Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 12, Problem 12D.7P

(a)

Interpretation Introduction

Interpretation:

The proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital has to be stated.  The proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital has to be stated.  The total probability that the electron is found on N atoms has to be stated.

Concept introduction:

EPR stands for electronic paramagnetic resonance.  It is used to study species that contain unpaired electrons.  EPR can be used to study both solids and liquids.  However, the study of the gas phase samples is difficult due to the free rotation of the molecules.  The g value represents the ease with which the applied field can generate currents through the molecule.

(a)

Expert Solution
Check Mark

Answer to Problem 12D.7P

The proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital is 10.3%_.

The proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital is 38.2%_.

The total probability that the electron is found on N atoms is 48.5%_.

Explanation of Solution

The given hyperfine interaction of 2s orbital of N atom with nucleus is 55.2mT.

The given isotropic hyperfine interaction of 2s orbital of N atom with nucleus is 5.7mT.

So, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital is calculated by the expression given below.

    P(N2s)=Isotropic hyperfine interactionof 2sorbitalHyperfine interaction of 2sorbital

Substitute the values of hyperfine interaction of 2s orbital and isotropic hyperfine interaction of 2s orbital in the above expression.

    P(N2s)=5.7mT55.2mT=0.103=0.103×100%=10.3%_

Therefore, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital is 10.3%_.

The given hyperfine interaction of 2p orbital of N atom with nucleus is 3.4mT.

The given anisotropic hyperfine interaction of 2p orbital of N atom with nucleus is 1.3mT.

So, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital is calculated by the expression given below.

    P(N2p)=Anisotropic hyperfine interactionof 2porbitalHyperfine interaction of 2porbital

Substitute the values of hyperfine interaction of 2p orbital and isotropic hyperfine interaction of 2p orbital in the above expression.

    P(N2p)=1.3mT3.4mT=0.382=0.382×100%=38.2%_

Therefore, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital is 38.2%_.

The total probability that the electron is found on N atoms is calculated by adding P(N2p) and P(N2s) as follow.

    P(N)=P(N2s)+P(N2p)

Substitute the values of P(N2p) and P(N2s) in the above expression

    P(N)=0.103+0.382=0.485=0.485×100%=48.5%_

Therefore, the total probability that the electron is found on N atoms is 48.5%_.

(b)

Interpretation Introduction

Interpretation:

The total probability that the electron is found on O atoms has to be stated. The hybridization ratio of N atom has to be stated. Whether the hybridization predicts that NO2 is angular or not has to be stated.

Concept introduction:

EPR stands for electronic paramagnetic resonance.  It is used to study species that contain unpaired electrons.  EPR can be used to study both solids and liquids.  However, the study of the gas phase samples is difficult due to the free rotation of the molecules.  The g value represents the ease with which the applied field can generate currents through the molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 12D.7P

The total probability that the electron is found on O atoms is 51.5%_.

The hybridization ratio of N atom is 3.7_.

The hybridization predicts that NO2 is angular in shape.

Explanation of Solution

The total probability that the electron is found on N atoms is 0.485.

The total probability that the electron is found on O atoms is calculated by subtracting P(N) from 1  as follow.

    P(O)=1P(N)

Substitute the value of P(N) in the above expression

    P(O)=10.485=0.515=0.515×100%=51.5%_

Therefore, the total probability that the electron is found on O atoms is 51.5%_.

The expression to calculate the hybridization ratio of N atom is given below.

    HybridizationratioofNatom=P(N2p)P(N2s)

Substitute the values of P(N2p) and P(N2s) in the above expression.

    HybridizationratioofNatom=0.3820.103=3.7_

Therefore, the hybridization ratio of N atom is 3.7_.

As the shape of NO2 radical is non-linear, thus, NO2 is angular in shape.

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 12 Solutions

Atkins' Physical Chemistry

Ch. 12 - Prob. 12A.2BECh. 12 - Prob. 12A.3AECh. 12 - Prob. 12A.3BECh. 12 - Prob. 12A.4AECh. 12 - Prob. 12A.4BECh. 12 - Prob. 12A.5AECh. 12 - Prob. 12A.5BECh. 12 - Prob. 12A.6AECh. 12 - Prob. 12A.6BECh. 12 - Prob. 12A.7AECh. 12 - Prob. 12A.7BECh. 12 - Prob. 12A.8AECh. 12 - Prob. 12A.8BECh. 12 - Prob. 12A.9AECh. 12 - Prob. 12A.9BECh. 12 - Prob. 12A.1PCh. 12 - Prob. 12A.3PCh. 12 - Prob. 12B.1DQCh. 12 - Prob. 12B.2DQCh. 12 - Prob. 12B.3DQCh. 12 - Prob. 12B.4DQCh. 12 - Prob. 12B.5DQCh. 12 - Prob. 12B.1AECh. 12 - Prob. 12B.1BECh. 12 - Prob. 12B.2AECh. 12 - Prob. 12B.2BECh. 12 - Prob. 12B.3AECh. 12 - Prob. 12B.3BECh. 12 - Prob. 12B.4AECh. 12 - Prob. 12B.4BECh. 12 - Prob. 12B.5AECh. 12 - Prob. 12B.5BECh. 12 - Prob. 12B.6AECh. 12 - Prob. 12B.6BECh. 12 - Prob. 12B.7AECh. 12 - Prob. 12B.7BECh. 12 - Prob. 12B.8AECh. 12 - Prob. 12B.8BECh. 12 - Prob. 12B.9AECh. 12 - Prob. 12B.9BECh. 12 - Prob. 12B.10AECh. 12 - Prob. 12B.10BECh. 12 - Prob. 12B.11AECh. 12 - Prob. 12B.11BECh. 12 - Prob. 12B.12AECh. 12 - Prob. 12B.12BECh. 12 - Prob. 12B.13AECh. 12 - Prob. 12B.13BECh. 12 - Prob. 12B.14AECh. 12 - Prob. 12B.14BECh. 12 - Prob. 12B.1PCh. 12 - Prob. 12B.2PCh. 12 - Prob. 12B.3PCh. 12 - Prob. 12B.5PCh. 12 - Prob. 12B.6PCh. 12 - Prob. 12B.7PCh. 12 - Prob. 12B.8PCh. 12 - Prob. 12B.9PCh. 12 - Prob. 12C.1DQCh. 12 - Prob. 12C.2DQCh. 12 - Prob. 12C.3DQCh. 12 - Prob. 12C.4DQCh. 12 - Prob. 12C.5DQCh. 12 - Prob. 12C.1AECh. 12 - Prob. 12C.1BECh. 12 - Prob. 12C.2AECh. 12 - Prob. 12C.2BECh. 12 - Prob. 12C.3AECh. 12 - Prob. 12C.3BECh. 12 - Prob. 12C.4AECh. 12 - Prob. 12C.4BECh. 12 - Prob. 12C.5AECh. 12 - Prob. 12C.5BECh. 12 - Prob. 12C.4PCh. 12 - Prob. 12C.5PCh. 12 - Prob. 12C.6PCh. 12 - Prob. 12C.10PCh. 12 - Prob. 12D.1DQCh. 12 - Prob. 12D.2DQCh. 12 - Prob. 12D.1AECh. 12 - Prob. 12D.1BECh. 12 - Prob. 12D.2AECh. 12 - Prob. 12D.2BECh. 12 - Prob. 12D.3AECh. 12 - Prob. 12D.3BECh. 12 - Prob. 12D.4AECh. 12 - Prob. 12D.4BECh. 12 - Prob. 12D.5AECh. 12 - Prob. 12D.5BECh. 12 - Prob. 12D.6AECh. 12 - Prob. 12D.6BECh. 12 - Prob. 12D.1PCh. 12 - Prob. 12D.2PCh. 12 - Prob. 12D.4PCh. 12 - Prob. 12D.5PCh. 12 - Prob. 12D.6PCh. 12 - Prob. 12D.7PCh. 12 - Prob. 12D.8PCh. 12 - Prob. 12.3IACh. 12 - Prob. 12.4IA
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
The Bohr Model of the atom and Atomic Emission Spectra: Atomic Structure tutorial | Crash Chemistry; Author: Crash Chemistry Academy;https://www.youtube.com/watch?v=apuWi_Fbtys;License: Standard YouTube License, CC-BY