28. Chemistry of transition elements

A transition element is a d-block element that forms one or more stable ions with an incomplete d subshell. The definition concerns ions, not merely the neutral atom’s position in the periodic table.

Check the electron configuration of the common ion. Zinc is d-block but Zn²⁺ is 3d¹⁰, so it is not a transition element under this definition.

Fe²⁺ is 3d⁶ and Fe³⁺ is 3d⁵, so iron qualifies; Sc³⁺ is d⁰ and does not meet the incomplete-d-ion criterion.

“d-block” and “transition element” are not synonyms in every syllabus definition.

The dxy orbital has four lobes lying between the x and y axes in the xy plane. The dz² orbital has two lobes along z with a torus around the centre.

All five d orbitals have the same general four-lobed/axial angular family in an isolated atom, but their orientations differ. In a ligand field, their energies can split.

Sketch axes first, then place dxy lobes between axes and dz² lobes on the z axis with the doughnut in the xy plane.

Do not put dxy lobes on the axes or draw dz² as a simple p orbital.

Partially filled d subshells allow transition elements to access several oxidation states, bond to ligands in complex ions, participate in catalytic cycles and absorb visible light in many compounds.

These are linked but not identical properties: oxidation-state flexibility supports catalysis, ligand-field splitting helps explain colour, and complex formation depends on donor ligands and geometry.

Iron forms Fe²⁺/Fe³⁺, catalyses redox processes, forms [Fe(H₂O)₆]²⁺ and [Fe(SCN)]²⁺, and gives coloured compounds.

Not every d-block ion is coloured or catalytic; d⁰/d¹⁰ configurations and ligand environment matter.

The 3d and 4s subshell energies are close enough that electrons from both can be removed or involved in bonding. This supports multiple stable oxidation states.

The 4s electrons are removed before 3d when forming ions, but once the ion forms the relative energies and ligand environment influence stability. Use actual configurations for the species named.

Fe can form Fe²⁺ ([Ar]3d⁶) and Fe³⁺ ([Ar]3d⁵); both are common because the energy difference is not prohibitive.

Do not say 4s always fills and empties independently of 3d or assume every oxidation state is equally stable.

Transition-metal catalysts can provide alternative pathways because they access multiple oxidation states and have energetically accessible vacant or partially occupied d orbitals that form dative bonds to reactants.

A catalyst may bind, activate and release a substrate, or shuttle electrons between oxidation states. It is regenerated overall and lowers activation energy without changing equilibrium.

Fe²⁺/Fe³⁺ can catalyse a redox chain by accepting an electron in one step and donating it in another.

The catalyst is not consumed permanently, and “vacant d orbital” does not mean every d orbital is empty.

A complex ion contains a central metal ion surrounded by ligands. Empty or energetically accessible orbitals on the metal accept lone pairs to form coordinate (dative) bonds.

The metal charge, size and electron configuration influence ligand binding. Count donor atoms and show the overall charge rather than describing the complex as an ordinary ionic lattice.

[Cu(H₂O)₆]²⁺ forms when six water lone pairs coordinate to Cu²⁺.

A dative bond has both electrons from the ligand but is still a covalent bond once formed; it is not a simple ion pair.

Ligands can replace one another around a metal ion. Water, ammonia, hydroxide, chloride and other ligands donate lone pairs and alter charge, colour and geometry.

Balance the metal, ligands and charge in the equation. Excess ligand or precipitating hydroxide can drive a substitution and may produce a different coordination number.

Cu²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺ changes the ligand environment and colour; adding OH⁻ can instead form Cu(OH)₂ precipitate.

A colour change does not by itself prove oxidation-state change; ligand substitution can change colour while Cu remains +2.

A ligand is an ion or molecule with at least one lone pair that forms a dative covalent bond to a metal atom or ion. The donor atom is the atom directly attached to the metal.

Water and ammonia donate oxygen or nitrogen lone pairs; chloride and cyanide donate anionic lone pairs. The ligand’s charge contributes to the complex charge but does not change the donor definition.

In [Co(NH₃)₆]³⁺, each NH₃ donates one nitrogen lone pair and the six ligands are neutral.

A ligand is not any surrounding ion, and a coordinate bond is not formed by the metal donating both electrons.

A monodentate ligand binds through one donor atom, a bidentate ligand through two, and a polydentate ligand through several donor atoms.

H₂O, NH₃, Cl⁻ and CN⁻ are monodentate; en and ethanedioate are bidentate; EDTA⁴⁻ is polydentate. Denticity affects coordination number and chelate stability.

One en ligand occupies two coordination sites, so three en ligands can surround a metal in an octahedral complex.

Denticity is not the same as the ligand’s charge or the number of ligands written in the formula.

A complex is a molecule or ion in which a central metal atom or ion is bonded to surrounding ligands through coordinate bonds. The bracketed species has its own overall charge.

Separate the complex ion from counter-ions outside the brackets. Count ligand donor atoms to determine coordination number, then check total charge.

In [CoCl₄]²⁻, Co is central, four chloride ligands coordinate, and the bracketed species carries −2 charge.

The counter-ions are not ligands inside the complex, and coordination number is not always equal to the number of written ligand molecules for polydentate ligands.

Common complex geometries are linear (2 sites, 180°), square planar (4, 90°), tetrahedral (4, about 109.5°) and octahedral (6, 90° between adjacent sites).

Geometry depends on coordination number, ligand size, metal and electronic configuration. Use the stated complex rather than assigning a shape from charge alone.

[Ag(NH₃)₂]⁺ is linear; [PtCl₄]²⁻ square planar; [CoCl₄]²⁻ tetrahedral; [Cu(H₂O)₆]²⁺ octahedral (often distorted).

Four-coordinate does not uniquely mean tetrahedral—square-planar complexes are also possible.

Coordination number is the number of donor atoms directly bonded to the central metal, not always the number of ligand molecules. The complex charge is metal charge plus all ligand charges.

Choose ligands to fill the stated coordination number, then add charges algebraically. Denticity matters: one bidentate ligand occupies two sites.

A +2 metal with four neutral NH₃ ligands gives [M(NH₃)₄]²⁺; a +3 metal with two oxalate ions (2− each) gives [M(C₂O₄)₂]⁻.

Do not use ligand count as coordination number for en or EDTA, and do not confuse oxidation state with overall complex charge.

Ligand exchange is a substitution at a metal centre: one ligand leaves as another donates a lone pair. The metal oxidation state may stay the same while colour and geometry change.

Use the ligand and stoichiometry to balance the equation. Copper(II) and cobalt(II) complexes show exchange with water, ammonia, hydroxide or chloride.

Adding excess NH₃ to [Cu(H₂O)₆]²⁺ forms a deep-blue ammine complex; adding OH⁻ can produce Cu(OH)₂ instead. The observations depend on ligand amount and medium.

Ligand exchange is not automatically redox and a colour change does not prove the metal changed oxidation state.

Write the relevant reduction half-equations, select the more positive reduction as the cathode, and calculate E°cell = E°cathode − E°anode. A positive value supports the written standard direction.

Transition-metal ions can act as oxidising or reducing agents depending on their pair and oxidation state. Balance electrons after choosing the direction.

If MnO₄⁻/Mn²⁺ is more positive than Fe³⁺/Fe²⁺ in acid, permanganate can oxidise Fe²⁺ to Fe³⁺ while being reduced.

Do not rank ions without specifying the half-equations or assume a reaction is feasible under non-standard concentrations from E° alone.

In acid, MnO₄⁻ is reduced to Mn²⁺; oxalate, Fe²⁺ or I⁻ are oxidised according to their half-equations. Combine half-equations so electrons, atoms and charge balance.

Use stoichiometric ratios to calculate titres or amounts. For Cu²⁺ + I⁻, iodine forms and can be titrated with thiosulfate in the full analytical method.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O; Fe²⁺ → Fe³⁺ + e⁻, so one permanganate reacts with five Fe²⁺.

Do not balance oxygen with OH⁻ in an acidic equation without converting correctly, and do not ignore the reaction medium.

For any supplied redox pair, identify oxidation and reduction, balance each half-equation, cancel electrons and use the resulting mole ratio for calculations.

Use E° or the stated conditions to choose direction, then verify mass and charge conservation. Unfamiliar species do not require a new method.

If X²⁺ + 2e⁻ → X and Y → Y³⁺ + 3e⁻, the least common electron number is six, giving 3X²⁺ + 2Y → 3X + 2Y³⁺.

Do not force a familiar coefficient pattern onto a new oxidation state or calculate mass before finding the electron ratio.

Degenerate orbitals have the same energy. In an isolated transition-metal ion the five d orbitals are degenerate; surrounding ligands create an electric field that splits them into non-degenerate energy levels.

The splitting pattern depends on geometry: octahedral ligands approach along axes, while tetrahedral approaches lie between axes. The energy gap affects colour and magnetic behaviour.

In an octahedral complex, dxy, dxz and dyz form one set and dz², dx²−y² another, with different energies.

Degenerate does not mean identical shapes, and ligand-field splitting is not caused by changing the principal quantum number.

Ligand-field splitting removes the equality of the five d-orbital energies. In an octahedral complex, two orbitals point towards ligands and rise in energy while three lie between axes and are lower; in a tetrahedral complex the pattern is reversed, with three higher and two lower.

The separation is ΔE. It is not a new electron shell: it is an energy difference within the same d subshell, created by the ligand arrangement around the metal.

For an octahedral ion, the dxy, dxz and dyz orbitals form the lower set, while dz² and dx²−y² form the upper set. A tetrahedral diagram must show the opposite 3:2 ordering.

Do not copy the octahedral diagram for a tetrahedral complex, or treat “higher” and “lower” as absolute energies independent of the complex.

A transition-metal complex can absorb visible light when a d electron is promoted from a lower split d level to a higher one. The absorbed photon has energy ΔE = hf, so its frequency is set by the size of the splitting.

The colour seen is not the colour absorbed: it is the complementary mixture of wavelengths that pass through or are reflected. A transition is possible only when the metal has an appropriate partially filled d set.

If a complex absorbs mainly orange-red light, the transmitted or observed colour is toward the blue-green complement. Changing ligands can change ΔE and therefore shift the observed colour.

“The d orbitals emit the colour” is the wrong model for the syllabus explanation; the key event is selective absorption during promotion between non-degenerate levels.

Different ligands create different electric fields around the same metal ion. That changes ΔE; because ΔE = hf, it changes the frequency and wavelength of light absorbed, so the complementary colour observed can change.

Compare complexes only after holding the metal oxidation state and geometry in view. A stronger ligand-field effect gives a larger splitting and absorption at higher frequency (shorter wavelength).

Replacing water ligands around a metal with ammonia can alter the splitting and shift the absorption band. The solution may therefore change colour even though the central metal and its oxidation state are unchanged.

Do not say a ligand “has a colour that it transfers” to the complex. The observed colour comes from the new energy gap and selective absorption.

When a ligand is replaced at a transition-metal centre, the geometry and ligand field can change. The new ΔE changes which visible wavelength is absorbed, so ligand exchange can produce a different solution colour.

Use the ligand, stoichiometry and medium to identify the complex actually present. Water, ammonia, hydroxide and chloride provide contrasting examples for Cu(II) and Co(II).

Adding excess ammonia to a hydrated Cu(II) solution produces a deep-blue ammine complex; hydroxide can instead precipitate Cu(OH)₂. These are different chemical outcomes, not simply “more blue” of the same species.

A colour change does not by itself prove redox. Ligand exchange can leave the oxidation state at +2 while changing coordination, geometry and absorption.

Stereoisomers have the same connectivity but different three-dimensional arrangements. Complexes may show geometrical cis/trans isomerism, or optical isomerism when a pair of arrangements are non-superimposable mirror images.

For cis/trans isomers, identical ligands are adjacent (cis) or opposite (trans). Bidentate ligands can create chirality by wrapping around the metal, so the mirror image cannot be rotated onto the original.

Square-planar [Pt(NH₃)₂Cl₂] has cis and trans forms. Octahedral [Ni(en)₃]²⁺ is a standard optical-isomer example; [Co(NH₃)₄(H₂O)₂]²⁺ illustrates octahedral cis/trans arrangements.

Different colours or conformations are not automatically stereoisomers. First check whether connectivity is unchanged and whether the spatial arrangements are genuinely non-superimposable.

A complex is overall polar when its bond or ligand dipoles do not cancel as vectors. The metal–ligand bonds may be polar individually, but symmetry and three-dimensional arrangement determine the net dipole.

Draw or inspect the geometry before deciding. Opposite equal dipoles can cancel in a symmetric arrangement; cis/trans changes or different ligands can destroy that cancellation.

A trans square-planar complex with matching opposite ligands can have cancellation that its cis isomer lacks. An octahedral complex with six identical ligands is more symmetric than one containing different ligand types.

Do not infer polarity from the overall ionic charge or from one polar bond. Charge and molecular dipole are different properties, and a charged complex can still have zero net dipole.

The stability constant, Kstab, is the equilibrium constant for forming a complex ion from its constituent metal ion and ligands in a specified solvent.

A larger Kstab means the equilibrium lies further towards the coordinated complex under the stated conditions. It is an equilibrium measure, not a statement that formation is infinitely fast.

For M²⁺ + 4NH₃ ⇌ [M(NH₃)₄]²⁺, Kstab compares the equilibrium amount of the ammine complex with the free metal ion and ammonia concentrations.

Do not confuse Kstab with a rate constant or with the overall charge of the complex; a stable complex can still undergo ligand exchange.

For a formation equation, Kstab is the concentration of the complex divided by the concentrations of the free metal ion and ligands, each raised to its stoichiometric coefficient.

Write the balanced equilibrium first, then omit pure solids, pure liquids and the solvent. Water is therefore not included when it is the solvent.

For M²⁺ + 4NH₃ ⇌ [M(NH₃)₄]²⁺, Kstab = [[M(NH₃)₄]²⁺]/([M²⁺][NH₃]⁴).

Do not put ligand count in front of a concentration, forget the power of four, or include [H₂O] merely because water appears in the hydrated starting ion.

A Kstab calculation links equilibrium concentrations through the formation expression. Rearrange the expression only after identifying the stoichiometry of the complex.

Substitute concentrations with consistent units, keep powers attached to the correct ligand, and check that the calculated value is chemically plausible relative to the starting amounts.

If Kstab = [[ML]⁺]/([M²⁺][L⁻]), then [[ML]⁺] = Kstab[M²⁺][L⁻]. For ML₂, the ligand concentration is squared instead.

A large Kstab does not mean every metal is complexed regardless of concentration; dilution and competing ligands still change the equilibrium composition.

When two ligands compete for a metal ion, the larger formation constant corresponds to the more thermodynamically favoured complex under comparable conditions.

Write the competing formation or exchange equation and combine Kstab expressions if needed. The resulting equilibrium constant tells which complex is favoured, while the ratio also depends on ligand concentrations.

If [M(NH₃)₄]²⁺ has a larger formation constant than [M(H₂O)₆]²⁺, adding enough ammonia can shift hydrated metal ions towards the ammine complex.

Do not infer that the largest Kstab always gives the largest concentration: a ligand present in tiny concentration may lose the competition.