K^{+},Cl^{-},Ca^{2+},and S^{2-} ions are isoe | vedclass

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Question

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases.The atomic numbers are: $S$ $(Z=16)$,$Cl$ $(Z=17)$,$K$

Vedclass · Accepted Answer

$S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$

Vedclass · Answer

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases.The atomic numbers are: $S$ $(Z=16)$,$Cl$ $(Z=17)$,$K$ $(Z=19)$,and $Ca$ $(Z=20)$.Since all these ions have the same number of electrons ($18$ electrons),the effective nuclear charge $(Z_{eff})$ increases with the increase in the number of protons $(Z)$.As $Z_{eff}$ increases,the attraction between the nucleus and the electrons increases,causing the ionic radius to decrease.Therefore,the decreasing order of ionic radii is: $S^{2-} (Z=16) > Cl^{-} (Z=17) > K^{+} (Z=19) > Ca^{2+} (Z=20)$.

$K^{+}$,$Cl^{-}$,$Ca^{2+}$,and $S^{2-}$ ions are isoelectronic. The decreasing order of their ionic radii is:

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