Transition elements — OCR A-Level Chemistry

This heading introduces the typical properties of transition elements that the specification expects you to describe and explain. Read the surrounding section to identify each property: formation of compounds with variable oxidation states, formation of complex ions, use as catalysts, and characteristic coloured ions. Link each property to the electronic structure of transition elements, particularly the presence of partially filled 3d subshells and the small energy gap between 3d and 4s. Use the course text to select examples of each property and to check which elements count as transition elements. Treat the heading as a reading guide: gather the properties, the explanation for each, and a worked example, then test yourself by explaining why a given element shows or lacks the property.

For Period 4 d-block atoms (Sc–Zn), fill orbitals in order 1s, 2s, 2p, 3s, 3p, 4s, 3d, but note the two exceptions: chromium is [Ar] 3d⁵ 4s¹ and copper is [Ar] 3d¹⁰ 4s¹, because a half-filled or full 3d subshell is more stable. For ions, remove electrons from the highest-energy occupied shell first, which for d-block cations means losing 4s electrons before 3d electrons. For example, Fe (Z = 26) is [Ar] 3d⁶ 4s², so Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵. Use the atomic number to count electrons and the charge to add or remove electrons, then write the configuration.

A transition element is defined here as a d-block element that forms at least one ion with an incomplete d-sub-shell. Apply the test across Ti to Cu. Titanium forms Ti³⁺ with [Ar] 3d¹, so it qualifies. Vanadium, chromium, manganese, iron, cobalt and nickel all form ions with partly filled 3d subshells. Copper forms Cu²⁺ with [Ar] 3d⁹, so it qualifies despite the full 3d¹⁰ of the Cu⁺ ion. Zinc does not qualify because Zn²⁺ is [Ar] 3d¹⁰, a complete d-sub-shell, and scandium does not qualify because Sc³⁺ is [Ar] 3d⁰, an empty d-sub-shell. The key skill is writing the ion configuration and checking whether the 3d count lies between 1 and 9.

This row is a lead-in to the sub-points that follow it, so read it as an instruction about how to illustrate transition-element chemistry rather than as a standalone fact to memorise. The requirement is to illustrate the chemistry of at least two different transition elements, drawing on the sub-points listed beneath it (such as variable oxidation states, catalytic behaviour, coloured ions and ligand substitution). Choose two elements that let you show the widest range of behaviour, for example iron and copper, or chromium and manganese. For each element, select a specific compound or ion and describe what it does: name the species, give its formula and oxidation state, and state the observation. Compare the two elements so the illustration is genuinely about two elements, not one element described twice.

Transition elements commonly form compounds in more than one oxidation state, and this row asks you to recognise that pattern for each element. Iron shows 2⁺ in iron(II) compounds such as FeCl₂ and 3⁺ in iron(III) compounds such as FeCl₃. Copper shows 1⁺ in copper(I) oxide, Cu₂O, and 2⁺ in copper(II) sulfate, CuSO₄. Manganese ranges widely, for example 2⁺ in MnSO₄, 4⁺ in MnO₂ and 7⁺ in KMnO₄. Chromium shows 3⁺ in CrCl₃ and 6⁺ in K₂Cr₂O₇. In an MCQ, work out the oxidation state from the formula using the usual rules: oxygen is usually 2⁻, hydrogen 1⁺, and the sum of oxidation states equals the overall charge. Then check whether the element appears in a second compound with a different value. The cross-reference to 5.3.1 k links this to redox behaviour, because a change in oxidation state is what makes these elements useful as oxidising or reducing agents.

Transition metal ions often form coloured solutions because the d subshell splits into slightly different energy levels when ligands bond to the ion. An electron absorbs visible light of one energy and is promoted to a higher d level; the light not absorbed is transmitted, so the solution shows the complementary colour. For example, hydrated copper(II) ions appear blue because red-orange light is absorbed. The size of the splitting, and so the colour, depends on the metal, its oxidation state, the identity of the ligands and their arrangement. Changing the ligand set, as when water is replaced by ammonia or chloride, changes the splitting and therefore the observed colour. This links to the ligand substitution and complex ion work in 5.3.1 h, j–k.

Transition metals and their compounds act as catalysts because they can change oxidation state and provide a surface or intermediate species that lowers the activation energy of a reaction. Heterogeneous catalysts, such as iron in the Haber process for making ammonia, provide a surface on which reactant molecules adsorb, weaken their bonds and react. Homogeneous catalysts, such as transition metal ions in solution, form intermediate complexes that react further and regenerate the catalyst. Their industrial importance lies in increasing reaction rates so that chemicals such as ammonia, sulfuric acid and methanol can be manufactured economically at lower temperatures and pressures. This links to the rate and catalysis work in 3.2.2 d.

This row is a course heading, not an assessed answer target. It tells you that the specification develops two linked ideas: ligands and complex ions. Read the following specification statements in order and treat each one as a separate learning point. A ligand is a species that donates a lone pair into a vacant orbital on a metal ion or metal atom, forming a coordinate (dative covalent) bond; monodentate ligands such as H₂O and NH₃ donate one pair, while bidentate ligands such as H₂NCH₂CH₂NH₂ donate two pairs from two donor atoms. A complex ion is a central metal ion surrounded by ligands, for example [Cu(H₂O)₆]²⁺ or [Cu(NH₃)₄(H₂O)₂]²⁺. Use the heading to organise your notes: define the terms, then practise drawing and naming complexes, counting coordination number and charge.

A ligand is an ion or molecule that donates one or more lone pairs of electrons into vacant orbitals on a central metal ion or metal atom, forming coordinate (dative covalent) bonds. In a coordinate bond both shared electrons come from the ligand. Monodentate ligands such as H₂O, NH₃, Cl⁻ and CN⁻ donate one lone pair from one donor atom, so each forms one coordinate bond. Bidentate ligands such as ethane-1,2-diamine (H₂NCH₂CH₂NH₂) and the ethanedioate ion (C₂O₄²⁻) have two donor atoms, each with a lone pair, so one bidentate ligand forms two coordinate bonds to the same metal ion and occupies two coordination sites. For example, [Cu(H₂O)₆]²⁺ is octahedral with six monodentate water ligands, while [Cu(H₂NCH₂CH₂NH₂)₂(H₂O)₂]²⁺ contains two bidentate ligands and two water molecules.

A complex ion is a central transition-metal ion surrounded by ligands bonded by coordinate (dative covalent) bonds; each ligand donates a lone pair into a vacant orbital on the metal. The coordination number is the number of coordinate bonds from ligands to the central metal ion, so it counts donor atoms, not ligands: ethanedioate (oxalate) is bidentate and contributes 2, and ethane-1,2-diamine (en) is bidentate and contributes 2, while water, ammonia and chloride are monodentate and contribute 1 each. For example, [Cu(H₂O)₆]²⁺ has coordination number 6, [Cu(NH₃)₄(H₂O)₂]²⁺ has coordination number 6, and [Ag(NH₃)₂]⁺ has coordination number 2. The overall charge equals the metal ion charge plus the sum of ligand charges, so neutral ligands leave the charge unchanged.

Six-fold coordination means six coordinate bonds from ligands to the central metal ion, giving coordination number 6. These six donor atoms arrange themselves as far apart as possible, producing an octahedral shape with the metal at the centre and the six ligands at the vertices of an octahedron; bond angles between adjacent ligands are 90°. Common examples include [Cu(H₂O)₆]²⁺, [Fe(H₂O)₆]³⁺, [Fe(CN)₆]⁴⁻ and [Cr(NH₃)₆]³⁺. A bidentate ligand such as ethane-1,2-diamine can also give six-fold coordination with three bidentate ligands, as in [Cr(en)₃]³⁺, where the octahedral arrangement is retained. Recognise the shape by counting six coordinate bonds, not by counting the number of ligands.

Four-fold coordination means a central metal ion is bonded to four donor atoms. Those four donors can lie in one plane at the corners of a square, giving a square planar shape with bond angles of 90°, or they can sit at the corners of a tetrahedron, giving a tetrahedral shape with bond angles of about 109.5°. The shape adopted depends on the metal, its oxidation state and the ligands: for example, [Cu(NH₃)₄]²⁺ and cisplatin, [Pt(NH₃)₂Cl₂], are square planar, whereas [CuCl₄]²⁻ and [CoCl₄]²⁻ are tetrahedral. The cross-reference to 2.2.2 g–h reminds you to apply the same VSEPR-style shape reasoning and bond-angle vocabulary used for simple molecules. When answering, count the donor atoms bonded to the metal, decide whether they are coplanar or three-dimensional, and state the shape and angle together.

Stereoisomerism occurs when complexes have the same molecular formula and the same bonds but a different spatial arrangement of atoms. Complexes show two main types. Geometric (cis–trans) isomerism arises when two identical ligands can be either adjacent (cis) or opposite (trans), as in square planar [Pt(NH₃)₂Cl₂] or octahedral [Co(NH₃)₄Cl₂]⁺. Optical isomerism arises when a complex and its mirror image are non-superimposable, which happens with certain bidentate and multidentate ligands: the tris(ethanedioate) complex [Fe(C₂O₄)₃]³⁻ and the tris(1,2-diaminoethane) complex [Co(en)₃]³⁺ each exist as a pair of optical isomers. Recognise that bidentate and multidentate ligands create chelate rings that can wind in two mirror-image senses. To answer, identify the ligand type, decide whether cis–trans or optical isomerism (or both) is possible, and justify your choice.

cis–trans isomerism arises in square-planar complexes of the type Pt(NH₃)₂Cl₂ because the four ligands occupy positions in one plane around the Pt²⁺ centre and two different ligands are each present twice. In the cis isomer the two NH₃ ligands sit adjacent at 90°, and the two Cl⁻ ligands also sit adjacent; in the trans isomer the two NH₃ ligands lie opposite at 180°, with the two Cl⁻ ligands opposite each other. Both isomers share the same molecular formula, so isomerism is a difference in spatial arrangement, not in composition. To decide which label applies, pick one ligand type and check whether its two copies are next to each other (cis) or on opposite corners (trans). The two forms can differ in properties such as colour, dipole moment and biological activity, which is why the distinction matters. Link this to the broader idea from 4.1.

Optical isomerism occurs when a complex and its mirror image are non-superimposable, so the two forms rotate plane-polarised light in opposite directions. The example [Ni(NH₂CH₂CH₂NH₂)₃]²⁺ contains a Ni²⁺ centre bonded to three bidentate ethane-1,2-diamine (en) ligands, giving an octahedral complex in which the three chelate rings wind around the metal like a propeller. The two propeller senses, conventionally labelled Δ and Λ, are mirror images that cannot be superimposed, so they are optical isomers (enantiomers). Each en ligand binds through two N donor atoms, so the coordination number is six. To decide whether a complex can be optically active, look for a geometry and ligand arrangement whose mirror image cannot be rotated to coincide with the original; a complex with a plane of symmetry is not optically active. Link this to 6.2.

cis-Platin is a square-planar platinum(II) complex with the formula [Pt(NH₃)₂Cl₂], in which the two chloride ligands sit adjacent to each other (cis) rather than opposite (trans). The Pt centre is a transition-metal ion and the ligands are two neutral ammonia molecules and two chloride ions, giving an overall neutral complex. In the body, the relatively low concentration of chloride means water can replace the chloride ligands in a ligand-substitution step, producing a reactive aqua species that can bind to DNA. The platinum binds to adjacent guanine bases on the same DNA strand, forming cross-links that distort the double helix. This distortion prevents the DNA from being replicated and transcribed, so the cell cannot divide and eventually dies. Because cancer cells divide rapidly, they are affected more than most normal cells, which is why cis-platin is used as an anti-cancer drug.

Ligand substitution is the replacement of one ligand in a complex ion by another ligand. In aqueous solution, [Cu(H₂O)₆]²⁺ is a pale blue, six-coordinate complex with water acting as a monodentate ligand. Adding excess ammonia solution replaces water ligands stepwise, giving [Cu(NH₃)₄(H₂O)₂]²⁺, a deep blue complex that is still six-coordinate because two water ligands remain. Adding concentrated hydrochloric acid to [Cu(H₂O)₆]²⁺ replaces water ligands with chloride ions, giving the green-yellow, four-coordinate [CuCl₄]²⁻. The coordination number changes only when the number of ligands bonded to the central ion changes, so compare the ligand count in the reactant and product before claiming a change.

Ligand substitution occurs when one ligand in a complex is replaced by another. In aqueous solution the hexaaqua ion [Cu(H₂O)₆]²⁺ is pale blue; adding excess ammonia solution replaces water ligands to give [Cu(NH₃)₄(H₂O)₂]²⁺, a deep blue solution, and adding concentrated hydrochloric acid replaces water ligands to give [CuCl₄]²⁻, a yellow-green solution. The colour change arises because the ligands alter the energy gap between the d orbitals, so a different wavelength of visible light is absorbed and the complementary colour is seen. The coordination number can change: six-coordinate octahedral becomes four-coordinate tetrahedral with chloride.

Starting from the pale blue complex [Cu(H₂O)₆]²⁺, two ligand substitutions are required. Adding excess ammonia solution replaces four water ligands, giving [Cu(NH₃)₄(H₂O)₂]²⁺, a deep blue complex that is six-coordinate because two water ligands remain. Adding concentrated hydrochloric acid replaces the water ligands with chloride ions, giving [CuCl₄]²⁻, a green-yellow complex that is four-coordinate. Check the overall charge of each product: [Cu(NH₃)₄(H₂O)₂]²⁺ carries a 2⁺ charge, while [CuCl₄]²⁻ carries a 2⁻ charge because four chloride ligands each contribute 1⁻ around a Cu²⁺ centre.

This item tests ligand substitution at a chromium(III) centre. In [Cr(H₂O)₆]³⁺ the chromium ion is surrounded by six water ligands acting as monodentate ligands through their oxygen atoms, giving an octahedral complex with an overall 3⁺ charge. When excess ammonia is added, each water ligand is replaced by an ammonia ligand, which bonds through its lone pair on nitrogen. Six substitutions give [Cr(NH₃)₆]³⁺, still octahedral and still 3⁺ because both ligands are neutral and the chromium oxidation state stays +3. The coordination number remains six, so the shape and charge are unchanged; only the identity of the ligands alters. Recognise that this is a stepwise substitution and that the metal ion's oxidation state does not change during ligand substitution.

Haemoglobin carries oxygen in the blood because each haem group holds an iron(II) ion, Fe²⁺, at the centre of a porphyrin ring. The iron acts as a transition-metal centre that binds an O₂ molecule by ligand substitution, replacing a weakly bound water ligand in the deoxygenated form. Because O₂ binds reversibly, haemoglobin picks up oxygen in the lungs, where oxygen concentration is high, and releases it in respiring tissues, where it is low; this transport depends on the iron centre. Carbon monoxide binds to the same iron site far more strongly and is not released readily, so it blocks the site and prevents oxygen transport, which is why CO is toxic. Explain the role of the Fe²⁺ centre, the reversible ligand substitution with O₂, and the competing substitution by CO.

A precipitation reaction in transition-element chemistry forms an insoluble solid when two aqueous solutions are mixed. In this topic the metal ion solution is treated with aqueous sodium hydroxide or aqueous ammonia, and the hydroxide of the metal separates as a coloured precipitate. For example, adding NaOH(aq) to Cu²⁺(aq) gives a blue precipitate of Cu(OH)₂, written as Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s). The ionic equation shows only the ions that change, so the sodium ion and any spectator ions are omitted. The colour of the precipitate, and whether it dissolves in excess reagent, helps identify the metal ion. Learn to write the state symbols and balance the charge as well as the atoms.

This topic covers how five aqueous transition-metal ions behave when a little, then excess, aqueous NaOH or aqueous NH₃ is added. For each ion you must know the hydroxide precipitate formula, its colour, and whether it dissolves in excess reagent, and you must write balanced ionic equations. Cu²⁺ gives a blue precipitate of Cu(OH)₂, insoluble in excess NaOH but dissolving in excess NH₃ to give a deep blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺. Fe²⁺ gives a green precipitate of Fe(OH)₂, insoluble in both excesses. Fe³⁺ gives a brown precipitate of Fe(OH)₃, insoluble in both excesses. Mn²⁺ gives a pale brown precipitate of Mn(OH)₂, insoluble in both excesses. Cr³⁺ gives a grey-green precipitate of Cr(OH)₃ that dissolves in excess NaOH to a green solution of [Cr(OH)₆]³⁻, and dissolves slightly in excess NH₃. Write each as an ionic equation, for example Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s).

Precipitation reactions of transition metal ions in aqueous solution are tested by adding a small volume of aqueous sodium hydroxide or aqueous ammonia and observing the hydroxide precipitate that forms. For example, Cu²⁺(aq) with OH⁻(aq) gives a blue gelatinous precipitate of Cu(OH)₂, while Fe²⁺(aq) gives a green precipitate of Fe(OH)₂ and Fe³⁺(aq) gives a reddish-brown precipitate of Fe(OH)₃. The identity of the metal ion is inferred from the colour of the precipitate and, where relevant, its behaviour in excess reagent. A multiple-choice question may ask you to select the correct ion–precipitate colour pairing or the correct ionic equation, so learn each colour and balance the charges and atoms in the equation.

When excess aqueous sodium hydroxide or excess aqueous ammonia is added, some transition metal hydroxide precipitates dissolve as ligand substitution forms soluble complex ions. For example, adding excess aqueous ammonia to Cu²⁺(aq) first gives a blue precipitate of Cu(OH)₂, which dissolves in excess ammonia to give a deep blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺. Similarly, Cr³⁺(aq) gives a green precipitate with a little hydroxide that dissolves in excess to give [Cr(OH)₆]³⁻. A multiple-choice question may ask which complex forms, its colour, or its charge, so learn the ligand, coordination number and overall charge for each case.

This row is a heading-level guide within the transition elements topic, so treat it as a signpost rather than a standalone examinable statement. Read it as telling you that redox behaviour runs through the whole section: you will meet it when transition metals change oxidation state in solution, when their ions act as oxidising or reducing agents, and when ligand substitution or precipitation is accompanied by a colour change. Build a working method: assign oxidation states to the metal before and after a change, identify which species is oxidised and which is reduced, then link the electron transfer to the visible colour. Keep a running table of ion, oxidation state and colour as you read, and check each entry against your specification and class notes rather than memorising this heading as a fact to be tested on its own.

This is an introductory stem: it announces that the following sub-statements list specific redox reactions, each with a colour change you must know. Read it as an instruction to work through those listed reactions one at a time, and for each one record the reactant ions with their colours, the oxidation state of the metal before and after, which species is oxidised and which is reduced, and the final colour. A reliable method is a four-column table: reaction, metal oxidation state before and after, species oxidised and reduced, colour change observed. Because the stem ends with a colon, the examinable detail sits in the sub-statements that follow, so do not treat this line alone as a complete fact to memorise; use it to organise your notes and check you have covered every listed reaction.

Iron exists in aqueous solution mainly as pale green Fe²⁺ (iron(II), +2 oxidation state) and yellow-orange Fe³⁺ (iron(III), +3 oxidation state). Oxidation converts Fe²⁺ to Fe³⁺ and needs an oxidising agent; reduction converts Fe³⁺ back to Fe²⁺ and needs a reducing agent. A common laboratory method uses acidified potassium manganate(VII): MnO₄⁻ oxidises Fe²⁺ to Fe³⁺ and is itself decolourised, so the purple colour fades. Adding zinc, or passing hydrogen sulfide, reduces Fe³⁺ to Fe²⁺. Aqueous sodium hydroxide gives a green precipitate with Fe²⁺ and a red-brown precipitate with Fe³⁺, which helps identify each ion. In each interconversion the oxidation state of iron changes by one, so one electron is transferred per iron ion. Write the half-equations and balance charge and atoms before combining them.

Chromium in aqueous solution appears as green Cr³⁺ (chromium(III), +3 oxidation state) and orange Cr₂O₇²⁻ (dichromate(VI), in which each chromium is +6). Oxidation converts Cr³⁺ to Cr₂O₇²⁻ and needs a strong oxidising agent, for example hydrogen peroxide in alkaline conditions followed by acidification. Reduction converts Cr₂O₇²⁻ back to Cr³⁺ and needs a reducing agent such as zinc and acid, when the orange colour turns green. The half-equation for the oxidation is 2Cr³⁺ + 7H₂O → Cr₂O₇²⁻ + 14H⁺ + 6e⁻, so six electrons are lost per dichromate ion formed. Balance atoms, then oxygen with water, then hydrogen with H⁺, then charge with electrons. Note that the chromium oxidation state rises from +3 to +6, a change of three per chromium atom.

Copper shows two common aqueous oxidation states, Cu²⁺ (blue, [Cu(H₂O)₆]²⁺) and Cu⁺ (colourless, often [Cu(H₂O)₂]⁺ or as an insoluble salt). A reducing agent such as iodide ions, I⁻, reduces Cu²⁺ to Cu⁺: 2Cu²⁺ + 4I⁻ → 2CuI + I₂, giving a white precipitate of copper(I) iodide and brown iodine. Cu⁺ is unstable in aqueous solution because it can disproportionate: 2Cu⁺ → Cu²⁺ + Cu. One Cu⁺ is oxidised (oxidation number rises from +1 to +2) while another is reduced (falls from +1 to 0), so copper appears as a pink solid. The E° values make this favourable. In an MCQ, identify the oxidation-state changes, the disproportionation definition and the correct equation or observation.

This criterion asks you to apply known transition-metal chemistry to reactions you have not met before. For ligand substitution, identify the incoming ligand, decide whether it replaces water ligands, and state the change in colour and formula while checking whether the coordination number stays the same. For example, adding excess NH₃ to [Cu(H₂O)₆]²⁺ replaces water ligands to give [Cu(NH₃)₄(H₂O)₂]²⁺, a deep blue solution that is still six-coordinate. For precipitation, use solubility rules and the charges on the ions to predict the precipitate formula and colour. For redox, assign oxidation states before and after, identify which species is oxidised and which is reduced, and write a balanced equation. Explain each prediction by reference to the species involved rather than reciting a memorised result.