Question

The atomic number of cerium (Ce) is 58. The correct electronic configuration of $C{{e}^{3+}}$ is
a.) $[Xe]4{{f}^{1}}$
b.) $[Kr]4{{f}^{1}}$
c.) $[Xe]4{{f}^{13}}$
d.) $[Kr]4{{d}^{1}}$

Answer 1

Hint: Nearest noble gas element to cerium is xenon and the atomic number of xenon is 56.
Atomic number is the number of protons present in the molecule. For every element in a periodic table it is different.
Atomic number of cerium is 58.

Complete answer:
So, for writing the noble gas configuration of an element symbol of the last noble gas that came before the element which is followed by the configuration of the electrons left.
Last inert gas element that comes before cerium is xenon.
Symbol of xenon is Xe.
Atomic number of xenon is 56 and of cerium is 58.
So, the difference of electrons between cerium and xenon is 2.
Therefore, electronic configuration of cerium is \[[Xe]4{{f}^{1}}5{{d}^{1}}6{{s}^{2}}\]
but the electron from \[5{{d}_{1}}\] moves to \[4{{f}_{1}}~\,\] therefore its electronic configuration becomes\[[Xe]4{{f}^{2}}5{{d}^{0}}6{{s}^{2}}\]
For the formation of$C{{e}^{3+}}$ion, cerium loses three electrons to form $C{{e}^{3+}}$.
So, the electron will be removed from outer orbitals that is 6s and 4f
Therefore, the electronic configuration of $C{{e}^{3+}}$comes out be \[[Xe]4{{f}^{1}}5{{d}^{0}}6{{s}^{0}}\] or simply\[[Xe]4{{f}^{1}}\]

Correct answer is option A.

Note: The noble gas configuration of an element is written by writing the symbol of the last noble gas that came before the element which is followed by the configuration of the electrons left.