ChemistryISC Class 12 · Chapter 4

d- and f-Block Elements ISC Class 12 Chemistry: Notes

Updated 15 October 2026 · Fact-checked

d- and f-Block Elements is Chapter 4 of ISC Class 12 Chemistry. It covers the position and configurations of transition metals, trends in radii, ionisation enthalpy and oxidation states, colour, magnetism, catalysis, alloys and interstitial compounds, lanthanoids, lanthanoid contraction, actinoids, and the preparation and oxidising reactions of KMnO₄ and K₂Cr₂O₇.

What d- and f-Block Elements is about

The d-block (groups 3-12) lies between the s- and p-blocks; its elements fill the (n−1)d orbitals. The three complete series are 3d (Sc-Zn), 4d (Y-Cd) and 5d (La, Hf-Hg). A transition element has an incompletely filled d subshell in the atom or in a common ion; so Zn, Cd and Hg ( $d^{10}$ in both atom and ions) are d-block elements but not transition elements. The f-block contains the lanthanoids (Ce-Lu, filling 4f) and actinoids (Th-Lr, filling 5f). The chapter shares the Inorganic Chemistry group (14 of the 70 theory marks) with Coordination Compounds.

Most ISC questions are 'give a reason' items built on a few ideas: partly filled d orbitals (colour by d-d transitions, paramagnetism, variable oxidation states, catalysis), small size and high charge with vacant d orbitals (complex formation), similar atomic sizes (alloys), and **extra stability of $d^{0}$ , $d^{5}$ and $d^{10}$ ** (Mn²⁺ more stable than Fe²⁺; Fe³⁺ more stable than Fe²⁺). The spin-only magnetic moment $μ = n (n + 2)$ BM is the one formula.

For the f-block, the key idea is lanthanoid contraction: the steady decrease in size from La to Lu because 4f electrons shield poorly. It makes Zr and Hf almost the same size and very hard to separate. You also need the reactions of lanthanoids with oxygen, hydrogen, halogens, sulphur, nitrogen, carbon and water, and a comparison of actinoids with lanthanoids.

Finally, two important compounds: potassium permanganate (from pyrolusite, MnO₂) and potassium dichromate (from chromite, FeCr₂O₄): their structures, extraction equations, oxidising reactions in acidic, basic and neutral media, use in redox titrations and the chromate-dichromate equilibrium.

Weightage: d- and f-Block Elements is part of the Inorganic Chemistry group, which carries 14 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Position, configuration and definition of transition elements: General configuration $(n - 1) d^{1 - 10} n s^{1 - 2}$ . 3d series: Sc $3 d^{1} 4 s^{2}$ to Zn $3 d^{10} 4 s^{2}$ , with two exceptions from the extra stability of half-filled and full d subshells: Cr $3 d^{5} 4 s^{1}$ and Cu $3 d^{10} 4 s^{1}$ . On forming ions, 4s electrons are lost first: Fe $3 d^{6} 4 s^{2}$ → Fe²⁺ $3 d^{6}$ → Fe³⁺ $3 d^{5}$ . Transition element: an element whose atom or a common ion has an incompletely filled d subshell. Cu is transition (Cu²⁺ $3 d^{9}$ ); Sc is transition (Sc atom $3 d^{1}$ ). Zn, Cd and Hg have completely filled $d^{10}$ in the atom and in their ions (M²⁺), so they are not regarded as transition elements and lack typical properties (they are colourless and diamagnetic, and show essentially one oxidation state). Occurrence: mostly as oxides, sulphides and carbonates (e.g. haematite Fe₂O₃, pyrolusite MnO₂, chromite FeCr₂O₄).; Worked example
Write the configurations of Mn²⁺ and Fe²⁺ (Z: Mn 25, Fe 26) and explain why Mn²⁺ is more stable than Fe²⁺ towards oxidation.
Try it first, then show the solutionSolution
Mn²⁺: [Ar] $3 d^{5}$ ; Fe²⁺: [Ar] $3 d^{6}$ .
Mn²⁺ already has a half-filled, extra-stable $d^{5}$ configuration; removing another electron would destroy it.
Fe²⁺ loses one electron to become Fe³⁺ ( $3 d^{5}$ ), gaining that stability, so Fe²⁺ is easily oxidised.
Answer: Mn²⁺ is $d^{5}$ (half-filled); Fe²⁺ ( $d^{6}$ ) readily becomes $d^{5}$ .
Trends: metallic character, radii, ionisation enthalpy and oxidation states: Metallic character: all are hard, lustrous metals with high melting and boiling points and high enthalpies of atomisation, because both ns and (n−1)d electrons take part in metallic bonding; melting points peak near the middle ( $d^{5}$ ) of each series. Atomic and ionic radii decrease from Sc to about Cr, stay nearly constant in the middle (increased nuclear charge is balanced by shielding from added d electrons) and rise slightly at Cu and Zn. Radii of 4d and 5d elements are almost equal (lanthanoid contraction). Ionisation enthalpy increases gradually and irregularly across a series; the irregularities reflect the stability of $d^{5}$ and $d^{10}$ (e.g. Cu has a high second IE because Cu⁺ is $d^{10}$ ). Oxidation states: variable, because ns and (n−1)d electrons are close in energy. Number of states rises to a maximum at Mn (+2 to +7) then falls; Sc only +3, Zn only +2. Lower states are common in ionic compounds; the highest appear in oxides and oxoanions with O or F (MnO₄⁻, CrO₄²⁻). Variable valency appears in pairs differing by one unit (Fe²⁺/Fe³⁺, Cu⁺/Cu²⁺), unlike the p-block where they differ by two.; Worked example
Find the oxidation state of the metal in (a) $M n O_{4}^{-}$ , (b) $C r_{2} O_{7}^{2 -}$ , (c) $[F e (C N)_{6}]^{4 -}$ .
Try it first, then show the solutionSolution
(a) x + 4(−2) = −1, x = +7
(b) 2x + 7(−2) = −2, x = +6
(c) x + 6(−1) = −4, x = +2
Answer: +7, +6, +2
Colour and magnetic properties: Colour: in a transition metal ion with partly filled d orbitals, the ligands (water molecules in solution) split the d orbitals into two sets of slightly different energy. The ion absorbs visible light of a particular wavelength to promote an electron from the lower to the higher set (d-d transition); the colour seen is the complementary colour. Ions with $d^{0}$ (Sc³⁺, Ti⁴⁺) or $d^{10}$ (Zn²⁺, Cu⁺) have no possible d-d transition and are colourless. Examples: Ti³⁺ ( $d^{1}$ ) purple, V³⁺ green, Mn²⁺ pale pink, Fe³⁺ yellow, Co²⁺ pink, Ni²⁺ green, Cu²⁺ blue. (MnO₄⁻ and Cr₂O₇²⁻ are $d^{0}$ but coloured by charge transfer, not d-d.) Magnetism: substances with unpaired electrons are paramagnetic (attracted by a magnetic field); with all electrons paired, diamagnetic. Spin-only magnetic moment $μ = n (n + 2)$ BM, n = number of unpaired electrons. Paramagnetism is maximum for $d^{5}$ (Mn²⁺, Fe³⁺, 5.92 BM).; Worked example
Calculate the spin-only magnetic moments of Ti³⁺ (Z = 22), Cr³⁺ (Z = 24) and Fe²⁺ (Z = 26). Which is colourless among Ti⁴⁺ and Ti³⁺?
Try it first, then show the solutionSolution
Ti³⁺ $3 d^{1}$ : n = 1, μ = √3 = 1.73 BM. Cr³⁺ $3 d^{3}$ : n = 3, μ = √15 = 3.87 BM. Fe²⁺ $3 d^{6}$ : n = 4, μ = √24 = 4.90 BM.
Ti⁴⁺ is $3 d^{0}$ : no d electron for a d-d transition, so it is colourless; Ti³⁺ ( $d^{1}$ ) is coloured (purple).
Answer: 1.73, 3.87, 4.90 BM; Ti⁴⁺ is colourless
Catalysis, complex formation, interstitial compounds and alloys: Catalytic property: transition metals and their compounds catalyse many reactions because they (1) show variable oxidation states and can form unstable intermediates, and (2) provide a surface where reactants are adsorbed and bonds weakened. Examples: Fe (with Mo) in the Haber process; V₂O₅ in the contact process ( $2 S O_{2} + O_{2} \to 2 S O_{3}$ ); finely divided Ni in hydrogenation of oils; Pt in oxidation of NH₃. Complex formation: due to small size and high charge of the ions, and vacant d orbitals of suitable energy to accept lone pairs from ligands, e.g. $[F e (C N)_{6}]^{4 -}$ , $[C u (N H_{3})_{4}]^{2 +}$ . Interstitial compounds: small atoms (H, B, C, N) occupy the holes in the metal lattice (e.g. TiC, Fe₃H, Mn₄N). They are non-stoichiometric, very hard, have high melting points, keep metallic conductivity and are chemically inert. Alloy formation: transition metals have similar atomic sizes, so atoms of one metal can easily replace atoms of another in the lattice (e.g. brass Cu-Zn, bronze Cu-Sn, stainless steel Fe-Cr-Ni). Alloys are harder and have higher melting points than the parent metals.; Worked example
Give reasons: (a) Transition metals form alloys easily. (b) Transition metals have a great tendency to form complex compounds.
Try it first, then show the solutionSolution
(a) Their atomic radii are very similar, so one metal's atoms can substitute for another's in the crystal lattice.
(b) Their ions are small with high charge density and have vacant d orbitals of suitable energy to accept electron pairs from ligands.
Answer: (a) Similar atomic size. (b) Small, highly charged ions with vacant d orbitals.
Lanthanoids and lanthanoid contraction: Configuration: [Xe] $4 f^{1 - 14} 5 d^{0 - 1} 6 s^{2}$ . Oxidation state: +3 is characteristic (all form Ln³⁺); some show +2 (Eu²⁺ $4 f^{7}$ , Yb²⁺ $4 f^{14}$ ) or +4 (Ce⁴⁺ $4 f^{0}$ , Tb⁴⁺ $4 f^{7}$ ) when an empty, half-filled or full f subshell results. Colour: many Ln³⁺ ions are coloured by f-f transitions; La³⁺ ( $4 f^{0}$ ) and Lu³⁺ ( $4 f^{14}$ ) are colourless. Lanthanoid contraction: the steady decrease in atomic and ionic radii from La to Lu. Cause: each added electron enters the 4f subshell, and 4f orbitals shield the outer electrons poorly (diffuse shape), so the effective nuclear charge increases. Consequences: (1) the 4d and 5d elements of a group have almost the same size, e.g. Zr 160 pm and Hf 159 pm, so they have very similar properties and are difficult to separate; (2) lanthanoids themselves are hard to separate; (3) basic strength of hydroxides decreases from La(OH)₃ to Lu(OH)₃ (smaller ion, more covalent); (4) 5d elements have higher ionisation enthalpies and densities. Chemical reactivity: fairly reactive metals (like Ca); tarnish in air. With oxygen: $4 L n + 3 O_{2} \to 2 L n_{2} O_{3}$ ; hydrogen (heated): $L n + H_{2} \to L n H_{2}$ ; halogens: $2 L n + 3 C l_{2} \to 2 L n C l_{3}$ ; sulphur: $2 L n + 3 S \to L n_{2} S_{3}$ ; nitrogen: $2 L n + N_{2} \to 2 L n N$ ; carbon (2773 K): $L n + 2 C \to L n C_{2}$ ; water: $2 L n + 6 H_{2} O \to 2 L n (O H)_{3} + 3 H_{2}$ . Alloys: misch metal (about 95% lanthanoids, 5% Fe) is used in lighter flints.; Worked example
Zr (Z = 40) and Hf (Z = 72) occur together in nature and are very difficult to separate. Explain.
Try it first, then show the solutionSolution
Hf follows the 14 lanthanoids, whose poorly shielding 4f electrons cause a steady contraction in size.
This cancels the size increase expected on going from period 5 to period 6, so Hf (159 pm) is nearly the same size as Zr (160 pm).
Same group, same valence configuration and same size give almost identical chemical properties, so separation is hard.
Answer: Lanthanoid contraction makes their atomic radii almost equal.
Actinoids and comparison with lanthanoids: Actinoids (Th to Lr) fill the 5f subshell: [Rn] $5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}$ . All are radioactive; those beyond uranium are man-made. Oxidation states: +3 is common, but they show a much wider range (+3 to +7, e.g. U +6, Np and Pu +7) because 5f, 6d and 7s orbitals are close in energy. Actinoid contraction is greater from element to element than lanthanoid contraction, as 5f electrons shield even more poorly. Comparison: *similarities*: both fill f orbitals, +3 is common, both show contraction, both form coloured ions by f-f transitions and are electropositive. *Differences*: actinoids show more oxidation states (lanthanoids mainly +3); actinoids are radioactive (only Pm among lanthanoids); actinoids form complexes more readily; actinoid contraction is larger; actinoids form oxocations such as $U O_{2}^{2 +}$ , lanthanoids do not; magnetic behaviour of actinoids is more complex.; Worked example
Why do actinoids show a greater range of oxidation states than lanthanoids?
Try it first, then show the solutionSolution
In actinoids the 5f, 6d and 7s levels have comparable energies, so varying numbers of electrons can be used in bonding.
In lanthanoids the 4f orbitals are buried deep inside and are much lower in energy than 5d and 6s, so only +3 (occasionally +2, +4) is common.
Answer: Small energy gap between 5f, 6d and 7s orbitals.
Potassium permanganate (KMnO₄): Structure: the permanganate ion $M n O_{4}^{-}$ is tetrahedral; Mn is +7 ( $d^{0}$ ) and the purple colour is due to charge transfer. Preparation from pyrolusite: (1) fuse MnO₂ with KOH in air (or with KNO₃) to form green potassium manganate: $2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$ . (2) Oxidise manganate to permanganate electrolytically ( $M n O_{4}^{2 -} \to M n O_{4}^{-} + e^{-}$ at the anode) or by passing CO₂ (disproportionation): $3 K_{2} M n O_{4} + 2 C O_{2} \to 2 K M n O_{4} + M n O_{2} + 2 K_{2} C O_{3}$ . Oxidising action: *acidic medium* (dil. H₂SO₄), Mn +7 → +2, gains 5e⁻: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ . With FeSO₄: $2 K M n O_{4} + 8 H_{2} S O_{4} + 10 F e S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O$ . With oxalic acid (warm, 333 K): $2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$ . With KI: $2 K M n O_{4} + 8 H_{2} S O_{4} + 10 K I \to 6 K_{2} S O_{4} + 2 M n S O_{4} + 5 I_{2} + 8 H_{2} O$ . *Basic medium* (Mn +7 → +4, MnO₂), KI is oxidised to iodate: $2 K M n O_{4} + K I + H_{2} O \to 2 M n O_{2} + 2 K O H + K I O_{3}$ . *Neutral medium:* H₂S is oxidised to sulphur: $2 K M n O_{4} + 3 H_{2} S \to 2 K O H + 2 M n O_{2} + 3 S + 2 H_{2} O$ . In titrations KMnO₄ acts as its own indicator: its pink colour vanishes while reductant remains; the first permanent pale pink marks the end point. HCl is not used to acidify (it is oxidised to Cl₂); HNO₃ is not used (it is itself an oxidant). On heating: $2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$ .; Worked example
What volume of 0.020 M KMnO₄ is needed to oxidise 25.0 mL of 0.050 M oxalic acid in acidic medium?
Try it first, then show the solutionSolution
Moles of oxalic acid = 0.0250 × 0.050 = $1.25 \times 1 0^{- 3}$
From the equation, 2 mol KMnO₄ react with 5 mol oxalic acid: moles KMnO₄ = $\frac{2}{5} \times 1.25 \times 1 0^{- 3} = 5.0 \times 1 0^{- 4}$
Volume = $\frac{5.0 \times 1 0 ^{- 4}}{0.020}$ = 0.025 L = 25.0 mL
Answer: 25.0 mL
Potassium dichromate (K₂Cr₂O₇) and the chromate-dichromate equilibrium: Structure: chromate $C r O_{4}^{2 -}$ is tetrahedral; dichromate $C r_{2} O_{7}^{2 -}$ is two tetrahedra sharing one corner oxygen (Cr-O-Cr bond angle about 126°). Cr is +6 in both. Preparation from chromite: (1) roast chromite with sodium carbonate in air: $4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2} \to 8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}$ . (2) Acidify the yellow sodium chromate: $2 N a_{2} C r O_{4} + H_{2} S O_{4} \to N a_{2} C r_{2} O_{7} + N a_{2} S O_{4} + H_{2} O$ . (3) Add KCl (K₂Cr₂O₇ is less soluble and crystallises): $N a_{2} C r_{2} O_{7} + 2 K C l \to K_{2} C r_{2} O_{7} + 2 N a C l$ . Effect of pH: in acid, yellow chromate changes to orange dichromate: $2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$ ; in alkali the reverse: $C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O$ . Cr stays +6: no redox. Oxidising action in acid (Cr +6 → +3, green; gains 6e⁻ per Cr₂O₇²⁻): $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ . With FeSO₄: $K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O$ . With KI: $K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 K I \to 4 K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 I_{2} + 7 H_{2} O$ . With H₂S: $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 H_{2} S \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 S + 7 H_{2} O$ . *In basic medium* dichromate becomes chromate, a weak oxidant. *In neutral solution* it oxidises H₂S slowly, precipitating chromium(III) hydroxide: $K_{2} C r_{2} O_{7} + 3 H_{2} S + H_{2} O \to 2 K O H + 2 C r (O H)_{3} + 3 S$ . Titrations: K₂Cr₂O₇ is a primary standard (pure, non-hygroscopic), used to estimate Fe²⁺ with diphenylamine as the internal indicator; it is not decomposed by HCl, so can be used in HCl solution.; Worked example
A yellow solution of potassium chromate turns orange when dilute sulphuric acid is added, and back to yellow with NaOH. Explain with equations. Is this a redox change?
Try it first, then show the solutionSolution
Acid: $2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$ (yellow to orange).
Alkali: $C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O$ (orange to yellow).
Cr is +6 in both ions, so it is an acid-base (condensation) equilibrium, not redox.
Answer: pH-controlled chromate-dichromate equilibrium; not redox.

Chemistry in real life: d- and f-Block Elements

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Permanganate titration of iron

Suppose a quality-control chemist checks an iron tablet by dissolving it in dilute sulphuric acid and titrating with 0.020 M KMnO₄; 18.0 mL are needed.

How many milligrams of iron(II) does the tablet contain? (Fe = 56)

Try it first, then show the solutionSolution

Moles $M n O_{4}^{-}$ = 0.0180 × 0.020 = $3.6 \times 1 0^{- 4}$
$M n O_{4}^{-} : F e^{2 +}$ = 1 : 5, so moles Fe²⁺ = $1.8 \times 1 0^{- 3}$ ; mass = $1.8 \times 1 0^{- 3} \times 56 = 0.1008$ g

Answer: About 101 mg of iron(II).

Memory hook: One permanganate eats five ferrous ions.

Dichromate breath test

Imagine an old roadside breath tester in which orange acidified dichromate crystals turn green when a driver who has been drinking blows through them.

What causes the colour change, and what happens to the ethanol?

Try it first, then show the solutionSolution

Ethanol is a reducing agent; it reduces $C r_{2} O_{7}^{2 -}$ (Cr +6, orange) to $C r^{3 +}$ (+3, green).
Ethanol is oxidised to ethanal and then ethanoic acid.

Answer: Cr(+6) orange → Cr(+3) green as ethanol is oxidised.

Memory hook: Orange to green means alcohol has been seen.

Misch metal and lanthanoid reactivity

Suppose a student wonders why the flint in a gas lighter, made of misch metal (mostly cerium and lanthanum with some iron), throws sparks when scratched.

Explain using the reactivity of lanthanoids.

Try it first, then show the solutionSolution

Lanthanoids are reactive, electropositive metals that combine readily with oxygen: $4 L n + 3 O_{2} \to 2 L n_{2} O_{3}$ .
Scratching produces tiny fresh particles which heat up by friction and burn in air, giving sparks.

Answer: The finely divided lanthanoid metal ignites in air as it is scraped off.

Memory hook: Lanthanoids burn bright: that's the lighter's spark.

Topics in this chapter

d- and f-Block Elements formula sheet

Spin-only magnetic moment: $μ = n (n + 2) B M$; n = 1, 2, 3, 4, 5 gives 1.73, 2.83, 3.87, 4.90, 5.92 BM.
Manganate from pyrolusite: $2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$; Fusion in air; green manganate.
Manganate to permanganate: $3 K_{2} M n O_{4} + 2 C O_{2} \to 2 K M n O_{4} + M n O_{2} + 2 K_{2} C O_{3}$; Or electrolytic oxidation at the anode.
KMnO₄ + FeSO₄ (acidic): $2 K M n O_{4} + 8 H_{2} S O_{4} + 10 F e S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O$; Mn +7 → +2; Fe +2 → +3.
KMnO₄ + oxalic acid (acidic): $2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$; Warm to about 333 K; CO₂ bubbles.
KMnO₄ + KI (acidic): $2 K M n O_{4} + 8 H_{2} S O_{4} + 10 K I \to 6 K_{2} S O_{4} + 2 M n S O_{4} + 5 I_{2} + 8 H_{2} O$; Iodine liberated.
KMnO₄ + KI (basic): $2 K M n O_{4} + K I + H_{2} O \to 2 M n O_{2} + 2 K O H + K I O_{3}$; Iodide to iodate; Mn +7 → +4.
KMnO₄ + H₂S (neutral): $2 K M n O_{4} + 3 H_{2} S \to 2 K O H + 2 M n O_{2} + 3 S + 2 H_{2} O$; Sulphur precipitated.
Chromite roasting: $4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2} \to 8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}$; Then acidify and add KCl.
Chromate to dichromate: $2 N a_{2} C r O_{4} + H_{2} S O_{4} \to N a_{2} C r_{2} O_{7} + N a_{2} S O_{4} + H_{2} O$; Yellow to orange.
K₂Cr₂O₇ + FeSO₄ (acidic): $K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O$; Orange to green.
K₂Cr₂O₇ + H₂S (acidic): $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 H_{2} S \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 S + 7 H_{2} O$; Milky sulphur.
K₂Cr₂O₇ + KI (acidic): $K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 K I \to 4 K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 I_{2} + 7 H_{2} O$; Used in iodometry.
Lanthanoids with water: $2 L n + 6 H_{2} O \to 2 L n (O H)_{3} + 3 H_{2}$; Also with O₂ → Ln₂O₃, halogens → LnX₃, N₂ → LnN, C → LnC₂.

d- and f-Block Elements: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Structures of permanganate, chromate and dichromate ions

'Draw the structure of MnO₄⁻ / Cr₂O₇²⁻' or 'state the shape'.

Draw

MnO₄⁻: Mn at the centre of a tetrahedron, three Mn=O and one Mn-O⁻ (show resonance or the charge on the ion).
CrO₄²⁻: tetrahedral Cr with four O, charge 2−.
Cr₂O₇²⁻: two CrO₄ tetrahedra joined through one bridging O; mark the Cr-O-Cr angle about 126°.

Label

Mn (+7)
tetrahedral
Cr (+6)
bridging O
126°
terminal O

Watch out: State the oxidation state of the metal alongside the shape; give both in your answer.

Trend in atomic radii across the 3d series and down a group

Explaining lanthanoid contraction and its effect on Zr and Hf.

Draw

x-axis: group number 3 to 12; y-axis: atomic radius.
Three curves for 3d, 4d and 5d series: each falling at first, flat in the middle, rising slightly at the end.
Show the 4d and 5d curves almost overlapping, well above the 3d curve.

Label

3d
4d
5d
Zr ≈ Hf
atomic radius / pm

Watch out: One sentence: 4d ≈ 5d because of lanthanoid contraction.

How the board asks d- and f-Block Elements

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Give reasons	Colour of Cu²⁺ vs Cu⁺, Ti³⁺ vs Ti⁴⁺; Zn not transition; complex formation; alloys; Mn²⁺ stability; catalysis.	Name the property-causing feature (d-d transition, $d^{5}$ / $d^{10}$ , similar size, vacant d orbitals) and the configuration.
Magnetic moment	Spin-only μ of a given ion; which ion is most paramagnetic.	Write the ion's configuration (4s lost first), count unpaired electrons, use √n(n+2).
Lanthanoid contraction	Cause and consequences; Zr/Hf; basicity of Ln(OH)₃; colourless La³⁺/Lu³⁺.	Cause: poor shielding by 4f. Always add 'similar size, so similar properties'.
KMnO₄ and K₂Cr₂O₇ equations	Extraction from ores; balanced equations with FeSO₄, oxalic acid, KI, H₂S; pH effect.	Learn complete molecular equations; check atom balance; state the change in oxidation state.
Actinoids vs lanthanoids	Two differences; reason for more oxidation states in actinoids.	Use energy gap 5f/6d/7s; radioactivity; oxocations.
Titration calculations	Volume or concentration of KMnO₄ or K₂Cr₂O₇ for a given reductant.	Mole ratio from the balanced equation (MnO₄⁻:Fe²⁺ = 1:5; Cr₂O₇²⁻:Fe²⁺ = 1:6).

Tricks to remember d- and f-Block Elements

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cr and Cu cheat
Chromium and Copper Cheat the order: $3 d^{5} 4 s^{1}$ and $3 d^{10} 4 s^{1}$ .
Use it when: Electronic configurations.
First in, first out? No: 4s out first
Ionise from 4s before 3d.
Use it when: Configurations of ions and magnetic moments.
Zero or full = no colour
$d^{0}$ and $d^{10}$ : colourless and diamagnetic (Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺).
Use it when: Colour and magnetism reasons.
Permanganate's five, dichromate's six
Acidic: $M n O_{4}^{-}$ gains 5e⁻; $C r_{2} O_{7}^{2 -}$ gains 6e⁻.
Use it when: Balancing and titration ratios.
Zr-Hf twins
Lanthanoid contraction makes Zr and Hf twins: same size, same chemistry.
Use it when: Consequences of lanthanoid contraction.

d- and f-Block Elements: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Colour and moment of one ion · 1 mark · Very short answerBoard level

Fill in the blanks: The V²⁺ ion (Z of V = 23) is ______ in aqueous solution and has a spin-only magnetic moment of ______ BM.

Try it first, then show the solutionSolution

V²⁺ is $3 d^{3}$ : three unpaired electrons, so a d-d transition is possible (coloured) and μ = √15 = 3.87 BM.

Answer: coloured; 3.87

Example 2: Oxidation state change in acidic permanganate · 1 mark · MCQBoard level

When acidified KMnO₄ oxidises potassium iodide, the oxidation state of manganese changes from:
(a) +7 to +6
(b) +7 to +4
(c) +7 to +2
(d) +6 to +2

Try it first, then show the solutionSolution

In acid, MnO₄⁻ gains 5e⁻ and becomes Mn²⁺; the change to +4 (MnO₂) happens in basic or neutral media.

Answer: (c) +7 to +2

Example 3: Colour and the definition of a transition element · 2 marks · Short answerBoard level

Account for: (a) Cr³⁺ salts are coloured but Sc³⁺ salts are colourless (Z: Sc 21, Cr 24). (b) Silver ( $4 d^{10} 5 s^{1}$ ) is regarded as a transition element although Ag⁺ salts are colourless. [2]

Try it first, then show the solutionSolution

(a) Cr³⁺ is $3 d^{3}$ with unpaired electrons, so a d-d transition absorbs visible light; Sc³⁺ is $3 d^{0}$ : no d electron, no d-d transition.
(b) In its +2 state (as in AgF₂) silver is $4 d^{9}$ , an incompletely filled d subshell, so it meets the definition; Ag⁺ itself is $4 d^{10}$ and colourless.

Answer: (a) $d^{3}$ vs $d^{0}$ (b) Ag²⁺ is $4 d^{9}$

Example 4: Most paramagnetic ion · 2 marks · Short answerHarder board level

Which of V³⁺, Cr³⁺, Mn²⁺ and Ni²⁺ (Z = 23, 24, 25, 28) has the highest magnetic moment? Calculate it, and that of Ni²⁺. [2]

Try it first, then show the solutionSolution

V³⁺ $d^{2}$ (2 unpaired), Cr³⁺ $d^{3}$ (3), Mn²⁺ $d^{5}$ (5), Ni²⁺ $d^{8}$ (2).
Mn²⁺: μ = √35 = 5.92 BM; Ni²⁺: μ = √8 = 2.83 BM

Answer: Mn²⁺, 5.92 BM; Ni²⁺ 2.83 BM

Example 5: Balancing permanganate reactions · 3 marks · Short answerHarder board level

Complete and balance: (a) KMnO₄ + H₂SO₄ + FeSO₄ → (b) KMnO₄ + KI + H₂O → (basic) (c) KMnO₄ + H₂S → (neutral). [3]

Try it first, then show the solutionSolution

(a) $2 K M n O_{4} + 8 H_{2} S O_{4} + 10 F e S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O$
(b) $2 K M n O_{4} + K I + H_{2} O \to 2 M n O_{2} + 2 K O H + K I O_{3}$
(c) $2 K M n O_{4} + 3 H_{2} S \to 2 K O H + 2 M n O_{2} + 3 S + 2 H_{2} O$

Answer: As above; Mn goes to +2 in acid and +4 in basic/neutral media

Example 6: Dichromate from chromite · 5 marks · Long answerHarder board level

(a) Write balanced equations for the preparation of K₂Cr₂O₇ from chromite ore. [3]
(b) What happens when K₂Cr₂O₇ is treated with acidified H₂S? Give the equation and the colour change. [2]

Try it first, then show the solutionSolution

(a) $4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2} \to 8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}$
$2 N a_{2} C r O_{4} + H_{2} S O_{4} \to N a_{2} C r_{2} O_{7} + N a_{2} S O_{4} + H_{2} O$
$N a_{2} C r_{2} O_{7} + 2 K C l \to K_{2} C r_{2} O_{7} + 2 N a C l$
(b) $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 H_{2} S \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 S + 7 H_{2} O$
Orange solution turns green (Cr³⁺) and pale yellow sulphur precipitates.

Answer: Equations as above; orange → green with milky sulphur

Example 7: Lanthanoid contraction and basicity · 3 marks · Short answerHarder board level

(a) What is lanthanoid contraction and what causes it? (b) Explain why La(OH)₃ is more basic than Lu(OH)₃. [3]

Try it first, then show the solutionSolution

(a) The steady decrease in atomic and ionic radii from La to Lu.
Cause: added electrons enter 4f orbitals, which shield poorly, so effective nuclear charge increases and pulls the electron cloud in.
(b) Lu³⁺ is smaller than La³⁺, so it polarises the OH⁻ more and the M-OH bond is more covalent; it releases OH⁻ less readily.

Answer: Poor shielding by 4f; smaller Lu³⁺ gives a more covalent, less basic hydroxide

Example 8: Why permanganate titrations need H₂SO₄ · 3 marks · Short answerHOTS / competency

In the estimation of Fe²⁺ with KMnO₄, why is the solution acidified with dilute H₂SO₄ rather than HCl or HNO₃? How is the end point recognised without an indicator? [3]

Try it first, then show the solutionSolution

HCl is oxidised by KMnO₄ to Cl₂, consuming extra permanganate and giving high results.
HNO₃ is itself an oxidising agent and would oxidise some Fe²⁺, giving low results. Dilute H₂SO₄ is neither oxidised nor oxidising.
KMnO₄ is decolourised as long as Fe²⁺ remains; the first drop in excess gives a permanent pale pink: it is a self-indicator.

Answer: H₂SO₄ is inert to redox here; KMnO₄ acts as its own indicator

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

The traffic-light test tube

Suppose a teacher adds dilute sulphuric acid to a yellow solution of potassium chromate in one test tube, and passes H₂S through an acidified orange dichromate solution in another, which turns green and cloudy.

(i) · 1 mark
Why does the yellow solution turn orange?
Answer
Answer: Chromate changes to dichromate in acid.
$2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$
(ii) · 1 mark
What is the oxidation state of Cr in chromate and in dichromate?
Answer
Answer: +6 in both.
So the first change is not a redox reaction.
(iii) · 1 mark
What causes the green colour and the cloudiness in the second tube?
Answer
Answer: Cr³⁺ ions (green) and colloidal sulphur.
H₂S reduces Cr(+6) to Cr(+3) and is oxidised to S.
(iv) · 2 marks
Write the balanced equation for the second reaction.
Answer
Answer: $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 H_{2} S \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 S + 7 H_{2} O$
Each Cr₂O₇²⁻ gains 6e⁻; each H₂S loses 2e⁻; ratio 1 : 3.

Choosing a catalyst and an alloy

Imagine a chemical plant that uses V₂O₅ to make SO₃, iron to make ammonia and nickel to harden vegetable oils, and a cutlery maker who uses stainless steel (Fe with Cr and Ni).

(i) · 1 mark
Give two reasons why transition metals are good catalysts.
Answer
Answer: Variable oxidation states and a large surface for adsorption.
They form unstable intermediates and weaken bonds of adsorbed reactants.
(ii) · 1 mark
Write the reaction catalysed by V₂O₅.
Answer
Answer: $2 S O_{2} + O_{2} \to 2 S O_{3}$
Contact process for sulphuric acid.
(iii) · 1 mark
Why can Cr and Ni atoms replace Fe atoms in the lattice of steel?
Answer
Answer: They have similar atomic sizes.
Similar radii allow substitutional alloys without distorting the lattice much.
(iv) · 2 marks
Steel also contains carbon atoms in the holes of the iron lattice. Name this type of compound and give two properties.
Answer
Answer: Interstitial compound.
Hard, high melting, retains metallic conductivity, chemically inert, often non-stoichiometric.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Give reasons: (a) V⁴⁺ salts are coloured but V⁵⁺ salts are colourless (Z of V = 23). (b) Transition metals and their compounds act as catalysts. (2 marks)

Step	Marks
(a) V⁴⁺ $3 d^{1}$ allows a d-d transition; V⁵⁺ $3 d^{0}$ has no d electron	1
(b) Variable oxidation states let them form unstable intermediates, and their surfaces adsorb reactants and weaken bonds	1

Explain why the pink colour of KMnO₄ disappears when it is added to acidified FeSO₄, and write the ionic equation. (3 marks)

Step	Marks
Mn is reduced from +7 (purple $M n O_{4}^{-}$ ) to +2 (almost colourless Mn²⁺)	1
Fe²⁺ is oxidised to Fe³⁺	1
$M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$	1

(a) Describe the preparation of KMnO₄ from pyrolusite with equations. (b) State its structure and shape. (c) Write its reaction with acidified oxalic acid. (5 marks)

Step	Marks
$2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$	1
Manganate oxidised to permanganate (CO₂ disproportionation or electrolysis) with equation	1
Tetrahedral $M n O_{4}^{-}$ with Mn +7	1
Correct products of oxalic acid reaction (K₂SO₄, MnSO₄, CO₂, H₂O)	1
Balanced: $2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$	1

Where students lose marks in d- and f-Block Elements

The slips examiners see most often, and what to write instead.

Mistake: Vague reasons for colour
CISCE examiners' comments, 2024 and 2025: candidates did not mention d-d transition for the colour of Cu²⁺ or Ti³⁺, confused empty, partly filled and completely filled d orbitals and paired with unpaired electrons, and in a fill-in item wrote 'incompletely filled' and 'paramagnetic' for colourless ions.
Wrong working
Cu²⁺ is coloured because it has a high charge.
Right working
Cu²⁺ ( $3 d^{9}$ ) has an unpaired electron; absorption of visible light causes a d-d transition; Cu⁺ ( $3 d^{10}$ ) cannot.
Do this: Write the configuration, then 'd-d transition possible / not possible'.
Usually costs: 1 mark each.
Mistake: Naming lanthanoid contraction without explaining it
CISCE examiners' comments, 2024 and 2025: for Zr and Hf, candidates wrote only 'lanthanoid contraction' without stating their similar size and properties, some wrote 'actinoid contraction', and in an assertion-reason item many chose (b) though the reason was the correct explanation.
Wrong working
Due to lanthanoid contraction.
Right working
Lanthanoid contraction makes Hf almost the same size as Zr, so their properties are almost identical and separation is difficult.
Do this: Always finish with the consequence: same size, same properties.
Usually costs: 1 mark.
Mistake: Wrong reasons for complex formation, alloys and Mn²⁺ stability
CISCE examiners' comments, 2024 and 2025: candidates gave variable valency as the reason for complex formation instead of small size, high nuclear charge and vacant d orbitals; gave 'small size' instead of 'similar size' for alloys; could not write configurations of Mn²⁺ and Fe²⁺ or omitted the stability of half-filled orbitals; and confused d-d with f-f transitions for La³⁺ and Lu³⁺.
Wrong working
Transition metals form alloys because they are small.
Right working
They form alloys because their atomic sizes are similar.
Do this: Learn one exact key phrase for each property.
Usually costs: 1 mark each.
Mistake: Unbalanced or wrong permanganate and dichromate equations
CISCE examiners' comments, 2024 and 2025: candidates chose SO₂ as the gas from KMnO₄ and oxalic acid, did not state the change of Mn from +7 to +2 when the pink colour disappears with Mohr's salt, wrote correct products but an unbalanced KMnO₄-oxalic acid equation, and wrote wrong product formulae for K₂Cr₂O₇ with H₂S.
Wrong working
$K M n O_{4} + H_{2} S O_{4} + H_{2} C_{2} O_{4} \to K_{2} S O_{4} + M n S O_{4} + C O_{2} + H_{2} O$ (unbalanced).
Right working
$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$
Do this: Balance electrons (5 per Mn, 2 per oxalate) then atoms; count K, S, O last.
Usually costs: 1-2 marks.

Practice questions on d- and f-Block Elements

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which ion is colourless in aqueous solution?
(a) Ti³⁺
(b) Cu²⁺
(c) Sc³⁺
(d) Fe³⁺
Hint
$d^{0}$ .
Answer
(c) Sc³⁺
Q2 · 1 mark · MCQBoard level
The shape of the dichromate ion is best described as:
(a) two tetrahedra sharing a corner
(b) two octahedra sharing an edge
(c) square planar
(d) linear
Hint
Cr-O-Cr bridge.
Answer
(a)
Q3 · 1 mark · Very short answerBoard level
Fill in the blanks: Lanthanoids show ______ transitions, and the ions La³⁺ and Lu³⁺ are ______.
Hint
$4 f^{0}$ and $4 f^{14}$ .
Answer
f-f; colourless
Q4 · 1 mark · Very short answerBoard level
Why do transition metals have high enthalpies of atomisation?
Hint
Which electrons bond?
Answer
Both ns and (n−1)d electrons, including many unpaired d electrons, take part in strong metallic bonding.
Q5 · 2 marks · Short answerHarder board level
Calculate the spin-only magnetic moment of Co²⁺ (Z = 27) and predict whether [CoCl₄]²⁻ is coloured.
Hint
$3 d^{7}$ .
Answer
Co²⁺ $3 d^{7}$ : 3 unpaired, μ = √15 = 3.87 BM. Partly filled d subshell allows d-d transitions, so it is coloured (blue).
Q6 · 2 marks · Short answerHarder board level
Why is Fe³⁺ more stable than Fe²⁺, while Mn³⁺ is less stable than Mn²⁺?
Hint
Count d electrons.
Answer
Fe³⁺ is $3 d^{5}$ (half-filled, extra stable) whereas Fe²⁺ is $3 d^{6}$ . Mn²⁺ is $3 d^{5}$ while Mn³⁺ is $3 d^{4}$ ; so Mn²⁺ is the stable one. Each pair favours the $d^{5}$ ion.
Q7 · 2 marks · Short answerHarder board level
Write balanced equations for the reaction of acidified K₂Cr₂O₇ with (a) FeSO₄ and (b) KI.
Hint
Cr₂O₇²⁻ gains 6e⁻.
Answer
(a) $K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O$ (b) $K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 K I \to 4 K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 I_{2} + 7 H_{2} O$
Q8 · 2 marks · Short answerHarder board level
Give two consequences of lanthanoid contraction other than the Zr-Hf similarity.
Hint
Basicity; separation.
Answer
Basic strength of Ln(OH)₃ decreases from La to Lu; lanthanoids are difficult to separate from one another (similar sizes); 5d elements have high densities and ionisation enthalpies.
Q9 · 3 marks · Short answerHarder board level
Write the equations for the reactions of a lanthanoid (Ln) with (a) nitrogen, (b) carbon at 2773 K, (c) chlorine.
Hint
Ln is +3 except in the carbide.
Answer
(a) $2 L n + N_{2} \to 2 L n N$ (b) $L n + 2 C \to L n C_{2}$ (c) $2 L n + 3 C l_{2} \to 2 L n C l_{3}$
Q10 · 3 marks · Short answerHarder board level
25.0 mL of an acidified Fe²⁺ solution needs 20.0 mL of 0.0200 M K₂Cr₂O₇. Find the molarity of Fe²⁺.
Hint
1 : 6.
Answer
Moles Cr₂O₇²⁻ = 4.00 × 10⁻⁴; Fe²⁺ = 2.40 × 10⁻³ mol; M = 2.40 × 10⁻³/0.0250 = 0.096 M.
Q11 · 5 marks · Long answerHarder board level
(a) Give two differences between lanthanoids and actinoids. [2]
(b) Why does the number of oxidation states rise to a maximum at Mn in the 3d series? [2]
(c) Name the oxidation state of Mn in the green compound formed when MnO₂ is fused with KOH in air. [1]
Hint
Count 4s + 3d electrons.
Answer
(a) Actinoids show many oxidation states (+3 to +7), lanthanoids mainly +3; actinoids are all radioactive; actinoids form complexes and oxocations (UO₂²⁺) more readily. (b) Mn ( $3 d^{5} 4 s^{2}$ ) has seven electrons in 4s and 3d that can all be used in bonding (up to +7); after Mn, pairing in d orbitals makes fewer electrons available, so the maximum falls. (c) +6 (K₂MnO₄).
Q12 · 2 marks · Short answerHOTS / competency
A student says 'KMnO₄ is purple because Mn has unpaired d electrons'. Correct the statement.
Hint
Mn +7.
Answer
In MnO₄⁻, Mn is +7 ( $3 d^{0}$ ): no d electrons, so no d-d transition and it is diamagnetic. The intense purple colour is due to charge transfer, an electron moving from oxygen to the metal on absorbing light.
Q13 · 2 marks · Short answerHOTS / competency
Explain why Cu (Z = 29), which has a $3 d^{10}$ configuration in the atom, is regarded as a transition element while Zn is not.
Hint
Look at the common ion.
Answer
Cu forms Cu²⁺ ( $3 d^{9}$ ), a common ion with an incompletely filled d subshell, so it meets the definition. Zn and its only common ion Zn²⁺ are both $d^{10}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define a transition element.Show answer
An element whose atom or a common ion has an incompletely filled d subshell.
2.Configuration of Cr and Cu?Show answer
Cr [Ar] $3 d^{5} 4 s^{1}$ ; Cu [Ar] $3 d^{10} 4 s^{1}$ .
3.Spin-only formula?Show answer
$μ = n (n + 2)$ BM
4.Cause of lanthanoid contraction?Show answer
Poor shielding by 4f electrons, so effective nuclear charge increases from La to Lu.
5.Ore of KMnO₄ and of K₂Cr₂O₇?Show answer
Pyrolusite (MnO₂); chromite (FeCr₂O₄).
6.Electrons gained by MnO₄⁻ in acid and by Cr₂O₇²⁻ in acid?Show answer
5 and 6.
7.What are interstitial compounds?Show answer
Compounds formed when small atoms (H, C, N, B) occupy holes in a metal lattice.
8.Colour change on acidifying chromate?Show answer
Yellow to orange (dichromate).

Best order to study it

Learn configurations of 3d elements and ions (Cr, Cu exceptions; 4s lost first).
Learn trends in radii, ionisation enthalpy, oxidation states and metallic properties.
Learn colour, magnetism (with μ calculations), catalysis, complexes, interstitial compounds and alloys.
Learn lanthanoids: configuration, +3 state, contraction, its consequences and reactions.
Compare actinoids with lanthanoids.
Learn KMnO₄ and K₂Cr₂O₇: structure, extraction and every listed reaction, balanced.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain why Zn, Cd and Hg are not transition elements?
Can you write configurations of any 3d ion and find μ?
Can you explain colour by d-d transition and say which ions are colourless?
Can you state the cause and three consequences of lanthanoid contraction?
Can you write the lanthanoid reactions with O₂, H₂, X₂, S, N₂, C and H₂O?
Can you give two differences between lanthanoids and actinoids?
Can you write the extraction equations for KMnO₄ and K₂Cr₂O₇?
Can you balance KMnO₄ with FeSO₄, oxalic acid, KI and H₂S in the right medium?

d- and f-Block Elements: frequently asked questions

Which KMnO₄ reactions does the ISC 2027 syllabus name?

In acidic medium with FeSO₄, oxalic acid and KI; in basic medium with KI; and in neutral medium with H₂S. You also need its structure, extraction from pyrolusite and use in redox titration.

Do I need the structures of chromate and dichromate?

Yes. The syllabus asks for the structure and shape of K₂Cr₂O₇, its extraction from chromite, its oxidising nature in different media and the effect of pH on chromate and dichromate.

Is Zn a d-block element?

Yes, it is in the d-block (group 12), but it is not a transition element because its atom and its common ion Zn²⁺ have completely filled d orbitals.

Why are KMnO₄ and K₂Cr₂O₇ coloured if the metal is $d^0$?

Their colour comes from charge transfer (an electron moving from oxygen to the metal on absorbing light), not from d-d transitions.

How much of the lanthanoid chemistry is needed?

Configuration, radii, oxidation states, colour, complexes, alloys, lanthanoid contraction and its consequences, and reactions with oxygen, hydrogen, halogens, sulphur, nitrogen, carbon and water. Actinoids: mainly oxidation states and comparison with lanthanoids.

How many marks does this chapter carry?

It is part of the Inorganic Chemistry group, 14 of the 70 theory marks, shared with Coordination Compounds.

Based on: CISCE ISC 2027 Chemistry syllabus; CISCE ISC 2027 Chemistry specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Chemistry, 2025; CISCE Analysis of Pupil Performance, ISC Chemistry, 2024