Chapter 13, Problem 13.72E

Interpretation Introduction

Interpretation:

The effect of the involvement of orbitals for sulfur on the symmetry-adapted linear combination of molecular orbitals for hydrogen sulfide is to be stated.

Concept introduction:

Symmetry-adapted linear combination is referred to as the wavefunction of the symmetry equivalent orbitals. The symmetry-adapted linear combination is used to determine the binding schemes and symmetries of the molecule. The SALC’s are determined with the help of character tables and great orthogonality theorem.

Answer to Problem 13.72E

The symmetry-adapted linear combinations of molecular orbitals for hydrogen sulfide involving the core orbitals for sulfur are shown below.

$\begin{array}{rll}{\Psi }_{{\text{A}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& 1s_{\text{S}}\\ {\Psi }_{{\text{A}}_1}& =& 2s_{\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{A}}_1}=2p_{z,\text{S}}\\ {\Psi }_{{\text{A}}_1}=3s_{\text{S}}\\ {\Psi }_{{\text{A}}_1}=3p_{z,\text{S}}\end{array}$

$\begin{array}{rll}{\Psi }_{{\text{B}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{{\text{B}}_1}& =& \frac{1}{2}\left(1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{{\text{B}}_1}& =& 2p_{x,\text{S}}\\ {\Psi }_{{\text{B}}_1}& =& 3p_{x,\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{B}}_2}=2p_{y,\text{S}}\\ {\Psi }_{{\text{B}}_2}=3p_{y,\text{S}}\end{array}$

Explanation of Solution

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 given 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. Therefore, the point group of the given 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 oxygen atom which involves 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 $3p_{y,\text{S}}$. The symmetry operation $E$ has no effect on the orbitals. The symmetry has no effect on the hydrogen $1s$ orbitals but can reverse $2p_{y,\text{S}}$ and $3p_{y,\text{S}}$ orbitals. The symmetry labels are multiplied with the character for the symmetry operation of that irreducible representation. The table obtained for ${\text{A}}_1$ symmetry is shown below.

$\begin{array}{cccccccccccc}& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ E& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ C_2& 1s_{\text{H2}}& 1s_{\text{H1}}& 1s_{\text{S}}& 2s_{\text{S}}& 1\cdot -2p_{x,\text{S}}& 1\cdot -2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 1\cdot -3p_{x,\text{S}}& 1\cdot -3p_{y,\text{S}}& 3p_{z,\text{S}}\\ {\sigma }_{\text{v}}& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 1\cdot -2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 1\cdot -3p_{y,\text{S}}& 3p_{z,\text{S}}\\ {\sigma }_{\text{v}}^{\text{'}}& 1s_{\text{H2}}& 1s_{\text{H1}}& 1s_{\text{S}}& 2s_{\text{S}}& 1\cdot -2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 1\cdot -3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\end{array}$

The summation of each column in the table gives the eleven linear combinations as shown below.

$\begin{array}{rll}{\Psi }_{{\text{A}}_1}& =& \frac{1}{4}\left(1s_{\text{H1}}+1s_{\text{H2}}+1s_{\text{H1}}+1s_{\text{H2}}\right)=\frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& \frac{1}{4}\left(1s_{\text{H2}}+1s_{\text{H1}}+1s_{\text{H2}}+1s_{\text{H1}}\right)=\frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& \frac{1}{4}\left(1s_{\text{S}}+1s_{\text{S}}+1s_{\text{S}}+1s_{\text{S}}\right)=1s_{\text{S}}\\ {\Psi }_{{\text{A}}_1}& =& \frac{1}{4}\left(2s_{\text{S}}+2s_{\text{S}}+2s_{\text{S}}+2s_{\text{S}}\right)=2s_{\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{A}}_1}=\frac{1}{4}\left(2p_{x,\text{S}}-2p_{x,\text{S}}+2p_{x,\text{S}}-2p_{x,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_1}=\frac{1}{4}\left(2p_{y,\text{S}}-2p_{y,\text{S}}-2p_{y,\text{S}}+2p_{y,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_1}=\frac{1}{4}\left(2p_{z,\text{S}}+2p_{z,\text{S}}+2p_{z,\text{S}}+2p_{z,\text{S}}\right)=2p_{z,\text{S}}\\ {\Psi }_{{\text{A}}_1}=\frac{1}{4}\left(3s_{\text{S}}+3s_{\text{S}}+3s_{\text{S}}+3s_{\text{S}}\right)=3s_{\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{A}}_1}=\frac{1}{4}\left(3p_{x,\text{S}}-3p_{x,\text{S}}+3p_{x,\text{S}}-3p_{x,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_1}=\frac{1}{4}\left(3p_{y,\text{S}}-3p_{y,\text{S}}-3p_{y,\text{S}}+3p_{y,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_1}=\frac{1}{4}\left(3p_{z,\text{S}}+3p_{z,\text{S}}+3p_{z,\text{S}}+3p_{z,\text{S}}\right)=3p_{z,\text{S}}\end{array}$

Similarly the table obtained for ${\text{A}}_2$ symmetry is shown below.

$\begin{array}{cccccccccccc}& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ E& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ C_2& 1s_{\text{H2}}& 1s_{\text{H1}}& 1s_{\text{S}}& 2s_{\text{S}}& 1\cdot -2p_{x,\text{S}}& 1\cdot -2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 1\cdot -3p_{x,\text{S}}& 1\cdot -3p_{y,\text{S}}& 3p_{z,\text{S}}\\ {\sigma }_{\text{v}}& -1s_{\text{H1}}& -1s_{\text{H2}}& -1s_{\text{S}}& -2s_{\text{S}}& -2p_{x,\text{S}}& -1\cdot -2p_{y,\text{S}}& -2p_{z,\text{S}}& -3s_{\text{S}}& -3p_{x,\text{S}}& -1\cdot -3p_{y,\text{S}}& -3p_{z,\text{S}}\\ {\sigma }_{\text{v}}^{\text{'}}& -1s_{\text{H2}}& -1s_{\text{H1}}& -1s_{\text{S}}& -2s_{\text{S}}& -1\cdot -2p_{x,\text{S}}& -2p_{y,\text{S}}& -2p_{z,\text{S}}& -3s_{\text{S}}& -1\cdot -3p_{x,\text{S}}& -3p_{y,\text{S}}& -3p_{z,\text{S}}\end{array}$

The summation of each column in the table gives the eleven linear combinations as shown below.

$\begin{array}{rllll}{\Psi }_{{\text{A}}_2}& =& \frac{1}{4}\left(1s_{\text{H1}}+1s_{\text{H2}}-1s_{\text{H1}}-1s_{\text{H2}}\right)& =& 0\\ {\Psi }_{{\text{A}}_2}& =& \frac{1}{4}\left(1s_{\text{H2}}+1s_{\text{H1}}-1s_{\text{H2}}-1s_{\text{H1}}\right)& =& 0\\ {\Psi }_{{\text{A}}_2}& =& \frac{1}{4}\left(1s_{\text{S}}+1s_{\text{S}}-1s_{\text{S}}-1s_{\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_2}& =& \frac{1}{4}\left(2s_{\text{S}}+2s_{\text{S}}-2s_{\text{S}}-2s_{\text{S}}\right)=0\end{array}$

$\begin{array}{l}{\Psi }_{{\text{A}}_2}=\frac{1}{4}\left(2p_{x,\text{S}}-2p_{x,\text{S}}-2p_{x,\text{S}}+2p_{x,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_2}=\frac{1}{4}\left(2p_{y,\text{S}}-2p_{y,\text{S}}+2p_{y,\text{S}}-2p_{y,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_2}=\frac{1}{4}\left(2p_{z,\text{S}}+2p_{z,\text{S}}-2p_{z,\text{S}}-2p_{z,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_2}=\frac{1}{4}\left(3s_{\text{S}}+3s_{\text{S}}-3s_{\text{S}}-3s_{\text{S}}\right)=0\end{array}$

$\begin{array}{l}{\Psi }_{{\text{A}}_2}=\frac{1}{4}\left(3p_{x,\text{S}}-3p_{x,\text{S}}-3p_{x,\text{S}}+3p_{x,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_2}=\frac{1}{4}\left(3p_{y,\text{S}}-3p_{y,\text{S}}+3p_{y,\text{S}}-3p_{y,\text{S}}\right)=0\\ {\Psi }_{{\text{A}}_2}=\frac{1}{4}\left(3p_{z,\text{S}}+3p_{z,\text{S}}-3p_{z,\text{S}}-3p_{z,\text{S}}\right)=0\end{array}$

Similarly the table obtained for ${\text{B}}_1$ symmetry is shown below.

$\begin{array}{cccccccccccc}& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ E& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ C_2& -1s_{\text{H2}}& -1s_{\text{H1}}& -1s_{\text{S}}& -2s_{\text{S}}& -1\cdot -2p_{x,\text{S}}& -1\cdot -2p_{y,\text{S}}& -2p_{z,\text{S}}& -3s_{\text{S}}& -1\cdot -3p_{x,\text{S}}& -1\cdot -3p_{y,\text{S}}& -3p_{z,\text{S}}\\ {\sigma }_{\text{v}}& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 1\cdot -2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 1\cdot -3p_{y,\text{S}}& 3p_{z,\text{S}}\\ {\sigma }_{\text{v}}^{\text{'}}& -1s_{\text{H2}}& -1s_{\text{H1}}& -1s_{\text{S}}& -2s_{\text{S}}& -1\cdot -2p_{x,\text{S}}& -2p_{y,\text{S}}& -2p_{z,\text{S}}& -3s_{\text{S}}& -1\cdot -3p_{x,\text{S}}& -3p_{y,\text{S}}& -3p_{z,\text{S}}\end{array}$

The summation of each column in the table gives the eleven linear combinations as shown below.

$\begin{array}{rllll}{\Psi }_{{\text{B}}_1}& =& \frac{1}{4}\left(1s_{\text{H1}}-1s_{\text{H2}}+1s_{\text{H1}}-1s_{\text{H2}}\right)& =& \frac{1}{2}\left(1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{{\text{B}}_1}& =& \frac{1}{4}\left(1s_{\text{H2}}-1s_{\text{H1}}+1s_{\text{H2}}-1s_{\text{H1}}\right)& =& \frac{1}{2}\left(1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{{\text{B}}_1}& =& \frac{1}{4}\left(1s_{\text{S}}-1s_{\text{S}}+1s_{\text{S}}-1s_{\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_1}& =& \frac{1}{4}\left(2s_{\text{S}}-2s_{\text{S}}+2s_{\text{S}}-2s_{\text{S}}\right)=0\end{array}$

$\begin{array}{l}{\Psi }_{{\text{B}}_1}=\frac{1}{4}\left(2p_{x,\text{S}}+2p_{x,\text{S}}+2p_{x,\text{S}}+2p_{x,\text{S}}\right)=2p_{x,\text{S}}\\ {\Psi }_{{\text{B}}_1}=\frac{1}{4}\left(2p_{y,\text{S}}+2p_{y,\text{S}}-2p_{y,\text{S}}-2p_{y,\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_1}=\frac{1}{4}\left(2p_{z,\text{S}}-2p_{z,\text{S}}+2p_{z,\text{S}}-2p_{z,\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_1}=\frac{1}{4}\left(3s_{\text{S}}-3s_{\text{S}}+3s_{\text{S}}-3s_{\text{S}}\right)=0\end{array}$

$\begin{array}{l}{\Psi }_{{\text{B}}_1}=\frac{1}{4}\left(3p_{x,\text{S}}+3p_{x,\text{S}}+3p_{x,\text{S}}+3p_{x,\text{S}}\right)=3p_{x,\text{S}}\\ {\Psi }_{{\text{B}}_1}=\frac{1}{4}\left(3p_{y,\text{S}}+3p_{y,\text{S}}-3p_{y,\text{S}}-3p_{y,\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_1}=\frac{1}{4}\left(3p_{z,\text{S}}-3p_{z,\text{S}}+3p_{z,\text{S}}-3p_{z,\text{S}}\right)=0\end{array}$

Similarly the table obtained for ${\text{B}}_2$ symmetry is shown below.

$\begin{array}{cccccccccccc}& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ E& 1s_{\text{H1}}& 1s_{\text{H2}}& 1s_{\text{S}}& 2s_{\text{S}}& 2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\\ C_2& -1s_{\text{H2}}& -1s_{\text{H1}}& -1s_{\text{S}}& -2s_{\text{S}}& -1\cdot -2p_{x,\text{S}}& -1\cdot -2p_{y,\text{S}}& -2p_{z,\text{S}}& -3s_{\text{S}}& -1\cdot -3p_{x,\text{S}}& -1\cdot -3p_{y,\text{S}}& -3p_{z,\text{S}}\\ {\sigma }_{\text{v}}& -1s_{\text{H1}}& -1s_{\text{H2}}& -1s_{\text{S}}& -2s_{\text{S}}& -2p_{x,\text{S}}& -1\cdot -2p_{y,\text{S}}& -2p_{z,\text{S}}& -3s_{\text{S}}& -3p_{x,\text{S}}& -1\cdot -3p_{y,\text{S}}& -3p_{z,\text{S}}\\ {\sigma }_{\text{v}}^{\text{'}}& 1s_{\text{H2}}& 1s_{\text{H1}}& 1s_{\text{S}}& 2s_{\text{S}}& 1\cdot -2p_{x,\text{S}}& 2p_{y,\text{S}}& 2p_{z,\text{S}}& 3s_{\text{S}}& 1\cdot -3p_{x,\text{S}}& 3p_{y,\text{S}}& 3p_{z,\text{S}}\end{array}$

The summation of each column in the table gives the eleven linear combinations as shown below.

$\begin{array}{rllll}{\Psi }_{{\text{B}}_2}& =& \frac{1}{4}\left(1s_{\text{H1}}-1s_{\text{H2}}-1s_{\text{H1}}+1s_{\text{H2}}\right)& =& 0\\ {\Psi }_{{\text{B}}_2}& =& \frac{1}{4}\left(1s_{\text{H2}}-1s_{\text{H1}}-1s_{\text{H2}}+1s_{\text{H1}}\right)& =& 0\\ {\Psi }_{{\text{B}}_2}& =& \frac{1}{4}\left(1s_{\text{S}}-1s_{\text{S}}-1s_{\text{S}}+1s_{\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_2}& =& \frac{1}{4}\left(2s_{\text{S}}-2s_{\text{S}}-2s_{\text{S}}+2s_{\text{S}}\right)=0\end{array}$

$\begin{array}{l}{\Psi }_{{\text{B}}_2}=\frac{1}{4}\left(2p_{x,\text{S}}+2p_{x,\text{S}}-2p_{x,\text{S}}-2p_{x,\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_2}=\frac{1}{4}\left(2p_{y,\text{S}}+2p_{y,\text{S}}+2p_{y,\text{S}}+2p_{y,\text{S}}\right)=2p_{y,\text{S}}\\ {\Psi }_{{\text{B}}_2}=\frac{1}{4}\left(2p_{z,\text{S}}-2p_{z,\text{S}}-2p_{z,\text{S}}+2p_{z,\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_2}=\frac{1}{4}\left(3s_{\text{S}}-3s_{\text{S}}-3s_{\text{S}}+3s_{\text{S}}\right)=0\end{array}$

$\begin{array}{l}{\Psi }_{{\text{B}}_2}=\frac{1}{4}\left(3p_{x,\text{S}}+3p_{x,\text{S}}-3p_{x,\text{S}}-3p_{x,\text{S}}\right)=0\\ {\Psi }_{{\text{B}}_2}=\frac{1}{4}\left(3p_{y,\text{S}}+3p_{y,\text{S}}+3p_{y,\text{S}}+3p_{y,\text{S}}\right)=3p_{y,\text{S}}\\ {\Psi }_{{\text{B}}_2}=\frac{1}{4}\left(3p_{z,\text{S}}-3p_{z,\text{S}}-3p_{z,\text{S}}+3p_{z,\text{S}}\right)=0\end{array}$

The total unique combinations are shown below.

$\begin{array}{rll}{\Psi }_{{\text{A}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& 1s_{\text{S}}\\ {\Psi }_{{\text{A}}_1}& =& 2s_{\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{A}}_1}=2p_{z,\text{S}}\\ {\Psi }_{{\text{A}}_1}=3s_{\text{S}}\\ {\Psi }_{{\text{A}}_1}=3p_{z,\text{S}}\end{array}$

$\begin{array}{rll}{\Psi }_{{\text{B}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{{\text{B}}_1}& =& \frac{1}{2}\left(1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{{\text{B}}_1}& =& 2p_{x,\text{S}}\\ {\Psi }_{{\text{B}}_1}& =& 3p_{x,\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{B}}_2}=2p_{y,\text{S}}\\ {\Psi }_{{\text{B}}_2}=3p_{y,\text{S}}\end{array}$

Conclusion

The symmetry-adapted linear combinations of molecular orbitals for hydrogen sulfide involving the core orbitals for sulfur are shown below.

$\begin{array}{rll}{\Psi }_{{\text{A}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}+1s_{\text{H2}}\right)\\ {\Psi }_{{\text{A}}_1}& =& 1s_{\text{S}}\\ {\Psi }_{{\text{A}}_1}& =& 2s_{\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{A}}_1}=2p_{z,\text{S}}\\ {\Psi }_{{\text{A}}_1}=3s_{\text{S}}\\ {\Psi }_{{\text{A}}_1}=3p_{z,\text{S}}\end{array}$

$\begin{array}{rll}{\Psi }_{{\text{B}}_1}& =& \frac{1}{2}\left(1s_{\text{H1}}-1s_{\text{H2}}\right)\\ {\Psi }_{{\text{B}}_1}& =& \frac{1}{2}\left(1s_{\text{H2}}-1s_{\text{H1}}\right)\\ {\Psi }_{{\text{B}}_1}& =& 2p_{x,\text{S}}\\ {\Psi }_{{\text{B}}_1}& =& 3p_{x,\text{S}}\end{array}$

$\begin{array}{l}{\Psi }_{{\text{B}}_2}=2p_{y,\text{S}}\\ {\Psi }_{{\text{B}}_2}=3p_{y,\text{S}}\end{array}$