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d-and f-Block Elements - Lanthanoid Contraction

Grade 12CBSEChemistry

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

🔑Concepts

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Lanthanoid Contraction is the steady decrease in the atomic and ionic radii (specifically for M3+M^{3+} ions) of lanthanoid elements with an increase in atomic number from Lanthanum (Z=57Z = 57) to Lutetium (Z=71Z = 71).

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The primary cause is the poor shielding effect of 4f4f electrons. As the atomic number increases, the nuclear charge increases by unity at each step, and the new electron enters the 4f4f subshell. Due to the diffused shape of 4f4f orbitals, they shield the outer electrons from the nucleus very effectively, leading to an increase in the effective nuclear charge (ZeffZ_{eff}).

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One major consequence is the similarity in atomic and ionic radii of the second (4d4d) and third (5d5d) transition series. For example, Zirconium (ZrZr) and Hafnium (HfHf) have almost identical sizes, making their separation difficult.

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The basic strength of lanthanoid hydroxides decreases from La(OH)3La(OH)_3 to Lu(OH)3Lu(OH)_3. As the size of the M3+M^{3+} ion decreases, the covalent character of the M−OHM-OH bond increases (according to Fajan's Rule), leading to a decrease in basicity.

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The separation of lanthanoid elements in their pure state is difficult due to their very similar chemical properties resulting from the contraction, though it can be achieved through ion-exchange methods.

📐Formulae

[Xe]4f1−145d0−16s2[Xe] 4f^{1-14} 5d^{0-1} 6s^2

Zeff=Z−σZ_{eff} = Z - \sigma

Shielding Effect Order: s>p>d>f\text{Shielding Effect Order: } s > p > d > f

💡Examples

Problem 1:

Explain why ZrZr (Atomic radius ≈160\approx 160 pm) and HfHf (Atomic radius ≈159\approx 159 pm) have almost identical sizes despite HfHf being in the period below ZrZr.

Solution:

This is due to the phenomenon of Lanthanoid Contraction.

Explanation:

Normally, the size increases down a group (4d4d to 5d5d). However, before HfHf (Z=72Z=72) in the 5d5d series, there are 14 lanthanoid elements (Z=58Z=58 to 7171) where the 4f4f shell is being filled. The poor shielding of these 14 4f4f electrons results in a contraction in size that almost exactly cancels the expected increase in size due to the addition of a new energy shell. Thus, ZrZr and HfHf have nearly identical radii.

Problem 2:

Arrange the following in increasing order of basic strength: La(OH)3La(OH)_3, Gd(OH)3Gd(OH)_3, Lu(OH)3Lu(OH)_3.

Solution:

Lu(OH)3<Gd(OH)3<La(OH)3Lu(OH)_3 < Gd(OH)_3 < La(OH)_3

Explanation:

According to lanthanoid contraction, the size of the M3+M^{3+} ion decreases from La3+La^{3+} to Lu3+Lu^{3+}. As the size of the cation decreases, its polarizing power increases, which increases the covalent character of the M−OHM-OH bond. A more covalent bond releases OH−OH^- ions less easily in solution, making it a weaker base.