Question

Consider the isoelectronic series: $K^{+}$ , $S^{2-}$ , $C{l}^{-}$ and $C{a}^{2+}$ ,the radii of the ions decrease as

• A. $C{a}^{2+}>K^{+}>C{l}^{-}>S^{2-}$
• B. $C{l}^{-}>S^{2-}>K^{+}>C{a}^{2+}$
• C. $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$
• D. $K^{+}>C{a}^{2+}>S^{2-}>C{l}^{-}$

Answer 1

Answer: (C)

Amongst isoelectronic ions, ionic radii of anions is more than that of cations. Further size of the anion increases with increase in negative charge and size of cation decreases with increase in positive charge. $\therefore$ The decreasing order of ionic radii is $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}.$