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
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Chapter 10, Problem 10B.3P

(a)

Interpretation Introduction

Interpretation:

The relation C2σv=σv and σvσv=C2 has to be confirmed using an explicit matrix multiplication method.

Concept introduction:

A symmetry operation is defined as an action on an object to reproduce an arrangement using a symmetry argument. The spatial arrangement of the object remains identical after a symmetry operation. The point of reference through which a symmetry operation takes place is termed as a symmetry element.

(a)

Expert Solution
Check Mark

Answer to Problem 10B.3P

The relation C2σv=σv and σvσv=C2 is confirmed from the explicit matrix multiplication method.

Explanation of Solution

The water (H2O) molecule taking two H1s orbital and the four valence orbitals of the O atom are shown below.

Atkins' Physical Chemistry, Chapter 10, Problem 10B.3P , additional homework tip  1

Figure 1

The two H1s orbital taken as a basis are 1S(H1) and 1S(H2).  The four valence orbitals of the O atom taken as a basis are 2S(O), 2px(O), 2py(O) and 2pz(O).

The E operator is applied on the orbitals as shown below.

Atkins' Physical Chemistry, Chapter 10, Problem 10B.3P , additional homework tip  2

Figure 2

The changes taking place in the above operation are shown below

  (2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))

The above result is expressed as a matrix multiplication as shown below.

  (2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))(100000010000001000000100000010000001)=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))D(E)

The identity operation (E) does not affect the basis and is represented in a 6×6 matrix as shown below.

  D(E)=(100000010000001000000100000010000001)

The C2 operator is applied on the orbitals as shown below.

Atkins' Physical Chemistry, Chapter 10, Problem 10B.3P , additional homework tip  3

Figure 3

The changes taking place in the above operation are shown below

  (2S(O)2px(O)2py(O)2pz(O)1S(H2)1S(H1))(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))

The above result is expressed as a matrix multiplication as shown below.

(2S(O)2px(O)2py(O)2pz(O)1S(H2)1S(H1))=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))(100000010000001000000100000001000010)=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))D(C2)

The C2 operation is represented in a 6×6 matrix as shown below.

  D(C2)=(100000010000001000000100000001000010)

The σv operator is applied on the orbitals as shown below.

Atkins' Physical Chemistry, Chapter 10, Problem 10B.3P , additional homework tip  4

Figure 4

The changes taking place in the above operation are shown below

  (2S(O)2px(O)2py(O)2pz(O)1S(H2)1S(H1))(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))

The above result is expressed as a matrix multiplication as shown below.

(2S(O)2px(O)2py(O)2pz(O)1S(H2)1S(H1))=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))(100000010000001000000100000001000010)=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))D(σv)

The σv operation is represented in a 6×6 matrix as shown below.

  D(σv)=(100000010000001000000100000001000010)

The σv operator is applied on the orbitals as shown below.

Atkins' Physical Chemistry, Chapter 10, Problem 10B.3P , additional homework tip  5

Figure 5

The changes taking place in the above operation are shown below

  (2S(O)2px(O)2py(O)2pz(O)1S(H2)1S(H1))(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))

The above result is expressed as a matrix multiplication as shown below.

(2S(O)2px(O)2py(O)2pz(O)1S(H2)1S(H1))=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))(100000010000001000000100000010000001)=(2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2))D(σv)

The σv operation is represented in a 6×6 matrix as shown below.

  D(σv)=(100000010000001000000100000010000001)

The explicit matrix multiplication C2σv is calculated as shown below.

  C2σv=D(C2)×D(σv)        (1)

Substitute the value of D(C2) and D(σv) in equation (1) as shown below.

  C2σv=(100000010000001000000100000001000010)(100000010000001000000100000001000010)=(100000010000001000000100000010000001)=σ

Therefore, the relation C2σv=σv is confirmed.

The explicit matrix multiplication σvσv is calculated as shown below.

  σvσv=D(σv)×D(σv)        (2)

Substitute the value of D(σv) and D(σv) in equation (2) as shown below.

  σvσv=(100000010000001000000100000001000010)(100000010000001000000100000010000001)=(100000010000001000000100000001000010)=C2

Therefore, the relation σvσv=C2 is confirmed.

(b)

Interpretation Introduction

Interpretation:

The corresponding representation is reducible and span 3A1+B1+2B2 has to be proved.

Concept introduction:

As mentioned in the concept of introduction in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 10B.3P

The corresponding representation is reducible because the characters are not similar to any irreducible representation.  The corresponding representation span 3A1+B1+2B2 because the character for the sum 3A1+B1+2B2 is the same as the basis (2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2)).

Explanation of Solution

The characters of a representative are the sum of their diagonal elements as shown below.

RECσvσvD(R)(100000010000001000000100000010000001)(100000010000001000000100000001000010)(100000010000001000000100000001000010)(100000010000001000000100000010000001)χ(R)6024

The characters shown above are not similar to any irreducible representation.  Therefore, the representation is reducible.

The character for the sum 3A1+B1+2B2 is shown below.

  RECσvσv3A13333B111112B222223A1+B1+2B26024

The character for the sum 3A1+B1+2B2 is the same as the basis (2S(O)2px(O)2py(O)2pz(O)1S(H1)1S(H2)).  Therefore, the corresponding representation spans 3A1+B1+2B2.

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Chapter 10 Solutions

Atkins' Physical Chemistry

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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