Chapter 14, Problem 14.82E

Interpretation Introduction

Interpretation:

The cubane, ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$, 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 $3N-6$.

Answer to Problem 14.82E

The cubane, ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$, 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.

Figure 1

The point group of cubane is $O_{\text{h}}$. It is non-linear.

The character table for point group $O_{\text{h}}$ with the number of stationary atom after the operation and angle of rotation by The symmetry element is shown below.

$O_{\text{h}}$	$E$	$8C_3$	$3C_2$	$6C_4$	$6C_2^{\text{'}}$	$i$	$8S_6$	$3{\sigma }_{\text{h}}$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$16$	$1$	$0$	$0$	$4$	$8$	$0$	$0$	$0$	$0$
$\theta$	$0°$	$120°$	$180°$	$90°$	$180°$	$90°$	$60°$	$180°$	$90°$	$180°$

The expression $\left(1+2\cos \theta \right)$ is evaluate for $\theta =0°$ as shown below.

$\begin{array}{rll}1+2\cos \theta & =& 1+2\cos 0°\\ & =& 1+\left(2\right)\left(1\right)\\ & =& 3\end{array}$

Similarly, the value of $\left(1+2\cos \theta \right)$ is evaluate for all value of angle as shown in the table below.

$O_{\text{h}}$	$E$	$8C_3$	$3C_2$	$6C_4$	$6C_2^{\text{'}}$	$i$	$8S_6$	$3{\sigma }_{\text{h}}$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$16$	$1$	$0$	$0$	$4$	$8$	$0$	$0$	$0$	$0$
$\theta$	$0°$	$120°$	$180°$	$90°$	$180°$	$90°$	$60°$	$180°$	$90°$	$180°$
$\left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$

The expression ${\chi }_{\text{tot}}$ is shown below.

${\chi }_{\text{tot}}=\pm N_{\text{stationary}}\left(1+2\cos \theta \right)$

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

Substitute the value of $N_{\text{stationary}}=16$ and $\theta =0°$ in the above equation.

$\begin{array}{rll}{\chi }_{\text{tot}}& =& \left(16\right)\left(1+2\cos 0°\right)\\ & =& \left(16\right)\left(3\right)\\ & =& 48\end{array}$

Similarly, the value of ${\chi }_{\text{tot}}$ is evaluate for all symmetry elements as shown in the table below.

$O_{\text{h}}$	$E$	$8C_3$	$3C_2$	$6C_4$	$6C_2^{\text{'}}$	$i$	$8S_6$	$3{\sigma }_{\text{h}}$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$16$	$1$	$0$	$0$	$4$	$8$	$0$	$0$	$0$	$0$
$\theta$	$0°$	$120°$	$180°$	$90°$	$180°$	$90°$	$60°$	$180°$	$90°$	$180°$
$\left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$
${\chi }_{\text{tot}}$	$48$	$0$	$0$	$0$	$-4$	$-8$	$0$	$0$	$0$	$0$

The expression ${\chi }_{\text{r}}$ is shown below.

${\chi }_{\text{r}}=\pm \left(1+2\cos \theta \right)$

Substitute the value of $\theta =0°$ in the above equation.

$\begin{array}{rll}{\chi }_{\text{r}}& =& \left(1+2\cos 0°\right)\\ & =& 3\end{array}$

Similarly, the value of ${\chi }_{\text{r}}$ is evaluate for all symmetry elements as shown in the table below.

$O_{\text{h}}$	$E$	$8C_3$	$3C_2$	$6C_4$	$6C_2^{\text{'}}$	$i$	$8S_6$	$3{\sigma }_{\text{h}}$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$16$	$1$	$0$	$0$	$4$	$8$	$0$	$0$	$0$	$0$
$\theta$	$0°$	$120°$	$180°$	$90°$	$180°$	$90°$	$60°$	$180°$	$90°$	$180°$
$\left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$
${\chi }_{\text{tot}}$	$48$	$0$	$0$	$0$	$-4$	$-8$	$0$	$0$	$0$	$0$
${\chi }_{\text{r}}$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$

The expression ${\chi }_{\text{t}}$ is shown below.

${\chi }_{\text{t}}=\pm \left(1+2\cos \theta \right)$

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

Substitute the value of $\theta =0°$ in the above equation.

$\begin{array}{rll}{\chi }_{\text{t}}& =& \left(1+2\cos 0°\right)\\ & =& 3\end{array}$

Similarly, the value of ${\chi }_{\text{t}}$ is evaluate for all symmetry elements as shown in the table below.

$O_{\text{h}}$	$E$	$8C_3$	$3C_2$	$6C_4$	$6C_2^{\text{'}}$	$i$	$8S_6$	$3{\sigma }_{\text{h}}$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$16$	$1$	$0$	$0$	$4$	$8$	$0$	$0$	$0$	$0$
$\theta$	$0°$	$120°$	$180°$	$90°$	$180°$	$90°$	$60°$	$180°$	$90°$	$180°$
$\left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$
${\chi }_{\text{tot}}$	$48$	$0$	$0$	$0$	$-4$	$-8$	$0$	$0$	$0$	$0$
${\chi }_{\text{r}}$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$
${\chi }_{\text{t}}$	$3$	$0$	$-1$	$1$	$-1$	$-1$	$-2$	$1$	$-1$	$1$

The expression ${\chi }_{\text{v}}$ is shown below.

${\chi }_{\text{v}}={\chi }_{\text{tot}}-{\chi }_{\text{t}}-{\chi }_{\text{r}}$

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

Substitute the value of ${\chi }_{\text{tot}}$, ${\chi }_{\text{r}}$ and ${\chi }_{\text{t}}$ for $E$ in the above equation.

$\begin{array}{rll}{\chi }_{\text{v}}& =& 48-3-3\\ & =& 42\end{array}$

Similarly, the value of ${\chi }_{\text{v}}$ is evaluate for all symmetry elements as shown in the table below.

$O_{\text{h}}$	$E$	$8C_3$	$3C_2$	$6C_4$	$6C_2^{\text{'}}$	$i$	$8S_6$	$3{\sigma }_{\text{h}}$	$6S_4$	$6{\sigma }_{\text{d}}$
$N_{\text{stationary}}$	$16$	$1$	$0$	$0$	$4$	$8$	$0$	$0$	$0$	$0$
$\theta$	$0°$	$120°$	$180°$	$90°$	$180°$	$90°$	$60°$	$180°$	$90°$	$180°$
$\left(1+2\cos \theta \right)$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$
${\chi }_{\text{tot}}$	$48$	$0$	$0$	$0$	$-4$	$-8$	$0$	$0$	$0$	$0$
${\chi }_{\text{r}}$	$3$	$0$	$-1$	$1$	$-1$	$1$	$2$	$-1$	$1$	$-1$
${\chi }_{\text{t}}$	$3$	$0$	$-1$	$1$	$-1$	$-1$	$-2$	$1$	$-1$	$1$
${\chi }_{\text{v}}$	$42$	$0$	$2$	$-2$	$-2$	$-8$	$0$	$0$	$0$	$0$

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 term ${\chi }_{\text{v}}$ is taken in place of ${\chi }_{\text{linear}\text{combo}}$.

The irreducible representation for point group $O_{\text{h}}$ is shown below.

$O_{\text{h}}$	$E$	$8C_3$	$3C_2$	$6C_4$	$6C_2^{\text{'}}$	$i$	$8S_6$	$3{\sigma }_{\text{h}}$	$6S_4$	$6{\sigma }_{\text{d}}$
$\Gamma$	$3$	$0$	$-1$	$-1$	$1$	$-3$	$0$	$1$	$1$	$-1$

Substitute the $h$, ${\chi }_{\Gamma }$, ${\chi }_{\text{v}}$, and $N$ in the above equation.

$\begin{array}{rll}{a}_{\Gamma }& =& \frac{1}{48}\left(\begin{array}{l}\left(1\right)\left(3\right)\left(42\right)+\left(8\right)\left(0\right)\left(0\right)+\left(3\right)\left(-1\right)\left(2\right)+\left(6\right)\left(-1\right)\left(-2\right)+\\ \left(6\right)\left(1\right)\left(-1\right)+\left(1\right)\left(-3\right)\left(-8\right)+\left(8\right)\left(0\right)\left(0\right)+\left(3\right)\left(1\right)\left(0\right)+\\ \left(6\right)\left(1\right)\left(0\right)+\left(6\right)\left(-1\right)\left(0\right)\end{array}\right)\\ & =& \frac{1}{48}\left(126+0-6+12-6+24+0+0+0+0\right)\\ & =& \frac{1}{48}\left(144\right)\\ & =& 3\end{array}$

Therefore, the cubane, ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$, has only three IR-active vibrational. The degeneracies of each IR-active vibrational is triply degenerated due to the presence of label $\left(x,y,z\right)$. 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, ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$, 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 $9$.