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)

We need to identify which set of ions contains species that are all isoelectronic (have the same number of electrons).

Recall the concept of isoelectronic species

Isoelectronic species are atoms or ions that have the same number of electrons. To find the number of electrons in an ion, we use: Electrons = Atomic number - Charge.

Check Option (4): $$K^+, Cl^-, Ca^{2+}, Sc^{3+}$$

Using the given atomic numbers:

- $$K^+$$: Atomic number 19, charge +1, so electrons = 19 - 1 = 18

- $$Cl^-$$: Atomic number 17, charge -1, so electrons = 17 + 1 = 18

- $$Ca^{2+}$$: Atomic number 20, charge +2, so electrons = 20 - 2 = 18

- $$Sc^{3+}$$: Atomic number 21, charge +3, so electrons = 21 - 3 = 18

All four ions have 18 electrons. This is an isoelectronic set (all have the electron configuration of Argon).

Verify by checking the other options

Option (1): $$Li^+$$ has 2 electrons, $$Na^+$$ has 10, $$Mg^{2+}$$ has 10, $$Ca^{2+}$$ has 18. Not isoelectronic.

Option (2): $$Ba^{2+}$$ has 54 electrons, $$Sr^{2+}$$ has 36, $$K^+$$ has 18, $$Ca^{2+}$$ has 18. Not isoelectronic.

Option (3): $$N^{3-}$$ has 10, $$O^{2-}$$ has 10, $$F^-$$ has 10, $$S^{2-}$$ has 18. Not isoelectronic ($$S^{2-}$$ breaks the pattern).

The correct answer is Option (4): $$K^+, Cl^-, Ca^{2+}, Sc^{3+}$$.