Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter U2.16, Problem 4E

(a)

Interpretation Introduction

Interpretation : Whether the given molecule N2 is a polar covalent or a nonpolar covalent is to be identified.

Concept Introduction :

A covalent bond is formed when there is sharing of electrons between the atoms to fulfill the noble gas configuration. There are two types of covalent bonds; polar covalent bonds and nonpolar covalent bonds. In a polar molecule the bonded pair of electrons is not shared equally among the atoms. If in a molecule the electronegativity of both the atoms is equal or identical, there is no pulling of electrons and they form nonpolar bond.

(a)

Expert Solution
Check Mark

Answer to Problem 4E

N2has nonpolar covalent bond.

Explanation of Solution

If in a molecule the electronegativity of both the atoms is equal or identical, there is no pulling of electrons and they share the electrons equally between them. There is no formation of partial charges and this type of bond is called nonpolar covalent bond. In nitrogen molecule both atoms are same so they have no difference in electronegativity. Hence the bond is nonpolar covalent.

(b)

Interpretation Introduction

Interpretation : Whether the given molecule HF is a polar covalent or a nonpolar covalent is to be identified.

Concept Introduction :

A covalent bond is formed when there is sharing of electrons between the atoms to fulfill the octet configuration. There are two types of covalent bonds; polar covalent bonds and nonpolar covalent bonds. In a polar molecule the bonded pair of electrons is not shared equally among the atoms. If in a molecule the electronegativity of both the atoms is equal or identical, there is no pulling of electrons and they form nonpolar bond.

(b)

Expert Solution
Check Mark

Answer to Problem 4E

HF has polar covalent bond between hydrogen atom and fluorine atom.

Explanation of Solution

In a polar molecule the bonded pair of electrons is not shared equally among the atoms. In HF molecule, the hydrogen and fluorine atom do not share the bonded pair of electrons equally. Fluorine atom with higher electronegativity attracts more electrons than hydrogen atom with low electronegativity. Fluorine has partial negative charge and hydrogen has positive charge. This results in formation of a polar covalent bond.

(c)

Interpretation Introduction

Interpretation : Whether the given molecule F2 is a polar covalent or a nonpolar covalent is to be identified.

Concept Introduction :

A covalent bond is formed when there is sharing of electrons between the atoms to fulfill the noble gas configuration. There are two types of covalent bonds; polar covalent bonds and nonpolar covalent bonds. In a polar molecule the bonded pair of electrons is not shared equally among the atoms.If in a molecule the electronegativity of both the atoms is equal or identical, there is no pulling of electrons and they form nonpolar bond.

(c)

Expert Solution
Check Mark

Answer to Problem 4E

F2has nonpolar covalent bond between two fluorine atoms.

Explanation of Solution

If in a molecule the electronegativity of both the atoms is equal or identical, there is no pulling of electrons and they share the electrons equally between them. There is no formation of partial charges and this type of bond is called nonpolar covalent bond. In fluorine molecule both atoms are same so they have no difference in electronegativity. Hence the bond is nonpolar covalent.

(d)

Interpretation Introduction

Interpretation : Whether the given molecule NO is a polar covalent or a nonpolar covalent is to be identified.

Concept Introduction :

There are two types of covalent bonds; polar covalent bonds and nonpolar covalent bonds. In polar covalent molecule there is partial charge on atoms and in nonpolar covalent molecule ther is no partial charge on atoms.

(d)

Expert Solution
Check Mark

Answer to Problem 4E

NO is a polar covalent molecule.

Explanation of Solution

In NO molecule oxygen is more electronegative than nitrogen atom. It pulls the electrons towards itself and so develops a partial negative charge on itself. Nitrogen has a partial positive charge, so the molecule is polar covalent.

(e)

Interpretation Introduction

Interpretation : Whether the given molecule FCl is a polar covalent or a nonpolar covalent is to be identified.

Concept Introduction :

A covalent bond is formed when there is sharing of electrons between the atoms to fulfill the octet configuration. There are two types of covalent bonds; polar covalent bonds and nonpolar covalent bonds.

(e)

Expert Solution
Check Mark

Answer to Problem 4E

FCl has polar covalent bond.

Explanation of Solution

When two atoms with different electronegativities form a covalent bond with each other, the resultant bond is called polar covalent bond. FCl has polar covalent bond between fluorine and chlorine atom as fluorine is more electronegative than chlorine. So fluorine develops a partial negative charge and chlorine develops a partial positive charge.

Chapter U2 Solutions

Living By Chemistry: First Edition Textbook

Ch. U2.2 - Prob. 3ECh. U2.2 - Prob. 4ECh. U2.2 - Prob. 5ECh. U2.2 - Prob. 6ECh. U2.2 - Prob. 7ECh. U2.3 - Prob. 1TAICh. U2.3 - Prob. 1ECh. U2.3 - Prob. 2ECh. U2.3 - Prob. 3ECh. U2.3 - Prob. 4ECh. U2.3 - Prob. 5ECh. U2.4 - Prob. 1TAICh. U2.4 - Prob. 1ECh. U2.4 - Prob. 2ECh. U2.4 - Prob. 3ECh. U2.4 - Prob. 4ECh. U2.4 - Prob. 5ECh. U2.4 - Prob. 6ECh. U2.4 - Prob. 7ECh. U2.4 - Prob. 8ECh. U2.5 - Prob. 1TAICh. U2.5 - Prob. 1ECh. U2.5 - Prob. 2ECh. U2.5 - Prob. 3ECh. U2.5 - Prob. 4ECh. U2.5 - Prob. 5ECh. U2.5 - Prob. 6ECh. U2.5 - Prob. 7ECh. U2.5 - Prob. 8ECh. U2.5 - Prob. 9ECh. U2.6 - Prob. 1TAICh. U2.6 - Prob. 1ECh. U2.6 - Prob. 2ECh. U2.6 - Prob. 3ECh. U2.6 - Prob. 4ECh. U2.6 - Prob. 5ECh. U2.6 - Prob. 6ECh. U2.6 - Prob. 7ECh. U2.6 - Prob. 8ECh. U2.6 - Prob. 9ECh. U2.6 - Prob. 10ECh. U2.6 - Prob. 11ECh. U2.7 - Prob. 1TAICh. U2.7 - Prob. 1ECh. U2.7 - Prob. 2ECh. U2.7 - Prob. 3ECh. U2.7 - Prob. 4ECh. U2.7 - Prob. 5ECh. U2.8 - Prob. 1TAICh. U2.8 - Prob. 1ECh. U2.8 - Prob. 2ECh. U2.8 - Prob. 3ECh. U2.8 - Prob. 4ECh. U2.8 - Prob. 5ECh. U2.8 - Prob. 6ECh. U2.9 - Prob. 1TAICh. U2.9 - Prob. 1ECh. U2.9 - Prob. 2ECh. U2.9 - Prob. 3ECh. U2.9 - Prob. 4ECh. U2.9 - Prob. 5ECh. U2.9 - Prob. 6ECh. U2.9 - Prob. 7ECh. U2.10 - Prob. 1TAICh. U2.10 - Prob. 1ECh. U2.10 - Prob. 2ECh. U2.10 - Prob. 3ECh. U2.10 - Prob. 4ECh. U2.10 - Prob. 5ECh. U2.10 - Prob. 6ECh. U2.10 - Prob. 7ECh. U2.10 - Prob. 8ECh. U2.10 - Prob. 9ECh. U2.11 - Prob. 1TAICh. U2.11 - Prob. 1ECh. U2.11 - Prob. 2ECh. U2.11 - Prob. 3ECh. U2.11 - Prob. 4ECh. U2.11 - Prob. 5ECh. U2.11 - Prob. 6ECh. U2.11 - Prob. 7ECh. U2.12 - Prob. 1TAICh. U2.12 - Prob. 1ECh. U2.12 - Prob. 2ECh. U2.12 - Prob. 3ECh. U2.12 - Prob. 4ECh. U2.12 - Prob. 5ECh. U2.12 - Prob. 6ECh. U2.12 - Prob. 7ECh. U2.13 - Prob. 1TAICh. U2.13 - Prob. 1ECh. U2.13 - Prob. 2ECh. U2.13 - Prob. 3ECh. U2.13 - Prob. 4ECh. U2.13 - Prob. 5ECh. U2.14 - Prob. 1TAICh. U2.14 - Prob. 1ECh. U2.14 - Prob. 2ECh. U2.14 - Prob. 3ECh. U2.14 - Prob. 4ECh. U2.14 - Prob. 5ECh. U2.14 - Prob. 6ECh. U2.14 - Prob. 7ECh. U2.14 - Prob. 8ECh. U2.15 - Prob. 1TAICh. U2.15 - Prob. 1ECh. U2.15 - Prob. 2ECh. U2.15 - Prob. 4ECh. U2.15 - Prob. 5ECh. U2.15 - Prob. 6ECh. U2.16 - Prob. 1TAICh. U2.16 - Prob. 1ECh. U2.16 - Prob. 2ECh. U2.16 - Prob. 3ECh. U2.16 - Prob. 4ECh. U2.16 - Prob. 6ECh. U2.16 - Prob. 7ECh. U2.17 - Prob. 1TAICh. U2.17 - Prob. 1ECh. U2.17 - Prob. 2ECh. U2.17 - Prob. 3ECh. U2.17 - Prob. 4ECh. U2.17 - Prob. 5ECh. U2.17 - Prob. 6ECh. U2.17 - Prob. 7ECh. U2.17 - Prob. 8ECh. U2.17 - Prob. 9ECh. U2.18 - Prob. 1TAICh. U2.18 - Prob. 1ECh. U2.18 - Prob. 2ECh. U2.18 - Prob. 3ECh. U2.18 - Prob. 4ECh. U2.18 - Prob. 5ECh. U2.18 - Prob. 6ECh. U2.18 - Prob. 7ECh. U2.19 - Prob. 1TAICh. U2.19 - Prob. 1ECh. U2.19 - Prob. 2ECh. U2.19 - Prob. 3ECh. U2.19 - Prob. 4ECh. U2.19 - Prob. 5ECh. U2.19 - Prob. 6ECh. U2.19 - Prob. 7ECh. U2.19 - Prob. 8ECh. U2.19 - Prob. 9ECh. U2.20 - Prob. 1TAICh. U2.20 - Prob. 1ECh. U2.20 - Prob. 2ECh. U2.20 - Prob. 3ECh. U2.20 - Prob. 4ECh. U2.20 - Prob. 5ECh. U2.20 - Prob. 6ECh. U2.20 - Prob. 7ECh. U2.20 - Prob. 8ECh. U2.20 - Prob. 9ECh. U2.21 - Prob. 1TAICh. U2.21 - Prob. 1ECh. U2.21 - Prob. 2ECh. U2.21 - Prob. 3ECh. U2.21 - Prob. 4ECh. U2.21 - Prob. 6ECh. U2 - Prob. SI1RECh. U2 - Prob. SI2RECh. U2 - Prob. SI3RECh. U2 - Prob. SI4RECh. U2 - Prob. SI5RECh. U2 - Prob. SII1RECh. U2 - Prob. SII2RECh. U2 - Prob. SII3RECh. U2 - Prob. SII4RECh. U2 - Prob. SII5RECh. U2 - Prob. SIII1RECh. U2 - Prob. SIII2RECh. U2 - Prob. SIII3RECh. U2 - Prob. SIII4RECh. U2 - Prob. SIII5RECh. U2 - Prob. SIII6RECh. U2 - Prob. SIV1ECh. U2 - Prob. SIV2ECh. U2 - Prob. SIV3ECh. U2 - Prob. SIV4ECh. U2 - Prob. SIV5ECh. U2 - Prob. 1RECh. U2 - Prob. 2RECh. U2 - Prob. 3RECh. U2 - Prob. 4RECh. U2 - Prob. 5RECh. U2 - Prob. 6RECh. U2 - Prob. 7RECh. U2 - Prob. 8RE
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Calorimetry Concept, Examples and Thermochemistry | How to Pass Chemistry; Author: Melissa Maribel;https://www.youtube.com/watch?v=nSh29lUGj00;License: Standard YouTube License, CC-BY