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Q.What is Lanthanide contraction? Explain its reasons and consequences. OR

(a) Complete the following chemical reaction equations (1×3=3):
(i) [not rendered]
(ii) [not rendered]
(iii) [not rendered].
(b) Transition metals and their compounds show paramagnetic behaviour. Explain. (2 marks)
Haryana BsehBSEH Intermediate Board 2020Subjective· 5mImportance★★★★★
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Lanthanide contraction — the steady shrinkage in size across La to Lu — arises from poor shielding by the diffuse 4f electrons, and its main consequence is that second- and third-row transition metals end up nearly the same size.

What it is: As you move across the lanthanide series (La, Ce, ..., Lu; Z = 57 to 71), the atomic and ionic (M³⁺) radii show a small but steady, cumulative decrease, rather than the increase you might expect from rising atomic number alone.

Reason: Each successive lanthanide adds one more electron to the deeply-buried 4f subshell rather than to an outer shell. The 4f orbitals are very diffuse and shield the increasing nuclear charge from the outer (5s, 5p, 6s) electrons very poorly (worse than d or p electrons would). So as the nuclear charge (number of protons) increases across the series, the outer electrons feel an increasingly strong effective nuclear pull that is only weakly screened — pulling the outer shells inward. This 'contraction' happens at every step, and although each individual step is small, the cumulative effect across all 14 elements is significant.

Consequences:

  1. Near-identical radii of 4d and 5d transition elements: Because of the lanthanide contraction occurring in between them, the second (4d) and third (5d) transition series elements of the same group end up with almost the same atomic/ionic radii (e.g. Zr and Hf, Nb and Ta) — normally you'd expect the third-row element to be noticeably larger due to an extra shell. This makes such pairs chemically very similar and notoriously difficult to separate.
  2. Decreasing basicity of lanthanide hydroxides: As the ionic radius of M³⁺ decreases from La³⁺ to Lu³⁺, the M–OH bond becomes more covalent and less ionic, so the basic strength of Ln(OH)₃ steadily decreases from La(OH)₃ (most basic) to Lu(OH)₃ (least basic). …

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