d- and f-Block Elements

1. Position, Configuration and General Trends

• Location: d-block occupies Groups 3–12, bridging electropositive s-block metals and covalent p-block elements. General configuration (n−1)d^1–10ns^1–2. Transition element = has an incompletely filled d-subshell in the ground state OR in any common oxidation state. Four series: 3d (Sc→Zn, Period 4), 4d (Y→Cd, Period 5), 5d (La, Hf→Hg, Period 6), 6d (Ac, Rf onward, Period 7).
• Group 12 (Zn, Cd, Hg) is d-block but NOT transition: d¹⁰ in both the ground state and the common +2 state, so no incomplete d-subshell anywhere. Pd is the odd one out: [Kr]4d¹⁰5s⁰ (skips 5s entirely) for maximum subshell stability.
• Atomic/ionic radius across a period: falls initially (rising nuclear charge), nearly flat Cr→Cu (added d-electrons screen the outer s-electrons), slight RISE at Zn (inter-electronic repulsion in filled d¹⁰). Down a group: 4d > 3d, but 5d ≈ 4d (not bigger!) — this is the lanthanoid contraction's direct consequence.

• Lanthanoid contraction: the 4f orbitals fill just before the 5d elements; 4f electrons shield poorly, so nuclear charge rises faster than the electron cloud can compensate, steadily shrinking the lanthanoids. Net effect: 4d and 5d congeners (Zr/Hf, Nb/Ta) end up nearly IDENTICAL in size and chemistry.
• Melting/boiling points: very high (strong metallic bonding), rising to a maximum around Group 6 (Cr, Mo, W — maximum unpaired d-electrons available for bonding), then falling as electrons pair up. Mn and Tc dip abnormally low: their stable half-filled d⁵ configuration resists the electron delocalisation that metallic bonding needs.
• Density: high (small atomic volume + high nuclear charge); rises steadily toward the end of the series. Os (22.57 g/cm³) and Ir (22.61 g/cm³) are the densest elements known. Zn is the exception (larger atomic volume, lower density).
• Ionization enthalpy: intermediate between s- and p-block. IE₁ of 5d elements > 3d and 4d (weak shielding by the 4f electrons beneath them, same cause as lanthanoid contraction).

2. Oxidation States and Ionization Enthalpy Anomalies

• Why oxidation states vary: (n−1)d and ns orbitals are close in energy, so multiple numbers of electrons can be lost at comparable energy cost. Lowest state = ns electrons only; highest state = ns + all unpaired (n−1)d electrons.
• Sc shows only +3 (minimum variability); Mn shows +2 to +7 (maximum variability, mid-series). Step difference between successive transition-metal oxidation states is 1 unit (V^II, V^III, V^IV, V^V), unlike the p-block's 2-unit steps (inert pair effect).
• Highest oxidation state overall: +8, shown only by Os (OsO₄) and Ru (RuO₄).
• Group stability trend (opposite of p-block): heavier transition elements stabilize their HIGHER oxidation state better — Mo^VI and W^VI are stable, while Cr^VI (in CrO₄²⁻/Cr₂O₇²⁻) is strongly oxidising and readily falls to Cr³⁺.
• IE anomalies (all from 3d⁵/3d¹⁰ stability): IE₁ of Cr < Mn (Cr loses a lone 4s¹ electron; Mn must break a filled 4s², plus Mn has higher nuclear charge) — Cr = 653, Mn = 717 kJ/mol. IE₂ of Cr > Mn (Cr⁺ has the stable [Ar]3d⁵; Mn⁺ only has to lose an ordinary 4s¹) — Cr = 1592, Mn = 1509 kJ/mol. IE₃ of Mn > Cr (Mn²⁺ is the stable 3d⁵, so removing a THIRD electron from it costs more than from Cr²⁺'s 3d⁴) — Mn = 3248, Cr = 2987 kJ/mol. The same logic gives IE₂ of Cu > Zn (Cu⁺ is stable 3d¹⁰).
• Ni(II) vs Pt(IV) thermodynamic stability: Σ(IE₁+IE₂) is lower for Ni than Pt, so Ni(II) compounds are more stable than Pt(II). But Σ(IE₁...IE₄) is lower for Pt than Ni, so Pt(IV) compounds (K₂PtCl₆) are MORE stable than any Ni(IV) — this is why platinum chemistry favours +4 while nickel chemistry favours +2.

3. Electrode Potentials and Redox Behaviour

• What sets E°(M²⁺/M): ΔH_total = Δ_subH° + IE₁ + IE₂ + Δ_hydH° — a balance of sublimation, ionization and hydration energies, not any single one alone.
• Copper is the one positive E° (+0.34 V) among first-row M²⁺/M couples: Cu cannot displace H₂ from dilute acids because the energy needed to convert Cu(s) → Cu²⁺(aq) is not repaid by Cu²⁺'s hydration enthalpy. Only oxidising acids (HNO₃, hot conc. H₂SO₄) dissolve copper.
• Most negative E° peaks: Mn²⁺ (3d⁵) and Zn²⁺ (3d¹⁰) — both stable configurations resist further oxidation, making the METAL comparatively easy to oxidise (very negative E°). Ni²⁺ is exceptionally negative for a different reason: an unusually high (negative) hydration enthalpy.
• M³⁺/M²⁺ couples: Mn³⁺/Mn²⁺ is exceptionally HIGH (+1.57 V) because Mn²⁺ (3d⁵) resists further oxidation strongly. Fe³⁺/Fe²⁺ is comparatively low (+0.77 V) because Fe³⁺ itself reaches the stable 3d⁵. Sc³⁺/Sc²⁺ is very low (Sc³⁺ has a noble-gas-like configuration, extremely hard to reduce further). V³⁺/V²⁺ is negative (−0.26 V, V²⁺'s half-filled t_2g³ in an octahedral field is comparatively stable).
• Reducing power of early +2 ions: Ti²⁺, V²⁺ and Cr²⁺ are all strong enough reducing agents to liberate H₂ gas from dilute acid (2Cr²⁺ + 2H⁺ → 2Cr³⁺ + H₂), unlike Cu which cannot.

4. Colour, Magnetism, Catalysis and Alloys

• Colour from d-d transitions: under a ligand field, d-orbitals split into t_2g and e_g sets; visible light promotes an electron between them, and the complementary wavelength is transmitted/reflected as colour. d⁰ (Sc³⁺, Ti⁴⁺) and d¹⁰ (Cu⁺, Zn²⁺) are colourless — no d-electron to promote, or no vacancy to promote it into.
• Spin-only magnetic moment: μ = √[n(n+2)] BM, n = number of unpaired electrons. Maximum at high-spin d⁵ (Mn²⁺, Fe³⁺): μ = 5.92 BM. Mn²⁺/Fe³⁺ are technically coloured (weak, spin-forbidden d-d transitions) but are treated as effectively colourless in JEE-level reasoning.
• Interstitial compounds: small atoms (H, B, C, N) trapped in the metal lattice's voids. Non-stoichiometric (TiH_1.7, VH_0.56), higher melting point than the pure metal, extremely hard (some borides near diamond hardness), retain metallic conductivity, chemically inert.
• Catalysis: driven by variable oxidation states and the ability to form unstable intermediates that lower activation energy — V₂O₅ in the Contact Process, Fe (with Mo promoter) in the Haber Process.
• Alloys: form readily when atomic radii are within ~15% of each other (Brass = Cu–Zn, Bronze = Cu–Sn, most steels).

5. Key Compounds of Fe, Cu, Zn and Ag

• Ferrous sulphate (FeSO₄·7H₂O, green vitriol): thermal decomposition 2FeSO₄·7H₂O → FeSO₄·H₂O (140°C) → Fe₂O₃ + SO₂ + SO₃ (high T). Fenton's reagent = FeSO₄ + H₂O₂, used to oxidise organic compounds. Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O) crystallisation needs dilute H₂SO₄ added BEFORE dissolving, specifically to prevent Fe²⁺ hydrolysis.
• Ferric chloride (FeCl₃): anhydrous form made by passing dry Cl₂ over heated Fe; exists as the dimer Fe₂Cl₆ in vapour. Heating hydrated FeCl₃·6H₂O does NOT give anhydrous FeCl₃ — it hydrolyses instead, to Fe₂O₃ + HCl.
• Zinc compounds: ZnO (philosopher's wool) is amphoteric, turns yellow on heating and white again on cooling (lattice-defect effect, not decomposition). Heating with Co(NO₃)₂ gives Rinmann's Green (ZnCoO₂) — a qualitative test for Zn²⁺. Lithopone = BaSO₄ + ZnS, a white pigment.
• Copper sulphate (CuSO₄·5H₂O, blue vitriol): 4 H₂O molecules coordinate directly to Cu²⁺; the 5th is hydrogen-bonded to SO₄²⁻. Stepwise dehydration: → CuSO₄·H₂O (100°C) → CuSO₄ white (250°C) → CuO + SO₂ + O₂ (750°C). Cu²⁺ + I⁻: 2Cu²⁺ + 4I⁻ → Cu₂I₂↓ (white ppt) + I₂ (brown solution) — used in iodometric estimation of Cu²⁺.
• Cu(I) vs Cu(II) stability: Cu(II) (3d⁹) is MORE stable in aqueous/ionic chemistry than Cu(I) (3d¹⁰), because Cu²⁺'s high hydration enthalpy compensates for its extra ionization energy. But Cu(I) is stabilised whenever a very strong, low-charge-density ligand is present: e.g. Cu²⁺ + excess CN⁻ forms the extremely stable complex [Cu(CN)₄]³⁻, driving Cu down to +1 (CuCN) and releasing cyanogen gas — so little free Cu⁺ remains that passing H₂S afterward precipitates NO CuS at all.
• Silver compounds: AgNO₃ (lunar caustic) stains skin black (reduced to metallic Ag by organic tissue) and forms explosive silver acetylide (Ag₂C₂) with acetylene. Photography: exposure gives a latent image (2AgBr → 2Ag + Br₂, photochemical); developing (hydroquinone reduces exposed grains fully to Ag); fixing (hypo, Na₂S₂O₃, dissolves UNEXPOSED AgBr as a soluble complex: AgBr + 2Na₂S₂O₃ → Na₃[Ag(S₂O₃)₂] + NaBr).

6. Potassium Permanganate and Potassium Dichromate

• KMnO₄ preparation: pyrolusite fusion 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ (dark green) + 2H₂O, then oxidation (by Cl₂/CO₂, or electrolytically at the anode: MnO₄²⁻ → MnO₄⁻ + e⁻) to purple KMnO₄.
• MnO₄²⁻ (manganate, Mn⁶⁺) disproportionates in acidic/neutral medium: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (Mn simultaneously oxidised +6→+7 and reduced +6→+4).
• Both manganate and permanganate are tetrahedral (sp³), but manganate (MnO₄²⁻, Mn⁶⁺, d¹) is PARAMAGNETIC while permanganate (MnO₄⁻, Mn⁷⁺, d⁰) is DIAMAGNETIC — easy to mix up since they look alike structurally.
• Acidified permanganate OXIDISES (never reduces) oxalate, nitrite and iodide ions.
• K₂Cr₂O₇ preparation: chromite ore roasting 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂; acidify to Na₂Cr₂O₇, then treat with KCl to crystallise K₂Cr₂O₇ (lower solubility than NaCl).

• Chromyl chloride test (identifies chloride specifically): heating a chloride salt with K₂Cr₂O₇ and conc. H₂SO₄ gives deep-red CrO₂Cl₂ vapour (Cr stays +6): 4Cl⁻ + Cr₂O₇²⁻ + 6H⁺ → 2CrO₂Cl₂ + 3H₂O. Confirmed further with NaOH (→ yellow CrO₄²⁻) then Pb²⁺ (→ yellow PbCrO₄ precipitate). Bromide and iodide salts do NOT give this test (their analogous chromyl halides are unstable).

7. f-Block: Lanthanoids and Actinoids

• Filling: Lanthanoids fill 4f (Ce→Lu); Actinoids fill 5f (Th→Lr). Both series are predominantly +3, but actinoids show a much WIDER spread of oxidation states (+4, +5, +6, +7 early in the series) because 5f, 6d and 7s are all close in energy.
• Radioactivity: lanthanoids are non-radioactive (except Promethium, Pm); ALL actinoids are radioactive.
• Contraction: lanthanoid contraction (steady, ~1 pm/element) vs actinoid contraction (larger, more irregular per element) — both from poor f-electron shielding, worse for 5f than 4f.
• Exceptional lanthanoid oxidation states (all driven by f⁰/f⁷/f¹⁴ stability): Ce⁴⁺ ([Xe]4f⁰) is a strong OXIDISING agent, reverting to stable Ce³⁺. Eu²⁺ ([Xe]4f⁷, half-filled) is a strong REDUCING agent, going to the common Eu³⁺. Yb²⁺ ([Xe]4f¹⁴, fully-filled) is also reducing. Tb⁴⁺ ([Xe]4f⁷) is oxidising, mirroring Eu²⁺'s logic from the other direction.
• Colourless lanthanoid ions: only La³⁺ (4f⁰) and Lu³⁺ (4f¹⁴) are colourless — every other Ln³⁺ has partially-filled f-orbitals and at least weak colour. (Compare: in the d-block, BOTH d⁰ and d¹⁰ are colourless the same way.)
• Mischmetall: ~95% lanthanoid metal (mainly Ce, La) + ~5% Fe + traces of S, C, Ca, Al. Used in lighter flints and Mg-based alloys.

8. How JEE Frames d- and f-Block Questions

• Two/three/four-statement identification questions: verify each statement independently against a specific fact (an E° value, a configuration, a colour) rather than pattern-matching against "usual" trends.
• Spin-only magnetic moment numericals: identify the ion/oxidation state first (often via a clue like "highest atomisation enthalpy" or "least metallic radius"), then count unpaired electrons and apply μ = √[n(n+2)].
• KMnO₄/K₂Cr₂O₇ reaction-sequence questions: track the oxidation state of Mn or Cr through each step; most "identify A, B, C" chains are just the standard preparation or chromyl-chloride sequence in disguise.
• Mixed-oxide counting: an oxide is "mixed" only if it can be split into two oxides of the SAME element in different oxidation states (Fe₃O₄ = FeO·Fe₂O₃, Mn₃O₄ = MnO·Mn₂O₃, Pb₃O₄ = 2PbO·PbO₂) — check each oxide individually.
• Colour/oxidising-agent identification from a spin-only clue: a "strong oxidising agent" among a list of ions is usually the one with a stable LOWER oxidation state waiting to be reached (Co³⁺→Co²⁺, Mn³⁺→Mn²⁺); compute its configuration and unpaired-electron count to get μ.

9. Quick Sheet and Checklist

Before the exam, check you can:

• Identify whether an element is d-block-but-not-transition, and explain Pd's exceptional configuration.
• Reproduce the Cr/Mn IE1, IE2, IE3 flip and explain each step from configuration.
• Rank E°(M2+/M) anomalies (Mn, Zn negative; Cu positive) and M3+/M2+ trends (Mn3+/Mn2+ high, Fe3+/Fe2+ lower) from 3d5/3d10 stability.
• Compute spin-only magnetic moment for any given ion and identify colourless vs coloured/paramagnetic species.
• Pick the correct KMnO4 half-reaction and equivalent weight for a stated medium.
• State which lanthanoid ions are exceptionally oxidising or reducing (Ce4+, Eu2+, Yb2+, Tb4+) and why.