B
In lanthanide series ionic radius of $Ln^{3+}$ ions decreases
C
La is actually an element of transition series rather than lanthanide
D
Atomic radii of Zr and Hf are same because of lanthanide contraction
The answer is A.
Solution
Due to the lanthanide contraction, the size of $Ln^{3+}$ ions decreases steadily from La to Lu; correct. As the size of the cation decreases, its polarising power increases and the covalent character of the M–OH bond increases. So basicity decreases from $La(OH)_3$ to $Lu(OH)_3$; $La(OH)_3$ is the most basic. So the statement '$La(OH)_3$ is less basic than $Lu(OH)_3$' is not correct. La ($5d^1\,6s^2$, no 4f electron) is often regarded as a d-block (transition) element; and Zr and Hf have almost the same atomic radius due to lanthanide contraction; both statements are correct.
ConceptBasicity of lanthanoid hydroxides
Lanthanoid contraction in past papers
6 questions from this chapter have appeared across 6 exam years.
