/*! 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 25 Give the electron-domain and mol... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 25

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

Short Answer

Expert verified
(a) Electron-domain geometry: tetrahedral; Molecular geometry: tetrahedral. (b) Electron-domain geometry: trigonal bipyramidal; Molecular geometry: T-shaped. (c) Electron-domain geometry: octahedral; Molecular geometry: square pyramidal. (d) Electron-domain geometry: octahedral; Molecular geometry: square planar.

Step by step solution

01

(a) Four bonding domains and no nonbonding domains

In this case, since there are four bonding domains and no nonbonding domains, the electron-domain geometry will be a tetrahedral shape. Since the molecular geometry only accounts for the positions of bonded atoms and all atoms are bonded, the molecular geometry will also be tetrahedral.
02

(b) Three bonding domains and two nonbonding domains

In this situation, with three bonding domains and two nonbonding domains, the electron-domain geometry is trigonal bipyramidal. For the molecular geometry, we must consider only the positions of the bonded atoms. The nonbonding domains take up equatorial positions, so the molecular geometry will be T-shaped, as there will be three bonded atoms along with a 90-degree bond angle.
03

(c) Five bonding domains and one nonbonding domain

In this case, five bonding domains and one nonbonding domain add up to a total of six electron domains. Therefore, the electron-domain geometry is octahedral. The molecular geometry must account only for the positions of the bonded atoms. The nonbonding domain will occupy one position, while the bonded atoms will occupy the other five positions, giving the molecular geometry a square pyramidal shape.
04

(d) Four bonding domains and two nonbonding domains

In this scenario, four bonding domains and two nonbonding domains together make up a total of six electron domains, so the electron-domain geometry will be octahedral, similar to the case of (c). Regarding the molecular geometry of this molecule, the two nonbonding domains will be located on opposite sides of the central atom, while the four bonding domains form a square plane. Consequently, the molecular geometry will be square planar.

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

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

Predict whether each of the following molecules is polar or nonpolar: (a) IF, (b) \(\mathrm{CS}_{2}\) (c) \(\mathrm{SO}_{3}\) (d) \(\mathrm{PCl}_{3}\) (e) \(\mathrm{SF}_{6}\) (f) \(\mathrm{IF}_{5}\)

(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?

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}\) is a planar molecule that has the following carbocarbon bond lengths: (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(C-\) bond length in butadiene \((1.48\) A) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length \((1.54 \hat{\mathrm{A}}) .\) Does this imply that the middle \(\mathrm{C}-\mathrm{Cbond}\) in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\)? (\mathbf{d} ) Based on your answer for part ( c ),discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\) C bond.

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on \(\mathrm{F}\) interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

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.