The correct order of ionic radii of $\mathrm{Y}^{3+}$, $\mathrm{La}^{3+}$ and $\mathrm{Lu}^{3+}$ is: (Atomic…

The correct order of ionic radii of $\mathrm{Y}^{3+}$, $\mathrm{La}^{3+}$ and $\mathrm{Lu}^{3+}$ is: (Atomic nos $\mathrm{Y}=39, \mathrm{La}=57, \mathrm{Eu}=63 \mathrm{Lu}$ $=71$ )
  1. $\mathrm{Y}^{3+} < \mathrm{La}^{3+} < \mathrm{Eu}^{3+} < \mathrm{Lu}^{3+}$
  2. $\mathrm{Y}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+}$
  3. $\mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+} < \mathrm{Y}^{3+}$
  4. $\mathrm{La}^{3+} < \mathrm{Eu}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Y}^{3+}$

Solution

Eu and Lu are the members of lanthanide series (so they show lanthanide contraction) and $\mathrm{La}$ is the representative element of all elements of such series and $\mathrm{Y}^{3+}$ ion has lower radii as comparison to $\mathrm{La}^{3+}$ because it lies immediately above it in the periodic table. $\mathrm{Y}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+}$

Asked in: NEET 2003

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