Question

Which of the following ions are isoelectronic?

• A. $K^+, Al^{3+}, Ca^{2+}, Sc^{2+}$
• B. $O^{2-}, Al^{3+}, Cl^{-}, Sc^{2+}$
• C. $K^+, S^{2-}, Ca^{2+}, Cl^{-}$
• D. $Na^+, Mg^{2+}, Ca^{2+}, Sc^{2+}$
• E. $Mg^{2+}, Al^{3+}, Ca^{2+}, S^{2-}$

Hint:

Noble gas configurations help quickly identify isoelectronic species. Here all ions have Argon configuration (18 electrons).

Answer 1

The Correct Option is C

Concept: Chemistry - Isoelectronic species. Species having same number of electrons are called isoelectronic.
Step 1: Find electrons in each ion of option (C). Atomic numbers: * $K = 19 \Rightarrow K^+ = 18$ electrons * $S = 16 \Rightarrow S^{2-} = 18$ electrons * $Ca = 20 \Rightarrow Ca^{2+} = 18$ electrons * $Cl = 17 \Rightarrow Cl^- = 18$ electrons
Step 2: Compare all values. All four ions contain: [ 18 \text{ electrons} ] Hence they are isoelectronic.
Step 3: Final answer. [ \boxed{K^+, S^{2-}, Ca^{2+}, Cl^-} ]
Hence, correct option is (C).