CLASS 12 CHEMISTRY · CHAPTER 4 · STUDY NOTES

The d- and f-Block Elements

This chapter explores the chemistry of transition and inner transition elements, focusing on their electronic configurations, variable oxidation states, magnetic properties, and the significant roles of lanthanoid and actinoid contractions in periodic trends.

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Position and Definition of Transition Elements

Locate the d-block, then distinguish position from the transition-element definition.

Locate the block, then test the definition

The compact map marks groups 3–12 in the centre as the d-block and the detached inner-transition rows below. Its tiles represent periodic-table positions, not atom size. This chapter follows NCERT’s placement convention; group-3 layout conventions can differ.

Transition element: An element whose atom has an incomplete d subshell, or which forms a cation with an incomplete d subshell.

Groups 3–12 occupy the centre of the chapter’s periodic-table map. The 3d, 4d, 5d and 6d series involve the (n−1)d subshell. The detached inner-transition rows belong within the long periods; they are not extra periods below the table.

Sc qualifies through its d¹ atom even though Sc³⁺ is d⁰. Copper qualifies through Cu²⁺, which is d⁹, despite its d¹⁰ atom and Cu⁺. Zinc has d¹⁰ in its atom and common Zn²⁺ ion, so NCERT treats it as a d-block member rather than a transition element.

Let’s look a little closer.

Go deeper: Use the definition rather than a colour checklist

Magnetism, colour and variable oxidation states are useful patterns, not tests every transition compound must pass. A colourless d⁰ ion can belong to a transition element, while a coloured compound need not owe its colour to a d–d transition.

NCERT places Ce–Lu after La and Th–Lr after Ac in its discussion. Treat the displayed periodic-table layout as the chapter convention; row labels and the formal electronic definition answer different questions.

Take a moment. What do you think?

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Question

An element has a d¹⁰ atom, a d¹⁰ singly charged ion and a d⁹ doubly charged ion. Does it satisfy the transition-element definition?

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NCERT reference: chapter PDF pages 1, 2, 3.

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Electronic Configurations of d-Block Elements

Follow the actual ground state, then remove outer s electrons before d electrons for these cations.

Fill orbitals, then remove the outer electrons

Fe atom: d occupancy 6, outer s occupancy 2. Each horizontal line denotes one orbital and each arrow one electron’s spin projection, not a trajectory. Fill separate degenerate d orbitals before pairing in this free-ion bookkeeping. In forming the displayed 3d-metal cations, remove 4s before 3d. Pd uses 4d¹⁰5s⁰; labels retain that exception.

d-electron configuration: A statement of d-subshell occupancy for a specified atom or ion, after accounting for outer-electron removal.

Element	Atomic Number (Z)	4s Electrons	3d Electrons
Scandium (Sc)	21	2	1
Titanium (Ti)	22	2	2
Vanadium (V)	23	2	3
Chromium (Cr)	24	1	5
Manganese (Mn)	25	2	5
Iron (Fe)	26	2	6
Cobalt (Co)	27	2	7
Nickel (Ni)	28	2	8
Copper (Cu)	29	1	10
Zinc (Zn)	30	2	10

The useful general pattern is (n−1)d¹–¹⁰ns⁰–². Chromium is [Ar]3d⁵4s¹ and copper [Ar]3d¹⁰4s¹; palladium is [Kr]4d¹⁰5s⁰. These exceptions show why the simple filling mnemonic is not a complete energy calculation.

For Fe, [Ar]3d⁶4s² becomes Fe²⁺ [Ar]3d⁶ and Fe³⁺ [Ar]3d⁵. Cu²⁺ is [Ar]3d⁹. Orbital arrows represent electron spin projections; orbitals are probability distributions rather than paths.

Go deeper: Configurations explain patterns without replacing energetics

Similar s and d orbital energies and electron interactions contribute to ground-state arrangements. “Half-filled or full-filled stability” is a summary of some cases, not a universal prediction rule.

The occupancy drawn for an isolated ion follows Hund’s rule across degenerate d orbitals. Ligands split their energies and can change pairing. Specify the ion and environment before using its d count to predict magnetism.

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Question

Write the d counts of Fe²⁺, Fe³⁺ and Cu²⁺. Which removal step produces Fe³⁺ from Fe²⁺?

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NCERT reference: chapter PDF pages 2, 3.

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Physical Properties and Enthalpy of Atomisation

Use measured values to connect metallic bonding with physical properties.

Metallic bonding has measurable consequences

The plotted first-series atomisation enthalpies are chapter Table 4.2 data in kJ/mol: Sc 326, Ti 473, V 515, Cr 397, Mn 281, Fe 416, Co 425, Ni 430, Cu 339 and Zn 126. The Mn dip shows why “maximum at d⁵” is not a strict curve law. Metallic bonding and structure matter; the graph is not a direct count of free-atom unpaired electrons.

Enthalpy of atomisation: The enthalpy change for forming one mole of gaseous atoms from an element in its stated standard form.

Transition metals are generally hard, dense, electrically conducting solids, often with high melting points. Zinc, cadmium and mercury are useful exceptions to broad bonding trends. Crystal structure and electronic bonding both matter, so no single d-electron count predicts all physical properties.

Atomisation M(s)→M(g) requires energy for these metals. The first-series graph uses chapter data: vanadium 515 kJ/mol, manganese 281 and zinc 126. The conspicuous Mn dip contradicts a simplistic maximum-at-d⁵ rule.

Go deeper: Separate solid bonding from gaseous ionisation

Atomisation separates neighbours in the metal; ionisation removes electrons from gaseous species. First and third ionisation curves therefore measure different changes from the atomisation bars.

Third ionisation of Mn²⁺ disrupts a d⁵ configuration and costs 3260 kJ/mol, whereas Fe²⁺→Fe³⁺ gives d⁵ and costs 2962 kJ/mol in the source table. Aqueous-ion stability also includes hydration and other thermodynamic contributions.

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Question

Why does the largest number of unpaired electrons in a free atom not guarantee the largest atomisation enthalpy?

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NCERT reference: chapter PDF pages 4, 5, 6, 7.

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Atomic Sizes and Lanthanoid Contraction

Compare the same kind of radius before explaining a trend.

Compare like radii and see contraction

These are metallic radii in pm from the chapter’s Table 4.2, not ionic radii. The modest overall contraction has real deviations, including Mn and the late-series rise. Nuclear attraction, shielding and bonding all contribute. Compare the same radius definition rather than mixing metallic and ionic values.

Lanthanoid contraction: The progressive decrease in comparable atomic or ionic sizes across the lanthanoid series as nuclear charge increases with incomplete shielding.

Same radius convention	Zr	Hf
Chapter quoted radii / pm	160	159

The first-series graph plots metallic radii, not ionic radii. Its contraction is irregular: manganese and the late-series metals deviate. Nuclear attraction, shielding and metal bonding all contribute.

NCERT quotes zirconium at 160 pm and hafnium at 159 pm. Filling the intervening 4f series contributes to contraction, largely offsetting the extra period’s expected size increase. Their near-equal sizes help explain closely similar chemistry and difficult separation.

Go deeper: Radius is a model-dependent comparison

A radius depends on the definition, charge and, for ions, coordination environment. Comparing metallic Zr with an unrelated Hf ionic radius would not demonstrate contraction.

Poor shielding is relative: added 4f electrons do not fully compensate the increased nuclear attraction. The visual envelopes isolate competing influences; they are not literal electron-shell boundaries.

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Question

Calculate the percentage difference between the chapter’s Zr and Hf radii using Zr as the reference. What explains the unexpectedly small difference?

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NCERT reference: chapter PDF pages 5, 6.

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Variability of Oxidation States

Account for ligands and total charge, then compare which states occur.

One metal can have several oxidation states

Sc: the dots list selected chapter oxidation states. Oxidation numbers are formal bookkeeping in compounds, not a claim that every state is an isolated aqueous ion or equally stable. Manganese spans +2 to +7 in the chapter; near-edge elements show fewer common states.

Oxidation state: A formal charge assigned by electron bookkeeping in a compound; it need not be a real isolated-ion charge.

Similar outer s and d energies allow several oxidation states for many transition metals. Mn spans +2 to +7 in the chapter, while Sc commonly gives +3 and Zn +2. The occurrence of a state does not imply equal stability in every medium.

In MnO₄⁻, x+4(−2)=−1 gives Mn(VII); in CrO₄²⁻, chromium is +6. CO is neutral in Ni(CO)₄, giving nickel state zero. Oxygen and fluorine can stabilise high states through favourable bonding and lattice effects.

Go deeper: Formal state and chemical behaviour are related, not identical

MnO is a low-state, predominantly ionic basic oxide; Mn₂O₇ is a high-state covalent acidic oxide. Cr₂O₃ and V₂O₅ illustrate amphoteric behaviour, so do not force every oxide onto a simple linear acidity rule.

A formula containing Mn(VII) does not establish a free aqueous Mn⁷⁺ ion. High formal states often occur in covalent oxo compounds; reduction products depend on the medium and available reactants.

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Question

Find chromium’s oxidation state in Cr₂O₇²⁻ and nickel’s in Ni(CO)₄. Explain why these are bookkeeping results.

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NCERT reference: chapter PDF pages 8, 9.

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Standard Electrode Potentials (E°)

Aqueous redox tendencies depend on more than an isolated d configuration.

A redox tendency includes the whole environment

Standard aqueous M²⁺/M reduction potentials are chapter values relative to SHE, in volts. More positive values favour reduction relative to a partner under standard conditions. Cu²⁺/Cu is +0.34 V, so Cu does not liberate H₂ by reducing H⁺ in a simple non-oxidising acid under the standard comparison; oxidising acids and kinetics require separate consideration.

Standard reduction potential: The potential of a specified reduction half-cell relative to a reference electrode under standard conditions.

The M²⁺/M ladder plots chapter standard reduction potentials relative to SHE. Cu²⁺/Cu at +0.34 V is more positive than Zn²⁺/Zn at −0.76 V. A compatible standard Zn/Cu cell has E°cell=0.34−(−0.76)=1.10 V; half-reaction stoichiometric multiplication does not multiply E°.

Cu does not displace hydrogen from a simple non-oxidising acid by the standard comparison. Oxidising acids, complexation and kinetic barriers need separate analysis. Cr²⁺(d⁴) is commonly reducing towards Cr³⁺(d³), whereas Mn³⁺(d⁴) is oxidising towards Mn²⁺(d⁵).

Go deeper: Include the complete energetic accounting

For Cu, atomisation +339 and ionisation +745,+1958 combine with hydration −2121 kJ/mol to give an enthalpy bookkeeping sum +921 kJ/mol. This is not directly the Gibbs energy of a complete electrochemical cell.

Electrode potential depends on Gibbs energy and the reference comparison. Dividing an isolated enthalpy sum by 2F does not produce a justified E°. Hydration, entropy, ligands and concentration qualify configuration-based explanations.

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Question

Can the +921 kJ/mol copper enthalpy bookkeeping sum be divided by 2F to obtain its standard reduction potential? Explain.

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NCERT reference: chapter PDF pages 10, 11, 13.

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Magnetic Properties

Count unpaired electrons and state the pairing assumptions.

Count unpaired electrons before predicting a moment

d-electron count: 5

For the displayed high-spin d5 bookkeeping, n=5 unpaired electrons and the spin-only estimate μ=√[n(n+2)]=5.92 BM. Arrows show spin projections, not moving particles. Orbital contributions and the ligand environment can cause measured values to differ from this estimate.

Spin-only magnetic moment: The estimate μ=√[n(n+2)] BM based on n unpaired electrons, neglecting orbital contributions.

For high-spin d⁵, five unpaired electrons give √35≈5.92 BM. High-spin d⁶ has four and gives √24≈4.90 BM. Paired electrons cancel their spin contribution in this model; measured moments can differ because orbital and collective effects matter.

In an octahedral d⁶ complex, weak splitting can favour a high-spin arrangement; sufficiently strong splitting can favour three paired lower orbitals and no unpaired electrons. Paramagnetic samples tend towards stronger regions of a non-uniform field; diamagnetic samples weakly favour weaker regions.

Go deeper: Magnetism includes several distinct models

The d count alone does not settle the measured moment without the ligand environment and pairing. The five equal levels shown in the free-ion view are not the split levels of every complex.

Ferromagnetism involves collective ordering and domains. It is not simply a large independent-ion spin-only moment. Nickel, cobalt and iron metals illustrate why metallic magnetism needs a different description from an isolated ion.

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Question

A high-spin d⁷ ion has three unpaired electrons. Calculate its spin-only moment and identify one reason an experimental value could differ.

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NCERT reference: chapter PDF pages 13, 14, 15.

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Formation of Coloured Ions and Complexes

The colour seen is the transmitted or reflected light, rather than the light absorbed.

The light absorbed differs from the colour seen

Ti³⁺ has d1 and the aqueous-colour example purple. Colours are illustrative and depend on ligand environment, concentration and lighting; colour alone does not identify an unknown ion. Zn²⁺ is d¹⁰ with no ordinary d–d excitation, but d⁰/d¹⁰ compounds can still be coloured through charge-transfer processes.

Picture from the NCERT book

NCERT Chapter 4, PDF page 15: coloured solutions in a specific aqueous environment. Ligands, concentration and lighting affect appearance.Source: NCERT Class XII Chemistry · Chapter 4 · chapter PDF page 15.

Ligand-field excitation: An electronic excitation between metal-centred levels whose energies depend on surrounding ligands.

Aqueous example	d count	Illustrative colour
Ti³⁺	1	Purple
V³⁺	2	Green
Mn²⁺	5	Pale pink
Co²⁺	7	Pink
Ni²⁺	8	Green
Cu²⁺	9	Blue
Zn²⁺	10	Colourless

Surrounding ligands split d-orbital energies. An absorbed photon matching an allowed energy gap can excite an electron. A solution absorbing an orange/red portion may transmit blue-biased light; real spectra contain ranges of wavelengths.

Aqueous-ion colours vary with ligands, concentration and lighting. Zn²⁺ is d¹⁰ and has no ordinary d–d excitation, yet some d⁰ or d¹⁰ compounds are coloured through charge transfer. Purple permanganate is an important d⁰ example, so “all colour means unpaired d electrons” is false.

Go deeper: Complexes have real neighbours and an environment

A coordination complex binds donor atoms of ligands to a central metal. The six-donor drawing is octahedral perspective, not a count of all atoms within the ligands. Small size, charge density and favourable bonding help explain complex formation.

Colour alone cannot identify an unknown substance. Even one metal oxidation state can appear differently with changed ligand coordination. The chapter’s source photograph is an observation of particular solutions, not a universal reference palette.

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Question

Permanganate contains formally d⁰ Mn(VII), yet is purple. What is wrong with explaining its colour as an ordinary d–d transition?

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NCERT reference: chapter PDF pages 15, 16.

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Catalytic Properties and Interstitial Compounds

A catalytic cycle, an alloy site and surface adsorption describe different relationships.

A surface, a redox cycle and a crystal gap are different ideas

In the chapter’s simplified Fe³⁺/Fe²⁺ catalytic cycle, iodide reduces Fe³⁺ to Fe²⁺ and peroxodisulphate restores Fe³⁺. Adding 2Fe³⁺+2I⁻→2Fe²⁺+I₂ and 2Fe²⁺+S₂O₈²⁻→2Fe³⁺+2SO₄²⁻ gives the uncatalysed net stoichiometry. The cycle shows regeneration, not a measured timing model.

Interstitial compound: A material in which sufficiently small atoms occupy gaps in a metal structure rather than replace host metal atoms.

Transition metals and compounds can catalyse reactions through variable states, adsorption or both. The Fe³⁺/Fe²⁺ illustration shows iodide reducing Fe³⁺ and peroxodisulphate restoring it. The catalyst participates in intermediate steps but cancels from the net stoichiometry.

On a schematic nickel surface, H₂ is split and its two H atoms add across an alkene’s C=C, leaving C–C and two new C–H bonds. Small H, C or N atoms can also occupy crystal gaps in interstitial phases. These phases often become harder while retaining metallic conductivity.

Go deeper: Conserve atoms and distinguish structural roles

An interstitial atom occupies a gap; a substitutional alloy atom replaces a host site. Similar atomic radii favour substitutional alloy formation, whereas suitably small atoms can enter gaps. Interstitial phases can have variable stoichiometry and reactivity.

A catalyst changes the pathway rather than supplying a net reactant. Surface illustrations slow and simplify events for learning; they do not specify the unique mechanism, actual trajectories or an at-home preparation.

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Question

Adding the two iron-catalysed steps gives 2I⁻+S₂O₈²⁻→I₂+2SO₄²⁻. Why is iron omitted from this net equation?

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NCERT reference: chapter PDF page 16.

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Potassium Dichromate (K₂Cr₂O₇)

Count shared atoms and distinguish acid–base speciation from reduction.

Chromate and dichromate join without changing Cr(VI)

Chromate has one Cr with four O neighbours in a tetrahedral arrangement. Dichromate has two such tetrahedra sharing a single O corner, so it contains seven O atoms rather than eight. The displayed Cr–O–Cr bridge angle is schematic; the chapter gives about 126°. All Cr atoms remain +6. Solid versus dashed bonds indicate perspective, not alternating chemical identities.

Dichromate ion: Cr₂O₇²⁻, containing two tetrahedral chromium–oxygen units sharing one oxygen corner.

Two separate CrO₄ units would contain eight oxygen atoms; sharing one leaves seven. The chapter gives a Cr–O–Cr bridge angle of about 126°. In both chromate and dichromate, chromium has oxidation state +6.

The equilibrium 2CrO₄²⁻+2H⁺⇌Cr₂O₇²⁻+H₂O shifts towards orange dichromate on acidification and yellow chromate on raising pH in the simplified comparison. Actual solutions can contain additional protonated species. This colour change is not chromium reduction.

Go deeper: Preparation and oxidising action answer different questions

Chromite undergoes oxidative fusion with carbonate and air to chromate. Acidification forms dichromate; adding KCl permits the less soluble potassium salt to crystallise in the chapter route. This is industrial chemistry, not student laboratory guidance.

Acidic reduction is Cr₂O₇²⁻+14H⁺+6e⁻→2Cr³⁺+7H₂O. Six Fe²⁺ ions can supply the six electrons by becoming Fe³⁺. Potassium ions are spectators; verify oxygen, hydrogen and charge as well as electron count.

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Question

Why does converting two chromate ions to one dichromate ion require no electron transfer, whereas reducing one dichromate ion to two Cr³⁺ requires six electrons?

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NCERT reference: chapter PDF pages 17, 18.

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Potassium Permanganate (KMnO₄)

Choose the medium before balancing a permanganate reduction.

The medium changes the reduction product

Both MnO₄²⁻ and MnO₄⁻ have tetrahedral oxygen coordination. Manganate(VI) is green with one unpaired electron; permanganate(VII) is purple and formally d⁰. Its intense colour therefore cannot be explained as an ordinary d–d excitation; charge-transfer transitions matter. Ball colours identify atoms, not the solution’s absorption colour.

Disproportionation: A reaction in which one oxidation state of an element is both oxidised and reduced.

Manganate MnO₄²⁻ has Mn(VI), d¹ and green colour; permanganate MnO₄⁻ has Mn(VII), d⁰ and purple colour. Both have tetrahedral oxygen coordination. The intense permanganate colour involves charge transfer rather than ordinary d–d excitation.

Typical chapter reduction products are Mn²⁺ in acid, MnO₂ in neutral/faintly alkaline medium, and MnO₄²⁻ in strongly alkaline comparisons. The corresponding electron counts per Mn are five, three and one. Medium, reductant, concentration and kinetics qualify these patterns.

Go deeper: Balance half-reactions and the preparation route

In acid: MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O. In the neutral/alkaline representation: MnO₄⁻+2H₂O+3e⁻→MnO₂+4OH⁻. In strongly alkaline manganate formation: MnO₄⁻+e⁻→MnO₄²⁻.

Oxidative fusion of MnO₂ in alkali yields manganate, which can be oxidised electrolytically to permanganate. Alternatively 3MnO₄²⁻+4H⁺→2MnO₄⁻+MnO₂+2H₂O disproportionates Mn(VI): two atoms each lose one electron, and one gains two.

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Question

One mole of manganate undergoes the written disproportionation completely. How many moles of Mn(VII) and Mn(IV) products result?

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NCERT reference: chapter PDF pages 18, 19, 20.

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The Lanthanoids (4f series)

Follow comparable trivalent radii, then examine exceptions to the common +3 state.

4f occupancy and a shrinking series

The plot uses La³⁺ through Yb³⁺ ionic radii from chapter Table 4.9: 106 to 86 pm for comparable trivalent ions. Lu³⁺ is intentionally omitted because that table leaves its radius blank; no endpoint is invented. These are a stated tabulated radius convention, not universal coordination-independent sizes.

Lanthanoids: The chapter’s fourteen elements Ce–Lu following lanthanum, associated with progressive 4f occupancy.

Comparable Ln³⁺ radii contract across the series. The plotted Table 4.9 values run La³⁺ 106 pm to Yb³⁺ 86 pm; Lu³⁺ is not plotted because the source table leaves it blank. Increasing nuclear attraction with imperfect 4f shielding explains the broad trend.

The predominant oxidation state is +3, with important +2 and +4 cases. Ce⁴⁺ is 4f⁰, Eu²⁺ 4f⁷ and Yb²⁺ 4f¹⁴. These configurations help explain exceptions without guaranteeing their stability in every medium. The source’s La(II) line conflicts with its table; La(III) is intended.

Go deeper: Similar sizes affect separation and applications

Closely similar +3 chemistry makes separation difficult; contraction also reduces the basicity of comparable lanthanoid hydroxides across the series. Metallic reactivity produces oxides, halides and hydroxides under suitable conditions.

Rare-earth alloy additions, oxide catalysts and phosphors illustrate applications. NCERT describes mischmetall as roughly 95% rare-earth metals and 5% iron with minor additional elements. Luminescence is light emission, not identical to the absorption process in coloured transition-ion solutions.

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Question

Use the tabulated La³⁺ 106 pm and Yb³⁺ 86 pm to find the percentage contraction relative to La³⁺. Is the result universal for all coordination environments?

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NCERT reference: chapter PDF pages 20, 21, 22, 23.

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The Actinoids (5f series)

Compare oxidation-state patterns while keeping nuclear decay separate.

Compare 5f chemistry without forcing a smooth trend

Selected chapter actinoid oxidation states span more possibilities in the early series than the predominantly +3 lanthanoid comparison. The plot shows a selection of documented chapter states, not their relative abundance or a strict maximum for all modern compounds. Oxidation state is a chemical property, not nuclear decay rate.

Actinoids: The chapter’s fourteen elements Th–Lr following actinium, whose chemistry involves 5f and nearby orbital energies.

Comparison	Lanthanoids	Actinoids
Characteristic orbitals	4f occupancy	5f and nearby energies
Oxidation states	Common +3; +2/+4 exceptions	Several early states; later +3 prominent
Nuclear point	Radioactive isotopes exist; Pm has no stable isotope	All elements have radioactive isotopes

All actinoid elements have radioactive isotopes; half-life belongs to a specified isotope. Early actinoids show several higher oxidation states, whereas +3 becomes more prominent later. Nearby 5f, 6d and 7s energies and bonding contribute; their states are not forced into a smooth curve.

Both series show contraction, but early 5f orbitals can participate more in bonding than relatively buried 4f orbitals. The nucleus illustration represents radioactive change, not electron loss to form a chemical ion. Oxidation state and nuclear stability answer different questions.

Go deeper: Correct broad historical generalisations

The chapter’s old Lr ground-state configuration is superseded by NIST’s [Rn]5f¹⁴7s²7p¹. This does not remove Lr from the named actinoid series. Likewise, “only promethium is radioactive” overstates the lanthanoid comparison: other lanthanoid elements also have radioactive isotopes.

Uranium and plutonium illustrate isotope-specific nuclear applications; they are not interchangeable materials simply because both are actinoids. Transition/inner-transition chemistry also supports structural alloys, catalysts and pigments: the common white titanium pigment is TiO₂, not TiO.

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Question

A uranium ion changes oxidation state from +4 to +6 without changing its isotope. Is this radioactive decay? Explain.

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NCERT reference: chapter PDF pages 23, 24, 25.

The d- and f-Block Elements Class 12 Notes

Chapter notes

The d- and f-Block Elements

Position and Definition of Transition Elements

Locate the block, then test the definition

Electronic Configurations of d-Block Elements

Fill orbitals, then remove the outer electrons

Physical Properties and Enthalpy of Atomisation

Metallic bonding has measurable consequences

Atomic Sizes and Lanthanoid Contraction

Compare like radii and see contraction

Variability of Oxidation States

One metal can have several oxidation states

Standard Electrode Potentials (E°)

A redox tendency includes the whole environment

Magnetic Properties

Count unpaired electrons before predicting a moment

Formation of Coloured Ions and Complexes

The light absorbed differs from the colour seen

Catalytic Properties and Interstitial Compounds

A surface, a redox cycle and a crystal gap are different ideas

Potassium Dichromate (K₂Cr₂O₇)

Chromate and dichromate join without changing Cr(VI)

Potassium Permanganate (KMnO₄)

The medium changes the reduction product

The Lanthanoids (4f series)

4f occupancy and a shrinking series

The Actinoids (5f series)

Compare 5f chemistry without forcing a smooth trend

Now practise The d- and f-Block Elements, free