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
The electron-domain and molecular geometries for the given molecules are as follows: a) Tetrahedral electron-domain geometry and tetrahedral molecular geometry. b) Trigonal bipyramidal electron-domain geometry and T-shaped molecular geometry. c) Octahedral electron-domain geometry and square pyramidal molecular geometry. d) Octahedral electron-domain geometry and square planar molecular geometry.

Step by step solution

01

Identify the electron-domain geometry.

With four bonding domains and no nonbonding domains, the central atom has a total of four electron domains. According to the VSEPR model, the electron-domain geometry will be tetrahedral.
02

Determine the molecular geometry.

Since all electron domains are bonding domains, the molecular geometry will be the same as the electron-domain geometry. Hence, the molecular geometry is also tetrahedral. #b) Three bonding domains and two nonbonding domains.#
03

Identify the electron-domain geometry.

With three bonding domains and two nonbonding domains, the central atom has a total of five electron domains. According to the VSEPR model, the electron-domain geometry will be trigonal bipyramidal.
04

Determine the molecular geometry.

As there are two nonbonding electron domains, the molecular geometry will differ from the electron-domain geometry. The nonbonding electron domains will occupy the equatorial positions, while the three bonding domains occupy the remaining positions forming a "T" shape. Hence, the molecular geometry is T-shaped. #c) Five bonding domains and one nonbonding domain.#
05

Identify the electron-domain geometry.

With five bonding domains and one nonbonding domain, the central atom has a total of six electron domains. According to the VSEPR model, the electron-domain geometry will be octahedral.
06

Determine the molecular geometry.

As there is one nonbonding electron domain, the molecular geometry will differ from the electron-domain geometry. The nonbonding electron domain will occupy one position, while the five bonding domains occupy the remaining positions sharing vertices with the nonbonding domain. The shape formed by the five bonding domains is a square pyramid. Hence, the molecular geometry is square pyramidal. #d) Four bonding domains and two nonbonding domains.#
07

Identify the electron-domain geometry.

With four bonding domains and two nonbonding domains, the central atom has a total of six electron domains. According to the VSEPR model, the electron-domain geometry will be octahedral.
08

Determine the molecular geometry.

As there are two nonbonding electron domains, the molecular geometry will differ from the electron-domain geometry. These nonbonding electron domains will occupy two positions, while the four bonding domains occupy the remaining positions, forming a square plane. Hence, the molecular geometry is 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 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) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right),\) ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule?

There are two compounds of the formula Pt \(\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} :\) The compound on the right is called cisplatin, and the compound on the left is called transplatin. (a) Which compound has a nonzero dipole moment? (b) One of these compounds is an anticancer drug, and one is inactive. The anticancer drug works by its chloride ions undergoing a substitution reaction with nitrogen atoms in DNA that are close together, forming a \(\mathrm{N}-\mathrm{Pt}-\mathrm{N}\) angle of about \(90^{\circ} .\) Which compound would you predict to be the anticancer drug?

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2},\) is a fragrant substance used both as a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(I_{2} ?\) Explain your answer.

In the formate ion, \(\mathrm{HCO}_{2}^{-}\) , the carbon atom is the central atom with the other three atoms attached to it. (a) Draw a Lewis structure for the formate ion. (b) What hybridization is exhibited by the C atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

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