/*! 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 52 Show how a hydrogen \(1 s\) atom... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 52

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 (HF) molecule, the hydrogen 1s atomic orbital and the fluorine half-filled 2pz atomic orbital overlap to form bonding and antibonding molecular orbitals, which are both \(\sigma\) (sigma) molecular orbitals. This overlap is symmetric around the bond axis, resulting in a strong covalent bond that stabilizes the hydrogen fluoride molecule.

Step by step solution

01

Understand orbital overlap

Orbital overlap is the process by which atomic orbitals of two atoms combine to form molecular orbitals in a new molecule. In the formation of a covalent bond, the atomic orbitals of the bonding atoms overlap, leading to the sharing of the electron pair between atoms. This results in a region of increased electron density between the atoms, creating a stable molecule.
02

Hydrogen 1s atomic orbital and Fluorine 2p atomic orbital

The hydrogen atom has only one electron, which occupies the lowest energy 1s atomic orbital. The fluorine atom has nine electrons, with two electrons occupying the 1s atomic orbital, two electrons occupying the 2s atomic orbital, and five electrons occupying the 2p atomic orbital. In the hydrogen fluoride molecule (HF), one electron from the hydrogen 1s orbital and one electron from the fluorine half-filled 2pz atomic orbital will combine to form a new molecular orbital.
03

Formation of bonding and antibonding molecular orbitals

When two atomic orbitals overlap, two new molecular orbitals are formed: a bonding molecular orbital and an antibonding molecular orbital. The bonding molecular orbital is formed when there's constructive interference between overlapping orbitals, resulting in a region of higher electron density and lower energy. Electrons in the bonding orbital stabilize the molecule. The antibonding molecular orbital is formed when there's destructive interference between overlapping orbitals, resulting in a region of lower electron density and higher energy. Electrons in the antibonding orbital destabilize the molecule.
04

Type of molecular orbitals formed in hydrogen fluoride

When hydrogen's 1s atomic orbital and fluorine's 2p atomic orbital overlap, a \(\sigma\) (sigma) type molecular orbital is formed. The reason for this is that a \(\sigma\) molecular orbital is symmetric around the bond axis, while a \(\pi\) (pi) molecular orbital is asymmetric with respect to the bond axis. In the hydrogen fluoride molecule, since the overlapping orbitals are symmetric around the bond axis, the resulting molecular orbitals will be \(\sigma\) molecular orbitals.
05

Conclusion

In the hydrogen fluoride (HF) molecule, the hydrogen 1s atomic orbital and the fluorine 2p atomic orbital overlap to form bonding and antibonding molecular orbitals, which are both \(\sigma\) (sigma) molecular orbitals. This overlap is symmetric around the bond axis, resulting in a strong covalent bond that stabilizes 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.

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

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?

As compared with \(\mathrm{CO}\) and \(\mathrm{O}_{2}, \mathrm{CS}\) and \(\mathrm{S}_{2}\) are very unstable molecules. Give an explanation based on the relative abilities of the sulfur and oxygen atoms to form \(\pi\) bonds.

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.

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{aligned} \mathrm{NCl}_{3}(g) & \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g) & \Delta H &=375 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{ONCl}(g) & \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g) & \Delta H &=158 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Rationalize the difference in the values of \(\Delta H\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double and triple bond cannot rotate about the internuclear axis unless the bond is broken. Why?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic 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.