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 Concept:
Lanthanoids generally show a +3 oxidation state. Some show +2 or +4 to achieve stable empty ($f^0$), half-filled ($f^7$), or completely filled ($f^{14}$) configurations.

Step 2: Formula Application:
Cerium ($Z=58$): $[Xe] 4f^1 5d^1 6s^2$.

Step 3: Explanation:
When Cerium loses 4 electrons (two from $6s$, one from $5d$, and one from $4f$), it forms $Ce^{4+}$. This ion has a noble gas configuration ($[Xe] 4f^0$), which provides extra stability.

Step 4: Final Answer:
Cerium ($Ce$) exhibits the +4 state with $f^0$ configuration.