Question

The correct order of ionic radii of ${\text{}}{{\text{Y}}^{{\text{3 + }}}}{\text{, L}}{{\text{a}}^{{\text{3 + }}}}{\text{, E}}{{\text{u}}^{{\text{3 + }}}}{\text{and L}}{{\text{u}}^{{\text{3 + }}}}$ is
(At. no. Y = 39, La = 57, Eu = 63, Lu = 71)
(A) ${\text{ L}}{{\text{u}}^{{\text{3 + }}}}{\text{ < E}}{{\text{u}}^{{\text{3 + }}}}{\text{ < L}}{{\text{a}}^{{\text{3 + }}}}{\text{ < }}{{\text{Y}}^{{\text{3 + }}}}$
(B) ${\text{ L}}{{\text{a}}^{{\text{3 + }}}}{\text{ < E}}{{\text{u}}^{{\text{3 + }}}}{\text{ < L}}{{\text{u}}^{{\text{3 + }}}}{\text{ < }}{{\text{Y}}^{{\text{3 + }}}}$
(C) ${\text{}}{{\text{Y}}^{{\text{3 + }}}}{\text{ < L}}{{\text{a}}^{{\text{3 + }}}}{\text{ < E}}{{\text{u}}^{{\text{3 + }}}}{\text{ < L}}{{\text{u}}^{{\text{3 + }}}}$
(D) ${\text{}}{{\text{Y}}^{{\text{3 + }}}}{\text{ < L}}{{\text{u}}^{{\text{3 + }}}}{\text{ < E}}{{\text{u}}^{{\text{3 + }}}}{\text{ < L}}{{\text{a}}^{{\text{3 + }}}}$

Answer 1

Ionic radii is defined as the distance between the nucleus and the electron in the outermost shell of an ion. Due to lanthanide contraction, there is a steady decrease in size of atoms or ions of the rare earth metals with increasing atomic number.

Complete answer:
${\text{L}}{{\text{a}}^{{\text{3 + }}}}{\text{, E}}{{\text{u}}^{{\text{3 + }}}}{\text{and L}}{{\text{u}}^{{\text{3 + }}}}$ belong to the lanthanide series elements. ${{\text{Y}}^{{\text{3 + }}}}$ is a transition element. Generally, across the periodic table the ionic radii decrease and down the group the ionic radii increase. Since ${{\text{Y}}^{{\text{3 + }}}}$ lies above the lanthanide series it has the smaller ionic radii comparatively.
The atomic radius or ionic radii, as according to the Lanthanide Contraction, of these elements decreases as the atomic number increases. It occurs due to the poor shielding effect of the 4f electrons. As we move down the group from left to right in a lanthanide series, the atomic no. increases and for every proton in the nucleus the extra electron goes to 4f orbital.
The lanthanide elements are arranged in the order (according to their atomic number): ${\text{L}}{{\text{a}}^{{\text{3 + }}}}{\text{,E}}{{\text{u}}^{{\text{3 + }}}}{\text{andL}}{{\text{u}}^{{\text{3 + }}}}$ .
Thus, ionic radii decrease along the lanthanide series in the following order:
${\text{L}}{{\text{a}}^{{\text{3 + }}}}{\text{ > E}}{{\text{u}}^{{\text{3 + }}}}{\text{ > L}}{{\text{u}}^{{\text{3 + }}}}$
Hence, the order of ionic radii is
${\text{L}}{{\text{a}}^{{\text{3 + }}}}{\text{ > E}}{{\text{u}}^{{\text{3 + }}}}{\text{ > L}}{{\text{u}}^{{\text{3 + }}}} > {\text{}}{{\text{Y}}^{{\text{3 + }}}}$
In other way, we can write it as
${\text{}}{{\text{Y}}^{{\text{3 + }}}}{\text{ < L}}{{\text{u}}^{{\text{3 + }}}}{\text{ < E}}{{\text{u}}^{{\text{3 + }}}}{\text{ < L}}{{\text{a}}^{{\text{3 + }}}}$
So, the correct option is D.

Note:
Lanthanide contraction decreases ionic radii and it also increases the ionisation energy and electronegativity, thereby separation of lanthanides is very difficult.