Question

The most common oxidation state of Cerium are :
( A ) $+ 2$ and $+ 4$
( B )$+ 3$ and $+ 4$
( C )$+ 3$ and $+ 5$
( D ) $+ 2$and $+ 3$

Answer 1

Cerium is the important member of Lanthanide . Lanthanides exhibit different oxidation states. Stability of $4f$ orbital is greater than $5d$ and $6s$.

Complete step-by-step answer: We know that Cerium is the member of Lanthanide.
Electronic Configuration of Cerium $($ $Z = 58$ $)$ is ${[Xe]^{54}}4{f^1}5{d^1}6{s^2}$. So to attain the stability of noble gas Xenon $(Z = 54)$ , Cerium has to lose $4$ electron to attain noble gas stability .
Lanthanoids exhibit different oxidation states i.e. $+ 2, + 3,and + 4$.
When Cerium shows $+ 2$ oxidation state then the electronic configuration is $4{f^1}5{d^1}6{s^0}$.
When Cerium shows $+ 3$ oxidation state the electronic configuration is $4{f^1}5{d^0}6{s^0}$.
When Cerium shows $+ 4$ oxidation state the electronic configuration is $4{f^0}5{d^0}6{s^0}$.
The $+ 3$ oxidation state of Cerium is most stable than the $+ 4$ oxidation state due to the greater stabilization of $4f$ orbital than$5d$and $6s$orbitals i.e. $4f > 5d > 6s$.
As $4f$ orbital is closest to the nucleus, attraction of electrons is more in $4f$ orbital and thus penetration of electrons from $4f$ is difficult. As it requires more Ionization energy for penetration of electrons from $4f$ orbital.
That’s why Cerium$({\rm I}V)$ acts as an oxidising agent .
While penetrating electrons from $5d$ and $6s$ orbitals require less ionization energy . And because of the high penetration energy of $4f$ orbital , the $+ 3$ oxidation state is the most stable one among all other oxidation states .
So from the above explanation we can say that the most common oxidation state of Cerium is $+ 3$ and $+ 4$.

Hence option (B) is correct.

Note: For attaining a noble gas configuration which is a stable configuration, Cerium has to lose electrons and among all the oxidation states $+ 3$and $+ 4$ are the common ones. And $+ 3$ is the most stable one .