Question

Although \(+3\) oxidation state is most common in lanthanoids, cerium still shows \(+4\) oxidation state because:

• A. After losing one more electron, it acquires \(4d^4\) electronic configuration.
• B. Its atomic number is \(61\).
• C. After losing one more electron, it acquires \(4f^0\) electronic configuration.
• D. Its common inert gas is radon.

Hint:

Cerium shows \(+4\) oxidation state because \(Ce^{4+}\) has stable \(4f^0\) configuration.

Answer 1

The Correct Option is C

Step 1: Understanding the Topic:
This problem belongs to the "f-Block Elements," specifically the Lanthanoid series. While the $+3$ oxidation state is the most common and stable for all lanthanoids, some elements exhibit $+2$ or $+4$ states. These anomalous oxidation states are almost always driven by the extra stability that comes from achieving an empty ($f^0$), half-filled ($f^7$), or completely filled ($f^{14}$) $4f$ subshell. Achieving these specific configurations mimics the stability found in noble gases.
Step 2: Key Formulas and Approach:
The approach involves checking the electronic configuration of Cerium in its neutral state and its various ionic states to see which one provides maximum stability.
Step 3: Detailed Explanation:

Ground State: Cerium (Atomic Number 58) has the ground-state configuration: $[Xe] 4f^1 5d^1 6s^2$.
$+3$ State: Removing three electrons (from $6s$ and $5d$) gives $Ce^{3+}$ with the configuration $[Xe] 4f^1$.
$+4$ State: Removing one more electron (the $4f$ electron) results in $Ce^{4+}$.
The Result: The electronic configuration of $Ce^{4+}$ is simply $[Xe]$, or $4f^0$.
An empty $4f$ subshell is highly stable. In the case of Cerium, the $+4$ state is favored because it leaves the atom with the stable electronic core of the noble gas Xenon. Although $Ce^{4+}$ is a strong oxidizing agent (it wants to get back to the more common $+3$ state), it is stable enough to exist in many solid compounds and solutions.
Step 4: Final Answer:
Cerium shows the $+4$ state due to the attainment of the stable $4f^0$ configuration, which is option (D).