibrium to the left and the blue colour returns. Adding silver nitrate to blue [CoCl₄]²⁻ precipitates white AgCl, lowers [Cl⁻], and the solution turns pink as [Co(H₂O)₆]²⁺ re-forms. With ammonia the two equilibria compete: a little NH₃ raises [OH⁻] and pushes the precipitation equilibrium to the right; excess NH₃ pushes the ligand-exchange equilibrium to the right, lowering the concentration of the aqua ion so that the precipitation equilibrium shifts left and the precipitate dissolves.

Past-paper practice · Set 28B · Ligands, complexes and ligand exchange

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 28B.1[9]

Answer and marking guidance

(a)(i) a molecule or ion formed by a central metal atom/ion surrounded by (bonded to) one or more ligands ✔. (ii) NaOH: blue precipitate ✔; [Co(H₂O)₆]²⁺ + 2OH⁻ → Co(OH)₂ + 6H₂O ✔; precipitation / acid–base ✔. Excess NH₃: the mark scheme records "blue solution" ✔ (the colour of [Co(NH₃)₆]²⁺ is described differently in some sources — see section 8); [Co(H₂O)₆]²⁺ + 6NH₃ → [Co(NH₃)₆]²⁺ + 6H₂O ✔; ligand exchange / substitution ✔. (b) any two for ✔, three for ✔✔: the solution turns from blue to pink; a white precipitate of AgCl forms; [Cl⁻] decreases so the equilibrium shifts to the left. (No examiner report was published for this session.)

Question 28B.2[4]

Parts (a) and (d) of the question; parts (b) and (c) are 28E.4 and 28F.2.

Answer and marking guidance

(a)(i) [Cu(H₂O)₆]²⁺ + 2OH⁻ → Cu(OH)₂(H₂O)₄ + 2H₂O ✔. (ii) [Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O ✔ (or with 4HCl, giving 4H⁺). (d) Ru³⁺ with two bidentate phen (four bonds) and two Cl⁻: [Ru(C₁₂H₈N₂)₂Cl₂]⁺ ✔; Fe³⁺ with three C₂O₄²⁻: [Fe(C₂O₄)₃]³⁻ ✔. Examiner insight: common errors were [Cu(OH)₂(H₂O)₄]²⁺ and Na instead of Na⁺; equations unbalanced for charge, or with H₂ instead of 4H⁺; in (d) [Ru(phen)₂(C₂O₄)₂]⁻, [Ru(phen)₂Cl₄]³⁺ and [Fe(C₂O₄)₆]⁹⁻ were seen.

Question 28B.3[4]

Answer and marking guidance

A [Cu(H₂O)₆]²⁺, (pale) blue; B Cu(H₂O)₄(OH)₂ or Cu(OH)₂, (pale) blue; C [Cu(NH₃)₄(H₂O)₂]²⁺, dark blue; D [CuCl₄]²⁻, yellow. Two correct for ✔, four ✔✔, six ✔✔✔, eight ✔✔✔✔. Examiner insight: most scored some or all marks; [Cu(NH₃)₆]²⁺ was a common error for C.

Question 28B.4[2]

Answer and marking guidance

(i) adding NH₃ increases [OH⁻] (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻), shifting equilibrium 1 to the right, so Ni(OH)₂ precipitates ✔. (ii) a large excess of NH₃ shifts equilibrium 2 to the right, lowering [[Ni(H₂O)₆]²⁺], so equilibrium 1 shifts to the left and the precipitate dissolves ✔. (No examiner report was published for this session.)

Question 28B.5[4]

Part (a) only.

Answer and marking guidance

(i) NaOH(aq): blue (or pink) solid ✔; excess conc. HCl: blue, aqueous ✔ (two cells per mark). (ii) [Co(H₂O)₆]²⁺ + 2OH⁻ → Co(OH)₂ + 6H₂O ✔. (iii) [Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O ✔. Examiner insight: the states are solid and aqueous; equations were sometimes unbalanced (displaced water omitted) or carried wrong charges, such as CoCl₄⁻, or showed Na instead of Na⁺.

Predicting redox reactions with E^⦵28.2.8

Variable oxidation states make transition elements rich in redox chemistry. Whether one species will oxidise another under standard conditions is predicted exactly as in topic 24: write the two relevant half-equations as reductions, and the one with the more positive E^⦵ proceeds as a reduction while the other runs in reverse as an oxidation. The reaction is feasible if

Vanadium and zinc

Ammonium vanadate(V) in acid gives yellow VO₂⁺. Zinc and acid reduce it step by step: yellow → blue (VO²⁺) → green (V³⁺) → violet (V²⁺). A green colour is also seen early on as yellow and blue mix. The E^⦵ values (Figure 28.8) explain why zinc goes all the way to V²⁺ but no further.

Ligands change E^⦵

The E^⦵ of a metal-ion couple depends on the ligands. For iron, Fe³⁺/Fe²⁺ as aqua ions is +0.77 V, but [Fe(CN)₆]³⁻/[Fe(CN)₆]⁴⁻ is +0.36 V: with cyanide ligands the iron(III) complex is harder to reduce (the equilibrium lies further to the left), because cyanide stabilises the +3 state relative to +2. The E^⦵ values of a series of complexes of the same metal therefore compare the relative stabilities of the two oxidation states.

The limits of the prediction

E^⦵ values apply to standard conditions (1 mol dm⁻³, 298 K) and say nothing about rate. A reaction predicted to be feasible may be too slow to observe; one predicted to be not feasible may occur if concentrations are far from standard. The reaction of Cu²⁺ with I⁻ is the classic case: from Cu²⁺/Cu⁺ (+0.15 V) and I₂/I⁻ (+0.54 V), E^⦵_cell = −0.39 V, yet copper(II) oxidises iodide readily, because the copper(I) formed is removed as a precipitate of CuI. With [Cu⁺] tiny, the Cu²⁺/Cu⁺ electrode potential becomes much more positive (Nernst equation, topic 24) and the reaction becomes feasible.

Redox titrations28.2.9, 28.2.10

Transition-metal oxidising agents make excellent titrants because their colour changes signal the end-point. Every calculation follows the same route: balanced equation → moles of titrant → mole ratio → moles of analyte → scale up for any dilution → answer.

Manganate(VII) with iron(II)

Acidified potassium manganate(VII) is added from the burette. While Fe²⁺ remains, the purple MnO₄⁻ is decolorised as it is reduced to almost colourless Mn²⁺; the end-point is the first permanent pale pink colour. No indicator is needed — the titration is self-indicating. The acid must be dilute sulfuric acid: hydrochloric acid would be oxidised to chlorine by manganate(VII), using up titrant, and nitric acid is itself an oxidising agent.

Manganate(VII) with ethanedioate

The half-equation for ethanedioate is C₂O₄²⁻ → 2CO₂ + 2e⁻, so ten electrons link two MnO₄⁻ with five C₂O₄²⁻. The reaction is slow at room temperature, so the flask is warmed (to about 60 °C) before titrating; once some Mn²⁺ has formed the reaction speeds up, because Mn²⁺ catalyses it. The end-point is again the first permanent pale pink.

Copper(II) with iodide, then thiosulfate

Excess potassium iodide is added to the copper(II) solution: an off-white precipitate of copper(I) iodide forms in a brown solution of iodine. The iodine is titrated with sodium thiosulfate. As the brown colour fades to pale yellow, starch is added, giving a blue-black colour; the end-point is when the blue-black colour disappears. Starch is added near the end because at high iodine concentration the starch–iodine complex forms too strongly and the end-point is less sharp. Combining the equations, 2Cu²⁺ ≡ I₂ ≡ 2S₂O₃²⁻, so moles of Cu²⁺ = moles of thiosulfate.

Other redox systems

Any balanced redox equation can be used in the same way (28.2.10). Two common extensions:

• Dichromate(VI): Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O (ratio 1 : 6); orange to green, so a redox indicator is needed. For example, if 25.0 cm³ of an Fe²⁺ solution needs 23.60 cm³ of 0.00250 mol dm⁻³ K₂Cr₂O₇, moles of Fe²⁺ = 6 × 5.90 × 10⁻⁵ = 3.54 × 10⁻⁴ mol.
• Back titration: when the analyte cannot be titrated directly, a known excess of oxidant is added and the excess is titrated. The moles that reacted with the analyte = initial moles − excess moles. Each stage has its own ratio, and both must be applied.

Past-paper practice · Set 28C · Redox chemistry and titrations

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 28C.1[2]

Answer and marking guidance

moles VO₂⁺ = 0.02500 × 0.0300 = 7.50 × 10⁻⁴ ✔; mean titre 28.125 cm³, so moles C₂O₄²⁻ = 0.028125 × 0.0400 = 1.13 × 10⁻³; ratio 7.50 × 10⁻⁴ : 1.13 × 10⁻³ = 1 : 1.5 = 2 : 3, as in equation 2 ✔. Examiner insight: most answered well with a clear justification.

Question 28C.2[8]

Part (c)(iii) uses the Nernst equation from topic 24.

Answer and marking guidance

(b) n(S₂O₃²⁻) = 0.02835 × 0.500 = 0.0142 mol ✔ = n(Cu²⁺) ✔; mass Cu = 0.014175 × 63.5 = 0.90 g; % Cu = 0.90 ÷ 1.50 × 100 = 60% ✔. (c)(i) E^⦵_cell = 0.15 − 0.54 = −0.39 V ✔ (Cu²⁺/Cu⁺, not Cu²⁺/Cu). (ii) E^⦵_cell is negative, so the reaction is not feasible under standard conditions ✔. (iii) E = E^⦵ + 0.059 log([Cu²⁺]/[Cu⁺]) = 0.15 + 0.059 log(1.0 ÷ 1.3 × 10⁻⁶) ✔ = +0.50 V ✔ — the removal of Cu⁺ as CuI raises the potential. (iv) E^⦵_cell is very negative (0.15 − 1.36 = −1.21 V), so even this shift cannot make the reaction with chloride feasible ✔. Examiner insight: many fully correct answers to (b); errors were 30% (2 : 1 ratio) and 59.3% (rounding); in (c)(i) many used Cu²⁺/Cu (+0.34 V); the Nernst equation was not well known; (iv) needed a reference to E^⦵ data.

Question 28C.3[4]

Part (c)(ii) only: a back titration.

Answer and marking guidance

moles MnO₄⁻ initially = 0.0200 × 0.0500 = 1.00 × 10⁻³; moles Fe²⁺ = 0.0500 × 0.03040 = 1.52 × 10⁻³ ✔; MnO₄⁻ unreacted = 1.52 × 10⁻³ ÷ 5 = 3.04 × 10⁻⁴, so MnO₄⁻ reacted with ethanedioate = 6.96 × 10⁻⁴ ✔; moles C₂O₄²⁻ = 6.96 × 10⁻⁴ × 5/2 = 1.74 × 10⁻³ ✔; mass BaC₂O₄ = 1.74 × 10⁻³ × 225.3 = 0.392 g; % = 0.392 ÷ 0.500 × 100 = 78.4% ✔. Examiner insight: weaker answers did not calculate the initial moles of manganate(VII); common wrong answers were 12.5% (2/5 used instead of 5/2), 31.4% (step omitted) and 34.2%.

Question 28C.4[4]

Part (c) only.

Answer and marking guidance

(i) 2MnO₄⁻ + 6H⁺ + 5SO₃²⁻ → 2Mn²⁺ + 3H₂O + 5SO₄²⁻ ✔. (ii) moles MnO₄⁻ = 0.02240 × 0.0250 = 5.60 × 10⁻⁴; moles SO₃²⁻ = × 5/2 = 1.40 × 10⁻³ in 25.0 cm³; × 10 = 1.40 × 10⁻² mol; mass K₂SO₃ = 1.40 × 10⁻² × 158.3 = 2.216 g ✔✔ (two bullets per mark); purity = 2.216 ÷ 3.40 × 100 = 65.2% ✔. Examiner insight: a common error in (i) was cancellable H₂O or H⁺ on both sides; in (ii) errors were 6.52 (no × 10), 10.4 (2/5 ratio), 26.1 (5/2 omitted) and 49.1 (M_r 119.1).

Question 28C.5[1]

Answer and marking guidance

[Fe(CN)₆]³⁻, because its E^⦵ (+0.36 V) is the least positive, so its equilibrium lies furthest to the left ✔. Examiner insight: generally well answered; some gave [Fe(CN)₆]⁴⁻, which is not an iron(III) complex.

Degenerate and non-degenerate d orbitals28.3.1, 28.3.2

In an isolated gaseous transition-metal ion the five 3d orbitals have exactly the same energy: they are degenerate. When ligands approach to form a complex, their lone pairs repel the electrons in the d orbitals and raise the energy of all five. But the orbitals point in different directions (Figure 28.1), so they are not affected equally. Orbitals whose lobes point directly towards the ligands are raised more than those whose lobes point between them. The d orbitals are split into two sets of different energy — they become non-degenerate — separated by an energy gap, ΔE.

• In an octahedral complex the six ligands lie along the x, y and z axes. The d_z² and d_x²−y² orbitals, whose lobes lie along the axes, are raised most: two higher and three lower orbitals (d_xy, d_xz, d_yz).
• In a tetrahedral complex the four ligands lie between the axes. The pattern is inverted: three higher and two lower. With only four ligands, none pointing straight at an orbital, ΔE is smaller than in an octahedral complex of the same ion and ligand.

Why transition-element compounds are coloured28.3.3

White light contains all visible frequencies. When it passes through a solution of a complex, an electron in a lower-energy d orbital can absorb a photon whose energy exactly matches ΔE and be promoted to a higher-energy d orbital:

For most complexes ΔE corresponds to a frequency in the visible region. That band of frequencies is removed from the light; the rest is transmitted (or reflected), and the eye sees the complementary colour of the light absorbed (Figure 28.6). Hydrated copper(II) ions absorb in the orange–red region, and appear blue.

The explanation requires a partly filled d sub-shell: there must be an electron in a lower orbital and a space for it in a higher one. That is why:

• Sc³⁺ (3d⁰) and Ti⁴⁺ compounds are colourless — no d electron to promote;
• Zn²⁺ and Cu⁺ (3d¹⁰) compounds are white or colourless — every d orbital is full, so no electron can move to a higher d orbital (copper(I) iodide is white);
• compounds of transition-element ions with partly filled d sub-shells are coloured.

How the ligand changes ΔE and the colour28.3.4, 28.3.5

ΔE depends on the metal, its oxidation state, the geometry and — the syllabus focus — the ligand. Ligands that interact more strongly with the metal's d orbitals split them more. For the ligands in the syllabus the order of increasing ΔE is, approximately,

A larger ΔE means a photon of higher frequency (shorter wavelength) is absorbed, so the absorption moves from the red end towards the violet end of the spectrum, and the colour seen shifts accordingly. When ligand exchange changes the ligands around a metal ion, ΔE changes, a different frequency of light is absorbed, and the colour changes.

Two different metal ions with the same ligands also have different colours, because their ΔE values differ: Cr³⁺(aq) and Fe³⁺(aq) are both hexaaqua ions with a 3+ charge, but different numbers of d electrons and different nuclear charges give different ΔE, so different frequencies are absorbed.

Past-paper practice · Set 28D · Colour of complexes

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 28D.1[3]

Answer and marking guidance

in a complex (in the presence of ligands) the d orbitals are split into two sets of different energy ✔; an electron is promoted (excited) to a higher d orbital ✔; visible light is absorbed AND the colour seen is the complementary colour ✔. Examiner insight: performed well — splitting, absorption of photons of visible light, excitation of electrons and the complementary colour were all described clearly.

Question 28D.2[2]

Answer and marking guidance

any two for ✔, three for ✔✔: five 3d orbitals of the isolated Fe²⁺ ion drawn at the same energy; in the octahedral complex two higher and three lower; both sets of non-degenerate orbitals at higher energy than the degenerate orbitals of the isolated ion. Examiner insight: the splitting into three lower and two upper orbitals was well appreciated; their energy compared with the isolated ion was not.

Question 28D.3[5]

Answer and marking guidance

(a) [Ar] 3d: ↑↓ ↑↓ ↑↓ ↑ ↑; 4s empty ✔. (b) the d orbitals split into two levels ✔; an electron is promoted to a higher d orbital ✔; a frequency of (visible) light is absorbed ✔; the colour observed is the complement of the light absorbed ✔. (No examiner report was published for this session.)

Question 28D.4[2]

Answer and marking guidance

ΔE is different in the two complexes ✔; so a different frequency (wavelength) of light is absorbed ✔. Examiner insight: candidates who began by thinking about d-orbital splitting did well; stronger answers stated that "the two complexes have different ΔE", since ΔE is the conventional symbol for the energy gap.

Question 28D.5[5]

Part (a) of the same question as 28C.2.

Answer and marking guidance

(i) (1s²2s²2p⁶)3s²3p⁶3d¹⁰4s¹ ✔. (ii) d orbitals are split into two levels by the approaching ligands ✔; light is absorbed and the complementary colour observed ✔; a d electron is promoted to a higher d orbital ✔; in Cu(I) all the d orbitals are full (3d¹⁰), so no promotion is possible ✔. Examiner insight: 3d⁹4s² was a common error; answers to (ii) often lacked clarity — some said light is emitted, did not mention splitting of the d orbitals, or omitted that Cu(I) is 3d¹⁰.

Geometrical (cis/trans) isomerism28.4.1(a)

Stereoisomers have the same molecular formula and the same bonds, but a different arrangement of atoms in space. In complexes the arrangement is fixed by the geometry around the metal: ligands cannot rotate from one position to another without breaking bonds. When a complex contains two or more different ligands, they can often be placed in more than one way.

In geometrical (cis/trans) isomerism two identical ligands are either next to each other — cis, at 90° — or opposite each other — trans, at 180°. The syllabus names two examples (Figure 28.7):

• square planar Pt(NH₃)₂Cl₂: cis-platin, with the two Cl on the same side, and trans-platin. Only the cis isomer is an effective anticancer drug: it binds to DNA, the two chloride positions being replaced by nitrogen atoms of adjacent bases, which prevents the DNA from replicating. The trans isomer cannot bind in the same way.
• octahedral [Co(NH₃)₄(H₂O)₂]²⁺: the two water ligands are either adjacent (cis) or opposite (trans).

With bidentate ligands, [Ni(en)₂(H₂O)₂]²⁺ also shows cis/trans isomerism: in the trans isomer the two en ligands lie in one plane with the waters above and below; in the cis isomer the waters are adjacent.

Optical isomerism28.4.1(b)

Optical isomers are non-superimposable mirror images of each other. In organic chemistry optical isomerism arises from a chiral carbon atom; in complexes it arises from the arrangement of bidentate ligands round an octahedral metal ion, which gives the complex a "propeller" shape that can twist either way.

• [Ni(en)₃]²⁺: three en ligands, each spanning two cis positions. The complex and its mirror image cannot be superimposed however they are rotated (Figure 28.7), so there are two optical isomers.
• [Ni(en)₂(H₂O)₂]²⁺: the cis isomer is chiral and exists as a pair of optical isomers; the trans isomer has a plane of symmetry (the plane of the two en ligands) and is not optically active. This complex therefore has three stereoisomers in all.

Like organic enantiomers, optical isomers of a complex rotate the plane of plane-polarised light by equal amounts in opposite directions.

The polarity of complexes28.4.2

A complex is polar if the individual metal–ligand bond dipoles do not cancel. The reasoning is the same as for molecules (topic 3): in a symmetrical arrangement, dipoles in opposite directions cancel.

• trans-Pt(NH₃)₂Cl₂: each Pt–Cl dipole is opposite another Pt–Cl dipole, and each Pt–N opposite another Pt–N. The dipoles cancel: non-polar.
• cis-Pt(NH₃)₂Cl₂: the two Pt–Cl bonds are on the same side; their resultant is not cancelled: polar.
• The same applies to octahedral trans complexes such as trans-[Co(NH₃)₄(H₂O)₂]²⁺ (non-polar) and their cis isomers (polar).
• Complexes with a single type of ligand in a regular geometry — linear [Ag(NH₃)₂]⁺, octahedral [Fe(C₂O₄)₃]³⁻ — are non-polar.

"Polar" here refers to the distribution of charge within the complex, not to its overall charge: a complex ion with a 2+ charge can still be non-polar.

Past-paper practice · Set 28E · Stereoisomerism and polarity of complexes

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 28E.1[2]

Answer and marking guidance

two 3D octahedral diagrams: one with the three Cl on one face (fac), one with the three Cl in a plane including the Co (mer) ✔; type: geometric (cis–trans) ✔. (No examiner report was published for this session.)

Question 28E.2[2]

Answer and marking guidance

two correct square-planar structures, one with the two Br adjacent (cis) and one with them opposite (trans) ✔; type: cis–trans / geometrical ✔. (No examiner report was published for this session.)

Question 28E.3[5]

Answer and marking guidance

(i) a species that donates two lone pairs ✔ to form dative bonds to a metal atom or ion ✔. (ii) one correct 3D diagram with three bipy ligands, each spanning two cis positions ✔; both diagrams correct as non-superimposable mirror images ✔; optical isomerism ✔. Examiner insight: (i) was generally well answered; many accurate drawings using the correct 3D conventions were seen in (ii).

Question 28E.4[2]

Answer and marking guidance

(i) cis and trans isomers drawn, with the two H₂O ligands adjacent in one and opposite in the other ✔. (ii) cis is polar AND trans is non-polar ✔. Examiner insight: performed well; errors included O₂H for OH₂ and drawing the same isomer twice. Most identified cis as polar and trans as non-polar.

Question 28E.5[3]

Answer and marking guidance

(i) vacant d orbitals that are energetically accessible, which can accept lone pairs to form dative bonds ✔. (ii) cis-[Pt(en)Cl₂]: square planar, 4, polar; [Ag(NH₃)₂]⁺: linear, 2, non-polar; [Fe(C₂O₄)₃]³⁻: octahedral, 6, non-polar — three cells ✔, six ✔✔. Examiner insight: (i) was less well known — many gave the definition of a transition element; in (ii) coordination numbers were generally right but shape and polarity were less well known.

Question 28E.6[2]

This part follows 28C.1 (the same question).

Answer and marking guidance

a 3D octahedral structure with three bidentate C₂O₄²⁻ ligands, each across two cis positions ✔; charge 4− (V²⁺ + 3 × (2−)) ✔. Examiner insight: many drew a suitable structure; the charge was often omitted.

The stability constant, K_stab28.5.1, 28.5.2

Ligand exchange reactions are equilibria. In aqueous solution the "free" metal ion is really the aqua complex, so the formation of a complex with another ligand is written as replacement of water:

The equilibrium constant for this formation reaction measures how far it goes, and so how stable the new complex is compared with the aqua complex.

K_stab is written like any K_c (topic 7), with one important convention: [H₂O] is not included. Water is the solvent, present in huge excess, and its concentration is effectively constant.

The aqua ion may also be written simply as [Cu²⁺]. The units follow from the powers: here there are (mol dm⁻³)¹ on top and (mol dm⁻³)⁵ underneath, giving mol⁻⁴ dm¹². In general, for a complex formed with n monodentate ligands the units are mol⁻ⁿ dm³ⁿ.

Calculations with K_stab28.5.3

K_stab values for transition-metal complexes are often very large (10¹⁰ to 10⁴⁰), which means that when enough ligand is present, almost none of the aqua ion remains. Calculations substitute equilibrium concentrations into the expression, exactly as for K_c.

Ligand exchange explained by K_stab28.5.4

A large K_stab means that the position of equilibrium lies far to the right: the complex formed is stable relative to the aqua complex. Comparing K_stab values of two complexes of the same metal ion predicts which ligand will replace which:

• when a ligand is added that forms a complex with a larger K_stab than the existing complex, ligand exchange occurs and the more stable complex forms;
• when several ligands compete for the same metal ion, the complex with the largest K_stab is present in the highest concentration at equilibrium, and the one with the smallest K_stab in the lowest.

Multidentate ligands and the chelate effect

Complexes of bidentate and polydentate ligands are generally much more stable than those of similar monodentate ligands. The reason is largely entropic. When one EDTA⁴⁻ ion replaces six water molecules,

the number of particles in solution increases, so the entropy change is positive and ΔG is more negative. The enthalpy change is small, because six Cu–O or Cu–N dative bonds are broken and six are formed. The larger ΔS makes K_stab very large. This is why EDTA is used to remove toxic metal ions such as Pb²⁺ from the body (as its calcium complex, so that calcium is not stripped from the blood), and why chelating agents are added to foods and detergents to lock up metal ions.

Past-paper practice · Set 28F · Stability constants

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 28F.1[4]

Answer and marking guidance

(i) the equilibrium constant for the formation of the complex ion in a solvent (from its constituent ions or molecules) ✔. (ii) K_stab = [[Ni(en)₃]²⁺] / ([[Ni(H₂O)₆]²⁺][en]³) ✔; units mol⁻³ dm⁹ ✔. (iii) both equilibria lie to the right (equilibrium 1 further) AND [Ni(en)₃]²⁺ is more stable ✔. Examiner insight: most did not identify K_stab as an equilibrium constant; errors in (ii) included (H₂O)₆ and the charge written outside the brackets; in (iii) many did not answer both parts.

Question 28F.2[5]

Answer and marking guidance

(i) the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules ✔. (ii) [Cu(NH₃)₄(H₂O)₂]²⁺ is more stable, as K_stab is large ✔. (iii) K_stab = [[Cu(NH₃)₄(H₂O)₂]²⁺] / ([[Cu(H₂O)₆]²⁺][NH₃]⁴) ✔; units mol⁻⁴ dm¹² ✔. (iv) [[Cu(H₂O)₆]²⁺] = 0.0074 ÷ (1.40 × 10¹³ × 0.57⁴) = 5.0 × 10⁻¹⁵ mol dm⁻³ ✔. Examiner insight: some did not identify K_stab as an equilibrium constant; (H₂O)₆ and charges outside the brackets were common errors; (iv) was usually correct.

Question 28F.3[6]

Part (d) of the same question as 28B.3 and 28D.4.

Answer and marking guidance

(i) a species that donates more than two lone pairs ✔ to form dative bonds to a metal atom or ion ✔. (ii) six atoms circled: the two N and one O of each of four different CO₂⁻ groups ✔. (iii) the number of coordinate (dative) bonds formed by the metal ion ✔. (iv) ligand exchange ✔. (v) [FeEDTA]⁻ > [CrEDTA]⁻ > [PbEDTA]²⁻, because [FeEDTA]⁻ has the largest K_stab ✔. Examiner insight: stronger candidates knew the standard definition of polydentate; many did not attempt (ii); some described denticity rather than coordination number in (iii); "redox" was a common wrong answer in (iv); in (v) references to "the position of equilibrium" could not be credited because no reversible equation had been given.

Question 28F.4[3]

Answer and marking guidance

(i) K_stab = [[Hg(CN)₄]²⁻] / ([Hg²⁺][CN⁻]⁴) ✔. (ii) highest concentration [Hg(CN)₄]²⁻, lowest [HgCl₄]²⁻ ✔; because their K_stab values (stabilities) are the highest and lowest ✔. Examiner insight: great care was needed with species and with charges inside the square brackets; most identified the highest and lowest and explained by K_stab or stability.

Misconceptions and how the topic is assessed28.1–28.5

The misconceptions below recur in examiner reports on this topic. Each is set out as the incorrect idea, why it fails, the correct model and what it costs in an examination.

How the topic is assessed

Transition-element questions are among the longest on the structured paper and very often combine several of these families with topics 23–26: a lattice-energy part for a Group 2 nitrate before a copper(II) reaction scheme, a kinetics part on cisplatin hydrolysis, an EDTA titration inside a K_stab question.

Self-test

Twelve questions across the whole unit, each with the reasoning behind the answer.

Definitions to learn

Data used in this chapter

K_stab values quoted in Worked examples 28.9–28.10 and Figure 28.9 are those printed in examination questions. The E^⦵ values in Table 28.4 and in the models are reference values of the kind printed in the data booklet; use the values in your own data booklet in an examination. Titration data in Worked examples 28.4–28.6 are illustrative. Colours are those normally quoted for dilute aqueous solutions; descriptions of some colours (for example [Co(NH₃)₆]²⁺, and green mixtures of [Cu(H₂O)₆]²⁺ and [CuCl₄]²⁻) vary between sources.

The isomer model counts distinct arrangements of ligands on an ideal octahedron or square, identifying two arrangements as the same if one can be rotated into the other, and as optical isomers if they are mirror images but not rotations of each other. The colour model maps an absorption wavelength to its complementary colour using the six-colour wheel of Figure 28.6; it is a guide, not a spectrum.

Summary

28.1 General properties

• Transition element: d-block element forming one or more stable ions with incomplete d orbitals; Sc and Zn are excluded. 4s electrons are removed before 3d.
• Properties: variable oxidation states, catalysis, complex ions, coloured compounds.
• Variable oxidation states: 3d and 4s close in energy. Complexes: vacant, energetically accessible d orbitals accept ligand lone pairs. Catalysis: more than one stable oxidation state and vacant accessible d orbitals.

28.2 Complexes and redox

• Ligands: monodentate (H₂O, NH₃, Cl⁻, CN⁻), bidentate (en, C₂O₄²⁻), polydentate (EDTA⁴⁻). Coordination number counts dative bonds.
• Shapes: linear 180° (2), square planar 90° (4), tetrahedral 109.5° (4), octahedral 90° (6). Charge = metal oxidation state + ligand charges.
• Cu(II): pale blue → OH⁻ pale blue ppt → excess NH₃ deep blue [Cu(NH₃)₄(H₂O)₂]²⁺; conc. HCl yellow [CuCl₄]²⁻. Co(II): pink → blue ppt → excess NH₃ [Co(NH₃)₆]²⁺; conc. HCl blue [CoCl₄]²⁻.
• E^⦵_cell > 0 → feasible under standard conditions. Titrations: MnO₄⁻ : Fe²⁺ 1 : 5; MnO₄⁻ : C₂O₄²⁻ 2 : 5; Cu²⁺ : S₂O₃²⁻ 1 : 1 (starch).

28.3 Colour

• Ligands split the degenerate d orbitals: octahedral two higher/three lower; tetrahedral three higher/two lower.
• A d electron absorbs visible light of energy ΔE and is promoted; the complementary colour is seen. d⁰ and d¹⁰ ions are colourless.
• Changing the ligand changes ΔE, the frequency absorbed and the colour.

28.4–28.5 Isomerism and stability

• Cis/trans: square planar Pt(NH₃)₂Cl₂, octahedral [Co(NH₃)₄(H₂O)₂]²⁺; optical: [Ni(en)₃]²⁺, cis-[Ni(en)₂(H₂O)₂]²⁺. Trans isomers are non-polar; cis are polar.
• K_stab = [[complex]]/([[M(H₂O)_n]][L]^x), no [H₂O]; larger K_stab = more stable; ligand exchange favours the complex with the larger K_stab; chelate complexes are especially stable (entropy).

Examination checklist

• Can I write configurations of Ti–Cu atoms and ions, removing 4s electrons first, including Cr and Cu?
• Can I define transition element, ligand, complex and coordination number precisely, and explain why Sc and Zn are excluded?
• Can I sketch 3d_xy and 3d_z² on labelled axes?
• Do I match each property to its explanation (3d/4s energies; vacant accessible d orbitals; both for catalysis)?
• Can I deduce the formula, charge, shape and coordination number of a complex from a description?
• Can I write balanced equations, with colours, states and reaction types, for Cu(II) and Co(II) with OH⁻, NH₃ and Cl⁻?
• Can I use E^⦵ values to predict feasibility and products, and know the limits of the prediction?
• Can I do titration calculations with MnO₄⁻, Cr₂O₇²⁻, C₂O₄²⁻, Fe²⁺ and Cu²⁺/I⁻/S₂O₃²⁻, including back titrations?
• Can I explain colour in four steps and draw the octahedral and tetrahedral splitting diagrams, above the isolated-ion level?
• Can I draw 3D cis/trans and optical isomers and deduce polarity?
• Can I write K_stab with units, calculate with it and explain ligand exchange from K_stab values?

Knowledge organiser