Question

Consider following statements.
$\left(I\right)$ The size of the lanthanide ions $\left(M^{3+}\right)$ decreases as the atomic number of $M$ increases.
$\left(II\right)$ Electronic spectra of lanthanides show very broad bands.
$\left(III\right)$ As with transition metals, coordination number six is very common in lanthanide complexes.
Which of the statements given above is/are correct?

• A. $I$ only
• B. $I$ and $II$
• C. $I$ and $III$
• D. $III$ only

Answer 1

Answer: (B)

$I$ and $II$

In lanthanides, the additional electron enters $4f$ sub-shell but not in the valence shell namely sixth shell. The shielding effect of one electron in $4f$ sub-shell by another in the same sub-shell (i.e. mutual shielding effect of $4f$ electrons) is very little, being even smaller than that of d-electrons because the shape of f-sub shell is very much diffused.

The nuclear charge (i.e. atomic number), however, increases by unity at each step. Thus, the nuclear charge increases at each step, while there is no comparable increase in the mutual shielding effect of $4f$ electrons. This results in that electrons in the outermost shell experience increasing nuclear attraction from the growing nucleus. Consequently, the atomic and ionic radii go on decreasing as we move from ${}^{57}La$ and ${}^{71}Lu$.
Due to higher number of electrons, electronic spectra of lanthanides show very broad bands.