ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 10, Problem 10B.4AE
Interpretation Introduction

Interpretation:

Whether all the irreducible representations are orthonormal for the point group C2h has to be confirmed.

Concept introduction:

The systematic discussion of symmetry is known as group theory.  The table showing all the characters of the operations of a group are known as character tables.  The character of an operation in a particular matrix is the sum of the diagonal elements of the representative of that operation.

Expert Solution & Answer
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Answer to Problem 10B.4AE

All the irreducible representations are orthonormal for the point group C2h.

Explanation of Solution

The condition for orthonormality of irreducible representation is given below.

    1hcN(C)χΓ(i)(C)χΓ(j)(C)={0forij1fori=j        (1)

Where,

  • h is the number of operations of the group.
  • N(C) is the number of classes.
  • χΓ(i)(C),χΓ(j)(C) are irreducible representations.

The character table for C2h point group is given below.

 EC2iσh  
Ag1111Rzx2,y2,z2,xy
Bg1111Rx,Ryxz,yz
Au1111z 
Bu1111x,y 

The number of operation, h is 4.

The number of classes, N(C) is 1.

The condition in equation (1) is applied to Ag and Bg to check orthogonality as shown below.

    14(χΓ(Ag)χΓBg)=14((1×1)+(1×(1))+(1×1)+(1×(1)))=14(1+(1)+1+(1))=0

The normality of Ag is checked as shown below.

    14(χΓ(Ag)χΓAg)=14((1×1)+(1×1)+(1×1)+(1×1))=14(1+1+1+1)=1

The normality of Bg is checked as shown below.

    14(χΓ(Bg)χΓBg)=14((1×1)+(1×(1))+(1×1)+(1×(1)))=14(1+1+1+1)=1

Therefore, the irreducible representations, Ag and Bg are orthonormal.

The condition in equation (1) is applied to Ag and Au to check orthogonality as shown below.

    14(χΓ(Ag)χΓ(Au))=14((1×1)+(1×1)+(1×(1))+(1×(1)))=14(1+1+(1)+(1))=0

The normality of Ag is checked as shown below.

    14(χΓ(Ag)χΓ(Ag))=14((1×1)+(1×1)+(1×1)+(1×1))=14(1+1+1+1)=1

The normality of Au is checked as shown below.

    14(χΓ(Au)χΓ(Au))=14((1×1)+(1×1)+(1×(1))+(1×(1)))=14(1+1+1+1)=1

Therefore, the irreducible representations, Ag and Au are orthonormal.

The condition in equation (1) is applied to Ag and Bu to check orthogonality as shown below.

    14(χΓ(Ag)χΓ(Bu))=14((1×1)+(1×(1))+(1×(1))+(1×1))=14(1+(1)+(1)+1)=0

The normality of Ag is checked as shown below.

    14(χΓ(Ag)χΓ(Ag))=14((1×1)+(1×1)+(1×1)+(1×1))=14(1+1+1+1)=1

The normality of Bu is checked as shown below.

    14(χΓ(Bu)χΓ(Bu))=14((1×1)+(1×(1))+(1×(1))+(1×1))=14(1+1+1+1)=1

Therefore, the irreducible representations, Ag and Bu are orthonormal.

The condition in equation (1) is applied to Au and Bg to check orthogonality as shown below.

    14(χΓ(Au)χΓ(Bg))=14((1×1)+(1×(1))+((1)×1)+((1)×(1)))=14(1+(1)+(1)+1)=0

The normality of Au is checked as shown below.

    14(χΓ(Au)χΓ(Au))=14((1×1)+(1×1)+(1×(1))+(1×(1)))=14(1+1+1+1)=1

The normality of Bg is checked as shown below.

    14(χΓ(Bg)χΓ(Bg))=14((1×1)+(1×(1))+(1×1)+(1×(1)))=14(1+1+1+1)=1

Therefore, the irreducible representations, Au and Bg are orthonormal.

The condition in equation (1) is applied to Bu and Bg to check orthogonality as shown below.

    14(χΓ(Bu)χΓ(Bg))=14((1×1)+((1)×(1))+((1)×1)+(1×(1)))=14(1+1+(1)+(1))=0

The normality of Bu is checked as shown below.

    14(χΓ(Bu)χΓ(Bu))=14((1×1)+(1×(1))+(1×(1))+(1×1))=14(1+1+1+1)=1

The normality of Bg is checked as shown below.

    14(χΓ(Bg)χΓ(Bg))=14((1×1)+(1×(1))+(1×1)+(1×(1)))=14(1+1+1+1)=1

Therefore, the irreducible representations, Bu and Bg are orthonormal.

The condition in equation (1) is applied to Au and Bu to check orthogonality as shown below.

    14(χΓ(Au)χΓ(Bg))=14((1×1)+(1×(1))+((1)×(1))+((1)×1))=14(1+(1)+1+(1))=0

The normality of Au is checked as shown below.

    14(χΓ(Au)χΓ(Au))=14((1×1)+(1×1)+(1×(1))+(1×(1)))=14(1+1+1+1)=1

The normality of Bu is checked as shown below.

    14(χΓ(Bu)χΓ(Bu))=14((1×1)+(1×(1))+(1×(1))+(1×1))=14(1+1+1+1)=1

Therefore, the irreducible representations, Au and Bu are orthonormal.

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Chapter 10 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 10 - Prob. 10A.2AECh. 10 - Prob. 10A.2BECh. 10 - Prob. 10A.3AECh. 10 - Prob. 10A.3BECh. 10 - Prob. 10A.4AECh. 10 - Prob. 10A.4BECh. 10 - Prob. 10A.5AECh. 10 - Prob. 10A.5BECh. 10 - Prob. 10A.6AECh. 10 - Prob. 10A.6BECh. 10 - Prob. 10A.7AECh. 10 - Prob. 10A.1PCh. 10 - Prob. 10A.2PCh. 10 - Prob. 10A.3PCh. 10 - Prob. 10A.4PCh. 10 - Prob. 10A.5PCh. 10 - Prob. 10B.1DQCh. 10 - Prob. 10B.2DQCh. 10 - Prob. 10B.3DQCh. 10 - Prob. 10B.4DQCh. 10 - Prob. 10B.5DQCh. 10 - Prob. 10B.1AECh. 10 - Prob. 10B.1BECh. 10 - Prob. 10B.2AECh. 10 - Prob. 10B.2BECh. 10 - Prob. 10B.3AECh. 10 - Prob. 10B.3BECh. 10 - Prob. 10B.4AECh. 10 - Prob. 10B.4BECh. 10 - Prob. 10B.5AECh. 10 - Prob. 10B.5BECh. 10 - Prob. 10B.6AECh. 10 - Prob. 10B.6BECh. 10 - Prob. 10B.7AECh. 10 - Prob. 10B.7BECh. 10 - Prob. 10B.1PCh. 10 - Prob. 10B.2PCh. 10 - Prob. 10B.3PCh. 10 - Prob. 10B.4PCh. 10 - Prob. 10B.5PCh. 10 - Prob. 10B.6PCh. 10 - Prob. 10B.7PCh. 10 - Prob. 10B.8PCh. 10 - Prob. 10B.9PCh. 10 - Prob. 10B.10PCh. 10 - Prob. 10C.1DQCh. 10 - Prob. 10C.2DQCh. 10 - Prob. 10C.1AECh. 10 - Prob. 10C.1BECh. 10 - Prob. 10C.2AECh. 10 - Prob. 10C.2BECh. 10 - Prob. 10C.3AECh. 10 - Prob. 10C.3BECh. 10 - Prob. 10C.4AECh. 10 - Prob. 10C.4BECh. 10 - Prob. 10C.5AECh. 10 - Prob. 10C.6AECh. 10 - Prob. 10C.6BECh. 10 - Prob. 10C.7AECh. 10 - Prob. 10C.7BECh. 10 - Prob. 10C.8AECh. 10 - Prob. 10C.8BECh. 10 - Prob. 10C.9AECh. 10 - Prob. 10C.9BECh. 10 - Prob. 10C.1PCh. 10 - Prob. 10C.2PCh. 10 - Prob. 10C.3PCh. 10 - Prob. 10C.4PCh. 10 - Prob. 10C.5PCh. 10 - Prob. 10C.6P
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