Which of the following is the incorrect order | vedclass

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Question

(D) For isoelectronic species,the ionic radius decreases as the effective nuclear charge ($Z/e$ ratio) increases.In option $A$,$Sr^{2+}$,$Rb^{+}$,$Br^

Vedclass · Accepted Answer

$Ba^{2+} < Cs^{+} < Se^{2-} < As^{3-}$

Vedclass · Answer

(D) For isoelectronic species,the ionic radius decreases as the effective nuclear charge ($Z/e$ ratio) increases.In option $A$,$Sr^{2+}$,$Rb^{+}$,$Br^{-}$,and $Se^{2-}$ are isoelectronic ($36$ electrons). The order of radii is $Sr^{2+} In option $B$,$Nb^{5+}$,$Zr^{4+}$,and $Y^{3+}$ are isoelectronic ($36$ electrons). The order of radii is $Nb^{5+} In option $C$,for the same element,the size decreases as the positive charge increases: $Co > Co^{2+} > Co^{3+} > Co^{4+}$,which is correct.In option $D$,$Ba^{2+}$ and $Cs^{+}$ belong to the $6^{th}$ period,while $Se^{2-}$ and $As^{3-}$ belong to the $4^{th}$ period. Ions in the $6^{th}$ period are significantly larger than those in the $4^{th}$ period. Thus,the correct order is $As^{3-} < Se^{2-} < Ba^{2+} < Cs^{+}$. Therefore,the given order is incorrect.

Which of the following is the incorrect order of ionic/atomic radius?

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