ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 11, Problem 11.5IA
Interpretation Introduction

Interpretation:

The bond lengths of HgCl2,HgBr2 and HgI2 have to be stated.

Concept introduction:

A rotational spectrum is a spectrum in which only the rotation of molecule changes.  For a molecule to exhibit pure rotational spectrum, the molecule must possess a permanent electric dipole moment.  The molecules possessing dipole moment is considered as a handle that stirs the electromagnetic field into oscillation.  The energy for a rotational spectrum relates the quantum number and rotational constant to give the expression, E=B˜J(J+1).  The rotational constant further gives a relation with the moment of inertia of the molecule.

Expert Solution & Answer
Check Mark

Answer to Problem 11.5IA

s

The value of bond lengths of HgCl2,HgBr2 and HgI2 are 3.486×1010 m_,4.0077×1010 m_ and 4.57×1010 m_.

Explanation of Solution

The Placzek-Teller relation is shown below.

  δ=(32B˜kThc)1/2        (1)

Where,

  • B˜ is the rotational constant.
  • k is the Boltzmann constant.
  • T is the temperature.
  • h is the Planck’s constant.
  • c is the speed of light.

The value of rotational constant helps in determining the moment of inertia of a molecule as shown below.

  B˜=h2Ic        (2)

Substitute equation (2) in equation (1).

  δ=(32kThc×(h8π2Ic))1/2=(4kTπ2Ic2)1/2

Rearrange the above equation for value of inertia of molecule.

  δ=(4kTπ2Ic2)1/2δ2=(4kTπ2Ic2)

The equation for moment of inertia is shown below.

  I=4kTπ2δ2c2        (3)

For HgCl2.

Convert 282°C in K as shown below.

T(K)=T(°C)+273=282+273=555K

Substitute the value of δ as 23.8 cm1, c as 2.998×1010 cm s1 T as 555K and k as 1.38×1023JK1 in equation (3)

  I=4×1.38×1023J K1×555K(3.14)2×(23.8cm1)2×(2.998×1010cms1)2=3.06×10209.8596×566.44×8.988×1020kgm2=6.089×1045kgm2

The moment of inertia is given by the equation as shown below.

  I=μR2        (4)

Where,

  • μ is the reduced mass
  • R is the bond length of the molecule.

From the masses of the molecules, the reduced mass is calculated as shown below.

  μ=mambma+mb        (5)

Where,

  • ma is the mass of first atom
  • mb is the mass of second atom

For HgCl2, substitute the value of ma as 200.59 g/mol and mb as 35.5g/mol and in equation (5).

  μ=200.59 g/mol×35.5 g/mol200.59g/mol+35.5 g/mol=7120.945236.09 g/mol=30.161g/mol

Convert the reduced mass in kg as shown below.

  1 g/mol =103 kg6.022×1023/mol30.161 g/mol=30.161 g/mol×103 kg6.022×1023 /mol=5.008×1026 kg

Substitute μ as 5.008×1026 kg and I as 6.089×1045 kg m2 in equation (4) as shown below.

  6.089×1045kg m2=5.008×1026 kg×R2R2=1.21×1019 m2R=3.486×1010 m_

Therefore, the value of bond length of HgCl2 is 3.486×1010 m_.

For HgBr2.

Convert 282°C in K as shown below.

T(K)=T(°C)+273=292+273=565K

Substitute the value of δ as 15.2 cm1, c as 2.998×1010 cm s1 T as 565K and k as 1.38×1023JK1 in equation (3)

  I=4×1.38×1023JK1×565K(3.14)2×(15.2cm1)2×(2.998×1010cms1)2=3.118×10209.8596×231.04×8.988×1020kgm2=1.524×1044kgm2

The moment of inertia is given by the equation as shown below.

  I=μR2        (4)

Where,

  • μ is the reduced mass
  • R is the bond length of the molecule.

From the masses of the molecules, the reduced mass is calculated as shown below.

  μ=mambma+mb        (5)

Where,

  • ma is the mass of first atom
  • mb is the mass of second atom

For HgBr2, substitute the value of ma as 200.59g/mol and mb as 79.9g/mol and in equation (5).

  μ=200.59 g/mol×79.9 g/mol200.59g/mol+79.9 g/mol=16027.141280.49 g/mol=57.139g/mol

Convert the reduced mass in kg as shown below.

  1 g/mol =103 kg6.022×1023/mol57.139 g/mol=57.139 g/mol×103 kg6.022×1023 /mol=9.488×1026 kg

Substitute μ as 9.488×1026 kg and I as 1.524×1044 kg m2 in equation (4) as shown below.

  1.524×1044kg m2=9.488×1026 kg×R2R2=1.606×1019 m2R=4.0077×1010 m_

Therefore, the value of bond length of HgBr2 is 4.0077×1010 m_.

For HgI2.

Convert 282°C in K as shown below.

T(K)=T(°C)+273=292+273=565K

Substitute the value of δ as 11.4 cm1, c as 2.998×1010 cm s1 T as 565K and k as 1.38×1023JK1 in equation (3)

  I=4×1.38×1023JK1×565K(3.14)2×(11.4cm1)2×(2.998×1010cms1)2=3.118×10209.8596×129.96×8.988×1020kgm2=2.704×1044kgm2

The moment of inertia is given by the equation as shown below.

  I=μR2        (4)

Where,

  • μ is the reduced mass
  • R is the bond length of the molecule.

From the masses of the molecules, the reduced mass is calculated as shown below.

  μ=mambma+mb        (5)

Where,

  • ma is the mass of first atom
  • mb is the mass of second atom

For HgI2, substitute the value of ma as 200.59g/mol and mb as 126.9g/mol and in equation (5).

  μ=200.59 g/mol×126.9 g/mol200.59g/mol+126.9 g/mol=25454.871327.49 g/mol=77.72g/mol

Convert the reduced mass in kg as shown below.

  1 g/mol =103 kg6.022×1023/mol77.72 g/mol=77.72 g/mol×103 kg6.022×1023 /mol=1.29×1025 kg

Substitute μ as 1.29×1025 kg and I as 2.704×1044 kg m2 in equation (4) as shown below.

  2.704×1044kg m2=1.29×1025 kg×R2R2=2.096×1019 m2R=4.57×1010 m_

Therefore, the value of bond length of HgI2 is 4.57×1010 m_.

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 11 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 11 - Prob. 11E.2STCh. 11 - Prob. 11F.2STCh. 11 - Prob. 11A.1DQCh. 11 - Prob. 11A.2DQCh. 11 - Prob. 11A.3DQCh. 11 - Prob. 11A.1AECh. 11 - Prob. 11A.1BECh. 11 - Prob. 11A.2AECh. 11 - Prob. 11A.2BECh. 11 - Prob. 11A.3AECh. 11 - Prob. 11A.3BECh. 11 - Prob. 11A.4AECh. 11 - Prob. 11A.4BECh. 11 - Prob. 11A.5AECh. 11 - Prob. 11A.5BECh. 11 - Prob. 11A.6AECh. 11 - Prob. 11A.6BECh. 11 - Prob. 11A.7AECh. 11 - Prob. 11A.7BECh. 11 - Prob. 11A.8AECh. 11 - Prob. 11A.8BECh. 11 - Prob. 11A.9AECh. 11 - Prob. 11A.9BECh. 11 - Prob. 11A.10AECh. 11 - Prob. 11A.10BECh. 11 - Prob. 11A.11AECh. 11 - Prob. 11A.11BECh. 11 - Prob. 11A.12AECh. 11 - Prob. 11A.12BECh. 11 - Prob. 11A.1PCh. 11 - Prob. 11A.2PCh. 11 - Prob. 11A.3PCh. 11 - Prob. 11A.4PCh. 11 - Prob. 11A.5PCh. 11 - Prob. 11A.7PCh. 11 - Prob. 11A.8PCh. 11 - Prob. 11A.9PCh. 11 - Prob. 11A.11PCh. 11 - Prob. 11B.1DQCh. 11 - Prob. 11B.2DQCh. 11 - Prob. 11B.3DQCh. 11 - Prob. 11B.4DQCh. 11 - Prob. 11B.5DQCh. 11 - Prob. 11B.6DQCh. 11 - Prob. 11B.7DQCh. 11 - Prob. 11B.8DQCh. 11 - Prob. 11B.1AECh. 11 - Prob. 11B.1BECh. 11 - Prob. 11B.3AECh. 11 - Prob. 11B.3BECh. 11 - Prob. 11B.4BECh. 11 - Prob. 11B.5AECh. 11 - Prob. 11B.5BECh. 11 - Prob. 11B.6AECh. 11 - Prob. 11B.6BECh. 11 - Prob. 11B.7AECh. 11 - Prob. 11B.7BECh. 11 - Prob. 11B.8AECh. 11 - Prob. 11B.8BECh. 11 - Prob. 11B.9AECh. 11 - Prob. 11B.9BECh. 11 - Prob. 11B.10AECh. 11 - Prob. 11B.10BECh. 11 - Prob. 11B.11AECh. 11 - Prob. 11B.11BECh. 11 - Prob. 11B.12AECh. 11 - Prob. 11B.12BECh. 11 - Prob. 11B.13AECh. 11 - Prob. 11B.13BECh. 11 - Prob. 11B.14AECh. 11 - Prob. 11B.14BECh. 11 - Prob. 11B.1PCh. 11 - Prob. 11B.2PCh. 11 - Prob. 11B.3PCh. 11 - Prob. 11B.4PCh. 11 - Prob. 11B.5PCh. 11 - Prob. 11B.6PCh. 11 - Prob. 11B.7PCh. 11 - Prob. 11B.8PCh. 11 - Prob. 11B.9PCh. 11 - Prob. 11B.10PCh. 11 - Prob. 11B.11PCh. 11 - Prob. 11B.12PCh. 11 - Prob. 11B.13PCh. 11 - Prob. 11B.14PCh. 11 - Prob. 11C.1DQCh. 11 - Prob. 11C.2DQCh. 11 - Prob. 11C.3DQCh. 11 - Prob. 11C.4DQCh. 11 - Prob. 11C.1AECh. 11 - Prob. 11C.1BECh. 11 - Prob. 11C.2AECh. 11 - Prob. 11C.2BECh. 11 - Prob. 11C.3AECh. 11 - Prob. 11C.3BECh. 11 - Prob. 11C.4AECh. 11 - Prob. 11C.4BECh. 11 - Prob. 11C.5AECh. 11 - Prob. 11C.5BECh. 11 - Prob. 11C.6AECh. 11 - Prob. 11C.6BECh. 11 - Prob. 11C.7AECh. 11 - Prob. 11C.7BECh. 11 - Prob. 11C.8AECh. 11 - Prob. 11C.8BECh. 11 - Prob. 11C.2PCh. 11 - Prob. 11C.3PCh. 11 - Prob. 11C.4PCh. 11 - Prob. 11C.5PCh. 11 - Prob. 11C.6PCh. 11 - Prob. 11C.7PCh. 11 - Prob. 11C.8PCh. 11 - Prob. 11C.9PCh. 11 - Prob. 11C.10PCh. 11 - Prob. 11C.11PCh. 11 - Prob. 11C.12PCh. 11 - Prob. 11C.13PCh. 11 - Prob. 11C.15PCh. 11 - Prob. 11C.17PCh. 11 - Prob. 11C.18PCh. 11 - Prob. 11C.19PCh. 11 - Prob. 11D.1DQCh. 11 - Prob. 11D.2DQCh. 11 - Prob. 11D.3DQCh. 11 - Prob. 11D.1AECh. 11 - Prob. 11D.1BECh. 11 - Prob. 11D.2AECh. 11 - Prob. 11D.2BECh. 11 - Prob. 11D.3AECh. 11 - Prob. 11D.3BECh. 11 - Prob. 11D.4AECh. 11 - Prob. 11D.4BECh. 11 - Prob. 11D.5AECh. 11 - Prob. 11D.5BECh. 11 - Prob. 11D.6AECh. 11 - Prob. 11D.6BECh. 11 - Prob. 11D.7AECh. 11 - Prob. 11D.7BECh. 11 - Prob. 11D.2PCh. 11 - Prob. 11E.1DQCh. 11 - Prob. 11E.1AECh. 11 - Prob. 11E.1BECh. 11 - Prob. 11E.2AECh. 11 - Prob. 11E.2BECh. 11 - Prob. 11E.3AECh. 11 - Prob. 11E.3BECh. 11 - Prob. 11E.1PCh. 11 - Prob. 11E.2PCh. 11 - Prob. 11F.1DQCh. 11 - Prob. 11F.2DQCh. 11 - Prob. 11F.3DQCh. 11 - Prob. 11F.4DQCh. 11 - Prob. 11F.5DQCh. 11 - Prob. 11F.6DQCh. 11 - Prob. 11F.1AECh. 11 - Prob. 11F.1BECh. 11 - Prob. 11F.2AECh. 11 - Prob. 11F.2BECh. 11 - Prob. 11F.3AECh. 11 - Prob. 11F.3BECh. 11 - Prob. 11F.4AECh. 11 - Prob. 11F.4BECh. 11 - Prob. 11F.5AECh. 11 - Prob. 11F.5BECh. 11 - Prob. 11F.6AECh. 11 - Prob. 11F.6BECh. 11 - Prob. 11F.7AECh. 11 - Prob. 11F.7BECh. 11 - Prob. 11F.8AECh. 11 - Prob. 11F.8BECh. 11 - Prob. 11F.9AECh. 11 - Prob. 11F.9BECh. 11 - Prob. 11F.10AECh. 11 - Prob. 11F.10BECh. 11 - Prob. 11F.11AECh. 11 - Prob. 11F.11BECh. 11 - Prob. 11F.12AECh. 11 - Prob. 11F.12BECh. 11 - Prob. 11F.13AECh. 11 - Prob. 11F.13BECh. 11 - Prob. 11F.1PCh. 11 - Prob. 11F.2PCh. 11 - Prob. 11F.3PCh. 11 - Prob. 11F.4PCh. 11 - Prob. 11F.5PCh. 11 - Prob. 11F.6PCh. 11 - Prob. 11F.7PCh. 11 - Prob. 11F.8PCh. 11 - Prob. 11F.9PCh. 11 - Prob. 11F.10PCh. 11 - Prob. 11F.11PCh. 11 - Prob. 11F.12PCh. 11 - Prob. 11G.1DQCh. 11 - Prob. 11G.2DQCh. 11 - Prob. 11G.3DQCh. 11 - Prob. 11G.4DQCh. 11 - Prob. 11G.5DQCh. 11 - Prob. 11G.1AECh. 11 - Prob. 11G.1BECh. 11 - Prob. 11G.2AECh. 11 - Prob. 11G.2BECh. 11 - Prob. 11G.1PCh. 11 - Prob. 11G.2PCh. 11 - Prob. 11G.3PCh. 11 - Prob. 11G.4PCh. 11 - Prob. 11G.5PCh. 11 - Prob. 11G.6PCh. 11 - Prob. 11.1IACh. 11 - Prob. 11.5IACh. 11 - Prob. 11.8IA
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