Write ground-state electron configurations for these ions: (a) \(\mathrm{Li}^{+},\) (b) \(\mathrm{H}^{-}\), (c) \(\mathrm{N}^{3-},\) (d) \(\mathrm{F}^{-},\) (e) \(\mathrm{S}^{2-}\) (f) \(\mathrm{Al}^{3+},(\mathrm{g}) \mathrm{Se}^{2-}\) (h) \(\mathrm{Br}^{-}\) (i) \(\mathrm{Rb}^{+},(\mathrm{j}) \mathrm{Sr}^{2+},(\mathrm{k}) \mathrm{Sn}^{2+}\)

Short Answer

Expert verified
The electron configurations are: (a) \(\mathrm{Li}^{+}\) = [He], (b) \(\mathrm{H}^{-}\) = [He], (c) \(\mathrm{N}^{3-}\) = [Ne], (d) \(\mathrm{F}^{-}\) = [Ne], (e) \(\mathrm{S}^{2-}\) = [Ar], (f) \(\mathrm{Al}^{3+}\) = [Ne], (g) \(\mathrm{Se}^{2-}\) = [Kr], (h) \(\mathrm{Br}^{-}\) = [Kr], (i) \(\mathrm{Rb}^{+}\) = [Kr], (j) \(\mathrm{Sr}^{2+}\) = [Kr], (k) \(\mathrm{Sn}^{2+}\) = [Xe] 4f14 5d10.

Step by step solution

01

Understand Basic Concepts

Electron configurations describe where electrons are around a nucleus of an atom. Elements in the periodic table follow a pattern, in order, as they gain electrons. This is known as the Aufbau Principle. An ion is an atom that has lost or gained electrons. A cation, denoted by a plus sign, has lost electrons. An anion denoted by a minus sign, has gained electrons.
02

Determine Electron Configurations

Determine the electron configuration for each ion. Here, electron configurations are written as noble gas configuration which is a shortcut method where the last noble gas that is smaller than the ion in question is written first, followed by the remaining electron configuration. (a) \(\mathrm{Li}^{+}\) goes to Helium's configuration: [He]. (b) \(\mathrm{H}^{-}\), hydrogen gains an electron so its configuration is the same as Helium: [He]. and so on for the remaining ions.
03

Complete Remaining Ions

(c) \(\mathrm{N}^{3-}\): As Nitrogen gains three electrons, it goes to Neon's configuration: [Ne]. (d) \(\mathrm{F}^{-}\): Fluorine gains one electron so its configuration will be the same as Neon: [Ne]. (e) \(\mathrm{S}^{2-}\): Sulphur gains two electrons so its configuration will be as Argon: [Ar]. (f) \(\mathrm{Al}^{3+}\): Aluminium loses three electrons making it go to Neon's configuration: [Ne]. (g) \(\mathrm{Se}^{2-}\): Selenium gains two electrons so it will be as Krypton: [Kr]. (h) \(\mathrm{Br}^{-}\): Bromine gains one electron making it be like Krypton: [Kr]. (i) \(\mathrm{Rb}^{+}\): Rubidium loses an electron making it go to Krypton’s configuration: [Kr]. (j) \(\mathrm{Sr}^{2+}\): Strontium loses two electrons making it go to Krypton's configuration: [Kr]. (k) \(\mathrm{Sn}^{2+}\): Tin loses two electrons making it go to Xenon's previous configuration: [Xe] 4f14 5d10.

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