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d-and f-Block Elements - The Lanthanoids

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The Lanthanoids consist of 14 elements from Cerium (Z=58Z=58) to Lutetium (Z=71Z=71), following Lanthanum (Z=57Z=57).

General Electronic Configuration: The general configuration is represented as [Xe]4f1145d016s2[Xe] 4f^{1-14} 5d^{0-1} 6s^2.

Lanthanoid Contraction: The steady decrease in atomic and ionic radii (Ln3+Ln^{3+}) with increasing atomic number. This is due to the imperfect shielding of one 4f4f electron by another, leading to an increase in effective nuclear charge.

Oxidation States: The most common and stable oxidation state is +3+3. Elements like Cerium (Ce4+Ce^{4+}) and Europium (Eu2+Eu^{2+}) show other states to achieve stable f0,f7,f^0, f^7, or f14f^{14} configurations.

Consequences of Lanthanoid Contraction: It results in the similarity of atomic radii between 4d and 5d transition series (e.g., ZrZr and HfHf), making their separation difficult.

Chemical Reactivity: Lanthanoids are highly electropositive. They react with acids to liberate H2H_2, with oxygen to form Ln2O3Ln_2O_3, and with water to form Ln(OH)3Ln(OH)_3.

Color and Magnetic Properties: Many Ln3+Ln^{3+} ions are colored due to fff-f transitions. They exhibit paramagnetism (except La3+La^{3+} and Lu3+Lu^{3+}) because of the presence of unpaired electrons in 4f4f orbitals.

📐Formulae

[Xe]4f1145d016s2[Xe] 4f^{1-14} 5d^{0-1} 6s^2

2Ln+6H2O2Ln(OH)3+3H22Ln + 6H_2O \rightarrow 2Ln(OH)_3 + 3H_2

2Ln+3X22LnX3 (where X=Halogen)2Ln + 3X_2 \rightarrow 2LnX_3 \text{ (where } X = \text{Halogen)}

Ln+N2ΔLnNLn + N_2 \xrightarrow{\Delta} LnN

μ=n(n+2) B.M. (where n=number of unpaired electrons)\mu = \sqrt{n(n+2)} \text{ B.M. (where } n = \text{number of unpaired electrons)}

💡Examples

Problem 1:

Explain why Ce4+Ce^{4+} is a strong oxidizing agent even though it has a stable [Xe]4f0[Xe] 4f^0 configuration.

Solution:

In aqueous solution, Ce4+Ce^{4+} tends to revert to the more stable +3+3 oxidation state.

Explanation:

Although Ce4+Ce^{4+} achieves a noble gas configuration (4f04f^0), the +3+3 state is the most thermodynamically stable oxidation state for all lanthanoids in water. Therefore, Ce4+Ce^{4+} acts as an oxidizing agent (E=+1.74VE^{\circ} = +1.74 V) to be reduced to Ce3+Ce^{3+}.

Problem 2:

Name the lanthanoid element that exhibits the +4+4 oxidation state and acts as a good analytical reagent.

Solution:

Cerium (CeCe)

Explanation:

Cerium exhibits the +4+4 oxidation state (Ce4+Ce^{4+}). It is widely used in volumetric analysis (Cerimetry) as a strong oxidizing agent.

Problem 3:

Why do Zirconium (Zr,Z=40Zr, Z=40) and Hafnium (Hf,Z=72Hf, Z=72) possess almost identical chemical and physical properties?

Solution:

Due to Lanthanoid Contraction.

Explanation:

Normally, atomic size increases down a group. However, the intervention of the lanthanoids and the resulting lanthanoid contraction causes the atomic radius of HfHf (a 5d element) to be almost the same as ZrZr (a 4d element). This makes their properties nearly identical.

The Lanthanoids - Revision Notes & Key Formulas | CBSE Class 12 Chemistry