28. Chemistry A2 of transition elements

Classify the element by the electron configuration of its stable ion or ions, not by the neutral atom alone. Only one stable ion with a partly filled d subshell is required.

Put the nucleus at the origin and label the relevant axes. For 3dxy the axes pass between the lobes; for 3dz² the two main lobes lie on the z axis and the ring lies perpendicular to it.

These behaviours arise because 3d and 4s levels are close enough in energy for several electron arrangements and bonding interactions to be accessible.

When a first-row transition metal forms a positive ion, its 4s electrons are removed before its 3d electrons, even though 4s fills before 3d in the neutral atom.

In either route, the catalyst participates in intermediate steps but is regenerated by the end. The alternative pathway has a lower activation energy, so a larger fraction of collisions is successful at the same temperature.

Water is a neutral ligand, so the complex charge is the Ti³⁺ charge plus six zeros: +3. Once a dative bond has formed, it behaves as a covalent bond; its name records the origin of the shared electron pair.

For each product, conserve the metal and ligands and add charges algebraically. Neutral H₂O/NH₃ retain the metal's +2 overall charge; two OH⁻ give a neutral complex; four Cl⁻ give an overall 2− charge.

Three bidentate en ligands provide six donor atoms, so [M(en)₃]ⁿ⁺ has coordination number 6 even though only three ligand molecules are present.

First count coordinated donor atoms, then use the identity of the stated complex: coordination number 4 can be either tetrahedral or square planar, so donor count alone does not decide between them.

M²⁺ with coordination number 4 and four neutral NH₃ ligands gives [M(NH₃)₄]²⁺. M³⁺ with coordination number 4 and two bidentate C₂O₄²⁻ ligands gives [M(C₂O₄)₂]⁻ because +3 + 2(−2) = −1.

Exchange occurs when an entering ligand donates a lone pair and replaces one or more existing ligands. Larger Cl⁻ favours four-coordinate tetrahedral products, whereas small H₂O/NH₃ commonly give six-coordinate octahedral environments.

For Fe³⁺/Fe²⁺, +0.77 V, and Cu²⁺/Cu⁺, +0.15 V: Fe³⁺ is reduced and Cu⁺ is oxidised, so E°cell = 0.77 − 0.15 = +0.62 V.

Convert the known concentration and volume to moles, apply the coefficient ratio from the balanced equation, then convert the target moles to the requested concentration, mass or percentage. Track any aliquot or dilution factor separately.

If X²⁺ + 2e⁻ → X and Y → Y³⁺ + 3e⁻, the least common electron count is 6. The combined ratio is 3X²⁺ : 2Y, giving 3X²⁺ + 2Y → 3X + 2Y³⁺.

Equal energy does not mean equal shape or orientation. The five d orbitals can have different spatial orientations while remaining degenerate before the ligand field is applied.

Orbitals directed more toward the approaching ligand lone pairs experience greater repulsion and lie higher in energy. Octahedral ligands approach along the axes; tetrahedral ligands approach between the axes, so the 2:3 ordering reverses.

A valid text-first energy diagram has energy increasing upward, two horizontal levels separated by a labelled ΔE, and the correct 2/3 degeneracy on each level for the stated geometry.

For a fair ligand comparison, keep track of metal identity, oxidation state and geometry as well as ligand identity; each can affect the splitting. A ligand exchange can change colour while the metal remains Cu(II) or Co(II).

The incoming ligand changes the ligand field and sometimes coordination geometry, changing ΔE and the light absorbed. Adding water to concentrated chloride equilibria favours the hexaaqua complex and reverses the characteristic colour shift.

Each en ligand, H₂NCH₂CH₂NH₂, uses two nitrogen donor atoms and therefore occupies two adjacent octahedral sites. Keep the two donor atoms joined as one chelate when constructing or comparing structures.

To test a proposed pair: keep the formula and metal–ligand bonds unchanged; place all six octahedral or four square-planar sites; mark cis/trans by 90°/180° separation; for optical isomers reflect the entire chelate arrangement and check that no rotation superimposes it.

Polarity, ionic charge and chirality answer different questions. Charge is the algebraic total on the complex; polarity is a vector resultant; chirality asks whether the mirror image is superimposable. One does not determine either of the others.

A larger Kstab means the formation equilibrium lies further toward the complex under the stated conditions, so the complex is thermodynamically more stable relative to the separated constituents used in that definition.

Write the balanced formation equation first. Put the complex concentration in the numerator and each dissolved constituent concentration in the denominator, raised to its stoichiometric coefficient. Omit H₂O when water is the solvent because its activity is effectively constant.

Derive units from the final expression rather than memorising one unit. In the copper example, one concentration divided by five concentration factors gives (mol dm⁻³)⁻⁴ = dm¹² mol⁻⁴.

Use the balanced coefficients to construct concentration changes, calculate every equilibrium concentration, write the expression with water omitted, substitute equilibrium—not initial—values, then derive units and interpret the magnitude.

Here Kstab = 40, so products are favoured for this step under the stated conditions, although appreciable reactant remains. A large value does not mean the equilibrium goes literally to completion.

Adding sufficient NH₃ therefore favours replacement of chloride by ammonia because formation of the ammine complex has the larger Kstab. The actual equilibrium mixture still depends on incoming and outgoing ligand concentrations through the reaction quotient.