Question

The correct increasing order of the ionic radii is

• A. $Ca ^{2+}< K ^{+}< Cl ^{-}< S ^{2-}$
• B. $S ^{2-}< Cl ^{-}< Ca ^{2+}< K ^{+}$
• C. $K ^{+}< S ^{2-}< Ca ^{2+}< Cl ^{-}$
• D. $Cl ^{-}< Ca ^{2+}< K ^{+}< S ^{2-}$

Hint:

In isoelectronic ions, as the nuclear charge increases, the ionic radius decreases.

Answer 1

The Correct Option is A

The question asks us to arrange the given ions in the increasing order of their ionic radii. The ions provided are \(Ca^{2+}\), \(K^+\), \(Cl^-\), and \(S^{2-}\). Let's analyze these ions based on their positions in the periodic table and their electronic configurations: 

1. General Trends of Ionic Radii:
   • Anions (negatively charged ions) are larger than neutral atoms due to the addition of electrons, which increases electron-electron repulsion.
   • Cations (positively charged ions) are smaller than their neutral counterparts because the loss of electrons reduces electron-electron repulsion, and the effective nuclear charge draws remaining electrons closer.
   • Within the same group, ionic size increases as we move down the group.
   • Across a period, the effective nuclear charge increases and ion size decreases for cations, while it increases for anions after reaching the noble gases.
2. Individual Analysis of Given Ions:
   • \(Ca^{2+}\): It loses two electrons, having the configuration of the noble gas \(Ar\). Thus, it is relatively small.
   • \(K^+\): It loses one electron, also having the configuration of \(Ar\), but it is slightly larger than \(Ca^{2+}\) since \(Ca\) has an additional proton pulling electrons closer.
   • \(Cl^-\): Gains one electron to achieve \(Ar\) configuration, larger than both cations due to electron-electron repulsion.
   • \(S^{2-}\): Gains two electrons reaching the configuration of \(Ar\), slightly larger than \(Cl^-\) due to increased electron-electron repulsion.
3. Correct Order: Arranging these ions from smallest to largest based on ionic radii, we have:
   • \(Ca^{2+} < K^+ < Cl^- < S^{2-}\)

Thus, the increasing order of ionic radii is \(Ca^{2+} < K^+ < Cl^- < S^{2-}\), matching the correct option provided in the question.