Identify the correct order of the size of the following:
A
$Ca^{2+} \lt K^+ \lt Ar \lt Cl^- \lt S^{2-}$
B
$Ar \lt Ca^{2+} \lt K^+ \lt Cl^- \lt S^{2-}$
C
$Ca^{2+} \lt Ar \lt K^+ \lt Cl^- \lt S^{2-}$
D
$Ca^{2+} \lt K^+ \lt Ar \lt S^{2-} \lt Cl^-$
The answer is A.
Solution
All five species, $Ca^{2+}$, $K^+$, $Ar$, $Cl^-$ and $S^{2-}$, have 18 electrons; they are isoelectronic. For isoelectronic species the number of electrons is the same, so the size is decided by the nuclear charge: greater the nuclear charge, stronger the pull on the electrons and smaller the size. Nuclear charges: $Ca^{2+}$ = 20, $K^+$ = 19, $Ar$ = 18, $Cl^-$ = 17, $S^{2-}$ = 16. So the size of an anion increases as the negative charge increases, whereas the size of a cation decreases with increase in positive charge; the radii of the anions are more than those of the cations. Thus the correct order is: $Ca^{2+} \lt K^+ \lt Ar \lt Cl^- \lt S^{2-}$
ConceptSize of isoelectronic species
Ionic Radii of Isoelectronic Species in past papers
5 questions from this chapter have appeared across 5 exam years.
