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

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. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)

Short Answer

Expert verified
The ions with noble gas electron configurations are \(\mathrm{Sc}^{3+}\), \(\mathrm{Tl}^{+}\), \(\mathrm{Te}^{2-}\), and \(\mathrm{Ba}^{2+}\).

Step by step solution

01

Determine the electron configuration of each ion#a.

To find the electron configuration, subtract the charge of the ion from the atomic number to determine the number of electrons. Then, fill the atomic orbitals according to the Aufbau principle (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.): 1. \(\mathrm{Fe}^{2+}\): 26 - 2 = 24 electrons, configuration: [Ar] 3d^6 4s^0 2. \(\mathrm{Fe}^{3+}\): 26 - 3 = 23 electrons, configuration: [Ar] 3d^5 4s^0 3. \(\mathrm{Sc}^{3+}\): 21 - 3 = 18 electrons, configuration: [Ar] 3d^0 4s^0 4. \(\mathrm{Co}^{3+}\): 27 - 3 = 24 electrons, configuration: [Ar] 3d^6 4s^0
02

Compare ion electron configurations to noble gases#a.

Among these ions, only \(\mathrm{Sc}^{3+}\) has the electron configuration of a noble gas, Argon: [Ar]. b. \(\mathrm{Tl}^{+}, \mathrm{Te}^{2-}, \mathrm{Cr}^{3+}\)
03

Determine the electron configuration of each ion#b.

Same process as before: 1. \(\mathrm{Tl}^{+}\): 81 - 1 = 80 electrons, configuration: [Xe] 4f^14 5d^10 6s^2 6p^0 2. \(\mathrm{Te}^{2-}\): 52 + 2 = 54 electrons, configuration: [Kr] 4d^10 5s^2 5p^6 3. \(\mathrm{Cr}^{3+}\): 24 - 3 = 21 electrons, configuration: [Ar] 3d^3 4s^0
04

Compare ion electron configurations to noble gases#b.

In this set, both \(\mathrm{Tl}^{+}\) and \(\mathrm{Te}^{2-}\) have electron configurations that match noble gases: Xenon and Krypton, respectively. c. \(\mathrm{Pu}^{4+}, \mathrm{Ce}^{4+}, \mathrm{Ti}^{4+}\) We'll skip these ions, as their electron configurations are more complex and beyond the scope of high school chemistry. d. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)
05

Determine the electron configuration of each ion#d.

Same process as before: 1. \(\mathrm{Ba}^{2+}\): 56 - 2 = 54 electrons, configuration: [Kr] 4d^10 5s^2 5p^6 2. \(\mathrm{Pt}^{2+}\): 78 - 2 = 76 electrons, configuration: [Xe] 4f^14 5d^8 6s^0 3. \(\mathrm{Mn}^{2+}\): 25 - 2 = 23 electrons, configuration: [Ar] 3d^5 4s^0
06

Compare ion electron configurations to noble gases#d.

In this set, only \(\mathrm{Ba}^{2+}\) has the electron configuration of a noble gas, Krypton: [Kr]. In summary, the ions with noble gas electron configurations are \(\mathrm{Sc}^{3+}\), \(\mathrm{Tl}^{+}\), \(\mathrm{Te}^{2-}\), and \(\mathrm{Ba}^{2+}\).

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

Use the following data for potassium chloride to estimate \(\Delta E\) for the reaction: $$\mathrm{K}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s) \quad \Delta E=?$$ Lattice energy Ionization energy for \(\mathbf{K}\) Electron affinity of Cl Bond energy of \(\mathrm{Cl}_{2}\) Energy of sublimation for \(\mathrm{K}\) \(-2913 \mathrm{kJ} / \mathrm{mol}\) \(735 \mathrm{kJ} / \mathrm{mol}\) \(1445 \mathrm{kJ} / \mathrm{mol}\) -328 kJ/mol \(154 \mathrm{kJ} / \mathrm{mol}\) 150\. kJ/mol

Use the following data formagnesium fluoride to estimate \(\Delta E\) for the reaction: $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s) \quad \Delta E=?$$ Lattice energy First ionization energy of \(\mathrm{Mg}\) Second ionization energy of \(\mathbf{M g}\) Electron affinity of \(\mathbf{F}\) Bond energy of \(\mathrm{F}_{2}\) Energy of sublimation for \(\mathrm{Mg}\) \(-2913 \mathrm{kJ} / \mathrm{mol}\) \(735 \mathrm{kJ} / \mathrm{mol}\) \(1445 \mathrm{kJ} / \mathrm{mol}\) \(-328 \mathrm{kJ} / \mathrm{mol}\) \(154 \mathrm{kJ} / \mathrm{mol}\) 150\. kJ/mol

Name each of the following compounds: a. \(\mathrm{BaSO}_{3}\) b. \(\mathrm{NaNO}_{2}\) c. \(\mathrm{KMnO}_{4}\) \(\mathbf{d .} \mathbf{K}_{2} \mathrm{Cr}_{2} \mathbf{O}_{7}\)

Arrange the following in order of increasing radius and increasing ionization energy. a. \(\mathrm{N}^{+}, \mathrm{N}, \mathrm{N}^{-}\) b. \(\mathrm{Se}, \mathrm{Se}^{-}, \mathrm{Cl}, \mathrm{Cl}^{+}\) c. \(\mathrm{Br}^{-}, \mathrm{Rb}^{+}, \mathrm{Sr}^{2+}\)

Consider the following: $$\mathrm{Li}(s)+\frac{1}{2} \mathrm{I}_{2}(s) \longrightarrow \mathrm{LiI}(s) \quad \Delta E=-272 \mathrm{kJ} / \mathrm{mol}$$ LiI(s) has a lattice energy of -753 kJ/mol. The ionization energy of \(\mathrm{Li}(g)\) is \(520 . \mathrm{kJ} / \mathrm{mol},\) the bond energy of \(\mathrm{I}_{2}(g)\) is \(151 \mathrm{kJ} /\) mol, and the electron affinity of \(\mathrm{I}(g)\) is \(-295 \mathrm{kJ} / \mathrm{mol} .\) Use these data to determine the energy of sublimation of Li(s).

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