What are the most common oxidation states of Ce?

+3, +4

The correct answer is option 1. +3, +4.

The electronic configuration of Cerium (Z=58) is \([Xe] \, 4f^{15} 5d^{16} 6s^2\). To attain the stability of the noble gas Xenon (Z=54), Cerium needs to lose 4 electrons.

Cerium exhibits different oxidation states, including +2, +3, and +4. When Cerium shows the +2 oxidation state, its electronic configuration is \(4f^1 5d^1 6s^0\). In the +3 oxidation state, the electronic configuration is \(4f^1 5d^0 6s^0\). Finally, in the +4 oxidation state, the electronic configuration is \(4f^0 5d^0 6s^0\).

Among these oxidation states, the +3 oxidation state of Cerium is more stable than the +4 oxidation state. This is due to the greater stabilization of the 4f orbital compared to the 5d and 6s orbitals, with the order of stability being \(4f > 5d > 6s\). The 4f orbital is closer to the nucleus, resulting in stronger electron attraction and making it more difficult for electrons to penetrate this orbital. As a result, it requires more ionization energy for electrons to penetrate the 4f orbital.

Cerium(IV) (in the +4 oxidation state) acts as an oxidizing agent because of the high ionization energy required for electron penetration from the 4f orbital. On the other hand, electrons from the 5d and 6s orbitals require less ionization energy for penetration. The high penetration energy of the 4f orbital contributes to the stability of the +3 oxidation state, making it the most stable among all the oxidation states of Cerium.

Based on this explanation, we can conclude that the most common oxidation states of Cerium are +3 and +4.