Chapter 10, Problem 10C.9BE

(i)

Interpretation Introduction

Interpretation:

The states of anthracene that might be reached by electric dipole transitions from its ground state have to be stated.

Concept introduction:

The selection rules for molecular transitions can be made on the basis of group theory by knowing the symmetry species of the wavefunctions.  The intensity of transition depends on the electric transition dipole moment.

Answer to Problem 10C.9BE

The states of anthracene that might be reached by electric dipole transitions from its ground state are ${\text{B}}_{\text{1u}}$, ${\text{B}}_{\text{2u}}$, and ${\text{B}}_{\text{3u}}$.

Explanation of Solution

The structure of anthracene is shown below.

Figure 1

The anthracene molecule belongs to the $D_{\text{2h}}$ point group.

The dipole integral of $D_{\text{2h}}$ point group is ${\text{B}}_{\text{1u}}+{\text{B}}_{\text{2u}}+{\text{B}}_{\text{3u}}$.

The character table for $D_{6\text{h}}$ point group is given below.

$D_{\text{2h}}$	$E$	$C_2\left(x\right)$	$C_2\left(y\right)$	$C_2\left(z\right)$	$i$	$\sigma \left(xy\right)$	$\sigma \left(yz\right)$	$\sigma \left(zx\right)$
${\text{A}}_{\text{g}}$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$
${\text{B}}_{\text{1u}}$	$1$	$1$	$-1$	$-1$	$-1$	$-1$$1$	$1$	$1$
${\text{B}}_{\text{2u}}$	$1$	$-1$	$1$	$-1$	$-1$	$1$	$1$	$-1$
${\text{B}}_{\text{3u}}$	$1$	$-1$	$-1$	$1$	$-1$	$1$	$-1$	$1$

The electric transition dipole moment is calculated using the formula given below.

    ${\mu }_{q\text{,fi}}=-e{\displaystyle \int {\psi }_{\text{f}}^{*}q{\psi }_i\text{d}\tau }$        (1)

Where,

• $q$ is the component in $x,y$ or $z$ direction.
• ${\psi }_i$ is the initial wavefunction.
• ${\psi }_{\text{f}}^{*}$ is the conjugate of the final wavefunction.
• ${\mu }_{q\text{,fi}}$ is the electric transition dipole moment.
• $e$ is the charge of an electron.

The ground state is ${\text{A}}_{\text{g}}$.

The transition ${\psi }_{{\text{A}}_{\text{g}}}\to {\psi }_{{\text{B}}_{\text{1u}}}$ is considered.

This gives the product in the integral to be ${\text{A}}_{\text{g}}\times q\times {\text{B}}_{\text{1u}}$.

The product of ${\text{A}}_{\text{g}}\times {\text{A}}_{\text{2u}}={\text{B}}_{\text{1u}}$.  Therefore, the product in the integral can be written as ${\text{B}}_{\text{1u}}\times q$.

The product in the integral will be equal to ${\text{A}}_{\text{g}}$ and is allowed only when the $q$ component is equal to ${\text{B}}_{\text{1u}}$.

The $q$ component has symmetry species ${\text{B}}_{\text{1u}}$, therefore, the transition ${\psi }_{{\text{A}}_{\text{g}}}\to {\psi }_{{\text{B}}_{\text{1u}}}$ is allowed and polarized at $z$-axis.

The transition ${\psi }_{{\text{A}}_{\text{g}}}\to {\psi }_{{\text{B}}_{\text{2u}}}$ is considered.

This gives the product in the integral to be ${\text{A}}_{\text{g}}\times q\times {\text{B}}_{\text{2u}}$.

The product of ${\text{A}}_{\text{g}}\times {\text{A}}_{\text{2u}}={\text{B}}_{\text{2u}}$.  Therefore, the product in the integral can be written as ${\text{B}}_{\text{2u}}\times q$.

The product in the integral will be equal to ${\text{A}}_{\text{g}}$ and is allowed only when the $q$ component is equal to ${\text{B}}_{\text{2u}}$.

The $q$ component has symmetry species ${\text{B}}_{\text{2u}}$, therefore, the transition ${\psi }_{{\text{A}}_{\text{g}}}\to {\psi }_{{\text{B}}_{\text{2u}}}$ is allowed and polarized at $y$-axis.

The transition ${\psi }_{{\text{A}}_{\text{g}}}\to {\psi }_{{\text{B}}_{\text{3u}}}$ is considered.

This gives the product in the integral to be ${\text{A}}_{\text{g}}\times q\times {\text{B}}_{\text{3u}}$.

The product of ${\text{A}}_{\text{g}}\times {\text{A}}_{\text{2u}}={\text{B}}_{\text{3u}}$.  Therefore, the product in the integral can be written as ${\text{B}}_{\text{3u}}\times q$.

The product in the integral will be equal to ${\text{A}}_{\text{g}}$ and is allowed only when the $q$ component is equal to ${\text{B}}_{\text{3u}}$.

The $q$ component has symmetry species ${\text{B}}_{\text{3u}}$, therefore, the transition ${\psi }_{{\text{A}}_{\text{g}}}\to {\psi }_{{\text{B}}_{\text{3u}}}$ is allowed and polarized at $x$-axis.

(b)

Interpretation Introduction

Interpretation:

The states of coronene that might be reached by electric dipole transitions from its ground state have to be stated.

Concept introduction:

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

Answer to Problem 10C.9BE

The states of coronene that might be reached by electric dipole transitions from its ground state are ${\text{A}}_{\text{2u}}$ and ${\text{E}}_{\text{1u}}$.

Explanation of Solution

The structure of coronene is shown below.

Figure 2

The anthracene molecule belongs to the $D_{6\text{h}}$ point group.

The dipole integral of $D_{6\text{h}}$ point group is ${\text{A}}_{\text{2u}}+{\text{E}}_{\text{1u}}$.

The character table for $D_{6\text{h}}$ point group is given below.

$D_{\text{6h}}$	$E$	$2C_6$	$2C_3$	$C_2$	$3C_2^{\text{'}}$	$3C_2^{"}$	$i$	$2S_3$	$2S_6$	${\sigma }_{\text{h}}$	$3{\sigma }_{\text{d}}$	$3{\sigma }_{\text{v}}$
${\text{A}}_{1\text{g}}$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$
${\text{A}}_{\text{2u}}$	$1$	$1$	$1$	$1$	$-1$	$-1$	$-1$	$-1$	$-1$	$-1$	$1$	$1$
${\text{E}}_{\text{1u}}$	$2$	$1$	$-1$	$-2$	$0$	$0$	$-2$	$-1$	$1$	$2$	$0$	$0$

The electric transition dipole moment is calculated using the formula given below.

    ${\mu }_{q\text{,fi}}=-e{\displaystyle \int {\psi }_{\text{f}}^{*}q{\psi }_i\text{d}\tau }$        (1)

Where,

• $q$ is the component in $x,y$ or $z$ direction.
• ${\psi }_i$ is the initial wavefunction.
• ${\psi }_{\text{f}}^{*}$ is the conjugate of the final wavefunction.
• ${\mu }_{q\text{,fi}}$ is the electric transition dipole moment.
• $e$ is the charge of an electron.

The ground state is ${\text{A}}_{\text{1g}}$.

The transition ${\psi }_{{\text{A}}_{\text{1g}}}\to {\psi }_{{\text{A}}_{\text{1u}}}$ is considered.

This gives the product in the integral to be ${\text{A}}_{\text{1g}}\times q\times {\text{A}}_{\text{2u}}$.

The product of ${\text{A}}_{\text{1g}}\times {\text{A}}_{\text{2u}}={\text{A}}_{\text{2u}}$.  Therefore, the product in the integral can be written as ${\text{A}}_{\text{2u}}\times q$.

The product in the integral will be equal to ${\text{A}}_{\text{1g}}$ and is allowed only when the $q$ component is equal to ${\text{A}}_{\text{2u}}$.

The $q$ component has symmetry species ${\text{A}}_{\text{2u}}$, therefore, the transition ${\psi }_{{\text{A}}_{\text{1g}}}\to {\psi }_{{\text{A}}_{\text{1u}}}$ is allowed and polarized at $z$-axis.

The transition ${\psi }_{{\text{A}}_{\text{1g}}}\to {\psi }_{{\text{E}}_{\text{1u}}}$ is considered.

This gives the product in the integral to be ${\text{A}}_{\text{1g}}\times q\times {\text{E}}_{\text{1u}}$.

The product of ${\text{A}}_{\text{1g}}\times {\text{A}}_{\text{2u}}={\text{E}}_{\text{1u}}$.  Therefore, the product in the integral can be written as ${\text{E}}_{\text{1u}}\times q$.

The product in the integral will be equal to ${\text{A}}_{\text{1g}}$ and is allowed only when the $q$ component is equal to ${\text{E}}_{\text{1u}}$.

The $q$ component has symmetry species ${\text{E}}_{\text{1u}}$, therefore, the transition ${\psi }_{{\text{A}}_{\text{1g}}}\to {\psi }_{{\text{E}}_{\text{1u}}}$ is allowed and polarized at $x$-axis and $y-$ axis.