/*! 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 75 Using an MO energy-level diagram... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 75

Using an MO energy-level diagram, would you expect \(\mathrm{F}_{2}\) to have a lower or higher first ionization energy than atomic fluorine? Why?

Short Answer

Expert verified
Using the MO energy-level diagram, F2 would have a lower first ionization energy than atomic fluorine. This is because F2's highest occupied molecular orbital (HOMO) is the 蟺*2p orbital, which has a higher energy than fluorine's 2p orbital. Consequently, it takes less energy to remove an electron from F2.

Step by step solution

01

Recall the Molecular Orbital (MO) theory

The Molecular Orbital theory helps us understand the bonding, anti-bonding, and non-bonding energies in a molecule. In an MO energy-level diagram, the x-axis represents the energy level of the orbitals, and the y-axis represents the positions of the atomic nuclei.
02

Draw the MO energy-level diagram for F2

Consider the electron configurations of fluorine atoms: Fluorine (F): 1s虏 2s虏 2p鈦� Since there are two fluorine atoms in F2, the total number of electrons is 14. The energy order for molecular orbitals in the second period of the periodic table is: 蟽1s < 蟽*1s < 蟽2s < 蟽*2s < 蟺2p < 蟺*2p < 蟽2p < 蟽*2p Start filling the molecular orbitals based on the energy levels, and follow the Pauli Exclusion Principle and Hund's rule: F2 MO electron configuration: (蟽1s)虏 (蟽*1s)虏 (蟽2s)虏 (蟽*2s)虏 (蟺2p)鈦� (蟺*2p)虏
03

Compare the ionization energies

To determine whether F2 has a lower or higher first ionization energy than atomic fluorine, consider the valence electrons in each species: - F2: The highest occupied molecular orbital is 蟺*2p and this is the orbital where electrons would be removed from F2. - Atomic fluorine: The electron configuration of atomic fluorine is 1s虏 2s虏 2p鈦�, so an electron will be removed from the 2p orbital. Now, let's compare the energies: - 蟺*2p in F2: In F2, 蟺*2p is an anti-bonding orbital with higher energy than the bonding 蟺2p orbital. - 2p in atomic Fluorine: In atomic fluorine, the 2p orbital is a valence shell electron with intermediate energy.
04

Conclusion

Since the HOMO (highest occupied molecular orbital) of F2 is the 蟺*2p orbital that has higher energy than fluorine's 2p orbital, it will require less energy to remove an electron from F2. Therefore, F2 has a lower first ionization energy than atomic fluorine.

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

Show how a hydrogen \(1 s\) atomic orbital and a fluorine \(2 p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lower-energy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of \(\mathrm{OH}\). Explain.

Describe the bonding in the first excited state of \(\mathrm{N}_{2}\) (the one closest in energy to the ground state) using the molecular orbital model. What differences do you expect in the properties of the molecule in the ground state as compared to the first excited state? (An excited state of a molecule corresponds to an electron arrangement other than that giving the lowest possible energy.)

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity. a. \(\mathrm{CF}_{4}\) e. \(\mathrm{BeH}_{2}\) i. \(\mathrm{KrF}_{4}\) b. \(\mathrm{NF}_{3}\) f. \(\mathrm{TeF}_{4}\) j. \(\mathrm{SeF}_{6}\) c. \(\mathrm{OF}_{2}\) g. \(\mathrm{AsF}_{5}\) k. \(\mathrm{IF}_{5}\) d. \(\mathrm{BF}_{3}\) h. \(\mathrm{KrF}_{2}\) 1\. \(\mathrm{IF}_{3}\)

In the hybrid orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

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.