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d & f Block Elements – JEE Main Formula Sheet & Class 12 Notes | Transition & Inner Transition Metals
This is the complete JEE Main Formula Sheet and Class 12 Formula Sheet for d and f Block Elements — Chapter 21 from the Aakash Rapid Revision & Formula Bank. The chapter covers the characteristic properties of transition metals (d-block, Groups 3–12) — their electronic configurations, variable oxidation states, magnetic behaviour, coloured compounds, catalytic activity, alloy formation, and interstitial compounds — along with the important potassium dichromate (K₂Cr₂O₇) and potassium permanganate (KMnO₄) reactions. The f-block section covers lanthanoids and actinoids, the lanthanoid contraction, and key comparisons between the two series. These topics contribute 4–6 questions in JEE Main every year, with a strong emphasis on electronic configuration exceptions, magnetic moment calculations, and KMnO₄/K₂Cr₂O₇ reaction applications. Download the Free PDF below for all formulas, reaction details, and trend explanations in one exam-ready reference.
Topics Covered in This Formula Sheet
JEE Main Formula Sheet PDF Preview
Scroll to explore all d & f Block Elements formulas — JEE Main & Class 12 Formula Sheet
Introduction: Why d & f Block Is a Consistent High-Yield Chapter in JEE Main
Transition metals occupy the ten groups in the middle of the periodic table and are the workhorses of both industrial chemistry and biological systems. Their unique ability to exist in multiple oxidation states, form intensely coloured compounds, act as catalysts, and exhibit paramagnetism through unpaired d-electrons makes them chemically unlike any other group of elements. For JEE Main, this translates into a chapter that is both conceptually rich and consistently examined — electronic configuration exceptions, magnetic moment calculations, and the reactions of KMnO₄ and K₂Cr₂O₇ appear in virtually every examination session.
The f-block elements — lanthanoids and actinoids — bring a different dimension. The lanthanoid contraction is one of the most consequence-rich single phenomena in inorganic chemistry: it explains why 4d and 5d transition metals have nearly identical sizes, why Zr and Hf are virtually inseparable, and why the properties of 5d transition metals are more similar to 4d metals than to 3d metals. Understanding lanthanoid contraction once gives you the reasoning tool to answer a whole class of JEE Main questions.
Download the Free PDF to access all electronic configurations, magnetic moment values, KMnO₄ and K₂Cr₂O₇ reaction tables, and lanthanoid/actinoid comparison charts in one structured revision reference.
Overview: Structure of This Chapter
The chapter divides into two major sections. The first covers the d-block (transition metals) — primarily the first transition series (Sc to Zn, atomic numbers 21–30) and their characteristic properties: variable oxidation states, magnetic behaviour based on unpaired electrons, coloured ions due to d-d transitions, catalytic activity explained through multiple oxidation states and surface adsorption, and the specific chemistry of chromium and manganese through their industrially important potassium salts. The second section covers the f-block — lanthanoids (La to Lu, 4f filling) and actinoids (Ac to Lr, 5f filling) — with emphasis on lanthanoid contraction and its consequences, and a comparison of lanthanoid vs. actinoid oxidation state behaviour and radioactivity.
Throughout both sections, the key analytical tool is electronic configuration. Every characteristic property of transition metals — variable oxidation states, paramagnetism, colour, catalysis — can be traced directly to the partially filled d-subshell. Students who keep this connection in mind find that the chapter becomes logically consistent rather than a collection of isolated facts. Download the Free PDF for the complete chapter summary with all configurations and properties mapped.
Key Concepts Covered in This Chapter
Electronic Configuration of d-Block Elements and Exceptions
Why Configuration Exceptions Are the Most Directly Tested Content in JEE Main
The d-block elements fill the (n-1)d subshell. The general configuration for the first transition series is [Ar] 3d¹⁻¹⁰ 4s¹⁻². However, the expected filling order is overridden in several cases where a half-filled (3d⁵) or completely filled (3d¹⁰) d-subshell provides extra stability. These exceptions are direct JEE Main one-mark questions.
First transition series (Sc to Zn) configurations: Sc: [Ar]3d¹4s²; Ti: [Ar]3d²4s²; V: [Ar]3d³4s²; Cr: [Ar]3d⁵4s¹ (exception — half-filled 3d and half-filled 4s, not 3d⁴4s²); Mn: [Ar]3d⁵4s²; Fe: [Ar]3d⁶4s²; Co: [Ar]3d⁷4s²; Ni: [Ar]3d⁸4s²; Cu: [Ar]3d¹⁰4s¹ (exception — completely filled 3d, not 3d⁹4s²); Zn: [Ar]3d¹⁰4s².
Similar exceptions in second transition series: Mo ([Kr]4d⁵5s¹, like Cr) and Ag ([Kr]4d¹⁰5s¹, like Cu). In the third series: W follows regular filling; Au ([Xe]4f¹⁴5d¹⁰6s¹, like Cu/Ag). When transition metals form ions, the 4s electrons are lost first (before 3d electrons), even though 4s fills before 3d during the build-up of the atom. So Fe²⁺ = [Ar]3d⁶ (not [Ar]3d⁴4s²), and Fe³⁺ = [Ar]3d⁵. Zn, Cd, and Hg are technically not transition metals by the strict definition (they have completely filled d-subshells in all common oxidation states — Zn²⁺ = [Ar]3d¹⁰) but are studied with the d-block due to their position. Download the Free PDF for the complete first series configuration table.
Variable Oxidation States of Transition Metals
Why Oxidation State Trends Are a Multi-Mark JEE Main Topic
The most characteristic property of transition metals is their ability to exhibit variable oxidation states in compounds. This arises because the 3d and 4s electrons are close in energy and can both be involved in bonding. The energy difference between successive ionisation enthalpies is small enough that multiple electrons can be removed under appropriate chemical conditions.
Trend in maximum oxidation state across first series: The maximum oxidation state increases from Sc (+3) to Mn (+7) as more d-electrons become available for bonding. After Mn, the maximum oxidation state decreases because the d-electrons become increasingly paired and the nuclear charge holds them more tightly: Fe(+6 in ferrate FeO₄²⁻ — rare), Co(+4 rare), Ni(+4 rare), Cu(+2 common, +3 rare), Zn(+2 only). The general trend is that maximum oxidation state = group number for Groups 3–7 (Sc=+3, Ti=+4, V=+5, Cr=+6, Mn=+7). For Groups 8–10 the highest observed states are lower than the group number.
Stability of +2 oxidation state: The +2 state is shown by almost all first series transition metals (from Ti to Cu) because it corresponds to removal of the two 4s electrons. The stability of the +2 state relative to higher oxidation states changes across the series — Mn²⁺ is particularly stable (half-filled 3d⁵), which is why Mn³⁺ is a stronger oxidising agent than Fe³⁺ (Fe³⁺ itself being half-filled 3d⁵ and hence stable, making Fe²⁺ a weaker reducing agent than Mn²⁺). Cr²⁺ is a strong reducing agent (tends to be oxidised to Cr³⁺, which is very stable — 3d³ configuration). Cu²⁺ is more stable than Cu⁺ in aqueous solution due to higher hydration enthalpy of Cu²⁺ (small 2+ ion) outweighing the second ionisation enthalpy cost.
Magnetic Properties — Spin-Only Formula and Unpaired Electrons
Why Magnetic Moment Calculations Are a Guaranteed JEE Main Numerical Topic
Transition metal compounds exhibit paramagnetism when unpaired electrons are present. The magnetic moment (μ) can be calculated from the number of unpaired electrons (n) using the spin-only formula: μ = √[n(n+2)] Bohr Magnetons (BM). This formula assumes that only spin angular momentum contributes to magnetic moment (orbital contribution is quenched in most first-series transition metal complexes).
Values for common ion configurations: 1 unpaired electron: μ = √3 ≈ 1.73 BM (Cu²⁺: 3d⁹, Ti³⁺: 3d¹). 2 unpaired: μ = √8 ≈ 2.83 BM (V³⁺: 3d², Cu⁰ in some). 3 unpaired: μ = √15 ≈ 3.87 BM (Cr³⁺: 3d³, Co²⁺ in high-spin: 3d⁷). 4 unpaired: μ = √24 ≈ 4.90 BM (Fe²⁺: 3d⁶ high-spin, Cr²⁺: 3d⁴). 5 unpaired: μ = √35 ≈ 5.92 BM (Mn²⁺: 3d⁵, Fe³⁺: 3d⁵ — maximum for first series). Zn²⁺ (3d¹⁰) and Sc³⁺ (3d⁰) have 0 unpaired electrons: μ = 0 BM — diamagnetic. JEE Main approach: write the electronic configuration of the ion (remember: remove 4s first, then 3d), count unpaired electrons, substitute into the formula. Example: Fe²⁺ = [Ar]3d⁶ — draw 3d box: ↑↓ ↑ ↑ ↑ ↑ → 4 unpaired electrons → μ = √[4×6] = √24 ≈ 4.90 BM. Download the Free PDF for the complete magnetic moment table for all first series ions.
Colour of Transition Metal Compounds — d-d Transitions
Why Coloured Ion Identification Is a Direct JEE Main Question Type
Most transition metal compounds are intensely coloured. The colour arises from d-d transitions — when visible light falls on a transition metal ion with a partially filled d-subshell, electrons absorb photons of specific wavelengths and jump from lower-energy d-orbitals (t₂g) to higher-energy d-orbitals (eg) in the presence of ligands (crystal field splitting). The colour observed is the complementary colour of the absorbed wavelength.
Conditions required for colour: The d-subshell must be partially filled — ions with d⁰ (Ti⁴⁺, Sc³⁺) or d¹⁰ (Zn²⁺, Cu⁺) configurations have no d-d transition possible and are colourless. This is a critical JEE Main fact: Zn²⁺ compounds are white/colourless despite zinc being a d-block element. Ligands must also be present to create the crystal field splitting (ΔCF) — in the gaseous ion without ligands, all d-orbitals are degenerate and no d-d transition is possible.
Common ion colours to know for JEE Main: Ti³⁺ (d¹) — purple/violet; V³⁺ (d²) — green; Cr³⁺ (d³) — violet; Cr²⁺ (d⁴) — blue; Mn²⁺ (d⁵) — very pale pink (d⁵ is half-filled — transition is spin-forbidden → very weak colour); Fe³⁺ (d⁵) — pale yellow to yellow; Fe²⁺ (d⁶) — pale green; Co²⁺ (d⁷) — pink; Ni²⁺ (d⁸) — green; Cu²⁺ (d⁹) — blue; Zn²⁺ (d¹⁰) — colourless. The deep blue colour of [Cu(NH₃)₄]²⁺ (tetraamminecopper(II)) is a classic qualitative test for Cu²⁺ ions. KMnO₄ is intensely purple-violet — this colour is NOT due to d-d transition (Mn in +7 has d⁰ configuration) but due to charge transfer transition (ligand-to-metal electron transfer). Similarly, K₂Cr₂O₇ is orange (charge transfer, Cr is d⁰ in +6). Download the Free PDF for the complete colour table.
Catalytic Activity, Interstitial Compounds, and Alloy Formation
Why These Properties Are Tested as Conceptual JEE Main Questions
Catalytic activity: Transition metals and their compounds are outstanding catalysts. The catalytic ability arises from two properties working in combination. First, variable oxidation states allow the metal to form reactive intermediates with reactants at different oxidation levels, providing alternative reaction pathways with lower activation energies (homogeneous catalysis mechanism — e.g., Fe²⁺/Fe³⁺ cycle in the Haber process region). Second, transition metals have partially filled d-orbitals that can form weak bonds with reactant molecules, adsorbing them on the metal surface (heterogeneous catalysis — chemisorption). The adsorbed molecules are activated and positioned to react more easily, then the product desorbs. Key examples: Fe in Haber process (N₂ + 3H₂ → 2NH₃), Pt/Rh in Ostwald process (4NH₃ + 5O₂ → 4NO + 6H₂O), V₂O₅ in Contact process (2SO₂ + O₂ → 2SO₃), Ni in hydrogenation of oils (addition of H₂ to C=C bonds), Pd and Pt in catalytic converters. MnO₂ as catalyst in laboratory preparation of O₂ from KClO₃.
Interstitial compounds: Small non-metal atoms (H, B, C, N) can occupy the interstitial (vacant) spaces in the close-packed metallic lattice of transition metals. The resulting interstitial compounds are non-stoichiometric (variable composition, e.g., TiH₁.₇₃, VH₀.₅₆, PdH₀.₇). Properties: harder and less ductile than the parent metal, higher melting point than the parent metal, retain metallic conductivity and lustre, chemically inert. Examples: steel (Fe with interstitial C), WC (tungsten carbide, extremely hard cutting tool), TiN (very hard golden coating). The compounds are distinct from ionic carbides or hydrides — they have no fixed formulae and are best described as insertion compounds.
Alloy formation: Transition metals form alloys easily with other transition metals and with non-metals because their atomic radii are similar and their electronic structures allow intermixing. Alloys are generally harder, stronger, and more resistant to corrosion than the component pure metals. Examples: steel (Fe + C, sometimes + Cr, Ni, Mn), stainless steel (Fe + Cr + Ni), brass (Cu + Zn), bronze (Cu + Sn), nichrome (Ni + Cr, used in heating elements). Download the Free PDF for a complete list of catalytic applications and alloy compositions.
Potassium Dichromate (K₂Cr₂O₇) — Preparation and Reactions
Why K₂Cr₂O₇ Reactions Are Direct JEE Main Equations Questions
Potassium dichromate is one of the most important oxidising agents in inorganic and analytical chemistry. Chromium is in the +6 oxidation state in Cr₂O₇²⁻ (d⁰ — hence the orange colour is from charge transfer, not d-d). It is a strong oxidising agent, particularly in acidic medium.
Preparation of K₂Cr₂O₇: Step 1 — roasting of chromite ore with sodium carbonate in air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ (sodium chromate, yellow). Step 2 — acidification converts chromate to dichromate: 2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O. Step 3 — treatment with KCl: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl (K₂Cr₂O₇ is less soluble and crystallises out).
Chromate–dichromate equilibrium: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. In acidic medium (add H⁺): chromate (yellow, CrO₄²⁻) converts to dichromate (orange, Cr₂O₇²⁻). In basic medium (add OH⁻): dichromate converts back to chromate. This pH-dependent interconversion is a very common JEE Main question.
Oxidising reactions of K₂Cr₂O₇ in acidic medium (Cr goes from +6 to +3, orange to green): Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (half-reaction). Key reactions: K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3Fe₂(SO₄)₃ + 7H₂O (oxidises Fe²⁺ to Fe³⁺). K₂Cr₂O₇ + 6KI + 7H₂SO₄ → Cr₂(SO₄)₃ + 4K₂SO₄ + 3I₂ + 7H₂O (oxidises I⁻ to I₂). K₂Cr₂O₇ + 3H₂S + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3S + 7H₂O (oxidises H₂S to S). K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O (oxidises SO₂ to SO₄²⁻ — used in breathalyser test for alcohol: C₂H₅OH + K₂Cr₂O₇ + H₂SO₄ → CH₃COOH + Cr₂(SO₄)₃ + K₂SO₄ + H₂O, orange to green). Download the Free PDF for all balanced K₂Cr₂O₇ equations.
Potassium Permanganate (KMnO₄) — Preparation and Reactions in Different Media
Why KMnO₄ Reactions Are the Most Frequently Tested Equations in This Chapter
Potassium permanganate (KMnO₄) is the most powerful and versatile oxidising agent among inorganic compounds studied at this level. Manganese is in the +7 oxidation state in MnO₄⁻ (d⁰ — the intense purple colour is due to charge transfer, not d-d transitions). The product of KMnO₄ reduction depends on the medium of the reaction — this medium-dependence is the most tested concept about KMnO₄ in JEE Main.
Preparation of KMnO₄: Step 1 — alkaline fusion of MnO₂ with KOH in the presence of an oxidising agent (KNO₃ or atmospheric O₂): 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O (potassium manganate, dark green, Mn = +6). Step 2 — oxidation of manganate to permanganate, either by chlorine: 2K₂MnO₄ + Cl₂ → 2KMnO₄ + 2KCl (Mn: +6 → +7), or by electrolytic oxidation at the anode: MnO₄²⁻ → MnO₄⁻ + e⁻. In acidic solution, manganate disproportionates: 3K₂MnO₄ + 2H₂SO₄ → 2KMnO₄ + MnO₂ + 2K₂SO₄ + 2H₂O (Mn: +6 → +7 and +4).
KMnO₄ reduction products by medium — the most important JEnE Main content here:
Acidic medium (H₂SO₄): MnO₄⁻ → Mn²⁺ (colourless/pale pink). Half-reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Purple colour → colourless (intense to colourless is a visual indicator of complete reaction). Reactions: 2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 5O₂ + 8H₂O (H₂O₂ oxidised to O₂, KMnO₄ is reduced); 2KMnO₄ + 5FeSO₄ + 8H₂SO₄ → 2MnSO₄ + K₂SO₄ + 5Fe₂(SO₄)₃/2 + 8H₂O (Fe²⁺ → Fe³⁺); 2KMnO₄ + 5C₂H₄ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 5C₂H₄(OH)₂ + 3H₂O (Baeyer's test — alkene decolourises KMnO₄ solution, forming diol); 2KMnO₄ + 5Na₂SO₃ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 5Na₂SO₄ + 3H₂O (SO₃²⁻ → SO₄²⁻).
Neutral/very weakly alkaline medium: MnO₄⁻ → MnO₂ (brown precipitate). Half-reaction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Reaction: 2KMnO₄ + 3MnSO₄ + 2H₂O → 5MnO₂ + K₂SO₄ + 2H₂SO₄ (brown precipitate forms).
Strongly alkaline medium (excess KOH): MnO₄⁻ → MnO₄²⁻ (manganate, green). Half-reaction: MnO₄⁻ + e⁻ → MnO₄²⁻ (Mn: +7 → +6). Reaction: 2KMnO₄ + 2KOH → 2K₂MnO₄ + H₂O + [O] (also given as 2KMnO₄ + I₂ + 2KOH → 2K₂MnO₄ + 2HIO₃ in alkaline medium). Download the Free PDF for all KMnO₄ equations in all three media with balancing steps.
Lanthanoids — Electronic Configuration, Properties, and Lanthanoid Contraction
Why Lanthanoid Contraction Is One of the Most Consequence-Rich Concepts in JEE Main
The lanthanoids are the 14 elements from La (57) to Lu (71) in which the 4f subshell is progressively filled. Their general electronic configuration is [Xe] 4f⁰⁻¹⁴ 5d⁰⁻¹ 6s². The predominant oxidation state is +3 (formed by losing 2 electrons from 6s and 1 from 4f or 5d). A few lanthanoids show +2 (Eu, Sm, Yb) or +4 (Ce, Pr, Tb) states when the resulting 4f subshell has an empty, half-filled, or completely filled configuration.
Lanthanoid contraction: Across the lanthanoid series (La to Lu), as electrons are added to the inner 4f subshell, the nuclear charge increases by one proton with each element. The 4f electrons provide very poor shielding to the outer 5d and 6s electrons (due to the diffuse and deeply penetrating nature of f-orbitals). As a result, the effective nuclear charge experienced by the outer electrons increases steadily across the series, causing a progressive decrease in atomic and ionic radii from La to Lu. This steady decrease in size across 14 elements is the lanthanoid contraction.
Consequences of lanthanoid contraction — a critical JEE Main topic: (1) The radii of lanthanoids are so small by Lu that the 5d transition metals of the third series (Hf, Ta, W, Re, Os, Ir, Pt, Au) have nearly the same atomic radii as their 4d counterparts in the second series (Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag) immediately above them. This makes Zr–Hf and Nb–Ta pairs nearly inseparable by chemical means — a major industrial challenge. (2) The similarity in size of 4d and 5d elements makes their chemistry and properties much more alike than would normally be expected for elements two periods apart. (3) Lanthanoids are separated from each other with great difficulty (ion exchange chromatography) because they have such similar sizes and hence similar chemical properties. Download the Free PDF for the complete lanthanoid property table and contraction diagram.
Actinoids and Comparison with Lanthanoids
Why the Lanthanoid–Actinoid Comparison Is a Direct JEE Main Table Question
The actinoids are the 14 elements from Ac (89) to Lr (103) in which the 5f subshell is progressively filled. Their general electronic configuration is [Rn] 5f⁰⁻¹⁴ 6d⁰⁻¹ 7s². The actinoids show much greater variation in oxidation states than lanthanoids — this is because the 5f, 6d, and 7s electrons are closer in energy than 4f, 5d, and 6s in lanthanoids, making it easier to involve all three subshells in bonding. Actinoids from Th to Np (Z=90–93) show high oxidation states (+3 to +6 and beyond): Th(+4), Pa(+5), U(+4, +5, +6 in UO₂²⁺ — uranyl ion), Np(+3 to +7). Heavier actinoids (from Am onwards) increasingly favour +3, more like lanthanoids.
Key differences between lanthanoids and actinoids: All actinoids are radioactive (unstable nuclei), while lanthanoids are generally stable (except Pm, which is radioactive). Actinoids show a larger range of oxidation states than lanthanoids. Actinoids form more stable complexes than lanthanoids. Actinoids form oxo cations like UO₂²⁺ (uranyl) and NpO₂²⁺ which are not known for lanthanoids. The magnetic properties of actinoids are more complex than lanthanoids because both spin and orbital contributions are significant (not quenched as in lanthanoids). Actinoid contraction (across 5f series) is larger per element than lanthanoid contraction because 5f electrons shield less effectively than 4f electrons.
Similarities between lanthanoids and actinoids: Both series involve filling of inner f-subshells. Both exhibit +3 as a common oxidation state. Both show lanthanoid/actinoid contraction. Both are largely metallic with high melting points. Both form coloured compounds (due to 4f–4f or 5f–5f transitions, though these are Laporte-forbidden and give weak colours). All f-block elements are placed in two separate rows at the bottom of the periodic table. Download the Free PDF for the complete lanthanoid vs. actinoid comparison table.
Download Free PDF — d & f Block Elements Formula Sheet
All electronic configuration tables with exceptions, magnetic moment values, coloured ion identification, KMnO₄ reactions in all three media with balanced equations, K₂Cr₂O₇ reactions, chromate–dichromate equilibrium, lanthanoid contraction consequences, and lanthanoid vs. actinoid comparison tables are compiled in the Aakash Rapid Revision & Formula Bank PDF — structured specifically for JEE Main, CBSE boards, and NEET.
Why This Chapter Is Important for Students and Exams
Four clear reasons make d & f block chemistry one of the most reward-per-revision-hour chapters for JEE Main.
Electronic configuration exceptions are guaranteed marks. Cr and Cu exceptions appear in virtually every JEE Main paper in some form — either as a direct configuration question or as the basis for an oxidation state or magnetic property question. Memorising [Ar]3d⁵4s¹ for Cr and [Ar]3d¹⁰4s¹ for Cu, and understanding why, converts two potentially tricky questions into certain marks.
Magnetic moment calculations are fast and formulaic. Given the formula μ = √n(n+2) BM and the ability to write an ion's configuration (removing 4s before 3d), calculating the magnetic moment of any transition metal ion takes under 30 seconds. These are among the fastest-answerable numerical questions in the entire JEE Main chemistry paper.
KMnO₄ reactions in different media are a complete, learnable set. Three products, three half-reactions, and a handful of specific reactions with H₂O₂, Fe²⁺, SO₃²⁻, and alkenes — this is a finite and well-defined topic. Students who systematically revise these equations can answer every KMnO₄ reaction question in JEE Main confidently.
Lanthanoid contraction explains a cascade of facts with one concept. Understanding why 5d metals have the same size as 4d metals (because lanthanoid contraction compensates for the expected size increase) explains the inseparability of Zr–Hf and Nb–Ta, the similar chemistry of 4d and 5d transition metals, and several periodic anomalies across the sixth period. This is the kind of conceptual investment that pays returns across multiple question types. Download the Free PDF for all of these revision essentials in one place.
Who Should Use This Formula Sheet?
Learning Outcomes After Completing This Chapter
After working through this chapter using the formula sheet and notes above, a student should be able to accomplish the following confidently.
For electronic configuration: write the correct configuration for all 10 first series transition metals including Cr and Cu exceptions. Write the configuration of their ions (removing 4s first, then 3d). Identify whether Zn is a true transition metal and explain why not.
For properties: explain why transition metals show variable oxidation states. State the trend in maximum oxidation state across the first series. Explain why Mn²⁺ is more stable than Mn³⁺ using half-filled 3d⁵ configuration. Calculate the magnetic moment of any transition metal ion using the spin-only formula. Identify which ions are coloured and which are colourless, with reasons. Explain d-d transition mechanism for colour. Explain why KMnO₄ is intensely coloured despite Mn being in d⁰ state.
For KMnO₄ and K₂Cr₂O₇: write the three-step preparation of KMnO₄ from MnO₂. Write the half-reactions and complete balanced equations for KMnO₄ in acidic, neutral, and strongly alkaline media with at least two examples each. Write the preparation of K₂Cr₂O₇ from chromite ore. Write and balance K₂Cr₂O₇ reactions with FeSO₄, KI, H₂S, and SO₂. Write the chromate–dichromate equilibrium with the effect of pH.
For f-block: explain lanthanoid contraction and its consequences. State the common oxidation states of lanthanoids and actinoids. Give examples of lanthanoids that show +2 and +4 states with electronic configuration reasoning. Write four differences and three similarities between lanthanoids and actinoids. Download the Free PDF to test all outcomes before your exam.
Get the Free PDF for Quick Revision
Whether you are preparing for JEE Main, CBSE Class 12 boards, or NEET, having a focused formula sheet for this chapter ensures that every configuration exception, every magnetic moment value, every KMnO₄ reaction, and every lanthanoid fact is accessible during revision. The Aakash Rapid Revision & Formula Bank PDF for d and f Block Elements is built precisely for that purpose.
Conclusion — Partial d-Filling Explains Everything
The entire chemistry of transition metals flows from one structural feature: partially filled d-orbitals. Variable oxidation states exist because 3d and 4s electrons are close in energy and both can participate in bonding. Paramagnetism arises because partially filled d-orbitals contain unpaired electrons. Colour arises because partially filled d-orbitals split in the presence of ligands, allowing visible-light absorption through d-d transitions. Catalytic activity arises because partial d-filling gives the metal the ability to form weak, reversible bonds with reactant molecules. Even the exceptions to regular electronic configuration (Cr, Cu) are explained by the extra stability of the half-filled or fully filled d-subshell.
For the f-block, lanthanoid contraction is the central concept — and it is best understood not as a memorised fact but as a logical outcome of poor 4f shielding. Once the mechanism is clear, all its consequences (size similarity of 4d and 5d metals, inseparability of Zr–Hf, difficulty of separating lanthanoids from each other) follow naturally. Use this page, the concept boxes, and the Free PDF Download as your structured revision foundation for this chapter in JEE Main.
Frequently Asked Questions
Why do chromium and copper have anomalous electronic configurations?
The expected configurations based on the Aufbau principle would be Cr: [Ar]3d⁴4s² and Cu: [Ar]3d⁹4s². However, the actual configurations are Cr: [Ar]3d⁵4s¹ and Cu: [Ar]3d¹⁰4s¹. The reason is that completely half-filled (3d⁵) and completely filled (3d¹⁰) d-subshells have extra thermodynamic stability — this stability arises from two sources: symmetrical electron distribution (which reduces electron-electron repulsion) and maximum exchange energy (the quantum mechanical energy lowering when electrons of the same spin can exchange positions). For chromium, promoting one electron from 4s to 3d achieves the half-filled 3d⁵ configuration, and the resulting energy lowering exceeds the cost of the promotion. For copper, one electron moves to 3d to achieve the completely filled 3d¹⁰, again with a net energy benefit. Similar anomalies occur in the second series for Mo and Ag for the same reasons.
How do you calculate the magnetic moment of a transition metal ion?
To calculate the magnetic moment of any transition metal ion, follow these three steps. First, write the electronic configuration of the neutral metal atom. Second, remove electrons from the 4s subshell first (before 3d) to get the ion's configuration — this is important because during ionisation, 4s electrons are lost before 3d electrons even though 4s fills first. Third, count the number of unpaired electrons (n) in the 3d subshell by applying Hund's rule to the d electrons. Then apply the spin-only formula: μ = √[n(n+2)] Bohr Magnetons. Example for Fe³⁺: Fe = [Ar]3d⁶4s² → remove 3 electrons (2 from 4s, 1 from 3d) → Fe³⁺ = [Ar]3d⁵ → Hund's rule: ↑ ↑ ↑ ↑ ↑ in five separate d-orbitals → n = 5 → μ = √[5×7] = √35 ≈ 5.92 BM. Example for Cu²⁺: Cu = [Ar]3d¹⁰4s¹ → remove 2 electrons (1 from 4s, 1 from 3d) → Cu²⁺ = [Ar]3d⁹ → one unpaired electron → n = 1 → μ = √3 ≈ 1.73 BM.
Why does the colour of KMnO₄ and K₂Cr₂O₇ not arise from d-d transitions?
The deep purple colour of KMnO₄ and the orange colour of K₂Cr₂O₇ are both due to charge transfer transitions, not d-d transitions. In KMnO₄, manganese is in the +7 oxidation state (MnO₄⁻ ion). Mn⁷⁺ has the configuration [Ar]3d⁰ — there are no d-electrons at all. Since d-d transitions require at least one electron in the d-subshell and at least one empty d-orbital for it to transition into, a d⁰ configuration makes d-d transitions impossible. Similarly, in K₂Cr₂O₇, chromium is in the +6 state (d⁰). Instead, the intense colour arises from charge transfer — an electron is transferred from the O²⁻ ligand (which is a good π-donor) to the high-oxidation-state metal. This electron transfer occurs at visible wavelengths and produces intense colours because the transitions are fully allowed (unlike most d-d transitions which are partially Laporte-forbidden). The charge transfer transition absorbs specific visible wavelengths and the transmitted/reflected light appears as the complementary colour (purple for KMnO₄, orange for K₂Cr₂O₇).
What are the products of KMnO₄ in acidic, neutral, and strongly alkaline media?
The reduction product of permanganate ion (MnO₄⁻) depends on the medium. In acidic medium (H₂SO₄ or HCl added), MnO₄⁻ is reduced to Mn²⁺ (manganous ion, colourless or very pale pink). The half-reaction is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. This involves a change of +5 in oxidation number per Mn (+7 → +2) and is the most powerful oxidation — KMnO₄ is a 5-electron oxidant in acid. The solution changes from intense purple to colourless, making KMnO₄ its own indicator in acid permanganometry. In neutral or very weakly alkaline medium, MnO₄⁻ is reduced to MnO₂ (manganese dioxide, a brown/black precipitate). Half-reaction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. This is a 3-electron oxidant. In strongly alkaline medium (with excess NaOH or KOH), MnO₄⁻ is reduced only to MnO₄²⁻ (manganate ion, green). Half-reaction: MnO₄⁻ + e⁻ → MnO₄²⁻. This is only a 1-electron oxidant — the weakest oxidising action of the three media.
What is the chromate–dichromate equilibrium and how does pH affect it?
Chromate (CrO₄²⁻) and dichromate (Cr₂O₇²⁻) are interconvertible depending on the pH of the solution. The equilibrium is: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. In acidic conditions (adding H⁺, lowering pH), the equilibrium shifts to the right — chromate (yellow) is converted to dichromate (orange). In alkaline conditions (adding OH⁻, increasing pH), the equilibrium shifts to the left — dichromate (orange) is converted back to chromate (yellow). Both chromium species have Cr in the +6 oxidation state — this is NOT a redox reaction, only a condensation–hydrolysis equilibrium. Practical consequence: if you need to use K₂Cr₂O₇ as an oxidising agent, you acidify the solution — this ensures maximum dichromate concentration and best oxidising action. If you add base to a dichromate solution, it turns yellow. This colour change (orange ↔ yellow with pH) is a direct one-mark JEE Main observation question.
What is lanthanoid contraction and what are its three major consequences?
Lanthanoid contraction refers to the progressive and steady decrease in atomic and ionic radii across the lanthanoid series from La (Z=57) to Lu (Z=71). As each successive lanthanoid is reached, one proton is added to the nucleus and one electron is added to the inner 4f subshell. The 4f electrons are very poor at shielding the outer 5s, 5p, 5d, and 6s electrons from the nuclear charge — f-electrons have complex angular distributions with many nodes, and they are deeply penetrating yet also highly diffuse, resulting in poor mutual shielding. The net effective nuclear charge experienced by the outer electrons therefore increases with each element, drawing the electron cloud inward and decreasing the radius. The three major consequences: (1) The atomic radii of 5d transition metals (Hf to Pt) are almost the same as those of the corresponding 4d metals (Zr to Pd) immediately above them — because the expected increase in going one period down is exactly offset by the lanthanoid contraction. (2) Pairs like Zr–Hf and Nb–Ta have nearly identical chemical properties and sizes, making them among the hardest pairs to separate in nature — they always occur together in ores. (3) The lanthanoid elements themselves are very difficult to separate from each other (done by ion-exchange chromatography) because their very similar sizes give them nearly identical chemistry.
Why is Zn not considered a true transition metal?
The strict definition of a transition metal is an element that, in its ground state or in any of its commonly occurring oxidation states, has a partially filled d-subshell. Zinc (Zn, atomic number 30) has the ground state configuration [Ar]3d¹⁰4s² — a completely filled 3d subshell. When zinc forms its only common ion Zn²⁺ (by losing both 4s electrons), the configuration becomes [Ar]3d¹⁰ — still a completely filled d-subshell. At no common oxidation state does zinc have a partially filled d-subshell. As a result, Zn²⁺ compounds are colourless (no d-d transitions possible), Zn²⁺ is diamagnetic (no unpaired electrons), and zinc does not show variable oxidation states (it is always +2). Despite being positioned in the d-block of the periodic table and studied alongside transition metals, zinc technically does not meet the definition of a transition element. Similarly, Cd and Hg are not true transition metals. Sc could also be questioned (Sc³⁺ is d⁰) but it is a transition metal by ground state configuration.
Why do actinoids show a wider range of oxidation states than lanthanoids?
In lanthanoids, the 4f, 5d, and 6s subshells are well separated in energy — the 4f orbitals are much more deeply buried inside the atom and have significantly lower energy than the 5d and 6s. This means only certain combinations of electrons can participate in bonding, and +3 (from 6s² and one 4f or 5d) is strongly favoured. The 4f electrons in most lanthanoids are largely non-bonding (except for a few cases like Ce⁴⁺ where loss of the 4f electron gives the stable empty-4f configuration). In actinoids, the 5f, 6d, and 7s orbitals are much closer in energy to each other — the 5f orbitals extend further from the nucleus relative to the atomic size and interact more with the outer orbitals. This means that 5f electrons can participate more readily in chemical bonding, and a wider range of oxidation states becomes accessible. Early actinoids (Th to Np) show states from +3 up to +7. As the series progresses, the 5f orbitals become more core-like and the chemistry starts resembling lanthanoids with +3 being more prevalent (from Am onwards).
How is potassium permanganate prepared from manganese dioxide?
Potassium permanganate (KMnO₄) is prepared from MnO₂ in two steps. In the first step, MnO₂ is fused with potassium hydroxide (KOH) in the presence of an oxidising agent — either potassium nitrate (KNO₃) or by passing air through the molten mixture. The reaction is: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O. This produces potassium manganate (K₂MnO₄, dark green solution), where Mn is in the +6 oxidation state. In the second step, K₂MnO₄ must be oxidised to KMnO₄ (Mn: +6 → +7). This can be done by: (a) chlorine gas oxidation: 2K₂MnO₄ + Cl₂ → 2KMnO₄ + 2KCl; or (b) electrolytic oxidation at the anode: MnO₄²⁻ → MnO₄⁻ + e⁻. Alternatively, in slightly acidic solution, K₂MnO₄ disproportionates: 3K₂MnO₄ + 2H₂SO₄ → 2KMnO₄ + MnO₂ + 2K₂SO₄ + 2H₂O (two-thirds of Mn goes to +7 as permanganate, one-third to +4 as MnO₂). The purple KMnO₄ crystals are obtained by evaporating the solution.
Why are transition metals good catalysts?
Transition metals are excellent catalysts for two complementary reasons related to their electronic structure. For heterogeneous catalysis (where the catalyst is in a different phase from the reactants, typically a solid metal with gaseous or liquid reactants), transition metals have partially filled d-orbitals that allow them to form weak, temporary bonds with reactant molecules (chemisorption). The reactant molecules adsorb on the metal surface, becoming activated — their bonds are weakened as electron density is redistributed between the molecule and the metal d-orbitals. The activated molecules are also brought into proximity and correct orientation on the surface, increasing the probability of reaction. The products then desorb, freeing the active site. For homogeneous catalysis (where the catalyst is in the same phase as reactants), the variable oxidation states of transition metals allow them to participate in electron-transfer reactions with reactants, forming reactive intermediates at different oxidation states and providing a lower-energy pathway to the same products. Classic examples: Fe (Haber process, heterogeneous), V₂O₅ (Contact process, heterogeneous), Pt/Rh (Ostwald process, heterogeneous), Ni (hydrogenation, heterogeneous).
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