Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

Short Answer

Expert verified
(a) The physical basis for the VSEPR model is the repulsion between negatively charged valence shell electron pairs, which arrange themselves to maintain a maximum distance between them, determining the three-dimensional shape of molecules. (b) Treating double and triple bonds as a single electron domain in the VSEPR model is justified because they have similar spatial constraints and repulsive force effects upon molecular geometry as a single bond does, despite having additional electron pairs.

Step by step solution

01

(a) Explanation of VSEPR Model's Physical Basis

The physical basis for the VSEPR model lies in the fact that electron pairs in the valence shell of an atom will repel each other due to their negative charges. Electrons being negatively charged, they will repel each other in order to minimize electrostatic force or repulsion. As a result, these valence shell electron pairs will arrange themselves in such a way to maintain a maximum distance between them, leading to distinct molecular geometries. Thus, the VSEPR model is built upon the concept of repulsion between valence shell electron pairs, which determines the three-dimensional shape of molecules.
02

(b) Justification for Treating Double and Triple Bonds as a Single Electron Domain

Although double and triple bonds involve more than one electron pair, they are still considered a single electron domain in the VSEPR model. There are two main reasons for this: 1. Molecular geometry is primarily influenced by repulsions between electron pairs, and since the electron pairs involved in bonding are usually closer to each other than to other nearby electron pairs, the repulsive force their interaction creates will tend to be more significant, so the actual repulsion due to other electron pairs in double or triple bonds becomes less critical. 2. In double or triple bonds, the additional electron pairs are part of the same bonding structure and are shared between the same two atoms, so they tend to be concentrated in the same region of space. This means that they effectively act as a single electron domain for the purposes of considering repulsions and molecular geometry determination. In summary, counting double or triple bonds as a single electron domain in the VSEPR model is justified because they have similar spatial constraints and repulsive force effects upon molecular geometry as a single bond does, despite having additional electron pairs.

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 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

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion, and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\) (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher- energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}^{-}\) localized or delocalized?

(a) Write a single Lewis structure for \(\mathrm{SO}_{3}\), and determine the hybridization at the \(\mathrm{S}\) atom. (b) Are there other equivalent Lewis structures for the molecule? (c) Would you expect \(\mathrm{SO}_{3}\) to exhibit delocalized \(\pi\) bonding? Explain.

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.27 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

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