d- and f-Block Elements JEE Notes PDF: Download Now

d- and f-Block Elements JEE Notes: Important Concepts

NCERT coverage: Class XII Chemistry, Unit 8.
Official JEE syllabus tags: General characteristics of transition elements (d-block), lanthanides and actinides (f-block), oxidation states, magnetic properties, coordination tendencies, alloy formation and interstitial compounds.

• Location in periodic table: d-block = Groups 3-12 (periods 4-7). f-block = inner transition series placed separately below the main table.
• Transition element definition: An element whose atoms or ions have an incompletely filled d-subshell in at least one common oxidation state.
• Inner transition definition: Incompletely filled f-subshell in the ground or common oxidation state.
• Standard abbreviations: $$\mu_{\text{eff}}$$ for spin-only magnetic moment, $$\Delta_{o}$$ for octahedral CFSE, $$\lambda_{\max}$$ for absorption peak in coloured ions.

Key periodic trends across the d-block rows

Atomic and ionic radii: Large contraction from Sc to Cu in period 4 due to ineffective shielding of 3d electrons. A second contraction, lanthanide contraction, compresses the 5d series.

Ionisation enthalpy: Increases irregularly. Notable dips at Cr and Cu because of added stability of half-filled and completely filled 3d subshells.

Electronegativity: Low-to-moderate (1.2–1.9 Pauling) yet higher than s-block, enabling covalent bonding in higher oxidation states.

Density & melting point: Very high owing to strong metallic bonding and close packing of small atoms.

General characteristics of f-block series

• Two rows: 4f = lanthanides (La 57 to Lu 71) and 5f = actinides (Ac 89 to Lr 103).
• Oxidation state: +3 dominates for lanthanides while actinides show +3 to +6 (Uranium even +7 in $$\mathrm{UO_{4}^{-}}$$).
• Lanthanide contraction: Steady, 0.2 pm electron-shell shrink per element. Responsible for similarity in radii of Zr/Hf, Nb/Ta etc.
• Almost all actinides are radioactive; first four (Th, Pa, U, Np) occur naturally, rest are synthetic.

Electronic Configuration, Trends and Reactivity of Transition Metals

Ground-state configurations

Write $$\mathrm{(n-1)d^{1-10}\,ns^{0-2}}$$ for periods 4-6. Two famous exceptions maximise exchange energy:

Variable oxidation states

• Lower states use 4s electrons, higher states draw from 3d.
• Maximum shown by Mn (+7 in $$\mathrm{MnO_{4}^{-}}$$) in first transition series.
• E° values: $$\mathrm{Cr^{3+}/Cr} = -0.74\,V$$, $$\mathrm{Cu^{2+}/Cu} = +0.34\,V$$ explain ease of redox conversions tested repeatedly in Mains.

Reactivity pointers

Complex formation: Small, highly charged $$\mathrm{M^{n+}}$$ ions with vacant d orbitals accept lone pairs from ligands, leading to high $$K_{\text{f}}$$. VBT and CFT questions often ask geometries of $$\mathrm{[Fe(CN)_{6}]^{4-}}$$, $$\mathrm{[Cu(NH_{3})_{4}]^{2+}}$$ etc.

Exam tip: After revising configs, solve previous Sc–Cu redox questions from the JEE Mains Previous Papers for accurate trend recall.

Solved Example 1 (Spin-only magnetic moment)

Calculate $$\mu_{\text{eff}}$$ for $$\mathrm{Fe^{2+}}$$ in high-spin $$\mathrm{FeSO_{4}}$$.

1. $$\mathrm{Fe^{2+}}: [Ar]\,3d^{6}$$ → high spin → 4 unpaired electrons.
2. Formula: $$\mu_{\text{eff}} = \sqrt{n(n+2)}\,\text{B}_{\mathrm{M}}$$ where $$n$$ is unpaired count.
3. Insert $$n=4$$: $$\mu_{\text{eff}} = \sqrt{4\times6} = \sqrt{24}=4.90\,\text{B}_{\mathrm{M}}$$.

Answer: $$4.90\,\text{B}_{\mathrm{M}}$$.

Solved Example 2 (Balancing a dichromate redox in acidic medium)

Balance $$\mathrm{Fe^{2+}+\,Cr_{2}O_{7}^{2-}\rightarrow Fe^{3+}+\,Cr^{3+}}$$ in $$\mathrm{H^{+}}$$ and find electrons transferred.

1. Oxidation: $$\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^{-}}$$ (1 e⁻).
2. Reduction: $$\mathrm{Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O}$$
3. LCM = 6: multiply Fe-reaction by 6. Net electrons 6.

Balanced ionic: $$6Fe^{2+}+\,Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 6Fe^{3+}+\,2Cr^{3+}+7H_{2}O$$. Electrons transferred: 6.

Properties of d-Block Elements

Colour of ions

Magnetic behaviour

• Spin-only formula: $$\mu_{\text{eff}}=\sqrt{n(n+2)}$$ (in Bohr magneton).
• Orbital contribution is quenched in octahedral fields, so the formula works for most first-row ions asked in exam.
• Diamagnetism is rare; $$\mathrm{Zn^{2+}}$$, $$\mathrm{Cd^{2+}}$$ and $$\mathrm{Hg^{2+}}$$ are classic cases.

Catalytic activity

d-orbitals can supply or accept electrons to weaken reactant bonds and form intermediates. Popular JEE examples: $$\mathrm{V_{2}O_{5}}$$ in contact process, finely divided $$\mathrm{Ni}$$ in hydrogenation, $$\mathrm{Fe}$$ in Haber process.

Alloy and interstitial compound formation

Close-packed lattices plus comparable radii let transition metals mix: brass (Cu + Zn), stainless steel (Fe + Cr + Ni). Carbon, B, N enter holes to form very hard interstitials such as $$\mathrm{Fe_{3}C}$$ (cementite).

For concise values such as hardness ranking and density, download our JEE Formula Sheets after finishing this section.

Solved Example 3 (CFSE comparison)

Which ion has greater octahedral CFSE: $$\mathrm{Co^{3+}\,(d^{6})}$$ low-spin or $$\mathrm{Ni^{2+}\,(d^{8})}$$ high-spin?

• Low-spin $$d^{6}: t_{2g}^{6}e_{g}^{0}\Rightarrow$$ CFSE = $$-2.4\Delta_{o}$$
• High-spin $$d^{8}: t_{2g}^{6}e_{g}^{2}\Rightarrow$$ CFSE = $$-1.2\Delta_{o}$$

Answer: $$\mathrm{Co^{3+}}$$ (low-spin) with larger magnitude CFSE.

f-Block Elements: Lanthanides and Actinides

Lanthanide series (4f)

1. Common oxidation state +3; +2 and +4 possible when they lead to empty, half-filled or full 4f subshells (Eu²⁺, Yb²⁺, Ce⁴⁺, Tb⁴⁺).
2. Magnetic moments are not spin-only; orbital contribution significant, so JEE rarely asks numerical here.
3. Lanthanides form coloured ions but colours are f-f transitions: lap-forbidden hence pale compared to d-ions.
4. Separation technique: Ion-exchange chromatography and solvent extraction exploit fractional differences in $$K_{\text{f}}$$ values.

Actinide series (5f)

• Show wider oxidation range: +3 to +6, sometimes +7 (e.g. $$\mathrm{Np^{7+}}$$ in $$\mathrm{NpO_{5}}$$).
• 5f orbitals extend outside; actinide contraction is less smooth, leading to greater variability in chemistry.
• All are radioactive; questions may ask half-life ordering of Th → U → Pu.

Complex multi-step oxidation-reduction balancing of actinides frequently finds space in Advanced papers, so after reading this block practise from the JEE Advanced Previous Papers set for integration with electrochemistry and nuclear equations.

Solved Example 4 (Lanthanide contraction consequence)

Explain why zirconium (Zr) and hafnium (Hf) have nearly identical radii though they belong to different periods.

Solution: The 14-element lanthanide contraction shrinks the entire 5d row. As 4f electrons poorly shield, 6s and 5d are pulled in, making Hf’s radius (159 pm) almost equal to Zr (160 pm) despite the extra shell.

Important Formulas and Results at a Glance

Magnetic moment and CFSE are the two most numerically tested formulas from this chapter. Memorise $$\mu_{\text{eff}}$$ and the table above.

JEE Important Points, Common Mistakes and Quick Revision

• Always check spin state before applying magnetic moment; students often plug n from ground-state instead of complex-state configuration.
• Remember colourless ions: $$\mathrm{Ti^{4+}, Zn^{2+}, Sc^{3+}}$$ lack d-electrons, hence no d-d transition.
• Disproportionation traps: $$\mathrm{Cu^{+}}$$ in aqueous acid gives $$\mathrm{Cu^{2+}+Cu}$$; check stability of +1 oxidation state.
• Lanthanide contraction effects: density and ionic radii parity used for tricky matching questions.
• For rapid mixed-concept drills, attempt topic-wise JEE Questions immediately after closing these notes.

30-second flashcards

Swipe through this checklist mentally:

1. Write $$\mu_{\text{eff}}$$ formula.
2. State the cause of colour in $$\mathrm{KMnO_{4}}$$ (charge-transfer, not d-d).
3. Recall which element shows +8 state (Os, Ru).
4. Name the hardest known metal carbide (WC).
5. Quote E° of $$\mathrm{Cr^{3+}/Cr}$$.

Quick solve triggers