Question

The correct order of ionic radii of $Y ^{3+}, La ^{3+}$ $Eu ^{3+}$, and $Lu ^{3+}$ is : - Atomic nos. $Y =39,\, La =57,\, Eu =63,\, Lu =71$.

• A. $Lu^{3+ } < Eu^{3+} < La^{3+} < Y^{3+} $
• B. $La^{3+ } < Eu^{3+} < Lu^{3+} < Y^{3+} $
• C. $Y^{3+ } < La^{3+} < Eu^{3+} < Lu^{3+} $
• D. $Y^{3+ } < Lu^{3+} < Eu^{3+} < La^{3+} $

Answer 1

The Correct Option is D

To determine the correct order of ionic radii for the given ions $Y^{3+}$, $La^{3+}$, $Eu^{3+}$, and $Lu^{3+}$, we need to understand the concept of lanthanide contraction. Lanthanide contraction refers to the gradual decrease in the ionic radii of the lanthanide series elements with increasing atomic number.

Let's examine the ions in the question:

1. $Y^{3+}$ (Yttrium, Atomic number = 39) - Although not a lanthanide, Yttrium is similar and located just before the series on the periodic table.
2. $La^{3+}$ (Lanthanum, Atomic number = 57) - Marks the beginning of the lanthanide series.
3. $Eu^{3+}$ (Europium, Atomic number = 63) - A member of the lanthanide series.
4. $Lu^{3+}$ (Lutetium, Atomic number = 71) - Represents the end of the lanthanide series.

Due to lanthanide contraction, as we move from $La^{3+}$ to $Lu^{3+}$, the ionic size decreases. Therefore, we should expect $Lu^{3+}$ to have the smallest ionic radius among the lanthanides given.

Thus, comparing the radii based on the concept of lanthanide contraction and placement in the periodic table:

1. $Y^{3+}$ has a larger ionic radius than $Lu^{3+}$ because it is positioned before the lanthanide series and is less affected by the contraction.
2. $La^{3+}$ has the largest ionic radius among the lanthanides listed, as it is the first element in the series.
3. $Eu^{3+}$ is between $La^{3+}$ and $Lu^{3+}$ in terms of atomic number and size.

Therefore, the correct order of decreasing ionic radii is:

$Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$