Consider the isoelectronic series K⊕,S2−,Cl⊖,Ca2+ the radii of the ions decrease as
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a
Ca2+>K⊕>Cl⊖>S2−
b
Cl⊖>S2−>K⊕>Ca2+
c
S2−>Cl⊖>K⊕>Ca2+
d
K⊕>Ca2+>S2−>Cl⊖
answer is C.
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Detailed Solution
Among the isoelectronic species, the ionic radii decreases as ' More negative charge > Less negative charge > Neutral atom > Less positive charge > More positive charge. Hence, the answer is S2−>Cl⊖>K⊕>Ca2+