Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 70

What hybridization do you expect for the atom indicated in red in each of the following species? (a) \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-} ;\) (b) \(\mathrm{PH}_{4}^{+}\) (c) \(\mathrm{AlF}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}^{+}\)

Short Answer

Expert verified
The hybridizations for the atoms indicated in red in each species are: (a) sp2 (b) sp3 (c) sp2 (d) sp2

Step by step solution

01

(a) CH3CO2-

For this species, we need to determine the hybridization of the carbon atom in red. To do this, we need to consider the electron domains around the carbon atom. The central carbon atom is bonded to 3 atoms (one carbon and two oxygen atoms) and has no lone pairs of electrons. This results in 3 electron domains. Thus, the hybridization of this carbon atom is sp2.
02

(b) PH4+

For this species, we need to determine the hybridization of the phosphorus atom. To do this, we need to consider the electron domains around the phosphorus atom. The central phosphorus atom is bonded to 4 atoms (four hydrogen atoms) and has no lone pairs of electrons. This results in 4 electron domains. Therefore, the hybridization of the phosphorus atom is sp3.
03

(c) AlF3

For this species, we need to determine the hybridization of the aluminum atom. To do this, we need to consider the electron domains around the aluminum atom. The central aluminum atom is bonded to 3 atoms (three fluorine atoms) and has no lone pairs of electrons. This leads to 3 electron domains. Consequently, the hybridization of the aluminum atom is sp2.
04

(d) H2C=CH-CH2+

For this species, we need to determine the hybridization of the carbon atom in red. To do this, we need to consider the electron domains around the carbon atom. The carbon atom in red is bonded to 3 atoms (two carbon atoms and one hydrogen atom) and has no lone pairs of electrons. There are 3 electron domains around it. Therefore, the hybridization of this carbon atom is sp2.

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

(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}^{2-} ?\) For those ions that are paramagnetic, determine the number of unpaired electrons.

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$\mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6}\), you run into a problem. What is it? (c) What could you do to resolve the difficulty in part \((\mathrm{b}) ?(\mathbf{d})\) The molecule \(\mathrm{IF}_{7}\) has a pentagonalbipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\).

(a) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+} ?(\mathbf{b})\) What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger O \(-\) O bond than \(\mathrm{O}_{2}\) itself.

(a) Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{BF}_{3}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Are any valence atomic orbitals of \(\mathrm{B}\) left unhybridized? If so, how are they oriented relative to the hybrid orbitals?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Kinetics

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