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What makes a transition element?

Part of the UNEB UACE (Uganda) study roadmap. Chemistry topic chemis-004 of Chemistry.

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What makes a transition element?

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Transition elements (d-block) have a partially filled (n−1)d subshell in the ground state or in a common ion; Zn²⁺ (d¹⁰) is the classic exception often re-tested.

Their five signature properties — variable oxidation states, coloured complex ions, catalytic activity, magnetic behaviour, and complex formation — all trace back to how easily d-electrons are lost or excited.

  • Spin-only magnetic moment: μ_so = √(n(n+2)) BM, where n = unpaired electrons.
  • Colour rule: d¹⁰ (Zn²⁺), d⁰ (Sc³⁺, Ti⁴⁺) → colourless; intermediate dn → coloured.
  • MnO₄⁻ / Cr₂O₇²⁻ oxidising power changes with pH (E° = +1.51 V and +1.33 V respectively in acid).

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

What makes a transition element?

A d-block element is a transition element when it (or one of its common ions) has an incomplete d-subshell. Zn and Cu sit on the boundary: Zn²⁺ is d¹⁰ and therefore not a true transition ion, while Cu²⁺ (d⁹) is.

Why variable oxidation states?

The (n−1)d and ns orbitals are very close in energy, so different numbers of d-electrons can be removed with little energetic cost. Manganese for instance spans +2, +4, +6, +7, and chromium shows +3, +6 most commonly.

Complex ions and coordination number

Ligands donate lone pairs into vacant hybrid orbitals on the metal. The most frequent coordination numbers are 6 (octahedral) and 4 (tetrahedral or square-planar).

Common ionGeometryColour
[Cr(NH₃)₆]³⁺octahedralviolet
[Cu(NH₃)₄]²⁺square-planardeep blue
[Fe(H₂O)₆]²⁺octahedralpale green
[Zn(OH)₄]²⁻tetrahedralcolourless

Colour, magnetism and catalysis

Colour comes from d–d transitions; the wavelength absorbed depends on Δ_o, which is set by the spectrochemical series (I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < CN⁻). Magnetic moment is read directly from the count of unpaired electrons using the spin-only formula. Catalysis works because the metal can cycle between oxidation states, supplying a low-energy pathway (heterogeneous, e.g. Fe in the Haber process) or acting in solution (homogeneous, e.g. MnO₄⁻ titrations).

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

pH-dependent redox interconversion

Dichromate(VI) and chromate(VI) interconvert on pH change:

2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O

In alkaline solution Cr³⁺ is oxidised to yellow CrO₄²⁻; in acid it becomes orange Cr₂O₇²⁻. Failing to flip the equation when pH changes is the most common half-equation error.

Worked micro-example — spin-only moment

For [Fe(H₂O)₆]²⁺ (Fe²⁺, d⁶, high-spin in water), n = 4 unpaired electrons.

μ_so = √(4 × 6) = √24 ≈ 4.90 BM, matching literature ≈ 5.0 BM once orbital contribution is added. Predict the moment for [Ni(CN)₄]²⁻: strong-field CN⁻ gives a low-spin square-planar d⁸ with 0 unpaired electrons → 0 BM (diamagnetic).

Identification tests and pitfalls

NH₄OH added dropwise to Cu²⁺ gives a pale-blue Cu(OH)₂ precipitate dissolving in excess to [Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue). Adding OH⁻ to Zn²⁺ gives amphoteric Zn(OH)₂ which redissolves in excess forming [Zn(OH)₄]²⁻ — a favourite UACE Paper 1 structured item.

Common mistakes

  • Writing Cr(VI) as CrO₄²⁻ in acid (should be Cr₂O₇²⁻).
  • Citing Cu²⁺ as [Cu(NH₃)₆]²⁺ instead of the square-planar tetraammine.
  • Forgetting that d⁰ and d¹⁰ species are colourless — no d–d transitions exist.
  • Treating all catalysts like enzymes; heterogeneous and homogeneous require different justifications.

Practice prompts

  1. Explain why [Co(H₂O)₆]²⁺ is pink but [CoCl₄]²⁻ is blue, using Δ_o and the spectrochemical series.
  2. Balance MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ in acidic solution and calculate the volume of 0.02 M KMnO₄ needed to react with 25.0 cm³ of 0.10 M Na₂C₂O₄.

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