# The Quantum Molecule

By

Tee Kai Ze

,

Guan Xin

,

Jensie Low

,

Seow Ryan

●

Published 2022-08-27

In the previous chapter, we have established what is a chemical bond and how electron transitions between energy levels of a molecule are observable. Building upon the principles of linear combination of atomic orbitals, one might expect discrete energy levels and transitions leading to discrete peaks in molecular spectra. Is it really so?

Let us take a look at a molecule you should see and eat everyday: chlorophyll :) There are 2 types of chlorophyll (Chl), Chl a and Chl b . Their molecular structure and (solvent-free) absorption spectra are shown in the figure below.

One notable difference between atomic and molecular (UV-Vis) spectra is that while the former is composed of distinct discrete peaks, the latter usually exhibit broader peaks or bands. A fine example of a molecular spectrum is that of Chl a and b shown above. What is the reason for the broader bands as opposed to narrow peaks?

One major contributing factor to the broad bands is the existence of vibrational and rotational energy levels within an (molecular) electronic state. For a molecule at a certain electronic orbital state, it can wobble and rotate. The vibrations and rotations changes the energy of the molecule slightly. As many vibrational and rotational states are possible for each molecular orbital state, the electronic state no longer possesses a single well-defined discrete energy but a range of energies. Depending on the resolution of the spectrometer used, the band of energy may be resolved such that the finer structures can be seen (see spectrum of Chl a and b above). Very often a lower resolution spectrometer is used to probe the molecule , in which case the finer details will smear out to give smooth and broad peaks (google image for chlorophyll sprectrum).

The energies associated with transitions across electronic energy states are orders of magnitude larger than that of the vibrational states, which are in turn orders of magnitude larger than transitions across rotational states.

\begin{equation}
\Delta E_{\text{MO}}\approx10^{3}\Delta E_{\text{vib}}\approx10^{6}\Delta E_{\text{rot}}
\end{equation}

A simplistic representation of this is given in the following diagram:

In order to simplify the quantum description of the molecule, Born and Oppenheimer assumed that the motion of the electrons and nuclei in molecules can be considered separately. Since the mass of the nuclei is of several orders of magnitude heavier than the electrons, the motion of nuclei is negligible to that of the electrons. The energies of the electron can then be calculated at fixed internuclei distances. The energies of electronic, vibrational and rotational states can be considered to be independent of each other. This is known as the Born-Oppenheimer approximation.

In this course we will examine the vibrational and rotational energy levels and their corresponding transitions in greater details.

Let us consider the rotation of a diatomic molecule. We assume the bond between the two atoms is rigid, meaning that the internuclear distance does not change. As shown in section 4.2.1, the kinetic and potential energies of the system is

\begin{equation}
T=\frac{L^{2}}{2I}
\end{equation}

and

\begin{equation}
V=0
\end{equation}

respectively. The Schrödinger equation is

\begin{equation}
-\frac{\hbar^{2}}{2I}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\theta^{2}}\right]\psi=E\psi
\end{equation}

It is out of the scope of the course to derive Eq.(4). For the interested reader, see the Appendix on a note about operators.

The procedure to solve the above SE is in essence similar to how it was done in Chapter 3. The allowed energies of the system are

\begin{equation}
E=\frac{\hbar^{2}J(J+1)}{2I}=BJ(J+1)
\end{equation}

where J=0,1,2,3,\dots is the rotational quantum number and B=\hbar^{2}/2I is called the rotational constant.

The rotational constant depends on the moment of inertia of the diatomic system, which in turn depends on the masses of the nuclei as well as the bond length. Different molecules will have different rotational constants thus making the set of rotational energies unique to each molecule.

The rotational constant above has units of energy. Some (eg. radio astronomers) like to state the rotational constant in units of frequency (Hz). In this case, divide the above B by h. Other (eg. chemists) prefer to state the rotational constant in units of wavenumber (\text{cm}^{-1}). In this case divide the above by hc where c=3.00\times10^{10}\text{cm s}^{-1} is the speed of light.

Define P(J) as the probability of a molecule being in the J^{\text{th }} energy state, we have

\begin{equation}
P(J)=(2J+1)P(0)\exp\left(-\frac{BJ(J+1)}{kT}\right)
\end{equation}

where P(0) is the probability of finding the particle at ground state (J=0).

The diatomic molecule with a permanent dipole moment can transit from one rotational energy state to another by absorbing or emitting a photon.¹ By examining the probability of transitions, it can be shown that only transitions across more than one energy level is highly improbable. Thus transitions across rotational energy levels are restricted to the following selection rule

\begin{equation}
\Delta J=\pm1
\end{equation}

This means that transitions such as J=3\rightarrow2 and J=3\rightarrow4 are allowed while J=5\rightarrow3 is highly unlikely.

The energy of a photon emitted in a J+1\rightarrow J transition is given by

\begin{align}
hf=\Delta E & =B(J+1)(J+2)-BJ(J+1)\nonumber \\
hf & =2B(J+1)
\end{align}

The probablility of particle residing in the J^{\text{th}} energy state given in Eq.(6) gives the relative intensity of the spectral line.

Try it yourself!

The figure below is the rotation spectrum of CO. Explain qualitatively why different spectral lines have different intensities.

In the previous section, the bond between atoms are treated as a rigid rod. In this section, we shall allow the bonds to stretch and compress. This means that the molecule can vibrate when the atoms are displaced from their equilibrium position.

Let us consider the case of a diatomic molecule. We model its vibrational motion with a harmonic oscillator described in section 4.2.2, i.e. a 1D system with effective mass (see Eq.(Rot.18))

\begin{equation}
\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}
\end{equation}

where m_{1} and m_{2} are the masses of the atoms. The potential energy is (see Eq.(Rot.18))

\begin{equation}
V=\frac{1}{2}\mu\omega^{2}x^{2}
\end{equation}

The Schrödinger equation is

\begin{equation}
-\frac{\hbar^{2}}{2\mu}\frac{d^{2}\psi}{dx^{2}}+\frac{1}{2}\mu\omega^{2}x^{2}\psi=E\psi
\end{equation}

and the boundary conditions are

\begin{equation}
x\rightarrow\pm\infty,\psi(x)\rightarrow0
\end{equation}

Without going into the details of solving the quantum mechanics for a harmonic oscillator, we just state here the results. The molecules display discrete vibrational energy states given by

\begin{equation}
E=\hbar\omega\left(v+\frac{1}{2}\right)\,,\;v=0,1,2,3,\dots
\end{equation}

where v is the vibrational quantum number which takes on integer values.

For transitions across vibrational energy levels, a similar rule to rotational levels holds:

\begin{equation}
\Delta v=\pm1
\end{equation}

This means that all allowed transitions across vibrational energy levels will have energy

\begin{equation}
\Delta E_{\text{vib}}=E_{v+1}-E_{v}=\frac{h\omega}{2\pi}\left(\left(v+1+\frac{1}{2}\right)-\left(v+\frac{1}{2}\right)\right)=\frac{h\omega}{2\pi}
\end{equation}

Note the the lowest energy state is non-zero.

Although it seems from Eq.(15) that the vibrational spectrum is a simple single line for a molecule, it is not so after taking rotational energies into account. In each vibrational energy band, the molecule can be at many different rotational energy states. For a heterogenous diatomic molecule, a transition across the vibrational energy will be accompanied by a change in rotational levels with the following rule:

\begin{equation}
\Delta v=\pm1,\;\Delta J=\pm1
\end{equation}

The vibrational spectrum of a molecule thus exhibits 2 bands. The higher energy band (R-branch) comprises of \Delta J=+1 transitions, while the lower energy band (P-branch) comprises of \Delta J=-1.

The image above is a ro-vibrational absorbance spectrum for HCl, and we can clearly see that there is a peak near the centre that is missing. That missing peak belongs to the pure vibrational transition, \Delta E=\frac{1}{2}\hbar\omega. To the left, is the P branch (as in P for poor), where \Delta J=-1, and to the right is the R branch (as in R for rich), where \Delta J=+1. The pure vibrational (\Delta J=0) peak do not show up as the transition is much less favourable than the \Delta J=\pm1 transitions.