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 equations corresponding to the following. a. the fourth ionization energy of Se b. the electron affinity of \(\mathrm{S}^{-}\) c. the electron affinity of \(\mathrm{Fe}^{3+}\) d. the ionization energy of \(\mathrm{Mg}\)

Short Answer

Expert verified
a. \(Se^{3+} + IE_{4} \rightarrow Se^{4+} + e^{-}\) b. \(S^{-} + e^{-} \rightarrow S^{2-} + EA\) c. \(Fe^{3+} + e^{-} \rightarrow Fe^{2+} + EA\) d. \(Mg + IE \rightarrow Mg^{+} + e^{-}\)

Step by step solution

01

a. The fourth ionization energy of Se

To find the fourth ionization energy of Se, we need to determine the initial and final states of the selenium atom in this process. After three ionizations, we have the Se³⁺ ion (three electrons removed). The fourth ionization implies the removal of another electron from this ion. The general formula for ionization energy is: Atom/ion + Ionization Energy → Atom/ion⁺ + e⁻ Applying this to Se: Se³⁺ + IE₄ → Se⁴⁺ + e⁻
02

b. The electron affinity of \(\mathrm{S}^{-}\)

Electron affinity refers to the energy change that occurs when an electron is added to a gaseous atom or ion. For this case, we are considering the addition of an electron to a sulfur ion, S⁻. The general formula for electron affinity is: Atom/ion + e⁻ → Atom/ion⁻ + Electron Affinity Applying this to S⁻: S⁻ + e⁻ → S²⁻ + EA
03

c. The electron affinity of \(\mathrm{Fe}^{3+}\)

In this case, we are considering the electron affinity of an iron ion, Fe³⁺. We need to add an electron to this ion. Using the general formula for electron affinity: Fe³⁺ + e⁻ → Fe²⁺ + EA
04

d. The ionization energy of \(\mathrm{Mg}\)

For the ionization energy of magnesium, Mg, we need to remove one electron from the gaseous atom. Using the general formula for ionization energy: Mg + IE → Mg⁺ + e⁻

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

Complete and balance the equations for the following reactions. a. \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow\) b. \(\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow\)

Three elements have the electron configurations \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}\), \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\), and \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}\). The first ionization energies of these elements (not in the same order) are \(0.419,0.735\), and \(1.527 \mathrm{MJ} / \mathrm{mol}\). The atomic radii are \(1.60,0.98\), and \(2.35 \AA\). Identify the three elements, and match the appropriate values of ionization energy and atomic radius to each configuration.

Francium, Fr, is a radioactive element found in some uranium minerals and is formed as a result of the decay of actinium. a. What are the electron configurations of francium and its predicted most common ion? b. It has been estimated that at any one time, there is only one (1.0) ounce of francium on earth. Assuming this is true, what number of francium atoms exist on earth? c. The longest-lived isotope of francium is \({ }^{223} \mathrm{Fr}\). What is the total mass in grams of the neutrons in one atom of this isotope?

Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

Identify the following elements. a. An excited state of this element has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{5} 3 s^{1}\). b. The ground-state electron configuration is \([\mathrm{Ne}] 3 s^{2} 3 p^{4}\). c. An excited state of this element has the electron configuration \([\mathrm{Kr}] 5 s^{2} 4 d^{6} 5 p^{2} 6 s^{1}\) d. The ground-state electron configuration contains three unpaired \(6 p\) electrons.

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