Question

The atomic radii of cerium and promethium are quite similar because
A ) they belong to the same period in the periodic table
B ) they have similar electronic configuration
C ) f-orbitals have very poor shielding effect
D ) nuclear charge is higher on cerium than promethium

Answer 1

Hint: Atomic radius is the distance between the center of the nucleus and outermost (valence) electron.

Complete answer:
> The Lanthanide series has elements from Ce-58 to Lu-71 . Lanthanide series shows lanthanoid contraction. With increase in the atomic number of lanthanides, the atomic size decreases. For the lanthanide series, the atomic size decreases from 187 pm to 172 pm. For each successive element in the series, the decrease in size is around 1 pm. This decrease in size is continuous, but not regular. $$4f$$ subshell electrons have imperfect shielding effects. $$4f$$electrons do not properly shield the nuclear charge. Due to this the attraction between nucleus and $$6s$$electrons, pulls $$6s$$electrons towards nucleus. This decreases atomic radius.
> Cerium and promethium are lanthanoids. The atomic numbers of promethium and cerium are 61 and 62 respectively. Since cerium has a higher atomic number than promethium, the nuclear charge on cerium is higher than that on promethium.
Hence, cerium and promethium have quite similar radii.
> When the atomic number increases by 1, the additional electron enters the same $$4f$$ sub-shell. Due to this poor shielding, the atomic radius should decrease. But when an additional electron is added to the same $$4f$$orbital, the electron electron repulsion increases as now more electrons repel each other. Due to this the atomic radius should increase. When these two effects balance each other, there is little (if any) change in the atomic radius.

Hence the option C ) is the correct answer.

Note: Even though the electron configurations are similar, the nuclear charges and the number of electrons are different. The general electronic configuration of lanthanides is $$\left[\mathrm{X}\mathrm{e}\right]4f^{1-14}5d^{0-1}6s^2$$. The general electronic configuration of actinides is $$\left[\mathrm{R}\mathrm{n}\right]5f^{1-14}6d^{0-1}7s^2$$.