/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Does the minimization of electro... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 33

Does the minimization of electron-electron repulsions correlate with Hund's rule?

Short Answer

Expert verified
Yes, the minimization of electron-electron repulsions correlates with Hund's rule. Hund's rule ensures that electrons fill degenerate orbitals one at a time with the same spin orientation, thereby maximizing the total spin of the system before pairing up electrons. This arrangement effectively minimizes electron-electron repulsions, as two electrons in the same orbital have increased repulsion due to their close proximity. Distributing electrons among the available orbitals before pairing them reduces these repulsive forces, resulting in the lowest energy and most stable configuration for the atom.

Step by step solution

01

Understand electron-electron repulsions and Hund's rule

Electron-electron repulsions occur between negatively charged electrons in the same atom. In order to minimize these repulsions, electrons should be distributed among available orbitals in an organized manner. Hund's rule is one of the guiding principles in electron configuration, which states that for a given electron configuration, the term with the maximum spin multiplicity has the lowest energy.
02

Electron configuration in an atomic system

According to the Aufbau principle, electrons populate orbitals in order of increasing energy levels. Each orbital can accommodate at most 2 electrons, each with opposite spins denoted by +1/2 and -1/2. When there are multiple degenerate orbitals available (orbitals with the same energy level), electrons start to fill the orbitals one at a time with the same spin orientation. This arrangement abides by Hund's rule as it maximizes the total spin of the system before pairing up electrons in the same orbital.
03

Explanation of Hund's rule's correlation with minimization of electron-electron repulsions

As electrons fill degenerate orbitals one at a time with the same spin orientation, they remain unpaired for as long as possible, which effectively minimizes electron-electron repulsions. Two electrons in the same orbital have increased electron-electron repulsion due to their close proximity. By distributing the electrons among the available orbitals before pairing them, the electron-electron repulsions are reduced. So, the minimization of electron-electron repulsions does correlate with Hund's rule, as Hund's rule ensures that the electrons are arranged in a way that reduces repulsive forces between them, ultimately resulting in the lowest energy (and most stable) configuration for the atom.

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 91影视!

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 ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ $$ \begin{array}{ll} \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\). c. Which one of the four ions has the greatest electron affinity? Explain. d. List the four aluminum ions given in order of increasing size, and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{~s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms? a. The principal quantum number completely determines the energy of a given electron. b. The angular momentum quantum number, \(\ell\), determines the shapes of the atomic orbitals. c. The magnetic quantum number, \(m_{\ell}\), determines the direction that the atomic orbitals point in space.

Predict the atomic number of the next alkali metal after francium and give its ground-state electron configuration.

How many electrons in an atom can have the designation \(1 p\), \(6 d_{x^{2}-y^{2}}, 4 f, 7 p_{y}, 2 s, n=3 ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

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.