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d and f Block Elements - Lanthanoid contraction

Grade 12ICSEChemistry

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

🔑Concepts

Lanthanoid Contraction is the steady decrease in the atomic and ionic radii (especially for M3+M^{3+} ions) of the elements in the lanthanoid series (58Ce_{58}Ce to 71Lu_{71}Lu) with the increase in atomic number.

The primary cause of Lanthanoid Contraction is the poor shielding effect of the 4f4f electrons. As the nuclear charge increases by +1+1 at each step, the 4f4f electrons, being in diffused orbitals, fail to shield the outer electrons effectively from the increasing nuclear pull.

One major consequence is the similarity in the atomic and ionic radii of the 4d4d (second transition series) and 5d5d (third transition series) elements. For instance, ZrZr and HfHf have nearly identical sizes, making them 'chemical twins'.

Basicity of hydroxides decreases across the series. Due to the decrease in size of the M3+M^{3+} ion, the covalent character of the MOHM-OH bond increases (according to Fajans' rules), making La(OH)3La(OH)_3 the most basic and Lu(OH)3Lu(OH)_3 the least basic.

The contraction makes the separation of lanthanoids in their pure state very difficult because their chemical properties, which depend on ionic size, are extremely similar.

📐Formulae

General Electronic Configuration: [Xe]4f1145d016s2\text{General Electronic Configuration: } [Xe] 4f^{1-14} 5d^{0-1} 6s^2

Ionic Radius Trend: r(La3+)>r(Ce3+)>>r(Lu3+)\text{Ionic Radius Trend: } r(La^{3+}) > r(Ce^{3+}) > \dots > r(Lu^{3+})

Basicity Order: La(OH)3>Ce(OH)3>>Lu(OH)3\text{Basicity Order: } La(OH)_3 > Ce(OH)_3 > \dots > Lu(OH)_3

Radius Comparison: r(Zr)r(Hf)160 pm\text{Radius Comparison: } r(Zr) \approx r(Hf) \approx 160\text{ pm}

💡Examples

Problem 1:

Explain why ZrZr (Atomic number 4040) and HfHf (Atomic number 7272) possess almost similar properties.

Solution:

ZrZr belongs to the 4d4d series and HfHf belongs to the 5d5d series. Normally, the size should increase down the group. However, the 1414 lanthanoid elements (58Ce_{58}Ce to 71Lu_{71}Lu) occur between ZrZr and HfHf. Due to the lanthanoid contraction, the expected increase in size from ZrZr to HfHf is cancelled out.

Explanation:

The atomic radius of ZrZr is 160 pm160\text{ pm} and that of HfHf is 159 pm159\text{ pm}. Because their sizes and valence electronic configurations are similar, their chemical and physical properties are nearly identical.

Problem 2:

Which is more basic: Ce(OH)3Ce(OH)_3 or Lu(OH)3Lu(OH)_3? Justify.

Solution:

Ce(OH)3Ce(OH)_3 is more basic than Lu(OH)3Lu(OH)_3.

Explanation:

According to Lanthanoid contraction, the size of the M3+M^{3+} ion decreases from Ce3+Ce^{3+} to Lu3+Lu^{3+}. As the size of the cation decreases, the polarising power increases, which increases the covalent character of the MOHM-OH bond. Therefore, the tendency to release OHOH^- ions decreases, leading to a decrease in basicity.

Lanthanoid contraction - Revision Notes & Key Formulas | ICSE Class 12 Chemistry