Chapter 12, Problem 12.24E

(a) Construct Slater determinant wavefunctions for $\text{Be}$ and $\text{B}$. (Hint: Although you need only include one $p$ orbital for $\text{B}$, you should recognize that up to six possible determinants can be constructed.)

(b) How many different Slater determinants can be constructed for $\text{C}$, assuming that the $p$ electrons spread out among the available $p$ orbitals and have the same spin? How many different Slater determinants are there for $\text{F}$?

Interpretation Introduction

(a)

Interpretation:

The slater determinant wavefunctions for $\text{Be}$ and $\text{B}$ is to be constructed.

Concept introduction:

The wavefunctions can be represented in the form of Slater determinants. The terms in the wavefunction is equal to $n!$. The $n$ is the number of electrons present in the given atom. Also, the $n$ is equal to the number of rows or columns. The normalization factor in the slater determinant is $1/\sqrt{n!}$.

Answer to Problem 12.24E

The slater determinant wavefunctions for $\text{Be}$ and $\text{B}$ is,

${\Psi }_{\text{Be}}=\frac{1}{\sqrt{24}}\left|\begin{array}{cccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta \end{array}\right|$

${\Psi }_{\text{B}}=\frac{1}{\sqrt{120}}\left|\begin{array}{ccccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta & 2p_{x,1}\alpha \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta & 2p_{x,2}\alpha \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta & 2p_{x,3}\alpha \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta & 2p_{x,4}\alpha \\ 1s_5\alpha & 1s_5\beta & 2s_5\alpha & 2s_5\beta & 2p_{x,5}\alpha \end{array}\right|$

The total slater determinants possible for boron is $6$.

Explanation of Solution

The number of electrons in $\text{Be}$ is $4$. Thus, there are four rows and four columns and the value of n in the normalization constant is also $4$. The rows represent electrons $1$, $2$, $3$ and $4$. The columns represent the spin orbitals $1s\alpha$, $1s\beta$, $2s\alpha$ and $2s\beta$. The slater determinant for the $\text{Be}$ is,

$\begin{array}{l}{\Psi }_{\text{Be}}=\frac{1}{\sqrt{4!}}\left|\begin{array}{cccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta \end{array}\right|\\ {\Psi }_{\text{Be}}=\frac{1}{\sqrt{24}}\left|\begin{array}{cccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta \end{array}\right|\end{array}$

Where,

• $1$ and $2$ represents the electrons.

• $\alpha$ and $\beta$ are the spin parts of the wavefunction.

The number of electrons in $\text{B}$ is $5$. Thus, there are five rows and five columns and the value of n in the normalization constant is also $5$.The rows represent electrons $1$, $2$, $3$, $4$ and $5$. The column represent the spin orbitals $1s\alpha$, $1s\beta$, $2s\alpha$, $2s\beta$ and $2p_x\alpha$. The slater determinant for $\text{B}$ is,00000000000000000000000000000

$\begin{array}{l}{\Psi }_{\text{B}}=\frac{1}{\sqrt{5!}}\left|\begin{array}{ccccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta & 2p_{x,1}\alpha \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta & 2p_{x,2}\alpha \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta & 2p_{x,3}\alpha \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta & 2p_{x,4}\alpha \\ 1s_5\alpha & 1s_5\beta & 2s_5\alpha & 2s_5\beta & 2p_{x,5}\alpha \end{array}\right|\\ {\Psi }_{\text{B}}=\frac{1}{\sqrt{120}}\left|\begin{array}{ccccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta & 2p_{x,1}\alpha \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta & 2p_{x,2}\alpha \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta & 2p_{x,3}\alpha \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta & 2p_{x,4}\alpha \\ 1s_5\alpha & 1s_5\beta & 2s_5\alpha & 2s_5\beta & 2p_{x,5}\alpha \end{array}\right|\end{array}$

Where,

• $1$ and $2$ represents the electrons.

• $\alpha$ and $\beta$ are the spin part of the wavefunction.

In boron, the electron can be placed in any of the three $2p$ orbitals, that is, $2p_x$, $2p_y$ and $2p_z$. Also the electron can take any spin value $\alpha$ or $\beta$. Thus, the total slater determinants possible for boron is $3\times 2=6$.

Conclusion

The slater determinant wavefunctions for $\text{Be}$ and $\text{B}$ is,

${\Psi }_{\text{Be}}=\frac{1}{\sqrt{24}}\left|\begin{array}{cccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta \end{array}\right|$

${\Psi }_{\text{B}}=\frac{1}{\sqrt{120}}\left|\begin{array}{ccccc}1s_1\alpha & 1s_1\beta & 2s_1\alpha & 2s_1\beta & 2p_{x,1}\alpha \\ 1s_2\alpha & 1s_2\beta & 2s_2\alpha & 2s_2\beta & 2p_{x,2}\alpha \\ 1s_3\alpha & 1s_3\beta & 2s_3\alpha & 2s_3\beta & 2p_{x,3}\alpha \\ 1s_4\alpha & 1s_4\beta & 2s_4\alpha & 2s_4\beta & 2p_{x,4}\alpha \\ 1s_5\alpha & 1s_5\beta & 2s_5\alpha & 2s_5\beta & 2p_{x,5}\alpha \end{array}\right|$

The total slater determinants possible for boron is $6$.

Interpretation Introduction

(b)

Interpretation:

The number of different Slater determinants that can be constructed for $\text{C}$, assuming that the $p$ electrons spread out among the available $p$ orbitals and have the same spin is to be predicted. Also the number of different slater determinants for $\text{F}$ is to be predicted.

Concept introduction:

The wavefunctions can be represented in the form of Slater determinants. The terms in the wavefunction is equal to $n!$. The $n$ is the number of electrons present in the given atom. Also, the $n$ is equal to the number of rows or columns. The normalization factor in the slater determinant is $1/\sqrt{n!}$.

Answer to Problem 12.24E

The number of different Slater determinants that can be constructed for $\text{C}$, assuming that the $p$ electrons spread out among the available $p$ orbitals and have the same spin is $6$. Also the number of different slater determinants for $\text{F}$ is $6$.

Explanation of Solution

The number of electrons in $\text{C}$ is $6$. Thus, there are six rows and six columns and the value of n in the normalization constant is also $6$.The rows represent electrons $1$, $2$, $3$, $4$, $5$ and $6$.

In carbon, the two electrons can be placed in any of the three $2p$ orbitals, that is, $2p_x$, $2p_y$ and $2p_z$. The possible combination is, $2p_y^{1}2p_z^1$, $2p_x^{1}2p_y^1$ and $2p_x^{1}2p_z^1$. Also the electron can take any spin value $\alpha$ or $\beta$. It is assumed that that the two electrons have same spin. Thus, the total slater determinants possible for boron is $3\times 2=6$.

The number of electrons in $\text{F}$ is $9$. Thus, there are nine rows and nine columns and the value of n in the normalization constant is also $9$.

In fluorine, the one unpaired electron can be placed in any of the three $2p$ orbitals, that is, $2p_x$, $2p_y$ and $2p_z$. Also the electron can take any spin value $\alpha$ or $\beta$. Thus, the total slater determinants possible for boron is $3\times 2=6$.

Conclusion

The number of different Slater determinant that can be constructed for $\text{C}$, assuming that the $p$ electrons spread out among the available $p$ orbitals and have the same spin is $6$. Also the number of different slater determinants for $\text{F}$ is $6$.