Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 37

(a) Does SCl\(_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does BeCl\(_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

Short Answer

Expert verified
(a) Yes, SCl$_{2}$ has a dipole moment, with the net dipole pointing towards the more electronegative chlorine atoms. (b) No, BeCl$_{2}$ does not have a dipole moment due to its linear geometry causing symmetrical bond dipoles that cancel each other.

Step by step solution

01

Draw the Lewis structure for SCl2

Start by drawing the Lewis structure for SCl2: 1. Count the total number of valence electrons: S (6 valence electrons) + 2 Cl (2 x 7 valence electrons) = 20 valence electrons. 2. Write the central atom (S) followed by its surrounding atoms (2 Cl). 3. Distribute the valence electrons between the atoms to form a stable octet (fulfilling the octet rule for S and each Cl atom). 4. The Lewis structure for SCl2 is: Cl - S - Cl | | . . where "-" represents a single bond, and "." represents a lone pair of electrons.
02

Determine the molecular geometry using VSEPR

The number of electron regions around sulfur is 4 (2 bonding regions from the 2 Cl atoms and 2 lone pairs of electrons). According to the VSEPR theory, this gives the molecular geometry a bent/angular shape with a bond angle of approximately 104.5°, similar to that of water (H2O).
03

Determine the polarity and dipole moment of SCl2

SCl2 has a bent molecular geometry, and the electronegativity difference between sulfur (S) and chlorine (Cl) atoms indicates polar bonds. As a result, there is an uneven distribution of electron density that creates a net dipole moment pointing towards the more electronegative chlorine atoms.
04

Draw the Lewis structure for BeCl2

Now, draw the Lewis structure for BeCl2: 1. Count the total number of valence electrons: Be (2 valence electrons) + 2 Cl (2 x 7 valence electrons) = 16 valence electrons. 2. Write the central atom (Be) followed by its surrounding atoms (2 Cl). 3. Distribute the valence electrons between the atoms to fulfill the octet rule (Be can have less than 8 electrons though). 4. The Lewis structure for BeCl2 is: Cl - Be - Cl where "-" represents a single bond.
05

Determine the molecular geometry using VSEPR

There are only 2 electron regions around beryllium, which gives the molecular geometry a linear shape with a bond angle of 180°.
06

Determine the polarity and dipole moment of BeCl2

BeCl2 has a linear molecular geometry. Even though there is an electronegativity difference between beryllium (Be) and chlorine (Cl) atoms, indicating polar bonds, the linear geometry of the molecule causes the bond dipoles to be symmetrical and cancel each other out. Therefore, BeCl2 does not have a net dipole moment. To summarize the results: (a) SCl2 has a dipole moment, with the net dipole pointing towards the more electronegative chlorine atoms. (b) BeCl2 does not have a dipole moment due to its linear geometry which causes symmetrical bond dipoles that cancel each other.

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

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

Indicate the hybridization of the central atom in \((\mathbf{a}) \mathrm{BCl}_{3}$$(\mathbf{b}) \mathrm{AlCl}_{4}^{-},(\mathbf{c}) \mathrm{CS}_{2},(\mathbf{d}) \mathrm{GeH}_{4}\)

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_ \(_{4} )\) according to the following unbalanced reaction: \begin{equation}\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) \end{equation} The O atom and the four \(\mathrm{F}\) atoms in OSF \(_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of OSF_ in which the formal charges of all atoms are zero.(c) Use average bond enthalpies (Table 8.3 ) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

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

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 \(\mathrm{P}-\mathrm{F}\) 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 \((\mathrm{c}) ?(\mathbf{d})\) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

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