Question

Which lanthanoid from following may exhibit +4 oxidation state with $f^{0}$ configuration?

• A. Eu
• B. Tb
• C. Ce
• D. Lu

Hint:

Anomalous Lanthanoid Oxidation States:
$\text{Ce}^{4+} \rightarrow 4f^0$ (Empty, highly stable).
$\text{Tb}^{4+} \rightarrow 4f^7$ (Half-filled, stable).
$\text{Eu}^{2+} \rightarrow 4f^7$ (Half-filled, stable).
$\text{Yb}^{2+} \rightarrow 4f^{14}$ (Fully-filled, stable).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The dominant oxidation state for all lanthanoids is +3. However, certain elements can exhibit a +4 state if losing that fourth electron results in a highly stable, completely empty $4f^0$, exactly half-filled $4f^7$, or completely filled $4f^{14}$ configuration. We need to find the element that achieves $f^0$.

Step 2: Detailed Explanation:
Let's evaluate the electronic configuration of the given lanthanoid elements:
(a) Europium (Eu, Z=63): $[\text{Xe}] 4f^7 6s^2$. Readily loses 2 electrons to form $\text{Eu}^{2+}$ ($4f^7$ stable half-filled state).
(b) Terbium (Tb, Z=65): $[\text{Xe}] 4f^9 6s^2$. Readily loses 4 electrons to form $\text{Tb}^{4+}$ ($4f^7$ stable half-filled state).
(d) Lutetium (Lu, Z=71): $[\text{Xe}] 4f^{14} 5d^1 6s^2$. Readily loses 3 electrons to form $\text{Lu}^{3+}$ ($4f^{14}$ stable fully-filled state).
(c) Cerium (Ce, Z=58): The first element in the lanthanoid series. Its ground state electronic configuration is $[\text{Xe}] 4f^1 5d^1 6s^2$.
When Cerium loses its three standard valence electrons ($5d^1$ and $6s^2$) to form $\text{Ce}^{3+}$, its configuration becomes $[\text{Xe}] 4f^1$.
Because an entirely empty orbital shell is exceptionally thermodynamically stable, Cerium is highly eager to lose its one remaining $4f$ electron.
By losing this 4th electron, it achieves the $+4$ oxidation state ($\text{Ce}^{4+}$) with the highly stable, noble gas core configuration of Xenon ($[\text{Xe}] 4f^0$).
Because it desperately wants to gain that electron back to return to the universal $+3$ lanthanoid state, $\text{Ce}^{4+}$ is an exceptionally strong oxidizing agent used widely in analytical chemistry (Ceric ammonium nitrate).

Step 3: Final Answer:
Cerium may exhibit a +4 oxidation state with $f^{0}$ configuration, matching option (c).