Class 12Chemistry · Inorganic ChemistryFull chapter

The d- and f-Block Elements

The whole chapter in one place — read it, then test yourself. Clear notes, key facts, a practice quiz, and worked NCERT solutions & PYQs.

Position and Electronic Configuration of d-Block Elements

Quick answer Transition (d-block) elements occupy groups 3-12 between the s- and p-blocks and are defined by partially filled (n-1)d orbitals in the atom or a common ion, giving anomalous configurations for elements like Cr and Cu.

The d-block lies between the s-block and p-block, spanning groups 3 to 12. It contains four series, each corresponding to the filling of a different set of d orbitals: the 3d series (Sc to Zn, Z = 21-30), the 4d series (Y to Cd, Z = 39-48), the 5d series (La, then Hf to Hg, Z = 57, 72-80), and the largely incomplete 6d series (Ac, then Rf onwards).

The general electronic configuration of d-block atoms is (n-1)d¹⁻¹⁰ns⁰⁻², reflecting the fact that the (n-1)d and ns subshells have very close orbital energies, so electrons occupy both as the series is traversed.

Strictly, a transition element is one whose atom (in the ground state) or one of whose commonly formed ions has an incompletely filled d subshell. By this definition, Zn, Cd and Hg are excluded from the "typical" transition elements: their atoms are (n-1)d¹⁰ns², and in their only common oxidation state (+2) they lose the ns electrons to leave a still-complete (n-1)d¹⁰ ion.

Two configurations break the simple (n-1)d¹⁻⁹ns² pattern: chromium and copper. A half-filled or completely filled d subshell combined with a half-filled s subshell is more stable than the "expected" configuration, because of extra exchange energy when electrons of parallel spin occupy separate orbitals. Hence Cr (Z = 24) is [Ar]3d⁵4s¹ (not 3d⁴4s²), and Cu (Z = 29) is [Ar]3d¹⁰4s¹ (not 3d⁹4s²). The same reasoning extends to Mo, W, Ag, Au and several actinoids.

Worked example - configuration of an ion: Find the ground-state configuration of Fe (Z = 26) and hence of Fe³⁺, and count its unpaired electrons.

  1. Fe (Z = 26): [Ar] 3d⁶4s².
  2. When a transition metal forms a cation, the ns electrons are always removed before any (n-1)d electrons. Removing both 4s electrons gives Fe²⁺: [Ar]3d⁶.
  3. Removing one more electron (from 3d) gives Fe³⁺: [Ar]3d⁵.
  4. 3d⁵ is exactly half-filled, so by Hund's rule all five electrons are unpaired: Fe³⁺ has 5 unpaired electrons, which is why it is markedly paramagnetic.
General configuration of d-block atoms (n−1)d¹⁻¹⁰ ns⁰⁻² n = principal quantum number of the outer shell; applies across the four transition series
Chromium (anomalous) Cr (Z=24): [Ar] 3d⁵ 4s¹ Half-filled 3d⁵ + half-filled 4s¹ gives extra exchange stabilisation
Copper (anomalous) Cu (Z=29): [Ar] 3d¹⁰ 4s¹ A completely filled 3d¹⁰ is more stable than 3d⁹4s²
Remember
  • d-block spans groups 3-12, across four series: 3d, 4d, 5d and 6d.
  • General configuration (n-1)d¹⁻¹⁰ns⁰⁻² arises because (n-1)d and ns subshells are close in energy.
  • Transition element = atom or common ion has a partially filled d subshell; excludes Zn, Cd, Hg (always d¹⁰).
  • Cr and Cu adopt (n-1)d⁵ns¹ and (n-1)d¹⁰ns¹ configurations for extra exchange-energy stability.
  • When 3d metals form cations, ns electrons are always removed before (n-1)d electrons.

General Properties of Transition Elements

Quick answer Transition metals show characteristic trends in size, ionisation enthalpy, oxidation states, colour, magnetism and catalytic behaviour, all traceable to their partly filled d orbitals.

Across any transition series, atomic and ionic radii fall gradually rather than sharply. Increasing nuclear charge pulls the electron cloud inward, but the added electrons enter the inner (n-1)d subshell, which screens the outer ns electrons reasonably well; radii level off towards the middle of a series and creep up slightly near the end where d-d electron repulsion becomes significant. A striking outcome (covered fully in the lanthanoid section) is that corresponding 4d and 5d elements - e.g. Zr and Hf - end up with almost the same radius.

Because unpaired d electrons contribute to metallic bonding in addition to the s electrons, elements near the middle of a series (Cr, Mo, W; Fe) have the greatest number of unpaired electrons and hence the strongest metallic bonding, the highest melting points, and the highest enthalpies of atomisation. Tungsten has the highest melting point of any metal. Elements at the ends of a series (Sc, Zn) have fewer unpaired d electrons and correspondingly lower atomisation enthalpies.

Ionisation enthalpies generally increase across a series but with irregularities, because successive electrons are removed from configurations of differing relative stability. Removing an electron from a stable half-filled (d⁵) or fully filled (d¹⁰) ion is harder than the general trend predicts - this is why the third ionisation enthalpy of manganese (breaking into the stable Mn²⁺, 3d⁵ configuration) is unusually high.

Transition metals show variable oxidation states that typically differ by units of 1 (e.g. Fe²⁺/Fe³⁺, Mn²⁺/Mn³⁺/Mn⁴⁺), unlike main-group elements, whose states usually differ by 2. Both the ns and several (n-1)d electrons lie close enough in energy to take part in bonding. Manganese shows the widest range in the 3d series, from +2 up to +7 (as in MnO₄⁻); the number of accessible states rises from Sc to Mn and falls again towards Zn.

Most transition metal ions are coloured because their partially filled d orbitals allow electrons to be promoted between split d-orbital energy levels by absorbing visible light (a "d-d transition"). Ions with no d electrons (d⁰, e.g. Sc³⁺, Ti⁴⁺) or a completely filled set (d¹⁰, e.g. Zn²⁺, Cu⁺) have no d-d transition available and are colourless.

Because of unpaired d electrons, most transition metal ions are paramagnetic; the number of unpaired electrons can be estimated from the measured magnetic moment.

Worked example: Estimate the spin-only magnetic moment of Mn²⁺.

  1. Mn (Z = 25): [Ar]3d⁵4s². Mn²⁺ is formed by removing the two 4s electrons: [Ar]3d⁵.
  2. 3d⁵ is exactly half-filled, so all 5 electrons are unpaired: n = 5.
  3. μ = √(n(n+2)) BM = √(5 × 7) BM = √35 BM ≈ 5.92 BM, consistent with the strong paramagnetism of Mn²⁺ salts.

Transition metals are also widely used as catalysts (variable oxidation states let them form reactive intermediates), readily form interstitial compounds (small atoms such as H, C or N occupy gaps in the metal lattice, giving hard, high-melting, conducting solids), and form alloys easily with one another because their atomic sizes are similar.

Spin-only magnetic moment μ = √(n(n+2)) BM Bohr Magneton (BM) · n = number of unpaired electrons in the ion
Remember
  • Atomic/ionic radii shrink gradually across a series and are nearly equal for corresponding 4d/5d elements.
  • Melting points and atomisation enthalpies peak near the middle of each series (max unpaired d electrons).
  • Oxidation states change in steps of 1; Mn shows the widest range (+2 to +7).
  • Colour arises from d-d transitions; d⁰ and d¹⁰ ions (Sc³⁺, Ti⁴⁺, Zn²⁺) are colourless.
  • Paramagnetism is proportional to unpaired d electrons; μ = √(n(n+2)) BM.
  • Variable oxidation states plus intermediate formation make transition metals and their compounds good catalysts.

Important Compounds: Potassium Dichromate and Potassium Permanganate

Quick answer K2Cr2O7 and KMnO4 are the two industrially and analytically important oxoanion compounds of the d-block, made from their ores and valued as strong oxidising agents in acidic medium.

Potassium dichromate (K₂Cr₂O₇) is manufactured from chromite ore (FeCr₂O₄) in three stages. First, the ore is fused with sodium carbonate in free access of air, converting the chromium into water-soluble sodium chromate while iron is oxidised to insoluble Fe₂O₃: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂.

The filtered yellow sodium chromate solution is acidified, converting chromate (CrO₄²⁻) into the more stable orange dichromate (Cr₂O₇²⁻): 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O. Finally, treating the sodium dichromate solution with potassium chloride precipitates the less soluble potassium dichromate on cooling: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl. The dichromate ion consists of two CrO₄ tetrahedra (Cr in +6) sharing one corner oxygen.

Chromate and dichromate interconvert reversibly with pH: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. Acid shifts the equilibrium towards orange dichromate; base shifts it back towards yellow chromate.

In acidic solution, dichromate is a powerful oxidant, reduced to green Cr³⁺: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (E° = +1.33 V). It oxidises iodide to iodine and Fe²⁺ to Fe³⁺, and is used in volumetric analysis.

Potassium permanganate (KMnO₄) is manufactured from pyrolusite (MnO₂). Fusing MnO₂ with KOH in the presence of an oxidant (air or KNO₃) gives the green manganate ion: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O. The manganate is then oxidised (commercially by electrolytic oxidation at the anode) to deep purple permanganate, MnO₄⁻ (Mn in +7, tetrahedral).

KMnO₄ is an even stronger oxidant than dichromate in acid, reduced all the way to Mn²⁺: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (E° = +1.51 V). In neutral/faintly alkaline solution the product is brown MnO₂ (Mn reduced only to +4), and in strongly alkaline solution the product is green manganate, MnO₄²⁻ (Mn reduced only to +6) - the depth of reduction, and hence oxidising strength, depends on pH.

Worked example - redox titration: What volume of 0.02 M KMnO₄ is required to completely oxidise the Fe²⁺ in 25.0 mL of 0.1 M FeSO₄ solution in acidic medium?

  1. Balanced equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, so 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺.
  2. Moles of Fe²⁺ = 0.1 mol L⁻¹ × 0.0250 L = 2.5 × 10⁻³ mol.
  3. Moles of MnO₄⁻ needed = (2.5 × 10⁻³)/5 = 5.0 × 10⁻⁴ mol.
  4. Volume of 0.02 M KMnO₄ = (5.0 × 10⁻⁴ mol)/(0.02 mol L⁻¹) = 0.025 L = 25.0 mL.
Chromite ore fusion 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ Roasting chromite with sodium carbonate in air
Chromate-dichromate equilibrium 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O pH-dependent; acid favours orange dichromate, base favours yellow chromate
Dichromate as oxidant (acidic) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O E° = +1.33 V · Six-electron change per dichromate ion
Permanganate as oxidant (acidic) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O E° = +1.51 V · Five-electron change per permanganate ion; strongest oxidising action of KMnO4 occurs in acidic medium
Remember
  • K2Cr2O7 is made from chromite ore via sodium chromate → sodium dichromate → potassium dichromate (which crystallises preferentially, being less soluble).
  • Chromate (yellow, CrO4²⁻) and dichromate (orange, Cr2O7²⁻) interconvert reversibly with pH.
  • Cr2O7²⁻ is a strong oxidant in acid (E° = +1.33 V), reduced to Cr³⁺.
  • KMnO4 is made from pyrolusite (MnO2) via potassium manganate, oxidised electrolytically to permanganate.
  • MnO4⁻ is a strong oxidant in acid (E° = +1.51 V) reduced to Mn²⁺; the product changes with pH (MnO2 in neutral, MnO4²⁻ in strongly alkaline medium).
  • Both compounds are widely used as volumetric (redox-titration) reagents in quantitative analysis.

The Lanthanoids

Quick answer The 14 elements from cerium to lutetium follow lanthanum by filling the 4f orbitals; they are chemically similar, show the lanthanoid contraction, and are dominated by the +3 oxidation state.

The lanthanoids are the 14 elements from cerium (Ce, Z = 58) to lutetium (Lu, Z = 71) that follow lanthanum (La, Z = 57), corresponding to progressive filling of the 4f orbitals. Their general configuration is [Xe]4f¹⁻¹⁴5d⁰⁻¹6s² (La itself, often studied alongside them, is [Xe]4f⁰5d¹6s²).

Because 4f orbitals lie deep inside the atom, shielded by the filled 5s and 5p subshells, they take almost no part in bonding. All lanthanoids are therefore chemically similar, and +3 is overwhelmingly the characteristic oxidation state. A few elements also show +2 or +4 when this produces a specially stable f⁰, f⁷ (half-filled) or f¹⁴ (fully filled) configuration.

Worked example - anomalous oxidation states:

  1. Ce (Z = 58): [Xe]4f¹5d¹6s². Losing all four outer electrons gives Ce⁴⁺: [Xe]4f⁰ - an empty f subshell, isoelectronic with the preceding noble gas Xe. This extra stability makes Ce⁴⁺ compounds useful oxidising agents, unusual for a lanthanoid.
  2. Eu (Z = 63): [Xe]4f⁷6s². Losing only the two 6s electrons gives Eu²⁺: [Xe]4f⁷ - an exactly half-filled f subshell with maximum exchange stabilisation, so Eu commonly appears as Eu²⁺ alongside the "normal" Eu³⁺.

The defining structural feature of the series is the lanthanoid contraction: atomic and ionic radii decrease steadily from La to Lu. Each additional 4f electron shields the rising nuclear charge poorly, because 4f orbitals are diffuse with a complex radial shape, so the effective nuclear charge felt by outer electrons rises steadily, pulling the electron cloud inward.

This contraction has consequences beyond the lanthanoids themselves: it almost exactly cancels the size increase normally expected on descending a group, so the 4d and 5d transition elements of a given group end up nearly identical in radius (e.g. Zr ≈ Hf, Nb ≈ Ta), making such pairs difficult to separate and chemically very alike. Within the lanthanoids, the steadily shrinking, increasingly charge-dense Ln³⁺ ions become progressively less basic, so the base strength of Ln(OH)₃ falls steadily from La(OH)₃ to Lu(OH)₃.

Most Ln³⁺ ions are coloured, from f-f transitions among partly filled 4f orbitals; ions with f⁰, f⁷ or f¹⁴ configurations (La³⁺, Gd³⁺, Lu³⁺) are essentially colourless. Practically, lanthanoids are used as mischmetal (a lanthanoid-iron alloy used in lighter flints), as catalysts in petroleum cracking, and in glass and ceramic formulations.

General configuration of lanthanoids [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s² Applies to Ce (Z=58) through Lu (Z=71); La (Z=57) is 4f⁰5d¹6s²
Ce4+ (special stability) Ce (Z=58): [Xe]4f¹5d¹6s² → Ce⁴⁺: [Xe]4f⁰ Loses all 4 outer electrons to reach a noble-gas-like f⁰ core; makes Ce⁴⁺ a useful oxidising agent
Eu2+ (special stability) Eu (Z=63): [Xe]4f⁷6s² → Eu²⁺: [Xe]4f⁷ Retains the exactly half-filled, exchange-stabilised f⁷ core
Remember
  • Lanthanoids = Ce(58) to Lu(71); general configuration [Xe]4f¹⁻¹⁴ 5d⁰⁻¹ 6s².
  • +3 is the characteristic oxidation state; +2/+4 appear only for extra f⁰, f⁷ or f¹⁴ stability (Ce⁴⁺, Eu²⁺, Tb⁴⁺, Yb²⁺).
  • Lanthanoid contraction: steady fall in atomic/ionic radii across the series due to poor shielding by diffuse 4f electrons.
  • Consequence: 4d and 5d elements of the same group (e.g. Zr/Hf) have almost identical radii and properties.
  • Ln³⁺ ions are often coloured (f-f transitions); f⁰, f⁷, f¹⁴ ions (La³⁺, Gd³⁺, Lu³⁺) are colourless.
  • Basicity of Ln(OH)3 decreases steadily from La(OH)3 to Lu(OH)3 as the ion shrinks.

The Actinoids

Quick answer The 14 elements from thorium to lawrencium fill the 5f orbitals; all are radioactive, show a wider range of oxidation states than lanthanoids, and undergo an even larger actinoid contraction.

The actinoids are the 14 elements from thorium (Th, Z = 90) to lawrencium (Lr, Z = 103) that follow actinium (Ac, Z = 89), corresponding to filling of the 5f orbitals. Their general configuration is [Rn]5f¹⁻¹⁴6d⁰⁻¹7s², but with more frequent irregularities than the lanthanoids because 5f, 6d and 7s orbitals lie unusually close together in energy.

Worked example - configuration irregularities:

  1. Th (Z = 90): no 5f electron at all in the ground state - [Rn]6d²7s² (analogous to the La anomaly among lanthanoids).
  2. U (Z = 92): [Rn]5f³6d¹7s² - both 5f and 6d orbitals are occupied.
  3. Because 5f, 6d and 7s electrons are all comparably easy to remove or share, uranium can adopt oxidation states from +3 up to +6 (the +6 state occurs as the uranyl ion, UO₂²⁺), unlike lanthanoids, which are almost always +3.

Every actinoid is radioactive; only thorium and uranium occur in nature in appreciable quantity, while the rest are obtained in trace amounts or synthesised artificially. This radioactivity, together with the very short half-lives of the heaviest members, makes some of their chemical properties difficult to study experimentally.

Actinoids show a much wider range of oxidation states than lanthanoids, especially among earlier members: +3 is common throughout, but +4 (Th, U, Pu), +5 and +6 (U, Np, Pu, as oxo-cations like UO₂²⁺) and even +7 (Np, Pu under strongly oxidising conditions) are known. This reflects the closely spaced 5f/6d/7s orbital energies, allowing more electrons to be involved in bonding than the deeply buried, largely non-bonding 4f electrons of lanthanoids.

Like lanthanoids, actinoids show a steady contraction in atomic and ionic radii across the series (the "actinoid contraction"), caused by poor shielding of the increasing nuclear charge by 5f electrons. Because 5f orbitals are even more diffuse and penetrate less effectively than 4f orbitals, the actinoid contraction per element is somewhat larger than the lanthanoid contraction. Actinoid ions are also frequently coloured, from 5f-5f (and sometimes 5f-6d) transitions, and actinoids show a greater tendency to form complexes than lanthanoids, since their 5f, 6d and 7s orbitals are all energetically accessible for bonding.

General configuration of actinoids [Rn] 5f¹⁻¹⁴ 6d⁰⁻¹ 7s² Th (Z=90) through Lr (Z=103); exceptions exist (e.g. Th is 6d²7s² with no 5f electron)
Uranium ground-state configuration U (Z=92): [Rn] 5f³ 6d¹ 7s² Comparable 5f/6d/7s energies allow U to show +3 to +6 oxidation states
Remember
  • Actinoids = Th(90) to Lr(103); general configuration [Rn]5f¹⁻¹⁴ 6d⁰⁻¹ 7s², with several irregular fillings.
  • All actinoids are radioactive; only Th and U are found in nature in significant amounts.
  • Oxidation states range more widely than lanthanoids: +3 is common, but +4, +5, +6 (and even +7) occur for early actinoids (U, Np, Pu).
  • 5f, 6d and 7s orbitals lie close in energy, so more electrons can be involved in bonding than in lanthanoids.
  • Actinoid contraction (per element) is larger than lanthanoid contraction because 5f orbitals shield the nuclear charge even less effectively than 4f orbitals.
  • Actinoid ions are frequently coloured and show a greater tendency to form complexes than lanthanoids.

Comparing Lanthanoids and Actinoids, and Applications of d- and f-Block Elements

Quick answer Lanthanoids and actinoids share a family resemblance but differ in radioactivity, oxidation-state range and orbital energetics, while transition and inner-transition metals together underpin catalysts, alloys, pigments and nuclear fuel.

Lanthanoids and actinoids are both "f-block" families and share several resemblances: both show a steady contraction in radius across the series, both are dominated (though to different extents) by the +3 oxidation state, and within each family the elements are chemically very similar to one another. The differences are equally important for understanding their chemistry.

  • All actinoids are radioactive; only a few lanthanoids (like promethium) are radioactive, the rest are stable.
  • Actinoids show a much wider spread of oxidation states (+3 to +7 for some elements) because 5f, 6d and 7s orbital energies are close together; lanthanoids are overwhelmingly +3, with only occasional +2/+4 exceptions.
  • The actinoid contraction (per element) is greater than the lanthanoid contraction, since 5f orbitals shield the nuclear charge even less effectively than 4f orbitals.
  • Actinoids have a distinctly greater tendency to form coordination complexes than lanthanoids, because more of their orbitals are available for bonding.

Both the d-block and f-block families underpin many practical applications. Transition metals and their compounds are used extensively as catalysts - iron in the Haber process for ammonia, vanadium(V) oxide in the Contact process for sulfuric acid, and finely divided nickel or platinum/palladium for hydrogenation reactions. Their ability to form alloys readily (owing to similar atomic sizes) gives us steel and stainless steel (Fe with C, Cr, Ni). Transition-metal compounds are also valued for their colour and redox behaviour - titanium dioxide as a white pigment, and manganese dioxide as the oxidant in ordinary dry-cell batteries.

f-block elements have their own distinctive applications: mischmetal (a lanthanoid alloy) in lighter flints, mixed lanthanoid oxides as catalysts and in glass-polishing/decolourising formulations, and - most significantly for actinoids - thorium, uranium and plutonium as nuclear fuels, exploiting the very properties (variable oxidation states, large diffuse f orbitals, radioactivity) that define this family.

Remember
  • Both lanthanoids and actinoids show contraction and a dominant +3 state, but actinoids are radioactive and show far wider oxidation-state ranges.
  • Actinoids form complexes more readily than lanthanoids because 5f, 6d and 7s orbitals are all energetically accessible.
  • Transition metals and compounds are used as catalysts (Fe, Ni, V2O5, Pt/Pd), in alloys (steel, stainless steel), as pigments (TiO2), and in batteries (MnO2).
  • f-block elements are used in mischmetal alloys, glass/ceramic formulations, and as nuclear fuel (U, Pu, Th).
  • Studying the two inner-transition families together shows how (n-2)f orbital filling shapes size, colour, magnetism and oxidation-state chemistry.

Key facts & terms

Every formula in this chapter, in one place — screenshot it before your exam.

(n−1)d¹⁻¹⁰ ns⁰⁻²
General configuration of d-block atoms
Cr (Z=24): [Ar] 3d⁵ 4s¹
Chromium (anomalous)
Cu (Z=29): [Ar] 3d¹⁰ 4s¹
Copper (anomalous)
μ = √(n(n+2)) BM
Spin-only magnetic momentBohr Magneton (BM)
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
Chromite ore fusion
2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O
Chromate-dichromate equilibrium
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Dichromate as oxidant (acidic)E° = +1.33 V
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Permanganate as oxidant (acidic)E° = +1.51 V
[Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s²
General configuration of lanthanoids
Ce (Z=58): [Xe]4f¹5d¹6s² → Ce⁴⁺: [Xe]4f⁰
Ce4+ (special stability)
Eu (Z=63): [Xe]4f⁷6s² → Eu²⁺: [Xe]4f⁷
Eu2+ (special stability)
[Rn] 5f¹⁻¹⁴ 6d⁰⁻¹ 7s²
General configuration of actinoids
U (Z=92): [Rn] 5f³ 6d¹ 7s²
Uranium ground-state configuration

Test yourself

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0 correct · 0/12 answered
Q1 Position and definition of d-block elements easy

Which of the following is NOT regarded as a typical transition element?

Q2 Electronic configuration easy

The general electronic configuration of d-block elements is

Q3 Magnetic properties medium

Which pair of ions has the same number of unpaired 3d electrons?

Q4 Magnetic moment calculation medium

What is the spin-only magnetic moment of high-spin Fe²⁺ (3d⁶)?

Q5 Chromate-dichromate equilibrium medium

An orange aqueous solution of K2Cr2O7 turns yellow on adding NaOH. This happens because

Q6 Redox titration numerical medium

20.0 mL of 0.5 M FeSO4 solution (acidified) is exactly oxidised by 20.0 mL of KMnO4 solution (MnO4⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H2O). What is the molarity of the KMnO4 solution?

Q7 Lanthanoid contraction medium

The lanthanoid contraction is primarily caused by

Q8 Consequences of lanthanoid contraction medium

Which of the following is a direct consequence of the lanthanoid contraction?

Q9 Oxidation states of lanthanoids hard

Cerium readily forms the Ce⁴⁺ ion, which is a useful oxidising agent in volumetric analysis, mainly because

Q10 Properties of actinoids hard

Which statement about the actinoids is INCORRECT?

Q11 Magnetic moment - inverse calculation hard

A first-row transition metal ion M²⁺ has a measured spin-only magnetic moment of 3.87 BM. How many unpaired electrons does it have?

Q12 Standard reduction potentials hard

Which of the following half-reactions (all in acidic aqueous medium) has the most positive standard reduction potential, i.e. represents the strongest oxidising agent?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Silver has a completely filled d subshell (4d¹⁰) in its ground state ([Kr]4d¹⁰5s¹). How can it be classified as a transition element?Definition of transition elements

A transition element is formally defined as an element whose atom, in its ground state or in any of its commonly encountered oxidation states, has an incompletely filled d subshell.

Silver (Z = 47) has the ground-state configuration [Kr]4d¹⁰ 5s¹, which does have a completely filled 4d subshell. However, silver also shows the +2 oxidation state in compounds such as AgF₂. In the Ag²⁺ ion, one electron is removed from the filled 4d shell, giving [Kr]4d⁹ 5s⁰ - a partially filled d subshell.

Because this partially filled configuration appears in one of silver's known oxidation states, silver satisfies the definition of a transition element, even though its more common Ag⁺ state (4d¹⁰) is not itself transition-like.

2 In the 3d series (Sc to Zn), zinc has the lowest enthalpy of atomisation. Why?Enthalpy of atomisation

Enthalpy of atomisation of a metal depends on the strength of metallic bonding, which in transition metals is reinforced by delocalisation of unpaired d electrons in addition to the s electrons.

Across the 3d series, elements with more unpaired d electrons (e.g. Cr, 3d⁵ 4s¹) form stronger metallic bonds and have very high atomisation enthalpies.

Zinc (Z = 30) has the configuration [Ar]3d¹⁰ 4s². Its 3d subshell is completely filled, so it has no unpaired d electrons available for metallic bonding - only the two 4s electrons contribute. This gives comparatively weak metallic bonding and hence the lowest enthalpy of atomisation in the series.

3 Write the electronic configurations of the elements with atomic numbers 61, 91, 101 and 109.Electronic configurations of f-block and 6d elements

Using the filling pattern for the sixth and seventh periods, where 5f/6d/7s (and 4f/5d/6s) orbitals lie close in energy:

  • Z = 61 (Promethium, Pm): [Xe] 4f⁵ 6s² - a lanthanoid with 5 electrons in the 4f subshell.
  • Z = 91 (Protactinium, Pa): [Rn] 5f² 6d¹ 7s² - an actinoid; both 5f and 6d orbitals are partly occupied because their energies are very close.
  • Z = 101 (Mendelevium, Md): [Rn] 5f¹³ 7s² - a late actinoid with no 6d electron.
  • Z = 109 (Meitnerium, Mt): [Rn] 5f¹⁴ 6d⁷ 7s² - a 6d-series (group 9) element; this configuration is based on periodic extrapolation, as Mt is a synthetic, extremely short-lived element.
4 Why is Cu⁺ ion not stable in aqueous solution?Stability of oxidation states

Cu⁺ (3d¹⁰) disproportionates in aqueous solution: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s).

Although the second ionisation enthalpy of copper (Cu⁺ → Cu²⁺) is higher than the first, the much greater (more negative) hydration enthalpy of the smaller, higher-charged Cu²⁺ ion more than compensates for this extra ionisation energy. The overall free energy change for the disproportionation is negative, so Cu⁺ is thermodynamically unstable relative to Cu²⁺ and Cu(s) in water, and only Cu²⁺ salts are commonly encountered in aqueous solution.

5 Calculate the 'spin-only' magnetic moment of a divalent ion in aqueous solution if its atomic number is 27.Magnetic moment calculation

Step 1 - identify the ion: Z = 27 is cobalt. The divalent ion Co²⁺ is formed by removing the two 4s electrons from [Ar]3d⁷ 4s², giving [Ar] 3d⁷.

Step 2 - count unpaired electrons: Distributing 7 electrons among the five 3d orbitals by Hund's rule (high-spin, as normal for Co²⁺(aq)): two orbitals get paired electrons and three remain singly occupied, giving 3 unpaired electrons.

Step 3 - apply the spin-only formula: μ = √(n(n+2)) BM = √(3 × 5) BM = √15 BM ≈ 3.87 BM.

So the spin-only magnetic moment of the Co²⁺(aq) ion is approximately 3.87 Bohr Magnetons.

6 What is lanthanoid contraction? What are its consequences?Lanthanoid contraction

Lanthanoid contraction is the steady, cumulative decrease in atomic and ionic radii of the lanthanoid elements (Ce to Lu) with increasing atomic number as the 4f orbitals are progressively filled.

Cause: as each additional electron enters a 4f orbital, it shields the increasing nuclear charge very poorly because 4f orbitals are diffuse and radially extended. The effective nuclear charge felt by outer electrons therefore rises steadily across the series, pulling the outer shell inward.

Consequences:

  • Radii of the second (4d) and third (5d) transition series elements of a given group become almost identical (e.g. Zr ≈ Hf, Nb ≈ Ta), because the lanthanoid contraction offsets the normal radius increase down a group. This makes such pairs of elements very similar in chemical properties and hard to separate.
  • The basic strength of the lanthanoid hydroxides, Ln(OH)3, decreases steadily from La(OH)3 to Lu(OH)3 as the ions get smaller and more polarising.
  • Because the lanthanoids shrink only gradually and their +3 ions are chemically very similar, separating individual lanthanoids requires methods such as ion-exchange chromatography.

Previous-year board questions 4

Q1 Account for the following: (i) Zn, Cd and Hg are not regarded as transition elements although they belong to the d-block. (ii) The E° value for the Mn³⁺/Mn²⁺ couple (+1.57 V) is much more positive than that for the Cr³⁺/Cr²⁺ couple (−0.41 V). 2023 3 marks

(i) Zn, Cd and Hg have the configuration (n−1)d¹⁰ns² in the free metal, and in their only common oxidation state (+2) they lose the two ns electrons to give an ion with configuration (n−1)d¹⁰ - still completely filled. Since neither the atom nor its common ion has a partially filled d subshell, these elements do not satisfy the definition of a transition element and are excluded, even though they lie in the d-block.

(ii) Mn²⁺ has the configuration 3d⁵, exceptionally stable because of its symmetrical, exactly half-filled arrangement (maximum exchange energy). Converting Mn³⁺ (3d⁴) to Mn²⁺ (3d⁵) therefore releases extra stabilisation energy, making the reduction strongly favourable (highly positive E°); Mn³⁺ is consequently a good oxidising agent.

Cr²⁺ (3d⁴) is comparatively less stable, while Cr³⁺ (3d³, an exactly half-filled t2g set in an octahedral field) is unusually stable. Oxidation of Cr²⁺ to Cr³⁺ is energetically favourable, corresponding to a negative E° for the Cr³⁺/Cr²⁺ couple - i.e. Cr²⁺ is a strong reducing agent rather than Cr³⁺ being a strong oxidant.

Q2 Calculate the spin-only magnetic moment of the Ni²⁺ ion. (Atomic number of Ni = 28) 2022 4 marks

Step 1: Ni (Z = 28): [Ar] 3d⁸ 4s². Removing the two 4s electrons to form Ni²⁺ gives [Ar] 3d⁸.

Step 2: Distributing 8 electrons in the five d orbitals by Hund's rule leaves 2 unpaired electrons (three orbitals doubly occupied, two singly occupied).

Step 3: μ = √(n(n+2)) BM = √(2 × 4) BM = √8 BM ≈ 2.83 BM.

This value (~2.8-2.9 BM) matches the magnetic moment observed experimentally for octahedral Ni²⁺ complexes, confirming 2 unpaired d electrons.

Q3 Describe the preparation of potassium dichromate starting from chromite ore. Write the ionic equations for its reaction with (i) KI and (ii) FeSO4 in acidic medium. 2024 5 marks

Preparation from chromite ore (FeCr2O4):

Step 1 - Fusion with sodium carbonate in air: 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2. This converts chromium into water-soluble yellow sodium chromate, while iron is left behind as insoluble Fe2O3.

Step 2 - Acidification: the filtered sodium chromate solution is acidified, converting chromate into the more stable orange dichromate: 2Na2CrO4 + 2H⁺ → Na2Cr2O7 + 2Na⁺ + H2O.

Step 3 - Conversion to the potassium salt: the sodium dichromate solution is treated with potassium chloride. Potassium dichromate is less soluble than sodium dichromate, so it crystallises out on cooling and is separated by filtration: Na2Cr2O7 + 2KCl → K2Cr2O7↓ + 2NaCl.

Reactions of acidified dichromate (Cr2O7²⁻ → 2Cr³⁺):

(i) With iodide: Cr2O7²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I2 + 7H2O

(ii) With Fe²⁺: Cr2O7²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H2O

In both cases dichromate (Cr in the +6 state) is reduced to Cr³⁺, while iodide is oxidised to iodine and Fe²⁺ is oxidised to Fe³⁺ respectively.

Q4 The third ionisation enthalpy of manganese is exceptionally high compared to its neighbouring elements in the 3d series. Explain, and state its consequence for the stability of the +2 oxidation state of manganese. 2023 3 marks

Manganese (Z = 25) has the ground-state configuration [Ar] 3d⁵ 4s². Removing two electrons to form Mn²⁺ gives [Ar] 3d⁵ - a symmetrical, exactly half-filled d subshell that is extra stable due to maximum exchange energy among the five parallel-spin electrons.

Removing a third electron (Mn²⁺ → Mn³⁺) means breaking this especially stable half-filled configuration to give 3d⁴. This costs significantly more energy than the general trend predicts, so the third ionisation enthalpy of Mn is unusually high - higher than that of both its neighbours Cr and Fe.

Consequence: the +2 oxidation state (3d⁵) is the most stable and most commonly encountered oxidation state of manganese in simple aqueous salts, while Mn³⁺ is comparatively unstable in solution and behaves as a fairly strong oxidising agent, readily gaining an electron to revert to the stable Mn²⁺ state.

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