Chemistry of transition elements

Defining a transition element, and the shapes of the 3d orbitals

A transition element is a d-block element which forms one or more stable ions with an incomplete d sub-shell. Every transition element sits in the d-block, but the reverse is not true. Scandium forms Sc3+, which is [Ar]3d0, and zinc forms Zn2+, which is [Ar]3d10. In both ions the d sub-shell is either empty or completely full, so scandium and zinc are d-block elements but not transition elements. The elements you meet in this topic are the first row, titanium to copper: titanium, vanadium, chromium, manganese, iron, cobalt, nickel and copper.

Everything in this topic grows out of the 3d sub-shell. It is close in energy to the 4s sub-shell, it holds up to ten electrons, and in the common ions of these elements it is partially filled. Because the 3d electrons are available and the sub-shell is not full, the elements show variable oxidation states, act as catalysts, form complex ions and form coloured compounds.

In an isolated atom or ion the five 3d orbitals all have exactly the same energy: they are degenerate. You need to be able to sketch two of them. The 3dxy orbital has four lobes that lie in the xy plane, in the four regions between the x and the y axes; the axes themselves pass between the lobes, so there is no electron density on the axes. The 3dz2 orbital is the odd one out: it has two lobes pointing along the z axis, one each way, and a ring of electron density (a doughnut) around the middle in the xy plane.

Both shapes come from the same idea. An orbital is a region of space in which a pair of electrons is most likely to be found, and the d orbitals are the ones that point either between the axes or along one axis and around it. That difference matters later, because the orbitals that point at the ligands are pushed up in energy when ligands approach, and the ones that point between the ligands are not.

• Transition element: a d-block element that forms one or more stable ions with an incomplete d sub-shell.
• Sc3+ is 3d0 and Zn2+ is 3d10, so scandium and zinc are d-block but not transition elements.
• Transition elements: variable oxidation states, catalysts, complex ions, coloured compounds.
• 3dxy: four lobes between the x and y axes. 3dz2: two lobes along z with a ring of density in the xy plane.

Variable oxidation states and why they occur

Transition elements have variable oxidation states because the 3d and 4s sub-shells are very close in energy. There is no large jump between removing an electron from 4s and removing one from 3d, so electrons can be lost from both sub-shells and the energies needed to remove successive electrons stay similar. An element with a small energy gap between one sub-shell and the next can settle at several different positive oxidation states; an element with a large gap, such as sodium, is stuck at one.

Manganese is the extreme case. It is found as Mn(II) in manganese(II) sulphate, Mn(IV) in manganese(IV) oxide, Mn(VI) in potassium manganate(VI) and Mn(VII) in potassium manganate(VII), KMnO4. Iron is found as Fe(II) in iron(II) sulphate and Fe(III) in iron(III) chloride. Chromium is found as Cr(III) in chromium(III) sulphate and Cr(VI) in potassium dichromate(VI), K2Cr2O7. Copper is found as Cu(I) in copper(I) oxide and Cu(II) in copper(II) sulphate.

There is a pattern worth knowing. The highest oxidation states of the later elements in the row are usually found only in oxides and oxo-ions such as MnO4- and Cr2O7^2-, where the oxygen atoms carry away electron density, while the simple ions in aqueous solution settle at +2 or +3. That is why a solution of manganese(II) sulphate and a solution of potassium manganate(VII) look nothing like each other even though both contain manganese.

The oxidation state also decides the chemistry. Mn(VII) in acidic solution is a strong oxidising agent and is reduced; Mn(II) in acidic solution is stable and can be oxidised only by something stronger. Whenever an element can slide between two oxidation states, both an oxidation and a reduction become possible, which is exactly why these elements make such good catalysts.

Catalysis by transition elements and their compounds

Transition elements and their compounds are catalysts because they have more than one stable oxidation state and because they have vacant d orbitals that are energetically accessible and can form dative bonds with ligands, including with the molecules of the reactants. A metal ion with two accessible oxidation states can accept electrons from one reactant in one step and hand them to the other reactant in the next step. The reaction is broken into two easier steps, each with a lower activation energy than the single uncatalysed step, and the catalyst finishes in the oxidation state it started in.

The vacant d orbitals do a second job. They can accept lone pairs from reactant molecules and hold them in position on the metal, which weakens bonds inside those molecules and lines them up for reaction. That is why many transition metal surfaces, such as iron in the Haber process and vanadium(V) oxide in the Contact process, are such effective heterogeneous catalysts: the reactant molecules are adsorbed onto the metal or metal oxide surface, held by dative bonds, and converted there.

Heterogeneous examples worth naming: iron in the Haber process, N2 + 3H2 <-> 2NH3; vanadium(V) oxide in the Contact process, 2SO2 + O2 <-> 2SO3; manganese(IV) oxide in the decomposition of hydrogen peroxide, 2H2O2 -> 2H2O + O2; and platinum, palladium and rhodium in a catalytic converter.

Homogeneous examples are just as important. Iron(II) ions catalyse the reaction between peroxodisulphate(VI) ions and iodide ions, shuttling between Fe(II) and Fe(III). The reaction between manganate(VII) ions and ethanedioate ions in acid is autocatalysed by one of its own products: at the start the purple colour fades slowly because no Mn2+ has been formed, then the rate increases as Mn2+ builds up, then it slows again as the reactants run out.

• Two accessible oxidation states: the catalyst takes electrons from one reactant and gives them to the other.
• Vacant d orbitals: form dative bonds with reactant molecules, holding them and weakening their bonds.
• Heterogeneous: Fe (Haber), V2O5 (Contact), MnO2 (decomposition of H2O2), Pt/Pd/Rh (catalytic converter).
• Homogeneous: Fe2+/Fe3+ with peroxodisulphate(VI) and iodide ions; Mn2+ autocatalysing the manganate(VII)/ethanedioate reaction.

Complex ions: ligands, coordination number and geometry

A ligand is a species that contains a lone pair of electrons and 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. Water, ammonia, the chloride ion and the cyanide ion each have one lone pair to give, so each attaches to the metal at one point only: they are monodentate ligands. The bond is drawn with an arrow from the ligand to the metal, because both electrons come from the ligand.

When a ligand has two or more lone pairs, it can attach to the same metal ion at two or more points. 1,2-diaminoethane, H2NCH2CH2NH2, often written en, has a lone pair on each nitrogen; the ethanedioate ion, C2O4^2-, has a lone pair on each of the two oxygen atoms that bond to the metal. Both are bidentate and form a five-membered ring with the metal, which makes the complex more stable than one held by the same number of separate monodentate ligands. EDTA4- is a polydentate ligand: it has six donor atoms and wraps around a metal ion, forming six dative bonds at once.

The coordination number is the total number of dative covalent bonds from ligands to the central metal ion, not the number of ligands. The geometry of the complex then follows from the coordination number, though it is not fixed by it alone: a coordination number of 4 can give a square planar ion such as [Ni(CN)4]2- or a tetrahedral ion such as [CuCl4]2-.

You can predict the formula and charge of a complex by adding the charges. Copper(II) with six water molecules gives [Cu(H2O)6]2+: Cu2+ plus six neutral ligands is 2+. Copper(II) with four chloride ions gives [CuCl4]2-: 2+ plus four lots of 1- is 2-. Nickel(II) with four cyanide ions gives [Ni(CN)4]2-. Silver(I) with two ammonia molecules gives [Ag(NH3)2]+, which is linear with a bond angle of 180 degrees.

Ligand exchange in copper(II) and cobalt(II) complexes

Ligand exchange happens when one ligand in a complex is replaced by another. Water is the default ligand in aqueous solution, because every metal ion is surrounded by water molecules, but a ligand with a stronger attraction for the metal ion can push water out. The exchanges you must know involve copper(II) and cobalt(II).

Copper(II) in water is the pale blue hexaaquacopper(II) ion, [Cu(H2O)6]2+. Adding a little aqueous ammonia gives a pale blue precipitate of copper(II) hydroxide, because ammonia solution contains hydroxide ions: Cu2+(aq) + 2OH-(aq) -> Cu(OH)2(s). Adding excess ammonia dissolves that precipitate and gives a deep blue solution, because four of the six water ligands have been replaced by ammonia: [Cu(H2O)6]2+ + 4NH3 <-> [Cu(NH3)4(H2O)2]2+ + 4H2O. The copper(II) hydroxide precipitate is not soluble in excess dilute sodium hydroxide, and that is the difference between adding hydroxide and adding excess ammonia. Adding concentrated hydrochloric acid to copper(II) ions replaces the water with chloride and gives the yellow-green [CuCl4]2- ion, in which the coordination number has fallen from 6 to 4: [Cu(H2O)6]2+ + 4Cl- <-> [CuCl4]2- + 6H2O. Adding water reverses that change.

Cobalt(II) in water is the pink hexaaquacobalt(II) ion, [Co(H2O)6]2+. Adding excess ammonia first gives a blue-green precipitate of cobalt(II) hydroxide, which dissolves in the excess ammonia to give the straw-coloured [Co(NH3)6]2+; that ion is oxidised quite quickly by oxygen in the air to the yellow-brown [Co(NH3)6]3+ containing cobalt(III), which is a useful reminder that cobalt(II) is easily oxidised. Adding concentrated hydrochloric acid to cobalt(II) gives the blue tetrahedral [CoCl4]2- ion: [Co(H2O)6]2+ + 4Cl- <-> [CoCl4]2- + 6H2O.

Which ligand wins is decided by the stability constant, Kstab, defined as the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules. For the copper(II) ammonia complex you write Kstab = [Cu(NH3)4(H2O)2^2+] / ([Cu^2+][NH3]^4); water is left out of the expression because it is the solvent and its concentration is effectively constant. A large Kstab means the equilibrium lies far over to the right, so the complex forms readily and is stable. Ammonia gives a much larger Kstab with copper(II) than water does, which is why the deep blue ammonia complex forms as soon as excess ammonia is added, and chloride gives a larger Kstab with cobalt(II) than water does, which is why the blue [CoCl4]2- ion appears in concentrated hydrochloric acid.

Colour: the splitting of degenerate d orbitals

In an isolated transition element ion the five d orbitals are degenerate: they all have the same energy. In a complex they are not, because the ligands have lone pairs that repel the d electrons. The d orbitals that point straight at the ligands are repelled more than the orbitals that point between them, so the five degenerate d orbitals split into two non-degenerate sets separated by an energy gap, usually written as dE.

In an octahedral complex the six ligands approach along the x, y and z axes. The two orbitals that point along those axes, dz2 and dx2-y2, are raised in energy, and the three that point between the axes, dxy, dxz and dyz, are lowered: two higher, three lower. In a tetrahedral complex the four ligands sit between the axes, so the pattern is the other way round: three higher, two lower. The gap is smaller in a tetrahedral complex than in an octahedral one.

Why the compounds are coloured: an electron in a lower d orbital can be promoted to a higher one by absorbing a photon whose energy is exactly dE. Because dE for most first-row complexes matches the energy of visible light, the complex absorbs part of the visible spectrum and lets the rest through. The colour you see is the complementary colour, the light that was not absorbed. A solution that absorbs orange light looks blue; one that absorbs red light looks blue-green.

Different ligands split the d orbitals by different amounts. Ligands can be arranged in the spectrochemical series, I- < Br- < Cl- < F- < OH- < H2O < NH3 < 1,2-diaminoethane < CN- < CO. A ligand further to the right causes a larger dE, so a higher frequency of light is absorbed and a different complementary colour is seen. This is exactly what happens when excess ammonia is added to copper(II) ions: the pale blue [Cu(H2O)6]2+ has a small gap, so it absorbs only low-frequency red light and looks pale blue, while the deep blue [Cu(NH3)4(H2O)2]2+ has a larger gap, absorbs more strongly in the visible region and looks a much deeper blue. The metal ion matters too: a more highly charged or smaller metal ion pulls the ligands closer, increases dE and changes the colour.

Stereoisomerism in complexes, and overall polarity

Stereoisomers contain the same atoms joined in the same way but arranged differently in space, and complexes show both of the main types.

Geometrical (cis/trans) isomerism appears when two ligands can sit either next to each other or opposite each other. In the square planar complex [Pt(NH3)2Cl2], the two chloride ligands can be cis, on the same side and 90 degrees apart, or trans, opposite each other and 180 degrees apart. These are two different compounds with different properties: the cis isomer is cisplatin, a drug used in cancer treatment, while the trans isomer has no such activity. Octahedral complexes show the same effect. In [Co(NH3)4(H2O)2]2+, the two water ligands can be cis (90 degrees apart) or trans (180 degrees apart), and the same is true of [Ni(H2NCH2CH2NH2)2(H2O)2]2+, where the two remaining water ligands are cis or trans to each other across the two bidentate ligands.

Optical isomerism appears when a complex and its mirror image cannot be superimposed. [Ni(H2NCH2CH2NH2)3]2+ has three bidentate ligands arranged like the three blades of a propeller, and the two possible propeller directions give a pair of mirror-image isomers. The cis form of [Ni(H2NCH2CH2NH2)2(H2O)2]2+ is also optically active, because it has no plane of symmetry, while the trans form is not, because a plane through the middle of the ion cuts it into two identical halves.

Polarity follows from the same pictures. In trans-[Co(NH3)4(H2O)2]2+ the two water ligands point in exactly opposite directions and the four ammonia ligands sit opposite each other in a plane, so the individual bond dipoles cancel and the ion is non-polar. In the cis isomer the two water ligands are at 90 degrees to each other, their dipoles add instead of cancel and the ion is polar. The square planar pair behaves in the same way: trans-[Pt(NH3)2Cl2] is non-polar and cis-[Pt(NH3)2Cl2] is polar.

Redox reactions of transition elements and E values

Standard electrode potentials tell you whether a redox reaction is feasible. The couple with the more positive E is the one that gains electrons: it is the stronger oxidising agent, and the half-equation of the other couple is reversed. The two couples combine in the direction that gives a positive cell potential, Ecell = E of the more positive couple - E of the less positive couple, and a positive Ecell means the reaction is feasible.

MnO4- with Fe2+ in acid. Manganate(VII) is a strong oxidising agent: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O, E = +1.52 V. The purple colour of MnO4- disappears as the almost colourless Mn2+ ion forms, which is what makes the titration self-indicating. Iron(III)/iron(II) is +0.77 V, so MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 5Fe3+ + 4H2O has Ecell = 1.52 - 0.77 = +0.75 V. A positive value means the reaction goes, and that is the basis of the manganate(VII) titration of iron(II).

MnO4- with C2O4^2- in acid. The ethanedioate ion is oxidised to carbon dioxide: 2CO2 + 2e- -> C2O4^2-, E = -0.49 V. Combining the two half-equations gives 2MnO4- + 16H+ + 5C2O4^2- -> 2Mn2+ + 10CO2 + 8H2O, with Ecell = 1.52 - (-0.49) = +2.01 V, so again the reaction is feasible. The mixture is warmed because the reaction is slow at room temperature, and it is autocatalysed by the Mn2+ that is formed.

Cu2+ with I-. Here the E values alone seem to say no: E(Cu2+/Cu+) = +0.15 V and E(I2/I-) = +0.54 V, so copper(II) is not a strong enough oxidising agent to oxidise iodide ions and Ecell would be 0.15 - 0.54 = -0.39 V. In practice 2Cu2+ + 4I- -> 2CuI + I2 does happen, because copper(I) iodide is very insoluble. Removing Cu+ from solution as a precipitate pulls the copper(I) concentration down, which shifts the effective potential of the copper couple to a much more positive value, and the reaction becomes feasible. This is a favourite exam point: E values predict feasibility, but a precipitation or a complexation that removes a product can change the effective potential. The liberated iodine is then titrated with sodium thiosulphate: 2S2O3^2- + I2 -> S4O6^2- + 2I-.