Bundle: Physical Chemistry, 2nd + Student Solutions Manual
Bundle: Physical Chemistry, 2nd + Student Solutions Manual
2nd Edition
ISBN: 9781285257594
Author: David W. Ball
Publisher: Cengage Learning
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Chapter 14, Problem 14.82E
Interpretation Introduction

Interpretation:

The cubane, C8H8, has only three IR-active vibrational is to be verified. The degeneracies of each IR-active vibrational are to be determined. The number of total vibrations out of the 42 vibrational degrees are represented by the three IR-active vibrational motions is to be stated.

Concept introduction:

The complex vibrations exhibit by the polyatomic molecule is known as normal modes of vibrations. The vibrational modes of a molecule are IR or Raman active. If a molecule has centre of symmetry, then the modes which are IR-active will be Raman inactive and the modes that are IR-inactive will be Raman active. The total number of vibrational degrees of freedom for nonlinear molecule is represented by 3N6.

Expert Solution & Answer
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Answer to Problem 14.82E

The cubane, C8H8, has only three IR-active vibrational. The degeneracies of each IR-active vibrational is 3. The number of total vibrations out of the 42 vibrational degrees are represented by the three IR-active vibrational motions is 9.

Explanation of Solution

The structure of cubane is shown below.

Bundle: Physical Chemistry, 2nd + Student Solutions Manual, Chapter 14, Problem 14.82E

Figure 1

The point group of cubane is Oh. It is non-linear.

The character table for point group Oh with the number of stationary atom after the operation and angle of rotation by The symmetry element is shown below.

Oh E 8C3 3C2 6C4 6C2' i 8S6 3σh 6S4 6σd
Nstationary 16 1 0 0 4 8 0 0 0 0
θ 0° 120° 180° 90° 180° 90° 60° 180° 90° 180°

The expression (1+2cosθ) is evaluate for θ=0° as shown below.

1+2cosθ=1+2cos0°=1+(2)(1)=3

Similarly, the value of (1+2cosθ) is evaluate for all value of angle as shown in the table below.

Oh E 8C3 3C2 6C4 6C2' i 8S6 3σh 6S4 6σd
Nstationary 16 1 0 0 4 8 0 0 0 0
θ 0° 120° 180° 90° 180° 90° 60° 180° 90° 180°
(1+2cosθ) 3 0 1 1 1 1 2 1 1 1

The expression χtot is shown below.

χtot=±Nstationary(1+2cosθ)

The positive sign is taken for the proper rotation and negative sign is taken for improper axis.

Substitute the value of Nstationary=16 and θ=0° in the above equation.

χtot=(16)(1+2cos0°)=(16)(3)=48

Similarly, the value of χtot is evaluate for all symmetry elements as shown in the table below.

Oh E 8C3 3C2 6C4 6C2' i 8S6 3σh 6S4 6σd
Nstationary 16 1 0 0 4 8 0 0 0 0
θ 0° 120° 180° 90° 180° 90° 60° 180° 90° 180°
(1+2cosθ) 3 0 1 1 1 1 2 1 1 1
χtot 48 0 0 0 4 8 0 0 0 0

The expression χr is shown below.

χr=±(1+2cosθ)

Substitute the value of θ=0° in the above equation.

χr=(1+2cos0°)=3

Similarly, the value of χr is evaluate for all symmetry elements as shown in the table below.

Oh E 8C3 3C2 6C4 6C2' i 8S6 3σh 6S4 6σd
Nstationary 16 1 0 0 4 8 0 0 0 0
θ 0° 120° 180° 90° 180° 90° 60° 180° 90° 180°
(1+2cosθ) 3 0 1 1 1 1 2 1 1 1
χtot 48 0 0 0 4 8 0 0 0 0
χr 3 0 1 1 1 1 2 1 1 1

The expression χt is shown below.

χt=±(1+2cosθ)

The positive sign is taken for the proper rotation and negative sign is taken for improper axis.

Substitute the value of θ=0° in the above equation.

χt=(1+2cos0°)=3

Similarly, the value of χt is evaluate for all symmetry elements as shown in the table below.

Oh E 8C3 3C2 6C4 6C2' i 8S6 3σh 6S4 6σd
Nstationary 16 1 0 0 4 8 0 0 0 0
θ 0° 120° 180° 90° 180° 90° 60° 180° 90° 180°
(1+2cosθ) 3 0 1 1 1 1 2 1 1 1
χtot 48 0 0 0 4 8 0 0 0 0
χr 3 0 1 1 1 1 2 1 1 1
χt 3 0 1 1 1 1 2 1 1 1

The expression χv is shown below.

χv=χtotχtχr

The positive sign is taken for the proper rotation and negative sign is taken for improper axis.

Substitute the value of χtot, χr and χt for E in the above equation.

χv=4833=42

Similarly, the value of χv is evaluate for all symmetry elements as shown in the table below.

Oh E 8C3 3C2 6C4 6C2' i 8S6 3σh 6S4 6σd
Nstationary 16 1 0 0 4 8 0 0 0 0
θ 0° 120° 180° 90° 180° 90° 60° 180° 90° 180°
(1+2cosθ) 3 0 1 1 1 1 2 1 1 1
χtot 48 0 0 0 4 8 0 0 0 0
χr 3 0 1 1 1 1 2 1 1 1
χt 3 0 1 1 1 1 2 1 1 1
χv 42 0 2 2 2 8 0 0 0 0

The great orthogonality theorem for the reducible representation can be represented as,

aΓ=1hallclassesofpointgroupNχΓχlinearcombo

Where,

aΓ is the number of times the irreducible representation appears in a linear combination.

h is the order of the group.

χΓ is the character of the class of the irreducible representation.

χlinearcombo is the character of the class linear combination.

N is the number of symmetry operations.

The term χv is taken in place of χlinearcombo.

The irreducible representation for point group Oh is shown below.

Oh E 8C3 3C2 6C4 6C2' i 8S6 3σh 6S4 6σd
Γ 3 0 1 1 1 3 0 1 1 1

Substitute the h, χΓ, χv, and N in the above equation.

aΓ=148((1)(3)(42)+(8)(0)(0)+(3)(1)(2)+(6)(1)(2)+(6)(1)(1)+(1)(3)(8)+(8)(0)(0)+(3)(1)(0)+(6)(1)(0)+(6)(1)(0))=148(126+06+126+24+0+0+0+0)=148(144)=3

Therefore, the cubane, C8H8, has only three IR-active vibrational. The degeneracies of each IR-active vibrational is triply degenerated due to the presence of label (x,y,z). Therefore, the number of total vibrations out of the 42 vibrational degrees are represented by the three IR-active vibrational motions is 9.

Conclusion

The cubane, C8H8, has only three IR-active vibrational. The degeneracies of each IR-active vibrational is 3. The number of total vibrations out of the 42 vibrational degrees are represented by the three IR-active vibrational motions is 9.

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Chapter 14 Solutions

Bundle: Physical Chemistry, 2nd + Student Solutions Manual

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