Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 42

Predict whether each of the following molecules is polar or nonpolar: \((\mathbf{a}){CCl}_{4},(\mathbf{b}) \mathrm{NH}_{3},(\mathbf{c}) \mathrm{SF}_{4},(\mathbf{d}) \mathrm{XeF}_{4},(\mathbf{e}) \mathrm{CH}_{3} \mathrm{Br}\)

Short Answer

Expert verified
In summary: - \(CCl_{4}\) is nonpolar - \(NH_{3}\) is polar - \(SF_{4}\) is polar - \(XeF_{4}\) is nonpolar - \(CH_{3}Br\) is polar

Step by step solution

01

Determine the Shape and Geometry of Each Molecule

To determine whether the molecules are polar or nonpolar, we first need to know their shape and geometry. We can find the shape and geometry by considering the central atom's valence electrons and creating a Lewis structure for each molecule.
02

Analyzing \((\mathbf{a}){CCl}_{4}\)

For CCl4, the central atom is Carbon (C) and is surrounded by four Chlorine atoms. The CCl4 molecule has four C-Cl bonds with identical bond dipoles, and it has a tetrahedral shape. Since the bond dipoles cancel out in this symmetrical shape, it has no net molecular dipole moment, and CCl4 is a nonpolar molecule.
03

Analyzing \(\mathrm{NH}_{3}\)

For NH3, the central atom is Nitrogen (N) and is surrounded by three Hydrogen atoms and has one lone pair of electrons. The NH3 molecule has three N-H bonds with non-zero bond dipoles, and it has a trigonal pyramidal shape. In this case, the bond dipoles do not cancel, leading to a net molecular dipole moment. Therefore, NH3 is a polar molecule.
04

Analyzing \(\mathrm{SF}_{4}\)

For SF4, the central atom is Sulfur (S), surrounded by four Fluorine atoms and has one lone pair of electrons. The SF4 molecule has four S-F bonds with non-zero bond dipoles, and it has a seesaw shape. In this case, the bond dipoles do not cancel, resulting in a net molecular dipole moment. Therefore, SF4 is a polar molecule.
05

Analyzing \(\mathrm{XeF}_{4}\)

For XeF4, the central atom is Xenon (Xe), surrounded by four Fluorine atoms and has two lone pairs of electrons. The XeF4 molecule has four Xe-F bonds with non-zero bond dipoles, and it has a square planar shape. In this case, bond dipoles cancel out due to the symmetrical shape, and there is no net molecular dipole moment, making XeF4 a nonpolar molecule.
06

Analyzing \(\mathrm{CH}_{3}\mathrm{Br}\)

For CH3Br, the central atom is Carbon (C), surrounded by three Hydrogen atoms and one Bromine atom. The CH3Br molecule has three C-H bonds and one C-Br bond, with different bond dipoles. Its geometry is tetrahedral. However, due to the presence of the C-Br bond, the bond dipoles do not cancel, resulting in a net molecular dipole moment, making CH3Br a polar molecule.
07

Final Results

Based on our analysis: - CCl4 is nonpolar - NH3 is polar - SF4 is polar - XeF4 is nonpolar - CH3Br is polar

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) Is the molecule BF \(_{3}\) polar or nonpolar? (b) If you react BF \(_{3}\) to make the ion \(\mathrm{BF}_{3}^{2-}\) , is this ion planar? (c) Does the molecule BF\(_{2}\)Cl have a dipole moment?

Consider the following \(\mathrm{XF}_{4}\) ions: \(\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+},\) and \(\mathrm{AlF}_{4}^{-}\) (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electron-domain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry (d) Which of the ions will exhibit a see-saw molecular geometry?

(a) An AB \(_{2}\) molecule is linear. How many non bonding electron pairs are around the A atom from this information? (b) How many non bonding electrons surround the Xe in \(\mathrm{XeF}_{2} ?(\mathbf{c})\) Is XeF \(_{2}\) linear?

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

Would you expect the nonbonding electron-pair domain in \(\mathrm{NH}_{3}\) to be greater or less in size than the corresponding one in \(\mathrm{PH}_{3}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical 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.

Sign-up for free