Correct order of basic strength of metallic hydroxides

Correct order of basic strength of metallic hydroxides
  1. $\mathrm{Ce}(\mathrm{OH})_3 < \mathrm{Lu}(\mathrm{OH})_3 < \mathrm{Eu}(\mathrm{OH})_3$
  2. $\mathrm{Ce}(\mathrm{OH})_3 < \mathrm{Eu}(\mathrm{OH})_3 < \mathrm{Lu}(\mathrm{OH})_3$
  3. $\mathrm{Lu}(\mathrm{OH})_3 < \mathrm{Eu}(\mathrm{OH})_3 < \mathrm{Ce}(\mathrm{OH})_3$
  4. $\mathrm{Lu}(\mathrm{OH})_3 < \mathrm{Ce}(\mathrm{OH})_3 < \mathrm{Eu}(\mathrm{OH})_3$

Solution

The atomic numbers of the given lanthanides are:- $\mathrm{Ce}=58, \mathrm{Eu}=63, \mathrm{Lu}=71$ Since the metallic character decreases from left to right in a period, the order of metallic character for the given metal will be $\mathrm{Lu} < \mathrm{Eu} < \mathrm{Ce}$. Thus, the basic strength of their hydroxides will be:- $\mathrm{Lu}$ $(\mathrm{OH})_3 < \mathrm{Eu}(\mathrm{OH})_3 < \mathrm{Ce}(\mathrm{OH})_3$.

Asked in: AP EAMCET 2023 (16 May Shift 1)

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