The correct order of decreasing ionic radii among the following isoelectronic species is

Options

A. `Cl^(-) gt S^(2-) gt Ca^(2+) gt K^(+)`
B. `S^(2-) gt Cl^(-) gt K^(+) gt Ca^(2+)`
C. `K^(+) gt Ca^(2+) gt Cl^(-) gt S^(2-)`
D. `Ca^(2+) gt K^(+) gt S^(2-) gt Cl^(-)`

Answer

B

Explanation

In the formation of a cation, the electrons are lost from the outer valence shell and the remaining electrons experience of greater force of attraction by the nucleus . In other words, nucleus hold the remaining electrons more tightly and this results in decreases radii. In case of anion formation, the addition of electron occurs in the same outer shell thus the hold of nucleus on the electron of outer shell decreases and results in increased ionic radii.
`:.` Order of ionic radii is `S^(2-) gt Cl^(-) gtK^(+) gt Ca^(2+)` these are isoelectronic species with 18 electrons each. more than nuclear charge, smaller is the size of the ion . nuclear charge are S(16),Cl(17),K(19),Ca(20).