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(B) To determine the ionic radius,consider the following rules:$1$. For isoelectronic species,the radius decreases as the nuclear charge (atomic numbe

Vedclass · Accepted Answer

$Te^{2-} > Br^{-} > Cl^{-} > K^{+}$

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(B) To determine the ionic radius,consider the following rules:$1$. For isoelectronic species,the radius decreases as the nuclear charge (atomic number) increases.$2$. For ions with different electron shells,the radius increases as the number of shells increases.$K^{+}$ ($18$ electrons,$Z=19$) and $Cl^{-}$ ($18$ electrons,$Z=17$) are isoelectronic. Since $Cl^{-}$ has a lower nuclear charge,$Cl^{-} > K^{+}$.$Br^{-}$ ($36$ electrons,$Z=35$) has $4$ shells,while $Cl^{-}$ and $K^{+}$ have $3$ shells. Thus,$Br^{-} > Cl^{-} > K^{+}$.$Te^{2-}$ ($54$ electrons,$Z=52$) has $5$ shells,which is more than $Br^{-}$. Thus,$Te^{2-} > Br^{-} > Cl^{-} > K^{+}$.The correct order is $Te^{2-} > Br^{-} > Cl^{-} > K^{+}$.

Which of the following represents the correct order of ionic radius?

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