Chapter 13, Problem 13.69E

Interpretation Introduction

Interpretation:

The symmetry labels of the $\text{H}$-like $p$ orbitals in $O_{\text{h}}$ symmetry is to be determined.

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 13.69E

The symmetry labels of the $\text{H}$-like $p$ orbitals in $O_{\text{h}}$ symmetry is $T_{\text{1u}}$.

Explanation of Solution

The formula to calculate the value of $\chi C_3$ is,

$\chi C_3=& 1+2\cos \theta$ …(1)

Substitute the value of $\theta =& 120°$ in equation (1).

$\begin{array}{rll}\chi C_3& =& 1+2\cos 120°\\ & =& 0\end{array}$

The formula to calculate the value of $\chi C_2$ is,

$\chi C_2=& 1+2\cos \theta$ …(2)

Substitute the value of $\theta =& 180°$ in equation (2).

$\begin{array}{rll}\chi C_2& =& 1+2\cos 180°\\ & =& -1\end{array}$

The formula to calculate the value of $\chi C_4$ is,

$\chi C_4=& 1+2\cos \theta$ …(3)

Substitute the value of $\theta =& 90°$ in equation (3).

$\begin{array}{rll}\chi C_4& =& 1+2\cos 90°\\ & =& 1\end{array}$

The formula to calculate the value of $\chi S_4$ is,

$\chi S_4=& -1+2\cos \theta$ …(4)

Substitute the value of $\theta =& 90°$ in equation (4).

$\begin{array}{rll}\chi S_4& =& -1+2\cos 90°\\ & =& -1\end{array}$

The formula to calculate the value of $\chi S_6$ is,

$\chi S_6=& -1+2\cos \theta$ …(5)

Substitute the value of $\theta =& 60°$ in equation (5).

$\begin{array}{rll}\chi S_6& =& -1+2\cos 60°\\ & =& 0\end{array}$

Therefore, the character table for $p$ orbital is shown below.

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

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 $48$.

Substitute the value of order of the group, character of the class of the irreducible representation from character table of $O_{\text{h}}$ point group, character of the class linear combination and number of symmetry operations for $A_{1g}$.

$\begin{array}{rll}{a}_{{\text{A}}_{1g}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 1\cdot 3\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot -1\right)+\left(6\cdot 1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(1\cdot 1\cdot -3\right)\\ +\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot 1\cdot 1\right)\end{array}\right]\\ & =& \frac{1}{48}\left[0\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_{1g}$ appears in a linear combination is $0$.

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

$\begin{array}{rll}{a}_{{\text{A}}_{\text{2g}}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 1\cdot 3\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot -1\right)+\left(6\cdot -1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(1\cdot 1\cdot -3\right)\\ +\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(6\cdot -1\cdot 1\right)\end{array}\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_{2g}$ appears in a linear combination is $0$.

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

$\begin{array}{rll}{a}_{{\text{E}}_{\text{g}}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 2\cdot 3\right)+\left(8\cdot -1\cdot 0\right)+\left(3\cdot 2\cdot -1\right)+\left(6\cdot 0\cdot 1\right)+\left(6\cdot 0\cdot -1\right)+\left(1\cdot 2\cdot -3\right)\\ +\left(8\cdot -1\cdot 0\right)+\left(3\cdot 2\cdot 1\right)+\left(6\cdot 0\cdot -1\right)+\left(6\cdot 0\cdot 1\right)\end{array}\right]\\ & =& 0\end{array}$

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

Similarly, for $T_{1g}$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $O_{\text{h}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{T}}_{\text{1g}}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 3\cdot 3\right)+\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot -1\right)+\left(6\cdot 1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(1\cdot 3\cdot -3\right)\\ +\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot -1\cdot 1\right)\end{array}\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $T_{1g}$ appears in a linear combination is $0$.

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

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

The number of times the irreducible representation for $T_{2g}$ appears in a linear combination is $0$.

Similarly, for $A_{\text{1u}}$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $O_{\text{h}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{T}}_{A_{\text{1u}}}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 1\cdot 3\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot -1\right)+\left(6\cdot 1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(1\cdot -1\cdot -3\right)\\ +\left(8\cdot -1\cdot 0\right)+\left(3\cdot -1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(6\cdot -1\cdot 1\right)\end{array}\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_{\text{1u}}$ appears in a linear combination is $0$.

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

$\begin{array}{rll}{a}_{{\text{T}}_{A_{\text{2u}}}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 1\cdot 3\right)+\left(8\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot -1\right)+\left(6\cdot -1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(1\cdot -1\cdot -3\right)\\ +\left(8\cdot -1\cdot 0\right)+\left(3\cdot -1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot 1\cdot 1\right)\end{array}\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $A_{\text{2u}}$ appears in a linear combination is $0$.

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

$\begin{array}{rll}{a}_{{\text{E}}_u}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 2\cdot 3\right)+\left(8\cdot -1\cdot 0\right)+\left(3\cdot 2\cdot -1\right)+\left(6\cdot 0\cdot 1\right)+\left(6\cdot 0\cdot -1\right)+\left(1\cdot -2\cdot -3\right)\\ +\left(8\cdot 1\cdot 0\right)+\left(3\cdot -2\cdot 1\right)+\left(6\cdot 0\cdot -1\right)+\left(6\cdot 0\cdot 1\right)\end{array}\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $E_{\text{u}}$ appears in a linear combination is $0$.

Similarly, for $T_{\text{1u}}$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $O_{\text{h}}$ point group, character of the class linear combination and number of symmetry operations.

$\begin{array}{rll}{a}_{{\text{T}}_{1u}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 3\cdot 3\right)+\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot -1\right)+\left(6\cdot 1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(1\cdot -3\cdot -3\right)\\ +\left(8\cdot 0\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot -1\cdot -1\right)+\left(6\cdot 1\cdot 1\right)\end{array}\right]\\ & =& \frac{1}{48}\left[48\right]\\ & =& 1\end{array}$

The number of times the irreducible representation for $T_{\text{1u}}$ appears in a linear combination is $1$.

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

$\begin{array}{rll}{a}_{{\text{T}}_{1u}}& =& \frac{1}{48}\left[\begin{array}{l}\left(1\cdot 3\cdot 3\right)+\left(8\cdot 0\cdot 0\right)+\left(3\cdot -1\cdot -1\right)+\left(6\cdot -1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(1\cdot -3\cdot -3\right)\\ +\left(8\cdot 0\cdot 0\right)+\left(3\cdot 1\cdot 1\right)+\left(6\cdot 1\cdot -1\right)+\left(6\cdot -1\cdot 1\right)\end{array}\right]\\ & =& 0\end{array}$

The number of times the irreducible representation for $T_{\text{2u}}$ appears in a linear combination is $0$.

Conclusion

The symmetry labels of the $\text{H}$-like $p$ orbitals in $O_{\text{h}}$ symmetry is $T_{\text{1u}}$.