Question

Name a member of the lanthanoid series
(I) which exhibits +4 oxidation state
(II) which exhibits +2 oxidation state.

Hint:

Lanthanoid oxidation states are driven by the stability of empty (\(f^0\)), half-filled (\(f^7\)), and completely filled (\(f^{14}\)) f-subshells.

Answer 1

Step 1: Understanding the Concept:
The most common oxidation state for lanthanoids is +3. However, some elements exhibit +2 or +4 states to achieve stable configurations like \(f^0\), \(f^7\), or \(f^{14}\).
Step 2: Detailed Explanation:
(I) +4 Oxidation State: Cerium (\(Ce\), \(Z = 58\)) has the configuration \([Xe] 4f^1 5d^1 6s^2\). By losing 4 electrons, it achieves the stable noble gas configuration of Xenon (\(f^0\)). Thus, \(Ce^{4+}\) is well-known.
(II) +2 Oxidation State: Europium (\(Eu\), \(Z = 63\)) has the configuration \([Xe] 4f^7 6s^2\). By losing 2 electrons, it achieves a stable half-filled \(f^7\) configuration. Thus, \(Eu^{2+}\) is a common ion.
Step 3: Final Answer:
(I) Cerium exhibits +4.
(II) Europium exhibits +2.