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AITS 03Chemistry · Chapter 1 of 6

Chemistry · last-minute revision

The d- and f-Block Elements

Configuration anomalies, oxidation states and E° logic, magnetic moment and colour, the two oxidising agents K₂Cr₂O₇ and KMnO₄, and lanthanoid contraction.

(n−1)d¹⁻¹⁰ ns¹⁻²every property traces back to partly filled d orbitals

Position and electronic configurationMust

d-Block = groups 3–12. General configuration (n−1)d1–10 ns1–2; Pd is the lone exception with 4d10 5s0. A transition element has an incompletely filled d subshell in the atom or in a common ion.

3d series configurations with the chromium and copper anomalies21Sc3d¹4s²22Ti3d²4s²23V3d³4s²24Cr3d⁵4s¹25Mn3d⁵4s²26Fe3d⁶4s²27Co3d⁷4s²28Ni3d⁸4s²29Cu3d¹⁰4s¹30Zn3d¹⁰4s²Zn: d¹⁰ in atom and ion → not a transition elementCr (Z = 24)3d4sCu (Z = 29)3d4sExpected 3d⁴4s² → actual 3d⁵4s¹half-filled d⁵: extra exchange energyExpected 3d⁹4s² → actual 3d¹⁰4s¹completely filled d¹⁰: extra stability
The brass dot is the electron that shifts from 4s into 3d. Paused, the plate shows the actual ground-state configuration.
AnomalyConfigurationReason / exam use
Cr (24)3d5 4s1half-filled d5 stability
Cu (29)3d10 4s1fully filled d10
Mo (42), Nb (41)4d5 5s1, 4d4 5s14d series irregularities
Pd (46)4d10 5s0only element with empty outer s
Ag (47), Au (79)4d105s1, 5d106s1coinage metals like Cu
Pt (78)5d9 6s1favourite odd one out
  • Zn, Cd, Hg: d10 in atom and in common ions → not transition elements (studied with d-block for completeness).
  • Cu, Ag, Au are transition elements: Cu2+ is 3d9, Au3+ is 5d8.
  • Forming ions: remove 4s before 3d. Fe2+ = [Ar]3d6, Fe3+ = 3d5, Mn2+ = 3d5, Cu+ = 3d10.
  • Four series: 3d (Sc–Zn), 4d (Y–Cd), 5d (La, Hf–Hg), 6d (Ac, Rf–Cn).

Physical trends of the 3d seriesHigh yield

PropertyTrendWhy
Metallic characterAll are metals: hard, lustrous, high tensile strength, good conductorsmetallic bonding with d electrons
Melting pointRises to a maximum near the middle (d5 region), then falls; Mn and Tc anomalously low; W highest mp of all; Hg liquidmore unpaired d electrons → stronger metallic bonds
Enthalpy of atomisationHigh; peaks mid-series; Zn lowest. Very high ΔaH° → metal tends to be noblesame unpaired-electron logic
Atomic radiusDecreases Sc → Cr, nearly constant Cr → Cu, slight rise at Znextra d electron shields poorly, then d–d repulsion balances
DensityIncreases Sc → Cumass rises, radius shrinks
Ionisation enthalpyGradual rise across; irregular3d and 4s close in energy; exchange-energy effects
  • 2nd IE unusually high for Cr (breaks d5) and Cu (breaks d10).
  • 3rd IE unusually high for Mn (Mn2+ is d5) and Zn (d10).
  • 4d and 5d atoms of the same group have almost equal radii (Zr 160 pm, Hf 159 pm) → lanthanoid contraction (see below).

Oxidation states and electrode potentialsMust

Variable oxidation states because (n−1)d and ns energies are close. Maximum number of states occurs in the middle: Mn shows +2 to +7. Sc shows only +3; Zn only +2. In the 4d/5d series Ru and Os reach +8 (OsO4).

ElementScTiVCrMnFeCoNiCuZn
Common states+3+4+5, +4+6, +3+7, +2+3, +2+2, +3+2+2, +1+2
Highest+3+4+5+6+7+6+4+4+2+2
  • Highest states appear in oxides and fluorides; oxygen stabilises higher states than F by forming multiple bonds (Mn2O7 has Mn +7, but highest fluoride is MnF4).
  • Low states stabilised by π-acceptor ligands: Ni(CO)4, Fe(CO)5 have metal in 0 state.
  • Oxide character rises with oxidation state: CrO basic, Cr2O3 amphoteric, CrO3 acidic; MnO basic, Mn2O7 acidic (covalent green oil); V2O5 amphoteric but mainly acidic.
  • Cu+ disproportionates in water: 2Cu+ → Cu2+ + Cu, because hydration enthalpy of Cu2+ is far more negative. CuI2 does not exist: 2Cu2+ + 4I− → Cu2I2 + I2.

Standard electrode potentials that decide answers

CoupleTiVCrMnFeCoNiCuZn
E°(M2+/M) / V−1.63−1.18−0.90−1.18−0.44−0.28−0.25+0.34−0.76
E°(M3+/M2+) / V−0.37−0.26−0.41+1.57+0.77+1.97———
Cu has the only positive E°(M2+/M)
High ΔaH + IE not repaid by hydration → Cu does not liberate H2 from non-oxidising acids.
Mn, Zn, Ni more negative than trend
Mn2+ d5 and Zn2+ d10 stable; Ni2+ has very negative hydration enthalpy.
Mn3+ strong oxidant, Cr2+ strong reductant
Both d4. Mn3+ → Mn2+ reaches d5; Cr2+ → Cr3+ reaches half-filled t2g3.
Co3+ very high +1.97 V
Co3+ unstable in water (oxidises water) but stable in complexes.
Fe3+ more stable than Fe2+
Fe3+ is d5.

Magnetic moment, colour, catalysis, complexesMust

Spin-only magnetic moment against number of unpaired electronsμ = √[n(n+2)] BM0.00n = 0Sc³⁺ Ti⁴⁺Cu⁺ Zn²⁺1.73n = 1Ti³⁺Cu²⁺2.83n = 2V³⁺Ni²⁺3.87n = 3Cr³⁺ V²⁺Co²⁺4.90n = 4Mn³⁺ Cr²⁺Fe²⁺5.92n = 5Mn²⁺Fe³⁺
Spin-only moment depends only on unpaired electrons n. Ions under each bar share that n in the free-ion (high-spin) state.
Spin-only moment
μ = √[n(n+2)] BM
Unpaired count for 3d ion
d electrons = Z − 18 − charge; fill 5 singly, then pair
Diamagnetic
n = 0: Sc3+, Ti4+, Cu+, Zn2+

Colour of hydrated ions

Colour comes from d–d transitions: the ion absorbs part of visible light, we see the complementary colour. d0 and d10 ions are colourless.

IondnColourIondnColour
Sc3+3d0colourlessMn3+3d4violet
Ti3+3d1purpleFe2+3d6green
Ti4+3d0colourlessFe3+3d5yellow
V3+3d2greenCo2+3d7pink
Cr3+3d3violetNi2+3d8green
Mn2+3d5pinkCu2+3d9blue
Zn2+3d10colourless
  • Catalysts (variable oxidation states + surface adsorption): V2O5 in Contact process, finely divided Fe in Haber process, Ni in hydrogenation. Fe3+ catalyses 2I− + S2O82− → I2 + 2SO42− by cycling Fe3+ ⇌ Fe2+.
  • Complexes: small, highly charged ions with vacant d orbitals → [Fe(CN)6]3−, [Cu(NH3)4]2+.
  • Interstitial compounds: small H, C, N trapped in metal lattice — TiC, Mn4N, Fe3H, VH0.56, TiH1.7 (non-stoichiometric). High mp (above pure metal), very hard, retain metallic conductivity, chemically inert.
  • Alloys: similar atomic radii let atoms substitute. Brass = Cu + Zn; bronze = Cu + Sn.

Potassium dichromate, K2Cr2O7Must

Preparation from chromite ore, FeCr2O4

1. Fuse with Na2CO3 in air
4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2
2. Acidify (yellow → orange)
2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O
3. Add KCl
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

K2Cr2O7 is preferred because sodium dichromate is hygroscopic; potassium dichromate is a primary standard in volumetric analysis.

Chromate and dichromate interconversion with pH and structuresOOOOCr2−OOOOOOOCrCr126°2−add H⁺ (acid)add OH⁻ (base)CrO₄²⁻ chromate, tetrahedralCr₂O₇²⁻ dichromate, two tetrahedra share one Oyelloworangesame +6 state:not a redox change
pH switches between the two ions. Cr stays +6 on both sides, so this is an acid–base equilibrium, not redox. Cr–O–Cr bridge angle is 126°.
Equilibrium
2CrO42− + 2H+ ⇌ Cr2O72− + H2O
Acidic oxidant
Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O (E° = 1.33 V, n = 6, orange → green)
Oxidises
I− → I2; Fe2+ → Fe3+; Sn2+ → Sn4+; H2S → S

Potassium permanganate, KMnO4Must

Preparation

Fuse MnO2 with KOH + oxidant
2MnO2 + 4KOH + O2 → 2K2MnO4 (green) + 2H2O
Manganate disproportionates
3MnO42− + 4H+ → 2MnO4− + MnO2 + 2H2O
Commercial
alkaline oxidative fusion of MnO2, then electrolytic oxidation of manganate
Laboratory
Mn2+ oxidised by peroxodisulphate (S2O82−)

Dark purple crystals, isostructural with KClO4. Manganate (green) is paramagnetic with one unpaired electron; permanganate (purple) is diamagnetic. Both ions are tetrahedral. On heating at 513 K: 2KMnO4 → K2MnO4 + MnO2 + O2.

Reduction products of permanganate in different media+2+3+4+5+6+7MnO₄⁻purple, d⁰, colour from charge transferMnO₄²⁻strongly alkalinegreen · gains 1 e⁻ · n-factor 1MnO₂neutral / faintly alkalinebrown ppt · gains 3 e⁻ · n-factor 3Mn²⁺acidic (dil. H₂SO₄)nearly colourless · gains 5 e⁻ · n-factor 5
Same starting ion, three products. The electrons gained fix the n-factor and therefore every mole ratio in titrations.
Acidic (n = 5)
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (E° = 1.52 V)
Acidic reactions
5Fe2+, 5/2 C2O42− (at 333 K), I− → I2, H2S → S, SO32− → SO42−, NO2− → NO3−
Neutral / faintly alkaline (n = 3)
MnO4− + 2H2O + 3e− → MnO2 + 4OH−
Neutral reactions
I− → IO3−; S2O32− → SO42−; Mn2+ → MnO2 (ZnSO4/ZnO catalyst)
Mole ratios in acid
MnO4− : Fe2+ = 1 : 5; MnO4− : C2O42− = 2 : 5
Why H2SO4, not HCl, for acidifying

KMnO4 oxidises Cl− to Cl2, so part of the titrant is wasted and the reading is wrong. HNO3 is avoided because it is itself an oxidant.

Lanthanoids and actinoidsHigh yield

Lanthanoid contraction: trivalent ion radius decreasing from La to Lu86909498102Ln³⁺ radius / pm (approx.)LaCePrNdPmSmEuGdTbDyHoErTmYbLuSteady shrink: 4f shields nuclear charge poorlyConsequence: Zr 160 pm ≈ Hf 159 pm(4d and 5d radii nearly equal; Zr/Hf hard to separate)
Radius drops across La → Lu. Because Hf comes after the 4f series, its radius is pulled back to that of Zr above it.
PointLanthanoids (4f)Actinoids (5f)
Configuration[Xe] 4f1–14 5d0–1 6s2[Rn] 5f1–14 6d0–1 7s2
Common state+3+3; wider range because 5f, 6d, 7s are close
Other statesCe4+ (f0, oxidant, E° = +1.74 V); Tb4+ (f7); Eu2+ (f7) and Yb2+ (f14) reductantsTh +4, Pa +5, U +6, Np and Pu up to +7
Contractionlanthanoid contraction: poor shielding by 4factinoid contraction greater element to element: 5f shields even worse
Radioactivityonly Pmall radioactive
Chemistryearly members react like Ca, later ones behave more like Al; Ln + H2O → Ln(OH)3 + H2; Ln2O3 and Ln(OH)3 basicmore reactive, especially finely divided; HNO3 barely attacks (protective oxide)
  • Consequences of lanthanoid contraction: Zr/Hf near-identical radii and chemistry; basicity of hydroxides falls La(OH)3 → Lu(OH)3; 4d and 5d series resemble each other more than 3d and 4d.
  • Ln3+ ions are coloured from f–f transitions except f0 (La3+, Ce4+) and f14 (Yb2+, Lu3+); these are also diamagnetic.
  • Lanthanoids react with C at 2773 K → Ln3C, Ln2C3, LnC2; burn in halogens → LnX3; heated with N2 → LnN.
Gap content

Mischmetall ≈ 95% lanthanoid metal + 5% Fe with traces of S, C, Ca, Al; used in Mg-based alloy for bullets, shells and lighter flint. From the older NCERT applications section; still seen in coaching papers.

Standard question patternsMust

Spin-only magnetic moment of an ion
Remove 4s electrons first. Count d electrons, find n, use √[n(n+2)]. Memorise 1.73, 2.83, 3.87, 4.90, 5.92.
Pick the colourless / diamagnetic ion
Look for d0 (Sc3+, Ti4+) or d10 (Cu+, Zn2+).
Moles of KMnO4 or K2Cr2O7 needed
Equate equivalents: moles × n-factor. MnO4− acid n = 5, Cr2O72− n = 6, Fe2+ n = 1, C2O42− n = 2.
Which is the strongest oxidant in water
Highest E°(M3+/M2+): Co3+ > Mn3+ > Fe3+.
Why Zr and Hf are similar
Lanthanoid contraction.
Colour of KMnO4 or K2Cr2O7
Charge transfer (ligand → metal), not d–d; both are d0.
Maximum oxidation state in 3d series
Mn +7; for the whole d-block Ru/Os +8.

Traps that cost marksMust

Trap · Ratio reversal

Acidic MnO4− oxidises Fe2+ in the ratio 1 : 5, so 1 mol Fe2+ needs 0.2 mol KMnO4 — not 5 mol. Write the equivalents equation before dividing.

Trap · Right content, wrong arrangement

Colour pairs are swapped easily: Mn2+ pink but Mn3+ violet; Fe2+ green but Fe3+ yellow; Co2+ pink. Match one ion at a time using its dn, then check the last two rows independently.

Trap · Configuration of ions

Fe2+ is 3d6 4s0, never 3d4 4s2. The 4s electrons leave first.

Trap · Assertion–reason

'Cu+ is unstable in water because Cu+ is d10' — the assertion is true but this reason argues the opposite (d10 should be stable). The real reason is hydration enthalpy of Cu2+.

Trap · Chromate ⇌ dichromate

Not a redox reaction: Cr is +6 on both sides. Adding alkali turns orange dichromate yellow.

Ten-question checkMust

Attempt all ten before opening any answer. Aim for 9/10 in under 8 minutes.

Q1.The spin-only magnetic moment of which ion is 3.87 BM?

  1. (A)Fe2+
  2. (B)Cu2+
  3. (C)Cr3+
  4. (D)Mn2+
Show answer

Answer (C). 3.87 BM means n = 3. Cr3+ is 3d3 → 3 unpaired. Fe2+ (3d6) has 4 → 4.90 BM; Cu2+ (3d9) has 1 → 1.73 BM; Mn2+ (3d5) has 5 → 5.92 BM.

Q2.Which of these is not regarded as a transition element?

  1. (A)Zn
  2. (B)Cu
  3. (C)Sc
  4. (D)Mn
Show answer

Answer (A). Zn is 3d10 in the atom and in Zn2+. Cu counts because Cu2+ is 3d9; Sc is 3d1.

Q3.In neutral or faintly alkaline solution, MnO4− is reduced to

  1. (A)MnO2
  2. (B)Mn2+
  3. (C)MnO42−
  4. (D)Mn2O3
Show answer

Answer (A). Neutral medium: MnO4− + 2H2O + 3e− → MnO2 + 4OH−. Mn2+ is the acidic product; MnO42− forms in strongly alkaline solution.

Q4.When NaOH solution is added to aqueous K2Cr2O7, the colour changes from

  1. (A)yellow to orange
  2. (B)orange to green
  3. (C)orange to yellow
  4. (D)purple to colourless
Show answer

Answer (C). OH− converts Cr2O72− (orange) to CrO42− (yellow). Orange → green happens only on reduction to Cr3+.

Q5.Zirconium and hafnium have almost identical atomic radii because of

  1. (A)actinoid contraction
  2. (B)inert pair effect
  3. (C)d–d repulsion in 5d orbitals
  4. (D)lanthanoid contraction
Show answer

Answer (D). Hf follows the 4f series; the steady contraction across the lanthanoids cancels the size increase expected from Zr to Hf.

Q6.The highest oxidation state shown by any element of the 3d series is shown by

  1. (A)Cr
  2. (B)Fe
  3. (C)V
  4. (D)Mn
Show answer

Answer (D). Mn uses all seven 3d + 4s electrons: +7 in MnO4− and Mn2O7. Cr reaches +6; V +5; Fe +6 only rarely.

Q7.How many moles of KMnO4 are needed to oxidise 1 mol of Fe2+ in acidic medium?

  1. (A)5 mol
  2. (B)0.6 mol
  3. (C)0.2 mol
  4. (D)0.4 mol
Show answer

Answer (C). Equivalents: n(MnO4−) × 5 = 1 × 1 → 0.2 mol. 5 mol is the upside-down ratio; 0.4 mol is the value for 1 mol oxalate.

Q8.Assertion (A): Cu+ ion is unstable in aqueous solution.
Reason (R): The more negative hydration enthalpy of Cu2+ more than compensates for the second ionisation enthalpy of Cu.

  1. (A)Both A and R are true, and R is the correct explanation of A
  2. (B)Both A and R are true, but R is not the correct explanation of A
  3. (C)A is true, but R is false
  4. (D)A is false, but R is true
Show answer

Answer (A). A is true (2Cu+ → Cu2+ + Cu). R is true and is exactly why the disproportionation is favoured.

Q9.Which ion is the strongest oxidising agent in aqueous solution?

  1. (A)Mn3+
  2. (B)Co3+
  3. (C)Fe3+
  4. (D)Cr3+
Show answer

Answer (B). E°(M3+/M2+): Co +1.97 V > Mn +1.57 V > Fe +0.77 V > Cr −0.41 V. Cr3+ has negative E°; Cr2+ is the reductant.

Q10.Which of the following ions is colourless in aqueous solution?

  1. (A)Ti3+
  2. (B)Sc3+
  3. (C)Cr3+
  4. (D)Cu2+
Show answer

Answer (B). Sc3+ is 3d0, so no d–d transition. Ti3+ (d1) purple, Cr3+ (d3) violet, Cu2+ (d9) blue.

60-second recap before the paperMust

  1. Cr 3d5 4s1, Cu 3d10 4s1, Pd 4d10 5s0. Zn/Cd/Hg are not transition elements.
  2. Ions lose 4s first. μ = √[n(n+2)]: 1.73, 2.83, 3.87, 4.90, 5.92.
  3. Mn has most oxidation states (+2 to +7). Cu is the only 3d metal with positive E°(M2+/M).
  4. Mn3+ oxidant, Cr2+ reductant (both d4). Co3+/Co2+ = +1.97 V.
  5. d0 and d10 ions colourless. KMnO4 and K2Cr2O7 colour = charge transfer.
  6. KMnO4: acid n = 5 → Mn2+; neutral n = 3 → MnO2; strong alkali n = 1 → MnO42−. Acidify with H2SO4.
  7. K2Cr2O7: n = 6, orange → green; chromate ⇌ dichromate by pH (not redox); Cr–O–Cr 126°.
  8. Lanthanoid contraction → Zr ≈ Hf, basicity falls La → Lu. Ce4+ oxidant; Eu2+, Yb2+ reductants.
  9. Actinoids: more oxidation states (up to +7 at Np/Pu), all radioactive, bigger contraction.