Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 9, Problem 9.98QP

(a)

Interpretation Introduction

Interpretation: The diatomic molecule whose bond order increases with the loss of two electrons is to be found from the given molecules.

Concept introduction: When two atomic orbitals come close to each other they lose their identity and form new pair of orbitals knows as molecular orbitals. Among the two molecular orbitals formed one has energy lower than the atomic orbitals is known as bonding molecular orbital and the other has energy higher than the atomic orbitals and is known as antibonding molecular orbital. The filling electrons in molecular orbitals follow Aufbau’s principle and Hund’s rule.

To determine: If the bond order of diatomic molecule B2 increases after the loss of two electrons.

(a)

Expert Solution
Check Mark

Answer to Problem 9.98QP

Solution

The bond order of B2 decreases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of B2 is,

2s)2*2s)22p)2

The bond order for B2 is calculated by using formula,

BondorderofB2=12(NumberofbondingelectronsinB2NumberofofantibondongelctronsinB2)

The number of bonding electrons in B2=4 .

The number of antibonding electrons in B2=2

Substitute the value of number of electrons in bonding and antibonding orbitals in B2 in the above equation.

Bondorder=12(42)=1

The bond order of B2=1

The B22+ ion is formed by the loss of two electrons from B2 . The electronic configuration of B22+ is,

2s)2*2s)2

The bond order of B22+ is calculated by formula,

BondorderofB22+12(NumberofbondingelectronsinB22+NumberofofantibondongelctronsinB22+)

The number of bonding electrons in B22+=2

The number of antibonding electrons in B22+=2

Substitute the value of number of electrons in bonding and antibonding orbitals in B22+ in the above equation.

Bondorder=12(22)=0

The bond order of B22+=0 .

Hence, the bond order of B2 molecule decreases with the loss of two electrons.

(b)

Interpretation Introduction

To determine: If the bond order of diatomic molecule C2 increases after the loss of two electrons.

(b)

Expert Solution
Check Mark

Answer to Problem 9.98QP

Solution

The bond order of C2 decreases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of C2 is,

2s)2*2s)22p)4

The bond order for C2 is calculated by using formula,

BondorderofC2=12(NumberofbondingelectronsinC2NumberofofantibondongelctronsinC2)

The number of bonding electrons in C2=6

The number of antibonding electrons in C2=2

Substitute the value of number of electrons in bonding and antibonding orbitals in C2 in the above equation.

Bondorder=12(62)=2

The bond order of C2=2

The C22+ ion is formed by the loss of two electrons from C2 . The electronic configuration of C22+ is,

2s)2*2s)22p)2

The bond order for C22+ is calculated by using formula,

BondorderofC22+=12(NumberofbondingelectronsinC22+NumberofofantibondongelctronsinC22+)

The number of bonding electrons in C22+=4

The number of antibonding electrons in C22+=2

Substitute the value of number of electrons in bonding and antibonding orbitals in C22 in the above equation.

Bondorder=12(42)=1

The bond order of C22+=1

Hence, the bond order of C2 molecule decreases with the loss of two electrons.

(c)

Interpretation Introduction

To determine: If the bond order of diatomic molecule N2 after the loss of two electrons.

(c)

Expert Solution
Check Mark

Answer to Problem 9.98QP

Solution

The bond order of N2 decreases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of N2 is,

2s)2*2s)22p)42p)2

The bond order for N2 is calculated by using formula,

BondorderofN2=12(NumberofbondingelectronsinN2NumberofofantibondongelctronsinN2)

The number of bonding electrons in N2=8

The number of antibonding electrons in N2=2

Substitute the value of number of electrons in bonding and antibonding orbitals in N2 in the above equation.

Bondorder=12(82)=3

The bond order of N2=3

The N22+ ion is formed by the loss of two electrons from N2 . The electronic configuration of N22+ is,

2s)2*2s)22p)4

The bond order for N22+   is calculated by using formula,

BondorderofN22+=12(NumberofbondingelectronsofN22+NumberofofantibondongelctronsinN22+)

The number of bonding electrons in N22+=6

The number of antibonding electrons in N22+=2

Substitute the value of number of electrons in bonding and antibonding orbitals in N22 in the above equation.

Bondorder=12(62)=2

The bond order of N22+=2

Hence, the bond order of N2 molecule decreases with the loss of two electrons.

(d)

Interpretation Introduction

To determine: The bond order of diatomic molecule O2 after the loss of two electrons.

(d)

Expert Solution
Check Mark

Answer to Problem 9.98QP

Solution

The bond order of O2 increases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of O2 is,

2s)2*2s)22p)22p)42p*)2

The bond order for O2 is calculated by using formula,

BondorderofO2=12(NumberofbondingelectronsinO2NumberofofantibondongelctronsinO2)

The number of bonding electrons in O2=8

The number of antibonding electrons in O2=4

Substitute the value of number of electrons in bonding and antibonding orbitals in O2 in the above equation.

Bondorder=12(84)=2

The bond order of O2=2

he O22+ ion is formed by the loss of two electrons from N2 . The electronic configuration of O22 is,

2s)2*2s)22p)22p)4

The bond order for O22+ is calculated by using formula,

Bondorderof O22+=12(NumberofbondingelectronsofO22+NumberofofantibondongelctronsinO22+)

The number of bonding electrons in O22+=8 .

The number of antibonding electrons in O22+=2 .

Substitute the value of number of electrons in bonding and antibonding orbitals in O22+ in the above equation.

Bondorder=12(82)=3

The bond order of O22+=3

Hence, the bond order of O2 molecule increases with the loss of two electrons.

Conclusion

Higher the bond order of a molecule higher will be its bond strength. The bond order of oxygen increases while as the bond order of boron, carbon and nitrogen molecules decreases with the loss of two electrons.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 9 Solutions

Chemistry

Ch. 9 - Prob. 9.1VPCh. 9 - Prob. 9.2VPCh. 9 - Prob. 9.3VPCh. 9 - Prob. 9.4VPCh. 9 - Prob. 9.5VPCh. 9 - Prob. 9.6VPCh. 9 - Prob. 9.7VPCh. 9 - Prob. 9.8VPCh. 9 - Prob. 9.9QPCh. 9 - Prob. 9.10QPCh. 9 - Prob. 9.11QPCh. 9 - Prob. 9.12QPCh. 9 - Prob. 9.13QPCh. 9 - Prob. 9.14QPCh. 9 - Prob. 9.15QPCh. 9 - Prob. 9.16QPCh. 9 - Prob. 9.17QPCh. 9 - Prob. 9.18QPCh. 9 - Prob. 9.19QPCh. 9 - Prob. 9.20QPCh. 9 - Prob. 9.21QPCh. 9 - Prob. 9.22QPCh. 9 - Prob. 9.23QPCh. 9 - Prob. 9.24QPCh. 9 - Prob. 9.25QPCh. 9 - Prob. 9.26QPCh. 9 - Prob. 9.27QPCh. 9 - Prob. 9.28QPCh. 9 - Prob. 9.29QPCh. 9 - Prob. 9.30QPCh. 9 - Prob. 9.31QPCh. 9 - Prob. 9.32QPCh. 9 - Prob. 9.33QPCh. 9 - Prob. 9.34QPCh. 9 - Prob. 9.35QPCh. 9 - Prob. 9.36QPCh. 9 - Prob. 9.37QPCh. 9 - Prob. 9.38QPCh. 9 - Prob. 9.39QPCh. 9 - Prob. 9.40QPCh. 9 - Prob. 9.41QPCh. 9 - Prob. 9.42QPCh. 9 - Prob. 9.43QPCh. 9 - Prob. 9.44QPCh. 9 - Prob. 9.45QPCh. 9 - Prob. 9.46QPCh. 9 - Prob. 9.47QPCh. 9 - Prob. 9.48QPCh. 9 - Prob. 9.49QPCh. 9 - Prob. 9.50QPCh. 9 - Prob. 9.51QPCh. 9 - Prob. 9.52QPCh. 9 - Prob. 9.53QPCh. 9 - Prob. 9.54QPCh. 9 - Prob. 9.55QPCh. 9 - Prob. 9.56QPCh. 9 - Prob. 9.57QPCh. 9 - Prob. 9.58QPCh. 9 - Prob. 9.59QPCh. 9 - Prob. 9.60QPCh. 9 - Prob. 9.61QPCh. 9 - Prob. 9.62QPCh. 9 - Prob. 9.63QPCh. 9 - Prob. 9.64QPCh. 9 - Prob. 9.65QPCh. 9 - Prob. 9.66QPCh. 9 - Prob. 9.67QPCh. 9 - Prob. 9.68QPCh. 9 - Prob. 9.69QPCh. 9 - Prob. 9.70QPCh. 9 - Prob. 9.71QPCh. 9 - Prob. 9.72QPCh. 9 - Prob. 9.73QPCh. 9 - Prob. 9.74QPCh. 9 - Prob. 9.75QPCh. 9 - Prob. 9.76QPCh. 9 - Prob. 9.77QPCh. 9 - Prob. 9.78QPCh. 9 - Prob. 9.79QPCh. 9 - Prob. 9.80QPCh. 9 - Prob. 9.81QPCh. 9 - Prob. 9.82QPCh. 9 - Prob. 9.83QPCh. 9 - Prob. 9.84QPCh. 9 - Prob. 9.85QPCh. 9 - Prob. 9.86QPCh. 9 - Prob. 9.87QPCh. 9 - Prob. 9.88QPCh. 9 - Prob. 9.89QPCh. 9 - Prob. 9.90QPCh. 9 - Prob. 9.91QPCh. 9 - Prob. 9.92QPCh. 9 - Prob. 9.93QPCh. 9 - Prob. 9.94QPCh. 9 - Prob. 9.95QPCh. 9 - Prob. 9.96QPCh. 9 - Prob. 9.97QPCh. 9 - Prob. 9.98QPCh. 9 - Prob. 9.99QPCh. 9 - Prob. 9.100QPCh. 9 - Prob. 9.101APCh. 9 - Prob. 9.102APCh. 9 - Prob. 9.103APCh. 9 - Prob. 9.104APCh. 9 - Prob. 9.105APCh. 9 - Prob. 9.106APCh. 9 - Prob. 9.107APCh. 9 - Prob. 9.108APCh. 9 - Prob. 9.109APCh. 9 - Prob. 9.110APCh. 9 - Prob. 9.111APCh. 9 - Prob. 9.112APCh. 9 - Prob. 9.113APCh. 9 - Prob. 9.114APCh. 9 - Prob. 9.115APCh. 9 - Prob. 9.116APCh. 9 - Prob. 9.117APCh. 9 - Prob. 9.118APCh. 9 - Prob. 9.119APCh. 9 - Prob. 9.120APCh. 9 - Prob. 9.121APCh. 9 - Prob. 9.122APCh. 9 - Prob. 9.123APCh. 9 - Prob. 9.124APCh. 9 - Prob. 9.125APCh. 9 - Prob. 9.126APCh. 9 - Prob. 9.127APCh. 9 - Prob. 9.128AP
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.
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
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY