Chapter 9, Problem 9B.2AE

Interpretation Introduction

Interpretation:

The linear combination of the orbitals $\text{A}$ and $\text{B}$ which is orthogonal to the linear combination $\left(0.145A+0.844B\right)$ has to be predicted.  The normalization constants of both combinations have to be determined by the use of $S=0.250$.

Concept introduction:

For the orthogonality of the two different wave functions, the product of the wave functions is integrated over the entire limits.  It is expressed by the equation as given below.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\psi }_m{\psi }_n}d\tau =0$

Where,

• ${\psi }_m$ and ${\psi }_n$ are two different wave functions.

The essential condition for two wave functions to be orthogonal to each other is $m\ne n$.

Answer to Problem 9B.2AE

The linear combination of the orbitals $\text{A}$ and $\text{B}$ which is orthogonal to the linear combination $\left(0.145A+0.844B\right)$ is shown below.

    ${\psi }_2=N_2\left(0.88025A-0.356B\right)$

The normalization constant of the linear combination $\left(0.145A+0.844B\right)$ is $\underset{\_}{1.12186}$.  The normalization constant of the linear combination $\left(0.88025A-0.356B\right)$ is $\underset{\_}{1.15865}$.

Explanation of Solution

The normalized molecular orbital is shown below.

  ${\psi }_1=N\left(0.88025A-0.356B\right)$                                                                         (1)

Where,

• $N$ is the normalization constant.
• $A$ is the wavefunction of the atomic orbital of $\text{A}$.
• $B$ is the wavefunction of the atomic orbital of $\text{B}$.

The linear combination of the orbitals $\text{A}$ and $\text{B}$ which is orthogonal to the above linear combination can be given by the expression as shown below.

  ${\psi }_2=aA+bB$                                                                                                  (2)

Where,

• $a$ and $b$ are constants.

The overlap integral for atom $\text{A}$ and atom $\text{B}$ is given by the expression shown below.

    $S={\displaystyle \int AB\text{d}\tau }$                                                                                                    (3)

The value of overlap integral $\left(S\right)$ is $0.250$.

The integral of square of orbital is given by the expression shown below.

    $\begin{array}{l}{\displaystyle \int A^2\text{d}\tau }=1\\ {\displaystyle \int B^2\text{d}\tau }=1\end{array}$

The relation between two orthogonal wavefunctions is shown below.

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\psi }_m{\psi }_n}d\tau =0$                                                                                                (4)

Where,

• ${\psi }_m$ and ${\psi }_n$ are two different wave functions.

Substitute the value of ${\psi }_1$ and ${\psi }_2$ from equation (1) and equation (2) in equation (4).

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\psi }_1{\psi }_2}d\tau ={\displaystyle \underset{0}{\overset{\infty }{\int }}N\left(0.145A+0.844B\right)\left(aA+bB\right)}\text{d}\tau \\ =N{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(0.145a{A}^2+0.145bAB+0.844aAB+0.844b{B}^2\right)}\text{d}\tau \\ =N\left(\begin{array}{l}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(0.145a{A}^2\right)}\text{d}\tau +{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(0.145bAB\right)}\text{d}\tau \\ +{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(0.844aAB\right)}\text{d}\tau +{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(0.844b{B}^2\right)}\text{d}\tau \end{array}\right)\end{array}$

The above equation can be further simplified as shown below.

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\psi }_1{\psi }_2}d\tau =N\left(\begin{array}{l}0.145a{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(A^2\right)}\text{d}\tau +0.145b{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(AB\right)}\text{d}\tau \\ +0.844a{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(AB\right)}\text{d}\tau +0.844b{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(B^2\right)}\text{d}\tau \end{array}\right)$

Substitute the value of integration terms in the above expression.

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\psi }_1{\psi }_2}d\tau =N\left(0.145a\left(1\right)+0.145b\left(S\right)+0.844a\left(S\right)+0.844b\left(1\right)\right)$

Substitute the value of equation (4) in the above equation.

    $0=N\left(0.145a\left(1\right)+0.145b\left(S\right)+0.844a\left(S\right)+0.844b\left(1\right)\right)$

Rearrange the above equation for the value of $a$.

    $a=\frac{0.145Sb+0.844b}{0.145+0.844S}$

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

  $\begin{array}{c}a=-\frac{0.145\left(0.250\right)b+0.844b}{0.145+0.844\left(0.250\right)}\\ =-\frac{\left(0.03625+0.844\right)b}{0.145+0.211}\\ =-\frac{0.88025b}{0.356}\end{array}$

Therefore, the new orbital will be normalized for the condition shown below.

    $\begin{array}{l}a=0.88025N_2\\ b=-0.356N_2\end{array}$

Where,

• $N_2$ is the normalization constant for the new orbital.

Therefore, the linear combination of the orbitals $\text{A}$ and $\text{B}$ which is orthogonal to the linear combination $\left(0.145A+0.844B\right)$ is shown below.

    ${\psi }_2=N_2\left(0.88025A-0.356B\right)$                                                                       (5)

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

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(\psi \right)}^{*}\psi }\text{d}\tau =1$                                                                                              (6)

Substitute the value of ${\psi }_1$ in the equation (6).

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(N\left(0.145A+0.844B\right)\right)}^{*}N\left(0.145A+0.844B\right)}\text{d}\tau =1\\ {\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(N\left(0.145A+0.844B\right)\right)}^2}\text{d}\tau =1\\ N^2{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(0.145A+0.844B\right)}^2}\text{d}\tau =1\end{array}$

The above equation is further simplified as shown below.

    $\begin{array}{c}{N}^2{\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\left(0.145A\right)}^2+{\left(0.844B\right)}^2+2\left(0.145A\right)\left(0.844B\right)\right)}\text{d}\tau =1\\ N^2\left(0.021025{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(A^2\right)}\text{d}\tau +0.712336{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(B^2\right)}\text{d}\tau +0.24476{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(AB\right)}\text{d}\tau \right)=1\end{array}$

Substitute the value of integration terms in the above expression.

    $\begin{array}{c}{N}^2\left(0.021025\left(1\right)+0.712336\left(1\right)+0.24476S\right)=1\\ N^2\left(0.733361+0.24476S\right)=1\\ N^2=\frac{1}{\left(0.733361+0.24476S\right)}\\ N={\left(\frac{1}{\left(0.733361+0.24476S\right)}\right)}^{1/2}\end{array}$

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

    $\begin{array}{c}N={\left(\frac{1}{\left(0.733361+0.24476\left(0.250\right)\right)}\right)}^{1/2}\\ ={\left(\frac{1}{\left(0.794551\right)}\right)}^{1/2}\\ =\underset{\_}{1.12186}\end{array}$

Therefore, the normalization constant of the linear combination $\left(0.145A+0.844B\right)$ is $\underset{\_}{1.12186}$.

Substitute the value of ${\psi }_2$ in the equation (6).

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(N_2\left(\left(0.88025A-0.356B\right)\right)\right)}^{*}{N}_2\left(\left(0.88025A-0.356B\right)\right)}\text{d}\tau =1\\ {\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(N_2\left(\left(0.88025A-0.356B\right)\right)\right)}^2}\text{d}\tau =1\\ N_2^2{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(\left(0.88025A-0.356B\right)\right)}^2}\text{d}\tau =1\end{array}$

The above equation is further simplified as shown below.

    $\begin{array}{c}{N}_2^2{\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\left(0.88025A\right)}^2+{\left(0.356B\right)}^2-2\left(0.88025A\right)\left(0.356B\right)\right)}\text{d}\tau =1\\ N_2^2\left(0.77484{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(A^2\right)}\text{d}\tau +0.126736{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(B^2\right)}\text{d}\tau -0.626738{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(AB\right)}\text{d}\tau \right)=1\end{array}$

Substitute the value of integration terms in the above expression.

    $\begin{array}{c}{N}_2^2\left(0.77484\left(1\right)+0.126736\left(1\right)-0.626738\left(S\right)\right)=1\\ N_2^2\left(0.901576-0.626738S\right)=1\\ N_2^2=\frac{1}{\left(0.901576-0.626738S\right)}\\ N_2={\left(\frac{1}{0.901576-0.626738S}\right)}^{1/2}\end{array}$

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

    $\begin{array}{c}{N}_2={\left(\frac{1}{0.901576-0.626738\left(0.250\right)}\right)}^{1/2}\\ ={\left(\frac{1}{\left(0.74489\right)}\right)}^{1/2}\\ =\underset{\_}{1.15865}\end{array}$

Therefore, the normalization constant of the linear combination $\left(0.88025A-0.356B\right)$ is $\underset{\_}{1.15865}$.