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Answer: Draw a Lewis structure for each molecule and determine its molecular geometry

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 98P Chapter 12

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 98P

PROBLEM 98P

Draw a Lewis structure for each molecule and determine its molecular geometry. What kind of intermolecular forces are present in each substance?

(a) BCl3 (remember that B is a frequent exception to the octet rule)

(b) HCOH (carbon is central; each H and O bonded directly to C)

(c) CS2

(d) NCl3

Step-by-Step Solution:
Step 1 of 3

Solution 98P

Here , we have to draw a Lewis structure for each molecule and determine its molecular geometry. And mention the kind of intermolecular forces that are present in each substance.

(a) BCl3 (remember that B is a frequent exception to the octet rule) :

Lewis structure :

        

Molecular geometry :

The molecular geometry of BCl3 is trigonal planar with symmetric charge distribution around the central atom.

Intermolecular forces :

BCl3 has three polar B-Cl bonds but because of the symmetry of the molecule these bond moments cancel and the BCl3 molecule is nonpolar. Hence the intermolecular forces acting in BCl3 are Dispersion(London) forces.

(b) HCOH (carbon is central; each H and O bonded directly to C) :

Lewis structure :

                

Molecular geometry :

The molecular geometry of HCOH is trigonal planar with asymmetric charge distribution.

Intermolecular forces :

 There is a carbon with two hydrogen and the carbon is also double bonded to an oxygen atom which points straight up on its molecular structure creating a strong polarity and the oxygen is very electronegative. So the intermolecular forces in HCOH is dispersion and dipole-dipole forces.

(c) CS2 :

Lewis structure :

                

Molecular geometry :

In the Lewis structure we see that there are two atoms attached to the central Carbon (C)atom and that there are no lone pairs of electrons (on the central C). the electron clouds of atoms around the C will repel each other. As a result they will be pushed apart giving the CS2 molecule a linear molecular geometry or shape.

Intermolecular forces :

        since the two dipoles between the C and S molecules(C=S) cancel out, giving no net dipole moment, the intermolecular forces present in CS2 is Dispersion(London) forces only.

(d) NCl3 :

Lewis structure :

Molecular geometry :

The molecule has a central nitrogen(N) bonded to three chlorine(Cl) atoms each by a single covalent bond.The other two valence of the nitrogen are a non-bonding pair and give the molecule, trigonal pyramidal shape.

Intermolecular forces :

The intermolecular forces present in the NCl3 molecule are Dipole-dipole and London dispersion force.

Step 2 of 3

Chapter 12, Problem 98P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The full step-by-step solution to problem: 98P from chapter: 12 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. This full solution covers the following key subjects: bcl, Bonded, carbon, central, determine. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The answer to “Draw a Lewis structure for each molecule and determine its molecular geometry. What kind of intermolecular forces are present in each substance?(a) BCl3 (remember that B is a frequent exception to the octet rule)(b) HCOH (carbon is central; each H and O bonded directly to C)(c) CS2(d) NCl3” is broken down into a number of easy to follow steps, and 48 words. Since the solution to 98P from 12 chapter was answered, more than 1912 students have viewed the full step-by-step answer.

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Answer: Draw a Lewis structure for each molecule and determine its molecular geometry