Answer» - The common oxidation state of the Lanthanoids is 3 + due to the loss of 2 electrons from outermost 6s orbital and one electron from the penultimate 5d sub-shell.
- Gd3+ and Lu3+ show extra stability due to their half-filled and completely filled f-orbitals, Gd3+ = [Xe]4f7, Lu3+ = [Xe]4f14
- Ce and Tb attain the 4f° and 4f7 configurations in the 4 + oxidation states. Eu and Yb attain the 4f7 and 4f14 configurations in the 2 + oxidation states. Sm and Tm also show the 2+ oxidation state although their stability can be explained based on thermodynamic factors.
- Some lanthanoids show 2 + and 4 + oxidation states even though they do not have stable electronic configuration of 4f°, 4f7 or 4f14. E.g. Pr4+ (4f1), Nd2+ (4f4), Sm2+(4f6), Dy4+(4f8) etc
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