Linearly polarized light can be assigned a specific irreducible representation of a symmetry point group. If the electronic ground state of benzene has A 1 g symmetry and y - polarized light has the label E 1 u , what are the symmetry label(s) of allowed excited electronic states? Use the D 6h character table in Appendix 3.

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Answer 1

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Chapter 15, Problem 15.1E

Linearly polarized light can be assigned a specific irreducible representation of a symmetry point group. If the electronic ground state of benzene has $A 1 g$ symmetry and $y -$ polarized light has the label $E 1 u$ , what are the symmetry label(s) of allowed excited electronic states? Use the $D 6h$ character table in Appendix 3.

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The symmetry label(s) of allowed excited electronic states for the electronic ground state of benzene as $A1g$ symmetry and $y-$ polarized light is to be predicted.

Concept introduction:

The electronic transitions should have a nonzero transition moment, given by the expression shown below.

$M=∫Ψfinal*μ^Ψinitialdτ$

Answer to Problem 15.1E

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Explanation of Solution

The electronic ground state of benzene has $A1g$ symmetry and $y-$ polarized light has $E1u$ symmetry.

The character table for point group $D6h$ is shown below.

$D6hE2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g111111111111E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is calculated as follows:

$E2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g⊗E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is the same as the representations of $E1u$ . Therefore, the symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Conclusion

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

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Chapter 15, Problem 15.1E

Linearly polarized light can be assigned a specific irreducible representation of a symmetry point group. If the electronic ground state of benzene has $A 1 g$ symmetry and $y -$ polarized light has the label $E 1 u$ , what are the symmetry label(s) of allowed excited electronic states? Use the $D 6h$ character table in Appendix 3.

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The symmetry label(s) of allowed excited electronic states for the electronic ground state of benzene as $A1g$ symmetry and $y-$ polarized light is to be predicted.

Concept introduction:

The electronic transitions should have a nonzero transition moment, given by the expression shown below.

$M=∫Ψfinal*μ^Ψinitialdτ$

Answer to Problem 15.1E

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Explanation of Solution

The electronic ground state of benzene has $A1g$ symmetry and $y-$ polarized light has $E1u$ symmetry.

The character table for point group $D6h$ is shown below.

$D6hE2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g111111111111E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is calculated as follows:

$E2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g⊗E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is the same as the representations of $E1u$ . Therefore, the symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Conclusion

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

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Answer 3

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Chapter 15, Problem 15.1E

Linearly polarized light can be assigned a specific irreducible representation of a symmetry point group. If the electronic ground state of benzene has $A 1 g$ symmetry and $y -$ polarized light has the label $E 1 u$ , what are the symmetry label(s) of allowed excited electronic states? Use the $D 6h$ character table in Appendix 3.

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The symmetry label(s) of allowed excited electronic states for the electronic ground state of benzene as $A1g$ symmetry and $y-$ polarized light is to be predicted.

Concept introduction:

The electronic transitions should have a nonzero transition moment, given by the expression shown below.

$M=∫Ψfinal*μ^Ψinitialdτ$

Answer to Problem 15.1E

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Explanation of Solution

The electronic ground state of benzene has $A1g$ symmetry and $y-$ polarized light has $E1u$ symmetry.

The character table for point group $D6h$ is shown below.

$D6hE2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g111111111111E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is calculated as follows:

$E2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g⊗E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is the same as the representations of $E1u$ . Therefore, the symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Conclusion

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

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Answer 4

Textbook Question

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Chapter 15, Problem 15.1E

Linearly polarized light can be assigned a specific irreducible representation of a symmetry point group. If the electronic ground state of benzene has $A 1 g$ symmetry and $y -$ polarized light has the label $E 1 u$ , what are the symmetry label(s) of allowed excited electronic states? Use the $D 6h$ character table in Appendix 3.

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The symmetry label(s) of allowed excited electronic states for the electronic ground state of benzene as $A1g$ symmetry and $y-$ polarized light is to be predicted.

Concept introduction:

The electronic transitions should have a nonzero transition moment, given by the expression shown below.

$M=∫Ψfinal*μ^Ψinitialdτ$

Answer to Problem 15.1E

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Explanation of Solution

The electronic ground state of benzene has $A1g$ symmetry and $y-$ polarized light has $E1u$ symmetry.

The character table for point group $D6h$ is shown below.

$D6hE2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g111111111111E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is calculated as follows:

$E2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g⊗E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is the same as the representations of $E1u$ . Therefore, the symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Conclusion

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

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Answer 5

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Answer 6

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Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The symmetry label(s) of allowed excited electronic states for the electronic ground state of benzene as $A1g$ symmetry and $y-$ polarized light is to be predicted.

Concept introduction:

The electronic transitions should have a nonzero transition moment, given by the expression shown below.

$M=∫Ψfinal*μ^Ψinitialdτ$

Answer to Problem 15.1E

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Explanation of Solution

The electronic ground state of benzene has $A1g$ symmetry and $y-$ polarized light has $E1u$ symmetry.

The character table for point group $D6h$ is shown below.

$D6hE2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g111111111111E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is calculated as follows:

$E2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g⊗E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is the same as the representations of $E1u$ . Therefore, the symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Conclusion

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The symmetry label(s) of allowed excited electronic states for the electronic ground state of benzene as $A1g$ symmetry and $y-$ polarized light is to be predicted.

Concept introduction:

The electronic transitions should have a nonzero transition moment, given by the expression shown below.

$M=∫Ψfinal*μ^Ψinitialdτ$

Answer to Problem 15.1E

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Explanation of Solution

The electronic ground state of benzene has $A1g$ symmetry and $y-$ polarized light has $E1u$ symmetry.

The character table for point group $D6h$ is shown below.

$D6hE2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g111111111111E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is calculated as follows:

$E2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g⊗E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is the same as the representations of $E1u$ . Therefore, the symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Conclusion

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The symmetry label(s) of allowed excited electronic states for the electronic ground state of benzene as $A1g$ symmetry and $y-$ polarized light is to be predicted.

Concept introduction:

The electronic transitions should have a nonzero transition moment, given by the expression shown below.

$M=∫Ψfinal*μ^Ψinitialdτ$

Answer 12

Answer to Problem 15.1E

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Answer 13

Explanation of Solution

The electronic ground state of benzene has $A1g$ symmetry and $y-$ polarized light has $E1u$ symmetry.

The character table for point group $D6h$ is shown below.

$D6hE2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g111111111111E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is calculated as follows:

$E2C62C3C23C2'3C2''i2S32S6σh3σd3σvA1g⊗E1u21−1−200−2−11200$

The representations of $A1g⊗E1u$ is the same as the representations of $E1u$ . Therefore, the symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Answer 14

Conclusion

The symmetry label of allowed excited state for the $y−polarized$ light is $E1u$ .

Answer 15

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Answer to Problem 15.1E

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