Question

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

1. \(K^+, \, Ca^{2+}, \, Cl^-, \, S^{2-}\)
2. \(K^+, \, Cl^-, \, Ca^{2+}, \, S^{2-}\)
3. \(S^{2-}, \, Ca^{2+}, \, Cl^-, \, K^+\)
4. \(S^{2-}, \, Cl^-, \, K^+, \, Ca^{2+}\)

Choose the correct answer from the options given below:

• (A) > (B) > (C) > (D)
• (A) > (C) > (B) > (D)
• (D) > (B) > (C) > (A)
• (D) > (C) > (A) > (B)

Hint:

In isoelectronic species, ionic radii are inversely proportional to the effective nuclear charge (Zeff).

Answer 1

The Correct Option is D

For isoelectronic species, ionic radii decrease as the nuclear charge increases. Since \( \text{S}^{2-} \) has the least nuclear charge and \( \text{Ca}^{2+} \) has the highest nuclear charge, their ionic radii follow the order: \[ \text{S}^{2-} > \text{Cl}^- > \text{K}^+ > \text{Ca}^{2+}. \]