Part 1: Electronic structure, oxidation states and colour

Distinguish d-block position from the transition-metal definition, write configurations correctly and use ligand-induced d-orbital splitting to explain colour.

Remove 4s electrons before 3d electrons

The Period 4 d-block runs from scandium to zinc. Write the atomic configuration from the atomic number, then remove 4s electrons first when forming the usual positive ions. Chromium and copper are atomic exceptions to the simple filling pattern: Cr is [Ar]3d⁵4s¹ and Cu is [Ar]3d¹⁰4s¹. Orbital energies depend on electron occupation, so ‘4s fills first’ does not imply that it is retained during ion formation.

Worked example: chromium has Z = 24. Cr has six electrons beyond [Ar], arranged 3d⁵4s¹. Cr²⁺ loses the 4s electron and one 3d electron, becoming [Ar]3d⁴. Cr³⁺ loses one more, becoming [Ar]3d³. Check totals: 18 + 4 = 22 = 24 − 2 for Cr²⁺. If asked for a full configuration, expand [Ar] to 1s²2s²2p⁶3s²3p⁶.

An incomplete d subshell in a stable ion is the defining feature

A transition metal is a d-block element that forms one or more stable ions with an incompletely filled d subshell. ‘Incomplete’ means partially filled: d¹ through d⁹. Sc³⁺ is d⁰ and Zn²⁺ is d¹⁰, so scandium and zinc do not meet the course definition. Copper does because Cu²⁺ is d⁹, even though Cu⁺ is d¹⁰.

Coloured compounds, variable oxidation states, complexes and catalytic activity are characteristic behaviours, but they are not substitutes for the definition. A metal can form a colourless ion and still be a transition metal; the criterion needs only one suitable stable ion. Conversely, making a complex alone does not prove an element is a transition metal.

Several oxidation states are energetically accessible

The 3d and 4s electrons have sufficiently similar energies that different numbers can participate in bonding or be removed. Energy released by forming bonds, hydration or a crystal lattice can compensate for different ionisation-energy costs. Consequently several oxidation states can be stable in different chemical environments. This is more informative than saying transition metals ‘lose any number of electrons’.

Iron commonly changes between +2 and +3, copper between +1 and +2, and vanadium has important +2, +3, +4 and +5 states. A high oxidation state in an oxoanion is not a bare aqueous cation of that charge: Mn in MnO₄⁻ is +7, but the ion’s total charge is −1. Use oxygen at −2 and the total ion charge to deduce the metal oxidation number.

Worked check: in Cr₂O₇²⁻, 2x + 7(−2) = −2, so x = +6. In [Co(NH₃)₆]³⁺ the neutral ammonia ligands contribute zero, so cobalt is +3. Changes in ligand identity can alter stability without changing oxidation number; always calculate the number rather than inferring it from colour.

Ligands split d orbitals into different energies

The five d orbitals of an isolated ion have the same energy. Ligands approach in particular directions and their electron pairs interact differently with differently oriented d orbitals. In an octahedral complex the d orbitals split into a lower group of three and a higher group of two. Detailed orbital names and high-spin/low-spin arrangements are not required here; the energy separation is the key model.

If there is a suitable occupied lower d orbital and an available higher d orbital, an electron can absorb a photon whose energy matches the separation and become excited. ΔE = hf = hc/λ. If this absorption lies in the visible range, some wavelengths are removed from white light. The transmitted or reflected light gives the observed colour, which is complementary to the absorbed region. A blue solution is not blue because it absorbs blue light.

Changing the metal’s oxidation number, the ligand or the coordination number can change d-orbital splitting, so a different wavelength is absorbed and the observed colour changes. Colour is evidence about the electronic environment, not proof of which one of these changes occurred. In the chloride substitutions of Cu²⁺ and Co²⁺, the oxidation state stays +2 even though the colour and shape change.

Swipe horizontally to view the whole diagram.

Explain colourless ions without overgeneralising

Sc³⁺ has no d electrons to excite; Zn²⁺ and Cu⁺ have full d subshells without an available higher d state within the split d set. They therefore cannot give the usual visible d–d absorption described above and their simple aqueous ions are colourless. A d-electron transition must also have an energy in the visible range to produce a visible colour.

The d–d explanation is not the only way a compound can absorb light. MnO₄⁻ is intensely purple despite manganese(VII) being formally d⁰; ligand-to-metal charge-transfer absorption explains this exception at a more advanced level. This extension prevents the incorrect universal claim ‘every d⁰ compound is colourless’. Likewise, the colour of a precipitate or mixed solution need not be the pure hexaaqua-ion colour.

For typical solutions remember Cu²⁺ blue, Fe²⁺ pale green and Co²⁺ pink. A pure chromium(III) hexaaqua ion is violet, although aqueous chromium(III) mixtures commonly appear green because other coordinated species are present. Iron(III) solutions commonly appear yellow to brown through hydrolysis/speciation. Record the observed shade and conditions rather than forcing every solution into one idealised colour.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources