Question

Which of the following oxidation state configurations represents the most common, energetically stable, and dominant oxidation state exhibited by nearly all the Lanthanoid elements across the f-block series?

• A. \( +2 \)
• B. \( +4 \)
• C. \( +3 \)
• D. \( +5 \)

Hint:

If an f-block lanthanoid question asks for the most stable or dominant oxidation state without naming a specific element, \( +3 \) is always the correct choice.

Answer 1

The Correct Option is C

Concept: The electronic configurations of the lanthanoids generally follow the pattern \( [\text{Xe}] \, 4\text{f}^{0-14} \, 5\text{d}^{0-1} \, 6\text{s}^2 \). Their oxidation stability patterns depend on how easily they can lose their outer \(6\text{s}\) and \(5\text{d}\) electrons along with inner \(4\text{f}\) electrons.

Step 1: Evaluate the general ionic stabilization pattern of lanthanoids.
Across the lanthanoid series, the \( +3 \) oxidation state is the most common and dominant oxidation state. Forming the trivalent aquated ions (\(\text{Ln}^{3+}\)) involves removing the two outer \(6\text{s}\) valence electrons along with one additional electron from either the \(5\text{d}\) or \(4\text{f}\) subshells. This state is highly stable across the entire series due to favorable hydration and lattice energies.

Step 2: Analyze exceptions to the general rule.
While certain elements can exhibit alternative \( +2 \) or \( +4 \) oxidation states, they do so only when the loss of electrons yields an exceptionally stable empty (\(4\text{f}^0\)), half-filled (\(4\text{f}^7\)), or fully filled (\(4\text{f}^{14}\)) subshell configuration. For example:

• \(\text{Ce}^{4+}\) is stable because it achieves an empty \(4\text{f}^0\) noble gas core.
• \(\text{Eu}^{2+}\) is stable because it achieves a half-filled \(4\text{f}^7\) subshell.

However, these alternative ions act as strong oxidizing or reducing agents, readily reverting back to the dominant \( +3 \) state.