NCERT Plus Two (Class 12) Chemistry • Chapter 4

The d- and f-Block Elements

Transition Metal Trends, Variable Oxidation States, Magnetic Moments, Coloured Ions, K2Cr2O7 & KMnO4, Lanthanoid Contraction & Actinoids

📄 Official NCERT Textbook PDF

Topic 4.1 – 4.2: Position in the Periodic Table & Electronic Configurations of d-Block Elements

Picture a bustling suspension bridge linking two massive cliffs: on the left sit the reactive alkali & alkaline-earth metals (s-block), and on the right stand the diverse non-metals & metalloids (p-block). Spanning the wide middle valley (Groups 3–12) is the d-block—our metallic bridge of iron, copper, titanium, silver, and gold that gives human civilisation its steel, wires, jewels, and catalysts! Even more fascinating: inside this bridge, Chromium and Copper pull a classic "electron robbery" from their outer 4s shell to achieve symmetrical 3d5 and 3d10 perfection.

The d-block occupies Groups 3 to 12 in the middle of the periodic table, where (n–1)d orbitals of the penultimate shell are progressively filled across four series: 3d (Sc to Zn), 4d (Y to Cd), 5d (La and Hf to Hg), and 6d (Ac and Rf to Cn).

General Outer Electronic Configuration: (n–1)d1–10 ns1–2   (Notable exception: Pd = [Kr] 4d10 5s0)

1. Why Cr (Z = 24) and Cu (Z = 29) Break the Rule

Because the energy gap between (n–1)d and ns orbitals is very small, and because half-filled (d5) and completely filled (d10) subshells enjoy extra stability (high exchange energy & symmetrical electron distribution):

  • Chromium (Cr, Z = 24): [Ar] 3d5 4s1 (instead of 3d4 4s2).
  • Copper (Cu, Z = 29): [Ar] 3d10 4s1 (instead of 3d9 4s2).

2. IUPAC Definition: Transition vs Non-Transition Elements

According to IUPAC, a transition metal is an element whose neutral atom or at least one common ion has an incompletely filled d-subshell (d1 to d9).

ElementGround StateCommon Ion(s)Transition Element?
Sc (21)[Ar] 3d1 4s2Sc3+: 3d0YES (incomplete 3d1 in ground state)
Cu (29)[Ar] 3d10 4s1Cu2+: 3d9YES (incomplete 3d9 in Cu2+)
Ag (47)[Kr] 4d10 5s1Ag2+: 4d9YES (incomplete 4d9 in Ag2+)
Zn (30), Cd (48), Hg (80)(n–1)d10 ns2M2+: (n–1)d10NO (full d10 in both M and M2+!)
NCERT Example 4.1 & Intext 4.1: Why is Scandium (Z = 21) a transition element whereas Zinc (Z = 30) is not? Sc has an incompletely filled 3d1 orbital in its ground state, whereas Zn has completely filled 3d10 orbitals in both ground state and Zn2+. Similarly, Silver (Z = 47) has 4d10 in ground state, yet is a transition element because Ag2+ has an incomplete 4d9 configuration!
When forming cations from any d-block atom, electrons are ALWAYS removed from the outer ns orbital BEFORE the inner (n–1)d orbital! Thus Fe ([Ar] 3d6 4s2) → Fe2+ is [Ar] 3d6 (NOT 3d4 4s2), and Cu+ is 3d10 while Cu2+ is 3d9.

Topic 4.3.1 – 4.3.3: Physical Trends, Enthalpy of Atomisation, Atomic Radii & Ionisation Enthalpies

Why does Tungsten (W) glow white-hot inside a bulb filament at 3400 °C without melting, while Zinc melts easily on a stove and Mercury pours like water at room temperature? The secret lies in how many unpaired d-electrons each atom commits to the metallic "electron glue" holding the crystal together: Tungsten uses all its valence electrons, whereas Zinc keeps its 3d10 electrons locked tightly at home!

1. Metallic Properties, Melting Points & Enthalpy of Atomisation (ΔaH°)

  • Transition metals are hard, ductile, malleable, high-density conductors with high melting/boiling points and high enthalpies of atomisation (ΔaH°) because both (n–1)d unpaired electrons and ns electrons form strong interatomic metallic bonds (NCERT Example 4.2).
  • Mid-Series Peaks & Anomalies: Melting points peak around d5 (Cr, Mo, W), except Mn and Tc which show anomalously low m.p. (Mn also has low ΔaH° = 281 kJ mol-1). Zn has the lowest ΔaH° in the 3d series (126 kJ mol-1) because no 3d electrons participate in metallic bonding (Intext 4.2).
  • 4d and 5d metals have higher ΔaH° than 3d metals, causing frequent metal–metal bonding in heavy transition metal compounds.

2. Atomic & Ionic Radii Trends (and the 4d ≈ 5d Surprise!)

  • Across a period: Radii decrease initially (rising nuclear charge outweighs poor d-shielding), stay nearly flat mid-series, and increase slightly at Cu/Zn (inter-electronic repulsion of filled d-orbitals). Density rises steadily from Sc (3.43) and Ti (4.1) to Cu (8.9 g cm-3).
  • Down a group (3d → 4d → 5d): Size increases from 3d to 4d, but 4d and 5d metals have virtually identical radii — e.g., Zr = 160 pm and Hf = 159 pm! This is due to Lanthanoid Contraction (imperfect shielding by intervening 4f electrons).

3. Ionisation Enthalpies (ΔiH°) & Exchange Energy Breaks

  • First IE rises gently across 3d (inner 3d electrons shield 4s moderately well).
  • 2nd IE (ΔiH2°) spikes: Unusually high for Cr (1592) and Cu (1958 kJ mol-1) because removing the second electron breaks stable 3d5 (Cr+) and 3d10 (Cu+).
  • 3rd IE (ΔiH3°) break between Mn & Fe: ΔiH3° of Fe (2962 kJ mol-1) is LOWER than Mn (3260 kJ mol-1)! Why? Removing a 3rd electron from Fe2+ (3d6) produces the extra-stable half-filled Fe3+ (3d5), whereas Mn2+ (3d5) loses high exchange energy when forced into Mn3+ (3d4)!
Do NOT expect a smooth straight-line increase in IE₂ or IE₃ across the 3d row! Exams love asking why ΔiH3° of Fe < Mn (Fe2+ → Fe3+ gains 3d5 stability) and why Cu has the highest ΔiH2° (breaks full 3d10 of Cu+).

Topic 4.3.4 – 4.3.8: Variable Oxidation States, Electrode Potentials (E°) & Disproportionation

Drop zinc or iron nails into dilute hydrochloric acid and vigorous hydrogen bubbles fizz right off. Drop a shiny copper wire into the very same dilute acid and—nothing happens! Copper refuses to dissolve. Why is Copper the odd noble hero with a positive reduction potential E°(Cu2+/Cu) = +0.34 V while every other 3d metal sits negative? Let's follow the energy money trail!

1. Variable Oxidation States (Differing by Unity!)

  • Because (n–1)d and ns energies are close, both participate in bonding, producing oxidation states differing by 1 (e.g., VII, VIII, VIV, VV)—unlike p-block elements which differ by 2 (inert pair effect).
  • Maximum states in the middle: Mn (Z = 25) exhibits the largest number of oxidation states (+2 to +7; Intext 4.3). Sc exhibits only +3 (no variable state; Example 4.3); Zn shows only +2.
  • Heavier d-metals favour HIGHER states (opposite to p-block!): Mo(VI) and W(VI) are stable, whereas Cr(VI) in dichromate is a strong oxidising agent.
  • Zero oxidation state: With π-acceptor ligands like CO in Ni(CO)4 and Fe(CO)5.

2. Standard Electrode Potentials: E°(M2+/M) & Why Copper is +0.34 V

Converting solid M(s) → M2+(aq) depends on endothermic ΔaH° + ΔiH1° + ΔiH2° versus exothermic ΔhydH°:

  • Across Ti → Zn, E° generally becomes less negative, except Mn (–1.18 V), Zn (–0.76 V) (stable d5 / d10 ions + lower atomisation enthalpy), and Ni (–0.25 V) (highest negative hydration enthalpy –2121 kJ mol-1).
  • Positive Cu value (+0.34 V; Intext 4.4): The high energy needed to convert Cu(s) → Cu2+(g) (ΔaH° 339 + ΔiH1 745 + ΔiH2 1958 = +3042 kJ mol-1) is NOT compensated by its hydration enthalpy (–2121 kJ mol-1). Thus Cu cannot liberate H2 from dilute acids (only reacts with oxidising acids like HNO3 and hot conc. H2SO4).

3. E°(M3+/M2+) & Disproportionation

NCERT Example 4.4 & Intext 4.7: Both Cr2+ and Mn3+ are d4, yet Cr2+ is strongly reducing (E° Cr3+/Cr2+ = –0.41 V) because d4 → d3 gives a half-filled t2g3 level in water, whereas Mn3+ is strongly oxidising (E° Mn3+/Mn2+ = +1.57 V) because d4 → d5 gains the extra stability of a half-filled 3d5 subshell! Cr2+ is even stronger reducing than Fe2+ (E° Fe3+/Fe2+ = +0.77 V).
  • Highest states in Oxides & Fluorides (Intext 4.6): Small size & high electronegativity of O and F stabilise high states (e.g., CrF6, VF5, Mn2O7). Oxygen beats fluorine via multiple pπ–dπ bonding (highest Mn fluoride is MnF4, highest oxide is Mn2O7 with Mn in +7!).
  • Increasing oxidising power (Example 4.5): VO2+ < Cr2O72– < MnO4– (increasing stability of lower reduced species).
  • Disproportionation (Example 4.9 & Intext 4.9): One species simultaneously oxidised and reduced:
    • 2Cu+(aq) → Cu2+(aq) + Cu(s) (Cu2+ is stable in water due to much more negative ΔhydH° than Cu+!)
    • 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O
Remember the opposite group trend! In p-block (Pb, Bi), lower oxidation states are more stable down the group (inert pair effect). In d-block, higher oxidation states become MORE stable down a group: Mo(VI) and W(VI) are stable, unlike oxidising Cr(VI). Also, CuI2 does not exist because Cu2+ oxidises I– to I2 (2Cu2+ + 4I– → Cu2I2 + I2)!

Topic 4.3.9 – 4.3.10: Spin-Only Magnetic Moment & Formation of Coloured Ions

Walk past a chemistry shelf of transition salts and you are walking through an art gallery: blue CuSO4, green FeSO4, pink MnCl2, violet CrCl3, and crystal-clear colourless ZnSO4! Even better, place those coloured crystals next to a magnet and almost every single coloured one feels a gentle pull. Why are colour and magnetism twins born from the same unpaired 3d electrons?

1. Magnetic Properties & Spin-Only Magnetic Moment (μ)

  • Diamagnetic: Repelled by a magnetic field (all electrons paired, n = 0).
  • Paramagnetic: Attracted into a magnetic field due to unpaired electrons (n > 0). Extreme paramagnetism (Fe, Co, Ni) is ferromagnetism.
  • For first-row (3d) transition ions, orbital angular momentum is effectively quenched by surrounding ligands, so magnetic moment is given by the Spin-Only Formula:
μ = √[n(n + 2)]  BM   (where n = number of unpaired electrons; unit = Bohr Magneton, BM)
NCERT Example 4.8 & Intext 4.8:
• M2+(aq) with Z = 25 (Mn2+): [Ar] 3d5 → 5 unpaired electrons (n = 5) → μ = √(5 × 7) = √35 = 5.92 BM!
• M2+(aq) with Z = 27 (Co2+): [Ar] 3d7 → 3 unpaired electrons (n = 3) → μ = √(3 × 5) = √15 = 3.87 BM!

2. Formation of Coloured Ions (d–d Transitions)

When ligands (like H2O) approach a transition metal ion, its five d-orbitals split into lower and higher energy sets (Unit 5). An electron from a lower d-orbital absorbs visible light energy and jumps to a higher d-orbital (d–d transition). We see the complementary colour of the absorbed light!

  • Why d0 and d10 ions are COLOURLESS: Ions with empty d-orbitals (Sc3+, Ti4+ → 3d0) or completely filled d-orbitals (Cu+, Zn2+ → 3d10) have NO available d–d electron transition and appear colourless!
  • Notice that the frequency of absorbed light—and thus colour—depends on the nature of the ligand (e.g., [Cu(H2O)6]2+ is sky blue; [Cu(NH3)4]2+ is deep royal blue).
Config (n)Ion ExampleCalc. μ (BM)Hydrated Colour (Table 4.8)
3d0 (n = 0)Sc3+, Ti4+0.00Colourless
3d1 (n = 1)Ti3+ / V4+1.73Purple (Ti3+) / Blue (V4+)
3d3 (n = 3)V2+, Cr3+3.87Violet
3d5 (n = 5)Mn2+ / Fe3+5.92Pink (Mn2+) / Yellow (Fe3+)
3d6 (n = 4)Fe2+ / Co3+4.90Green (Fe2+) / Blue (Co3+)
3d7 (n = 3)Co2+3.87Pink
3d8 (n = 2)Ni2+2.84Green
3d9 (n = 1)Cu2+1.73Blue
3d10 (n = 0)Cu+, Zn2+0.00Colourless
A quick mental trick for μ = √[n(n+2)]: the answer is ALWAYS just under (n + 0.9)! Check: n=1 → 1.73, n=2 → 2.84, n=3 → 3.87, n=4 → 4.90, n=5 → 5.92 BM! Also never forget: d6 has 4 unpaired electrons (same n = 4 as d4), NOT 6!

Topic 4.3.11 – 4.3.14 & 4.7: Complex Formation, Catalysis, Interstitial Compounds & Alloys

Why is pure iron too soft for bridges, yet mixing just a pinch of carbon locks its atomic layers into rock-hard steel? Imagine stacking identical marbles in a tray—they roll smoothly past one another. Now wedge tiny ball-bearings (carbon, hydrogen, or nitrogen atoms) into the hollow gaps between those marbles! Suddenly the layers jam tight—welcome to interstitial compounds and alloys!

1. Tendency to Form Complex Compounds (4.3.11)

Transition metals form vast numbers of coordination complexes (e.g., [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+, [PtCl4]2–) due to three reasons:

  1. Comparatively smaller sizes of metal ions,
  2. Their high ionic charges (high charge-to-size ratio), and
  3. Availability of vacant d-orbitals to accept electron pairs from ligands.

2. Catalytic Properties (4.3.12 & 4.7)

Transition metals and their compounds act as outstanding catalysts because they can (i) adopt multiple oxidation states and (ii) form complexes / surface bonds using (n–1)d & ns electrons, concentrating reactants and lowering activation energy.

  • Key NCERT Industrial Catalysts: V2O5 (Contact process: SO2 → SO3 for H2SO4), finely divided Fe (Haber's process for NH3), Ni (hydrogenation of fats/oils), TiCl4 + Al(CH3)3 (Ziegler-Natta catalyst for polythene), PdCl2 (Wacker process: ethyne → ethanal). (Plus AgBr in photography).
  • Homogeneous Redox Example: Fe3+ catalyses 2I– + S2O82– → I2 + 2SO42– via two fast steps:
    2Fe3+ + 2I– → 2Fe2+ + I2   followed by   2Fe2+ + S2O82– → 2Fe3+ + 2SO42–.

3. Interstitial Compounds (4.3.13)

Formed when small non-metal atoms like H, C, B, or N are trapped inside the crystal lattice voids of transition metals. Usually non-stoichiometric and neither purely ionic nor covalent (e.g., TiC, Mn4N, Fe3H, VH0.56, TiH1.7). Four classic properties:

  • (i) Higher melting points than pure metals; (ii) Very hard (some borides approach diamond); (iii) Retain metallic conductivity; (iv) Chemically inert.

4. Alloy Formation (4.3.14)

Blends (homogeneous solid solutions) form readily when metallic radii of atoms lie within ~15% of each other. Examples: Ferrous alloys (Cr, V, W, Mo, Mn in steels & stainless steel) and non-ferrous alloys like Brass (Cu + Zn) and Bronze (Cu + Sn).

Students often mistakenly think interstitial compounds lose conductivity like ionic ceramics—NO! According to NCERT, interstitial compounds retain metallic conductivity while gaining higher hardness, higher melting points, and chemical inertness.

Topic 4.4.1 (K₂Cr₂O₇): Potassium Dichromate (K₂Cr₂O₇): Preparation, pH Equilibrium & Oxidising Action

Pour acid into a sunny yellow potassium chromate beaker and—like magic—it flashes into fiery sunset orange! Pour alkali back in and the orange instantly snaps back to yellow! You are watching two chromium(+6) geometries—single pyramids vs twin shared pyramids—playing tug-of-war with H+ ions, yet neither atom changed its oxidation state!

1. 3-Step Industrial Preparation from Chromite Ore (FeCr2O4)

Step 1 (Roasting/Fusion with Na2CO3 in air):
4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 (yellow) + 2Fe2O3 + 8CO2

Step 2 (Acidification with H2SO4):
2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O

Step 3 (Double decomposition with KCl — K2Cr2O7 is less soluble & crystallises out):
Na2Cr2O7 + 2KCl → K2Cr2O7 (orange crystals) + 2NaCl

2. Chromate ↔ Dichromate pH Equilibrium & Structures

  • In Acid (pH < 7): 2CrO42– (Yellow) + 2H+ ⇌ Cr2O72– (Orange) + H2O
  • In Alkali (pH > 7): Cr2O72– (Orange) + 2OH– ⇌ 2CrO42– (Yellow) + H2O
  • Geometry: Chromate (CrO42–) is tetrahedral. Dichromate (Cr2O72–) consists of two tetrahedra sharing one oxygen corner with a Cr–O–Cr bond angle of 126°. In BOTH ions, Cr is in the +6 oxidation state!

3. Oxidising Action in Acidic Medium (Primary Standard in Volumetric Analysis)

In acid, orange Cr2O72– is reduced to green Cr3+ (E° = +1.33 V):

Cr2O72– + 14H+ + 6e– → 2Cr3+ (green) + 7H2O
  • Oxidises Fe2+ to Fe3+: Cr2O72– + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O
  • Oxidises I– to I2: Cr2O72– + 14H+ + 6I– → 2Cr3+ + 3I2 + 7H2O
  • Oxidises H2S to S: Cr2O72– + 8H+ + 3H2S → 2Cr3+ + 3S↓ + 7H2O
  • Oxidises Sn2+ to Sn4+: Cr2O72– + 14H+ + 3Sn2+ → 2Cr3+ + 3Sn4+ + 7H2O
When asked "What happens to the oxidation state of Cr when pH of dichromate increases?", NEVER say it changes! Increasing pH converts orange Cr2O72– to yellow CrO42–, and Cr stays at +6 in both (non-redox equilibrium)! Also remember: we use Na2CO3 + KCl instead of K2CO3 directly, and K2Cr2O7 precipitates because it is less soluble than Na2Cr2O7.

Topic 4.4.1 (KMnO₄): Potassium Permanganate (KMnO₄): Preparation, Structure & Medium-Dependent Redox

Drop one tiny purple grain of KMnO4 into water and it blooms into deep royal violet—yet Mn(+7) has zero 3d electrons (3d0)! How does an ion with no d-electrons shine more vividly purple than almost any compound on Earth, and why does it act like a chemical chameleon—turning colourless in acid, brown in water, and emerald green in strong alkali?

1. Preparation from Pyrolusite Ore (MnO2) & Laboratory Route

Step 1 (Alkaline oxidative fusion with KOH + air / KNO3):
2MnO2 + 4KOH + O2 → 2K2MnO4 (dark green) + 2H2O
Step 2 (Disproportionation in neutral/acidic medium OR Commercial Electrolytic Oxidation in alkali):
3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O   |   MnO42– (electrolytic oxidation) → MnO4– + e–
Lab Method (Mn2+ oxidised by peroxodisulphate): 2Mn2+ + 5S2O82– + 8H2O → 2MnO4– + 10SO42– + 16H+

2. Physical Properties, Structure & Magnetism

  • Dark purple (almost black) crystals isostructural with KClO4; decomposes on heating at 513 K: 2KMnO4 → K2MnO4 + MnO2 + O2.
  • Both MnO42– and MnO4– are tetrahedral (pπ–dπ overlap of O 2p and Mn 3d).
  • MnO42– (Manganate, +6): Dark green and paramagnetic (1 unpaired d1 electron).
  • MnO4– (Permanganate, +7): Intense purple and diamagnetic (0 unpaired d0 electrons; colour explained via Molecular Orbital / charge-transfer, NOT d–d transition!).

3. Oxidising Power Controlled by pH (Medium)

MediumMnO4– Half-ReactionFinal State & ColourKey Reactions
Acidic (dil. H2SO4)MnO4– + 8H+ + 5e– → Mn2+ + 4H2O (E° = +1.52 V)+2 (Mn2+) ColourlessI– → I2; Fe2+ → Fe3+; C2O42– → CO2 (333 K); H2S → S; SO32– → SO42–; NO2– → NO3–
Neutral / Faintly AlkalineMnO4– + 2H2O + 3e– → MnO2 + 4OH– (E° = +1.69 V)+4 (MnO2) Brown pptI– → IO3– (Iodate!); S2O32– → SO42–; Mn2+ → MnO2 (ZnO/ZnSO4 cat.)
Strongly AlkalineMnO4– + e– → MnO42– (E° = +0.56 V)+6 (MnO42–) GreenReversible 1e– reduction to manganate
Must-Know Balanced Equations:
• Acidic with Fe2+: MnO4– + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
• Acidic with Oxalate (333 K): 2MnO4– + 5C2O42– + 16H+ → 2Mn2+ + 10CO2 + 8H2O
• Acidic with I–: 2MnO4– + 10I– + 16H+ → 2Mn2+ + 5I2 + 8H2O
• Neutral/Faintly Alkaline with I–: 2MnO4– + H2O + I– → 2MnO2 + 2OH– + IO3–
NEVER use HCl to acidify KMnO4 during titrations! KMnO4 oxidises HCl into Cl2 gas, giving false titration volumes—only dilute H2SO4 is used! Also watch the iodide trap: acidic MnO4– oxidises I– to I2, whereas neutral/faintly alkaline MnO4– oxidises I– to IO3– (iodate)!

Topic 4.5 – 4.6: The f-Block: Lanthanoids, Lanthanoid Contraction, Actinoids & Comparison

Every smartphone speaker, electric car motor, and lighter flint hides elements from the twin rows tucked cleanly beneath the periodic table—the f-block (inner transition elements)! Though their outer shells look nearly identical, sliding across the 14 lanthanoids triggers a mysterious shrinkage called Lanthanoid Contraction—which secretly makes Zirconium and Hafnium almost identical twins, while down in the 5f actinoid row, Uranium and Plutonium unlock nuclear energy!

1. The Lanthanoids (4f series: Ce 58 to Lu 71, studied with La 57; symbol Ln)

  • Electronic Configuration: [Xe] 4f1–14 5d0–1 6s2 (5d1 appears only in La, Ce, Gd [4f75d16s2], and Lu [4f145d16s2]). All Ln3+ ions have [Xe] 4fn (n = 0 to 14).
  • Oxidation States: Principal and most stable state is +3. Occasional +4 and +2 occur to reach stable f0, f7, or f14 shells:
    • Ce4+ (4f0) & Tb4+ (4f7) — strong oxidants (E° Ce4+/Ce3+ = +1.74 V; slow kinetics lets Ce(IV) serve as an analytical reagent).
    • Eu2+ (4f7) & Yb2+ (4f14) — strong reductants (revert to +3).
  • Colour & Magnetism: Ln3+ ions are coloured & paramagnetic except f0 (La3+, Ce4+) and f14 (Yb2+, Lu3+).
  • Important Alloy: Mischmetall (~95% lanthanoid metal + ~5% Fe + traces of S, C, Ca, Al) — used in Mg-based alloys for bullets, shells, and lighter flints!
Lanthanoid Contraction (Most Repeated Board Question!):
• Definition: Steady decrease in atomic and ionic (Ln3+) radii from La3+ (106 pm) to Lu3+ (~85 pm; Yb3+ = 86 pm) with increasing atomic number.
• Cause: Increasing nuclear charge along the series while electrons enter inner 4f orbitals, where shielding of one 4f electron by another is very imperfect (even poorer than d-electrons).
• Consequences: (1) 4d and 5d metals have nearly identical radii (e.g., Zr = 160 pm, Hf = 159 pm) and similar chemistry, making separation hard; (2) Separation of lanthanoids requires ion-exchange methods; (3) Covalent character of M(OH)3 increases → basic strength decreases from La(OH)3 (most basic) to Lu(OH)3 (least basic).

2. The Actinoids (5f series: Th 90 to Lr 103, studied with Ac 89)

  • All are radioactive; later elements have short half-lives (Lr: 3 min) and are made only in nanogram amounts.
  • Actinoid Contraction: Greater from element to element than lanthanoid contraction because 5f electrons provide even poorer shielding than 4f electrons (Intext 4.10).
PropertyLanthanoids (4f)Actinoids (5f)
Oxidation StatesMainly +3; rarely +2, +4+3 plus wide range +4, +5, +6, +7 (comparable 5f, 6d, 7s energies!)
Orbital Shielding4f deeply buried; poor shielding5f less buried, even poorer shielding → greater contraction & greater bonding participation
RadioactivityNon-radioactive (except Promethium, Pm)All radioactive
Oxocations & ComplexesDo NOT form oxocations; low complex tendencyForm oxocations (UO22+, PuO22+); strong complex tendency
Chemical ReactivityModerately reactive; oxides/hydroxides less basicMore reactive (lower early IE, tarnished/attacked by boiling H2O & HCl); more basic
Remember TWO big exceptions in f-block questions: (1) Only Promethium (Pm, Z=61) is radioactive among lanthanoids, whereas ALL actinoids are radioactive; (2) Ce4+ (4f0) is an oxidising agent (gains e– to become +3), whereas Eu2+ (4f7) is a reducing agent (loses e– to become +3)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated) • 1 Marks): Why are Zn, Cd, and Hg NOT regarded as transition elements, whereas Silver (Z = 47, 4d¹⁰ 5s¹) IS a transition element?

Answer: According to IUPAC, transition elements must have an incompletely filled d-subshell in their neutral atom or common ions. Zn, Cd, and Hg have completely filled (n–1)d10 configurations in both their ground states and common +2 oxidation states, so they are excluded. Silver has 4d105s1 in its ground state, but exhibits a +2 oxidation state (Ag2+) with an incomplete 4d9 orbital, making it a transition element.

Q2 (Kerala HSE SAY • 1 Marks): What is 'Mischmetall'? Mention one important use of it.

Answer: Mischmetall is a lanthanoid alloy containing about 95% lanthanoid metal, 5% iron, and traces of S, C, Ca, and Al. It is used in Mg-based alloys to make bullets, shells, and lighter flints.

Q3 (Kerala HSE March (Most Asked) • 2 Marks): Calculate the 'spin-only' magnetic moment of (a) Mn2+ (Z = 25) and (b) Co2+ (Z = 27) ions in aqueous solution.

Formula: μ = √[n(n + 2)] BM

  • (a) Mn2+ (Z = 25): Config = [Ar] 3d5 → 5 unpaired electrons (n = 5) → μ = √(5 × 7) = √35 = 5.92 BM.
  • (b) Co2+ (Z = 27): Config = [Ar] 3d7 → 3 unpaired electrons (n = 3) → μ = √(3 × 5) = √15 = 3.87 BM.

Q4 (Kerala HSE March (High Yield) • 2 Marks): Explain why (a) E°(Cu2+/Cu) is positive (+0.34 V), and (b) Cu+ ion is unstable in aqueous solution.

(a) The high energy needed to convert Cu(s) → Cu2+(g) (high ΔaH° and high ΔiH1° + ΔiH2°) is not compensated by the hydration enthalpy (ΔhydH°) of Cu2+(aq). Hence E° is positive (+0.34 V).

(b) In aqueous solution, Cu+ undergoes disproportionation: 2Cu+(aq) → Cu2+(aq) + Cu(s), because the much more negative hydration enthalpy of Cu2+ compensates for the second ionisation enthalpy of Cu.

Q5 (Kerala HSE Model / NCERT • 2 Marks): Account for the following: (i) Of the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising. (ii) Manganese exhibits the largest number of oxidation states in the 3d series.

(i) Cr2+ (d4) oxidises to Cr3+ (d3), which has a stable half-filled t2g3 level in aqueous complexes. Conversely, Mn3+ (d4) reduces to Mn2+ (d5), gaining the extra stability of a half-filled 3d5 subshell.

(ii) Mn ([Ar] 3d54s2) has the maximum number of unpaired electrons (five 3d + two 4s) available for bond formation, exhibiting all oxidation states from +2 to +7.

Q6 (Kerala HSE SAY • 2 Marks): What are interstitial compounds? Give two examples and list any two characteristic properties.

Definition: Non-stoichiometric compounds formed when small atoms like H, C, or N get trapped inside the interstitial voids of transition metal crystal lattices (e.g., TiC, Mn4N, Fe3H, VH0.56).

Properties: (1) Higher melting points than pure metals; (2) Very hard (some borides approach diamond); (3) Retain metallic conductivity; (4) Chemically inert.

Q7 (Kerala HSE March • 2 Marks): Give reasons why transition metals (i) form a large number of complex compounds and (ii) act as good catalysts.

(i) Complex formation: Comparatively smaller size of metal ions, high ionic charge, and availability of vacant d-orbitals for accepting ligand electron pairs.

(ii) Catalytic activity: Ability to adopt multiple/variable oxidation states (forming intermediates with lower Ea) and ability to adsorb reactants on their surface using (n–1)d and ns electrons.

Q8 (Kerala HSE March (Guaranteed!) • 3 Marks): What is Lanthanoid Contraction? What is its cause? Write any two important consequences of lanthanoid contraction.

Definition: The steady overall decrease in atomic and ionic (Ln3+) radii from Lanthanum (106 pm) to Lutetium (~85 pm) with increasing atomic number.

Cause: Increasing nuclear charge along the series while electrons enter inner 4f orbitals, where shielding of one 4f electron by another is very imperfect.

Consequences:
1. 4d and 5d transition metals have virtually identical radii (e.g., Zr = 160 pm, Hf = 159 pm) and similar properties, making separation very difficult.
2. Basic strength of hydroxides decreases from La(OH)3 to Lu(OH)3 due to rising covalent character.

Q9 (Kerala HSE March / SAY • 3 Marks): Describe the preparation of Potassium Dichromate (K2Cr2O7) from chromite ore (FeCr2O4). What is the effect of increasing pH on an aqueous solution of K2Cr2O7?

Preparation Steps:
1. Fuse chromite ore with Na2CO3 in air: 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2.
2. Acidify yellow Na2CrO4 with H2SO4: 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O.
3. Treat with KCl to crystallise less-soluble orange K2Cr2O7: Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl.

Effect of increasing pH: Adding alkali shifts equilibrium from orange dichromate to yellow chromate (Cr remains +6):
Cr2O72– (Orange) + 2OH– ⇌ 2CrO42– (Yellow) + H2O.

Q10 (Kerala HSE March • 3 Marks): Write the steps in the preparation of KMnO4 from pyrolusite ore (MnO2). Write balanced ionic equations for the reaction of MnO4– with (i) Fe2+ in acidic medium and (ii) I– in neutral/faintly alkaline medium.

Preparation:
1. Alkaline oxidative fusion: 2MnO2 + 4KOH + O2 → 2K2MnO4 (green) + 2H2O.
2. Disproportionation in acid/neutral medium (3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O) or commercial electrolytic oxidation of MnO42– in alkaline medium → MnO4–.

Ionic Equations:
(i) MnO4– + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
(ii) 2MnO4– + H2O + I– → 2MnO2 + 2OH– + IO3–

Q11 (Kerala HSE Model / NCERT • 3 Marks): Compare Lanthanoids and Actinoids with respect to: (i) Oxidation states, (ii) Atomic/ionic contraction, and (iii) Radioactivity & chemical reactivity.

  1. Oxidation states: Lanthanoids mainly exhibit +3 (rarely +2, +4), whereas actinoids exhibit +3 plus a wide range (+4, +5, +6, +7) because 5f, 6d, and 7s energies are comparable.
  2. Contraction: Actinoid contraction is greater from element to element than lanthanoid contraction due to poorer shielding by 5f electrons compared to 4f electrons.
  3. Radioactivity & Reactivity: All actinoids are radioactive and chemically more reactive (and form oxocations like UO22+), whereas only Promethium (Pm) is radioactive among lanthanoids.

Q12 (Kerala HSE March (4-Mark Cluster) • 4 Marks): Give scientific reasons for the following: (i) Zinc has the lowest enthalpy of atomisation in the 3d series. (ii) Transition metal ions generally form coloured solutions. (iii) Third ionisation enthalpy of Fe is lower than that of Mn. (iv) Hydrochloric acid is not used to acidify KMnO4 in volumetric titrations.

  1. Zn lowest ΔaH° (126 kJ/mol): Completely filled 3d10 electrons do NOT take part in metallic bonding, unlike other 3d metals.
  2. Coloured ions: Splitting of partially filled d-orbitals allows absorption of visible light for d–d transitions, showing the complementary colour.
  3. ΔiH3° of Fe < Mn: Fe2+ (3d6) readily loses a 3rd electron to attain the extra-stable half-filled Fe3+ (3d5) configuration, whereas Mn2+ (3d5) resists losing an electron.
  4. HCl not used with KMnO4: KMnO4 oxidises HCl into Cl2 gas (2MnO4– + 16H+ + 10Cl– → 2Mn2+ + 5Cl2 + 8H2O), consuming extra KMnO4.