Question

The electronic configuration of cerium is:
A.$\left[\text{Xe}\right]4f^{0}5d^{1}6s^2$
B.$\left[\text{Xe}\right]4f^{1}5d^{1}6s^2$
C.$\left[\text{Xe}\right]4f^{2}5d^{0}6s^2$
D.Both A and B

Answer 1

Hint: The arrangement of electrons in the energy levels around the nucleus of an atom is known as the electronic configuration. The electrons fill up in the energy levels according to the Aufbau’s principle.

Complete step by step answer:
The atomic number of cerium is 58.
The Aufbau’s principle states that in the ground state of the atoms, the orbitals are filled with electrons in order of the increasing energies. The order of energy of different orbitals in an atom is as follows:
$1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d$ and so on.
The maximum number of electrons that can be accommodated is s-orbital are 2, p-orbital are 6, d-orbital are 10 and f-orbital is 14.
Thus, the electronic configuration of cerium having atomic number 58 is as follows:
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{1}5d^1$
One electron fills in the 4f and 5d orbitals each because the 4f and 5d orbitals are quite similar in case of cerium.
The abbreviated electron configuration of an element is written by writing the nearest noble gas in the parenthesis.
The nearest noble gas to cerium is xenon (Xe). The electronic configuration of xenon is as follows:
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$
Thus, the abbreviated electron configuration of cerium is as follows:
$\left[\text{Xe}\right]6s^{2}4f^{1}5d^1$
Thus, the correct option is option (B).


Note:
Other rules that explain the electronic configuration are as follows:
Pauli’s exclusion principle: The Pauli’s exclusion principle states that the two electrons in an atom cannot have the same set of all four quantum numbers.
Hund’s rule of maximum multiplicity: The Hund’s rule of maximum multiplicity states that when several orbitals of equal energy are available, the electrons first fill all the orbitals singly before pairing in any of these orbitals.