Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write electron configurations for the most stable ion formed by each of the elements Al, Ba, Se, and I (when in stable ionic compounds).

Short Answer

Expert verified
The electron configurations for the most stable ions formed by the elements Al, Ba, Se, and I are: Al³⁺: \(1s^2 2s^2 2p^6\) Ba²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\) Se²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6\) I⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\)

Step by step solution

01

Find the atomic numbers of the elements and their most common oxidation states

Look up the atomic numbers and the most common oxidation states for the elements Al, Ba, Se, and I. For reference purposes, we can use a periodic table. Al (Aluminium) - Atomic number: 13, Most common oxidation state: +3 Ba (Barium) - Atomic number: 56, Most common oxidation state: +2 Se (Selenium) - Atomic number: 34, Most common oxidation state: -2 I (Iodine) - Atomic number: 53, Most common oxidation state: -1
02

Write the ground state electron configuration for each element

Using the atomic numbers, write the electron configurations for the neutral atoms of Al, Ba, Se, and I. Al: \(1s^2 2s^2 2p^6 3s^2 3p^1\) Ba: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 6s^2\) Se: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^4\) I: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^5\)
03

Write the electron configurations for the most stable ions

Based on the most common oxidation states, we can determine the electron configurations for the most stable ions formed by each element. Al: As it tends to lose 3 electrons, the most stable ion is Al³⁺ (Al+3). Its electron configuration is: \(1s^2 2s^2 2p^6\) Ba: As it tends to lose 2 electrons, the most stable ion is Ba²⁺ (Ba+2). Its electron configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\) Se: As it tends to gain 2 electrons, the most stable ion is Se²⁻ (Se-2). Its electron configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6\) I: As it tends to gain 1 electron, the most stable ion is I⁻ (I-1). Its electron configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\) In conclusion, the electron configurations for the most stable ions formed by the elements Al, Ba, Se, and I are: Al³⁺: \(1s^2 2s^2 2p^6\) Ba²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\) Se²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6\) I⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\)

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

Consider the following reaction: $$ \mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB} \quad \Delta E=-285 \mathrm{kJ} $$ The bond energy for \(A_{2}\) is one-half the amount of the AB bond energy. The bond energy of \(\mathbf{B}_{2}=432 \mathrm{kJ} / \mathrm{mol} .\) What is the bond energy of \(\mathrm{A}_{2} ?\)

Each of the following compounds has three possible names listed for it. For each compound, what is the correct name and why aren't the other names used? a. \(\mathrm{N}_{2} \mathrm{O}\) : nitrogen oxide, nitrogen(I) oxide, dinitrogen monoxide b. \(\mathrm{Cu}_{2} \mathrm{O}:\) copper oxide, copper(I) oxide, dicopper monoxide c. \(\mathrm{Li}_{2} \mathrm{O}:\) lithium oxide, lithium(I) oxide, dilithium monoxide

For each of the following, write an equation that corresponds to the energy given. a. lattice energy of \(\mathrm{NaCl}\) b. lattice energy of \(\mathrm{NH}_{4} \mathrm{Br}\) c. lattice energy of \(\mathrm{MgS}\) d. \(O=O\) double bond energy beginning with \(O_{2}(g)\) as a reactant

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3}\) Give a possible explanation for why these two reactions occur.

Which of the following ions have noble gas electron configurations? a. \(\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Sc}^{3+}, \mathrm{Co}^{3+}\) b. \(\mathrm{Tl}^{+}, \mathrm{Te}^{2-}, \mathrm{Cr}^{3+}\) c. \(\mathrm{Pu}^{4+}, \mathrm{Ce}^{4+}, \mathrm{Ti}^{4+}\) d. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)

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