Explain why the +2 oxidation state is relatively more stable for the heavier lanthanides (such as Yb and Lu) compared to the lighter lanthanides. Include in your answer the role of 4f‑electron shielding and ionisation energies.
Answer
Because 4f electrons poorly shield the nuclear charge, the effective nuclear attraction on the outer 6s electrons increases across the series. For heavier lanthanides, removal of the two 6s electrons (forming +2) requires relatively less energy than removing a 4f electron to achieve +3. Hence the +2 state becomes comparatively stable for Yb, Lu, etc.
Step-by-step solution
- State that lanthanides normally exhibit the +3 oxidation state, which corresponds to loss of the two 6s electrons and one 4f electron.
- Explain that 4f electrons are poor shields; as atomic number increases, the effective nuclear charge felt by the outer electrons rises (lanthanide contraction).
- Because of this increased attraction, the ionisation energy to remove a 4f electron (to go from +2 to +3) becomes significantly higher for the heavier lanthanides.
- Consequently, for elements like Yb and Lu, losing only the two 6s electrons (giving +2) is energetically more favorable than also removing a 4f electron.
- Therefore, the +2 oxidation state is unusually stable for the heavier lanthanides.
Common mistake: Attributing the stability of +2 solely to the inert‑pair effect, ignoring the poor shielding of 4f electrons.
Concept tested: Oxidation states of lanthanides and 4f shielding effect
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