d- and f-Block Elements

Inorganic Chemistry Weightage: 7 Marks CBSE Unit 4

“Hi there! The d- and f-Block Elements chapter is one of the most predictable in CBSE Class 12. It carries 7 marks (about 10% of the theory paper) and the board loves three things from it: give-reasons questions (Cr and Cu configuration, Cu+ instability, Zr and Hf), the spin-only magnetic moment, and balanced equations for KMnO4 and K2Cr2O7. Nothing here needs heavy calculation. If you understand the stability of half-filled d5 and full d10 shells, the pattern behind most answers becomes obvious. The red trap boxes are the exact places where students lose marks.”

— SCORECHEM ACADEMIC TEAM

1. Transition Elements & Their Configurations

The d-block is groups 3-12 (four series: 3d, 4d, 5d, 6d). The f-block (lanthanoids 4f, actinoids 5f) sits at the bottom.

IUPAC definition: a transition element has an incomplete d subshell either in the neutral atom or in its common ions.

Sc (3d¹4s²) qualifies. Cu (Cu²⁺ = 3d⁹) and Ag (Ag²⁺ = 4d⁹) qualify.
Zn, Cd, Hg (d¹⁰s²) have full d orbitals in both the atom and the common +2 ion, so they are not transition metals (they are studied along with them as end members).

General configuration: $(n-1)d^{1-10}\,ns^{1-2}$ .

Two exceptions in the 3d series (learn these):

Element	Expected	Actual	Reason
Cr (24)	$3d^4\,4s^2$	$3d^5\,4s^1$	half-filled d⁵ extra stable
Cu (29)	$3d^9\,4s^2$	$3d^{10}\,4s^1$	fully filled d¹⁰ extra stable

The 4s and 3d levels are very close in energy, so an electron easily moves to reach d⁵ or d¹⁰. When ions form, 4s electrons leave first: $\text{Fe}^{2+}=3d^6$ , $\text{Cr}^{3+}=3d^3$ , $\text{Cu}^{2+}=3d^9$ .

⚠️ Board Exam Trap: Removing electrons from d before s. For M²⁺ ions, remove the two 4s electrons first. Mn²⁺ is [Ar]3d⁵ (not 3d³4s²). Getting this wrong ruins every magnetic-moment and colour answer.

2. Trends in the 3d Series

Property	Trend	Why (exam answer)
Melting point / hardness	Rises to a maximum near the middle (V, Cr, Mo, W), then falls. Mn is low, Zn low	more unpaired (n-1)d electrons take part in metallic bonding; d⁵ is favourable
Enthalpy of atomisation	High; Zn lowest (126 kJ mol⁻¹); 4d, 5d higher than 3d	Zn has no unpaired d electrons (3d¹⁰), so weak metallic bonding
Atomic/ionic radius	Decreases slightly Sc → Cr, nearly constant Mn–Ni, rises slightly at Cu, Zn	d electrons shield poorly; at the end, repulsion between d¹⁰ electrons
Ionisation enthalpy	Rises slowly and irregularly	the added electron goes into inner (n-1)d, which shields the 4s

Ionisation enthalpy exceptions:

Third IE is highest for Mn among Cr, Mn, Fe, V: removing an electron from stable half-filled Mn²⁺ (d⁵) is hardest. Fe²⁺ (d⁶ → d⁵) is easier (2024 board MCQ).
Second IE of Cr and Cu is unusually high because M⁺ is d⁵ or d¹⁰.
First IE of Cr is lower than Zn's: Cr loses one electron to reach a stable d⁵ (3d⁵4s⁰), whereas Zn has a fully filled 3d¹⁰4s² (practice-set question).

⚠️ Board Exam Trap: Atomisation enthalpy of Zn. The reason is not "filled orbitals are stable". Say: in zinc no 3d electrons take part in metallic bonding (all paired), so metallic bonds are weak and Δ_aH is lowest (2023 board).

3. Oxidation States & Their Stability

Oxidation states of the 3d series. The number of states rises to a maximum at Mn (mid-series, most unpaired electrons) then falls. States differ by one unit (V^II, V^III, V^IV, V^V), unlike p-block elements where they differ by two.

Variable oxidation states because (n-1)d and ns electrons have nearly the same energy; states differ by one unit (unlike p-block's two).
Mn shows the most states (+2 to +7) because it has the most unpaired electrons (3d⁵4s²). Sc (+3 only) has too few electrons; Zn (+2 only) has d¹⁰ so no d electrons are used.
Highest state = sum of s and d electrons up to Mn (Ti⁴⁺, VO₂⁺, CrO₄²⁻, MnO₄⁻).
+2 state becomes more stable left to right (IE₃ rises); +3 becomes less stable after Mn.
Down a group, higher states get more stable (Mo(VI), W(VI) > Cr(VI)). This is the opposite of the p-block inert-pair effect. So Cr₂O₇²⁻ (Cr(VI)) is a strong oxidant but MoO₃ and WO₃ are not.
Highest states appear in oxides and fluorides (O, F small and highly electronegative; O forms multiple bonds). Hence MnF₄ but Mn₂O₇; oxoanions MnO₄⁻, CrO₄²⁻, FeO₄²⁻.
Low (zero) oxidation states occur with π-acceptor ligands: Ni(CO)₄, Fe(CO)₅.
As oxidation number rises, the oxide changes basic → amphoteric → acidic: MnO (basic), Mn₃O₄ (amphoteric), Mn₂O₇ (acidic, covalent green oil).

Disproportionation: a state that is unstable relative to one higher and one lower state.

$2\text{Cu}^+\rightarrow\text{Cu}^{2+}+\text{Cu}$ (aqueous)
$3\text{MnO}_4^{2-}+4\text{H}^+\rightarrow2\text{MnO}_4^-+\text{MnO}_2+2\text{H}_2\text{O}$

Standard electrode potentials E°(M²⁺/M) for Ti to Zn. The general trend is towards less negative values (rising sum of first and second ionisation enthalpies); Mn, Ni and Zn are the irregular ones. Cu (+0.34 V) is the only positive value because its high atomisation and ionisation enthalpies are not paid back by hydration enthalpy.

Making sense of E° values:

Cu is the only positive E°(M²⁺/M) (+0.34 V): its high atomisation + ionisation enthalpy is not compensated by hydration enthalpy. So Cu does not liberate H₂ from acids.
Cu⁺ is unstable in water: Cu²⁺ has a much more negative hydration enthalpy that more than pays for the second ionisation enthalpy (2023 board).
Cr²⁺ is reducing (d⁴ → d³, half-filled t₂g); Mn³⁺ is oxidising (d⁴ → d⁵, half-filled). Same d⁴ configuration, opposite behaviour (2023 board 5-marker).
Mn²⁺ is more stable to oxidation than Fe²⁺: Mn²⁺ is d⁵, whereas Fe²⁺ (d⁶) loses one electron to become d⁵ Fe³⁺.
Co(II) is stable in water, but easily oxidised to Co(III) when strong-field ligands (NH₃, CN⁻) are present.

⚠️ Board Exam Trap: Cr²⁺ vs Fe²⁺. Cr²⁺ is the stronger reducing agent: its change d⁴ → d³ gives the stable half-filled t_2g³ level, and in water d³ is more stable than d⁵. Fe²⁺ (d⁶ → d⁵) is a weaker reductant (NCERT 4.7). Do not answer "half-filled" alone.

4. Magnetism, Colour, Catalysis, Interstitial Compounds & Alloys

Magnetic moment (spin-only):

\mu=\sqrt{n(n+2)}\ \text{BM}

Unpaired e⁻ (n)	1	2	3	4	5
$\mu$ (BM)	1.73	2.84	3.87	4.90	5.92
Example ion	Ti³⁺, Cu²⁺	Ni²⁺, V³⁺	Cr³⁺, V²⁺, Co²⁺	Fe²⁺, Cr²⁺	Mn²⁺, Fe³⁺

Worked (2025 board): Cr³⁺ = [Ar]3d³, n = 3, $\mu=\sqrt{15}=3.87$ BM. Paramagnetism comes from unpaired electrons; ions with d⁰ or d¹⁰ are diamagnetic.

Colour: caused by d-d transitions (an electron jumps between split d levels, absorbing visible light; we see the complementary colour). No unpaired/partly filled d = colourless: Sc³⁺, Ti⁴⁺ (d⁰), Zn²⁺, Cu⁺ (d¹⁰). Coloured: Ti³⁺ purple, V³⁺ green, Cr³⁺ violet, Mn²⁺ pale pink, Fe²⁺ green, Fe³⁺ yellow, Co²⁺ pink, Ni²⁺ green, Cu²⁺ blue.

Catalysis: (i) variable oxidation states give easy reaction intermediates; (ii) vacant d orbitals/large surface adsorb reactants and weaken bonds, lowering E_a.

Catalyst	Process
V₂O₅	Contact process (SO₂ → SO₃)
Fe (finely divided)	Haber process (NH₃)
Ni	Hydrogenation of oils
PdCl₂	Wacker process (ethyne → ethanal)
TiCl₄ + Al(CH₃)₃	Ziegler-Natta polymerisation
Fe³⁺	$2\text{I}^-+\text{S}_2\text{O}_8^{2-}\rightarrow\text{I}_2+2\text{SO}_4^{2-}$

Complex formation: small ions, high charge, vacant d orbitals (e.g. [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺).

Interstitial compounds: small atoms (H, C, N) trapped in the metal lattice (TiC, Mn₄N, Fe₃H). Non-stoichiometric; hard, high melting, conduct electricity, chemically inert.

Alloys: transition metals have similar radii (within ~15%), so they mix easily. Steels (Cr, V, W, Mo, Mn), brass (Cu-Zn), bronze (Cu-Sn). Misch metal (about 95% lanthanoids + 5% Fe) is used for lighter flints and Mg alloys.

⚠️ Board Exam Trap: "Coloured because of unpaired electrons" is incomplete. Write "because of d-d transitions in partly filled d orbitals". Zn²⁺ is colourless because 3d¹⁰ has no vacant d orbital to jump into (2025 board).

5. Potassium Dichromate and Potassium Permanganate

K₂Cr₂O₇ from chromite ore

Fuse chromite with Na₂CO₃ in excess air: $4\text{FeCr}_2\text{O}_4+8\text{Na}_2\text{CO}_3+7\text{O}_2\rightarrow8\text{Na}_2\text{CrO}_4+2\text{Fe}_2\text{O}_3+8\text{CO}_2$ (yellow sodium chromate).
Acidify: $2\text{Na}_2\text{CrO}_4+2\text{H}^+\rightarrow\text{Na}_2\text{Cr}_2\text{O}_7+2\text{Na}^++\text{H}_2\text{O}$ (orange).
Add KCl: $\text{Na}_2\text{Cr}_2\text{O}_7+2\text{KCl}\rightarrow\text{K}_2\text{Cr}_2\text{O}_7+2\text{NaCl}$ (K salt is less soluble and crystallises).

Chromate and dichromate are interconverted by changing pH, not by redox. Acid gives orange dichromate; alkali gives yellow chromate.

Oxidising action (acidic medium): $\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6e^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$ ( $E^\circ=1.33$ V). Orange changes to green Cr³⁺.

Reductant	Balanced ionic equation
Iodide	$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{I}^-\rightarrow2\text{Cr}^{3+}+3\text{I}_2+7\text{H}_2\text{O}$
Iron(II)	$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{Fe}^{2+}\rightarrow2\text{Cr}^{3+}+6\text{Fe}^{3+}+7\text{H}_2\text{O}$
H₂S	$\text{Cr}_2\text{O}_7^{2-}+8\text{H}^++3\text{H}_2\text{S}\rightarrow2\text{Cr}^{3+}+3\text{S}+7\text{H}_2\text{O}$
Tin(II)	$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++3\text{Sn}^{2+}\rightarrow2\text{Cr}^{3+}+3\text{Sn}^{4+}+7\text{H}_2\text{O}$

Uses: primary standard in volumetric analysis, leather tanning, azo dyes.

KMnO₄ from pyrolusite

Fuse MnO₂ with KOH in air (or KNO₃): $2\text{MnO}_2+4\text{KOH}+\text{O}_2\rightarrow2\text{K}_2\text{MnO}_4+2\text{H}_2\text{O}$ (green manganate).
Oxidise manganate: electrolytic oxidation in alkaline solution (industrial), or disproportionation in acid: $3\text{MnO}_4^{2-}+4\text{H}^+\rightarrow2\text{MnO}_4^-+\text{MnO}_2+2\text{H}_2\text{O}$ .
Lab: $2\text{Mn}^{2+}+5\text{S}_2\text{O}_8^{2-}+8\text{H}_2\text{O}\rightarrow2\text{MnO}_4^-+10\text{SO}_4^{2-}+16\text{H}^+$ .

Properties: dark purple crystals; on heating at 513 K, $2\text{KMnO}_4\rightarrow\text{K}_2\text{MnO}_4+\text{MnO}_2+\text{O}_2$ . Both MnO₄⁻ and MnO₄²⁻ are tetrahedral; manganate (1 unpaired e⁻) is paramagnetic, permanganate (d⁰) is diamagnetic. Its intense colour is due to charge transfer, not d-d.

KMnO₄ is made from pyrolusite in two stages. Its reduction product depends on pH: Mn²⁺ (5 e⁻) in acid, MnO₂ (3 e⁻) in neutral or weakly alkaline solution, MnO₄²⁻ (1 e⁻) in strong alkali.

Acidic-medium reactions (all give Mn²⁺):

Reductant	Balanced ionic equation
Iodide	$10\text{I}^-+2\text{MnO}_4^-+16\text{H}^+\rightarrow2\text{Mn}^{2+}+8\text{H}_2\text{O}+5\text{I}_2$
Iron(II)	$5\text{Fe}^{2+}+\text{MnO}_4^-+8\text{H}^+\rightarrow\text{Mn}^{2+}+4\text{H}_2\text{O}+5\text{Fe}^{3+}$
Oxalate	$5\text{C}_2\text{O}_4^{2-}+2\text{MnO}_4^-+16\text{H}^+\rightarrow2\text{Mn}^{2+}+8\text{H}_2\text{O}+10\text{CO}_2$
Sulphite	$5\text{SO}_3^{2-}+2\text{MnO}_4^-+6\text{H}^+\rightarrow2\text{Mn}^{2+}+3\text{H}_2\text{O}+5\text{SO}_4^{2-}$
Nitrite	$5\text{NO}_2^-+2\text{MnO}_4^-+6\text{H}^+\rightarrow2\text{Mn}^{2+}+5\text{NO}_3^-+3\text{H}_2\text{O}$
H₂S	$5\text{H}_2\text{S}+2\text{MnO}_4^-+6\text{H}^+\rightarrow2\text{Mn}^{2+}+5\text{S}+8\text{H}_2\text{O}$

Neutral/weakly alkaline: $2\text{MnO}_4^-+\text{H}_2\text{O}+\text{I}^-\rightarrow2\text{MnO}_2+2\text{OH}^-+\text{IO}_3^-$ .

⚠️ Board Exam Trap: Balance electrons first, then H⁺ and H₂O. Use the ratio of electrons: MnO₄⁻ takes 5 e⁻ (so 2 MnO₄⁻ = 10 e⁻); Fe²⁺ gives 1, C₂O₄²⁻ gives 2, SO₃²⁻ gives 2. Never use HCl to acidify KMnO₄ (it oxidises Cl⁻ to Cl₂); use dilute H₂SO₄.

6. Lanthanoids & Actinoids

Lanthanoids (Ce to Lu, 4f filling)

Configuration: [Xe]4f¹⁻¹⁴5d⁰⁻¹6s². Ln³⁺ is 4fⁿ.
Oxidation state: +3 is the most stable and common (2023 board MCQ). Extra +2/+4 arise from empty, half-filled or full 4f:
- Ce⁴⁺ (4f⁰) strong oxidant, E° Ce⁴⁺/Ce³⁺ = +1.74 V; Tb⁴⁺ (4f⁷) oxidant.
- Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴) are reductants (change to +3).

Lanthanoid contraction: steady decrease of atomic and ionic radii from La to Lu.

Lanthanoid contraction: Ln³⁺ radius falls from 106 pm (La) to 86 pm (Yb) because each added 4f electron shields the others poorly. Result: 4d and 5d elements of a group have almost equal radii (Zr 160 pm, Hf 159 pm), so they are chemically alike and very hard to separate.

Cause: poor shielding of one 4f electron by another, so the effective nuclear charge keeps rising.
Consequences: (i) 4d and 5d elements have nearly the same size (Zr 160 pm, Hf 159 pm), so Zr/Hf are hard to separate and occur together; (ii) the basic strength of Ln(OH)₃ decreases from La to Lu; (iii) the very similar chemistry of lanthanoids makes their mutual separation difficult.
Ln³⁺ ions are coloured (f-f transitions) except La³⁺ (4f⁰) and Lu³⁺ (4f¹⁴).
Uses: misch metal for lighter flints; Mg alloys; phosphors in TV screens; catalysts in petroleum cracking.

Actinoids (Th to Lr, 5f filling)

All are radioactive. 5f, 6d and 7s energies are comparable, so they show a wide range of oxidation states (+3 to +7); +3 is common (2026 board AR).
Actinoid contraction is larger than lanthanoid contraction (5f electrons shield even more poorly).

Property	Lanthanoids	Actinoids
Orbital filling	4f	5f (5f electrons take part in bonding more readily)
Oxidation states	Mainly +3 (+2, +4 occasionally)	+3 to +7, wide range
Radioactivity	Non-radioactive (except Pm)	All radioactive
Contraction	Lanthanoid contraction (regular)	Actinoid contraction (larger, irregular)
Complex formation	Less tendency	Greater tendency
Basic character of hydroxides	Higher (fall from La to Lu)	Less basic
Magnetic behaviour	Simpler, explained by 4fⁿ	More complex

⚠️ Board Exam Trap: Assertion-Reason on actinoids. "Actinoids show a wide range of oxidation states" and "actinoids are radioactive" are both true, but radioactivity does not explain the variable oxidation states. The reason is the small energy gap between 5f, 6d and 7s (2026 board: option B).

7. Quick Sheet & Last-Minute Checklist

Concept	Remember
Spin-only moment	$\mu=\sqrt{n(n+2)}$ BM (n = unpaired e⁻)
Exceptions	Cr 3d⁵4s¹, Cu 3d¹⁰4s¹
Not transition	Zn, Cd, Hg
Most states	Mn (+2 to +7)
Highest IE₃	Mn (from Mn²⁺ d⁵)
Only positive E°	Cu (+0.34 V)
Disproportionation	Cu⁺, MnO₄²⁻
Reducing / oxidising (d⁴)	Cr²⁺ reducing, Mn³⁺ oxidising
Cr₂O₇²⁻ half-reaction	+14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
MnO₄⁻ half-reaction (acid)	+8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Common Ln state	+3 (Ce⁴⁺ oxidant; Eu²⁺, Yb²⁺ reductants)
Lanthanoid contraction	Zr ≈ Hf; hard separation

Before the exam, check:

Can I write 3d configurations for Cr, Cu, Cr³⁺, Mn²⁺, Cu⁺ and Cu²⁺?
Can I give reasons for: Zn not a TM, Cu positive E°, Cu⁺ unstable, Zn lowest atomisation enthalpy?
Can I calculate μ for any 3d ion and name paramagnetic or diamagnetic?
Can I write and balance six KMnO₄/K₂Cr₂O₇ equations without notes?
Can I explain lanthanoid contraction and two consequences in three lines?