Molecules

2.0 Energy conversion table

2.1 Born-Oppenheimer approximation

2.2 Molecular orbital Hamiltonian

2.3 Molecular orbitals

2.4 The molecular ion H2+

2.5 Homonuclear diatomic molecules

2.6 Linear combination of atomic orbitals

2.7 Molecular databases

2.8 Conjugated rings and conjugated chains

2.9 Cyclopropene

2.10 H2 and He2

2.11 Effective nuclear charge

2.12 Bonds

2.13 Water

2.14 Bond potentials

2.15 Molecular vibrations

2.16 A 6-12 bond potential

2.17 Lennard-Jones potential

2.18 Lennard-Jones bond

2.19 CO molecule

2.20 Rotational energy states

2.21 The rotational spectrum of BrF

2.22 Estimating vibrational energies from the Morse potential

2.23 Vibrational modes of CO2 and H2O

2.24 C-O bond

2.25 Thermal occupation of rotational states

2.26 HD and D2

2.27 Vibrational modes

2.28 LCAO - water

2.29 Ionic bonds

2.30 Molecular orbitals

2.31 Morse potential

2.32 Carbon hybridization

2.33 Bond length and bond angle

2.34 He2 molecule

2.35 Effective spring constant

2.0 Energy conversion table

 Electronvolts (eV) 

 Joule (J) 

 Wave number [cm-1] 

 Frequency [Hz] 

0.235 (Vibrational mode) 

2710 (Vibrational mode)

12 (Ionization energy)

25104 (6 kcal)

13 kHz (Audio)

1201 (XPS)

Molecular orbital Hamiltonian
The starting point for the quantum description of any molecule or crystal is the many-electron Hamiltonian,

\[ \begin{equation} H_{\text{me}}= -\sum\limits_i \frac{\hbar^2}{2m_e}\nabla^2_i -\sum\limits_a \frac{\hbar^2}{2m_a}\nabla^2_a -\sum\limits_{a,i} \frac{Z_ae^2}{4\pi\epsilon_0 |\vec{r}_i-\vec{r}_a|}+\sum\limits_{i< j} \frac{e^2}{4\pi\epsilon_0 |\vec{r}_i-\vec{r}_j|}+\sum\limits_{a< b} \frac{Z_aZ_be^2}{4\pi\epsilon_0 |\vec{r}_a-\vec{r}_b|} . \end{equation} \]

The first sum describes the kinetic energy of the electrons. The electrons are labeled with the subscript $i$. The second sum describes the kinetic energy of all of the atomic nuclei. The atoms are labeled with the subscript $a$. The third sum describes the attractive Coulomb interaction between the positively charged nuclei and the negatively charged electrons. $Z_a$ is the atomic number (the number of protons) of nucleus $a$. The fourth sum describes the repulsive electron-electron interactions. Notice the plus sign before the sum for repulsive interactions. The fifth sum describes the repulsive nuclei-nuclei interactions.

This Hamiltonian neglects some small details like the spin-orbit interaction and the hyperfine interaction. These effects will be ignored in this discussion. If they are relevant, they could be included as perturbations later. Remarkably, this Hamiltonian can tell us the shape of every molecule and the energy released (or absorbed) in a chemical reaction. Any observable quantity of any solid can also be calculated from this Hamiltonian. It turns out, however, that solving the Schrödinger equation associated with this Hamiltonian is usually terribly difficult. Here we will present a standard approach to determine approximate solutions to the Schrödinger equation.

An approximate solution of the many-electron Hamiltonian for a molecule can be found by applying the Born-Oppenheimer approximation and neglecting the electron-electron interactions. The resulting Hamiltonian is called the reduced Hamiltonian. The reduced Hamiltonian is the sum of $N_e$ identical molecular orbital Hamiltonians where $N_e$ is the number of electrons in the molecule.

\begin{equation} H_{\text{red}}= \sum \limits_{i=1}^{N_e} H_{\text{mo}}. \end{equation}

2.1 Born-Oppenheimer approximation

This question is designed to test your conceptual understanding of the Born-Oppenheimer approximation. Consider three quantum particles confined to move only in the $x$-direction. There are two heavy particles with mass $M$ and $M$ and one light particle with mass $m \ll M$. This is similar to an H2+ molecule with two heavy protons and one light electron. The potential energy term of the total Hamiltonian reads

$$ V(X_1,X_2,x) = -\frac{1}{2}K_1(X_1-X_2)^2 + \frac 1 2 K_2(X_1-X_2)^2\left(x-\frac{X_1+X_2}{2}\right)^2,$$

where $X_1$ and $X_2$ are the positions of the heavy particles and $x$ is the position of the light particle. The first term represents a repulsive interaction between the two heavy particles and the second term is an attractive term between the the light particle and the average position of the heavy particles.

(a) Show that this potential is translationally invariant. If you add $\delta$ to all three coordinates, the potential does not change. In the following it is convenient to choose origin to be at the center of mass of the heavy atoms.

(b) Draw the potential energy of the light particle for a fixed distance between the heavy particles.

(c) Use the Born-Oppenheimer approximation to eliminate the light particle degree of freedom and obtain the effective potential $U(X_1,X_2)$ between the heavy particles. Discuss the stability of this potential.

2.2 Molecular orbital Hamiltonian

(a) Explain the Born-Oppenheimer approximation.

(b) Explain Slater's rules.

(c) What are the molecular orbital Hamiltonians $H_{\text{mo}}$ for H2O, O2, and CH4?

Linear Combination of Atomic Orbitals (LCAO)
Molecular orbitals are used in a similar way as hydrogen wave functions are used to construct the many-electron wave functions of atoms. In the simplest approximation, the molecular orbitals are the wave function of a single electron moving in a potential created by all of the positively charged nuclei in a molecule. The molecular orbital Hamiltonian can be written,

\[ \begin{equation} H_{\text{mo}}= - \frac{\hbar^2}{2m_e}\nabla^2 -\sum\limits_{a} \frac{Z_ae^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_a|}. \end{equation} \]

Here $\vec{r}_a$ are the positions of the nuclei in the molecule and $\vec{r}$ is the position of the single electron. The molecular orbital Hamiltonian is often solved by a method called the Linear Combination of Atomic Orbitals (LCAO). It is assumed that the wave function can be written in terms of atomic orbitals that are centered around the nuclei,

\begin{equation} \psi_{\text{mo}}(\vec{r})= \sum\limits_{a} \sum\limits_{ao} c_{ao,a}\phi^{Z_a}_{ao}\left(\vec{r}-\vec{r}_a\right). \end{equation}

where $ao$ labels the atomic orbitals ($ao=1$: 1s, $ao=2$: 2s, $ao=3$: 2px, $\cdots$). The number of molecular orbitals that we calculate will be equal to the number of unknown coefficients. We call this number $N$. We have to decide how many atomic orbitals should be included for each atom. A reasonable choice is to take all of the occupied atomic orbitals of the isolated atoms. For instance, for water one might use the 1s, 2s, and 2p orbitals of oxygen and the 1s orbitals of the two hydrogen atoms. In that case, there would be $N=7$ terms in the wave function for the molecular orbital. There is no strict rule as to which atomic orbitals to include. Including more atomic orbitals leads to a higher accuracy but makes the numerical calculation more difficult.

At this point it is convenient to relabel the atomic orbitals used in the wave function with integers $p=1\cdots N$. For water we might choose $\phi_1 = \phi^{Z=1}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{H1}}\right)$, $\phi_2 = \phi^{Z=1}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{H2}}\right)$, $\phi_3 = \phi^{Z=8}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{O}}\right)$, $\phi_4 = \phi^{Z=8}_{\text{2s}}\left(\vec{r}-\vec{r}_{\text{O}}\right)$, etc. The trial wave function can then be written more compactly as,

\begin{equation} \psi_{\text{mo}} = \sum\limits_{p=1}^N c_p\phi_p. \end{equation}

The time independent Schrödinger equation is,

\[ \begin{equation} H_{\text{mo}}\psi_{\text{mo}}=E\psi_{\text{mo}} . \end{equation} \]

Multiply the Schrödinger equation from the left by each of the atomic orbitals and integrate over all space. This results in a set of $N$ algebraic equations called the Roothaan equations.

\[ \begin{equation} \begin{matrix} \large \langle \phi_1 | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_1|\psi_{\text{mo}} \rangle \\ \large \langle \phi_2 | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_2|\psi_{\text{mo}} \rangle\\ \large \vdots \\ \large \langle \phi_N | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_N|\psi_{\text{mo}} \rangle \end{matrix} \end{equation} \]

By substituting in the form for $\psi_{\text{mo}}$ from above, the Roothaan equations can be written in matrix form,

\[ \begin{equation} \left[ \begin{matrix} H_{11} & H_{12} & \cdots & H_{1N} \\ H_{21} & H_{22} & \cdots & H_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ H_{N1} & H_{N2} & \cdots & H_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right] = E\left[ \begin{matrix} S_{11} & S_{12} & \cdots & S_{1N} \\ S_{21} & S_{22} & \cdots & S_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ S_{N1} & S_{N2} & \cdots & S_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right]. \end{equation} \]

Here the elements of the Hamiltonian matrix and the overlap matrix are,

\[ \begin{equation} H_{pq}= \langle\phi_{p}|H_{\text{mo}}|\phi_{q}\rangle \hspace{1cm}\text{and}\hspace{1cm} S_{pq}= \langle\phi_{p}|\phi_{q}\rangle . \end{equation} \]

The Roothaan equations can be solved numerically to find $N$ solutions for the energy $E$ along with the corresponding coefficients that describe the $N$ wave functions which are the molecular orbitals.

The Hückel model is an approximation that is often made to simplify the Roothaan equations. The overlap matrix is nearly the identity matrix. For normalized atomic orbitals, $S_{pp}= 1$ so all of the diagonal elements equal $1$. If two wave functions $\phi_p$ and $\phi_q$ are not the same but are centered on the same nucleus then $S_{pq}= 0$ because the hydrogen atomic orbitals are orthogonal to each other. If two wave functions $\phi_p$ and $\phi_q$ are centered on different nuclei that are far apart in the molecule, then $S_{pq}\approx 0$. This only leaves off-diagonal elements of the matrix that correspond to two wave functions $\phi_p$ and $\phi_q$ that are centered on nearby atoms. Although these elements may not really be zero, they will be small compared to $1$ and we make the approximation,

\[ \begin{equation} S_{pp}= 1 \hspace{1cm} \text{and} \hspace{1cm} S_{pq}= 0 \hspace{1cm} \text{for }p\ne q. \end{equation} \]

The equations that need to be solved reduce to an eigenvalue problem,

\[ \begin{equation} \left[ \begin{matrix} H_{11} & H_{12} & \cdots & H_{1N} \\ H_{21} & H_{22} & \cdots & H_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ H_{N1} & H_{N2} & \cdots & H_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right] = E \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right]. \end{equation} \]

2.3 Molecular orbitals

(a) What are molecular orbitals? How do you calculate them?

(b) What are the Roothaan equations? What is the Hückel model?

2.4 The molecular ion H2+

The molecular ion H2+ consists of one electron and two protons.

(a) Explain the Born-Oppenheimer approximation. What is the Schrödinger equation for the electron when the Born-Oppenheimer approximation is used?

(b) Assuming that the wave function can be expressed as a linear combination of atomic orbitals of the form $\psi=c_1\phi_{\text{1s}}(\vec{r}-\vec{r}_A) +c_2\phi_{\text{1s}}(\vec{r}-\vec{r}_B)$, construct the two molecular orbitals and plot them.

(c) Consider the case that the protons are moved far apart. What are the values of the energies of the two molecular orbitals in this limit?

(d) Draw the potential energy of the electron along the interatomic axis under the assumption that the bond length is 1.06 Å.

2.5 Homonuclear diatomic molecules

Homonuclear diatomic molecules consist of two atoms of the same type like H2, N2 or O2. The molecular orbitals of all homonuclear diatomic molecules have the same form as for H2; only the effective nuclear charge differs. The energies of the first few molecular orbitals for homonuclear diatomic molecules are ordered $1\sigma_g \lt 1\sigma_u \lt 2\sigma_g \lt 2\sigma_u \lt 3\sigma_g \approx 1\pi_u \lt 1\pi_g \lt 3\sigma_u$. The $\sigma$ orbitals are singly degenerate (they can hold 2 electrons) but the $\pi$ orbitals are doubly degenerate (they can hold 4 electrons). $1\pi_u$ is lower in energy than $3\sigma_g$ up to and including N2 but $3\sigma_g$ is lower in energy for atoms with more electrons than nitrogen.

Specify the ground state of H2, N2, Li2+, Be2, C2, and O2.

H2: $1\sigma_g^2$

N2: $1\sigma_g^2 \, 1\sigma_u^2 \, 2\sigma_g^2 \, 2\sigma_u^2 \, 1\pi_u^4 \, 3\sigma_g^2$

Li2+: $1\sigma_g^2 \, 1\sigma_u^2 \, 2\sigma_g^1 $

Be2: $1\sigma_g^2 \, 1\sigma_u^2 \, 2\sigma_g^2 \, 2\sigma_u^2 $

C2: $1\sigma_g^2 \, 1\sigma_u^2 \, 2\sigma_g^2 \, 2\sigma_u^2 \, 1\pi_u^4 $

O2: $1\sigma_g^2 \, 1\sigma_u^2 \, 2\sigma_g^2 \, 2\sigma_u^2 \, 3\sigma_g^2\, 1\pi_u^4 \, 1\pi_g^2$

2.6 Linear combination of atomic orbitals

A linear combination of atomic orbitals used to find the molecular orbitals of a He2 molecule contains four atomic orbitals,

\[ \begin{equation} \psi(\vec{r})= c_1\phi^{Z=2}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{He1}}\right)+c_2\phi^{Z=2}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{He2}}\right)+c_3\phi^{Z=2}_{\text{2s}}\left(\vec{r}-\vec{r}_{\text{He1}}\right)+c_4\phi^{Z=2}_{\text{2s}}\left(\vec{r}-\vec{r}_{\text{He2}}\right). \end{equation} \]

What is the integral that needs to be evaluated to determine the Hamiltonian matrix element $H_{12}$ for this molecule? What is the integral that needs to be evaluated to determine the matrix element of the overlap matrix $S_{13}$ in the Roothaan equations? This integral is easy to evaluate. What is $S_{13}$?

2.7 Molecular databases

Use a database (such as the one embedded in jmol) to determine the Si-O bond length and the angle formed by the atoms Si-O-C in tetraethyl orthosilicate (TEOS).

2.8 Conjugated rings and conjugated chains

Aromatic molecules are substances which are chemically unreactive. In this exercise, we will compare the prototypical aromatic molecule benzene with its linear chain equivalent, 1,3,5-hexatriene (shown in the picture below).

(a) How many $p_z$ electrons are in these two molecules?

(b) Set up the Roothaan-equations for both molecules assuming that only the $p_z$ orbitals need to be included. Look at the discussion of molecular orbitals of a conjugated rings and the molecular orbitals of conjugate chains. Use $H_{11} = - 12$ eV, $H_{12} = -3.8$ eV, $S_{11} = 1$, and $S_{12} = 0.27$. These are not the values that the simple model gives but are close to the observed values.

(c) Which molecular orbitals will be occupied for benzene and for 1,3,5-hexatriene? Which molecular orbitals are degenerate? (Degenerate molecular orbitals have the same energy).

(d) The total energy of the molecules can be approximated as the sum over all occupied molecular orbitals. Which molecule has the lower energy?

(e) Calculate and compare the energy difference between the highest occupied molecular orbital and the lowest unoccupied molecular orbital for both molecules. This energy difference could be measured spectroscopically.

2.9 Cyclopropene

The cation of cyclopropene is the smallest aromatic molecule. The Roothaan equations can be used to express three molecular orbitals in terms of the three $2p_z$ orbitals. Use $H_{11} = - 12$ eV, $H_{12} = -3.8$ eV, $S_{11} = 1$, and $S_{12} = 0.27$.

(a) The cation of cyclopropene has one less electron than neutral cyclopropene. How many electrons are in the cation in total? How many electrons are in the molecular $\pi$-system and have to be accounted for in the Roothaan equations?

(b) Calculate the energies of the orbitals. What is the degeneracy of the unoccupied orbitals?

(c) The molecule can absorb a photon to promote one electron from the highest occupied molecular orbital to the lowest unoccupied molecular orbital. What is the wavelength of the photon?

2.10 H2 and He2

Atomic orbitals are often used to construct trial wavefunctions that solve a molecular orbital Hamiltonian. In the simplest approximation where we neglect the electron-electron interactions, we can use the molecular orbitals of H2+ to estimate the ground state energies of H2 and He2.

The energies of the two lowest molecular orbitals of H2+ are shown in the plot below.

If the atoms are moved far apart, we obtain the energy of two isolated atoms with an energy of zero.

(a) How many molecular orbitals are occupied for H2 and for He2?

(b) Compare the energies of H2 and He2 as a function of interatomic distance. Explain why one of these molecules is more stable than the other.

2.11 Effective nuclear charge

Atomic orbitals are often used to construct trial wavefunctions that solve a molecular orbital Hamiltonian. To include the effects of electron shielding, an effective nuclear charge is often introduced. The form of the 1s orbital with an effective nuclear charge is,

$$\phi_{\text{1s}}=\sqrt{\frac{Z_{eff}^3}{\pi a_0^3}}\exp\left(-\frac{Z_{eff}r}{a_0}\right),$$

where $Z_{eff}$ is the effective charge of the nucleus. The appropriate effective charge to use is given in the table of Slater's rules.

Consider H2 and He2. How many molecular orbitals are occupied for each of these molecules? The molecular orbital energies are calculated as for H2+,

\begin{equation} E_{+}= \frac{H_{11}+ H_{12}}{1+ S_{12}}, \qquad E_{-}= \frac{H_{11}- H_{12}}{1- S_{12}}. \end{equation}

Use the programs of the page Molecular orbitals of the molecular ion H2+ to calculate $H_{11}$, $H_{12}$, $S_{12}$, $E_+$ and $E_-$ for the protons far apart and at the bond length of H2, 0.074 nm. When the atoms are far apart, the energy of the two isolated H atoms is $2H_{11}$. Compare this to the energy of both electrons in the $\psi_{+}$ orbital at the bond length of H2. Do the same for He2 $(Z_{eff} = 1.7)$. The bond length of He2 is 0.3 nm. What does this say about the bond strengths of H2 and He2? By including electron-electron interactions this calculation could be performed more accurately. This, however, would require a significant numerical effort and you don't have to do this.

Energy of an many-electron wave function
Many-electron wave functions consisting of any product of molecular orbitals solve the reduced electronic Hamiltonian. However, not all such solutions are physically relevant because in addition to being a solution to the Schrödinger equation, multielectron wave functions must also be antisymmetric. The common way to construct a properly antisymmetrized multi-electron wave function is to use a Slater determinant. An estimation of the energy of a multi-electron molecular state can be made by calculating the energy,

\[ \begin{equation} E= \frac{ \langle \Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})|H_{\text{elec}}|\Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})\rangle}{\langle \Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})|\Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})\rangle} \end{equation} \]

Here $\Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})$ is an antisymmetrized wave function with $N_e$ electrons and $H_{\text{elec}}$ is the electronic Hamiltonian which includes the electron-electron interactions. The integrals that have to be performed to calculate the energy are defined in $3N_e$ dimensions. When comparing the energies of two molecular states, only include the valence molecular orbitals so that integrals do not become too computationally intensive. In practice, a numerical approach such as density functional theory or the Hartree_Fock method is often used to find the molecular orbitals. We will not discuss these advanced methods in this course.

Being able to calculate the energy of a molecular state is very powerful because molecules will go to a minimum of the energy. The bond potential of a chemical bond can be calculated by changing the distance between the atoms that form that bond, finding the molecular orbitals and then using the molecular orbitals to calculate the energy of the multi-electron state for each distance. Don't forget to include the change in energy of the nuclear-nuclear interaction as the distance between the nuclei changes. This calculation yields the equilibrium bond length and the bond strength. Similarly, bond angles can be calculated by determining which angle gives the lowest energy. The energy released in a chemical reaction can be found by subtracting the energies of the products from the energies of the reactants.

2.12 Bonds

(a) The bond length and the bond energy of a C-C bond depends on the other atoms in the molecule. How do you calculate the bond potential for a particular C-C bond in a molecule?

(b) How could the H-N-H bond angle be calculated for NH3?

(c) How can you calculate the energy released when one methane molecule burns in oxygen to form water and carbon dioxide?

2.13 Water

Molecular orbitals for water are calculated using the LCAO method by assuming a molecular orbital that is a linear combination of the two hydrogen $1\text{s}$ orbitals plus the oxygen $1\text{s}$, $2\text{s}$, $2\text{p}_x$, $2\text{p}_y$, and $2\text{p}_z$ orbitals. How many molecular orbitals will be calculated? How many of the molecular orbitals are filled in the ground state of the molecule? The energy is of the many-electron ground state calculated including all of the electrons of the molecule in the Slater determinant, how many dimensions is the integral $\langle \Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})|H_{\text{elec}}|\Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})\rangle$ defined in?

2.14 Bond potentials

(a) How do you calculate the bond potential for a

(b) How can you calculate the energy released when one methane molecule burns in oxygen to form water and carbon dioxide?

(c) How could the H-N-H bond angle be calculated for NH3?

Molecular vibrations and rotations
Although the bond potentials of any bond is calculated the same way, they are often fit to different bond potentials. For covalent bonds, the Morse potential is often used,

\begin{equation} U(r)= U_0\left(e^{-2a(r-r_0)}-2e^{-a(r-r_0)}\right). \end{equation}

For van der Waals bonds, a Lennard-Jones potential is often used,

\begin{equation} U(r)= U_0\left(\left(\frac{r_0}{r}\right)^{12}-2\left(\frac{r_0}{r}\right)^{6}\right). \end{equation}

Here $r$ is the distance between the atoms and $r_0$ is the minimum of the potential. For diatomic molecules, the bond potentials can be used to find the vibrational and rotational spectra of a molecule. Diatomic molecules only have one stretching mode where the two atoms in a molecule vibrate with respect to each other. In the simplest approximation, we imagine that the atoms are attached to each other by a linear spring. The spring constant can be determined from the bond potential. Near the minimum of the bond potential at $r_0$, the potential can be approximated by a parabola. This is the same potential as for a harmonic oscillator where the effective spring constant is $k_{\text{eff}}=\frac{d^2U}{dr^2}|_{r=r_0}$ and the reduced mass is $m_r=(1/m_a+1/m_b)^{-1}$. Here $m_a$ and $m_b$ are the masses of the two atoms. The angular frequency of this vibration is $\omega=\sqrt{k_{\text{eff}}/m_r}$. The vibrational modes of the molecule have the same energy level spectrum as a harmonic oscillator,

\[ \begin{equation} E_{\nu}=\hbar\omega(\nu+1/2) \hspace{1.5cm} \nu=0,1,2,\cdots. \end{equation} \]

The rotational levels of a diatomic molecule can be estimated by assuming that the atoms remain at their equilibrium spacing $r_0$ during rotation. In this can, the quantized energy levels of a rigid rotator can be used,

\[ \begin{equation} E_{\mathcal{l}}= \frac{\hbar^2}{2I}\mathcal{l}(\mathcal{l}+1), \end{equation} \]

where $I$ is the moment of inertia and $\mathcal{l}$ is orbital quantum number, $\mathcal{l}=0,1,2,\cdots$. For a diatomic molecule with an equilibrium spacing of the atoms of $r_0$ and atomic masses $m_a$ and $m_b$, the moment of inertia is $I=\frac{m_am_b}{m_a+m_b}r_0^2$.

2.15 Molecular vibrations

The $\nu = 0$ to $\nu = 1$ vibrational transition of an HI molecule occurs at a frequency of $6.69\times 10^{13}$ Hz. The same transition for the NO molecule occurs at a frequency of $5.63 \times 10^{13}$ Hz. The atomic masses (in atomic units) are: I = 126.9, H = 1.01, N = 14.01, O = 16. Calculate:

(a) The effective force constants.

(b) The classical amplitude of vibration for each molecule for $\nu = 10$. The amplitude can be calculated with $E=kA^2/2$ where $E$ is the energy of the vibrational mode, $k$ is the effective spring constant, and $A$ is the amplitude.

(c) Explain why the force constant of the NO molecule is so much larger than that of the HI molecule.

2.16 A 6-12 bond potential

The potential energy of a diatomic molecule is $U(R)=-\frac{\alpha}{ R^6}+\frac{\beta}{ R^{12}}$, with $\alpha$ and $\beta$ being characteristic constants for the molecule.

(a) Find the interaction force between the two atoms as a function of distance $R$.

(b) Find the equilibrium spacing $R_0$ between the two atoms.

2.17 Lennard-Jones potential

The bond potential of a diatomic molecule has a Lennard-Jones form, $U(R)=-\frac{A}{ R^6}+\frac{B}{ R^{12}}$. This form is typical of a Van der Waals bond. The bond length of this molecule is 3 Å and the minmum of the bond potential is at -70 meV.

Calculate $A$ and $B$.

$A=$  eV Å6   $B=$  eV Å12

A force is needed to hold the atoms further apart than the equilibrium distance. This force is $F=-\frac{dU}{dR}$. What is the maximum value this force can have?

$|F|=$  N

A force is applied that separates the molecule. As this force is slowly increased, the atoms move further apart until a critical distance where the molecule dissociates. What is the critical distance?

$R_c=$  Å

2.18 Lennard-Jones bond

Van der Waals bonds are often described by a Lennard-Jones potential with the form, $U(R)= 4\epsilon\left[\left(\frac{\sigma}{R}\right)^{12}-\left(\frac{\sigma}{R}\right)^{6}\right]$, where $\epsilon=$  meV and $\sigma=$  Å are characteristic constants for the bond and $R$ is the bond length (the distance between the atoms).

(a) Find the interaction force between the two atoms at a function of distance $R=$  Å. A positive force pushes the atoms apart and a negative force pulls them together.

$F=$  N.

(b) Find the equilibrium spacing $R_0$ between the two atoms.

$R_0=$  Å

2.19 CO molecule

What are the rotational energy levels of an CO molecule? The bond length is 112.8 pm.

2.20 Rotational energy states

A molecule can store energy in electronic excitations (where an electron moves from the ground state to an excited state), vibrational excitations, and rotational excitations. Linear molecules (where all of the atoms are in a line) have two rotational degrees of freedom and non-linear molecules have three rotational degrees of freedom. The rotational degrees of freedom are characterized by their moments of inertia. Generally, the moment of inertia is a matrix,

\begin{equation} \bf{I}=\left[\begin{matrix} \sum\limits_a m_a(y_{a}^2+z_a^2) & \sum\limits_a m_ax_{a}y_{a} & \sum\limits_a m_ax_{a}z_{a}\\ \sum\limits_a m_ay_{a}x_{a} & \sum\limits_a m_a(x_{a}^2+z_a^2) & \sum\limits_a m_ay_{a}z_{a}\\ \sum\limits_a m_az_{a}x_{a} & \sum\limits_a m_az_{a}y_{a} & \sum\limits_a m_a(x_{a}^2+y_a^2)\\ \end{matrix}\right]. \end{equation}

Here $a$ sums over the atoms in a molecule, $m_a$ is the mass of atom $a$ and $(x_a,y_a,z_a)$ is the position of atom $a$ measured from the center of mass of the molecule. The three rotational degrees of freedom correspond to rotations about the principle axes of the moment of inertia tensor. These principle axes are found by diagonalizing the moment of inertia tensor. If it is assumed that the atoms are rigidly attached to each other then the rotational energy levels are described by the rigid rotator model. The energy levels of a rotational degree of freedom are

$$ E_J= \frac{\hbar^2}{2I}J(J+1), $$

where $I$ is the eigenvalue of the moment of inertia tensor for that degree of freedom, and $J$ is the orbital quantum number, $J=0,1,2,\cdots$. For a diatomic molecule with an equilibrium spacing of the atoms of $r_0$ and atomic masses $m_a$ and $m_b$, the moment of inertia is $I=\frac{m_am_b}{m_a+m_b}r_0^2$.

The wave functions for a rigid rotor are the the spherical harmonics $Y_{Jm}(\theta,\phi)$. Here, $m$ is the quantum number that characterizes the $z$-component of the molecule's angular momentum and takes on the values $m=-J,...,J$. Each rotational energy level has a degeneracy of $(2J+1)$.

Since photons have a spin of $1$, $J$ can only change by $\pm 1$ when photons are emitted or absorbed by a molecule.

(a) Calculate the moment of inertia of CHCl3 for rotation around the C-H bond. Use an online database such as JSmol molecule viewer to determine the bond lengths and bond angles. Type CHCl3 into the search box to retrieve the data for this molecule. Double-click on an atom and then double-click on another atom to find the distance between the atoms. To determine an angle: double click on an atom, single click on a second atom, double click on a third atom.

(b) Calculate the three eigenvalues of the moment of inertia tensor of a water molecule.

The moment of inertia depends on the isotopes of the atoms. Use the isotopes 1H 16O, 12C, and 35Cl.

2.21 The rotational spectrum of BrF

The rotational spectrum of 79Br19F has a uniform line spacing of 0.71433 cm-1. In the rigid rotator model, the rotational energies are given by,

$$E_{J}= \frac{\hbar^2}{2I}J(J+1)=BJ(J+1)\qquad J= 0,1,2,\cdots,$$

where $B$ is the rotation constant and $I$ is the moment of inertia of the molecule. For a diatomic molecule,

$$I=\frac{m_am_b}{m_a+m_b}r_0^2,$$

where $r_0$ is the bond length. Rotational transitions are allowed for $\Delta J = \pm 1$.

Calculate the rotation constant $B$, the moment of inertia, and the bond length of the molecule. Determine the wave number of the transition $J = 9 \rightarrow 10$. For the state $J = 10$, calculate the classical number of rotations per second.

2.22 Estimating vibrational energies from the Morse potential

A Morse potential is often used to approximate the bond potential of a covalent bond.

$$U(r)= U_0\left(e^{-2a(r-r_0)}-2e^{-a(r-r_0)}\right)$$

For a Morse potential with $U_0=1.76\,\text{eV}$, $r_0=1.1$ Å, and $a=9\times 10^{10}\,\text{1/m}$, determine the spacing of the vibrational levels by approximating the minimum of the potential using a parabola. For the mass, use the reduced mass of N2, $m=1.16\times 10^{-26}$ kg.

Linearizing the bond potentials to find the normal mode vibratins
For molecules with more than two atoms, a multidimensional potential energy can be constructed by calculating the energy for the atoms of the molecule being in different positions. This potential energy has a minimum when all of the atoms are in their equilibrium positions. For small displacements from the minimum, it is possible to model this as a collection of masses attached by linear springs. A molecule with $n$ atoms has $3n$ degrees of freedom. Let $u_1$ be the displacement of the first atom from its equilibrium position in the $x$-direction, $u_2$ be the displacement of the first atom from its equilibrium position in the $y$-direction, $u_3$ be the displacement of the first atom from its equilibrium position in the $z$-direction, $u_4$ be the displacement of the second atom from its equilibrium position in the $x$-direction, etc. Newton's law, in this case, takes the form of $3n$ second-order differential equations,

\[ \begin{eqnarray} m_1\frac{d^2u_1}{dt^2} = k_{12}(u_2-u_1)+ k_{13}(u_3-u_1)+\cdots+k_{1,3n}(u_{3n}-u_1)\\ m_1\frac{d^2u_2}{dt^2} = k_{12}(u_1-u_2)+ k_{23}(u_3-u_2)+\cdots+k_{2,3n}(u_{3n}-u_2)\\ m_1\frac{d^2u_3}{dt^2} = k_{13}(u_1-u_3)+ k_{23}(u_2-u_3)+\cdots+k_{3,3n}(u_{3n}-u_3)\\ m_2\frac{d^2u_4}{dt^2} = k_{12}(u_1-u_4)+ k_{23}(u_2-u_4)+\cdots+k_{4,3n}(u_{3n}-u_4)\\ m_2\frac{d^2u_5}{dt^2} = k_{12}(u_1-u_5)+ k_{23}(u_2-u_5)+\cdots+k_{5,3n}(u_{3n}-u_5)\\ m_2\frac{d^2u_6}{dt^2} = k_{12}(u_1-u_6)+ k_{23}(u_2-u_6)+\cdots+k_{6,3n}(u_{3n}-u_6)\\ \end{eqnarray} \] \[ \begin{equation} \vdots \end{equation} \] \[ \begin{eqnarray} m_n\frac{d^2u_{3n-2}}{dt^2} = k_{1,3n-2}(u_1-u_{3n-2})+ k_{2,3n-2}(u_2-u_{3n-2})+\cdots+k_{3n-2,3n}(u_{3n}-u_{3n-2})\\ m_n\frac{d^2u_{3n-1}}{dt^2} = k_{1,3n-1}(u_1-u_{3n-1})+ k_{2,3n-1}(u_2-u_{3n-1})+\cdots+k_{3n-1,3n}(u_{3n}-u_{3n-1})\\ m_n\frac{d^2u_{3n}}{dt^2} = k_{1,3n}(u_1-u_{3n})+ k_{2,3n}(u_2-u_{3n})+\cdots+k_{3n-1,3n}(u_{3n-1}-u_{3n})\\ \end{eqnarray} \]

For a normal mode solution, all of the atoms move with the same frequency $u_p=A_pe^{i\omega t}$, where $A_p$ is the amplitude of displacement $p=1,2,\cdots,3n$. Substituting this into the equations above results in an eigenvalue problem.

\[ \begin{equation} \left[ \begin{matrix} \sum\limits_{p\neq 1} \frac{k_{1p}}{m_1} & -\frac{k_{12}}{m_1} & -\frac{k_{13}}{m_1} & \cdots & -\frac{k_{1,3n}}{m_1} \\ -\frac{k_{21}}{m_1} &\sum\limits_{p\neq 2} \frac{k_{2p}}{m_1}& -\frac{k_{23}}{m_1} & \cdots & -\frac{k_{2,3n}}{m_1} \\ -\frac{k_{31}}{m_1} & -\frac{k_{32}}{m_1} & \sum\limits_{p\neq 3} \frac{k_{3p}}{m_1} & \cdots & -\frac{k_{3,3n}}{m_1} \\ & \vdots & & \\ -\frac{k_{1,3n-2}}{m_n} & \cdots & \sum\limits_{p\neq 3n-2} \frac{k_{p,3n-2}}{m_n} & -\frac{k_{3n-1,3n-2}}{m_n} & -\frac{k_{3n,3n-2}}{m_n} \\ -\frac{k_{1,3n-1}}{m_n} & \cdots & -\frac{k_{3n-2,3n-1}}{m_n} & \sum\limits_{p\neq 3n-1} \frac{k_{p,3n-1}}{m_n} & -\frac{k_{3n,3n-1}}{m_n} \\ -\frac{k_{1,3n}}{m_n} & \cdots & -\frac{k_{3n-2,3n}}{m_n} & -\frac{k_{3n-1,3n}}{m_n} & \sum\limits_{p\neq 3n} \frac{k_{p,3n}}{m_n} \\ \end{matrix} \right] \left[ \begin{matrix} A_{1} \\ A_{2} \\ A_{3} \\ \vdots \\ A_{3n-1} \\ A_{3n-2} \\ A_{3n} \end{matrix} \right]=\omega^2\left[ \begin{matrix} A_{1} \\ A_{2} \\ A_{3} \\ \vdots \\ A_{3n-1} \\ A_{3n-2} \\ A_{3n} \end{matrix} \right] \end{equation} \]

Here $k_{pq}=k_{qp}$ is the effective spring constant between describing how the energy increases as $u_p-u_q$ increases. Of the $3n$ degrees of freedom, three correspond to the translation of the center of mass of the molecule. The potential energy of the mass-spring system does not change if the whole molecule is translated. The mathematical consequence of this is that there are three eigenvectors that describe the center-of-mass motion which have an eigenvalue of zero. The potential energy of the molecule also does not depend on the orientation of the molecule. This means that the eigenvectors that correspond to the rotational degrees of freedom also have eigenvalues of zero. There are two rotational degrees of freedom for linear molecules where the atoms are all in a line and three rotational degrees of freedom for nonlinear molecules. The rest of the eigenvectors correspond to vibrational normal modes. There are $N_{\text{vib}}=3n-5$ vibrational modes for linear molecules and $N_{\text{vib}}=3n-6$ vibrational modes for nonlinear molecules.

The state of a molecule is described by its electronic state (how the electrons occupy the molecular orbitals) plus its vibrational state plus its rotational state. A molecule can change states by interacting with electromagnetic waves. Molecules can absorb a photon of energy $\hbar \omega$ if that photon has the energy to move an electron from an occupied state to an unoccupied state, $\hbar \omega = \Delta E_{\text{elec}}+\Delta E_{\text{vib}}+\Delta E_{\text{rot}}$. If a molecule is in a state other than the ground state, it can also emit a photon such that $\hbar \omega = \Delta E_{\text{elec}}+\Delta E_{\text{vib}}+\Delta E_{\text{rot}}$. Transitions that involve a change in the electronic state typically result in the emission of a visible photon. Transitions where the electronic configuration of the electrons stays the same but the vibrational state changes typically result in the emission of an infrared photon. Transitions where the electronic configuration of the electrons stays the same and the vibrational state stays the same but the rotational state changes typically result in the emission of a microwave photon. Although the transition from any state to any other state is possible, some transitions occur at a very low rate. The absorption spectrum or emission spectrum of a molecule tends to be dominated by the transitions that occur at a high rate. The adsorption and emission spectra of a molecule are unique for that molecule and can be used as a means of measuring the presence of a specific molecule.

2.23 Vibrational modes of CO2 and H2O

(a) How many vibrational modes do CO2 and H2O have?

(b) An N2 molecule is in its electronic ground state and its vibronic ground state. Which rotational states would be occupied at room temperature? (The app on Rotational and vibrational energy levels of diatomic molecules includes N2. Room temperature corresponds to an energy of about 0.025 eV. )

2.24 C-O bond

Calculate the force constant of the C-O bond from the oscillation frequency of the CO molecule (6.42 × 1013 Hz).

The frequency is related to the spring constant by,

$$f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{k_{\text{eff}}/m_r},$$

where $m_r$ is the reduced mass, $m_r=(1/m_a+1/m_b)^{-1}$. This can be solved for the spring constant.

2.25 Thermal occupation of rotational states

The energy levels of the vibrational and rotational states of some molecules can be determined with the app Rotational and vibrational energy levels of diatomic molecules.

The ratio of the occupation probabilities of state $j$ to state $i$ is given by a Boltzmann factor,

$$\frac{P_j}{P_i} = \frac{g_j\exp\left(\frac{-E_j}{k_BT}\right)}{g_i\exp\left(\frac{-E_i}{k_BT}\right)}.$$

Here $g_i$ is the degeneracy of state $i$. For the rotational energy levels, $g_J=(2J+1)$.

An N2 molecule is in its electronic ground state and its vibronic ground state. What are the occupation probabilities of the first 40 rotational states compared to the rotational ground state at 300 K? Copy the rotational energy levels from the app and paste them into a spreadsheet program to calculate the probabilities.

2.26 HD and D2

The energy needed to separate two hydrogen atoms from their equilibrium distance in a hydrogen molecule until they are far apart is $U_0=$ 4.46 eV. The vibrational energy of the ground state vibrational mode of H2 is $\hbar\omega/2 =$ 0.269 eV, and the rotational constant of H2 is $B=7.6\times 10^{-3}$ eV. The rotational energy levels are given by,

$$E_J= \frac{\hbar^2}{2I}J(J+1)= BJ(J+1)\qquad J=0,\,1,\,2,\cdots$$

Consider what would happen if one or both of the hydrogen atoms were replaced by deuterium.

(a) Explain why the mass of the isotope is irrelevant if the Born-Oppenheimer approximation is used to calculate the bond potential.

(b) Calculate $U_0$, $\hbar\omega/2$, and $B$ for HD and D2.

2.27 Vibrational modes

(a) How many vibrational modes do CO2 and H2O have?

(b) Look up the vibrational modes of CO2 and H2O online and make sketches of how the atoms move. Which modes create an oscillating electric dipole moment? Modes where the dipole moment oscillate are IR-active.

2.28 LCAO - water

Molecular orbitals are used in a similar way as hydrogen wave functions are used to construct the many-electron wave functions of atoms. In the simplest approximation, the molecular orbitals are the wave function of a single electron moving in a potential created by all of the positively charged nuclei in a molecule. The molecular orbital Hamiltonian can be written,

\[ \begin{equation} H_{\text{mo}}= - \frac{\hbar^2}{2m_e}\nabla^2 -\sum\limits_{a} \frac{Z_ae^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_a|}. \end{equation} \]

Here $\vec{r}_a$ are the positions of the nuclei in the molecule and $\vec{r}$ is the position of the single electron. The molecular orbital Hamiltonian is often solved by a method called the Linear Combination of Atomic Orbitals (LCAO). It is assumed that the wave function can be written in terms of atomic orbitals that are centered around the nuclei,

\begin{equation} \psi_{\text{mo}}(\vec{r})= \sum\limits_{a} \sum\limits_{ao} c_{ao,a}\phi^{Z_a}_{ao}\left(\vec{r}-\vec{r}_a\right). \end{equation}

where $ao$ labels the atomic orbitals ($ao=1$: 1s, $ao=2$: 2s, $ao=3$: 2px, $\cdots$). The number of molecular orbitals that we calculate will be equal to the number of unknown coefficients. We call this number $N$. We have to decide how many atomic orbitals should be included for each atom. A reasonable choice is to take all of the occupied atomic orbitals of the isolated atoms. For instance, for water one might use the 1s, 2s, and 2p orbitals of oxygen and the 1s orbitals of the two hydrogen atoms. In that case, there would be $N=7$ terms in the wave function for the molecular orbital. There is no strict rule as to which atomic orbitals to include. Including more atomic orbitals leads to a higher accuracy but makes the numerical calculation more difficult.

At this point it is convenient to relabel the atomic orbitals used in the wave function with integers $p=1\cdots N$. For water we might choose $\phi_1 = \phi^{H}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{H1}}\right)$, $\phi_2 = \phi^{H}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{H2}}\right)$, $\phi_3 = \phi^{O}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{O}}\right)$, $\phi_4 = \phi^{O}_{\text{2s}}\left(\vec{r}-\vec{r}_{\text{O}}\right)$, etc. The trial wave function can then be written more compactly as,

\begin{equation} \psi_{\text{mo}} = \sum\limits_{p=1}^N c_p\phi_p. \end{equation}

The time independent Schrödinger equation is,

\[ \begin{equation} H_{\text{mo}}\psi_{\text{mo}}=E\psi_{\text{mo}} . \end{equation} \]

Multiply the Schrödinger equation from the left by each of the atomic orbitals and integrate over all space. This results in a set of $N$ algebraic equations called the Roothaan equations.

\[ \begin{equation} \begin{matrix} \large \langle \phi_1 | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_1|\psi_{\text{mo}} \rangle \\ \large \langle \phi_2 | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_2|\psi_{\text{mo}} \rangle\\ \large \vdots \\ \large \langle \phi_N | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_N|\psi_{\text{mo}} \rangle \end{matrix} \end{equation} \]

By substituting in the form for $\psi_{\text{mo}}$ from above, the Roothaan equations can be written in matrix form,

\[ \begin{equation} \left[ \begin{matrix} H_{11} & H_{12} & \cdots & H_{1N} \\ H_{21} & H_{22} & \cdots & H_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ H_{N1} & H_{N2} & \cdots & H_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right] = E\left[ \begin{matrix} S_{11} & S_{12} & \cdots & S_{1N} \\ S_{21} & S_{22} & \cdots & S_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ S_{N1} & S_{N2} & \cdots & S_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right]. \end{equation} \]

Here the elements of the Hamiltonian matrix and the overlap matrix are,

\[ \begin{equation} H_{pq}= \langle\phi_{p}|H_{\text{mo}}|\phi_{q}\rangle \hspace{1cm}\text{and}\hspace{1cm} S_{pq}= \langle\phi_{p}|\phi_{q}\rangle . \end{equation} \]

The Roothaan equations can be solved numerically to find $N$ solutions for the energy $E$ along with the corresponding coefficients that describe the $N$ wave functions which are the molecular orbitals. Multiplying from the left with the inverse of the overlap matrix results in an eigenvalue problem,

$$\textbf{S}^{-1}\textbf{H}\,\vec{c}=E\,\vec{c}.$$

Molecular orbitals for water are calculated using the LCAO method by assuming a molecular orbital that is a linear combination of the two hydrogen $1\text{s}$ orbitals plus the oxygen $2\text{s}$, $2\text{p}_x$, $2\text{p}_y$, and $2\text{p}_z$ orbitals. Following Slater's rules, the effective charge of the oxygen atom is $Z_{eff} = 4.55$. How many molecular orbitals will be calculated? How many of the molecular orbitals are filled in the ground state of the molecule? The energy of the many-electron ground state is calculated including all of the electrons of the molecule in the Slater determinant, how many dimensions is the integral $\langle \Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})|H_{\text{elec}}|\Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_{N_e})\rangle$ defined in?

2.29 Ionic bonds

In an ionic bond one binding partner loses an electron to the other binding partner. For a large separation between the atoms, energy is required to transfer an electron from one atom to the other. For a small separation between the atoms, the energy is reduced by transferring an electron from one atom to the other. Consider a neutral Na atom separated by a long distance from a neutral Cl atom. We assign this configuration an energy of zero. An electron is then transferred from the Na atom to the Cl atom while they remain well separated. This will increase the energy by the ionization energy of Na, 5.14 eV, minus the electron affinity of Cl, 3.61 eV. The increase in energy is 1.53 eV. The ions are then slowly brought together and as they get closer, the energy of the system decreases like a Coulomb potential,

$$ U_{\text{Coulomb}}=\frac{e^2}{4\pi\epsilon_0R},$$

where $\epsilon_0$ is the permeativity constant. At first, the energy is greater than zero but for a critical separation $R_{\text{crit}}$, the energy of the two ions becomes lower than the energy of the isolated neutral atoms. Calculate this critical separation for NaCl, and CsCl. You will have to look up the ionization energies and electron affinities.

2.30 Molecular orbitals

Draw the bonding and antibonding orbitals that can be constructed from

  1. 2 1s orbitals
  2. 2 2pz orbitals
  3. 2 2px orbitals

The interatomic axis is the $x$-axis. Indicate the nodal planes.

2.31 Morse potential

Covalent bonds are often described by a Morse potential of the form,

\begin{equation} U(r)= U_0\left(e^{-2(r-r_0)/a}-2e^{-(r-r_0)/a}\right). \end{equation}

For one particular covalent bond, $U_0 = $  eV, $r_0 = $ 0.1 nm, and $a = $ 0.0 nm.

For small deviations around $r_0$, the force between the two atoms can be described by Hooke's law $F=-k(r - r_0)$. Here $k$ is the effective spring constant. The relationship between the potential and the force is $F = -\frac{dU}{dr}$. What is the effective spring constant for this bond?

$k=$  N/m.

The atoms form a crystal 1 cm high, 1 cm wide, and 6 cm long. The crystal has a simple cubic structure and the bond between each pair of atoms can be described by the Morse potential given above. The crystal is stretched along the long axis. What is the spring constant of the crystal?

$k_{\text{crystal}}=$  N/m.

2.32 Carbon hybridization

Carbon forms different hybrid orbitals depending on which molecule it is in. In ethyne (acetylene) carbon is present in sp hybridization, in graphene the carbon atom is sp² hybridized and in methane C is sp³ hybridized.

The sp, sp², and sp³ hybrid orbitals are defined in terms of the 2s and 2p atomic orbitals.

Sketch the hybrid orbitals $\psi_1$, $\psi_2$, $\psi_3$, and $\psi_4$ for sp, sp², and sp³ hybridization.

2.33 Bond length and bond angle

Use the NCI database in the jsmol app to determine the N-C bond length and the angle formed by the atoms C-C-N in Glycine.

Other databases might give slightly different results for the bond lengths and bond angles.

2.34 He2 molecule

A linear combination of atomic orbitals used to find the molecular orbitals of a He2 molecule contains four atomic orbitals,

\[ \begin{equation} \psi(\vec{r})= c_1\phi^{Z_{eff}}_{\text{1s}}\left(\vec{r}-\vec{r}_A\right)+c_2\phi^{Z_{eff}}_{\text{1s}}\left(\vec{r}-\vec{r}_B\right)+c_3\phi^{Z_{eff}}_{\text{2s}}\left(\vec{r}-\vec{r}_A\right)+c_4\phi^{Z_{eff}}_{\text{2s}}\left(\vec{r}-\vec{r}_B\right). \end{equation} \]

What is the integral that needs to be evaluated to determine the Hamiltonian matrix element $H_{12}$ for this molecule? Following Slater's rules, use $Z_{eff} = 1.7$ to describe the effective positive charge of the helium nuclei.

What is the integral that needs to be evaluated to determine the matrix element of the overlap matrix $S_{13}$ in the Roothaan equations? This integral is easy to evaluate. What is $S_{13}$?

2.35 Effective spring constant

Common models for bond potentials are:

Covalent bond - Morse potential

$$U(R)= U_0\left(e^{-2a(R-R_E)}-2e^{-a(R-R_E)}\right) +U_{\infty}.$$

Van der Waals bond - Lennard-Jones potential,

$$U(R)= 4\epsilon\left[\left(\frac{\sigma}{R}\right)^{12}-\left(\frac{\sigma}{R}\right)^{6}\right].$$

Find expressions for the effective spring constants near the minima of the bond potentials. Check that the expression that you derive has the correct units.