Skip to notes
9476 / 2027
Transition Elements

Full chapter

Transition Elements

Connect d-electron configurations to physical trends, variable oxidation states, coloured complexes and catalysis.

A-Level 9476 (2026-2027)

01

Which elements count as transition elements?

Use the d subshell of the atom and its cations, not just its position on the table.

A transition element is a d-block element whose atom has an incomplete d subshell, or which forms a cation with an incomplete d subshell. Incomplete means between one and nine d electrons. The definition tests electron configuration; being coloured or forming a complex is supporting behaviour, not the definition.

Three useful boundary cases
ElementConfiguration evidenceConclusion
ScSc is [Ar]3d14s2; Sc3+ is [Ar].Included by its atom, although its usual +3 ion is d0.
CuCu is [Ar]3d104s1; Cu2+ is [Ar]3d9.Included by the incomplete d subshell in Cu2+.
ZnZn is [Ar]3d104s2; its usual Zn2+ ion is [Ar]3d10.A d-block element, but excluded from the transition elements in this course.

For first-row transition-metal cations, write the neutral atom first, then remove 4s electrons before 3d electrons. The order used to build a neutral-atom configuration is not a universal order of removal. Chromium and copper are the familiar neutral-atom exceptions to a simple 4s2 pattern.

First-row configurations: the [Ar] core is omitted in every entry
ElementAtom after [Ar]Example ion after [Ar]
Sc3d14s2Sc3+: 3d0
Ti3d24s2Ti3+: 3d1
V3d34s2V3+: 3d2
Cr3d54s1Cr3+: 3d3
Mn3d54s2Mn2+: 3d5
Fe3d64s2Fe2+: 3d6; Fe3+: 3d5
Co3d74s2Co2+: 3d7
Ni3d84s2Ni2+: 3d8
Cu3d104s1Cu+: 3d10; Cu2+: 3d9

Check your understanding

Cobalt has proton number 27. What is the configuration of Co3+?

  1. [Ar]3d7, after losing the 4s electrons and no more
  2. [Ar]3d4 4s2, after losing three of the 3d electrons
  3. [Ar]3d6, after losing both 4s electrons and then one 3d
Show the answer

[Ar]3d6, after losing both 4s electrons and then one 3d

Co is [Ar]3d7 4s2. Ions form by losing 4s electrons first, then 3d. 27 - 3 = 24 electrons, [Ar]3d6.

02

Why the physical trends are comparatively gentle

Increasing nuclear attraction is partly offset by added 3d shielding.

From scandium towards copper, nuclear charge rises while electrons are added mainly to 3d. These added electrons partly shield the outer electrons from the increasing nuclear charge, and the atoms do not acquire a new principal shell. The two changes largely offset one another: atomic radii and first ionisation energies vary relatively little compared with a full main-group period. Radii generally contract modestly and first ionisation energies show small, irregular changes; neither quantity is exactly constant.

Qualitative comparison with calcium
PropertyFirst-row transition metals compared with CaExplanation
Melting pointGenerally higher.More extensive participation of 3d and 4s electrons in metallic bonding, together with relatively small atoms, generally produces stronger bonding than in calcium. The precise trend is irregular.
DensityGenerally higher.Mass increases while atomic volumes remain relatively small. More mass packed into a similar or smaller volume gives a higher density.

Do not explain a high melting point using strong intermolecular forces: these are metallic structures. Do not explain density only by atomic mass; the volume occupied is also essential.

Worked exampleExplain a supplied trend

A data table shows only a small change in atomic radius across Sc to Cu, despite eight extra protons. What balances the increased nuclear attraction?

Try it first. Then show the worked stepsHide the worked steps
  1. The added electrons mainly enter the inner 3d subshell rather than a new outer shell.
  2. Their additional shielding partly offsets the increase in nuclear charge experienced by outer electrons.
  3. A small net increase in effective attraction can still cause a modest contraction.
Answer

The radius is relatively invariant because nuclear charge and 3d shielding increase together. This is not a claim of perfect cancellation.

Check your understanding

Why do atomic radii change only a little from Sc to Cu?

  1. Each element adds a new outer shell, which cancels the rising nuclear charge.
  2. Nuclear charge stays the same across the series, so the pull is unchanged.
  3. Extra electrons go into the inner 3d subshell and shield the outer 4s electrons.
Show the answer

Extra electrons go into the inner 3d subshell and shield the outer 4s electrons.

Nuclear charge rises across the series, but the added 3d electrons partly shield the 4s electrons. The two effects largely cancel, and no new shell is added.

03

Close 3d and 4s energies allow several oxidation states

Configuration suggests possibilities; the reaction environment decides which are stable.

Transition elements commonly have variable oxidation states because their 3d and 4s electrons have comparable energies and different numbers can participate in bonding. Fe commonly forms +2 and +3 compounds; copper commonly forms +1 and +2 compounds. Calcium, by contrast, usually forms +2: removing more electrons would disrupt a much more tightly bound noble-gas core.

A configuration can suggest likely oxidation states by accounting for 4s electrons and, where chemically feasible, additional 3d electrons. Early members can use more of these electrons in high oxidation states. This is a prediction of possibilities, not proof that every intermediate state is equally stable in water.

Worked exampleFrom configuration to likely states

Vanadium has [Ar]3d34s2. Explain why +2 and +5 are plausible oxidation states.

Try it first. Then show the worked stepsHide the worked steps
  1. Using the two 4s electrons gives a +2 state and a d3 metal centre.
  2. Using three 3d electrons as well gives the formal +5 state. Strong bonding to oxygen can stabilise this high state.
  3. The formula VO2+ has V at +5 because x + 2(-2) = +1.
Answer

Both +2 and +5 are plausible; V(III) and V(IV) also occur. VO2+ does not contain a free V5+ ion in water.

Watch out for this

A student says: "Mn is +7 in MnO4-, so the solution contains bare Mn7+ ions." What is wrong with this?

Show the answer

In MnO4-, Mn has oxidation number +7; this does not mean the ion contains a bare aqueous Mn7+ cation. Ligand exchange can change colour without changing oxidation number.

Check your understanding

Why does iron form both Fe2+ and Fe3+, while calcium forms only Ca2+?

  1. Iron has more protons than calcium, so it can afford to lose more electrons each time.
  2. Iron's 3d and 4s electrons are close in energy; calcium's next electron is in a tightly held inner shell.
  3. Calcium's 3d subshell is full, so it cannot lose any more electrons than two.
Show the answer

Iron's 3d and 4s electrons are close in energy; calcium's next electron is in a tightly held inner shell.

Iron can lose its two 4s electrons and then a 3d electron for little extra energy. Removing a third electron from calcium would break into its noble-gas core, which costs far too much energy.

04

Use three redox systems confidently

Write the correct acidic half-equation before combining electrons or comparing potentials.

Reduction half-equations in acidic aqueous solution
SystemHalf-equationUseful observation
Fe(III)/Fe(II)Fe3+(aq) + e- → Fe2+(aq)Common Fe(III) solutions are yellow and Fe(II) solutions pale green; intensity and ligands affect the appearance.
Manganate(VII)/Mn(II)MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)Purple disappears to a very pale pink, often effectively colourless, dilute Mn(II) solution.
Dichromate(VI)/Cr(III)Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)Orange changes to green under the usual reaction conditions.

The oxidising agent is reduced. Acidified manganate(VII) and dichromate(VI) accept electrons from reducing agents such as Fe2+, which becomes Fe3+. Acid is a reactant in both oxyanion half-equations, not just a label above the arrow. Changing pH can change both the potential and the manganese or chromium product.

Worked exampleWill dichromate oxidise iron(II)?

Use E°(Cr2O72-/Cr3+) = +1.33 V and E°(Fe3+/Fe2+) = +0.77 V. Predict the reaction under standard conditions and balance it.

Try it first. Then show the worked stepsHide the worked steps
  1. Dichromate has the more positive reduction potential: reduce it and reverse the iron reduction half-equation.
  2. Multiply Fe2+ → Fe3+ + e- by six to cancel six electrons. Do not multiply its electrode potential.
  3. E°cell = 1.33 - 0.77 = +0.56 V.
  4. Add the half-equations: Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+.
Answer

The positive standard cell potential predicts thermodynamic feasibility for the written reaction. It does not predict its speed. Atoms and total charge (+24 on each side) balance.

Check your understanding

How many moles of Fe2+ react with 0.00400 mol of acidified MnO4-?

  1. 0.0200 mol, since MnO4- : Fe2+ = 1 : 5
  2. 0.00400 mol, since the ions react 1 : 1
  3. 0.000800 mol, since Fe2+ : MnO4- = 1 : 5
Show the answer

0.0200 mol, since MnO4- : Fe2+ = 1 : 5

MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+. Each MnO4- gains five electrons and each Fe2+ gives one, so 5 x 0.00400 = 0.0200 mol.

05

Ligands donate electron pairs to a metal centre

A complex can be positive, neutral or negative.

Make a guess. You are not marked.

Excess concentrated hydrochloric acid is added to blue aqueous copper(II) sulfate. What do you see?

  1. The solution turns yellow-green as [CuCl4]2- forms
  2. A blue precipitate of copper(II) chloride forms
  3. No change, because chloride is a spectator ion here
Show the answer

The solution turns yellow-green as [CuCl4]2- forms

Chloride ions replace the water ligands to form [CuCl4]2-, which absorbs light differently from [Cu(H2O)6]2+.

A ligand is an ion or molecule that donates a lone pair to a central metal atom or ion to form a coordinate bond. A complex contains that central metal bonded to its surrounding ligands. H2O and NH3 are neutral donors through O and N respectively; Cl- is a negatively charged donor. The metal accepts the pair: it does not supply both electrons in that coordinate bond.

Three copper(II) complexes
ComplexLigands and coordinationOverall charge
[Cu(H2O)6]2+Six water donors; coordination number 6.+2: all water ligands are neutral.
[Cu(NH3)4(H2O)2]2+Four ammonia and two water donors; coordination number 6. Often abbreviated [Cu(NH3)4]2+ in equations.+2: ammonia and water are neutral.
[CuCl4]2-Four chloride donors; coordination number 4.+2 + 4(-1) = -2.

Coordination number counts donor atoms directly bonded to the metal. For these one-donor ligands it equals the number of ligands. Determine oxidation state separately from overall complex charge: the negative chloride complex still contains copper in the +2 state.

Two ligand changes, the same copper oxidation state

Water gives the blue copper(II) aqua complex. Excess ammonia produces a deep-blue ammine complex; concentrated chloride produces a yellow chloro complex. Both retain copper(II).

These are final complex colours. A small ammonia addition first gives a hydroxide precipitate; a mixture of blue aqua and yellow chloro complexes may appear green.

Excess ammonia: [Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l). Ammonia replaces four coordinated water molecules; the solution becomes deep blue.

Concentrated chloride: [Cu(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CuCl4]2-(aq) + 6H2O(l). High chloride concentration favours the yellow complex. Dilution favours the blue aqua form. These are ligand exchanges, not redox reactions.

Check your understanding

What is the oxidation state of chromium in [Cr(OH)6]3-?

  1. -3, the overall charge of the ion
  2. +6, one for each hydroxide ligand
  3. +3, since x + 6(-1) = -3
Show the answer

+3, since x + 6(-1) = -3

Each OH- ligand is -1. The overall charge includes the ligands, so x - 6 = -3 and x = +3.

06

Separate precipitation, ligand exchange and oxidation

The sequence of additions explains the qualitative-analysis observation.

A few drops of aqueous ammonia supply OH- through NH3 + H2O ⇌ NH4+ + OH-. Copper(II) first forms blue Cu(OH)2(s): Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s). In excess ammonia, complex formation removes dissolved copper ions and the precipitate dissolves to the deep-blue ammine solution. NaOH does not provide that ammonia ligand, so the precipitate remains in excess NaOH.

Transition-metal entries in the official qualitative-analysis notes
IonNaOH(aq), then excessNH3(aq), then excess
Cr3+Grey-green Cr(OH)3 precipitate; dissolves in excess to dark-green [Cr(OH)6]3-.Grey-green precipitate; insoluble in excess.
Cu2+Pale-blue Cu(OH)2 precipitate; insoluble in excess.Blue precipitate; dissolves in excess to a deep-blue complex.
Fe2+Green Fe(OH)2 precipitate; insoluble in excess; turns brown in air.Same precipitate and excess-reagent behaviour.
Fe3+Red-brown Fe(OH)3 precipitate; insoluble in excess.Same precipitate and excess-reagent behaviour.
Mn2+Off-white Mn(OH)2 precipitate; insoluble in excess; rapidly turns brown in air.Same precipitate and excess-reagent behaviour.

Chromium(III) hydroxide dissolves in excess hydroxide by forming a soluble hydroxo complex: Cr(OH)3(s) + 3OH-(aq) ⇌ [Cr(OH)6]3-(aq). Chromium remains +3. In contrast, the browning of freshly formed iron(II) or manganese(II) hydroxide in air involves oxidation. Record the initial colour and subsequent change separately.

Watch out for this

A student says: "The solution changed colour, so the metal ion must have been oxidised or reduced." What is wrong with this?

Show the answer

Cr(III) solution colours vary with the ligands present; the usual green reaction mixture is not proof that every pure hexaqua chromium(III) solution must be green. A colour change can come from redox, ligand exchange or precipitation. Use the reagent, formula and phase as well as the colour. Complex formation is also possible for non-transition species, including zinc and silver ions.

In haemoglobin, oxygen binds reversibly to an iron-containing centre. Carbon monoxide competes for the binding site and binds much more strongly, displacing oxygen and reducing oxygen transport. A simplified exchange is HbO2 + CO ⇌ HbCO + O2. This illustrates ligand competition; it is not a reaction in which CO burns with oxygen.

Check your understanding

A blue solution gives a blue precipitate with a little ammonia, then a deep-blue solution with excess. What happens in each stage?

  1. First ammonia forms the complex; in excess the extra ammonia dissolves the copper metal.
  2. First Cu2+ is reduced to a Cu+ precipitate; in excess it is oxidised back to Cu2+.
  3. First NH3 acts as a base, making Cu(OH)2; in excess it acts as a ligand, making a soluble complex.
Show the answer

First NH3 acts as a base, making Cu(OH)2; in excess it acts as a ligand, making a soluble complex.

A little ammonia supplies OH- and precipitates Cu(OH)2. Excess ammonia replaces ligands to form the deep-blue copper(II) ammine complex. Copper stays +2 throughout.

07

Octahedral ligands split the five d orbitals

Visible-light absorption can promote an electron across the energy gap.

Make a guess. You are not marked.

Zinc ions in water, [Zn(H2O)6]2+, have the configuration 3d10. What colour is the solution?

  1. Colourless, as its 3d subshell is full
  2. Green, like aqueous iron(II) ions are
  3. Blue, like aqueous copper(II) ions are
Show the answer

Colourless, as its 3d subshell is full

With a full 3d subshell, no electron can be promoted from a lower d orbital into a higher one, so no visible light is absorbed.

For an isolated metal ion, the five d orbitals have equal energy: they are degenerate. In an octahedral complex, six ligands approach along the positive and negative x, y and z axes. Their electron pairs repel d-electron density unevenly because the orbitals point in different directions.

Orientation determines the amount of repulsion
OrbitalsShape and orientationOctahedral result
dx2-y2 and dz2The first has four lobes along x and y. The second has two z-axis lobes and a ring around its middle. Both put substantial density towards axial ligands.Greater repulsion; the higher-energy pair.
dxy, dxz and dyzFour lobes between the indicated axes, in the xy, xz or yz plane.Less direct repulsion; the lower-energy set of three.

Octahedral splitting and a d-d transition

The two d orbitals directed towards axial ligands are higher in energy than the three oriented between axes. Absorbing a photon with energy equal to the gap can promote a d electron from the lower to the upper set.

The diagram shows relative energies within an octahedral complex. Splitting is caused by ligand interactions; light then supplies the energy for an electronic transition.

If a suitable lower d level is occupied and a higher one can receive an electron, light of energy equal to the gap can cause a d-d transition. Absorbing selected visible wavelengths leaves the complementary mixture to be transmitted or reflected, giving the observed colour. A blue solution transmits blue light; it does not look blue because it absorbs blue most strongly.

Changing the ligand, metal or oxidation state can change the splitting and the wavelength absorbed. This explains why ligand exchange can change colour while oxidation state remains fixed. You do not need a memorised ranking of ligand field strengths for this syllabus.

Check your understanding

Why are Cu2+ compounds often coloured, while Zn2+ compounds are usually white?

  1. Cu2+ is d9, so a d-d electron jump can absorb visible light; Zn2+ is d10, with no d-d jump.
  2. Zn2+ is not a metal ion, so it has no d electrons that can absorb any light.
  3. Cu2+ has an empty d subshell, so light passes straight through and makes it look blue.
Show the answer

Cu2+ is d9, so a d-d electron jump can absorb visible light; Zn2+ is d10, with no d-d jump.

Ligands split the d orbitals. In Cu2+ an electron can move to a higher d orbital by absorbing visible light. Zn2+ has a full d subshell, so no d-d transition is possible.

08

Catalysis uses accessible bonding and oxidation states

A catalyst participates in steps and is regenerated overall.

Transition metals and their compounds can offer a pathway with lower activation energy. Their ability to interact with reactant electron density helps surface catalysis; accessible oxidation states help electron-transfer cycles. These are chemical reasons for catalytic activity, not a claim that every transition metal catalyses every reaction.

Iron in the Haber process: heterogeneous catalysis
  1. Adsorb

    N2 and H2 attach to active sites on solid iron. Interactions with the surface weaken their bonds.

  2. React

    Surface species are brought together and react through steps with lower activation barriers than the uncatalysed route.

  3. Desorb

    NH3 leaves, freeing sites for another cycle. Iron is regenerated; N2 + 3H2 ⇌ 2NH3.

In a catalytic converter, transition-metal surfaces also help CO react with NO: 2CO(g) + 2NO(g) → 2CO2(g) + N2(g). Adsorption holds reactants near each other and facilitates bond changes. The catalyst does not change the reaction enthalpy or equilibrium constant.

Worked exampleAn iron-ion cycle transfers electrons in solution

Explain how Fe2+ can catalyse S2O82- + 2I- → 2SO42- + I2. All reactants and the catalyst are aqueous.

Try it first. Then show the worked stepsHide the worked steps
  1. First: S2O82- + 2Fe2+ → 2SO42- + 2Fe3+. Iron(II) donates electrons.
  2. Next: 2Fe3+ + 2I- → 2Fe2+ + I2. Iron(III) accepts electrons.
  3. Adding the steps cancels both iron species and gives the required overall reaction. Fe2+ is regenerated.
  4. Oppositely charged reactants can meet in each catalysed step, replacing the direct encounter between two anions with a lower-barrier route.
Answer

This is homogeneous catalysis using the Fe(II)/Fe(III) pair. A positive overall cell potential establishes feasibility, while the catalyst addresses the kinetic barrier.

Check your understanding

Why are transition metals and their ions often good catalysts?

  1. They have variable oxidation states and can adsorb reactants onto their surfaces.
  2. They shift the equilibrium to the right, so more product forms from the reactants.
  3. They are very reactive metals, so they react with and use up one of the reactants.
Show the answer

They have variable oxidation states and can adsorb reactants onto their surfaces.

Changing oxidation state lets an ion pass electrons in a cycle and be regenerated. A metal surface can hold reactants close and weaken their bonds.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Transition elementA d-block element whose atom has an incomplete d subshell, or which forms a stable cation with an incomplete d subshell.
LigandA molecule or ion with at least one lone pair of electrons that it donates to a central metal atom or ion to form a dative bond.
Complex ionA central metal atom or ion surrounded by ligands bonded to it by dative bonds.
Coordination numberThe number of dative bonds formed between the ligands and the central metal ion.
One question for each property
QuestionUseful reasoning
Does it qualify?Test the atom and its cations for an incomplete d subshell.
What is the ion configuration?Remove 4s electrons before 3d; keep Cr and Cu exceptions in the neutral atoms.
Why are trends gentle?Increasing nuclear charge is partly offset by added 3d shielding.
Why several oxidation states?3d and 4s energies are close; different numbers of electrons can participate.
Why a colour change?Distinguish ligand exchange, redox and precipitation from the actual species.
Why visible colour?Unequal ligand repulsion splits d levels; a suitable d-d excitation absorbs selected wavelengths.
Why catalysis?A lower-barrier surface or oxidation-state cycle regenerates the catalyst.

Named redox electron counts: Fe3+/Fe2+: 1; MnO4-/Mn2+: 5; Cr2O72-/Cr3+: 6. The latter two equations require acid. Copper: water blue; excess ammonia deep blue; concentrated chloride favours yellow complex, often green in a mixture. Copper remains +2 in those exchanges.

Scope

Learning outcomes

13. An Introduction to the Chemistry of Transition Elements. Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map
  1. 13(a) Apply the transition-element definition.

    • d-block atom with incomplete d subshell or cation with incomplete d subshell
    • Sc and Cu inclusion; Zn distinction

    Which elements count as transition elements?

  2. 13(b) Write first-row atom and ion configurations.

    • Sc to Cu
    • Cr and Cu neutral configurations
    • Remove 4s before 3d on forming cations
    • Cross-reference atomic structure 1(h)

    Which elements count as transition elements?

  3. 13(c) Explain relatively small radius and first-ionisation-energy changes.

    • Increasing nuclear charge
    • Added 3d shielding
    • Relatively invariant does not mean exactly constant

    Why the physical trends are comparatively gentle

  4. 13(d) Compare transition-metal physical properties with calcium.

    • Qualitative melting-point comparison
    • Qualitative density comparison
    • Metallic bonding and mass per volume

    Why the physical trends are comparatively gentle

  5. 13(e) Explain variable oxidation states.

    • Similar 3d/4s energies
    • Fe and Cu examples
    • Contrast with usual calcium +2

    Close 3d and 4s energies allow several oxidation states

  6. 13(f) Suggest likely oxidation states from configuration.

    • 4s and 3d electron participation
    • Worked vanadium +2 and +5 deduction
    • Formal oxidation number distinguished from a free aqueous ion

    Close 3d and 4s energies allow several oxidation states

  7. 13(g) Explain the three specified redox systems.

    • Fe3+/Fe2+
    • MnO4-/Mn2+ in acid
    • Cr2O7^2-/Cr3+ in acid
    • Balanced half-equations, roles, observations and electron ratios

    Use three redox systems confidently

  8. 13(h) Predict redox likelihood from standard potentials.

    • Reduction-potential comparison
    • Positive Ecell standard thermodynamic criterion
    • Do not multiply potentials by equation coefficients
    • Cross-reference electrochemistry limitations

    Use three redox systems confidently

  9. 13(i) Define ligands and complexes using required examples.

    • Cu(II) with water, ammonia and chloride
    • Electron-pair donation and charge accounting
    • Transition-metal complexes in official QA notes, including chromium hydroxo complex

    Ligands donate electron pairs to a metal centreSeparate precipitation, ligand exchange and oxidation

  10. 13(j) Explain ligand exchange and associated colours.

    • Cu aqua/ammine/chloro equilibria
    • Initial copper hydroxide versus excess-ammonia complex
    • CO/O2 exchange in haemoglobin

    Ligands donate electron pairs to a metal centreSeparate precipitation, ligand exchange and oxidation

  11. 13(k) Explain octahedral d-orbital splitting.

    • Five initially degenerate d orbitals
    • Shapes and orientation relative to six axial ligands
    • Higher pair and lower group of three

    Octahedral ligands split the five d orbitals

  12. 13(l) Explain colour through d-d transitions.

    • Visible-light absorption across the splitting
    • Observed versus absorbed colour
    • Occupancy limits for d0/d10
    • Ligand field-strength ranking not required

    Octahedral ligands split the five d orbitals

  13. 13(m) Explain catalytic action of transition elements and compounds.

    • Surface adsorption, reaction and desorption
    • Iron Haber and catalytic-converter examples
    • Homogeneous Fe2+/Fe3+ peroxodisulfate/iodide cycle
    • Catalyst regeneration and lower activation barrier; cross-reference 8(j)

    Catalysis uses accessible bonding and oxidation states