Question

Correct order of basic strength of metallic hydroxides

• A. \( \text{Ce(OH)}_3<\text{Lu(OH)}_3<\text{Eu(OH)}_3 \)
• B. \( \text{Ce(OH)}_3<\text{Eu(OH)}_3<\text{Lu(OH)}_3 \)
• C. \( \text{Lu(OH)}_3<\text{Eu(OH)}_3<\text{Ce(OH)}_3 \)
• D. \( \text{Lu(OH)}_3<\text{Ce(OH)}_3<\text{Eu(OH)}_3 \)

Hint:

The basic strength of lanthanoid hydroxides decreases as we move from left to right across the lanthanoid series due to lanthanoid contraction, which leads to a decrease in ionic size and an increase in the covalent character of the \( M-OH \) bond.

Answer 1

The Correct Option is C

The basic strength of metallic hydroxides of lanthanoids is primarily determined by the size of the metal ion \( M^{3+} \).
As we move across the lanthanoid series from left to right, the ionic radii of \( M^{3+} \) ions gradually decrease due to lanthanoid contraction.
This decrease in ionic size leads to an increase in the polarizing power of the \( M^{3+} \) ion.
A higher polarizing power increases the covalent character of the \( M-OH \) bond, making it more difficult to release \( OH^- \) ions in aqueous solution.
Consequently, the basic strength of the hydroxide \( M(OH)_3 \) decreases as the size of the \( M^{3+} \) ion decreases.
The order of the given lanthanoids in the series is Cerium (Ce), Europium (Eu), and Lutetium (Lu).
Due to lanthanoid contraction, the ionic radii follow the order: $$ r(Ce^{3+})>r(Eu^{3+})>r(Lu^{3+}) $$ Since the basic strength of the hydroxide is inversely related to the polarizing power of the metal ion, which in turn is related to the inverse of the ionic size, the basic strength order will be the reverse of the ionic size order: $$ \text{Basic strength of } Ce(OH)_3>\text{Basic strength of } Eu(OH)_3>\text{Basic strength of } Lu(OH)_3 $$ Therefore, the correct order of basic strength of the metallic hydroxides is: $$ \text{Lu(OH)}_3<\text{Eu(OH)}_3<\text{Ce(OH)}_3 $$ This corresponds to option (C).