Question

The electronic configuration of Cerium is:
(A) $\left[ Xe \right] { 4f }^{ 0 }{ 5d }^{ 1 }{ 6s }^{ 2 }$
(B) $\left[ Xe \right] { 4f }^{ 1 }{ 5d }^{ 1 }{ 6s }^{ 2 }$
(C) $\left[ Xe \right] { 4f }^{ 2 }{ 5d }^{ 0 }{ 6s }^{ 2 }$
(D) Both (b) and (c)

Answer 1

Hint: Cerium is a lanthanide element with an atomic number of 58. It is soft enough to be cut with a knife like sodium and tarnishes when exposed to air.

Complete step by step solution:
Cerium is an element with atomic number 58. It is the second element in the lanthanide series, and shows +3 and +4 oxidation states. It is one of the rare-earth elements. The first 54 electrons will be placed in exactly the same manner as is placed in Xenon since its atomic number is 54.
Rules for filling the electrons:
Rule 1 –Aufbau Principle- Lowest energy orbitals are filled first. This can be determined by using the formula n+l where n is the principal quantum number and l is the orbital angular momentum quantum number. If the n+l value is the same then look at their n value to determine their order of filling. Therefore, the filling pattern is 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc. The orbitals within a subshell are degenerate which implies that the entire subshell of a particular orbital type is filled before moving to the next subshell of higher energy.
Rule 2 – Pauli’s Exclusion Principle – “Only two electrons are permitted per orbital and they must be of opposite spin. If one electron within an orbital possesses a clockwise spin, then the second electron within that orbital will possess a counter clockwise spin”.
Rule 3- Hund's Rule of maximum multiplicity – “All orbitals within a subshell are of equal energy. Electrons are repulsive to one another and only pair after all of the orbitals have been singly filled”.
Keeping all these rules in mind, the electronic configuration up to 54 electrons will be equal to that of the Xenon.
$\left[ Xe \right] ={ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 3d }^{ 10 }{ 4s }^{ 2 }{ 4p }^{ 6 }{ 4d }^{ 10 }{ 5s }^{ 2 }{ 5p }^{ 6 }$
Now 4 more electrons remain. After the filling of the 5p orbital, these 4 electrons will end up in 4f, 5d and 6s orbitals. According to the n+l rule, the lowest energy orbital among them is 6s. Therefore the 2 electrons will enter the 6s orbital. Now the remaining 2 electrons remain. Now according to the n+l rule both 4f and 5d orbitals are equal in energy therefore their n value has to be taken into account. Therefore the 4f orbital will be lower in energy than the 5d orbital. But one electron occupies the 4f orbital and the other one is present in the 5d orbital. This seems to violate the Aufbau principle and Hund’s rule of maximum multiplicity. In Cerium, the 4f and 5d orbitals are almost similar in energy. Therefore one electron resides in the 4f orbital and the other in the 5d orbital.
Therefore the correct configuration is (b) $\left[ Xe \right] { 4f }^{ 1 }{ 5d }^{ 1 }{ 6s }^{ 2 }$.

Note: Cerium belongs to the lanthanide series. Therefore it has intervening 4f orbitals that are poor shielders of the effective nuclear charge. Due to this the effective nuclear charge on the outermost shell orbital i.e. 6s increases than what is expected due to which there is a dramatic decrease in the ionic radii as we move along the series which is termed as ‘Lanthanide contraction’.