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 (atomic number) increases.Atomic numbers are: $S = 16$,$Cl = 17$,$K = 1

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.Atomic numbers are: $S = 16$,$Cl = 17$,$K = 19$,$Ca = 20$.Since all these ions have $18$ electrons,the ionic size depends on the effective nuclear charge.As the nuclear charge increases,the electrons are pulled more strongly towards the nucleus,resulting in a smaller size.Thus,the decreasing order of size is: $S^{2-} (16) > Cl^{-} (17) > K^{+} (19) > Ca^{2+} (20)$.

Ion	Radius $(pm)$
$Q^{a+}$	$53$
$X^{b+}$	$66$
$Y^{c+}$	$40$
$Z^{d+}$	$100$

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

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