Which of the following statements is not correct?

\(La(OH)_3\) is less basic than \(Lu(OH)_3\)

The correct answer is option 2. \(La(OH)_3\) is less basic than \(Lu(OH)_3\).

The incorrect statement is: (2) \(La(OH)_3\) is less basic than \(Lu(OH)_3\)

Let us analyze each statement:

(1) La is an element of transition series rather than Lanthanoid series - Correct. Lanthanum (La) is the first element in the lanthanoid series, not a transition metal.

(2) \(La(OH)_3\) is less basic than \(Lu(OH)_3\)- Incorrect. This statement is not necessarily correct. Basicity in hydroxides generally increases as you move down a group in the periodic table. This is because the size of the metal cation increases, leading to weaker binding of hydroxide ions and thus greater availability of hydroxide ions in solution.

Lanthanum (\(La\)) is at the beginning of the lanthanoid series, and Lutetium (\(Lu\)) is at the end. Since basicity tends to increase down a group, \(Lu(OH)_3\) would typically be expected to be less basic than \(La(OH)_3\), not the other way around.

Therefore, option 2 seems incorrect.

(3) In Lanthanoid series, ionic radius of \(Ln^{3+}\) ion decreases - Correct. This phenomenon is called lanthanide contraction. As you move across the series, the additional electrons filling the \4f\) subshell don't effectively shield the increasing nuclear charge. This pulls the outer electrons closer, resulting in a decrease in ionic radii.

(4) Atomic radii of \(Zr\) and \(Hf\) are same because of Lanthanide contraction - Correct. Lanthanide contraction does contribute to the very close atomic radii of Zirconium \((Zr)\) and Hafnium \((Hf)\), but it's not the sole reason. Other factors like effective nuclear charge and electron configuration also play a role.