Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 76

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?

Short Answer

Expert verified
In the hydrogen fluoride molecule, the hydrogen 1s atomic orbital and the fluorine 2p atomic orbital overlap along the internuclear axis, resulting in the formation of σ bonding and σ* antibonding molecular orbitals. This is because their overlap is symmetric with respect to the internuclear axis. The σ bonding molecular orbital has increased electron density between the hydrogen and fluorine nuclei, while the σ* antibonding molecular orbital has decreased electron density between the hydrogen and fluorine nuclei.

Step by step solution

01

Understand Atomic Orbitals and Molecular Orbitals

Atomic orbitals represent the regions in space where electrons are most likely to be found in atoms. Molecular orbitals represent the regions in space where electrons are most likely to be found in a molecule. When two atoms combine to form a molecule, their atomic orbitals overlap to create molecular orbitals.
02

Recognize Bonding and Antibonding Orbitals

When atomic orbitals overlap, two types of molecular orbitals are created: bonding and antibonding orbitals. Bonding molecular orbitals are created when the overlap of atomic orbitals results in an increase in electron density between the nuclei of the atoms involved. This increased electron density leads to a stronger attractive force between the nuclei, resulting in a stable bound state. Antibonding molecular orbitals are created when the overlap of atomic orbitals results in a decrease in electron density between the nuclei of the atoms involved. This decreased electron density leads to a weaker attractive force between the nuclei, resulting in an unstable, high-energy state.
03

Understand σ and π Molecular Orbitals

Molecular orbitals can be classified as either σ (sigma) or π (pi) based on their symmetry with respect to the internuclear axis - the line connecting the nuclei of the two atoms involved in the bond. σ molecular orbitals have symmetric electron distributions around the internuclear axis, and the orbital can freely rotate around the axis. In contrast, π molecular orbitals have asymmetric electron distributions around the internuclear axis, and the orbital cannot rotate around the axis without breaking the bond.
04

Overlap of Hydrogen 1s and Fluorine 2p Atomic Orbitals

In the hydrogen fluoride molecule, the hydrogen 1s atomic orbital and the fluorine 2p atomic orbital will overlap. Since the hydrogen 1s orbital is spherical and the fluorine 2p orbital is dumbbell-shaped, their overlap can be visualized along the internuclear axis. The resulting molecular orbitals will be symmetric with respect to this axis.
05

Determine the Type of Molecular Orbitals Formed

As the overlap of hydrogen 1s and fluorine 2p atomic orbitals is symmetric with respect to the internuclear axis, the resulting molecular orbitals are σ orbitals. When these orbitals combine, they form a σ bonding molecular orbital with increased electron density between the hydrogen and fluorine nuclei, and a σ* antibonding molecular orbital with decreased electron density between the hydrogen and fluorine nuclei. In conclusion, the hydrogen 1s and fluorine 2p atomic orbitals overlap to form σ bonding and σ* antibonding molecular orbitals in the hydrogen fluoride molecule.

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

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

Draw the Lewis structures for \(\mathrm{SO}_{2}, \mathrm{PCl}_{3}, \mathrm{NNO}, \mathrm{COS},\) and \(\mathrm{PF}_{3} .\) Which of the compounds are polar? Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit \(s p^{3}\) hybridization by the central atom? Which of the compounds have a linear molecular structure?

Determine the molecular structure and hybridization of the central atom \(X\) in the polyatomic ion \(X Y_{3}^{+}\) given the following information: A neutral atom of X contains 36 electrons, and the element Y makes an anion with a \(1-\) charge, which has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6}\).

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and \(\mathrm{NO}\) than the hybrid orbital model?

As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, \(\mathbf{B}_{2}\) a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2} ?\) b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part b) as follows: $$\mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta E=6455 \mathrm{kJ} / \mathrm{mol}$$ How much energy is needed to ionize \(1.5 \mathrm{kg} \mathrm{B}_{2}\) to the desired isoelectronic species?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

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.

Sign-up for free