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: Understand common oxidation state of lanthanoids.
Lanthanoids usually show: \[ +3 \] oxidation state. This is because they commonly lose: \[ 2 \] electrons from \(6s\) orbital and: \[ 1 \] electron from \(5d\) or \(4f\) orbital.

Step 2: Consider cerium.
Cerium has atomic number: \[ 58 \] Its electronic configuration is approximately: \[ [Xe]\,4f^1\,5d^1\,6s^2 \]

Step 3: Cerium in \(+3\) state.
In \(Ce^{3+}\), cerium loses three electrons. The configuration becomes: \[ [Xe]\,4f^1 \]

Step 4: Cerium in \(+4\) state.
If cerium loses one more electron, it becomes: \[ Ce^{4+} \] The configuration becomes: \[ [Xe]\,4f^0 \] This is especially stable because the \(4f\) subshell becomes empty.

Step 5: Final conclusion.
Therefore, cerium shows \(+4\) oxidation state because it attains: \[ 4f^0 \] electronic configuration.