Animated Solution for Chemistry - d and f-Block Elements: Given below are two statements. Statement I The $E^{\circ }$ value of $C{e}^{4+}/C{e}^{3+}$ is $+1.74\text{ V}$. Statement II Ce is more stable in $C{e}^{4+}$ state than $C{e}^{3+}$ state. In the light of the above statements, choose the most appropriate answer from the options given below.

The chemistry of lanthanides is a fascinating interplay of electronic configurations, thermodynamic stability, and redox potentials. In this problem, we are tasked with evaluating two statements regarding Cerium ($Ce$), a prominent member of the lanthanide series. Let's break down the concepts step-by-step to uncover the truth.

To understand the behavior of Cerium, we must first look at its ground state electronic configuration. Cerium has an atomic number of $Z=58$.

Its electronic configuration is:

When Cerium loses its four outermost electrons, it forms the $C{e}^{4+}$ ion. The configuration becomes:

Notice that $C{e}^{4+}$ achieves the completely empty $4f^0$ state, which corresponds to the highly stable noble gas configuration of Xenon. Intuitively, one might assume that this makes $C{e}^{4+}$ the most stable state for Cerium. However, aqueous chemistry has a different story to tell.

Despite the apparent stability of the noble gas core in $C{e}^{4+}$, the $+3$ oxidation state is universally the most stable state for all lanthanides, including Cerium.

When Cerium is in the $+3$ state, its configuration is:

The stability of the $+3$ state is governed by a complex balance of ionization enthalpies and hydration enthalpies in aqueous solutions. Because $C{e}^{3+}$ is thermodynamically more stable, any Cerium ion in the $+4$ state will have a strong natural tendency to gain an electron and revert to the $+3$ state.

This brings us to Statement I, which provides the standard reduction potential:

A positive $E^{\circ }$ value indicates that the reduction reaction is spontaneous. A value as high as $+1.74\text{ V}$ means that $C{e}^{4+}$ is exceptionally eager to gain an electron:

Because $C{e}^{4+}$ forces other substances to lose electrons so it can be reduced, it acts as a powerful oxidizing agent. This high positive value is a well-documented factual data point, making Statement I absolutely correct.

Now, let's evaluate Statement II, which claims that Cerium is more stable in the $C{e}^{4+}$ state than the $C{e}^{3+}$ state.

As we just established, the strong tendency of $C{e}^{4+}$ to reduce to $C{e}^{3+}$ (evidenced by the high $E^{\circ }$ value) directly proves that $C{e}^{3+}$ is the lower-energy, more stable state. If $C{e}^{4+}$ were truly more stable, it wouldn't act as such a strong oxidizing agent to escape that state!

Therefore, Statement II is incorrect.

By combining these insights, we can confidently conclude that Statement I is correct, but Statement II is incorrect, leading us to our final answer.