Question

(a) What is an isoelectronic series? (b) Which neutral atom is isoelectronic with each of the following ions: \(\mathrm{Al}^{3+}, \mathrm{Ti}^{4+}, \mathrm{Br}^{-}, \mathrm{Sn}^{2+}\)

Short answer

(a) An isoelectronic series is a group of atoms or ions with the same number of electrons but different atomic numbers and charges. (b) The neutral atoms isoelectronic with the given ions are: \(\mathrm{Al}^{3+}\) - Ne (neon); \(\mathrm{Ti}^{4+}\) - Ar (argon); \(\mathrm{Br}^{-}\) - Kr (krypton); \(\mathrm{Sn}^{2+}\) - Cd (cadmium).

Step-by-step solution

Part (a): Definition of an Isoelectronic Series

An isoelectronic series is a group of atoms or ions that have the same number of electrons (or the same electron configuration) but different atomic numbers and different charges. These species will have similar chemical and physical properties because their electron configurations are the same.

Part (b): Finding neutral atoms isoelectronic with given ions

1. \(\mathrm{Al}^{3+}\): Aluminum has an atomic number of 13. Since it has lost 3 electrons to form the \(\mathrm{Al}^{3+}\) ion, it has 10 electrons remaining. A neutral atom with 10 electrons is neon (Ne), so the neutral isoelectronic atom with \(\mathrm{Al}^{3+}\) is Ne. 2. \(\mathrm{Ti}^{4+}\): Titanium has an atomic number of 22. Since it has lost 4 electrons to form the \(\mathrm{Ti}^{4+}\) ion, it has 18 electrons remaining. A neutral atom with 18 electrons is argon (Ar), so the neutral isoelectronic atom with \(\mathrm{Ti}^{4+}\) is Ar. 3. \(\mathrm{Br}^{-}\): Bromine has an atomic number of 35. Since it has gained 1 electron to form the \(\mathrm{Br}^{-}\) ion, it has 36 electrons in total. A neutral atom with 36 electrons is krypton (Kr), so the neutral isoelectronic atom with \(\mathrm{Br}^{-}\) is Kr. 4. \(\mathrm{Sn}^{2+}\): Tin has an atomic number of 50. Since it has lost 2 electrons to form the \(\mathrm{Sn}^{2+}\) ion, it has 48 electrons remaining. A neutral atom with 48 electrons is cadmium (Cd), so the neutral isoelectronic atom with \(\mathrm{Sn}^{2+}\) is Cd.