Chapter 13, Problem 13.71E

Construct the symmetry-adapted linear combination molecular orbitals for hydrogen sulfide, ${\text{H}}_{\text{2}}\text{S}$.

Interpretation Introduction

Interpretation:

The symmetry-adapted linear combination molecular orbitals for hydrogen sulfide are to be constructed.

Concept introduction:

A symmetry operation is defined as an action on an object to reproduce an arrangement that is identical to its original spatial arrangement. The spatial arrangement of the object remains identical after a symmetry operation. The point of reference through which a symmetry operation takes place is termed as a symmetry element.

Answer to Problem 13.71E

The symmetry-adapted linear combination molecular orbitals for hydrogen sulfide with $A_{\text{1}}$ are shown below.

$\begin{array}{l}{\Psi }_{A_1}=\frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{A_1}=3s_{\text{S}}\\ {\Psi }_{A_1}=3p_{z,\text{S}}\end{array}$

The symmetry-adapted linear combination molecular orbitals for hydrogen sulfide with $B_{\text{1}}$ are shown below.

$\begin{array}{rll}{\Psi }_{B_{\text{1}}}& =& \frac{1}{2}\left(1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{B_{\text{1}}}& =& \frac{1}{2}\left(1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{B_{\text{1}}}& =& 3p_{x,\text{S}}\end{array}$

The symmetry-adapted linear combination molecular orbitals for hydrogen sulfide with $B_{\text{2}}$ are shown below.

${\Psi }_{B_2}=3p_{y,\text{S}}$

Explanation of Solution

The character table for $C_{\text{2v}}$ symmetry is shown below.

$C_{\text{2v}}$	$E$	$C_2$	${\sigma }_{\text{v}}$	${\sigma }_{\text{v}}{}^{\text{'}}$
$A_{\text{1}}$	$1$	$1$	$1$	$1$
$A_2$	$1$	$1$	$-1$	$-1$
$B_{\text{1}}$	$1$	$-1$	$1$	$-1$
$B_2$	$1$	$-1$	$-1$	$1$

The structure of ${\text{H}}_{\text{2}}\text{S}$ is shown below.

Figure 1

The geometry of hydrogen selenide is bent due to presence of two lone pair of electrons. The $180°$ out of plane rotation of the molecule produces same structure. Hence, it contains $C_2$ symmetry operation. It contains vertical plane of symmetry, ${\sigma }_{\text{v}}$ and ${\sigma }_{\text{v}}\text{'}$ in two different planes. Hence, the point group of the molecule is $C_{2\text{v}}$.

The atomic orbitals of $\text{H}$ which participates in bonding are $1s_{\text{H1}}$ and $1s_{\text{H2}}$. The atomic orbitals of sulfur atom which are involved in the bond formation are $1s_{\text{S}}$, $2s_{\text{S}}$, $2p_{x,\text{S}}$, $2p_{y,\text{S}}$, $2p_{z,\text{S}}$, $3s_{\text{S}}$, $3p_{x,\text{S}}$, $3p_{y,\text{S}}$ and $3p_{y,\text{S}}$. Only outer orbitals are to be considered. The table for $A_{\text{1}}$ is shown below.

	$1s_{\text{H1}}$	$1s_{\text{H2}}$	$3s_{\text{S}}$	$3p_{x,\text{S}}$	$3p_{y,\text{S}}$	$3p_{z,\text{S}}$
$E$	$1\cdot 1s_{\text{H1}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 3s_{\text{S}}$	$1\cdot 3p_{x,\text{S}}$	$1\cdot 3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
$C_2$	$1\cdot 1s_{\text{H2}}$	$1\cdot 1s_{\text{H1}}$	$1\cdot 3s_{\text{S}}$	$1\cdot -3p_{x,\text{S}}$	$1\cdot -3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}$	$1\cdot 1s_{\text{H1}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 3s_{\text{S}}$	$1\cdot 3p_{x,\text{S}}$	$1\cdot -3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}^{\text{'}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 1s_{\text{H1}}$	$1\cdot 3s_{\text{S}}$	$1\cdot -3p_{x,\text{S}}$	$1\cdot 3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$

The six linear combinations can be obtained from the above table as shown below.

$\begin{array}{l}{\Psi }_{A_1}=\frac{1}{4}\left(1s_{\text{H1}}+1s_{\text{H2}}+1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{A_1}=\frac{1}{4}\left(1s_{\text{H2}}+1s_{\text{H1}}+1s_{\text{H2}}+1s_{\text{H1}}\right)\\ {\Psi }_{A_1}=\frac{1}{4}\left(3s_{\text{S}}+3s_{\text{S}}+3s_{\text{S}}+3s_{\text{S}}\right)\\ {\Psi }_{A_1}=\frac{1}{4}\left(3p_{x,\text{S}}-3p_{x,\text{S}}+3p_{x,\text{S}}-3p_{x,\text{S}}\right)\end{array}$

The linear combination for $3p_{y,\text{S}}$ and $3p_{z,\text{S}}$ are shown below.

$\begin{array}{l}{\Psi }_{A_1}=\frac{1}{4}\left(3p_{y,\text{S}}-3p_{y,\text{S}}-3p_{y,\text{S}}+3p_{y,\text{S}}\right)\\ {\Psi }_{A_1}=\frac{1}{4}\left(3p_{z,\text{S}}+3p_{z,\text{S}}+3p_{z,\text{S}}+3p_{z,\text{S}}\right)\end{array}$

The first and second combinations are the same. The fourth and fifth combinations are exactly zero. Therefore, the unique wavefunctions after the simplification for $A_{\text{1}}$ are shown below.

$\begin{array}{l}{\Psi }_{A_1}=\frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{A_1}=3s_{\text{S}}\\ {\Psi }_{A_1}=3p_{z,\text{S}}\end{array}$

The table for $A_{\text{2}}$ is shown below.

	$1s_{\text{H1}}$	$1s_{\text{H2}}$	$3s_{\text{S}}$	$3p_{x,\text{S}}$	$3p_{y,\text{S}}$	$3p_{z,\text{S}}$
$E$	$1\cdot 1s_{\text{H1}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 3s_{\text{S}}$	$1\cdot 3p_{x,\text{S}}$	$1\cdot 3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
$C_2$	$1\cdot 1s_{\text{H2}}$	$1\cdot 1s_{\text{H1}}$	$1\cdot 3s_{\text{S}}$	$1\cdot -3p_{x,\text{S}}$	$1\cdot -3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}$	$-1\cdot 1s_{\text{H1}}$	$-1\cdot 1s_{\text{H2}}$	$-1\cdot 3s_{\text{S}}$	$-1\cdot 3p_{x,\text{S}}$	$-1\cdot -3p_{y,\text{S}}$	$-1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}^{\text{'}}$	$-1\cdot 1s_{\text{H2}}$	$-1\cdot 1s_{\text{H1}}$	$-1\cdot 3s_{\text{S}}$	$-1\cdot -3p_{x,\text{S}}$	$-1\cdot 3p_{y,\text{S}}$	$-1\cdot 3p_{z,\text{S}}$

The six linear combinations can be obtained from the above table as shown below.

$\begin{array}{l}{\Psi }_{A_2}=\frac{1}{4}\left(1s_{\text{H1}}+1s_{\text{H2}}-1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{A_2}=\frac{1}{4}\left(1s_{\text{H2}}+1s_{\text{H1}}-1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{A_2}=\frac{1}{4}\left(3s_{\text{S}}+3s_{\text{S}}-3s_{\text{S}}-3s_{\text{S}}\right)\\ {\Psi }_{A_2}=\frac{1}{4}\left(3p_{x,\text{S}}-3p_{x,\text{S}}-3p_{x,\text{S}}+3p_{x,\text{S}}\right)\end{array}$

The linear combination for $3p_{y,\text{S}}$ and $3p_{z,\text{S}}$ are shown below.

$\begin{array}{l}{\Psi }_{A_2}=\frac{1}{4}\left(3p_{y,\text{S}}-3p_{y,\text{S}}+3p_{y,\text{S}}-3p_{y,\text{S}}\right)\\ {\Psi }_{A_2}=\frac{1}{4}\left(3p_{z,\text{S}}+3p_{z,\text{S}}-3p_{z,\text{S}}-3p_{z,\text{S}}\right)\end{array}$

All combinations are exactly zero.

The table for $B_{\text{1}}$ is shown below.

	$1s_{\text{H1}}$	$1s_{\text{H2}}$	$3s_{\text{S}}$	$3p_{x,\text{S}}$	$3p_{y,\text{S}}$	$3p_{z,\text{S}}$
$E$	$1\cdot 1s_{\text{H1}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 3s_{\text{S}}$	$1\cdot 3p_{x,\text{S}}$	$1\cdot 3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
$C_2$	$-1\cdot 1s_{\text{H2}}$	$-1\cdot 1s_{\text{H1}}$	$-1\cdot 3s_{\text{S}}$	$-1\cdot -3p_{x,\text{S}}$	$-1\cdot -3p_{y,\text{S}}$	$-1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}$	$1\cdot 1s_{\text{H1}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 3s_{\text{S}}$	$1\cdot 3p_{x,\text{S}}$	$1\cdot -3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}^{\text{'}}$	$-1\cdot 1s_{\text{H2}}$	$-1\cdot 1s_{\text{H1}}$	$-1\cdot 3s_{\text{S}}$	$-1\cdot -3p_{x,\text{S}}$	$-1\cdot 3p_{y,\text{S}}$	$-1\cdot 3p_{z,\text{S}}$

The six linear combinations can be obtained from the above table as shown below.

$\begin{array}{l}{\Psi }_{B_{\text{1}}}=\frac{1}{4}\left(1s_{\text{H1}}-1s_{\text{H2}}+1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{B_{\text{1}}}=\frac{1}{4}\left(1s_{\text{H2}}-1s_{\text{H1}}+1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{B_{\text{1}}}=\frac{1}{4}\left(3s_{\text{S}}-3s_{\text{S}}+3s_{\text{S}}-3s_{\text{S}}\right)\\ {\Psi }_{B_{\text{1}}}=\frac{1}{4}\left(3p_{x,\text{S}}+3p_{x,\text{S}}+3p_{x,\text{S}}+3p_{x,\text{S}}\right)\end{array}$

The linear combination for $3p_{y,\text{S}}$ and $3p_{z,\text{S}}$ are shown below.

$\begin{array}{l}{\Psi }_{B_{\text{1}}}=\frac{1}{4}\left(3p_{y,\text{S}}+3p_{y,\text{S}}-3p_{y,\text{S}}-3p_{y,\text{S}}\right)\\ {\Psi }_{B_{\text{1}}}=\frac{1}{4}\left(3p_{z,\text{S}}-3p_{z,\text{S}}+3p_{z,\text{S}}-3p_{z,\text{S}}\right)\end{array}$

The third, fifth and sixth combinations are exactly zero. Therefore, the unique wavefunctions after the simplification for $B_{\text{1}}$ are shown below.

$\begin{array}{rll}{\Psi }_{B_{\text{1}}}& =& \frac{1}{2}\left(1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{B_{\text{1}}}& =& \frac{1}{2}\left(1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{B_{\text{1}}}& =& 3p_{x,\text{S}}\end{array}$

The table for $B_2$ is shown below.

	$1s_{\text{H1}}$	$1s_{\text{H2}}$	$3s_{\text{S}}$	$3p_{x,\text{S}}$	$3p_{y,\text{S}}$	$3p_{z,\text{S}}$
$E$	$1\cdot 1s_{\text{H1}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 3s_{\text{S}}$	$1\cdot 3p_{x,\text{S}}$	$1\cdot 3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$
$C_2$	$-1\cdot 1s_{\text{H2}}$	$-1\cdot 1s_{\text{H1}}$	$-1\cdot 3s_{\text{S}}$	$-1\cdot -3p_{x,\text{S}}$	$-1\cdot -3p_{y,\text{S}}$	$-1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}$	$-1\cdot 1s_{\text{H1}}$	$-1\cdot 1s_{\text{H2}}$	$-1\cdot 3s_{\text{S}}$	$-1\cdot 3p_{x,\text{S}}$	$-1\cdot -3p_{y,\text{S}}$	$-1\cdot 3p_{z,\text{S}}$
${\sigma }_{\text{v}}^{\text{'}}$	$1\cdot 1s_{\text{H2}}$	$1\cdot 1s_{\text{H1}}$	$1\cdot 3s_{\text{S}}$	$1\cdot -3p_{x,\text{S}}$	$1\cdot 3p_{y,\text{S}}$	$1\cdot 3p_{z,\text{S}}$

The six linear combinations can be obtained from the above table as shown below.

$\begin{array}{l}{\Psi }_{B_2}=\frac{1}{4}\left(1s_{\text{H1}}-1s_{\text{H2}}-1s_{\text{H1}}+1s_{\text{H2}}\right)=0\\ {\Psi }_{B_2}=\frac{1}{4}\left(1s_{\text{H2}}-1s_{\text{H1}}-1s_{\text{H2}}+1s_{\text{H1}}\right)=0\\ {\Psi }_{B_2}=\frac{1}{4}\left(3s_{\text{S}}-3s_{\text{S}}-3s_{\text{S}}+3s_{\text{S}}\right)=0\\ {\Psi }_{B_2}=\frac{1}{4}\left(3p_{x,\text{S}}+3p_{x,\text{S}}-3p_{x,\text{S}}-3p_{x,\text{S}}\right)=0\end{array}$

The linear combination for $3p_{y,\text{S}}$ and $3p_{z,\text{S}}$ are shown below.

$\begin{array}{l}{\Psi }_{B_2}=\frac{1}{4}\left(3p_{y,\text{S}}+3p_{y,\text{S}}+3p_{y,\text{S}}+3p_{y,\text{S}}\right)=3p_{y,\text{S}}\\ {\Psi }_{B_2}=\frac{1}{4}\left(3p_{z,\text{S}}-3p_{z,\text{S}}-3p_{z,\text{S}}+3p_{z,\text{S}}\right)=0\end{array}$

All combinations except the fifth combination are exactly zero. Therefore, the unique wavefunctions after the simplification for $B_2$ are shown below.

${\Psi }_{B_2}=3p_{y,\text{S}}$

Conclusion

The symmetry-adapted linear combination molecular orbitals for hydrogen sulfide with $A_{\text{1}}$ are shown below.

$\begin{array}{l}{\Psi }_{A_1}=\frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{A_1}=3s_{\text{S}}\\ {\Psi }_{A_1}=3p_{z,\text{S}}\end{array}$

The symmetry-adapted linear combination molecular orbitals for hydrogen sulfide with $B_{\text{1}}$ are shown below.

$\begin{array}{rll}{\Psi }_{B_{\text{1}}}& =& \frac{1}{2}\left(1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{B_{\text{1}}}& =& \frac{1}{2}\left(1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{B_{\text{1}}}& =& 3p_{x,\text{S}}\end{array}$

The symmetry-adapted linear combination molecular orbitals for hydrogen sulfide with $B_{\text{2}}$ are shown below.

${\Psi }_{B_2}=3p_{y,\text{S}}$