Transition and Inner Transition Elements — Chemistry STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryTransition and Inner Transition Elements3 Marks
Question
Explain oxidation states of lanthanoids.
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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.
$Gd^{3+}$ and $Lu^{3+}$ show extra stability due to their half$-$filled and completely filled $f-$orbitals, $Gd^{3+} = [Xe]4f^7, Lu^{3+} = [Xe]4f^{14}$
$Ce$ and $Tb$ attain the $4f^\circ$ 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.
Some lanthanoids show $2+$ and $4+$ oxidation states even though they do not have stable electronic configuration of $4f^\circ , 4f^7$ or $4f^{14}.$ E.g. $Pr^{4+} (4f^1), Nd^{2+} (4f^4), Sm^{2+} (4f^6), Dy^{4+} (4f^8)$ etc
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