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Explain oxidation states of lanthanoids. |
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Answer» Solution :(1) The common oxidation state of the Lanthanoids is 3+ due to the loss of 2 electrons from OUTERMOST 6s orbital amd one electron from the penultimate 5d-sub-shell. (2) `Gd^(3+)` and `Lu^(3+)`e showextrastabilitydue to theirhalffilledandcompletely filled f-sub-shells. `Gd^(3+) = [Xe]4f^(7)` `Lu^(3+) = [Xe]4f^(14)` (3) Ce and Tb ATTAIN the `4f^(0)` and `4f^(7)` configurations in the 4+ oxidation states. Eu and Yb attain the `4f^(7)` and `4f^(14)` 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. (4) Somelanthnoidsshow 2+ and 4+ oxidationstateseven THOUGH theydo nothavestableelectronicconfigurationof `4f^(0)` , `4f^(7)` or `4f^(14)` E.g. `Pr^(4+) (4f^(1)),Nd^(2+),Sm^(2+)(4f^(6)),DY^(4+)(4f^(8)),` etc |
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