Chapter 14, Problem 14.94E

Interpretation Introduction

(a)

Interpretation:

The number of Raman-active vibrations for the ${\text{CH}}_{\text{4}}$ molecule is to be predicted. The answer is to be compared with the answers of exercise 14.79.

Concept introduction:

The characters of the irreducible representations of the given point group can be multiplied by each other. The only condition is the characters of the same symmetry operations are multiplied together. The multiplication of the characters is commutative.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

Answer to Problem 14.94E

The number of Raman-active vibrations for the ${\text{CH}}_{\text{4}}$ molecule is four. The number of Raman-active vibrations for the ${\text{CH}}_{\text{4}}$ is more than the IR-active vibrations.

Explanation of Solution

The symmetry of ${\text{CH}}_{\text{4}}$ molecule is tetrahedral.

The character table for point group $T_{\text{d}}$ is shown below.

operations	$E$	$8C_3$	$3C_2$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$5$	$2$	$1$	$1$	$3$
$\theta (°)$	$0$	$120$	$180$	$90$	$180$
${\chi }_{\text{r}}=\left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$1$	$-1$
${\chi }_{\text{tot}}=\pm N_{\text{stationary}}\left(1+2\cos \theta \right)$	$15$	$0$	$-1$	$-1$	$3$
${\chi }_l=\pm \left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$-1$	$1$
${\chi }_{\text{v}}={\chi }_{\text{tot}}-{\chi }_l-{\chi }_{\text{r}}$	$9$	$0$	$1$	$-1$	$3$

This reducible representation reduced using great orthogonality theorem as shown below.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

The order of the group is $24$.

The great orthogonality theorem orthogonality of the irreducible representation of $A_1$, $A_2$$,E$, $T_1$ and $T_2$ is shown below.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations for $A_1$.

$\begin{array}{rll}{a}_{{\text{A}}_1}& =& \frac{1}{24}\left[\left(1\cdot 1\cdot 9\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[9+0+3-6+18\right]\\ & =& \frac{1}{24}\left[24\right]\\ & =& 1\end{array}$

The number of times the irreducible representation for $A_1$ appears in a linear combination is $1$.

Similarly, for $A_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{A}}_2}& =& \frac{1}{24}\left[\left(1\cdot 1\cdot 9\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(6\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[9+0+3+6-18\right]\\ & =& \frac{1}{24}\left[0\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_2$ appears in a linear combination is $0$.

Similarly, for $E$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{\text{E}}& =& \frac{1}{24}\left[\left(1\cdot 2\cdot 9\right)+\left(8\cdot -1\cdot 0\right)+\left(3\cdot 2\cdot 1\right)+\left(6\cdot 0\cdot -1\right)+\left(6\cdot 0\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[18+0+6+0+0\right]\\ & =& \frac{1}{24}\left[24\right]\\ & =& 1\end{array}$

The number of times the irreducible representation for $E$ appears in a linear combination is $1$.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations for $T_1$.

$\begin{array}{rll}{a}_{{\text{T}}_1}& =& \frac{1}{24}\left[\left(1\cdot 3\cdot 9\right)+\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[37+0-3-6-18\right]\\ & =& \frac{1}{24}\left[24\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $T_1$ appears in a linear combination is $0$.

Similarly, for $T_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{T}}_2}& =& \frac{1}{24}\left[\left(1\cdot 3\cdot 9\right)+\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(6\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[27+0-3+6+18\right]\\ & =& \frac{1}{24}\left[48\right]\\ & =& 2\end{array}$

The number of times the irreducible representation for $T_2$ appears in a linear combination is $2$.

The character of $E$ symmetry species is $2$. This shows that the $E$ is doubly degenerate. The character of $T_2$ symmetry species is $3$. This shows that the $T_2$ is triply degenerate.

Therefore, $A_1$ irreducible representation contains $x^2+y^2+z^2$, this indicates that it will give one Raman-active vibration. The $E$ irreducible representation contains $\left(x^2-y^2,2z^2-x^2-y^2\right)$, this indicates that it will give one Raman-active vibration. This will be doubly degenerate. The $T_2$ irreducible representation contains $\left(xy,yz,zx\right)$ and $\left(x,y,z\right)$, this indicates that it will give two Raman-active vibration and IR-active vibrations.

Therefore, there are four Raman-active vibrations and two IR-active vibrations would be observed by ${\text{CH}}_{\text{4}}$. From the exercise 14.79, the number of IR-active vibrations for ${\text{CH}}_{\text{4}}$ is $2$.

Therefore, the number of Raman-active vibrations for the ${\text{CH}}_{\text{4}}$ is more than the IR-active vibrations.

Conclusion

The number of Raman-active vibrations for the ${\text{CH}}_{\text{4}}$ molecule is four. The number of Raman-active vibrations for the ${\text{CH}}_{\text{4}}$ is more than the IR-active vibrations.

Interpretation Introduction

(b)

Interpretation:

The number of Raman-active vibrations for the ${\text{CH}}_{\text{3}}\text{Cl}$ molecule is to be predicted. The answer is to be compared with the answers of exercise 14.79.

Concept introduction:

The characters of the irreducible representations of the given point group can be multiplied by each other. The only condition is the characters of the same symmetry operations are multiplied together. The multiplication of the characters is commutative.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

Answer to Problem 14.94E

The number of Raman-active vibrations for the ${\text{CH}}_{\text{3}}\text{Cl}$ molecule is six. The number of Raman-active vibrations for the ${\text{CH}}_{\text{3}}\text{Cl}$ are equal to the IR-active vibrations.

Explanation of Solution

The symmetry of ${\text{CH}}_{\text{3}}\text{Cl}$ molecule is $C_{\text{3V}}$.

The character table for point group $C_{\text{3V}}$ is shown below.

operations	$E$	$2C_3$	$3{\sigma }_{\text{v}}$
$N_{\text{stationary}}$	$5$	$2$	$3$
$\theta (°)$	$0$	$120$	$180$
${\chi }_{\text{r}}=\left(1+2\cos \theta \right)$	$3$	$0$	$-1$
${\chi }_{\text{tot}}=\pm N_{\text{stationary}}\left(1+2\cos \theta \right)$	$15$	$0$	$3$
${\chi }_l=\pm \left(1+2\cos \theta \right)$	$3$	$0$	$1$
${\chi }_{\text{v}}={\chi }_{\text{tot}}-{\chi }_l-{\chi }_{\text{r}}$	$9$	$0$	$3$

This reducible representation reduced using great orthogonality theorem as shown below.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

The order of the group is $6$.

The great orthogonality theorem orthogonality of the irreducible representation of $A_1$, $A_2$ and $E$ is shown below.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{3V}}$ point group, character of the class linear combination and number of symmetry operations for $A_1$.

$\begin{array}{rll}{a}_{{\text{A}}_1}& =& \frac{1}{6}\left[\left(1\cdot 1\cdot 9\right)+\left(2\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{6}\left[9+0+9\right]\\ & =& \frac{1}{6}\left[18\right]\\ & =& 3\end{array}$

The number of times the irreducible representation for $A_1$ appears in a linear combination is $3$.

Similarly, for $A_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{3V}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{A}}_2}& =& \frac{1}{6}\left[\left(1\cdot 1\cdot 9\right)+\left(2\cdot 1\cdot 0\right)+\left(3\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{6}\left[9+0-9\right]\\ & =& \frac{1}{6}\left[0\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_2$ appears in a linear combination is $0$.

Similarly, for $E$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{3V}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{\text{E}}& =& \frac{1}{6}\left[\left(1\cdot 2\cdot 9\right)+\left(2\cdot -1\cdot 0\right)+\left(3\cdot 0\cdot 3\right)\right]\\ & =& \frac{1}{6}\left[18+0+0\right]\\ & =& 3\end{array}$

The number of times the irreducible representation for $E$ appears in a linear combination is $3$.

The character of $E$ symmetry species is $2$. This shows that the $E$ is doubly degenerate. The character of $T_2$ symmetry species is $3$. This shows that the $T_2$ is triply degenerate.

Therefore, $A_1$ irreducible representation contains second order labels $x^2+y^2$, $z^2$ and first order label $z$, this indicates that it will give three Raman-active vibration and three IR-active vibrations.

The $E$ irreducible representation contains second order labels $\left(x^2-y^2,xy\right)$, $\left(xy,yz\right)$ and first order labels $\left(x,y\right)$, this indicates that it will give three Raman-active vibrations and three IR-active vibrations.

Therefore, there are six Raman-active vibrations and six IR-active vibrations would be observed by ${\text{CH}}_{\text{3}}\text{Cl}$. From the exercise 14.79, the number of IR-active vibrations for ${\text{CH}}_{\text{3}}\text{Cl}$ is $6$.

Therefore, the number of Raman-active vibrations for the ${\text{CH}}_{\text{3}}\text{Cl}$ are equal to the IR-active vibrations.

Conclusion

The number of Raman-active vibrations for the ${\text{CH}}_{\text{3}}\text{Cl}$ molecule is six. The number of Raman-active vibrations for the ${\text{CH}}_{\text{3}}\text{Cl}$ are equal to the IR-active vibrations.

Interpretation Introduction

(c)

Interpretation:

The number of Raman-active vibrations for the ${\text{CH}}_2{\text{Cl}}_2$ molecule is to be predicted. The answer is to be compared with the answers of exercise 14.79.

Concept introduction:

The characters of the irreducible representations of the given point group can be multiplied by each other. The only condition is the characters of the same symmetry operations are multiplied together. The multiplication of the characters is commutative.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

Answer to Problem 14.94E

The number of Raman-active vibrations for the ${\text{CH}}_2{\text{Cl}}_2$ molecule is nine. The number of Raman-active vibrations for the ${\text{CH}}_2{\text{Cl}}_2$ are greater than the IR-active vibrations.

Explanation of Solution

The symmetry of ${\text{CH}}_2{\text{Cl}}_2$ molecule is $C_{\text{2V}}$.

The character table for point group $C_{\text{2V}}$ is shown below.

operations	$E$	$C_2$	${\sigma }_{\text{v}}$	${\sigma }^{\text{'}}{}_{\text{v}}$
$N_{\text{stationary}}$	$5$	$1$	$3$	$3$
$\theta (°)$	$0$	$180$	$180$	$180$
${\chi }_{\text{r}}=\left(1+2\cos \theta \right)$	$3$	$-1$	$-1$	$-1$
${\chi }_{\text{tot}}=\pm N_{\text{stationary}}\left(1+2\cos \theta \right)$	$15$	$-1$	$3$	$3$
${\chi }_l=\pm \left(1+2\cos \theta \right)$	$3$	$-1$	$1$	$1$
${\chi }_{\text{v}}={\chi }_{\text{tot}}-{\chi }_l-{\chi }_{\text{r}}$	$9$	$1$	$3$	$3$

This reducible representation reduced using great orthogonality theorem as shown below.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

The order of the group is $4$.

The great orthogonality theorem orthogonality of the irreducible representation of $A_1$, $A_2$, $B_1$ and $B_2$ is shown below.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{2V}}$ point group, character of the class linear combination and number of symmetry operations for $A_1$.

$\begin{array}{rll}{a}_{{\text{A}}_1}& =& \frac{1}{4}\left[\left(1\cdot 1\cdot 9\right)+\left(1\cdot 1\cdot 1\right)+\left(1\cdot 1\cdot 3\right)+\left(1\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{4}\left[9+1+3+3\right]\\ & =& \frac{1}{4}\left[16\right]\\ & =& 4\end{array}$

The number of times the irreducible representation for $A_1$ appears in a linear combination is $4$.

Similarly, for $A_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{2V}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{A}}_2}& =& \frac{1}{4}\left[\left(1\cdot 1\cdot 9\right)+\left(1\cdot 1\cdot 1\right)+\left(1\cdot -1\cdot 3\right)+\left(1\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{4}\left[9+1-3-3\right]\\ & =& \frac{1}{4}\left[4\right]\\ & =& 1\end{array}$

The number of times the irreducible representation for $A_2$ appears in a linear combination is $1$.

Similarly, for $B_1$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{2V}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{B}}_{\text{1}}}& =& \frac{1}{4}\left[\left(1\cdot 1\cdot 9\right)+\left(1\cdot -1\cdot 1\right)+\left(1\cdot 1\cdot 3\right)+\left(1\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{4}\left[9-1+3-3\right]\\ & =& \frac{1}{4}\left[8\right]\\ & =& 2\end{array}$

The number of times the irreducible representation for $B_1$ appears in a linear combination is $2$.

Similarly, for $B_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{2V}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{B}}_{\text{2}}}& =& \frac{1}{4}\left[\left(1\cdot 1\cdot 9\right)+\left(1\cdot -1\cdot 1\right)+\left(1\cdot -1\cdot 3\right)+\left(1\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{4}\left[9-1-3+3\right]\\ & =& \frac{1}{4}\left[8\right]\\ & =& 2\end{array}$

The number of times the irreducible representation for $B_2$ appears in a linear combination is $2$.

Therefore, $A_1$ irreducible representation contains second order labels $x^2$, $y^2$, $z^2$ and first order label $z$, this indicates that it will give four Raman-active vibration and four IR-active vibrations.

The $A_2$ irreducible representation contains second order labels $\left(xy\right)$, this indicates that it will give one Raman-active vibrations.

The $B_1$ irreducible representation contains second order labels $\left(xz\right)$ and first order label $x$, this indicates that it will give two Raman-active vibrations and two IR-active vibrations.

The $B_2$ irreducible representation contains second order labels $\left(yz\right)$ and first order label $y$, this indicates that it will give two Raman-active vibrations and two IR-active vibrations.

Therefore, there are nine Raman-active vibrations and eight IR-active vibrations would be observed by ${\text{CH}}_2{\text{Cl}}_2$. From the exercise 14.79, the number of IR-active vibrations for ${\text{CH}}_{\text{2}}{\text{Cl}}_2$ is $8$.

Therefore, the number of Raman-active vibrations for the ${\text{CH}}_2{\text{Cl}}_2$ are greater than the IR-active vibrations.

Conclusion

The number of Raman-active vibrations for the ${\text{CH}}_2{\text{Cl}}_2$ molecule is nine. The number of Raman-active vibrations for the ${\text{CH}}_2{\text{Cl}}_2$ are greater than the IR-active vibrations.

Interpretation Introduction

(d)

Interpretation:

The number of Raman-active vibrations for the ${\text{CHCl}}_3$ molecule is to be predicted. The answer is to be compared with the answers of exercise 14.79.

Concept introduction:

The characters of the irreducible representations of the given point group can be multiplied by each other. The only condition is the characters of the same symmetry operations are multiplied together. The multiplication of the characters is commutative.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

Answer to Problem 14.94E

The number of Raman-active vibrations for the ${\text{CHCl}}_3$ molecule is six. The number of Raman-active vibrations for the ${\text{CHCl}}_3$ are equal to the IR-active vibrations.

Explanation of Solution

The symmetry of ${\text{CHCl}}_3$ molecule is $C_{\text{3V}}$.

The character table for point group $C_{\text{3V}}$ is shown below.

operations	$E$	$2C_3$	$3{\sigma }_{\text{v}}$
$N_{\text{stationary}}$	$5$	$2$	$3$
$\theta (°)$	$0$	$120$	$180$
${\chi }_{\text{r}}=\left(1+2\cos \theta \right)$	$3$	$0$	$-1$
${\chi }_{\text{tot}}=\pm N_{\text{stationary}}\left(1+2\cos \theta \right)$	$15$	$0$	$3$
${\chi }_l=\pm \left(1+2\cos \theta \right)$	$3$	$0$	$1$
${\chi }_{\text{v}}={\chi }_{\text{tot}}-{\chi }_l-{\chi }_{\text{r}}$	$9$	$0$	$3$

This reducible representation reduced using great orthogonality theorem as shown below.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

The order of the group is $6$.

The great orthogonality theorem orthogonality of the irreducible representation of $A_1$, $A_2$ and $E$ is shown below.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{3V}}$ point group, character of the class linear combination and number of symmetry operations for $A_1$.

$\begin{array}{rll}{a}_{{\text{A}}_1}& =& \frac{1}{6}\left[\left(1\cdot 1\cdot 9\right)+\left(2\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{6}\left[9+0+9\right]\\ & =& \frac{1}{6}\left[18\right]\\ & =& 3\end{array}$

The number of times the irreducible representation for $A_1$ appears in a linear combination is $3$.

Similarly, for $A_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{3V}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{A}}_2}& =& \frac{1}{6}\left[\left(1\cdot 1\cdot 9\right)+\left(2\cdot 1\cdot 0\right)+\left(3\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{6}\left[9+0-9\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_2$ appears in a linear combination is $0$.

Similarly, for $E$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{3V}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{\text{E}}& =& \frac{1}{6}\left[\left(1\cdot 2\cdot 9\right)+\left(2\cdot -1\cdot 0\right)+\left(3\cdot 0\cdot 3\right)\right]\\ & =& \frac{1}{6}\left[18+0+0\right]\\ & =& 3\end{array}$

The number of times the irreducible representation for $E$ appears in a linear combination is $3$.

The character of $E$ symmetry species is $2$. This shows that the $E$ is doubly degenerate. The character of $T_2$ symmetry species is $3$. This shows that the $T_2$ is triply degenerate.

Therefore, $A_1$ irreducible representation contains second order labels $x^2+y^2$, $z^2$ and first order label $z$, this indicates that it will give three Raman-active vibration and three IR-active vibrations.

The $E$ irreducible representation contains second order labels $\left(x^2-y^2,xy\right)$, $\left(xy,yz\right)$ and first order labels $\left(x,y\right)$, this indicates that it will give three Raman-active vibrations and three IR-active vibrations.

Therefore, there are six Raman-active vibrations and six IR-active vibrations would be observed by ${\text{CHCl}}_3$. From the exercise 14.79, the number of IR-active vibrations for ${\text{CH}}_{\text{3}}\text{Cl}$ is $6$.

Therefore, the number of Raman-active vibrations for the ${\text{CHCl}}_3$ are equal to the IR-active vibrations.

Conclusion

The number of Raman-active vibrations for the ${\text{CHCl}}_3$ molecule is six. The number of Raman-active vibrations for the ${\text{CHCl}}_3$ are equal to the IR-active vibrations.

Interpretation Introduction

(e)

Interpretation:

The number of Raman-active vibrations for the ${\text{CCl}}_{\text{4}}$ molecule is to be predicted. The answer is to be compared with the answers of exercise 14.79.

Concept introduction:

The characters of the irreducible representations of the given point group can be multiplied by each other. The only condition is the characters of the same symmetry operations are multiplied together. The multiplication of the characters is commutative.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

Answer to Problem 14.94E

The number of Raman-active vibrations for the ${\text{CCl}}_{\text{4}}$ molecule is four. The number of Raman-active vibrations for the ${\text{CCl}}_{\text{4}}$ is more than the IR-active vibrations.

Explanation of Solution

The symmetry of ${\text{CCl}}_{\text{4}}$ molecule is tetrahedral.

The character table for point group $T_{\text{d}}$ is shown below.

operations	$E$	$8C_3$	$3C_2$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$5$	$2$	$1$	$1$	$3$
$\theta (°)$	$0$	$120$	$180$	$90$	$180$
${\chi }_{\text{r}}=\left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$1$	$-1$
${\chi }_{\text{tot}}=\pm N_{\text{stationary}}\left(1+2\cos \theta \right)$	$15$	$0$	$-1$	$-1$	$3$
${\chi }_l=\pm \left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$-1$	$1$
${\chi }_{\text{v}}={\chi }_{\text{tot}}-{\chi }_l-{\chi }_{\text{r}}$	$9$	$0$	$1$	$-1$	$3$

This reducible representation reduced using great orthogonality theorem as shown below.

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

$a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.

• $h$ is the order of the group.

• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.

• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.

• $N$ is the number of symmetry operations.

The order of the group is $24$.

The great orthogonality theorem orthogonality of the irreducible representation of $A_1$, $A_2$$E$, $T_1$ and $T_2$ is shown below.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations for $A_1$.

$\begin{array}{rll}{a}_{{\text{A}}_1}& =& \frac{1}{24}\left[\left(1\cdot 1\cdot 9\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[9+0+3-6+18\right]\\ & =& \frac{1}{24}\left[24\right]\\ & =& 1\end{array}$

The number of times the irreducible representation for $A_1$ appears in a linear combination is $1$.

Similarly, for $A_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{A}}_2}& =& \frac{1}{24}\left[\left(1\cdot 1\cdot 9\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(6\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[9+0+3+6-18\right]\\ & =& \frac{1}{24}\left[0\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_2$ appears in a linear combination is $0$.

Similarly, for $E$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{\text{E}}& =& \frac{1}{24}\left[\left(1\cdot 2\cdot 9\right)+\left(8\cdot -1\cdot 0\right)+\left(3\cdot 2\cdot 1\right)+\left(6\cdot 0\cdot -1\right)+\left(6\cdot 0\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[18+0+6+0+0\right]\\ & =& \frac{1}{24}\left[24\right]\\ & =& 1\end{array}$

The number of times the irreducible representation for $E$ appears in a linear combination is $1$.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations for $T_1$.

$\begin{array}{rll}{a}_{{\text{T}}_1}& =& \frac{1}{24}\left[\left(1\cdot 3\cdot 9\right)+\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot -1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[27+0-3-6-18\right]\\ & =& \frac{1}{24}\left[0\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $T_1$ appears in a linear combination is $0$.

Similarly, for $T_2$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $T_{\text{d}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{T}}_2}& =& \frac{1}{24}\left[\left(1\cdot 3\cdot 9\right)+\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(6\cdot 1\cdot 3\right)\right]\\ & =& \frac{1}{24}\left[27+0-3+6+18\right]\\ & =& \frac{1}{24}\left[48\right]\\ & =& 2\end{array}$

The number of times the irreducible representation for $T_2$ appears in a linear combination is $2$.

The character of $E$ symmetry species is $2$. This shows that the $E$ is doubly degenerate. The character of $T_2$ symmetry species is $3$. This shows that the $T_2$ is triply degenerate.

Therefore, $A_1$ irreducible representation contains $x^2+y^2+z^2$, this indicates that it will give one Raman-active vibration. The $E$ irreducible representation contains $\left(x^2-y^2,2z^2-x^2-y^2\right)$, this indicates that it will give one Raman-active vibration. This will be doubly degenerate. The $T_2$ irreducible representation contains $\left(xy,yz,zx\right)$ and $\left(x,y,z\right)$, this indicates that it will give two Raman-active vibration and IR-active vibrations.

Therefore, there are four Raman-active vibrations and two IR-active vibrations would be observed by ${\text{CCl}}_{\text{4}}$. From the exercise 14.79, the number of IR-active vibrations for ${\text{CCl}}_{\text{4}}$ is $2$.

Therefore, the number of Raman-active vibrations for the ${\text{CCl}}_{\text{4}}$ is more than the IR-active vibrations.

Conclusion

The number of Raman-active vibrations for the ${\text{CCl}}_{\text{4}}$ molecule is four. The number of Raman-active vibrations for the ${\text{CCl}}_{\text{4}}$ is more than the IR-active vibrations.