Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 27

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?

Short Answer

Expert verified
A hybrid orbital differs from a pure atomic orbital in that it is a combination of two or more atomic orbitals, resulting in an orbital with a different characteristic shape and energy level. Yes, two \(2p\) orbitals can hybridize to give two hybridized orbitals.

Step by step solution

01

Defining Pure Atomic Orbital

A pure atomic orbital is the region in an atom where an electron with a given energy is likely to be found. They have characteristic shapes, denoted as \(s\), \(p\), \(d\), and \(f\) orbitals, each corresponding to a different shape and energy level.
02

Defining Hybrid Orbital

A hybrid orbital is a kind of atomic orbital that results from the mixing or 'hybridization' of two or more different pure atomic orbitals, resulting in a new, different orbital. They have characteristic shapes different from the pure atomic orbitals, and are often involved in covalent bonding.
03

Explaining the Difference

The key difference therefore between a pure atomic orbital and a hybrid orbital is that while the former exists alone and defines the space in which an electron is likely to be found, the latter is a combination of two or more atomic orbitals that results in an orbital with a different characteristic shape and energy level.
04

Hybridization of \(2p\) Orbitals

Indeed, two \(2p\) orbitals can hybridize to give two hybridized orbitals. In the process known as 'hybridization', different atomic orbitals mix to form new, equivalent hybrid orbitals. Therefore, two \(2p\) orbitals can mix to create two hybrid orbitals.

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

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) has a tendency to lose two protons \(\left(\mathrm{H}^{+}\right)\) and form the carbide ion \(\left(\mathrm{C}_{2}^{2-}\right),\) which is present in a number of ionic compounds, such as \(\mathrm{CaC}_{2}\) and \(\mathrm{MgC}_{2}\). Describe the bonding scheme in the \(\mathrm{C}_{2}^{2}-\) ion in terms of molecular orbital theory. Compare the bond order in \(\mathrm{C}_{2}^{2-}\) with that in \(\mathrm{C}_{2}\)

The molecules cis-dichloroethylene and transdichloroethylene shown on p. 324 can be interconverted by heating or irradiation. (a) Starting with cis-dichloroethylene, show that rotating the \(\mathrm{C}=\mathrm{C}\) bond by \(180^{\circ}\) will break only the pi bond but will leave the sigma bond intact. Explain the formation of trans- dichloroethylene from this process. (Treat the rotation as two, stepwise \(90^{\circ}\) rotations.) (b) Account for the difference in the bond enthalpies for the pi bond (about \(270 \mathrm{~kJ} / \mathrm{mol}\) ) and the sigma bond (about \(350 \mathrm{~kJ} / \mathrm{mol}\) ). (c) Calculate the longest wavelength of light needed to bring about this conversion.

Assume that the third-period element phosphorus forms a diatomic molecule, \(\mathrm{P}_{2}\), in an analogous way as nitrogen does to form \(\mathrm{N}_{2}\). (a) Write the electronic configuration for \(\mathrm{P}_{2}\). Use \(\left[\mathrm{Ne}_{2}\right]\) to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

Draw the Lewis structure of ketene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}\right)\) and describe the hybridization states of the \(\mathrm{C}\) atoms. The molecule does not contain \(\mathrm{O}-\mathrm{H}\) bonds. On separate diagrams, sketch the formation of sigma and pi bonds.

Predict the geometry of these molecules and ion using the VSEPR method: (a) \(\operatorname{HgBr}_{2}\), (b) \(\mathrm{N}_{2} \mathrm{O}\) (arrangement of atoms is NNO), (c) SCN \(^{-}\) (arrangement of atoms is SCN).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

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