Question

Which element among the following exhibits electronic configuration as \([Xe]4f^0\) in +4 oxidation state?

• A. Neodymium (z=60)
• B. Praseodymium (z=59)
• C. Cerium (z = 58)
• D. Dysprosium (z = 66)

Hint:

Cerium is the only element in the given list that exhibits the \([Xe]4f^0\) electronic configuration in its +4 oxidation state.

Answer 1

The Correct Option is C

Step 1: Understanding electronic configuration.
The electronic configuration for elements in the +4 oxidation state of the lanthanides typically follows the pattern \([Xe]4f^0\), where the 4f orbital is empty. Cerium (Z=58) exhibits this configuration in the +4 oxidation state.
Step 2: Analyzing the options.
(A) Neodymium (z=60): Incorrect. Neodymium has an electron configuration of \([Xe]4f^4\) in its +4 oxidation state, not \([Xe]4f^0\).
(B) Praseodymium (z=59): Incorrect. Praseodymium has an electron configuration of \([Xe]4f^3\) in its +4 oxidation state.
(C) Cerium (z = 58): Correct — Cerium has the electronic configuration \([Xe]4f^0\) in its +4 oxidation state.
(D) Dysprosium (z = 66): Incorrect. Dysprosium has an electron configuration of \([Xe]4f^10\) in its +4 oxidation state.
Step 3: Conclusion.
The correct answer is (C) Cerium (z = 58), as it exhibits the electronic configuration \([Xe]4f^0\) in the +4 oxidation state.