Animated Solution for Chemistry - d and f-Block Elements: Cerium ($Z=58$) is an important member of the lanthanides. Which of the following statements about cerium is incorrect?

To understand the chemistry of Cerium ($Z=58$), we must first look at its place in the periodic table. As a prominent member of the lanthanide series, Cerium follows the general trend of its family. For all lanthanides, the $+3$ oxidation state is the most common and thermodynamically stable state.

When Cerium is in its ground state, its electronic configuration is $[\text{Xe}]4f^{1}5d^{1}6s^2$. By losing three electrons (the two $6s$ electrons and the one $5d$ electron), it forms the ${\text{Ce}}^{3+}$ ion with a configuration of $[\text{Xe}]4f^1$. This $+3$ state is the bedrock of lanthanide chemistry.

However, Cerium is special. It has a trick up its sleeve. If it loses just one more electron—the lone electron in the $4f$ orbital—it forms the ${\text{Ce}}^{4+}$ ion.

The electronic configuration of ${\text{Ce}}^{4+}$ is $[\text{Xe}]4f^0$. By emptying its valence shell completely, Cerium achieves the highly stable noble gas core of Xenon. Because of this exceptional electronic stability, Cerium frequently and readily exhibits the $+4$ oxidation state, making it unique among the early lanthanides.

Here is where the plot thickens. While the noble gas configuration of ${\text{Ce}}^{4+}$ provides kinetic and electronic stability, thermodynamics dictates the ultimate fate of ions in solution. The hydration enthalpy and the sum of ionization energies make the $+3$ state the undisputed king of stability in aqueous mediums.

Because the $+3$ state is thermodynamically favored, ${\text{Ce}}^{4+}$ is always "hungry" for an electron to revert back to ${\text{Ce}}^{3+}$.

This strong tendency to gain an electron means that ${\text{Ce}}^{4+}$ acts as a powerful oxidising agent.

Now, let's evaluate the statements given in the question.

We know that Cerium commonly shows both $+3$ and $+4$ states, and that the $+3$ state is ultimately more stable. We also established that ${\text{Ce}}^{4+}$ acts as an oxidising agent.

What about its existence in solutions? ${\text{Ce}}^{4+}$ is actually incredibly famous in analytical chemistry! Compounds like Ceric Ammonium Nitrate (CAN) are widely used as oxidising agents in aqueous titrations and organic synthesis. Therefore, the claim that the $+4$ oxidation state of cerium is not known in solutions is blatantly false.

Thus, statement (c) is the incorrect one.