Chemistry of transition elements

All five subtopics here are printed under "A Level subject content" in the 9701 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

With twenty-seven objectives this is the largest topic in the A Level chemistry course. Everything in it follows from one structural fact, that the 3d and 4s subshells are close in energy and the 3d subshell is only partly filled, so it is worth reading the first section carefully before the rest.

Syllabus points

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

• Define a transition element as a d-block element which forms one or more stable ions with incomplete d orbitals.
• Sketch the shape of a 3d_xy orbital and a 3d_z² orbital.
• Understand that transition elements have the following properties: variable oxidation states; behaviour as catalysts; formation of complex ions; formation of coloured compounds.
• Explain why transition elements have variable oxidation states in terms of the similarity in energy of the 3d and the 4s subshells.
• Explain why transition elements behave as catalysts in terms of having more than one stable oxidation state, and vacant d orbitals that are energetically accessible and can form dative bonds with ligands.
• Explain why transition elements form complex ions in terms of vacant d orbitals that are energetically accessible.

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

• Describe and explain the reactions of transition elements with ligands to form complexes, including the complexes of copper(II) and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions.
• Define the term ligand as a species that contains a lone pair of electrons that forms a dative covalent bond to a central metal atom or ion.
• Understand and use the terms monodentate ligand, including H₂O, NH₃, Cl⁻ and CN⁻; bidentate ligand, including 1,2-diaminoethane (en) and the ethanedioate ion, C₂O₄²⁻; polydentate ligand, including EDTA⁴⁻.
• Define the term complex as a molecule or ion formed by a central metal atom or ion surrounded by one or more ligands.
• Describe the geometry (shape and bond angles) of transition element complexes which are linear, square planar, tetrahedral or octahedral.
• State what is meant by coordination number, and predict the formula and charge of a complex ion, given the metal ion, its charge or oxidation state, the ligand and its coordination number or geometry.
• Explain qualitatively that ligand exchange can occur, including the complexes of copper(II) ions and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions.
• Predict, using E° values, the feasibility of redox reactions involving transition elements and their ions.
• Describe the reactions of, and perform calculations involving: MnO₄⁻/C₂O₄²⁻ in acid solution; MnO₄⁻/Fe²⁺ in acid solution; Cu²⁺/I⁻, given suitable data.
• Perform calculations involving other redox systems given suitable data.

28.3 Colour of complexes

• Define and use the terms degenerate and non-degenerate d orbitals.
• Describe the splitting of degenerate d orbitals into two non-degenerate sets of d orbitals of higher energy, and the use of ΔE in octahedral complexes (two higher and three lower d orbitals) and tetrahedral complexes (three higher and two lower d orbitals).
• Explain why transition elements form coloured compounds in terms of the frequency of light absorbed as an electron is promoted between two non-degenerate d orbitals.
• Describe, in qualitative terms, the effects of different ligands on ΔE, on the frequency of light absorbed, and hence on the complementary colour that is observed.
• Use the complexes of copper(II) ions and cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions as examples of ligand exchange affecting the colour observed.

28.4 Stereoisomerism in transition element complexes

• Describe the types of stereoisomerism shown by complexes, including those associated with bidentate ligands: geometrical (cis/trans) isomerism, for example square planar such as [Pt(NH₃)₂Cl₂] and octahedral such as [Co(NH₃)₄(H₂O)₂]²⁺ and [Ni(H₂NCH₂CH₂NH₂)₂(H₂O)₂]²⁺; optical isomerism, for example [Ni(H₂NCH₂CH₂NH₂)₃]²⁺ and [Ni(H₂NCH₂CH₂NH₂)₂(H₂O)₂]²⁺.
• Deduce the overall polarity of such complexes.

28.5 Stability constants, K_stab

• Define the stability constant, K_stab, of a complex as the equilibrium constant for the formation of the complex ion in a solvent, from its constituent ions or molecules.
• Write an expression for a K_stab of a complex ([H₂O] should not be included).
• Use K_stab expressions to perform calculations.
• Describe and explain ligand exchanges in terms of K_stab values, and understand that a large K_stab is due to the formation of a stable complex ion.

What a transition element is

A transition element is a d-block element which forms one or more stable ions with an incomplete d subshell.

The definition is worded that way to exclude two elements that sit in the d block and are not transition elements:

• Scandium forms only Sc³⁺, which is 3d⁰, an empty d subshell.
• Zinc forms only Zn²⁺, which is 3d¹⁰, a full d subshell.

Neither has an ion with a partly filled d subshell, so neither shows the characteristic properties. Being asked why zinc is not a transition element is a standard question and the answer is exactly that.

Two electron configurations are irregular and must be learned, in each case because a half-filled or a full 3d subshell is more stable than the arrangement with a filled 4s.

• Chromium: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹, with a half-filled 3d.
• Copper: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹, with a full 3d.

When a transition element forms an ion, the 4s electrons are lost first, before the 3d. So Fe is 3d⁶4s² and Fe²⁺ is 3d⁶, not 3d⁴4s².

The shapes of the d orbitals

Two must be sketched.

• 3d_xy: four lobes lying in the xy plane, pointing between the x and y axes, like a four-leaf clover.
• 3d_z²: two lobes along the z axis with a doughnut-shaped ring of electron density round the middle in the xy plane.

The difference between them is the reason the orbitals split in a complex, so the sketches are not decoration.

The four characteristic properties, and why

All four come from the same source: the 3d and 4s subshells are very close in energy, and the 3d subshell is only partly filled with vacant orbitals available.

Variable oxidation states. Because 3d and 4s are similar in energy, the energy needed to remove a 3d electron is not much greater than for a 4s electron, so different numbers of electrons can be lost without a large energy penalty. Manganese runs from +2 to +7. Contrast a Group 2 metal, where removing a third electron means breaking into an inner shell and is prohibitively costly.

Catalytic behaviour. Two mechanisms, and the syllabus wants both:

• Having more than one stable oxidation state allows the element to be oxidised in one step and reduced back in a later step, which is homogeneous catalysis, as with Fe²⁺ and Fe³⁺ in the peroxodisulfate and iodide reaction from topic 26.
• Having vacant d orbitals that are energetically accessible allows reactant molecules to form dative bonds to the metal, which is adsorption on a surface, and this weakens the bonds within them. That is heterogeneous catalysis, as with iron in the Haber process.

Complex ion formation. The vacant, energetically accessible d orbitals can accept lone pairs from ligands to form dative covalent bonds.

Coloured compounds. Explained in section 28.3 below.

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 covalent bonds from ligands to the central metal ion. Note that it counts bonds, not ligands, which matters as soon as a bidentate ligand appears.

So [Ni(en)₃]²⁺ has three ligands but a coordination number of six, and [Cu(EDTA)]²⁻ has one ligand and a coordination number of six.

Shapes

The useful rule of thumb is that small ligands such as water and ammonia give six-coordinate octahedral complexes, while large ligands such as chloride give four-coordinate tetrahedral ones, because only four of the larger ligands fit around the metal ion.

Working out the charge on a complex

The overall charge is the oxidation state of the metal plus the sum of the charges on the ligands.

Worked examples.

• Cu²⁺ with six neutral water molecules: 2 + 0 = 2, so [Cu(H₂O)₆]²⁺.
• Cu²⁺ with four chloride ions: 2 - 4 = -2, so [CuCl₄]²⁻.
• Fe³⁺ with six cyanide ions: 3 - 6 = -3, so [Fe(CN)₆]³⁻.
• Co²⁺ with four hydroxide ions: 2 - 4 = -2, so [Co(OH)₄]²⁻.

Ligand exchange, with the colours

Ligand exchange is the replacement of one ligand by another in a complex, and it usually changes the colour and sometimes the shape. The copper(II) and cobalt(II) series must be known.

Copper(II)

Adding a little aqueous ammonia to copper(II) sulfate solution gives the pale blue precipitate of copper(II) hydroxide, because ammonia acts as a base. Adding excess ammonia dissolves the precipitate to give the deep blue solution, because ammonia now acts as a ligand and replaces four of the six water molecules. Being able to explain the same reagent doing two different things is a favourite question, and the answer is that ammonia is both a base and a ligand.

Adding concentrated hydrochloric acid replaces all six water molecules with four chloride ions, so both the colour and the shape change, from octahedral to tetrahedral. The change is reversible: adding water turns the yellow solution blue again.

Cobalt(II)

The pink to blue change on adding concentrated hydrochloric acid is the one to remember, and it reverses on dilution.

Note the pattern across both: water and ammonia are similar in size, so an exchange between them keeps the octahedral shape, whereas chloride is much larger, so an exchange to chloride reduces the coordination number to four and the shape to tetrahedral.

Why complexes are coloured

Degenerate orbitals are orbitals of equal energy. In an isolated gaseous ion the five 3d orbitals are degenerate.

When ligands approach, their lone pairs repel the d electrons, and they repel some d orbitals more than others because the orbitals point in different directions. So the five orbitals split into two non-degenerate sets separated by an energy gap ΔE:

• In an octahedral complex, two orbitals are raised and three are lowered, because two of them point directly at the ligands.
• In a tetrahedral complex, the pattern is reversed: three are raised and two are lowered, and ΔE is smaller.

The colour then arises in three steps:

1. An electron absorbs energy from visible light and is promoted from the lower set to the higher set.
2. The energy absorbed equals ΔE, and the frequency absorbed is given by ΔE = hf, so only light of that frequency is absorbed.
3. The light transmitted or reflected is the original white light minus the absorbed frequency, so the colour seen is the complementary colour of the light absorbed.

That is why a compound with an empty or a full d subshell is colourless: there is no electron to promote, or no vacancy to promote it into. Sc³⁺, Zn²⁺ and Cu⁺ are all colourless, and this is the neatest evidence that the colour comes from d-d transitions.

What changes ΔE. Different ligands split the d orbitals by different amounts, and so do a different oxidation state of the metal and a different coordination number.

• A ligand that causes a larger ΔE means a higher frequency of light absorbed, and therefore a different complementary colour observed.
• Ammonia causes a larger splitting than water, which is why replacing water with ammonia in a copper complex shifts the colour from pale blue to deep blue.
• Chloride causes a smaller splitting than water, and it also changes the shape to tetrahedral, where ΔE is smaller again, so a lower frequency is absorbed.

Two lines carry most of the marks in an explanation question: name the d-d transition between non-degenerate orbitals, and say that the observed colour is the complementary colour of the frequency absorbed.

Stereoisomerism in complexes

Geometrical (cis/trans) isomerism occurs where two identical ligands can be arranged adjacent or opposite:

• Square planar, as in [Pt(NH₃)₂Cl₂]. The cis isomer has the two chlorides adjacent at 90°, the trans isomer has them opposite at 180°. Only the cis isomer is the anticancer drug cisplatin, which is a useful demonstration that stereochemistry can decide biological activity.
• Octahedral, as in [Co(NH₃)₄(H₂O)₂]²⁺. The two water molecules can be adjacent (cis) or opposite (trans).

Optical isomerism occurs where a complex has no plane of symmetry, so it exists as two non-superimposable mirror images. It arises most often with bidentate ligands:

• [Ni(en)₃]²⁺, with three bidentate ligands arranged like the blades of a propeller, exists as two enantiomers.
• The cis isomer of [Ni(en)₂(H₂O)₂]²⁺ is also optically active, while the trans isomer is not, because the trans arrangement has a plane of symmetry.

Polarity

The trans isomer of a square planar complex such as [Pt(NH₃)₂Cl₂] is non-polar: the two identical ligands are opposite one another, so the bond dipoles cancel. The cis isomer is polar, because the two chlorides are on the same side and the dipoles reinforce rather than cancel.

The same reasoning applies to an octahedral complex with two identical ligands: trans cancels, cis does not.

Redox reactions of transition elements

Because transition elements have several oxidation states, they take part in redox reactions, and the feasibility of any of them is judged from E° values exactly as in topic 24: the reaction is feasible if E°_cell is positive.

Three systems must be known.

Manganate(VII) and iron(II) in acid.

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

The ratio is 1 mole of MnO₄⁻ to 5 moles of Fe²⁺. The manganate(VII) is self-indicating: the purple colour is decolourised until the end point, at which the first permanent faint pink appears. Use dilute sulfuric acid to acidify, never hydrochloric, since chloride would itself be oxidised, and never nitric, since it is an oxidising agent.

Manganate(VII) and ethanedioate in acid.

2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂

The ratio is 2 : 5. This reaction is slow at first and then speeds up, because the Mn²⁺ produced catalyses it, which is an example of autocatalysis and a good illustration of a transition element's two oxidation states at work.

Copper(II) and iodide.

2Cu²⁺ + 4I⁻ → 2CuI + I₂

Copper(II) oxidises iodide to iodine, and is itself reduced to a white precipitate of copper(I) iodide, so the mixture looks like a brown suspension. The iodine liberated is then titrated with sodium thiosulfate:

I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

with starch added near the end point, when the colour has faded to pale straw, and the end point is the disappearance of the blue-black colour. The overall ratio is 2 moles of Cu²⁺ to 1 mole of I₂ to 2 moles of thiosulfate, so moles of thiosulfate equals moles of copper(II).

Worked example. 25.0 cm³ of a copper(II) solution liberates iodine that requires 22.4 cm³ of 0.100 mol dm⁻³ sodium thiosulfate. Find the concentration of the copper(II) solution.

Moles of thiosulfate:

22.4 × 0.100 / 1000 = 0.00224

Since moles of Cu²⁺ equals moles of thiosulfate, that is also 0.00224 mol in 25.0 cm³:

0.00224 × 1000 / 25.0 = 0.0896

so the concentration is 0.0896 mol dm⁻³.

Worked example, manganate(VII) and iron(II). 25.0 cm³ of an iron(II) solution requires 18.6 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII).

Moles of MnO₄⁻:

18.6 × 0.0200 / 1000 = 0.000372

Moles of Fe²⁺, five times as many:

0.000372 × 5 = 0.00186

Concentration:

0.00186 × 1000 / 25.0 = 0.0744

so 0.0744 mol dm⁻³.

Stability constants

The stability constant K_stab of a complex is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.

For a ligand exchange in aqueous solution, the water being displaced is the solvent and is present in vast excess, so [H₂O] is not included in the expression.

For [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O:

K_stab = [[Cu(NH₃)₄(H₂O)₂]²⁺] / ([[Cu(H₂O)₆]²⁺] [NH₃]⁴)

A large K_stab means the equilibrium lies far to the right, so the complex formed is more stable than the aqua complex it replaced, and the substituting ligand binds more strongly. Comparing two K_stab values therefore predicts which ligand will displace the other: the ligand giving the larger K_stab wins.

Values are often very large, so they are frequently quoted as lg K_stab, and a difference of one unit in lg K_stab is a factor of ten in stability.

Bidentate and polydentate ligands give particularly large stability constants. EDTA⁴⁻ occupies all six coordination positions with a single ligand, which is why it is used to remove metal ions from solution, in water treatment, in analysis and as an antidote in cases of heavy-metal poisoning.

Worked example. For the reaction [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ ⇌ [Cu(EDTA)]²⁻ + 6H₂O, K_stab = 6.3 × 10¹⁸. At equilibrium [[Cu(EDTA)]²⁻] = 0.0100 mol dm⁻³ and [EDTA⁴⁻] = 0.0500 mol dm⁻³. Find the concentration of free aqua copper(II) ions.

Rearranging the expression:

0.0100 / (6.3 × 10¹⁸ × 0.0500) = 3.175 × 10⁻²⁰

so about 3.2 × 10⁻²⁰ mol dm⁻³, which is effectively zero. That is what a large stability constant means in practice.

Common mistakes

• Defining a transition element as any d-block element, so scandium and zinc are wrongly included.
• Writing the electron configuration of Fe²⁺ as 3d⁴4s². The 4s electrons are removed first.
• Forgetting the irregular configurations of chromium and copper.
• Confusing coordination number with the number of ligands. A bidentate ligand forms two bonds.
• Saying ammonia forms a precipitate with copper(II) in excess. A little gives the precipitate; excess dissolves it.
• Giving the ammonia complex of copper as [Cu(NH₃)₆]²⁺. Only four water molecules are replaced, giving [Cu(NH₃)₄(H₂O)₂]²⁺.
• Saying a complex is coloured because it contains a transition metal, without naming the d-d transition between non-degenerate orbitals.
• Saying the colour observed is the colour absorbed. It is the complementary colour.
• Saying zinc compounds are coloured. Zn²⁺ has a full d subshell, so no transition is possible.
• Reversing the splitting pattern for a tetrahedral complex. Octahedral is two up and three down; tetrahedral is three up and two down.
• Saying the trans isomer of a square planar complex is polar. The dipoles cancel.
• Acidifying a manganate(VII) titration with hydrochloric acid.
• Getting the 1 : 5 ratio for manganate(VII) and iron(II) the wrong way round, or forgetting the 2 : 5 ratio for ethanedioate.
• Including [H₂O] in a K_stab expression.