Which ion has the largest radius?

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(A) To determine the ionic radius of isoelectronic or related species,we consider the effective nuclear charge $(Z_{eff})$. For ions with different el

Vedclass · Accepted Answer

$Se^{2-}$

Vedclass · Answer

(A) To determine the ionic radius of isoelectronic or related species,we consider the effective nuclear charge $(Z_{eff})$. For ions with different electron configurations,the radius generally increases as we move down a group and decreases as we move across a period. Comparing the given ions: $O^{2-}$ ($Z=8$,$10$ electrons) $F^{-}$ ($Z=9$,$10$ electrons) $Se^{2-}$ ($Z=34$,$36$ electrons) $Rb^{+}$ ($Z=37$,$36$ electrons) Among these,$Se^{2-}$ and $Rb^{+}$ are isoelectronic ($36$ electrons). For isoelectronic species,the radius is inversely proportional to the atomic number $(Z)$. Since $Z_{Se} = 34$ and $Z_{Rb} = 37$,$Se^{2-}$ has a larger radius than $Rb^{+}$. Comparing $O^{2-}$ and $F^{-}$,they are smaller than $Se^{2-}$ because they belong to lower periods. Thus,$Se^{2-}$ has the largest radius.

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