Which of the following statements are true regarding Lanthanoid elements?

i. \(La(OH)_3\) is the basic hydroxide of Lanthanoids

ii. \(Zr^{4+}\) and \(Hf^{4+}\) ions have almost similar ionic sizes

iii. \(Ce^{4+}\) act as an oxidizing agent.

iv. Trivalent lanthanoid ions are not coloured.

Choose the correct answer from the options given below:

(ii) and (iii)

The correct answer is option 3. (ii) and (iii).

Let us go through each statement in detail to understand why certain statements about lanthanoids are true or false.

Statement i:La(OH)\(_3\) is the basic hydroxide of Lanthanoid

Lanthanum hydroxide (\(La(OH)_3\)) is indeed a basic compound. Lanthanides, which are also called lanthanoids, typically form basic hydroxides. The basicity of lanthanoid hydroxides decreases as we move across the series from lanthanum to lutetium due to the lanthanoid contraction. This contraction occurs because the 4f orbitals poorly shield the increasing nuclear charge, leading to a decrease in atomic and ionic radii across the series. \(La(OH)_3\) is one of the most basic hydroxides among the lanthanoids, and as we move to heavier lanthanoids, the hydroxides become less basic but still remain basic in nature.

Conclusion: This statement is true because \(La(OH)_3\) is indeed a basic hydroxide typical of lanthanoids.

Statement ii: Zr\(^{4+}\) and Hf\(^{4+}\) ions have almost similar ionic sizes

Zirconium (Zr) and Hafnium (Hf) are in Group 4 of the periodic table, and they have the same charge state (\(Zr^{4+}\) and \(Hf^{4+}\)). Due to the lanthanoid contraction (a significant reduction in size that occurs across the lanthanide series), Hafnium, which comes after the lanthanoids, has an atomic and ionic radius that is very similar to that of Zirconium, despite Hf having more protons. The lanthanoid contraction causes the \(4f\) orbitals to poorly shield the outer electrons, making the overall atomic and ionic size of Hf much closer to Zr.

Conclusion: This statement is true because the ionic sizes of \(Zr^{4+}\) and \(Hf^{4+}\) are indeed almost identical due to the lanthanoid contraction.

Statement iii: Ce\(^{4+}\) acts as an oxidizing agen

Cerium (Ce) can exist in the oxidation states of \(+3\) and \(+4\). The \(+4\) oxidation state of cerium is relatively unstable and has a strong tendency to gain an electron and revert to the more stable \(+3\) oxidation state. This makes \(Ce^{4+}\) a strong oxidizing agent. \(Ce^{4+}\) in compounds like ceric ammonium nitrate (\((NH_4)_2[Ce(NO_3)_6]\)) is widely used as an oxidizing agent in organic and inorganic chemistry.

Conclusion: This statement is true because \(Ce^{4+}\) is a well-known oxidizing agent.

Statement iv:Trivalent lanthanoid ions are not coloured

Trivalent lanthanoid ions (\(Ln^{3+}\)) typically exhibit colors due to electronic transitions within the partially filled \(4f\) orbitals. The \(4f\) orbitals in lanthanoids are shielded by the outer 5s and 5p orbitals, which results in sharp, specific \(f\) to \(f\) transitions. These transitions are often responsible for the characteristic colors of the lanthanoid ions. For example, \(Pr^{3+}\) (Praseodymium) is green, \(Nd^{3+}\) (Neodymium) is violet, and \(Sm^{3+}\) (Samarium) is yellow. Since many of the lanthanoid ions are colored, the statement that they are not colored is incorrect.

Conclusion: This statement is false because trivalent lanthanoid ions are generally colored due to \(f\) to \(f\) electronic transitions.

The statements that are true are ii and iii:

ii: Zr\(^{4+}\) and Hf\(^{4+}\) ions have almost similar ionic sizes.

iii: Ce\(^{4+}\) acts as an oxidizing agent.