Q.(a) What is Lanthanoid contraction ? What are the causes and effects of Lanthanoid contraction ? OR

(a) The lanthanoid contraction — the cumulative radius shrinkage caused by poor 4f shielding — makes the second- and third-row transition elements in the same group nearly identical in size. OR (b) Valence bond theory explains the differing geometries of two nickel complexes by how strongly each ligand forces the pairing of nickel's d-electrons: no pairing needed gives $sp^3$/tetrahedral, while forced pairing by the strong-field $CN^-$ ligand gives $dsp^2$/square planar.

(a) Lanthanoid contraction — causes and effects: as atomic number increases across the lanthanoid series (La to Lu), each additional electron enters an inner 4f orbital. Because 4f orbitals are diffuse and shield the outer (5s, 5p, 6s) electrons very poorly from the increasing nuclear charge, the effective nuclear charge experienced by the outer electrons rises steadily across the series, causing a slow, steady, cumulative decrease in atomic and ionic radii — the lanthanoid contraction. Effects: (1) it makes the atomic/ionic radii of the second-row (4d) and third-row (5d) transition elements in the same group (e.g. Zr & Hf, Nb & Ta, Mo & W) almost identical, since the expected size increase from period 5 to period 6 is nearly cancelled by the intervening contraction — making these element pairs very difficult to separate chemically; (2) it causes a gradual decrease in the basicity of lanthanide hydroxides, $Ln(OH)_3$, across the series; (3) it makes the lanthanoids themselves very similar in chemical properties, complicating their mutual separation.

OR (b) VBT explanation of geometry — $[Ni(CO)_4]$ (tetrahedral) vs $[Ni(CN)_4]^{2-}$ (square planar):

In $[Ni(CO)_4]$, nickel is in the 0 oxidation state (neutral $CO$ ligands), retaining its full 10 valence electrons ($3d^84s^2$ as a free atom). $CO$ is a weak-field-type ligand here in the sense that it does not force any electron pairing beyond what already occurs; the nickel atom's electrons rearrange so that all 10 electrons pair up within the five 3d orbitals ($3d^{10}$ configuration), leaving the $4s$ and three $4p$ orbitals completely empty and available for hybridisation. These four empty orbitals ($4s + 4p^3$) undergo $sp^3$ hybridisation, giving four equivalent hybrid orbitals directed tetrahedrally, into which the four $CO$ ligands donate their lone pairs — giving $[Ni(CO)_4]$ a tetrahedral geometry (and it is diamagnetic, since all electrons are paired).

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