Question

Radius order of \(Yb^{3+}, La^{3+}\), \(Ce^{3+}\), \(Pm^{3+}\)?

• A. Yb$^{3+}$ $>$ La$^{3+}$ $>$ Ce$^{3+}$ $>$ Pm$^{3+}$
• B. Yb$^{3+}$ $<$ La$^{3+}$ $<$ Ce$^{3+}$ $<$ Pm$^{3+}$
• C. Yb$^{3+}$ $>$ Ce$^{3+}$ $>$ Pm$^{3+}$ $>$ La$^{3+}$
• D. Yb$^{3+}$ $<$ Ce$^{3+}$ $<$ Pm$^{3+}$ $<$ La$^{3+}$

Hint:

When analyzing the ionic radii of lanthanides, remember that the radii decrease as the atomic number increases, but the trend can be interrupted by the unique electronic configurations of certain lanthanides.

Answer 1

The Correct Option is A

The ionic radii of lanthanide ions decrease with increasing atomic number. This is due to the progressive filling of the 4f orbitals and the increasing nuclear charge, which pulls the electrons closer to the nucleus, resulting in smaller ionic radii. However, this trend is not always perfectly monotonic due to the unique electronic configurations of individual lanthanides. Specifically:

Yb$^{3+}$: Ytterbium has a small ionic radius because it has a fully filled 4f orbital configuration, making it more stable and compact.
La$^{3+}$: Lanthanum is the first element in the lanthanide series, and it has a relatively larger ionic radius compared to the later lanthanides
Ce$^{3+}$: Cerium has a slightly larger ionic radius than La$^{3+}$ due to its unique electron configuration.
Pm$^{3+}$: Promethium has a larger ionic radius compared to both La$^{3+}$ and Ce$^{3+}$ due to its electron configuration.

Thus, the correct order of ionic radii is Yb$^{3+}$ > La$^{3+}$ > Ce$^{3+}$ > Pm$^{3+}$.