Must-memorise formulas

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Transition elements are d-block metals whose atoms (or common ions) have partially filled d-orbitals, giving them characteristic properties: variable oxidation states, coloured complex ions, paramagnetism, and catalytic activity. The lanthanide contraction is the gradual decrease in Ln³⁺ ionic radii from La³⁺ (103 pm) to Lu³⁺ (86.1 pm) caused by poor 4f-electron shielding, which makes 5d elements (Hf vs Zr) unusually similar in size.

Property	Cause	Typical Question
Colour of complexes	d–d transitions within crystal field splitting Δₒ	Why is [Ti(H₂O)₆]³⁺ purple?
Paramagnetism	Unpaired d-electrons (n)	Calculate μ for Fe²⁺ (high-spin)
Variable oxidation states	Similar successive IE values, d-electron participation	List oxidation states of Mn
Catalysis	Variable oxidation states + surface adsorption	Why is V₂O₅ used in Contact process?

Must-memorise formulas:

Spin-only magnetic moment: μ = √(n(n+2)) BM (n = unpaired electrons)
Octahedral CFSE: CFSE = (−0.4n_t₂g + 0.6n_eg)Δₒ
Δₒ–λ relation: Δₒ = hc/λ (energy of light absorbed equals splitting energy)

Mnemonic for Cr/Cu anomalies: “Chromium Shifts One Up; Copper Shifts One Down” — Cr is 3d⁵4s¹ not 3d⁴4s²; Cu is 3d¹⁰4s¹ not 3d⁹4s².

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Electronic configuration of the 3d series

Each 3d metal is [Ar] 3dⁿ 4s², except Cr = [Ar] 3d⁵ 4s¹ and Cu = [Ar] 3d¹⁰ 4s¹, where half-filled and fully-filled d-subshells gain extra stability. The 4s electron is removed first during ionisation because 4s penetrates less effectively in the cationic field.

Variable oxidation states

The successive first ionization energies of the 3d series differ by less than ~150 kJ mol⁻¹ across the row, so removing 1, 2, or even 7 electrons (Mn: +2 to +7) costs comparably. The d-electrons participate directly in bonding, allowing oxidation states from Sc(+3) to Mn(+7) in MnO₄⁻. Maximum oxidation state rises to the middle of the series, then falls as the ionisation cost of higher d-electron removal increases.

Colour and crystal field theory

In an octahedral ligand field, the five degenerate d-orbitals split into a lower t₂g set (3 orbitals) and an upper e_g set (2 orbitals) separated by Δₒ. When Δₒ matches the energy of visible light, electrons absorb one colour and the complex transmits the complementary hue.

Spin-only magnetic moment

For a complex with n unpaired electrons: μ = √(n(n+2)) BM. Compare observed μ to predict high-spin vs low-spin geometry — tetrahedral complexes are almost always high-spin because Δ_t ≈ ⁴⁄₉ Δₒ (small splitting relative to pairing energy).

Lanthanide contraction

Across the 4f row, each added proton is poorly shielded by f-electrons, so Z_eff rises gradually. The total contraction (~17 pm from La to Lu) makes post-lanthanide 5d elements (Hf, Ta, W) nearly identical in radius to their 4d predecessors (Zr, Nb, Mo) — explaining why Hf is harder to separate from Zr.

Typical Makerere exam question types

5-mark: “Account for the colour of [Cu(H₂O)₆]²⁺ using crystal field theory.”
10-mark: Compare 3d, 4d, 5d transition series — oxidation state ranges, M–M bonding, low-spin tendencies.
Numerical: “Calculate the spin-only magnetic moment of Fe²⁺ in [Fe(CN)₆]⁴⁻ (low-spin, n = 0).” Answer: μ = 0 BM.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Worked micro-example: CFSE of [Fe(H₂O)₆]²⁺

Fe²⁺ is d⁶. H₂O is a weak-field ligand (early in the spectrochemical series), so [Fe(H₂O)₆]²⁺ is high-spin: t₂g⁴ e_g².

Quantity	Calculation	Result
n_t₂g	4	—
n_eg	2	—
CFSE (in Δₒ)	(−0.4×4) + (0.6×2) = −1.6 + 1.2	−0.4 Δₒ
Pairing energy penalty	2 extra pairs (vs free ion d⁶)	+2P
Unpaired electrons	n = 4	μ = √(4×6) = 4.90 BM

Compare with low-spin [Fe(CN)₆]⁴⁻ (t₂g⁶, n = 0): CFSE = −2.4 Δₒ, μ = 0 — illustrates why CN⁻ > H₂O in the spectrochemical series.

Edge cases and frequent traps

Zn, Cd, Hg are d¹⁰ — no unpaired electrons, no d–d colour, often excluded from “transition elements” in Makerere CEM/BCS marking schemes despite being d-block.
Δ_t = ⁴⁄₉ Δₒ — never write CFSE for a tetrahedral ion using octahedral coefficients.
Catalysis distinction: heterogeneous (Fe in Haber, V₂O₅ in Contact, Pt in catalytic converters — surface adsorption) vs homogeneous (Ni, Pd, Rh organometallics in cross-coupling).
Interstitial carbides/nitrides (TiC, Fe₃C, TiN) are not true stoichiometric salts — small atoms sit in metal lattice voids and confer hardness and metallic conductivity.

Connection to adjacent topics

Builds on atomic structure (Topic 2–3), periodic trends (Topic 4), and chemical bonding; leads directly to coordination chemistry (Topic 12) and organometallic catalysis in Years 2–3 inorganic modules.

Common mistakes at Makerere exams

Confusing Δₒ (energy absorbed) with emitted wavelength — Δₒ corresponds to the wavelength the complex absorbs, not the colour it appears.

Treating “high oxidation state” as automatic for oxidising strength — MnO₄⁻ (Mn +7) is a strong oxidiser; VO₂⁺ (V +5) is much weaker; only E° values decide.

Practice prompts

Calculate the CFSE, in Δₒ, of a high-spin d⁵ octahedral ion and predict whether an additional electron would be oxidised or reduced first.
Explain, with two piece of evidence, why the ionic radius of Zr (4d) is almost the same as Hf (5d), while Nb (4d) and V (3d) differ markedly.

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Must-memorise formulas