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For each polar molecule in the molecule and indicate the

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

Solution for problem 87P Chapter 10

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 87P

PROBLEM 87P

For each polar molecule in Problem draw the molecule and indicate the positive and negative ends of the dipole moment.

Problem

Classify each diatomic molecule as polar or nonpolar.

(a) CO

(b) O2

(c) F2

(d) HBr

Step-by-Step Solution:

Solution 87P

Here, we are going to classify the molecules as polar or nonpolar. Polarity of a molecule can be determined as follows.

Determining the geometry of the molecule.A molecule is said to be polar when the two bonding atoms have sufficiently different electronegativities. Determination of the dipole moment of the molecule.

(a) CO

Step 1: The Lewis structure of the molecule is as follows

The molecule contains three bonding electron pair and 2 lone pair of electrons. Therefore, CO is a linear molecule.

Step 2: Determination of the polarity of the bond.

The electronegativity of C and O are 2.55 and 3.44 respectively

The electronegativity difference between C-O is 1.11. Therefore C-O is polar

Step 3: The molecule has net dipole moment.

Thus, CO is a polar molecule. Thus the structure for CO by depicting the positive and negative ends of the dipole moment.

(b) O2

Step 1: The Lewis structure of the molecule is as follows

The molecule contains two bonding electron pair and 4 lone pair of electrons. Therefore, O2 is a linear molecule.

Step 2: Determination of the polarity of the bond.

The electronegativity of O is 3.44 respectively

The electronegativity difference between O-O is 0.0. Therefore O-O is non polar.

Step 3: The molecule has no net dipole moment.

Thus, O2 is a non polar molecule.

(c) F2

Step 1: The Lewis structure of the molecule is as follows

The molecule contains one bonding electron pair and 6 lone pair of electrons. Therefore, F2 is a linear molecule.

Step 2: Determination of the polarity of the bond.

The electronegativity of F is 4.0 respectively

The electronegativity difference between F-F is 0.0. Therefore F-F is non polar.

Step 3: The molecule has no net dipole moment.

Thus, F2 is a non polar molecule.

(d) HBr

Step 1: The Lewis structure of the molecule is as follows

The molecule contains one bonding electron pair and 3 lone pair of electrons. Therefore, HBr is a linear molecule.

Step 2: Determination of the polarity of the bond.

The electronegativity of H and Br are 2.2 and 2.96  respectively

The electronegativity difference between H-Br is 0.76. Therefore H-Br bond is polar.

Step 3 of 3

Chapter 10, Problem 87P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

The answer to “For each polar molecule in the molecule and indicate the positive and negative ends of the dipole moment. each diatomic molecule as polar or nonpolar.(a) CO(b) O2(c) F2(d) HBr” is broken down into a number of easy to follow steps, and 29 words. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. This full solution covers the following key subjects: molecule, polar, indicate, draw, ends. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. The full step-by-step solution to problem: 87P from chapter: 10 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 87P from 10 chapter was answered, more than 299 students have viewed the full step-by-step answer.

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