Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 18

a) Based on its position in the periodic table, predict the valence shell, core charge, and number of valence electrons for \(\mathrm{Rb}\) and add these values to Table 1 . b) Using the shell model and referring to the Coulombic Potential Energy relationship (equation in Model 1, CA 3), explain clearly how the IE \(_{1}\) for \(\mathrm{Rb}\) is consistent with your answer to part a.

Short Answer

Expert verified
The valence shell for Rubidium (Rb) is the 5th shell, its core charge is 1, and the number of valence electrons is 1. In reference to the Coulombic Potential Energy, Rb's ionization energy aligns with these values as the larger valence shell and low core charge mean less energy is required to remove an electron.

Step by step solution

01

Determine the Valence Shell, Core Charge, and Number of Valence Electrons

Rubidium (Rb) is located in Group 1, period 5 of the periodic table. Its electron configuration is \(2,8,18,8,1\). The outermost shell with electrons, in this case, is the 5th shell and this is the valence shell for Rb. The core charge is the atomic number minus the inner shell electrons, which in this case is \(37-36=1\). The number of valence electrons for elements in Group 1 is 1, hence Rb has 1 valence electron.
02

Understand Ionization Energy (IE) and Coulombic Potential Energy Relationship

First Ionization Energy (IE) is the energy required to remove 1 valence electron from an atom, turning it into a positive ion. The Coulombic Potential Energy relationship states that the potential energy of two charged particles is directly proportional to the product of their charges and inversely proportional to the distance between them. The greater the charge and the closer the distance, the greater the interaction and the higher the energy.
03

Explain the Consistency of Ionization with part A

For Rubidium, the ionization energy corresponds with the Valence shell which is the 5th shell, a larger shell number means the electron is further from the nucleus, thus less ionization energy will be required to remove the electron. This aligns with the Coulombic Potential Energy relationship where potential energy is inversely proportional to the distance. And given that Rubidium has a core charge of 1, the positive charge on the nucleus is relatively low, which means there is less attraction holding the valence electron close, hence less ionization energy is required to remove it.

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.

Start your free trial

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

Based on its position in the periodic table, what is the valence shell and what is the core charge for C? Explain your reasoning.

How does the core charge for Na compare to the core charge for \(\mathrm{Li} ?\)

Assuming that the valence shells of \(\mathrm{Li}\) and \(\mathrm{Be}\) are at approximately the same distance from their nuclei, explain how the core charges of \(\mathrm{Li}\) and \(\mathrm{Be}\) are consistent with the \(\mathrm{IE}_{1}\) values for \(\mathrm{Li}(0.52 \mathrm{MJ} / \mathrm{mole})\) and \(\mathrm{Be}(0.90 \mathrm{MJ} / \mathrm{mole})\).

Within our model and referring to the Coulombic Potential Energy expression, explain why the \(\mathrm{IE}_{1}\) increases from left to right across a row of the periodic table.

Complete the following table: $$ \begin{array}{|c|c|c|c|c|} \hline \text { Atom } & \begin{array}{c} \text { Total number } \\ \text { of electrons } \end{array} & \begin{array}{c} \text { Number of } \\ \text { valence shell } \\ \text { electrons } \end{array} & \begin{array}{c} \text { Number of } \\ \text { inner shell } \\ \text { electrons } \end{array} & \text { Core Charge } \\ \hline \text { H } & 1 & & & \\ \hline \text { He } & & & & \\ \hline \text { Li } & & & & \\ \hline \end{array} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation
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