Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 67

Briefly compare the VSEPR and hybridization approaches to the study of molecular geometry.

Short Answer

Expert verified
The VSEPR and hybridization theories both tackle molecular geometry, albeit using different lenses. VSEPR theory is based on the repulsion between the pairs of electrons surrounding a central atom, thus predicting the 3D shape of the molecules. In contrast, hybridization is concerned with how atomic orbitals mix and form newly hybridized orbitals, exerting influence over the shape and properties of molecules. Hence, while VSEPR is more spatial, hybridization provides a more atomic perspective.

Step by step solution

01

Define VSEPR

The VSEPR theory is a model used to predict the geometry of individual molecules from the number of electron pairs surrounding central atoms. This theory posits that electron pairs, bound or unbound, in the outermost shells of atoms, or valence shells, extend outwards from the central atom and repel each other. These repulsions result in the unique 3D shape or geometry of a molecule.
02

Define Hybridization

Hybridization in chemistry refers to the idea that atomic orbitals fuse to form newly hybridized orbitals, which influence the shape and reactivity of the molecules. These new orbitals result from the mixing of at least two nonequivalent atomic orbitals (s-orbitals, p-orbitals, d-orbitals) from the same atom.
03

Compare both approaches

While both theories are used in the explanation of molecular geometry, their approaches are different. The VSEPR theory is more about spatial considerations. It looks at electron pairs as they repel each other, leading to the formation of the molecular shape. The Hybridization Theory, on the other hand, is more atomic-level focused. It is concerned with the blending or mixing of individual atomic orbitals to form new hybrid orbitals. In essence, it helps to explain how atomic orbitals blend to accommodate the shape and bonding of molecules.

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

Does the molecule OCS have a higher or lower dipole moment than \(\mathrm{CS}_{2} ?\)

Draw the Lewis structure of mercury(II) bromide. Is this molecule linear or bent? How would you establish its geometry?

Aluminum trichloride \(\left(\mathrm{AlCl}_{3}\right)\) is an electron-deficient molecule. It has a tendency to form a dimer (a molecule made of two \(\mathrm{AlCl}_{3}\) units): $$ \mathrm{AlCl}_{3}+\mathrm{AlCl}_{3} \longrightarrow \mathrm{Al}_{2} \mathrm{Cl}_{6} $$ (a) Draw a Lewis structure for the dimer. (b) Describe the hybridization state of \(\mathrm{Al}\) in \(\mathrm{AlCl}_{3}\) and \(\mathrm{Al}_{2} \mathrm{Cl}_{6}\). (c) Sketch the geometry of the dimer. (d) Do these molecules possess a dipole moment?

What is the hybridization state of \(\mathrm{Si}\) in \(\mathrm{SiH}_{4}\) and in \(\mathrm{H}_{3} \mathrm{Si}-\mathrm{SiH}_{3} ?\)

How does a hybrid orbital differ from a pure atomic orbital? Can two \(2 p\) orbitals of an atom hybridize to give two hybridized orbitals?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

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