The d- and f-Block ElementsNEET Chemistry · Class 12 · NCERT Chapter 4

Introduction: What Are d- and f-Block Elements?

The d-block elements (also called transition metals) are those in which the last electron enters the d-subshell of the penultimate energy level. They occupy Groups 3 to 12 of the periodic table. The 3d series runs from Sc (Z=21) to Zn (Z=30), the 4d series from Y (Z=39) to Cd (Z=48), and the 5d series from La/Hf (Z=57/72) to Hg (Z=80).

The f-block elements are those in which the last electron enters the f-subshell. They are divided into:

• Lanthanoids: La (Z=57) to Lu (Z=71) — filling the 4f-subshell
• Actinoids: Ac (Z=89) to Lr (Z=103) — filling the 5f-subshell

The term transition metals strictly refers to elements whose atoms or ions have an incomplete d-subshell. By this definition, Zn (3d10 4s2) and Sc (only +3 ion, 3d0) are borderline — Zn2+ has a complete 3d10 shell, and Sc3+ has no d-electrons. For NEET, treat Sc through Zn as the 3d series.

Electronic Configuration and Properties of 3d Metals

The general configuration of 3d elements is $[\text{Ar}]3d^{1-10}4s^{1-2}$. Two exceptions break the expected pattern because half-filled and fully filled d-subshells are extra stable:

• Cr (Z=24): $[\text{Ar}]3d^{5}4s^1$ (not $3d^{4}4s^2$) — half-filled 3d5 is more stable
• Cu (Z=29): $[\text{Ar}]3d^{10}4s^1$ (not $3d^{9}4s^2$) — fully filled 3d10 is more stable

Full Table: 3d Series Electronic Configurations

Electronic Configuration of Ions

When a transition metal loses electrons to form an ion, the 4s electrons are lost before the 3d electrons. This is because 4s is higher in energy for the cation. So Fe (3d6 4s2) loses 4s2 to give Fe2+ (3d6), then loses one more 3d to give Fe3+ (3d5).

Key ion configurations to memorise:

• Fe2+: [Ar] 3d6 (4 unpaired electrons)
• Fe3+: [Ar] 3d5 (5 unpaired electrons)
• Cu2+: [Ar] 3d9 (1 unpaired electron)
• Mn2+: [Ar] 3d5 (5 unpaired electrons)
• Cr3+: [Ar] 3d3 (3 unpaired electrons)

Characteristic Properties of Transition Metals

1. Variable Oxidation States

Transition metals show multiple oxidation states because both 4s and 3d electrons have similar energies and can participate in bonding. The number of oxidation states increases to a maximum near the middle of the series (Mn shows the most: +2 to +7) then decreases.

At the ends: Sc shows only +3; Zn shows only +2. This is because Sc has only one 3d electron (so +2 would leave 3d1, which it loses completely for +3), and Zn has 3d10 (fully stable, not used in bonding).

2. Colour of Transition Metal Ions

Transition metal ions in solution are often coloured because of d-d electronic transitions. When white light falls on an ion with a partially filled d-subshell, an electron absorbs a specific wavelength and jumps from a lower d-orbital to a higher d-orbital. The colour you see is the complementary colour to what is absorbed.

Ions with 3d0 (empty) or 3d10 (full) d-subshells are colourless because no d-d transition is possible. MnO4- is intensely purple because of charge transfer (O to Mn), not a d-d transition.

3. Magnetic Properties

Substances with unpaired electrons are paramagnetic (attracted by a magnetic field). Substances with no unpaired electrons are diamagnetic (weakly repelled). The spin-only magnetic moment formula for NEET is:

where n is the number of unpaired electrons.

Example: Fe3+ has 3d5 with 5 unpaired electrons: $\mu =\sqrt{5\times 7}=\sqrt{35}\approx 5.92$ BM.

4. Catalytic Properties

Transition metals are excellent catalysts because:

• Variable oxidation states: they can form unstable intermediate compounds that break down to give the product, providing an alternative lower-energy pathway.
• Surface adsorption: as metals or metal oxides, they adsorb reactant molecules on their surface. Bonds in reactant molecules weaken, lowering activation energy.

Key examples: Fe in the Haber process (N2 + 3H2 → 2NH3), V2O5 in the Contact process (2SO2 + O2 → 2SO3), Pt/Pd in catalytic converters, Ni in hydrogenation of oils, MnO2 as the catalyst in KClO3 decomposition.

5. Interstitial Compounds

Transition metals can trap small non-metal atoms (H, C, N, B) in the holes (interstices) of their crystal lattice to form interstitial compounds. Examples: steel (Fe + C), metal hydrides (TiH2), metal nitrides (Fe2N).

These compounds are: harder and less malleable than the pure metal; have higher melting points; retain metallic conductivity; and their composition is non-stoichiometric (variable ratio).

6. Alloy Formation

Transition metals readily form alloys (solid solutions with other metals) because they have similar atomic radii and can substitute for each other in the crystal lattice. Examples: stainless steel (Fe + Cr + Ni), brass (Cu + Zn), bronze (Cu + Sn).

Important Compounds: KMnO4 and K2Cr2O7

Potassium Permanganate, KMnO4

KMnO4 is a dark purple crystalline solid. In aqueous solution it gives an intense purple colour. Mn is in the +7 oxidation state (check: K = +1, O = -2; so +1 + x + 7(-2) = 0 gives x = +7).

Preparation: MnO2 is first fused with KOH in the presence of an oxidising agent (KNO3 or air) to give K2MnO4 (green), which is then oxidised (by Cl2 or electrolysis) to KMnO4.

KMnO4 Reactions in Different Media

Key reactions of KMnO4 in acidic medium:

• With FeSO4: oxidises Fe2+ to Fe3+
  
  $${\text{MnO}}_4^{-}+5{\text{Fe}}^{2+}+8{\text{H}}^{+}\to {\text{Mn}}^{2+}+5{\text{Fe}}^{3+}+4{\text{H}}_2\text{O}$$
• With oxalic acid:
  
  $$2{\text{KMnO}}_4+5{\text{H}}_2{\text{C}}_2{\text{O}}_4+3{\text{H}}_2{\text{SO}}_4\to 2{\text{MnSO}}_4+{\text{K}}_2{\text{SO}}_4+10{\text{CO}}_2+8{\text{H}}_2\text{O}$$
• With H2O2 (acting as reducing agent here):
  
  $$2{\text{KMnO}}_4+5{\text{H}}_2{\text{O}}_2+3{\text{H}}_2{\text{SO}}_4\to 2{\text{MnSO}}_4+{\text{K}}_2{\text{SO}}_4+5{\text{O}}_2+8{\text{H}}_2\text{O}$$

Potassium Dichromate, K2Cr2O7

K2Cr2O7 is an orange crystalline solid. Cr is in the +6 oxidation state. In the Cr2O72- ion, two CrO4 tetrahedra share one oxygen atom.

Interconversion of CrO42- and Cr2O72-:

Chromate (CrO42-, yellow) in basic medium; dichromate (Cr2O72-, orange) in acidic medium. This reversible equilibrium is a classic NEET question on the effect of pH on chromium chemistry.

K2Cr2O7 as an oxidising agent in acidic medium:

Colour change: orange (Cr2O72-) to green (Cr3+). K2Cr2O7 oxidises: I- to I2, Fe2+ to Fe3+, SO32- to SO42-, and is used in volumetric analysis (dichromate titrations).

Lanthanoids and Lanthanoid Contraction

Lanthanoids are the 15 elements from La (Z=57) to Lu (Z=71). They fill the 4f-subshell. Their general configuration is $[\text{Xe}]4f^{1-14}5d^{0-1}6s^2$.

Electronic Configurations of Select Lanthanoids

Lanthanoid Contraction

As you move from La to Lu, one proton and one 4f-electron are added at each step. The 4f-orbitals are poor at shielding (they have many nodes and a complex shape, so they do not effectively screen the nucleus). The effective nuclear charge on the outer 6s electrons increases steadily. This pulls in the electron cloud, causing the atomic and ionic radii to decrease progressively from La to Lu. This effect is called lanthanoid contraction.

Consequences of lanthanoid contraction:

• 4d and 5d metals in the same group have nearly equal radii: The expected size increase from period 4 to period 5 (4d to 5d) is cancelled by lanthanoid contraction. So Zr and Hf, Nb and Ta, Mo and W are almost the same size. This makes them chemically very similar and very hard to separate industrially.
• Density and hardness of 5d metals: Because the radii do not increase as expected, 5d metals are denser than 4d metals in the same group. Osmium (Os) is one of the densest metals known.
• Basicity decreases La to Lu: As the ionic radius of Ln3+ decreases from La to Lu, the metal-oxygen bond becomes stronger. This means the hydroxides become less basic from La(OH)3 (strongly basic) to Lu(OH)3 (weakly basic).

Oxidation States of Lanthanoids

All lanthanoids show the +3 oxidation state (most stable). Exceptions:

• Ce4+: stable because Ce4+ achieves empty 4f0 shell. Ce4+ is a mild oxidising agent used in cerium titrations.
• Eu2+ and Sm2+: stable because losing one fewer electron gives 4f7 (half-filled, Eu) and 4f6 (near-half-filled, Sm).
• Yb2+: has 4f14 (fully filled), hence extra stable.

Actinoids

Actinoids are the 15 elements from Ac (Z=89) to Lr (Z=103). They fill the 5f-subshell. Their general configuration is $[\text{Rn}]5f^{1-14}6d^{0-1}7s^2$.

Key Differences: Actinoids vs Lanthanoids

Why Actinoids Show More Oxidation States

In lanthanoids, the 4f-orbitals are deeply embedded below the 5s and 5p-orbitals and have very high ionisation enthalpies. The 4f, 5d, and 6s energy levels differ enough that only 6s (and occasionally 5d) electrons participate in bonding. In actinoids, the 5f, 6d, and 7s orbitals are much closer in energy and spatially more extended. More electrons can participate in bonding, so actinoids like U (can be +3 to +6) and Np (can reach +7) show many oxidation states.

Worked NEET Problems

Summary Cheat Sheet

d-Block Quick Reference

f-Block Quick Reference