Which is the correct order of ionic sizes (At | vedclass

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(A) The ionic size depends on the charge and the shell number. $Sn^{4+}$ has a configuration of $[Kr] 4d^{10}$,which is a pseudo-noble gas configurati

Vedclass · Accepted Answer

$Sn^{4+} > Ce^{4+} > Yb^{3+} > Lu^{3+}$

Vedclass · Answer

(A) The ionic size depends on the charge and the shell number. $Sn^{4+}$ has a configuration of $[Kr] 4d^{10}$,which is a pseudo-noble gas configuration,and it is in the $5th$ period,making it larger than the lanthanide ions. $Ce^{4+}$ $(Z=58)$ is a lanthanide ion with a $4f^0$ configuration. $Yb^{3+}$ $(Z=70)$ and $Lu^{3+}$ $(Z=71)$ are lanthanide ions. Due to lanthanide contraction,the ionic size decreases as the atomic number increases from $Ce$ to $Lu$. Thus,the order is $Sn^{4+} > Ce^{4+} > Yb^{3+} > Lu^{3+}$.

Which is the correct order of ionic sizes $(At. \, No. : Ce = 58, Sn = 50, Yb = 70, Lu = 71)$?

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The atomic number of cerium $(Ce)$ is $58$. The correct electronic configuration of $Ce^{3+}$ ion is

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