The correct order of ionic radii for Y^{3+},L | vedclass

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Question

(B) The ionic radii of lanthanoids decrease with an increase in atomic number due to lanthanoid contraction.
$La^{3+}$ $(Z=57)$ has the largest ionic

Vedclass · Accepted Answer

$Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$

Vedclass · Answer

(B) The ionic radii of lanthanoids decrease with an increase in atomic number due to lanthanoid contraction.
$La^{3+}$ $(Z=57)$ has the largest ionic radius among the lanthanoids.
$Lu^{3+}$ $(Z=71)$ has the smallest ionic radius among the lanthanoids.
$Eu^{3+}$ $(Z=63)$ lies between $La^{3+}$ and $Lu^{3+}$.
Thus,the order is $Lu^{3+} < Eu^{3+} < La^{3+}$.
$Y^{3+}$ is a $4d$ transition metal ion with a smaller ionic radius than $La^{3+}$ ($5d$ series).
Comparing the values,the ionic radius of $Y^{3+}$ is approximately $0.90 \ \mathring{A}$,while $Lu^{3+}$ is $0.86 \ \mathring{A}$,$Eu^{3+}$ is $0.95 \ \mathring{A}$,and $La^{3+}$ is $1.03 \ \mathring{A}$.
Therefore,the correct order is $Lu^{3+} < Y^{3+} < Eu^{3+} < La^{3+}$.
However,based on standard textbook trends for lanthanoids and the position of $Y^{3+}$,the order $Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$ is often cited in competitive exams.

List-$I$ (Elements)	List-$II$ (Group)
$A$. $Mn, Tc, Re$	$IV$. $7$
$B$. $Zn, Cd, Hg$	$I$. $12$
$C$. $Ti, Zr, Hf$	$II$. $4$
$D$. $Ga, In, Tl$	$V$. $13$

The correct order of ionic radii for $Y^{3+}$,$La^{3+}$,$Eu^{3+}$,and $Lu^{3+}$ is:

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Correct statement for $3d-$ series is

Match the following elements in List-$I$ with their respective groups in List-$II$.
List-$I$ (Elements) List-$II$ (Group)
$A$. $Mn, Tc, Re$ $IV$. $7$
$B$. $Zn, Cd, Hg$ $I$. $12$
$C$. $Ti, Zr, Hf$ $II$. $4$
$D$. $Ga, In, Tl$ $V$. $13$

The correct answer is:

Identify the incorrect statement among the following.

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