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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The ionic radii of lanthanoids decrease with an increase in atomic number due to lanthanide contraction. $La^{3+}$ $(Z=57)$ has the largest size,while $Lu^{3+}$ $(Z=71)$ has the smallest size among the lanthanoids. $Y^{3+}$ $(Z=39)$ has an ionic radius comparable to the heavier lanthanoids (like $Ho^{3+}$ or $Er^{3+}$),but it is smaller than $La^{3+}$ and $Eu^{3+}$. Comparing the given ions: $La^{3+} (1.03 \ \mathring{A}) > Eu^{3+} (0.95 \ \mathring{A}) > Y^{3+} (0.90 \ \mathring{A}) > Lu^{3+} (0.86 \ \mathring{A})$. Therefore,the correct order is $Lu^{3+} < Y^{3+} < Eu^{3+} < La^{3+}$.

The correct order of ionic radii of $Y^{3+}$,$La^{3+}$,$Eu^{3+}$ and $Lu^{3+}$ is
(Atomic No. $Y = 39$,$La = 57$,$Eu = 63$,$Lu = 71$)

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The correct order of ionic radii of $Y^{3+}$,$La^{3+}$,$Eu^{3+}$ and $Lu^{3+}$ is (Atomic No. $Y = 39$,$La = 57$,$Eu = 63$,$Lu = 71$)