Question

Given,$Ce\left( {58} \right)$ is a member of
A) s-block
B) p-block
C) d-block
D) f-block

Answer 1

We need to know that the cerium is a chemical element with the symbol $Ce$ and atomic number 58. Cerium is a soft, ductile, and silvery-white metal that tarnishes when exposed to air, and it is soft enough to be cut with a steel kitchen knife. Cerium is the second element in the lanthanide series, and while it often shows the $+ 3$ oxidation state characteristic of the series.

Complete answer:
It also has a stable $+ 4$ state that does not oxidize water. It is also considered one of the rare-earth elements. Cerium has a variable electronic structure. The energy of the 4f electron is nearly the same as that of the outer 5d and 6s electrons that are delocalized in the metallic state, and only a small amount of energy is required to change the relative occupancy of these electronic levels. This gives rise to dual valence states.
Cerium is having atomic number $58$
Electronic configuration of Cerium = $\left[ {Xe} \right]6{s^2}5{d^1}4{f^1}$
As the last electron goes to the f-orbital, hence the valence shell is f-shell.
Therefore, the element is a member of the f-block of the periodic table. This element is the second element of the lanthanide series.

The correct answer is Option D

Note:
We have to remember that cerium is the second element of the lanthanide series. In the periodic table, it appears between the lanthanides lanthanum to its left and praseodymium to its right, and above the actinide thorium. It is a ductile metal with hardness similar to that of silver. At lower temperatures the behavior of cerium is complicated by the slow rates of transformation. Transformation temperatures are subject to substantial hysteresis and values quoted here are approximate.