The correct order of ionic radii of Ce,La,Pm | vedclass

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(D) In the lanthanoid series,as we move from $La$ $(Z=57)$ to $Lu$ $(Z=71)$,the ionic radii of the $+3$ ions decrease regularly. This phenomenon is kn

Vedclass · Accepted Answer

$Yb^{+3} < Pm^{+3} < Ce^{+3} < La^{+3}$

Vedclass · Answer

(D) In the lanthanoid series,as we move from $La$ $(Z=57)$ to $Lu$ $(Z=71)$,the ionic radii of the $+3$ ions decrease regularly. This phenomenon is known as lanthanoid contraction.This decrease occurs because the $4f$ electrons are added,which provide poor shielding for the increasing nuclear charge,leading to a stronger attraction between the nucleus and the outer electrons.The atomic numbers are: $La$ $(57)$,$Ce$ $(58)$,$Pm$ $(61)$,and $Yb$ $(70)$.Therefore,the order of ionic radii is $La^{+3} > Ce^{+3} > Pm^{+3} > Yb^{+3}$,which is equivalent to $Yb^{+3} < Pm^{+3} < Ce^{+3} < La^{+3}$.

List-$I$ (Property)	List-$II$ (Element)
$(A)$ Lanthanoid showing $+4$ oxidation state	$(i)$ $Pm$
$(B)$ Lanthanoid showing $+2$ oxidation state	$(ii)$ $Ce$
$(C)$ Radioactive lanthanoid	$(iii)$ $Lu$
$(D)$ Lanthanoid with $4f^7$ configuration in $+3$ state	$(iv)$ $Eu$
$(E)$ Lanthanoid with $4f^{14}$ configuration in $+3$ state	$(v)$ $Gd$
-	$(vi)$ $Dy$

The correct order of ionic radii of $Ce$,$La$,$Pm$ and $Yb$ in $+3$ oxidation state is

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