The d-and f-Block Elements

What this chapter is about

The d-block and f-block elements occupy the central and lower portions of the periodic table. The d-block includes elements where the last electron enters a d-orbital (groups 3–12), while the f-block contains the lanthanoids and actinoids where the last electron enters an f-orbital. These elements are commonly called transition elements because they form a bridge between the highly reactive s-block metals and the less metallic p-block elements.

This chapter builds on your Class 11 knowledge of electronic configuration and periodic properties. You will learn why transition metals show variable oxidation states, form coloured compounds, exhibit catalytic activity, and readily form complex ions. The lanthanoids and actinoids present special features owing to the gradual filling of f-orbitals, including the famous lanthanoid contraction.

After studying this chapter, you should be able to write electronic configurations of d- and f-block elements, predict their oxidation states, explain trends in properties like atomic radii and ionisation enthalpy, describe the preparation and properties of important compounds (potassium dichromate, potassium permanganate), and understand why these elements behave differently from s- and p-block elements.

Key ideas

• Definition of transition elements: Elements whose atoms or common ions have incompletely filled d-orbitals. Zinc, cadmium and mercury are d-block but not true transition elements because their d-orbitals are completely filled in both atomic and common ionic states.
• Variable oxidation states: Transition metals show multiple oxidation states because the (n−1)d and ns electrons have similar energies and can both participate in bonding. For example, iron shows +2 and +3, manganese shows +2 to +7.
• Colour of compounds: Incompletely filled d-orbitals allow d–d transitions when visible light is absorbed. The colour seen is complementary to the wavelength absorbed. Compounds of Zn²⁺ (d¹⁰) are colourless.
• Catalytic activity: Transition metals act as catalysts because they can adopt multiple oxidation states and provide a surface for reactant adsorption (e.g., iron in the Haber process, vanadium(V) oxide in the Contact process).
• Formation of complex ions: Small size and high charge density allow transition metal ions to accept electron pairs from ligands, forming coordination compounds such as [Fe(CN)₆]⁴⁻.
• Lanthanoid contraction: As the 4f orbitals fill across the lanthanoid series, the poor shielding by f-electrons causes a steady decrease in atomic and ionic radii. This makes the sizes of 4d and 5d series elements (e.g., Zr and Hf) almost identical.
• Actinoids: These 5f elements are all radioactive. They show more variable oxidation states than lanthanoids because the 5f, 6d and 7s energy levels are closer together.

Formulas and facts to remember

1. General electronic configuration of d-block: (n−1)d¹⁻¹⁰ ns¹⁻². For the first transition series (Sc to Zn), n = 4.
2. Exceptional configurations: Chromium is [Ar] 3d⁵ 4s¹ and copper is [Ar] 3d¹⁰ 4s¹ owing to the extra stability of half-filled and fully filled d-orbitals.
3. Oxidation state trend (first series): Maximum oxidation state increases from Sc (+3) to Mn (+7), then decreases towards Zn (+2). The +2 state arises from loss of two 4s electrons.
4. Standard electrode potential: E° values become less negative across the series, meaning later elements are weaker reducing agents. Copper has a positive E° for Cu²⁺/Cu, so it does not liberate H₂ from dilute acids.
5. Dichromate ion in acid: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O. Orange dichromate is reduced to green Cr³⁺.
6. Permanganate ion in acid: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O. Purple permanganate becomes nearly colourless Mn²⁺.
7. Lanthanoid configuration: [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s². The +3 state is most stable because removal of two 6s and one 4f or 5d electron gives a stable configuration.
8. Actinoid configuration: [Rn] 5f¹⁻¹⁴ 6d⁰⁻¹ 7s². Uranium commonly shows +6 (as in UO₂²⁺).

Worked examples

Example 1: Writing electronic configuration and predicting oxidation states

Problem: Write the electronic configuration of Fe (Z = 26) and Fe³⁺. Explain why Fe³⁺ is more stable than Fe²⁺ in some reactions.

Solution:

1. Ground-state Fe: [Ar] 3d⁶ 4s².
2. When Fe loses electrons, the 4s electrons are removed first (they are farther from the nucleus).
3. Fe²⁺: [Ar] 3d⁶ (loss of two 4s electrons).
4. Fe³⁺: [Ar] 3d⁵ (loss of one more electron from the 3d subshell).
5. The 3d⁵ configuration is half-filled and has extra stability due to exchange energy.
6. Therefore, Fe³⁺ is often more stable, especially in oxidising conditions.

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Example 2: Balancing a redox equation involving dichromate

Problem: Balance the reaction of potassium dichromate with ferrous sulphate in acidic medium.

Solution:

1. Identify changes: Cr in Cr₂O₇²⁻ goes from +6 to +3 (gain of 3 e⁻ per Cr, total 6 e⁻). Fe²⁺ goes to Fe³⁺ (loss of 1 e⁻ each).
2. To balance electrons, 6 Fe²⁺ ions are needed per Cr₂O₇²⁻.
3. Half-reactions:
   • Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
   • 6 Fe²⁺ → 6 Fe³⁺ + 6 e⁻
4. Add them: Cr₂O₇²⁻ + 14 H⁺ + 6 Fe²⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O
5. In terms of salts: K₂Cr₂O₇ + 7 H₂SO₄ + 6 FeSO₄ → Cr₂(SO₄)₃ + 3 Fe₂(SO₄)₃ + K₂SO₄ + 7 H₂O.

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Example 3: Explaining lanthanoid contraction

Problem: The atomic radius of Zr (160 pm) is almost equal to that of Hf (159 pm), even though Hf is in the next period. Explain.

Solution:

1. Between Zr (period 5) and Hf (period 6), the 14 lanthanoid elements intervene.
2. Across the lanthanoids, 4f electrons are added. These f-electrons shield the outer electrons poorly from nuclear charge.
3. Effective nuclear charge increases steadily, causing a contraction of about 10–20 pm across the series.
4. This lanthanoid contraction almost cancels the expected size increase from adding a new shell.
5. Hence Zr and Hf end up with nearly identical radii and very similar chemical properties.

Common mistakes

• Assuming Zn, Cd and Hg are transition elements → They are d-block but not transition elements because their d-orbitals are fully filled in both atom and common ion.
• Removing 3d electrons before 4s when forming cations → Always remove 4s electrons first; they are higher in energy in the cation.
• Thinking all transition-metal compounds are coloured → Compounds with d⁰ or d¹⁰ configuration (e.g., Sc³⁺, Zn²⁺) are colourless because no d–d transition is possible.
• Confusing lanthanoids with actinoids regarding radioactivity → All actinoids are radioactive; most lanthanoids are not.
• Forgetting to balance oxygen with water and hydrogen with H⁺ in acidic-medium redox equations → Use the ion-electron method systematically.

Quick revision

• Transition element: incomplete d-orbitals in atom or common ion.
• Variable oxidation states arise because (n−1)d and ns electrons have similar energies.
• Colour results from d–d electronic transitions; d⁰ and d¹⁰ ions are colourless.
• Lanthanoid contraction makes 4d and 5d congeners nearly equal in size.
• KMnO₄ and K₂Cr₂O₇ are strong oxidising agents; learn their half-reactions in acid.
• Actinoids are all radioactive and show wider range of oxidation states than lanthanoids.