/*! 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 40 (a) The \(\mathrm{PH}_{3}\) mole... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 40

(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
(a) The polarity of the \(\mathrm{PH}_{3}\) molecule alone does not offer experimental proof that the molecule cannot be planar. It is the presence of a lone pair on the central phosphorus atom that makes the molecule non-planar, which in turn contributes to its polarity. (b) Ozone, \(\mathrm{O}_{3}\), can have a small dipole moment even though all the atoms are the same because of the difference in formal charges resulting from varying bond types (single and double bonds) between the oxygen atoms. This difference in bond types leads to an uneven distribution of charges and, as a result, a molecular dipole moment.

Step by step solution

01

(a) Polarity of \(\mathrm{PH}_{3}\) molecule

In order to determine if the polarity of the \(\mathrm{PH}_{3}\) molecule can be considered as proof for it being non-planar, we first need to understand what causes a molecule to be polar. A molecule is considered polar when there is an uneven distribution of charges, leading to an overall molecular dipole moment.
02

(a) Molecular geometry of \(\mathrm{PH}_{3}\)

The central atom in \(\mathrm{PH}_{3}\) molecule is phosphorus (P) which forms bonds with three hydrogen (H) atoms. Phosphorus has five valence electrons, and it shares one electron with each hydrogen atom to form three single covalent bonds. This leaves a lone pair of electrons on the phosphorus atom, which will affect the shape of the molecule. The electron repulsion due to the lone pair will push the hydrogen atoms downward, resulting in a tetrahedral geometry. The bond angle between the hydrogen atoms is approximately 93.5掳 which deviates from the ideal angle of 109.5掳 in a perfect tetrahedron due to the presence of the lone pair.
03

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

Now that we know the molecular geometry of \(\mathrm{PH}_{3}\), we can analyze its polarity. Since the molecule has a lone pair on the central atom, there will be an uneven distribution of charge, and the molecule will be polar. However, the fact that the molecule is polar alone does not necessarily mean that it cannot be planar. There are polar molecules that are planar as well (e.g., water). In the case of \(\mathrm{PH}_{3}\), it is the tetrahedral geometry due to the presence of a lone pair on the central atom that makes it non-planar, not the fact that it is polar.
04

(a) Conclusion for \(\mathrm{PH}_{3}\)

The polarity of the \(\mathrm{PH}_{3}\) molecule itself cannot offer experimental proof that the molecule cannot be planar. It is the presence of the lone pair on the central atom that makes the molecule non-planar, which in turn contributes to its polarity.
05

(b) Molecular geometry of ozone, \(\mathrm{O}_{3}\)

Ozone, \(\mathrm{O}_{3}\), consists of three oxygen atoms. Each oxygen atom forms a bond with one other oxygen atom, and one of the oxygen atoms is doubly bonded to the central oxygen atom, forming a bent or V-shaped molecular geometry with an approximate bond angle of 116掳.
06

(b) Polarity of ozone, \(\mathrm{O}_{3}\)

In ozone, there is a formal charge on both the terminal oxygen atoms: one being negative and the other positive due to the difference in electronegativity between the singly bonded and doubly bonded oxygen atoms. This difference creates a molecular dipole moment along the axis connecting the central oxygen atom to the singly bonded oxygen atom, making the molecule polar.
07

(b) Conclusion for ozone, \(\mathrm{O}_{3}\)

It is possible for ozone, \(\mathrm{O}_{3}\), to have a small dipole moment even though all the atoms are the same because of the difference in formal charges due to varying bond types (single and double bonds) between the oxygen atoms. This difference in bond types leads to an uneven distribution of charges and, as a result, a molecular 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 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

What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\), (b) HCN, (d) \(\mathrm{TeCl}_{2}\) (c) \(\mathrm{SO}_{3}\)

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

(a) If you combine two atomic orbitals on two different atoms to make a new orbital, is this a hybrid orbital or a molecular orbital? (b) If you combine two atomic orbitals on one atom to make a new orbital, is this a hybrid orbital or a molecular orbital? (c) Does the Pauli exclusion principle (Section 6.7\()\) apply to MOs? Explain.

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) What conditions must be met if a molecule with polar bonds is nonpolar? (b) What geometries will signify nonpolar molecules for \(\mathrm{AB}_{2}, \mathrm{AB}_{3},\) and \(\mathrm{AB}_{4}\) geometries?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

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.