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

The correct option is Option 1: K⁺, Ca²⁺, Sc³⁺, Cl^–.

An isoelectronic species refers to a group of atoms, molecules, or ions that have the same number of electrons. To determine which set represents a collection of isoelectronic species, we calculate the total number of electrons in each ion of the chosen set using their atomic numbers.

The atomic number (Z) of an element represents the number of protons, which is equal to the number of electrons in a neutral atom. For ions:
• A positive charge (cation) indicates a loss of electrons: $\text{Number of electrons}=Z-\text{charge}$
• A negative charge (anion) indicates a gain of electrons: $\text{Number of electrons}=Z+\text{magnitude of charge}$

Let us calculate the number of electrons for each ion in the correct set:
1. Potassium ion (${\text{K}}^{+}$): The atomic number of K is 19. It has a +1 charge, meaning it has lost 1 electron.
$\text{Number of electrons}=19-1=18$
2. Calcium ion (${\text{Ca}}^{2+}$): The atomic number of Ca is 20. It has a +2 charge, meaning it has lost 2 electrons.
$\text{Number of electrons}=20-2=18$
3. Scandium ion (${\text{Sc}}^{3+}$): The atomic number of Sc is 21. It has a +3 charge, meaning it has lost 3 electrons.
$\text{Number of electrons}=21-3=18$
4. Chloride ion (${\text{Cl}}^{–}$): The atomic number of Cl is 17. It has a –1 charge, meaning it has gained 1 electron.
$\text{Number of electrons}=17+1=18$

Since each of the ions in this set (${\text{K}}^{+}$, ${\text{Ca}}^{2+}$, ${\text{Sc}}^{3+}$, and ${\text{Cl}}^{–}$) contains exactly 18 electrons, they form a collection of isoelectronic species.