Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) The \(\mathrm{PH}_{3}\) molecule is polar. Does this offer experimental proof that the molecule cannot be planar? Explain. (b) It turns out that ozone, \(\mathrm{O}_{3}\), has a small dipole moment. How is this possible, given that all the atoms are the same?

Short Answer

Expert verified
The polarity of the $\mathrm{PH}_{3}$ molecule suggests that it cannot be planar, as planar and symmetric molecules have their individual dipole moments canceling each other, resulting in nonpolar molecules. Ozone ($\mathrm{O}_{3}$) has a small dipole moment despite having the same type of atom due to its bent structure, which causes an unsymmetrical charge distribution within the molecule. The electronic difference between the single and double bonds creates overall molecular polarity, leading to a small dipole moment.

Step by step solution

01

Understanding molecular polarity

Polarity comes from the distribution of electric charge in the molecule. A molecule has a dipole moment if there is an overall charge separation across the atoms in the molecule, creating one side with an overall partial positive charge and another side with a partial negative charge.
02

Molecule structure and plane

A molecule is considered planar if all its atoms lie in the same plane. In the case of \(\mathrm{PH}_{3}\), the central atom is Phosphorus (P), which is bonded to three Hydrogen (H) atoms. If the molecule is planar, then these atoms should be positioned in one plane.
03

(a) Polarity and planar structure of \(\mathrm{PH}_{3}\)

The \(\mathrm{PH}_{3}\) molecule is polar, which means there is a net dipole moment. If the molecule were planar and symmetric, the dipole moments of individual bonds would cancel each other out, making the molecule nonpolar. The fact that \(\mathrm{PH}_{3}\) is polar suggests that the molecule cannot be planar because, in a planar configuration, the dipole moments would cancel each other out, and the molecule would be nonpolar.
04

(b) Ozone (\(\mathrm{O}_{3}\)) molecule's structure and polarity

In the case of ozone, the molecule is composed of three Oxygen atoms. It has a bent structure with a bond angle of approximately 116° and a bond length of about 128 pm. Despite having the same type of atom, ozone has a small dipole moment because of its bent structure. The central Oxygen atom shares a coordinate bond with one neighboring Oxygen atom and forms a double bond with the other Oxygen atom. This creates an unsymmetrical charge distribution within the molecule. The electronic difference between the single and double bonds causes overall molecular polarity, leading to a small dipole moment.

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

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN}\), and \(\mathrm{CN}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

Suppose that silicon could form molecules that are precisely the analogs of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). How would you describe the bonding about \(\mathrm{Si}\) in terms of hydrid orbitals? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

Consider the molecule \(\mathrm{PF}_{4} \mathrm{Cl}\). (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a P \(-F\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl}\). How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\), (b) \(\mathrm{HCN}\), (c) \(\mathrm{H}_{2} \mathrm{C}_{2}\), (d) \(\mathrm{CH}_{3} \mathrm{~F}\) ?

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3}\), (b) \(\mathrm{CH}_{3}^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{3}^{-}\), (e) \(\mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

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