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

You can expect 2 to 4 questions from this chapter every year. The most tested topics are electronic configurations (especially Cr and Cu exceptions), colour and magnetic properties of transition metal ions, KMnO4 reactions in different media, and lanthanoid contraction. Knowing the magnetic moment formula and a few key oxidation state examples will cover most of what NEET asks.

A half-filled d-subshell (3d5) is extra stable because all five d-orbitals are singly occupied. This symmetric arrangement lowers electron-electron repulsion and gives extra exchange energy. So one electron moves from 4s to 3d, giving Cr the configuration [Ar] 3d5 4s1 instead of [Ar] 3d4 4s2. Similarly, Cu is [Ar] 3d10 4s1 rather than [Ar] 3d9 4s2 because a fully filled 3d10 is also extra stable. These two exceptions (Cr and Cu) are almost always asked in NEET.

Transition metal ions have partially filled d-orbitals. When white light falls on the ion, electrons absorb specific wavelengths and jump to a higher d-orbital (this is called a d-d transition). The colour you see is the complementary colour of what was absorbed. For example, Cu2+ absorbs orange-red light and appears blue. If all d-orbitals are empty (like Sc3+ or Ti4+) or completely filled (like Cu+ or Zn2+), no d-d transition is possible and the ion is colourless. This is why Sc3+ and Zn2+ solutions are colourless.

As you move from La (Z=57) to Lu (Z=71), protons are added to the nucleus and electrons fill the 4f-subshell. The 4f-orbitals have poor shielding ability, so the effective nuclear charge felt by the outer electrons increases steadily across the lanthanoid series. This causes a gradual decrease in atomic and ionic size from La to Lu, called lanthanoid contraction. The main consequences are: (1) the atomic radii of 4d and 5d transition metals in the same group are nearly equal (e.g., Zr and Hf), making them very hard to separate; (2) the 5d metals (W, Re, Os, Ir, Pt, Au) are denser and harder than expected.

Use the spin-only formula: μ = √(n(n+2)) BM, where n is the number of unpaired electrons. Steps: (1) write the ground-state electronic configuration of the ion, (2) count unpaired electrons in the d-orbitals. For example, Fe3+ is [Ar] 3d5 (5 unpaired electrons), so μ = √(5 × 7) = √35 = 5.92 BM. Fe2+ is [Ar] 3d6 (4 unpaired), so μ = √(4 × 6) = √24 = 4.90 BM. You only need the spin-only formula for NEET; orbital contributions are ignored.

In KMnO4, manganese is in the +7 oxidation state. This is very high and unstable, so Mn readily accepts electrons and gets reduced to a lower, more stable state. In acidic medium, Mn7+ is reduced to Mn2+ (colourless): MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. In neutral or faintly alkaline medium, it is reduced to MnO2 (brown precipitate): MnO4- + 2H2O + 3e- → MnO2 + 4OH-. In strongly alkaline medium, it is reduced to MnO42- (green): MnO4- + e- → MnO42-. This versatility makes KMnO4 one of the most important oxidising agents in chemistry.

Lanthanoids (La to Lu, Z=57-71) fill the 4f-subshell and are non-radioactive (except promethium). They mainly show the +3 oxidation state with a few exceptions (+2 for Eu, Sm; +4 for Ce, Tb). Actinoids (Ac to Lr, Z=89-103) fill the 5f-subshell and are all radioactive. They show a much wider range of oxidation states (up to +7 for some) because the 5f, 6d, and 7s orbitals are very close in energy. Actinoids like U and Th are used in nuclear reactors; others like Pu in nuclear weapons.

Transition metals make good catalysts for two main reasons: (1) Variable oxidation states allow them to provide alternative reaction pathways by forming intermediate compounds that break down more easily than the direct pathway. For example, Fe acts as a catalyst in the Haber process. (2) Large surface area (when used as fine powders or supported metals) allows reactant molecules to adsorb onto the metal surface, which weakens bonds in the reactants and lowers the activation energy. Examples: Pt and Pd catalyse hydrogenation, V2O5 catalyses the oxidation of SO2 to SO3 in the Contact process.