Animated Solution for Chemistry - d and f-Block Elements: Given below are two statements: Statement I ${\text{CeO}}_2$ can be used for oxidation of aldehydes and ketones. Statement II Aqueous solution of ${\text{EuSO}}_4$ is a strong reducing agent. In the light of the above statements, choose the correct answer from the options given below.

Welcome, future engineers and doctors! Today, we are going to unravel a beautiful mystery from the d and f-block elements. Inorganic chemistry often feels like a massive list of exceptions, but once you understand the underlying thermodynamic principles, it transforms into a logical and predictable science.

In this problem, we are evaluating two statements regarding the redox behavior of specific lanthanide compounds. To solve this, we don't need to memorize a hundred reactions; we just need to understand one fundamental rule of the f-block elements.

The +3 Comfort Zone

The lanthanide series consists of elements from atomic number 58 (Cerium) to 71 (Lutetium). As we move across this series, electrons are progressively filled into the deep-seated 4f orbitals.

The most crucial, non-negotiable fact about lanthanides is that their most stable oxidation state is +3. Why? Because the thermodynamic energy required to remove the first three electrons (usually two from the 6s orbital and one from the 5d or 4f orbital) is perfectly balanced by the high hydration energy (in aqueous solutions) or lattice energy (in solid state) of the resulting ${\text{Ln}}^{3+}$ ions.

Think of the +3 state as a cozy armchair. If a lanthanide finds itself in a +2 or +4 state, it feels uncomfortable and will aggressively try to return to the +3 state by either losing or gaining an electron. This simple desire dictates their entire redox chemistry.

The Case of Cerium Dioxide

Let's look at Statement I, which claims that ${\text{CeO}}_2$ can be used for the oxidation of aldehydes and ketones.

First, we need to determine the oxidation state of Cerium (Ce) in ${\text{CeO}}_2$. Since oxygen typically exhibits an oxidation state of -2, the two oxygen atoms contribute a total charge of -4. To maintain neutrality, Cerium must be in the +4 oxidation state (${\text{Ce}}^{4+}$).

Now, recall our golden rule. Cerium is a lanthanide, and it desperately wants to be in the +3 state. Being in the +4 state means it is deficient in electrons. To achieve stability, ${\text{Ce}}^{4+}$ will aggressively snatch an electron from any available source:

Because ${\text{Ce}}^{4+}$ forces another substance to lose an electron (so it can gain one), it acts as a powerful oxidising agent. It undergoes reduction itself while oxidising the other species. This is precisely why ${\text{CeO}}_2$ (and other ${\text{Ce}}^{4+}$ salts like Ceric Ammonium Nitrate) are widely used in organic chemistry to oxidise alcohols, aldehydes, and ketones. Therefore, Statement I is absolutely true.

The Case of Europium Sulphate

Now, let's shift our focus to Statement II, which states that an aqueous solution of ${\text{EuSO}}_4$ is a strong reducing agent.

What is the oxidation state of Europium (Eu) in ${\text{EuSO}}_4$? The sulphate ion (${\text{SO}}_4^{2-}$) carries a -2 charge, which means Europium is sitting in the +2 oxidation state (${\text{Eu}}^{2+}$).

Europium (Atomic number 63) has an interesting electronic configuration: $[\text{Xe}]4{\text{f}}^{7}6{\text{s}}^2$. When it loses its two 6s electrons, it forms ${\text{Eu}}^{2+}$ with a exactly half-filled $4{\text{f}}^7$ subshell. This half-filled state provides some extra kinetic stability, allowing ${\text{Eu}}^{2+}$ compounds to exist.

However, thermodynamic stability always wins in the end. The +3 state is still the ultimate comfort zone. To get from +2 to +3, Europium must throw away an electron:

By losing an electron, ${\text{Eu}}^{2+}$ provides that electron to another species in the solution, causing that species to be reduced. Because Europium facilitates this reduction, it acts as a strong reducing agent. It undergoes oxidation itself to help another molecule reduce. Thus, Statement II is also perfectly true.

Both Cerium and Europium, despite starting from opposite ends of the stability spectrum (+4 and +2 respectively), are driven by the exact same thermodynamic destiny: the overwhelming stability of the +3 oxidation state.

Since both statements correctly describe the chemical consequences of this drive, the correct option is (a) Both statement I and statement II are true.