d- and f-block elements for NEET

The d-block is where chemistry gets colourful, literally. Transition metals form coloured ions, stick to magnets, switch between oxidation states and speed up industrial reactions. Nearly every one of these properties comes from a partly filled d subshell, so if you understand the d electrons, you understand the chapter.

The NMC syllabus asks for general trends across the first row (Sc to Zn), the preparation, properties and uses of K₂Cr₂O₇ and KMnO₄, and, for the f-block, configurations, oxidation states and lanthanoid contraction. NEET questions are usually a spin-only magnetic moment, "which ion is colourless", the highest oxidation state in the series, a balanced dichromate or permanganate reaction, or a consequence of lanthanoid contraction.

What counts as a transition element

A transition element has an incompletely filled d subshell in the atom or in one of its common ions. The general configuration is (n−1)d¹⁻¹⁰ ns¹⁻².

• Exceptions in the first row: Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹, because half-filled and fully filled d subshells are extra stable.
• Zn, Cd and Hg are not typical transition elements: they are d¹⁰ in the ground state and in their common +2 ions.
• Copper counts, because Cu²⁺ is 3d⁹.
• When a transition metal forms a cation, the 4s electrons leave first. Fe is 3d⁶4s², so Fe²⁺ is 3d⁶ and Fe³⁺ is 3d⁵.

Trends across the first row

Physical properties. Transition metals are hard, dense and have high melting points, because both d and s electrons take part in metallic bonding. Melting points peak near the middle of the row. Zinc, with no unpaired d electrons, has a low melting point and the lowest enthalpy of atomisation in the row. Mercury is a liquid at room temperature.

Atomic radii decrease at first across the row, then stay almost constant in the middle, because the added d electrons shield the added nuclear charge fairly well.

Ionisation enthalpy rises only gradually and irregularly. Cr and Cu have unusually high second ionisation enthalpies, because removing a second electron breaks a stable d⁵ or d¹⁰.

Oxidation states:

• The number of oxidation states rises to Mn, which reaches +7, the highest in the row, then falls.
• Mn²⁺ (3d⁵) is especially stable, so Mn³⁺ is a strong oxidising agent. Cr³⁺ (t₂g³) is stable, so Cr²⁺ is a reducing agent.
• In high oxidation states, metals bond to oxygen or fluorine: MnO₄⁻, CrO₄²⁻, VO₂⁺.
• E°(Cu²⁺/Cu) is positive (+0.34 V), which is why copper does not release hydrogen from dilute acids.

Colour comes from d–d transitions: in a complex, the d orbitals split in energy, and an electron jumping between them absorbs visible light. The colour we see is the complementary colour.

Magnetic properties. Unpaired electrons make an ion paramagnetic. The spin-only magnetic moment is:

μ = √[n(n + 2)] BM, where n is the number of unpaired electrons.

Catalytic behaviour. Transition metals and their compounds are good catalysts because they have variable oxidation states and can adsorb reactants on their surfaces. Examples: iron in the Haber process, V₂O₅ in the contact process, nickel in the hydrogenation of oils.

Other properties in the syllabus:

• Complex formation: small, highly charged ions with vacant d orbitals, for example [Fe(CN)₆]⁴⁻ and [Cu(NH₃)₄]²⁺. See coordination compounds.
• Interstitial compounds: small atoms such as H, C or N trapped in the metal lattice, as in TiC, Mn₄N and Fe₃H. They are hard, have high melting points, keep metallic conductivity and are chemically inert. They are usually non-stoichiometric.
• Alloys form easily because transition metal radii are similar: brass (Cu + Zn), bronze (Cu + Sn), stainless steel (Fe with Cr and Ni).

Potassium dichromate, K₂Cr₂O₇

Preparation from chromite ore (FeCr₂O₄):

1. Fuse with sodium carbonate in air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2. Acidify the yellow chromate: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
3. Add KCl, since K₂Cr₂O₇ is less soluble: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Chromate and dichromate interconvert with pH: 2CrO₄²⁻ (yellow) + 2H⁺ ⇌ Cr₂O₇²⁻ (orange) + H₂O. Acid pushes it to dichromate; alkali pushes it back to chromate. Chromate is tetrahedral; dichromate is two tetrahedra sharing one oxygen (Cr–O–Cr).

As an oxidising agent in acid: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. The colour changes from orange to green. It oxidises Fe²⁺ to Fe³⁺, I⁻ to I₂, H₂S to S, and Sn²⁺ to Sn⁴⁺. It is a primary standard in volumetric analysis.

Potassium permanganate, KMnO₄

Preparation from pyrolusite (MnO₂):

1. Fuse with KOH in air or with an oxidant such as KNO₃: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O. This gives green potassium manganate.
2. Manganate disproportionates in neutral or acidic solution: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.

Commercially, manganate is oxidised to permanganate electrolytically.

The permanganate ion is tetrahedral and diamagnetic (Mn +7 is d⁰). The manganate ion is green and paramagnetic, with one unpaired electron (Mn +6 is d¹). In acid, KMnO₄ oxidises Fe²⁺, oxalate, I⁻ and H₂S. It is acidified with H₂SO₄, not HCl, because it would oxidise chloride to chlorine.

The f-block: lanthanoids and actinoids

Lanthanoids (Ce to Lu) fill the 4f subshell. Their general configuration is 4f¹⁻¹⁴ 5d⁰⁻¹ 6s².

• The characteristic oxidation state is +3.
• Ce⁴⁺ (4f⁰) is a good oxidising agent. Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴) are reducing agents. These states are stable because they reach an empty, half-filled or filled f subshell.

Lanthanoid contraction is the steady decrease in atomic and ionic radii from La to Lu. 4f electrons shield the nucleus poorly, so each added proton pulls the outer electrons in a little more.

Consequences:

• Zr (160 pm) and Hf (159 pm) have almost the same radius, so they occur together and are hard to separate. The same holds for the whole second and third transition series.
• Basic strength of the hydroxides decreases from La(OH)₃ to Lu(OH)₃, as the ions get smaller and more covalent.

Actinoids (Th to Lr) fill the 5f subshell. All are radioactive. They show a wider range of oxidation states than the lanthanoids, because 5f, 6d and 7s orbitals are close in energy. Actinoid contraction is greater than lanthanoid contraction, because 5f electrons shield even more poorly.

Worked NEET-style questions

Example 1: spin-only magnetic moment

Find μ for Mn²⁺, Cr³⁺ and Cu²⁺.

• Mn²⁺ is 3d⁵: 5 unpaired, μ = √35 ≈ 5.92 BM.
• Cr³⁺ is 3d³: 3 unpaired, μ = √15 ≈ 3.87 BM.
• Cu²⁺ is 3d⁹: 1 unpaired, μ = √3 ≈ 1.73 BM.

Example 2: identify the ion from μ

An ion of the first row has μ = 4.90 BM. Which could it be: Fe²⁺, Fe³⁺, Co²⁺ or Ni²⁺?

• √[n(n + 2)] = 4.90 gives n = 4.
• Fe²⁺ (3d⁶) has 4 unpaired; Fe³⁺ has 5, Co²⁺ has 3, Ni²⁺ has 2.
• Answer: Fe²⁺.

Example 3: a colourless ion

Which is colourless in water: Ti³⁺, V³⁺, Sc³⁺ or Ni²⁺?

• Colour needs d electrons that can jump; d⁰ and d¹⁰ ions have none.
• Sc³⁺ is 3d⁰, so the answer is Sc³⁺.

Example 4: dichromate stoichiometry

How many moles of Fe²⁺ does 1 mol of K₂Cr₂O₇ oxidise in acid?

• Cr₂O₇²⁻ gains 6 electrons (two Cr going from +6 to +3).
• Each Fe²⁺ gives 1 electron.
• Answer: 6 mol.

Practice MCQs

1. The highest oxidation state shown by any first-row transition element is: (a) +6 (b) +7 (c) +8 (d) +5
2. Which ion has a spin-only moment of 2.83 BM? (a) Ti³⁺ (b) Zn²⁺ (c) Mn²⁺ (d) Ni²⁺
3. The colour of KMnO₄ is due to: (a) charge transfer (b) d–d transitions (c) unpaired electrons (d) f–f transitions
4. Zr and Hf have almost identical radii because of: (a) lanthanoid contraction (b) inert pair effect (c) actinoid contraction (d) d–d transition
5. The most common oxidation state of lanthanoids is: (a) +2 (b) +3 (c) +4 (d) +6
6. When an acidified dichromate solution oxidises a reducing agent, the colour changes from: (a) yellow to orange (b) purple to colourless (c) green to orange (d) orange to green
7. The manganate ion MnO₄²⁻ is: (a) green and paramagnetic (b) purple and diamagnetic (c) green and diamagnetic (d) colourless
8. Which is not regarded as a typical transition element? (a) Cu (b) Fe (c) Zn (d) Cr

Answers

1. (b) Manganese reaches +7 in MnO₄⁻.
2. (d) Ni²⁺ is 3d⁸, with 2 unpaired electrons: √8 ≈ 2.83.
3. (a) Mn(+7) is d⁰, so there are no d–d transitions.
4. (a) Poor 4f shielding cancels the expected size increase.
5. (b) +3 is characteristic; +2 and +4 occur for a few.
6. (d) Cr₂O₇²⁻ (orange) goes to Cr³⁺ (green).
7. (a) Mn(+6) is d¹, with one unpaired electron.
8. (c) Zn is 3d¹⁰ in the atom and in Zn²⁺.

What to do next

• Write the configurations of every first-row ion from Sc³⁺ to Zn²⁺ and their spin-only moments.
• Learn the colour table and the charge-transfer exception.
• Write the preparation steps for K₂Cr₂O₇ and KMnO₄ from memory, balancing each equation.
• Revise redox reactions for the permanganate and dichromate half-reactions, then move on to coordination compounds.