Question

Which one of the following sets of ions represents a collection of isoelectronic species ? 
(Given : Atomic Number : \(F : 9, Cl : 17, Na =11, Mg =12, Al =13, K =19, Ca =20, Sc =21\))

• A. \(N ^{3-}, O ^{2-}, F ^{-}, S ^{2-}\)
• B. \(K ^{+}, Cl ^{-}, Ca ^{2+}, Sc ^{3+}\)
• C. \(Ba ^{2+}, Sr ^{2+}, K ^{+}, Ca ^{2+}\)
• D. \(Li ^{+}, Na ^{+}, Mg { }^{2+}, Ca ^{2+}\)

Hint:

For isoelectronic species, the number of electrons should be the same. Calculate the number of electrons for each ion by considering the atomic number and the charge.

Answer 1

The Correct Option is B

To determine which set of ions represents a collection of isoelectronic species, we need to understand the concept of isoelectronicity. Isoelectronic species are atoms or ions that have the same number of electrons, thus the same electronic configuration.

1. Calculate the number of electrons in each ion:
   • \(K^+\): Potassium (K) with an atomic number of 19 loses one electron to become \(K^+\), so the number of electrons = 19 - 1 = 18.
   • \(Cl^−\): Chlorine (Cl) with an atomic number of 17 gains one electron to become \(Cl^−\), so the number of electrons = 17 + 1 = 18.
   • \(Ca^{2+}\): Calcium (Ca) with an atomic number of 20 loses two electrons to become \(Ca^{2+}\), so the number of electrons = 20 - 2 = 18.
   • \(Sc^{3+}\): Scandium (Sc) with an atomic number of 21 loses three electrons to become \(Sc^{3+}\), so the number of electrons = 21 - 3 = 18.
2. All the ions above have 18 electrons and thus are isoelectronic with each other.

The correct set of isoelectronic species is \(K^+, Cl^−, Ca^{2+}, Sc^{3+}\).

Let’s rule out the other options:

• \(N^{3−}\), \(O^{2−}\), \(F^{−}\), \(S^{2−}\): They have different numbers of electrons (10, 10, 10, and 18, respectively).
• \(Ba^{2+}\), \(Sr^{2+}\), \(K^{+}\), \(Ca^{2+}\): Different electron counts (54, 36, 18, and 18, respectively).
• \(Li^{+}\), \(Na^{+}\), \(Mg^{2+}\), \(Ca^{2+}\): Different electron counts (2, 10, 10, and 18, respectively).

Based on this analysis, the correct answer is indeed:

\(K^+, Cl^−, Ca^{2+}, Sc^{3+}\)