Chapter 9, Problem 9B.1ST

Interpretation Introduction

Interpretation:

The molecular orbital ${\psi }_{-}$ has to be normalized to $1$.  The value of $N_{-}$ has to be calculated when $S=0.59$.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\psi$.  The wavefunction contains all the information about the state of the system.  The wavefunction is the function of the coordinates of particles and time.  The normalized wavefunction is given by the wavefunction shown below.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\psi }^{*}\psi }\text{d}\tau =1$

Answer to Problem 9B.1ST

The normalized molecular orbital is shown below.

    $\psi ={\left(\frac{1}{2\left(1-S\right)}\right)}^{1/2}\left({\psi }_{\text{A}}+{\psi }_{\text{B}}\right)$

The value of $N_{-}$ for $S=0.59$ is $\underset{\_}{1.10}$.

Explanation of Solution

The molecular orbital ${\psi }_{-}$ is shown below.

  ${\psi }_{-}=N_{-}\left({\psi }_{\text{A}}-{\psi }_{\text{B}}\right)$                                                                                        (1)

Where,

• ${\psi }_{\text{A}}$ is the wavefunction of atom $\text{A}$.
• ${\psi }_{\text{B}}$ is the wavefunction of atom $\text{B}$.
• $\lambda$ is the parameter.

The overlap integral is given by the expression as shown below.

    $S={\displaystyle \int {\psi }_{\text{A}}{\psi }_{\text{B}}\text{d}\tau }$                                                                                                (2)

The value of $S$ is $0.59$.

The normalization of the wavefunction is given by the expression as shown below.

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(\psi \right)}^{*}\psi }\text{d}\tau =1$                                                                                               (3)

The wavefunction ${\psi }_{\text{A}}$ and ${\psi }_{\text{B}}$ are normalized.  Therefore, the integral of square of these wavefunction can be given by the expression as shown below.

    $\begin{array}{l}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\psi }_{\text{A}}^2\right)}\text{d}\tau =1\\ {\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\psi }_{\text{B}}^2\right)}\text{d}\tau =1\end{array}$

Substitute the value of $\psi$ from equation (1) in equation (3).

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(N_{-}\left({\psi }_{\text{A}}-{\psi }_{\text{B}}\right)\right)}^2}\text{d}\tau =1\\ N_{-}^2{\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\psi }_{\text{A}}^2+{\psi }_{\text{B}}^2-2{\psi }_{\text{A}}{\psi }_{\text{B}}\right)}\text{d}\tau =1\\ N_{-}^2\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\psi }_{\text{A}}^2\right)}\text{d}\tau +{\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\psi }_{\text{B}}^2\right)}\text{d}\tau -2{\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\psi }_{\text{A}}{\psi }_{\text{B}}\right)}\text{d}\tau \right)=1\end{array}$

Substitute the value of integration terms in the above expression.

    $\begin{array}{c}{N}_{-}^2\left(\left(1\right)+\left(1\right)-2\left(S\right)\right)=1\\ N_{-}^2=\frac{1}{2-2S}\\ N_{-}={\left(\frac{1}{2\left(1-S\right)}\right)}^{1/2}\end{array}$

Substitute the value of $S$ in the above expression.

    $\begin{array}{c}{N}_{-}={\left(\frac{1}{2\left(1-\left(0.59\right)\right)}\right)}^{1/2}\\ ={\left(\frac{1}{0.82}\right)}^{1/2}\\ =\underset{\_}{1.10}\end{array}$

Therefore, the value of $N_{-}$ for $S=0.59$ is $\underset{\_}{1.10}$.

The normalized molecular orbital is shown below.

    $\psi ={\left(\frac{1}{2\left(1-S\right)}\right)}^{1/2}\left({\psi }_{\text{A}}+{\psi }_{\text{B}}\right)$