The d and f-Block Elements: CBSE Class 12 Chemistry Chapter Notes

Key Terms and Definitions

Transition Elements (d-block)

Elements that have an incompletely filled d-orbital in their ground state or in any one of their common oxidation states. General electronic configuration: [Noble Gas] (n-1)d¹⁻¹⁰ ns¹⁻².

Inner-Transition Elements (f-block)

Elements in which the last electron enters the f-orbital of the ante-penultimate shell (n-2). They consist of the Lanthanoid series (4f) and Actinoid series (5f).

Lanthanoid Contraction

The steady decrease in the atomic and ionic radii of lanthanoid elements with increasing atomic number. It is caused by the poor shielding effect of 4f electrons.

Actinoid Contraction

The gradual decrease in atomic and ionic radii across the actinoid series. It is more pronounced than lanthanoid contraction due to the poorer shielding effect of 5f electrons compared to 4f electrons.

Interstitial Compounds

Compounds formed when small atoms like H, C, or N are trapped inside the crystal lattices of transition metals. They are non-stoichiometric, hard, and have high melting points.

Mischmetal

An alloy consisting of a lanthanoid metal (~95%), iron (~5%), and traces of S, C, Ca, and Al. It is used in making bullets, shells, and lighter flints.

Disproportionation

A redox reaction where an element in an intermediate oxidation state is simultaneously oxidized and reduced. For example, 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.

Characteristic Properties of Transition Elements

Transition elements exhibit several unique properties that distinguish them from s-block and p-block elements. A key feature is their ability to show variable oxidation states. This arises because the energy difference between the (n-1)d and ns orbitals is very small. As a result, electrons from both subshells can participate in bonding. For example, Manganese (Mn) can show oxidation states from +2 to +7. This variability is fundamental to another significant property: their catalytic activity. Transition metals and their compounds act as excellent catalysts (e.g., V₂O₅ in the Contact Process, Fe in Haber's Process). They can provide a large surface area for reactants and, more importantly, can form unstable intermediate compounds by adopting multiple oxidation states. This provides an alternative reaction pathway with lower activation energy, thereby increasing the reaction rate. At the end of the reaction, the catalyst is regenerated in its original form.

Must-Remember Facts

• Zn, Cd, and Hg are d-block elements but NOT transition elements because they have completely filled d-orbitals (d¹⁰) in their ground state and common oxidation state (+2).
• The color of transition metal ions is generally due to d-d transitions. However, compounds like KMnO₄ and K₂Cr₂O₇ are intensely colored due to Ligand-to-Metal Charge Transfer (LMCT), not d-d transitions.
• Lanthanoid contraction is the reason why the atomic radii of the second (4d) and third (5d) transition series are very similar (e.g., Zr and Hf).
• Magnetic moment (μ) is calculated using the spin-only formula: μ = √n(n+2) B.M., where 'n' is the number of unpaired electrons.
• Transition metals form alloys readily because their atomic sizes are very similar, allowing atoms of one metal to substitute atoms of another in the crystal lattice.
• Most transition metals have high melting points and boiling points due to the involvement of (n-1)d electrons in addition to ns electrons in strong metallic bonding.
• The oxidizing action of KMnO₄ and K₂Cr₂O₇ is strongest in acidic medium.
• In neutral or faintly alkaline medium, KMnO₄ is reduced to MnO₂, while in acidic medium it is reduced to Mn²⁺.

Lanthanoids vs. Actinoids

Worked Examples

• {"title":"Calculating Magnetic Moment","bodyMarkdown":"Question: Calculate the 'spin-only' magnetic moment of the M²⁺(aq) ion (Z = 27).\n\nSolution:\n1. Identify the element: Z = 27 is Cobalt (Co).\n2. Ground state configuration of Co: [Ar] 3d⁷ 4s².\n3. Configuration of the ion Co²⁺: Electrons are removed from the outermost shell (4s) first. So, Co²⁺ is [Ar] 3d⁷.\n4. Count unpaired electrons (n): In the 3d⁷ configuration, there are 5 orbitals. Filling them gives ↑↓, ↑↓, ↑, ↑, ↑. Thus, there are n = 3 unpaired electrons.\n5. Apply the formula: μ = √n(n+2) = √3(3+2) = √15 ≈ 3.87 B.M."}
• {"title":"Predicting Colour in Ions","bodyMarkdown":"Question: Why is Sc³⁺ colorless while Ti³⁺ is colored?\n\nSolution:\n- Scandium (Sc): Z = 21. Configuration: [Ar] 3d¹ 4s². For Sc³⁺, the configuration is [Ar] 3d⁰. Since there are no electrons in the d-orbital, d-d transition is not possible. Hence, Sc³⁺ is colorless.\n- Titanium (Ti): Z = 22. Configuration: [Ar] 3d² 4s². For Ti³⁺, the configuration is [Ar] 3d¹. It has one unpaired electron in its d-orbital. This electron can undergo d-d transition by absorbing light from the visible region, and thus Ti³⁺ is colored (purple)."}

Board Exam Traps & Tips

Pay close attention to 'Reasoning' based questions, which are very common from this chapter.

1. Transition vs. d-block: Always remember the precise definition. Questions like "Why is Zinc not a transition element?" are frequent. Your answer must mention its filled d¹⁰ configuration in both ground state and its common +2 oxidation state.
2. Lanthanoid Contraction Consequences: Don't just define it; learn its consequences. The key ones are: a) Similarity in size of 4d and 5d series elements (e.g., Zr/Hf, Nb/Ta), and b) Difficulty in separating lanthanoids due to similar properties.
3. KMnO₄/K₂Cr₂O₇ Reactions: You must know the products in acidic, basic, and neutral media. For KMnO₄, remember: Acidic → Mn²⁺, Alkaline/Neutral → MnO₂. For K₂Cr₂O₇, it's Cr³⁺ in acidic medium. Balancing the redox equations is a high-scoring skill here.

Quick Revision Check

• Q: Why are d-block elements good catalysts? A: Because they can exhibit variable oxidation states, which allows them to form unstable intermediate compounds, and they provide a suitable surface for the reaction to occur.
• Q: What is the primary consequence of lanthanoid contraction on the elements of the 3rd transition series? A: The atomic radii of elements in the 3rd transition series (5d) are very similar to their corresponding elements in the 2nd transition series (4d), e.g., Hf has a radius almost identical to Zr.
• Q: Why is Cu²⁺ ion coloured but Zn²⁺ ion is colourless? A: Cu²⁺ has a d⁹ configuration with one unpaired electron, allowing d-d transitions. Zn²⁺ has a filled d¹⁰ configuration, so no d-d transition is possible.
• Q: Give the balanced ionic equation for the reaction of acidified potassium permanganate with oxalate ion (C₂O₄²⁻). A: 2MnO₄⁻(aq) + 16H⁺(aq) + 5C₂O₄²⁻(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g)

Frequently Asked Questions

Frequently Asked Questions