HomeLearning HubA Level ChemistryA2 28: Transition elements
A2 28

Chemistry of transition elements

A Level · Inorganic chemistry · Papers 4 and 5

🎯What you need to be able to do

  • Define a transition element, sketch the 3dxy and 3d orbitals, and explain variable oxidation states, catalytic activity and complex formation.
  • Define ligand and complex; use monodentate, bidentate and polydentate; describe linear, square planar, tetrahedral and octahedral complexes; state coordination number; predict a complex’s formula and charge.
  • Describe ligand exchange in the complexes of Cu2+ and Co2+ with water, ammonia, hydroxide and chloride, including the colour changes.
  • Predict feasibility of transition-metal redox reactions from E, and do calculations on MnO4/C2O42−, MnO4/Fe2+ and Cu2+/I.
  • Explain colour by the splitting of degenerate d orbitals in octahedral and tetrahedral complexes, and how the ligand affects ΔE and the colour seen.
  • Describe geometrical and optical isomerism in complexes, and deduce their polarity.
  • Define, write and calculate with stability constants, and use them to explain ligand exchange.

📚The chemistry

28.1 What makes a transition element

A transition element is a d-block element that forms one or more stable ions with incomplete d orbitals. In the first row that means titanium to copper. Scandium and zinc are in the d-block but are not transition elements: scandium’s only ion, Sc3+, has no d electrons (3d0), and zinc’s only ion, Zn2+, has a full 3d sub-shell (3d10). Copper qualifies because Cu2+ is 3d9.

d orbital shapes. The 3dxy orbital has four lobes lying in the xy plane, pointing between the x and y axes (like a four-leaf clover set at 45° to the axes). The 3d orbital has two lobes along the z axis with a ring (doughnut) around the middle in the xy plane.

The characteristic properties, and why:

  • Variable oxidation states. The 3d and 4s sub-shells are close in energy, so different numbers of electrons can be lost (or used in bonding) for similar energy costs. Manganese shows every state from +2 to +7; vanadium +2 to +5; chromium commonly +3 and +6; iron +2 and +3.
  • Catalysis. Transition elements and their compounds are good catalysts because they have more than one stable oxidation state (so they can accept and donate electrons, being oxidised and reduced in turn, as Fe2+/Fe3+ does in the I/S2O82− reaction, topic 26) and vacant d orbitals that are energetically accessible, which can form dative bonds with reactant molecules and hold them on a surface.
  • Complex ions. The same vacant, energetically accessible d orbitals can accept lone pairs from ligands, forming dative bonds.
  • Coloured compounds — explained in 28.3 below.

28.2 Complexes and ligands

  • A ligand is a species that contains a lone pair of electrons that forms a dative covalent bond to a central metal atom or ion.
  • A complex is a molecule or ion formed by a central metal atom or ion surrounded by one or more ligands.
  • The coordination number is the number of dative bonds from ligands to the central metal ion.

Ligands are classified by how many dative bonds each one forms:

Monodentate (one bond): H2O, NH3, Cl, CN, OH
Bidentate (two): 1,2-diaminoethane, en, H2NCH2CH2NH2; ethanedioate, C2O42−
Polydentate (several): EDTA4−, which forms six bonds

Shapes

Linear, 180°, coordination number 2: [Ag(NH3)2]+
Tetrahedral, 109.5°, CN 4: [CuCl4]2−, [CoCl4]2−
Square planar, 90°, CN 4: [Pt(NH3)2Cl2]
Octahedral, 90°, CN 6: [Cu(H2O)6]2+, [Co(NH3)6]2+

Large ligands such as Cl often give four-coordinate complexes, because fewer fit around the metal. With bidentate ligands, count the bonds, not the ligands: [Ni(en)3]2+ has three ligands but coordination number 6.

Formula and charge: overall charge = charge on the metal ion + the sum of the ligand charges. Fe3+ with six CN: 3 − 6 = −3, so [Fe(CN)6]3−. Co3+ with six NH3 (neutral): [Co(NH3)6]3+.

Ligand exchange: copper(II) and cobalt(II)

A ligand exchange reaction replaces some or all of the ligands in a complex with different ones. In aqueous solution the starting complex is the hexaaqua ion.

Copper(II), starting from blue [Cu(H2O)6]2+:

  • + OH (NaOH): pale blue precipitate, insoluble in excess. \[ \mathrm{[Cu(H_2O)_6]^{2+} + 2OH^{-} \rightarrow Cu(OH)_2(H_2O)_4 + 2H_2O} \]
  • + NH3: a little gives the same pale blue precipitate (ammonia solution contains OH); excess dissolves it to a deep (dark) blue solution: \[ \mathrm{[Cu(H_2O)_6]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O} \]
  • + Cl (concentrated HCl): a yellow tetrahedral complex; mixtures with the blue aqua ion look green. Reversible on adding water. \[ \mathrm{[Cu(H_2O)_6]^{2+} + 4Cl^{-} \rightleftharpoons [CuCl_4]^{2-} + 6H_2O} \]

Cobalt(II), starting from pink [Co(H2O)6]2+:

  • + OH: a blue precipitate of cobalt(II) hydroxide, which turns pink on standing and slowly brown in air as it is oxidised. \[ \mathrm{[Co(H_2O)_6]^{2+} + 2OH^{-} \rightarrow Co(OH)_2(H_2O)_4 + 2H_2O} \]
  • + NH3: a little gives the blue precipitate; excess concentrated ammonia dissolves it to a pale yellow-brown (straw) solution, which darkens in air as Co(II) is oxidised to Co(III). \[ \mathrm{[Co(H_2O)_6]^{2+} + 6NH_3 \rightarrow [Co(NH_3)_6]^{2+} + 6H_2O} \]
  • + Cl (concentrated HCl): a blue tetrahedral complex. Reversible; the pink colour returns on dilution. \[ \mathrm{[Co(H_2O)_6]^{2+} + 4Cl^{-} \rightleftharpoons [CoCl_4]^{2-} + 6H_2O} \]

Notice the change in coordination number and shape with chloride: six water ligands (octahedral) are replaced by four larger chloride ligands (tetrahedral).

Redox and titrations

Use E values (topic 24) to predict whether a transition-metal redox reaction is feasible: it is if Ecell is positive. Three titrations are named in the syllabus:

  • Manganate(VII) and iron(II), in acid; purple MnO4 is decolourised until the end-point, when a permanent pale pink appears (self-indicating). Ratio 1 : 5. \[ \mathrm{MnO_4^{-} + 8H^{+} + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O} \]
  • Manganate(VII) and ethanedioate, in acid, warmed to about 60 °C (the reaction is slow at first, then catalysed by the Mn2+ it produces — autocatalysis). Ratio 2 : 5. \[ \mathrm{2MnO_4^{-} + 16H^{+} + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O} \]
  • Copper(II) and iodide: Cu2+ oxidises I to iodine, forming a white precipitate of copper(I) iodide; the iodine is then titrated with sodium thiosulfate, using starch near the end-point (blue-black → colourless). \[ \mathrm{2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_2} \] \[ \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}} \] Overall ratio Cu2+ : S2O32− = 1 : 1.

Manganate(VII) titrations are acidified with dilute sulfuric acid, never hydrochloric acid, because MnO4 can oxidise Cl (E +1.52 against +1.36 V).

28.3 Colour

In an isolated transition metal ion, the five 3d orbitals have the same energy: they are degenerate. When ligands approach, their lone pairs repel the d electrons, but not equally: orbitals pointing towards the ligands are raised in energy more than those pointing between them. The d orbitals split into two non-degenerate sets separated by an energy gap, ΔE:

  • Octahedral complexes: two orbitals higher (d and dx²−y², which point at the ligands on the axes) and three lower.
  • Tetrahedral complexes: the reverse — three higher and two lower, with a smaller gap.

Why complexes are coloured: an electron in a lower d orbital can absorb a photon and be promoted to a higher d orbital. The photon’s energy must equal ΔE, so only light of one particular frequency (ΔE = hf) is absorbed. The remaining light is transmitted or reflected, and we see the complementary colour of what was absorbed. [Cu(H2O)6]2+ absorbs in the red-orange, so it looks blue.

Ions with no d electrons (Sc3+, Ti4+) or a full d sub-shell (Zn2+, Cu+) have no possible d–d transition, so their compounds are colourless (white as solids).

Different ligands split the d orbitals by different amounts, so they change ΔE, the frequency absorbed, and the colour seen. Replacing water by ammonia around Cu2+ increases ΔE: a higher frequency is absorbed and the colour changes from pale blue to deep blue. Changes of coordination number and shape (octahedral to tetrahedral with chloride) change ΔE even more — pink to blue for cobalt. That is why every ligand exchange above comes with a colour change.

28.4 Stereoisomerism in complexes

Geometrical (cis/trans) isomerism occurs when two identical ligands can be next to each other (cis, 90° apart) or opposite (trans, 180°):

  • Square planar [Pt(NH3)2Cl2]: the cis isomer is the anticancer drug cisplatin; the trans isomer is inactive.
  • Octahedral [Co(NH3)4(H2O)2]2+: the two water ligands cis or trans.
  • Octahedral with bidentate ligands [Ni(en)2(H2O)2]2+: the two water ligands cis or trans.

Optical isomerism occurs in complexes with no plane of symmetry, which exist as non-superimposable mirror images. The examples involve bidentate ligands, which wrap around the metal like propeller blades:

  • [Ni(en)3]2+ — three en ligands; two enantiomers.
  • cis-[Ni(en)2(H2O)2]2+ — the cis form is chiral and has two enantiomers; the trans form has a plane of symmetry and is not. So this complex has three stereoisomers in total.

Polarity: a complex is polar if the bond dipoles do not cancel. In trans isomers identical ligands are opposite each other, so their dipoles cancel and the complex is non-polar (trans-[Pt(NH3)2Cl2]). In cis isomers they are on the same side, so the complex is polar (cisplatin).

28.5 Stability constants

The stability constant, Kstab, of a complex is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules. Water is the solvent, so [H2O] is not included:

\[ \mathrm{Cu^{2+}(aq) + 4NH_3(aq) \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}(aq)} \] \[ K_\mathrm{stab} = \frac{[\mathrm{[Cu(NH_3)_4(H_2O)_2]^{2+}}]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4} \]

A large Kstab means the equilibrium lies far to the right: the complex is stable. Values are often quoted as log Kstab. In ligand exchange, a ligand forming a complex with a larger Kstab replaces one with a smaller Kstab. For copper(II): the ammine complex (log Kstab about 13) is far more stable than the chloro complex (about 5.6), and EDTA complexes are more stable still (about 18.8). Adding EDTA to almost any aqua or ammine complex therefore replaces the other ligands, which is why EDTA is used to treat metal poisoning and to measure metal ions by titration.

✏️Worked example

1.20 g of brass is dissolved in nitric acid and the solution is neutralised and made up to 250 cm3. A 25.0 cm3 portion is added to excess potassium iodide, and the iodine released needs 12.6 cm3 of 0.100 mol dm−3 sodium thiosulfate. (a) Write the two equations and deduce the Cu2+ : S2O32− ratio. (b) Calculate the percentage by mass of copper in the brass. [Ar Cu 63.5] (c) Explain why zinc, the other metal in brass, does not interfere. (d) Describe what you would see if concentrated ammonia were added, in excess, to another portion of the copper solution, and write an equation.

(a)

\[ \mathrm{2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_2} \] \[ \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}} \]

2 Cu2+ give 1 I2, which reacts with 2 S2O32−: so Cu2+ : S2O32− = 1 : 1.

(b)

\[ n(\mathrm{S_2O_3^{2-}}) = 0.100 \times \frac{12.6}{1000} = 1.26 \times 10^{-3}\ \mathrm{mol} = n(\mathrm{Cu^{2+}})\ \text{in 25.0 cm}^3 \] \[ n(\mathrm{Cu})\ \text{in 250 cm}^3 = 1.26 \times 10^{-2}\ \mathrm{mol} \qquad m = 1.26 \times 10^{-2} \times 63.5 = 0.800\ \mathrm{g} \] \[ \%\ \mathrm{Cu} = \frac{0.800}{1.20} \times 100 = 66.7\% \]

(c) Zn2+ has a full 3d10 sub-shell and only one oxidation state; it cannot be reduced by iodide (E Zn2+/Zn is −0.76 V), so it releases no iodine and does not affect the titre.

(d) A pale blue precipitate forms at first, then dissolves in excess ammonia to give a deep blue solution:

\[ \mathrm{[Cu(H_2O)_6]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O} \]
Check it. Brass is typically 60–70% copper, so 66.7% is realistic. And remember the scale-up: the titration used one tenth of the solution (25.0 of 250 cm3), so the moles of copper in the whole sample are ten times the titration value.
Using a 1 : 2 ratio for copper and thiosulfate. Chaining the two equations gives 2Cu2+ → I2 → 2S2O32−, i.e. 1 : 1. Stopping after the first equation and reading “2 Cu : 1 I2” as the answer halves the copper content. The other common loss is forgetting the factor of 10 for the aliquot.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain why zinc is not classed as a transition element, although it is in the d-block.
A transition element must form at least one stable ion with an incomplete d sub-shell. Zinc forms only the Zn2+ ion, [Ar] 3d10, in which the 3d sub-shell is full. So zinc does not meet the definition. (Consequently its compounds are white and it does not show variable oxidation states.)
2. Give the formula, charge, coordination number and shape of the complex formed from (a) Fe3+ and six CN; (b) Ni2+ and three en ligands; (c) Co2+ and four Cl.
(a) [Fe(CN)6]3− (3 − 6 = −3); coordination number 6; octahedral. (b) [Ni(en)3]2+ (en is neutral); each en forms two dative bonds, so coordination number 6; octahedral. (c) [CoCl4]2− (2 − 4 = −2); coordination number 4; tetrahedral.
3. Explain why [Cu(H2O)6]2+ is coloured but [Zn(H2O)6]2+ is not.
In an octahedral complex the ligands split the d orbitals into two higher and three lower energy orbitals, separated by ΔE. Cu2+ is 3d9: there is a vacancy in the upper set, so an electron in a lower orbital can absorb visible light of frequency f (ΔE = hf) and be promoted. Light of that frequency (red-orange) is removed, and the complementary colour, blue, is seen. Zn2+ is 3d10: every d orbital is full, so no d–d promotion is possible, no visible light is absorbed, and the solution is colourless.
4. 20.0 cm3 of acidified potassium manganate(VII) reacts exactly with 0.1340 g of sodium ethanedioate, Na2C2O4 (Mr 134.0). Calculate the concentration of the manganate(VII).
n(C2O42−) = 0.1340 / 134.0 = 1.000 × 10−3 mol. Ratio MnO4 : C2O42− = 2 : 5, so n(MnO4) = 1.000 × 10−3 × 2/5 = 4.00 × 10−4 mol. c = 4.00 × 10−4 / 0.0200 = 0.0200 mol dm−3.
5. Explain why [Pt(NH3)2Cl2] has two isomers, and which is polar.
The complex is square planar. The two Cl ligands can be adjacent (cis, 90°) or opposite (trans, 180°), giving geometrical isomers that cannot be interconverted without breaking bonds. In the trans isomer the Pt–Cl dipoles point in opposite directions and cancel (as do the Pt–N dipoles), so it is non-polar. In the cis isomer (cisplatin) they are on the same side and do not cancel, so it is polar.
6. For [Ag(NH3)2]+, Kstab = 1.7 × 107 mol−2 dm6. A solution has [Ag(NH3)2+] = 0.050 and [NH3] = 0.10 mol dm−3. Calculate [Ag+].
Kstab = [Ag(NH3)2+] / ([Ag+][NH3]2). Rearranging: [Ag+] = 0.050 / (1.7 × 107 × 0.102) = 0.050 / 1.7 × 105 = 2.9 × 10−7 mol dm−3. The large Kstab means almost all the silver is in the complex; only a tiny concentration of free Ag+ remains.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemguide (Jim Clark) — the transition metals section, with photographs of the copper and cobalt ligand-exchange colours
  • ChemTube3D (University of Liverpool) — rotatable d orbitals and complex ions, including the enantiomers of [Ni(en)3]2+
  • Royal Society of Chemistry — practical guides for manganate(VII) and iodine–thiosulfate titrations