Question

Increasing ionic radii of lanthanoids having +2 oxidation state?

• A. La$^{2+}$ $<$ Ce$^{2+}$ $<$ Pr$^{2+}$ $<$ Nd$^{2+}$
• B. Eu$^{2+}$ $<$ Sm$^{2+}$ $<$ Gd$^{2+}$ $<$ Tb$^{2+}$
• C. La$^{2+}$ $<$ Nd$^{2+}$ $<$ Pr$^{2+}$ $<$ Eu$^{2+}$
• D. Eu$^{2+}$ $<$ Gd$^{2+}$ $<$ Tb$^{2+}$ $<$ Dy$^{2+}$

Hint:

When considering ionic radii of lanthanides, remember that as the ionic charge decreases, the ionic radius increases. The +2 state of lanthanides typically exhibits a gradual increase in radius as you move across the series.

Answer 1

The Correct Option is B

The lanthanides have two oxidation states: +3 and +2. The ionic radii of lanthanides increase when the oxidation state decreases from +3 to +2. This is because the +2 state has fewer protons holding the electrons, allowing the radius to expand. The correct order of increasing ionic radii for the lanthanides in the +2 oxidation state is as follows:
\(Eu^{2+}\) has the smallest ionic radius because it is the first to form the +2 state in the lanthanide series.
\(Sm^{2+}, Gd^{2+},\) and \(Tb^{2+}\) follow, with each having progressively larger ionic radii as you move down the series.

Thus, the correct order is \(Eu^{2+} < Sm^{2+} < Gd^{2+} < Tb^{2+}\).