Write the ground-state electron configurations of the following ions: (a) \(\mathrm{Li}^{+},(\mathrm{b}) \mathrm{H}^{-},(\mathrm{c}) \mathrm{N}^{3-},(\mathrm{d}) \mathrm{F}^{-}\) (e) \(\mathrm{S}^{2-}\), (f) \(\mathrm{Al}^{3+},(\mathrm{g}) \mathrm{Se}^{2-},(\mathrm{h}) \mathrm{Br}^{-},(\mathrm{i}) \mathrm{Rb}^{+},(\mathrm{j}) \mathrm{Sr}^{2+}\) (k) \(\mathrm{Sn}^{2+},\) (1) \(\mathrm{Te}^{2-},(\mathrm{m}) \mathrm{Ba}^{2+},(\mathrm{n}) \mathrm{Pb}^{2+},(\mathrm{o}) \mathrm{In}^{3+}\) (p) \(\mathrm{Tl}^{+},(\mathrm{q}) \mathrm{Tl}^{3+}\)

Short Answer

Expert verified
The ground-state electron configurations are: (a) \(\mathrm{Li}^{+}: 1s^{2}\), (b) \(\mathrm{H}^{-}: 1s^{2}\), (c) \(\mathrm{N}^{3-}: 1s^{2}2s^{2}2p^{6}\), (d) \(\mathrm{F}^{-}: 1s^{2}2s^{2}2p^{6}\),(e) \(\mathrm{S}^{2-}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\), (f) \(\mathrm{Al}^{3+}: 1s^{2}2s^{2}2p^{6}\), (g) \(\mathrm{Se}^{2-}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\), (h) \(\mathrm{Br}^{-}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}\), (i) \(\mathrm{Rb}^{+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}\), (j) \(\mathrm{Sr}^{2+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}\), (k) \(\mathrm{Sn}^{2+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{2}\), (1) \(\mathrm{Te}^{2-}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}\),(m) \(\mathrm{Ba}^{2+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}\), (n) \(\mathrm{Pb}^{2+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}\), (o) \(\mathrm{In}^{3+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}\), (p) \(\mathrm{Tl}^{+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{1}\), (q) \(\mathrm{Tl}^{3+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}\).

Step by step solution

01

Identify the ion type

Identify whether each ion is a cation or an anion. Cations have a positive charge and fewer electrons than protons, meaning electrons are removed in relation to the atom. Anions have a negative charge and more electrons than protons, so electrons are added.
02

Adjust the atomic number

For each ion, adjust the atomic number (from the periodic table) according to its charge. Subtract the charge for cations (since they lose electrons) and add the charge for anions (since they gain electrons). For example, for \(\mathrm{Li}^{+}\), the atomic number for uncharged Lithium is 3. Since it is a \(+\) charge, one electron is removed, leaving 2 electrons.
03

Write the electron configuration

Write the electron configuration for the calculated number of electrons. For instance, for \(\mathrm{Li}^{+}\), with 2 electrons, the configuration is \(1s^2\). Repeat these steps for all ions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free