Transition Elements

Learn and apply Transition Elements in the published Chemistry course sequence.

GCE A-Level H2 Chemistry 9476-2027

Learning goals

Transition Elements: Definition and Electron Configurations
Variable Oxidation States and Redox Systems
Complex Ions, Ligands, Ligand Exchange
d-Orbital Splitting and Colour
Transition-metal Catalysis

What You’ll Learn

Apply the two-branch transition-element definition—an incomplete d subshell in the atom or in at least one cation—and write first-row atom and ion configurations.
Explain the relatively small changes in radius and first ionisation energy, then compare melting point and density with calcium.
Predict variable oxidation states and use the Fe³⁺/Fe²⁺, MnO₄⁻/Mn²⁺ and Cr₂O₇²⁻/Cr³⁺ redox systems.
Explain copper(II) ligand exchange with water, ammonia and chloride ions, including colour evidence and haemoglobin competition.
Describe octahedral d-orbital splitting and connect visible-light absorption to complex colour.
Explain how transition elements or their compounds provide catalytic pathways.

Transition Elements is prescribed in H2 9476. H1 8873 uses electron configurations and redox reasoning elsewhere but does not include this transition-element, complex-ion and d-orbital-colour sequence.

Use the map as a causal overview: identity and ligand environment set the d-level splitting, which controls absorbed and observed colour.

Lessons (Recommended Order)

Definition and Electron Configurations — classify elements, form ions and explain physical trends.
Variable Oxidation States and Redox Systems — predict oxidation states, balance the named systems and use E⦵ data.
Complex Ions, Ligands and Ligand Exchange — follow the prescribed copper(II) and haemoglobin exchanges.
d-Orbital Splitting and Colour — connect octahedral orientation, absorption and observed colour.
Transition-metal Catalysis — distinguish surface and solution pathways and prove regeneration.

On a first pass, keep this order: configuration supplies oxidation-state and d-electron evidence; those ideas then support redox, complexes, colour and catalysis.

Quick Reference

Evidence in the question	First move	What to check
“transition element?”	inspect the atom and any cation supplied	either may qualify if its d subshell is incomplete; d⁰/d¹⁰ does not qualify that species
first-row cation	write the atom, then remove 4s before 3d	retain the Cr/Cu atom exceptions
trends across the series	balance rising nuclear charge against added 3d shielding	say relatively invariant, not constant
oxidation-state prediction	remove accessible 4s then 3d electrons	label it a prediction unless chemistry is supplied
acidic permanganate/dichromate	write the memorised reduction half-equation	include H⁺, H₂O and colour evidence
two E⦵ values	more positive couple is reduced	positive E⦵_cell is a standard-state prediction
copper ligand exchange	write complete bracketed complexes	conserve ligand count, atoms and charge
colour explanation	octahedral approach → splitting → absorption	observed colour is complementary
catalytic cycle	add the steps	catalyst must cancel and be regenerated

The 9476 colour model requires octahedral d-orbital orientation, but not memorisation of the relative order of ligand field strengths.

Common Exam Traps

Defining a transition element as any d-block element, or using an ion-only shortcut that wrongly excludes scandium. Test the atom and any cation supplied.
Removing 3d electrons before 4s when forming a cation.
Saying radius or first ionisation energy is constant across the series rather than relatively invariant.
Explaining high density only with “strong bonding”; density also depends on mass per volume.
Predicting an oxidation state as guaranteed without considering supplied chemical evidence.
Using acidic permanganate or dichromate equations without H⁺ and H₂O.
Treating a positive E⦵_cell as proof of a fast reaction.
Calling precipitation or redox a ligand exchange without tracking the central metal’s oxidation state and ligands.
Memorising non-prescribed complex geometries or ligand-strength rankings as though they were required.
Saying a complex reflects its own colour without describing selective absorption and a d–d transition.
Claiming a catalyst increases equilibrium yield rather than the rates of both directions.

Hub Quiz

Complete the Transition Elements hub quiz after all five lessons. Then use structured Transition Elements practice for longer redox, colour, ligand-exchange and catalysis explanations.

Practise

Work through questions with marking and feedback as you learn.

About 10 minutes

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Which set of electronic configurations is correct?

Cr: [Ar]3d⁵4s¹; Cu: [Ar]3d¹⁰4s¹; Cu²⁺: [Ar]3d⁹
Cr: [Ar]3d⁴4s²; Cu: [Ar]3d⁹4s²; Cu²⁺: [Ar]3d⁹
Cr: [Ar]3d⁵4s¹; Cu: [Ar]3d¹⁰4s¹; Cu²⁺: [Ar]3d⁸4s¹
Cr: [Ar]3d⁵4s¹; Cu: [Ar]3d¹⁰4s¹; Cu²⁺: [Ar]3d¹⁰

E⦵ values: [Co(H₂O)₆]³⁺/[Co(H₂O)₆]²⁺ +1.82 V; [Co(NH₃)₆]³⁺/[Co(NH₃)₆]²⁺ +0.10 V; O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O +1.23 V; O₂ + 2H₂O + 4e⁻ ⇌ 4OH⁻ +0.40 V. Which statement is correct?

In ammoniacal (alkaline) solution, air oxidises [Co(NH₃)₆]²⁺ (E⦵cell = +0.30 V), but in acidic solution it does not oxidise [Co(H₂O)₆]²⁺ (E⦵cell = −0.59 V); the ligand changes the relative stability of cobalt(II) and cobalt(III).
Air oxidises both cobalt(II) complexes, because oxygen is a strong oxidising agent.
Air oxidises neither complex, because E⦵cell for the ammine complex is −0.30 V.
Air oxidises [Co(H₂O)₆]²⁺ easily, with E⦵cell = +3.05 V.

In haemoglobin, O₂ is bonded as a ligand to Fe²⁺. Why is carbon monoxide poisonous, and how does breathing pure oxygen help a patient?

CO replaces O₂ as the ligand on Fe²⁺ because it bonds much more strongly, so less oxygen is carried; a high concentration of O₂ shifts the exchange equilibrium back towards oxyhaemoglobin.
CO oxidises Fe²⁺ to Fe³⁺, which cannot bond to O₂.
CO bonds to Fe²⁺ more weakly than O₂ does, but it blocks the lungs.
Breathing pure oxygen cannot help, because ligand exchange is irreversible.

Aqueous solutions containing [Zn(H₂O)₆]²⁺ and [Sc(H₂O)₆]³⁺ are colourless. Which explanation is correct?

Zn²⁺ has a full d subshell and Sc³⁺ an empty one, so no electron can move between the split d orbitals, even though the ligands split them.
Water ligands do not split the d orbitals of these ions.
The ions absorb all wavelengths of visible light equally.
Zinc and scandium atoms contain no d electrons.

How does iron act as the catalyst in the Haber process?

N₂ and H₂ are adsorbed onto the iron surface, using its partially filled 3d orbitals; this weakens the N≡N and H–H bonds, the atoms react on the surface, and NH₃ desorbs, freeing the active sites.
Fe²⁺ and Fe³⁺ ions in solution transfer electrons to N₂ molecules.
Iron adsorbs N₂ so strongly that it stays bound, removing N₂ from the equilibrium.
Iron shifts the equilibrium to the right, increasing the yield of ammonia.

Practise after feedback

After a check, practise the skills it showed you need to work on.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

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No completed attempts are saved yet.

This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Which set of electronic configurations is correct?

Cr: [Ar]3d⁵4s¹; Cu: [Ar]3d¹⁰4s¹; Cu²⁺: [Ar]3d⁹
Cr: [Ar]3d⁴4s²; Cu: [Ar]3d⁹4s²; Cu²⁺: [Ar]3d⁹
Cr: [Ar]3d⁵4s¹; Cu: [Ar]3d¹⁰4s¹; Cu²⁺: [Ar]3d⁸4s¹
Cr: [Ar]3d⁵4s¹; Cu: [Ar]3d¹⁰4s¹; Cu²⁺: [Ar]3d¹⁰

In ammoniacal (alkaline) solution, air oxidises [Co(NH₃)₆]²⁺ (E⦵cell = +0.30 V), but in acidic solution it does not oxidise [Co(H₂O)₆]²⁺ (E⦵cell = −0.59 V); the ligand changes the relative stability of cobalt(II) and cobalt(III).
Air oxidises both cobalt(II) complexes, because oxygen is a strong oxidising agent.
Air oxidises neither complex, because E⦵cell for the ammine complex is −0.30 V.
Air oxidises [Co(H₂O)₆]²⁺ easily, with E⦵cell = +3.05 V.

In haemoglobin, O₂ is bonded as a ligand to Fe²⁺. Why is carbon monoxide poisonous, and how does breathing pure oxygen help a patient?

CO replaces O₂ as the ligand on Fe²⁺ because it bonds much more strongly, so less oxygen is carried; a high concentration of O₂ shifts the exchange equilibrium back towards oxyhaemoglobin.
CO oxidises Fe²⁺ to Fe³⁺, which cannot bond to O₂.
CO bonds to Fe²⁺ more weakly than O₂ does, but it blocks the lungs.
Breathing pure oxygen cannot help, because ligand exchange is irreversible.

Aqueous solutions containing [Zn(H₂O)₆]²⁺ and [Sc(H₂O)₆]³⁺ are colourless. Which explanation is correct?

Zn²⁺ has a full d subshell and Sc³⁺ an empty one, so no electron can move between the split d orbitals, even though the ligands split them.
Water ligands do not split the d orbitals of these ions.
The ions absorb all wavelengths of visible light equally.
Zinc and scandium atoms contain no d electrons.

How does iron act as the catalyst in the Haber process?

N₂ and H₂ are adsorbed onto the iron surface, using its partially filled 3d orbitals; this weakens the N≡N and H–H bonds, the atoms react on the surface, and NH₃ desorbs, freeing the active sites.
Fe²⁺ and Fe³⁺ ions in solution transfer electrons to N₂ molecules.
Iron adsorbs N₂ so strongly that it stays bound, removing N₂ from the equilibrium.
Iron shifts the equilibrium to the right, increasing the yield of ammonia.

Check what I know

Start here to see which parts you already know.

About 8 minutes

Answer 5 short questions. It shows what to work on next and doesn't count towards mastery.

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Which statement about iron, zinc and scandium is correct?

Fe²⁺ is [Ar]3d⁶, because the 4s electrons are lost before the 3d electrons; zinc is not a transition element, because neither its atom nor its Zn²⁺ ion has an incomplete d subshell.
Fe²⁺ is [Ar]3d⁴4s², because the 3d electrons are lost first; zinc is not a transition element.
Fe²⁺ is [Ar]3d⁶; zinc is a transition element because it is in the d-block.
Fe²⁺ is [Ar]3d⁶; scandium is not a transition element, because Sc³⁺ has no d electrons.

E⦵ values: VO₂⁺/VO²⁺ +1.00 V; VO²⁺/V³⁺ +0.34 V; V³⁺/V²⁺ −0.26 V; Sn²⁺/Sn −0.14 V; Zn²⁺/Zn −0.76 V. An acidified solution containing VO₂⁺ is shaken with excess tin. What is the final vanadium species?

V³⁺
V²⁺
VO²⁺
Vanadium metal

Concentrated hydrochloric acid is added to aqueous copper(II) sulfate. Which statement is correct?

[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O: the blue solution turns yellow-green, and the coordination number falls from 6 to 4 because Cl⁻ ligands are larger than H₂O.
[Cu(H₂O)₆]²⁺ + 6Cl⁻ ⇌ [CuCl₆]⁴⁻, with the coordination number staying at 6.
HCl reduces Cu²⁺ to Cu⁺, and the solution turns colourless.
[CuCl₄]²⁻ forms, and adding water afterwards cannot restore the blue colour, because ligand exchange is irreversible.

Aqueous [Ti(H₂O)₆]³⁺ is purple. Which explanation is correct?

Ti³⁺ has one 3d electron; the water ligands split the 3d orbitals into two sets, and the electron absorbs visible light (mainly yellow-green) to move to the higher set, so the transmitted light looks purple.
The complex absorbs purple light, so the solution looks purple.
The 3d electron is promoted to the 4s orbital, absorbing visible light.
Excited electrons fall back to lower orbitals, emitting purple light.

Fe²⁺ ions catalyse S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂. E⦵ values: S₂O₈²⁻/SO₄²⁻ +2.01 V; Fe³⁺/Fe²⁺ +0.77 V; I₂/I⁻ +0.54 V. Which statement is correct?

S₂O₈²⁻ oxidises Fe²⁺ to Fe³⁺ (E⦵cell = +1.24 V), then Fe³⁺ oxidises I⁻ (E⦵cell = +0.23 V), regenerating Fe²⁺; each step is between oppositely charged ions, so it is faster than the direct reaction between two anions.
Fe²⁺ adsorbs S₂O₈²⁻ and I⁻ onto its surface, weakening their bonds.
S₂O₈²⁻ oxidises Fe²⁺ with E⦵cell = +2.78 V.
Fe²⁺ increases the equilibrium yield of I₂.

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Connections and targeted revision

Transition Elements consolidates The Periodic Table and Electrochemistry, then supplies inorganic evidence used in Practical Skills and Qualitative Analysis.

If an error persists, return to the matching lesson:

definition, configuration or physical trend → Foundations;
oxidation state, half-equation or E⦵ direction → Redox Systems;
ligand identity, exchange equation or haemoglobin → Complex Ions;
orbital orientation, absorbed wavelength or hue → Colour;
surface steps, intermediate or regeneration → Catalysis.

Transition Elements

What You’ll Learn

Lessons (Recommended Order)

Quick Reference

Hub Quiz

Practise

Practice complete

Your answers

Practise after feedback

Practice complete

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Check what I know

Starting check complete

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Check my progress

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Check again

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Review

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Connections and targeted revision