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Learning Objectives
• Understand the utility and limits of using the quantum harmonic oscillator as a model for molecular vibrations
The quantum harmonic oscillator is one of the most important model systems in quantum mechanics. This is due in partially to the fact that an arbitrary potential curve $V(x)$ can usually be approximated as a harmonic potential at the vicinity of a stable equilibrium point. Furthermore, it is one of the few quantum-mechanical systems for which an exact, analytical solution exists. Solving other potentials typically require either approximations or numerical approaches to identify the corresponding eigenstates and eigenvalues (i.e., wavefunctions and energies).
A general potential energy ($V(x)$) curve for a molecular vibration can be expanded as a Taylor series
$V(x) = V(x_0) + \left. \dfrac {d V(x)}{d x} \right|_{x_0}^{x} (x - x_0) + \left. \dfrac {1}{2!} \dfrac {d^2 V(x)}{d x^2} \right|_{x_0}^{x} (x - x_0)^2 + \ldots + \left. \dfrac {1}{n!} \dfrac {d^n V(x)}{d x^n} \right|_{x_0}^{x} (x - x_0)^n \label{5.3.1}$
It is important to note that this approximation is only good for $x$ near $x_0$, and that $x_0$ stands for the equilibrium bond distance. $V(x)$ is often (but not always) shortened to the cubic term and can be rewritten as
$V(x) = \dfrac {1}{2} kx^2 + \dfrac {1}{6} \gamma x^3 \label{5.3.2}$
where $V(x_0) = 0$, $k$ is the harmonic force constant (harmonic term), and $\gamma$ is the first anharmonic term (i.e., cubic). As Figure 5.3.2 demonstrates, the harmonic oscillator (red curve) is a good approximation for the exact potential energy of a vibration (blue curve).
Adding anharmonic perturbations to the harmonic oscillator (Equation $\ref{5.3.2}$) better describes molecular vibrations. Anharmonic oscillation is defined as the deviation of a system from harmonic oscillation, or an oscillator not oscillating in simple harmonic motion. Anharmonic oscillation is described as the restoring force is no longer proportional to the displacement. Adding the cubic term (Figure 5.3.2 ; green curve) improves the harmonic oscillation approximation especially under greater displacement from equilibrium.
Naturally, adding higher order anharmonic terms, like quartic terms (Figure $\PageIndex{2; right}$) improve the approximation. Almost all diatomics have experimentally determined potentials for their lowest energy states. $\ce{H2}$, $\ce{Li2}$, $\ce{O2}$, $\ce{N2}$, and $\ce{F2}$ with terms up to $n < 10$ determined of Equation $\ref{5.3.1}$.
Figure 5.3.1 shows the the general potential with (numerically) calculated energy levels ($E_0$, $E_1$ etc.). $D_o$ is the dissociation energy, which is different from the well depth $D_e$. These vibrational energy levels of this plot can be calculated using the harmonic oscillator model (i.e., Equation 5.3.1 with the Schrödinger equation) and have the general form
$E_v = \left(v + \dfrac{1}{2}\right) v_e - \left(v + \dfrac{1}{2}\right)^2 v_e x_e + \left(v + \dfrac{1}{2}\right)^3 v_e y_e + \text{higher terms} \label{5.3.7}$
where $v$ is the vibrational quantum number and $x_e$ and $y_e$ are the first and second anharmonicity constants, respectively.
The $v = 0$ level is the vibrational ground state. Because this potential is less confining than a parabola used in the harmonic oscillator, the energy levels become less widely spaced at high excitation (Figure 5.3.1 ; top of potential).
Limitations of the Harmonic Oscillator Model for Molecular Vibrations
The harmonic oscillation is a great approximation of a molecular vibration, but has key limitations:
• Due to equal spacing of energy, all transitions occur at the same frequency (i.e. single line spectrum). However experimentally many lines are often observed (called overtones).
• The harmonic oscillator does not predict bond dissociation; you cannot break it no matter how much energy is introduced.
Morse Potentials are better Approximations of Vibrational Motion
A more powerful approach than just "patching up" the harmonic oscillator solution with anharmonic corrections is to adopt a different potential ($V(x)$). One such approach is the Morse potential, named after physicist Philip M. Morse, and a better approximation for the vibrational structure of the molecule than the harmonic oscillator because it explicitly includes the effects of bond breaking and accounts for the anharmonicity of real bonds (Figure 5.3.4 ).
The Morse Potential is a good approximation to $V(x)$ and is best when looking for a general formula for all $x$ from 0 to $\infty$, not just applicable for the local region around the $x_o$:
$V(x) = D(1-e^{-\beta (x - x_0)})^2 \label{5.3.8}$
with $V(x = x_0) = 0$ and $V(x = \infty) = D$.
The Morse Potential (Figure 5.3.4 ) approaches zero at infinite $r_e$ and equals $-D_e$ at its minimum (i.e. $r=r_e$). It clearly shows that the Morse potential is the combination of a short-range repulsion term (small $r$ values) and a long-range attractive term (large $r$ values).
Solving the Schrödinger Equation with the Morse Potential (Equation \ref{5.3.8}) is not trivial, but can be done analytically.
$\hat{H}|\psi \rangle = E_n | \psi \rangle \nonumber$
with
\begin{align} \hat{H} &= \hat{T} + \hat{V} \[4pt] &= \dfrac{- \hbar ^2 d^2}{2m \;dx^2} + D(1-e^{-\beta (x - x_0)})^2 \label{5.3.9} \end{align}
The solutions and energies for the Morse potential will not be used in this course and will not be discussed in more detail.
Contributors and Attributions
• Peter Kelly (UCDavis)
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For a classical oscillator, we know exactly the position, velocity, and momentum as a function of time. The frequency of the oscillator (or normal mode) is determined by the reduced mass $\mu$ and the effective force constant $k$ of the oscillating system and does not change unless one of these quantities is changed. There are no restrictions on the energy of the oscillator, and changes in the energy of the oscillator produce changes in the amplitude of the vibrations experienced by the oscillator.
For the quantum mechanical oscillator, the oscillation frequency of a given normal mode is still controlled by the mass and the force constant (or, equivalently, by the associated potential energy function). However, the energy of the oscillator is limited to certain values. The allowed quantized energy levels are equally spaced and are related to the oscillator frequencies as given by Equation $\ref{5.4.1}$ and Figure 5.4.1 .
$E_v = \left ( v + \dfrac {1}{2} \right ) \hbar \omega = \left ( v + \dfrac {1}{2} \right ) h \nu \label {5.4.1}$
with
$v = 0, 1, 2, 3, \cdots \infty \nonumber$
In a quantum mechanical oscillator, we cannot specify the position of the oscillator (the exact displacement from the equilibrium position) or its velocity as a function of time; we can only talk about the probability of the oscillator being displaced from equilibrium by a certain amount. This probability is given by
$P_{Q \rightarrow Q + dQ} = \int_{Q}^{Q + dQ} \psi ^*_v (Q) \psi _v (Q) dQ \label {5.4.3}$
We can, however, calculate the average displacement and the mean square displacement of the atoms relative to their equilibrium positions. This average is just $\left \langle Q \right \rangle$, the expectation value for $Q$, and the mean square displacement is $\left \langle Q^2 \right \rangle$, the expectation value for $Q^2$. Similarly we can calculate the average momentum $\left \langle P_Q \right \rangle$, and the mean square momentum $\left \langle P^2_Q \right \rangle$, but we cannot specify the momentum as a function of time.
Physically what do we expect to find for the average displacement and the average momentum? Since the potential energy function is symmetric around $Q = 0$, we expect values of $Q > 0$ to be equally as likely as $Q < 0$. The average value of $Q$ therefore should be zero.
These results for the average displacement and average momentum do not mean that the harmonic oscillator is sitting still. As for the particle-in-a-box case, we can imagine the quantum mechanical harmonic oscillator as moving back and forth and therefore having an average momentum of zero. Since the lowest allowed harmonic oscillator energy, $E_0$, is $\dfrac{\hbar \omega}{2}$ and not 0, the atoms in a molecule must be moving even in the lowest vibrational energy state. This phenomenon is called the zero-point energy or the zero-point motion, and it stands in direct contrast to the classical picture of a vibrating molecule. Classically, the lowest energy available to an oscillator is zero, which means the momentum also is zero, and the oscillator is not moving.
Exercise 5.4.1
Compare the quantum mechanical harmonic oscillator to the classical harmonic oscillator at $v=1$ and $v=50$.
Answer
At v=1 the classical harmonic oscillator poorly predicts the results of quantum mechanical harmonic oscillator, and therefore reality. At v=1 the particle will be near the ground state and the classical model will predict the particle to spend most it's time on the outer edges when the KE goes to zero and PE is at a maximum, while the quantum model says the opposite and that the particle will be more likely to be found in the center. At v=50 the quantum model will begin to match the classical much more closely, with the particle most likely to be found at the edges. The quantum model looking more like the classical at higher quantum numbers can be referred to as the correspondence principle.
Since the average values of the displacement and momentum are all zero and do not facilitate comparisons among the various normal modes and energy levels, we need to find other quantities that can be used for this purpose. We can use the root mean square deviation (see also root-mean-square displacement) (also known as the standard deviation of the displacement) and the root-mean-square momentum as measures of the uncertainty in the oscillator's position and momentum.
For a molecular vibration, these quantities represent the standard deviation in the bond length and the standard deviation in the momentum of the atoms from the average values of zero, so they provide us with a measure of the relative displacement and the momentum associated with each normal mode in all its allowed energy levels. These are important quantities to determine because vibrational excitation changes the size and symmetry (or shape) of molecules. Such changes affect chemical reactivity, the absorption and emission of radiation, and the dissipation of energy in radiationless transitions.
The harmonic oscillator wavefunctions form an orthonormal set; this means that all functions in the set are normalized individually
$\int \limits _{-\infty}^{\infty} \psi ^*_v (x) \psi _v (x) dx = 1 \label {5.4.4}$
and are orthogonal to each other.
$\int \limits _{-\infty}^{\infty} \psi ^*_{v'} (x) \psi _v (x) dx = 0 \;\; \text {for} \;\; v' \ne v \label {5.4.5}$
The fact that a family of wavefunctions forms an orthonormal set is often helpful in simplifying complicated integrals. We will use these properties when we determine the harmonic oscillator selection rules for vibrational transitions in a molecule and calculate the absorption coefficients for the absorption of infrared radiation.
Finally, we can calculate the probability that a harmonic oscillator is in the classically forbidden region. What does this tantalizing statement mean? Classically, the maximum extension of an oscillator is obtained by equating the total energy of the oscillator to the potential energy, because at the maximum extension all the energy is in the form of potential energy. If all the energy weren't in the form of potential energy at this point, the oscillator would have kinetic energy and momentum and could continue to extend further away from its rest position. Interestingly, as we show below, the wavefunctions of the quantum mechanical oscillator extend beyond the classical limit, i.e. beyond where the particle can be according to classical mechanics.
The lowest allowed energy for the quantum mechanical oscillator is called the zero-point energy, $E_0 = \dfrac {\hbar \omega}{2}$. Using the classical picture described in the preceding paragraph, this total energy must equal the potential energy of the oscillator at its maximum extension. We define this classical limit of the amplitude of the oscillator displacement as $Q_0$. When we equate the zero-point energy for a particular normal mode to the potential energy of the oscillator in that normal mode, we obtain
$\dfrac {\hbar \omega}{2} = \dfrac {k Q^2_0}{2} \label {5.4.6}$
The zero-point energy is the lowest possible energy that a quantum mechanical physical system may have. Hence, it is the energy of its ground state.
Recall that $k$ is the effective force constant of the oscillator in a particular normal mode and that the frequency of the normal mode is given by Equation $\ref{5.4.1}$ which is
$\omega = \sqrt {\dfrac {k}{\mu}} \label {5.4.7}$
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Learning Objectives
• Understand how the quantum harmonic oscillator model can be used to interpret the infrared spectra of diatomic molecules
• Understand the origin of the transition moment integral and selection rules and how they are related
Infrared (IR) spectroscopy is one of the most common and widely used spectroscopic techniques employed mainly by inorganic and organic chemists due to its usefulness in determining structures of compounds and identifying them. Chemical compounds have different chemical properties due to the presence of different functional groups. IR spectroscopy is one of the most common and widely used spectroscopic techniques. Absorbing groups in the infrared region absorb within a certain wavelength region. The absorption peaks within this region are usually sharper when compared with absorption peaks from the ultraviolet and visible regions. In this way, IR spectroscopy can be very sensitive to determination of functional groups within a sample since different functional group absorbs different particular frequency of IR radiation. Also, each molecule has a characteristic spectrum often referred to as the fingerprint. A molecule can be identified by comparing its absorption peak to a data bank of spectra. IR spectroscopy is very useful in the identification and structure analysis of a variety of substances, including both organic and inorganic compounds. It can also be used for both qualitative and quantitative analysis of complex mixtures of similar compounds.
IR Spectroscopy
Transitions between vibrational energy levels can be induced about by absorption or emission of radiation. To understand this, knowledge of both the initial and final eigenstates is needed. The energy of the $v^{th}$ eigenstate of a harmonic oscillator can be written as
$E_v = \left(v+\dfrac{1}{2}\right) \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5.5.16}$
where $h$ is Planck's constant and $v$ is the vibrational quantum number and ranges from 0,1,2,3.... $\infty$. Equation \ref{5.5.16} is often rewritten as
$E_v = \left(v+\dfrac{1}{2}\right)h \nu_m \label{5.5.17}$
where $\nu_m$ is the vibrational frequency of the vibration. Equation \ref{5.5.17} is often written as
$E_v = \left(v + \dfrac{1}{2}\right)\hbar\omega \label{5.5.17b}$
where $\omega$ is the angular frequency (i.e., $2\pi \nu$).
Transitions in vibrational energy levels can be brought about by absorption of radiation, provided the energy of the radiation ($h\nu_{photon}$) exactly matches the difference in energy ($\Delta E_{vv^\prime}$) between the vibrational quantum state $v$ to quantum state $v^\prime$. This can be expressed as
\begin{align} h \nu_{photon} &= \Delta E_{vv^\prime} \nonumber \[4pt] &= E_{v^\prime} - E_{v} \nonumber \[4pt] &= \left(v^\prime + \dfrac{1}{2}\right)h \nu _m- \left(v +\dfrac{1}{2}\right)h \nu_m \[4pt] &= \left(v^\prime - v\right) h \nu_m \label{5.5.18} \end{align}
Let's consider only transitions between adjacent eigenstates (discussed in more details below) so
$v^\prime - v = \pm 1 \nonumber$
which is positive if an IR photon is absorbed and negative if it is emitted. For the absorption of a IR photon, Equation \ref{5.5.18} simplifies to
\begin{align} h \nu_{photon} &= h \nu_m \nonumber \[4pt] &= \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5.5.19} \end{align}
The frequency of radiation $\nu_{photon}$ that will bring about this change is identical to the classical vibrational frequency of the bond $\nu$.
The Wavenumber as a Unit of Energy
The cm-1 is the wavenumber scale and it can also be defined as 1/wavelength in cm. A wavenumber is often used due to its direct relationship with both frequency and energy. The frequency of the absorbed radiation causes the molecular vibrational frequency for the absorption process:
\begin{align} \bar{\nu}(cm^{-1}) &= \dfrac{1}{\lambda(\mu m)} \times 10^4 \left(\dfrac{\mu m}{cm}\right) \[4pt] &= \dfrac{\nu\,(Hz)}{c\,(cm/s)} \label{5.5.28} \end{align}
Equation \ref{5.5.18} can be modified so that the radiation can be expressed in wavenumbers
$\widetilde{\nu}_m = \dfrac{1}{2\pi c} \sqrt{\dfrac{k}{\mu}} \label{5.5.23}$
where $c$ is the velocity of light (cm/s) and $\widetilde{\nu}$ is the wavenumber of an absorption maximum (cm-1).
Example 5.5.1 : Hydrogen Halides
The force constants for typical diatomic molecules are in the range between 400 to 2000 $N \cdot m^{-1}$.
Molecule HF HCl HBr HI CO NO
Force constant, $k$ (N.m-1) 970 480 410 320 1860 1530
For the diatomic molecules listed above, calculate the following:
1. angular frequency ($\text{rad} \cdot s^{-1}$)
2. natural frequency (Hz)
3. period (s)
4. separation between energy levels
5. wavelength $\lambda$ of the electromagnetic radiation absorbed in the transition $v=0 \rightarrow v=1$.
Solution
For $\ce{HCl}$:
1. angular frequency $\omega = 5.45 \times 10^{14} \text{rad} \cdot s^{-1} \nonumber$
2. natural frequency: $\nu = \dfrac{\omega}{2\pi} = 8.68 \times 10^{13}\nonumber$
3. period (s): $T =\dfrac{1}{\nu} = 1.15 \times 10^{-14}\nonumber$
4. separation between energy levels: $\Delta E = E_{v=1} - E_{v=0} = \hbar \omega = 5.75 \times 10^{-20} \; J\nonumber$
5. wavelength $\lambda$ of the electromagnetic radiation absorbed in the transition $v=0 \rightarrow v=1$: $3.46 \times 10^{-6} \, m\nonumber$
The electromagnetic radiation released (and absorbed) for vibrations is primarily in the infrared (IR) part of the spectrum. Calculating the above properties for the other molecules remains as an exercise.
Note: We did not designate the specific isotopes in the hydrogen halides above because that information was not required to solve these questions.
IR deals with the interaction between a molecule and radiation from the electromagnetic region ranging (4000- 40 cm-1). The IR region of the electromagnetic spectrum ranges in wavelength from 2 -15 µm. Conventionally the IR region is subdivided into three regions (Table 5.5.1 ): near IR, mid IR and far IR. Most of the IR used originates from the mid IR region.
Table 5.5.1 : Regions of the IR Spectrum
Region Wavelength Wavenumbers ($\widetilde{\nu}$), cm-1 Frequencies (v), HZ
Near 0.78 -2.5 12800 - 4000 3.8 x 1014 - 1.2 x 1014
Middle 2.5 - 50 4000 - 200 3.8 x 1014 - 1.2 x 1014
Far 50 -100 200 -10 3.8 x 1014 - 1.2 x 1014
Most Used 2.5 -15 4000 -670 3.8 x 1014 - 1.2 x 1014
Example 5.5.2 : Hydrogen Chloride
The $\ce{H^{35}Cl}$ equilibrium bond length is 0.127 nm and the $v = 0$ to $v = 1$ transition is observed in the infrared at 2,886 cm-1. Compute the vibrational energy of $\ce{H^{35}Cl}$ in its lowest state.
1. Compute the classical limit for the stretching of the $\ce{HCl}$ bond from its equilibrium length in this state.
2. What percent of the equilibrium bond length is this extension?
Solution
The $\ce{H-Cl}$ bond length is $0.127\,nm = 1.27 \times 10^{-10}m$ and the IR transition is observed at $2886 \,cm^{-1}$. We first convert this to Hertz
$2886\, cm^{-1} (3 \times 10^{10} cm/s) = 8.646 \times 10^{13} s^{-1} = 8.646 \times 10^{13}\, Hz \nonumber$
The equations for the energy of the vth eigenstate of a harmonic oscillator is
$E_v = (v + 1/2)\hbar\omega \nonumber$
So the ground state ($v=0$) energy is
$E_0 = \dfrac{1}{2} \hbar \omega \nonumber$
with
\begin{align*} \omega &= 2\pi \nu \[4pt] &= 5.54 \times 10^{14}\, rad/s \end{align*} \nonumber
We can then extra the ground state vibrational energy (aka zero-point energy) from
\begin{align*} E_o &= \dfrac{1}{2}\hbar \omega \[4pt] &= \dfrac{1}{2} \hbar (5.54 \times 10^{14} rad/s) \[4pt] &= 2.916 \times 10^{-20}\, J\end{align*} \nonumber
The classical limit of the stretch is denoted as $Q_0$, this can be equated as potential energy in relation to the total $E_0$ found above as, at $E_0$, all of the energy would be potential energy in the form of the stretch. In comparison to the classic spring potential
$V = \dfrac{1}{2}k Q_0^2 \nonumber$
As described above, we can relate the two as
$\dfrac{1}{2}\hbar \omega = \dfrac{1}{2} k Q_0^2 \nonumber$
or,
$2.916 \times 10^{-20} J = \dfrac{1}{2}481 Q_0^2 \nonumber$
where $Q_0 = 1.10 \times 10^{-11} m = 0.0110\, nm$.
Lastly, this classical limit to length can be compared to the equilibrium bond length by a simple relation of
\begin{align*} \text{Percent bond length} &= \dfrac{Q_0}{x_{eq}} \times 100\% \[4pt] &= \dfrac{0.110\,nm}{0.127\,nm} \times 100\% \[4pt] &= 86.6\% \end{align*} \nonumber
Selection Rules for IR Transitions
Photons can be absorbed or emitted, and the harmonic oscillator can transition from one vibrational energy state to another. Which transitions between vibrational states are allowed? When discussing the Bohr hydrogen atom in Chapter 1, we identified multiple transitions between different states as being responsible for the line spectra emission of the hydrogen atom. No limitations were identified so that any transition between any starting state and ending state was possible and this resulted in a complex pattern in the spectra. As we will identify later on, when considering mulit-electron atoms, there are selection rules that limits which transitions will be observed (i.e., are allowed).
A similar situation exist for the harmonic oscillator and IR transitions and a set of selection rules must be satisfied to ensure a particular transition is allowed. Selection rules result from evaluating a transition moment integral that expresses the probability of a transition from the $v$ to the $v'$ eigenstates:
\begin{align} \mu_T &= \langle \psi_{v'} | \hat {\mu} (x) | \psi_{v} \rangle \[4pt] &= \int \limits _{-\infty}^{\infty} \psi_{v'}^* (x) \hat {\mu} (x) \psi _v (x) dx \label {5.5.29} \end{align}
When the transition moment integral is zero for two specific eigenstates (i.e., $\psi _v (x)$ and $\psi'_v (x)$), then the probability of observing that transition is zero. In this case, this transition is called a forbidden transition (i.e., it will not be obversed). Transitions with low probabilities are call "weakly allowed transitions" since they will have small amplitudes in spectra and transitions with high probabilities are "strongly allowed transitions" and will exhibit high amplitudes in spectra.
To evaluate Equation \ref{5.5.29}, we need to express the dipole moment operator, $\hat {\mu}$, in terms of the magnitude of the vibration $x$. The dipole moment operator is defined as
$\hat {\mu} = \sum _{electrons} e \vec{r} + \sum _{nuclei} q \vec{R} \label {5.5.30}$
where the two sums are over all the electrons and nuclei and involve the particle charge ($-e$ or $q$) multiplying the position vector ($\vec{r}$ or $\vec{R}$, respectively). We can obtain this dipole moment operator in terms of the magnitude of the displacement coordinate, $x$, in a simple way by using a Taylor series expansion for the dipole moment around the equlibrium position ($x=0$).
$\hat{\mu} (x) = \mu _{x=0} + \left. \dfrac {d \mu (x)}{dx} \right |_{x=0} x + \left. \dfrac {d^2 \mu (x)}{dx^2} \right |_{x=0} x^2 + \cdots \label {5.5.31}$
Retaining only the first two terms and substituting into Equation $\ref{5.5.29}$ produces
$\mu _T = \mu _{x=0} \int \limits _{-\infty}^{\infty} \psi _{v'} (x) \psi _v (x) dx + \left . \dfrac {d \mu (x)}{dx} \right |_{x=0} \int \limits _{-\infty}^{\infty} x\psi _{v'}^* (x) \psi _v (x) dx \label {5.5.32}$
where $\mu _{x=0} = 0$ is the dipole moment of the molecule when the nuclei are at their equilibrium positions, and
$\left.\dfrac {d\mu (x)}{dx} \right|_{x=0} \nonumber$
is the linear change in the dipole moment due to the displacement of the nuclei in the normal mode. The derivative is the linear change because it multiplies $x$ and not a higher power of $x$ in Equation $\ref{5.5.31}$. Both $\mu$ and
$\left.\dfrac {d\mu (x)}{dx}\right |_{x=0} \nonumber$
are moved outside of the integral because they are constants that no longer depend on $x$ because they are evaluated at $x = 0$.
The integral in the first term in Equation $\ref{5.5.32}$ is 0 because any two harmonic oscillator wavefunctions are orthogonal. The integral in the second term of Equation \ref{5.5.32} is zero except when $v' = v \pm 1$ as demonstrated in Exercise 5.5.1 . Also note that the second term is zero if
$\left .\dfrac {d\mu (x)}{dx}\right |_{x=0} = 0 \label {5.5.33}$
Hence we can identify two "rules" that need to be satisfied for an IR photon to be absorb (or emitted) by a vibrating molecule.
Rules for IR Absorption
For IR absorption to occur two conditions must be met:
1. There must be a change in the dipole moment of the molecule as a result of a molecular vibration (or rotation). The change (or oscillation) in the dipole moment allows interaction with the alternating electrical component of the IR radiation. Symmetric molecules (or bonds) do not absorb IR radiation since there is no dipole moment.
2. If the frequency of the radiation $\nu_{photon}$ matches the natural frequency of the vibration ($\nu_m$), the IR photon can be absorbed and the amplitude of the vibration increases.
Exercise 5.5.1
Use one of the Hermite polynomial recursion relations to verify that the second integral in Equation \ref{5.5.32} is 0 unless $v' = v \pm 1$.
If we are to observe absorption of infrared radiation due to a vibrational transition in a molecule, the transition moment cannot be zero. This condition requires that the dipole moment derivative Equation $\ref{5.5.33}$ cannot be zero and that the vibrational quantum number change by one unit. The normal coordinate motion must cause the dipole moment of the molecule to change in order for a molecule to absorb infrared radiation. If the normal coordinate oscillation does not cause the dipole moment to change then $\mu _T = 0$ and no infrared absorption is observed. So we can
$\underbrace{ \Delta v = \pm 1}_{\text{For allowed transitions}} \label {5.5.34}$
Consider oxygen and nitrogen molecules. Because they are symmetrical, their dipole moments are zero, $\mu = 0$. Since the vibrational motion (only bond stretching for a diatomic molecule) preserves this symmetry, the change in the dipole moment due to the vibrational motion also is zero, $\dfrac {d\mu (x)}{dx} = 0$. Consequently, oxygen and nitrogen do not absorb infrared radiation as a result of vibrational motion.
Exercise 5.5.2
Explain why the molar absorptivity or molar extinction coefficients ($\epsilon$) in Beer's Law for the IR absorption of some vibrations are greater than others.
Answer
Qualitatively, if the probability of transition is large, then the molar absorptivity is large. And similar if the transition were not allowed, then there will be no intensity and no observed peak in the spectrum. Transitions can be "partially allowed" as well, and these bands appear with a lower intensity than the full allowed transitions.
When looking at the Beer's Law
$A=\epsilon c l \nonumber$
where $A$ is absorbance. $\epsilon$ is the molar absorptivity, $c$ is the molar concentration and $l$ is the optical path length
We are only looking at the change in the molar absorptivity as the IR absorptions of vibrations vary. To do this we have to look at the transition moment integral
$\mu_{T}=\left\langle\psi_{v^{\prime}}|\not \hat{\mu}(x)| \psi_{v}\right\rangle$
When the transition moment integral is zero, there is no transition as it is not allowed under the selection rules. This means that the IR is not absorbing any vibrations and therefore the molar absorptivity is zero, which means that absorbance according to Beer's Law is zero. As the transition moment integral increases, the molar absorptivity also increases and the overall absorbance increases.
The case $v' = v + 1$ corresponds to going from one vibrational state to a higher energy one by absorbing a photon with energy $hν$. The case $v' = v − 1$ corresponds to a transition that emits a photon with energy $hν$. In the harmonic oscillator model infrared spectra are very simple; only the fundamental transitions, $\Delta = \pm 1$, are allowed. The associated transition energy is $\hbar \omega$, according to Equation \ref{5.5.19}. The transition energy is the change in energy of the oscillator as it moves from one vibrational state to another, and it equals the photon energy.
\begin{align} \Delta E &= E_{final} - E_{initial} \[4pt] &= hv_{photon} \[4pt] &= \hbar \omega _{oscillator} \label {5.5.35} \end{align}
For perfect harmonic oscillators, the only possible allowed transitions are $\Delta = \pm 1$ with all other transitions forbidden (Figure 5.5.1 ). This conclusion predicts that the vibrational absorption spectrum of a diatomic molecule consists of a single line since the energy levels are equally spaced in the harmonic oscillator model (Figure 5.5.1 ). If the vibration were anharmonic, then the levels would not be equally spaced and then transitions from $v = 0$ to $v=1$ and from $v = 1$ to $v=2$, etc. would occur at different frequencies.
Only the fundamental transitions, $\Delta = \pm 1$, are observed in infrared spectra within harmonic oscillator model.
The actual IR spectrum is more complex, especially at high resolution. There is a fine structure due to the rotational states of the molecule. These states will be discussed in the next chapter. The spectrum is enriched further by the appearance of lines due to transitions corresponding to $\Delta = \pm n$ where $n > 1$. These transitions are called overtone transitions and their appearance in spectra despite being forbidden in the harmonic oscillator model is due to the anharmonicity of molecular vibrations. Anharmonicity means the potential energy function is not strictly the harmonic potential. The first overtone, $Δv = 2$, generally appears at a frequency slightly less than twice that of the fundamental, i.e. the frequency due to the $Δv = 1$ transition.
Exercise 5.5.3 : Hydrogen Chloride
Compute the approximate transition frequencies in wavenumber units for the first and second overtone transitions in $\ce{HCl}$ given that the fundamental is at 2,886 cm-1.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra.txt
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Learning Objectives
• The Equation for a Harmonic-Oscillator Model of a Diatomic Molecule Contains the Reduced Mass of the Molecule
For a diatomic molecule, there is only one vibrational mode, so there will be only a single set of vibrational wavefunctions with associated energies for this system. For polytatomic molecules, there will be a set of wavefunctions with associated energy associated with each vibrational mode.
The Hamiltonian operator, the general quantum mechanical operator for energy, includes both a kinetic energy term, $\hat {T}$, and a potential energy term, $\hat {V}$.
$\hat {H} = \hat {T} + \hat {V} \label {5.6.2}$
For the free particle and the particle in a box, the potential energy term used in the Hamiltonian was zero. As shown in Equation $\ref{5.6.2}$, the classical expression for the energy of a harmonic oscillator includes both a kinetic energy term and the harmonic potential energy term. Transforming this equation into the corresponding Hamiltonian operator gives,
$\hat {H} (q) = \dfrac {1}{2 \mu} \hat {P}^2_q + \dfrac {k}{2} \hat {q}^2 \label {5.6.3}$
where $\hat {q}$ is the operator for the length of the normal coordinate, and $\hat {P}_q$ is the momentum operator associated with the normal coordinate. $\mu$ is an effective (reduced) mass, and $k$ is an effective force constant, and these quantities will be different for each of the normal modes (vibrations).
Substituting the definitions for the operators yields
$\hat {H} (q) = -\dfrac {\hbar ^2}{2\mu} \dfrac {d^2}{dq^2} + \dfrac {k}{2} q^2 \label {5.6.4}$
since the operator for position or displacement is just the position or displacement. The time-independent Schrödinger Equation then becomes
$- \dfrac {\hbar ^2}{2\mu} \dfrac {d^2 \psi _v (q)}{dq^2} + \dfrac {k}{2} q^2 \psi _v (q) = E_v \psi _v (q) \label {15.6.5}$
or upon rearranging
$\dfrac {d^2 \psi _v (q)}{dq^2} + \dfrac {2 \mu}{\hbar ^2} \left ( E_v - \dfrac {k}{2} q^2 \right ) \psi _v (q) = 0 \label {15.6.6}$
This differential equation is not straightforward to solve. Rather than fully develop the details of the solution, we will outline the method used because it represents a common strategy for solving differential equations. The steps taken to solve Equation $\ref{15.6.6}$ are to simplify the equation by collecting constants in the parameter $\beta$
$\beta ^2 = \dfrac {\hbar}{\sqrt { \mu k}} \label {15.6.7}$
and then changing the variable from $q$ to $x$ where
$x = \dfrac {q}{\beta} \label{scale}$
so that
$\dfrac {d^2}{dq^2} = \dfrac {1}{\beta^2} \dfrac {d^2}{dx^2} \label {15.6.8}$
After substituting Equations $\ref{15.6.7}$ and $\ref{15.6.8}$ into Equation $\ref{15.6.6}$, the differential equation for the harmonic oscillator becomes
$\dfrac {d^2 \psi _v (x)}{dx^2} + \left ( \dfrac {2 \mu \beta ^2 E_v}{\hbar ^2} - x^2 \right ) \psi _v (x) = 0 \label {15.6.9}$
Exercise 5.6.1
Make the substitutions given in Equations $\ref{15.6.7}$ and $\ref{15.6.8}$ into Equation $\ref{15.6.6}$ to get Equation $\ref{15.6.9}$.
Solving for the Quantum Wavefunctions
A common strategy for solving differential equations, which is employed here, is to find a solution that is valid for large values of the variable and then develop the complete solution as a product of this asymptotic solution and a power series. Since the potential energy approaches infinity as $x$ and the coordinate $q$ approach infinity, the wavefunctions must approach zero (this is, the wavefunctions must converge to zero):
$\lim_{x \rightarrow \infty} \psi_v(x) = 0 \label {15.6.10a}$
The function that has this property and satisfies the differential equation for large values of $x$ is the exponential function, i.e.,
$\lim_{x \rightarrow \infty} \phi_v(x) \exp \left ( \dfrac {-x^2}{2} \right ) = 0 \label {15.6.10}$
where the wavefunctions are
$\psi_v(x) \propto \phi_v(x) \exp \left ( \dfrac {-x^2}{2} \right ) \nonumber$
The general expression for a power series of $\phi_v(x)$ is
\begin{align*}\phi_v(x) &= \sum _{n=0}^\infty c_n(v) x^n \[4pt] &=c_n(v) x + c_n(v) x^2 + c_n(v) x^3 + \cdots\label {15.6.11} \end{align*}
which can be truncated after the first term, after the second term, after the third term, etc. to produce a set of polynomials. There is one polynomial for each value of $v$ where $v$ can be equal to any integer value including zero.
$\sum _{n=0}^v c_n x^n \label {15.6.12}$
Each of the truncations of the power series in Equation $\ref{15.6.12}$ can be multiplied by the exponential function in Equation $\ref{15.6.10}$ to create a family of valid solutions to the differential equation in Equation \ref{15.6.9}.
$\psi _v (x) = \sum _{n=0}^v c_n x^n \exp \left ( \dfrac {-x^2}{2} \right ) \label {15.6.13}$
Exercise 5.6.2
Write the first four polynomials, $v=0$ to $v=1$, $v=12$, $v=13$, $v=14$ for Equation $\ref{15.6.12}$ and use suitable software to prepare plots of these polynomials. Identify the curves in the plots.
Exercise 5.6.3
Confirm that the wavefunction given by Equation \ref{15.6.13} is a solution to the harmonic oscillator Schrödinger Equation in Equation \ref{15.6.9} for $v=0$ and $v=1$.
Hermite Polynomials
While polynomials in general approach $∞$ (or $-∞$) as $x$ approaches $∞$, the decreasing exponential term overpowers the polynomial term so that the overall wavefunction exhibits the desired approach to zero at large values of $x$ or $-x$. The exact forms of polynomials that solve Equation $\ref{15.6.9}$ are the Hermite polynomials, which are standard mathematical functions known from the work of Charles Hermite. The first eight Hermite polynomials, $H_v(x)$, are given below.
• $H_0 = 1$
• $H_1 = 2x$
• $H_2 = -2 + 4x^2$
• $H_3 = -12x + 8x^3$
• $H_4 = 12 - 48x^2 +16x^4$
• $H_5 = 120x - 160x^3 + 32x^5$
• $H_6 = -120 + 720x^2 - 480 x^4 + 64x^6$
• $H_7 = -1680x + 3360 x^3 - 1344 x^5 + 128 x^7$
The first six Hermite polynomials are plotted in Figure 5.6.1 . Hermite polynomials will be discussed in more detail in the following Section.
Exercise 5.6.4
Determine the units of $β$ and the units of $x$ in the Hermite polynomials.
Because of the association of the wavefunction with a probability density, it is necessary for the wavefunction to include a normalization constant, $N_v$.
$N_v = \dfrac {1}{(2^v v! \sqrt {\pi} )^{1/2}} \label {5.6.15}$
The final form of the harmonic oscillator wavefunctions is thus
$\psi _v (x) = N_v H_v (x) e^{-x^2/2} \label {5.6.16}$
Alternative and More Common Formulation of Harmonic Oscillator Wavefunctions
The harmonic oscillator wavefunctions are often written in terms of $Q$, the unscaled displacement coordinate (Equation $\ref{scale}$) and a different constant $\alpha$:
$\alpha =1/\sqrt{\beta} = \sqrt{\dfrac{k \mu}{\hbar ^2}} \nonumber$
so Equation $\ref{5.6.16}$ becomes
$\psi _v (x) = N_v'' H_v (\sqrt{\alpha} Q) e^{-\alpha Q^2/ 2} \nonumber$
with a slightly different normalization constant
$N_v'' = \sqrt {\dfrac {1}{2^v v!}} \left(\dfrac{\alpha}{\pi}\right)^{1/4} \nonumber$
Exercise 5.6.5
Compute the normalization factor for $\psi_v(x)$ where $v = 0$ and $v = 4$. What is the purpose of $N_v$?
The energy eigenvalues for a quantum mechanical oscillator also are obtained by solving the Schrödinger equation. The energies are restricted to discrete values
$E_v = \left ( v + \dfrac {1}{2} \right ) \hbar \omega \label {5.6.17}$
with $v = 0, 1, 2, 3, \cdots$.
The energies depend both on the quantum number, $v$, and the oscillator frequency
$\omega = \sqrt {\dfrac {k}{\mu}} \nonumber$
which in turn depends on the spring constant $k$ and the reduced mass of the vibration $\mu$.
Exercise 5.6.6
Determine the energy for the first ten harmonic oscillator energy levels in terms of $\hbar \omega$. Sketch an energy level diagram of these energies.
1. What insights do you gain from Equation $\ref{5.6.17}$, your calculations, and your diagram?
2. Is it possible to have a molecule that is not vibrating?
3. In terms of $\hbar \omega$, what is the energy of the photon required to cause a transition from one vibrational state to the next higher one?
4. If a transition from energy level $v = 9$ to $v = 10$ were observed in a spectrum, where would that spectral line appear relative to the one for the transition from level $v = 0$ to $v = 1$?
5. If a vibrational transition is observed at 3000 cm-1 in an infrared spectrum, what is the value of $\hbar \omega$ for the normal mode?
6. Identify all the possible meanings of $ΔE = hν$ and the definition of the frequency, $ν$, in each case.
The normalized wavefunctions for the first four states of the harmonic oscillator are shown in Figure 5.6.2 , and the corresponding probability densities are shown in Figure 5.6.3 . You should remember the mathematical and graphical forms of the first few harmonic oscillator wavefunctions, and the correlation of $v$ with $E_v$. The number of nodes in the wavefunction will help you to remember these characteristics. Also note that the functions fall off exponentially and that the symmetry alternates. For $v$ equal to an even number, $\psi_v$ is gerade; for $v$ equal to an odd number, $\psi_v$ is ungerade.
Exercise 5.6.7
Write a few sentences describing and comparing the plots in Figure 5.6.2 . How many nodes are there as a function of $v$? Do the wavefunctions converge at extreme displacement? Where is the most likely displacements for the oscillator to be found?
Exercise 5.6.8
Explain how Figure 5.6.3 is related to Figure 5.6.2 . Explain the physical significance of the plots in Figure 5.6.3 in terms of the magnitude of the normal coordinate $Q$. Couch your discussion in terms of the $\ce{HCl}$ molecule. How would you describe the location of the atoms in each of the states? How does the oscillator position correspond to the energy of a particular level?
Answer
Figure 5.6.2 is simply the wavefunction in Figure 5.6.1 squared. The normal coordinate is the linear combination of the atomic cartesian coordinates. As Q is often in relation to the energy (kinetic and potential), they would be displaced by a certain amount dependent on Q (energy) along with an increase in nodes. This displacement is apparent when comparing the ascending energy levels of each of the wavefunctions. In the n=0 (first) energy state, it is most probable to be found between -2, 2. (in a range of -4, 4) In the second energy state, it is likely to be between -2.5, 2.5 (range -5, 5), third level: (-3,3) (range -6,6), fourth level (-4,4) (range -6,6).
Exercise 5.6.9
Plot the probability density for energy level 10 of the harmonic oscillator. How many nodes are present? Plot the probability density for energy level 20. Compare the plot for level 20 with that of level 10 and level 1. Compare these quantum mechanical probability distributions to those expected for a classical oscillator. What conclusion can you draw about the probability of the location of the oscillator and the length of a chemical bond in a vibrating molecule? Extend your analysis to include a very high level, like level 50.
In completing Exercise 5.6.9 , you should have noticed that as the quantum number increases and becomes very large, the probability distribution approaches that of a classical oscillator. This observation is very general. It was first noticed by Bohr, and is called the Bohr Correspondence Principle. This principle states that classical behavior is approached in the limit of large values for a quantum number. A classical oscillator is most likely to be found in the region of space where its velocity is the smallest. This situation is similar to walking through one room and running through another. In which room do you spend more time? Where is it more likely that you will be found?
Examination of the quantum mechanical wavefunction for the lowest-energy state reveals that the wavefunction $\psi_0(x)$ extends beyond the classical limit (i.e., outside of the harmonic oscillator well, albeit slightly). Higher energy states have higher total energies, so the classical limits to the amplitude of the displacement will be larger for these states.
Tunneling in the Quantum Harmonic Oscillator
The observation that the wavefunctions are not zero at the classical limit means that the quantum mechanical oscillator has a finite probability of having a displacement that is larger than what is classically possible. The oscillator can be in a region of space where the potential energy is greater than the total energy. Classically, when the potential energy equals the total energy, the kinetic energy and the velocity are zero, and the oscillator cannot pass this point. A quantum mechanical oscillator, however, has a finite probability of passing this point. For a molecular vibration, this property means that the amplitude of the vibration is larger than what it would be in a classical picture. In some situations, a larger amplitude vibration could enhance the chemical reactivity of a molecule.
Exercise 5.6.10
Plot the probability density for $v = 0$ and $v = 1$ states. Mark the classical limits on each of the plots, since the limits are different because the total energy is different for $v = 0$ and $v = 1$. Shade in the regions of the probability densities that extend beyond the classical limit.
We should be able to calculate the probability that the quantum mechanical harmonic oscillator is in the classically forbidden region for the lowest energy state of the harmonic oscillator, the state with $v = 0$. The classically forbidden region is shown by the shading of the regions beyond $Q_0$ in the graph you constructed for Exercise 5.6.3 . The area of this shaded region gives the probability that the bond oscillation will extend into the forbidden region (Figure 5.6.3 ). To calculate this probability, we use
$P [ \text {forbidden}] = 1 - P [ \text {allowed}] \label {5.4.9}$
because the integral from 0 to $Q_0$ for the allowed region can be found in integral tables and the integral from $Q_0$ to $\infty$ cannot. The form of the integral, $P[ \text{allowed}]$, to evaluate is
$P[ \text {allowed}] = 2 \int \limits _0^{Q_0} \psi _0^* (Q) \psi _0 (Q) dQ \label {5.4.10}$
The factor 2 appears in Equation $\ref{5.4.10}$ from the chancing the limits of integration from $-Q_0$ to $+Q_0$ into $0$ to $+Q_0$; we can do this since the integrand is an even function, i. e., $f(-x)=f(x)$. To evaluate the integral in Equation $\ref{5.4.10}$, use the wavefunction and do the integration in terms of $x$. Recall that for $v = 0$, $Q = Q_0$ corresponds to $x = 1$. Including the normalization constant, Equation $\ref{5.4.10}$ produces
$P[ \text {allowed}] = \dfrac {2}{\sqrt {\pi}} \int \limits _0^1 \exp (-x^2) dx \label {5.4.11}$
The integral in Equation $\ref{5.4.11}$ is called an error function (ERF) and can only be evaluated numerically. Values can be found in books of mathematical tables. When the limit of integration is 1, ERF(l) = 0.843 and P[forbidden] = 0.157. This result means that the quantum mechanical oscillator can be found in the forbidden region 16% of the time. This effect is substantial and leads to the phenomenon called quantum mechanical tunneling.
Exercise 5.6.11
Numerically verify that $P[ \text{allowed}]$ in Equation $\ref{5.4.11}$ equals 0.843. To obtain a value for the integral do not use symbolic integration or symbolic equals.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.06%3A_The_Harmonic-Oscillator_Wavefunctions_involve_Hermite_Polynomials.txt
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Learning Objectives
• Understand key properties of the Hermite polynomials including orthogonality and symmetry.
• Be proficient at using symmetries of integrands to quickly solve integrals.
Hermite polynomials were defined by Laplace (1810) though in scarcely recognizable form, and studied in detail by Chebyshev (1859). Chebyshev's work was overlooked and they were named later after Charles Hermite who wrote on the polynomials in 1864 describing them as new. They were consequently not new although in later 1865 papers Hermite was the first to define the multidimensional polynomials. The first six Hermite polynomial are plotted in Figure 5.7.1 .
Generating Formula
Any Hermite polynomial $H_n(x)$ can be generated from a previous one $H_{n-1}(x)$ via the following using the recurrence relation
$H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x). \label{5.7.2}$
Hermite Polynomials are Symmetric
Let $f(x)$ be a real-valued function of a real variable.
• Then $f$ is even if the following equation holds for all x and -x in the domain of f $f(x) = f(-x) \nonumber$
• Then $f$ is odd if the following equation holds for all x and -x in the domain of f $-f(x) = f(-x) \nonumber$
Even and odd are terms used to describe particularly well-behaved functions. An even function is symmetric about the y-axis (Figure 5.7.2 ; left). That is, if we reflect the graph of the function in the $y$-axis, then it does not change. Formally, we say that $f$ is even if, for all $x$ and $−x$ in the domain of $f$, we have
$f(-x)=f(x) \nonumber$
Two examples of even functions are $f(x)=x^2$ and $f(x)=\cos x$.
An odd function has rotational symmetry of order two about the origin (Figure 5.7.2 ; middle). That is, if we rotate the graph of the function 180° about the origin, then it does not change. Formally, we say that ff is odd if, for all $x$ and $−x$ in the domain of $f$, we have
$f(-x)=-f(x) \nonumber$
Examples of odd functions are $f(x)=x^3$ and $f(x)=\sin x$.
Naturally, not all functions can be classified as even or odd. For example $f=x^3+1$ shown in the right side of Figure 5.7.2 , is neither.
You can also think of these properties as symmetry conditions at the origin. More symmetries in 3D space are discussed in Group Theory.
Without proof, we can identify several key features involving multiplication properties of even and odd functions:
• The product of two even functions is an even function.
• The product of two odd functions is an even function.
• The product of an even function and an odd function is an odd function.
This can be shown graphically as a product table like that in Table 5.7.1 .
Table 5.7.1 : Product table of 1D Functions
Product table Odd Function (anti-symmetric) Even Function (symmetric) No symmetry (neither)
Odd Function (anti-symmetric) Even Function (symmetric) Odd Function (anti-symmetric) who knows
Even Function (symmetric) Odd Function (anti-symmetric) Even Function (symmetric) who knows
No symmetry (neither) who knows who knows who knows
Notice that the Hermite polynomials in Figure 5.7.1 oscillate from even to odd. We can take advantage of that aspect in our calculation of Harmonic Oscillator wavefunctions. Hermite Polynomial is an even or odd function depends on its degree $n$. Based on
$H_n(-x) = (-1)^n H_n(x) \label{5.7.3}$
• $H_n(x)$ is an even function, when $n$ is even.
• $H_n(x)$ is an odd function, when $n$ is odd.
Integration over Symmetric Functions
You often consider integrals of the form
$I=\int_{-a}^a f(x)\,\mathrm{d}x \nonumber$
If $f$ is odd or even, then sometimes you can make solving this integral easier. For example, we can rewrite that integral in the following way:
\begin{align} I=\int_{-a}^a f(x)\,\mathrm{d}x &= \int_{-a}^0 f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \ &= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \label{intsym} \end{align}
For an even function, we have $f(-x)=f(x)$ and Equation \ref{intsym} can be simplified
\begin{align*} I &= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \[4pt] &=2\int_0^a f(x)\,\mathrm{d}x \end{align*} \nonumber
For an odd function, we have $f(-x)=-f(x)$ and Equation \ref{intsym} can be simplified
\begin{align*} I &= -\int_0^a f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \[4pt] &=0 \end{align*} \nonumber
That’s what it means to simplify the integration: the integral of an odd or even function over the interval $[−L,L]$ can be put into a nicer form (and sometimes we can see that it vanishes without ever computing an integral).
Example 5.7.1
Technically, evaluating the orthogonality of Hermite polynomials requires integrating over the $\exp(-x^2)$ weight function (Equations $\ref{1}$ and $\ref{2}$).
Solution
For the Hermite polynomials $H_n(x)$, the relevant inner product (using Dirac Notation)
$\langle f,g \rangle=\int_{-\infty}^\infty f(x)g(x)\color{red}{\exp(-x^2)}\,\mathrm dx \nonumber$
While the $H_2(x)H_3(x)$ product is indeed an odd function (Table 5.7.1 ), while $exp(−x^2)$ is even. Their product is odd, and thus $\langle f,g \rangle$ certainly ought to be zero.
Symmetry is an important aspect of quantum mechanics and mathematics, especially in calculating integrals. Using this symmetry, integrals can be identified to be equal to zero without explicitly solving them. For example, the integral of an odd integrand over all possible values will always be zero irrespective of the exact nature of the function:
$\int_{-\infty}^{\infty} f(x) \, dx= 0 \nonumber$
This simplifies calculations greatly as demonstrated in the following chapters.
Hermite Polynomials are Orthogonal
Hermite polynomials $H_n(x)$ are nth-degree polynomials for n = 0, 1, 2, 3 and form an orthogonal set of functions for the weight function $e^{-x^2/2}$. The exact relation is:
$\int_{-\infty}^{\infty} H_m(x)H_n(x) e^{-x^2/2} dx = 0 \label{1}$
if $m \neq n$ and
$\int_{-\infty}^{\infty} H_m(x)H_n(x) e^{-x^2/2} dx = 2^n n! \sqrt{\pi} \label{2}$
if $m = n$.
This will not be proved, but can the demonstrated using any of the Hermite polynomials listed in the previous section. The orthogonality property becomes important when solving the Harmonic oscillator problems. Note that the integral of Equation \ref{2} is important for normalizing the quantum harmonic oscillator wavefunctions discussed in last Section.
Example 5.7.2 : Hermite Polynomials are Orthogonal
Demonstrate that $H_2(x)$ and $H_3(x)$ are orthogonal.
Solution
We need to confirm
$\int_{-\infty}^{\infty}H_2(x)H_3(x) dx=0 \nonumber$
or when substituted
$\int_{-\infty}^{\infty} (4x^2-2)(8x^3-12x) dx=0 \nonumber$
because it says I need to show it's orthogonal on $[ -\infty, \infty ]$ or we can just evaluate it on a finite interval $[−L,L]$, where $L$ is a constant.
\begin{align*} \int_{-L}^{L} (4x^2-2)(8x^3-12x) dx &=\left. 8 \left(\dfrac{2 x^6}{3}-2 x^4+\dfrac{3 x^2}{2}\right)\right|_{-L}^{L} \[4pt] &= 8 \left(\dfrac{2 L^6}{3}-2 L^4+\frac{3 L^2}{2}\right)-8 \left(\dfrac{2 (-L)^6}{3}-2 (-L)^4+\dfrac{3 (-L)^2}{2}\right) \[4pt] &=0. \end{align*} \nonumber
math.stackexchange.com/questi...are-orthogonal
Concluding
Hermite polynomials are a component in the harmonic oscillator wavefunction that dictates the symmetry of the wavefunctions. If your integration interval is symmetric around 0, then the integral over any integrable odd function is zero, no exception. Therefore as soon as you've found that your integrand is odd and your integration interval is symmetric, you're done. Also, for general functions, if you can easily split them into even and odd parts, you only have to consider the integral over the even part for symmetric integration intervals.
Another important property is that the product of two even or of two odd functions is even, and the product of an even and an odd function is odd. For example, if ff is even, $x↦f(x)\sin(x)$ is odd, and therefore the integral over it is zero (provided it is well defined).
Contributors and Attributions
• StackExchange: alexwlchan
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.07%3A_Hermite_Polynomials_are_either_Even_or_Odd_Functions.txt
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Learning Objectives
• Compare the classical and quantum rigid rotor in three dimensions
• Demonstrate how to use the Separation of Variable technique to solve the 3D rigid rotor Schrödinger Equation
• Identify and interpret the two quantum numbers for a 3D quantum rigid rotor including the range of allowed values
• Describe the wavefunctions of the 3D quantum rigid rotor in terms of nodes, average displacements and most probable displacements
• Describe the energies of the 3D quantum rigid rotor in terms of values and degeneracies
Rigid rotor means when the distance between particles do not change as they rotate. A rigid rotor only approximates a rotating diatomic molecular if vibration is ignored.
The Classical Rigid Rotor in 3D
The rigid rotor is a mechanical model that is used to explain rotating systems. The linear rigid rotor model consists of two point masses located at fixed distances from their center of mass. The fixed distance between the two masses and the values of the masses are the only characteristics of the rigid model. However, for many actual diatomics this model is too restrictive since distances are usually not completely fixed and corrections on the rigid model can be made to compensate for small variations in the distance. Even in such a case the rigid rotor model is a useful model system to master.
For a rigid rotor, the total energy is the sum of kinetic ($T$) and potential ($V$) energies
$E_{tot} = T + V \label{5.8.2}$
The potential energy, $V$, is set to $0$ because the distance between particles does not change within the rigid rotor approximation. However, In reality, $V \neq 0$ because even though the average distance between particles does not change, the particles still vibrate. The rigid rotor approximation greatly simplifys our discussion.
Since $V=0$ then $E_{tot} = T$ and we can also say that:
$T = \dfrac{1}{2}\sum{m_{i}v_{i}^2} \label{5.8.3}$
However, we have to determine $v_i$ in terms of rotation since we are dealing with rotation motion. Since,
$\omega = \dfrac{v}{r} \label{5.8.4}$
where $\omega$ is the angular velocity, we can say that:
$v_{i} = \omega{X}r_{i} \label{5.8.5}$
Thus we can rewrite Equation $\ref{5.8.3}$ as:
$T = \dfrac{1}{2}\sum{m_{i}v_{i}\left(\omega{X}r_{i}\right)} \label{5.8.6}$
Since $\omega$ is a scalar constant, we can rewrite Equation \ref{5.8.6} as:
$T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2} \label{5.8.7}$
where $l_i$ is the angular momentum of the ith particle, and $L$ is the angular momentum of the entire system. Also, we know from physics that,
$L = I\omega \label{5.8.9}$
where $I$ is the moment of inertia of the rigid body relative to the axis of rotation. We can rewrite Equation $\ref{5.8.3}$ as
$T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2 \label{5.8.10}$
Equation \ref{5.8.10} shows that the energy of the rigid rotor scales with increasing angular frequency (i.e., the faster is rotates) and with increasing moment of inertia (i.e, the inertial resistance to rotation). Also, as expected, the classical rotational energy is not quantized (i.e., all possible rotational frequencies are possible).
The Quantum Rigid Rotor in 3D
It is convenient to discuss rotation with in the spherical coordinate system rather than the Cartesian system (Figure 5.8.1 ).
To solve the Schrödinger equation for the rigid rotor, we will separate the variables and form single-variable equations that can be solved independently. Only two variables $\theta$ and $\varphi$ are required in the rigid rotor model because the bond length, $r$, is taken to be the constant $r_0$. We first write the rigid rotor wavefunctions as the product of a theta-function depending only on $\theta$ and a phi-function depending only on $\varphi$
$| \psi (\theta , \varphi ) \rangle = | \Theta (\theta ) \Phi (\varphi) \rangle \label {5.8.11}$
We then substitute the product wavefunction and the Hamiltonian written in spherical coordinates into the Schrödinger Equation $\ref{5.8.12}$
$\hat {H} | \Theta (\theta ) \Phi (\varphi) \rangle = E | \Theta (\theta ) \Phi (\varphi) \rangle \label {5.8.12}$
to obtain
$-\dfrac {\hbar ^2}{2\mu r^2_0} \left [ \dfrac {\partial}{\partial r_0} r^2_0 \dfrac {\partial}{\partial r_0} + \dfrac {1}{\sin \theta} \dfrac {\partial}{\partial \theta } \sin \theta \dfrac {\partial}{\partial \theta } + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] | \Theta (\theta ) \Phi (\varphi) \rangle = E | \Theta (\theta) \Phi (\varphi) \rangle \label {5.8.13}$
Since $r = r_0$ is constant for the rigid rotor and does not appear as a variable in the functions, the partial derivatives with respect to $r$ are zero; i.e. the functions do not change with respect to $r$. We also can substitute the symbol $I$ for the moment of inertia, $\mu r^2_0$ in the denominator of the left hand side of Equation $\ref{5.8.13}$, to give
$-\dfrac {\hbar ^2}{2I} \left [ \dfrac {1}{\sin \theta} \dfrac {\partial}{\partial \theta } \sin \theta \dfrac {\partial}{\partial \theta } + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2}\right ] | \Theta (\theta ) \Phi (\varphi) \rangle = E | \Theta (\theta) \Phi (\varphi) \rangle \label {5.8.14}$
To begin the process of the Separating of Variables technique, multiply each side of Equation $\ref{5.8.14}$ by $\dfrac {2I}{\hbar ^2}$ and $\dfrac {-\sin ^2 \theta}{\Theta (\theta) \Phi (\varphi)}$ to give
$\dfrac {1}{\Theta (\theta) \psi (\varphi)} \left [ \sin \theta \dfrac {\partial}{\partial \theta } \sin \theta \dfrac {\partial}{\partial \theta } + \dfrac {\partial ^2}{\partial \varphi ^2}\right ] \Theta (\theta ) \Phi (\varphi) = \dfrac {-2IE \sin ^2 \theta}{\hbar ^2} \label {5.8.15}$
Simplify the appearance of the right-hand side of Equation $\ref{5.8.15}$ by defining a parameter $\lambda$:
$\lambda = \dfrac {2IE}{\hbar ^2}. \label {5.8.16}$
Note that this $\lambda$ has no connection to a wavelength; it is merely being used as an algebraic symbol for the combination of constants shown in Equation $\ref{5.8.16}$.
Inserting $\lambda$, evaluating partial derivatives, and rearranging Equation $\ref{5.8.15}$ produces
$\dfrac {1}{\Theta (\theta)} \left [ \sin \theta \dfrac {\partial}{\partial \theta } \left (\sin \theta \dfrac {\partial}{\partial \theta } \right ) \Theta (\theta) + \left ( \lambda \sin ^2 \theta \right ) \Theta (\theta) \right ] = - \dfrac {1}{\Phi (\varphi)} \dfrac {\partial ^2}{\partial \varphi ^2} \Phi (\varphi) \label {5.8.17}$
Exercise 5.8.1
Carry out the steps leading from Equation $\ref{5.8.15}$ to Equation $\ref{5.8.17}$. Keep in mind that, if $y$ is not a function of $x$,
$\dfrac {dy}{dx} = y \dfrac {d}{dx} \nonumber$
Equation $\ref{5.8.17}$ says that the function on the left, depending only on the variable $\theta$, always equals the function on the right, depending only on the variable $\varphi$, for all values of $\theta$ and $\varphi$. The only way two different functions of independent variables can be equal for all values of the variables is if both functions are equal to a constant (review separation of variables). We call this constant $m_J^2$ because soon we will need the square root of it. The two differential equations to solve are the $\theta$-equation
$\sin \theta \dfrac {d}{d \theta} \left ( \sin \theta \dfrac {d}{d \theta} \right ) \Theta (\theta ) + \left ( \lambda \sin ^2 \theta - m_J^2 \right ) \Theta (\theta ) = 0 \label {5.8.18}$
and the $\varphi$-equation
$\dfrac {d^2}{d \varphi ^2} \Phi (\varphi ) + m_J^2 \Phi (\varphi) = 0 \label {5.8.21}$
The partial derivatives have been replaced by total derivatives because only a single variable is involved in each equation.
Often $m_J$ is referred to as just $m$ for convenience.
Exercise 5.8.2
Show how Equations $\ref{5.8.18}$ and $\ref{5.8.21}$ are obtained from Equation $\ref{5.8.17}$.
Solving the $\varphi$ Equation
The $\varphi$-equation is similar to the Schrödinger Equation for the free particle. Since we already solved this previously, we immediately write the solutions:
$\Phi _m (\varphi) = N e^{\pm im_J \varphi} \label {5.8.22}$
where we introduce the number $m$ to track how many wavelengths of the wavefunction occur around one rotation (similar to the wavelength description of the Bohr atom).
Exercise 5.8.3
Substitute Equation $\ref{5.8.22}$ into Equation $\ref{5.8.21}$ to show that it is a solution to that differential equation.
Answer
Substitute
$\Phi_m(\varphi)= \mathrm{N} e^{\pm \mathrm{i} m_{J} \varphi} \nonumber$
into
$\frac{d^{2}}{d \varphi^{2}} \Phi(\varphi)+m_{J}^{2} \Phi(\varphi)=0 \nonumber$
\begin{aligned} \frac{d^{2}}{d \varphi^{2}} \Phi_{\mathrm{m}}(\varphi)+m_{J}^{2} \Phi_{\mathrm{m}}(\varphi)=& \frac{d}{d \varphi}\left(\mathrm{N}\left(\pm \mathrm{i} m_{J}\right) e^{\pm \mathrm{i} m_{J} \varphi}\right)+m_{J}^{2} \Phi_{\mathrm{m}}(\varphi) \ &\left.=\mathrm{N}\left(\pm \mathrm{i} m_{J}\right)^{2} e^{\pm i m_{J} \varphi}\right)+m_{J}^{2}\left(\mathrm{N} e^{\pm \mathrm{i} m_{J} \varphi}\right) \ &=-\mathrm{N} m_{J}^{2} e^{\pm i m_{J} \varphi}+\mathrm{N} m_{J}^{2} e^{\pm i m_{J} \varphi}=0 \end{aligned} \nonumber
Thus
$\Phi_{m}(\varphi)=N e^{\pm i m_{J} \varphi} \nonumber$
is a solution to the differential equations.
The normalization condition, Equation $\ref{5.8.23}$ is used to find a value for $N$ that satisfies Equation $\ref{5.8.22}$.
$\int \limits ^{2 \pi} _0 \Phi ^*(\varphi) \Phi (\varphi) d \varphi = 1 \label {5.8.23}$
The range of the integral is only from $0$ to $2π$ because the angle $\varphi$ specifies the position of the internuclear axis relative to the x-axis of the coordinate system and angles greater than $2π$ do not specify additional new positions.
Exercise 5.8.4
Use the normalization condition in Equation $\ref{5.8.23}$ to demonstrate that $N = 1/\sqrt{2π}$.
Answer
We need to evaluate Equation \ref{5.8.23} with $\psi(\varphi)=N e^{\pm i m J \varphi}$
\begin{align*} \psi^{*}(\varphi) \psi(\varphi) &= N e^{+i m J \varphi} N e^{-i m J \varphi} \[4pt] &=N^{2} \[4pt] 1=\int_{0}^{2 \pi} N^* N d \varphi=1 & \[4pt] N^{2} (2 \pi) =1 \[4pt] N=\sqrt{1 / 2 \pi} \end{align*} \nonumber
Values for $m$ are found by using a cyclic boundary condition. The cyclic boundary condition means that since $\varphi$ and $\varphi + 2\varphi$ refer to the same point in three-dimensional space, $\Phi (\varphi)$ must equal $\Phi (\varphi + 2 \pi )$, i.e.
\begin{align} e^{im_J \varphi} &= e^{im_J (\varphi + 2\pi)} \label{5.8.24} \[4pt] &= e^{im_J\varphi} e^{im_J2\pi} \label {5.8.25} \end{align}
For the equality in Equation $\ref{5.8.25}$ to hold, $e^{i m_J 2 \pi}$ must equal 1, which is true only when
$m_J = \cdots , -3, -2, -1, 0, 1, 2, 3, \cdots \label {5.8.26}$
In other words $m_J$ can equal any positive or negative integer or zero.
Exercise 5.8.5 : Cyclic Boundary Conditions
Use Euler’s Formula to show that $e^{im_J2\pi}$ equals 1 for $m_J$ equal to zero or any positive or negative integer.
Thus, the $Φ$ function is
$\Phi _{m_J} (\varphi) = \sqrt{\dfrac{1}{2\pi}} e^{\pm i m_J \varphi} \nonumber$
with
$m_J = 0, \pm 1, \pm 2, \cdots \nonumber$
Solving the $\Theta (\theta)$ Equation
Finding the $\Theta (\theta)$ functions that are solutions to the $\theta$-equation (Equation $\ref{5.8.18}$) is a more complicated process. Solutions are found to be a set of power series called Associated Legendre Functions (Table M2), which are power series of trigonometric functions, i.e., products and powers of sine and cosine functions. The $\Theta (\theta)$ functions, along with their normalization constants, are shown in the third column of Table 5.8.1 .
Table 5.8.1 : Spherical Harmonic Wavefunctions
$m_J$
$J$
$\Theta ^{m_J}_J (\theta)$
$\Phi (\varphi)$
$Y^{m_J}_J (\theta , \varphi)$
0 0 $\dfrac {1}{\sqrt {2}}$ $\dfrac {1}{\sqrt {2 \pi}}$ $\dfrac {1}{\sqrt {4 \pi}}$
0 1 $\sqrt {\dfrac {3}{2}}\cos \theta$ $\dfrac {1}{\sqrt {2 \pi}}$ $\sqrt {\dfrac {3}{4 \pi}}\cos \theta$
1 1 $\sqrt {\dfrac {3}{4}}\sin \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{i \varphi}$ $\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{i \varphi}$
-1 1 $\sqrt {\dfrac {3}{4}}\sin \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{-i\varphi}$ $\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{-i \varphi}$
0 2 $\sqrt {\dfrac {5}{8}}(3\cos ^2 \theta - 1)$ $\dfrac {1}{\sqrt {2 \pi}}$ $\sqrt {\dfrac {5}{16\pi}}(3\cos ^2 \theta - 1)$
1 2 $\sqrt {\dfrac {15}{4}} \sin \theta \cos \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{i \varphi}$ $\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{i\varphi}$
-1 2 $\sqrt {\dfrac {15}{4}} \sin \theta \cos \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{-i\varphi}$ $\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{-i\varphi}$
2 2 $\sqrt {\dfrac {15}{16}} \sin ^2 \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{2i\varphi}$ $\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{2i\varphi}$
-2 2 $\sqrt {\dfrac {15}{16}} \sin ^2 \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{2i\varphi}$ $\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{-2i\varphi}$
The solution to the $\theta$-equation requires that $λ$ in Equation $\ref{5.8.17}$ be given by
$\lambda = J (J + 1) \label {5.8.28}$
where
$J \ge |m_J| \label {5.8.29}$
$J$ can be 0 or any positive integer greater than or equal to $m_J$. Each pair of values for the quantum numbers, $J$ and $m_J$, identifies a rotational state with a wavefunction (Equation $\ref{5.8.11}$) and energy (below). Equation $\ref{5.8.29}$ means that $J$ controls the allowed values of $m_J$.
Each pair of values for the quantum numbers, $J$ and $m_J$, identifies a rotational state and hence a specific wavefunction with associated energy.
The combination of Equations $\ref{5.8.16}$ and $\ref{5.8.28}$ reveals that the energy of this system is quantized.
$E = \dfrac {\hbar ^2 \lambda}{2I} = J(J + 1) \dfrac {\hbar ^2}{2I} \label {5.8.30}$
Using Equation $\ref{5.8.30}$, you can construct a rotational energy level diagram (Figure 5.8.2 ). For simplicity, use energy units of $\dfrac {\hbar ^2}{2I}$.
• $J=0$: The lowest energy state has $J = 0$ and $m_J = 0$. This state has an energy $E_0 = 0$. There is only one state with this energy, i.e. one set of quantum numbers, one wavefunction, and one set of properties for the molecule.
• $J=1$: The next energy level is $J = 1$ with energy $\dfrac {2\hbar ^2}{2I}$. There are three states with this energy because $m_J$ can equal +1, 0, or ‑1. These different states correspond to different orientations of the rotating molecule in space. States with the same energy are said to be degenerate. The degeneracy of an energy level is the number of states with that energy. The degeneracy of the $J = 1$ energy level is 3 because there are three states with the energy $\dfrac {2\hbar ^2}{2I}$.
• $J=2$: The next energy level is for $J = 2$. The energy is $\dfrac {6\hbar ^2}{2I}$, and there are five states with this energy corresponding to $m_J = +2, \,+1,\, 0,\, ‑1,\, ‑2$. The energy level degeneracy is five. Note that the spacing between energy levels increases as J increases. Also note that the degeneracy increases. The degeneracy is always $2J+1$ because $m_J$ ranges from $+J$ to $‑J$ in integer steps, including 0.
Each allowed energy of rigid rotor is $(2J+1)$-fold degenerate. Hence, there exist $(2J+1)$ different wavefunctions with that energy.
Exercise 5.8.6
Compute the energy levels for a rotating molecule for $J = 0$ to $J = 5$ using units of $\dfrac {\hbar ^2}{2I}$.
Answer
This rotating molecule can be assumed to be a rigid rotor molecule. From solving the Schrödinger equation for a rigid rotor we have the relationship for energies of each rotational eigenstate (Equation \ref{5.8.30}):
$E = J(J+1)(ħ^2/2I) \nonumber$
Using this equation, we can plug in the different values of the $J$ quantum number so that
For J=0, $E = (0)(1)(ħ^2/2I) = 0$ For J=1, $E = (1)(2)(ħ^2/2I) = 2(ħ^2/2I)$
This shows that as $J$ increases, the energy levels get farther apart (Figure 5.8.2 ).
• For J=2, $E = (2)(3)(ħ^2/2I) = 6(ħ^2/2I)$
• For J=3, $E = (3)(4)(ħ^2/2I) = 12(ħ^2/2I)$
• For J=4, $E = (4)(5)(ħ^2/2I) = 20(ħ^2/2I)$
• For J=5, $E = (5)(6)(ħ^2/2I) = 30(ħ^2/2I)$
Exercise 5.8.7
For $J = 0$ to $J = 5$, identify the degeneracy of each energy level and the values of the $m_J$ quantum number that go with each value of the $J$ quantum number. Construct a rotational energy level diagram including $J = 0$ through $J=5$. Label each level with the appropriate values for the quantum numbers $J$ and $m_J$. Describe how the spacing between levels varies with increasing $J$.
Answer
This rotating molecule can be assumed to be a rigid rotor molecule. From solving the Schrödinger equation for a rigid rotor, we have: $λ = 2IE/ħ^2 \nonumber$
Where λ is an arbitrary parameter not related to wavelength. Additionally, λ is assigned a quantum relation so that $λ = J(J+1) \nonumber$
Thus, combining the two equations and solving for E yields $E = J(J+1)(ħ^2/2I) \nonumber$
Using this equation, we can plug in the different values of J so that
• For J=0, $E = (0)(1)(ħ^2/2I) = 0$
• For J=1, $E = (1)(2)(ħ^2/2I) = 2(ħ^2/2I)$
• For J=2, $E = (2)(3)(ħ^2/2I) = 6(ħ^2/2I)$
• For J=3, $E = (3)(4)(ħ^2/2I) = 12(ħ^2/2I)$
• For J=4, $E = (4)(5)(ħ^2/2I) = 20(ħ^2/2I)$
• For J=5, $E = (5)(6)(ħ^2/2I) = 30(ħ^2/2I)$
This shows that as $J$ increases, the energy levels get closer and closer together.
Interpretation of Quantum Numbers for a Rigid Rotor
The $m_J$ quantum number reflects the component of the angular momentum along the $z$ direction (and hence is sometimes called the azimuthal quantum number). For a fixed value of $J$, the different values of $m_J$ reflect the different directions the angular momentum vector could be pointing – for large, positive $m_J$ the angular momentum is mostly along +z; if $m_J$ is zero the angular momentum is orthogonal to $z$. Physically, the energy of the rotation does not depend on the direction, which is reflected in the fact that the energy depends only on $J$ (Equation $\ref{5.8.30}$), which measures the length of the vector, not its direction given mb $m_J$.
Example 5.8.7 : Molecular Oxygen
Calculate $J = 0$ to $J = 1$ rotational transition of the $\ce{O2}$ molecule with a bond length of 121 pm.
Solution
$E = \dfrac {\hbar^2}{I} = \dfrac {\hbar^2}{\mu r^2} \nonumber$
$\mu_{O2} = \dfrac{m_{O} m_{O}}{m_{O} + m_{O}} = \dfrac{(15.9994)(15.9994)}{15.9994 + 15.9994} = 7.9997 \nonumber$
convert from atomic units to kilogram using the conversion: 1 au = 1.66 x 10-27 kg. Plug and chug.
$E = 5.71 \times 10^{-27} \;Joules \nonumber$
Spherical Harmonics
A wavefunction that is a solution to the rigid rotor Schrödinger Equation (Equation $\ref{5.8.11}$) can be written as a single function $Y(\theta, \varphi)$, which is called a spherical harmonic function.
$Y^{m_J} _J (\theta , \varphi ) = \Theta ^{|m_J|}_J (\theta) \Phi _{m_J} (\varphi) \label {5.8.31}$
The spherical harmonic wavefunction is labeled with $m_J$ and $J$ because its functional form depends on both of these quantum numbers. These functions are tabulated above for $J = 0$ through $J = 2$ and for $J = 3$ in the Spherical Harmonics Table (M4) Polar plots of some of the $\theta$-functions are shown in Figure 5.8.3 .
The two-dimensional space for a rigid rotor is defined as the surface of a sphere of radius $r_0$, as shown in Figure 5.8.2 .
The probability of finding the internuclear axis at specific coordinates $\theta _0$ and $\varphi _0$ within an infinitesimal area $ds$ on this curved surface is given by
$Pr \left [ \theta _0, \varphi _0 \right ] = Y^{m_{J*}}_J (\theta _0, \varphi _0) Y^{m_J}_J (\theta _0, \varphi _0) ds \label {5.8.32}$
where the area element $ds$ is centered at $\theta _0$ and $\varphi _0$.
Within the Copenhagen interpretation of wavefunctions, the absolute square (or modulus squared) of the rigid rotor wavefunction $Y^{m_{J*}}_J (\theta, \varphi) Y^{m_J}_J (\theta, \varphi)$ gives the probability density for finding the internuclear axis oriented at $\theta$ to the z-axis and $\varphi$ to the x-axis. In spherical coordinates the area element used for integrating $\theta$ and $\varphi$ is
$ds = \sin \theta\, d \theta \,d \varphi \label {5.8.33}$
Exercise 5.8.8
Use calculus to evaluate the probability of finding the internuclear axis of a molecule described by the $J = 1$, $m_J = 0$ wavefunction somewhere in the region defined by a range in $\theta$ of 0° to 45°, and a range in of 0° to 90°. Note that a double integral will be needed. Sketch this region as a shaded area on Figure 5.8.1 .
Consider the significance of the probability density function by examining the $J = 1$, $m_J = 0$ wavefunction. The Spherical Harmonic for this case is
$Y^0_1 = \sqrt{ \dfrac {3}{4 \pi}} \cos \theta \label {5.8.34}$
The polar plot of $( Y^0_1)^2$ is shown in Figure 5.8.1 . For $J = 1$ and $m_J = 0$, the probability of finding the internuclear axis is independent of the angle $\varphi$ from the x-axis, and greatest for finding the internuclear axis along the z‑axis, but there also is a probability for finding it at other values of $\theta$ as well. So, although the internuclear axis is not always aligned with the z-axis, the probability is highest for this alignment. Also, since the probability is independent of the angle $\varphi$, the internuclear axis can be found in any plane containing the z-axis with equal probability.
The $J = 1$, $m_J = 0$ function is 0 when $\theta$ = 90°. Therefore, the entire xy-plane is a node. This fact means the probability of finding the internuclear axis in this particular horizontal plane is 0 in contradiction to our classical picture of a rotating molecule. In the classical picture, a molecule rotating in a plane perpendicular to the xy‑plane must have the internuclear axis lie in the xy‑plane twice every revolution, but the quantum mechanical description says that the probability of being in the xy-plane is zero. This conclusion means that molecules are not rotating in the classical sense, but they still have some, but not all, of the properties associated with classical rotation. The properties they retain are associated with angular momentum.
Exercise 5.8.9
For each state with $J = 0$ and $J = 1$, use the function form of the $Y$ spherical harmonics and Figure 5.8.1 to determine the most probable orientation of the internuclear axis in a diatomic molecule, i.e., the most probable values for $\theta$ and $\theta$.
Exercise 5.8.10
Write a paragraph describing the information about a rotating molecule that is provided in the polar plot of $Pr [\theta, \theta ]$ for the $J = 1$, $m_J = \pm 1$ state in Figure 5.8.1 . Compare this information to the classical picture of a rotating object.
Summary
There are two quantum numbers that describe the quantum behavior of a rigid rotor in three-deminesions: $J$ is the total angular momentum quantum number and $m_J$ is the z-component of the angular momentum. The spherical harmonics called $Y_J^{m_J}$ are functions whose probability $|Y_J^{m_J}|^2$ has the well known shapes of the s, p and d orbitals etc learned in general chemistry.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.08%3A_The_Energy_Levels_of_a_Rigid_Rotor.txt
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Learning Objectives
• Demonstrate how to use the 3D regid rotor to describe a rotating diatomic molecules
• Demonstate how microwave spectroscopy can get used to characterize rotating diatomic molecules
• Interprete a simple microwave spectrum for a diatomic molecule
To develop a description of the rotational states, we will consider the molecule to be a rigid object, i.e. the bond lengths are fixed and the molecule cannot vibrate. This model for rotation is called the rigid-rotor model. It is a good approximation (even though a molecule vibrates as it rotates, and the bonds are elastic rather than rigid) because the amplitude of the vibration is small compared to the bond length.
The rotation of a rigid object in space is very simple to visualize. Pick up any object and rotate it. There are orthogonal rotations about each of the three Cartesian coordinate axes just as there are orthogonal translations in each of the directions in three-dimensional space (Figures 5.9.1 and 5.9.2 ). These rotations are said to be orthogonal because one can not describe a rotation about one axis in terms of rotations about the other axes just as one can not describe a translation along the x-axis in terms of translations along the y- and z-axes. For a linear molecule, the motion around the interatomic axis (x-axis) is not considered a rotation.
In this section we examine the rotational states for a diatomic molecule by comparing the classical interpretation of the angular momentum vector with the probabilistic interpretation of the angular momentum wavefunctions. We want to answer the following types of questions. How do we describe the orientation of a rotating diatomic molecule in space? Is the molecule actually rotating? What properties of the molecule can be physically observed? In what ways does the quantum mechanical description of a rotating molecule differ from the classical image of a rotating molecule?
Introduction to Microwave Spectroscopy
The permanent electric dipole moments of polar molecules can couple to the electric field of electromagnetic radiation. This coupling induces transitions between the rotational states of the molecules. The energies that are associated with these transitions are detected in the far infrared and microwave regions of the spectrum. For example, the microwave spectrum for carbon monoxide spans a frequency range of 100 to 1200 GHz, which corresponds to 3 - 40 $cm^{-1}$.
The selection rules for the rotational transitions are derived from the transition moment integral by using the spherical harmonic functions and the appropriate dipole moment operator, $\hat {\mu}$.
$\mu _T = \int Y_{J_f}^{m_f*} \hat {\mu} Y_{J_i}^{m_i} \sin \theta \,d \theta\, d \varphi \label{5.9.1a}$
or in braket notation
$\mu _T = \langle Y_{J_f}^{m_f} | \hat {\mu} | Y_{J_i}^{m_i} \rangle \label{5.9.1b}$
Evaluating the transition moment integral involves a bit of mathematical effort. This evaluation reveals that the transition moment depends on the square of the dipole moment of the molecule, $\mu ^2$ and the rotational quantum number, $J$, of the initial state in the transition,
$\mu _T = \mu ^2 \dfrac {J + 1}{2J + 1} \label {5.9.2}$
and that the selection rules for rotational transitions are
$\Delta J = \pm 1 \label {5.9.3}$
and
$\Delta m_J = 0, \pm 1 \label {5.9.4}$
A photon is absorbed for $\Delta J = +1$ and emitted for $\Delta J = -1$.
Exercise 5.9.1
Explain why your microwave oven heats water, but not air. Hint: draw and compare Lewis structures for components of air and for water.
The energies of the $J^{th}$ rotational levels are given by
$E_J = J(J + 1) \dfrac {\hbar ^2}{2I} \label{energy}$
with each $J^{th}$ energy level having a degeneracy of $2J+1$ due to the different possible $m_J$ values.
Microwave Transition Energies
The transition energies for absorption of radiation are given by
\begin{align} E_{photon} &= \Delta E \[4pt] &= E_f - E_i \label{5.9.5A} \[4pt] &= h \nu \[4pt] &= hc \bar {\nu} \label {5.9.5} \end{align}
Substituting the relationship for energy (Equation \ref{energy}) into Equation \ref{5.9.5A} results in
\begin{align} E_{photon} &= E_f - E_i \[4pt] &= J_f (J_f +1) \dfrac {\hbar ^2}{2I} - J_i (J_i +1) \dfrac {\hbar ^2}{2I} \label {5.9.6} \end{align}
with $J_i$ and $J_f$ representing the rotational quantum numbers of the initial (lower) and final (upper) levels involved in the absorption transition.
Since microwave spectroscopists use frequency units and infrared spectroscopists use wavenumber units when describing rotational spectra and energy levels, both $\nu$ and $\bar {\nu}$ are important to calculate. When we add in the constraints imposed by the selection rules to identify possible transitions, $J_f$ in Equation \ref{5.9.6} can be replaced by $J_i + 1$, since the selection rule requires $J_f – J_i = 1$ for the absorption of a photon (Equation \ref{5.9.3}). The equation for absorption transitions (Equation \ref{5.9.6}) then can be written in terms of the only the quantum number $J_i$ of the initial state.
\begin{align} E_{photon} &= h \nu \[4pt] &= hc \bar {\nu} \[4pt] &= 2 (J_i + 1) \dfrac {\hbar ^2}{2I} \label {5.9.7} \end{align}
Equation \ref{5.9.7} can be rewritten as
$E_{photon} = 2B (J_i+1) \nonumber$
where $B$ is the rotational constant for the molecule and is defined in terms of the energy of the absorbed photon
$B = \dfrac {\hbar ^2}{2I} \label {5.9.9}$
Often spectroscopists want to express the rotational constant in terms of frequency of the absorbed photon and do so by dividing Equation $\ref{5.9.9}$ by $h$
\begin{align} B (\text{in freq}) &= \dfrac{B}{h} \[4pt] &= \dfrac {h}{8\pi^2 \mu r_0^2} \end{align} \nonumber
More often, spectroscopists want to express the rotational constant in terms of wavenumbers ($\bar{\nu}$) of the absorbed photon by dividing Equation $\ref{5.9.9}$ by $hc$,
$\tilde{B} = \dfrac{B}{hc} = \dfrac {h}{8\pi^2 c \mu r_0^2} \label {5.9.8}$
The rotational constant depends on the distance ($R$) and the masses of the atoms (via the reduced mass) of the nuclei in the diatomic molecule.
Exercise 5.9.2
Construct a rotational energy level diagram for $J = 0$, $1$, and $2$ and add arrows to show all the allowed transitions between states that cause electromagnetic radiation to be absorbed or emitted.
Exercise 5.9.3
Complete the steps going from Equation $\ref{5.9.6}$ to Equation $\ref{5.9.9}$ and identify the units of $B$ at the end.
Answer
\begin{align*} \Delta E_{photon} &= E_{f} - E{i}\ E_{r.rotor} &= J(J+1)\frac{\hbar^2}{2I}\ E_{photon} = h_{\nu} = hc\widetilde{\nu} &= J_f(J_f+1)\frac{\hbar^2}{2I} - J_i(J_i+1)\frac{\hbar^2}{2I}\ J_f - J_i &= 1\ J_f &= 1 + J_i\ E_{photon} = h_{\nu} = hc\widetilde{\nu} &= (1+J_i)(2+J_i)\frac{\hbar^2}{2I} - J_i(J_i+1)\frac{\hbar^2}{2I} \ &= \frac{\hbar^2}{2I}[2 + 3J_i + J_i^2 -J_i^2 - J_i]\ &= \frac{\hbar^2}{2I}2(J_i+1)\ &= 2B(J_i + 1) \end{align*} \nonumber
Now we do a standard dimensional analysis
\begin{align*} B &= \frac{\hbar^2}{2I} \equiv \left[\frac{kg m^2}{s^2}\right] = [J]\ \frac{B}{h} = B(in freq.) &= \frac{h}{8 \pi^2\mu r_o^2} \equiv \left[\frac{1}{s}\right]\ \frac{B}{hc} = \widetilde{B} &= \frac{h}{8 \pi^2\mu c r_o^2} \equiv \left[\frac{s}{m}\right]\ \end{align*} \nonumber
Exercise 5.9.4
Infrared spectroscopists use units of wavenumbers. Rewrite the steps going from Equation $\ref{5.9.6}$ to Equation $\ref{5.9.9}$ to obtain expressions for $h\nu$ and $B$ in units of wavenumbers. Note that to convert $B$ in Hz to $B$ in $cm^{-1}$, you simply divide the former by $c$.
Figure 5.9.3 shows the rotational spectrum as a series of nearly equally spaced lines. The line positions $\nu _J$, line spacings, and the maximum absorption coefficients ( $\gamma _{max}$, the absorption coefficients associated with the specified line position) for each line in this spectrum are given here in Table 5.9.1 .
Table 5.9.1 : Rotational Transitions in $^{12}C^{16}O$ at 40 K
J
$\nu _J$ (MHz)
Spacing from previous line (MHz)
$\gamma _{max}$
$0 \rightarrow 1$ 115,271.21 0 0.0082
$1 \rightarrow 2$ 230,538.01 115,266.80 0.0533
$2 \rightarrow 3$ 345,795.99 115,257.99 0.1278
$3 \rightarrow 4$ 461,040.76 115,244.77 0.1878
$4 \rightarrow 5$ 576,267.91 115,227.15 0.1983
$6 \rightarrow 6$ 691,473.03 115,205.12 0.1618
$6 \rightarrow 7$ 806,651.78 115,178.68 0.1064
$7 \rightarrow 8$ 921,799.55 115,147.84 0.0576
$8 \rightarrow 9$ 1,036,912.14 115,112.59 0.0262
$9 \rightarrow 10$ 1,151,985.08 115,072.94 0.0103
Let’s try to reproduce Figure 5.9.3 from the data in Table 5.9.1 by using the quantum theory that we have developed so far. Equation $\ref{5.9.8}$ predicts a pattern of exactly equally spaced lines. The lowest energy transition is between $J_i = 0$ and $J_f = 1$ so the first line in the spectrum appears at a frequency of $2B$. The next transition is from $J_i = 1$ to $J_f = 2$ so the second line appears at $4B$. The spacing of these two lines is $2B$. In fact the spacing of all the lines is $2B$, which is consistent with the experimental data in Table 5.9.1 showing that the lines are very nearly equally spaced. The difference between the first spacing and the last spacing is less than 0.2%.
Exercise 5.9.5
Use Equation $\ref{5.9.8}$ to prove that the spacing of any two lines in a rotational spectrum is $2B$, i.e. derive:
$\nu _{J_i + 1} - \nu _{J_i} = 2B \nonumber$
Answer
To prove the relationship, evaluate the LHS. First, define the terms:
$\nu_{J_{i}}=2B(J_{i}+1),\nu_{J_{i}+1}=2B((J_{i}+1)+1) \nonumber$
Substitute into the equation and evaluate:
$2B((J_{i}+1)+1)-2B(J_{i}+1)=2B \nonumber$
$2B(J_{i}+1)+2B-2B(J_{i}+1)=2B \nonumber$
$2B=2B \nonumber$
LHS equals RHS.Therefore, the spacing between any two lines is equal to $2B$.
Example 5.9.1 : Rotation of Sodium Hydride
The molecule $\ce{NaH}$ undergoes a rotational transition from $J=0$ to $J=1$ when it absorbs a photon of frequency $2.94 \times 10^{11} \ Hz$. What is the equilibrium bond length of the molecule?
Solution
We use $J=0$ in the formula for the transition frequency
$\nu =2B=\dfrac{\hbar}{2\pi I}=\dfrac{\hbar}{2\pi \mu R_{e}^{2}} \nonumber$
Solving for $R_e$ gives
$R_e = \sqrt{\dfrac{\hbar}{2\pi \mu \nu}} \nonumber$
The reduced mass is given by
\begin{align*}\mu &= \dfrac{m_{Na}m_H}{m_{Na}+m_H} \[4pt] &=\dfrac{(22.989)(1.0078)}{22.989+1.0078}\[4pt] &=0.9655\end{align*} \nonumber
which is in atomic mass units or relative units. To convert to kilograms, we need the conversion factor $1 \ au = 1.66\times 10^{-27} \ kg$. Multiplying this by $0.9655$ gives a reduced mass of $1.603\times 10^{-27} \ kg$. Substituting in for $R_e$ gives
\begin{align*} R_e &= \sqrt{\dfrac{(1.055 \times 10^{-34} \ J\cdot s)}{2\pi (1.603\times 10^{-27} \ kg)(2.94\times 10^{11} \ Hz)}}\[4pt] &= 1.899\times 10^{-10} \ m \[4pt] &=1.89 \ \stackrel{\circ}{A}\end{align*} \nonumber
Example 5.9.6
Use the frequency of the $J = 0$ to $J = 1$ transition observed for carbon monoxide to determine a bond length for $^{12}C^{16}O$.
Solution
• J=0: $v_{0}=115271.21\, MHz$
• J=1: $v_{1}=230538.01\, MHz$
\begin{align*} \Delta v &=230538.01 M H z-115271.21\, MHz \[4pt] &=115266.8 MHz \[4pt] &=1.153 \times 10^{11} Hz \[4pt] &=\dfrac{\hbar}{2 \pi \mu R_{e}^2} \end{align*} \nonumber
The reduced mass is
\begin{align*} \mu &=\dfrac{m_{C} m_{O}}{m_{C}+m_{O}} \[4pt] &=\frac{12.01 \times 16.00}{12.01+16.00} \[4pt] &=6.86\, amu \end{align*} \nonumber
Convert to kg
$6.86 amu \, \left( \frac{1.661 \times 10^{-27} k g}{12 m u} \right) =1.139 \times 10^{-26} kg \nonumber$
\begin{align*} R_{e} &=\sqrt{\frac{\hbar}{2 \pi \mu \Delta v}} \[4pt] &= \sqrt{\frac{1.055 \times 10^{-34} J \cdot s}{2 \pi \cdot 1.139 \times 10^{-26} k g \cdot 1.153 \times 10^{11} H z}} \[4pt] &=1.131 \times 10^{-10} \mathrm{m} \[4pt] &=1.131\, \stackrel{\circ}{A} \end{align*} \nonumber
Advanced: Non-Rigid Rotors
Centrifugal stretching of the bond as $J$ increases causes the decrease in the spacing between the lines in an observed spectrum (Table 5.9.1 ). This decrease shows that the molecule is not really a rigid rotor. As the rotational angular momentum increases with increasing $J$, the bond stretches. This stretching increases the moment of inertia and decreases the rotational constant (Figure 5.9.5 ).
The effect of centrifugal stretching is smallest at low $J$ values, so a good estimate for $B$ can be obtained from the $J = 0$ to $J = 1$ transition. From $B$, a value for the bond length of the molecule can be obtained since the moment of inertia that appears in the definition of $B$ (Equation $\ref{5.9.9}$) is the reduced mass times the bond length squared. When the centrifugal stretching is taken into account quantitatively, the development of which is beyond the scope of the discussion here, a very accurate and precise value for B can be obtained from the observed transition frequencies because of their high precision. Rotational transition frequencies are routinely reported to 8 and 9 significant figures.
As we have just seen, quantum theory successfully predicts the line spacing in a rotational spectrum. An additional feature of the spectrum is the line intensities. The lines in a rotational spectrum do not all have the same intensity, as can be seen in Figure 5.9.3 and Table 5.9.1 . This is related to the populations of the initial and final states. This aspect of spectroscopy will be discussed in more detail in the following chapters
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.09%3A_The_Rigid_Rotator_is_a_Model_for_a_Rotating_Diatomic_Molecule.txt
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Solutions to select questions can be found online.
5.7
Calculate the reduced mass of HCl molecule given that the mass of H atom is 1.0078 amu and the mass of Cl atom is 34.9688 amu. Note that 1 amu = 1.660565*10-27 kg.
Solution
$\mu = \dfrac{m_1m_2}{m_1+m_2} \nonumber$
$\mu = \dfrac{1.0078\; amu \times 34.9688\; amu}{1.0078\; amu +34.9688\; amu} = 0.9796 \;amu \nonumber$
$\mu =0.9796 \;amu \times \dfrac{1.660565 \cdot 10^{-27} \;kg}{1\;amu} =1.627 \times 10^{-27} kg \nonumber$
5.8
Calculate the reduced mass for the Br2, Cl2, and I2 diatomics.
Solution
From the periodic table, the atomic masses for Br, Cl, and I are 79.904, 35.453, and 126.904 respectively.
Covert the atomic mass to kg.
$Br= (79.904 \, amu)(1.6606 \times 10^{-27} \,amu/kg)=1.327 \times 10^{-25}\;kg$ $Cl= (35.453 \, amu )(1.6606 \times 10^{-27} \,amu/kg )=5.887 \times 10^{-26}\;kg$ $I= (126.904 \, amu )(1.6606 \times 10^{-27} \,amu/kg )=2.107 \times 10^{-25}\;kg$
$\mu=\dfrac{m}{2} \nonumber$
therefore
μBr2=1.327x10-25 kg/2=6.635x10-26 kg μCl2= 5.887x10-26kg/2 = 2.9435x10-26kg μI2=2.107x10-25kg=1.0535x10-25kg
The equation for a reduced mass ($\mu$) of a diatomic is
$\mu= \dfrac{m_1m_2}{m_1+m_2} \nonumber$
for a diatomic molecule with identical atoms ($m_1=m_2=m$) so
5.14
79Br79Br has a force constant of $240\,N \cdot m^{-1}$. Given this information:
1. Calculate the fundamental vibrational frequency and
2. Calculate the 79Br79Br zero point energy.
Solution
We must first know which formula to use which is
$\nu_{obs}=\dfrac{1}{2π} \sqrt{\dfrac{k}{\mu}} \nonumber$
calculate the reduced mass
$\mu=\dfrac{(79\; amu)^2}{79 \;amu+ 79 \;amu}=39.5 amu \nonumber$
and convert to Kg:
1.66e-27 kg•amu-1
substitute the given values
$\nu=\dfrac{1}{2π} \sqrt{\dfrac{ 240 kg \;m \;s^{-2} \;s }{39.5 amu \times 1.66 \times 10^{-27} kg \;amu^{-1}}} = 9.63 \times 10^{12} s^{-1} \nonumber$
It can also be convert to wavenumber (inverse centimeter $cm^{-1}$):
$\nu_{cm^{-1}}=\dfrac{1}{\lambda}=\dfrac{\nu}{c}=\dfrac{9.63\times 10^{12} s^{-1}}{3.0\times 10^{10} cm\;s^{-1}}=321 cm^{-1} \nonumber$
Zero Point Energy:
$E_{0}=\dfrac{1}{2}h\nu= \dfrac{1}{2}hc\nu_{cm^{-1}} \nonumber$ (formula to use)
E0=1/2(6.626e-34J•s)(2.998e10cm•s-1)(321cm-1)
E0= 3.19e-21J
5.19
Prove that the second derivative of an even function is even and odd function is odd.
Solution
This is an example..not a proof
The following is an even function:
$y(x) = a + bx^2 + cx^4 + dx^6 \nonumber$
so
$\dfrac{dy}{dx}= 2bx + 4cx^3 + 6dx^5 \nonumber$
and
$\dfrac{d^2y}{dx^2}= 2b + 12cx^2 + 30dx^4 \nonumber$
which is an even function.
The following is an odd function:
$f(x)= ax + bx^3 + cx^5 \nonumber$
so
$\dfrac{df}{dx}= a + 3bx^2 + 5cx^4 \nonumber$
and
$\dfrac{d^2f}{dx^2}= 6bx + 10cx^3 \nonumber$
which is an odd function.
5.27
The Harmonic oscillator Hamiltonian obeys the reflective property:
$\hat{H}(x) = \hat{H}(-x) \nonumber$
What does this say about the nature of the harmonic oscillator wave function?
Solution
The harmonic oscillator switches from odd to even due to the fact that the reflective property will alternate.
5.28
If $\langle x \rangle$ is an odd function, what does that say about $p_x$?
Hint: use
$\dfrac{d \langle p_x \rangle }{dt} = \left\langle\dfrac{-dV}{dx} \right \rangle \nonumber$
also known as Ehrenfest's Theorem, where $V$ is the potential of a one dimensional harmonic oscillator.
Hence, $\langle p_x \rangle$ does not depend on time.
5.32
Convert $\nabla^2$ from Cartesian coordinates to cylindrical coordinates.
Solution
We have to start with the conversion of Cartesian coordinates $\{x, y, z\}$ to cylindrical coordinates $\{r, \theta , z\}$
$x = r\cos\theta$ $y = r\sin\theta$ $z = z$
Now putting it all together
$\nabla^2 = \dfrac{d^2}{dr^2} + \dfrac{1}{r}\dfrac{d}{dr} + \dfrac{1}{r^2}\dfrac{d^2}{d\theta^2} + \dfrac{d^2}{dz^2}$
$r = \sqrt{x^2+y^2}$
$\cos\theta = \dfrac{x}{\sqrt{x^2+y^2}}$
$\sin\theta = \dfrac{x}{\sqrt{x^2+y^2}}$
Now by chain rule we get
$\dfrac{d}{dx} = \dfrac{dr}{dx}\dfrac{d}{dr} + \dfrac{d\theta}{dx}\dfrac{d}{d\theta}$
$\dfrac{d}{dy} = \dfrac{dr}{dy}\dfrac{d}{dr} + \dfrac{d\theta}{dy}\dfrac{d}{d\theta}$
$\dfrac{dr}{dx} = \dfrac{x}{r} = \cos\theta$
$\dfrac{dr}{dx} = \sin\theta$
using implicit differentiation and taking the second derivatives will yield
$\dfrac{d^2}{dx^2} = \left(\cos\theta\dfrac{d}{dr} - \dfrac{\sin\theta}{r} \dfrac{d}{d\theta}\right)\left(\cos\theta\dfrac{d}{dr} - \dfrac{\sin\theta}{r} \dfrac{d}{d\theta}\right)$
$\dfrac{d^2}{dy^2} = \left(\sin\theta\dfrac{d}{dr} + \dfrac{\cos\theta}{r} \dfrac{d}{d\theta}\right)\left(\sin\theta\dfrac{d}{dr} + \dfrac{\cos\theta}{r} \dfrac{d}{d\theta}\right)$
$\dfrac{d^2}{dz^2} = \dfrac{d^2}{dz^2}$
5.37
Find the magnitude of angular momentum and the $z$ component of angular momentum for electrons in a hydrogen-like species with
1. quantum numbers $n \ = \ 1$, $l \ = \ 0$, $m \ = \ 0$; and
2. $n \ = \ 2$, $l \ = \ 0$, $m \ = \ 0$.
Compare your answers and explain your results.
Solution
The wave function for this problem is given by:
$\psi_{100}=R(r)_{10}Y(\theta,\phi)_{00}=2\left(\frac{Z}{2a_0}\right)^\frac{3}{2}e^\frac{-Zr}{a_0} \nonumber$
Using that:
$\hat{L}^{2}Y_{lm}(\theta, \phi)=\hbar^{2}l(l+1)Y_{lm}(\theta,\phi), \nonumber$
and
$\hat{L_z}=m\hbar \nonumber$
Then $\hat{L}^{2}=0$ and $\hat{L_z}=0$. Given that the values for $l$ and $m$ are the same as above, the answers would also be the same.
The reason why both answers are the same is that the operators for angular momentum only act on the angular part of the wave function. Since only the quantum number $n$ varied between these two states, the angular momentum eigenvalues did not change.
5.38
Apply the angular momentum operator in the x direction to the following functions ($Y(\theta,\phi)$).
1. $\dfrac{5\pi}{4} + 7\exp(\pi^2)$
2. $3\pi \sin(\theta)$
3. $\dfrac{3}{2}\cos(\theta)\exp(i\phi)$
Solution
Let us begin by stating the angular momentum operator in terms of $\theta$ and $\phi$.
$\hat{L_x} = i\hbar\Big(\sin(\phi)\dfrac{\partial}{\partial \theta} + \cot(\theta)\cos(\phi)\dfrac{\partial}{\partial \phi}\Big) \nonumber$
a) $Y(\theta,\phi) = \dfrac{5\pi}{4} + 7exp(\pi^2)$
$\hat{L_x}(\dfrac{5\pi}{4} + 7exp(\pi^2)) = i\hbar\Big(\sin(\phi)\dfrac{\partial \dfrac{5\pi}{4} + 7exp(\pi^2)}{\partial \theta} + \cot(\theta)\cos(\phi)\dfrac{\partial \dfrac{5\pi}{4} + 7exp(\pi^2)}{\partial \phi}\Big) \nonumber$
$= 0 \nonumber$
The function does not depend on $\theta$ or $\phi$ so when the angular momentum operator is applied to the function, it equals 0.
b) $Y(\theta,\phi) = 3\pi \sin(\theta)$
$\hat{L_x}(3\pi \sin(\theta)) = i\hbar\Big(\sin(\phi)\dfrac{\partial}{\partial \theta}3\pi \sin(\theta) + \cot(\theta)\cos(\phi)\dfrac{\partial}{\partial \phi}3\pi \sin(\theta)\Big) \nonumber$
$= 3i\pi\hbar \sin(\phi)\cos(\theta) \nonumber$
c) $Y(\theta,\phi) = \dfrac{3}{2}\cos(\theta)exp(i\phi)$
$\hat{L_x}(3\pi \sin(\theta)) = i\hbar\Big(\sin(\phi)\dfrac{\partial}{\partial \theta}\dfrac{3}{2}\cos(\theta)exp(i\phi) + \cot(\theta)\cos(\phi)\dfrac{\partial}{\partial \phi}\dfrac{3}{2}\cos(\theta)exp(i\phi)\Big) \nonumber$
$= i\hbar\Big( \dfrac{-3}{2}\sin(\phi)\sin(\theta)exp(i\phi) + \dfrac{3i}{2}\cot(\theta)\cos(\phi)\cos(\theta)exp(i\phi)\Big) \nonumber$
$= \dfrac{3i\hbar exp(i\phi)}{2}(i\cot(\theta)\cos(\phi)\cos(\theta) - \sin(\phi)\sin(\theta)) \nonumber$
5.41
Use the fact that $\hat x$ and $\hat p$ are Hermitian in the number operator
$\hat a_- = \dfrac{1}{\sqrt{2}}(\hat x +i\hat p) \nonumber$
$\hat a_+ = \dfrac{1}{\sqrt{2}}(\hat x -i\hat p) \nonumber$
and
$\hat H=\dfrac{\hbar w}{2}(\hat a_-\hat a_+ + \hat a_+\hat a_-) \nonumber$
Show that
$\int \psi^*_v \hat{v} \psi dx \geq 0 \nonumber$
5.43
Determine the unnormalized wave function $\psi_\circ \big(x\big)$ given that $\hat{a}_- = 2^{-1/2}\big(\hat{x}+i\hat{p}\big)$ and that $\hat{a_-}\psi_\circ = 0$ Then find the unnormalized wave function for $\psi_1\big(x\big)$ using $\hat{a}_+$.
Solution
It was given that $\hat{a}_-\psi_\circ = 0$, so substituting in $\hat{a}_-$ so we know
$\hat{a_-} = 2^{-1/2}\big(\hat{x}+i\hat{p}\big) \psi_\circ = 0 \nonumber$
We can expand and simplify this expression to a first order partial differential equation
$x\psi_\circ + \dfrac{d\psi_\circ}{dx} = 0 \nonumber$
Solve by separating like terms
$\dfrac{d\psi_\circ}{\psi_\circ} = -xdx \nonumber$
Solving this equation for $\psi_\circ \big(x\big)$ we find that
$\psi_\circ = e^\dfrac{-x^2}{2} \nonumber$
To solve for $\psi_1$ we understand that $\psi_1\backsim\hat{a_+}\psi_\circ\backsim\hat{x}-i\hat{p}\psi_\circ$, as well as that
$\hat{x}-i\hat{p}\psi_\circ = x\psi_\circ - \dfrac{d\psi_\circ}{dx} = 2xe^\dfrac{-x^2}{2}=2x\psi_\circ \nonumber$
So then we can say
$\boxed{\psi_1\backsim xe^\dfrac{-x^2}{2}} \nonumber$
5.46
Find the reduced mass of an electron in a Tritium atom. Set the mass of the Tritium to be $5.008267 \times 10^{-27}\, kg$. Then find the value of the Rydberg constant for the Tritium atom.
Solution
To solve, use the reduced mass equation, and for mass 1 enter the mass of the electron, and for mass 2 enter the mass of the Tritium atom:
$\mu = \dfrac{m_1m_2}{m_1+m_2} \nonumber$
For which one attains a value of $\mu = 9.1077 x 10^{-31} kg$
5.46
The mass of a deuterium atom is $3.343586 \times 10^{-27}\; kg$. First calculate the reduce mass of the deuterium atom. Then using the reduced mass calculated find the Rydberg constant for a deuterium atom.
Solution
$\mu$ = reduced mass
$\mu_{deuterium} = \dfrac{(9.109390 \times 10^{-31} kg) ( 3.343586\times 10^{-27}kg)}{(9.109390 \times 10^{-31} kg + 3.343586\times 10^{-27}kg )} \nonumber$
$\mu_{deuterium} = 9.106909 \times 10^{-31}kg = 0.9997277 m_e \nonumber$
$R_H$ = Rydberg constant
$R_H = (109,737.2 cm^{-1}) (0.9997277 m_e) = 109,707.3 cm^{-1} \nonumber$
5.47
What is the ratio of the frequency of spectral lines of C-14 that has been ionized 5 times and C-12 that has been ionized 5 times?
Solution
Carbon that has been ionized 5 times is a hydrogen like ion, so we can use the Bohr model to find the desired ratio.
$E = \dfrac{uZ^2e^4n^2}{8ε_0^2h^3c} \nonumber$
gives the placement of spectral lines. The coefficient of n2 is proportional to the frequency of these lines, so the ratio of EC-14/Ec-12 will give the ratio of frequency of the lines. The only difference between these two isotopes is the reduced mass u. So the problem reduces to uC-14/ uC-12. Mass in amu is used below. me = mass of electron = 5.4858*10-4 amu.
$\mu_{C-14} = \dfrac{m_em_{c-14}}{m_e + m_{c-14}} = \dfrac{(14.003)(5.4858 \times 10^{-4})}{14.003 + 5.4858 \times 10^{-4}} = 5.485585 \times 10^{-4} \nonumber$
$\mu_{C-12} = \dfrac{m_em_{c-12}}{m_e + m_{c-12}}= \dfrac{(12)(5.4858 \times 10^{-4})}{12 + 5.4858 \times 10^{-4}} = 5.485549 \times 10^{-4} \nonumber$
$\dfrac{\mu_{C-14}}{\mu_{C-12}} = 1.0000065 \nonumber$
5.47
Calculate the Rydberg constant for a deuterium atom and atomic hydrogen given the reduced mass of a deuterium atom is $9.106909 \times 10^{-31} kg$ and the reduced mass of hydrogen is $9.104431 \times 10^{-31} kg$. Compare both of these answers with the experimental result ($109677.6 cm^{-1}$). Then determine the ratio of the frequencies of the lines in the spectra of atomic hydrogen and atomic deuterium.
Solution
The Rydberg constant is found using
$R_H=\dfrac{me^4}{8\epsilon_o^2 ch^3} \nonumber$
For a deuterium atom
$R_H=\dfrac{(9.104431 \times 10^{-31}kg)(1.602 \times 10^{-19} C)^4}{8(8.854 \times 10^{-12} \dfrac{F}{m})^2(2.998 \times 10^{8} \dfrac{m}{s})(6.626 \times 10^{-34} J \cdot s)^3} \nonumber$
$R_H=109707.3 cm^{-1} \nonumber$
This is different by $2.7 \times 10^{-2}\%$.
The ratio of the frequencies of the lines in the spectra of atomic hydrogen and atomic deuterium is equivalent to the ratio of the Rydberg constants we just found.
$\dfrac{109707.3\; cm^{-1}}{109677.5\; cm^{-1}}=1.000272 \nonumber$
$R_H=\dfrac{(9.106909 \times 10^{-31}kg)(1.602 \times 10^{-19} C)^4}{8(8.854 \times 10^{-12} \dfrac{F}{m})^2(2.998 \times 10^{8} \dfrac{m}{s})(6.626 \times 10^{-34} J \cdot s)^3} \nonumber$
$R_H=109677.5 cm^{-1} \nonumber$
This is different by $9.1\times 10^{-5}\%$.
For a hydrogen atom
5.46
Find the reduced mass of HCl where the mass of hydrogen in 1 amu and the mass of chloride is 35 amu.
Solution
$\mu = \dfrac{m_1 \times m_2}{m_1 + m_2} \nonumber$
$\mu = \dfrac{(1.00)(35.00)}{36.00} 1.603 \times 10^{-27} kg = 1.558 \times 10^{-27} kg \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.E%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor_%28Exercises%29.txt
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The solution of the Schrödinger equation (wave equation) for the hydrogen atom uses the fact that the Coulomb potential produced by the nucleus is isotropic (it is radially symmetric in space and only depends on the distance to the nucleus). Although the resulting energy eigenfunctions (the orbitals) are not necessarily isotropic themselves, their dependence on the angular coordinates follows completely generally from this isotropy of the underlying potential: the eigenstates of the Hamiltonian (that is, the energy eigenstates) can be chosen as simultaneous eigenstates of the angular momentum operator. This corresponds to the fact that angular momentum is conserved in the orbital motion of the electron around the nucleus. Therefore, the energy eigenstates may be classified by two angular momentum quantum numbers, ℓ and m (both are integers). The angular momentum quantum number ℓ = 0, 1, 2, ... determines the magnitude of the angular momentum. The magnetic quantum number m = −ℓ, ..., +ℓ determines the projection of the angular momentum on the (arbitrarily chosen) z-axis.
• 6.1: The Schrodinger Equation for the Hydrogen Atom Can Be Solved Exactly
The solution of the Schrödinger equation (wave equation) for the hydrogen atom uses the fact that the Coulomb potential produced by the nucleus is isotropic (it is radially symmetric in space and only depends on the distance to the nucleus). Although the resulting energy eigenfunctions (the orbitals) are not necessarily isotropic themselves, their dependence on the angular coordinates follows completely generally from this isotropy.
• 6.2: The Wavefunctions of a Rigid Rotator are Called Spherical Harmonics
The solutions to the hydrogen atom Schrödinger equation are functions that are products of a spherical harmonic functions and a radial function.
• 6.3: The Three Components of Angular Momentum Cannot be Measured Simultaneously with Arbitrary Precision
The angular momentum operator is one of several related operators analogous to classical angular momentum. The angular momentum operator plays a central role in the theory of atomic physics and other quantum problems involving rotational symmetry. Two orthogonal components of angular momentum (e.g., \(L_x\) and \(L_y\)) are complementary and cannot be simultaneously known or measured. It is, however, possible to simultaneously measure or specify \(L^2\) and any one component of \(L\).
• 6.4: Hydrogen Atomic Orbitals Depend upon Three Quantum Numbers
In solving the Schrödinger equation of the hydrogen atom, we have encountered three quantum numbers. The quantum numbers are not independent; the choice of nn limits the choice of ll, which in turn limits the choice of mm. A fourth quantum number, ss, does not follow directly from solving the Schrödinger equation but is to do with spin (discussed later).
• 6.5: s-orbitals are Spherically Symmetric
The hydrogen atom wavefunctions are called atomic orbitals. An atomic orbital is a function that describes one electron in an atom. The radial probability distribution is introduced in this section.
• 6.6: Orbital Angular Momentum and the p-Orbitals
The physical quantity known as angular momentum plays a dominant role in the understanding of the electronic structure of atoms.
• 6.7: The Helium Atom Cannot Be Solved Exactly
The second element in the periodic table provides our first example of a quantum-mechanical problem which cannot be solved exactly. Nevertheless, as we will show, approximation methods applied to helium can give accurate solutions in perfect agreement with experimental results. In this sense, it can be concluded that quantum mechanics is correct for atoms more complicated than hydrogen. By contrast, the Bohr theory failed miserably in attempts to apply it beyond the hydrogen atom.
• 6.E: The Hydrogen Atom (Exercises)
These are homework exercises to accompany Chapter 6 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Thumbnail: Hydrogen atom. (Public Domain; Bensaccount via Wikipedia)
06: The Hydrogen Atom
The hydrogen atom, consisting of an electron and a proton, is a two-particle system, and the internal motion of two particles around their center of mass is equivalent to the motion of a single particle with a reduced mass. This reduced particle is located at $r$, where $r$ is the vector specifying the position of the electron relative to the position of the proton. The length of $r$ is the distance between the proton and the electron, and the direction of $r$ and the direction of $r$ is given by the orientation of the vector pointing from the proton to the electron. Since the proton is much more massive than the electron, we will assume throughout this chapter that the reduced mass equals the electron mass and the proton is located at the center of mass.
Exercise 6.1.1
1. Assuming the Bohr radius gives the distance between the proton and electron, calculate the distance of the proton from the center of mass, and calculate the distance of the electron from the center of mass.
2. Calculate the reduced mass of the electron-proton system.
3. In view of your calculations in (a) and (b), comment on the validity of a model in which the proton is located at the center of mass and the reduced mass equals the electron mass.
Since the internal motion of any two-particle system can be represented by the motion of a single particle with a reduced mass, the description of the hydrogen atom has much in common with the description of a diatomic molecule discussed previously. The time-independent Schrödinger Equation for the hydrogen atom
$\hat {H} (r , \theta , \varphi ) \psi (r , \theta , \varphi ) = E \psi ( r , \theta , \varphi) \label {6.1.1}$
employs the same kinetic energy operator, $\hat {T}$, written in spherical coordinates. For the hydrogen atom, however, the distance, $r$, between the two particles can vary, unlike the diatomic molecule where the bond length was fixed and the rigid rotor model was applicable. The hydrogen atom Hamiltonian also contains a potential energy term, $\hat {V}$, to describe the attraction between the proton and the electron. This term is the Coulomb potential energy,
$\hat {V} (r) = - \dfrac {e^2}{4 \pi \epsilon _0 r } \label {6.1.2}$
where r is the distance between the electron and the proton. The Coulomb potential energy depends inversely on the distance between the electron and the nucleus and does not depend on any angles. Such a potential is called a central potential.
The full expression for $\hat {H}$ in spherical coordinates is
$\hat {H} (r , \theta , \varphi ) = - \dfrac {\hbar ^2}{2 \mu r^2} \left [ \dfrac {\partial}{\partial r} \left (r^2 \dfrac {\partial}{\partial r} \right ) + \dfrac {1}{\sin \theta } \dfrac {\partial}{\partial \theta } \left ( \sin \theta \dfrac {\partial}{\partial \theta} \right ) + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] - \dfrac {e^2}{4 \pi \epsilon _0 r } \label {6.1.3}$
The contributions from rotational and radial components of the motion become clearer if we write out the complete Schrödinger equation,
$\left \{ -\dfrac {\hbar ^2}{2 \mu r^2} \left [ \dfrac {\partial}{\partial r} \left (r^2 \dfrac {\partial}{\partial r} \right ) + \dfrac {1}{\sin \theta } \dfrac {\partial}{\partial \theta } \left ( \sin \theta \dfrac {\partial}{\partial \theta} \right ) + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] - \dfrac {e^2}{4 \pi \epsilon _0 r } \right \} \psi (r , \theta , \varphi ) = E \psi (r , \theta , \varphi ) \label {6.1.4}$
multiply both sides of Equation $\ref{6.1.4}$ by $2 \mu r^2$, and rearrange to obtain
$\hbar ^2 \dfrac {\partial}{\partial r} \left ( r^2 \dfrac {\partial}{\partial r} \psi (r , \theta , \varphi ) \right ) + 2 \mu r^2 \left [ E + \dfrac {e^2}{4 \pi \epsilon _0 r } \right ] \psi (r , \theta , \varphi ) = \nonumber$
$- \hbar^2 \left [ \dfrac {1}{\sin \theta } \dfrac {\partial}{\partial \theta } \left ( \sin \theta \dfrac {\partial}{\partial \theta} \right ) + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] \psi (r , \theta , \varphi ) \label {6.1.5}$
Manipulating the Schrödinger equation in this way helps us recognize the square of the angular momentum operator in Equation $\ref{6.1.5}$. The square of the angular momentum operator in Equation $\ref{6.1.6}$.
$\hat {M} ^2 = -\hbar ^2 \left [\dfrac {1}{\sin \theta } \dfrac {\partial}{\partial \theta } \left ( \sin \theta \dfrac {\partial}{\partial \theta} \right ) + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] \label {6.1.6}$
Substituting Equation $\ref{6.1.6}$ into Equation $\ref{6.1.5}$ produces
$\hbar ^2 \dfrac {\partial}{\partial r } \left ( r^2 \dfrac {\partial}{\partial r} \psi (r , \theta , \varphi ) \right ) + 2 \mu r^2 [ E - \hat {V} ] \psi (r , \theta , \varphi ) = \hat {M} ^2 \psi (r, \theta , \varphi ) \label {6.1.7}$
Exercise 6.1.2
Show the algebraic steps going from Equation $\ref{6.1.4}$ to $\ref{6.1.5}$ and finally to $\ref{6.1.7}$. Justify the statement that the rotational and radial motion are separated in Equation $\ref{6.1.7}$.
Since the angular momentum operator does not involve the radial variable, $r$, we can separate variables in Equation $\ref{6.1.7}$ by using a product wavefunction, as we did previously for rigid rotors. We know that the eigenfunctions of the angular momentum operator are the Spherical Harmonic functions (Table M4), $Y (\theta , \varphi )$, so a good choice for a product function is
$\psi (r , \theta , \varphi ) = R (r) Y (\theta , \varphi ) \label {6.1.8}$
The Spherical Harmonic functions provide information about where the electron is around the proton, and the radial function $R(r)$ describes how far the electron is away from the proton.
To separate variables, substitute the product function, Equation $\ref{6.1.8}$ into Equation $\ref{6.1.7}$, evaluate partial derivatives, divide each side by R(r) $Y (\theta, \varphi )$, and set each side of that resulting equation equal to a constant $\lambda$.
\begin{align} \dfrac {\hbar ^2}{R (r)} \dfrac {\partial}{\partial r} r^2 \dfrac {\partial}{\partial r} R(r) + \dfrac {2 \mu r^2}{R (r)} [ E - V ] R (r) &= \lambda \label {6.1.9} \[4pt] \dfrac {1}{Y (\theta , \varphi )} \hat {M} ^2 Y (\theta , \varphi ) &= \lambda \label {6.1.10} \end{align}
Equations $\ref{6.1.9}$ and $\ref{6.1.10}$ represent the radial differential equation and the angular differential equation, respectively. As we describe below, they are solved separately to give the $Y (\theta , \varphi )$ angular functions and the R(r) radial functions.
Exercise 6.1.3
Complete the steps leading from Equations $\ref{6.1.7}$ to $\ref{6.1.9}$ and $\ref{6.1.10}$.
Rearranging Equation $\ref{6.1.10}$ yields
$\hat {M} ^2 Y^{m_l}_l (\theta , \varphi ) = \lambda Y^{m_l}_l (\theta , \varphi ) \label {6.1.11}$
where we have added the indices $l$ and $m_l$ to identify a particular spherical harmonic function. Note that the notation has changed from that used with the Rigid Rotor; it is customary to use $J$ and $m_J$ to represent the angular momentum quantum numbers for rotational states, but for electronic states, it is customary to use $l$ and $m_l$ to represent the same thing. Furthermore, the electronic angular momentum is designated by $L$ and the corresponding operator is called $\hat {L}$. In complete electronic notation, Equation $\ref{6.1.11}$ is
$\hat {L} ^2 Y^{m_l}_l (\theta , \varphi ) = \lambda Y^{m_l}_l (\theta , \varphi ) \label {6.1.12}$
Equation $\ref{6.1.12}$ says that $Y^{m_l}_l (\theta , \varphi )$ must be an eigenfunction of the angular momentum operator $\hat {L} ^2$ with eigenvalue λ. We know from the discussion of the Rigid Rotor that the eigenvalue $λ$ is $J(J+1)ħ^2$, or in electronic notation, $l (l + 1) \hbar ^2$. Consequently, Equation $\ref{6.1.12}$ becomes
$\hat {L} ^2 Y^{m_l}_l (\theta , \varphi ) = l (l + 1) \hbar ^2 Y^{m_l}_l (\theta , \varphi ) \label{6.1.13}$
Using this value for λ and rearranging Equation \ref{6.1.9}, we obtain
$- \dfrac {\hbar ^2}{2 \mu r^2} \dfrac {\partial}{\partial r} r^2 \dfrac {\partial}{\partial r} R(r) + \left [ \dfrac {l(l +1) \hbar ^2}{2 \mu r^2} + V (r) - E \right ] R (r) = 0 \label {6.1.14}$
Exercise 6.1.4
Write the steps leading from Equation $\ref{6.1.9}$ to Equation $\ref{6.1.14}$.
The details for solving Equation $\ref{6.1.14}$ are provided elsewhere, but the procedure and consequences are similar to previously examined cases. As for the harmonic oscillator, an asymptotic solution (valid at large $r$) is found, and then the complete solutions are written as products of the asymptotic solution and polynomials arising from sequential truncations of a power series expansion.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/06%3A_The_Hydrogen_Atom/6.01%3A_The_Schrodinger_Equation_for_the_Hydrogen_Atom_Can_Be_Solved_Exactly.txt
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The solutions to the hydrogen atom Schrödinger equation are functions that are products of a spherical harmonic function and a radial function:
$\psi _{n, l, m_l } (r, \theta , \phi) = R_{n,l} (r) Y^{m_l}_l (\theta , \phi) \label {6.2.20}$
The wavefunctions for the hydrogen atom depend upon the three variables $r$, $\theta$, and $\phi$ and the three quantum numbers n, $l$, and $m_l$. The variables give the position of the electron relative to the nucleus in spherical coordinates. The absolute square of the wavefunction, $| \psi (r, \theta , \phi )|^2$, evaluated at $r$, $\theta$, and $\phi$ gives the probability density of finding the electron inside a differential volume $d \tau$, centered at the position specified by $r$, $\theta$, and $\phi$.
Exercise 6.2.1
What is the value of the integral (in braket notation)
$\langle \psi (r, \theta , \phi ) | \psi (r, \theta , \phi ) \rangle \;? \nonumber$
or expanded in integral notation (in Cartesian coordinates)
$\int \limits_{-\infty}^{+\infty} \int \limits_{-\infty}^{+\infty} \int \limits_{-\infty}^{+\infty} | \psi (x, y , z)|^2 \,dx\,dy\,dz \nonumber$
or expanded in integral notation (in spherical coordinates)
$\int \limits_{0}^{+\infty} \int \limits_{0}^{\pi} \int \limits_{0}^{+2\pi} | \psi (r, \theta , \phi)|^2 r^2\sin \theta \,d\theta\, d\phi\,dr \nonumber$
Answer
No need to know the functional form of the hydrogen atom eigenstates in any coordinate system. If the eigenstates (i.e., solutions to the hydrogen atom Schrödinger equation) are normalized - as we normally like them- then this integral will be 1.
The quantum numbers have names:
• $n$ is called the principal quantum number,
• $l$ is called the angular momentum quantum number, and
• $m_l$ is called the magnetic quantum number because (as we will see, the energy in a magnetic field depends upon $m_l$).
Often $l$ is called the azimuthal quantum number because it is a consequence of the $\theta$-equation, which involves the azimuthal angle $\Theta$, referring to the angle to the zenith. These quantum numbers have specific values that are dictated by the physical constraints or boundary conditions imposed upon the Schrödinger equation: $n$ must be an integer greater than 0, $l$ can have the values 0 to $n‑1$, and $m_l$ can have $2l + 1$ values ranging from $-l$ to $+l$ in unit or integer steps. The total number of orbitals with a particular value of $n$ is $n^2$ - i.e., this is the degeneracy of the system.
Wavefunction Nomenclature
The values of the quantum number $l$ usually are coded by a letter: s means 0, p means 1, d means 2, f means 3; the next codes continue alphabetically (e.g., g means $l = 4$). The quantum numbers specify the quantization of physical quantities. The discrete energies of different states of the hydrogen atom are given by $n$, the magnitude of the angular momentum is given by $l$, and one component of the angular momentum (usually chosen by chemists to be the z‑component) is given by $m_l$.
The $\Phi$ function is found to have the quantum number $m$, where
$\Phi_m (\phi) = A_m e^{im\phi} \nonumber$
and $A_m$ is the normalization constant and $m = 0, \pm1, \pm2 ... \pm\infty$. The $\Theta$ function was solved and is known as Legendre polynomials, which have quantum numbers $m$ and $\ell$. When $\Theta$ and $\Phi$ are multiplied together, the product is known as spherical harmonics with labeling $Y_{J}^{m} (\theta, \phi)$.
Figure 6.2.1 shows the spherical harmonics $Y_J^M$, which are solutions of the angular Schrödinger equation. These are explicitly written in Table 6.2.1 . Notice that these functions are complex in nature.
Table 6.2.1 : spherical harmonics $Y_J^M$
$m_J$
$J$
$\Theta ^{m_J}_J (\theta)$
$\Phi (\varphi)$
$Y^{m_J}_J (\theta , \varphi)$
0 0 $\dfrac {1}{\sqrt {2}}$ $\dfrac {1}{\sqrt {2 \pi}}$ $\dfrac {1}{\sqrt {4 \pi}}$
0 1 $\sqrt {\dfrac {3}{2}}\cos \theta$ $\dfrac {1}{\sqrt {2 \pi}}$ $\sqrt {\dfrac {3}{4 \pi}}\cos \theta$
1 1 $\sqrt {\dfrac {3}{4}}\sin \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{i \varphi}$ $\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{i \varphi}$
-1 1 $\sqrt {\dfrac {3}{4}}\sin \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{-i\varphi}$ $\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{-i \varphi}$
0 2 $\sqrt {\dfrac {5}{8}}(3\cos ^2 \theta - 1)$ $\dfrac {1}{\sqrt {2 \pi}}$ $\sqrt {\dfrac {5}{16\pi}}(3\cos ^2 \theta - 1)$
1 2 $\sqrt {\dfrac {15}{4}} \sin \theta \cos \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{i \varphi}$ $\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{i\varphi}$
-1 2 $\sqrt {\dfrac {15}{4}} \sin \theta \cos \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{-i\varphi}$ $\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{-i\varphi}$
2 2 $\sqrt {\dfrac {15}{16}} \sin ^2 \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{2i\varphi}$ $\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{2i\varphi}$
-2 2 $\sqrt {\dfrac {15}{16}} \sin ^2 \theta$ $\dfrac {1}{\sqrt {2 \pi}}e^{-2i\varphi}$ $\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{-2i\varphi}$
Exercise 6.2.2
Consider several values for $n$, and show that the number of orbitals for each $n$ is $n^2$.
Exercise 6.2.3
Construct a table summarizing the allowed values for the quantum numbers $n$, $l$, and $m_l$. for energy levels 1 through 7 of hydrogen.
Exercise 6.2.4
The notation 3d specifies the quantum numbers for an electron in the hydrogen atom.
1. What are the values for $n$ and $l$?
2. What are the values for the energy and angular momentum?
3. What are the possible values for the magnetic quantum number?
4. What are the possible orientations for the angular momentum vector?
Answer
1. $n=3$ and $l=2$
2. Energy: $E_{n}=\frac{E_{1}}{n^{2}}=\frac{-13.6 \mathrm{eV}}{9}=-1.51 \mathrm{eV} \nonumber$ and angular momentum $\sqrt{l(l+1)} \hbar=\sqrt{2 \times 3} \hbar=2.584 \times 10^{-34} J \cdot S \nonumber$
3. Magnetic quantum number: $m_{l}=-2,-1,0,1,2 \nonumber$
Atomic Orbitals
The hydrogen atom wavefunctions, $\psi (r, \theta , \phi )$, are called atomic orbitals. An atomic orbital is a function that describes one electron in an atom. The wavefunction with $n = 1$, $\ell = 0$, and ($m_{ell}=0$ is called the 1s orbital, and an electron that is described by this function is said to be “in” the 1s orbital, i.e. have a 1s orbital state. The constraints on $n$, $\ell$, and $m_\ell$ that are imposed during the solution of the hydrogen atom Schrödinger equation explain why there is a single 1s orbital, why there are three 2p orbitals, five 3d orbitals, etc. We will see when we consider multi-electron atoms, these constraints explain the features of the Periodic Table. In other words, the Periodic Table is a manifestation of the Schrödinger model and the physical constraints imposed to obtain the solutions to the Schrödinger equation for the hydrogen atom.
Visualizing the variation of an electronic wavefunction with $r$, $\theta$, and $\phi$ is important because the absolute square of the wavefunction depicts the charge distribution (electron probability density) in an atom or molecule. The charge distribution is central to chemistry because it is related to chemical reactivity. For example, an electron-deficient part of one molecule is attracted to an electron-rich region of another molecule, and such interactions play a major role in chemical interactions ranging from substitution and addition reactions to protein folding and the interaction of substrates with enzymes.
Visualizing wavefunctions and charge distributions is challenging because it requires examining the behavior of a function of three variables in three-dimensional space. This visualization is made easier by considering the radial and angular parts separately, but plotting the radial and angular parts separately do not reveal the shape of an orbital very well. The shape can be revealed better in a probability density plot. To make such a three-dimensional plot, divide space up into small volume elements, calculate $\psi^* \psi$ at the center of each volume element, and then shade, stipple or color that volume element in proportion to the magnitude of $\psi^* \psi$. Do not confuse such plots with polar plots, which look similar. Probability densities also can be represented by contour maps, as shown in Figure 6.2.2 .
Methods for separately examining the radial portions of atomic orbitals provide useful information about the distribution of charge density within the orbitals. Graphs of the radial functions, $R(r)$, for the 1s, 2s, and 2p orbitals plotted in Figure 6.2.3 .
The 1s function in Figure 6.2.3 starts with a high positive value at the nucleus and exponentially decays to essentially zero after 5 Bohr radii. The high value at the nucleus may be surprising, but as we shall see later, the probability of finding an electron at the nucleus is vanishingly small.
Next notice how the radial function for the 2s orbital, Figure 6.2.3 , goes to zero and becomes negative. This behavior reveals the presence of a radial node in the function. A radial node occurs when the radial function equals zero other than at $r = 0$ or $r = ∞$. Nodes and limiting behaviors of atomic orbital functions are both useful in identifying which orbital is being described by which wavefunction. For example, all of the s functions have non-zero wavefunction values at $r = 0$, but p, d, f and all other functions go to zero at the origin. It is useful to remember that there are $n-1-l$ radial nodes in a wavefunction, which means that a 1s orbital has no radial nodes, a 2s has one radial node, and so on.
Exercise 6.2.5
Examine the mathematical forms of the radial wavefunctions. What feature in the functions causes some of them to go to zero at the origin while the s functions do not go to zero at the origin?
Exercise 6.2.6
What mathematical feature of each of the radial functions controls the number of radial nodes?
Answer
The Laguerre polynomial controls the radial nodes with the number of roots for the Laguerre polynomial is the number of radial nodes.
Exercise 6.2.7 : Radial Node
At what value of $r$ does the 2s radial node occur?
Answer
A node exists when the radial portion of the wavefunction equals 0. The radial portion of the 2s wavefunction is:
$R(r) = \left(\frac{1}{2 \sqrt{2}}\right)\left(\frac{Z}{\alpha_{0}}\right)^{\frac{3}{2}}(2-\rho) e^{\frac{-\rho}{2}} \nonumber$
where $\rho=\frac{Z r}{\alpha_{0}}$, $Z=1$ for the hydrogen atom and $\alpha_{0}$ is the Bohr radius.
Therefore:
$R(r) = \left(\frac{1}{2 \sqrt{2}}\right)\left(\frac{1}{\alpha_{0}}\right)^{\frac{3}{2}}\left(2-\frac{r}{\alpha_{0}}\right) e^{\frac{-2 r}{\alpha_{0}}} \nonumber$
For this wavefunction:
• $\left(\frac{1}{\alpha_{0}}\right)^{\frac{3}{2}}$ is a constant and will never equal 0.
• $e^{\frac{-2 r}{\alpha_{0}}}$ is an exponential, and will also never equal 0.
Therefore, our node is when $\left(2-\frac{r}{\alpha_{0}}\right)=0$ and $r=2 \alpha_{0} \nonumber$
Exercise 6.2.8
Make a table that provides the energy, number of radial nodes, and the number of angular nodes and total number of nodes for each function with $n = 1$, $n=2$, and $n=3$. Identify the relationship between the energy and the number of nodes. Identify the relationship between the number of radial nodes and the number of angular nodes.
Answer
Wavefunction Energy Total Nodes Radial Nodes Angular Nodes
1s 13.6 eV 0 0 0
2s 3.4 eV 1 1 0
2p 3.4 eV 1 0 1
3s 1.5 eV 2 2 0
3p 1.5 eV 2 1 1
3d 1.5 eV 2 0 2
The energy of the electron in each wavefunction is $E = \dfrac{Z^2E_h}{2n^2} \nonumber$
The number of total nodes is (N-1\), the number of radial nodes is $N-L-1$ and the number of angular nodes is $L$.
The quantity $R(r)^* R(r)$ gives the radial probability density; i.e., the probability density for the electron to be at a point located the distance $r$ from the proton. Radial probability densities for three atomic orbitals are plotted in Figure 6.2.4 .
Probabilities and Distribution Functions
When the radial probability density for every value of $r$ is multiplied by the area of the spherical surface represented by that particular value of r, we get the radial distribution function. The radial distribution function gives the probability density for an electron to be found anywhere on the surface of a sphere located a distance r from the proton. Since the area of a spherical surface is $4 \pi r^2$, the radial distribution function is given by
$\underbrace{4 \pi r^2 R(r) ^* R(r)}_{\text{radial probability density}} \nonumber$
Radial distribution functions are shown in Figure 6.2.5 . At small values of $r$, the radial distribution function is low because the small surface area for small radii modulates the high value of the radial probability density function near the nucleus. As we increase $r$, the surface area associated with a given value of r increases, and the $r^2$ term causes the radial distribution function to increase even though the radial probability density is beginning to decrease. At large values of $r$, the exponential decay of the radial function outweighs the increase caused by the $r^2$ term and the radial distribution function decreases.
Exercise 6.2.9
Write a quality comparison of the radial function and radial distribution function for the 2s orbital. See Figure 6.2.6 .
Answer
$R(r)$ is the radial function of the eigenstate and $4\pi r^2 R(r)^* R(r)$ gives the radial distribution function of the distance of electron at distance $r$ from the nucleus (i.e., the probability integrated over all angles).
The radial probability function is low at small values of $r$ because of a small surface area near nucleus, for example at 2s at a small value of $r$ the radial probability function is low. At higher values of $r$ the surface area increases while radial probability density decreases, this causes the radial distribution function to increase. In contrast the radial probability density is high at small surface area and when $r$ is near the nucleus, i.e low values of $r$.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/06%3A_The_Hydrogen_Atom/6.02%3A_The_Wavefunctions_of_a_Rigid_Rotator_are_Called_Spherical_Harmonics.txt
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Learning Objectives
• Understand how to measure the orbital angular momentum of an electron around a nucleus.
• Understand how the Heisenburgh Uncertainty Principle extends to orbital angular momenta.
• Manipulate the angular momenta cyclic permutations that allow two of the three projects to be simultaneous measured
Consider a particle described by the Cartesian coordinates $(x, y, z)\equiv \vec{r}$ and their conjugate momenta $(p_x, p_y, p_z)\equiv \vec{p}$. The classical definition of the orbital angular momentum of such a particle about the origin is (i.e., via the vector cross product):
$\vec{L} = \vec{r} \times \vec{p} \nonumber$
which can be separated into projections into each of the primary axes :
\begin{align*} L_x &= y\, p_z - z\, p_y, \label{6.3.1a} \[4pt] L_y &= z\, p_x - x\, p_z \label{6.3.1b} \[4pt] L_z &= x\,p_y - y \,p_x \label{6.3.1c} \end{align*}
Extending this discussion to the quantum mechanics, we can assume that the operators $(\hat{L}_x, \hat{L}_y, \hat{L}_z)\equiv \vec{L}$ - that represent the components of orbital angular momentum in quantum mechanics - can be defined in an analogous manner to the corresponding components of classical angular momentum. In other words, we are going to assume that the above equations specify the angular momentum operators in terms of the position and linear momentum operators.
In Cartesian coordinates, the three operators for the orbital angular momentum components can be written as
$\hat{L}_x = -{\rm i}\,\hbar\left(y\,\dfrac{\partial}{\partial z} - z\,\dfrac{\partial} {\partial y}\right) \label{6.3.2a}$
$\hat{L}_y = -{\rm i}\,\hbar\left(z\,\dfrac{\partial}{\partial x} - x\,\dfrac{\partial} {\partial z}\right) \label{6.3.2b}$
$\hat{L}_z = -{\rm i}\,\hbar\left(x\,\dfrac{\partial}{\partial y} - y\,\dfrac{\partial} {\partial x}\right) \label{6.3.2c}$
These can be transforming to operators in standard spherical polar coordinates,
\begin{align*} x &= r \,\sin\theta\, \cos\varphi \label{6.3.3a} \[4pt] y &= r\, \sin\theta\, \sin\varphi \label{6.3.3b} \[4pt] z &=r \cos \theta \label{ 6.3.3c} \end{align*}
we obtain
\begin{align*} \hat{L}_x &= {\rm i}\,\hbar\,\left(\sin\varphi\, \dfrac{\partial}{\partial \theta} + \cot\theta \cos\varphi\,\dfrac{\partial}{\partial \varphi}\right) \label{6.3.4a} \[4pt] \hat{L}_y &= -{\rm i} \,\hbar\,\left(\cos\varphi\, \dfrac{\partial}{\partial\theta} -\cot\theta \sin\varphi \,\dfrac{\partial}{\partial \varphi}\right) \label{6.3.4b} \[4pt] \hat{L}_z &= -{\rm i}\,\hbar\,\dfrac{\partial}{\partial\varphi} \label{6.3.4c} \end{align*}
We can introduce a new operator $\hat{L^2}$:
\begin{align} \hat{L^2} &= \hat{L}_x^{\,2}+\hat{L}_y^{\,2}+\hat{L}_z^{\,2} \label{6.3.5} \[4pt] &= - \hbar^2\left( \dfrac{1}{\sin\theta}\dfrac{\partial}{\partial \theta} \sin \theta \dfrac{\partial}{\partial \theta} + \dfrac{1}{\sin^2\theta}\dfrac{\partial^2} {\partial\varphi^2}\right) \label{6.3.6} \end{align}
The eigenvalue problem for $\hat{L^2}$ takes the form
$\hat{L^2} | \psi \rangle = \lambda \,\hbar^2 | \psi \rangle \label{6.3.6a}$
where $\psi(r, \theta, \varphi)$ is the wavefunction, and $\lambda$ is a number. Let us write
$\psi(r, \theta, \varphi) = R(r) \,Y(\theta, \varphi) \label{6.3.6b}$
By definition,
$\boxed{L^2 \,Y_{l}^{m_l} = l\,(l+1)\,\hbar^2\,Y_{l}^{m_l}} \label{6.3.9}$
where $l$ is an integer. This is an important conclusion that argues the angular momentum is quantized with the square of the magnitude of the angular momentum only capable of assume one of the discrete set of values (Equation $\ref{6.3.9}$). From this, the amplitude of angular momentum can be expressed
$\boxed{ |\vec{L}| =\sqrt{L^2} = \sqrt{l(l+1)} \hbar }\label{6.3.10}$
Warning
We often refer to a particle in a state with angular momentum quantum number $l$ as having angular momentum $l$, rather than saying that it has angular momentum of $\sqrt{l(l+1)} \hbar$ magnitude, primarily since it is awkward to say quickly.
The properties of spherical harmonics that the z-component of the angular momentum ($L_z$) is also quantized and can only assume a one of a discrete set of values
$L_z \,Y_{l}^{m_l} = m\,\hbar\,Y_l^{m_l} \label{6.3.11}$
where $m_l$ is an integer lying in the range $-l\leq m_l \leq l$.
• $l$ is sometimes called "azimuthal quantum number" or "orbital quantum number"
• $m_l$ is sometimes called "magnetic quantum number"
Simultaneous Measurements
Note that observables associated with $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$ can, in principle, be measured. However, to determine if they can be measured simultaneously with infinite precision, the corresponding operators must commute. Remember that the fundamental commutation relations satisfied by the position and linear momentum operators are:
\begin{align*} [\hat{x}_i, \hat{x}_j] &=0 \label{6.3.12} \[4pt] [\hat{p}_i, \hat{p}_j] &=0 \label{6.3.13} \[4pt] [\hat{x}_i, \hat{p}_j] &= {\rm i}\,\hbar \,\delta_{ij} \label{6.3.14} \end{align*}
where $i$ and $j$ stand for either $x$, $y$, or $z$. Consider the commutator of the operators $\hat{L}_x$ and $\hat{L}_z$ :
\begin{align*} [\hat{L}_x, \hat{L}_y] & = [(y\,p_z-z\,p_y), (z\,p_x-x \,p_z)] \[4pt] &= y\,[p_z, z]\,p_x + x\,p_y\,[z, p_z] \label{6.3.15} \[4pt] &= {\rm i}\,\hbar\,(-y \,p_x+ x\,p_y) \[4pt] &= {\rm i}\,\hbar\, \hat{L}_z \label{6.3.16} \end{align*}
The cyclic permutations of the above result yield the fundamental commutation relations satisfied by the components of an orbital angular momentum:
$[\hat{L}_x, \hat{L}_y] = {\rm i}\,\hbar\, \hat{L}_z \label{6.3.17a}$
$[\hat{L}_y, \hat{L}_z] = {\rm i}\,\hbar\, \hat{L}_x \label{6.3.17b}$
$[\hat{L}_z, \hat{L}_x] = {\rm i}\,\hbar\, \hat{L}_y \label{6.3.17c}$
The three commutation relations (Equations $\ref{6.3.17a}$ - $\ref{6.3.17c}$) are the foundation for the whole theory of angular momentum in quantum mechanics. Whenever we encounter three operators having these commutation relations, we know that the dynamical variables that they represent have identical properties to those of the components of an angular momentum (which we are about to derive). In fact, we shall assume that any three operators that satisfy the commutation relations (Equations $\ref{6.3.17a}$ - $\ref{6.3.17c}$) represent the components of some sort of angular momentum.
Any three operators that satisfy the cyclic commutation relations represent the components of some sort of angular momentum.
Example 6.3.1 : Commutators
Show that the $\hat{L^2}$ and $\hat{L}_x$ operators commute.
Solution
We want to confirm that $[\hat{L^2}, \hat{L}_x] = 0$ that from Equation $\ref{6.3.5}$ this can be expanded
$[\hat{L^2}, \hat{L}_x] = [\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2, \hat{L}_x] \nonumber$
from the properties of commutators, this can be expanded
$[\hat{L^2}, \hat{L}_x] = [\hat{L}_x^2, \hat{L}_x] + [\hat{L}_y^2, \hat{L}_x] + [\hat{L}_z^2 , \hat{L}_x] \nonumber$
However,
$[\hat{L}_x^2, \hat{L}_x] = \hat{L}_x^2 \hat{L}_x - \hat{L}_x \hat{L}_x^2 = \hat{L}_x \hat{L}_x \hat{L}_x - \hat{L}_x \hat{L}_x \hat{L}_x = 0\nonumber$
So
\begin{align*} [\hat{L^2}, \hat{L}_x] &= [\hat{L}_y^2, \hat{L}_x] + [\hat{L}_z^2, \hat{L}_x]\nonumber \ &= \hat{L}_y^2 \hat{L}_x - \hat{L}_x \hat{L}_y^2 + \hat{L}_z^2 \hat{L}_x - \hat{L}_x \hat{L}_z^2 \nonumber \ &= \hat{L}_y \hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y \hat{L}_y + \hat{L}_z \hat{L}_z \hat{L}_x - \hat{L}_x \hat{L}_z \hat{L}_z \nonumber \end{align*} \nonumber
Lets look at some related forms which can be used to simplify the above expression. The first two terms can and final two terms can be rewritten as different commutators
\begin{align*} [\hat{L}_y , \hat{L}_x] &= \hat{L}_y + \hat{L}_y [\hat{L}_y , \hat{L}_x] \nonumber \ &= (\hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y) \hat{L}_y + \hat{L}_y (\hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y) \nonumber \ &= \cancel{\hat{L}_y \hat{L}_x \hat{L}_y} - \hat{L}_x \hat{L}_y \hat{L}_y + \hat{L}_y \hat{L}_y \hat{L}_x - \cancel{\hat{L}_y \hat{L}_x \hat{L}_y} \nonumber \end{align*} \nonumber
The first & fourth terms cancel, giving
$[\hat{L}_y , \hat{L}_x] \hat{L}_y + \hat{L}_y [\hat{L}_y , \hat{L}_x] = \hat{L}_y \hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y \hat{L}_y \nonumber$
Similarly,
$[Lz , \hat{L}_x] \hat{L}_z + \hat{L}_z [\hat{L}_z, \hat{L}_x] = \hat{L}_z \hat{L}_z \hat{L}_x - \hat{L}_x \hat{L}_z \hat{L}_z \nonumber$
So,
\begin{align*} [\hat{L^2} , \hat{L}_x] &= [\hat{L}_y , \hat{L}_x] \hat{L}_y + \hat{L}_y [\hat{L}_y , \hat{L}_x] + [\hat{L}_z , \hat{L}_x] \hat{L}_z + \hat{L}_z [\hat{L}_z , \hat{L}_x]\nonumber \[4pt] &= - i\hbar \hat{L}_z \hat{L}_y - i\hbar \hat{L}_y \hat{L}_z + i\hbar \hat{L}_y \hat{L}_z + ih \hat{L}_z \hat{L}_y \[4pt] &= 0 \nonumber \end{align*} \nonumber
One can also show similarly that
$[\hat{L^2}, \hat{L}_y] = [\hat{L^2}, \hat{L}_z] = 0 \nonumber$
Example 6.3.1 shows that while $L_z$ can be known with certainty, $L_x$ and $L_y$ would unknown. This means that every vector with the appropriate length and z-component can drawn to represent $\vec{L}$, which forms a cone (Figure 6.3.1 ). The expected value of the angular momentum for a given ensemble of systems in the quantum state characterized by $l$ and $m_l$ could be somewhere on this cone while it cannot be defined for a single system (since the components of $L$ do not commute with each other).
The Meaning of Commutation of Two Operators
The mathematics of commutation relations is relatively straightforward, but what does it physically mean for an observable (Hermitian operator) to commute with another observable (Hermitian operator) in quantum mechanics?
If two operators $\hat{A}$ and $\hat{B}$ commute with each other then
$\hat A \hat B - \hat B \hat A = 0, \nonumber$
which can be rearranged to
$\hat A \hat B = \hat B \hat A. \nonumber$
This is not a trivial statement and many operations do not commute and hence the end-result depends on how you have ordered the operations.
If you recall that operators act on quantum mechanical states and give you a new state in return, then this means that with $\hat{A}$ and $\hat{B}$ commuting, the state you obtain from letting first $\hat{A}$ act and then $\hat{B}$ act on some initial state is the same as if you let first $\hat{B}$ and then $\hat{A}$ act on that state, i.e.,
$\hat A \hat B | \psi \rangle = \hat B \hat A | \psi \rangle. \nonumber$
Recall that when you perform a quantum mechanical measurement, you will always measure an eigenvalue of your operator, and after the measurement your state is left in the corresponding eigenstate. The eigenstates to the operator are precisely those states for which there is no uncertainty in the measurement: You will always measure the eigenvalue.
Therefore, $\hat{B} |a \rangle$ must be an eigenfunction of $\hat{A}$ with eigenvalue $a$ just like $|a \rangle$ itself is. That is essentially saying that $|a \rangle$ is an eigenfunction of $\hat{B}$.
A key example of this is since $\hat{L^2}$ and $\hat{L}_x$ commute (Example 6.3.1 ) then both operators share the same eigenstates. Hence, we do no need to solve two eignevalue problems:
$\hat{L^2} | \psi \rangle = \lambda | \psi \rangle \nonumber$
and
$\hat{L}_x | \psi \rangle = \beta| \psi \rangle \nonumber$
If we solve one, we then know the eigenvalues ($| \psi \rangle$) for the other!
What does it mean when some observable $\hat{A}$ commutes with the Hamiltonian $\hat{H}$? First, we get all the result from above: There is a simultaneous eigenbasis of the energy-eigenstates and the eigenstates of $\hat{A}$. This can yield a tremendous simplification of the task of solving Schrödinger equations. For example, the Hamiltonian of the hydrogen atom commutes with $\hat{L}$, the angular momentum operator, and with $\hat{L}_z$, its z-component. This tells you that you can classify the eigenstates by an angular- and magnetic quantum number $l$ and $m$.
Summary
In the quantum world, angular momentum is quantized. The square of the magnitude of the angular momentum (determined by the eigenvalues of the $\hat{L^2}$ operator) can only assume one of the discrete set of values
$L^2 = l(l + 1)\hbar^2 \nonumber$
or the magnitude of the angular momentum
$L = \sqrt{l(l + 1)}\hbar \nonumber$
with $l = 0, 1, 2, ... \nonumber$
The z-component of the angular momentum (i.e., projection of $L$ onto the $z$-axis) is also quantized with
$L_z= m_{l} \hbar \nonumber$
with $m_l = -l, 0-1, ..., 0, ... +l +1, l \nonumber$ for a given value of $l$. Hence, $l$ and $m_l$ are the angular momentum quantum number and the magnetic quantum number, respectively.
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Learning Objectives
• Recognize how the hydrogen atomic orbitals vary as a function of the three primary quantum numbers
The solutions to the hydrogen atom Schrödinger equation discussed previously are functions that are products of a spherical harmonic function and a radial function.
$\psi _{n, l, m_l } (r, \theta , \varphi) = \underbrace{R_{n,l} (r)}_{radial} \underbrace{ Y^{m_l}_l (\theta , \varphi)}_{angular} \label {6.1.14}$
The wavefunctions for the hydrogen atom depend upon the three variables $r$, $\theta$, and $\varphi$ and the three quantum numbers $n$, $l$, and $m_l$. The variables give the position of the electron relative to the proton in spherical coordinates. The absolute square of the wavefunction, $| \psi (r, \theta , \varphi )|^2$, evaluated at $r$, $\theta$, and $\varphi$ gives the probability density of finding the electron inside a differential volume $d \tau$, centered at the position specified by $r$, $\theta$, and $\varphi$.
Exercise 6.4.1
Evaluate the following integrals
1. $\langle \psi (r, \theta, \varphi )| \psi (r, \theta , \varphi ) \rangle \nonumber$
2. $\langle \psi (r, \theta, \varphi )| \psi (r', \theta' , \varphi' ) \rangle \nonumber$
Answer
a. This integral is equal to one since $\psi(r, \theta, \varphi)$ are normalized eigenstates.
b. However, we can explicitly evaluate this integral for any arbitrary pair of eigenstates
\begin{align*} \langle\psi(r,\theta,\varphi)|\psi(r',\theta',\varphi')\rangle & = \int\limits_{all space}\psi^*(r,\theta,\varphi)\psi(r',\theta',\varphi')d\tau \[4pt] &=\int\limits_{0}^{\infty} dr \int\limits_{0}^{\pi}d\theta\int\limits_{0}^{2\pi}d\varphi(r^2\sin(\theta))\overbrace{\psi*(r,\theta,\varphi)}^{R_{n,l}(r)Y_{l}^{m_l}(\theta,\varphi)}\overbrace{\psi(r,\theta,\varphi)}^{R_{n',l'}(r)Y_{l'}^{m'_l}(\theta,\varphi)} \[4pt] &=\int \limits_{0}^{\infty} dr \int\limits_{0}^{\pi}d\theta \int \limits_{0}^{2\pi} d\varphi(r^2\sin(\theta))[R_{n,l}(r)Y_{l}^{m_l}(\theta,\varphi)][R_{n',l'}(r)Y_{l'}^{m'_l}(\theta,\varphi)] \[4pt] &=\left[\int\limits_{0}^{\infty}r^2[R_{n,l}(r)R_{n',l'}(r)]dr\right]\left[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\sin(\theta)[Y_{l}^{m_l}(\theta,\varphi)Y_{l'}^{m'_l}(\theta,\varphi)]d\theta d\varphi \right] \[4pt] &=\langle R_{n,l}(r)|R_{n',l'}(r)\rangle\langle Y_{l}^{m_l}(\theta,\varphi)|Y_{l'}^{m'_l}(\theta,\varphi)\rangle \[4pt] &=(\delta_{nn'}\delta_{ll'})(\delta_{ll'}\delta_{mm'}) =\delta_{nn'}\delta_{ll'}\delta_{mm'} \end{align*} \nonumber
While part a demonstrates normality of the eigenstates, part b demonstrates the orthogonality of the eigenstate (and normality too).
The quantum numbers have names:
• $n$ is called the principal quantum number,
• $l$ is called the angular momentum quantum number, and
• $m_l$ is called the magnetic quantum number because the energy in a magnetic field depends upon $m_l$.
Often $l$ is called the azimuthal quantum number because it is a consequence of the $\theta$-equation, which involves the azimuthal angle $\Theta$, referring to the angle to the zenith.
Radial Part of the Wavefunction
The asymptotic behavior (i.e., far away from the nucleus) to the radial part of the wavefunction is
$R_{asymptotic} (r) \sim \exp \left(-\dfrac {r}{n} a_0 \right) \label {6.1.15}$
where $n$ will turn out to be a quantum number and $a_0$ is the Bohr radius (~52.9 pm). Note that this function decreases exponentially with distance, in a manner similar to the decaying exponential portion of the harmonic oscillator wavefunctions, but with a different distance dependence, $r$ vs. $r^2$.
Exercise 6.4.2
What happens to the magnitude of $R_{asymptotic}(r)$ as the distance $r$ from the proton approaches infinity? Sketch a graph of the function, $R_{asymptotic}(r)$. Why might this behavior be expected for an electron in a hydrogen atom?
Answer
$R(r)=e^{-\frac{c}{n} a_{0}} \nonumber$
As $r$ approaches infinity, the exponential decay goes to zero, this is to be expected as the likelihood of an electron being found at an infinite distance away is almost zero too.
The polynomials produced by the truncation of the power series are related to the associated Laguerre polynomials, $L_n , _l(r)$, where the set of $c_i$ are constant coefficients.
$L_{n, l} (r) = \sum _{r=0}^{n-l-1} c_i r^i \label {6.1.16}$
These polynomials are identified by two indices or quantum numbers, $n$ and $l$. Physically acceptable solutions require that $n$ must be greater than or equal to $l +1$. The smallest value for $l$ is zero, so the smallest value for $n$ is 1. The angular momentum quantum number affects the solution to the radial equation because it appears in the radial differential equation, (Equation $\ref{6.1.14}$).
The $R(r)$ functions that solve the radial differential Equation $\ref{6.1.14}$, are products of the associated Laguerre polynomials and the exponential factor, multiplied by a normalization factor $(N_{n,l})$ and $\left (\dfrac {r}{a_0} \right ) ^l$.
$R (r) = N_{n,l} \left ( \dfrac {r}{a_0} \right ) ^l L_{n,l} (r) e^{-\frac {r}{n {a_0}}} \label {6.1.17}$
The decreasing exponential term overpowers the increasing polynomial term so that the overall wavefunction exhibits the desired approach to zero at large values of $r$. The first six radial functions are provided in Table 6.4.1 . Note that the functions in the table exhibit a dependence on $Z$, the atomic number of the nucleus. As discussed later in this chapter, other one electron systems have electronic states analogous to those for the hydrogen atom, and inclusion of the charge on the nucleus allows the same wavefunctions to be used for all one-electron systems. For hydrogen, $Z = 1$.
Table 6.4.1 : Radial functions for one-electron atoms and ions. $Z$ is the atomic number of the nucleus, and $\rho = \frac {Zr}{a_0}$, where $a_0$ is the Bohr radius and r is the radial variable.
n
$l$
$R_{n,l} (\rho)$
1 0 $2 \left (\dfrac {Z}{a_0} \right ) ^{3/2} e^{-\rho}$
2 0 $\dfrac {1}{2 \sqrt {2}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} (2 - \rho) e^{-\rho/2}$
2 1 $\dfrac {1}{2 \sqrt {6}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} \rho e^{-\rho/2}$
3 0 $\dfrac {2}{81 \sqrt {3}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} (27 - 18 \rho + 2\rho ^2) e^{-\rho/3}$
3 1 $\dfrac {1}{81 \sqrt {6}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} (6 \rho + \rho ^2) e^{-\rho/3}$
3 2 $\dfrac {1}{81 \sqrt {30}}\left (\dfrac {Z}{a_0} \right ) ^{3/2} \rho ^2 e^{-\rho/3}$
The constraint that $n$ be greater than or equal to $l +1$ also turns out to quantize the energy, producing the same quantized expression for hydrogen atom energy levels that was obtained from the Bohr model of the hydrogen atom.
$E_n = - \dfrac {\mu e^4}{8 \epsilon ^2_0 h^2 n^2} \nonumber$
Bohr Theory vs. Schrödinger Theory
It is interesting to compare the results obtained by solving the Schrödinger equation with Bohr’s model of the hydrogen atom. There are several ways in which the Schrödinger and Bohr models differ.
1. First, and perhaps most strikingly, the Schrödinger model does not produce well-defined orbits for the electron. The wavefunctions only give us the probability for the electron to be at various directions and distances from the proton.
2. Second, the quantization of angular momentum is different from that proposed by Bohr. Bohr proposed that the angular momentum is quantized in integer units of $\hbar$, while the Schrödinger model leads to an angular momentum of $\sqrt{(l (l +1)} \hbar$.
3. Third, the quantum numbers appear naturally during solution of the Schrödinger equation while Bohr had to postulate the existence of quantized energy states. Although more complex, the Schrödinger model leads to a better correspondence between theory and experiment over a range of applications that was not possible for the Bohr model.
Exercise 6.4.3
Explain how the Schrödinger equation leads to the conclusion that the angular momentum of the hydrogen atom can be zero, and explain how the existence of such states with zero angular momentum contradicts Bohr's idea that the electron is orbiting around the proton in the hydrogen atom.
The Three Quantum Numbers
These quantum numbers have specific values that are dictated by the physical constraints or boundary conditions imposed upon the Schrödinger equation: n must be an integer greater than 0, $l$ can have the values 0 to $n‑1$, and $m_l$ can have $2l + 1$ values ranging from $-l$ ‑ to $+l$ in unit or integer steps. The values of the quantum number $l$ usually are coded by a letter: s means 0, p means 1, d means 2, f means 3; the next codes continue alphabetically (e.g., g means $l = 4$). The quantum numbers specify the quantization of physical quantities. The discrete energies of different states of the hydrogen atom are given by n, the magnitude of the angular momentum is given by $l$, and one component of the angular momentum (usually chosen by chemists to be the z‑component) is given by $m_l$. The total number of orbitals with a particular value of $n$ is $n^2$.
Exercise 6.4.4
Consider several values for $n$, and show that the number of orbitals for each $n$ is $n^2$.
Exercise 6.4.5
Construct a table summarizing the allowed values for the quantum numbers $n$, $l$, and $m_l$ for energy levels 1 through 7 of hydrogen.
Exercise 6.4.6
The notation 3d specifies the quantum numbers for an electron in the hydrogen atom. What are the values for $n$ and $l$ ? What are the values for the energy and angular momentum? What are the possible values for the magnetic quantum number? What are the possible orientations for the angular momentum vector?
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The hydrogen atom wavefunctions, $\psi (r, \theta , \varphi )$, are called atomic orbitals. An atomic orbital is a function that describes one electron in an atom. The wavefunction with $n = 1$, $l$ = 0 is called the 1s orbital, and an electron that is described by this function is said to be “in” the ls orbital, i.e. have a 1s orbital state. The constraints on n, $l$, and $m_l$ that are imposed during the solution of the hydrogen atom Schrödinger equation explain why there is a single 1s orbital, why there are three 2p orbitals, five 3d orbitals, etc. We will see when we consider multi-electron atoms, these constraints explain the features of the Periodic Table. In other words, the Periodic Table is a manifestation of the Schrödinger model and the physical constraints imposed to obtain the solutions to the Schrödinger equation for the hydrogen atom.
Visualizing the variation of an electronic wavefunction with r, $\theta$, and $\varphi$ is important because the absolute square of the wavefunction depicts the charge distribution (electron probability density) in an atom or molecule. The charge distribution is central to chemistry because it is related to chemical reactivity. For example, an electron deficient part of one molecule is attracted to an electron rich region of another molecule, and such interactions play a major role in chemical interactions ranging from substitution and addition reactions to protein folding and the interaction of substrates with enzymes.
We can obtain an energy and one or more wavefunctions for every value of $n$, the principal quantum number, by solving Schrödinger's equation for the hydrogen atom. A knowledge of the wavefunctions, or probability amplitudes $\psi_n$, allows us to calculate the probability distributions for the electron in any given quantum level. When n = 1, the wavefunction and the derived probability function are independent of direction and depend only on the distance r between the electron and the nucleus. In Figure 6.5.1 , we plot both $\psi_1$ and $P_1$ versus $r$, showing the variation in these functions as the electron is moved further and further from the nucleus in any one direction. (These and all succeeding graphs are plotted in terms of the atomic unit of length, $a_0 = 0.529 \times 10^{-8}\, cm$.)
Two interpretations can again be given to the $P_1$ curve. An experiment designed to detect the position of the electron with an uncertainty much less than the diameter of the atom itself (using light of short wavelength) will, if repeated a large number of times, result in Figure 6.5.1 for $P_1$. That is, the electron will be detected close to the nucleus most frequently and the probability of observing it at some distance from the nucleus will decrease rapidly with increasing $r$. The atom will be ionized in making each of these observations because the energy of the photons with a wavelength much less than 10-8 cm will be greater than $K$, the amount of energy required to ionize the hydrogen atom. If light with a wavelength comparable to the diameter of the atom is employed in the experiment, then the electron will not be excited but our knowledge of its position will be correspondingly less precise. In these experiments, in which the electron's energy is not changed, the electron will appear to be "smeared out" and we may interpret $P_1$ as giving the fraction of the total electronic charge to be found in every small volume element of space. (Recall that the addition of the value of Pn for every small volume element over all space adds up to unity, i.e., one electron and one electronic charge.)
Visualizing wavefunctions and charge distributions is challenging because it requires examining the behavior of a function of three variables in three-dimensional space. This visualization is made easier by considering the radial and angular parts separately, but plotting the radial and angular parts separately does not reveal the shape of an orbital very well. The shape can be revealed better in a probability density plot. To make such a three-dimensional plot, divide space up into small volume elements, calculate $\psi ^* \psi$ at the center of each volume element, and then shade, stipple or color that volume element in proportion to the magnitude of $\psi ^* \psi$.
We could also represent the distribution of negative charge in the hydrogen atom in the manner used previously for the electron confined to move on a plane (Figure 6.5.2 ), by displaying the charge density in a plane by means of a contour map. Imagine a plane through the atom including the nucleus. The density is calculated at every point in this plane. All points having the same value for the electron density in this plane are joined by a contour line (Figure 6.5.2 ). Since the electron density depends only on r, the distance from the nucleus, and not on the direction in space, the contours will be circular. A contour map is useful as it indicates the "shape" of the density distribution.
When the electron is in a definite energy level we shall refer to the $P_n$ distributions as electron density distributions, since they describe the manner in which the total electronic charge is distributed in space. The electron density is expressed in terms of the number of electronic charges per unit volume of space, e-/V. The volume V is usually expressed in atomic units of length cubed, and one atomic unit of electron density is then e-/a03. To give an idea of the order of magnitude of an atomic density unit, 1 au of charge density e-/a03 = 6.7 electronic charges per cubic Ångstrom. That is, a cube with a length of $0.52917 \times 10^{-8}\; cm$, if uniformly filled with an electronic charge density of 1 au, would contain 6.7 electronic charges.
For every value of the energy En, for the hydrogen atom, there is a degeneracy equal to $n^2$. Therefore, for n = 1, there is but one atomic orbital and one electron density distribution. However, for n = 2, there are four different atomic orbitals and four different electron density distributions, all of which possess the same value for the energy, E2. Thus for all values of the principal quantum number n there are n2 different ways in which the electronic charge may be distributed in three-dimensional space and still possess the same value for the energy. For every value of the principal quantum number, one of the possible atomic orbitals is independent of direction and gives a spherical electron density distribution which can be represented by circular contours as has been exemplified above for the case of n = 1. The other atomic orbitals for a given value of n exhibit a directional dependence and predict density distributions which are not spherical but are concentrated in planes or along certain axes. The angular dependence of the atomic orbitals for the hydrogen atom and the shapes of the contours of the corresponding electron density distributions are intimately connected with the angular momentum possessed by the electron.
Methods for separately examining the radial portions of atomic orbitals provide useful information about the distribution of charge density within the orbitals. Graphs of the radial functions, $R(r)$, for the 1s and 2s orbitals plotted in Figure 6.5.3 . The 1s function in Figure $\PageIndex{3; left}$ starts with a high positive value at the nucleus and exponentially decays to essentially zero after 5 Bohr radii. The high value at the nucleus may be surprising, but as we shall see later, the probability of finding an electron at the nucleus is vanishingly small.
Next notice how the radial function for the 2s orbital, Figure $\PageIndex{3; right}$, goes to zero and becomes negative. This behavior reveals the presence of a radial node in the function. A radial node occurs when the radial function equals zero other than at $r = 0$ or $r = ∞$. Nodes and limiting behaviors of atomic orbital functions are both useful in identifying which orbital is being described by which wavefunction. For example, all of the s functions have non-zero wavefunction values at $r = 0$.
Exercise 6.5.1
Examine the mathematical forms of the radial wavefunctions. What feature in the functions causes some of them to go to zero at the origin while the s functions do not go to zero at the origin?
Exercise 6.5.2
What mathematical feature of each of the radial functions controls the number of radial nodes?
Exercise 6.5.3 : Radial Nodes
At what value of $r$ does the 2s radial node occur?
Exercise 6.5.4
Make a table that provides the energy, number of radial nodes, and the number of angular nodes and total number of nodes for each function with $n = 1$, $n=2$, and $n=3$. Identify the relationship between the energy and the number of nodes. Identify the relationship between the number of radial nodes and the number of angular nodes.
Answer
Energy
1.Particle in a Box (h2n2/8meL2)
2.Harmonic Oscillator((n+0.5)ℏω)
3.Hydrogen (-13.6eV/n2)
Number of Radial Nodes
(n-l-1)
Number of Angular Nodes
l = (n-1)
l : s =0
p = 1
d = 2
Total Number of Nodes
n = 1
1. 6.02 *10-38 J/L2
2. 1.5ℏω
3. -13.6 eV
- 0 0
n = 2
1. 6.02 *10-38 J/L2
2. 2.5ℏω
3. -3.4 eV
for s: 1
for p: 0
for s: 0
for p: 1
1
n = 3
1. 6.02 *10-38 J/L2
2. 3.5ℏω
3. 1.51 eV
for s: 2
for p: 1
for d: 0
for s: 0
for p:1
for d: 2
2
For a particle in a box the energy is equivalent to $E_n= 6.02 \times 10^{-38} n^2L^2$ where $n$ is any value greater than and not equal to 0 and L is the length of the box.
Radial probability densities for the 1s and 2s atomic orbitals are plotted in Figure 6.5.4 .
Radial Distribution Functions
Rather than considering the amount of electronic charge in one particular small element of space, we may determine the total amount of charge lying within a thin spherical shell of space. Since the distribution is independent of direction, consider adding up all the charge density which lies within a volume of space bounded by an inner sphere of radius $r$ and an outer concentric sphere with a radius only infinitesimally greater, say $r + \Delta r$. The area of the inner sphere is $4\pi r^2$ and the thickness of the shell is $\Delta r$. Thus the volume of the shell is $4\pi r^2 \Delta r$ and the product of this volume and the charge density P1(r), which is the charge or number of electrons per unit volume, is therefore the total amount of electronic charge lying between the spheres of radius $r$ and $r + \Delta r$. The product $4\pi r^2P_n$ is given a special name, the radial distribution function.
Volume Element for a Shell in Spherical Coordinates
The reader may wonder why the volume of the shell is not taken as:
$\dfrac{4}{3} \pi \left[ (r + \Delta r)^3 -r^3 \right] \nonumber$
the difference in volume between two concentric spheres. When this expression for the volume is expanded, we obtain
$\dfrac{4}{3} \pi \left(3r^2 \Delta r + 3r \Delta r^2 + \Delta r^3\right) \nonumber$
and for very small values of $\Delta r$ the $3r \Delta r^2$ and $\Delta r^3$ terms are negligible in comparison with $3r^2\Delta r$. Thus for small values of $\Delta r$, the two expressions for the volume of the shell approach one another in value and when $\Delta r$ represents an infinitesimal small increment in $r$ they are identical.
The radial distribution function is plotted in Figure 6.5.5 for the ground state of the hydrogen atom.
The curve passes through zero at $r = 0$ since the surface area of a sphere of zero radius is zero. As the radius of the sphere is increased, the volume of space defined by $4 \pi r^2Dr$ increases. However, as shown in Figure 6.5.4 , the absolute value of the electron density at a given point decreases with $r$ and the resulting curve must pass through a maximum. This maximum occurs at $r_{max} = a_0$. Thus more of the electronic charge is present at a distance $a_o$, out from the nucleus than at any other value of $r$. Since the curve is unsymmetrical, the average value of $r$, denoted by $\bar{r}$, is not equal to $r_{max}$. The average value of $r$ is indicated on the figure by a dashed line. A "picture" of the electron density distribution for the electron in the $n = 1$ level of the hydrogen atom would be a spherical ball of charge, dense around the nucleus and becoming increasingly diffuse as the value of $r$ is increased.
The radial distribution function gives the probability density for an electron to be found anywhere on the surface of a sphere located a distance $r$ from the proton. Since the area of a spherical surface is $4 \pi r^2$, the radial distribution function is given by $4 \pi r^2 R(r) ^* R(r)$.
Radial distribution functions are shown in Figure 6.5.6 . At small values of $r$, the radial distribution function is low because the small surface area for small radii modulates the high value of the radial probability density function near the nucleus. As we increase $r$, the surface area associated with a given value of $r$ increases, and the $r^2$ term causes the radial distribution function to increase even though the radial probability density is beginning to decrease. At large values of $r$, the exponential decay of the radial function outweighs the increase caused by the $r^2$ term and the radial distribution function decreases.
Example 6.5.1 :
Calculate the probability of finding a 1s hydrogen electron being found within distance $2a_o$ from the nucleus.
Solution
Note the wavefunction of hydrogen 1s orbital which is
$ψ_{100}= \dfrac{1}{\sqrt{π}} \left(\dfrac{1}{a_0}\right)^{3/2} e^{-\rho} \nonumber$
with $\rho=\dfrac{r}{a_0}$.
The probability of finding the electron within $2a_0$ distance from the nucleus will be:
$prob= \underbrace{\int_{0}^{\pi} \sin \theta \, d\theta}_{over\, \theta} \, \overbrace{ \dfrac{1}{\pi a_0^3} \int_{0}^{2a_0} r^2 e^{-2r/a_0} dr}^{over\, r} \, \underbrace{ \int_{0}^{2\pi} d\phi }_{over\, \phi } \nonumber$
Since $\int_0^{\pi} \sin \theta d\theta=2$ and $\int_0^{2\pi} d\phi=2\pi$, we have
\begin{align*} prob &=2 \times 2\pi \times \dfrac{1}{\pi a_0^3} \int_0^2a_0 (-a_0/2)r^2 d e^{-2r/a_0} \[4pt]&=\dfrac{4}{a_0^3}\left(-\dfrac{a_0}{2}\right) (r^2 e^{-2r/a_0} |_0^{2a_0} - \int_0^{2a_0} 2r e^{-2r/a_0} dr) \[4pt]&= -\dfrac{2}{a_0^2} [(2a_0)^2 e^{-4}-0-2\int_0^{2a_0} r \left(-\dfrac{a_0}{2}\right) d e^{-2r/a_0} ] \[4pt]&=-\dfrac{2}{a_0^2}4a_0^2 e^{-4} +\dfrac{4}{a_0^2}(-\dfrac{a_0}{2}) (r e^{-2r/a_0} |_0^{2a_0}-\int_0^{2a_0} e^{-2r/a_0} dr ) \[4pt]&=-8e^{-4}-\dfrac{2}{a_0} \left[2a_0e^{-4}-0-(-\dfrac{a_0}{2})e^{-2r/a_0} |_0^{2a_0} \right] \[4pt]&=-8e^{-4}-4e^{-4}-e^{2r/a_0} |_0^{2a_0} \[4pt]&=-12 e^{-4}-(e^{-4}-1)=1-13e^{-4}=0.762 \end{align*} \nonumber
There is a 76.2% probability that the electrons will be within $2a_o$ of the nucleus in the 1s eigenstate.
Summary
This completes the description of the most stable state of the hydrogen atom, the state for which $n = 1$. Before proceeding with a discussion of the excited states of the hydrogen atom we must introduce a new term. When the energy of the electron is increased to another of the allowed values, corresponding to a new value for $n$, $y_n$ and $P_n$ change as well. The wavefunctions $y_n$ for the hydrogen atom are given a special name, atomic orbitals, because they play such an important role in all of our future discussions of the electronic structure of atoms. In general the word orbital is the name given to a wavefunction which determines the motion of a single electron. If the one-electron wavefunction is for an atomic system, it is called an atomic orbital.
Do not confuse the word orbital with the classical word and notion of an orbit. First, an orbit implies the knowledge of a definite trajectory or path for a particle through space which in itself is not possible for an electron. Secondly, an orbital, like the wavefunction, has no physical reality but is a mathematical function which when squared gives the physically measurable electron density distribution.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/06%3A_The_Hydrogen_Atom/6.05%3A_s-orbitals_are_Spherically_Symmetric.txt
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Learning Objectives
• To relate the classical orbital angular momentum for an particle to the quantum equivalent
• Characterize the mangnitude and orientation of orbital angular momentum for an electron in terms of quantum numbers
Classical Orbital Angular Momentum
The physical quantity known as angular momentum plays a dominant role in the understanding of the electronic structure of atoms. To gain a physical picture and feeling for the angular momentum it is necessary to consider a model system from the classical point of view. The simplest classical model of the hydrogen atom is one in which the electron moves in a circular orbit with a constant speed or angular velocity (Figure 6.6.1 ). Just as the linear momentum $m\vec{v}$ plays a dominant role in the analysis of linear motion, so angular momentum ($L$) plays the central role in the analysis of a system with circular motion as found in the model of the hydrogen atom.
In Figure 6.6.1 , $m$ is the mass of the electron, $\vec{v}$ is the linear velocity (the velocity the electron would possess if it continued moving at a tangent to the orbit) and $r$ is the radius of the orbit. The linear velocity $\vec{v}$ is a vector since it possesses at any instant both a magnitude and a direction in space. Obviously, as the electron rotates in the orbit the direction of $\vec{v}$ is constantly changing, and thus the linear momentum $m\vec{v}$ is not constant for the circular motion. This is so even though the speed of the electron (i.e, the magnitude of $\vec{v}$ which is denoted by $|\vec{v}|$) remains unchanged. According to Newton's second law, a force must be acting on the electron if its momentum changes with time. This is the force which prevents the electron from flying on tangent to its orbit. In an atom the attractive force which contains the electron is the electrostatic force of attraction between the nucleus and the electron, directed along the radius r at right angles to the direction of the electron's motion.
The angular momentum, like the linear momentum, is a vector and is defined as follows:
$|\vec{L}| = m \nu r \nonumber$
The angular momentum vector $\vec{L}$ is directed along the axis of rotation. From the definition it is evident that the angular momentum vector will remain constant as long as the speed of the electron in the orbit is constant ($|\vec{v}|$ remains unchanged) and the plane and radius of the orbit remain unchanged. Thus for a given orbit, the angular momentum is constant as long as the angular velocity of the particle in the orbit is constant. In an atom the only force on the electron in the orbit is directed along $r$; it has no component in the direction of the motion. The force acts in such a way as to change only the linear momentum. Therefore, while the linear momentum is not constant during the circular motion, the angular momentum is. A force exerted on the particle in the direction of the vector $\vec{v}$ would change the angular velocity and the angular momentum. When a force is applied which does change $\vec{L}$, a torque is said to be acting on the system. Thus angular momentum and torque are related in the same way as are linear momentum and force.
Quantum Angular Momentum
The important point of the above discussion is that both the angular momentum and the energy of an atom remain constant if the atom is left undisturbed. Any physical quantity which is constant in a classical system is both conserved and quantized in a quantum mechanical system. Thus both the energy and the angular momentum are quantized for an atom.
Any physical quantity which is constant in a classical system is both conserved and quantized in a quantum mechanical system.
There is a quantum number, denoted by $l$, which governs the magnitude of the angular momentum, just as the quantum number $n$ determines the energy. The magnitude of the angular momentum may assume only those values given by:
$|L| = \sqrt{l(l+1)} \hbar \label{4}$
with $l = 0, 1, 2, 3, ... n-1$.
Furthermore, the value of n limits the maximum value of the angular momentum as the value of l cannot be greater than n - 1. For the state n = 1 discussed above, $l$ may have the value of zero only. When n = 2, l may equal 0 or 1, and for n = 3, l = 0 or 1 or 2, etc. When l = 0, it is evident from Equation $\ref{4}$ that the angular momentum of the electron is zero. The atomic orbitals which describe these states of zero angular momentum are called s orbitals. The s orbitals are distinguished from one another by stating the value of n, the principal quantum number. They are referred to as the 1s, 2s, 3s, etc., atomic orbitals.
The preceding discussion referred to the 1s orbital since for the ground state of the hydrogen atom $n = 1$ and $l = 0$. This orbital, and all s orbitals in general, predict spherical density distributions for the electron as discussed previously.
It is common usage to refer to an electron as being "in" an orbital even though an orbital is, but a mathematical function with no physical reality. To say an electron is in a particular orbital is meant to imply that the electron is in the quantum state which is described by that orbital. For example, when the electron is in the 2s orbital the hydrogen atom is in a state for which $n = 2$ and $l = 0$.
Comparing these results with those for the 1s orbital in Figure 6.6.2 we see that as $n$ increases the average value of $r$ increases. This agrees with the fact that the energy of the electron also increases as $n$ increases. The increased energy results in the electron being on the average pulled further away from the attractive force of the nucleus. As in the simple example of an electron moving on a line, nodes (values of $r$ for which the electron density is zero) appear in the probability distributions. The number of nodes increases with increasing energy and equals $n - 1$.
When the electron possesses angular momentum the density distributions are no longer spherical. In fact for each value of $l$, the electron density distribution assumes a characteristic shapes in Figure 6.6.2 .
When $l = 1$, the orbitals are called p orbitals. In this case the orbital and its electron density are concentrated along a line (axis) in space. The 2p orbital or wavefunction is positive in value on one side and negative in value on the other side of a plane which is perpendicular to the axis of the orbital and passes through the nucleus. The orbital has a node in this plane, and consequently an electron in a 2p orbital does not place any electronic charge density at the nucleus. The electron density of a 1s orbital, on the other hand, is a maximum at the nucleus. The same diagram for the 2p density distribution is obtained for any plane which contains this axis. Thus in three dimensions the electron density would appear to be concentrated in two lobes, one on each side of the nucleus, each lobe being circular in cross section Figure 6.6.3 .
An electron possesses orbital angular momentum has a density distributions is no longer spherical.
The $m_l$ Quantum Number and Magnetic Fields
The magnetic quantum number, designated by the letter $m_l$, is the third quantum numbers which describe the unique quantum state of an electron. The magnetic quantum number distinguishes the orbitals available within a subshell, and is used to calculate the azimuthal component of the orientation of the orbital in space. As with our discussion of rigid rotors, the quantum number $m_l$ refers to the projection of the angular momentum in this arbitrarily chosen direction, conventionally called the $z$ direction or quantization axis. $L_z$, the magnitude of the angular momentum in the z direction, is given by the formula
$L_z = m_l \hbar \nonumber$
The quantum number $m_l$ refers, loosely, to the direction of the angular momentum vector. The magnetic quantum number $m_l$ only affects the electron's energy if it is in a magnetic field because in the absence of one, all spherical harmonics corresponding to the different arbitrary values of $m_l$ are equivalent. The magnetic quantum number determines the energy shift of an atomic orbital due to an external magnetic field (this is called the Zeeman effect) - hence the name magnetic quantum number. However, the actual magnetic dipole moment of an electron in an atomic orbital arrives not only from the electron angular momentum, but also from the electron spin, expressed in the spin quantum number, which is the fourth quantum number. $m_s$ and discussed in the next chapter.
Which $m_l$ Number Corresponds to which p-Orbital?
The answer is complicated; while $m_l=0$ corresponds to the $p_z$, the orbitals for $m_l=+1$ and $m_l=−1$ lie in the xy-plane (see Spherical Harmonics), but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the math easier, but graphs in the Cartesian coordinates make more intuitive sense for humans. The $p_x$ and $p_y$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $xy$ (this was discussed previously). Thus, it is not possible to directly correlate the values of $m_l=±1$ with specific orbitals. The notion that we can do so is sometimes presented in introductory courses to make a complex mathematical model just a little bit simpler and more intuitive, but it is incorrect.
The three wavefunctions for $n=2$ and $l=1$ are as follows.
\begin{align} |\psi_{2,1,0} \rangle &=r \cos θR(r) \[4pt] |\psi_{2,1,+1} \rangle &=−\dfrac{r}{2} \sinθ e^{iϕ} R(r) \[4pt] |\psi_{2,1,-1} \rangle &=+\dfrac{r}{2} \sinθ e^{-iϕ} R(r) \end{align} \nonumber
The notation is $|\psi_{n,l,m_l} \rangle$ with $R(r)$ is the radial component of this wavefuction, $θ$ is the angle with respect to the z-axis and $ϕ$ is the angle with respect to the $xz$-plane.
$R(r)=\sqrt{\dfrac{Z^5}{32\pi a_0^5}}\mathrm{e}^{-Zr/2a_0} \nonumber$
in which $Z$ is the atomic number (or probably better nuclear charge) and $a_0$ is the Bohr radius.
In switching from spherical to Cartesian coordinates, we make the substitution $z=r \cosθ$, so:
$|\psi_{2,1,0} \rangle =z R(r) \nonumber$
This is $\psi_{2p_z}$ since the value of $\psi$ is dependent on $z$: when $z=0$; $\psi =0$, which is expected since $z=0$ describes the $xy$-plane.
The other two wavefunctions are degenerate in the $xy$-plane. An equivalent statement is that these two orbitals do not lie on the x- and y-axes, but rather bisect them. Thus it is typical to take linear combinations of them to make the equation look prettier. If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution (Section 2.4). We can do this because of the linearity of the Schrödinger equation.
In the equations below, we're going to make use of some trigonometry, notably Euler's formula:
\begin{align} \mathrm{e}^{\mathrm{i}\phi} &=\cos{\phi}+\mathrm{i}\sin{\phi}\[4pt] \sin{\phi} &= \dfrac{\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi}}{2\mathrm{i}}\[4pt] \cos{\phi} &= \dfrac{\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi}}{2} \end{align} \nonumber
We're also going to use $x=\sin θ\cos ϕ$ and $y=\sin θ \sinϕ$.
\begin{align*} \psi_{2p_x} &=\dfrac{1}{\sqrt{2}}\left(\psi_{2,1,+1}-\psi_{2,1,-1}\right) \[4pt] &=\dfrac{1}{2}\left(\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta} f(r) \[4pt] &=r\sin{\theta}\cos{\phi}f(r)=xf(r) \[4pt] \psi_{2p_y} &=\dfrac{\mathrm{i}}{\sqrt{2}}\left(\psi_{2,1,+1}+\psi_{2,1,-1}\right)\[4pt] &=\dfrac{1}{2\mathrm{i}}\left(\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)\[4pt] &=r\sin{\theta}\sin{\phi}f(r)=yf(r)\ \end{align*}
So, while $m_l=0$ corresponds to $|p_z \rangle$, $m_l=+1$ and $m_l=−1$ cannot be directly assigned to either $|p_x \rangle$ or $|p_y \rangle$, but rather a combination of $|p_x \rangle$ and $|p_y \rangle$. An alternative description is that $m_l=+1$ might correspond to $(|p_x \rangle\ + |p_y \rangle )$ and $m_l=−1$ might correspond to $(|p_x \rangle\ - |p_y \rangle)$.
d-Orbitals (even higher angular momenta wavefunctions)
When $l = 2$, the orbitals are called d orbitals and Figure 6.6.4 shows the contours in a plane for a 3d orbital and its density distribution. Notice that the density is again zero at the nucleus and that there are now two nodes in the orbital and in its density distribution. As the angular momentum of the electron increases, the density distribution becomes increasingly concentrated along an axis or in a plane in space. Only electrons in $s$ orbitals with zero angular momentum give spherical density distributions and in addition place charge density at the position of the nucleus.
As with the p-orbitals, the only d-orbital that a specific $m_l$ can be ascribed is the $d_{z^2}$ orbitals with $m_l=0$. The rest are linear combinations of the hydrogen atom wavefunctions with complex spherical harmonic angular components.
There seems to be neither rhyme nor reason for the naming of the states corresponding to the different values of $\ell$ (s, p, d, f for l = 0, 1, 2, 3). This set of labels had its origin in the early work of experimental atomic spectroscopy. The letter s stood for sharp, p for principal, d for diffuse and f for fundamental in characterizing spectral lines. From the letter f onwards the naming of the orbitals is alphabetical $l = 4,5,6 \rightarrow g,h,i, ....$.
We have not as yet accounted for the full degeneracy of the hydrogen atom orbitals which we stated earlier to be $n^2$ for every value of $n$. For example, when $n = 2$, there are four distinct atomic orbitals. The remaining degeneracy is again determined by the angular momentum of the system. Since angular momentum like linear momentum is a vector quantity, we may refer to the component of the angular momentum vector which lies along some chosen axis. For reasons we shall investigate, the number of values a particular component can assume for a given value of $l$ is ($2l + 1$). Thus when $l = 0$, there is no angular momentum and there is but a single orbital, an s orbital. When $l = 1$, there are three possible values for the component ($2 \times 1 + 1$) of the total angular momentum which are physically distinguishable from one another. There are, therefore, three p orbitals. Similarly there are five d orbitals, ($2 \times 2+1$), seven f orbitals, ($2 \times 3 +1$), etc. All of the orbitals with the same value of $n$ and $l$, the three 2p orbitals for example, are similar but differ in their spatial orientations.
To gain a better understanding of this final element of degeneracy, we must consider in more detail what quantum mechanics predicts concerning the angular momentum of an electron in an atom.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/06%3A_The_Hydrogen_Atom/6.06%3A_Orbital_Angular_Momentum_and_the_p-Orbitals.txt
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Learning Objectives
• Adding electrons to the quantum hydrogen atom results in analytically unsolvable Schrödinger Equations (they exist, we just do not have analytical forms for them)
• A basic aspect of the corresponding multi-electron Hamiltonians is that they are NOT separable with respect to the spatial coordinate of each electron
• The solutions to multi-electron Schrödinger Equations are called multi-electron wavefunctions and they are often approximated as a product of single-electron wavefunctions (called the orbital approximation).
Multi-electron Hamiltonians
The second element in the periodic table provides our first example of a quantum-mechanical problem which cannot be solved exactly. Nevertheless, as we will show, approximation methods applied to helium can give accurate solutions in perfect agreement with experimental results. In this sense, it can be concluded that quantum mechanics is correct for atoms more complicated than hydrogen. By contrast, the Bohr theory failed miserably in attempts to apply it beyond the hydrogen atom.
Figure 6.7.1 shows a schematic representation of a helium atom with two electrons whose coordinates are given by the vectors $r_1$ and $r_2$. The electrons are separated by a distance $r_{12} = |r_1-r_2|$. The origin of the coordinate system is fixed at the nucleus. As with the hydrogen atom, the nuclei for multi-electron atoms are so much heavier than an electron that the nucleus is assumed to be the center of mass. Fixing the origin of the coordinate system at the nucleus allows us to exclude translational motion of the center of mass from our quantum mechanical treatment.
The Hamiltonian operator for the hydrogen atom serves as a reference point for writing the Hamiltonian operator for atoms with more than one electron. Start with the same general form we used for the hydrogen atom Hamiltonian
$\hat {H} = \hat {T} + \hat {V} \label {6.7.1}$
Include a kinetic energy term for each electron and a potential energy term for the attraction of each negatively charged electron for the positively charged nucleus and a potential energy term for the mutual repulsion of each pair of negatively charged electrons. The He atom Hamiltonian is
$\hat {H} = -\dfrac {\hbar ^2}{2m_e} (\nabla ^2_1 + \nabla ^2_2) + V_1 (r_1) + V_2 (r_2) + V_{12} (r_{12}) \label {6.7.2}$
where
$V_1(r_1) = -\dfrac {2e^2}{4 \pi \epsilon _0 r_1} \label {6.7.3}$
$V_2(r_2) = -\dfrac {2e^2}{4 \pi \epsilon _0 r_2} \label {6.7.4}$
$V_{12}(r_{12}) = \dfrac {e^2}{4 \pi \epsilon _0 r_{12}} \label {6.7.5}$
The two-electron Hamiltonian in Equation $\ref{6.7.2}$ can be extended to any atom or ion by replacing the He nuclear charge of +2 with a general charge $Z$; e.g.
$V_1(r_1) = -\dfrac {Ze^2}{4 \pi \epsilon _0 r_1} \label {6.7.6}$
and including terms for the additional electrons. The subsequent multi-electron atom with $n$ electron is
$\hat {H} = \underbrace{-\dfrac {\hbar ^2}{2m_e} \sum_i^n \nabla ^2_i}_{\text{Kinetic Energy}} + \underbrace{\sum_i^n V_i (r_i)}_{\text{Coulombic Attraction}} + \underbrace{ \sum_{i \ne j}^{n,n} V_{ij} (r_{ij})}_{\text{electron-electron Repulsion}} \label {6.7.7}$
This multi-electron Hamiltonian is qualitatively similar to the 2-electron Hamiltonian (Equation \ref{6.7.1}) with each electron having its own kinetic energy and nuclear potential energy terms (Equations $\ref{6.7.3}$ and $\ref{6.7.4}$). The other big difference between single electron systems and multi-electron systems is the presence of the $V_{ij}(r_{ij})$ terms which contain $1/r_{ij}$, where $r_{ij}$ is the distance between electrons $i$ and $j$. These terms account for the electron-electron repulsion that we expect between like-charged particles.
Exercise 6.7.1 : Multi-electron atom Hamiltonians
For the generalized multi-electron atom Hamiltonian (Equation $\ref{6.7.7}$):
1. Explain the origin of each of the three summations.
2. What do these summations over (i.e., what is the origin of the summing index)?
3. Write expressions for $V_i(r_i)$ and $V_{ij}(r_{ij})$.
Exercise 6.7.2 : Boron Atom
Boron is the fifth element of the periodic table (Z=5) and is located in Group 13.
1. Write the multi-electron Hamiltonian for a $\ce{^{11}B}$ atom.
2. Would it be any different for a $\ce{^{11}B^{+}}$ ion?
3. Would it be any different for a $\ce{^{10}B}$ atom?
Answer
a.
$\hat {H}_{\ce{B}}(\vec r_1,\vec r_2,\vec r_3,\vec r_4,\vec r_5) = -\dfrac {\hbar ^2}{2m_e} \sum_i^{5} \nabla ^2_i + \sum_i^{5} \dfrac {-5e^2}{4 \pi \epsilon _0 r_i} + \sum_{i \ne j}^{5,5} \dfrac {e^2}{4 \pi \epsilon _0 r_{ij}} \nonumber$
which expands to 20 terms
\begin{align*} \hat {H}_B(\vec r_1,\vec r_2,\vec r_3,\vec r_4,\vec r_5) = &-\dfrac {\hbar ^2}{2m_e} \nabla ^2_1 -\dfrac {\hbar ^2}{2m_e} \nabla ^2_2 -\dfrac {\hbar ^2}{2m_e} \nabla ^2_3 -\dfrac {\hbar ^2}{2m_e} \nabla ^2_4 -\dfrac {\hbar ^2}{2m_e} \nabla ^2_5 \[4pt] &- \dfrac {-5e^2}{4 \pi \epsilon _0 r_1} - \dfrac {-5e^2}{4 \pi \epsilon _0 r_2} - \dfrac {-5e^2}{4 \pi \epsilon _0 r_3} - \dfrac {-5e^2}{4 \pi \epsilon _0 r_4} - \dfrac {-5e^2}{4 \pi \epsilon _0 r_5} \[4pt] &+ \dfrac {e^2}{4 \pi \epsilon _0 r_{12}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{13}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{14}} +\dfrac {e^2}{4 \pi \epsilon _0 r_{15}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{23}} +\dfrac {e^2}{4 \pi \epsilon _0 r_{24}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{25}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{34}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{35}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{45}} \end{align*} \nonumber
b. Yes, $\ce{^{11}B^{+}}$ has one less electron than $\ce{^{11}B}$. Its Hamiltonian is
$\hat {H}_{\ce{B^+}}(\vec r_1,\vec r_2,\vec r_3,\vec r_4 ) = -\dfrac {\hbar ^2}{2m_e} \sum_i^{4} \nabla ^2_i + \sum_i^{4} \dfrac {-5e^2}{4 \pi \epsilon _0 r_i} + \sum_{i \ne j}^{4,4} \dfrac {e^2}{4 \pi \epsilon _0 r_{ij}} \nonumber$
or expanded to 14 terms
\begin{align*} \hat {H}_B(\vec r_1,\vec r_2,\vec r_3,\vec r_4) = &-\dfrac {\hbar ^2}{2m_e} \nabla ^2_1 -\dfrac {\hbar ^2}{2m_e} \nabla ^2_2 -\dfrac {\hbar ^2}{2m_e} \nabla ^2_3 -\dfrac {\hbar ^2}{2m_e} \nabla ^2_4 \[4pt] &- \dfrac {-5e^2}{4 \pi \epsilon _0 r_1} - \dfrac {-5e^2}{4 \pi \epsilon _0 r_2} - \dfrac {-5e^2}{4 \pi \epsilon _0 r_3} - \dfrac {-5e^2}{4 \pi \epsilon _0 r_4} \[4pt] &+ \dfrac {e^2}{4 \pi \epsilon _0 r_{12}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{13}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{14}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{23}} +\dfrac {e^2}{4 \pi \epsilon _0 r_{24}} + \dfrac {e^2}{4 \pi \epsilon _0 r_{34}} \end{align*} \nonumber
c. No effect. Changing the number of neutrons in the nucleus does not affect kinetic nor potential energies of the electrons. The Hamilitonian for $\ce{^{10}B}$ is identical to $\ce{^{11}B}$. This is technically correct for this level of discussion, but as we we will see in later, if we expand the Hamiltonian with hyperfine structure the number of neutrons can play a role.
Multi-electron Wavefunctions and the Orbital Approximation
Given what we have learned from the previous quantum mechanical systems we’ve studied, we predict that exact solutions to the multi-electron Schrödinger equation in Equation $\ref{6.7.7}$ would consist of a family of multi-electron wavefunctions, each with an associated energy eigenvalue. These wavefunctions and energies would describe the ground and excited states of the multi-electron atom, just as the hydrogen wavefunctions and their associated energies describe the ground and excited states of the hydrogen atom. We would predict quantum numbers to be involved, as well.
The fact that electrons interact through their Coulomb repulsion means that an exact wavefunction for a multi-electron system would be a single function that depends simultaneously upon the coordinates of all the electrons; i.e., a multi-electron wavefunction,
$\psi (\vec r _1, \vec r_2, \vec r_3, \vec r_4, \cdots \vec r_n) \nonumber$
The modulus squared of such a wavefunction would describe the probability of finding the electrons (though not specific ones) at a designated volume ($V$) in the atom.
$p(V)= \int_{V} | \psi (\vec r _1, \vec r_2, \vec r_3, \vec r_4, \cdots \vec r_n) |^2 d\tau \nonumber$
All of the electrons are described simultaneously by a multi-electron wavefunction, so the total amount of electron density represented by the wavefunction equals the number of electrons in the atom.
$\int_{\text{all space}} | \psi (\vec r _1, \vec r_2, \vec r_3, \vec r_4, \cdots \vec r_i) |^2 d\tau = n \nonumber$
Unfortunately, the Coulomb repulsion terms (Equation $\ref{6.7.5}$) make it impossible to find an exact solution to the Schrödinger equation for many-electron atoms and molecules even for two electrons atoms. We have to rely on approximations and the orbital approximation is central to basic chemistry concepts.
The Orbital Approximation
The most basic approximations to the exact solutions to a multi-electron atom Hamiltonian, $\hat{H}$, (Equation \ref{6.7.7}) involve writing a multi-electron wavefunction ($\psi (r_1, r_2, \cdots , r_n)$) as a simple product of single-electron wavefunctions ($\varphi _i (r_i)$ ):
$\psi (r_1, r_2, \cdots , r_n) \approx \varphi _1 (r_1) \varphi _2 (r_2) \cdots \varphi _n(r_n) \label {6.7.8}$
or in Dirac notation
$|\psi (r_1, r_2, \cdots , r_n) \rangle \approx | \varphi _1 (r_1) \rangle |\varphi _2 (r_2) \rangle \cdots |\varphi _n(r_n) \rangle\label {6.7.9}$
The energy of the atom in the state associated with a specific multi-electron wavefunction ($E$) is obtained from the multi-electron Schrödinger Equation
$\hat{H} \psi (r_1, r_2, \cdots , r_n) = E \psi (r_1, r_2, \cdots , r_n) \nonumber$
Within the approximation in Equation \ref{6.7.8}, $E$ can be expressed sum of the energies of the one-electron components ($\epsilon_i$).
$E \approx \sum_i \epsilon_i \nonumber$
This is called the orbital approximation.
By writing the multi-electron wavefunction as a product of single-electron functions (Equations \ref{6.7.8} or \ref{6.7.9}), we conceptually transform a multi-electron atom into a collection of individual electrons located in individual orbitals whose spatial characteristics and energies can be separately identified. For atoms, these single-electron wavefunctions are called atomic orbitals and resemble the wavefunctions for hydrogen-like atoms. For molecules, as we will see in the following chapters, these are called molecular orbitals. While a great deal can be learned from such an analysis, it is important to keep in mind that such a discrete, compartmentalized picture of the electrons is an approximation, albeit a powerful one.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/06%3A_The_Hydrogen_Atom/6.07%3A_The_Helium_Atom_Cannot_Be_Solved_Exactly.txt
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Solutions to select questions can be found online.
6.5
$\int_{-1}^{1} T_n(x)T_m(x) \frac{1}{\sqrt{1-x^2}}dx= \begin{cases} 0, & n \neq m \ \pi, & n=m=0 \ \pi/2, & n=m \neq 0 \end{cases} \nonumber$
First 6 Chebyshev Polynomials
\begin{align*} T_0(x) &=1 \[4pt] T_1(x)&=x \[4pt] T_2(x)&=2x^2-1 \[4pt] T_3(x) &=4x^3-3x \[4pt] T_4(x) &=8x^4-8x^2+1 \[4pt] T_5(x) &=16x^5-20x^3+5x \end{align*}
Use the orthogonality of Chebyshev polynomials to determine what the following polynomials are equal to
1. $\int_{-1}^{1} x^2 \frac{dx}{\sqrt{1-x^2}}$
2. $\int_{-1}^{1} 4x^3-2x \frac{dx}{\sqrt{1-x^2}}$
3. $\int_{-1}^{1} 1 \frac{dx}{\sqrt{1-x^2}}$
4. $\int_{-1}^{1} 4x^4-4x^2+1 \frac{dx}{\sqrt{1-x^2}}$
Solution
1. x^2= T1*T1; therefore the answer is π /2
2. here the following polynomial is not a product of either Chebyshev polynomials; therefore, answer is doesn't follow orthogonality conditions
3. 1=T0*T0; therefore, answer is π
4. x^4-4x^2+1= T2*T2; therefore the answer is π/2
6.6
Use Eq. 6.47 to generate the radial functions $R_{nl}\left(r\right)$ for $n=1,2$.
Solution
$R_{10}\left(r\right)={\left\{\dfrac{\left(1-0-1\right)!}{2\left(1\right){\left[\left(1+0\right)!\right]}^3}\right\}}^{\dfrac{1}{2}}{\left(\dfrac{2}{1a_0}\right)}^{\dfrac{0+3}{2}}r^0e^{-\dfrac{r}{1a_0}}L^1_1\left(\dfrac{2r}{1a_0}\right)\nonumber$
$R_{10}\left(r\right)=-{\left\{\dfrac{1}{2}\right\}}^{\dfrac{1}{2}}{\left(\dfrac{2}{a_0}\right)}^{\dfrac{3}{2}}e^{-\dfrac{r}{a_0}}\nonumber$
$R_{20}\left(r\right)={\left\{\dfrac{\left(2-0-1\right)!}{2\left(2\right){\left[\left(2+0\right)!\right]}^3}\right\}}^{\dfrac{1}{2}}{\left(\dfrac{2}{2a_0}\right)}^{\dfrac{0+3}{2}}r^0e^{-\dfrac{r}{2a_0}}L^1_2\left(\dfrac{2r}{2a_0}\right)\nonumber$
$R_{20}\left(r\right)={\left\{\dfrac{1}{32}\right\}}^{\dfrac{1}{2}}{\left(\dfrac{1}{a_0}\right)}^{\dfrac{3}{2}}e^{-\dfrac{r}{2a_0}}\left(-2!\left(2-\dfrac{r}{a_0}\right)\right)\nonumber$
$R_{20}\left(r\right)=-2{\left\{\dfrac{1}{32}\right\}}^{\dfrac{1}{2}}{\left(\dfrac{1}{a_0}\right)}^{\dfrac{3}{2}}e^{-\dfrac{r}{2a_0}}\left(\left(2-\dfrac{r}{a_0}\right)\right)\nonumber$
$R_{21}\left(r\right)={\left\{\dfrac{\left(2-1-1\right)!}{2\left(2\right){\left[\left(2+1\right)!\right]}^3}\right\}}^{\dfrac{1}{2}}{\left(\dfrac{2}{2a_0}\right)}^{\dfrac{1+3}{2}}r^1e^{-\dfrac{r}{2a_0}}L^3_3\left(\dfrac{2r}{2a_0}\right)\nonumber$
$R_{21}\left(r\right)=-6{\left\{\dfrac{1}{864}\right\}}^{\dfrac{1}{2}}{\left(\dfrac{1}{a_0}\right)}^2r^1e^{-\dfrac{r}{2a_0}}\nonumber$
6.29
Compare $\psi_{310}$ and $\psi_{311}$.
Hint: What do the subscripts tell you about the wave function? What do they denote?
Solution
The first subscript tells you the quantum number $n$. The second denotes the angular momentum $l$. The last denotes the magnetic spin number $m_l$. These two functions have the same $n$ values, and thus they are degenerate.
6.30
What is the probability density of the 3p orbital by evaluating
$\left (\sum_{m=-1}^{1}\psi_{31m}^{2}\right )\nonumber$
Solution
\begin{align*} \sum_{m=-1}^{1}\psi_{31m}^{2} &=\left (\dfrac{2}{6561\pi}\right )\left (\dfrac{z^{3}}{a_o^{3}}\right )\sigma^{3}\left (6-\sigma\right )^{2}\exp^{\dfrac{-2\sigma}{3}} \left (\cos^{2} \theta+\sin^{2} \theta \cos^{2} \phi + \sin^{2} \theta \sin^{2} \phi \right ) \[4pt] \sum_{m=-1}^{1}\psi_{31m}^{2} &=\left (\dfrac{2z^{3}\sigma^{2} \left (6-\sigma\right )^{2}\exp^{\dfrac{-2\sigma}{3}}}{6561\pi a_o^{3}}\right ) \left (\cos^{2}\theta+\sin^{2}\theta \left(\cos^{2}\phi+\sin^{2}\phi \right ) \right)\[4pt] \sum_{m=-1}^{1}\psi_{31m}^{2} &=\left (\dfrac{2z^{3}\sigma^{2}\left (6-\sigma\right )^{2}\exp^{\dfrac{-2\sigma}{3}}}{6561\pi a_o^{3}}\right )\end{align*}
6.34
Find the energy, and wavefunction for a single electron located in the 2p orbital of the hydrogen atom. Include all possible wavefunctions.
Solution
Identify the quantum numbers for the electron of interest (in our case, $n=2$; $l =1$). Energy of the electron can be defined as
$E_n = \dfrac{-m_ee^4}{8n^2\epsilon_o^2h^2} \nonumber$
this leads us to
$E_2 = \dfrac{-m_ee^4}{32\epsilon_o^2h^2} \nonumber$
we have two possible wave functions
$\Psi_{210}= \dfrac{1}{\sqrt{32}} (\dfrac{z}{a_o})^{3/2}\sigma/e^{-\sigma/2} \cos{\theta} \nonumber$
and
$\Psi_{21\pm1}= \dfrac{1}{\sqrt{32}} (\dfrac{z}{a_o})^{3/2}\sigma/e^{-\sigma/2} \sin{\theta}e^{\pm i\theta } \nonumber$
6.37
The Hamiltonian is given by $\hat{H} = \dfrac{-\hbar}{2m}\nabla^2 + V$ is an Hermitian Operator. Using this fact, show that
$\int{\psi^*[\hat{H},\hat{A}]\psi} d\tau = 0\nonumber$
where $\hat{A}$ is any operator.
Solution
Through the commutation relation
$\int{\psi^*\hat{H}\hat{A}\psi} d\tau - \int{\psi^*\hat{A}\hat{H}\psi} d\tau= 0\nonumber$
because $\hat{H}$ is a Hermitian operator, the above goes to
$\int{(\psi\hat{H})^*\hat{A}\psi} d\tau - \int{\psi^*\hat{A}(\hat{H}\psi)} d\tau= 0\nonumber$
$E\int{\psi^*\hat{A}\psi} d\tau - E\int{\psi^*\hat{A}\psi} d\tau= 0\nonumber$
6.38
Prove that $\langle{\hat{K}}\rangle \ = \ \langle{V}\rangle = E/2$ for a harmonic oscillator using the virial theorem
Solution
The virial theorem gives us,
$\Bigg\langle{x\dfrac{\partial V}{\partial x} + y\dfrac{\partial V}{\partial y} + z\dfrac{\partial V}{\partial z}}\Bigg\rangle = 2\langle{\hat{K}}\rangle\nonumber$
For a three-dimensional harmonic oscillator,
$V(x,y,z) = \dfrac{k_xx^2}{2} + \dfrac{k_yy^2}{2} + \dfrac{k_zz^2}{2}\nonumber$
Therefore,
$x\dfrac{\partial V}{\partial x} + y\dfrac{\partial V}{\partial y} + z\dfrac{\partial V}{\partial z} = k_xx^2 + k_yy^2 + k_zz^2 = 2V\nonumber$
and substituting into the equation given by the virial theorem gives us $2\langle{V}\rangle = 2\langle{\hat{K}}\rangle$. Because $\langle{\hat{K}}\rangle + \langle{V}\rangle = E$, we can also write
$\langle{\hat{K}}\rangle = \langle{V}\rangle = \dfrac{1}{2}E\nonumber$
6.41
Find the expected values of $1/r$ and $1/r^2$ for a hydrogenlike atom in the $2p_z$ orbital.
Solution
The $2p_z$ orbital:
$Ψ_{210} = \dfrac{1}{4\sqrt{2π}} (Z/a_0)^{3/2}\rho e^{-\rho} \sinθ \cos ϕ \,dθ \nonumber$
where $\rho=Zr/a_o$
$\langle 1/r \rangle _{Ψ210}= \int_{0}^2π \, dθ \int\limits_{0}^{π}\, \sin θ \cos 2ϕ dθ \int\limits_{0}^{∞}\ (Z^{3/2}/a_o^{3/2}4\sqrt{2π}^{2*}r^2p^*e^{-p*}(1/r) dr\nonumber$
$\int_0^{2π}\, dθ = 2π\nonumber$
$\int_0^π \, \sinθ \cos^2 \,ϕ \,dθ = \dfrac{2}{3}\nonumber$
$\int _ 0^∞ \ (Z^{3/2}/a_o^{3/2}4\sqrt{2π})^2 r^2 \phi e-{\rho}(1/r)dr =(Z^3/a_o^332π)^* [3!/(Z/ao)^4]\nonumber$
$\langle 1/r \rangle_{Ψ210} = (Z^3/a_o^332π)(2π)(2/3)[3!/(Z/ao)^4]\nonumber$
Simplify to get:
$\left \langle \dfrac{1}{r} \right \rangle _{Ψ_{210}} = \dfrac{Z}{4a_o}\nonumber$
For the hydrogen atom $Z=1$, therefore
$\langle \dfrac{1}{r} \rangle _{Ψ_{210}} = \dfrac{1}{4a_o}\nonumber$
For $\langle \dfrac{1}{r^2} \rangle$
$\langle \dfrac{1}{r^2} \rangle_{Ψ_{210}} = \int_{0}^2π \, dθ \int\limits_{0}^{π}\, \sin θ \cos^2\, ϕ\,dθ \int\limits_{0}^{∞}\ (Z^3/2/a_o^3/24\sqrt{2π})^2r^2pe^{-p}(1/r^2)dr\nonumber$
$\int\limits_{0}^{2π}\, dθ = 2π\nonumber$
$\int\limits_{0}^{π}\, \sinθ \cos^2ϕdθ = 2/3\nonumber$
$\int \limits_{0}^{∞}\ (Z^{3/2}/a_o^{3/2}4\sqrt{2π})^2r2pe^{-p}(1/r^2)dr = (Z^3/a_o^332π) [2!/(Z/ao)^3]\nonumber$
$\langle 1/r^2 \rangle Ψ_{210} =(Z^3/a_o^332π)(2π)(2/3)[2!/(Z/a_o)^3]\nonumber$
Simplify to get:
$\langle \dfrac{1}{r^2} \rangle _{Ψ_{210}} =\dfrac{Z^2}{12a_o^2} \nonumber$
where $Z=1$
$\langle \dfrac{1}{r^2} \rangle _{Ψ_{210}} = \dfrac{1}{12a_o^2} \nonumber$
6.43
Derive the classical magnetic moment of an electron orbiting a nucleus in terms of charge, mass and angular momentum.
Solution
We can begin by recalling the classical expression for a magnetic moment,
$\mu = IArea\nonumber$
Where $I$ is the current the electron makes by revolving around the nucleus. The definition of current is
$I = \dfrac{Q}{time}\nonumber$
In this case $Q$ is simply the charge $(q_e)$ of the electron and $time$ is the time it takes the electron to orbit the nucleus once. The area is the of loop that the electron takes when revolving around the nucleus. We also know from classical mechanics that $x=vt$. solving for $t$ and evaluating $x$ to be $2\pi r$ for a circle. We can figure out the time of revolution to be,
$t = \dfrac{x}{v}= \dfrac{2\pi r}{v}\nonumber$
Our current equations becomes,
$I = \dfrac{q_ev}{2\pi r}\nonumber$
To introduce angular momentum $L=m_evr$ we can multiply the right side of our current equation by $\dfrac{m_er}{m_er}$ to arrive at
$I = \dfrac{q_em_evr}{2\pi m_er^2} \ I = \dfrac{q_eL}{2\pi m_er^2}\nonumber$
Substituting in the area of a circle $(\pi r^2)$ we can show that,
$\boxed{\mu = IArea = \dfrac{q_eL}{2m_e}}\nonumber$
6.46
Find the magnitude of the splitting shown in figure below. The magnetic field in the figure is at 20 T.
Solution
We know from a previous problem that
$\Delta E = E_{2} - E_{1} = \beta _{e}B_{z}(m_2 - m_1)\nonumber$
In the 1\s\ state where m = 0 and in the 2\p\ state where m = 0, \pm\ 1. The condition will cause (m_{2} - m_{1}) become equal to 0, or \pm\ 1 which will affect the magnitude of splitting, calculated below
$\Delta E = (9.274 * 10^{-24} J*T^{-1}) (20T)(1) \nonumber$
$\Delta E = 1.8548 * 10^{-22} J*T^{-1}$ or 0
6.47
Consider the transition between the $l=1$ and the $l=2$ states for atomic hydrogen. Determine the total number of possible allowed transitions between these two states in an external magnetic field given the following selection rules
1. Light whose electric field vector is parallel to the external magnetic field's direction has a selection rule of $\Delta m=0$ for allowed transitions.
2. Light whose electric field vector is perpendicular to the external magnetic field's direction has a selection rule of $\Delta m=\pm 1$ for allowed transitions.
Solution
An external magnetic field splits a state with given values n and $l$ into $2l+1$ levels. So the $l=1$ state will be split into three states ($m=0, \pm 1$) and the $l=2$ state will be split into five states ($m=0, \pm 1, \pm 2$). This means that the $l=1 \rightarrow l=2$ transition will have a possible of 15 transitions (ignoring any selection rules that reduce this number).
Using the selection rule $\Delta m=0$, then three transitions are possible: $m=0$, $m=1$, $m=-1$ Using the selection rule $\Delta m= \pm 1$, then six transitions are possible:
$l=1$ $\rightarrow$ $l=2$ Relative Orientation of light Polarization to Magnetic field
m=0 m=1 parallel
m=0 m=-1 parallel
m=1 m=2 perpendicular
m=1 m=0 perpendicular
m=-1 m=-2 perpendicular
m=-1 m=0 perpendicular
6.49
Prove that
$\hat{L_+}\hat{L_-} - \hat{L_-}\hat{L_+} = 2\hbar\hat{L_z}\nonumber$
given that
$\hat{L_+} = \hat{L_x} + i\hat{L_y}\nonumber$
and
$\hat{L_-} = \hat{L_x} - i\hat{L_y}.\nonumber$
Solution
$\hat{L_+}\hat{L_-} = (\hat{L_x} + i\hat{L_y})(\hat{L_x} - i\hat{L_y}) = \hat{L^2_x} + \hat{L^2_y} - i\hat{L^2_x}\hat{L^2_y} + i\hat{L^2_y} \hat{L^2_x} = \hat{L^2_x} + \hat{L^2_y} +i[\hat{L_y},\hat{L_x}]\nonumber$
$\hat{L_+}\hat{L_-} = \hat{L^2} - \hat{L^2_z} +\hbar \hat{L_z}\nonumber$
and
$\hat{L_-}\hat{L_+} = (\hat{L_x} - i\hat{L_y})(\hat{L_x} + i\hat{L_y}) = \hat{L^2_x} + \hat{L^2_y} +i\hat{L^2_x}\hat{L^2_y} - i\hat{L^2_y} \hat{L^2_x} = \hat{L^2_x} + \hat{L^2_y} +i[\hat{L_x},\hat{L_y}]\nonumber$
$\hat{L_-}\hat{L_+} = \hat{L^2} - \hat{L^2_z} - \hbar \hat{L_z}\nonumber$
thus
$\hat{L_+}\hat{L_-} - hat{L_-}\hat{L_+} = \hat{L^2} - \hat{L^2_z} +\hbar \hat{L_z} - \hat{L^2} + \hat{L^2_z} - \hbar \hat{L_z}\nonumber$
$\hat{L_+}\hat{L_-} - \hat{L_-}\hat{L_+} = 2\hbar \hat{L_z}\nonumber$
6.49
Show that the commutative property applies to $\hat{L}_{-}\hat{L}_{+}\nonumber$
Solution
$\hat{L}_{-}\hat{L}_{+} = \hat{L}_{+}\hat{L}_{-}\nonumber$
$\hat{L}_{-} = \hat{L}_{x} - i \hat{L}_{y}\nonumber$
and
$\hat{L}_{+} = \hat{L}_{x} + i \hat{L}_{y}\nonumber$
so
$\hat{L}_{-}\hat{L}_{+}=[\hat{L}_x -i\hat{L}_y][\hat{L}_x + i \hat{L}_y]\nonumber$
$= \hat{L}_{x}^2 + i \hat{L}_{x} \hat{L}_{y} - i \hat{L}_{x}\hat{L}_{y} + \hat{L}_{y}^2\nonumber$
and
$\hat{L}_{+}\hat{L}_{-}= [\hat{L}_{x} + i\hat{L}_{y}][\hat{L}_{x}-i\hat{L}_{y}]\nonumber$
$= \hat{L}_{x}^2 -i \hat{L}_{y}\hat{L}_{x}+ i\hat{L}_{x}\hat{L}_{y}+ \hat{L}_{y}^2\nonumber$
which shows that
$\hat{L}_{-}\hat{L}_{+}= \hat{L}_{+}\hat{L}_{-}\nonumber$
Q7.29
Calculate the ground-state energy for particle in the box model using variational method.
Solution
Variational method equations is:
$E_\phi=\dfrac{\langle\phi | \hat{H}| \phi\rangle}{\langle \phi | \phi \rangle}\nonumber$
where the wavefunctions are unnormalized
The unnormalized Schrödinger equation for PIB:
$\phi(x)=A \sin (\dfrac{xn\pi}{L})\nonumber$
$\langle\phi | \phi\rangle\nonumber$ $= A = \sqrt[]{\dfrac{2}{L}}\nonumber$
and
$\langle \phi | \hat{H}| \phi\rangle\nonumber$ $= \dfrac{n^2 h^2}{8mL^2}\cdot \sqrt[]{\dfrac{2}{L}}\nonumber$ so
$E_\phi = \dfrac{\dfrac{n^2 h^2}{8mL^2}\cdot \sqrt[]{\dfrac{2}{L}}}{\sqrt[]{\dfrac{2}{L}}}\nonumber$
so
$E_\phi = \dfrac{n^2h^2}{8mL^2}\nonumber$
where n=1 we get
$E_\phi = \dfrac{h^2}{8mL^2}\nonumber$
6.50
If two functions commute, they have mutual eigenfunctions, such as $\hat{L}$2 and $\hat{L}$$z$. These mutual eigenfunctions are also known as spherical harmonics, $Y$$l$$m$($\theta$, $\phi$), however this information is not pertinent in this case. Let
$\psi$$\alpha$$\beta$ be a mutual eigenfunction of $\hat{L}$2 and $\hat{L}$$z$ so that
$\hat{L}$$z$ $\psi$$\alpha$$\beta$ = $\beta$2\$\alpha$$\beta$
and
$\hat{L}$$z$ $\psi$$\alpha$$\beta$ = $\alpha$$\psi$$\alpha$$\beta$
Now let
$\psi$$\alpha$$\beta$+1 = $\hat{L}$+$\psi$$\alpha$$\beta$
Show that
$\hat{L}$$z$$\psi$$\alpha$$\beta$+1 = ($\alpha$ + $\hbar$)$\psi$$\alpha$$\beta$+1
and
$\hat{L}$2$\psi$$\alpha$$\beta$+1 = $\beta$2$\psi$$\alpha$$\beta$+1
This proves that if $\alpha$ is an eigenvalue of $\hat{L}$$z$, then $\alpha$ + $\hbar$ also is an eigenvalue.
Solution
Solve this problem as given below:
$\psi$$\alpha$$\beta$+1 = $\hat{L}$+$\psi$$\alpha$$\beta$
$\hat{L}$$z$ $\psi$$\alpha$$\beta$+1 =$\hat{L}$$z$ $\hat{L}$+$\psi$$\alpha$$\beta$
= ($\hat{L}$$z$ $\hat{L}$$x$ + $i$$\hat{L}$$z$ $\hat{L}$$y$ )$\psi$$\alpha$$\beta$
$z$, $\hat{L}$$x$] + $\hat{L}$$x$ $\hat{L}$$z$ + $i$[$\hat{L}$$z$, $\hat{L}$$y$] + $i$$\hat{L}$$y$ $\hat{L}$$z$)$\psi$$\alpha$$\beta$
$y$ + $\hat{L}$$z$$\hat{L}$$x$ + $i$$\hbar$$\hat{L}$$x$ + $\hat{L}$$y$$\hat{L}$$z$) $\psi$$\alpha$$\beta$
$\hat{L}$$+$$\hat{L}$$z$ + $\hbar$$\hat{L}$$+$)$\psi$$\alpha$$\beta$
$+$($\alpha$ + $\hbar$)$\psi$$\alpha$$\beta$
$\alpha$$\beta$+1
Therefore proven.
Finally, you can write:
$\hat{L}$2$\psi$$\alpha$$\beta$+1 = $\hat{L}$2$\hat{L}$$+$$\psi$$\alpha$$\beta$
= ($\hat{L}$2$\hat{L}$$x$ + $i$$\hat{L}$2$\hat{L}$$y$)$\psi$$\alpha$$\beta$
=([$\hat{L}$2,$\hat{L}$$x$] + $\hat{L}$$x$$\hat{L}$2 + $i$[$\hat{L}$2,$\hat{L}$$y$] + $i$$\hat{L}$$y$$\hat{L}$2)$\psi$$\alpha$$\beta$
$x$$\hat{L}$2 + $i$$\hat{L}$$y$$\hat{L}$2)$\psi$$\alpha$$\beta$
$+$$\beta$2$\psi$$\alpha$$\beta$
2$\psi$$\alpha$$\beta$ +1
Therefore proven.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/06%3A_The_Hydrogen_Atom/6.E%3A_The_Hydrogen_Atom_%28Exercises%29.txt
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The Schrödinger equation for realistic systems quickly becomes unwieldy, and analytical solutions are only available for very simple systems - the ones we have described as fundamental systems in this module. Numerical approaches can cope with more complex problems, but are still (and will remain for a good while) limited by the available computer power. Approximations are necessary to cope with real systems. Within limits, we can use a pick and mix approach, i.e. use linear combinations of solutions of the fundamental systems to build up something akin to the real system. There are two mathematical techniques, perturbation and variation theory, which can provide a good approximation along with an estimate of its accuracy. These two approximation techniques are described in this chapter.
• 7.1: The Variational Method Approximation
In this section we introduce the powerful and versatile variational method and use it to improve the approximate solutions we found for the helium atom using the independent electron approximation.
• 7.2: Linear Variational Method and the Secular Determinant
A special type of variation widely used in the study of molecules is the so-called linear variation function, a linear combination of N linearly independent functions (often atomic orbitals). Quite often a trial wavefunction is expanded as a linear combination of other functions (not the eigenvalues of the Hamiltonian, since they are not known) .
• 7.3: Trial Functions Can Be Linear Combinations of Functions That Also Contain Variational Parameters
An alternative approach to the general problem of introducing variational parameters into wavefunctions is the construction of a wavefunction as a linear combination of other functions each with one or multiple parameters that can be varied
• 7.4: Perturbation Theory Expresses the Solutions in Terms of Solved Problems
Perturbation theory is the single most important method of solving problems in quantum mechanics, and is widely used in atomic physics, condensed matter and particle physics. Perturbation theory is another approach to finding approximate solutions to a problem, by starting from the exact solution of a related, simpler problem. A critical feature of the technique is a middle step that breaks the problem into "solvable" and "perturbation" parts.
• 7.E: Approximation Methods (Exercises)
These are homework exercises to accompany Chapter 7.
07: Approximation Methods
Learning Objectives
• Appreciate the complexity of solving muliti-electron atoms
• Characterize multi-electron interactions within shielding and penetration concepts
• Use the variational method as an approximation to study insolvable problems
• User variational method to evaluate the effective nuclear charge for a specific atom
In this section we introduce the powerful and versatile variational method and use it to improve the approximate solutions we found for the helium atom using the independent electron approximation.
The True (Experimentally Determined) Energy of the Helium Atom
The helium atom has two electrons bound to a nucleus with charge $Z = 2$. The successive removal of the two electrons can be considered stepwise:
$\ce{He} \xrightarrow {\textit{I}_1} \ce{He}^+ + e^-\xrightarrow {\textit{I}_2}\ce{He}^{++}+2e^-\label{7.1.1}$
The first ionization energy $I_1$ is the minimum energy required to remove the first electron from helium gas and is experimentally determined:
\begin{align*} \textit{I}_1=-\textit{E}_{1\textit{s}}(\ce{He}) = 24.59\;eV \end{align*} \nonumber
While the second ionization energy, $I_2$ can be experimentally determined, it can also be calculated exactly from the hydrogen atom solutions since $\ce{He^{+}}$ is a hydrogen-like ion with $Z=2$. Hence, we have
\begin{align*} \textit{I}_2 &=-\textit{E}_{ 1\textit{s}}(\ce{He}^+) \[4pt] &=\dfrac{Z^2}{2n^2} \[4pt] &=54.42\mbox{ eV} \end{align*} \nonumber
The energy of the three separated particles on the right side of Equation $\ref{7.1.1}$ is zero (by definition). Therefore the ground-state energy of helium atom is given by
\begin{align*} E_{true}&=-(\textit{I}_1+\textit{I}_2) \[4pt] &=-79.02\mbox{ eV}.\end{align*} \nonumber
which can be expressed in terms of the Rydberg constant ($R_H=13.6 \; eV$) that also describes the lowest energy of the hydrogen atom
$E_{true} = -5.8066\,R_H \nonumber$
We will attempt to reproduce this true value, as close as possible, by different theoretical approaches (all approximations).
The "Ignorance is Bliss" Approximation
The Hamiltonian for the Helium atom is:
$\hat{H} = -\dfrac{\hbar^2}{2m_e}\nabla_{el_{1}}^2 -\dfrac{\hbar^2}{2m_e}\nabla_{el_{2}}^2 - \dfrac {Ze^2}{4\pi\epsilon_0 r_1} - \dfrac {Ze^2}{4\pi\epsilon_0 r_2} + \cancel{ \dfrac {e^2}{4\pi \epsilon_0 r_{12}} } \label{7.1.3}$
If we simply ignore the electron-electron repulsion term, then Equation \ref{7.1.3} can be simplified to
\begin{align} \hat{H} & \approx -\dfrac{\hbar^2}{2m_e}\nabla_{el_{1}}^2 - \dfrac {Ze^2}{4\pi\epsilon_0 r_1} - \dfrac{\hbar^2}{2m_e}\nabla_{el_{2}}^2 - \dfrac {Ze^2}{4\pi\epsilon_0 r_2} \label{7.1.3B} \[4pt] &\approx h_1(r_1) + h_2(r_2) \label{7.1.3C} \end{align}
where $h_1$ and $h_2$ are one electron Hamiltonians for electron 1 and 2, respectively, and are just the hydrogen-like Hamiltonians. The approximation in Equation \ref{7.1.3C} is convenient since electron 1 is separable from electron 2, so that the two-electron wavefunction is approximated as a product to two one-electron wavefunctions:
$\Psi_{total} = \psi_{el_{1}}\psi_{el_{2}} \label{7.1.4a}$
or in braket notation
$| \Psi_{total} \rangle = \hat{H} | \psi_{el_1} \psi_{el_2} \rangle \label{7.1.4b}$
With some operator algebra, something important arises - the one electron energies are additive:
\begin{align*} \hat{H} \Psi_{total} &= (\hat{H}_{el_1} + \hat{H}_{el_2}) \psi_{n\ {el_1}} \psi_{n\ {el_2}} = (E_{n_1} + E_{n_2}) \psi_{n\ {el_1}} \psi_{n\ {el_2}} \end{align*} \nonumber
or in bra-ket notation
\begin{align*} \hat{H} | \Psi_{total} \rangle &= \hat{H} | \psi_{el_1} \psi_{el_2} \rangle \[4pt] &= (E_{n_1} + E_{n_2}) | \psi_{1} \psi_{2} \rangle \end{align*} \nonumber
The energy for a ground state Helium atom (both electrons in lowest state) is then
\begin{align*} E_{He_{1s}} &= \underset{\text{energy of single electron in helium}}{E_{n_1}} + \underset{\text{energy of single electron in helium}}{E_{n_2}} \[4pt] &= -R_H\left(\dfrac{Z^2}{1}\right) -R_H \left(\dfrac{Z^2}{1}\right) \[4pt] &= -8R_H \end{align*} \nonumber
This approximation significantly overestimates the true energy of the helium atom $E_{He_{1s}} = -5.8066\,R$. This is a poor approximation and we need to address electron-electron repulsion properly (or better at least).
Shielding and Penetration
One way to take electron-electron repulsion into account is to modify the form of the wavefunction. A logical modification is to change the nuclear charge, $Z$, in the wavefunctions to an effective nuclear charge ($Z_{eff}$), from +2 to a smaller value. The rationale for making this modification is that one electron partially shields the nuclear charge from the other electron, as shown in Figure 7.1.1 .
A region of negative charge density between one of the electrons and the +2 nucleus makes the potential energy between them more positive (decreases the attraction between them). We can effect this change mathematically by using $\zeta < 2$ in the wavefunction expression. If the shielding were complete, then $Z_{eff}$ would equal 1. If there is no shielding, then $Z_{eff}= 2$. So a way to take into account the electron-electron interaction is by saying it produces a shielding effect. The shielding is not zero, and it is not complete, so the effective nuclear charge varies between one and two.
In general, a theory should be able to make predictions in advance of knowledge of the experimental result. Consequently, a principle and method for choosing the best value for $Z_{eff}$ or any other adjustable parameter that is to be optimized in a calculation is needed. The Variational Principle provides the required criterion and method and says that the best value for any variable parameter in an approximate wavefunction is the value that gives the lowest energy for the ground state; i.e., the value that minimizes the energy. The variational method is the procedure that is used to find the lowest energy and the best values for the variable parameters.
A Better Approximation: The Variational Method
The variational method is one way of finding approximations to the lowest energy eigenstate or ground state, and some excited states. This allows calculating approximate wavefunctions and is the variational principle. The method consists in choosing a "trial wavefunction" depending on one or more parameters, and finding the values of these parameters for which the expectation value of the energy is the lowest possible. The wavefunction obtained by fixing the parameters to such values is then an approximation to the ground state wavefunction, and the expectation value of the energy in that state is an upper bound to the ground state energy.
The variational principle means that the expectation value for the binding energy obtained using an approximate wavefunction and the exact Hamiltonian operator will be higher than or equal to the true energy for the system. This idea is really powerful. When implemented, it permits us to find the best approximate wavefunction from a given wavefunction that contains one or more adjustable parameters, called a trial wavefunction. A mathematical statement of the variational principle is
$E_{trial} \ge E_{true} \label {7.1.7}$
where
\begin{align} E_{trial} &= \dfrac{ \langle \psi _{trial}| \hat {H} | \psi _{trial} \rangle}{\langle \psi _{trial} | \psi _{trial} \rangle} \[4pt] &= \dfrac {\displaystyle \int \psi _{trial} ^* \hat {H} \psi _{trial} d \tau}{\displaystyle \int \psi _{trial} ^* \psi _{trial} d\tau } \label {7.1.8} \end{align}
Equation $\ref{7.1.7}$ is called the variational theorem and states that for a time-independent Hamiltonian operator, any trial wavefunction will have an variational energy (i.e., expectation value) that is greater than or equal to the true ground state wavefunction corresponding to the given Hamiltonian (Equation \ref{7.1.7}). Because of this, the variational energy is an upper bound to the true ground state energy of a given molecule. The general approach of this method consists in choosing a "trial wavefunction" depending on one or more parameters, and finding the values of these parameters for which the expectation value of the energy is the lowest possible (Figure 7.1.2 ).
The variational energy $E_{trial}$ is only equal to the true energy $E_{true}$ when the the corresponding trial wavefunction $\psi_{trial}$ is equal to the true wavefunction $\psi_{true}$.
Application to the Helium atom Ground State
Often the expectation values (numerator) and normalization integrals (denominator) in Equation $\ref{7.1.8}$ can be evaluated analytically. For the case of the He atom, let's consider the trial wavefunction as the product wavefunction given by Equation $\ref{7-13}$ (this is called the orbital approximation),
$\psi (r_1 , r_2) \approx \varphi (r_1) \varphi (r_2) \label {7-13}$
The adjustable or variable parameter in the trial wavefunction is the effective nuclear charge $\zeta$, and the Hamiltonian is the complete form given below (Note: quantum calculations typically refer to effective nuclear charge as $\zeta$ rather than $Z_{eff}$ as we used previously).
$\hat {H} = -\dfrac {\hbar ^2}{2m} \nabla^2_1 - \dfrac {\zeta e^2}{4 \pi \epsilon _0 r_1} - \dfrac {\hbar ^2}{2m} \nabla ^2_2 - \dfrac {\zeta e^2}{4 \pi \epsilon _0 r_2} + \dfrac {e^2}{4 \pi \epsilon _0 r_{12}} \label {9-9}$
the adjustable or variable parameter in the trial wavefunction is the effective nuclear charge $\zeta$ (would be equal to $\zeta=2$ if fully unshielded), and the Hamiltonian is the complete form. When the expectation value for the trial energy (Equation \ref{7.1.8}) is evaluated for helium, the result is a variational energy that depends on the adjustable parameter, $\zeta$.
$E_{trial} (\zeta) = \dfrac {\mu e^4}{4 \epsilon ^2_0 h} \left ( \zeta ^2 - \dfrac {27}{8} \zeta \right ) \label {7.1.9}$
This function is plotted in Figure 7.1.3 as a function of $\zeta$. According to the variational principle (Equation \ref{7.1.7}), the minimum value of the energy on this graph is the best approximation of the true energy of the system, and the associated value of $\zeta$ is the best value for the adjustable parameter.
Using the mathematical function for the energy of a system, the minimum energy with respect to the adjustable parameter can be found by taking the derivative of the energy with respect to that parameter, setting the resulting expression equal to zero, and solving for the parameter, in this case $\zeta$. This is a standard method in calculus for finding maxima and minima.
Exercise 7.1.1
Find the value for $\zeta$ that minimizes the helium binding energy for the product trial wavefunction in Equation \ref{7-13} with the Hamiltonian in Equation \ref{9-9}. and compare the binding energy to the experimental value. What is the percent error in the calculated value?
Solution
The variational method requires following the workflow in Figure 7.1.2 .
• Step 1: Define the Hamiltonian - This is given by Equation \ref{9-9}.
• Step 2: Define the trial wavefunction as a function of at least one parameter - This is given by Equation \ref{7-13}.
• Step 3: Evaluate variational energy ($E_{trial}$ integral (Equation \ref{7.1.8}) - This procedure was already above in Equation \ref{7.1.9}.
• Step 4: Minimize the variational energy as a function of the parameter(s) - Following the standard approach to find extrema in calculus, evaluate the derivative of $E_{trial}$ with respect to $\zeta$ and set to zero: $\dfrac{dE_{trial}}{d\zeta} = \dfrac {\mu e^4}{4 \epsilon ^2_0 h} \left ( 2 \zeta - \dfrac {27}{8} \right ) =0 \nonumber$ then find solve for the roots of this polynomial $2 \zeta - \dfrac {27}{8}=0 \nonumber$ or $\zeta = \dfrac {27}{16} \approx 1.6875 \nonumber$
• Step 5-6: The question does not ask for the optimized wavefunction (Step 5) or to compare the result with the true value to evaluate the quality of the approximation (Step 6). We can skip these steps.
From Exercise 7.1.1 , the $\zeta = 1.6875$ and the approximate energy we calculate using this approximation method, Eapprox = -77.483 eV. Table 7.1.1 show that a substantial improvement in the accuracy of the computed binding energy is obtained by using shielding to account for the electron-electron interaction. Including the effect of electron shielding in the wavefunction reduces the error in the binding energy to about 2%. This idea is very simple, elegant, and significant.
Table 7.1.1 : Comparison of the results of three approximation methods to experiment.
Method
He binding energy (eV)
"Ignorance is Bliss" Approximation (neglect repulsion between electrons) -108.8
Variational method with variable effective charge -77.483
Experimental -79.0
The improvement we have seen in the total energy calculations using a variable parameter $\zeta$ indicates that an important contribution of electron-electron interaction or repulsion to the total binding energy arises from the fact that each electron shields the nuclear charge from the other electron. It is reasonable to assume the electrons are independent; i.e., that they move independently, but the shielding must be taken into account in order to fine-tune the wavefunctions. The inclusion of optimizable parameters in the wavefunction allows us to develop a clear physical image of the consequences of our variation calculation. Calculating energies correctly is important, and it is also important to be able to visualize electron densities for multi-electron systems. In the next two sections, we take a temporary break from our consideration of approximation methods in order to examine multi-electron wavefunctions more closely.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/07%3A_Approximation_Methods/7.01%3A_The_Variational_Method_Approximation.txt
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Learning Objectives
• Understand how the variational method can be expanded to include trial wavefunctions that are a linear combination of functions with coefficients that are the parameters to be varied.
• To be able to construct secular equations to solve the minimization procedure intrinsic to the variational method approach.
• To map the secular equations into the secular determinant
• To understand how the Linear Combination of Atomic Orbital (LCAO) approximation is a specific application of the linear variational method.
A special type of variation widely used in the study of molecules is the so-called linear variation function, where the trial wavefunction is a linear combination of $N$ linearly independent functions (often atomic orbitals) that not the eigenvalues of the Hamiltonian (since they are not known). For example
$| \psi_{trial} \rangle = \sum_{j=1}^N a_j |\phi_j \rangle \label{Ex1}$
and
$\langle \psi_{trial} | = \sum_{j=1}^N a_j^* \langle \phi_j | \label{Ex2}$
In these cases, one says that a 'linear variational' calculation is being performed.
Linear Variational Basis Functions
The set of functions {$\phi_j$} are called the 'linear variational' basis functions and are nothing more than members of a set of functions that are convenient to deal with. However, they are typically not arbitrary and are usually selected to address specific properties of the system:
• to obey all of the boundary conditions that the exact state $| \psi _{trial} \rangle$ obeys,
• to be functions of the the same coordinates as $| \psi _{trial} \rangle$,
• to be of the same symmetry as $| \psi _{trial} \rangle$, and
• to be convenient to evaluate Hamiltonian terms elements $\langle \phi_i|H|\phi_j \rangle$.
Beyond these conditions, nothing other than effort can limit the selection and number of such basis functions in the expansions in Equations $\ref{Ex1}$ and $\ref{Ex2}$.
As discussed in Section 7.1, the variational energy for a generalized trial wavefunction is
$E_{trial} = \dfrac{ \langle \psi _{trial}| \hat {H} | \psi _{trial} \rangle}{\langle \psi _{trial} | \psi _{trial} \rangle} \label{7.1.8}$
Substituting Equations \ref{Ex1} and \ref{Ex2} into Equation \ref{7.1.8} involves addressing the numerator and denominator individually. For the numerator, the integral can be expanded thusly:
\begin{align} \langle\psi_{trial} |H| \psi_{trial} \rangle &= \sum_{i}^{N} \sum_{j} ^{N}a_i^{*} a_j \langle \phi_i|H|\phi_j \rangle. \[4pt] &= \sum_{i,\,j} ^{N,\,N}a_i^{*} a_j \langle \phi_i|H|\phi_j \rangle. \label{MatrixElement}\end{align}
We can rewrite the following integral in Equation \ref{MatrixElement} as a function of the basis elements (not the trial wavefunction) as
$H_{ij} = \langle \phi_i|H|\phi_j \rangle \nonumber$
So the numerator of the right side of Equation \ref{7.1.8} becomes
$\langle\psi_{trial} |H| \psi_{trial} \rangle = \sum_{i,\,j} ^{N,\,N}a_i^{*} a_j H_{ij} \label{numerator}$
Similarly, the denominator of the right side of Equation \ref{7.1.8} can be expanded
$\langle \psi_{trial}|\psi_{trial} \rangle = \sum_{i,\,j} ^{N,\,N}a_i^{*} a_j \langle \phi_i | \phi_j \rangle \label{overlap}$
We often simplify the integrals on the right side of Equation \ref{overlap} as
$S_{ij} = \langle \phi_i|\phi_j \rangle \nonumber$
where $S_{ij}$ are overlap integrals between the different {$\phi_j$} basis functions. Equation \ref{overlap} is thus expressed as
$\langle \psi_{trial}|\psi_{trial} \rangle = \sum_{i,\,j} ^{N,\,N}a_i^{*} a_j S_{ij} \label{denominator}$
Orthonormality of Basis Functions
There is no explicit rule that the {$\phi_j$} functions have to be orthogonal or normalized functions, although they often are selected that way for convenience. Therefore, a priori, $S_{ij}$ does not have to be $\delta_{ij}$.
Substituting Equations \ref{numerator} and \ref{denominator} into the variational energy formula (Equation \ref{7.1.8}) results in
$E_{trial} = \dfrac{ \displaystyle \sum_{i,\,j} ^{N,\,N}a_i^{*} a_j H_{ij} }{ \displaystyle \sum_{i,\,j} ^{N,\,N}a_i^{*} a_j S_{ij} } \label{Var}$
For such a trial wavefunction as Equation \ref{Ex1}, the variational energy depends quadratically on the 'linear variational' $a_j$ coefficients. These coefficients can be varied just like the parameters in the trial functions of Section 7.1 to find the optimized trial wavefunction ($| \psi_{trial} \rangle$) that approximates the true wavefunction ($| \psi \rangle$) that we cannot analytically solve for.
Minimizing the Variational Energy
The expression for variational energy (Equation \ref{Var}) can be rearranged
$E_{trial} \sum_{i,\,j} ^{N,\,N} a_i^*a_j S_{ij} = \sum_{i,\,j} ^{N,\,N} a_i^* a_j H_{ij} \label{7.2.9}$
The optimum coefficients are found by searching for minima in the variational energy landscape spanned by varying the $\{a_i\}$ coefficients (Figure 7.2.1 ).
We want to minimize the energy with respect to the linear coefficients $\{a_i\}$, which requires that
$\dfrac{\partial E_{trial}}{\partial a_i}= 0 \nonumber$
for all $i$.
Differentiating both sides of Equation $\ref{7.2.9}$ for the $k^{th}$ coefficient gives,
$\dfrac{\partial E_{trial}}{\partial a_k} \sum_{i,\,j} ^{N,\,N} a_i^*a_j S_{ij}+ E_{trial} \sum_i \sum_j \left[ \dfrac{ \partial a_i^*}{\partial a_k} a_j + \dfrac {\partial a_j}{\partial a_k} a_i^* \right ]S_{ij} = \sum_{i,\,j} ^{N,\,N} \left [ \dfrac{\partial a_i^*}{\partial a_k} a_j + \dfrac{ \partial a_j}{\partial a_k}a_i^* \right] H_{ij} \label{7.2.10}$
Since the coefficients are independent
$\dfrac{\partial a_i^*}{ \partial a_k} = \delta_{ik} \nonumber$
and
$S_{ij} = S_{ji} \nonumber$
and also since the Hamiltonian is a Hermitian Operator (see below)
$H_{ij} =H_{ji} \nonumber$
then Equation $\ref{7.2.10}$ simplifies to
$\dfrac{\partial E_{trial}}{\partial a_k} \sum_i \sum_j a_i^*a_j S_{ij}+ 2E_{trial} \sum_i a_i S_{ik} = 2 \sum_i a_i H_{ik} \label{7.2.11}$
At the minimum variational energy, when
$\dfrac{\partial E_{trial}}{\partial a_k} = 0 \nonumber$
then Equation $\ref{7.2.11}$ gives
${\sum _i^N a_i (H_{ik}–E_{trial} S_{ik}) = 0} \label{7.2.12}$
for all $k$. The equations in $\ref{7.2.12}$ are called the Secular Equations.
Hermitian Operators
Hermitian operators are operators that satisfy the general formula
$\langle \phi_i | \hat{A} | \phi_j \rangle = \langle \phi_j | \hat{A} | \phi_i \rangle \label{Herm1}$
If that condition is met, then $\hat{A}$ is a Hermitian operator. For any operator that generates a real eigenvalue (e.g., observables), then that operator is Hermitian. The Hamiltonian $\hat{H}$ meets the condition of a Hermitian operator. Equation \ref{Herm1} can be rewriten as
$A_{ij} =A_{ji}^* \nonumber$
where
$A_{ij} = \langle \phi_i | \hat{A} | \phi_j \rangle \nonumber$
and
$A_{ji} = \langle \phi_j | \hat{A} | \phi_i \rangle \nonumber$
Therefore, when applied to the Hamiltonian operator
$H_{ij}^* =H_{ji}. \nonumber$
If the functions $\{|\phi_j\rangle \}$ are orthonormal, then the overlap matrix $S$ reduces to the unit matrix (one on the diagonal and zero every where else) and the Secular Equations in Equation \ref{7.2.12} reduces to the more familiar Eigenvalue form:
$\sum\limits_i^N H_{ij}a_j = E_{trial} a_i .\label{seceq2}$
Hence, the secular equation, in either form, have as many eigenvalues $E_i$ and eigenvectors {$C_{ij}$} as the dimension of the $H_{ij}$ matrix as the functions in $| \psi_{trail} \rangle$ (Example \ref{Ex1}). It can also be shown that between successive pairs of the eigenvalues obtained by solving the secular problem at least one exact eigenvalue must occur (i.e.,$E_{i+1} > E_{exact} > E_i$, for all i). This observation is referred to as 'the bracketing theorem'.
Variational methods, in particular the linear variational method, are the most widely used approximation techniques in quantum chemistry. To implement such a method one needs to know the Hamiltonian $H$ whose energy levels are sought and one needs to construct a trial wavefunction in which some 'flexibility' exists (e.g., as in the linear variational method where the $a_j$ coefficients can be varied). This tool will be used to develop several of the most commonly used and powerful molecular orbital methods in chemistry.
The Secular Determinant
From the secular equations with an orthonormal functions (Equation \ref{seceq2}), we have $k$ simultaneous secular equations in $k$ unknowns. These equations can also be written in matrix notation, and for a non-trivial solution (i.e. $c_i \neq 0$ for all $i$), the determinant of the secular matrix must be equal to zero.
${ | H_{ik}–ES_{ik}| = 0} \label{7.2.13}$
Properties of Determinants
• The determinant is a real number, it is not a matrix.
• The determinant can be a negative number.
• It is not associated with absolute value at all except that they both use vertical lines.
• The determinant only exists for square matrices ($2 \times 2$, $3 \times 3$, ..., $n \times n$). The determinant of a $1 \times 1$ matrix is that single value in the determinant.
• The inverse of a matrix will exist only if the determinant is not zero.
Expanding Determinants
The determinant can be evaluated using an expansion method involving minors and cofactors. Before we can use them, we need to define them. It is the product of the elements on the main diagonal minus the product of the elements off the main diagonal. In the case of a $2 \times 2$ matrix, the specific formula for the determinant is
{\displaystyle {\begin{aligned}|A|={\begin{vmatrix}a&b\c&d\end{vmatrix}}=ad-bc.\end{aligned}}} \nonumber
Similarly, suppose we have a $3 \times 3$ matrix $A$, and we want the specific formula for its determinant $|A|$:
{\displaystyle {\begin{aligned}|A|={\begin{vmatrix}a&b&c\d&e&f\g&h&i\end{vmatrix}}&=a\,{\begin{vmatrix}e&f\h&i\end{vmatrix}}-b\,{\begin{vmatrix}d&f\g&i\end{vmatrix}}+c\,{\begin{vmatrix}d&e\g&h\end{vmatrix}}\&=aei+bfg+cdh-ceg-bdi-afh.\end{aligned}}} \nonumber
To solve Equation \ref{7.2.13}, the determinate should be expanded and then set to zero. That generates a polynomial (called a characteristic equation) that can be directly solved with linear algebra methods or numerically.
Example 7.2.1 : A Simple Two Component Basis Set
If $|\psi_{trial} \rangle$ is a linear combination of two functions. In math terms,
$|\psi_{trial} \rangle= \sum_{n=1}^{N=2} a_n |f_n\rangle = a_1 |\phi_1 \rangle + a_2 | \phi_2 \rangle \nonumber$
then the secular determinant (Equation $\ref{7.2.13}$), in matrix formulation would look like this
$\begin{vmatrix} H_{11}-E_{trial}S_{11}&H_{12}-E_{trial}S_{12} \ H_{12}-E_{trial}S_{12}&H_{22}-E_{trial}S_{22}\end{vmatrix}=0 \nonumber$
Solution
Solving the secular equations is done by finding $E_{trial}$ and putting the value into the expansion of the secular determinant
$a_1^2 H_{11} + 2a_1 a_2 H_{12}+ a_2^2 H_{22}=0 \nonumber$
and
$a_1(H_{12} - E_{trial}S_{12}) + a_2(H_{22} - E_{trial}S_{22}) = 0 \nonumber$
Equation $\ref{7.2.13}$ can be solved to obtain the energies $E$. When arranged in order of increasing energy, these provide approximations to the energies of the first $k$ states (each having an energy higher than the true energy of the state by virtue of the variation theorem). To find the energies of a larger number of states we simply use a greater number of basis functions $\{\phi_i\}$ in the trial wavefunction (Example \ref{Ex1}). To obtain the approximate wavefunction for a particular state, we substitute the appropriate energy into the secular equations and solve for the coefficients $a_i$.
Using this method it is possible to find all the coefficients $a_1 \ldots a_k$ in terms of one coefficient; normalizing the wavefunction provides the absolute values for the coefficients.
Example 7.2.2 : Linear Combination of Atomic Orbitals (LCAO) Approximation
Trial wavefunctions that consist of linear combinations of simple functions
$| \psi(r) \rangle = \sum_i a_i | \phi_i(r) \rangle \nonumber$
form the basis of the Linear Combination of Atomic Orbitals (LCAO) method introduced by Lennard and Jones and others to compute the energies and wavefunctions of atoms and molecules. The functions $\{| \phi_i \rangle \}$ are selected so that matrix elements can be evaluated analytically. Two basis sets of atomic orbitals functions can be used: Slater type and Gaussian type:
Slater orbitals using Hydrogen-like wavefunctions
$| \phi_i \rangle = Y_{l}^{m}(\theta,\phi) e ^{-\alpha r} \nonumber$
and Gaussian orbitals of the form
$| \phi_i \rangle = Y_{l}^{m}(\theta,\phi) e ^{-\alpha r^2} \nonumber$
are the most widely used forms, where $Y_l^m(\theta,\phi)$ are the spherical harmonics that represent the angular part of the atomic orbitals. Gaussian orbitals form the basis of many quantum chemistry computer codes.
Because Slater orbitals give exact results for Hydrogen, we will use Gaussian orbitals to test the LCAO method on Hydrogen, following S.F. Boys, Proc. Roy. Soc. A 200, 542 (1950) and W.R. Ditchfield, W.J. Hehre and J.A. Pople, J. Chem. Phys. Rev. 52, 5001 (1970) with the basis set. Because products of Gaussians are also Gaussian, the required matrix elements are easily computed.
The linear variational method is used extensively in molecular orbitals of molecules and further examples will be postponed until that discussion in Chapters 9.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/07%3A_Approximation_Methods/7.02%3A_Linear_Variational_Method_and_the_Secular_Determinant.txt
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Learning Objectives
• Demonstrate that variational problems can include changing parameters within the elements (normal variational method) and changing coefficients of a basis set (linear variational method)
An alternative approach to the general problem of introducing variational parameters into wavefunctions is the construction of a wavefunction as a linear combination of other functions each with one or multiple parameters that can be varied
For hydrogen, the radial function decays, or decreases in amplitude, exponentially as the distance from the nucleus increases. For helium and other multi-electron atoms, the radial dependence of the total probability density does not fall off as a simple exponential with increasing distance from the nucleus as it does for hydrogen. More complex single-electron functions therefore are needed in order to model the effects of electron-electron interactions on the total radial distribution function. One way to obtain more appropriate single-electron functions is to use a sum of exponential functions in place of the hydrogenic spin-orbitals.
An example of such a wavefunction created from a sum or linear combination of exponential functions is written as
$\varphi _{1s} (r_1) = \sum _j c_j e^{-\zeta _j r_j / a_o} \label{9-37}$
The linear combination permits weighing of the different exponentials through the adjustable coefficients ($c_j$) for each term in the sum. Each exponential term has a different rate of decay through the zeta-parameter $\zeta _j$. The exponential functions in Equation $\ref{9-37}$ are called basis functions. Basis functions are the functions used in linear combinations to produce the single-electron orbitals that in turn combine to create the product multi-electron wavefunctions. Originally the most popular basis functions used were the STO’s, but today STO’s are not used in most quantum chemistry calculations. However, they are often the functions to which more computationally efficient basis functions are fitted.
Physically, the $\zeta _j$ parameters account for the effective nuclear charge (often denoted with $Z_{eff}$). The use of several zeta values in the linear combination essentially allows the effective nuclear charge to vary with the distance of an electron from the nucleus. This variation makes sense physically. When an electron is close to the nucleus, the effective nuclear charge should be close to the actual nuclear charge. When the electron is far from the nucleus, the effective nuclear charge should be much smaller. See Slater's rules for a rule-of-thumb approach to evaluate $Z_{eff}$ values.
A term in Equation $\ref{9-37}$ with a small $\zeta$ will decay slowly with distance from the nucleus. A term with a large $\zeta$ will decay rapidly with distance and not contribute at large distances. The need for such a linear combination of exponentials is a consequence of the electron-electron repulsion and its effect of screening the nucleus for each electron due to the presence of the other electrons.
Exercise 7.3.1
Make plots of $\varphi$ in Equation $\ref{9-37}$ using three equally weighted terms with $\zeta$ = 1.0, 2.0, and 5.0. Also plot each term separately.
Computational procedures in which an exponential parameter like $\zeta$ is varied are more precisely called the Nonlinear Variational Method because the variational parameter is part of the wavefunction and the change in the function and energy caused by a change in the parameter is not linear. The optimum values for the zeta parameters in any particular calculation are determined by doing a variational calculation for each orbital to minimize the ground-state energy. When this calculation involves a nonlinear variational calculation for the zetas, it requires a large amount of computer time. The use of the variational method to find values for the coefficients, $\{c_j\}$, in the linear combination given by Equation $\ref{9-37}$ above is called the Linear Variational Method because the single-electron function whose energy is to be minimized (in this case $\varphi _{1s}$) depends linearly on the coefficients. Although the idea is the same, it usually is much easier to implement the linear variational method in practice.
Nonlinear variational calculations are extremely costly in terms of computer time because each time a zeta parameter is changed, all of the integrals need to be recalculated. In the linear variation, where only the coefficients in a linear combination are varied, the basis functions and the integrals do not change. Consequently, an optimum set of zeta parameters were chosen from variational calculations on many small multi-electron systems, and these values, which are given in Table 7.3.1 , generally can be used in the STOs for other and larger systems.
Table 7.3.1 : Orbital Exponents for Slater Orbitals
Atom $\zeta _{1s}$ $\zeta _{2s,2p}$
H 1.24 -
He 1.69 -
Li 2.69 0.80
Be 3.68 1.15
B 4.68 1.50
C 5.67 1.72
N 6.67 1.95
O 7.66 2.25
F 8.56 2.55
Ne 9.64 2.88
Exercise 7.3.2
Compare the value $\zeta _{1s}$ = 1.24 in Table 7.3.1 for hydrogen with the value you obtained in Exercise 7.3.1 . and comment on possible reasons for any difference. Why are the zeta values larger for 1s than for 2s and 2p orbitals? Why do the $\zeta _{1s}$ values increase by essentially one unit for each element from He to Ne while the increase for the $\zeta _{2s, 2p}$ values is much smaller?
Answer
ζ values represent the rate of decay in the radial function of an orbital. ζ values are larger for 1s than 2s and 2p orbitals because 1s orbitals have a smaller radial function. As a result, 1s orbitals decrease faster in radial function as you move further from the nucleus, and have a larger ζ value to represent this faster decay. The ζ values for 1s increase essentially by one unit for each element from He to Ne because the 1s orbital is closest to the nucleus, and experiences the greatest effects from change in electronegativity as nuclear density increases from He to Ne. This increase in electronegativity causes the radial function to decay more and more rapidly as atomic number/nucleus density increase. The 2s and 2p orbitals don't experience as great a change in radial function decay rate because they are shielded by the 1s orbital.
The discussion above gives us some new ideas about how to write flexible, useful single-electron wavefunctions that can be used to construct multi-electron wavefunctions for variational calculations. Single-electron functions built from the basis function approach are flexible because they have several adjustable parameters, and useful because the adjustable parameters still have clear physical interpretations. Such functions will be needed in the Hartree-Fock method discussed elsewhere.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/07%3A_Approximation_Methods/7.03%3A_Trial_Functions_Can_Be_Linear_Combinations_of_Functions_That_Also_Contain_Variational_Parameters.txt
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Learning Objectives
• Use perturbation theory to approximate the energies of systems as a series of perturbation of a solved system.
• Use perturbation theory to approximate the wavefunctions of systems as a series of perturbation of a solved system.
It is easier to compute the changes in the energy levels and wavefunctions with a scheme of successive corrections to the zero-field values. This method, termed perturbation theory, is the single most important method of solving problems in quantum mechanics and is widely used in atomic physics, condensed matter and particle physics. Perturbation theory is another approach to finding approximate solutions to a problem, by starting from the exact solution of a related, simpler problem. A critical feature of the technique is a middle step that breaks the problem into "solvable" and "perturbation" parts. Perturbation theory is applicable if the problem at hand cannot be solved exactly, but can be formulated by adding a "small" term to the mathematical description of the exactly solvable problem.
We begin with a Hamiltonian $\hat{H}^0$ having known eigenkets and eigenenergies:
$\hat{H}^o | n^o \rangle = E_n^o | n^o \rangle \label{7.4.1}$
The task is to find how these eigenstates and eigenenergies change if a small term $H^1$ (an external field, for example) is added to the Hamiltonian, so:
$( \hat{H}^0 + \hat{H}^1 ) | n \rangle = E_n | n \rangle \label{7.4.2}$
That is to say, on switching on $\hat{H}^1$ changes the wavefunctions:
$\underbrace{ | n^o \rangle }_{\text{unperturbed}} \Rightarrow \underbrace{|n \rangle }_{\text{Perturbed}}\label{7.4.3}$
and energies (Figure 7.4.1 ):
$\underbrace{ E_n^o }_{\text{unperturbed}} \Rightarrow \underbrace{E_n }_{\text{Perturbed}} \label{7.4.4}$
The basic assumption in perturbation theory is that $H^1$ is sufficiently small that the leading corrections are the same order of magnitude as $H^1$ itself, and the true energies can be better and better approximated by a successive series of corrections, each of order $H^1/H^o$ compared with the previous one.
The strategy is to expand the true wavefunction and corresponding eigenenergy as series in $\hat{H}^1/\hat{H}^o$. These series are then fed into Equation $\ref{7.4.2}$, and terms of the same order of magnitude in $\hat{H}^1/\hat{H}^o$ on the two sides are set equal. The equations thus generated are solved one by one to give progressively more accurate results.
To make it easier to identify terms of the same order in $\hat{H}^1/\hat{H}^o$ on the two sides of the equation, it is convenient to introduce a dimensionless parameter $\lambda$ which always goes with $\hat{H}^1$, and then expand both eigenstates and eigenenergies as power series in $\lambda$,
\begin{align} | n \rangle &= \sum _ i^m \lambda ^i| n^i \rangle \label{7.4.5} \[4pt] E_n &= \sum_{i=0}^m \lambda ^i E_n^i \label{7.4.6} \end{align}
where $m$ is how many terms in the expansion we are considering. The ket $|n^i \rangle$ is multiplied by $\lambda^i$ and is therefore of order $(H^1/H^o)^i$.
$\lambda$ is purely a bookkeeping device: we will set it equal to 1 when we are through! It’s just there to keep track of the orders of magnitudes of the various terms.
For example, in first order perturbation theory, Equations $\ref{7.4.5}$ are truncated at $m=1$ (and setting $\lambda=1$):
\begin{align} | n \rangle &\approx | n^o \rangle + | n^1 \rangle \label{7.4.7} \[4pt] E_n &\approx E_n^o + E_n^1 \label{7.4.8} \end{align}
However, let's consider the general case for now. Adding the full expansions for the eigenstate (Equation $\ref{7.4.5}$) and energies (Equation $\ref{7.4.6}$) into the Schrödinger equation for the perturbation Equation $\ref{7.4.2}$ in
$( \hat{H}^o + \lambda \hat{H}^1) | n \rangle = E_n| n \rangle \label{7.4.9}$
we have
$(\hat{H}^o + \lambda \hat{H}^1) \left( \sum _ {i=0}^m \lambda ^i| n^i \rangle \right) = \left( \sum_{i=0}^m \lambda^i E_n^i \right) \left( \sum _ {i=0}^m \lambda ^i| n^i \rangle \right) \label{7.4.10}$
We’re now ready to match the two sides term by term in powers of $\lambda$. Note that the zeroth-order term, of course, just gives back the unperturbed Schrödinger Equation (Equation $\ref{7.4.1}$).
Let's look at Equation $\ref{7.4.10}$ with the first few terms of the expansion:
\begin{align} (\hat{H}^o + \lambda \hat{H}^1) \left( | n ^o \rangle + \lambda | n^1 \rangle \right) &= \left( E _n^0 + \lambda E_n^1 \right) \left( | n ^o \rangle + \lambda | n^1 \rangle \right) \label{7.4.11} \[4pt] \hat{H}^o | n ^o \rangle + \lambda \hat{H}^1 | n ^o \rangle + \lambda H^o | n^1 \rangle + \lambda^2 \hat{H}^1| n^1 \rangle &= E _n^0 | n ^o \rangle + \lambda E_n^1 | n ^o \rangle + \lambda E _n^0 | n ^1 \rangle + \lambda^2 E_n^1 | n^1 \rangle \label{7.4.11A} \end{align}
Collecting terms in order of $\lambda$ and coloring to indicate different orders
$\underset{\text{zero order}}{\hat{H}^o | n ^o \rangle} + \color{red} \underset{\text{1st order}}{\lambda ( \hat{H}^1 | n ^o \rangle + \hat{H}^o | n^1 \rangle )} + \color{blue} \underset{\text{2nd order}} {\lambda^2 \hat{H}^1| n^1 \rangle} =\color{black}\underset{\text{zero order}}{E _n^0 | n ^o \rangle} + \color{red} \underset{\text{1st order}}{ \lambda (E_n^1 | n ^o \rangle + E _n^0 | n ^1 \rangle )} +\color{blue}\underset{\text{2nd order}}{\lambda^2 E_n^1 | n^1 \rangle} \label{7.4.12}$
If we expanded Equation $\ref{7.4.10}$ further we could express the energies and wavefunctions in higher order components.
Zero-Order Terms ($\lambda=0$)
Collecting the zero order terms in the expansion (black terms in Equation $\ref{7.4.12}$) results in just the Schrödinger Equation for the unperturbed system
$\hat{H}^o | n^o \rangle = E_n^o | n^o \rangle \label{Zero}$
First-Order Expression of Energy ($\lambda=1$)
The summations in Equations $\ref{7.4.5}$, $\ref{7.4.6}$, and $\ref{7.4.10}$ can be truncated at any order of $\lambda$. For example, the first order perturbation theory has the truncation at $\lambda=1$. Matching the terms that linear in $\lambda$ (red terms in Equation $\ref{7.4.12}$) and setting $\lambda=1$ on both sides of Equation $\ref{7.4.12}$:
$\hat{H}^o | n^1 \rangle + \hat{H}^1 | n^o \rangle = E_n^o | n^1 \rangle + E_n^1 | n^o \rangle \label{7.4.13}$
Equation $\ref{7.4.13}$ is the key to finding the first-order change in energy $E_n^1$. Taking the inner product of both sides with $\langle n^o |$:
$\langle n^o | \hat{H}^o | n^1 \rangle + \langle n^o | \hat{H}^1 | n^o \rangle = \langle n^o | E_n^o| n^1 \rangle + \langle n^o | E_n^1 | n^o \rangle \label{7.4.14}$
since operating the zero-order Hamiltonian on the bra wavefunction (this is just the Schrödinger equation; Equation $\ref{Zero}$) is
$\langle n^o | \hat{H}^o = \langle n^o | E_n^o \label{7.4.15}$
and via the orthonormality of the unperturbed $| n^o \rangle$ wavefunctions both
$\langle n^o | n^o \rangle = 1 \label{7.4.16}$
and Equation $\ref{7.4.8}$ can be simplified
$\bcancel{E_n^o \langle n^o | n^1 \rangle} + \langle n^o | H^1 | n^o \rangle = \bcancel{ E_n^o \langle n^o | n^1 \rangle} + E_n^1 \cancelto{1}{\langle n^o | n^o} \rangle \label{7.4.14new}$
since the unperturbed set of eigenstates are orthogonal (Equation \ref{7.4.16}) and we can cancel the other term on each side of the equation, we find that
$E_n^1 = \langle n^o | \hat{H}^1 | n^o \rangle \label{7.4.17}$
The first-order change in the energy of a state resulting from adding a perturbing term $\hat{H}^1$ to the Hamiltonian is just the expectation value of $\hat{H}^1$ in the unperturbed wavefunctions.
That is, the first order energies (Equation \ref{7.4.13}) are given by
\begin{align} E_n &\approx E_n^o + E_n^1 \[4pt] &\approx \underbrace{ E_n^o + \langle n^o | H^1 | n^o \rangle}_{\text{First Order Perturbation}} \label{7.4.17.2} \end{align}
Example $\PageIndex{1A}$: A Perturbed Particle in a Box
Estimate the energy of the ground-state and first excited-state wavefunction within first-order perturbation theory of a system with the following potential energy
$V(x)=\begin{cases} V_o & 0\leq x\leq L \ \infty & x< 0 \;\text{and} \; x> L \end{cases} \nonumber$
Solution
The first step in any perturbation problem is to write the Hamiltonian in terms of a unperturbed component that the solutions (both eigenstates and energy) are known and a perturbation component (Equation $\ref{7.4.2}$). For this system, the unperturbed Hamilonian and solutions is the particle in an infiinitely high box and the perturbation is a shift of the potential within the box by $V_o$.
$\hat{H}^1 = V_o \nonumber$
Using Equation $\ref{7.4.17}$ for the first-order term in the energy of the ground-state
$E_n^1 = \langle n^o | H^1 | n^o \rangle \nonumber$
with the wavefunctions known from the particle in the box problem
$| n^o \rangle = \sqrt{\dfrac{2}{L}} \sin \left ( \dfrac {n \pi}{L} x \right) \nonumber$
At this stage we can do two problems independently (i.e., the ground-state with $| 1 \rangle$ and the first excited-state $| 2 \rangle$). However, in this case, the first-order perturbation to any particle-in-the-box state can be easily derived.
$E_n^1 = \int_0^L \sqrt{\dfrac{2}{L}} \sin \left ( \dfrac {n \pi}{L} x \right) V_o \sqrt{\dfrac{2}{L}} \sin \left ( \dfrac {n \pi}{L} x \right) dx \nonumber$
or better yet, instead of evaluating this integrals we can simplify the expression
$E_n^1 = \langle n^o | H^1 | n^o \rangle = \langle n^o | V_o | n^o \rangle = V_o \langle n^o | n^o \rangle = V_o \nonumber$
so via Equation $\ref{7.4.17.2}$, the energy of each perturbed eigenstate is
\begin{align*} E_n &\approx E_n^o + E_n^1 \[4pt] &\approx \dfrac{h^2}{8mL^2}n^2 + V_o \end{align*}
While this is the first order perturbation to the energy, it is also the exact value.
Example $\PageIndex{1B}$: An Even More Perturbed Particle in a Box
Estimate the energy of the ground-state wavefunction within first-order perturbation theory of a system with the following potential energy
$V(x)=\begin{cases} V_o & 0\leq x\leq L/2 \ \infty & x< 0 \; and\; x> L \end{cases} \nonumber$
Solution
As with Example 7.4.1 , we recognize that unperturbed component of the problem (Equation $\ref{7.4.2}$) is the particle in an infinitely high well. For this system, the unperturbed Hamiltonian and solution is the particle in an infinitely high box and the perturbation is a shift of the potential within half a box by $V_o$. This is essentially a step function.
Using Equation $\ref{7.4.17}$ for the first-order term in the energy of the any state
\begin{align*} E_n^1 &= \langle n^o | H^1 | n^o \rangle \[4pt] &= \int_0^{L/2} \sqrt{\dfrac{2}{L}} \sin \left ( \dfrac {n \pi}{L} x \right) V_o \sqrt{\dfrac{2}{L}} \sin \left ( \dfrac {n \pi}{L} x \right) dx + \int_{L/2}^L \sqrt{\dfrac{2}{L}} \sin \left ( \dfrac {n \pi}{L} x \right) 0 \sqrt{\dfrac{2}{L}} \sin \left ( \dfrac {n \pi}{L} x \right) dx \end{align*}
The second integral is zero and the first integral is simplified to
$E_n^1 = \dfrac{2}{L} \int_0^{L/2} V_o \sin^2 \left( \dfrac {n \pi}{L} x \right) dx \nonumber$
this is evaluated to
\begin{align*} E_n^1 &= \dfrac{2V_o}{L} \left[ \dfrac{-1}{2 \dfrac{\pi n}{a}} \cos \left( \dfrac {n \pi}{L} x \right) \sin \left( \dfrac {n \pi}{L} x \right) + \dfrac{x}{2} \right]_0^{L/2} \[4pt] &= \dfrac{2V_o}{\cancel{L}} \dfrac{\cancel{L}}{4} = \dfrac{V_o}{2} \end{align*}
The energy of each perturbed eigenstate, via Equation $\ref{7.4.17.2}$, is
\begin{align*} E_n &\approx E_n^o + \dfrac{V_o}{2} \[4pt] &\approx \dfrac{h^2}{8mL^2}n^2 + \dfrac{V_o}{2} \end{align*}
First-Order Expression of Wavefunction ($\lambda=1$)
The general expression for the first-order change in the wavefunction is found by taking the inner product of the first-order expansion (Equation $\ref{7.4.13}$) with the bra $\langle m^o |$ with $m \neq n$,
$\langle m^o | H^o | n^1 \rangle + \langle m^o |H^1 | n^o \rangle = \langle m^o | E_n^o | n^1 \rangle + \langle m^o |E_n^1 | n^o \rangle \label{7.4.18}$
Last term on right side of Equation $\ref{7.4.18}$
The last integral on the right hand side of Equation $\ref{7.4.18}$ is zero, since $m \neq n$ so
$\langle m^o |E_n^1 | n^o \rangle = E_n^1 \langle m^o | n^o \rangle \label{7.4.19}$
and
$\langle m^o | n^0 \rangle = 0 \label{7.4.20}$
First term on right side of Equation $\ref{7.4.18}$
The first integral is more complicated and can be expanded back into the $H^o$
$E_m^o \langle m^o | n^1 \rangle = \langle m^o|E_m^o | n^1 \rangle = \langle m^o | H^o | n^1 \rangle \label{7.4.21}$
since
$\langle m^o | H^o = \langle m^o | E_m^o \label{7.4.22}$
so
$\langle m^o | n^1 \rangle = \dfrac{\langle m^o | H^1 | n^o \rangle}{ E_n^o - E_m^o} \label{7.4.23}$
and therefore the wavefunction corrected to first order is:
\begin{align} | n \rangle &\approx | n^o \rangle + | n^1 \rangle \[4pt] &\approx \underbrace{| n^o \rangle + \sum _{m \neq n} \dfrac{|m^o \rangle \langle m^o | H^1| n^o \rangle }{E_n^o - E_m^o}}_{\text{First Order Perturbation Theory}} \label{7.4.24} \end{align}
Equation $\ref{7.4.24}$ is essentially is an expansion of the unknown wavefunction correction as a linear combination of known unperturbed wavefunctions $\ref{7.4.24.2}$:
\begin{align} | n \rangle &\approx | n^o \rangle + | n^1 \rangle \[4pt] &\approx | n^o \rangle + \sum _{m \neq n} c_{m,n} |m^o \rangle \label{7.4.24.2} \end{align}
with the expansion coefficients determined by
$c_{m,n} = \dfrac{\langle m^o | H^1| n^o \rangle }{E_n^o - E_m^o} \label{7.4.24.3}$
This is justified since the set of original zero-order wavefunctions forms a complete basis set that can describe any function.
Calculating the first order perturbation to the wavefunctions is in general an infinite sum of off diagonal matrix elements of $H^1$ (Figure 7.4.2 ).
• However, the denominator argues that terms in this sum will be weighted by states that are of comparable energy. That means in principle, these sum can be truncated easily based off of some criterion.
• Another point to consider is that many of these matrix elements will equal zero depending on the symmetry of the $\{| n^o \rangle \}$ basis and $H^1$ (e.g., some $\langle m^o | H^1| n^o \rangle$ integrals in Equation $\ref{7.4.24}$ could be zero due to the integrand having an odd symmetry; see Example 7.4.3 ).
The denominators in Equation $\ref{7.4.24}$ argues that terms in this sum will be preferentially dictated by states that are of comparable energy. That is, eigenstates that have energies significantly greater or lower than the unperturbed eigenstate will weakly contribute to the perturbed wavefunction.
Example 7.4.2 : A Harmonic Oscillator with a Cubic Perturbation
Estimate the energy of the ground-state wavefunction associated with the Hamiltonian using perturbation theory
$\hat{H} = \dfrac{-\hbar}{2m} \dfrac{d^2}{dx^2} + \dfrac{1}{2} kx^2 + \epsilon x^3 \nonumber$
Solution
The first step in a perturbation theory problem is to identify the reference system with the known eigenstates and energies. For this example, this is clearly the harmonic oscillator model.
Energy
The first steps in flowchart for applying perturbation theory (Figure 7.4.1 ) is to separate the Hamiltonian of the difficult (or unsolvable) problem into a solvable one with a perturbation. For this case, we can rewrite the Hamiltonian as
$\hat{H}^{o} + \hat{H}^{1} \nonumber$
where
• $\hat{H}^{o}$ is the Hamitonian for the standard Harmonic Oscillator with known eigenstates and eigenenergies $\hat{H}^{(0)}= \dfrac{-\hbar}{2m} \dfrac{d^2}{dx^2} + \dfrac{1}{2} kx^2 \nonumber$
• $\hat{H}^{1}$ is the pertubtiation $\hat{H}^{1} = \epsilon x^3 \nonumber$
The first order perturbation is given by Equation $\ref{7.4.17}$, which for this problem is
$E_n^1 = \langle n^o | \epsilon x^3 | n^o \rangle \nonumber$
Notice that the integrand has an odd symmetry (i.e., $f(x)=-f(-x)$) with the perturbation Hamiltonian being odd and the ground state harmonic oscillator wavefunctions being even. So
$E_n^1=0 \nonumber$
This means to first order pertubation theory, this cubic terms does not alter the ground state energy (via Equation $\ref{7.4.17.2})$. However, this is not the case if second-order perturbation theory were used, which is more accurate (not shown).
Wavefunction
Calculating the first order perturbation to the wavefunctions (Equation $\ref{7.4.24}$) is more difficult than energy since multiple integrals must be evaluated (an infinite number if symmetry arguments are not applicable). The harmonic oscillator wavefunctions are often written in terms of $Q$, the unscaled displacement coordinate:
$| \Psi _v (x) \rangle = N_v'' H_v (\sqrt{\alpha} Q) e^{-\alpha Q^2/ 2} \nonumber$
with $\alpha$
$\alpha =1/\sqrt{\beta} = \sqrt{\dfrac{k \mu}{\hbar ^2}} \nonumber$
and
$N_v'' = \sqrt {\dfrac {1}{2^v v!}} \left(\dfrac{\alpha}{\pi}\right)^{1/4} \nonumber$
Let's consider only the first six wavefunctions that use these Hermite polynomials $H_v (x)$:
• $H_0 = 1$
• $H_1 = 2x$
• $H_2 = -2 + 4x^2$
• $H_3 = -12x + 8x^3$
• $H_4 = 12 - 48x^2 +16x^4$
• $H_5 = 120x - 160x^3 + 32x^5$
The first order perturbation to the ground-state wavefunction (Equation $\ref{7.4.24}$)
$| 0^1 \rangle = \sum _{m \neq 0}^5 \dfrac{|m^o \rangle \langle m^o | H^1| 0^o \rangle }{E_0^o - E_m^o} \label{energy1}$
given these truncated wavefunctions (we should technically use the infinite sum) and that we are considering only the ground state with $n=0$:
$| 0^1 \rangle = \dfrac{ \langle 1^o | H^1| 0^o \rangle }{E_0^o - E_1^o} |1^o \rangle + \dfrac{ \langle 2^o | H^1| 0^o \rangle }{E_0^o - E_2^o} |2^o \rangle + \dfrac{ \langle 3^o | H^1| 0^o \rangle }{E_0^o - E_3^o} |3^o \rangle + \dfrac{ \langle 4^o | H^1| 0^o \rangle }{E_0^o - E_4^o} |4^o \rangle + \dfrac{ \langle 5^o | H^1| 0^o \rangle }{E_0^o - E_5^o} |5^o \rangle \nonumber$
We can use symmetry of the perturbation and unperturbed wavefunctions to solve the integrals above. We know that the unperturbed harmonic oscillator wavefunctions $\{|n^{0}\} \rangle$ alternate between even (when $v$ is even) and odd (when $v$ is odd) as shown previously. Since the perturbation is an odd function, only when $m= 2k+1$ with $k=1,2,3$ would these integrals be non-zero (i.e., for $m=1,3,5, ...$).
So of the original five unperturbed wavefunctions, only $|m=1\rangle$, $|m=3\rangle$, and $|m=5 \rangle$ mix to make the first-order perturbed ground-state wavefunction so
$| 0^1 \rangle = \dfrac{ \langle 1^o | H^1| 0^o \rangle }{E_0^o - E_1^o} |1^o \rangle + \dfrac{ \langle 3^o | H^1| 0^o \rangle }{E_0^o - E_3^o} |3^o \rangle + \dfrac{ \langle 5^o | H^1| 0^o \rangle }{E_0^o - E_5^o} |5^o \rangle \nonumber$
At this stage, the integrals have to be manually calculated using the defined wavefuctions above, which is left as an exercise. Notice that each unperturbed wavefunction that can "mix" to generate the perturbed wavefunction will have a reciprocally decreasing contribution (w.r.t. energy) due to the growing denominator in Equation \ref{energy1}.
Exercise 7.4.3 : Harmonic Oscillator with a Quartic Perturbation
Use perturbation theory to estimate the energy of the ground-state wavefunction associated with this Hamiltonian
$\hat{H} = \dfrac{-\hbar}{2m} \dfrac{d^2}{dx^2} + \dfrac{1}{2} kx^2 + \gamma x^4 \nonumber. \nonumber$
Answer
The model that we are using is the harmonic oscillator model which has a Hamiltonian
$H^{0}=-\frac{\hbar}{2 m} \frac{d^2}{dx^2}+\dfrac{1}{2} k x^2 \nonumber$
Making the perturbed Hamiltonian
$H^{1}=\gamma x^{4} \nonumber$
To find the perturbed energy we approximate it using Equation \ref{7.4.17.2}
$E^{1}= \langle n^{0}\left|H^{1}\right| n^{0} \rangle \nonumber$
where is the wavefunction of the ground state harmonic oscillator
$n^{0}=\left(\frac{a}{\pi}\right)^{\left(\frac{1}{4}\right)} e^{-\frac{ax^2}{2}} \nonumber$
When we substitute in the Hamiltonian and the wavefunction we get
$E^{1}=\left\langle\left(\frac{a}{\pi}\right)^{\left(\frac{1}{4}\right)} e^{-\frac{ax^2}{2}}\right|\gamma x^{4}\left|\left(\frac{a}{\pi}\right)^{\left(\frac{1}{4}\right)} e^{-\frac{ax^2}{2}} \right \rangle \nonumber$
Changing this into integral form, and combining the wavefunctions,
\begin{align*} E^{1} &=\int_{-\infty}^{\infty}\left(\frac{a}{\pi}\right)^{\left(\frac{1}{2}\right)} e^{\frac{-ax^2}{2}} \gamma x^{4} dx \[4pt] &=\gamma\left(\frac{a}{\pi}\right)^{\frac{1}{2}} \int_{-\infty}^{\infty} x^{4} e^{-a x^2} d x \end{align*}
Now we use the integral table value
$\int_{0}^{\infty} x^{2 \pi} e^{-a x^2} dx=\frac{1 \cdot 3 \cdot 5 \ldots (2 n-1)}{2^{m+1} a^{n}}\left(\frac{\pi}{a}\right)^{\frac{1}{2}} \nonumber$
Where we plug in $\mathrm{n}=2$ and $\mathrm{a}=\alpha$ for our integral
\begin{aligned} E^{1} &=2 \gamma\left(\frac{a}{\pi}\right)^{\left(\frac{1}{2}\right)} \int_{0}^{\infty} x^{4} e^{-a x^2} d x \[4pt] &=2 \gamma\left(\frac{a}{\pi}\right)^{\left(\frac{1}{2}\right)} \frac{1\cdot 3}{2^{3} a^2}\left(\frac{\pi}{a}\right)^{\frac{1}{2}}\end{aligned} \nonumber
This is our perturbed energy.
Now we have to find our ground state energy using the formula for the energy of a harmonic oscillator that we already know,
$E_{r}^{0}=\left(v+\dfrac{1}{2}\right) h \nu \nonumber$
Where in the ground state $v=0$ so the energy for the ground state of the quantum harmonic oscillator is
$E_{\mathrm{r}}^{0}=\frac{1}{2} h \nu \nonumber$
Putting both of our energy terms together gives us the ground state energy of the wavefunction of the given Hamiltonian,
\begin{align*} E &=E^{0}+E^{1} \[4pt] &=\frac{1}{2} h\nu + \gamma \frac{3}{4 a^2} \end{align*} \nonumber
Second-Order Terms ($\lambda=2$)
There are higher energy terms in the expansion of Equation $\ref{7.4.5}$ (e.g., the blue terms in Equation $\ref{7.4.12}$), but are not discussed further here other than noting the whole perturbation process is an infinite series of corrections that ideally converge to the correct answer. It is truncating this series as a finite number of steps that is the approximation. The general approach to perturbation theory applications is giving in the flowchart in Figure 7.4.1 .
Perturbation Theory Does not always Work
It should be noted that there are problems that cannot be solved using perturbation theory, even when the perturbation is very weak, although such problems are the exception rather than the rule. One such case is the one-dimensional problem of free particles perturbed by a localized potential of strength $\lambda$. Switching on an arbitrarily weak attractive potential causes the $k=0$ free particle wavefunction to drop below the continuum of plane wave energies and become a localized bound state with binding energy of order $\lambda^2$. However, changing the sign of $\lambda$ to give a repulsive potential there is no bound state, the lowest energy plane wave state stays at energy zero. Therefore the energy shift on switching on the perturbation cannot be represented as a power series in $\lambda$, the strength of the perturbation.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/07%3A_Approximation_Methods/7.04%3A_Perturbation_Theory_Expresses_the_Solutions_in_Terms_of_Solved_Problems.txt
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Solutions to select questions can be found online.
7.3
Calculate the ground state energy of Harmonic Oscillator using variation method with the following trial wavefunction
$\phi(x) = | \phi (x) \rangle = \dfrac{1}{(1+\beta x^2)^2}\nonumber$
You may require these definite integrals:
$\int _{-\infty}^{\infty} \dfrac{dx}{(1+\beta x^2)^n} = \dfrac{ (2n - 3)(2n - 5)(2n - 7)…(1)}{ (2n - 2)(2n - 4)(2n - 6)…(2) \cdot π / \beta ^{1/2}} \nonumber$
$\int _{-\infty}^{\infty} \dfrac{dx}{(1- \beta x^2)^n} = \dfrac{ (2n - 5)(2n - 7)…(1) }{ (2n - 2)(2n - 4)…(2) \cdot π / \beta^{3/2} }\nonumber$
Solution
First, we must know the Hamiltonian operator for the harmonic oscillator, which is
$\hat{H} =\dfrac{-\hbar}{2μ} \dfrac{d^2}{dx^2}+\dfrac{1}{2} kx^2\nonumber$
From this point on, the determination of $E_ø$ can be found using the trail function
$| \phi (x) \rangle = \dfrac{1}{(1+\beta x^2)^2}\nonumber$
which once substitute get the following equation for the numerator portion:
$\int_{-\infty}^{-\infty} \dfrac{1}{(1+ \betax^2)^2 [ \hbar ^2/µ*2 \beta/(1+ \betax^2)^3 - \hbar^2/µ*12 \beta^2x^2/(1+ \betax^2)^4 + \dfrac{kx^2}{2}(1+ \betax^2)^2]} \nonumber$
$=2 \beta \hbar ^2/µ*(7*5*3*1*π/8*6*4*2* \beta1/2) - 12 \beta^2 \hbar^2/µ*(7*5*3*1*π/ 10*8*6*4*2* \beta1/2) + k/2*(3*1*π/6*4*2* \beta^{3/2})\nonumber$
$= \dfrac{7π \beta^{1/2} \hbar }{32µ + kπ/32 \beta^{3/2}}\nonumber$
Now solving the denominator:
$\int _{\infty}^{\infty} \phi^*\phi\,dx= \int _{\infty}^{\infty} \dfrac{1}{(1+ \beta x^2)^4}= \dfrac{5*3*1*\pi}{6*4*2*\beta^{1/2}} = \dfrac{5π}{/16 \beta^{1/2}}\nonumber$
After this we will find
$E_\phi= \dfrac{7π \beta^{1/2} \hbar^2 }{ 32µ *(16 \beta^{1/2}/ 5π )} + \dfrac{kπ}{32 \beta^{3/2} * (16 \beta^{1/2}/ 5π )} = \dfrac{7/10* \beta \hbar 2}{µ} + \dfrac{1}{10}\dfrac{k}{\beta}\nonumber$
Then find minimum value
$\dfrac{dE_\phi}{d \beta}= \dfrac{7 \hbar^2}{10µ} -\dfrac{ k}{10 \beta^2_{min} = 0\nonumber$
therefore
$\beta_{min} = \sqrt{\dfrac{µk}{7\hbar ^2}}\nonumber$
$E_{min}= \dfrac{7^{1/2} \hbar }{5 * (k/µ)^{1/2}} + \dfrac{7^{1/2} \hbar }{5 (k/µ)^{1/2} }= 71/2 \hbar /5 (k/µ)1/2 = 0.53 \hbar *(k/µ)1/2\nonumber$
Therefore overall get
$E_{exact} =0.500 \hbar \sqrt{(k/µ)}\nonumber$
⇒ this value differs by 6%.
7.8
What is the variational (trial) energy of the trial function
$| \phi \rangle = e^{-ax^2}\nonumber$
for the ground-state of a harmonic oscillator? Just set up the integral, but do not evaluate. Use
$\hat{H} = \dfrac{\hbar^2}{2m}\nabla n^2+\dfrac{kx^2}{2}\nonumber$
Solution
The variational energy:
$E_{trial}(a) = \dfrac{\langle \phi (a) \vert\hat{H}\vert\phi (a) \rangle}{\langle\phi (a)| \phi (a)\rangle} \geq E_{true}\nonumber$
numerator:
$\langle\phi | \phi\rangle = \int_{-\infty}^{\infty} e^{-2ax^2}dx\nonumber$
All combined together to extract the trial energy as a function of $a$:
$E_{trial}(a)= \dfrac{\int_{-\infty}^{\infty} e^{-ax^2} \left[\dfrac{\hbar^2}{2m} \dfrac{d^2(e^{-ax^2})}{d x^2} + \dfrac{kx^2}{2} \right] e^{-ax^2} dx}{\int_{-\infty}^{\infty} e^{-2ax^2}dx}\nonumber$
Use the components of $\hat{H}$ to operate on $\phi$
$\langle \phi\vert\hat{H}\vert\phi\rangle = \int_{-\infty}^{\infty} e^{-ax^2} \left[ \dfrac{\hbar^2}{2m} \dfrac{d^2}{d x^2} + \dfrac{kx^2}{2} \right] e^{-ax^2}dx\nonumber$
denominator:
$\langle \phi\vert\phi\rangle = \int_{-\infty}^{\infty} e^{-ax^2} e^{-ax^2}dx\nonumber$
7.9
Use the trial function
$| \exp^{\frac{-\alpha x^{2}}{2}} \rangle \nonumber$
to set up the integrals to find the ground state energy of a anharmonic oscillator whose potential is $V(x)=cx^{5}$, but do not evaluate.
Solution
$E=\dfrac{\int_{-\infty}^{\infty}\phi^{*}\hat H\phi d\tau}{\int_{-\infty}^{\infty}\phi^{*}\phi d\tau}\nonumber$
$\int_{-\infty}^{\infty}\phi^{*}\phi d\tau = \int_{-\infty}^{\infty}\exp^{-\alpha x^{2}} dx\nonumber$
$\int_{-\infty}^{\infty}\phi^{*}\hat H\phi d\tau = \int_{-\infty}^{\infty} (\dfrac{\hbar^{2}}{2m}\nabla ^{2} + \dfrac{kx^{2}}{2} + cx^{5})\exp^{-\alpha x^{2}} dx \nonumber$
$E=\dfrac{\int_{-\infty}^{\infty} (\dfrac{\hbar^{2}}{2m}\nabla ^{2} + \dfrac{kx^{2}}{2} + cx^{5})\exp^{-\alpha x^{2}} dx}{\int_{-\infty}^{\infty}\exp^{-\alpha x^{2}} dx}\nonumber$
7.12
Consider a particle of mass $m$ in a box from $x=-a$ to $x=a$ with $V\left(x\right)=-{V_0}$ for $\left|x\right|\ge a$. Assume a trial function of the form
$| \phi (x)\rangle =l^2-x^2\nonumber$
for $-l<x<l$ and ${\psi}\left(x\right)=0$ otherwise. $l$ is the parameter. Does the trial function satisfy the requirements of a particle in a box wavefunction?
The result of the variational method was
$E_{\phi}(s)=\dfrac{5}{16}\dfrac{{\hbar }^2}{ma^2}\left[\dfrac{4}{s^2}+\dfrac{4}{5}\left(8-\dfrac{15}{s}+\dfrac{10}{s^3}-\dfrac{3}{s^5}\right)\right]\nonumber$
Where $s=\dfrac{l}{a}$ is a new variational parameter for convenience of expression. Derive a polynomial expression for $s$ that can be solved to obtain the value of $s$ that yields the ground state energy, but do not attempt to solve for this value of $s$ .
Solution
Yes, it is finite over all $x$ values, it's first and second derivatives are continuous, and it meets the boundary conditions ${\psi}\left(-a\right)={\psi}\left(a\right)=0$, and it is normalizable for a choice of $l$.
Taking the derivative of $E$ with respect to $s$,
$\dfrac{\partial E}{\partial s}=0=-\dfrac{8}{s^3}+\dfrac{4}{5}\left(\dfrac{15}{s^2}-\dfrac{30}{s^4}+\dfrac{15}{s^6}\right)\nonumber$
With some algebra, this becomes,
${3s}^4-2s^3-6s^2+3=0\nonumber$
With a calculator or other root finding procedure, $s$ can be solved for.
7.13
Given a trial wavefunction equal to $\sin \lambda (x)$, explain in words a stepwise procedure on how you would go about solving for the energy of this trial wavefunction as well as how to minimize the error.
Solution
1. Denote $\sin(\lambda (x)) = \phi_n$
2. Solve the integral $\langle \phi_n^*| \phi_n \rangle$
3. Solve the integral $\langle \phi_n^*| \hat{H} | \phi_n \rangle$
4. Now that you solved for steps 2 and 3, plug into the equation
$E_n= \dfrac{ \langle \phi_n^*| \hat{H} | \phi_n \rangle}{\langle \phi_n^*| \phi_n \rangle} \nonumber$
5. Take the derivative of $E_n$ with respect to $\lambda$ and set equal to 0.
$\dfrac{dE_n}{d \lambda }\nonumber$
6. Solve for $\lambda$ and plug back into equation in step 4.
7.16
Using the variational method approximation, find the ground state energy of a particle in a box using this trial function:
$| \phi \rangle = N\cos\left(\dfrac{\pi x}{L}\right) \nonumber$
How does is it compare to the true ground state energy?
Solution
The problem asks that we apply variational methods approximation to our trial wavefunction.
$E_{\phi} = \dfrac {\langle \phi | \hat{H} | \phi \rangle} { \langle \phi | \phi \rangle} \ge E_o \nonumber$
$\langle \phi | \phi \rangle = 1 = \int_{0}^{L}N^2\cos^2\Big(\dfrac{\pi x}{L}\Big)\nonumber$
Performing this integral and solving for N yields
$N = \sqrt{\dfrac{2}{L}}\nonumber$
The Hamiltonian for a particle in a one dimensional box is $\hat{H} = \dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}$
\begin{align*} \langle \phi | \hat{H} | \phi \rangle &= \langle N\cos\Big(\dfrac{\pi x}{L}\Big) \Big| \dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2} \Big| N\cos\Big(\dfrac{\pi x}{L}\Big) \rangle\nonumber \ &= \int_{0}^{L}N\cos\Big(\dfrac{\pi x}{L}\Big)\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}N\cos\Big(\dfrac{\pi x}{L}\Big)dx\nonumber \ &= \dfrac{N^2\pi^2 \hbar^2}{2mL^2}\int_{0}^{L}\cos^2\Big(\dfrac{\pi x}{L}\Big) dx\end{align*}
where $N = \sqrt{\dfrac{2}{L}}$. The above equation after the integral becomes
$\dfrac{\pi^2 \hbar^2}{mL^3}\Big(\dfrac{L}{2}\Big)\nonumber$
$E_\phi = \dfrac{\pi^2 \hbar^2}{2mL^2}\nonumber$
This is equal to the ground state energy of the particle in a box that we calculated from the Schrodinger equation using
$\psi = \sqrt{\dfrac{2}{L}}\sin(\dfrac{n\pi x}{L})\nonumber$
7.17
For the three-electron detrimental wavefunction
$\psi = \begin{vmatrix} \phi_A(1) & \phi_A(2) & \phi_A(3) \ \phi_B(1) & \phi_B(2) & \phi_B(3) \ \phi_C(1) & \phi_C(2) & \phi_C(3) \end{vmatrix}.\nonumber$
confirm that:
1. the interchange of two columns changes the sign of the wavefunction,
2. the interchange of two rows changes the sign of the wavefunction, and
3. the three electrons cannot have the same spin orbital.
Solution
First find the determinant
$\psi = \phi_A(1) \; \begin{vmatrix} \phi_B(2) & \phi_B(3) \ \phi_C(2) & \phi_C(3) \end{vmatrix} - \phi_A(2) \; \begin{vmatrix} \phi_B(1) & \phi_B(3) \ \phi_C(1) & \phi_C(3) \end{vmatrix} + \phi_A(3) \; \begin{vmatrix} \phi_B(1) & \phi_B(2) \ \phi_C(1) & \phi_C(2) \end{vmatrix}\nonumber$
$= \phi_A(1) \; (\phi_B(2) \phi_C(3) - \phi_C(2) \phi_B(3)) - \phi_A(2) \; (\phi_B(1) \phi_C(3) - \phi_C(1) \phi_B(3)) + \phi_A(3) \; (\phi_B(1) \phi_C(2) - \phi_C(1) \phi_B(2)) \nonumber$
$\psi= \phi_A(1)\phi_B(2)\phi_C(3) - \phi_A(1)\phi_C(2)\phi_B(3) - \phi_A(2)\phi_B(1)\phi_C(3) + \phi_A(2)\phi_C(1)\phi_B(3) + \phi_A(3)\phi_B(1)\phi_C(2) - \phi_A(3)\phi_C(1)\phi_B(2)\tag{5}\nonumber$
a) Switch column $1$ with column $2$
$\psi_{(a)} = \begin{vmatrix} \phi_A(2) & \phi_A(1) & \phi_A(3) \ \phi_B(2) & \phi_B(1) & \phi_B(3) \ \phi_C(2) & \phi_C(1) & \phi_C(3) \end{vmatrix}\nonumber$
Now find the determinant
$\phi_{(a)} = \phi_A(2) \; \begin{vmatrix} \phi_B(1) & \phi_B(3) \ \phi_C(1) & \phi_C(3) \end{vmatrix} - \phi_A(1) \; \begin{vmatrix} \phi_B(2) & \phi_B(3) \ \phi_C(2) & \phi_C(3) \end{vmatrix} + \phi_A(3) \; \begin{vmatrix} \phi_B(2) & \phi_B(1) \ \phi_C(2) & \phi_C(1) \end{vmatrix}\nonumber$
$\phi_{(a)}= \phi_A(2)\phi_B(1)\phi_C(3) - \phi_A(2)\phi_C(1)\phi_B(3) - \phi_A(1)\phi_B(2)\phi_C(3) + \phi_A(1)\phi_C(2)\phi_B(3) + \phi_A(3)\phi_B(2)\phi_C(1) - \phi_A(3)\phi_C(2)\phi_B(1)\tag{6}\nonumber$
Comparing equation $(5)$ with equation $(6)$ we see that $\phi = -\phi_{(a)}$
b) Switch row $2$ with row $3$
$\phi_{(b)} = \begin{vmatrix} \phi_A(1) & \phi_A(2) & \phi_A(3) \ \phi_C(1) & \phi_C(2) & \phi_C(3) \ \phi_B(1) & \phi_B(2) & \phi_B(3) \end{vmatrix}.\nonumber$
Now find the determinant
$\phi_{(b)} = \phi_A(1) \; \begin{vmatrix} \phi_C(2) & \phi_C(3) \ \phi_B(2) & \phi_B(3) \end{vmatrix} - \phi_A(2) \; \begin{vmatrix} \phi_C(1) & \phi_C(3) \ \phi_B(1) & \phi_B(3) \end{vmatrix} + \phi_A(3) \; \begin{vmatrix} \phi_C(1) & \phi_C(2) \ \phi_B(1) & \phi_B(2) \end{vmatrix}\nonumber$
$\phi_{(b)}= \phi_A(1)\phi_C(2)\phi_B(3) - \phi_A(1)\phi_B(2)\phi_C(3) - \phi_A(2)\phi_C(1)\phi_B(3) + \phi_A(2)\phi_B(1)\phi_C(3) + \phi_A(3)\phi_C(1)\phi_B(2) - \phi_A(3)\phi_B(1)\phi_C(2)\tag{7}\nonumber$
Comparing equation $(5)$ with equation $(7)$ we see that $\phi = -\phi_{(b)}$
c) Replace column $2$ with column $1$
$\phi_{(c)} = \begin{vmatrix} \phi_A(1) & \phi_A(1) & \phi_A(3) \ \phi_B(1) & \phi_B(1) & \phi_B(3) \ \phi_C(1) & \phi_C(1) & \phi_C(3) \end{vmatrix}\nonumber$
Now find the determinant
$\phi_{(c)} = \phi_A(1) \; \begin{vmatrix} \phi_B(1) & \phi_B(3) \ \phi_C(1) & \phi_C(3) \end{vmatrix} - \phi_A(1) \; \begin{vmatrix} \phi_B(1) & \phi_B(3) \ \phi_C(1) & \phi_C(3) \end{vmatrix} + \phi_A(3) \; \begin{vmatrix} \phi_B(1) & \phi_B(1) \ \phi_C(1) & \phi_C(1) \end{vmatrix}\nonumber$
The first two terms are identical but opposite so they cancel one another. The third has a determinant of zero.
$\phi_{(c)} = 0 + \phi_A(3) \cdot (0) = 0\nonumber$
7.20
1. What is $\hat{H}^{(0)}$, $\hat{H}^{(1)}$, $Ψ^{(0)}$, and $E^{(0)}$ for an oscillator that has a potential of $V(x) = (1/2)kx^2 + x^3 + x^4 + x^5?\nonumber$
2. What is $\hat{H}^{(0)}$, $\hat{H}^{(1)}$, $Ψ^{(0)}$, and $E^{(0)}$ for a particle in a box that has a potential of $V(x) = 0$ between 0<x<L?
3. What is $\hat{H}^{(0)}$, $\hat{H}^{(1)}$, $Ψ^{(0)}$, and $E^{(0)}$ for a hydrogenlike atom that has a potential of $V(x) = \dfrac{-e^2}{4πϵ_or} + \dfrac{1}{2}ϵ r \cos θ?\nonumber$
Solution
For an oscillator:
$\hat{H} = \dfrac{- \hbar 2}{2m} \dfrac{∂^2 }{∂x^2} - \dfrac{1}{2} k x^2 + x^3 + x^4 + x^6\nonumber$
$\hat{H}^{(0)}$ is the Hamiltonian for a simple harmonic oscillator, therefore
$\hat{H}^{(0)} = \dfrac{- ћ^ 2}{2m} \dfrac{∂^2}{ ∂x^2} - \dfrac{1}{2}kx^2\nonumber$
$\hat{H}^{(1)}$ is what is added to the Hamiltonian for a simple harmonic oscillator. therefore
$\hat{H}^{(1)} = x^3 + x^4 + x^5\nonumber$
Ψ(0) is the wave function for a simple harmonic oscillator, therefore
$Ψ^{(0)} = N_vH_v (α^{1/2}x) e^{-αx^2 /2}\nonumber$
E(0) is the energy for a simple harmonic oscillator, therefore
$E^{(0)} = h\nu \left(v+ \dfrac{1}{2} \right)\nonumber$
where v= 0,1, 2...
Particle in a box
Using this as an example, we find that for a particle in a box with potential V(x) = 0 between 0<x<L
$\hat{H} = - \dfrac{\hbar^2}{2m} \dfrac{∂^2}{∂x^2} \nonumber$
$\hat{H}^{(0)} = - \dfrac{ \hbar 2}{2m} \dfrac{ ∂^2}{∂x^2}\nonumber$
$\hat{H}^{(1)} = 0\nonumber$
Ψ(0) =B*sin(nπx/L)
E(0) = n2h2 / 8mL2 where n= 1, 2, 3 ...
Hydrogen like Atom
For a hydrogen like atom that has a potential of
$V(x) = -\dfrac{e^2}{4πϵ_or} + (1/2) ϵ r \cosθ\nonumber$
$\hat{H} = - ћ 2/2μ ∂2 /∂x2 2 -e2/(4πϵor) + (1/2)ϵrcosθ\nonumber$
$\hat{H}(0) = - ћ 2/2μ ∂2 /∂x2 2 -e2/(4πϵor)\nonumber$
$\hat{H}(1) =(1/2)ϵrcosθ\nonumber$
Ψ(0) = Ψn,l,m (r,θ,ϕ)
E(0) = μe4 / 8ϵo2h2n2
7.21
Using a harmonic oscillator as the unperturbed problem, calculate the first-order correction to the energy of the $v=0$ level for the system described as
$V(x) = \dfrac {k}{2}{x^2}+ \dfrac{m}{6}{x^3}+ \dfrac{b}{24}{x^4}\nonumber$
7.22
Using the first order perturbation theory for particle in a box, calculate the ground-state energy for the system
$V(x) = ax^{3}\qquad 0 < x < b\nonumber$
Solution
$\psi_{1} = \sqrt{\dfrac{2}{b}}\sin(\dfrac{\pi\text{x}}{b})\nonumber$
$\widehat{H} = \widehat{H}^{0}+\widehat{H}^{1}\nonumber$
$\widehat{H}^{1} = ax^{3}\nonumber$
$E_{1} = E_{1}^{0}+E_{1}^{1}\nonumber$
$E_{1}^{0} = \dfrac{h^{2}}{8mb^{2}}\nonumber$
$E_{1}^{1} = \langle\psi^{1}\vert\widehat{H}^{1}\vert\psi^{1}\rangle\nonumber$
$= \int_{0}^{b} \dfrac{2a}{b}x^{3}\sin^{2}(\dfrac{\pi\text{x}}{b})\text{dx}\nonumber$
$= \dfrac{2a}{b}\dfrac{(\pi^{2}-3)b^{4}}{8\pi^{2}}\nonumber$
$=\dfrac{(\pi^{2}-3)ab^{3}}{4\pi^{2}}\nonumber$
$E_{1} = \dfrac{h^{2}}{8mb^{2}}+\dfrac{(\pi^{2}-3)ab^{3}}{4\pi^{2}}\nonumber$
7.23
In your chemistry lab you were able to manipulate an external electric field to have the strength $\kappa$. You're supervisor wants you to figure out what the first-order correction to the ground state energy of a hydrogen like atom of charge N in this electric field.
Solution
You should remember, or look up the ground state wavefunction for a hydrogen atom and find that
$\psi_{100} = \dfrac{1}{\sqrt{\pi}}\left(\dfrac{1}{Z_\circ}\right)^\dfrac{3}{2}e^{-r/a_o}\nonumber$
Our change in energy equation has a familiar form
$\Delta E = \int\psi^{(0)*}\hat{H}^{(1)}\psi^{(0)}d\tau\nonumber$
For this problem you construct a Hamiltonian for a Hydrogen atom in an electron field with strength $\kappa$.
$\hat{H} = -\dfrac{\hbar^2}{2m_e} \nabla ^2-\dfrac{Ne^2}{4r\pi\kappa_\circ} + er\kappa\cos{\theta}\nonumber$
Luckily you have previously calculated $\hat{H}^{(1)}$ for this system in a previous experiment, simply allowing you to substitute your variables into your expressions to find that
$\Delta E = \dfrac{Ne\kappa}{\pi}\Bigg(\dfrac{1}{Z_\circ}\Bigg)^3 \int_{0}^{\infty}r^3e^\dfrac{-r}{a_\circ}dr\int_{0}^{2\pi}d\phi\int_{0}^{\pi}\sin{\theta}\cos{\theta}d\theta\nonumber$
Notice that the problem gets simplified by the fact that
$\int_{0}^{\pi}\sin{\theta}\cos{\theta}d\theta = 0\nonumber$
So your answer is a trivial solution.
$\boxed{\Delta E = 0}\nonumber$
7.25A
Use first-order perturbation theory to calculate ground-state energy of a harmonic oscillator with a $cx^7$ added to the end of the potential.
Solution
The Hamiltonian to the system can be formulated as
$\hat{H}= \dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}+\dfrac{1}{2} kx^2 + cx^7 \nonumber$
we then solve
$E^1 = \langle \psi_0|cx^7|\psi_0 \rangle \nonumber$
We know that the integral is of an odd function over a symmetric boundary is 0, so by symmetry we can conclude that the energy is 0.
7.25B
In order to calculate the first-order correction to the ground-state energy of the quartic oscillator, use first-order perturbation theory. The potential energy is $V$($x$) = $c$$x$4. For this potential use the harmonic oscillator as the unperturbed system. Solve for the perturbing potential as well.
Solution
The Hamiltonian operator is given below:
$\hat{H}=-\dfrac{\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+cx^4\nonumber$
To use a harmonic oscillator as the reference system, add and subtract $\dfrac {1}{2} kx^2$ from $\hat{H}$.
$\hat{H}=-\dfrac{\hbar^2}{2\mu}\dfrac{d}{dx^2}+\dfrac{1}{2}kx^2\ + cx^4 - \dfrac{1}{2}kx^2\nonumber$
Hence we get :
$\hat{H}^{(0)}=cx^4-\dfrac {1}{2} kx^2\nonumber$
Now we have:
∆E = $\int\psi^{(0)*}\hat{H}^{(1)}\psi^{(0)} \;d\tau$ . . . .
By putting the values in the equation above, we get:
$\Delta E = \left(\dfrac{\alpha}{\pi}\right)^{1/2} \int_{-\infty}^{\infty} dxe^{-x^2\alpha} \left(cx^4-\dfrac {1}{2} kx^2\right)\nonumber$
$=\left(\dfrac {\alpha}{\pi}\right)^{1/2} 2 \left[\dfrac {3c}{8\alpha^2} \left(\dfrac {\pi}{\alpha}\right)^{1/2} -\dfrac {k}{8\alpha}\left(\dfrac {\pi}{\alpha}\right)^{1/2}\right]\nonumber$
$=\dfrac {3c}{4\alpha^2} - \dfrac {k}{4\alpha}\nonumber$
7.26
Solve the following integrals using this trial wavefunction
$| \phi \rangle = c_1x(a-x) + c_2x^{2}(a-x)^{2}\nonumber$
For simplicity purposes, we can assume that a = 1.
$H_{11} = \dfrac{\hbar^{2}}{6m}$ $S = \dfrac{1}{30}$
$H_{12} = H_{22} = \dfrac{\hbar^{2}}{30m}$ $S_{12} = S_{21} = \dfrac{1}{140}$
$H_{22} = \dfrac{\hbar^2}{105m}$ $S_{22} = \dfrac{1}{630}$
Solution
We know that for a particle in a box $\hat{H} = \dfrac{-\hbar^{2}}{2m}\dfrac{d^{2}}{dx^{2}}\nonumber$
We also know the two components of the trial function that was given are
$\phi_{1} = x(a-x)\nonumber$ and $\phi_{2} = x^{2}(a-x)^{2}\nonumber$
Using this we will have
$\hat{H}\phi_{1} = \dfrac{\hbar^2}{2m}\nonumber$ and $\hat{H}\phi_{2} = \dfrac{\hbar^2}{m}(a^{2} - 6ax + 6x^{2})\nonumber$
Using this we can solve for $H_{ii}$ and $S_{ij}$ using this integral
$\int_{0}^{1} x^{m}(1-x)^{n}dx = \dfrac{m!n!}{(m+n+1)!}\nonumber$
Letting $a = 1$, we can now solve for
$H_{11} = \dfrac{\hbar^{2}}{m}\nonumber$ $\int_{0}^{1} x(1-x)dx = \dfrac{\hbar^{2}}{6m}\nonumber$
$H_{12} = \dfrac{\hbar^{2}}{m}\nonumber$ $\int_{0}^{1} x(1-x)(1-6x+6x^{2})dx = \dfrac{\hbar^{2}}{30m}\nonumber$
$H_{21} = \dfrac{\hbar^{2}}{m}\nonumber$ $\int_{0}^{1} x^{2}(1-x)^{2}dx = \dfrac{\hbar^{2}}{30m}\nonumber$
$H_{22} = \dfrac{\hbar^2}{105m}\nonumber$
$S_{11} = \int_{0}^{1} x^{2}(1-x)^{2}dx = \dfrac{4}{5!} = \dfrac{1}{30}\nonumber$
$S_{12} = S_{21} = \int_{0}^{1} x^{3}(1-x)^{3}dx = \dfrac{36}{7!}= \dfrac{1}{140}\nonumber$
$S_{22} = \int_{0}^{1} x^{4}(1-x)^{4}dx = \dfrac{576}{9!}= \dfrac{1}{630}\nonumber$
7.27
Use Perturbation Theory to add cubic and quartic perturbations to the SHO and find the first three SHO energy levels. Do this by expanding the Morse potential:
$V(x)=D(1 - e-^{Bx})^2\nonumber$
into polynomials (i.e., a Taylor expansion). Show that the Hamiltonian can be written as
$\dfrac{-h^2∇^2}{8π^2m} + ax^2 + bx^3 + cx^4\nonumber$
Note which terms can be associated with $H^0$ and which are the $H^1$ perturbation. What are the relationships between a, b, c, and D, B? How do the new energy levels compare to the old ones?
Solution
The $e^{-Bx}$ function can be expanded noting that
$e^x \approx 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + ... + Ox^n\nonumber$
So e-Bx will expand similarly, replacing x in the above expansion with -Bx, so
$e^{-Bx] = 1 - Bx + B^2x^2/2 - B^3x^3/6 + ... + Ox^n\nonumber$
The Morse Potential therefore is
$D(1 - (1 - Bx + B^2\dfrac{x^2}{2} - B^3\dfrac{x^3}{6})^2\nonumber$
The expansion is shortened to 4 terms only.
= D( Bx - B2x2/2 + B3x3/6)2
=D (B6x6/36 - B5x5/6 + 7B4x4/12 - B3x3 + B2x2)
= DB6x6/36 - DB5x5/6 + 7DB4x4/12 - DB3x3 + DB2x2
= 7DB4x4/12 - DB3x3 + DB2x2
(We have truncated above the quartic term)
Here, it is seen that DB2x2 corresponds to the H0 potential, and 7DB4x4/12 - DB3x3 is H1
We can also see that a = DB2, b = - DB3, c = 7DB4/12 in in the Hamiltonian potential: ax2 + bx3 + cx4
Perturbation theory states that
$E_n = E^0_n + E^1_n = E^0_n + ∫Ψ^0_nH^1Ψ^0_n \,dτ \nonumber$
Therefore, with E00 = hv/2 and Ψ00 = (α/π)1/4e-α(x^2)/2
E01 = 3hv/2 and Ψ01 = (4α3/π)1/4xe-α(x^2)/2
E02 =5 hv/2 and Ψ02 = (α/4π)1/4(2αx2 - 1)e-α(x^2)/2
H1 = bx2 + cx2
the first three energy levels are:
E0 = hv/2 + (α/π)1/4e-α(x^2)/2( bx3 + cx4)(α/π)1/4e-α(x^2)/2dx
= hv/2 + (α/π)1/2 e-α(x^2)( bx3 + cx4)dx
= hv/2 + (α/π)1/2 [e-α(x^2)bx3dx + e-α(x^2)cx4dx] (The cubic integral is odd so evaluates to 0)
= hv/2 + (α/π)1/2 e-α(x^2)cx4dx
We can use ∫x2ne-αx^2dx = n!/(2αn + 1) (This is true from 0 to infinity, so we must double it)
= hv/2 + 2 * c(α/π)1/2 * 3/(23α2) * (π/α)1/2
= hv/2 + 3c/(4α2)
E1 = 3hv/2 + (4α3/π)1/4xe-α(x^2)/2( bx3 + cx4)(4α3/π)1/4xe-α(x^2)/2dx
= 3hv/2 + (4α3/π)1/2 x2e-α(x^2)( bx3 + cx4)dx
= 3hv/2 + (4α3/π)1/2 [x2e-α(x^2) bx3dx + x2e-α(x^2) cx4dx ] (First integral evaulates to 0)
= 3hv/2 + c(4α3/π)1/2 x6e-α(x^2) dx
We can use ∫x2ne-αx^2dx = n!/(2αn + 1) (This is true from 0 to infinity, so we must double it)
= 3hv/2 + 2* c(4α3/π)1/2 * 15/(24α3) * (π/α)1/2
= 3hv/2 + 15c/(4α2)
E2o = 5hv/2 + (α/4π)1/4(2αx2 - 1)e-α(x^2)/2( bx3 + cx4) (α/4π)1/4(2αx2 - 1)e-α(x^2)/2dx
= 5hv/2 + (α/4π)1/2( bx3 + cx4)(2αx2 - 1)2e-α(x^2)dx
= 5hv/2 + (α/4π)1/2[bx3(2αx2 - 1)2e-α(x^2)dx + cx4e-α(x^2)(2αx2 - 1)2dx
= 5hv/2 + (α/4π)1/2cx4e-α(x^2)(2αx2 - 1)2dx (First integral evaluates to 0)
= 5hv/2 + (α/4π)1/2∫4cα2x8e-α(x^2) - 4αcx6e-α(x^2) + cx4e-α(x^2)dx
We can use ∫x2ne-αx^2dx = n!/(2αn + 1) (This is true from 0 to infinity, so we must double it)
= 5hv/2 + (α/4π)1/2[4cα2*2*(105/(32α4))* (π/α)1/2 - αc*2*15/(24α3) * (π/α)1/2 + c*2*3/(23α2) * (π/α)1/2]
= 5hv/2 + 39c/4α2
It is evident that as the energy levels increase, the perturbation to the energy increases as well, making the Hooke potential increasingly bad as an approximation of intramolecular potential.
7.27
Use the perturbation theory to calculate the first - order corrections to the ground state energy of
1. A harmonic oscillator that arises from a cubic and quartic term.
2. A quartic oscillator that arises from only using a quartic term $cx^4$
and compare the results.
Solution
A) The Hamiltonian for this problem is
$\hat{H}=\dfrac{-\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+ax^2+bx^3+cx^4\nonumber$
We use the harmonic oscillator Hamiltonian for $\hat{H}^{(0)}$
$\hat{H}^{(0)}=\dfrac{-\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+ax^2\nonumber$
$\hat{H}^{(1)}=bx^3+cx^4\nonumber$
$\psi^{(0)}=N_vH_v(\alpha^{1/2}x)e^{-\alpha x^2/2} \nonumber$
$E^{(0)}=h\mu (v+\dfrac{1}{2})\nonumber$
$E_0=E_0^{(0)}+\int\psi^{(0)*}\hat{H}^{(1)}\psi^{(0)} \;d\tau\nonumber$
$E_0 = \dfrac{\hbar \mu}{2}+b(\dfrac{\alpha}{\pi})^{1/2} \int_{-\infty}^{\infty} dxx^3e^{-x^2\alpha} +c(\dfrac{\alpha}{\pi})^{1/2} \int_{-\infty}^{\infty} dxx^4e^{-x^2\alpha}\nonumber$
$E_0 = \dfrac{\hbar \mu}{2}+0+2c \dfrac{\alpha}{\pi}^{1/2} \int_{0}^{\infty} dxx^4e^{-x^2\alpha}\nonumber$
$E_0=\dfrac{\hbar \mu}{2}+\dfrac{3c}{4\alpha^2}\nonumber$
B) The Hamiltonian for this problem is
$\hat{H}=\dfrac{-\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+cx^4\nonumber$
We use the harmonic oscillator Hamiltonian for $\hat{H}^{(0)}$
$\hat{H}^{(0)}=\dfrac{-\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+ax^2\nonumber$
$\hat{H}^{(1)}=cx^4-\dfrac{kx^2}{2}\nonumber$
$E= \int\psi^{(0)*}\hat{H}^{(1)}\psi^{(0)} \;d\tau\nonumber$
$E=(\dfrac{\alpha}{\pi})^{1/2} \int_{-\infty}^{\infty} dxe^{-x^2\alpha}(cx^4-\dfrac{kx^2}{2})\nonumber$
$E =(\dfrac{\alpha}{\pi})^{1/2} 2(\dfrac{3c}{8\alpha^2} (\dfrac{\alpha}{\pi})^{1/2}-\dfrac{k}{8\alpha}(\dfrac{\alpha}{\pi})^{1/2}) \nonumber$
$E=\dfrac{3c}{4\alpha^2}+\dfrac{k}{4\alpha}\nonumber$
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Electrons with more than one atom, such as Helium (He), and Nitrogen (N), are referred to as multi-electron atoms. Hydrogen is the only atom in the periodic table that has one electron in the orbitals under ground state. We will learn how additional electrons behave and affect a certain atom.
• 8.1: Atomic and Molecular Calculations are Expressed in Atomic Units
Atomic units form a system of natural units which is especially convenient for atomic physics calculations. There are two different kinds of atomic units, Hartree atomic units and Rydberg atomic units, which differ in the choice of the unit of mass and charge. This article deals with Hartree atomic units, where the numerical values of the following four fundamental physical constants are all unity by definition.
• 8.2: Perturbation Theory and the Variational Method for Helium
Both perturbation theory and variation method (especially the linear variational method) provide good results in approximating the energy and wavefunctions of multi-electron atoms. We address both approximations with respect to the helium atom.
• 8.3: Hartree-Fock Equations are Solved by the Self-Consistent Field Method
The Hartree method is used to determined the wavefunction and the energy of a quantum multi-electron system in a stationary state. The Hartree method often assumes that the exact, N-body wave function of the system can be approximated by a product of single-electron wavefucntions. By invoking the variational method, one can derive a set of N-coupled equations for the N spin orbitals.
• 8.4: An Electron has an Intrinsic Spin Angular Momentum
Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to the classical angular momentum of orbital revolution. The existence of spin angular momentum is inferred from the Stern–Gerlach experiment, in which particles are observed to possess angular momentum that cannot be accounted for by orbital angular momentum alone.
• 8.5: Wavefunctions must be Antisymmetric to Interchange of any Two Electrons
The probability |Ψ(r1, r2)|² should be identical to the probability |Ψ(r2, r1)|² because the electrons have no label and they cannot be told apart because of Heisenberg principle. You can naively think that Ψ(r1, r2)=±Ψ(r2, r1) but it turns out that the sign must always be minus for the electrons. This is an additional postulate of quantum mechanics.
• 8.6: Antisymmetric Wavefunctions can be Represented by Slater Determinants
John Slater introduced an idea of a Slater determinant that is a relatively simple scheme for constructing antisymmetric wavefunctions of multi-electron systems from a product of one-electron functions in the form of a determinant.
• 8.7: Hartree-Fock Calculations Give Good Agreement with Experimental Data
The Hartree–Fock method is a method of approximation for the determination of the wave function and the energy of quantum many-body systems. The Hartree–Fock method often assumes that the exact, N-body wave function of the system can be approximated by a single Slater determinant of N spin-orbitals. By invoking the variational method, one can derive a set of N-coupled equations for the N spin orbitals. A solution of these equations yields the Hartree–Fock wave function and energy of the system.
• 8.8: Term Symbols Gives a Detailed Description of an Electron Configuration
The term symbol is an abbreviated description of the (total) angular momentum quantum numbers in a multi-electron atom (however, even a single electron can be described by a term symbol). Each energy level of an atom with a given electron configuration is described by not only the electron configuration but also its own term symbol, as the energy level also depends on the total angular momentum including spin. The usual atomic term symbols assume LS coupling (Russell-Saunders coupling).
• 8.9: The Allowed Values of J - the Total Angular Momentum Quantum Number
The total angular momentum quantum number parameterizes the total angular momentum of a given particle, by combining its orbital angular momentum and its intrinsic angular momentum (i.e., its spin).
• 8.10: Hund's Rules Determine the Term Symbols of the Ground Electronic States
The allocation electrons among degenerate orbitals can be formalized by Hund’s rule: For an atom in its ground state, the term with the highest multiplicity has the lowest energy. Hund's first rule states that the lowest energy atomic state is the one that maximizes the total spin quantum number for the electrons in the open subshell. The orbitals of the subshell are each occupied singly with electrons of parallel spin before double occupation occurs.
• 8.11: Using Atomic Term Symbols to Interpret Atomic Spectra
The electronic states that result from these excited orbital configurations are characterized by term symbols and are essential in understanding the spectra and energy level structure of atoms, and the orbital electron configurations. The orbital configurations help us understand many of the general or coarse features of spectra and are necessary to produce a physical picture of how the electron density changes because of a spectroscopic transition.
• 8.E: Multielectron Atoms (Exercises)
These are homework exercises to accompany Chapter 8 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Thumbnail: Neon Atom. (CC BY 3.0 Unported; BruceBlaus via Wikipedia)
08: Multielectron Atoms
Learning Objectives
• Demonstrate how solving electron structure problems are less cluttered by switching to atomic units instead of SI units.
Atomic units (au or a.u.) form a system of natural units which is especially convenient for atomic physics calculations. Atomic units, like SI units, have a unit of mass, a unit of length, and so on. However, the use and notation is somewhat different from SI. Suppose a particle with a mass of m has 3.4 times the mass of electron. The value of mass $m$ can be written in three ways:
• $m=3.4\; m_e$: This is the clearest notation (but least common), where the atomic unit is included explicitly as a symbol.
• $m=3.4\; a.u.$: This notation is ambiguous, but is common. Here, it means that the mass $m$ is 3.4 times the atomic unit of mass. If considering a length $L$ of 3.4 times the atomic unit of length, the equation would look the same, $L= 3.4 \;a.u.$ The dimension needs to be inferred from context, which is sloppy.
• $m = 3.4$: This notation is similar to the previous one, and has the same dimensional ambiguity. It comes from formally setting the atomic units to 1 (Table 8.1.1 ).
This article deals with "Hartree type" of atomic units, where the numerical values of the following four fundamental physical constants are all unity by definition:
Dimension Name Symbol/Definition Value in SI units Value in Atomic Units
Table 8.1.1 : Fundamental atomic units
mass electron rest mass $m_e$ 9.109×10−31 kg 1
charge elementary charge $e$ 1.602×10−19 C 1
action reduced Planck's constant $\hbar = \dfrac{h}{2\pi}$ 1.054×10−34 J·s 1
electric constant−1 Coulomb force constant $\displaystyle k_e = \frac{1}{4 \pi \epsilon_o}$ 8.987 x 109 kg·m3·s−2·C−2 1
Example 8.1.1 : Simplifying the Hamiltonian
Use the atomic units definitions in Table 8.1.1 to contrast the Hamiltonian for a Helium atom in Si units and in atomic units.
Solution
In SI units, the Hamiltonian for a Helium atom is
$\hat {H} = -\dfrac {\hbar ^2}{2m_e} (\nabla ^2_1 + \nabla ^2_2) -\dfrac {2e^2}{4 \pi \epsilon _0 r_1} - \dfrac {2e^2}{4 \pi \epsilon _0 r_2} + \dfrac {e^2}{4 \pi \epsilon _0 r_{12}} \nonumber$
In atomic units, the same Hamiltonian
$\hat {H} = -\dfrac {1}{2} (\nabla ^2_1 + \nabla ^2_2) - \dfrac {2}{r_1} - \dfrac {2}{r_2} + \dfrac {1}{r_{12}} \nonumber$
All the units that make the SI version of the Hamiltonian disappear to emphasize the key aspects of the operator.
Atomic units are derived from certain fundamental properties of the physical world, and are free of anthropocentric considerations. It should be kept in mind that atomic units were designed for atomic-scale calculations in the present-day universe, with units normalize the reduced Planck constant and also mass and charge of the electron are set to 1, and, as a result, the speed of light in atomic units is a large value, $1/\alpha \approx 137$. For example, the orbital velocity of an electron around a small atom is of the order of 1 in atomic units. Table 8.1.2 give a few derived units. Some of them have proper names and symbols assigned, as indicated in the table.
Table 8.1.2 : Derived atomic units
Dimension Name Symbol Expression Value in SI units Value in more common units
length bohr $a_o$ $4\pi \epsilon_0 \hbar^2 / (m_\mathrm{e} e^2) = \hbar / (m_\mathrm{e} c \alpha)$ 5.291×10−11 m 0.052 nm = 0.529 Å
energy hartree $E_h$ $m_\mathrm{e} e^4/(4\pi\epsilon_0\hbar)^2 = \alpha^2 m_\mathrm{e} c^2$ 4.359×10−18 J 27.2 eV = 627.5 kcal·mol−1
time $\hbar / E_\mathrm{h}$ 2.418×10−17 s
velocity $a_0 E_\mathrm{h} / \hbar = \alpha c$ 2.187×106 m·s−1
Bohr model in atomic units
Atomic units are chosen to reflect the properties of electrons in atoms. This is particularly clear from the classical Bohr model of the hydrogen atom in its ground state. The ground state electron orbiting the hydrogen nucleus has (in the classical Bohr model):
• Orbital velocity = 1
• Orbital radius = 1
• Angular momentum = 1
• Orbital period = 2π
• Ionization energy = 12
• Electric field (due to nucleus) = 1
• Electrical attractive force (due to nucleus) = 1
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Learning Objectives
• Demonstrate that both perturbation theory and variational methods can be used to solve the electron structure of the helium atom.
Both perturbation theory and variation method (especially the linear variational method) provide good results in approximating the energy and wavefunctions of multi-electron atoms. Below we address both approximations with respect to the helium atom.
Perturbation Theory of the Helium Atom
We use perturbation theory to approach the analytically unsolvable helium atom Schrödinger equation by focusing on the Coulomb repulsion term that makes it different from the simplified Schrödinger equation that we have just solved analytically. The electron-electron repulsion term is conceptualized as a correction, or perturbation, to the Hamiltonian that can be solved exactly, which is called a zero-order Hamiltonian. The perturbation term corrects the previous Hamiltonian to make it fit the new problem. In this way the Hamiltonian is built as a sum of terms, and each term is given a name. For example, we call the simplified or starting Hamiltonian, $\hat {H} ^0$, the zero order term, and the correction term $\hat {H} ^1$.
$\hat {H} = \hat {H} ^0 + \hat {H} ^1 \label {9-17}$
The Hamilonian for the helium atom (in atomic units) is:
\begin{align} \hat {H} ^0 &= \underbrace{-\dfrac {1}{2} \nabla ^2_1 - \dfrac {2}{r_1}}_{\text{H atom Hamiltonian}} - \underbrace{\dfrac {1}{2} \nabla ^2_2 - \dfrac {2}{r_2}}_{\text{H atom Hamiltonian}} \label {9-18} \[4pt] \hat {H} ^1 &= \dfrac {1}{r_{12}} = \dfrac{1}{|r_1-r_2|} \label {9-19} \end{align}
The expression for the first-order correction to the energy is
\begin{align} E^1 &= \langle \psi ^{0} | \hat {H} ^1 | \psi ^{0} \rangle \nonumber \[4pt] &= \int \psi ^{0*} \hat {H} ^1 \psi ^0 \,d\tau \nonumber \end{align} \label {9-28}
Equation $\ref{9-28}$ is a general expression for the first-order perturbation energy, which provides an improvement or correction to the zero-order energy we already obtained. Hence, $E^1$ is the average interaction energy of the two electrons calculated using wavefunctions that assume there is no interaction.
The solution to $\hat{H}^{0}$ (Equation \ref{9-18}) is the product of two single-electron hydrogen wavefunctions (scaled by the increased nuclear charge) since $\hat{H}^{0}$ can be separated into independent functions of each electron (i.e., Separation of Variables).
$| \psi ^{0} \rangle = | \varphi _{1s} (r_1) \varphi _{1s} (r_2) \rangle \nonumber$
So the integral in Equation $\ref{9-28}$ is
$E^1 = \iint \varphi _{1s} (r_1) \varphi _{1s} (r_2) \dfrac {1}{r_{12}} \varphi _{1s} (r_1) \varphi _{1s} (r_2)\, d\tau _1 d\tau _2 \label {9-29}$
where the double integration symbol represents integration over all the spherical polar coordinates of both electrons $r_1, \theta _1, \varphi _1 , r_2 , \theta _2 , \varphi _2$. The evaluation of these six integrals is lengthy. When the integrals are done, the result is $E^1$ = +34.0 eV so that the total energy calculated using our second approximation method, first-order perturbation theory, is
$E_{approx2} = E^0 + E^1 = - 74.8 eV \label {9-30}$
The new approximate value for the binding energy represents a substantial (~30%) improvement over the zero-order energy:
$E^{0} = \dfrac{2}{n^2} + \dfrac{2}{n^2} = 4\, \underbrace{E_h}_{hartrees} = 108.8\, eV \nonumber$
so the interaction of the two electrons is an important part of the total energy of the helium atom. We can continue with perturbation theory and find the additional corrections, $E^2$, $E^3$, etc. For example,
$E^0 + E^1 + E^2 = -79.2\, eV. \nonumber$
So with two corrections to the energy, the calculated result is within 0.3% of the experimental value of -79.01 eV. It takes thirteenth-order perturbation theory (adding $E^1$ through $E^{13}$ to $E^0$) to compute an energy for helium that agrees with experiment to within the experimental uncertainty. Interestingly, while we have improved the calculated energy so that it is much closer to the experimental value, we learn nothing new about the helium atom wavefunction by applying the first-order perturbation theory to the energy above. He need to expand the wavefunctions to first order perturbation theory, which requires more effort. Below, we will employ the variational method approximation to modify zero-order wavefunctions to address one of the ways that electrons are expected to interact with each other.
The Hartree Unit of Energy
The hartree is the atomic unit of energy (named after the British physicist Douglas Hartree) and is defined as
$E_h=2R_Hhc \nonumber$
where $R_H$ is the Rydberg constant, $h$ is the Planck constant and $c$ is the speed of light.
\begin{align} E_h &= 4.359 \times 10^{−18} \,J \nonumber \[4pt] &= 27.21\, eV. \nonumber \end{align} \nonumber
The hartree is usually used as a unit of energy in atomic physics and computational chemistry. As discussed before for hydrogen emission, IR, and microwave spectroscopies, experimental measurements prefer the electronvolt ($eV$) or the wavenumber ($cm^{−1}$).
Variational Method Applied to the Helium Method
As discussed in Section 6.7, because of the electron-electron interactions, the Schrödinger's Equation cannot be solved exactly for the helium atom or more complicated atomic or ionic species. However, the ground-state energy of the helium atom can be estimated using approximate methods. One of these is the variational method which requires the minimizing of the following variational integral.
\begin{align} E_{trial} &= \dfrac{\langle \psi_{trial}| \hat{H} | \psi_{trial} \rangle }{\langle \psi_{trial}| \psi_{trial} \rangle}\label{7.3.1b} \[4pt] &= \dfrac{\displaystyle \int_0^{\infty} \psi_{trial}^* \hat{H} \psi_{trial} d\tau}{\displaystyle \int_0^{\infty} \psi_{trial}^2 \, d\tau} \label{7.3.1a} \end{align}
The five trial wavefunctions discussions below are equally "valid" trial wavefunctions that describe the probability of finding each electron (technically the wavefunction squared). What separates the "poor" approximations from the "good" approximation is whether the trial wavefunction predicts experimental results. Consequently, for all the approximations used for the rest of this TextMap, it is important to compare the theoretical results to the "true" (i.e., experimental) results. No matter how complicated an approximation is, it is only as good as the accuracy of its predicted values to experimental values.
Trial Wavefunction #1: Simple Orbital Approximation with One Parameter
As is clear from Equation $\ref{7.3.1b}$, the variational method approximation requires that a trial wavefunction with one or more adjustable parameters be chosen. A logical first choice for such a multi-electron wavefunction would be to assume that the electrons in the helium atom occupy two identical, but scaled, hydrogen 1s orbitals.
\begin{align} | \psi (1,2) \rangle_{trial} &= \phi (1) \phi (2) \[4pt] &= \exp\left[- \alpha (r_1 +r_2)\right] \label{7.3.2} \end{align}
The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.2}) by varying $\alpha$ is
$E_{trial} = -2.84766 \;E_h \nonumber$
and the experimentally determined ground-state energy for the helium atom is the sum of first and second ionization energies
$E_{\exp}= I_1 +I_2 = -2.90372 \;E_h \label{exp}$
The deviation of energy for the optimized trial wavefunction from the experimental value is
\begin{align} \left| \dfrac{E_{trial}(\alpha)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.84766 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \[4pt] &= 1.93 \% \label{trial1} \end{align}
The value of -2.8477 hartrees is within 2% of the known ground-state energy of the helium atom. The error in the calculation is attributed to the fact that the wavefunction is based on the orbital approximation and, therefore, does not adequately take electron-electron interactions into account. In other words, this wavefunction gives the electrons too much independence, given that they have like charges and tend to avoid one another.
Trial Wavefunction #2: Orbital Approximation with Two Parameters
Some electron-electron interactions can be built into the multi-electron wavefunction by assuming that each electron is in an orbital which is a linear combination of two different and scaled hydrogen 1s orbitals.
$\phi(r_1) = \exp(- \alpha r_1) + \exp(- \beta r_1) \label{7.3.3}$
Under the orbital approximation this assumption gives a trial wavefunction of the form
\begin{align} | \psi (1,2) \rangle_{trial} &= \phi (1) \phi (2) \label{7.3.4a} \[4pt] &= {\exp(- \alpha r_1 )\exp(- \alpha r_2)}+\exp(- \alpha r_1 )\exp(- \beta r_2)+\exp(- \beta r_1 )\exp(- \alpha r_2 )+\exp(- \beta r_1 )\exp(- \beta r_2 ) \label{7.3.4b} \end{align}
Inspection of this trial wavefunction indicates that 50% of the time the electrons are in different orbitals, while for the first trial wavefunction the electrons were in the same orbital 100% of the time. Notice the enormous increase in the complexity of the variational expression for the energy for this trial wavefunction (Equation $\ref{7.3.1a}$). However, the calculation is very similar to that using the previous trial wavefunction. The differences are that in this case the expression for the energy is more complex and that it is being minimized simultaneously with respect to two parameters ($\alpha$ and $\beta$) rather than just one ($\alpha$).
The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.4b}) by varying $\alpha$ and $\beta$ is
$E_{trial} =-2.86035 \;E_h \nonumber$
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\begin{align} \left| \dfrac{E_{trial}(\alpha, \beta)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.86035 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \[4pt] &= 1.49 \% \label{trial2} \end{align}
Clearly introducing some electron-electron interactions into the trial wavefunction has improved the agreement between theory and experiment (Equation \ref{trial1} vs. \ref{trial2}).
Trial Wavefunction #3: Orbital Approximation with Two Parameters
The extent of electron-electron interactions can be increased further by eliminating the first and last term in the second trial wavefunction (Equation $\ref{7.3.4b}$). This yields a multi-electron wavefunction of the form,
$| \psi (1,2) \rangle_{trial} = \exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \label{7.3.5}$
This trial wavefunction places the electrons in different scaled hydrogen 1s orbitals 100% of the time this adds further improvement in the agreement with the literature value of the ground-state energy is obtained. The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.5}) by varying $\alpha$ and $\beta$ is
$E_{trial} = -2.87566 \;E_h \nonumber$
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\begin{align} \left| \dfrac{ E_{trial} (\alpha,\beta)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.87566 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \[4pt] &= 0.97 \%\label{trial3} \end{align}
This result is within 1% of the actual ground-state energy of the helium atom.
Trial Wavefunction #4: Approximation with Two Parameters
The third trial wavefunction, however, still rests on the orbital approximation and, therefore, does not treat electron-electron interactions adequately. Hylleraas took the calculation a step further by introducing electron-electron interactions directly into the first trial wavefunction by adding a term, $r_{12}$, involving the inter-electron separation.
$| \psi_{trial} (1,2) \rangle = \left(\exp[- \alpha ( r_1 + r_2 )]\right) \left(1 + \beta r_{12} \right) \label{7.3.6}$
In the trial multi-electron wavefunction of Equation \ref{7.3.6}, if the electrons are far apart, then $r_{12}$ is large and the magnitude of the wavefunction increases to favor that configuration. The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.6}) by varying $\alpha$ and $\beta$ is
$E_{trial} = - 2.89112\; E_h \nonumber$
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\begin{align} \left| \dfrac{ E_{trial} (\alpha,\beta)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{- 2.89112 \;E_h + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \[4pt] &= 0.43 \% \label{trial4} \end{align}
This modification of the trial wavefunction has further improved the agreement between theory and experiment to within 0.5%.
Fifth Trial Wavefunction #5: Approximation with Three Parameters
Chandrasakar brought about further improvement by adding Hylleraas's $r_{12}$ term to the third trial wavefunction (Equation $\ref{7.3.5}$) as shown here.
$| \psi (1,2) \rangle_{trial} = \left[\exp(- \alpha r_1 )\exp(- \beta r_2 ) + \exp(- \beta r_1 )\exp(- \alpha r_2 ) \right][1 + \gamma r _{12} ] \label{7.3.7}$
Chandrasakar's three parameter wavefunction gives rise to a fairly complicated variational expression for ground-state energy. The variational energy obtained after minimizing Equation $\ref{7.3.1a}$ after substituting the trial wavefunction (Equation \ref{7.3.7}) by varying $\alpha$, $\beta$ and $\gamma$ is
$E_{trial} = -2.90143 \;E_h \nonumber$
The deviation of energy for the optimized trial wavefunction from the experimental value (Equation \ref{exp}) is
\begin{align} \left| \dfrac{ E_{trial} (\alpha, \beta, \gamma)-E_{\exp}}{E_{\exp}} \right| &= \left| \dfrac{-2.90143 + 2.90372 \;E_h}{-2.90372 \;E_h} \right| \[4pt] &= 0.0789 \% \label{trial5} \end{align}
Chandrasakar's wavefunction gives a result for helium that is within 0.07% of the experimental value for the ground-state energy.
Summary
The purpose of this section is to examine five trial wavefunctions for the helium atom used within the Perturbation Theory and Variational method approximation. For the Variational method approximation, the calculations begin with an uncorrelated wavefunction in which both electrons are placed in a hydrogenic orbital with scale factor $\alpha$. The next four trial functions use several methods to increase the amount of electron-electron interactions in the wavefunction. As the summary of results that is appended shows this gives increasingly more favorable agreement with the experimentally determined value for the ground-state energy of the species under study. The detailed calculations show that the reason for this improved agreement with experiment is due to a reduction in electron-electron repulsion.
Five variational method calculations that have been outlined above for the helium atom ($Z=2$) can be repeated for two-electron atoms (e.g., $\ce{H^-}$, $\ce{Li^+}$, $\ce{Be^{2+}}$, etc). The hydride anion is a particularly interesting case because the first two trial wavefunctions do not predict a stable ion (i.e., they are poor approximations). This indicates that electron-electron interactions is an especially important issue for atoms and ions with small nuclear charge.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.02%3A_Perturbation_Theory_and_the_Variational_Method_for_Helium.txt
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Learning Objectives
• Show how the Hartree approximation can be used to solve for the wavefunctions and energies of multi-electron atoms.
• Understand the orbital approach of independent orbitals is an approximation to a mutlti-electron system with the motions of all electrons coupled together.
• Demonstate how the Self-Consistant Field (SCF) calculation is needed to support the Hartree approximation
The Hartree method is used to approximate the wavefunction and the energy of a quantum multi-electron system in a stationary state. This approximation assumes that the exact $N$-body wavefunction of the system can be approximated by a product of single-electron wavefunctions. By invoking the variational method, one can derive a set of $N$-coupled equations for the N spin orbitals. A solution of these equations yields the Hartree wavefunction and energy of the system. It is one step better than the "Ignorance is Bliss" approach, discussed previously, but still far from modern state-off-the-art methods.
Unsolvable Systems
The Hamiltonian for a generic multielectron atom includes nucleus-electron attraction terms for the additional electrons with a general charge $Z$; e.g.
$V_{\text{nuclear-electron}}(r_1) = -\dfrac {Z}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} \label{8.3.1}$
in atomic units with $\left\vert\mathbf{r} - \mathbf{R}\right\vert$ is the distance between the electron and the nucleus, The Hamiltonian must also have terms for electron-electron repulsion (also in atomic units)
$V_{\text{electron-electron}}(r_{12}) = \dfrac {1}{\left\vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} \nonumber$
with $\left\vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert$ is the distance between electron 1 and electron 2. So the proper multi-electron Hamiltonian can be constructed
$\hat {H} (r_1, r_2, ... r_n) = -\dfrac {\hbar ^2}{2m_e} \sum _i \nabla ^2_i + \sum _i V_{\text{nuclear-electron}} (r_i) + \sum _{i \ne j} V_{\text{electron-electron}} (r_{ij}) \label{8.3.3}$
Given what we have learned from the previous quantum mechanical systems we’ve studied, we predict that exact solutions to the multi-electron Schrödinger equation would consist of a family of multi-electron wavefunctions, each with an associated energy eigenvalue. These wavefunctions and energies would describe the ground and excited states of the multi-electron atom, just as the hydrogen wavefunctions and their associated energies describe the ground and excited states of the hydrogen atom. We would predict quantum numbers to be involved, as well.
The fact that electrons interact through their electron-electron repulsion (final term in Equation $\ref{8.3.3}$) means that an exact wavefunction for a multi-electron system would be a single function that depends simultaneously upon the coordinates of all the electrons; i.e., a multi-electron wavefunction:
$|\psi (r_1, r_2, \cdots r_i) \rangle \label{8.3.4}$
Unfortunately, the electron-electron repulsion terms make it impossible to find an exact solution to the Schrödinger equation for many-electron atoms.
The Hartree Approximation
The method for finding best possible one-electron wavefunctions that was published by Douglas Hartree in 1948 and improved two years later by Vladimir Fock. For the Schrödinger equation to be analytically solvable, the variables must be separable - the variables are the coordinates of the electrons. To separate the variables in a way that retains information about electron-electron interactions, the electron-electron term (Equation $\ref{8.3.1}$) must be approximated so it depends only on the coordinates of one electron. Such an approximate Hamiltonian can account for the interaction of the electrons in an average way. The exact one-electron eigenfunctions of this approximate Hamiltonian then can be found by solving the Schrödinger equation. These functions are the best possible one-electron functions.
The Hartree approximation starts by invoking an initial ansatz that the multi-electron wavefunction in Equation $\ref{8.3.4}$ can be expanded as a product of single-electron wavefunctions (i.e., orbitals)
$| \psi(\mathbf{r}_1,\mathbf{r}_2, \ldots, \mathbf{r}_N) \rangle \approx \psi_{1}(\mathbf{r}_1)\psi_{2}(\mathbf{r}_2) \ldots \psi_{N}(\mathbf{r}_N) \nonumber$
from which it follows that the electrons are independent, and interact only via the mean-field Coulomb potential. This yields one-electron Schrödinger equations of the form
$-\dfrac{\hbar^{2}}{2m} \nabla^{2}\psi_{i}(\mathbf{r}) + V(\mathbf{r})\psi_{i}(\mathbf{r}) = \epsilon_{i}\psi_{i}(\mathbf{r}) \nonumber$
or
$H_e(r) \psi_{i}(\mathbf{r}) = \epsilon_{i}\psi_{i}(\mathbf{r}) \nonumber$
where $V(r)$ is the potential in which the electron moves; this includes both the nuclear-electron interaction
$V_{nucleus}(\mathbf{r}) = -Ze^{2}\sum_{R} \dfrac{1}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} \nonumber$
and the mean field arising from the $N-1$ other electrons. We smear the other electrons out into a smooth negative charge density $\rho(\mathbf{r}')$ leading to a potential of the form
$V_{electron}(\mathbf{r}) = -e\int d\mathbf{r}^{\prime} \rho(\mathbf{r}^{\prime}) \dfrac{1}{\left\vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} \nonumber$
where
$\rho(\mathbf{r}) = \sum_{i}^{\text{occupied}}\vert\psi(\mathbf{r})\vert^{2}. \nonumber$
The sum over runs over all occupied states; i.e., only the states of electrons that exist in the atom. The wavefunctions that from this approach with the Hamiltonian $H_e(r)$ involve possess three kinds of energies discussed below.
Three Energies within the Hartree Approximation
The total energy $\epsilon_j$ of the orbital $\phi_j$, is the sum of the above three contributions:
1. Kinetic Energy: The Kinetic energy of the electron has an average value is computed by taking the expectation value of the kinetic energy operator $\dfrac{- \hbar^2}{2m} \nabla^2 \nonumber$ with respect to any particular solution $\phi_j(r)$ to the Schrödinger equation: $KE = \langle\phi_j| \dfrac{- \hbar^2}{2m} \nabla^2 |\phi_j\rangle \nonumber$
2. Nuclear-Electron Coulombic Attraction Energy: Coulombic attraction energy with the nucleus of charge $Z$: $\langle\phi_j| \dfrac{-Z e^2}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} |\phi_j\rangle \nonumber$
3. Electron-Electron Coulombic Repulsion Energy: Coulomb repulsion energies with all of the $N-1$ other electrons, which are assumed to occupy other atomic orbitals denoted $\phi_K$, with this energy computed as $\sum_{j\neq k} \langle\phi_j(r) \phi_k(r’) |\dfrac{e^2}{|r-r’|} | \phi_j(r) \phi_k(r’)\rangle.\label{8.3.8}$ The Dirac notation $\langle\phi_j(r) \phi_k(r’) |\dfrac{e^2}{|r-r’|} | \phi_j(r) \phi_k(r’)\rangle$ is used to represent the two-electron (six-dimensional) Coulomb integral $J_{j,k} = \int |\phi_j(r)|^2 |\phi_k(r’)|^2 \dfrac{e^2}{r-r'} dr dr’ \nonumber$ that describes the Coulomb repulsion between the charge density $|\phi_j(r)|^2$ for the electron in $\phi_j$ and the charge density $|\phi_k(r’)|^2$ for the electron in $\phi_k$. Of course, the sum over $k$ must be limited to exclude $k=j$ to avoid counting a “self-interaction” of the electron in orbital $\phi_j$ with itself.
Adding these all together to get the total energy $\epsilon_j$ of the orbital $\phi_j$:
$\epsilon_J = \langle\phi_j| \dfrac{- \hbar^2}{2m} \nabla^2 |\phi_j\rangle + \langle\phi_j| \dfrac{-Z e^2}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} |\phi_j\rangle + \sum_{j\neq k} \langle\phi_j(r) \phi_k(r’) |\dfrac{e^2}{|r-r’|} | \phi_j(r) \phi_k(r’)\rangle. \nonumber$
This treatment of the electrons and their orbitals is referred to as the Hartree-level of theory.
When screened hydrogenic atomic orbitals are used to approximate the $\phi_j$ and $\phi_K$ orbitals, the resultant $\epsilon_J$ values do not produce accurate predictions. For example, the negative of $\epsilon_J$ should approximate the ionization energy for removal of an electron from the orbitals $\phi_j$. Such ionization potentials (IP s) can be measured, and the measured values do not agree well with the theoretical values when a crude screening approximation is made for the atomic orbitals.
The Self-Consistant Field (SCF) Approach to the Variational Method
The Hartree Equations are nonlinear and must be solved iteratively. This is because if particles interact, that interaction must be in the Hamiltonian. So until we know where the particles are, we cannot write down the Hamiltonian, but until we know the Hamiltonian, we cannot tell where the particles are.
The idea is to solve the Schrödinger equation for an electron moving in the potential of the nucleus and all the other electrons. We start with a guess for the trial electron charge density, solve N/2 one-particle Schrödinger equations (initially identical) to obtain N electron wavefunctions. Then we construct the potential for each wavefunction from that of the nucleus and that of all the other electrons, symmetrize it, and solve the N/2 Schrödinger equations again. This method is ideal for a computer, because it is easily written as an algorithm (Figure 8.3.1 ).
Although we are concerned here with atoms, the same methodology is used for molecules or even solids (with appropriate potential symmetries and boundary conditions). This is a variational method, so wherever we refer to wavefunctions, we assume that they are expanded in some appropriate basis set.
Fock improved on Hartree’s method by using proper "antisymmetrized wavefunctions" (called the Hartree-Fock method) instead of simple one-electron wavefunctions.
Shielding and Effective Charges Concepts are Useful
The hydrogen-like orbitals provide qualitative descriptions of orbitals of atoms with more than a single electron. By introducing the concept of screening as a way to represent the repulsive interactions among the electrons of an atom, an effective nuclear charge $Z_{\rm eff}$ can be used in place of $Z$ in the hydrogenic $\psi_{n,l,m}$ and $E_{n,l}$ formulas to generate approximate atomic orbitals to be filled by electrons in a many-electron atom. For example, in the crudest approximation of a carbon atom, the two $1s$ electrons experience the full nuclear attraction so $Z_{\rm eff} =6$ for them, whereas the $2s$ and $2p$ electrons are screened by the two $1s$ electrons, so $Z_{\rm eff}= 4$ for them. Within this approximation, one then occupies two $1s$ orbitals with $Z=6$, two $2s$ orbitals with $Z=4$ and two $2p$ orbitals with $Z=4$ in forming the full six-electron product wavefunction of the lowest-energy state of carbon
$| \psi(1, 2, …, 6) \rangle = | \psi_{1s}(1) \psi_{1s}(2) \psi_{2s}(3) \ldots \psi_{1p}(6) \rangle . \label{8.3.6}$
However, such approximate orbitals are not sufficiently accurate to be of use in quantitative simulations of atomic and molecular structure. In particular, their energies do not properly follow the trends in atomic orbital (AO) energies that are taught in introductory chemistry classes (Figure 8.3.2 ).
For example, the relative energies of the $3d$ and $4s$ orbitals are not adequately described in a model that treats electron repulsion effects in terms of a simple screening factor. So, now it is time to examine how we can move beyond the screening model and take the electron repulsion effects, which cause the inter-electronic couplings that render the Schrödinger equation insolvable, into account in a more reliable manner.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.03%3A_Hartree-Fock_Equations_are_Solved_by_the_Self-Consistent_Field_Method.txt
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Learning Objectives
• Understand the forth quantum number for electrons - spin.
• Understand how spin is connected to magnetic properties of the electrons and atoms.
• Understand how to break degeneracy via externally applied magnetic fields in electrons and atoms.
Imagine doing a hypothetical experiment that would lead to the discovery of electron spin. Your laboratory has just purchased a microwave spectrometer with variable magnetic field capacity. We try the new instrument with hydrogen atoms using a magnetic field of 104 Gauss and look for the absorption of microwave radiation as we scan the frequency of our microwave generator (Figure 8.4.1 ).
Finally we see absorption at a microwave photon frequency of $28 \times 10^9\, Hz$ (28 gigahertz). This result is really surprising from several perspectives. Each hydrogen atom is in its ground state, with the electron in a 1s orbital. The lowest energy electronic transition that we predict based on existing theory (the electronic transition from the ground state ($\psi _{100}$ to $\psi _{21m}$) requires an energy that lies in the vacuum ultraviolet (the lower Lyman line at 121 nm), not the microwave, region of the spectrum. Furthermore, when we vary the magnetic field we note that the frequency at which the absorption occurs varies in proportion to the magnetic field.
The Zeeman Effect: Breaking Degeneracies with Magnetic Fields
Magnetism results from the circular motion of charged particles. This property is demonstrated on a macroscopic scale by making an electromagnet from a coil of wire and a battery. Electrons moving through the coil produce a magnetic field, which can be thought of as originating from a magnetic dipole or a bar magnet. Electrons in atoms also are moving charges with angular momentum so they too produce a magnetic dipole, which is why some materials are magnetic. A magnetic dipole interacts with a magnetic field, and the energy of this interaction is given by the scalar product of the magnetic dipole moment, and the magnetic field, $\vec{B}$.
$E_B = -\vec{\mu}_m \cdot \vec{B} \label {8.4.0}$
Pieter Zeeman was one of the first to observe the splittings of spectral lines in a magnetic field caused by this interaction. Consequently such splittings are known as the Zeeman effect. (Figure 8.4.2 ). The expectation value calculated for the total energy in this case is the sum of the energy in the absence of the field, $E_n$, plus the Zeeman energy:
\begin{align} \left \langle E \right \rangle &= E_n + \dfrac {e \hbar B_z m_l}{2m_e} \[4pt] &= E_n + \mu _B B_z m_l \label {8.4.13} \end{align}
The factor
$\dfrac {e \hbar}{2m_e} = - \gamma _e \hbar = \mu _B \label {8.4.14}$
defines the constant $\mu _B$, called the Bohr magneton, which is taken to be the fundamental magnetic moment. It has units of $9.2732 \times 10^{-21}$ erg/Gauss or $9.2732 \times 10^{-24}$ Joule/Tesla. This factor will help you to relate magnetic fields, measured in Gauss or Tesla, to energies, measured in ergs or Joules, for any particle with a charge and mass the same as an electron.
Equation \ref{8.4.13} demonstrates that that $m_l$ quantum number degeneracy of the hydrogen atom is removed by the externally applied magnetic field. For example, the three hydrogen atom eigenstates $|\psi _{211} \rangle$, $|\psi _{21-1} \rangle$, and $|\psi _{210} \rangle$ are degenerate in zero magnetic field, but have different energies in an externally applied magnetic field (Figure 8.4.2 ).
The $m_l = 0$ state, for which the component of angular momentum and hence also the magnetic moment in the external field direction is zero, experiences no interaction with the magnetic field. The $m_l = +1$ state, for which the angular momentum in the z-direction is +ħ and the magnetic moment is in the opposite direction, against the field, experiences a raising of energy in the presence of a field. Maintaining the magnetic dipole against the external field direction is like holding a small bar magnet with its poles aligned exactly opposite to the poles of a large magnet. It is a higher energy situation than when the magnetic moments are aligned with each other.
Electron Spin and the Stern-Gerlach Experiment
To discover new things, experimentalists sometimes must explore new areas in spite of contrary theoretical predictions. Our theory of the hydrogen atom at this point gives no reason to look for absorption in the microwave region of the spectrum. By doing the crazy experiment outlines above, we discovered that when an electron is in the $|1s \rangle$ orbital of the hydrogen atom, there are two different states that have the same energy. When a magnetic field is applied, this degeneracy is removed, and microwave radiation can cause transitions between the two states. In the rest of this section, we see what can be deduced from this experimental observation. This experiment actually could be done with electron spin resonance spectrometers available today (Figure 8.4.1 ). To explain our observations, a new model for the hydrogen atom. Our original model for the hydrogen atom accounted for the motion of the electron and proton in our three-dimensional world; the new model needs something else that can give rise to an additional Zeeman-like effect. We need a charged particle with angular momentum to produce a magnetic moment, just like that obtained by the orbital motion of the electron. We can postulate that our observation results from a motion of the electron that was not considered in the last section - electron spin. We have a charged particle spinning on its axis. We then have charge moving in a circle, angular momentum, and a magnetic moment, which interacts with the magnetic field and gives us the Zeeman-like effect that we observed (Figure 8.4.2 ).
In 1920, Otto Stern and Walter Gerlach designed an experiment, which unintentionally led to the discovery that electrons have their own individual, continuous spin even as they move along their orbital of an atom. Today, this electron spin is indicated by the fourth quantum number, also known as the Electron Spin Quantum Number and denoted by $m_s$. In 1925, Samuel Goudsmit and George Uhlenbeck made the claim that features of the hydrogen spectrum that were unexamined might by explained by assuming electrons act as if it has a spin, which can be denoted by an arrow pointing up, which is +1/2, or an arrow pointing down, which is -1/2. The Stern and Gerlach experiment which demonstrated this was done with a beam of vaporized silver atoms that split into two beams after passing through a magnetic field (Figure 8.4.2 ). An explanation of this is that an electron has a magnetic field due to its spin. When electrons that have opposite spins are put together, there is no net magnetic field because the positive and negative spins cancel each other out. The silver atom used in the experiment has a total of 47 electrons, 23 of one spin type, and 24 of the opposite. Because electrons of the same spin cancel each other out, the one unpaired electron in the atom will determine the spin.
Spin Eigenstates and Eigenvalues
To describe electron spin from a quantum mechanical perspective, we must have spin wavefunctions and spin operators. The properties of the spin states are deduced from experimental observations and by analogy with our treatment of the states arising from the orbital angular momentum of the electron. The important feature of the spinning electron is the spin angular momentum vector, which we label $S$ by analogy with the orbital angular momentum $L$. We define spin angular momentum operators with the same properties that we found for the rotational and orbital angular momentum operators. After all, angular momentum is angular momentum, no matter if it is orbital or spin in nature.
We found that
$\hat {L}^2 | Y^{m_l} _l \rangle = l(l + 1) \hbar^2 | Y^{m_l}_l \rangle \nonumber$
so by analogy for the spin states, we must have
$\hat {S}^2 | \sigma ^{m_s} _s \rangle = s( s + 1) \hbar ^2 | \sigma ^{m_s}_s \rangle \nonumber$
where $\sigma$ is a spin wavefunction with quantum numbers $s$ and $m_s$ that obey the same rules as the quantum numbers $l$ and $m_l$ associated with the spherical harmonic wavefunction $Y$. We also found the project of the orbital angular momentum on the z-axis is
$\hat {L}_z | Y^{m_l}_l \rangle = m_l \hbar | Y^{m_l}_l \rangle \nonumber$
so by analogy, we must have a similar projection for the spin angular momentum:
$\hat {S}_z | \sigma ^{m_s}_s \rangle = m_s \hbar | \sigma ^{m_s}_s \rangle \label {8.4.4}$
Since $m_l$ ranges in integer steps from $-l$ to $+l$, also by analogy $m_s$ ranges in integer steps from $-s$ to $+s$. In our hypothetical experiment, we observed one absorption transition, which means there are two spin states. Consequently, the two values of $m_s$ must be $+s$ and $-s$, and the difference in $m_s$ for the two states, labeled f and i below, must be the smallest integer step, i.e., 1. The result of this logic is that
\begin{align} m_{s,f} - m_{s,i} &= 1 \nonumber\[4pt] (+s) - (-s) &= 1 \nonumber\[4pt] 2s &= 1 \nonumber\[4pt] s &= \dfrac {1}{2} \label {8.4.5} \end{align}
Therefore our conclusion is that the magnitude of the spin quantum number is 1/2 and the values for $m_s$ are +1/2 and -1/2. The two spin states correspond to spinning clockwise and counter-clockwise with positive and negative projections of the spin angular momentum onto the z-axis (Figure 8.4.3 ). The state with a positive projection, $m_s$ = +1/2, is called $\alpha$; the other is called $\beta$. These spin states are arbitrarily labeled $\alpha$ and $\beta$, and the associated spin wavefunctions also are designated by $|\alpha \rangle$ and $| \beta \rangle$.
From Equation $\ref{8.4.4}$, the magnitude of the z-component of spin angular momentum, $S_z$, is given by
$S_z = m_s \hbar \label {8.4.6}$
so the value of $S_z$ is $+ħ/2$ for spin state $\alpha$ and $-ħ/2$ for spin state $\beta$. Hence, we conclude that the $\alpha$ spin state, where the magnetic moment is aligned against the external field direction, has a greater energy than the $\beta$ spin state.
Electrons are not Actually Spinning
Electron's hypothetical surface would have to be moving faster than the speed of light for it to rotate quickly enough to produce the observed angular momentum. Hence, an electron is not simply a spinning ball or ring and electron spin appears to be an intrinsic angular moment of the particle rather than a consequence of the rotation of a charge particle like Figure 8.4.3 suggests. Despite this, the term "electron spin" persists in quantum vernacular.
Properties of Spin Eigenstates
Even though we do not know their functional forms, the spin wavefunctions are taken to be normalized and orthogonal to each other.
$\int \alpha ^* \alpha \,d \tau _s = \int \beta ^* \beta \,d \tau _s = 1 \label {8.4.7a}$
or in braket notation
$\langle \alpha | \alpha \rangle = \langle \beta | \beta \rangle =1 \label{8.4.7b}$
and
$\int \alpha ^* \beta\, d \tau _s = \int \beta ^* \alpha\, d \tau _s = 0 \label {8.4.8a}$
or in braket notation
$\langle \alpha | \beta \rangle = \langle \alpha | \beta \rangle = 0 \label{8.4.8b}$
where the integral is over the spin variable $\tau _s$.
Now let's apply these deductions to the experimental observations in our hypothetical microwave experiment in Figure 8.4.1 . We can account for the frequency of the transition ($\nu$= 28 gigahertz) that was observed in this hypothetical experiment in terms of the magnetic moment of the spinning electron and the strength of the magnetic field. The photon energy, $h \nu$, is given by the difference between the energies of the two states, $E_{\alpha}$ and $E_{\beta}$
\begin{align} \Delta E &= h \nu \[4pt] &= E_{\alpha} - E_{\beta} \label {8-49}\end{align}
The energies of these two states consist of the sum of the energy of an electron in a 1s orbital, $E_{1s}$, and the energy due to the interaction of the spin magnetic dipole moment of the electron, $\mu _s$, with the magnetic field, $B$.
The two states with distinct values for spin magnetic moment $\mu _s$ are denoted by the subscripts $\alpha$ and $\beta$ (the spin version of Equation \ref{8.4.0}.
\begin{align*} E_{\alpha} &= E_{1s} - \mu _{s,\alpha} \cdot B \[4pt] E_{\beta} &= E_{1s} - \mu _{s,\beta} \cdot B \end{align*}
Substituting the two equations above into the expression for the photon (Equation \ref{8-49}) energy gives
\begin{align} h \nu &= E_{\alpha} - E_{\beta} \[4pt] &= (E_{1s} - \mu _{s, \alpha} \cdot B) - (E_{1s} - \mu_{s,\beta} \cdot B) \label {8-52} \[4pt] &= ( \mu _{s, \beta} - \mu _{s, \alpha}) \cdot B \label{8-53} \end{align}
Again by analogy with the orbital angular momentum and magnetic moment discussed above, we take the spin magnetic dipole of each spin state, $\mu _{s, \alpha}$ and $\mu _{s, \beta}$, to be related to the total spin angular momentum of each state, $S_{\alpha}$ and $S_{\beta}$, by a constant spin gyromagnetic ratio, $\gamma _s$, as shown below.
$\mu _s = \gamma _s S \nonumber$
or each of the two states
$\mu _{s, \alpha} = \gamma _s S_\alpha \nonumber$
$\mu _{s, \beta} = \gamma _s S_\beta \nonumber$
With the magnetic field direction defined as $z$, the scalar product in Equation \ref{8-53} becomes a product of the z-components of the spin angular momenta, $S_{z, \alpha}$ and $S_{z, \beta}$, with the external magnetic field.
Inserting the values for $S_{z,\alpha} = +\dfrac {1}{2} \hbar$ and $S_{z, \alpha} = -\dfrac {1}{2} \hbar$ from Equation \ref{8.4.6} and rearranging Equation \ref{8-53} yields
$\dfrac {h \nu}{B} = - \gamma _s \hbar \nonumber$
Calculating the ratio $\dfrac {h \nu}{B}$ from our experimental results, $\nu = 28 \times 10^9\, Hz$ when $B = 10^4\, gauss$, gives us a value for
$- \gamma_s \hbar = 18.5464 \times 10^{-21}\, erg/gauss. \nonumber$
This value is about twice the Bohr magneton,$-\gamma _e \hbar$, found in Equation \ref{8.4.14} i.e. $\gamma _s \hbar = 2.0023, \gamma _e \hbar$, or
$\gamma _s = 2.0023 \gamma _e \label \nonumber$
The factor of 2.0023 is called the g-factor and accounts for the deviation of the spin gyromagnetic ratio from the value expected for orbital motion of the electron. In other words, it accounts for the spin transition being observed where it is instead of where it would be if the same ratio between magnetic moment and angular momentum held for both orbital and spin motions. The value 2.0023 applies to a freely spinning electron; the coupling of the spin and orbital motion of electrons can produce other values for $g$.
Exercise 8.4.1
Carry out the calculations that show that the g-factor for electron spin is 2.0023.
Interestingly, the concept of electron spin and the value g = 2.0023 follow logically from Dirac's relativistic quantum theory, which is beyond the scope of this discussion. Electron spin was introduced here as a postulate to explain experimental observations. Scientists often introduce such postulates parallel to developing the theory from which the property is naturally deduced.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.04%3A_An_Electron_Has_an_Intrinsic_Spin_Angular_Momentum.txt
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Learning Objectives
• Interprete the consequence of exchanging two electrons in a multi-electron atom on the wavelengths
• Connect the Pauli’s Exclusion Principle to the permutation symmetry of multi-electron atoms
Quantum mechanics allows us to predict the results of experiments. If we conduct an experiment with indistinguishable particles a correct quantum description cannot allow anything which distinguishes between them. For example, if the wavefunctions of two particles overlap, and we detect a particle, which one is it? The answer to this is not only that we do not know, but that we cannot know. Quantum mechanics can only tell us the probability of finding a particle in a given region. The wavefunction must therefore describe both particles.
The Schrödinger equation for the helium atom is then:
$\left[ - \dfrac{\hbar^2}{2m} ( \nabla^2_1 + \nabla^2_1 ) - V(\mathbf{r}_1) - V(\mathbf{r}_2) + V_{12}(\mathbf{r}_1,\mathbf{r}_2) \right ] \psi(\mathbf{r}_1, \mathbf{r}_2) = E\psi(\mathbf{r}_1, \mathbf{r}_2) \nonumber$
where the subscripts label each particle, and there are six coordinates, three for each particle. While $|\psi(\mathbf{r}_1, \mathbf{r}_2) \rangle$ is a six dimensional wavefunction (three for each electron) and contains all the information we can measure (a Postulate of Quantum Mechanics), it only provide the probability of finding the electron at a specific volume element, and does not tell us which particle is which (e.g., is it electon $1$ or electron $2$?).
What basis states would be appropriate for $|\psi \rangle$? If we consider the orbital approximation that uses a product wavefunction
$|\psi(\mathbf{r}_1, \mathbf{r}_2) \rangle = |\varphi_a (\mathbf{r}_1) \rangle | \varphi_ b(\mathbf{r}_2)\rangle \nonumber$
where $| \varphi_a(\mathbf{r}_1)\rangle$ and $| \varphi_b(\mathbf{r}_2)\rangle$ are one-particle spin-orbitals (with both spin and spatial components) of atoms 1 and 2. This approximation allows us to separate the two particle equation into two one-electron equations:
$\left[ - \dfrac{\hbar^2}{2m} \nabla^2_1 + V(\mathbf{r}_1) \right ] |\varphi_a(\mathbf{r}_1) \rangle = E|\varphi_a(\mathbf{r}_1) \rangle \nonumber$
$\left[ - \dfrac{\hbar^2}{2m} \nabla^2_2 + V(\mathbf{r}_2) \right ] |\varphi_b(\mathbf{r}_2) \rangle= E|\varphi_a(\mathbf{r}_2) \rangle \nonumber$
provided that the particles do not interact (e.g., $\nabla^2_1$ does not act on $|\varphi_b(\mathbf{r}_2)\rangle$ and $V_{12} = 0$).
Unfortunately, by doing this we have introduced unphysical labels to the indistinguishable particles. And this is wrong: the effect of it is that the particles do not interfere with each other because they are in different dimensions (six dimensional space - remember?). When we construct a two particle wavefunction out of two one-particle wavefunctions we must be ensure that the probability density (the measurable quantity $|\psi|^2$) is independent of the artificial labels.
The Exchange Operator
We can deepen our understanding of the quantum mechanical description of multi-electron atoms by examining the concepts of electron indistinguishability and the Pauli Exclusion Principle in detail. We will use the following statement as a guide to keep our explorations focused on the development of a clear picture of the multi-electron atom: “When a multi-electron wavefunction is built as a product of single-electron wavefunctions, the corresponding concept is that exactly one electron’s worth of charge density is described by each atomic spin-orbital.”
A subtle, but important part of the conceptual picture, is that the electrons in a multi-electron system are not distinguishable from one another by any experimental means. Since the electrons are indistinguishable, the probability density we calculate by squaring the modulus of our multi-electron wavefunction also cannot change when the electrons are interchanged (permuted) between different orbitals. In general, if we interchange two identical particles, the world does not change. As we will see below, this requirement leads to the idea that the world can be divided into two types of particles based on their behavior with respect to permutation or interchange.
In order for the probability density to remain unchanged when two particles are permuted, the wavefunction itself can change only by a factor of $e^{i\varphi}$, which represents a complex number, when the particles described by that wavefunction are permuted. As we will show below, the $e^{i\varphi}$ factor is possible because the probability density depends on the absolute square of the function and all expectation values involve $\psi \psi ^*$. Consequently $e^{i\varphi}$ disappears in any calculation that relates to the real world because $e^{i\varphi} e^{-i\varphi} = 1$.
We could symbolically write an approximate two-particle wavefunction as $\psi (\mathbf{r}_1, \mathbf{r}_2)$. This could be, for example, a two-electron wavefunction for helium. To exchange the two particles, we simply substitute the coordinates of particle 1 ($\mathbf{r}_1$) for the coordinates of particle 2 ($\mathbf{r}_2$) and vice versa, to get the new wavefunction $\psi (\mathbf{r}_1, \mathbf{r}_2)$. This new wavefunction must have the property that
$|\psi (\mathbf{r}_1, \mathbf{r}_2)|^2 = \psi (\mathbf{r}_2, \mathbf{r}_1)^*\psi (\mathbf{r}_2, \mathbf{r}_1) = \psi (\mathbf{r}_1, \mathbf{r}_2)^* \psi (\mathbf{r}_1, \mathbf{r}_2) \label {8.5.1}$
since the probability density of the electrons in the atom does not change upon permutation of the electrons.
Exercise 8.5.1
Permute the electrons the product function for He wavefunction:
$\psi (\mathbf{r}_1,\mathbf{r}_2) \rangle = | \varphi_{1s}(\mathbf{r}_1) \rangle | \varphi_{1s}(\mathbf{r}_2) \rangle \nonumber$
Equation $\ref{8.5.1}$ will be true only if the wavefunctions before and after permutation are related by a factor of $e^{i\varphi}$,
$\psi (\mathbf{r}_1, \mathbf{r}_2) = e^{i\varphi} \psi (\mathbf{r}_1, \mathbf{r}_2) \label {8.5.2}$
so that
$\left ( e^{-i\varphi} \psi (\mathbf{r}_1, \mathbf{r}_2) ^*\right ) \left ( e^{i\varphi} \psi (\mathbf{r}_1, \mathbf{r}_2) ^*\right ) = \psi (\mathbf{r}_1 , \mathbf{r}_2 ) ^* \psi (\mathbf{r}_1 , \mathbf{r}_2) \label {8.5.3}$
If we exchange or permute two identical particles twice, we are (by definition) back to the original situation. If each permutation changes the wavefunction by $e^{i \varphi}$, the double permutation must change the wavefunction by $e^{i\varphi} e^{i\varphi}$. Since we then are back to the original state, the effect of the double permutation must equal 1; i.e.,
$e^{i\varphi} e^{i\varphi} = e^{i 2\varphi} = 1 \label {8.5.4}$
which is true only if $\varphi = 0$ or an integer multiple of $\pi$. The requirement that a double permutation reproduce the original situation limits the acceptable values for $e^{i\varphi}$ to either +1 (when $\varphi = 0$) or -1 (when $\varphi = \pi$). Both possibilities are found in nature.
Exercise 8.5.2
Use Euler’s Equality
$e^{i\pi} + 1=0 \nonumber$
to show that $e^{i 2\varphi} = 1$ when $\varphi = 0$ or $n \pi$ and consequently $e^{i \varphi} = \pm 1$.
We can introduce the exchange operator $\hat{P}_{12}$: an operator which permutes the labels of the particles in a multi-particle wavefucntion. This is a rather strange operator, because it only changes the unphysical labels which we have attached to the one-particle wavefunctions in order to make the maths more easy. For a meaningful solution we must have a wavefunction which has a probability amplitude unchanged by $\hat{P}_{12}$: it must be either symmetric or antisymmetric with respect to exchange:
$\hat{P}_{12} |\psi(\mathbf{r}_1, \mathbf{r}_2) \rangle = \pm |\psi(\mathbf{r}_2, \mathbf{r}_1) \rangle \label{exchange}$
Bosons and femions
Physical solutions must be eigenfunctions of $\hat{P}_{12}$ (i.e., $\hat{H}$ and $\hat{P_{12}}$ commute). Moveover, Equation \ref{exchange} argues that the eigenvalues of the Exchange Operator are either +1 (bosons) or −1 (fermions).
Bosons
Wavefunctions for which $e^{i \varphi} = +1$ are defined as symmetric with respect to permutation, because the wavefunction is identical before and after a single permutation. Wavefunctions that are symmetric with respect to interchange of the particles obey the following mathematical relationship,
$\hat{P}_{12} |\psi(\mathbf{r}_1, \mathbf{r}_2) \rangle = + |\psi(\mathbf{r}_2, \mathbf{r}_1) \rangle \nonumber$
The behavior of some particles requires that the wavefunction be symmetric with respect to permutation. These particles are called bosons and have integer spin such as deuterium nuclei, photons, and gluons.
Fermions
The behavior of other particles requires that the wavefunction be antisymmetric with respect to permutation $(e^{i\varphi} = -1)$. A wavefunction that is antisymmetric with respect to electron interchange is one whose output changes sign when the electron coordinates are interchanged, as shown below.
$\hat{P}_{12} |\psi(\mathbf{r}_1, \mathbf{r}_2) \rangle = - |\psi(\mathbf{r}_2, \mathbf{r}_1) \rangle \nonumber$
These particles are called fermions and have half-integer spin and include electrons, protons, and neutrinos. Since electrons are fermion, any wavefunction used to describe multiple electrons must be antisymmetric with respect to permutation of the electrons. The requirement that the wavefunction be antisymmetric applies to all multi-electron functions $\psi (\mathbf{r}_1, \mathbf{r}_2, \cdots r_i)$, including those approximated as products of single electron functions
$| \psi (\mathbf{r}_1, \mathbf{r}_2, \cdots r_i) \rangle \approx \varphi _a (\mathbf{r}_1) \varphi _b (\mathbf{r}_2) \cdots \varphi _i (r_i) \label{Product}$
Exercise 8.5.4
What is meant by the term permutation symmetry?
Exercise 8.5.5
Explain why the product function $\varphi (\mathbf{r}_1) \varphi (\mathbf{r}_2)$ could describe two bosons (deuterium nuclei), but can not describe two fermions (e.g. electrons).
Answer
Because if we switch $\mathbf{r}_1$ and $\mathbf{r}_2$, the product function becomes $φ(r_2)φ(r_1)$, which is equal to $φ(r_1)φ(r_2)$, which is consistent with the wavefunctions of bosons, because boson wavefunctions are symmetric. However, fermion wavefunctions are antisymmetric, which means that if switch r1 and r2, the result should be -1×the original wavefunction. Therefore, the product function $φ(r_1)φ(r_2)$ cannot describe fermions.
The Pauli’s Exclusion Principle
Any physically meaningful Hamiltonian must commute with $\hat{P}_{12}$, otherwise $\hat{H}$ and $\hat{P}_{12}$ could not have common eigenfunctions and the system could not remain in an eigenstate of exchange. A simple product wavefunction like that in Equation $\ref{Product}$ does not satisfy this (unless $\varphi_a = \varphi_b$). A linear combination of all permutations is required to satisfy indistinguishability constraints.
For a two particle system (e.g., Helium) there is the asymmetric combination
$|\psi^{−} \rangle = \dfrac{1}{\sqrt{2}} |\varphi_a(\mathbf{r}_1)\varphi_b(\mathbf{r}_2) − \varphi_a(\mathbf{r}_2)\varphi_b(\mathbf{r}_1) \rangle \label{ASym}$
and symmetric combination
$|\psi^{+} \rangle = C_{ab} |\varphi_a(\mathbf{r}_1)\varphi_b(\mathbf{r}_2) + \varphi_a(\mathbf{r}_2)\varphi_b(\mathbf{r}_1) \rangle + C_{aa} |\varphi_a(\mathbf{r}_2)\varphi_a(\mathbf{r}_1) \rangle + C_{bb}|\varphi_b(\mathbf{r}_2)\varphi_b(\mathbf{r}_1) \rangle \label{Sym}$
where the $C_{ab}$ terms are expansion and normalization parameters.
Note that the antisymmetric combination (Equation $\ref{ASym}$) cannot include terms where both particles are in the same state (spin-orbital), but there are three possibilities for the symmetric state (Equation $\ref{Sym}$).
Although any linear combinations of $C_{ab}$, $C_{bb}$, and $C_{aa}$ in Equation $\ref{Sym}$ is possible, there are three limiting expressions for possible symmetric combinations:
\begin{align} |\psi^{+i}_1 \rangle &= C_{ab} |\varphi_a(\mathbf{r}_1)\varphi_b(\mathbf{r}_2) + \varphi_a(\mathbf{r}_2)\varphi_b(\mathbf{r}_1) \rangle \[4pt] |\psi^{+i}_2 \rangle &=C_{aa} |\varphi_a(\mathbf{r}_2)\varphi_a(\mathbf{r}_1) \rangle \[4pt] |\psi^{+i}_3 \rangle &= C_{bb}|\varphi_b(\mathbf{r}_2)\varphi_b(\mathbf{r}_1) \rangle \end{align} \nonumber
If $\varphi_a(\mathbf{r}_1) = \varphi_a(\mathbf{r}_2)$, then $|\varphi^{−i} \rangle= 0$. Thus there is no possible antisymmetric combination involving electrons in the same state (spin-orbit). This is the Pauli exclusion principle.
The Pauli Exclusion Principle argues that two electrons could not be described by the same spin-orbital. To see the relationship between this statement and the requirement that the wavefunction be antisymmetric for electrons, try to construct an anti-symmetric wavefunction for two electrons that are described by the same spin-orbital
$|\varphi_b(\mathbf{r}_1) \rangle =\varphi_a(\mathbf{r}_2)\rangle \nonumber$
For example, if this were the case for the anti-symmetric combination for helium (Equation $\ref{ASym}$), then the wavefunction collapses to zero. We can only constructs wavefunctions that are antisymmetric with respect to permutation symmetry only if each electron is described by a different function.
The Pauli Exclusion Principle is simply the requirement that the wavefunction be antisymmetric for electrons, since they are fermions.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.05%3A_Wavefunctions_must_be_Antisymmetric_to_Interchange_of_any_Two_Electrons.txt
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Learning Objectives
• Understand how the Pauli Exclusion principle affects the electronic configuration of mulit-electron atoms
• Understand how determinantal wavefunctions (Slater determinents) ensure the proper symmetry to electron permutation required by Pauli Exclusion Principle.
• Connect the electron permutation symmetry requirement to multi-electron wavefunctions to the Aufbau principle taught in general chemistry courses
Let’s try to construct an antisymmetric function that describes the two electrons in the ground state of helium. Blindly following the first statement of the Pauli Exclusion Principle, then each electron in a multi-electron atom must be described by a different spin-orbital. For the ground-state helium atom, this gives a $1s^22s^02p^0$ configuration (Figure 8.6.1 ).
We try constructing a simple product wavefunction for helium using two different spin-orbitals. Both have the 1s spatial component, but one has spin function $\alpha$ and the other has spin function $\beta$ so the product wavefunction matches the form of the ground state electron configuration for He, $1s^2$.
$| \psi (\mathbf{r}_1, \mathbf{r}_2 ) \rangle = \varphi _{1s}\alpha (\mathbf{r}_1) \varphi _{1s}\beta ( \mathbf{r}_2) \label {8.6.1}$
After permutation of the electrons, this becomes
$| \psi ( \mathbf{r}_2,\mathbf{r}_1 ) \rangle = \varphi _{1s}\alpha ( \mathbf{r}_2) \varphi _{1s}\beta (\mathbf{r}_1) \label {8.6.2}$
which is different from the starting function since $\varphi _{1s\alpha}$ and $\varphi _{1s\beta}$ are different spin-orbital functions. Hence, the simple product wavefunction in Equation \ref{8.6.1} does not satisfy the indistinguishability requirement since an antisymmetric function must produce the same function multiplied by (–1) after permutation of two electrons, and that is not the case here. We must try something else.
To avoid getting a totally different function when we permute the electrons, we can make a linear combination of functions. A very simple way of taking a linear combination involves making a new function by simply adding or subtracting functions. The function that is created by subtracting the right-hand side of Equation $\ref{8.6.2}$ from the right-hand side of Equation $\ref{8.6.1}$ has the desired antisymmetric behavior. The constant on the right-hand side accounts for the fact that the total wavefunction must be normalized.
$| \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s}\alpha (\mathbf{r}_1) \varphi _{1s}\beta ( \mathbf{r}_2) - \varphi _{1s} \alpha( \mathbf{r}_2) \varphi _{1s} \beta (\mathbf{r}_1)] \label{8.6.3}$
In this orbital approximation, a single electron is held in a single spin-orbital with an orbital component (e.g., the $1s$ orbital) determined by the $n$, $l$, $m_l$ quantum numbers and a spin component determined by the $m_s$ quantum number.
The wavefunction in Equation \ref{8.6.3} can be decomposed into spatial and spin components:
$| \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \underbrace{[ \varphi _{1s}(1) \varphi _{1s}(2)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \beta( 2) - \alpha( 2) \beta(1)]}_{\text{spin component}} \label{8.6.3B}$
Example 8.6.1 : Symmetry to Electron Permutation
Show that the linear combination of spin-orbitals in Equation $\ref{8.6.3}$ is antisymmetric with respect to permutation of the two electrons.
Hint
Replace the minus sign with a plus sign (i.e. take the positive linear combination of the same two functions) and show that the resultant linear combination is symmetric.
Solution
First a reminder of permutation symmetries:
• If the wavefunction is symmetric with respect to permutation of the two electrons then $\left|\psi (\mathbf{r}_1, \mathbf{r}_2) \rangle=\right| \psi(\mathbf{r}_2, \mathbf{r}_1)\rangle \nonumber$
• If the wavefunction is antisymmetric with respect to permutation of the two electrons then $\left|\psi(\mathbf{r}_1, \mathbf{r}_2) \rangle= - \right| \psi(\mathbf{r}_2, \mathbf{r}_1)\rangle \nonumber$
We start with the original wavefunction
$| \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(\mathbf{r}_1) \varphi _{1s\beta}( \mathbf{r}_2) - \varphi _{1s\alpha}( \mathbf{r}_2) \varphi _{1s\beta}(\mathbf{r}_1)] \nonumber$
and flip the position of electron 1 with electron 2 and vice versa
$| \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(\mathbf{r}_2) \varphi _{1s\beta}( \mathbf{r}_1) - \varphi _{1s\alpha}( \mathbf{r}_1) \varphi _{1s\beta}(\mathbf{r}_2)] \label{permute1}$
We then we ask if we can rearrange the left side of Equation \ref{permute1} to either become $+ | \psi(\mathbf{r}_1, \mathbf{r}_2)\rangle$ (symmetric to permutation) or $- | \psi(\mathbf{r}_1, \mathbf{r}_2)\rangle$ (antisymmetric to permutation).
$| \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = \dfrac {1}{\sqrt {2}} [ - \varphi _{1s\alpha}( \mathbf{r}_1) \varphi _{1s\beta}(\mathbf{r}_2) + \varphi _{1s\alpha}(\mathbf{r}_2) \varphi _{1s\beta}( \mathbf{r}_1) ] \nonumber$
or
$| \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = - \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}( \mathbf{r}_1) \varphi _{1s\beta}(\mathbf{r}_2) - \varphi _{1s\alpha}(\mathbf{r}_2) \varphi _{1s\beta}( \mathbf{r}_1) ] \nonumber$
This is just the negative of the original wavefunction, therefore
$| \psi (\mathbf{r}_2, \mathbf{r}_1) \rangle = - | \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle \nonumber$
The wavefunction is antisymemtric.
Exercise 8.6.1 : Symmetry
Is this linear combination of spin-orbitals
$| \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(\mathbf{r}_1) \varphi _{1s\beta}( \mathbf{r}_2) + \varphi _{1s\alpha}( \mathbf{r}_2) \varphi _{1s\beta}(\mathbf{r}_1)] \nonumber$
symmetric or antisymmetric with respect to permutation of the two electrons?
Answer
Symmetric
The electronic configuration of the first excited state of He is $1s^12s^12p^0$ and we can envision four microstates for this configuration (Figure 8.6.2 ). As spected, the wavefunctions associated for of these microstate must satisfy indistinguishability requirement just like the ground state.
These electron configurations are used to construct four possible excited-state two-electron wavefunctions (but not necessarily in a one-to-one correspondence):
\begin{align} | \psi_1 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}2 \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2)+\varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \beta( 2) - \alpha( 2) \beta(1)]}_{\text{spin component}} \label{8.6.3C1} \[4pt] | \psi_2 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}{\sqrt {2}} \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2) - \varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \alpha( 2)]}_{\text{spin component}} \label{8.6.3C2} \[4pt] | \psi_3 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}{\sqrt {2}} \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2) - \varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \alpha(1) \beta( 2) + \alpha( 2) \beta(1)]}_{\text{spin component}} \label{8.6.3C3} \[4pt] | \psi_4 (\mathbf{r}_1, \mathbf{r}_2) \rangle &= \dfrac {1}2 \underbrace{[ \varphi _{1s}(1) \varphi _{2s}(2) - \varphi _{1s}(2) \varphi _{2s}(1)]}_{\text{spatial component}} \underbrace{[ \beta(1) \beta( 2)]}_{\text{spin component}} \label{8.6.3C4} \end{align}
All four wavefunctions are antisymmetric as required for fermionic wavefunctions (which is left to an exercise). Wavefunctions $| \psi_2 \rangle$ and $| \psi_4 \rangle$ correspond to the two electrons both having spin up or both having spin down (Configurations 2 and 3 in Figure 8.6.2 , respectively). Wavefunctions $| \psi_1 \rangle$ and $| \psi_3 \rangle$ are more complicated and are antisymmetric (Configuration 1 - Configuration 4) and symmetric combinations (Configuration 1 + 4). That is, a single electron configuration does not describe the wavefunction. For many electrons, this ad hoc construction procedure would obviously become unwieldy. However, there is an elegant way to construct an antisymmetric wavefunction for a system of $N$ identical particles.
Determinantal Wavefunctions
A linear combination that describes an appropriately antisymmetrized multi-electron wavefunction for any desired orbital configuration is easy to construct for a two-electron system. However, interesting chemical systems usually contain more than two electrons. For these multi-electron systems a relatively simple scheme for constructing an antisymmetric wavefunction from a product of one-electron functions is to write the wavefunction in the form of a determinant. John Slater introduced this idea so the determinant is called a Slater determinant.
John C. Slater introduced the determinants in 1929 as a means of ensuring the antisymmetry of a wavefunction, however the determinantal wavefunction first appeared three years earlier independently in Heisenberg's and Dirac's papers.
The Slater determinant for the two-electron ground-state wavefunction of helium is
$| \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) \end {vmatrix} \label {8.6.4}$
A shorthand notation for the determinant in Equation $\ref{8.6.4}$ is then
$| \psi (\mathbf{r}_1 , \mathbf{r}_2) \rangle = 2^{-\frac {1}{2}} Det | \varphi _{1s\alpha} (\mathbf{r}_1) \varphi _{1s\beta} ( \mathbf{r}_2) | \label {8.6.5}$
The determinant is written so the electron coordinate changes in going from one row to the next, and the spin orbital changes in going from one column to the next. The advantage of having this recipe is clear if you try to construct an antisymmetric wavefunction that describes the orbital configuration for uranium! Note that the normalization constant is $(N!)^{-\frac {1}{2}}$ for $N$ electrons.
The generalized Slater determinant for a multi-electrom atom with $N$ electrons is then
$\psi(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N)=\dfrac{1}{\sqrt{N!}} \left| \begin{matrix} \varphi_1(\mathbf{r}_1) & \varphi_2(\mathbf{r}_1) & \cdots & \varphi_N(\mathbf{r}_1) \ \varphi_1(\mathbf{r}_2) & \varphi_2(\mathbf{r}_2) & \cdots & \varphi_N(\mathbf{r}_2) \ \vdots & \vdots & \ddots & \vdots \ \varphi_1(\mathbf{r}_N) & \varphi_2(\mathbf{r}_N) & \cdots & \varphi_N(\mathbf{r}_N) \end{matrix} \right| \label{5.6.96}$
Example 8.6.2 : Helium Atom
Expand the Slater determinant in Equation $\ref{8.6.4}$ for the $\ce{He}$ atom.
Solution
To expand the Slater determinant of the Helium atom, the wavefunction in the form of a two-electron system:
$| \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) \end {vmatrix} \nonumber$
This is a simple expansion exercise of a $2 \times 2$ determinant
$| \psi (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \left[ \varphi _{1s} (1) \alpha (1) \varphi _{1s} (2) \beta (2) - \varphi _{1s} (2) \alpha (2) \varphi _{1s} (1) \beta (1) \right] \nonumber$
It is not unexpected that the determinant wavefunction in Equation \ref{8.6.4} is the same as the form for the helium wavefunction that is given in Equation \ref{8.6.3}.
Exercise 8.6.2 : Lithium Atom
Write and expand the Slater determinant for the ground-state $\ce{Li}$ atom.
Answer
Slater determinant for $\ce{Li}$ atom:
$\psi(1,2,3)=\frac{1}{\sqrt{6}} \operatorname{det}\left(\begin{array}{ccc} {\varphi _{1s} \alpha(1)} & {\varphi _{1s} \beta(1)} & {\varphi _{2s} \alpha(1)} \ \varphi _{1s} \alpha(2) & {\varphi _{1s} \beta(2)} & {\varphi _{2s} \alpha(2)} \ {\varphi _{1s} \alpha(3)} & {\varphi _{1s} \beta(3)} & {\varphi _{2s} \alpha(3)} \end{array}\right)\nonumber$
Expansion of Slater determinant:
$\psi(1,2,3)=\frac{1}{\sqrt{6}}[\varphi _{1s} \alpha(1) \varphi _{1s} \beta(2) \varphi _{2s} \alpha(3)-\varphi _{1s} \alpha(1) \varphi _{1s} \beta(3) \varphi _{2s} \alpha(2)+ \varphi _{1s} \alpha(3) \varphi _{1s} \beta(1) \varphi _{2s} \alpha(2) - \varphi _{1s} \alpha(3) \varphi _{1s} \beta(2) \varphi _{1s} \alpha(1)+ \varphi _{1s} \alpha(2) \varphi _{1s} \beta(3) \varphi _{2s} \alpha(3) ] \nonumber$
Note that this is also a valid ground state wavefunction
$\psi(1,2,3)=\frac{1}{\sqrt{6}} \operatorname{det}\left(\begin{array}{ccc} {\varphi _{1s} \alpha(1)} & {\varphi _{1s} \beta(1)} & {\varphi _{2s} \beta(1)} \ \varphi _{1s} \alpha(2) & {\varphi _{1s} \beta(2)} & {\varphi _{2s} \beta(2)} \ {\varphi _{1s} \alpha(3)} & {\varphi _{1s} \beta(3)} & {\varphi _{2s} \beta(3)} \end{array}\right)\nonumber$
What is the difference between these two wavefunctions?
Now that we have seen how acceptable multi-electron wavefunctions can be constructed, it is time to revisit the “guide” statement of conceptual understanding with which we began our deeper consideration of electron indistinguishability and the Pauli Exclusion Principle. What does a multi-electron wavefunction constructed by taking specific linear combinations of product wavefunctions mean for our physical picture of the electrons in multi-electron atoms? Overall, the antisymmetrized product function describes the configuration (the orbitals, regions of electron density) for the multi-electron atom. Because of the requirement that electrons be indistinguishable, we cannot visualize specific electrons assigned to specific spin-orbitals. Instead, we construct functions that allow each electron’s probability distribution to be dispersed across each spin-orbital. The total charge density described by any one spin-orbital cannot exceed one electron’s worth of charge, and each electron in the system is contributing a portion of that charge density.
The four configurations in Figure 8.6.2 for first-excited state of the helium atom can be expressed as the following Slater Determinants
$| \phi_a (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \beta(1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} \label {8.6.10A}$
$| \phi_b (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \alpha (1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \label {8.6.10B}$
$| \phi_c (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \alpha(1) \ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \label {8.6.10D}$
$| \phi_d (\mathbf{r}_1, \mathbf{r}_2) \rangle = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \beta (1) \ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} \label {8.6.10C}$
Slater determinants are constructed by arranging spinorbitals in columns and electron labels in rows and are normalized by dividing by $\sqrt{N!}$, where $N$ is the number of occupied spinorbitals. As you can imagine, the algebra required to compute integrals involving Slater determinants is extremely difficult. It is therefore most important that you realize several things about these states so that you can avoid unnecessary algebra:
• A Slater determinant corresponds to a single electron configuration diagram (Figure 8.6.2 ). Furthermore, recall that for the excited states of helium we had a problem writing certain stick diagrams as a (space)x(spin) product and had to make linear combinations of certain states to force things to separate (Equation \ref{8.6.3C2} and \ref{8.6.3C4}). Because of the direct correspondence of configuration diagrams and Slater determinants, the same pitfall arises here: Slater determinants sometimes may not be representable as a (space)x(spin) product, in which case a linear combination of Slater determinants must be used instead. This generally only happens for systems with unpaired electrons (like several of the Helium excited-states).
• A Slater determinant is anti-symmetric upon exchange of any two electrons. We recall that if we take a matrix and interchange two its rows, the determinant changes sign.
The wavefunctions in \ref{8.6.3C1}-\ref{8.6.3C4} can be expressed in term of the four determinants in Equations \ref{8.6.10A}-\ref{8.6.10C}.
\begin{align*} | \psi_2 \rangle &= |\phi_b \rangle \[4pt] &= \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \alpha (1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \end{align*} \nonumber
\begin{align*} | \psi_4 \rangle &= |\phi_d \rangle \[4pt] &= \dfrac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \beta (1) \ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} \end{align*} \nonumber
but the wavefunctions that represent combinations of spinorbitals and hence combinations of electron configurations (e.g., igure 8.6.2 ) are combinations of Slater determinants (Equation \ref{8.6.10A}-\ref{8.6.10D})
\begin{align*} | \psi_1 \rangle & = |\phi_a \rangle - |\phi_c \rangle \[4pt] &= \dfrac {1}{2} \left( \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \beta(1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} - \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \alpha(1) \ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \right) \end{align*} \nonumber
\begin{align*} | \psi_3 \rangle &= |\phi_a \rangle + |\phi_c \rangle \[4pt] &= \dfrac {1}{2} \left( \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{2s} (1) \beta(1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) \end {vmatrix} + \begin {vmatrix} \varphi _{1s} (1) \beta(1) & \varphi _{2s} (1) \alpha(1) \ \varphi _{1s} (2) \beta(2) & \varphi _{2s} (2) \alpha(2) \end {vmatrix} \right) \end{align*} \nonumber
Note the expected change in the normalization constants.
Example 8.6.3 : Carbon Atom
Write the Slater determinant for the ground-state carbon atom. If you expanded this determinant, how many terms would be in the linear combination of functions?
Solution
Carbon has 6 electrons which occupy the 1s 2s and 2p orbitals. Each row in the determinant represents a different electron and each column a unique spin-obital where the electron could be found. There are 6 rows, 1 for each electron, and 6 columns, with the two possible p orbitals both alpha (spin up), in the determinate. There are two columns for each s orbital to account for the alpha and beta spin possibilities. There are two different p orbitals because the electrons in their ground state will be in the different p orbitals and both spin up. N=6 so the normalization constant out front is 1 divided by the square-root of 6!
\begin{align*}\psi(1,2,3,4,5,6)=\frac{1}{6!^{1/2}}\begin{vmatrix}\varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) & \varphi _{2s} (1) \alpha (1) & \varphi _{2s} (1) \beta (1) & \varphi _{2px} (1) \alpha (1) & \varphi _{2py} (1) \alpha (1) \ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) & \varphi _{2s} (2) \alpha (2) & \varphi _{2s} (2) \beta (2) & \varphi _{2px} (2) \alpha (2) & \varphi _{2py} (2) \alpha (2) \ \varphi _{1s} (3) \alpha (3) & \varphi _{1s} (3) \beta (3) & \varphi _{2s} (3) \alpha (3) & \varphi _{2s} (3) \beta (3) & \varphi _{2px} (3) \alpha (3) & \varphi _{2py} (3) \alpha (3) \ \varphi _{1s} (4) \alpha (4) & \varphi _{1s} (4) \beta (4) & \varphi _{2s} (4) \alpha (4) & \varphi _{2s} (4) \beta (4) & \varphi _{2px} (4) \alpha (4) & \varphi _{2py} (4) \alpha (4)\ \varphi _{1s} (5) \alpha (5) & \varphi _{1s} (5) \beta (5) & \varphi _{2s} (5) \alpha (5) & \varphi _{2s} (5) \beta (5) & \varphi _{2px} (5) \alpha (5) & \varphi _{2py} (5) \alpha (5)\ \varphi _{1s} (6) \alpha (6) & \varphi _{1s} (6) \beta (6) & \varphi _{2s} (6) \alpha (6) & \varphi _{2s} (6) \beta (6) & \varphi _{2px} (6) \alpha (6) & \varphi _{2py} (6) \alpha (6)\end{vmatrix} \end{align*}
Expanding this determinant would result in a linear combination of functions containing 720 terms. An expanded determinant will contain N! factorial terms, where N is the dimension of the matrix.
Exercise $\PageIndex{3A}$: Excited-State of Helium Atom
Write the Slater determinant for the $1s^12s^1$ excited state orbital configuration of the helium atom.
Answer
Since there are 2 electrons in question, the Slater determinant should have 2 rows and 2 columns exactly. Additionally, this means the normalization constant is $1/\sqrt{2}$.
Each element of the determinant is a different combination of the spatial component and the spin component of the $1 s^{1} 2 s^{1}$ atomic orbitals
$\frac{1}{\sqrt{2}}\left[\begin{array}{cc} {\varphi _{1_s}(1) \alpha(1)} & {\varphi {2_s}(1) \beta(1)} \ {\varphi {1_s}(2) \alpha(2)} & {\varphi {2_s}(2) \beta(2)} \end{array}\right] \nonumber \nonumber$
Exercise $\PageIndex{3B}$
Critique the energy level diagram and shorthand electron configuration notation from the perspective of the indistinguishability criterion. Can you imagine a way to represent the wavefunction expressed as a Slater determinant in a schematic or shorthand notation that more accurately represents the electrons? (This is not a solved problem!)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.06%3A_Antisymmetric_Wavefunctions_can_be_Represented_by_Slater_Determinants.txt
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Learning Objectives
• Understand how the Hartree method is expanded to include symmetrized mulit-electron determential wavefunctions via the Hartree-Fock equations.
• Understand how to calculate the orbital energies from HF theory.
• Apply HF theory with Koopman's theory to estimate ionization energies and electron affinities.
The Hartree method discussed previously is useful as an introduction to the solution of many-particle system and to the concepts of self-consistency and of the self-consistent-field calculations, but its importance is confined to the history of physics. In fact the Hartree method is not just approximate, it is fundamentally wrong since its wavefunction is not antisymmetric to electron permutation! The Hartree-Fock approach discussed below is a better approach, which correctly takes into account the antisymmetric character of the trial wavefunctions.
Although the Hartree equations are numerically tractable via the self-consistent field method, it is not surprising that such a crude approximation fails to capture elements of the essential physics. The Pauli exclusion principle demands that the many-body wavefunction be antisymmetric with respect to interchange of any two electron coordinates, e.g.
$\Psi(\mathbf{r}_{1},\mathbf{r}_{2}, \ldots, \mathbf{r}_{N}) = - \Psi(\mathbf{r}_{2},\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) \label{2.7}$
which clearly cannot be satisfied by the multi-electron wavefunctions of the form used in the Hartree Approximation, i.e., the orbital approximation (Equation $\ref{2.3}$).
$\Psi(\mathbf{r}_1,\mathbf{r}_2, \ldots, \mathbf{r}_N) \approx \psi_{1}(\mathbf{r}_1)\psi_{2}(\mathbf{r}_2) \ldots \psi_{N}(\mathbf{r}_N) \label{2.3}$
This indistinguishability condition can be satisfied by forming a Slater determinant of single-particle orbitals
$\Psi(\mathbf{r}_{1}, \mathbf{r}_{2}, \ldots, \mathbf{r}_{N})= \dfrac{1}{\sqrt{N}} \left \vert\psi(\mathbf{r}_{1})\psi(\mathbf{r}_{2}) \ldots \psi(\mathbf{r}_{N}) \right\vert \label{2.8}$
This decouples the electrons resulting in $N$ single-particle Hartree-Fock equations:
$\underbrace{-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r})}_{\text{kinetic energy}} + \underbrace{V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r})}_{\text{electron-nucleus potential}} + \underbrace{V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r})}_{\text{Hartree Term}} - \sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} = \epsilon_{i}\psi_{i}(\mathbf{r}). \label{2.9}$
As with the Hartree equations, the first term is the kinetic energy of the $i^{th}$ electron
$-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r}) \nonumber$
and the second term is the electron-nucleus potential between the $i^{th}$ electron and nucleus
$V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r}) \nonumber$
The third term (sometimes called the “Hartree” term) is the electrostatic potential between the $i^{th}$ electron and the average charge distribution of the other $N-1$ electrons.
$V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r}) = J_{j,k} = \int |\phi_j(r)|^2 |\phi_k(r’)|^2 \dfrac{e^2}{r-r'} dr dr’ \label{8.3.9}$
These three terms are identical to Hartree Equations with the product wavefunction ansatz (i.e., orbital approximation). The fourth term of Equation $\ref{2.9}$ is not in the Hartree Equations:
$\sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} \nonumber$
and is the exchange term. This term resembles the direct Coulomb term, but for the exchanged indices. It is a manifestation of the Pauli exclusion principle, and acts so as to separate electrons of the same spin. This “exchange” term acts only on electrons with the same spin and comes from the Slater determinant form of the wavefunction (Equation \ref{2.8}). Physically, the effect of exchange is for like-spin electrons to avoid each other. The exchange term adds considerably to the complexity of these equations.
The Hartree-Fock Equations in Equation $\ref{2.9}$ can be recast as series of Schrödinger-like equations:
$\hat {F} | \varphi _i \rangle = \epsilon _i| \varphi _i \rangle \label {8.7.2}$
where $\hat {F}$ is called the Fock operator and $\{| \varphi_i \rangle \}$ are the Hatree-Fock orbitals with corresponding energies $\epsilon_i$.
The Fock operator is a one-electron operator and solving a Hartree-Fock equation gives the energy and Hartree-Fock orbital for one electron. For a system with 2N electrons, the variable i will range from 1 to N; i.e there will be one equation for each orbital. The reason for this is that only the spatial wavefunctions are used in Equation $\ref{8.7.2}$. Since the spatial portion of an orbital can be used to describe two electrons, each of the energies and wavefunctions found by solving Equation $\ref{8.7.2}$ will be used to describe two electrons.
The nature of the Fock operator reveals how the Hartree-Fock (HF) or Self-Consistent Field (SCF) Method accounts for the electron-electron interaction in atoms and molecules while preserving the idea of independent atomic orbitals. The wavefunction written as a Slater determinant of spin-orbitals is necessary to derive the form of the Fock operator, which is
\begin{align} \hat {F} &= \hat {H} ^0 + \sum _{j=1}^N ( 2 \hat {J} _j - \hat {K} _j ) \nonumber \[4pt] &= -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {Ze^2}{4 \pi \epsilon _0 r} + \sum _{j=1}^N (2\hat {J}_j - \hat {K} _j ) \label {8.7.3} \end{align}
As shown by the expanded version on the far right, the first term in this equation, $\hat {H}^0$, is the familiar hydrogen-like operator that accounts for the kinetic energy of an electron and the potential energy of this electron interacting with the nucleus. The next term accounts for the potential energy of one electron in an average field created by all the other electrons in the system. The $\hat {J}$ and $\hat {K}$ operators result from the electron-electron repulsion terms in the full Hamiltonian for a multi-electron system. These operators involve the one-electron orbitals as well as the electron-electron interaction energy.
The Fock operator (Equation $\ref{8.7.3}$) depends on all occupied orbitals (because of the exchange and Coulomb operators). Therefore, a specific orbital can only be determined if all the others are known. One must use iterative methods to solve the HF equations like the Self-consistent field method discussed previously for the Hartree Approximation.
Exchange Energy
The exchange interaction is a quantum mechanical effect that only occurs between identical particles. Despite sometimes being called an exchange force in analogy to classical force, it is not a true force, as it lacks a force carrier. The effect is due to the wavefunction of indistinguishable particles being subject to exchange symmetry, that is, either remaining unchanged (symmetric) or changing its sign (antisymmetric) when two particles are exchanged. Both bosons and fermions can experience the exchange interaction. For fermions, it is sometimes called Pauli repulsion and related to the Pauli exclusion principle. For bosons, the exchange interaction takes the form of an effective attraction that causes identical particles to be found closer together, as in Bose–Einstein condensation.
Example 8.7.1 : Hartree-Fock Energy of Helium
For example, the electron 1 in helium (with $Z=2$), then
$\hat {H}^0 (1) = - \dfrac {\hbar ^2}{2m} \nabla ^2_1 - \dfrac {2e^2}{4 \pi \epsilon _0 r_1} \nonumber$
The Fock operator is couched in terms of the coordinates of the one electron whose perspective we are taking (which we will call electron 1 throughout the following discussion), and the average field created by all the other electrons in the system is built in terms of the coordinates of a generic “other electron” (which we’ll call electron 2) that is considered to occupy each orbital in turn during the summation over the $N$ spatial orbitals. The best possible one-electron wavefunctions, by definition, will give the lowest possible total energy for a multi-electron system used with the complete multielectron Hamiltonian to calculate the expectation value for the total energy of the system. These wavefunctions are called the Hartree-Fock wavefunctions and the calculated total energy is the Hartree-Fock energy of the system.
As with the Hartree Equations, solving the Hartree-Fock Equations is mathematically equivalent to assuming each electron interacts only with the average charge cloud of the other electrons. This is how the electron-electron repulsion is handled. This also why this approach is also called the Self-Consistant Field (SCF) approach.
Hartree-Fock Energy
The Hartree-Fock equations $h_e \phi_i = \epsilon_i \phi_i$ imply that the orbital energies $\epsilon_i$ can be written as:
\begin{align*} \epsilon_i &= \langle \phi_i | h_e | \phi_i \rangle \[4pt] &= \langle \phi_i | T + V | \phi_i \rangle + \sum_{j({\rm occupied})} \langle \phi_i | J_j - K_j | \phi_i \rangle \label{8.7.6} \[4pt] &= \langle \phi_i | T + V | \phi_i \rangle + \sum_{j({\rm occupied})} [ J_{i,j} - K_{i,j} ],\label{8.7.7} \end{align*}
where $T + V$ represents the kinetic ($T$) and nuclear attraction ($V$) energies, respectively. Thus, $\epsilon_i$ is the average value of the kinetic energy plus Coulombic attraction to the nuclei for an electron in $\phi_i$ plus the sum over all of the spin-orbitals occupied in $\psi$ of Coulomb minus Exchange interactions of these electrons with the electron in $\phi_i$.
If $\phi_i$ is an occupied spin-orbital, the $j = i$ term $[ J_{i,i} - K_{i,i}]$ disappears in the above sum and the remaining terms in the sum represent the Coulomb minus exchange interaction of $\phi_i$ with all of the $N-1$ other occupied spin-orbitals. If $\phi_i$ is a virtual spin-orbital, this cancelation does not occur because the sum over $j$ does not include $j = i$. So, one obtains the Coulomb minus exchange interaction of $\phi_i$ with all $N$ of the occupied spin-orbitals in $\psi$. Hence the energies of occupied orbitals pertain to interactions appropriate to a total of $N$ electrons, while the energies of virtual orbitals pertain to a system with $N+1$ electrons. This difference is very important to understand and to keep in mind.
To give an idea of how well HF theory can predict the ground state energies of several atoms, consider Table 8.7.1 below:
Table 8.7.1 : Hartree-Fock Calculations of Ground Energies of Select Atoms (all energies are in $R_H$)
Atom Hartree-Fock Energy Experiment
$He$ $-5.72$ $-5.80$
$Li$ $-14.86$ $-14.96$
$Ne$ $-257.10$ $-257.88$
$Ar$ $-1053.64$ $-1055.20$
Koopmans' Theorem
Koopmans' theorem states that the first ionization energy is equal to the negative of the orbital energy of the highest occupied molecular orbital. Hence, the ionization energy required to generated a cation and detached electron is represented by the removal of an electron from an orbital without changing the wavefunctions of the other electrons. This is called the "frozen orbital approximation." Let us consider the following model of the detachment or attachment of an electron in an $N$-electron system.
1. In this model, both the parent molecule and the species generated by adding or removing an electron are treated at the single-determinant level.
2. The Hartree-Fock orbitals of the parent molecule are used to describe both species. It is said that such a model neglects orbital relaxation (i.e., the re-optimization of the spin-orbitals to allow them to become appropriate to the daughter species).
Within this model, the energy difference between the daughter and the parent can be written as follows ($\phi_k$ represents the particular spin-orbital that is added or removed:
• for electron detachment (vertical ionization energies) $E_{N-1} - E_N = - \epsilon_k \label{8.7.8}$
• and for electron attachment (electron affinities) $E_N - E_{N+1} = - \epsilon_k .\label{8.7.9}$
The Hartree-Fock equations deal with exchange exactly; however, the equations neglect more detailed correlations due to many-body interactions. The effects of electronic correlations are not negligible; indeed the failure of Hartree-Fock theory to successfully incorporate correlation leads to one of its most celebrated failures.
Example 8.7.2 : Electron Affinity
Let’s derive this result for the case in which an electron is added to the $N+1^{st}$ spin-orbital. The energy of the $N$-electron determinant with spin-orbitals $\phi_1$ through $f_N$ occupied is
$E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \sum_{i=1}^{N} [ J_{i,j} - K_{i,j} ] \nonumber$
which can also be written as
$E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N} [ J_{i,j} - K_{i,j} ].\nonumber$
Likewise, the energy of the $N+1$-electron determinant wavefunction is
$E_{N+1} = \sum_{i=1}^{N+1} \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N+1} [ J_{i,j} - K_{i,j} ]. \nonumber$
The difference between these two energies is given by
\begin{align*} E_{N} – E_{N+1} = &- \langle \phi_{N+1} | T + V | \phi_{N+1} \rangle - \frac{1}{2} \sum_{i=1}^{N+1} [ J_{i,N+1} - K_{i,N+1} ] \[4pt] &- \frac{1}{2} \sum_{j=1}^{N+1} [ J_{N+1,j} - K_{N+1,j} ] \[4pt] &= - \langle \phi_{N+1} | T + V | \phi_{N+1} \rangle - \sum_{i=1}^{N+1} [ J_{i,N+1} - K_{i,N+1} ] \[4pt] &= - \epsilon_{N+1}. \end{align*} \nonumber
That is, the energy difference is equal to minus the expression for the energy of the $N+1^{st}$ spin-orbital, which was given earlier.
Advanced: Electron Correlation and the "Exchange Hole"
In the Copenhagen Interpretation, the squared modulus of the wavefunction gives the probability of finding a particle in a given place. The many-body wavefunction gives the N-particle distribution function, i.e. $|Φ(r_1, ..., r_N )|^2$ is the probability density that particle 1 is at $r_1$, ..., and particle $N$ is at $r_N$. However, when trying to work out the interaction between electrons, what we want to know is the probability of finding an electron at $r$, given the positions of all the other electrons $\{r_i\}$. This implies that the electron behaves quantum mechanically when we evaluate its wavefunction, but as a classical point particle when it contributes to the potential seen by the other electrons.
The contributions of electron-electron interactions in N-electron systems within the Hartree and Hartree-Fock methods are shown in Figure 8.7.2 . The conditional electron probability distributions $n(r)$ of $N-1$ electrons around an electron with given spin situated at $r=0$. Within the Hartree approximation, all electrons are treated as independent, therefore $n(r)$ is structureless. However, within the Hartree-Fock approximation, the $N$-electron wavefunction reflects the Pauli exclusion principle and near the electron at $r=0$ the exchange hole can be seen where the the density of spins equal to that of the central electron is reduced. Electrons with opposite spins are unaffected (not shown).
Summary
So, within the limitations of the HF, frozen-orbital model, the ionization potentials (IPs) and electron affinities (EAs) are given as the negative of the occupied and virtual spin-orbital energies, respectively. This statement is referred to as Koopmans’ theorem; it is used extensively in quantum chemical calculations as a means of estimating ionization potentials (Equation $\ref{8.7.8}$) and Electron Affinities (Equation $\ref{8.7.9}$) and often yields results that are qualitatively correct (i.e., ± 0.5 eV). In general Hartree-Fock theory gives a great first order solution (99%) to describing multi-electron systems, but that last 1% is still too great for quantitatively describing many aspects of chemistry and more sophisticated approaches are necessary. These are discussed elsewhere.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.07%3A_Hartree-Fock_Calculations_Give_Good_Agreement_with_Experimental_Data.txt
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Learning Objectives
• Understand how electron configurations results in different manifestations of angular momenta (both oribital and spin)
• Describe the manifestations in atoms via atomic term symbols
Atoms have quantum numbers that are directly analogous to the electronic quantum numbers.
The Total Orbital Angular Momentum Quantum Number: L
One might naively think that you could get the total angular momentum of an atom ($L$) by simply adding up the $l$ values of the individual electrons. The problem with this idea is that the angular momenta of the various electrons are not necessarily pointing in the same direction. Let's consider the case of adding two 2p-orbitals together (i.e., with $l=1$ quantum numbers).
As discussed in Section 6.3, each of these electrons has $\sqrt{2} \hbar$ of angular momentum, but oriented in three different directions given by three different $m_l$ values of 1, 0, and -1 (Figure 8.8.1 ).
$L = \sum_i^n l_i \label{8.8.4A}$
However, we must recognize that while the magnitude of angular momentum is a scalar (represented by $l$), angular momenta are really vector quantities and to add them together will require vector addition (vs. scalar addition) to do properly and must address all possible orientations of $l$ for each electron.
For example, if two electrons are revolving in the same direction as each other (i.e., same $m_l$ values), one would add just their $l$ values together
$L(\text{same direction}) = \sqrt{2} \hbar + \sqrt{2} \hbar = 2 \hbar \nonumber$
If the two elections were revolving in opposites direction (e.., opposite $m_l$ values, you subtract their values.
$L(\text{same direction}) = \sqrt{2} \hbar - \sqrt{2} \hbar = 0 \nonumber$
If they are revolving at some off‐angle relative to each other (one revolving in a plane and one off plane), you would partially subtract them.
$L(\text{same direction}) = \sqrt{2} \hbar + 0 = \sqrt{2} \hbar \nonumber$
To figure out all of the possible combinations of $l$ for a pair of electrons, simply add them together to get the co‐aligned case, subtract them to get the opposing case, and then fill in all the numbers in between to get the off‐angle cases. For the two p-orbital case, there are no other possibilities (remember that $l$ and hence $L$ must be non-negative since they represent the magnitude of angular momenta (addressing the $m_l$ values is a different story as we discuss below).
If you prefer to have a formula for the possibilities of $L$, you can use this:
$L = |l_1 + l_2 | , |l_1 + l_2 ‐ 1| , \ldots, |l_1 ‐ l_2 | \label{8.8.5A}$
Technically, Equation \ref{8.8.5A} should include all electrons in a system, not just two, to account for all possible combinations of orbital angular momenta, however, there are some tricks that aid in applying this equations to address larger multi-electron atoms. First, electrons in s-orbitals have no orbital angular momentum, so they can be ignored. Second, if block of orbitals are completely filled (e.g., all six p-block spin-orbitals or all 10 d-block spin-orbitals), then the orbital angular momentum vector of each electron will be countered by another electron in the system; these are called closed-shell systems. Systems with partially unfilled blocks are called open-shell systems.
Hence, conveniently for chemists, an atom’s electronic state depends entirely on its unfilled sub shells. Because electrons distribute themselves in a symmetric manner, the inner shell electrons end up canceling out each other’s momenta. For an atom in the configuration $1s^2 2s^2 p^2$, only the two p‐electrons matter. For an atom in the configuration $1s^22s^12p^1$, we have to examine only the 2s and 2p electrons (and can ignore the s electrons too).
Example 8.8.1 : Total Orbital Angular Momentum of Carbon
What are the possible $L$ values for the electrons in the $1s^2 2s^2 2p^2$ configuration of carbon?
Solution
Both open‐shell electrons (i.e., the 2p electrons) are $l = 1$. The possible combinations are 2, 1, 0.
Example 8.8.2 : Total Orbital Angular Momentum of Unknown Species
What are the possible $L$ values for the electrons in the $[Xe]6s^2 4f^1 5d^1$ ?
Solution
We can ignore the electrons in the $[Xe]$ core and the electrons in the $6s$ block. So all we have to consider is the f electron $l=3$ and d electron $l=2$.
The two extremes possible (Equation \ref{8.8.5A}) are
$3+2 = 5 \nonumber$
and
$3 ‐ 2 = 1 \nonumber$
The possible combinations are thus 5, 4, 3, 2, and 1.
The Total Magnetic Quantum Number: $M_l$
The Total Magnetic Quantum Number $M_l$ is the total z‐component of all of the relevant electrons’ orbital momentum. Where $L$ told you how much total angular momentum there is in the system, $M_l$ tells you which direction it is pointing. Like $L$, a given configuration can have several possible values of $M_l$, depending on the electrons’ relative orientation. Unlike $L$, $M_l$ is allowed to have negative values. To list the possible $M_l$ values for a two electron system, take the case where both $m_l$ are positive, then take the case where they are both negative, and then fill in the numbers in between (Figure 8.8.1 ).
$M_l = m_{l1} + m_{l2}, m_{l1} + m_{l2} ‐ 1, ... , ‐m_{l1} ‐ m_{l2} \label{8.8.6}$
Note that there is no absolute value function in Equation \ref{8.8.6} like in Equation \ref{8.8.5A}.
Example 8.8.3 : Total Magnetic Quantum Number of the Zirconium Ground State
What are the possible values $M_l$ of a zirconium atom with the $[Kr] 5s^2 4d^2$ electron configuration?
Solution
Both open‐shell electrons (i.e., the 4d electrons) are $l=2$, so the values are 4, 3, 2, 1, 0, ‐1, ‐2, ‐3, ‐4.
The Total Spin Magnetic Quantum Number: $M_s$
$M_s$ is the sum total of the z‐components of the electrons’ inherent spin. Do not confuse it with $M_l$, which is the sum total of the z‐component of the orbital angular momentum. It is easily computed by finding all of the possible combinations of $m_s$. Since $m_s$ for each individual electron can only be +1/2 or ‐ 1/2, this isn’t too complicated.
$M_s = m_{s1} + m_{s2}, m_{s1} +m_{s2} ‐ 1 , ... , m_{s1} ‐ m_{s2} \label{8.8.7}$
Example 8.8.4 : Total Spin Magnetic Quantum Number of the Carbon Ground State
What are the M s values for $1s^2 2s^2 2p^2$ ?
Solution
$M_s = 1, 0, ‐1$
The Total Intrinsic Spin Quantum Number: $S$
The sum total of the spin vectors of all of the electrons is called $S$. The difference between $S$ and $M_s$ is subtle, but vital for understanding multiplicity. $M_s$ measures the total z‐component of the electrons’ spins, while $S$ measures the entire resultant vector. The values of $S$ are computing in a manner very similar to $M_s$. Because $S$ measures the magnitude of a vector, it cannot ever be negative.
$S = |s_1 + s_2 |, |s_1 +s_2 ‐ 1| , ... ,| s_1 ‐ s_2 | \label{8.8.8}$
Example 8.8.5 : The Hydrogen Ground State
Find $S$ for $1s^1$.
Solution
$S$ clearly has to be ½ since that’s the spin of a single electron and there’s only one electron to worry about.
Example 8.8.6 : The Beryllium Excited State
Find S for $1s^2 2s^1 2p^1$.
Solution
$S = 1, 0$
Example 8.8.7 : The Carbon Ground State
Find $S$ for carbon atoms with the $1s^2 2s^2 2p^2$ electron configuration.
Solution
$S=1,0$ This is the same as the previous problem. Notice that $S$ is not affected by which orbitals the electrons are in. $S$ only cares about how many open‐shell electrons (i.e., unpaired electrons) there are, not about where they are. This is because $S$ measures an inherent property of the electrons themselves
Example 8.8.8
Find $S$ for nitrogen atoms with the $1s^2 2s^2 2p^3$ electron configuration.
Solution
We have not done a three electron case yet, but they are not hard. Find all the combinations for a single pair first, and then factor in the third electron. For two electrons, we already know that the two possible S values are S=1,0. A third electron can either add or subtract ½ from these values, so the final $S$ can be $S=3/2, 1, 1/2$.
The Total Angular Momentum Quantum Number J
The total orbital angular momentum of an atom (measured in terms of $l$), and the total spin angular momentum of an atom (measure in $S$) combine to form total angular momentum, a number that is quantized by the number $J$. $L$ and $S$ do not necessarily have to be pointing in the same direction (Figure 8.8.1 ), so $J$ can range from $L + S$ to $|L – S|$.
Table 8.8.1 : Quantum Numbers and associated ranges
Symbol Name Allowed Range
$l$ Total orbital angular momentum $|l_1 + l_2 |, ..., |l_1 ‐ l_2 |$
$M_l$ Magnet Quantum number $[m_{l1} + m_{l2}, ..., ‐ m_{l1} ‐ m_{l2} ]$
$M_s$ Spin Magnetic Quantum Number $| m_{s1} + m_{s2} |,..., |m_{s1} ‐ m_{s2} |$
$S$ Inherent Spin Number $|s_1 + s_2 |,..., |s_1 ‐ s_2 |$
$M$ Multiplicity 2S+1
$J$ Total Angular Momentum $L+S,..., | L-S |$
Multiplicity
Multiplicity is a simple ‐ sounding concept that defies simple explanations. You know from your first ‐ year education that a singlet is when the net spin (S) is equal to zero (e.g. all the electrons are spin paired), and a triplet happens when the net spin is equal to 1 (e.g. two electrons are pointing in the same direction). They are called “singlet” and “triplet” because there are 3 ways to combine a pair of electron spins to get S=1, but only one way to get $S=0$. If you draw a picture of the possible ways that two electrons can arrange their spins, you get something like this:
While this picture is an improvement over the simple up‐down model, it is still misleading. The three spin axes of an electron share a Heisenberg Uncertainty Principle. The more you know about $S_x$, the less you can know about either $S_y$ or $S_z$. The same is true for all other combinations of x, y and z. Since we have defined $S_z$ as a known and fixed value, the values of $S_x$ and $S_y$ must be completely unknown. This causes the x and y orientations of the electrons to become smeared out across all possible values:
Constructing Term Symbols
Atomic term symbols contain two pieces of information. They tell you the total orbital angular momentum of the atom ($l$), and they tell you the multiplicity ($M$). $l$ is denoted by a simple code, similar to the code used to delineate the types of atomic orbitals:
• $L=0 \rightarrow S$
• $L=1 \rightarrow P$
• $L=2 \rightarrow D$
• $L=3 \rightarrow F$
Note that while the notation is similar, L does NOT say anything about what types of orbitals the electrons are in. A state that has the term symbol P does NOT necessarily have an open p‐shell. The multiplicity is indicated by appending a number to the upper left of the symbol. A $L=2$, $M=3$ state would be represented by $^3D$. The secret to writing the term symbols for an atom is to discover what combinations of $l$ and $M$ are possible for that atom with that specific electronic configuration. An atom that only has closed shells will always be $1S$.
Each term symbol represents a discrete energy level. We can place these levels in the correct order by using these simple rules:
• 1: High multiplicity values mean low energy
• 2: If there is a tie, high $l$ values mean low energy
• 3a: If there is still a tie and the shell is less than half full, then low J means low energy
• 3b: If the shell is more than half full, then high J means low energy
These rules reliably predict the ground state. They have only erratic agreement with experiment when ordering the other levels.
Example 8.8.9 : Hydrogen Ground State
What are the term symbols for the microstates possible for $1s^1$ electronic configuration of hydrogen?
Solution
Since there is only one electron, this is a simple problem. $L=0$ and $M=1$, so the only possible term symbol is $^2S$. With only one electron, $S = ½$, so $J = 0 + ½ = ½$. Only one microstate exists for this configuration and it has a term symbol of $^2S_{½}$.
Example 8.8.10 : Boron
What are the term symbols for the microstates possible for $1s^2 2s^2 2p^1$ electronic configuration of boron?
Solution
There still only one open shell electron, so $L=1$, $M=1$ and $S = ½$. We get a term symbol of the type $^2P$, which gets split into separate symbols because $J = 3/2$ and $1/2$. Two possible microstates exist for this system with term symbols of $^2P_{3/2}$ and $^2P_{1/2}$.
Example 8.8.11 : Beryllium Excited State
What are the term symbols for the microstates possible for the $1s^2 2s^1 2p^1$ excited-state electronic configuration of Beryllium?
Solution
Now we have two electrons to worry about. Since $l_1 = 0$ and $l_2 = 1$, the only possible combination is $L=1$. The possible combinations of S are: $S=1,0$. This means that $M=3$ or $M=1$. The term symbols will be of the form $^1P$ and $^3P$. For the $^1 P$ state, $L=1$ and $S=0$, so $J=1$. For the second state, $L=1$ and $S=1$, so $J=2,1,0$. There are four microstates for this configuration with term symbols of $^1 P_1$ and $^3 P_2$, $^3P_1$, and $^3P_0$.
Example 8.8.12 : Zirconium
What are the term symbols for the microstates possible for the $[Kr] 5s^2 4d^2$ ground-state electronic configuration of zirconium?
Solution
This is a much harder problem. We will need to use a special technique to disentangle all of the possible combinations of $l$ and $M$. Let’s start be listing the relevant quantum numbers for the two open-shell electrons:
$l_1 = 2$ $l_2 = 2$
$m_{l1} = 2, 1, 0, ‐ 1, ‐ 2$ $m_{l2} = 2, 1, 0, ‐ 1, ‐ 2$
$m_{s1} = ½, ‐½$ $m_{s1} = ½, ‐½$
Let’s combine these numbers to generate the atomic quantum numbers:
$L = 4,3,2,1,0 \nonumber$
$M_l = 4,3,2,1,0, ‐ 1, ‐ 2, ‐ 3, ‐ 4 \nonumber$
$M_s = 1,0 \nonumber$
We know that there will at least one each of S, P, D, F and G. It isn’t immediately clear which of these will be singlets and which will be triplets. To figure this out, we need to systematically examine the possible microstates. It turns out that there are 45 possible ways to put distribute two electrons between 5 d orbitals. That’s a lot! The easiest way to list the states is to organize them into a chart:
$M_s= -1$ $M_s= 0$ $M_s= +1$
Attacking the chart one row at a time. Ask yourself, how many ways can I arrange the two electrons to give me $M_l = 4$? It turns out there is only one possible combination that does this:
This state is has $M_s = 0$. This means that there is only 1 microstate that corresponds to $M_l =4$ and $M_s = 0$, and none that correspond to $M_l =4$ and $M_s = \pm ‐1$. We add this microstate to the chart like this:
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
4 0 1 0
Now, how many ways are there to get $M_l = 3$?
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
4 0 1 0
3 1 2 1
For $M_l =2$, we find the following states:
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
4 0 1 0
3 1 2 1
2 1 3 1
You should be able to draw the microstates on your own by now. You should find 8 states, four of which are singlet and four of which are triplets.
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
4 0 1 0
3 1 2 1
2 1 3 1
1 2 4 2
There are only nine possible ways to arrange the electrons to get $M_l =0$
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
4 0 1 0
3 1 2 1
2 1 3 1
1 2 4 2
0 2 5 2
The rest of the chart will be symmetric to the first half, so we do not need to do any more work:
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
4 0 1 0
3 1 2 1
2 1 3 1
1 2 4 2
0 2 5 2
-1 2 4 2
-2 1 3 1
-3 1 2 1
-4 0 1 0
Now that we have a listing of all of the microstates, we need to figure out how to divide them up between the term symbols. It turns out that each term symbol can have, at most, one microstate from each box on the chart. The term symbols always end up claiming a “box” of microstates, centered on the middle of the chart. This is easier shown than said.
Attacking the chart from the top, we can see that the $M_l = 4$ $M_s = 0$ state clearly belongs to a $^1G$ symbol. The $M_l = ‐4$ $M_s =0$ box also clearly belongs to this symbol. If I connect these states with a “box,” I get this:
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
4 0 1 0
3 1 2 1
2 1 3 1
1 2 4 2
0 2 5 2
-1 2 4 2
-2 1 3 1
-3 1 2 1
-4 0 1 0
The strickthough configurations all belong to the $^1 G$ state. Let's subtract them from the chart to indicate that they are not available for other term symbols.
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
3 1 2 1
2 1 3 1
1 2 4 2
0 2 5 2
-1 2 4 2
-2 1 3 1
-3 1 2 1
The next row indicates a $L=3$ state. Because there are three $M_s$ values available, this is a triplet. The term symbol will be $^3F$, which reduces the chart down to
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
1 2 4 2
0 2 5 2
-1 2 4 2
The next state will be $^1D$. Leaving us with
$M_l$ $M_s= -1$ $M_s= 0$ $M_s= +1$
1 1 1 1
0 1 2 1
-1 1 1 1
Next is a $^3P$ state. The chart is getting pretty small now wtih
$M_l$ $M_s= 0$
0
The last remaining microstate comprises the $^1S$ term symbol.
The total listing is
$^1G$, $^3F$, $^1D$, $^3P$, $^1S$.
Assigning $J$ values, we get
$^1G_4$, $^3F_4$, $^3F_3$, $^3F_2$, $^1D_2$, $^3P_2$, $^3P_1$, $^3P_0$, $^1S_0$
If you can do this problem, you can do almost any atomic term symbol.
Note
The secret to writing the term symbols for an atom is to discover what combinations of $l$ and $M$ are possible for that atom with that specific electronic configuration.
Shortcuts
There is a deep symmetry that connects different electronic configurations. It turns out that a $p^1$ configuration has the same term symbols as a $p^5$. Similarly, $p^2 = p^4$. A similar relationship can be used to figure out high electron number term symbols for the $d$ and $f$ orbitals.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.08%3A_Term_Symbols_Gives_a_Detailed_Description_of_an_Electron_Configuration.txt
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Learning Objectives
• Compate two spin-orbit coupling schemes that couple the total spin angular momenta and total orbital angular momenta of a multi-electron spectra
We need to be able to identify the electronic states that result from a given electron configuration and determine their relative energies. An electronic state of an atom is characterized by a specific energy, wavefunction (including spin), electron configuration, total angular momentum, and the way the orbital and spin angular momenta of the different electrons are coupled together. There are two descriptions for the coupling of angular momentum. One is called j-j coupling, and the other is called L-S coupling. The j-j coupling scheme is used for heavy elements (z > 40) and the L-S coupling scheme is used for the lighter elements. Only L-S coupling is discussed below.
L-S Coupling of Angular Momenta
L-S coupling also is called R-S or Russell-Saunders coupling. In L-S coupling, the orbital and spin angular momenta of all the electrons are combined separately
$L = \sum _i l_i \label{8.11.3}$
$S = \sum _i s_i \label{8.11.4}$
The total angular momentum vector then is the sum of the total orbital angular momentum vector and the total spin angular momentum vector.
$J = L + S \label{8.11.5}$
The total angular momentum quantum number parameterizes the total angular momentum of a given particle, by combining its orbital angular momentum and its intrinsic angular momentum (i.e., its spin). Due to the spin-orbit interaction in the atom, the orbital angular momentum no longer commutes with the Hamiltonian, nor does the spin.
However the total angular momentum $J$ does commute with the Hamiltonian and so is a constant of motion (does not change in time). The relevant definitions of the angular momenta are:
Orbital Angular Momentum
$|\vec{L}| = \hbar \sqrt{\ell(\ell+1)} \nonumber$
with its projection on the z-axis $L_z = m_\ell \hbar \nonumber$
Spin Angular Momentum
$|\vec{S}| = \hbar \sqrt{s(s+1)} \nonumber$
with its projection on the z-axis $S_z = m_s \hbar \nonumber$
Total Angular Momentum
$|\vec{J}| = \hbar \sqrt{j(j+1)} \nonumber$
with its projection on the z-axis $J_z = m_j \hbar \nonumber$
where
• $l$ is the azimuthal quantum number of a single electron,
• $s$ is the spin quantum number intrinsic to the electron,
• $j$ is the total angular momentum quantum number of the electron,
The quantum numbers take the values:
\begin{align} & m_\ell \in \{ -\ell, -(\ell-1) \cdots \ell-1, \ell \} , \quad \ell \in \{ 0,1 \cdots n-1 \} \& m_s \in \{ -s, -(s-1) \cdots s-1, s \} , \& m_j \in \{ -j, -(j-1) \cdots j-1, j \} , \& m_j=m_\ell+m_s, \quad j=|\ell+s|\\end{align} \nonumber
and the magnitudes are:
\begin{align} & |\textbf{J}| = \hbar\sqrt{j(j+1)} \& |\textbf{J}_1| = \hbar\sqrt{j_1(j_1+1)} \& |\textbf{J}_2| = \hbar\sqrt{j_2(j_2+1)} \\end{align} \nonumber
in which
$j \in \{ |j_1 - j_2|, |j_1 - j_2| - 1 \cdots j_1 + j_2 - 1, j_1 + j_2 \} \,\! \nonumber$
This process may be repeated for a third electron, then the fourth etc. until the total angular momentum has been found.
The result of these vector sums is specified in a code that is called a Russell-Saunders term symbol, and each term symbol identifies an energy level of the atom. Consequently, the energy levels also are called terms. A term symbol has the form $^{2s+1} L_J$ where the code letter that is used for the total orbital angular momentum quantum number L = 0, 1, 2, 3, 4, 5 is S, P, D, F, G, H, respectively. Note how this code matches that used for the atomic orbitals. The superscript $2S+1$ gives the spin multiplicity of the state, where S is the total spin angular momentum quantum number. The spin multiplicity is the number of spin states associated with a given electronic state. In order not to confuse the code letter S for the orbital angular momentum with the spin quantum number S, you must examine the context in which it is used carefully. In the term symbol, the subscript J gives the total angular momentum quantum number. Because of spin-orbit coupling, only $J$ and $M_j$ are valid quantum numbers, but because the spin-orbit coupling is weak $L$, $M_l$, $S$, and $m_s$ still serve to identify and characterize the states for the lighter elements.
For example, the ground state, i.e. the lowest energy state, of the hydrogen atom corresponds to the electron configuration in which the electron occupies the 1s spatial orbital and can have either spin $\alpha$ or spin $\beta$. The term symbol for the ground state is $^2 S_{1/2}$, which is read as “doublet S 1/2”. The spin quantum number is 1/2 so the superscript 2S+1 = 2, which gives the spin multiplicity of the state, i.e. the number of spin states equals 2 corresponding to $\alpha$ and $\beta$. The S in the term symbol indicates that the total orbital angular momentum quantum number is 0 (For the ground state of hydrogen, there is only one electron and it is in an s-orbital with $l = 0$ ). The subscript ½ refers to the total angular momentum quantum number. The total angular momentum is the sum of the spin and orbital angular momenta for the electrons in an atom. In this case, the total angular momentum quantum number is just the spin angular momentum quantum number, ½, since the orbital angular momentum is zero. The ground state has a degeneracy of two because the total angular momentum can have a z-axis projection of $+\frac {1}{2} \hbar$ or $-\frac {1}{2} \hbar$, corresponding to $m_J$ = +1/2 or -1/2 resulting from the two electron spin states $\alpha$ and $\beta$. We also can say, equivalently, that the ground state term or energy level is two-fold degenerate.
Exercise 8.9.1
Write the term symbol for a state that has 0 for both the spin and orbital angular momentum quantum numbers.
Exercise 8.9.2
Write the term symbol for a state that has 0 for the spin and 1 for the orbital angular momentum quantum numbers
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/08%3A_Multielectron_Atoms/8.09%3A_The_Allowed_Values_of_J_-_the_Total_Angular_Momentum_Quantum_Number.txt
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Learning Objectives
• Define Hund three rules
• Use Hunds three rules to predict the lowest energy configuration and term symbols for multi-electron systems
The Aufbau section discussed how that electrons fill the lowest energy orbitals first, and then move up to higher energy orbitals only after the lower energy orbitals are full. However, there a problem with this rule. Certainly, 1s orbitals should be filled before 2s orbitals, because the 1s orbitals have a lower value of n, and thus a lower energy. What about the three different 2p orbitals? In what order should they be filled? The answer to this question involves Hund's rule, which make a lot more sense in the context of generated term symbols that are used to combine the various \(L\) and \(S\) values represent vector additions of possible microstates.
Hund’s Rules
1. State with the largest value of \(S\) is most stable and stability decreases with decreasing \(S\).
2. For states with same values of \(S\), the state with the largest value of \(L\) is the most stable.
3. If states have same values of \(L\) and \(S\) then, for a subshell that is less than half filled, state with smallest \(J\) is most stable; for subshells that are more than half filled, state with largest value of \(J\) is most stable.
Example 8.10.1
Rank these terms associated with an electronic configuration of an atom based on energy (via Hund's rules):
\(^3D\), \(^3P\), \(^3S\), \(^1D\), \(^1P\), \(^1S\)
Hund's First Rule (Maximize Spin Multiplicity)
According to the first rule, electrons always enter an empty orbital before they pair up. Electrons are negatively charged and, as a result, they repel each other. Electrons tend to minimize repulsion by occupying their own orbitals, rather than sharing an orbital with another electron. Furthermore, quantum-mechanical calculations have shown that the electrons in singly occupied orbitals are less effectively screened or shielded from the nucleus.
There's a Coulomb repulsion between two electrons to put them in the same orbital (a spin pairing energy often discussed in Crystal Field Theory). However, there's also a quantum mechanical effect. The exchange energy (which is favorable) increases with the number of possible exchanges between electrons with the same spin and energy. In transitioning from the top state to the middle state of Figure 8.10.1 , we remove the Coulomb repulsion between electrons in the same orbital. Moreover, In transitioning from the middle state to the bottom state (most stable state predicted by Hund's first rule), we gain the exchange energy, because these two electrons are indistinguishable.
Hund's Second Rule (Maximize Orbital Angular Multiplicity)
What matters is the total (scalar) angular momentum, not the direction. The negative and positive signs refer only to the direction of the angular momentum, not the magnitude. The direction is furthermore arbitrary (except in, say, a magnetic or electric field). So is the spin direction, incidentally. By convention we usually draw the first electron in each orbital as "up" (positive spin). However we could just as easily draw it "down". It makes no difference - in the absence of an external EM field, the energy is the same, if only because molecules/atoms are rotating with respect to the lab frame anyway. "Up" and "down", in other words, is artificial. What matters is the relative momentum vectors of the various electrons in the system, and hence their sum total.
Hund's Third Rule (Minimize less than half filled or maximize greater than half filled shells)
A long time ago someone offering a reasonably simple explanation related to the fact that when the shell is more than half full, it's easier to visualize the system as an interaction between the spin and orbital momenta of holes rather than electrons, in which case the energetic stabilization term is reversed in sign. This would be because the spin angular momentum of a single hole would be opposite in sign compared to the spin angular momentum of a single electron. Taking as an example - the three p-orbitals. A situation with 1 electron and 5 electrons are functionally similar, except that one has a single electron and one has a single hole. All things being equal, the total spin angular momentum of the 1 electron system would be opposite in sign to whatever the total spin angular momentum of the 5 electron system is. So the expectations for Hund's rules would be switched. You can kind of see this if you draw out all the microstates of the 1-electron and 5-electron configurations: the everything is pretty much changed in sign in the latter case.
Example 8.10.2
What terms and levels can arise from an atom with the ground-state configuration of \(1s^22s^22p^6 3s^2 3p^6 4s^2 4p^1 3d^1\)? Which is the most stable (lowest in energy) state?
Solution
Possible states include:
\(^1F_3\), \(^1D_2\), \(^1P_1\), \(^3F_4\), \(^3F_3\), \(^3F_2\), \(^3D_3\), \(^3D_2\), \(^3D_1\), \(^3P_2\), \(^3P_1\), \(^3P_0\).
There are two unpaired electrons in this system from the electron configuration.
• Rule 1 predicts that the ground state will be a triplet with \(S=1\) so \(2S+1=3\). So the ground state is from this more narrowed list: \(^3F_4\), \(^3F_3\), \(^3F_2\), \(^3D_3\), \(^3D_2\), \(^3D_1\), \(^3P_2\), \(^3P_1\), \(^3P_0\).
• Rule 2 predicts a \(F\) state since that is the highest multiplicity with \(L= 3\): So the ground state is from this more narrowed list: \(^3F_4\), \(^3F_3\), \(^3F_2\)
• Rule 3 predicts the lowest \(J\) term since the d shell is less than half full. That is the \(J=2\) state.
Therefore for this system, the atom will have a ground-state structure of
\(^3F_2\)
Example 8.10.3 : Titanium cation
The ground configuration of a \(\ce{Ti^{2+}}\) ion is \([Ar]3d^2\). What is the term of lowest energy state?
Solution
• Rule 1: two unpaired electrons ⇒ highest S = 1 ⇒ 2S + 1 = 3
• Rule 2: two in d in parallel spin ⇒ highest L = 1 + 2 = 3 ⇒ 3F
• Rule 3:.L = 3, S = 1 ⇒ J = 4, 3, 2; less than half-filled ⇒ \(^3F_2\)
Exercise 8.10.3
What is the term of lowest energy state for the following atoms and ions.
• \(\ce{C}\): \([He]2s^2 2p^2\)
• \(\ce{N}\): \([He]2s^2 2p^3\)
• \(\ce{O}\): \([He]2s^2 2p^4\)
• \(\ce{Cr^{3+}}\): \([Ar]3d^3\)
• \(\ce{Mn^{3+}}\): \([Ar]3d^4\)
• \(\ce{Fe^{3+}}\): \([Ar]3d^5\)
Answer
\(^3P_0\), \(^4S_{3/2}\), \(^3P_2\), \(^4F_{3/2}\), \(^5D_0\), and \(^6S_{5/2}\), respectively.
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Learning Objectives
• Demonstrate how spin-orbit coupling is experimentally observed in atomic spectra
• Use atomic terms symbols to ascribe transitions to specific angular momenta states described by atomic term symbols
Around 1930, several spectroscopists using high resolution instruments found that lines in the hydrogen atom spectrum actually are not single lines but they are multiplets as shown for an isotopic mixture of hydrogen,$(H^1_{\alpha})$ and deuterium, ($H^2_{\alpha}$), in Figure 8.11.1 . A multiplet consists of two or more closely spaced lines. Two lines together form a doublet, three a triplet, etc. Multiplets also are called fine structure. The term fine structure means the lines are spaced close together, i.e. finely spaced. Such fine structure also was found in spectra of one-electron ions such as $\ce{He^{+}}$.
You should recall that the $H^1_{\alpha}$ line in the Balmer series at 656.279 nm was understood as resulting from a single transition of an electron from the n = 3 energy level to the n = 2 level. The observation of fine structure revealed that an orbital energy level diagram does not completely describe the energy levels of atoms. This fine structure also provided key evidence at the time for the existence of electron spin, which was used not only to give a qualitative explanation for the multiplets but also to furnish highly accurate calculations of the multiplet splittings.
Spin-Orbit Coupling
Specifying the orbital configuration of an atom does not uniquely identify the electronic state of the atom because the orbital angular momentum, the spin angular momentum, and the total angular momentum are not precisely specified. For example in the hydrogen 2p1 configuration, the electron can be in any of the three p-orbitals, $m_l$ = +1, 0, and –1, and have spins with $m_s$ = +1/2 or –1/2. Thus, there are 3 times 2 different possibilities or states. Also, the orbital and spin angular momentum of the electrons combine in multiple ways to produce angular momentum vectors that are characteristic of the entire atom not just individual electrons, and these different combinations can have different energies. This coupling of orbital and spin angular momentum occurs because both the electron spin and orbital motion produce magnetic dipole moments. As we have seen previously, the relationship between the angular momentum and the magnetic moment is given by the gyromagnetic ratio. These magnetic dipoles interact just like two tiny bar magnets attracting and repelling each other. This interaction is called spin-orbit interaction. The interaction energy is proportional to the scalar product of the magnetic dipole moments, which are proportional to the angular momentum vectors.
$E_{s-o} \propto S \cdot L \nonumber$
with the following terms added to the Hamiltonian
$\hat {H} _{s-o} \propto \hat {S} \cdot \hat {L} \nonumber$
where the constant of proportionality is called the spin-orbit coupling constant. The spin-orbit interaction couples the spin motion and orbital motion of all the electrons together. This coupling means that exact wavefunctions are not eigenfunctions of the spin and orbital angular momentum operators separately. Rather the total angular momentum $J = L+S$, the vector sum of the spin and orbital angular momentum, is required to be coupled for a completely accurate description of the system. Trying to describe the coupled system in terms of spin and orbital angular momentum separately is analogous to trying to describe the positions of two coupled bar magnets independently. It cannot be done; their interaction must be taken into account (Figure 8.11.1 ).
Atomic Spectroscopy
Higher energy or excited orbital configurations also exist. The hydrogen atom can absorb energy, and the electron can be promoted to a higher energy orbital. The electronic states that result from these excited orbital configurations also are characterized or labeled by term symbols. The details of how to determine the term symbols for multi-electron atoms and for cases where both the orbital and spin angular momentum differ from zero are given elsewhere, along with rules for determining the relative energies of the terms.
We have found that the selection rules for promoting a single electron moving from one atomic orbital to another via the absorption or emission of light are
\begin{align*} \Delta l &= \pm 1 \[4pt] \Delta m_l &= 0, \pm 1 \end{align*}
These selection rules arise from the conservation fo angular momentum during a spectroscopic transition and the fact that a photon has a spin 1. Within the limits of L-S coupling, these rules can be expressed in terms of atomic term symbols resulting in the resulting Russell-Saunders selection rules:
\begin{align} \Delta S &= 0 \label {8.11.8} \[4pt] \Delta L &= 0, \pm 1 \label {8.11.9} \[4pt] \Delta J &= 0, \pm 1,\label {8.11.10} \end{align}
but the $J =0$ to $J= 0$ transition is forbidden
$\Delta m_J = 0, \pm 1 \label {8.11.11}$
but the $m_J = 0$ to $m_J = 0$ transition is forbidden if $\Delta J = 0$.
These selection rules result from the general properties of angular momentum such as the conservation of angular momentum and commutation relations. The $\Delta L =0$ option in Equation $\ref{8.11.9}$ does not violate the conservation of angular momentum discussed previously, since $\Delta l = \pm 1$ is still required. The orbital angular momentum of an electron must change upon absorption, but this does not necessarily affect the overall momentum of the state given by Equation $\ref{8.11.9}$.
The selection rules apply only to atoms that can be described with Russell-Saunders (LS) coupling. These rules fail as the atomic number increases because the $S$ and $L$ quantum numbers become "bad" quantum numbers; this occurs when the jj-coupling coupling approach is more applicable. For example, the transition between single ($S=1/2$ and triplet $S=1$ states (violation of selection rule in Equation $\ref{8.11.8}$) are allowed and experimentally observed, in heavy atoms.
Example 8.11.1 : Sodium Atoms
An example of this fine structure is the emission of sodium atoms.
616.07 nm 615.42 nm 589.00 nm 589.59 nm 568.82 nm 568.26 nm
How can these transitions be described in terms of transitions between microstates
Solution
We need to discussed states in terms of not only electron configurations, but in terms of microstates (i.e., term symbols) and the principal quantum number of the valence electron, $n$:
• The ground state has a $(Ne]ns^1$ configuration, which has only one microstate $^2S_{1/2}$
• The excited state with the valence electron in the p-orbitals has an electron configuration of $[Ne]np^1$, which has two microstates: $^2P_{3/2}$ and $^2P_{1/2}$
• The excited state with the valence electron in the p-orbitals has an electron configuration of $[Ne]nd^1$, which has two microstates of $^2D_{5/2}$ and $^2D_{3/2}$
Observed lines can be explained:
• $5S \rightarrow 3P$ gives two lines since the initial configuration has two microstates: 616.07, 615.42 nm
• $3P \rightarrow 3S$ gives two lines since the terminal configuration has two microstates: 589.00, 589.59 nm
• $4D \rightarrow 3P$ gives two lines since the terminal configuration has two microstates: 568.82, 568.26 nm
Splitting of the Sodium D Line
One notable atomic spectral line of sodium vapor is the so-called D-line, which may be observed directly as the sodium flame-test line and also the major light output of low -pressure sodium lamps (these produce pressure sodium lamps (these produce an unnatural yellow). The D-line is one of the classified Fraunhofer lines in Sodium vapor in the upper layers of lines. Sodium vapor in the upper layers of the sun creates a dark line in the emitted spectrum of electromagnetic radiation by absorbing visible light in a band of wavelengths around 589.5 nm. This wavelength corresponds to transitions in atomic sodium in which the valence-electron transitions from a 3s to 3p electronic state.
Closer examination of the visible spectrum of atomic sodium reveals that the D-line actually consists of two lines called the $D_1$ and $D_2$ lines at 589 6 nm and 589.0 nm, respectively. The splitting between these lines arises because of spin-orbit coupling. Na has one unpaired electron ($S = ½$). If we consider the $S \rightarrow P$ transition, then for the excited state, $P$, we have $L = 1$. Thus, $J = 3/2$ or $J=1/2$.
Now we want to apply these ideas to understand why multiplet structure is found in the luminescence spectrum of hydrogen and single electron ions. As we have said, the $H_{\alpha}$ line in the Balmer series at 656.279 nm can be understood via a transition of an electron in a n = 3 atomic orbital to a n = 2 atomic orbital. When this spectral line was examined using high-resolution instruments, it was found actually to be a doublet, i.e. two lines separated by 0.326 cm-1.
There are 9 degenerate orbitals associated with the n = 3 level, and 4 associated with the n = 2 level. Since an electron can be in any orbital with any one of two spins, we expect the total number of states to be twice the number of orbitals. The number of orbitals is given by $n^2$ so there should be 8 states associated with n = 2 and 18 states associated with n = 3. Using the ideas of vector addition of angular momentum, the terms that result from having an electron in any one of these orbitals are given in Table 8.11.1 .
Orbital Configuration
Term Symbols
Degeneracy
Table 8.11.1 : H Atom Terms Originating from n = 1, 2, and 3
1s1 $^2S_{1/2}$ 2
2s1 $^2S_{1/2}$ 2
2p1 $^2P_{1/2}$, $^2P_{3/2}$ 2, 4
3s1 $^2S_{1/2}$ 2
3p1 $^2P_{1/2}$, $^2P_{3/2}$ 2, 4
3d1 $^2D_{3/2}$, $^2D_{5/2}$ 4, 6
Table 8.11.1 shows that there are three terms associated with n = 2, and 5 terms associated with n = 3. In principle, each term can have a different energy. The degeneracy of each term is determined by the number of projections that the total angular momentum vector has on the z-axis. These projections depend on the $m_J$ quantum number, which ranges from $+J$ to $–J$ in integer steps. J is the total angular momentum quantum number, which is given by the subscript in the term symbol. This relationship between $m_J$ and J ($m_J$ varies from $+J$ to $–J$ in integer steps) is true for any angular momentum vector.
Exercise 8.11.3
Confirm that the nine term symbols in Table 8.11.1 are correct.
Exercise 8.11.4
Confirm that the values for the degeneracy in Table 8.11.1 are correct and that the total number of states add up to 8 for n = 2 and 18 for n = 3.
The energies of the terms depend upon spin-orbit coupling and relativistic corrections that need to be included in the Hamiltonian operator in order to provide a more complete description of the hydrogen atom. As a consequence of these effects, all terms with the same $n$ and (J\) quantum numbers have the same energy, while terms with different values for $n$ or $J$ have different energies. The theoretical term splittings as given by H.E. White, Introduction to Atomic Spectra (McGraw-Hill, New York, 1934) pp. 132-137 and are shown in Figure 8.11.2 .
Figure 8.11.2 shows 5 allowed transitions for the electron in the states associated with n = 3 to the states associated with n = 2. Of these five, two are most intense and are responsible for the doublet structure. These two transitions are indicated by the wide black lines at the bottom of the figure to correspond to the lines observed in the photographic spectrum shown in Figure 8.11.2 . The other transitions contribute to the width of these lines or are not observed. The theoretical value for the doublet splitting is 0.328 cm-1, which is in excellent agreement with the measured value of 0.326 cm-1. The value of 0.328 cm-1 is obtained by taking the difference, 0.364 – 0.036 cm-1,in the term splittings.
As we have just seen, the electronic states, as identified by the term symbols, are essential in understanding the spectra and energy level structure of atoms, but it also is important to associate the term symbols and states with the orbital electron configurations. The orbital configurations help us understand many of the general or coarse features of spectra and are necessary to produce a physical picture of how the electron density changes because of a spectroscopic transition.
Exercise 8.11.5
Use the Russell-Saunders selection rules to determine which transitions contribute to the $H_{\alpha}$ line in the hydrogen spectrum.
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Solutions to select questions can be found online.
8.4
Prove the speed of electron in first Bohr orbit is $e^2/4\pi ϵ_0ℏ = 2.188 \times 10^{6} m•s^{-1}$. The speed is in atomic units.
Solution
Use following formula in order to make the calculations
$v=\dfrac{ℏ}{m_e a_0}\nonumber$
$v=\dfrac{ℏ}{m_e} \left(\dfrac{m_e e^2}{4πϵ_0ℏ^2}\right)\nonumber$
$v=\dfrac{e^2}{4πϵ_0ℏ}\nonumber$
Substituting values get the following:
$v=\dfrac{(1.6022 \times 10^{-19}\;C)^2}{(1.1127\times 10^{-10}C^2 \cdot J^{-1} \cdot m^{-1})(1.0546 \times 10^{-34}\;J \cdot s)}\nonumber$
$v=2.1877 \times 10^{6}\;m \cdot s^{-1} \nonumber$
This is the speed in atomic units. Are either of the two separate terms in the two term helium Hartree-Fock Orbital acceptable wavefunctions by themselves?
${\psi}\left(r\right)=0.81839e^{-1.44608r}+0.52072e^{-2.86222r}\nonumber$
8.15
Explain why:
$E = \dfrac{\displaystyle \int \Phi^*_2(1,2)\hat{H}\Phi_2(1,2)dr_1 dr_2 d\sigma_1 d\sigma_2 }{\int \Phi^*_2(1,2)\Phi_2(1,2)dr_1 dr_2 d\sigma_1 d\sigma_2}\nonumber$
can be rewritten as:
$E = \dfrac{\displaystyle \int \Phi^*_2(1,2)\hat{H}\Phi_2(1,2)dr_1 dr_2}{\int \Phi^*_2(1,2)\Phi_2(1,2)dr_1 dr_2}\nonumber$
Solution
The Hamiltonian does not depend on spin, so the spin integral can be factored out.
8.16
Why must you distinguish the two electrons in separated hydrogen atoms?
Solution
They must be distinguished from one another because they each belong to a separate nucleus, not to an individual one.
8.17
In both the Hartree-Fock approximation and hydrogen atomic wavefunction, why is the angular dependence the same?
Solution
Since the Hamiltonian used in the approximation only depends on $r$, the angular dependence is not affected in the Hartree-Fock approximation. So both the Hartree-Fock approximation and hydrogen atom will have the same angular dependence.
8.20
Given the two electron determinate wavefunction below, determine if the spin component of the system is symmetric, anti-symmetric, or neither.
$\begin{vmatrix} 1s\alpha(1) & 1s\beta(1) \ 1s\alpha(2) & 1s\alpha(2) \end{vmatrix}\nonumber$
Solution
We have to solve the determinate.
$1s{1}\alpha(2)1s{2}\beta(2)-1s{1}\beta(2)1s{1}\alpha(2)$
Now we have to factor out the spatial part from the spin part.
$1s{1}1s{2}[ \alpha(1)\beta(2)-\beta(1)\alpha(2)]$
We are focused on the spin part and by observation, we can tell that the spin component is anti-symmetric.
8.23
Given $\hat{S}_z\alpha = \dfrac{\hbar}{2}\alpha$ and $\hat{S}_z\alpha = \dfrac{-\hbar}{2}\beta$, show that $\Psi_{200}$ is an eigenfunction of $S_{z,total} = \hat{S}_{za} + \hat{S}_{zB}$.
Solution
Start with $\Psi_{200} = C(\psi_{2sa} + \psi_{2sB})$.
$\hat{S}_{z.total}\Psi = C(\hat{S}_{za} + \hat{S}_{zB})[\psi_{2sa} + \psi_{2sB}]\nonumber$
$= C \left(\dfrac{\hbar}{2}-\dfrac{\hbar}{2}\right) [\psi_{2sa} + \psi_{2sB}]\nonumber$
$=0\nonumber$
8.24
For the wavefunction:
$\psi =\begin{vmatrix} \psi _{A}(1)&\psi _{A}(2) \\psi_{B}(1) &\psi_{B}(2) \end{vmatrix}\nonumber$
discuss the effect on the wavefunction of (a) swapping rows of the matrix and (b) swapping columns of the matrix.
Solution
Let's expand the determinant to inspect the complete multi-electron wavefunction:
$\psi =\psi_{A}(1)\psi_{B}(2)-\psi_{A}(2)\psi_{B}(1)\nonumber$
If we swap rows: $\psi_{RowSwap} =\psi_{A}(2)\psi_{B}(1)-\psi_{A}(1)\psi_{B}(2)=-\psi\nonumber$
If we swap columns: $\psi_{ColumSwap} =\psi_{A}(2)\psi_{B}(1)-\psi_{A}(1)\psi_{B}(2)=-\psi\nonumber$
Conclusion: swapping the either two rows or two columns of a Slater determinant changes the sign of the wavefunction.
8.27
What are the term symbols for carbon and oxygen atoms in the ground state?
Solution
Carbon: The electron configuration for carbon at ground state (lowest energy) is $1s^2 2s^2 2p^2$
Since the 1s and 2s orbitals are completely filled, they can be neglected when calculating for S. For the lowest energy, you need to use the highest values of S and L, therefore we find that S= 1/2 +1/2=1 and L=1 which corresponds to the letter P. Since the 2p orbital is less than half-way filled, we use J=|L-S|=|1-1|=0.
The term symbol is $^{2S+1}L_J$ which is 3P0 for the ground-state of carbon atoms.
Oxygen: The electron configuration for oxygen at ground state (lowest energy) is $1s^2 2s^2 2p^4$
Again, the 1s and 2s orbitals may be neglected when calculating for $S$.
$S=1/2 +1/2 +1/2 -1/2 =1 \nonumber$
The last electron has a spin down due to the Pauli Exclusion Principle.
L=1 which corresponds to the letter P.
Since the 2p orbital for oxygen is more than half-filled,
$J= L+S = 2\nonumber$
The term symbol for the ground-state oxygen atom is 3P2
8.28
Show that the number of sets of magnetic quantum number ($m_l$) and spin quantum number ($m_s$) associated with any term symbol is equal to $(2L+1)(2S+1)$. Apply this result to the $np^2$ case and show that symbol $^1S$, $^3P$ and $^1D$ account for all the possible sets of magnetic quantum numbers and spin quantum numbers
8.29
Calculate all possible numbers of term symbol for an $np^{1}$ electron configuration.
Solution
$N = \dfrac{G!}{e! (G - e)!}\nonumber$
where G is the highest number of electrons that an orbital can hold and e is the highest number of electrons that a subshell can hold
$N = \dfrac{6!}{2!(6 - 2)!} = 15\nonumber$
8.30
Determine the ground state term symbol for the electron configuration of the Halogens.
Solution
Halogens have the electron configuration$np^5$ we can determine the term symbols for this configuration by mapping out all possible configurations the electrons can fit into the six spin orbitals. To quickly determine how many possible combinations there are we can use a statistical method of $N_{Comb.}= \binom{spin orbitals}{electrons}$
$N_{comb} = \binom{6}{5} = \dfrac{6!}{5!(6!-5!)} = 6\nonumber$
Now we know that there are 6 different configurations that we can map out to determine the term symbols. Since we are working with the $p$ orbital we know $l = 1$ and $m_s = -1, \ 0, \ 1$. I will denote spin using $\alpha$ as spin up $+\dfrac{1}{2}$ and $\beta$ as spin down $-\dfrac{1}{2}$. $M_L$ is the sum of the $m_s$ values corresponding to the number of electrons in that energy level. For example if you have $\alpha$$\beta$ in $m_s = +1$ only, you then have 2 electrons in $m_s = +1$ resulting in $M_L = +1+1=2$. $M_s$ is the sum of the spin up and spin down values.
+1 0 -1 $M_L$ $M_s$
$\alpha$$\beta$ $\alpha$$\beta$ $\alpha$ +1 $+\dfrac{1}{2}$
$\alpha$$\beta$ $\alpha$ $\alpha$$\beta$ 0 $+\dfrac{1}{2}$
$\alpha$ $\alpha$$\beta$ $\alpha$$\beta$ -1 $+\dfrac{1}{2}$
$\alpha$$\beta$ $\alpha$$\beta$ $\beta$ +1 $-\dfrac{1}{2}$
$\alpha$$\beta$ $\beta$ $\alpha$$\beta$ 0 $-\dfrac{1}{2}$
$\beta$ $\alpha$$\beta$ $\alpha$$\beta$ -1 $-\dfrac{1}{2}$
Now we need to determine the maximum value of $M_L$ and$M_s$. Looking at the table we see that
$max \ M_L = 1 \ max \ M_s = \dfrac{1}{2}\nonumber$
From this we know our maximum value of $L$ and $S$
$L_{max} = 1 \ S_{max} = \dfrac{1}{2}\nonumber$
Our possible values for $L$ and $S$ are
$L = 1, \ 0\ \ S = \dfrac{1}{2}\nonumber$
Since S is only $\dfrac{1}{2}$ we know we can only have doublet term symbols since $2\big(\dfrac{1}{2}\big)+1 = 2$. L ranges from 1 to 0 so our possible corresponding symbols will be P and S. This leaves us with the possibility of having
$^2P, \ ^2S\nonumber$
To figure out what is there we start with the term symbol that has the largest L value, with is the $P$. We see that for $P$, $L=1$ and $S = \dfrac{1}{2}$. For a value of $L=1$ our $m_l$ can be $+1, \ 0, \ -1$ and for an $S= \dfrac{1}{2}$ our $m_s = +\dfrac{1}{2}, \ -\dfrac{1}{2}$. In the table above all rows that contain these possible combination include every row. Therefore all of the configurations are contained in this doublet $P$ term symbol. Including values of $J$ we know that $L+S \geq J \geq |L-S|$. Since $L=1$ and $S = \dfrac{1}{2}$ our final term symbols are
$^2P_\dfrac{3}{2}, \ ^2P_\dfrac{1}{2}\nonumber$
Hund's rules say that when L and S are the same with a subshell more than half filled, you look to the largest $J$ value to be the most stable. Therefore our final answer and the ground state term symbol for halogens is
$\boxed{^2P_\dfrac{3}{2}}\nonumber$
8.33
$^2P$, $^2D$, and $^4S$ are the term symbols for an atom with the $np^{3}$ electron configuration. Using the term symbols for the $np^{3}$ electron configuration, calculate the $J$ values associated with each of the term symbol. Then find out which term symbol represent the ground state.
Solution
To calculate $J$ we use this equation
$J = L + S\nonumber$
This equation can be further expanded to be
$J = L + S, L + S - 1, L + S - 2, ...\left | L - S \right |\nonumber$
Term Symbol L S J Full Term Symbol
$^{2}P$ 1 $\dfrac{1}{2}$ $\dfrac{3}{2}$, $\dfrac{1}{2}$ $^2P_{3/2}$ and $^2P_{1/2}$
$^{2}D$ 2 $\dfrac{1}{2}$ $\dfrac{5}{2}\, \dfrac{3}{2}$ $^4D_{5/2}$ and $^4D_{3/2}$
$^{4}S$ 0 $\dfrac{3}{2}$ $\dfrac{3}{2}$ $^4S_{3/2}$
8.34
What are the ground state electron configuration and term symbol for Calcium?
Solution
1s22s22p63s23p64s2
or
[Ar]4s2
For the term symbol:
Spin Multiplicity: S = 0 (all electrons paired)
J = S + L = 0
There is only one valid value of $J$, so the term symbol for this configuration is
2S+1LJ = 1S0
2S + 1 = 1
Orbital Angular Multiplicity: L = 0 because we consider an s orbital. This corresponds to S.
8.34
Find the ground state term symbol for Ca.
Solution
$ns^2$ electron configurations have the term symbol $^1S_0$, so the term symbol for Ca in the ground state is $^1S_0$.
8.36
Write the electron configuration for vanadium and use this information to find the ground-state term symbol for V.
Solution
The electron configuration for vanadium is $[Ar]3d^3 4s^3$
Spin Multiplicity: The electron configuration predicts three unpaired electrons, so $S= 1/2+1/2+1/2=3/2\nonumber$ and the multiplicity of $2S+1$ predicts that this will be a quartet. Orbital Angular Momentum: The electron configuration predicts three electrons with $l=2$ and the rest so not contribute, so $L=1+1+1=3$, which is an $F$ state
The ground-state term symbol for vanadium is $^4F_{3/2}$ since vanadium has a half-filled 3d subshell.
• Total Angular Momentum: The shell is half full
8.36
What is the ground-state term symbol for $Ne$.
Solution
The term symbol is $^{2S + 1} L_{J}$
where
$S$ is the total electron spin $L$ is the total orbital angular momentum
so $2S+1 = 2\cdot 0+1 = 1$ and $J= 0+0=0$. So the term symbol for $Ne$ is $^{1}S_0$.
and $J= L+S\nonumber$
The electron configuration for:$Ne$ is $1s^2 2s^2 2p^6$ since Ne has spherical symmetry, we get:
$L= 0 + 0= S\nonumber$
$S = \dfrac{1}{2}+\dfrac{1}{2}+ \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}= 0\nonumber$
8.37
Looking at the 1$s$2$p$ electron configuration for He. Solve for the term symbols (states) of the Helium configuration and the degeneracies. If electron spin orbit coupling is included, what effect will this have?
Solution
There are two possible sets of $m$$l$ and $m$$s$ for the $ns$ electron and six possible sets of $m$$l$ and $m$$s$ for the $np$ electron, so there are 2 × 6 = 12 possible sets of $m$$l$ and $m$$s$ for the system. We can denote values for the electron in the $ns$ orbital as $m$1$j$ and those for the electron in the $np$ orbital as $m$2$j$. The allowed values are given below:
Microstate $m_{l} (1)$ $m_{s}(1)$ $m_{l}(2)$ $m_{s}(2)$ $M_{L}$ $M_S$ $M_J$
Electron 1 in $1s$ orbital Electron 2 in $2p$ orbital Combined Angular Momenta of both Electrons
1 0 + 1/2 1 + 1/2 1 1 2
2 0 - 1/2 1 + 1/2 1 0 0
3 0 + 1/2 1 - 1/2 1 0 0
4 0 - 1/2 1 - 1/2 1 -1 1
5 0 + 1/2 0 + 1/2 0 1 0
6 0 - 1/2 0 + 1/2 0 0 1
7 0 + 1/2 0 - 1/2 0 0 0
8 0 - 1/2 0 - 1/2 0 -1 0
9 0 + 1/2 -1 + 1/2 -1 1 0
10 0 - 1/2 -1 + 1/2 -1 0 -1
11 0 + 1/2 -1 - 1/2 -1 0 -2
12 0 - 1/2 -1 - 1/2 -1 -1 -2
Entries 1, 2, 4, 5, 6, 8, 9, 10, and 12 correspond to $L$ = 1 and $S$ = 1, or 3P term symbol, and entries 3, 7, and 11 correspond to $L$ = 1 and $S$ = 0, which is a 1P term symbol. The values of $J$ can be derived from the table or by using
$J$ = ($L$ + $S$), ($L$ + $S$ - 1), ($L$ + $S$ - 2), . . . . . . . , (|$L$ + $S$|),
The final results given the term symbols below:
3P2 3P1 3P0 1P1
($L$ + $S$) ($L$ + $S$ - 1) (|$L$ - $S$|) (|$L$ + $S$|)
The states corresponding to this electron configuration and their degeneracies are:
Term symbol: 3P2 3P1 3P0 1P1
Degeneracy: 5 3 1 3
According to Hund's rule, the ground state 3P0. Including the effect of spin-orbit coupling removes the degeneracy of the electronic states, and no spin orbit coupling splits the lines in an atomic spectra.
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Our basis for understanding chemical bonding and the structures of molecules is the electron orbital description of the structure and valence of atoms, as provided by quantum mechanics. We assume an understanding of the periodicity of the elements based on the nuclear structure of the atom and our deductions concerning valence based on electron orbitals.
• 9.1: The Born-Oppenheimer Approximation Simplifies the Schrödinger Equation for Molecules
The Born-Oppenheimer approximate is one of the most important and fundamental approximations in molecular quantum mechanics. This approximation separates the Schrödinger equation into two components, resulting in separate electronic and nuclear part of the wavefunction components. This separation is not exact, but approximate based on the separation of electronic and nuclear degrees of freedom we assume can be made based on the mass differential.
• 9.2: The H₂⁺ Prototypical Species
The simplest conceivable molecule would be made of two protons and one electron, namely $H_{2}^{+}$. This species actually has a transient existence in electrical discharges through hydrogen gas and has been detected by mass spectrometry and it also has been detected in outer space. The Schrödinger equation for $H_{2}^{+}$ can be solved exactly within the Born-Oppenheimer approximation. This ion consists of two protons held together by the electrostatic force of a single electron.
• 9.3: The Overlap Integral
Overlap integrals quantify the concentration of orbitals (often) on adjacent atoms in the same regions of space. Orbital overlap is a critical component in bond formation.
• 9.4: Chemical Bond Stability
From this LCAO-MO approach arises the Coulomb, Exchange (similar to HF calculations of atoms), and Overlap integrals. The concept of bonding and anti-bonding orbitals results.The application of LCAO toward molecular orbitals is demonstrated including linear variational theory and secular equations.
• 9.5: Bonding and Antibonding Orbitals
The spatial structure of the bonding and antibonding molecular orbitals are contrasted demonstrating features such as a node between the nuclei. The expansion of the LCAO MOs using a greater basis set than just the 1s atomic orbitals is discussed.
• 9.6: A Simple Molecular-Orbital Treatment of H₂ Places Both Electrons in a Bonding Orbital
To describe the electronic states of molecules, we construct wavefunctions for the electronic states by using molecular orbitals. These wavefunctions are approximate solutions to the Schrödinger equation. A mathematical function for a molecular orbital is constructed, $\psi _i$, as a linear combination of other functions, $\varphi _j$, which are called basis functions because they provide the basis for representing the molecular orbital.
• 9.7: Molecular Orbitals Can Be Ordered According to Their Energies
The linear combination of atomic orbitals always gives back the same number of molecular orbitals. So if we start with two atomic orbitals, we end up with two molecular orbitals. When atomic orbitals add in phase, we get constructive interference and a lower energy orbital. When they add out of phase, we get a node and the resulting orbital has higher energy. The lower energy MOs are bonding and higher energy MOs are antibonding.
• 9.8: Molecular-Orbital Theory Does not Predict a Stable Diatomic Helium Molecule
The occupied molecular orbitals (i.e., orbitals with electrons) are represented via an electron configuration like with atoms. For diatomics, these configurations are reflected at a "bond order" that is used to describe the strength and lengths of the bonds. They predict that stable molecules (i.e., observable) have bond orders that are > 0. For molecular orbitals consisting of only the 1s atomic orbitals, that suggests certain molecules will not exist. The typical example is the helium dimer.
• 9.9: Electrons Populate Molecular Orbitals According to the Pauli Exclusion Principle
The Pauli exclusion principle plays as important a role in the understanding of the electronic structure of molecules as it does in the case of atoms. We are now in a position to build up and determine the electronic configurations of the homonuclear diatomic molecules by adding electrons two at a time to the molecular orbitals with the spins of the electrons paired, always filling the orbitals of lowest energy first.
• 9.10: Molecular Orbital Theory Predicts that Molecular Oxygen is Paramagnetic
The molecular orbital configuration dictates the bond order of the bond. This in turns dictates the strength of the bond and the bond length with stronger bonds exhibiting small bond lengths. The molecular orbital configuration of molecular oxygen demonstrates that the ground-state neutral species has two unpaired electrons and hence is paramagnetic (attractive to external magnetic fields). This is a feature of MO theory that other theories do not predict.
• 9.11: Photoelectron Spectra Support the Existence of Molecular Orbitals
• 9.12: Molecular-Orbital Theory Also Applies to Heteronuclear Diatomic Molecules
• 9.13: SCF-LCAO-MO Wavefunctions are Molecular Orbitals formed from a Linear Combination of Atomic Orbitals and Whose Coefficients Are Determined Self-Consistently
• 9.14: Molecular Term Symbols Describe Electronic States of Molecules
Molecular term symbols specify molecular electronic energy levels. Term symbols for diatomic molecules are based on irreducible representations in linear symmetry groups, derived from spectroscopic notations. They usually consist of four parts: spin multiplicity, azimuthal angular momentum, total angular momentum and symmetry. All molecular term symbols discussed here are based on Russel-Saunders coupling.
• 9.15: Molecular Term Symbols Designate Symmetry
The quantum numbers for diatomic molecules are similar from the atomic quantum numbers. Be cautious, because the rules for finding the possible combinations are different
• 9.16: Most Molecules Have Excited Electronic States
• 9.E: Chemical Bond in Diatomic Molecules (Exercises)
These are homework exercises to accompany Chapter 9 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Thumbnail: A covalent bond forming $\ce{H2}$ where two hydrogen atoms share the two electrons. (CC BY-SA 3.0; Jacek FH via Wikipedia; modified by LibreTexts)
09: Chemical Bonding in Diatomic Molecules
Learning Objectives
• Understand the need to introduce an approximation like the Born-Oppenheimer approximation to solve multi-electron systems
• Understand the basis of parameterization involved in using the Born-Oppenheimer approximation
Using quantum mechanics to predict the chemical bonding patterns, optimal geometries, and physical and chemical properties of molecules is a large and active field of research known as molecular quantum mechanics or more commonly as quantum chemistry. The density functional theory referred to in the previous lecture, for which the chemistry Nobel prize was given in 1998, has had a tremendous impact in quantum chemistry, with some of the papers in this subject having acquired some 10,000 citations each since their publication. In fact, the 1998 chemistry Nobel prize was shared between Walter Kohn, one of the inventors of density functional theory and John Pople, the developer of a commonly used quantum chemistry software package.
Quantum chemistry calculations allow the geometries of molecules to be computed as well as a wide range of properties. Quantum chemistry can also be used in a novel way, in which the electrons are treated using quantum mechanics but the nuclei are treated as classical particles. We use quantum mechanics to calculate the internuclear forces but then use these forces in Newton's Second Law to study the motion of the nuclei during chemical reactions. This gives us a microscopic window into the specific motions, the complex dance, executed by the nuclei during a simple or complex chemical process.
The methods of quantum chemistry have become very sophisticated, and there are various software packages that can be downloaded for carrying out the calculations of quantum chemistry. It should be noted that these packages use a series of approximations to solve the Schrödinger equation because for all but the simplest of molecules, exact solutions are not available. We will discuss some of these methods, but first we need to introduce some of the underlying theory.
The Born-Oppenheimer Approximation
The Born-Oppenheimer approximation is one of the basic concepts underlying the description of the quantum states of molecules. This approximation makes it possible to separate the motion of the nuclei and the motion of the electrons. This is not a new idea for us. We already made use of this approximation in the particle-in-a-box model when we explained the electronic absorption spectra of cyanine dyes without considering the motion of the nuclei. Then we discussed the translational, rotational and vibrational motion of the nuclei without including the motion of the electrons. In this chapter we will examine more closely the significance and consequences of this important approximation. Note, in this discussion nuclear refers to the atomic nuclei as parts of molecules not to the internal structure of the nucleus.
The Born-Oppenheimer approximation neglects the motion of the atomic nuclei when describing the electrons in a molecule. The physical basis for the Born-Oppenheimer approximation is the fact that the mass of an atomic nucleus in a molecule is much larger than the mass of an electron (more than 1000 times). Because of this difference, the nuclei move much more slowly than the electrons. In addition, due to their opposite charges, there is a mutual attractive force of $Ze^2/r^2$ acting on an atomic nucleus and an electron. This force causes both particles to be accelerated. Since the magnitude of the acceleration is inversely proportional to the mass, a = F/m, the acceleration of the electrons is large and the acceleration of the atomic nuclei is small; the difference is a factor of more than 2000. Consequently, the electrons are moving and responding to forces very quickly, and the nuclei are not. You can imagine running a 100-yard dash against someone whose acceleration is a 2000 times greater than yours. That person could literally run circles around you.
Example 9.1.1 : Coupled Oscillators with Dissimilar Masses
If two particles interact in some way, and one is much heavier than the other, the light particle will move essentially as a "slave'' of the heavy particle. That is, it will simply follow the heavy particle wherever it goes, and, it will move rapidly in response to the heavy particle motion. As an illustration of this phenomenon, consider the simple mechanical system pictured below:
Considering this as a classical system, we expect that the motion will be dominated by the large heavy particle ($m_1$), which is attached to a fixed wall by a spring. The small, light particle ($m_2$, which is attached to the heavy particle by a spring will simply follow the heavy particle and execute rapid oscillations around it.
So a good approximation is to describe the electronic states of a molecule by thinking that the nuclei are not moving, i.e. that they are stationary. The nuclei, however, can be stationary at different positions so the electronic wavefunction can depend on the positions of the nuclei even though their motion is neglected.
Now we look at the mathematics to see what is done in solving the Schrödinger equation after making the Born-Oppenheimer approximation. For a diatomic molecule as an example, the Hamiltonian operator is grouped into three terms
$\hat {H} (r, R) = \hat {T}_{nuc} (R) + \dfrac {e^2}{4\pi \epsilon _0} \dfrac {Z_A Z_B}{R} + \hat {H} _{elec} (r,R) \label {9.1.1}$
where
$\hat{T}_{nuc} (R) = -\dfrac {\hbar^2}{2m_A} \nabla ^2_A - \dfrac {\hbar ^2}{2m_B} \nabla ^2_B \label {9.1.2}$
and
$\hat {H} _{elec} (\vec{r}, \vec{R}) = \dfrac {- \hbar ^2}{2m} \sum \limits _i \nabla ^2_i + \dfrac {e^2}{4 \pi \epsilon _0} \left ( -\sum \limits _i \dfrac {Z_A}{r_{Ai}} - \sum \limits _i \dfrac {Z_B}{r_{Bi}} + \dfrac {1}{2} \sum \limits _i \sum \limits _{j \ne i} \dfrac {1}{r_{ij}}\right ) \label{9.1.3}$
In Equation \ref{9.1.1}, the first term represents the kinetic energy of the nuclei, the second term represents the Coulomb repulsion of the two nuclei, and the third term represents the contribution to the energy from the electrons, which consists of their kinetic energy, mutual repulsion for each other, and attraction for the nuclei. $\vec{r}$ and $\vec{R}$ are vectors specifying the positions of all the electrons and all the nuclei, respectively.
Exercise 9.1.1
Define all the symbols in Equations \ref{9.1.1} through \ref{9.1.3}.
Answer
\begin{align*}
\hat{H}(r,R) &= \underbrace{\hat{T}_{nuc}(R)}_\text{Kinetic Energy Term for nuclei} + \underbrace{\frac{e^2}{4\pi\epsilon_o}\frac{Z_AZ_B}{R}}_\text{Repulsion Term for nuclei} + \underbrace{\hat{H}_{elec}(r,R)}_\text{Hamiltonian for electrons}\
\hat{T}_{nuc}(R) &= \underbrace{\frac{\hslash^2}{2m_A}\nabla^2_A}_\text{Kinetic Energy Term for nuclei A} - \underbrace{\frac{\hslash^2}{2m_b}\nabla^2_B}_\text{Kinetic Energy Term for nuclei B}\
\hat{H}_{elec}(\vec{r},\vec{R}) &= \underbrace{\frac{-\hslash^2}{2m}\sum_{i}\nabla_i^2}_\text{Kinetic Energy Term for electrons} + \frac{e^2}{4\pi\epsilon_o}\left(-\underbrace{\sum_{i}\frac{Z_A}{r_{Ai}}}_\text{Attraction Term between nuclei A and electron i} - \underbrace{\sum_{i}\frac{Z_B}{R_{Bi}}}_\text{Attraction Term between nuclei B and electron i} + \underbrace{\frac{1}{2}\sum_{i}\sum_{j \neq i}\frac{1}{r_{ij}}}_\text{Repulsion Term between electrons}\right)
\end{align*}
where, $Z_x$is the charge of particle x, $m_x$ is the mass of particle x and $r_{xz}$ is the distance between particle x and z.
Exercise 9.1.2
Explain why the factor of 1/2 appears in the last term in Equation \ref{9.1.3}.
Answer
The 1/2 term is there to make sure we do not double count the potential energies via the two summations. Otherwise, we would independently add the potential energy of electron 1 with electron 2 and the potential energy of electron 2 with electron 1. These are the same and hence one has to be removed.
The Born-Oppenheimer approximation says that the nuclear kinetic energy terms in the complete Hamiltonian, Equation \ref{9.1.1}, can be neglected in solving for the electronic wavefunctions and energies. Consequently, the electronic wavefunction $\varphi _e (r,R)$ is found as a solution to the electronic Schrödinger equation
$\hat {H} _{elec} (r, R) \varphi _e (r, R) = E_e (R) \varphi _e (r, R) \label {9.1.4}$
Even though the nuclear kinetic energy terms are neglected, the Born-Oppenheimer approximation still takes into account the variation in the positions of the nuclei in determining the electronic energy and the resulting electronic wavefunction depends upon the nuclear positions, $R$. As a result of the Born-Oppenheimer approximation, the molecular wavefunction can be written as a product
$\psi _{ne} (r, R) = X_{ne} (R) \varphi _e (r, R) \label {9.1.5}$
This product wavefunction is called the Born-Oppenheimer wavefunction. The function $X_{ne} (R)$ is the vibrational wavefunction, which is a function of the nuclear coordinates $R$ and depends upon both the vibrational and electronic quantum numbers or states, n and e, respectively. The electronic function, $\varphi _e (r, R)$, is a function of both the nuclear and electronic coordinates, but only depends upon the electronic quantum number or electronic state, e. Translational and rotational motion is not included here. The translational and rotational wavefunctions simply multiply the vibrational and electronic functions in Equation \ref{9.1.5} to give the complete molecular wavefunction when the translational and rotational motions are not coupled to the vibrational and electronic motion.
In the Crude Born-Oppenheimer Approximation, $R$ is set equal to $R_o$, the equilibrium separation of the nuclei, and the electronic wavefunctions are taken to be the same for all positions of the nuclei (i.e., the nuclei never move). The electronic energy, $E_e (R)$, in Equation \ref{9.1.4} combines with the repulsive Coulomb energy of the two nuclei, to form the potential energy function that controls the nuclear motion as shown in Figure 9.1.1 .
$V_e (R) = E_e (R) + \dfrac {e^2}{4\pi \epsilon _0} \dfrac {Z_A Z_B}{R} \label {9.1.6}$
Consequently the Schrödinger equation for the vibrational motion is
$( \hat {T} _{nuc} (R) + V (R) ) X_{ne} (R) = E_{ne} X_{ne} (R) \label {9.1.7}$
Potential Energy Curves and Surfaces
Previously, the potential energy was approximated as a harmonic potential or Morse potential depending on the displacement, $R$, of the nuclei from their equilibrium positions.
In practice the electronic Schrödinger equation is solved using approximations at particular values of $R$ to obtain the wavefunctions $\varphi _e (r,R)$ and potential energies $V_e (R)$. The potential energies can be graphed as illustrated in Figure 9.1.1 .
The graph in Figure 9.1.1 is the energy of a diatomic molecule as a function of internuclear separation, which serves as the potential energy function for the nuclei. When R is very large there are two atoms that are weakly interacting. As $R$ becomes smaller, the interaction becomes stronger, the energy becomes a large negative value, and we say a bond is formed between the atoms. At very small values of $R$, the internuclear repulsion is very large so the energy is large and positive. This energy function controls the motion of the nuclei. Previously, we approximated this function by a harmonic potential to obtain the description of vibrational motion in terms of the harmonic oscillator model. Other approximate functional forms could be used as well, e.g. the Morse potential. The equilibrium position of the nuclei is where this function is a minimum, i.e. at $R = R_0$. If we obtain the wavefunction at $R = R_0$ and use this function for all values of $R$, we have employed the Crude Born-Oppenheimer approximation.
Exercise 9.1.3
Relate Equation \ref{9.1.7} to the one previously used in our description of molecular vibrations in terms of the harmonic oscillator model.
While the potential energy function, $V_e (R)$, for a diatomic molecule is a 1-D curve (Figure 9.1.1 ), molecules with more than two atoms will have multi-dimensional potential energy surfaces with 3N-6 (or 3N-5 for linear molecule) dimensions for the number of internal degrees of freedom.
The potential energy surface concept can be used to theoretically explore properties of structures composed of atoms, for example, finding the minimum energy shape of a molecule or computing the rates of a chemical reaction. Qualitatively the reaction coordinate diagrams (one-dimensional slices through the potential energy surfaces) have numerous applications. Chemists use reaction coordinate diagrams as both an analytical and pedagogical aid for rationalizing and illustrating kinetic and thermodynamic events. The purpose of energy profiles and surfaces is to provide a qualitative representation of how potential energy varies with molecular motion for a given reaction or process.
Exercise 9.1.4
Explain the difference between the Born-Oppenheimer approximation and the Crude Born-Oppenheimer approximation.
Summary
In this section we started with the Schrödinger equation for a diatomic molecule and separated it into two equations, an electronic Schrödinger equation and a nuclear Schrödinger equation. In order to make the separation, we had to make an approximation. We had to neglect the effect of the nuclear kinetic energy on the electrons. The fact that this assumption works can be traced to the fact that the nuclear masses are much larger than the electron mass. We then used the solution of the electronic Schrödinger equation to provide the potential energy function for the nuclear motion. The solution to the nuclear Schrödinger equation provides the vibrational wavefunctions and energies.
Contributors and Attributions
David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski ("Quantum States of Atoms and Molecules")
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Molecular orbital theory is a conceptual extension of the orbital model, which was so successfully applied to atomic structure. As was once playfully remarked, "a molecule is nothing more than an atom with more nuclei." This may be overly simplistic, but we do attempt, as far as possible, to exploit analogies with atomic structure. Our understanding of atomic orbitals began with the exact solutions of a prototype problem – the hydrogen atom. We will begin our study of homonuclear diatomic molecules beginning with another exactly solvable prototype, the hydrogen molecule-ion $\ce{H2^{+}}$.
The Hydrogen Molecular Ion
The simplest conceivable molecule would be made of two protons and one electron, namely $\ce{H2^{+}}$. This species actually has a transient existence in electrical discharges through hydrogen gas and has been detected by mass spectrometry and it also has been detected in outer space. The Schrödinger equation for $\ce{H2^{+}}$ can be solved exactly within the Born-Oppenheimer approximation (i.e., fixed nuclei). This ion consists of two protons held together by the electrostatic force of a single electron. Clearly the two protons, two positive charges, repeal each other. The protons must be held together by an attractive Coulomb force that opposes the repulsive Coulomb force. A negative charge density between the two protons would produce the required counter-acting Coulomb force needed to pull the protons together. So intuitively, to create a chemical bond between two protons or two positively charged nuclei, a high density of negative charge between them is needed. We expect the molecular orbitals that we find to reflect this intuitive notion.
The electronic Hamiltonian for $\ce{H2^{+}}$ is
$\hat {H}_{elec} (r, R) = -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {e^2}{4 \pi \epsilon _0 r_A} - \dfrac {e^2}{4 \pi \epsilon _0 r_B} + \dfrac {e^2}{4 \pi \epsilon _0 R} \label{9.2.1}$
where $r_A$ and $r_B$ are the distances electron from the $A$ and $B$ hydrogen nuclei, respectively and $R$ is the distance between the two protons.
Although the Schrödinger equation for $\ce{H2^{+}}$ can be solved exactly (albeit within the Born-Oppenheimer approximation where the nuclei are fixed) because there is only one electron, we will develop approximate solutions in a manner applicable to other diatomic molecules that have more than one electron.
Linear Combination of Atomic Orbitals
For the case where the protons in H2+ are infinitely far apart, we have a hydrogen atom and an isolated proton when the electron is near one proton or the other. The electronic wavefunction would just be $1s_A(r)$ or $1s_B(r)$ depending upon which proton, labeled $A$ or $B$, the electron is near. Here $1s_A$ denotes a 1s hydrogen atomic orbital with proton A serving as the origin of the spherical polar coordinate system in which the position $r$ of the electron is specified. Similarly $1s_B$ has proton B as the origin. A useful approximation for the molecular orbital when the protons are close together therefore is a linear combination of the two atomic orbitals. The general method of using
$\psi (r) = C_A 1s_A (r) + C_B1s_B (r) \label{9.2.2}$
i.e. of finding molecular orbitals as linear combinations of atomic orbitals is called the Linear Combination of Atomic Orbitals - Molecular Orbital (LCAO-MO) Method. In this case we have two basis functions in our basis set, the hydrogenic atomic orbitals $1s_A$ and $1s_B$.
The Linear Combination of Atomic Orbitals (LCAO) Approximation and Interference
The LCAO approximation is an example of the linear variational method discussed previously with the true molecular orbital wavefunction approximated as an expansion of a basis set of atomic orbitals on each atom of the molecule with variable coefficients that can be optimized (e.g., via the secular equations). As discussed previously, the number of wavefunctions (solutions) extracted from solving the secular determinant is equal the number of elements in the expansion. So for the expansion in Equation \ref{9.2.2} with two atomic orbitals contributing result in two molecule orbitals.
This method yields a approximate picture of the molecular orbitals in a molecules. The figure below shows two atoms approaching along the axis of one of their $2p$ states. In the top row, the two lobes facing one another have the same sign; in the bottom row they have opposite sign. These are two different linear combinations of the same two atomic states, on different atoms and with difference phases (i.e., signs of $C_A$ vs. $C_B$ in the expansion). In the first example, the electron density increases between the nuclei and in the second example, a very steep-sided node between the two nuclei causes all the probability density to face away from the atom opposite.
For $\ce{H2^{+}}$, the simplest molecule, the starting function is given by Equation $\ref{9.2.2}$. We must determine the values for the coefficients, $C_A$ and $C_B$. We could use the variational method to find a value for these coefficients, but for the case of $\ce{H2^{+}}$ evaluating these coefficients is easy. Since the two protons are identical, the probability that the electron is near $A$ must equal the probability that the electron is near $B$. These probabilities are given by $|C_A|^2$ and $|C_B|^2$, respectively. Consider two possibilities that satisfy the condition $|C_A|^2 = |C_B|^2$; namely, $C_A = C_B = C_{+}$ and $C_A = -C_B = C_{-}$. These two cases produce two molecular orbitals:
$\psi _+ = C_+(1s_A + 1s_B) \label{9.2.3a}$
$\psi _{-} = C_{-}(1s_A - 1s_B) \label{9.2.3b}$
The probability density for finding the electron at any point in space is given by $|{\psi}^2|$ and the electronic charge density is just $|e{\psi}^2|$. The important difference between $\psi _+$ and $\psi _{-}$ is that the charge density for $\psi _+$ is enhanced (Figure 9.2.2 (bottom) between the two protons, whereas it is diminished for $\psi _{-}$ as shown in Figures 9.2.2 (top). $\psi _{-}$ has a node in the middle while $\psi _+$ corresponds to our intuitive sense of what a chemical bond must be like. The electronic charge density is enhanced in the region between the two protons.
So $\psi _+$ is called a bonding molecular orbital. If the electron were described by $\psi _{-}$, the low charge density between the two protons would not balance the Coulomb repulsion of the protons, so $\psi _{-}$ is called an antibonding molecular orbital.
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For $\ce{H2^{+}}$, the simplest molecule, we must determine the values for the coefficients, $C_A$ and $C_B$ for the LCAO approximation for the molecular orbital as a linear combination of the two atomic orbitals
$|\psi (r) \rangle = C_A 1s_A (r) + C_B1s_B (r) \nonumber$
We could use the linear variational method to find a value for these coefficients, but for the case of $\ce{H2^{+}}$ evaluating these coefficients is easy. Since the two protons are identical, the probability that the electron is near $A$ must equal the probability that the electron is near $B$. These probabilities are given by $|C_A|^2$ and $|C_B|^2$, respectively. Consider two possibilities that satisfy the condition
$|C_A|^2 = |C_B|^2 \nonumber$
so $C_A = C_B = C_{+}$ and $C_A = -C_B = C_{-}$. These two cases produce two molecular orbitals:
$\underbrace{| \psi _+ \rangle = C_+(1s_A + 1s_B)}_{\text{bonding molecular orbital}} \label{9.3.1a}$
and
$\underbrace{| \psi _{-} \rangle = C_{-}(1s_A - 1s_B)}_{\text{antibonding molecular orbital}} \label{9.3.1b}$
The probability density for finding the electron at any point in space is given by $|{\psi}^2|$ and the electronic charge density is just $|e{\psi}^2|$. The important difference between $\psi _+$ and $\psi _{-}$ is that the charge density for $| \psi_+ \rangle$ is enhanced between the two protons, whereas it is diminished for $|\psi _{-} \rangle$ as shown in Figures 9.3.1 . $\psi _{-}$ has a node in the middle while $| \psi_+ \rangle$ corresponds to our intuitive sense of what a chemical bond must be like. The electronic charge density is enhanced in the region between the two protons. So $|\psi_+ \rangle$ is called a bonding molecular orbital. If the electron were described by $|\psi _{-} \rangle$, the low charge density between the two protons would not balance the Coulomb repulsion of the protons, so $|\psi _{-} \rangle$ is called an antibonding molecular orbital.
Now we want to evaluate $C_+$ and $C_-$ and then calculate the energy. The bonding and antibonding character of $\psi _+$ and $\psi _{-}$ also should be reflected in the energy. If $|\psi _+ \rangle$ indeed describes a bonding orbital, then the energy of this state should be less than that of a proton and hydrogen atom that are separated. The calculation of the energy will tell us whether this simple theory predicts H2+ to be stable or not and also how much energy is required to dissociate this molecule.
The constants $C_+$ and $C_-$ are evaluated from the standard normalization condition:
\begin{align} \int \psi ^*_{\pm} \psi _{\pm} d\tau = \left \langle \psi _{\pm} | \psi _{\pm} \right \rangle &= 1 \label {9.3.1} \[4pt] \left \langle C_{\pm} ( 1s_A \pm 1s_B ) | C_{\pm} ( 1s_A \pm 1s_B ) \right \rangle &= 1 \label {9.3.2} \[4pt] |C_\pm|^2 [ \underbrace{\langle1s_A | 1s_A \rangle }_{1} + \underbrace{\langle 1s_B | 1s_B \rangle}_{1} \pm \underbrace{ \langle1s_B | 1s_A\rangle}_{S} \pm \underbrace{ \langle1s_A | 1s_B \rangle}_{S^*}] &= 1 \label {9.3.3} \end{align}
Since the atomic orbitals are normalized, the first two integrals are just 1. The last two integrals are called overlap integrals and are symbolized by $S$ and $S^*$, respectively, since one is the complex conjugate of the other.
The overlap integrals are telling us to take the value of 1sB at a point multiply by the value of 1sA at that point and sum (integrate) such a product over all of space (Figure 9.3.1 ). If the functions do not overlap, i.e. if one is zero when the other one is not and vice versa, these integrals then will be zero. It also is possible in general for such integrals to be zero even if the functions overlap because of the cancelation of positive and negative contributions.
If the overlap integral is zero, for whatever reason, the functions are said to be orthogonal. Notice that the overlap integral ranges from 0 to 1 as the separation between the protons varies from $R = ∞$ to $R = 0$. Clearly when the protons are infinite distance apart, there is no overlap, and when $R = 0$ both functions are centered on one nucleus and $\left \langle 1s_A | 1s_B \right \rangle$ becomes identical to $\left \langle 1s_B | 1s_A \right \rangle$, which is normalized to 1, because $1s_A = 1s_B$.
With these considerations and using the fact that $1s$ wavefunctions are real so
$\left \langle 1s_A | 1s_B \right \rangle = \left \langle 1s_B | 1s_A \right \rangle = S \label {9.3.4}$
Equation $\ref{9.3.3}$ becomes
$|C_{\pm}|^2 (2 \pm 2S ) = 1 \label {9.3.5}$
The solution to Equation $\ref{9.3.5}$ is given by
$C_{\pm} = \dfrac{1}{\sqrt{2(1 \pm S )}} \label {9.3.6}$
Hence, the normalized molecular orbitals in Equations $\ref{9.3.1a}$ and $\ref{9.3.1a}$ are
$| \psi _+\rangle = \dfrac{1}{\sqrt{2(1 + S )}} (1s_A + 1s_B) \label{9.3.7a}$
and
$|\psi _{-} \rangle = \dfrac{1}{\sqrt{2(1 - S )}} (1s_A - 1s_B) \label{9.3.7b}$
The energies associated with these wavefunctions requires a bit more effort to calculate though as demonstrated in the following section.
For the overlap integral of two 1s orbitals from the hydrogen dimer discussed above is difficult to evaluate analytically and is explained here. The final answer is:
\begin{align} S(R) &= \left \langle 1s_A | 1s_B \right \rangle \[4pt] &= e^{-R/a_o} \left( 1 +\dfrac{R}{a_o} + \dfrac{R^2}{3a_o^2} \right) \label{overlap} \end{align}
The overlap integral for two 1s atomic orbitals of hydrogen is graphically displayed below
Example 9.3.1
Calculate the difference in the electronic charge density (C/pm3) at a point halfway between the two nuclei for $\ce{H2^{+}}$ for an electron in the bonding molecular orbital compared to one in the antibonding molecular orbital.
Solution
The electronic charge density is calculated with the formula $|e{\psi}^2|$ where $e = 1.602 \times 10^{-19} C$
a: Bonding Molecular Orbital
First, squaring the bonding molecular orbital (Equation \ref{9.3.7a}) gives us:
\begin{align*} \psi_+^2 &=\dfrac{1}{2(1 + S )} (1s_A+1s_B)^2 \[4pt] &= \dfrac{1}{2(1 + S )} \dfrac{1}{\pi a_0^3}\Big(e^{-r_A/a_0}+e^{-r_B/a_0}\Big)^2 \[4pt] &= \dfrac{1}{2(1 + S )} \dfrac{1}{\pi a_0^3} \[4pt] &= (0.561)^2\times\left(\dfrac{1}{\pi\times(52.9\;\text{pm})^3}\right)=6.7672\times10^{-7}\;\text{pm}^{-3} \end{align*} \nonumber
(Note: The value of $C^2$ is obtained from equation 9.3.12 and the value of S is from Equation \ref{overlap})
To find the density at halfway between A and B for the bonding orbital evalulate at 1/2 R,
$r_A=\dfrac{1}{2}R$ and $r_B=\dfrac{1}{2}R$
$\psi_+^2=3.7\times10^{-7}\;\text{pm}^{-3} \nonumber$
Now multiply this answer by e to generate the electronic charge density,
$P = \Big(3.7\times10^{-7}\;\text{pm}^{-3}\Big)\times\Big(1.602\times10^{-19}\Big)\;\text{C} = 5.93\times10^{-26} C/pm^{3} \nonumber$
b: Antibonding Molecular Orbital
To find the halfway between A and B for the anti-bonding orbital at $r_A=\dfrac{1}{2}R$ and $r_B=\dfrac{1}{2}R$. However, simple inspection of Equation \ref{9.3.7b} or Figure 9.3.1 shows this will be zero.
$\text{e}\psi_-^2=0 \nonumber$
Obviously, when multiplied by e, the answer is still zero. This makes logical sense as there would be no charge density between the two atoms as there lies a node with no probability of finding an electron.
Exercise 9.3.1
Show that for two arbitrary functions $\left \langle \varphi _B | \varphi _A \right \rangle$ is the complex conjugate of $\left \langle \varphi _A | \varphi _B \right \rangle$ and that these two integrals are equal if the functions are real.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/09%3A_Chemical_Bonding_in_Diatomic_Molecules/9.03%3A_The_Overlap_Integral.txt
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Learning Objectives
• Identify the nature of the energy of molecular orbitals of a diatomic as a function of intermolecular distance
• Identify the three integrals involved in calculation the total Molecular Orbital Energy: coulomb Integral, exchange integral, and overlap integral
As shown previously, we can construct two molecular orbitals for the $\ce{H_2^{+}}$ system using the LCAO approximation with a basis set of two 1s atomic orbitals (i.e., the $1s$ orbitals on hydrogen $A$ ($1s_A$) and hydrogen $B$ ($1s_B$):
$| \psi _\pm\rangle = \dfrac{1}{\sqrt{2(1 \pm S )}} (1s_A \pm 1s_B) \label{9.3.7a}$
The energy of these two molecular orbitals can be calculated from the expectation value integral of the Hamiltonian,
$E_{\pm} = \left \langle \psi _{\pm} | \hat {H} _{elec} | \psi _{\pm} \right \rangle \label {9.4.1}$
which can be expanded using the expanded molecular orbital wavefunctions in Equations \ref{9.3.7a} to give
$E_{\pm} = \dfrac {1}{2(1 \pm S)} \left[ \underbrace{\left \langle 1s_A |\hat {H} _{elec} | 1s_A \right \rangle}_{H_{AA}} + \underbrace{\left \langle 1s_B |\hat {H} _{elec} | 1s_B \right \rangle }_{H_{BB}}\pm \underbrace{\left \langle 1s_A |\hat {H} _{elec} | 1s_B \right \rangle}_{H_{AB}} \pm \underbrace{\left \langle 1s_B |\hat {H} _{elec} | 1s_A \right \rangle}_{H_{BA}} \right] \label {9.4.2a}$
where $S$ is the overlap integral between the two atomic orbitals of the basis. The four integrals in Equation $\ref{9.4.2a}$ can be represented by $H_{AA}$, $H_{BB}$, $H_{AB}$, and $H_{BA}$, respectively.
$E_{\pm} = \dfrac {1}{2(1 \pm S)} \left[ H_{AA} + H_{BB} \pm H_{AB} \pm H_{BA} \right] \label {9.4.2b}$
Exercise 9.4.1
Show that Equation $\ref{9.4.1}$ expands to give Equation $\ref{9.4.2a}$ within the LCAO approximation that uses a basis set of only two 1s atomic orbitals.
Answer
Here we have the wavefunction within the LCAO approximation that uses a basis set of only two 1s atomic orbitals (Equation \ref{9.3.7a}).
$| \psi_{\pm} \rangle =\frac{1}{\sqrt{2(1 \pm S)}}\left(1_{S a} \pm 1_{S b}\right) \nonumber$
And our LCAO approximation is equivalent to this if we plug in the wavefunction directly.
$E_{\pm}= \langle \psi_{\pm}|\hat{H}| \psi_{\pm} \rangle =N^{2} \langle \left(1_{S a} \pm 1_{S b}\right)|\hat{H}|\left(1_{S a}-1_{S b}\right) \rangle \nonumber$
We can see form above that the normalization constant squared results in:
$N^{2}=\frac{1}{2(1 \pm S)} \nonumber$
We now FOIL the mulitple (i.e., expaned term by term):
$\langle 1_{S a}|\hat{H}| 1_{S a} \rangle + \langle1_{S b}|\hat{H}| 1_{S b} \rangle \pm \langle 1_{S a}|\hat{H}| 1_{S b} \rangle \pm \langle 1_{S b}|\hat{H}| 1_{S a} \rangle \nonumber$
Now we can see this is now equivalent to Equation \ref{9.4.2a} if the equation above is inserted (and adding a subscript to emphasize this only applies to the electronic wavefunction):
$E_{\pm} = \dfrac {1}{2(1 \pm S)} \left[ \underbrace{\left \langle 1s_A |\hat {H} _{elec} | 1s_A \right \rangle}_{H_{AA}} + \underbrace{\left \langle 1s_B |\hat {H} _{elec} | 1s_B \right \rangle }_{H_{BB}}\pm \underbrace{\left \langle 1s_A |\hat {H} _{elec} | 1s_B \right \rangle}_{H_{AB}} \pm \underbrace{\left \langle 1s_B |\hat {H} _{elec} | 1s_A \right \rangle}_{H_{BA}} \right] \nonumber$
Notice that $A$ and $B$ appear equivalently in the Hamiltonian operator for $\ce{H_2^{+}}$. This equivalence means that integrals involving $1s_A$ must be the same as corresponding integrals involving $1s_B$, i.e.
$H_{AA} = H_{BB} \label {9.4.3}$
and since the wavefunctions are real
$| A \rangle = \langle A | \nonumber$
so
$H_{AB} = H_{BA} \label {9.4.4}$
These two equalities simplify Equation \ref{9.4.2b}:
$E_{\pm} = \dfrac {1}{1 \pm S} (H_{AA} \pm H_{AB}) \label {9.4.5}$
Now examine the details of $H_{AA}$ after inserting the Hamiltonian operator for $\ce{H_2^{+}}$ (Equation 9.2.1):
$H_{AA} = \underbrace{ \left \langle 1s_A \left| - \dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {e^2}{4\pi \epsilon _0 r_A} \right| 1s_A \right \rangle }_{\color{red}{E_H}} + \dfrac {e^2}{4\pi \epsilon _0 R} \cancelto{ \color{red} {1}}{ \left \langle 1s_A | 1s_A \right \rangle} \underbrace{ - \left \langle 1s_A \left| \dfrac {e^2}{4 \pi \epsilon _0 r_B } \right| 1s_A \right \rangle}_{ \color{red} {J_{AB}}} \label {9.4.6}$
• The first term is just the integral for the energy of the hydrogen atom of the 1s orbital, $E_H$.
• The second integral is equal to 1 by normalization; the prefactor is just the Coulomb repulsion of the two protons.
• The last integral, including the minus sign, is represented by $J$ and is called the Coulomb integral.
Physically $J_{AB}$ is the potential energy of interaction of the electron located around proton $A$ with proton $B$. It is negative because it is an attractive interaction. It is the average interaction energy of an electron described by the $1s_A$ function with proton $B$.
The Coulomb Integral ($J$)
The Coulomb Integral is the potential energy of electrostatic repulsion between the electron with the electron density in $1S_A$ and the the electron with the electron density function $1S_B$
Now consider $H_{AB}$.
$H_{AB} = \underbrace{ \left \langle 1s_A \left| - \dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {e^2}{4\pi \epsilon _0 r_B} \right| 1s_B \right \rangle}_{\color{red}{E_HS}} + \dfrac {e^2}{4\pi \epsilon _0 R} \cancelto{\color{red}{S}}{\left \langle 1s_A | 1s_B \right \rangle} - \underbrace{ \left \langle 1s_A \left| \dfrac {e^2}{4 \pi \epsilon _0 r_A } \right| 1s_B \right \rangle }_ {\color{red} {K_{AB}}}\label {9.4.7S}$
• In the first integral we have the hydrogen atom Hamiltonian and the H atom function 1sB. The function 1sB is an eigenfunction of the operator with eigenvalue $E_H$. Since $E_H$ is a constant it factors out of the integral, which then becomes the overlap integral, $S$. The first integral therefore reduces to $E_HS$.
• The second term is just the Coulombic energy of the two protons times the overlap integral.
• The third term, including the minus sign, is given the symbol $K$ and is called the exchange integral because the electron is described by the 1sA orbital on one side and by the 1sB orbital on the other side of the operator. The electron changes or exchanges position in the molecule.
The Exchange Integral ($K$)
In a coulomb integral, the electron always is in the same orbital; whereas, in an Exchange Integral, the electron is in one orbital on one side of the operator and in a different orbital on the other side.
Using the expressions for $H_{AA}$ (Equation \ref{9.4.6}) and $H_{AB}$ (Equation \ref{9.4.7S}) and substituting into Equation $\ref{9.4.5}$ produces:
\begin{align} E_{\pm} &= \dfrac {1}{1 \pm S} \left[ \left(E_H + \dfrac {e^2}{4\pi \epsilon_0 R}\right) (1 \pm S ) + J \pm K \right] \label {9.4.8} \[4pt] &= E_H + \dfrac {e^2}{4\pi \epsilon _0 R} + \dfrac {J \pm K}{1 \pm S} \label {9.4.9} \end{align}
Equation $\ref{9.4.9}$ tells us that the energy of the $\ce{H_2^+}$ molecule is the energy of a hydrogen atom plus the repulsive energy of two protons plus some additional electrostatic interactions of the electron with the protons. These additional interactions are given by the last term
$\dfrac {J \pm K}{1 \pm S} \nonumber. \nonumber$
If the protons are infinitely far apart then only $E_H$ is nonzero, which we can set to zero by subtracting off:
\begin{align} \Delta E_{\pm} &= E_{\pm} - E_H \[4pt] &= \dfrac {e^2}{4\pi \epsilon _0 R} + \dfrac {J \pm K}{1 \pm S} \label {9.4.10} \end{align}
To get a chemical bond and a stable $\ce{H_2^+}$ molecule $\Delta E_{\pm}$ must be less than zero and have a minimum, i.e. $\dfrac {J \pm K}{1 \pm S}$ must be sufficiently negative to overcome the positive repulsive energy of the two protons $\dfrac {e^2}{4 \pi \epsilon _0R }$ for some value of $R$. For large $R$, these terms are zero, and for small $R$, the Coulomb repulsion of the protons rises to infinity.
Exercise 9.4.2
Show that Equation 9.2.1 follows from Equation $\ref{9.4.5}$
The Coulomb and Exchange Integrals
We will examine more closely how the Coulomb repulsion term and the integrals $J$, $K$, and $S$ depend on the separation of the protons, but first we want to discuss the physical significance of $J$, the Coulomb integral, and $K$, the exchange integral. $J$ and $K$ have been defined as
$J_{AB} = \left \langle 1s_A \left| \dfrac {-e^2}{4 \pi \epsilon _0 r_B } \right|1s_A \right \rangle = - \int \varphi ^*_{1s_A} (r) \varphi _{1s_A} (r) \dfrac {e^2}{4 \pi \epsilon _0 r_B } d\tau \label {9.4.11}$
$K_{AB} = \left \langle 1s_A \left| \dfrac {-e^2}{4 \pi \epsilon _0 r_A } \right|1s_B \right \rangle = - \int \varphi ^*_{1s_A} (r) \varphi _{1s_B} (r) \dfrac {e^2}{4 \pi \epsilon _0 r_A } d\tau \label {9.4.12}$
Figure 9.4.2 shows graphs of the four terms contributing to the energy of $\ce{H_2^{+}}$ (Equation $\ref{9.4.10}$). In Figure 9.4.2 , you can see that as the internuclear distance $R$ approaches zero,
• the Coulomb repulsion of the two protons goes from near zero to a large positive number,
• the overlap integral goes for zero to one, and
• $J$ and $K$ become increasingly negative.
Note that both $J$ and $K$ integrals are negative since all quantities in the integrands of Equation \ref{9.4.11} and \ref{9.4.12} are positive. In the Coulomb integral, $e \varphi ^*_{1s_A} (r) \varphi _{1a_A} (r)$ is the charge density of the electron around proton A, since r represents the coordinates of the electron relative to proton A. Since rB is the distance of this electron to proton B, the Coulomb integral gives the potential energy of the charge density around proton A interacting with proton B. $J$ can be interpreted as an average potential energy of this interaction because $e \varphi ^*_{1s_A} (r) \varphi _{1a_A} (r)$ is the probability density for the electron at point $r$, and $\dfrac {e^2}{4 \pi \epsilon _0 r_B }$ is the potential energy of the electron at that point due to the interaction with proton B. Essentially, J accounts for the attraction of proton B to the electron density of hydrogen atom A. As the two protons get further apart, this integral goes to zero because all values for rB become very large and all values for 1/rB become very small.
In the exchange integral, $K$, the product of the two functions is nonzero only in the regions of space where the two functions overlap. If one function is zero or very small at some point then the product will be zero or small. The exchange integral also approaches zero as internuclear distances increase because the both the overlap and the $1/r$ values become zero. The product $e \varphi ^*_{1s_A} (r) \varphi _{1a_B} (r)$ is called the overlap charge density. Since the overlap charge density is significant in the region of space between the two nuclei, it makes an important contribution to the chemical bond. The exchange integral, $K$, is the potential energy due to the interaction of the overlap charge density with one of the protons. While $J$ accounts for the attraction of proton $B$ to the electron density of hydrogen atom $A$, $K$ accounts for the added attraction of the proton due the build-up of electron charge density between the two protons.
Exercise 9.4.3
Write a paragraph describing in your own words the physical significance of the Coulomb and exchange integrals for $\ce{H_2^{+}}$.
Figure 9.4.3 shows the energy of H2+ relative to the energy of a separated hydrogen atom and a proton as given by Equation $\ref{9.4.9}$. For the electron in the $\psi_-$ orbital, the energy of the molecule, $E_{el}(R)$, always is greater than the energy of the separated atom and proton.
For the electron in the $\psi _+$ orbital, you can see that the big effect for the energy of the bonding orbital, $E_+(R)$, is the balance between the repulsion of the two protons $\dfrac {e^2}{4 \pi \epsilon _0R }$ and J and K, which are both negative. J and K manage to compensate for the repulsion of the two protons until their separation is less than 100 pm (i.e the energy is negative up until this point), and a minimum in the energy is produced at 134 pm. This minimum represents the formation of a chemical bond. The effect of S is small. It only causes the denominator in Equation $\ref{9.4.9}$ to increase from 1 to 2 as R approaches 0.
For the antibonding orbital, $-K$ is a positive quantity and essentially cancels J so there is not sufficient compensation for the Coulomb repulsion of the protons. The effect of the -K in the expression, Equation $\ref{9.4.9}$, for $E_-$ is to account for the absence of overlap charge density and the enhanced repulsion because the charge density between the protons for $\psi _-$ is even lower than that given by the atomic orbitals.
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Learning Objectives
• Characterize the bonding and anti-bonding molecular orbitals in $\ce{H^{+}}$
The two molecular orbitals of the $\ce{H^{+}}$ ion were created via the linear combinations of atomic orbitals (LCAOs) approximation were created from the sum and the difference of two atomic orbitals. Within this approximation, the jth molecular orbital can be expressed as a linear combination of many atomic orbitals {$\phi_i$}:
$| \psi_J \rangle = \sum_i^N c_{J,i} | \phi_i \rangle \label{9.5.12}$
A molecule will have as many molecular orbitals as there are atomic orbitals used in the basis set ($N$ in Equation $\ref{9.5.12}$). Adding two atomic orbitals corresponds to constructive interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is increased. The molecular orbital corresponding to the sum of the two H 1s orbitals is called a σ1s combination (parts (a) and (b) of Figure 9.5.1 ).
In the sigma ($σ$) orbital, the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1s denotes the atomic orbitals from which the molecular orbital was derived.
$| \sigma _{1s} \rangle = \dfrac{1}{\sqrt{2(1 + S )}} \left( | 1s_A \rangle + | 1s_B \rangle \right) \label{9.7.2}$
Conversely, subtracting one atomic orbital from another corresponds to destructive interference between two waves, which reduces their intensity and causes a decrease in the internuclear electron probability density (part (c) and part (d) in Figure 9.5.1 ). The resulting pattern contains a node where the electron density is zero. The molecular orbital corresponding to the difference is called $\sigma _{1s}^{*}$ and has a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis:
$| \sigma _{1s}^* \rangle = \dfrac{1}{\sqrt{2(1 - S )}} \left( | 1s_A \rangle - | 1s_B \rangle \right) \label{9.7.3}$
The electron density in the σ1s molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ1s orbital represents a bonding molecular orbital. A molecular orbital that forms when atomic orbitals or orbital lobes with the same sign interact to give increased electron probability between the nuclei due to constructive reinforcement of the wavefunctions. In contrast, electrons in the $\sigma _{1s}^{\star }$ orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the $\sigma _{1s}^{\star }$ orbital is an antibonding molecular orbital (a molecular orbital that forms when atomic orbitals or orbital lobes of opposite sign interact to give decreased electron probability between the nuclei due to destructive reinforcement of the wavefunctions).
Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not.
Because electrons in the σ1s orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ1s molecular orbital has a lower energy than either of the hydrogen 1s atomic orbitals. Conversely, electrons in the $\sigma _{1s}^{\star }$ orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1s atomic orbitals. Consequently, the $\sigma _{1s}^{\star }$ molecular orbital has a higher energy than either of the hydrogen 1s atomic orbitals. The σ1s (bonding) molecular orbital is stabilized relative to the 1s atomic orbitals, and the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is destabilized. The relative energy levels of these orbitals are shown in the energy-level diagram (a schematic drawing that compares the energies of the molecular orbitals (bonding, antibonding, and nonbonding) with the energies of the parent atomic orbitals) in Figure 9.5.2
A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable).
Expanding Beyond the 1s Orbital Basis Set
This picture of bonding in H2+ in the previous section is very simple, but gives reasonable results when compared to an exact calculation. The equilibrium bond distance is 134 pm compared to 106 pm (exact), and a dissociation energy is 1.8 eV compared to 2.8 eV (exact). To better describe chemical bonding we need to account for the increase in electron density between the two nuclei. The 1s orbitals alone are not particularly good for this purpose because they are spherically symmetric and show no preference for the space between the atomic nuclei. The use of additional atomic orbitals can correct this situation and provide additional parameters, which can be optimized by the linear variational method, to give a better function with a lower energy and more accurate description of the charge density.
The energy of the non-normalized molecular orbital can be calculated from the expectation value integral of the Hamiltonian,
$E_{J} = \dfrac{\left \langle \psi _{J} | \hat {H} _{elec} | \psi _{J} \right \rangle}{\left \langle \psi _{J} | \psi _{J} \right \rangle} \label {9.5.13}$
This is the variational energy using $| \psi _{J} \rangle$ as the trail wavefunction. After substituting the LCAO expansion for $| \psi _{J} \rangle$ (Equation \ref{9.5.12}) into the energy expression of Equation \ref{9.5.13} results in:
\begin{align} E_{J} &= \dfrac{\left \langle \displaystyle \sum_i c_{J,i}^* \phi_i \right | \hat {H} _{elec} \left | \displaystyle \sum_j c_{J,i} \phi_j \right \rangle}{\left \langle \displaystyle \sum_i c_{J,i}^* \phi_j | \displaystyle \sum_j c_{J,j} \phi_j\right \rangle} \label {9.5.14} \[4pt] &= \dfrac{ \displaystyle \sum_{i,j} c_{J,i}^* c_{J,j} \left \langle \phi_i \right| \hat {H} _{elec} \left| \phi_j \right \rangle}{ \displaystyle\sum_{i,j} c_{J,i}^* c_{J,j} \left \langle \phi_i | \phi_j\right \rangle} \label {9.5.15} \[4pt] &= \dfrac{ \displaystyle \sum_{i,j} c_{J,i}^* c_{J,j} H_{ij}}{ \displaystyle \sum_{i,j} c_{J,i}^* c_{J,j} S_{ij} } \label {9.5.16} \end{align}
where $H_{ij}$ is the Hamiltonian matrix element.
$H_{ij} = \langle \phi_i | \hat {H} _{elec} | \phi_j \rangle \nonumber$
Following the variational theorem, to determine the coefficients of the LCAO expansion $c_i$, we need to minimize $E_J$
$\dfrac{\partial E_J}{\partial c_k} = 0 \label{9.5.17}$
for all $k$. This requires solving $N$ linear equations to hold true (where $N$ is the number of atomic orbitals in the basis)
$\sum_{i=1}^{N} c_i (H_{ki} - ES_{ki}) = 0 \label{9.5.18}$
These equations are the secular equations and were discussed previously in the context of the linear variational method approximation. For the two basis set expansion ($N$) in Figure 9.5.1 , these are
$\begin{array}{rcl} c_1(H_{11} - ES_{11}) + c_2(H_{12} - ES_{12}) & = & 0 \ c_1(H_{12} - ES_{12}) + c_2(H_{22} - ES_{22}) & = & 0 \end{array} \label{9.5.19}$
where $c_1$ and $c_2$ are the coefficients in the linear combination of the atomic orbitals used to construct the molecular orbital. Writing this set of homogeneous linear equations in matrix form gives
$\begin{pmatrix} H_{11} - ES_{11} & H_{12} - ES_{12} \ H_{12} - ES_{12} & H_{22} - ES_{22} \end{pmatrix} \begin{pmatrix} c_1 \ c_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \label{9.5.20}$
Solving these secular equations with N different atomic orbitals in the expansion (Equation $\ref{9.5.12}$) requires finding the N roots of an N order polynomial.
$\left|\begin{array}{lcc} H_{11} - ES_{11} & H_{12} - ES_{12} & \ldots\ H_{12} - ES_{12} & H_{22} - ES_{22} &\ldots \ \ldots &\ldots &\ldots \end{array}\right|=0\label{23}$
Each molecular orbital ($| \psi_J \rangle$) from this treatment has an energy $E_J$ that is given by a different set of coefficients, $\{c_{ij}\}$ where $i$ runs over all $N$ functions in the basis (i.e., number of the atomic orbitals in the LCAO approximation of Equation $\ref{9.5.12}$), and $J$ runs over molecular orbitals. Solve the set of linear equations using that specific $E_J$ to determine $c_{ij}$ values.
Steps in a Solving the Secular Equations
1. Select a set of N basis functions
2. Determine all N( N –1)/2 values of both $H_{ij}$ and $S_{ij}$
3. Form the secular determinant; determine N roots $E_j$ of secular equation
4. For each $E_J$ solve the set of linear equations to determine the basis set coefficients $c_{ij\}) for the j-th molecular orbital For more information on solving the Secular equations check here. The greater the number of atomic orbitals \(N$ that combine to genera the molecular orbitals (Equation $\ref{9.5.12}$), the more accurate the LCAO approximation is. This is expected based on our discussions of the variational method examples. Hence, the $\psi_+$ and $\psi_-$ molecular orbitals for $H_2^+$ are better expressed with higher energy hydrogenic wavefunctions
$| \psi_J \rangle = c_{J,1} 1s_A + c_{J,2} 1s_B + c_{J,3} 2s_A + c_{J,4} 2s_B + c_{J,5} 2p_{z,A} + c_{J,6} 2p_{z,B} \label{9.5.24}$
The reasons that only the $p_z$ atomic orbitals are included in this expansion are discussed later.
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To describe the electronic states of molecules, we construct wavefunctions for the electronic states by using molecular orbitals. These wavefunctions are approximate solutions to the Schrödinger equation. A mathematical function for a molecular orbital is constructed, $\psi _i$, as a linear combination of other functions, $\varphi _j$, which are called basis functions because they provide the basis for representing the molecular orbital.
$| \psi _i \rangle = \sum _j c_{ij} \varphi _j \label{9.6.1}$
where
• $j$ is the index for the $j^{th}$ basis function (e.g., atomic orbital)
• $i$ is the $i^{th}$ molecular orbitals and
• $c_{ij}$ is the expansion coefficient of the $j^{th}$ basis function for the $i^{th}$ molecular orbital.
The variational method is used to find values for parameters in the basis functions and for the constant coefficients in the linear combination that optimize these functions, i.e. make them as good as possible. The criterion for quality in the variational method is making the ground state energy of the molecule as low as possible. Here and in the rest of this chapter, the following notation is used: $\sigma$ is a general spin function (can be either $\alpha$ or $\beta$), $\varphi$ is the basis function (this usually represents an atomic orbital), $\psi$ is a molecular orbital, and $\Psi$ is the electronic state wavefunction (representing a single Slater determinant or linear combination of Slater determinants).
The ultimate goal is a mathematical description of electrons in molecules that enables chemists and other scientists to develop a deep understanding of chemical bonding and reactivity, to calculate properties of molecules, and to make predictions based on these calculations. Just as for atoms, each electron in a molecule can be described by a product of spin-orbitals. Since electrons are fermions, the electronic wavefunction must be antisymmetric with respect to the permutation of any two electrons. A Slater determinant containing the molecular spin orbitals produces the antisymmetric wavefunction. For example for two electrons,
$\Psi (r_1, r_2) = \dfrac{1}{\sqrt{2}} \begin {vmatrix} \psi _A (r_1) \alpha (1) & \psi _B (r_1) \beta (1) \ \psi _A (r_2) \alpha (2) & \psi _B (r_2) \beta (2) \end {vmatrix} \label{9.6.2}$
Solving the Schrödinger equation in the orbital approximation will produce a set of spatial molecular orbitals, each with a specific energy, $\epsilon$. Following the Aufbau Principle, two electrons with different spins ( $\alpha$ and $\beta$, consistent with the Pauli Exclusion Principle discussed for muliti-electron atoms) are assigned to each spatial molecular orbital in order of increasing energy. For the ground state of the 2n electron molecule, the n lowest energy spatial orbitals will be occupied, and the electron configuration will be given as $\psi ^2_1 \psi ^2_2 \psi ^2_3 \dots \psi ^2_n$. The electron configuration also can be specified by an orbital energy level diagram as shown in Figure 9.6.1 . Higher energy configurations exist as well, and these configurations produce excited states of molecules. Some examples are shown in Figure 9.6.1 .
Symmetry
Molecular orbitals usually are identified by their symmetry or angular momentum properties. For example, a typical symbol used to represent an orbital in an electronic configuration of a diatomic molecule is $2\sigma ^2_g$. The superscript in symbol means that this orbital is occupied by two electrons; the prefix means that it is the second sigma orbital with gerade symmetry.
Diatomic molecules retain a component of angular momentum along the internuclear axis. The molecular orbitals of diatomic molecule therefore can be identified in terms of this angular momentum. A Greek letter, e.g. $\sigma$ or $\pi$, encodes this information, as well as information about the symmetry of the orbital. A $\sigma$ means the component of angular momentum is 0, and there is no node in any plane containing the internuclear axis, so the orbital must be symmetric with respect to reflection in such a plane. A $\pi$ means there is a node and the wavefunction is antisymmetric with respect to reflection in a plane containing the internuclear axis. For homonuclear diatomic molecules, a g or a u is added as a subscript to designate whether the orbital is symmetric or antisymmetric with respect to the center of inversion of the molecule.
A homonuclear diatomic molecule has a center of inversion in the middle of the bond. This center of inversion means that $\psi (x, y, z) = \pm \psi (-x, -y, -z)$ with the origin at the inversion center. Inversion takes you from $(x, y, z )$ to $(-x, -y, -z )$. For a heteronuclear diatomic molecule, there is no center of inversion so the symbols g and u are not used. A prefix 1, 2, 3, etc. simply means the first, second, third, etc. orbital of that type. We can specify an electronic configuration of a diatomic molecule by these symbols by using a superscript to denote the number of electrons in that orbital, e.g. the lowest energy configuration of N2 is
$1 \sigma ^2_g 1 \sigma ^2_u 2 \sigma ^2_g 2 \sigma ^2_u 1 \pi ^4_u 3 \sigma ^2_g \label{9.6.3}$
Contrasting LCAO with other Quantum Chemistry Methods
As we have already seen, the LCAO approach is very approximate, yielding only qualitative results. It should be noted that the Hartree-Fock method discussed earlier for atoms can also be used for molecules. For example, the molecule $He_{2}^{+}$ has three electrons, and the Li atom also has three electrons. As usual with Hartree-Fock, the idea is to optimize the shapes of the single-electron orbitals $\psi_1 (r)$, $\psi_2 (r)$ and $\psi_3 (r)$ by minimizing the guess to the ground state energy $E_g$. Of course, we will not get the same answer as for $Li$ because there is a different $V_{en}$ energy for $He_{2}^{+}$ due to the presence of two positively charge nuclei $(charge=+2e)$ separated by a distance $R$ in contrast to the single $+3e$ charged nucleus for $Li$. When the shapes of the orbitals are optimized, we also obtain three energy $\varepsilon_1$, $\varepsilon_2$ and $\varepsilon_3$. Not unexpectedly, we find that two of the HF orbitals resemble $1\sigma_g$ while the third resembles $1\sigma_{u}^{*}$ and the first two energies $\varepsilon_1$ and $\varepsilon_2$ will be nearly equal, while the third $\varepsilon_3$ will be noticeably higher. To contrast with the LCAO approach, in LCAO, we do not optimize the shapes of the orbitals (these are assumed a priori to be $1s$ shaped). All we do is choose the mixing coefficients so as to minimize the guess to the ground-state energy $E_g$.
We note, finally, that the density functional theory alluded to earlier can also be used for molecules. It is often the case that density functional theory yields a more accurate description than Hartree-Fock, but this depends on the molecule. In any case, both are more accurate than LCAO. There is also a hierarchy of methods called post Hartree-Fock methods, all of which are based on the wavefunction rather than the electron density, that can be used to improve upon the HF approximation systematically. The greater the accuracy that is desired for the calculation, the more costly computationally the post HF methods become, so quantum chemistry is often a trade-off between accuracy and efficiency, an issue that becomes more critical to take into account when calculations on large molecules must be carried out!
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The LCAO-MO method that we used for H2+ can be applied qualitatively to homonuclear diatomic molecules to provide additional insight into chemical bonding. A more quantitative approach also is helpful, especially for more complicated situations, like heteronuclear diatomic molecules and polyatomic molecules. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they are often ignored in constructing molecular orbitals. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals.
Molecular Orbitals Formed from ns Orbitals
The molecular orbitals diagrams formatted for the dihydrogen species are similar to the diagrams to any homonuclear diatomic molecule with two identical alkali metal atoms (Li2 and Cs2, for example) is shown in part (a) in Figure 9.7.1 , where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σns bonding molecular orbital and a σ*ns antibonding molecular orbital. Because each alkali metal (M) has an ns1 valence electron configuration, the M2 molecule has two valence electrons that fill the σns bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li2, Na2, K2, Rb2, and Cs2). The general features of these M2 diagrams are identical to the diagram for the H2 molecule. Experimentally, all are found to be stable in the gas phase, and some are even stable in solution.
Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be2), in which each metal atom has an ns2 valence electron configuration, resemble the diagram for the He2 molecule. As shown in Figure $\PageIndex{1b}$, this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σns bonding orbital and the σns* antibonding orbital and give a bond order of 0. Thus Be2, Mg2, Ca2, Sr2, and Ba2 are all expected to be unstable, in agreement with experimental data.In the solid state, however, all the alkali metals and the alkaline earth metals exist as extended lattices held together by metallic bonding. At low temperatures, $Be_2$ is stable.
Example 9.7.1 : Sodium Dimer Ion
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na2 ion.
Given: chemical species
Asked for: molecular orbital energy-level diagram, valence electron configuration, bond order, and stability
Strategy
1. Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system.
2. Determine the total number of valence electrons in the Na2 ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
3. Calculate the bond order and predict whether the species is stable.
Solution
A Because sodium has a [Ne]3s1 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1s atomic orbitals.
B The Na2 ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ3s molecular orbital, a half-filled σ3s* and a $\left ( \sigma _{3s} \right )^{2}\left ( \sigma _{3s}^{\star } \right )^{1}$ electron configuration.
(CC BY-SA-NC; anonymous).
C The bond order is (2-1)÷2=1/2 With a fractional bond order, we predict that the Na2 ion exists but is highly reactive.
Exercise 9.7.1 : Calcium Dimer Cation
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca2+ ion.
Answer
Ca2+ has a $\left ( \sigma _{4s} \right )^{2}\left ( \sigma _{4s}^{\star } \right )^{1}$ electron configurations and a bond order of 1/2 and should exist.
Molecular Orbitals Formed from np Orbitals
Atomic orbitals other than ns orbitals can also interact to form molecular orbitals. Because individual p, d, and f orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each np subshell, for example, there are npx, npy, and npz orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations.
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.1}$
Just as with ns orbitals, we can form molecular orbitals from np orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two npz atomic orbitals in part (a) in Figure 9.7.2 , it is the mathematical difference of their wavefunctions that results in constructive interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a $\sigma _{np_{z}}$ bonding molecular orbital because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the z-axis):
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.2}$
The other possible combination of the two npz orbitals is the mathematical sum:
$\sigma _{np_{z}}=np_{z}\left ( A \right )+np_{z}\left ( B \right ) \label{9.7.3}$
In this combination, shown in part (b) in Figure 9.7.2 , the positive lobe of one npz atomic orbital overlaps the negative lobe of the other, leading to destructive interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a $\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right )$ antibonding molecular orbital. Whenever orbitals combine, the bonding combination is always lower in energy (more stable) than the atomic orbitals from which it was derived, and the antibonding combination is higher in energy (less stable).
The remaining p orbitals on each of the two atoms, npx and npy, do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure 9.7.3 , we see that we have two pairs of np orbitals: the two npx orbitals lying in the plane of the page, and two npy orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the npx orbital on one atom can interact with only the npx orbital on the other, and the npy orbital on one atom can interact with only the npy on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital (a bonding molecular orbital formed from the side-to-side interactions of two or more parallel np atomic orbitals). The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital An antibonding molecular orbital formed from the difference of the side-to-side interactions of two or more parallel np atomic orbitals, creating a nodal plane perpendicular to the internuclear axis..
$\pi _{np_{x}}=np_{x}\left ( A \right )+np_{x}\left ( B \right ) \label{9.7.4}$
$\pi ^{\star }_{np_{x}}=np_{x}\left ( A \right )-np_{x}\left ( B \right ) \label{9.7.5}$
The two npy orbitals can also combine using side-to-side interactions to produce a bonding $\pi _{np_{y}}$ molecular orbital and an antibonding $\pi _{np_{y}}^{\star }$ molecular orbital. Because the npx and npy atomic orbitals interact in the same way (side-to-side) and have the same energy, the $\pi _{np_{x}}$ and $\pi _{np_{y}}$molecular orbitals are a degenerate pair, as are the $\pi _{np_{x}}^{\star }$ and $\pi _{np_{y}}^{\star }$ molecular orbitals.
Energies for Homonuclear Diatomic Molecules
We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N2, O2, and F2. When we draw a molecular orbital diagram for a molecule, there are four key points to remember:
1. The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them.
2. As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases.
3. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized.
4. The interaction between atomic orbitals is greatest when they have the same energy.
Figure 9.7.4 is an energy-level diagram that can be applied to two identical interacting atoms that have three np atomic orbitals each. There are six degenerate p atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals.
The number of molecular orbitals is always equal to the total number of atomic orbitals we started with.
We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F2. We use the diagram in Figure $\PageIndex{5a}$; the n = 1 orbitals (σ1s and σ1s*) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ2s and σ2s* molecular orbitals are much lower in energy than the molecular orbitals derived from the 2p atomic orbitals because of the large difference in energy between the 2s and 2p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2p orbitals on each F is $\sigma _{2p_{z}}$ and the next most stable are the two degenerate orbitals, $\pi _{2p_{x}}$ and $\pi _{2p_{y}}$. For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the $\sigma ^{\star }_{2p_{z}}$ orbital is higher in energy than either of the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star }$ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy.
Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rules. Two electrons each fill the σ2s and σ2s* orbitals, 2 fill the $\sigma _{2p_{z}}$ orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π* orbitals, for a total of 14 electrons.
For period 2 diatomic molecules to the left of N2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the $\sigma _{2p_{z}}$ molecular orbital is slightly higher in energy than the degenerate $\pi ^{\star }_{np_{x}}$ and $\pi ^{\star }_{np_{y}}$ orbitals. The difference in energy between the 2s and 2p atomic orbitals increases from Li2 to F2 due to increasing nuclear charge and poor screening of the 2s electrons by electrons in the 2p subshell. The bonding interaction between the 2s orbital on one atom and the 2pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ2s orbital and increases the energy of the $\sigma _{2p_{z}}$ orbital. Thus for Li2, Be2, B2, C2, and N2, the $\sigma _{2p_{z}}$ orbital is higher in energy than the $\sigma _{3p_{z}}$ orbitals, as shown in Figure 9.7.6 . Experimentally, the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure 9.7.6 ). Thus for example, the $\sigma _{2p_{z}}$ molecular orbital is at a lower energy than the $\pi _{2p_{x,y}}$ pair.
Example 9.7.2 : Diatomic Sulfur
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S2, a bright blue gas at high temperatures.
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons
Strategy:
1. Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S2. Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another.
2. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S2.
3. Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule.
4. Calculate the bond order and describe the bonding.
Solution:
A Sulfur has a [Ne]3s23p4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figures 9.7.6 and 9.7.7 , we need to know how close in energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the $\pi _{3p_{x,y}}$ or the $\sigma _{3p_{z}}$> molecular orbital is higher in energy. Because the nsnp energy gap increases as the nuclear charge increases, the $\sigma _{3p_{z}}$ molecular orbital will be lower in energy than the $\pi _{3p_{x,y}}$ pair.
B The molecular orbital energy-level diagram is as follows:
(CC BY-SA-NC; anonymous).
Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons.
C Ten valence electrons are used to fill the orbitals through $\pi _{3p_{x}}$ and $\pi _{3p_{y}}$, leaving 2 electrons to occupy the degenerate $\pi ^{\star }_{3p_{x}}$ and $\pi ^{\star }_{3p_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S2 is $\left ( \sigma _{3s} \right )^{2}\left ( \sigma ^{\star }_{3s} \right )^{2}\left ( \sigma _{3p_{z}} \right )^{2}\left ( \pi _{3p_{x,y}} \right )^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right )^{2}$ with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond.
Exercise 9.7.2
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O22−).
Answer
$\left ( \sigma _{2s} \right )^{2}\left ( \sigma ^{\star }_{2s} \right )^{2}\left ( \sigma _{2p_{z}} \right )^{2}\left ( \pi _{2p_{x,y}} \right )^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right )^{4}$ bond order of 1; no unpaired electrons
Molecular Orbitals Formed from ns with np Orbitals
Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an ns atomic orbital on one atom with an npz atomic orbital on another. As shown in Figure 9.7.7 , the sum of the two atomic wavefunctions (ns + npz) produces a σ bonding molecular orbital. Their difference (nsnpz) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis.
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Learning Objectives
• Using bond order as a metric for the existance of molecules
Bond Order
In the Lewis electron structures, the number of electron pairs holding two atoms together was called the bond order. Within the molecular orbital approach, bond order is defined as one-half the net number of bonding electrons:
$\text{bond order}=\dfrac{\text{number of bonding electrons} - \text{number of antibonding electrons}}{2} \label{9.8.1}$
To calculate the bond order of $H_2$, we know that the $σ_{1s}$ (bonding) molecular orbital contains two electrons, while the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is empty. The bond order of $H_2$ is therefore
$\text{bond order}=\dfrac{2-0}{2}=1 \label{9.8.2}$
This result corresponds to the single covalent bond; double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3.
We can use energy-level diagrams to describe the bonding in other pairs of atoms and ions where n = 1, such as the H2+ ion, the He2+ ion, and the He2 molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund's Rules.
Figure $\PageIndex{1a}$ shows the energy-level diagram for the H2+ ion, which contains two protons and only one electron. The single electron occupies the $σ_{1s}$ bonding molecular orbital, giving a (σ1s)1 electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is (via Equation \ref{9.8.1})
$\text{bond order}=\dfrac{1-0}{2}=1/2 \nonumber$
Because the bond order is greater than zero, the H2+ ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H2+. With a bond order of only 1/2 the bond in H2+ should be weaker than in the H2 molecule, and the H–H bond should be longer. As shown in Table 9.8.1 , these predictions agree with the experimental data.
Table 9.8.1 : Molecular Orbital Electron Configurations, Bond Orders, Bond Lengths, and Bond Energies for some Simple Homonuclear Diatomic Molecules and Ions
Species Electron Configuration Bond Order Bond Length (pm) Bond Energy (kJ/mol)
H2+ $(σ_{1s})^1$ 1/2 106 269
H2 $(σ_{1s})^2$ 1 74 436
He2+ $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ 1/2 108 251
He2 $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ 0 5,500 $4.6 \times 10^{−5}$
Figure $\PageIndex{1b}$ is the molecular orbital energy-level diagram for $\ce{He_2^{2+}}$. This ion has a total of three valence electrons. Because the first two electrons completely fill the $σ_{1s}$ molecular orbital, the Pauli principle states that the third electron must be in the $\sigma _{1s}^{\star}$ antibonding orbital, giving a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ electron configuration. This electron configuration gives a bond order (via Equation \ref{9.8.1}) of
$\text{bond order}=\dfrac{2-1}{2}=1/2 \nonumber$
As with H2+, the He2+ ion should be stable, but the He–He bond should be weaker and longer than in H2. In fact, the He2+ ion can be prepared, and its properties are consistent with our predictions (Table 9.8.1 ).
Example 9.8.1 : The $\ce{He_2^{2+}}$ ion
Use a molecular orbital energy-level diagrams to predict the bond order and stability of the $\ce{He_2^{2+}}$ ion.
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and stability
Strategy:
1. Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for the system.
2. Determine the total number of valence electrons in the He22+ ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
3. Calculate the bond order and predict whether the species is stable.
Solution:
A Two He 1s atomic orbitals combine to give two molecular orbitals: a $σ_{1s}$ bonding orbital at lower energy than the atomic orbitals and a $\sigma _{1s}^{\star }$ antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram:
B The He22+ ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He22+ as being formed from two He+ ions, each of which has a single valence electron in the 1s atomic orbital. We can now fill the molecular orbital diagram:
The two electrons occupy the lowest-energy molecular orbital, which is the bonding ($σ_{1s}$) orbital, giving a (σ1s)2 electron configuration. To avoid violating the Pauli principle, the electron spins must be paired.
C So the bond order is (via Equation \ref{9.8.1})
$\dfrac{2-0}{2} =1 \nonumber$
He22+ is therefore predicted to contain a single He–He bond. Thus it should be a stable species.
Exercise 9.8.1 : The $\ce{H_2^{2−}}$ Ion
Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the $\ce{H_2^{2−}}$ ion. Is this a stable species?
Answer
$\ce{H_2^{2−}}$ has a valence electron configuration of $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ with a bond order of 0. It is therefore predicted to be unstable.
The Helium Dimer
Finally, we examine the He2 molecule, formed from two He atoms with 1s2 electron configurations. Figure $\PageIndex{1c}$ is the molecular orbital energy-level diagram for He2. With a total of four valence electrons, both the $\sigma _{1s}$ bonding and $\sigma _{1s}^{\star }$ antibonding orbitals must contain two electrons. This gives a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ electron configuration, with a predicted bond order (via Equation \ref{9.8.1}) of
$\text{bond order}= \dfrac{2 − 2}{2} = 0 \label{heliumdimer}$
which indicates that the $He_2$ molecule has no net covalent bond and is not a stable species.
The ability to track bond order to bond strength is due to the fact that the energy difference between the anti-bonding $\sigma _{1s}^{\star }$ molecular orbital and the original $1s$ atomic orbital is larger than the energy difference between the bonding $\sigma _{1s}$ molecular orbital and the $1s$ atomic orbitals. This was derived previously where the stabilization energy ($\Delta E_{+}$) of the $\sigma _{1s}$ molecular orbital is less than the destabilization energy ($\Delta E_{-}$) of the anti-bonding $\sigma _{1s}^{\star }$ molecular orbital:
\begin{align} \Delta E_{\pm} &= E_{\pm} - E_H \[4pt] &= \dfrac {e^2}{4\pi \epsilon _0 R} + \dfrac {J \pm K}{1 \pm S} \label {10.31} \end{align}
Hence, the anti-bonding $\sigma _{1s}^{\star }$ molecular orbital is more destabilized relative to the atomic orbitals than the bonding $\sigma _{1s}$ molecular orbital is stabilized relative to the $1s$ atomic orbitals ($E_H$.
The fact that the anti-bonding MO energy difference is larger than the bonding $\sigma _{1s}$ molecular orbital energy difference is the true reason that helium dimer is not predicted to exist with a covalent bond. Of the four valence electrons in helium dimer, two will fill the bonding $\sigma _{1s}$ molecular orbital, and the other two will fill the anti-bonding $\sigma _{1s}^{\star }$ molecular orbital. The two electrons in the bonding $\sigma _{1s}$ molecular orbital will achieve some stabilization relative to the $1s$ atomic orbitals, but the two electrons in the anti-bonding $\sigma _{1s}^{\star }$ molecular orbital will achieve greater de-stabilization relative to their position in the atomic orbitals. The net result is a less stable molecule than if the electrons remained in their respective $1s$ atomic orbitals.
The electrons in antibonding orbitals cancel (and exceed) the stabilization resulting from electrons in bonding orbitals. Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature (at least as a covalently bonding complex).
However, Van der Waals Helium Dimers do exist
Based on molecular orbital theory discussed above, the $\ce{He_2}$ molecule should not exist since no covalent bond formed between the helium atoms (Equation \ref{heliumdimer}). However, the molecular orbital description above neglects the van der Waals force that exists between the atoms as demonstrated by the existence of liquid helium (at 4 K). So a "molecule" composed of two helium atoms bound by the van der Waals force may exist by this attractive force instead - and it does.
A helium dimer molecule bound by Van der Waals forces was first proposed by John Clarke Slater in 1928 and observed in 1993 by Gentry and coworkers. Interestingly, $\ce{He_2}$ is the largest known molecule of two atoms when in its ground state with an extremely long bond length with a separation of about 5,200 pm. The binding energy is only $4.6 \times 10^{−5}\, kJ/mol$, so the $\ce{He-He}$ bond is 5,000 times weaker than the covalent bond in the hydrogen molecule (Table 9.8.1 ).
Conclusion
The decrease in energy caused by the bonding orbital (constructive interference of the atomic orbitals) is canceled by the increase in energy caused by the antibonding orbital (destructive interference of the atomic orbitals), so it is not energetically favorable for the helium atoms to be in such proximity, so if that situation arises, they'll separate quickly since there's no force keeping them there.
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The Pauli exclusion principle plays as important a role in the understanding of the electronic structure of molecules as it does in the case of atoms. The end result of the Pauli principle is to limit the amount of electronic charge density that can be placed at any one point in space. For example, the Pauli principle prevents the 1s orbital in an atom from containing more than two electrons. Since the 1s orbital places most of its charge density in regions close to the nucleus, the Pauli principle, by limiting the occupation of the 1s orbital, limits the amount of density close to the nucleus. Any remaining electrons must be placed in orbitals which concentrate their charge density further from the nucleus.
In an earlier discussion we pointed out that the reason the electron doesn't fall onto the nucleus is because it must possess kinetic energy if Heisenberg's uncertainty principle is not to be violated. This is one reason why matter doesn't collapse. The Pauli principle is equally important in this regard. The electron density of the outer electrons in an atom cannot collapse and move closer to the nucleus since it can do so only if the electrons occupy an orbital with a lower n value. If, however, the inner orbital contains two electrons, then the Pauli principle states that the collapse cannot occur. We must be careful in our interpretation of this aspect of the Pauli principle. The density from a 2s orbital has a small but finite probability of being found well within the density of the 1s orbital. Do not interpret the Pauli principle as implying that the density from an occupied orbital has a clearly defined and distinct region in real space all to its own. This is not the case. The operation of the Pauli principle is more subtle than this. In some simple cases, such as the ones we wish to discuss below, the limiting effect of the Pauli principle on the density distribution can, however, be calculated and pictured in a very direct manner.
The Pauli principle demands that when two electrons are placed in the same orbital their spins must be paired. What restriction is placed on the spins of the electrons during the formation of a molecule, when two orbitals, each on a different atom, overlap one another? For example, consider the approach of two hydrogen atoms to form a hydrogen molecule. Consider atom A to have the configuration $1s^1 \alpha$ and atom B the configuration $1s^1 \beta$. Even when the atoms approach very close to one another the Pauli principle would be satisfied as the spins of the two electrons are opposed. This is the situation we have tacitly assumed in our previous discussion of the hydrogen molecule. However, what would occur if two hydrogen atoms approached one another and both had the same configuration and spin, say $1s^1 \alpha$? When two atoms are relatively close together the electrons become indistinguishable. It is no longer possible to say which electron is associated with which atom as both electrons move in the vicinity of both nuclei. Indeed this is the effect which gives rise to the chemical bond. In so far as we can still regard the region around each atom to be governed by its own atomic orbital, distorted as it may be, two electrons with the same spin will not be able to concentrate their density in the binding region. This region is common to the orbitals on both atoms, and since the electrons possess the same spin they cannot both be there simultaneously. In the region of greatest overlap of the orbitals, the binding region, the presence of one electron will tend to exclude the presence of the other if their spins are parallel. Instead of density accumulating in the binding region as two atoms approach, electron density is removed from this region and placed in the antibonding region behind each nucleus where the overlap of the orbitals is much smaller. Thus the approach of two hydrogen atoms with parallel spins does not result in the formation of a stable molecule. This repulsive state of the hydrogen molecule, in which both electrons have the same spin and atomic orbital quantum numbers, can be detected spectroscopically.
We can now give the general requirements for the formation of a chemical bond. Electron density must be accumulated in the region between the nuclei to an extent greater than that obtained by allowing the original atomic density distributions to overlap. In general, the increase in charge density necessary to balance the nuclear force of repulsion requires the presence of two electrons.
We are now in a position to build up and determine the electronic configurations of the homonuclear diatomic molecules by adding electrons two at a time to the molecular orbitals with the spins of the electrons paired, always filling the orbitals of lowest energy first. We shall, at the same time, discuss the effectiveness of each orbital in binding the nuclei and make qualitative predictions regarding the stability of each molecular configuration.
The Pauli Exclusion Principle in Hydrogen Dimer
The two electrons in the hydrogen molecule may both be accommodated in the 1sg orbital if their spins are paired and the molecular orbital configuration for H2 is 1sg2. Since the 1sg orbital is the only occupied orbital in the ground state of H2, the density distribution shown previously in Figure 9.9.2 for H2 is also the density distribution for the 1sg orbital when occupied by two electrons. The remarks made previously regarding the binding of the nuclei in H2 by the molecular charge distribution apply directly to the properties of the 1sg charge density.
The Pauli Exclusion Principle in Helium Dimer
The electronic configuration of He2 is 1sg2 1su2. A su orbital, unlike a sg orbital, possesses a node in the plane midway between the nuclei and perpendicular to the bond axis. The 1su orbital and all su orbitals in general, because of this nodal property, cannot concentrate charge density in the binding region. It is instead concentrated in the antibinding region behind each nucleus (Figure 9.9.3 ).
The su orbitals are therefore classified as antibonding. It is evident from the form of density distribution for the 1su orbital that the charge density in this orbital pulls the nuclei apart rather than drawing them together. Generally, the occupation of an equal number of sg and su orbitals results in an unstable molecule. The attractive force exerted on the nuclei by the charge density in the sg orbitals is not sufficient to balance both the nuclear force of repulsion and the antibinding force exerted by the density in the su orbitals. Thus molecular orbital theory ascribes the instability of He2 to the equal occupation of bonding and antibonding orbitals. Notice that the Pauli exclusion principle is still the basic cause of the instability. If it were not for the Pauli principle, all four electrons could occupy a sg-type orbital and concentrate their charge density in the region of low potential energy between the nuclei. It is the Pauli principle, and not a question of energetics, which forces the occupation of the 1su antibonding orbital.
The total molecular charge distribution is obtained by summing the individual molecular orbital densities for single or double occupation numbers as determined by the electronic configuration of the molecule. Thus the total charge distribution for He2 (Figure 9.9.3 ) is given by the sum of the 1sg and 1su orbital densities for double occupation of both orbitals. The adverse effect which the nodal property of the 1su orbital has on the stability of He2 is very evident in the total charge distribution. Very little charge density is accumulated in the central portion of the binding region. The value of the charge density at the mid-point of the bond in He2 is only 0.164 au compared to a value of 0.268 au for H2.
We should reconsider in the light of molecular orbital theory the stability of $He_2^+$ and the instability of the hydrogen molecule with parallel spins. He2+ will have the configuration 1sg2 1su1. Since the 1su orbital is only singly occupied in $He_2^+$, less charge density is accumulated in the antibinding regions than is accumulated in these same regions in the neutral molecule. Thus the binding forces of the doubly-occupied 1sg density predominate and $He_2^+$ is stable. The electron configuration of (triplet) $H_2$ is $1s_g^1(\alpha)1s_u^1(\alpha)$ when the electronic spins are parallel. The electrons must occupy separate orbitals because of the Pauli exclusion principle. With equal occupation of bonding and antibonding orbitals, the triplet $H_2$ species is predicted to be unstable.
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Learning Objectives
• To describe the connection between bond order, bond length and bond energy in diatomic molecules
• To explain the observed paramagnetic properties of molecular oxygen with Molecular Orbital theory
In general chemistry courses, students learn that covalent bonds can come as either single, double or triple bonds, which are identifies by their bond order. Both bond length and bond energy changes as the bond order increases and as the number of electrons shared between two atoms in a molecule increases, the bond order of a bond increases, the strength of the bond increases and the distance between nuclei decreases (Table 9.10.1 ).
Table 9.10.1 : General Correlation between Bond Strength, length and order in Covalent bonds
Bond Bond Order Bond Enthalpy (kJ/mol) Bond Length (Å)
$\ce{C-C}$ 1 348 1.54
$\ce{C=C}$ 2 614 1.34
$\ce{C#C}$ 3 839 1.20
$\ce{N-N}$ 1 163 1.47
$\ce{N=N}$ 2 418 1.24
$\ce{N#N}$ 3 941 1.10
The above trend can be observed in the first row diatomics in Figure 9.10.1 . The bond order can be determined directly form the molecular orbital electron configurations. For diatomics, the occupations can correlate to bond length, bond energies (Figure 9.10.1 ).
The trends in Figure 9.10.1 and Table 9.10.1 extend to molecular ions.
Example 9.10.1 : Molecular Oxygen
Arrange the following four molecular oxygen species in order of increasing bond length: $\ce{O_2^+}$, $\ce{O_2}$, $\ce{O_2^-}$, and $\ce{O_2^{2-}}$.
Solution
The bond length in the oxygen species can be explained by the positions of the electrons in molecular orbital theory. To obtain the molecular orbital energy-level diagram for $\ce{O2}$, we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in Figure 9.10.1 . We again fill the orbitals according to Hund’s rules and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ2s and σ2s* orbitals, two more to fill the $\sigma _{2p_{z}}$ orbital, and 4 to fill the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star}$ orbitals. According to Hund’s first rule, the last 2 electrons must be placed in separate $π^*$ orbitals with their spins parallel, giving a multiplicity of 3 (a triplet state) with two unpaired electrons. This leads to a predicted bond order of
$\dfrac{8 − 4}{2} = 2 \nonumber$
which corresponds to a double bond, in agreement with experimental data: the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K.
The bond order is determined from the the electron configurations. The electron configurations for the four species are contrasted below.
• $\ce{O_2}$: $σ_{1s}^2 {σ^*_{1s}}^2 σ_{2s}^2 {σ^*_{2s}}^2 σ_{2p}^2 π_{2p_y}^2 {π^*_{2p_y}}^2 π_{2p_x}^1 {π^*_{2p_y}}^1 \nonumber$ From Equation \ref{BO}, the bond order for $\ce{O_2}$ is 2 (i.e., a double bond).
• $\ce{O_2^{+}}$: $σ_{1s}^2 {σ^*_{1s}}^2 σ_{2s}^2 {σ^*_{2s}}^2 σ_{2p}^2 π_{2p_y}^2 {π^*_{2p_y}}^2 π_{2p_x}^1 {π^*_{2p_y}}^0 \nonumber$ From Equation \ref{BO}, the bond order for $\ce{O_2^{+}}$ is 2.5. An alternative and equally valid configuration is $σ_{1s}^2 {σ^*_{1s}}^2 σ_{2s}^2 {σ^*_{2s}}^2 σ_{2p}^2 π_{2p_y}^2 {π^*_{2p_y}}^2 π_{2p_x}^0 {π^*_{2p_y}}^1 \nonumber$
• $\ce{O_2^{-}}$: $σ_{1s}^2 {σ^*_{1s}}^2 σ_{2s}^2 {σ^*_{2s}}^2 σ_{2p}^2 π_{2p_y}^2 {π^*_{2p_y}}^2 π_{2p_x}^2 {π^*_{2p_y}}^1 \nonumber$ From Equation \ref{BO}, the bond order for $\ce{O_2^{-}}$ is 1.5. An alternative and equally valid configuration is $σ_{1s}^2 {σ^*_{1s}}^2 σ_{2s}^2 {σ^*_{2s}}^2 σ_{2p}^2 π_{2p_y}^2 {π^*_{2p_y}}^2 π_{2p_x}^1 {π^*_{2p_y}}^2 \nonumber$
• $\ce{O_2^{2-}}$: $σ_{1s}^2 {σ^*_{1s}}^2 σ_{2s}^2 {σ^*_{2s}}^2 σ_{2p}^2 π_{2p_y}^2 {π^*_{2p_y}}^2 π_{2p_x}^2 {π^*_{2p_y}}^2 \nonumber$ From Equation \ref{BO}, the bond order for $\ce{O_2^{2-}}$ is 1.
The bond order decreases and the bond length increases in the order. The predicted order of increasing bondlength then is $\ce{O_2^+}$ < $\ce{O_2}$ < $\ce{O_2^-}$ < $\ce{O_2^{2-}}$. This trend is confirmed experimentally with $\ce{O_2^+}$ (112.2 pm), $\ce{O_2}$ (121 pm), $\ce{O_2^-}$ (128 pm) and $\ce{O_2^{2-}}$ (149 pm).
Molecular Oxygen is Paramagnetic
We now turn to a molecular orbital description of the bonding in $\ce{O2}$. It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. None of the other bonding models (e.g., Valence Bond theory or Lewis bonding) can predict the presence of two unpaired electrons in $\ce{O_2}$. Chemists had long wondered why, unlike most other substances, liquid $\ce{O_2}$ is attracted into a magnetic field. As shown in Video 9.10.1 , it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for $\ce{O_2}$ to have unpaired electrons, making it paramagnetic. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches.
Advanced: Spin Barriers
The magnetic properties of $\ce{O2}$ are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H2O, CO2, and N2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H2O, CO2, and N2, have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of $\ce{O2}$ with organic compounds to give H2O, CO2, and N2 would require that at least one of the electrons on $\ce{O2}$ change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier.
9.11: Photoelectron Spectra Support the Existence of Molecular Orbitals
In the mid 1920's the German physicist Werner Heisenberg showed that if we try to locate an electron within a region $Δx$; e.g. by scattering light from it, some momentum is transferred to the electron, and it is not possible to determine exactly how much momentum is transferred, even in principle. Heisenberg showed that consequently there is a relationship between the uncertainty in position $Δx$ and the uncertainty in momentum $Δp$.
$\Delta p \Delta x \ge \frac {\hbar}{2} \label {5-22}$
You can see from Equation $\ref{5-22}$ that as $Δp$ approaches 0, $Δx$ must approach ∞, which is the case of the free particle discussed previously.
This uncertainty principle, which also is discussed in Chapter 4, is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space. Activity 1 at the end of this chapter illustrates that a reduction in the spatial extent of a wavefunction to reduce the uncertainty in the position of a particle increases the uncertainty in the momentum of the particle. This illustration is based on the ideas described in the next section.
Exercise $1$
Compare the minimum uncertainty in the positions of a baseball (mass = 140 gm) and an electron, each with a speed of 91.3 miles per hour, which is characteristic of a reasonable fastball, if the standard deviation in the measurement of the speed is 0.1 mile per hour. Also compare the wavelengths associated with these two particles. Identify the insights that you gain from these comparisons.
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Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χB > χA), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure $4$. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond.
A molecular orbital energy-level diagram is always skewed toward the more electronegative atom.
An Odd Number of Valence Electrons: NO
Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O2 with N2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O2 to produce NO2, which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals.
Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure $13$) shows that the general pattern is similar to that for the O2 molecule (Figure $11$). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2p atomic orbitals, the 11th electron must occupy one of the degenerate π* orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N2 and O2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot.
Note that electronic structure studies show the ground state configuration of $\ce{NO}$ to be $\left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{1}$ in order of increasing energy. Hence, the $\pi _{2p_{x,y}}$ orbitals are lower in energy than the $\sigma _{2p_{z}}$ orbital. This is because the $\ce{NO}$ molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond (Figure $11$).
Molecular orbital theory can also tell us something about the chemistry of $NO$. As indicated in the energy-level diagram in Figure $13$, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, $NO$ is easily oxidized to the $NO^+$ cation, which is isoelectronic with $N_2$ and has a bond order of 3, corresponding to an N≡O triple bond.
Nonbonding Molecular Orbitals
Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $6$ that the 1s orbital of atomic hydrogen is closest in energy to the 3p orbitals of chlorine. Consequently, the filled Cl 3s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1s and Cl 3p orbitals. Of the three p orbitals, only one, designated as 3pz, can interact with the H 1s orbital. The 3px and 3py atomic orbitals have no net overlap with the 1s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3s, 3px, and 3py orbitals do not change when HCl forms, they are called nonbonding molecular orbitals. A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3pz than to the H 1s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give $H^{\delta +} -- Cl^{\delta -}$.
Electrons in nonbonding molecular orbitals have no effect on bond order.
Example $4$: The Cyanide Ion
Use a “skewed” molecular orbital energy-level diagram like the one in Figure $4$ to describe the bonding in the cyanide ion (CN). What is the bond order?
Given: chemical species
Asked for: “skewed” molecular orbital energy-level diagram, bonding description, and bond order
Strategy:
1. Calculate the total number of valence electrons in CN. Then place these electrons in a molecular orbital energy-level diagram like Figure $4$ in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
2. Calculate the bond order and describe the bonding in CN.
Solution:
A The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure $4$ fills the five lowest-energy orbitals, as shown here:
Because $\chi_N > \chi_C$, the atomic orbitals of N (on the right) are lower in energy than those of C.
B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N2.
Exercise $4$: The Hypochlorite Ion
Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl). What is the bond order?
Answer
All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1.
Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods.
Summary
Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered homonuclear diatomic molecules, molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O2 molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules, using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry.
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Molecular term symbols specify molecular electronic energy levels. Term symbols for diatomic molecules are based on irreducible representations in linear symmetry groups, derived from spectroscopic notations. They usually consist of four parts: spin multiplicity, azimuthal angular momentum, total angular momentum and symmetry. All molecular term symbols discussed here are based on Russel-Saunders coupling.
Introduction
Molecular term symbols mark different electronic energy levels of a diatomic molecule. These symbols are similar to atomic term symbols, since both follow the Russell-Saunders coupling scheme. Molecular term symbols employ symmetry labels from group theory. The possibility of an electronic transition can be deducted from molecular term symbols following selection rules. For multi-atomic molecules, symmetry labels play most of term symbols' roles.
For homonuclear diatomics, the term symbol has the following form:
$^{2S+1}\Lambda_{\Omega,(g/u)}^{(+/-)}$
whereas Λ is the projection of the orbital angular momentum along the internuclear axis: Ω is the projection of the total angular momentum along the internuclear axis; g/u is the parity; and +/− is the reflection symmetry along an arbitrary plane containing the internuclear axis. Λ may be one of the greek letters in the sequence: Σ Π Δ Φ... when Λ = 0, 1, 2, 3..., respectively. For heteronuclear diatomics, the term symbol does not include the g/u part, for there is not inversion center in the molecule.
Determining term symbols of diatomics
Let's start with CO again. As we have seen before, the molecule has a close-shell configuration. Its ground state is a totally symmetric singlet, 1Σ+, since the only possible values of (S, Λ) are (0, 0). If one of the HOMO electrons on the 5σ+ orbital has jumped to the LUMO, this molecule will be in an excited state as follows.
Suppose a CO molecule is in the excited state shown above. In order to know the term symbol of this state, a direct product of the labels is required for the two MO's with unpaired electrons. The multiplication is such as $\Pi \times \Sigma^+ = \Pi$. According to Pauli's exclusion rule, these two unpaired electrons can never share the same set of quantum numbers, therefore the spin degeneracy S can reach its maximum 3. The resulting term symbols are 1Π and 3Π.
Now if we look at O2, it does not have a close-shell configuration at its ground state. There are two unpaired electrons each occupying one of the two degenerate 2π orbitals, which can be seen in the diagram below.
The term symbol for oxygen molecule at its ground state is therefore derived such as Π x Π = Σ+ + Σ- + [Δ], as the symbol in brackets does not allow the oxygen atoms to commute.
Transition between electronic states of diatomics
We'll focus on selection rules. Like atomic electronic states, different selection rules apply when differently incurred transitions occur. Usually for electric dipole field induced transitions, the selection rules are the same as for atoms.
1. ΔΛ = 0, ±1 except Λ = 0 ‡ Λ' = 0
2. ΔS = 0
3. ΔΩ = 0; ±1 except Ω = 0 ‡ Ω' = 0
9.15: Molecular Term Symbols Designate Symmetry
The quantum numbers for diatomic molecules are similar from the atomic quantum numbers. Be cautious, because the rules for finding the possible combinations are different. The total orbital angular momentum quantum number For the molecular case, this number is called $Λ$ instead of $L$. It follows the same naming convention as $L$, except that instead of using capital English letters, it uses capital Greek letters:
• $Λ = 0 \rightarrow Σ$
• $Λ = 1 \rightarrow Π$
• $Λ = 2 \rightarrow Δ$
• $Λ = 3 \rightarrow Φ$
Unlike $L$, there is not a general formula for finding the possible combinations of $Λ$. You have to examine the individual microstates. This is easier than it sounds.
• The total magnetic quantum number $M_L$: $M_L$ works like $M_l$ with atoms, except that there is no formula for finding the combinations.
• The total spin magnetic quantum number $M_S$: $M_S$ works exactly like $M_s$. Electrons can either point with or against the z ‐ axis, and being in a molecular orbital versus an atomic orbital doesn’t change this. $M_S$ can range from $m_{s1} + m_{s2}$ to $m_{s1} ‐ m_{s2}$.
Two New Symmetries: Parity and Reflection
Molecular orbitals are more complex than atomic ones and require more modifiers to completely define. Parity (sometimes called “inversion”) tells you if the orbital is symmetric or anti‐symmetric when an inversion operation is performed. The symmetry notation u and g are sometimes used when describing molecular orbitals. This refers to the operation of inversion, which requires starting at an arbitrary point in the orbital, traveling straight through the center, and then continuing outwards an equal distance from the center. The orbital is designated g (for gerade, even) if the phase is the same, and u (for ungerade, uneven) if the phase changes sign.
To determine whether or not a given state is $g$ or $u$, find the parity of each individual open‐shell electron and uses these simple (Laporte rules):
• $g + g \rightarrow g$
• $g + u \rightarrow u$
• $u + u \rightarrow g$
Example 9.15.1: Closed Shell Configuration
What is the parity of the state $1σ_g^21σ_u^22σ_g^22σ_u^22π_u^12π_u^1$ ?
Solution
Since both open shell electrons are ungerade, the overall parity is g. Helpful hint: bonding sigma orbitals and anti ‐ bonding pi orbitals are always gerade. Anti ‐ bonding sigmas and bonding pis are always ungerade. Draw them and see for yourself.
Reflection determines if a given orbital is symmetric or anti‐symmetric upon reflection through a plane that contains both nuclei. The choice of symmetry planes is arbitrary. As long as you pick a plane and stick with it, you will always get the right answer. When an orbital is symmetric, it is labeled +. When an orbital is anti ‐ symmetric, it is labeled ‐ . To find the overall reflection of a state, use these rules:
• (+)(+) \rightarrow +
• (+)(‐) \rightarrow ‐
• (‐)(‐) \rightarrow +
Reflection only applies to Σ states! For Λ > 0, there are no reflection labels! If you experiment with the rules, you will quickly realize why this is the case.
Example 9.15.2
What is the reflection of the state $1σ_g^21σ_u^22σ_g^22σ_u^22π_u^12π_u^1$ ?
Solution
You need to know what the orbitals look like. Draw a picture and then pick a plane. For this example, the plane of the page is selcted, but the orthogonal plane would have worked just as well.
The “vertical” orbital is + The “horizontal” orbital is ‐ Since one is + and one is ‐ , the overall reflection is ‐ . Try using the orthogonal plane and convince yourself that you still get the same answer.
Example 9.15.3: Oxygen
What are the term symbols for $O_2$ ?
Solution
The molecular orbital diagram for $O_2$ is
Where I chose arbitrary configurations for the last two electrons.
There are two open ‐ shell electrons occupying the anti‐bonding $π_g$ orbitals. These are the only electrons that matter. It is easiest to simply draw all of the permutations and figure out the bounds on $Λ$ and $M_L$ by inspection. If we do this, it is easy to see that $Λ = 2,0$ and that $M_L = 2,0, ‐ 2$
$M_s= -1$ $M_s= 0$ $M_s= 2$
0 1 0
1 2 1
0 1 0
The top row is a $Λ=2$ $M_S =0$ state, so it is $^1Δ$. Both electrons are in the rightmost orbital. This orbital is gerade, and (g)(g) = g, so the parity label is g. We do not assign reflection labels to non Σ states, so the term symbol is $^1Δ_g$. After removing the used up microstates, the chart becomes
$M_s= -1$ $M_s= 0$ $M_s= 2$
1 1 0
This is a Λ =0 state with three possible spin configurations, so it is $^3Σ$. We know that the electrons are in different sub‐orbitals (if you cannot see this, try drawing all of the possible combinations that give $Λ =0$). Both of the orbitals are gerade, so the overall parity is gerade. One of the orbitals will be +, the other will be ‐ . The final answer is $^1Δ_g \(^3Σ^+_g$
Exercise 9.15.1
Write the term symbols for $O_2^‐$
Solution
First draw the electron configuration diagram.
There are only two possible configurations. It should be easy to see that the term symbol is $^2\Pi_g$.
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Solutions to select questions can be found online.
9.3
The overlap integral and other integrals that arise in two-center systems like that of H2 which is called a two-center integral. The two center integrals are much easier being evaluated by simply using the coordinate system, elliptic coordinates. In the coordinate system, where there are simply two fixed points which are separated by distance $R$.
A point $P$ is given by three coordinates which are
• $λ= r_A + r_B / R$
• $µ= r_A + r_B / R$
the angle ø is the angle that ($r_A$, $r_B$, $R$) triangle makes the interfocal axis. The differential volume element (elliptic coordinates) $dr= R^3/ 8 (λ^2 - \mu^2) dλ\,dµ\,dø$
Given definitions $λ$, $\mu$, $\phi$ → prove that
• $1≤ λ < ∞$
• $-1≤ µ ≤ 1$
• $0 ≤ \phi ≤ 2π$
Use elliptic coordinates to evaluate the overlap integral.
Use Equation:
$S= \int dr \,1s_A1s_B = \dfrac{Z^3}{π} \int dr \,e^{-Zr_A} e^{-Zr_B} \nonumber$
Solution
From simple inspection of the figure:
• Have $r_A + r_B$ is never less than that of $R$. Therefore, $1≤ λ < ∞$.
• $r_A - r_B$ can never have a of magnitude greater that that of $R$, therefore have $-1≤ \mu ≤ 1$.
• The variable $\phi$ can undergo 1 complete revolution therefore have/get $0 ≤ \phi ≤ 2π$.
Knowing all of this information, we can now solve for this equation:
\begin{align*} S &= \dfrac{Z^3}{π} \int dr e^{-ZrA} e^{-Zr)B} \[4pt] &= \dfrac{Z^3}{π} ∫_0^{2\pi} d\phi \int_1^{∞} dλ \int _{-1}^{1} d\mu R^3/8 (λ^2 - µ^2) e^{-Z(r_A + r_B)} \[4pt] &= \dfrac{R^3Z^3}{4} ∫_1^∞ dλ \int_{-1}^{1} d\mu (λ^2 - \mu^2) e^{-ZR λ} \[4pt] &= \dfrac{R^3Z^3}{4} ∫_1^∞ d λ e^{-ZR λ} ∫_{-1}^1 dµ (λ^2 - µ^2) \[4pt] &= \dfrac{R^3Z^3}{4} ∫^∞_{1} dλ e^{-ZRλ} (2λ^2 - 2/3) \[4pt] &= \dfrac{R^3Z^3}{2} \left[(1/ZR + 2/Z^2R^2 + 2/Z^3R^3) e^{-ZR} - 1/3ZR e^{-ZR}\right] \[4pt] &= e^{-ZR}(1+ZR + Z^2R^2/3) \end{align*}
9.4
Repeat the calculation in 9.3 for ${\psi}_-={1s_A-1s_B}$
Solution
$E_-=\dfrac{\int{dr{\psi}^*_-H{\psi}_-}}{\int{dr{\psi}^*_-{\psi}_-}}\nonumber$
The denominator is
$\int{dr{\psi}^*_-{\psi}_-}=2(1-S)\nonumber$
The numerator can be evaluated using equations 9.18, 9.19, 9.20, 9.21 with a simple exchange of a minus sign in equations 9.18 and 9.21,
$\int{dr{\psi}^*_-H{\psi}_-}=2E_{1s}\left(1+S\right)+2J-2K\nonumber$
So
$E_-=\dfrac{2E_{1s}\left(1+S\right)+2J-2K}{2(1-S)}\nonumber$
9.9
Show that the $2p_x$ and the $2p_y$ orbitals do not overlap. Use $2p_x \equiv C\cos\theta$ and $2p_y \equiv C\sin\theta \cos\theta$ and $f(r)$ as the radial component.
Solution
If $\int \Psi_1 \Psi_2 dr \neq 0$ then the orbitals overlap.
Substituting in:
$\int C\cos^2\theta \sin\theta f(r) \nonumber$
$C\int_0^\pi cos^2\theta \sin\theta \int_0^\infty f(r)\neq 0 \nonumber$
This orbitals do not overlap.
9.10
what does the overlap integral represent?
Solution
The overlap integral represents the amount of overlap there is between the orbitals of 2 or more different nuclei. It is only significant for orbitals that have a large overlap. It is a component of the energy integral of the nuclei.
9.14
Which of the three species is the is the least stable due to bond order: $\ce{O2}$, $\ce{O2^{+}}$, $\ce{O_2^{2-}}$. Hint it may be helpful to draw molecular orbitals for each species although it is not required.
Solution
We have to use the bond order formula for each of the situations.
Bond order = $\dfrac{(\text{number of bonding electrons})- (\text{number of anti-bonding electrons})}{2}$
For O2: We have 12 valence electrons. Therefore
Bond order = $\dfrac{10 - 8}{2} = 1$
From the bond orders, we can see that O22- is the least stable.
Bond order = $\dfrac{10 - 5}{2}= \dfrac{5}{2}$
For O22-: We have 14 valence electrons. Therefore
Bond order =$\dfrac{10 - 6}{2} = 2$
For O2+: We have 11 valence electrons. Therefore
9.17
Write the electron configuration and bond order of carbon monoxide.
Solution
The electron configuration for carbon monoxide
$CO: KK(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4(\sigma2_{p_z})^2\nonumber$
The bond order of $\ce{CO}$ is three.
9.17
Determine the bond order diphosphorus $\ce{P2}$.
Solution
Each phosphorus atom will contribute 5 valence electrons, so there will be a total of 10 valence electrons. We use these ten valence electrons to fill up atomic orbitals and we learn that the ground state electron configuration has 8 electron in bonding orbitals and 2 electrons in antibonding orbitals. To find bond order we take bonding electrons(8 electrons) minus antibonding (2 electrons) and divide by 2 to find the bond order of 3.
9.18
Determine the bond order for the bond in nitrogen oxide cation, $\ce{NO^{+}}$.
Solution
The equation for bond order is
$\text{Bond Order} = \dfrac{1}{2}[(\text{# of electrons in bonding orbitals}) - (\text{# of electron in antibonding orbitals})]\nonumber$
There are 10 electrons total in the $2p$ level that have to fill the $\sigma_1, \sigma^{*}_1, \pi, \sigma_2, \pi^{*} \sigma^{*}_2$ in order of increasing energy, respectively. If one fills out the MO diagram, they will find that the number of bonding electrons is $8$ and the number of antibonding electrons is $2$.
Thus, the bond order is
$\dfrac{1}{2} [8 - 2] = 3\nonumber$
This indicates that there is triple bond for $NO^{+}$
9.18
What is the electron configuration and bond order of nitric oxide ion?
Solution
The electron configuration of $NO^+$ is
$NO^+: KK(\sigma2s)^2(\sigma^*2s)^2(\pi 2p)^4(\sigma2p_z)^2(\pi^*2p)^1\nonumber$
The bond order of $NO^+$ is $2\dfrac{1}{2}$
9.21
Determine the bond order for the $NO$ molecule.
Solution
Bond Order = [(# of binding electrons) - (# of anitbonding electrons)] /2
From drawing the Molecular Orbital Diagram, we find that there are 6 binding electrons and 1 antibonding electron, therefore,
Bond Order = (6-1)/2 =5/2
9.22
How is the energy-level diagram for diatomic OH radical differ from that of NH? What is the highest occupied molecular orbital of OH?
9.23
If a light source generates a light at 73.4 nm while a photoelectron spectrometer is measuring the kinetic energy of the electrons ionized when the molecule absorbs this light, what is the largest electron binding energy that can be measured? Also, is it possible to determine the energy of the occupied molecular orbitals of a molecule using the kinetic energy of the ionized electrons measured? If yes, explain. Hint : Recall the photoelectron effect discussed in lecture 2.
Solution
The energy of the source light is
$E = \dfrac{hc}{\lambda} = 2.71 \times 10^{-18} \;J\nonumber$
So 2.71 x 10-18 J is the largest electron binding energy that can be measured using this radiation source.
Using Einstein's explanation of the photoelectric effect, we know that
$\varphi + KE = h\nu\nonumber$
Hence, if we know the kinetic energy of the ionized electrons, as well as the energy absorbed by the electron v, we can find φ, which is the energy of the molecular orbital occupied by the electron being ionized.
9.24
Why would the ionization energy of a $3p$ electron from HCl molecule be lower than a $3p$ electron from a chlorine atom?
Solution
The electrons that are shared between the two atoms in HCl are pulled closer to the Chlorine atom because it has a greater electronegativity than hydrogen. This creates a dipole moment and the bonding electrons become localized in such a way that they shield the nonbonding $3p$ electrons on the chlorine atom more than the inner shells of just a chlorine atom. Since shielding is greater for HCl the electrons require less energy to be pulled away from the molecule. Less shielded valence electrons will experience a greater attractive force decreasing the radius and increasing ionization energy.
9.27
Photoelectron spectroscopy involves the measurement of kinetic energy of photoelectrons to determine the binding energy, intensity and angular distributions of these electrons and use the information obtained to examine the electronic structure of molecules. It differs from the conventional methods of spectroscopy in that it detects electrons rather than photons to study electronic structures of a material. The $O_2$ photoelectron spectrum emits two bands of 52.3898 MJ* mol^{-1} and 52.311 MJ*mol^{-1}. This emission corresponds to the 1$\s) ionization of the oxygen electron. In your own words explain the observation. Solution The two bands that oxygen is emitted corresponds to the spin \((+ \dfrac{1}{2}, - \dfrac{1}{2})$ ionization in the 1$s$ electron. The marginal difference in energy is due to the spin-orbit coupling, which is also known as spin-pairing. Spin-orbit coupling describes the weak magnetic interaction between the orbital motion and the particle spin in a particle
9.28
Would the ionization energies of non-bonding 1s electrons be different for HCl and Cl2?
Solution
The shielding effects for the 1s electrons are different for these two systems. In HCl, the bonding electron is localized on the more electronegative Cl, while in Cl2 the bonding electrons are right in between the two atoms. This means that the 1s electrons are more shielded in HCl than in Cl2 , and so the electron is easier to rip off in HCl. So the ionization energy for 1s electrons in HCl is smaller than in Cl2.
9.28
First write out the ground-state electron configuration for the homonuclear diatomic molecule $F_2$. Then given the experimental ionization energies for a 1s fluorine electron for HF and $F_2$ as 66.981 and 67.217 $MJ \centerdot mol^{-1}$, explain why even though the 1s electrons of fluorine are not involved in the chemical bonds their ionization energies are different.
Solution
The ground state configuration is
$(1\sigma_g)^2(1\sigma_u)^2(2\sigma_g)^2(2\sigma_u)^2(3\sigma_g)^2(1\pi_u)^2(1\pi_u)^2(1\pi_g)^2(1\pi_g)^2\nonumber$
The ionization energies are different because the 1s electrons affect the attraction of those electrons to the nucleus, even though they are not involved in the actual bond. The $F_2$ bonding electrons are equally distributed between two atoms, whereas the HF bonding electrons are localized on the fluorine atom. Therefore there is increased shielding of the 1s orbital on HF which causes a smaller ionization energy.
9.30
Try to solve for the ground-state term symbols for the diatomic molecules H2+ and H2, given that the ground-state electron configuration for H2+ is $(\sigma_g1s)^1$ and for H2 is $(\sigma_g1s)^2$.
Solution
(1σg)1 corresponds to ML=0 and Ms=1/2, meaning that there is an unpaired electron in the symmetric molecular orbitals of g and there is change in the 1σg wavefunction. The ground-state term symbol for H2+ will then be $^2\Sigma^+_g$
(1σg)2 corresponds to ML=0 and Ms=0, meaning that there is an unpaired electron in the molecular orbitals of g and there is change in the 1σg wavefunction. The ground-state term symbol for H2 will then be $^1\Sigma^+_g$
9.30
Find the ground-state term symbols for $B_{2}$.
Solution
The ground state electron configuration is
$(1\sigma_{g})^{2} (1\sigma_{u})^{2} (2\sigma_{g})^2 (1\pi_{u})^1 (1\pi _{u})^1\nonumber$
We do not care about the filled orbitals because their
$M_{L}=M_{S} =0\nonumber$
The two unfilled orbitals are
$(1\pi_{u})^1 (1\pi_{u})^1\nonumber$
all possible combinations of
$m_{l1}, m_{s1}, m_{l2}, m_{s2} \nonumber$
for $B_{2}$ are:
$M_{S}, M_{L}: \nonumber$
$(0, 2): 1^+, 1^-\nonumber$
$(0,0): 1^-, -1^+\nonumber$
$(0, -2): -1^+, -1^-\nonumber$
With these possible combinations we find the following to be possible ground state term symbols:
$^1\Delta, ^3\Sigma, ^1\Sigma\nonumber$
Following Hund's Rule we detect $^3\Sigma$ to be our ground-state term symbol because the largest spin multiplicity will be the ground state of $B_2$. Now we take the two unoccupied orbitals and take the product to get the symmetry of the molecular orbital.
$u \cdot u = g\nonumber$
So the ground-state term symbol for $B_2\nonumber$ is: $^3 \Sigma_{g}^-$
9.31
Solve for the ground-state molecular term symbols for O2, N2, N2+, and O2+.
Solution
The ground state molecular term symbols are given below:
O2
The molecular term symbol of O2 without the ± designation is 3$\Sigma$g. The electron configuration is (filled orbital) (1$\pi$g2$p$x)1(1$\pi$g2$p$y)1, so the symmetry with respect to a reflection through the $x$-$z$ plane is (+) (-) = (-); therefore, the complete molecular term symbol of O2 is 3$\Sigma$g-.
N2
(1$\sigma$g)2 (1$\sigma$u)2 (2$\sigma$g)2(2$\sigma$u)2(1$\pi$u)2(1$\pi$u)2(3$\sigma$g)2 Corresponds to | $M$$L$ | = 0 and | $M$$S$ | = 0 of a 1$\Sigma$ term symbol. The symmetry of the molecule is $g$. The complete ground state term symbol of N2 is 1$\Sigma$g+ because the molecular wave function does not change when reflected through a plane containing the two nuclei.
N2+
(1$\sigma$g)2 (1$\sigma$u)2 (2$\sigma$g)2(2$\sigma$u)2(1$\pi$u)2(1$\pi$u)2(3$\sigma$g)1 Corresponds to | $M$$L$ | = 0 and | $M$$S$ | = 0.5 of a 2$\Sigma$ term symbol. The symmetry of the molecule is $g$. The complete ground state term symbol of N2+ is 2$\Sigma$g+ because the molecular wave function does not change when reflected through a plane containing the two nuclei.
O2+
(1$\sigma$g)2 (1$\sigma$u)2 (2$\sigma$g)2(2$\sigma$u)2(3$\sigma$g)2(1$\pi$u)2(1$\pi$u)2(1$\pi$g)1 Corresponds to | $M$$L$ | = 1 and | $M$$S$ | = 0.5 of a 2$\Pi$ term symbol. The symmetry of the molecule is $g$, since the only unfilled molecular orbital has symmetry $g$, so the complete ground state term symbol of O2+ is 2$\Pi$$g$.
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The concept of a molecular orbital is readily extended to provide a description of the electronic structure of a polyatomic molecule. Indeed molecular orbital theory forms the basis for most of the quantitative theoretical investigations of the properties of large molecules. In general a molecular orbital in a polyatomic system extends over all the nuclei in a molecule and it is essential, if we are to understand and predict the spatial properties of the orbitals, that we make use of the symmetry properties possessed by the nuclear framework.
• 10.1: Hybrid Orbitals Account for Molecular Shape
The shape and bonding valecies of polyatomic molecules can be accounted for by hybrid orbitals. Molecular orbitals are formed from linear combinations of atomic orbitals which are similar in energy. These atomic orbitals could come from different atoms, or from the same atom. For example, the 2s and 2p atomic orbitals are very close energetically. When a linear combo of more than one atomic orbital from the same atom is formed, we have a hybrid orbital
• 10.2: Hybrid Orbitals in Water
The goal of applying Valence Bond Theory to water is to describe the bonding in H2O and account for its structure (i.e., appropriate bond angle and two lone pairs predicted from VSEPR theory). This means applying a localize two-atom bonding approach, which requires introducing hybrid orbitals to describe the experimentally observed bent structure.
• 10.3: BeH₂ is Linear and H₂O is Bent
Walsh correlation diagram is a plot of molecular orbital energy as a function of some systematic change in molecular geometry. For example, the correlation between orbital energies and bond angle for an $AH_2$ molecule. The geometry of a molecule is determined by which possible structure is lowest in energy. We can use the Walsh diagram to determine the energy trends based on which orbitals are occupied.
• 10.4: Photoelectron Spectroscopy
A photoelecton spectrum can show the relative energies of occupied molecular orbitals by ionization. (i.e. ejection of an electron). A photoelectron spectrum can also be used to determine energy spacing between vibrational levels of a given electronic state. Each orbital energy band has a structure showing ionization to different vibrational levels.
• 10.5: The pi-Electron Approximation of Conjugation
Molecular orbital theory has been very successfully applied to large conjugated systems, especially those containing chains of carbon atoms with alternating single and double bonds. An approximation introduced by Hückel in 1931 considers only the delocalized p electrons moving in a framework of $\pi$-bonds. This is, in fact, a more sophisticated version of a free-electron model.
• 10.6: Butadiene is Stabilized by a Delocalization Energy
Delocalization energy is intrinsic to molecular orbital theory, since it results from breaking the two-center bond concept. This is intrinsic to molecular orbital theory with the molecular orbitals spreading further than just one pair of atoms. However, within the two-center theory of valence bond theory, the delocalization energy results from a stabilization energy attributed to resonance.
• 10.7: Benzene and Aromaticity
The previous sections addressed the $\pi$ orbitals of linear conjugated system. Here we address conjugated systems of cyclic conjugated hydrocarons with the general formula of $C_nH_n$ where n is the number of carbon atoms in the ring. The molecule from this important class of organic molecule that you are most familiar with is benzene ($C_6H_6$) with n=6, although many other molecules exist like cyclobutadiene ($C_4H_4$ with n=4).
• 10.E: Bonding in Polyatomic Molecules (Exercises)
These are homework exercises to accompany Chapter 10 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
10: Bonding in Polyatomic Molecules
Learning Objectives
• Introduce hybrid orbital to explain non-linear molecular structure
Valence bond (VB) theory is one of two basic theories, along with molecular orbital (MO) theory, that were developed to use the methods of quantum mechanics to explain chemical bonding. It focuses on how the atomic orbitals of the dissociated atoms combine to give individual chemical bonds when a molecule is formed. In contrast, molecular orbital theory, which will be discussed elsewhere, predict wavefunctions that cover the entire molecule.
Review of Diatomics
Let us consider $H_2$. Recall that the Lewis structure for a single $H$ atom is $H\cdot$ and for $H_2$, it is $H:H$. Thus, each hydrogen brings one unpaired electron to the bond. Let the two protons be denoted A and B and the two electrons 1 and 2. Now, consider the potential energy
\begin{align} V &= V_{ee}+V_{en}+V_{nn}\[4pt] &=\dfrac{e^2}{4\pi \epsilon_0}\left [ \dfrac{1}{r_{12}}-\dfrac{1}{r_{1A}}-\dfrac{1}{r_{1B}}-\dfrac{1}{r_{2A}}-\dfrac{1}{r_{2B}}+\dfrac{1}{R_{AB}}\right ] \end{align} \nonumber
But as $R_{AB}\rightarrow \infty$, the $1/r_{12}$, $1/r_{1B}$, $1/r_{2A}$, and $1/R_{AB}$ terms vanish and the potential energy becomes simply that of two noninteracting hydrogen atoms
$V\rightarrow -\dfrac{e^2}{4\pi \epsilon_0}\left [ \dfrac{1}{r_{1A}}+\dfrac{1}{r_{2B}}\right ] \nonumber$
Since the potential energy becomes a simple sum of separate energies for electrons 1 and 2, the wavefunction should simply be a product $\psi_{1s}(r_1 -r_A)\psi_{1s}(r_2 -r_B)$. But as we let $R_{AB}\rightarrow R_e$, where $R_e$ is the equilibrium bond length, the electrons mix, and we can no longer tell if electron 1 belongs to atom A or atom B and the same for electron 2. Thus, we need to construct a combination of products that is consistent with the Pauli exclusion principle. If we just consider the coordinates $r_1$ and $r_2$ of the electrons, then the only wavefunction we can construct from a product of 1s orbitals is
$\psi_u (r_1 ,r_2)=C_u [\psi_{1s}^{A}(r_1)\psi_{1s}^{B} (r_2)-\psi_{1s}^{A}(r_2)\psi_{1s}^{B}(r_1)] \nonumber$
where the $u$ designator indicates that this is an odd function. The constant $C_u$ is the overall noramlization constant. Unfortunately, like in the LCAO method, such a wavefunction is antibonding and is not a good representation of the ground state. If, however, we construct the wavefunction
$\psi_g (r_1 ,r_2)=C_g [\psi_{1s}^{A} (r_1)\psi_{1s}^{B} (r_2)+\psi_{1s}^{A}(r_2) \psi_{1s}^{B} (r_1)] \nonumber$
(where $g$ designates that this is an even function), we violate the Pauli exclusion principle, even though such a wavefunction leads to a stable chemical bond.
What is missing here is the fact that we have not considered the spins of the electrons. Since the electrons are identical, if we exchange coordinates and spins, then the wavefunction should change sign. Thus, we can make both wavefunctions above consistent with the Pauli exclusion principle by multiplying by an appropriate spin wavefunction. We obtain
\begin{align}\psi_u (r_1 ,r_2 ,s_1 ,s_2 ) &=\psi_u(x_1 ,x_2)=C_u [\psi_{1s}^{A}(r_1)\psi_{1s}^{B}(r_2)-\psi_{1s}^{A}(r_2)\psi_{1s}^{B}(r_1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)+\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\[4pt] \psi_g (r_1 ,r_2 ,s_1 ,s_2) &= \psi_g(x_1 ,x_2)=C_g[\psi_{1s}^{A}(r_1)\psi_{1s}{B}(r_2)+\psi_{1s}^{A}(r_2)\psi_{1s}^{B}(r_1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\end{align} \nonumber
where $s_1$ and $s_2$ are the z-components of spin for electrons 1 and 2, respectively. We can now use $\psi_g$ as an approximate 2-electron wavefunction that leads to a stable chemical bond in $H_2$.
The fact that $\psi_u$ is antibonding can be easily determined by looking for a nodal plane between the two atoms, in this case, in the plane that exactly bisects the line joining the two atoms, midway between them. That this is, indeed, a nodal plane can be seen by considering two points $r_1$ and $r_2$ for the two electrons that are taken to lie in this plane. By symmetry, the functions $\psi_{1s}^{A}(r_1)$ and $\psi_{1s}^{B}(r_1)$ have the same value for $r_1$ in this plane, and the same for $\psi_{1s}^{B}(r_2)$ and $\psi_{1s}^{A}(r_2)$. Let us assign the following values:
\begin{align}\psi_{1s}^{A}(r_1) &= \psi_{1s}^{B}(r_1)=A\[4pt] \psi_{1s}^{B}(r_2) &= \psi_{1s}^{A}(r_2)=A' \end{align} \nonumber
Substituting these into $\psi_u (x_1 ,x_2)$, we obtain
$\psi_u (x_1 ,x_2)=C_u [AA' -A' A][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)+\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]=0 \nonumber$
Since the wavefunction has a node midway between the two atoms, it is clearly antibonding and should have a higher energy than the corresponding bonding wavefunction $\psi_g$.
A similar argument can be used for the molecule $F_2$. Each $F$ has an electronic configuration
$1s^2 2s^2 2p_{x}^{2}2p_{y}^{2}2p_{z}^{1} \nonumber$
and the Lewis structure of $F_2$ is
Most of the electrons are in lone pairs, but the $2p_z$ electrons, which are unpaired in each $F$ come together to form the bond. Thus, the bonding wavefunction should be a 2-electron wavefunction constructed from $2p_z$ orbitals. The bonding wavefunction takes the "gerade" form as in $H_2$:
$\psi_g (r_1 ,r_2 ,s_1 ,s_2)=C_g [\psi_{2p_z}^{A}(r_1)\psi_{2p_z}^{B}(r_2)+\psi_{2p_z}^{A}(r_2)\psi_{2p_z}^{B}(r_1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)] \nonumber$
For $HF$, the $2p_z$ orbital on $F$ and $1s$ orbital on $H$ come together to form the bonding wavefunction. To be consistent with the Pauli principle, we need a wavefunction of the form
$\psi (r_1 ,r_2 ,s_1 ,s_2)=[\psi_{1s}^{H}(1)\psi_{2p_z}^{F}(2)+\psi_{1s}^{H}(2)\psi_{2p_z}^{F}(1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)] \label{1})$
Looking at the $HF$ example, it becomes clear how much valence bond theory attempts to appear as a "quantum version'' of the Lewis dot structure model. Valence bond theory attempts to construct very approximate wavefunctions for the bonding electrons in a Lewis structure, leaving the orbitals unused in the construction of the valence bond wavefunctions for the lone pair electrons. In the case of $HF$, we use the $2p_z$ orbitals of $F$, which leaves the $2s$, $2p_x$ and $2p_y$ orbitals unused. Since there are three lone pairs, these three orbitals are sufficient to hold each of the lone pairs as spin-up/spin-down couples.
Hybridization
For polyatomic molecules, the valence bond theory becomes a very poor approximation because the directionalities of the $2s$ and $2p$ orbitals is too restrictive to describe molecules with steric numbers ranging between 2 and 4. The example considered above of $H_2 O$ illustrates this rather dramatically! Let us consider an even simpler molecule, $BeH_2$, which has a steric number of 2 and is linear. Let the atoms lie entirely along the z-axis in the arrangement $H-Be-H$.
Although $Be$ has a ground-state electronic configuration of $1s^2 2s^2$, but if we "promote" one of the $2s$ electrons to a state with higher energy and allow its electronic structure to be $1s^2 2s 2p_z$, then the unpaired electrons in the $2s$ and $2p_z$ orbitals can combine with the unpaired electrons in each of the hydrogen atoms to form bonds. The energy needed to excite the electron in Be would be repaid'' by the energy gained in the formation of stable bonds. The two valence-bond wavefunctions we would construct would be
\begin{align}\psi_1 (1,2) &= C_1 \left[\psi_{1s}^{H_1}(1)\psi_{2s}^{Be}(2)+\psi_{1s}^{H_1}(2)\psi_{2s}^{Be}(1) \right] \left[\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]\[4pt] \psi_2 (1,2) &= C_2 \left[\psi_{1s}^{H_2}(1)\psi_{2p_z}^{Be}(2)+\psi_{1s}^{H_2}(2)\psi_{2p_z}^{Be}(1)\right] \left[\psi_\uparrow (s_1)\psi_\downarrow (s_2)-\psi_\uparrow (s_2)\psi_\downarrow (s_1) \right]\end{align} \nonumber
Unfortunately, even this simple scheme does not work entirely because the two $Be-H$ bonds would be different due to their construction from different combinations of orbitals. By symmetry, however, we can see that the two $BeH$ bonds should be equivalent. A solution to this problem was proposed by Linus Pauling in the 30s in the form of orbital hybridization, a scheme that we still use today.
Pauling used the fact that in the first and second periods, the $2s$ and $2p$ orbitals have similar energies. Indeed, for $H$, the energies are exactly the same. Given that these energies are not that different, we can combine s and p orbitals and still have a valid solution of the Schrödinger equation. That is, a general orbital
$\chi (r)=C_1 \psi_{2s}(r)+C_2 \psi_{2p_x}(r)+C_3 \psi_{2p_y}(r)+C_4 \psi_{2p_z}(r) \nonumber$
is also a solution of the Schrödinger equation with the same energy as a $2s$ or $2p$ orbitals individually (this is exactly true for $H$). In the case of $BeH_2$, the external potential on the electrons in Be by the two hydrogens changes the energy levels and creates a near degeneracy between the $2s$ and $2p_z$ orbitals, hence, we are now free to combine the into linear combinations that are more suitable to the construction both of valence bond wavefunctions and MOs via the LCAO procedure.
$sp$ Hybrid Orbitals
For Be, we now allow the s and p orbitals to mix and create two hybrid orbitals known as $sp$ orbitals. The two new hybrid wavefunctions as linear combination of the functions for 2s and 2pz (using Dirac Notation):
$| \chi_i \rangle = a_1|2s \rangle + b_1 |2p_z \rangle \label{sp1}$
$| \chi_j \rangle = a_2|2s \rangle + b_2 |2p_z \rangle \label{sp2}$
These two wavefunctions must be orthogonal.
$\langle \chi_i | \chi_j \rangle = \delta_{ij} \nonumber$
Which can be separated into the following relationships:
$\langle \chi_i | \chi_i \rangle = 1 \label{norm1}$
and
$\langle \chi_j | \chi_j \rangle = 1 \label{norm2}$
and
$\langle \chi_i | \chi_j \rangle = \langle \chi_j | \chi_i \rangle = 0 \label{ortho}$
Equations $\ref{norm1}$ and $\ref{norm2}$ are the normality requirement and Equation $\ref{ortho}$ is the orthogonality requirement for the new hybrid wavefunctions. Substituting $\ref{sp1}$ into $\ref{norm1}$ results in
$\langle \chi_i | \chi_i \rangle = a_1^2 \cancelto{1}{\langle 2s | 2s \rangle } + a_1 b_1 \cancelto{0} {\langle 2s | 2p_z \rangle} + a_1 b_1 \cancelto{0} {\langle 2p_z | 2s} \rangle + b_1^2 \cancelto{1} {\langle 2p_z | 2p_z \rangle} =1 \nonumber$
and similarly for $\langle \chi_j | \chi_j \rangle$
$\langle \chi_j | \chi_j \rangle = a_2^2 \cancelto{1}{\langle 2s | 2s \rangle } + a_2 b_2 \cancelto{0} {\langle 2s | 2p_z \rangle} + a_2 b_2 \cancelto{0} {\langle 2p_z | 2s} \rangle + b_2^2 \cancelto{1} {\langle 2p_z | 2p_z \rangle} =1 \nonumber$
results in the following relationships
$\langle \chi_i | \chi_i \rangle = a_1^2 + b_1^2 = 1 \label{Con1}$
$\langle \chi_j | \chi_j \rangle = a_2^2 + b_2^2 = 1 \label{Con2}$
and
$\langle \chi_i | \chi_j \rangle = a_1a_2 + b_1b_2 = 0 \label{Con3}$
These are four unknowns and three equations. The fourth "constraint" is that we assume contribution of $|s\rangle$ is the same for both hybrid orbitals.
$a_1 = a_2 \nonumber$
Equations $\ref{Con1}$ to $\ref{Con3}$ revert to
$a_1^2 + b_1^2 = 1 \label{Con1a}$
$a_1^2 + b_2^2 = 1 \label{Con2a}$
$b_1^2 = b_2^2 \label{Con3a}$
Therefore
$b_1 = -b_2 \nonumber$
and
$a_1 = b_1 \label{eq231}$
Insert Equation \refPeq231} into $\ref{Con3a}$ to get
$a_1= \dfrac{1}{\sqrt{2}} \nonumber$
and the two hybrid orbitals are
\begin{align}\chi_1 (r) &= \dfrac{1}{\sqrt{2}}[\psi_{2s}(r)+\psi_{2p_z}(r)]\[4pt] \chi_2 (r) &= \dfrac{1}{\sqrt{2}}[\psi_{2s}(r)-\psi_{2p_z}(r)]\end{align} \nonumber
Note that these orbitals are both normalized and orthogonal:
$\int |\chi_1 (r)|^2 dV=1 \ ; \ \int |\chi_2 (r)|^2 dV=1 \ ; \ \int \chi_{1}^{*}(r)\chi_2 (r)dV=0 \nonumber$
These orbitals appear as shown in Figure 10.1.3
Given that the two $sp$ hybrid orbitals are mirror images of each other, they can overlap with the $1s$ orbital of $H$ (shown in the figure) and create two equal bonds, as needed for $BeH_2$. Using the valence bond formulation, now, one of the $BeH$ bonds will be described by a wavefunction of the form:
\begin{align}\psi_1 (1,2) &= C_1 [\psi_{1s}^{H_1}(1)\chi_{1}^{Be}(2)+\psi_{1s}^{H_1}(2)\chi_{1}^{Be}(1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\[4pt] \psi_2 (1,2) &= C_2 [\psi_{1s}^{H_2}(1)\chi_{2}^{Be}(2)+\psi_{1s}^{H_2}(2)\chi_{2}^{Be}(1)][\psi_{\uparrow}(s_1)\psi_\downarrow (s_2)-\psi_{\uparrow}(s_2)\psi_\downarrow (s_1)]\end{align} \nonumber
In the above wavefunctions, it is clear that $H_1$ is on the right and $H_2$ is on the left, based on the directionalities of $\chi_1$ and $\chi_2$.
$sp^2$ Hybrid Orbitals
For trigonal planar molecules such as $BH_3$, we start with the electronic configuration of $B$, which is $1s^2 2s^2 2p_x$, and we promote one of the $2s$ electrons to a $2p_y$ orbital, so that we have $1s^2 2s 2p_x 2p_y$. Suppose the geometry of $BH_3$ is such that one of the hydrogens lies along the positive x axis. The remaining hydrogens would be in the 3rd and 4th quadrants, respectively, as shown in Figure 10.1.4 .
If we simply combine the $2s$ with the $2p_x$ and $2p_y$ orbitals of boron, the resulting hybrid orbitals will not point in the correct direction. For this reason, we will create rotated versions of the $p_x$ and $p_y$ orbitals, which, as we will see are tantamount to taking new combinations of $2p_x$ and $2p_y$ orbitals to combine with the $2s$. Since the rotation occurs in the $xy$ plane, the coordinate that controls this is the azimuthal angle $\phi$. For the $p_x$ and $p_y$ orbitals, the $\phi$ dependence is
$\psi_{2p_x}\sim \cos\phi \ ; \ \psi_{2p_y}\sim \sin\phi \nonumber$
If we rotate $2p_y$ by $-30$ degrees (Figure 10.1.5 ; blue is positive and red is negative), the $\phi$ dependence becomes
$\psi_{2p_y}^{(rot)}\sim \sin(\phi +30) \nonumber$
Using the fact that
$\sin(\alpha \pm \beta)=\sin\alpha \cos\beta \pm cos\alpha sin\beta \nonumber$
this rotation gives
\begin{align}\psi_{2p_y}^{(rot,1)} &\sim sin\phi \cos30+\cos\phi \sin30\[4pt] &\sim \left [ \dfrac{\sqrt{3}}{2}\sin\phi +\dfrac{1}{2}\cos\phi \right ] \[4pt] &\sim \left [ \dfrac{\sqrt{3}}{2}\psi_{2p_y}+\dfrac{1}{2}\psi_{2p_x}\right ]\end{align} \nonumber
Similarly, consider rotating $-\psi_{2p_y}$ by $+30$ degrees (Figure 10.1.6 ). This gives
\begin{align}-\psi_{2p_y}^{(rot,2)} &\sim -\sin(\phi -30)\[4pt] &\sim -\left [ \dfrac{\sqrt{3}}{2}\sin\phi -\dfrac{1}{2}\cos\phi \right ] \[4pt] &\sim -\dfrac{\sqrt{3}}{2}\psi_{2p_y}+\dfrac{1}{2}\psi_{2p_x}\end{align} \nonumber
So, we now take the hybrid orbitals to be of the form
\begin{align}\chi_1 (r) &= a\psi_{2s}(r)-b\psi_{2p_x}\[4pt] \chi_2 (r) &= c\psi_{2s}(r)+d\psi_{2p_y}^{(rot,1)}(r)\[4pt] \chi_3 (r) &=c\psi_{2s}(r)-d\psi_{2p_y}^{(rot,2)}(r)\end{align} \nonumber
The coefficients $a$, $b$, and $c$ are determined by requiring that the orbitals are normalized and mutually orthogonal:
\begin{align}\int |\chi_1 (r)|^2 dV=1 \ &; \ \int \chi_{1}^{*}(r)\chi_2 (r)dV=0\[4pt] \int |\chi_2 (r)|^2 dV=1 \ &; \ \int \chi_{1}^{*}(r)\chi_3 (r)dV=0\[4pt] \int |\chi_3 (r)|^2 dV=1 \ &; \ \int \chi_{2}^{*}(r)\chi_3 (r)dV=0\end{align} \nonumber
Carrying out the algebra, we obtain the following $sp^2$ hybrid orbitals:
\begin{align}\chi_1 (r) &= \dfrac{1}{\sqrt{3}}[\psi_{2s}(r)-\sqrt{2}\psi_{2p_x}(r)]\[4pt] \chi_2 (r) &= \dfrac{1}{\sqrt{6}}[\sqrt{2}\psi_{2s}(r)+\psi_{2p_x}(r)+\sqrt{3}\psi_{2p_y} (r)]\[4pt] \chi_3 (r) &= \dfrac{1}{\sqrt{6}}[\sqrt{2}\psi_{2s}(r)+\psi_{2p_x}(r)-\sqrt{3}\psi_{2p_y}(r)]\end{align} \nonumber
The $sp^2$ hybrids allow bonding at $120^\circ$ degrees, and these orbitals appear as shown in Figure 10.1.7 :
The figure also shows the overlaps of these orbitals with the $1s$ orbitals of $H$.
$sp^3$ Hybrid Orbitals
Finally, we consider the case of methane $CH_4$. The electronic configuration of $C$ is $1s^2 2s^2 2p_x 2p_y$. We now promote one of the $2s$ orbitals to the $2p_z$ orbital and write $C$ as $1s^2 2s2p_x 2p_y 2p_z$. We can now hybridize the $2s$ orbital with each of the $2p$ orbitals to create four hybrids:
\begin{align}\chi_1 (r) &= \dfrac{1}{2} \left[\psi_{2s}(r)+\psi_{2p_x}(r)+\psi_{2p_y}(r)+\psi_{2p_z}(r)\right]\[4pt] \chi_2 (r) &= \dfrac{1}{2}\left[\psi_{2s}(r)-\psi_{2p_x}(r)-\psi_{2p_y}(r)+\psi_{2p_z}(r)\right]\[4pt] \chi_3 (r) &= \dfrac{1}{2}\left[\psi_{2s}(r)+\psi_{2p_x}(r)-\psi_{2p_y}(r)-\psi_{2p_z}(r)\right]\[4pt] \chi_4 (r) &= \dfrac{1}{2}\left[\psi_{2s}(r)-\psi_{2p_x}(r)+\psi_{2p_y}(r)-\psi_{2p_z}(r)\right]\end{align} \nonumber
The large lobes of the hybridized orbitals are oriented toward the vertices of a tetrahedron, with 109.5° angles between them (Figure 10.1.8 ). Like all the hybridized orbitals discussed earlier, the sp3 hybrid atomic orbitals are predicted to be equal in energy.
In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH2 or CF2), but these species are highly reactive, unstable intermediates that form in only certain chemical reactions.
Are hybrid orbitals Real?
Hybridization is an often misconceived concept. It only is a mathematical interpretation, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will result from various factors, such as steric and electronic interactions, and further more interactions with the surroundings like a solvent or external field. The geometric arrangement will not be formed because a molecule is hybridized in a certain way, it is the other way around, i.e. a result of the geometry or more precise and interpretation of the wavefunction for the given molecular arrangement.
The justification we gave for invoking hybridization in molecules such as BeH2, BF3 and CH4 was that the bonds in each are geometrically and chemically equivalent, whereas the atomic s- and p-orbitals on the central atoms are not. By combining these into new orbitals of sp, sp2 and sp3 types we obtain the required number of completely equivalent orbitals. This seemed easy enough to do on paper; we just drew little boxes and wrote “sp2” or whatever below them. But what is really going on here?
The full answer is beyond the scope of this course, so we can only offer the following very general explanation. First, recall what we mean by “orbital”: a mathematical function ψ having the character of a standing wave whose square ψ2 is proportional to the probability of finding the electron at any particular location in space. The latter, the electron density distribution, can be observed (by X-ray scattering, for example), and in this sense is the only thing that is “real”.
A given standing wave (ψ-function) can be synthesized by combining all kinds of fundamental wave patterns (that is, atomic orbitals) in much the same way that a color we observe can be reproduced by combining different sets of primary colors in various proportions. In neither case does it follow that these original orbitals (or colors) are actually present in the final product. So one could well argue that hybrid orbitals are not “real”; they simply turn out to be convenient for understanding the bonding of simple molecules at the elementary level, and this is why we use them.
Summary
The shape and bonding valecies of polyatomic molecules can be accounted for by hybrid orbitals. Molecular orbitals are formed from linear combinations of atomic orbitals which are similar in energy. These atomic orbitals could come from different atoms, or from the same atom. For example, the 2 sand 2patomic orbitals are very close energetically. When a linear combo of more than one atomic orbital from the same atom is formed, we have a hybrid orbital
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/10%3A_Bonding_in_Polyatomic_Molecules/10.01%3A_Hybrid_Orbitals_Account_for_Molecular_Shape.txt
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The goal of applying Valence Bond Theory to water is to describe the bonding in $H_2O$ and account for its structure (i.e., appropriate bond angle and two lone pairs predicted from VSEPR theory).
The ground state electronic configuration of atomic oxygen atom is $1s^2\,2s^2\,2p_x^2\,2p_y^1 \, 2p_z^1$ and of course the ground state electronic configuration of atomic hydrogen atom is $1s^1$, i.e., a spherical atomic orbital with no preferential orientation. If only the unfilled $2p_y$ and $2p_z$ atomic orbitals of the oxygen were used as bonding orbitals, then two bonds would be predicted. These bonding wavefunctions would be mixture of only $|2p_y \rangle$ and $|2p_z \rangle$ orbitals on oxygen and the $|1s \rangle$ orbitals on the hydrogens ($H_1$ and $H_2$):
$| \chi_1 \rangle = a_1 |1s \rangle_{H_1} + b_1 |2p_y \rangle_O \label{wrong1}$
$| \chi_2 \rangle = a_2 |1s \rangle_{H_2} + b_2 |2p_z \rangle_O \label{wrong2}$
However, with a H-O-H bond angle for these bonds would be expected to be 90° since $2p_y$ and $2p_z$ are oriented 90° with respect to each other. Note that $| \chi_1 \rangle$ and $| \chi_2 \rangle$ are two-center bonding orbitals common to Valence Bond theory.
Using the oxygen atomic orbitals directly is obviously not a good model for describing bonding in water, since we know from experiment that the bond angle for water is 104.45° (Figure 10.2.2 ), which is also in agreement with VSEPR theory. Since the $2s$ orbital is spherical, mixing some $2s$ character into the $2p_z$ orbitals can adjust the bond angle as discussed previously by creating new hybrid orbitals.
Historically, Valence Bond theory was used to explain bend angles in small molecules. Of course, it was only qualitatively correct in doing this, as the following example shows. Let us construct the Valence Bond wavefunctions for the two bonding pairs in $H_2O$ by mixing the $|2s \rangle$, $|2p_x \rangle$, $|2p_y \rangle$, and $|2p_z \rangle$ into four new $sp^3$ hybrid orbitals:
\begin{align*}\chi_1 (r) &= \dfrac{1}{2} \left[\psi_{2s}(r)+\psi_{2p_x}(r)+\psi_{2p_y}(r)+\psi_{2p_z}(r)\right]\ \chi_2 (r) &= \dfrac{1}{2}\left[\psi_{2s}(r)-\psi_{2p_x}(r)-\psi_{2p_y}(r)+\psi_{2p_z}(r)\right]\ \chi_3 (r) &= \dfrac{1}{2}\left[\psi_{2s}(r)+\psi_{2p_x}(r)-\psi_{2p_y}(r)-\psi_{2p_z}(r)\right]\ \chi_4 (r) &= \dfrac{1}{2}\left[\psi_{2s}(r)-\psi_{2p_x}(r)+\psi_{2p_y}(r)-\psi_{2p_z}(r)\right]\end{align*} \nonumber
Hence, the three $2p$ orbitals of the oxygen atom combined with the $2s$ orbitals of the oxygen to form four $sp^3$ hybrid orbitals (Figure 10.2.3 ).
The bond angle for four groups of electrons around a central atom is 109.5 degrees. However, for water the experimental bond angle is 104.45°. The VSPER picture (general chemistry) for this is that the smaller angle can be explained by the presence of the two lone-pairs of electrons on the oxygen atom. Since they take up more volume of space compared to a bonding pair of electrons the repulsions between lone pairs and bonding pairs is expected to be greater causing the H-O-H bond angle to be smaller than the ideal 109.5°.
We can rationalize this by thinking about the s and p characters of the hybrids. In a perfectly $sp^3$ hybridized set of hybrid orbitals, each $sp^3$ orbital should have: 25% s character and and 75% p character. Since the bond angle is not 109.5° in water, the hybrid orbitals cannot have exactly this ratio of s and p character. So there there is uneven distribution of s and p character between the 4 hybrid orbitals. First we will write down the wavefunction and see what this means and then we will rationalize it.
Note: A few cautionary words about hybridization.
Hybridization is an often misconceived concept. It only is a mathematical interpretation, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will result from various factors, such as steric and electronic interactions, and further more interactions with the surroundings like a solvent or external field. The geometric arrangement will not be formed because a molecule is hybridized in a certain way, it is the other way around, i.e., a result of the geometry or more precise and interpretation of the wavefunction for the given molecular arrangement.
Estimating Character of Hybrid Orbitals
The terminology we use for hybridization actually is just an abbreviation:
$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}} \nonumber$
In theory $x$ can have any value, hence any of the following combinations constitute valid hybridization schemes for 1 s orbital and 3 p orbitals:
\begin{align} 1\times\mathrm{s}, 3\times\mathrm{p} \nonumber &\leadsto 4\times\mathrm{sp}^3 \nonumber\ &\leadsto 3\times\mathrm{sp}^2, 1\times\mathrm{p} \nonumber \ &\leadsto 2\times\mathrm{sp}, 2\times\mathrm{p} \nonumber \ &\leadsto 2\times\mathrm{sp}^3, 1\times\mathrm{sp}, 1\times\mathrm{p} \nonumber \ &\leadsto \text{etc. pp.} \nonumber \ &\leadsto 2\times\mathrm{sp}^4, 1\times\mathrm{p}, 1\times\mathrm{sp}^{(2/3)} \nonumber \end{align} \nonumber
There are virtually infinite possibilities of combinations. Which one is "valid" is only determined by experiment (e.g., structure or spectroscopy). The generic $sp^x$ hybrid orbitals wavefunction can be roughly written in terms of atomic orbital character:
$|\chi_i \rangle = N ( p + \gamma s) \label{H1}$
where $N$ is a normalization constant and $\gamma$ is the relative contribution of s character to the hybrid orbital. For a pure $sp^3$ hybrid, $\gamma$ would be 0.25 and for a pure $sp$ hybrid, $\gamma$ would be 1. The question is how to determine $\gamma$ to get a better picture of the hybridization of water. Starting with the normalization criteria for wavefunctions:
$\langle \chi_i | \chi_i \rangle =1 \nonumber$
and substituting Equation $\ref{H1}$ into to get
$\langle N ( p + \gamma s) | N ( p + \gamma s) \rangle =1 \nonumber$
which in integral notation is
$\int N^2 ( p + \gamma s)^2 d\tau =1 \nonumber$
where $d\tau$ represents all space. This is then expanded to
$N^2 \cancelto{1}{\int p^2\; d\tau} + N^2 2 \gamma \cancelto{0} {\int sp\; d\tau} + N^2 \gamma^2 \cancelto{1} {\int s^2 \; d\tau} =1\label{H3}$
These terms simplify either due to orthogonality or normality of the constitute atomic orbitals. Equation $\ref{H3}$ simplifies to
$N^2 + N^2 \gamma^2 =1 \nonumber$
and thus the normalization factor can be expressed in terms of $\gamma$
$N = \dfrac{1}{\sqrt{1+\gamma^2}} \nonumber$
and the generic normalized $sp^x$ hybrid orbital (Equation $\ref{H1}$) is
$|\chi_i \rangle = \dfrac{1}{\sqrt{1+\gamma^2}} ( p + \gamma s) \label{H4}$
The s and p characters to a hybrid orbital are now easy to obtain by squaring $|\chi_i \rangle$
• The magnitude of p-character is $\left(\dfrac{1}{\sqrt{1+\gamma^2}} \right)^2 = \dfrac{1}{1+\gamma^2} \label{p}$ as $\gamma \rightarrow 0$, then the p character of the hybrid goes to 100%
• The magnitude of s-character is $\left(\dfrac{1}{\sqrt{1+\gamma^2}} \gamma ^2 \right)^2 = \dfrac{\gamma^2}{1+\gamma^2} \label{s}$ as $\gamma \rightarrow 1$, then the s character of the hybrid goes to 50%
As mentioned above, the geometric arrangement will not be formed because a molecule is hybridized in a certain way, it is the other way around. How do we choose the correct value of $\gamma$ for the hybrid orbitals? The mixing coefficient $\gamma$ is clearly related to the bond angle θ. Using some simple trigonometric relationships, it can be proven that:
$\cos θ = - \gamma^2 \label{angle}$
Equation $\ref{angle}$ is an important equation as it related experimentally determined structure to the nature of the bonding and specifically, the composition of the atomic orbitals that create the hybrid orbitals used in the bonding.
Example 10.2.1 : Carbon Dioxide
What is the s-character in the hybrid orbitals for $CO_2$.
Solution
We know from simple VSEPR theory that the geometry of $CO_2$ is a linear triatomic molecule.
Thus $\theta = 180°$ and via Equation $\ref{angle}$, $\gamma = 1$ since $\cos 180 ° = -1$. Hence, Equation $\ref{s}$ argues that the hybrid orbitals used in the bonding of $CO_2$ have 50% character; i.e., they are $sp$ hybrid orbitals
$| \chi_1 \rangle = \dfrac{1}{\sqrt{2}} ( s + p) \nonumber$
and
$| \chi_2 \rangle = \dfrac{1}{\sqrt{2}} ( s - p) \nonumber$
Now, let's apply Equation $\ref{s}$ to water to find the character of the hybrid orbitals in water. The bond angle in water is 104.45° (Figure 10.2.2 , hence
$\cos 104.5° = -0.25 \nonumber$
and
$\gamma = \sqrt{0.25} = 0.5 \nonumber$
From Equation $\ref{p}$, then the amount of p character in the hybrid orbitals are
$\dfrac{1}{1+\gamma^2} = \dfrac{1}{1+0.5^2} = 0.80\% \nonumber$
which leave 20% for s character (Equation $\ref{s}$).
$\dfrac{\gamma^2}{1+\gamma^2} = \dfrac{0.5^2}{1+0.5^ 2} = 0.20 \nonumber$
The two hybridized atomic orbitals of oxygen involved in bonding are each 80% p and 20% s character. This are not perfect $sp^3$ hybrid orbitals, as expected. Actually, the orbitals involved in the bonds would be better described as $sp^4$ hybridized. It does not mean that there are 4 p-orbitals in the hybrid orbital, but that each hybrid consists of 20% of s and 80% of p atomic orbitals.
Lone Pairs
Water has two sets of non-bonding electron pairs (Figure 10.2.4 ). Without a bond angle to start from, we cannot derive $\gamma$ that describes the nonbonding hybrid orbitals that they occupy. However, we do know that the O atom has three $p$ orbitals. So the TOTAL absolute p-character in all hybrid orbitals must be 3.
Let $x$ be the p-character in the lone pairs hybrid orbitals:
$0.8 + 0.8 + x + x = 3 \nonumber$
This is assuming the lone pairs are identical. Solving for this, x = 0.7 ( i.e. 70% p and 30% s ). From this we can estimate the angle between the lone pair using Equations $\ref{p}$ and $\ref{angle}$
• p-character:
$\dfrac{1}{1 +\gamma^2} = 0.7 \nonumber$
so
$-\gamma^2 = \dfrac{1}{0.7 -1} =-0.42 \nonumber$
and $\theta = 115°$.
The angle between the lone pairs is greater (115°) than the bond angle (104.5°). The $sp^3$ hybrid atomic orbitals of the lone pairs have > 25% s-character. These hybrid orbitals are less directional and held more tightly to the O atom. The $sp^3$ hybrid atomic orbitals of the bonding pairs have < 25% s-character. They are more directional (i.e., more p-character) and electron density found in the bonding region between O and H.
Warning
It should be noted that the valence bond theory application described above predicts that the two lone electron pairs are in the same hybrid orbitals and hence have the same energies. As discussed in the next sections, that is not experimentally observed in photoelectron spectroscopy, which is a shortcoming of valence bond theory's application to water.
Contributors and Attributions
• Andrew Wolff (Adjunct Professor of Chemistry)
• Martin (Stackexchange)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/10%3A_Bonding_in_Polyatomic_Molecules/10.02%3A_Hybrid_Orbitals_in_Water.txt
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In this section, we will construct approximate molecular orbitals for a water molecule by considering a simple linear triatomic of the general form $HXH$, where $X$ is a second row element. We will take a multi-centered molecular orbital approach instead of the two-centered valence bond/hybrid approach discussed previously. As with previous discussions of Molecular Orbitals, we approximate them as a linear combinations of atomic orbitals (LCAO). In molecular orbital theory linear combinations of all available (atomic) orbitals will form molecular orbitals. These are spread over the whole molecule, or delocalized, and in a quantum chemical interpretation they are called canonical orbitals. Since it is absolutely wrong to assume that there are only three types of $sp^x$ hybrid orbitals, it is possible, that there are multiple different types of orbitals involved in bonding for a certain atom.
AH₂ Molecules
We want to construct a reasonable argument for the energetic ordering and structure of the molecular orbitals. We first note that each $H$ will donate a $1s$ orbital in the LCAO scheme, and $A$ will likely donate at least $2s$ and possible $2p$ orbitals, depending on its chemical identity. In general, if we consider only the first row $A$ elements, the molecule orbitals (via the LCAO) can be expressed as combination of $1s$ orbitals on the two Hydrogens ($H_1$ and $H_2$) and the four n=2 orbitals ($2s\; 2p_x\; 2p_y; 2p_z$) on the $A$ atom:
$| \chi \rangle = a_1 | 1s \rangle _{H_1} + a_2 | 1s \rangle _{H_2} + a_3 | 2s \rangle _{A} + a_4 | 2p_x \rangle _{A} + a_5 | 2p_y \rangle _{A} + a_6 | 2p_z \rangle _{A} \label{MO1}$
These molecular orbitals were created with six atomic orbitals and hence six different $| \chi \rangle$ molecular orbitals can be created. As with previous molecular orbitals problems, the coefficients of this expansion ($\{a_i\}$) are determined by solving the secular determinant.
If we consider only linear $AH_2$ molecules, then Equation $\ref{MO1}$ can be simplified by ignoring $2p_x$ and $2p_y$ atomic orbitals since they are perpendicular to the bonds and are hence non-bonding (only for linear $AH_2$ molecules). Moreover, the molecule is symmetric about the center (the position of $A$), hence the orbitals have to have the same symmetry.
Only the $2p$ orbital of $A$ that will overlap with $1s$ of $H$ is the $2p_z$. Hence, Equation $\ref{MO1}$ can be simplified to consider the combination
$| \chi \rangle = a_1 | 1s \rangle _{H_1} + a_1 | 1s \rangle _{H_2} + a_3 | 2s \rangle _{A} + a_6 | 2p_z \rangle _{A} \label{MO2}$
Note that the two coefficients in front of the $1s$ orbitals of hydrogen are the same by symmetry. This since no hydrogen is "special" and they must have the same contribution to the molecular orbital.
Mixing Amplitudes
How big should the $2s$ orbital contributors of $A$ be compared to the $1s$ orbital of $H$?
This depends on several things. First, is the nuclear charge on $A$ and the second is the electronegativity difference between $H$ and $A$. The first determines how quickly the $2s$ orbitals, remembering that the exponential part is $exp(-Zr/a_0)$, and the electronegativity difference determines the relative magnitude of $H_1$ compared to $H_2$.
The six $| \chi \rangle$ molecular orbitals from Equation $\ref{MO1}$ are shown in Figure 10.3.1 .
The first molecular orbital $|\chi_1 \rangle$ constructed from Equation $\ref{MO1}$ is purely bonding because the $2s$ orbital is positive near the $A$ nucleus, but becomes negative as we go away from the nucleus. This orbital is also even (garade symmetry), so we can denote it as a $2\sigma_{g}$ orbital signifying that it is constructed from a $2s$ orbital of $A$ combined with the two $1s$ orbitals of $H$. The only other MO that can be constructed that has the right symmetry is $|\chi_6 \rangle$ which is denoted as $2\sigma_{u}$. This is an antibonding molecular orbital and is also even (garade symmetry). The corresponding wavefunctions are:
$| \chi_1 \rangle = a_1 | 1s \rangle _{H_1} + a_1 | 1s \rangle _{H_2} + a_3 | 2s \rangle _{A} \nonumber$
$| \chi_6 \rangle = a_1 | 1s \rangle _{H_1} + a_1 | 1s \rangle _{H_2} - a_3 | 2s \rangle _{A} \nonumber$
Next, if we combine a $2p_z$ orbital of $A$ with the $1s$ of $H$, there are two possibilities that have the right symmetry. The first is
$| \chi_2 \rangle = a_1 | 1s \rangle _{H_1} + a_1 | 1s \rangle _{H_2} + a_6 | 2p_z \rangle_{A} \nonumber$
which is a bonding orbital and denoted as $1\sigma_{u}$. This is purely antibonding and has an odd symmetry (ungarede). The other combination is
$| \chi_5 \rangle = a_1 | 1s \rangle _{H_1} - a_1 | 1s \rangle _{H_2} + a_6 | 2p_z \rangle_{A} \nonumber$
Hence, we denote this as $2\sigma_{g}$. The orbitals $2p_x$ and $2p_y$ from $X$ are nonbonding and become $\pi_{2p_x}$ and $\pi_{2p_y}$ nonbonding orbitals and designated as $1\pi_u$ orbitals:
$| \chi_3 \rangle = | 2p_x \rangle_{A} \nonumber$
$| \chi_4 \rangle = | 2p_y \rangle_{A} \nonumber$
Beryllium Hydride ($BeH_2$) is Linear
Consider the $BeF_2$ molecule: Be has a $1s^2\, 2s^2$ electron configuration with is no unpaired electrons available for bonding. From a perspective of using only atomic orbitals to generate the bonding orbitals, we would conclude that the molecule could not exist since no free orientals exist on $Be$ to bond. Clearly, atomic orbitals are not adequate to describe orbitals in molecules, but this can be solved by allowing the 2s and one 2p orbital on Be to mix to form $sp$ hybrid orbitals. The experimental H-Be-H bond angle is 180°. Presumably, one electron from Be is shared with each unpaired electrons from H. We could promote and electron from the 2s orbital on Be to the 2p orbital to get two unpaired electrons for bonding (predicting 90° bond angles, not 180°). Thus the geometry is still not explained with atomic orbitals alone.
Be has 2s and 2p orbitals, and it is in the middle. H has 1s orbitals; there are 2 H atoms on the outside. We initially make combinations of the H atomic orbitals that we previously used to make diatomic hydrogen, except there is no overlap (i.e., $S=0$). These combinations will mix with the 2s and 2pz on Be, as shown in Figure 10.3.2 .
Then we can put the Molecular Orbital diagram together, starting with the outside, drawing in bonding, non-bonding and anti-bonding MOs, and filling the electrons (Figure 10.3.3 ). The bond order is 2.
Walsh Correlation Diagrams
Walsh diagrams, often called angular coordinate diagrams or correlation diagrams, are representations of calculated orbital energies of a molecule versus a distortion coordinate, used for making quick predictions about the geometries of small molecules. By plotting the change in molecular orbital levels of a molecule as a function of geometrical change, Walsh diagrams explain why molecules are more stable in certain spatial configurations (i.e. why water adopts a bent conformation).
A major application of Walsh diagrams is to explain the regularity in structure observed for related molecules having identical numbers of valence electrons (i.e. why $\ce{H2O}$ and $\ce{H2S}$ look similar), and to account for how molecules alter their geometries as their number of electrons or spin state changes. Additionally, Walsh diagrams can be used to predict distortions of molecular geometry from knowledge of how the LUMO (Lowest Unoccupied Molecular Orbital) affects the HOMO (Highest Occupied Molecular Orbital) when the molecule experiences geometrical perturbation. Walsh's rule for predicting shapes of molecules states that a molecule will adopt a structure that best provides the most stability for its HOMO. If a particular structural change does not perturb the HOMO, the closest occupied molecular orbital governs the preference for geometrical orientation.
For the $AH_2$ molecular system, Walsh produced the first angular correlation diagram by plotting the orbital energy curves for the canonical molecular orbitals while changing the bond angle from 90° to 180° (Figure 10.3.4 ). As the bond angle is distorted, the energy for each of the orbitals can be followed along the lines, allowing a quick approximation of molecular energy as a function of conformation.
A typical prediction result for water is an bond angle of 90°, which is not even close to the experimental value of 104°. At best, the method is able to differentiate between a bent and linear molecule.
Walsh's rule for predicting shapes of molecules states that a molecule will adopt a structure that best provides the most stability for its HOMO. If a particular structural change does not perturb the HOMO, the closest occupied molecular orbital governs the preference for geometrical orientation.
Figure 10.3.4 illustrates the difference between the actual linear case we just analyzed and the truly bent molecule, e.g. $H_2 O$. The geometry changes the ordering somewhat, but the qualitative picture we obtain from the linear case makes it a useful construction. The oxygen atomic orbitals are labeled according to their symmetry (Figure 10.3.5 ) as $a_1$ for the 2s orbital and $b_1$ ($2p_x$), $b_2$ ($2p_y$) and $a_1$ ($2p_z$) for the three 2p orbitals. The two hydrogen 1s orbitals are premixed to form $a_1$ and $b_2$ molecular orbitals.
Mixing takes place between same-symmetry orbitals of comparable energy resulting a new set of MO's for water:
• 2a1 MO from mixing of the oxygen 2s atomic orbital and the hydrogen σ MO. Small oxygen 2pz atomic orbital admixture strengthens bonding and lowers the orbital energy.
• 1b2 MO from mixing of the oxygen 2py atomic orbital and the hydrogen σ* MO.
• 3a1 MO from mixing of the oxygen 2pz atomic orbital and the hydrogen σ MO. Small oxygen 2s atomic orbital admixture weakens bonding and raises the orbital energy.
• 1b1 nonbonding MO from the oxygen 2px atomic orbital (the p-orbital perpendicular to the molecular plane).
In the water molecule the highest occupied orbital, ($1b_1$) is non-bonding and highly localized on the oxygen atom, similar to the non-bonding orbitals of hydrogen fluoride. The next lowest orbital ($2a_1$) can be thought of as a non-bonding orbital, as it has a lobe pointing away from the two hydrogens. From the lower energy bonding orbitals, it is possible to see that oxygen also takes more than its "fair share" of the total electron density. The electronic configuration of water in the ground state (Figure 10.3.5 ) is therefore
$(a_1)^2(b_2)^2(a_1)^2(b_1)^2 \nonumber$
Table 10.3.1 list the respective LCAO coefficients for the six atomic orbitals. Table 10.3.1 combines the energy values with the description derived from the character table for molecules of point group C2v
Table 10.3.1 : LCAO Coefficients of first six molecular orbitals of water
Energy Symbol s(H) s(O) px(O) py(O) pz(O) s(H)
6.728
2b2
0.525
0
0
-0.669
0
-0.525
5.440
3a1
-0.553
0.306
0
0
-0.544
0.553
-12.191
1b1
0
0
-1.000
0
0
0
-14.467
2a1
-0.309
0.354
0
0
0.827
-0.309
-19.113
1b2
-0.473
0
0
-0.743
0
0.473
-40.032
1a1
0.315
0.884
0
0
-0.143
0.315
Note in contrast to the valence bond theory discussed previous for water, the two lone electron pairs are not in identical orbitals. The 1b1 MO is a lone pair, while the 3a1, 1b2 and 2a1 MO's can be localized to give two O−H bonds and an in-plane lone pair. This is in agreement with the experimentally measured photoelectron spectrum discussed in the next section.
Summary
Walsh correlation diagram is a plot of molecular orbital energy as a function of some systematic change in molecular geometry. For example, the correlation between orbital energies and bond angle for an $AH_2$ molecule. The geometry of a molecule is determined by which possible structure is lowest in energy. We can use the Walsh diagram to determine the energy trends based on which orbitals are occupied.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/10%3A_Bonding_in_Polyatomic_Molecules/10.03%3A_Why_is_BeH_Linear_and_HO_Bent.txt
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Learning Objectives
• Demonstrate how photoelectron spectroscopy can be used to resolve the absolute energies of molecular orbitals.
Photoelectron spectroscopy (PES) utilizes photo-ionization and analysis of the kinetic energy distribution of the emitted photoelectrons to study the composition and electronic state of the surface region of a sample.
• X-ray Photoelectron Spectroscopy (XPS) uses soft x-rays (with a photon energy of 200-2000 eV) to examine electrons in core-levels.
• Ultraviolet Photoelectron Spectroscopy (UPS) using vacuum UV radiation (with a photon energy of 10-45 eV) to examine electrons in valence levels.
Both photoelectron spectroscopies are based upon a single photon in/electron out process. The energy of a photon of all types of electromagnetic radiation is given by the Planck–Einstein relation:
$E = h \nu \label{5.3.1}$
where $h$ is Planck constant and $\nu$ is the frequency (Hz) of the radiation. UPS is a powerful technique to exam molecular electron structure since we are interested in the molecular orbitals from polyatomic molecules (especially the valence orbitals) and is the topic of this page.
Photoelectron spectroscopy uses monochromatic sources of radiation (i.e. photons of fixed energy). In UPS the photon interacts with valence levels of the molecule or solid, leading to ionization by removal of one of these valence electrons. The kinetic energy distribution of the emitted photoelectrons (i.e. the number of emitted photoelectrons as a function of their kinetic energy) can be measured using any appropriate electron energy analyzer and a photoelectron spectrum can thus be recorded.
The process of photoionization can be considered in several ways. One way is to look at the overall process for a species $A$:
$A + \text{photon} \rightarrow A^+ + e^- \label{5.3.2}$
Conservation of energy then requires that (after using Equation $\ref{5.3.1}$):
$E(A) + h\nu = E(A^+ ) + E(e^-) \label{5.3.3}$
Since the free electron's energy is present solely as kinetic energy ($KE$) (i.e., there is no internal energy in a free electron)
$E(e^-) = KE \nonumber$
Equation $\ref{5.3.3}$ can then be rearranged to give the following expression for the KE of the photoelectron:
$KE = h\nu - \left[ E(A^+ ) - E(A) \right] \label{5.3.4}$
The final term in brackets represents the difference in energy between the ionized and neutral species and is generally called the vertical ionization energy ($IE$) of the ejected electron; this then leads to the following commonly quoted equations:
$KE = h\nu - IE \label{5.3.5}$
or
$IE= h\nu - KE \label{Big}$
The vertical ionization energy is a direct measure of the energy required to just remove the electron concerned from its initial level to the vacuum level (i.e., a free electron). Photoelectron spectroscopy measures the relative energies of the ground and excited positive ion states that are obtained by removal of single electrons from the neutral molecule.
Note
Equation \ref{5.3.5} may look familiar to you as it the same equation Einstein used to describe the photoelectric effect except the vertical ionization energy ($IE$) is substituted for workfunction $\Phi$. Both vertical ionization energy and workfunctions are metrics for the binding energy of an electron in the sample.
At a fundamental level, ionization energies are well-defined thermodynamic quantities related to the heats of protonation, oxidation/reduction chemistry, and ionic and covalent bond energies. Ionization energies are closely related to the concepts of electronegativity, electron-richness, and the general reactivity of molecules. The energies and other characteristic features of the ionization bands observed in photoelectron spectroscopy provide some of the molecular orbitals detailed and specific quantitative information regarding the electronic structure and bonding in molecules.
Ionization is explicitly defined in terms of transitions between the ground state of a molecule and ion states as shown in Equation $\ref{Big}$ and as illustrated in the Figure 10.4.2 . Nonetheless, the information obtained from photoelectron spectroscopy is typically discussed in terms of the electronic structure and bonding in the ground states of neutral molecules, with ionization of electrons occurring from bonding molecular orbitals, lone pairs, antibonding molecular orbitals, or atomic cores. These descriptions reflect the relationship of ionization energies to the molecular orbital model of electronic structure.
Ionization energies are directly related to the energies of molecular orbitals by Koopmans' theorem, which states that the negative of the eigenvalue of an occupied orbital from a Hartree-Fock calculation is equal to the vertical ionization energy to the ion state formed by removal of an electron from that orbital (Figure 10.4.3 ), provided the distributions of the remaining electrons do not change (i.e., frozen).
$I_j = - \epsilon_j \label{Koopman}$
There are many limitations to Koopmans' theorem, but in a first order approximation each ionization of a molecule can be considered as removal of an electron from an individual orbital. The ionization energies can then be considered as measures of orbital stabilities, and shifts can be interpreted in terms of orbital stabilizations or destabilizations due to electron distributions and bonding. Koopmans' theorem is implicated whenever an orbital picture is involved, but is not necessary when the focus is on the total electronic states of the positive ions.
Koopmans' Theorem
Koopmans' theorem argues that the negative of the eigenvalue of an occupied orbital from a Hartree-Fock calculation is equal to the vertical ionization energy to the ion state formed by removal of an electron from that orbital.
Several different ionization energies can be defined, depending on the degree of vibrational excitation of the cations. In general, the following two types of ionization energies are considered (Figure 10.4.4 ):
• Adiabatic ionization energy corresponds to the ionization energy associated with this transition $M(X, v” = 0) + h\nu \rightarrow M^+(x, v’ = 0) + e^- \nonumber$ Adiabatic ionization energy that is, the minimum energy required to eject an electron from a molecule in its ground vibrational state and transform it into a cation in the lowest vibrational level of an electronic state x of the cation.
• Vertical ionization energy corresponds to the ionization energy associated with this transition $M(X, v” = 0) + h\nu \rightarrow M^+(x, v’ = n) + e^- \nonumber$ where, the value n of the vibrational quantum number v’ corresponds to the vibrational level whose wavefunction gives the largest overlap with the v” = 0 wavefunction. This is the most probable transition and usually corresponds to the vertical transition where the internuclear separations of the ionic state are similar to those of the ground state.
The geometry of an ion may be different from the neutral molecule. The measured ionzation energy in a PES experiment can refer to the vertical ionization energy, in which case the ion is in the same geometry as the neutral, or to the adiabatic ionzaiton energy, in which case the ion is in its lowest energy, relaxed geometry (mostly the former though). This is illustrated in the Figure 10.4.4 . For a diatomic the only geometry change possible is the bond length. The figure shows an ion with a slightly longer bond length than the neutral. The harmonic potential energy surfaces are shown in green (neutral) and red (ion) with vibrational energy levels. The vertical ionzation energy is always greater than the adiabatic ionzation energy.
Differing Ionization Energies
You have been exposed to three metrics of ionization energies already, which are similar, but with distinct differences:
• The ionization energy (also called adiabatic ionization energy) is the lowest energy required to effect the removal of an electron from a molecule or atom, and corresponds to the transition from the lowest electronic, vibrational and rotational level of the isolated molecule to the lowest electronic, vibrational and rotational level of the isolated ion.
• The binding energy (also called vertical ionization energy) is the energy change corresponding to an ionization reaction leading to formation of the ion in a configuration which is the same as that of the equilibrium geometry of the ground state neutral molecule.
• The workfunction is the minimum energy needed to remove an electron from a (bulk) solid to a point in the vacuum.
Example 10.4.1 : Molecular Hydrogen
As you remember, the molecular orbital description of hydrogen involves two $|1s \rangle$ atomic orbitals generating a bonding $1\sigma_g$ and antibonding $2\sigma_u^*$ molecular orbitals. The two electrons that are responsible for the $\ce{H_2}$ bond are occupied in the $1\sigma_g$.
The PES spectrum has a single band that corresponds to the ionization of a g electron. The multiple peaks are due to electrons ejecting from a range of stimulated vibrational energy levels. When extensive vibrational structure is resolved in a PES molecular orbital, then the removal of an electron from that molecular orbital induces a significant change in the bonding (in this case an increase in the bond length since the bond order has been reduced).
Example 10.4.2 : Molecular Nitrogen
Diatomic nitrogen is more complex than hydrogen since multiple molecular orbitals are occupied. Four molecular orbitals are occupied (the two $1\pi_u$ orbitals are both occupied). The UV photoelectron spectrum of $N_2$, has three bands corresponding to $3σ_g$, $1π_u$ and $2σ_u$ occupied molecular orbitals. Both $3σ_g$ and $2σ_u$ are weakly bonding and antibonding. The $1\sigma_g$ orbital is not resolved in this spectrum since the incident light $h\nu$ used did not have sufficient energy to ionize electrons in that deeply stabilized molecular orbital.
Note that extensive vibrational structure for the $1π_u$ band indicates that the removal of an electron from this molecular orbital causes a significant change in the bonding.
Hydrogen Chloride
The molecular energy level diagram for $\ce{HCl}$ is reproduced in Figure 10.4.5
Important aspects of molecular orbital diagram in Figure 10.4.5 :
• The H 1s energy lies well above the Cl 2s and 2p atomic orbitals;
• The valence electron configuration can be written 24;
• The H 1s orbital contributes only to the σ molecular orbitals, as does one of the Cl 2p orbitals (hence the lines in Figure 10.4.5 connecting these atomic orbitals and the and molecular orbitals);
• The remaining Cl 2p orbitals (ie those perpendicular to the bond axis) are unaffected by bonding, and these form the molecular orbitals;
• The orbitals are nonbonding - they are not affected energetically by the interaction between the atoms, and are hence neither bonding nor antibonding;
• The orbital is weakly bonding, and largely Cl 2p;
• The 3σ* orbital is antibonding, and primarily of H 1s character;
Figure 10.4.6 shows the analogous MO diagram and photoelectron spectrum for $\ce{HCl}$. The spectrum has two bands corresponding to non-bonding 1p (or $1\pi$) molecular orbitals (with negligible vibrational structure) and the 3s bonding molecular orbital (vibrational structure).
The higher energy (more stabilized) core molecular orbitals are not observed since the incident photon energy $h\nu$ is below their ionization energies.
Water
In the simplified valence bond theory perspective of the water molecule, the oxygen atom form four $sp^3$ hybrid orbitals. Two of these are occupied by the two lone pairs on the oxygen atom, while the other two are used for bonding. Within the molecular orbital picture, the electronic configuration of the $\ce{H_2O^{+}}$ molecule is $(1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^2 (1b_1)^2$ where the symbols $a_1$, $b_2$ and $b_1$ are orbital labels based on molecular symmetry that will be discussed later (Figure 10.4.7 ). Within Koopmans' theorem:
• The energy of the 1b1 HOMO corresponds to the ionization energy to form the $\ce{H_2O^{+}}$ ion in its ground state $(1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^2 (1b_1)^1$.
• The energy of the second-highest molecular orbitals $3a_1$ refers to the ion in the excited state $(1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^1 (1b_1)^2$.
The Hartree–Fock orbital energies (with sign changed) of these orbitals are tabulated below and compared to the experimental ionization energies.
Molecular orbital Hartree–Fock orbital Energies (eV) Experimental Ionization Energies (eV)
2a1 36.7 32.2
1b2 19.5 18.5
3a1 15.9 14.7
1b1 13.8 12.6
As explained above, the deviations between orbital energy and ionization energy is small and due to the effects of orbital relaxation as well as differences in electron correlation energy between the molecular and the various ionized states.
The molecular orbital perspective has the lone pair in different orbitals (one in a non-bonding orbital ($1b_1$ and one in the bonding orbitals). We tern to the photoelectron spectroscopy to help identify which theory is more accurate (i.e., describes reality better). The photoelectron spectrum of water in Figure 10.4.6 can be interpreted as having three major peaks with some fine structure arises from vibrational energy changes. The light source used in this experiment is not sufficiently energetic to ionize electrons from the lowest lying molecular orbitals.
If water was formed two identical O-H bonds and two lone pairs on the oxygen atom line valence bond theory predicts, then the PES in Figure 10.4.8 would have two (degenerate) peaks, one for the two bonds and one for the two lone pairs. The photoelectron spectrum clearly shows three peaks in the positions expected for the molecular orbitals in Figure 10.4.8 .
If the molecular orbitals in Figure 10.4.7 represent the real electronic structure, how do we view the bonding? These molecular orbitals are delocalized and bare little relationship to the familiar 2-center bonds used in valence bond theory. For example, the $2a_1$ $1b_1$ and $3a_1$ molecular orbitals all have contributions from all three atoms, they are really 3-centered molecular orbitals. The bonds however can be thought of as representing a build up of the total electron density which loosely put is a total of all the orbital contributions. Despite this, we keep the ideas of hybridization and 2-center bonds because they are useful NOT because they represent reality
Summary
A photoelecton spectrum can show the relative energies of occupied molecular orbitals by ionization. (i.e. ejection of an electron). A photoelectron spectrum can also be used to determine energy spacing between vibrational levels of a given electronic state. Each orbital energy band has a structure showing ionization to different vibrational levels.
Contributors and Attributions
• Roger Nix (Queen Mary, University of London)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/10%3A_Bonding_in_Polyatomic_Molecules/10.04%3A_Photoelectron_Spectroscopy.txt
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Learning Objectives
• Demonstrate how Hückel's theory approximates the full molecular orbital picture of molecules by treating the $\sigma$-bonding and $\pi$-bonding networks independently.
Molecular orbital theory has been very successfully applied to large conjugated systems, especially those containing chains of carbon atoms with alternating single and double bonds. An approximation introduced by Hückel in 1931 considers only the delocalized p electrons moving in a framework of $\pi$-bonds. This is, in fact, a more sophisticated version of a free-electron model.
The simplest hydrocarbon to consider that exhibits $\pi$ bonding is ethylene (ethene), which is made up of four hydrogen atoms and two carbon atoms. Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are sp2 hybridized, which means that a singly occupied sp2 orbital on one carbon overlaps with a singly occupied s orbital on each H and a singly occupied sp2 lobe on the other C. Thus each carbon forms a set of three $\sigma$ bonds: two C–H (sp2 + s) and one C–C (sp2 + sp2) (part (a) of Figure 10.5.1 ).
The Hückel approximation is used to determine the energies and shapes of the $\pi$ molecular orbitals in conjugated systems. Within the Hückel approximation, the covalent bonding in these hydrocarbones can be separated into two independent "frameworks": the $\sigma$-bonding framework and the the $\sigma$-bonding framework. The wavefunctions used to describe the bonding orbitals in each framework results from different combinations of atomic orbitals. The method limits itself to addressing conjugated hydrocarbons and specifically only $\pi$ electron molecular orbitals are included because these determine the general properties of these molecules; the sigma electrons are ignored. This is referred to as sigma-pi separability and is justified by the orthogonality of $\sigma$ and $\pi$ orbitals in planar molecules. For this reason, the Hückel method is limited to planar systems. Hückel approximation assumes that the electrons in the $\pi$ bonds “feel” an electrostatic potential due to the entire $\sigma$-bonding framework in the molecule (i.e. it focuses only on the formation of $\pi$ bonds, given that the $\sigma$ bonding framework has already been formed).
Conjugated Systems
A conjugated system has a region of overlapping p-orbitals, bridging the interjacent single bonds, that allow a delocalization of $\pi$ electrons across all the adjacent aligned p-orbitals. These $\pi$ electrons do not belong to a single bond or atom, but rather to a group of atoms.
Ethylene
Before considering the Hückel treatment for ethylene, it is beneficial to review the general bonding picture of the molecule. Bonding in ethylene involves the $sp^2$ hybridization of the $2s$, $2p_x$, and $2p_y$ atomic orbitals on each carbon atom; leaving the $2p_z$ orbitals untouched (Figure 10.5.2 ).
The use of hybrid orbitals in the molecular orbital approach describe here is merely a convenience and not invoking valence bond theory (directly). An identical description can be extracted using exclusively atomic orbitals on carbon, but the interpretation of the resulting wavefunctions is less intuitive. For example, the ith molecular orbital can be described via hybrid orbitals
$| \psi_1\rangle = c_1 | sp^2_1 \rangle + c_2 | 1s_a \rangle \nonumber \nonumber$
or via atomic orbitals.
$| \psi_1\rangle = a_1 | 2s \rangle + a_1 | 2p_x \rangle + a_1 | 2p_y \rangle + a_4| 1s_a \rangle \nonumber \nonumber$
where $\{a_i\}$ and $\{c_i\}$ are coefficients of the expansion. Either describe will work and both are identical approaches since
$| sp^2_1 \rangle = b_1 | 2s \rangle + b_1 | 2p_x \rangle + b_1 | 2p_y \rangle \nonumber \nonumber$
where $\{c_i\}$ are coefficients describing the hybridized orbital.
The bonding occurs via the mixing of the electrons in the $sp^2$ hybrid orbitals on carbon and the electrons in the $1s$ atomic orbitals of the four hydrogen atoms (Figure 10.5.1 ; left) resulting in the $\sigma$-bonding framework. The $\pi$-bonding framework results from the unhybridized $2p_z$ orbitals (Figure 10.5.2 ; right). The independence of these two frameworks is demonstrated in the resulting molecular orbital diagram in Figure 10.5.3 ; Hückel theory is concerned only with describing the molecular orbitals and energies of the $\pi$ bonding framework.
Hückel treatment is concerned only with describing the molecular orbitals and energies of the $\pi$ bonding framework.
Since Hückel theory is a special consideration of molecular orbital theory, the molecular orbitals $| \psi_i \rangle$ can be described as a linear combination of the $2p_z$ atomic orbitals $\phi$ at carbon with their corresponding $\{c_i\}$ coefficients:
$| \psi_i \rangle =c_1 | \phi_{1} \rangle +c_2 | \phi_2 \rangle \label{LCAO}$
This equation is substituted in the Schrödinger equation:
$\hat{H} | \psi_i \rangle =E_i | \psi_i \rangle \nonumber$
with $\hat{H}$ the Hamiltonian and $E_i$ the energy corresponding to the molecular orbital to give:
$\hat{H} c_{1} | \phi _{1} \rangle +\hat{H} c_{2} | \phi _{2} \rangle =E c_{1} | \phi _{1} \rangle +E c_{2} | \phi _{2} \rangle \label{SEq}$
If Equation $\ref{SEq}$ is multiplied by $\langle \phi _{1}|$ (and integrated), then
$c_1(H_{11} - ES_{11}) + c_2(H_{12} - ES_{12}) = 0 \label{Eq1}$
where $H_{ij}$ are the Hamiltonian matrix elements (see note below)
$H_{ij} = \langle \phi_i | \hat{H} | \phi_j \rangle = \int \phi _{i}H\phi _{j}\mathrm {d} v\nonumber$
and $S_{ij}$ are the overlap integrals.
$S_{ij}= \langle \phi_i | \phi_j \rangle = \int \phi _{i}\phi _{j}\mathrm {d} v\nonumber$
If Equation $\ref{SEq}$ is multiplied by $\langle \phi _{2} |$ (and integrated), then
$c_1(H_{21} - ES_{21}) + c_2(H_{22} - ES_{22}) = 0 \label{Eq2}$
Both Equations $\ref{Eq1}$ and $\ref{Eq2}$ can better represented in matrix notation,
${\begin{bmatrix}c_{1}(H_{11}-ES_{11})+c_{2}(H_{12}-ES_{12})\c_{1}(H_{21}-ES_{21})+c_{2}(H_{22}-ES_{22})\\end{bmatrix}}=0\nonumber$
or more simply as a product of matrices.
$\begin{bmatrix} H_{11} - ES_{11} & H_{12} - ES_{12} \ H_{21} - ES_{21} & H_{22} - ES_{22} \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \label{master}$
All diagonal Hamiltonian integrals $H_{ii}$ are called Coulomb integrals and those of type $H_{ij}$ are called resonance integrals. Both integrals are negative and the resonance integrals determines the strength of the bonding interactions. The equations described by Equation $\ref{master}$ are called the secular equations and will also have the trivial solution of
$c_1 = c_2 = 0 \nonumber$
Within linear algebra, the secular equations in Equation $\ref{master}$ will also have a non-trivial solution, if and only if, the secular determinant is zero
$\left| \begin{array} {cc} H_{11} - ES_{11} & H_{12} - ES_{12} \ H_{21} - ES_{21} & H_{22} - ES_{22} \ \end{array}\right| = 0 \label{SecDet}$
or in shorthand notation
$\text{det}(H -ES) =0\nonumber$
Everything in Equation $\ref{SecDet}$ is a known number except $E$. Since the secular determinant for ethylene is a $2 \times 2$ matrix, finding $E$, requires solving a quadratic equation (after expanding the determinant)
$( H_{11} - ES_{11} ) ( H_{22} - ES_{22} ) - ( H_{21} - ES_{21} )( H_{12} - ES_{12} ) = 0\nonumber$
There will be two values of $E$ which satisfy this equation and they are the molecular orbital energies. For ethylene, one will be the bonding energy and the other the antibonding energy for the $\pi$-orbitals formed by the combination of the two carbon $2p_z$ orbitals (Equation $\ref{LCAO}$). However, if more than two $| \phi \rangle$ atomic orbitals were used, e.g., in a bigger molecule, then more energies would be estimated by solving the secular determinant.
Solving the secular determinant is simplified within Hückel method via the following four assumptions:
1. All overlap integrals $S_{ij}$ are set equal to zero. This is quite reasonable since the $\pi-$ orbitals are directed perpendicular to the direction of their bonds (Figure 10.5.1 ). This assumption is often call neglect of differential overlap (NDO).
2. All resonance integrals $H_{ij}$ between non-neighboring atoms are set equal to zero.
3. All resonance integrals $H_{ij}$ between neighboring atoms are equal and set to $\beta$.
4. All coulomb integrals $H_{ii}$ are set equal to $\alpha$.
These assumptions are mathematically expressed as
$H_{11}=H_{22}=\alpha\nonumber$
$H_{12}=H_{21}=\beta\nonumber$
Assumptions 1 means that the overlap integral between the two atomic orbitals is 0
$S_{11}=S_{22}=1\nonumber$
$S_{12}=S_{21}=0\nonumber$
Matrix Representation of the Hamiltonian
The Coulomb integrals
$H_{ii}= \langle \phi _i|H| \phi _i \rangle \nonumber \nonumber$
and resonance integrals.
$H_{ij}= \langle \phi _i|H| \phi _j \rangle \,\,\, (i \neq i) \nonumber \nonumber$
are often described within the matrix representation of the Hamiltonian (specifically within the $| \phi \rangle$ basis):
$\hat{H} = \begin{bmatrix} H_{11} & H_{12} \ H_{21} & H_{22} \end {bmatrix} \nonumber \nonumber$
or within the Hückel assumptions
$\hat{H} = \begin{bmatrix} \alpha & \beta \ \beta & \alpha \end {bmatrix} \nonumber \nonumber$
The Hückel assumptions reduces Equation $\ref{master}$ in two homogeneous equations:
$\begin{bmatrix} \alpha - E & \beta \ \beta & \alpha - E \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \label{Eq12}$
if Equation $\ref{Eq12}$ is divided by $\beta$:
$\begin{bmatrix} \dfrac{\alpha - E}{\beta} & 1 \ 1 & \dfrac{\alpha - E}{\beta} \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0\nonumber$
and then a new variable $x$ is defined
$x = \dfrac {\alpha -E}{\beta} \label{new}$
then Equation $\ref{Eq12}$ simplifies to
$\begin{bmatrix} x & 1 \ 1 & x \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \label{seceq}$
The trivial solution gives both wavefunction coefficients equal to zero and the other (non-trivial) solution is determined by solving the secular determinant
$\begin{vmatrix}x&1\1&x\\end{vmatrix}=0\nonumber$
which when expanded is $x^{2}-1=0$ so $x=\pm 1$.
Knowing that $E=\alpha -x\beta$ from Equation $\ref{new}$, the energy levels can be found to be
$E=\alpha -\pm 1\times \beta \nonumber$
or
$E=\alpha \mp \beta \nonumber$
Since $\beta$ is negative, the two energies are ordered (Figure 10.5.4 )
• For $\pi_1$: $E_1 =\alpha + \beta$
• For $\pi_2$: $E_2 =\alpha - \beta$
To extract the coefficients attributed to these energies, the corresponding $x$ values can be substituted back into the Secular Equations (Equation $\ref{seceq}$). For the lower energy state ($x=-1$)
$\begin{bmatrix} -1 & 1 \ 1 & -1 \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \nonumber$
This gives $c_1=c_2$ and the molecular orbitals attributed to this energy is then (based off of Equation $\ref{LCAO}$):
$|\psi_1 \rangle = N_1 (\phi_1 \rangle + | \phi_2 \rangle ) \label{HOMO}$
where $N_1$ is the normalization constant for this molecular orbital; this is the bonding molecular orbital.
For the higher energy molecular orbital ($x=-1$) and then
$\begin{bmatrix} 1 & 1 \ 1 & 1 \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \nonumber$
This gives $c_1=-c_2$ and the molecular orbitals attributed to this energy is then (based off of Equation $\ref{LCAO}$):
$\psi_2 \rangle = N_2 (\phi_1 \rangle - | \phi_2 \rangle ) \label{LUMO}$
where $N_2$ is the normalization constant for this molecular orbital; this is the anti-bonding molecular orbital.
The normalization constants for both molecular orbitals can obtained via the standard normalization approach (i.e., $\langle \psi_i | \psi_i \rangle =1$) to obtain
$N_1 = N_2 = \dfrac{1}{\sqrt{2}}\nonumber$
These molecular orbitals form the $\pi$-bonding framework and since each carbon contributes one electron to this framework, only the lowest molecular orbital ($| \psi_1 \rangle$) is occupied (Figure 10.5.5 ) in the ground state. The corresponding electron configuration is then $\pi_1^2$.
HOMO and LUMO are acronyms for highest occupied molecular orbital and lowest unoccupied molecular orbital, respectively and are often referred to as frontier orbitals. The energy difference between the HOMO and LUMO is termed the HOMO–LUMO gap.
$\psi ={\$dfrac {1}{\sqrt {2}}}(\phi _{1}+\phi _{2})\,} The 3-D calculated $\pi$ molecular orbitals are shown in Figure 10.5.6 .
Limitations of Hückel Theory
Hückel theory was developed in the 1930's when computers were unavailable and a simple mathematical approaches were very important for understanding experiment. Although the assumptions in Hückel theory are drastic they enabled the early calculations of molecular orbitals to be performed with mechanical calculators or by hand. Hückel Theory can be extended to address other types of atoms in conjugated molecules (e.g., nitrogen and oxygen). Moreover, it can be extended to also treat $\sigma$ orbitals and this "Extended Hückel Theory" is still used today. Despite the utility of Hückel Theory, it is highly qualitative and we should remember the limitations of Hückel Theory:
• Hückel Theory is very approximate
• Hückel Theory cannot calculate energies accurately (electron-electron repulsion is not calculated)
• Hückel Theory typically overestimates predicted dipole moments
Hückel Theory is best used to provide simplified models for understanding chemistry and for a detailed understanding modern ab initio molecular methods discussed in Chapter 11 are needed.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/10%3A_Bonding_in_Polyatomic_Molecules/10.05%3A_The_%28pi%29-Electron_Approximation_of_Conjugation.txt
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Learning Objectives
• Apply Hückel theory to an extended $\pi$-bonding network
• Identify the origin of delocalization energy from Hückel theory and relate it to resonances structures in valence bond theory
1,3-Butadiene is a simple conjugated diene with the formula $\ce{C_4H_6}$ and can be viewed structurally as two vinyl groups ($\ce{CH_2=CH_2}$) joined together with a single bond. Butadiene can occupy either a cis or trans conformers and at room temperature, 96% of butadiene exists as the trans conformer, which is 2.3 kcal/mole more stable than the cis structure.
For the simple application of applying Hückel theory for understanding the electronic structure of butadiene, we will ignore the energetic differences between the two conformers. As discussed previously, the molecular orbitals are linear combination of the four $|p \rangle$ atomic orbitals on the carbon atoms that are not participating in the $\sigma$ bonding network:
$|\psi_i \rangle = \sum_j^4 c_{ij} |p_{i} \rangle \nonumber$
or explicitly
$|\psi_i \rangle =c_{i1} |p_1 \rangle+ c_ {i2} | p_2 \rangle + c_ {i3} | p_3 \rangle + c_ {i4} | p_4 \rangle \label{general}$
for the ith molecular orbital $|\psi_i \rangle$. The secular equations that need to be solved are
$\begin{bmatrix} H_{11} - ES_{11} & H_{12} - ES_{12} & H_{13} - ES_{13} & H_{14} - ES_{14} \ H_{12} - ES_{12} & H_{22} - ES_{22} & H_{23} - ES_{23} & H_{24} - ES_{24} \ H_{13} - ES_{13} & H_{23} - ES_{23} & H_{33} - ES_{33} & H_{34} - ES_{34} \ H_{14} - ES_{14} & H_{24} - ES_{24} & H_{34} - ES_{34} & H_{44} - ES_{44} \ \end{bmatrix} \times\begin{bmatrix} c_1 \ c_2 \ c_3 \ c_4 \ \end{bmatrix}= 0 \label{complete}$
If the standard Hückel theory approximations were used
$H_{ii}- ES_{ii} = \alpha\nonumber$
and
$H_{ij}- ES_{ij} = \beta\nonumber$
when $i=j\pm 1$, otherwise
$H_{ij}- ES_{ij} = 0\nonumber$
then the secular equations for butadiene in Equation $\ref{complete}$ become
$\begin{bmatrix} \alpha - E & \beta & 0 & 0 \ \beta & \alpha - E & \beta & 0 \ 0 & \beta & \alpha - E & \beta \ 0 & 0 & \beta & \alpha - E \ \end{bmatrix} \times\begin{bmatrix} c_1 \ c_2 \ c_3 \ c_4 \ \end{bmatrix}= 0 \label{Huckel1}$
Solving Equation $\ref{Huckel1}$ for $\{c_i\}$ coefficients and energy secular equation requires extracting the roots of the secular determinant:
$\left|\begin{array}{cccc}\alpha-E&\beta&0&0\\beta&\alpha-E&\beta&0\0&\beta&\alpha-E&\beta\0&0&\beta&\alpha-E\end{array}\right|=0\label{24}$
If both sides of Equation $\ref{24}$ were divided by $\beta^{4}$ and a new variable $x$ is defined
$x=\dfrac{\alpha-E}{\beta}\label{25}$
then Equation $\ref{24}$ simplifies further to
$\left|\begin{array}{cccc}x&1&0&0\1&x&1&0\0&1&x&1\0&0&1&x\end{array}\right|=0\label{26}$
This is essentially the connection matrix for the butadiene molecule. Each pair of connected atoms is represented by 1, each non-connected pair by 0 and each diagonal element by $x$. Expansion of the determinant in Equation $\ref{26}$ gives the 4th order polynomial equation
$x^{4}-3x^{2}+1=0 \label{27}$
While solving 4th order equations typically require numerical estimation, Equation $\ref{27}$ can be further simplified by recognizing that it is a quadratic equation in terms of $x^{2}$. Therefore, the roots are
$x^{2}= \dfrac{3\pm\sqrt{5}}{2} \nonumber$
or $x=\pm\; 0.618$ and $x= \pm\; 1.618$. Since $\alpha$ and $\beta$ are negative, these molecular orbital energies can ordered in terms of energy (from lowest to highest):
$E_1=\alpha+1.618\beta \label{E1}$
$E_2=\alpha+0.618\beta \label{E2}$
$E_3=\alpha-0.618\beta \label{E3}$
$E_4=\alpha-1.618\beta \label{E4}$
This sequence of energies is displayed in the energy diagram of Figure 10.6.1 .
Each p atomic orbital of carbon contributes a single electron to the $\pi$ manifold, so the ground-state occupation of the resulting four $\pi$ electrons have a $\pi_1^{2}\pi_2^{2}$ configuration (Figure 10.6.1 ). The the total $\pi$-electron energy is then determined by adding up the energies in Equations $\ref{E1}$-$\ref{E4}$ and scaling by their occupations to get
\begin{align} E_{\pi} (\text{butadiene}) &= 2 \times E_1 + 2 \times E_2 + 0 \times E_3 + 0 \times E_4 \nonumber \[4pt] &=2(\alpha+1.618\beta)+2(\alpha+0.618\beta) \nonumber \[4pt] &=4\alpha + 4.472\beta \label{29} \end{align}
If the bonding of butadiene were described only as two localized double bond as in its dominant valence-bond structure (Figure 10.6.1 ), then its $\pi$-electron energy would be given by twice the $E_{\pi}$ predicted for the ethlyene molecule:
\begin{align} E_{\pi} (\text{butadiene}) &= 2 \times E_{\pi} (\text{ethylene}) \nonumber \[4pt] &=2 \times 2(\alpha+\beta) \nonumber \[4pt] &= 4\alpha + 4\beta \label{ 30}\end{align}
Comparing Equation $\ref{29}$ with Equation $\ref{29}$, the total $\pi$ energy of butadiene lies lower than the total $\pi$ energy of two double bonds by $0.48\beta$ (the $\sigma$ bond does not contribute). This difference is known as the delocalization energy; a typical estimate of $\beta$ is around -75 kJ/mol, which results in a delocalization energy for butadiene of -35 kJ/mol.
The delocalization energy is the extra stabilization resulting from the electrons extending over the whole molecule.
Delocalization Energy in Valence Bond Theory
Delocalization energy is intrinsic to molecular orbital theory, since it results from breaking the two-center bond concept with the molecular orbitals that spread over more that just one pair of atoms. However, within the two-center theory of valence bond theory, the delocalization energy results from a stabilization energy attributed to resonance. Several conventional valence bond resonance structures that can be written for 1,3-butadiene, four of which are shown in Figure 10.6.2 . However, while structure $2a$ dominates, the other resonance structures also contribute to describing the total molecule and hence predict a corresponding stabilization energy akin to the delocalization energy in molecular orbital theory.
In general, the true description of the bonding within the valence bond theory is a superposition of resonance structures with amplitudes that are determined via a variational optimization to find the lowest possible energy for the valence bond wavefunctions.
The solving the secular equations (Euqation \ref{complete}) gives the $\{c_{ij}\}$ coefficients for the molecular orbitals in Equation $\ref{general}$ (not demonstrated):
$|\psi_1 \rangle =0.37 |p_1 \rangle + 0.60 | p_2 \rangle + 0.60 | p_3 \rangle + 0.37 | p_4 \rangle \label{MO1}$
$|\psi_3 \rangle =0.60 |p_1 \rangle + 0.37 | p_2 \rangle -0.37 | p_3 \rangle - 0.60 | p_4 \rangle \label{MO2}$
$|\psi_3 \rangle =0.60 |p_1 \rangle - 0.37 | p_2 \rangle -0.37 | p_3 \rangle + 0.60 | p_4 \rangle \label{MO3}$
$|\psi_4 \rangle =0.37 |p_1 \rangle - 0.60 | p_2 \rangle + 0.60 | p_3 \rangle - 0.37 | p_4 \rangle \label{MO4}$
These are depicted in Figure 10.6.3 .
Note the correlation of the energy of the $\pi$ molecular orbitals of butadiene to the number of nodes in the wavefunction; this is the general trend observed in previous systems like the particle in the box and atomic orbitals. The four 3-D calculated molecular orbitals are contrasted in Figure 10.6.1 .
Exercise 10.6.1
What would Hückel theory predict for the energy levels and $\pi$ molecular orbitals of cis-butadiene? Do you believe this approach adequately describes the energy levels of the cis-trans isomerization reaction of butadiene?
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Learning Objectives
• Apply Hückel theory to describing the pi bonding in cyclical conjugated system
• Identify the origin of aromaticity within Hückel theory to describe extra stabilization in certain cyclical conjugated systems
The previous sections addressed the $\pi$ orbitals of linear conjugated system. Here we address conjugated systems of cyclic conjugated hydrocarbons with the general formula of $C_nH_n$ where $n$ is the number of carbon atoms in the ring. The molecule from this important class of organic molecule that you are most familiar with is benzene ($C_6H_6$) with $n=6$, although many other molecules exist like cyclobutadiene ($C_4H_4$) with $n=4$ (Figure 10.7.1 ).
Structure of Benzene
The structure of benzene is an interesting historical topic. In 1865, the German chemist Friedrich August Kekulé published a paper suggesting that the structure of benzene contained a ring of six carbon atoms with alternating single and double bonds. Within this argument, two resonance structures can be formulated.
However, X-ray diffraction shows that all six carbon-carbon bonds in benzene are of the same length, at 140 pm. The C–C bond lengths are greater than a double bond (135 pm), but shorter than a typical single bond (147 pm). This means that neither structures Figure 10.7.2 are correct and the true 'structure' of benzene is a mixture of the two. As discussed previously, that such a valence bond perspective results in a delocalization energy within a molecular orbital approach.
Aromatic systems provide the most significant applications of Hückel theory. For benzene, we find the secular determinant
$\left|\begin{array}{cccccc}x&1&0&0&0&1\1&x&1&0&0&0\0&1&x&1&0&0\0&0&1&x&1&0\0&0&0&1&x&1\1&0&0&0&1&x\end{array}\right|=0\label{31}$
with the six roots $x=\pm2,\pm1,\pm1$. This corresponds to the following energies (ordered from most stable to least since $\beta < 0$):
• $E_1 = α + 2β$
• $E_2 = α + β$
• $E_3 = α + β$
• $E_4 = α − β$
• $E_5 = α − β$
• $E_6 = α − 2β$
The two pairs of $E=\alpha\pm\beta$ energy levels are two-fold degenerate (Figure 10.7.3 ).
The resulting wavefunctions are below (expanded in terms of carbon $| 2p\rangle$ atomic orbitals).
\begin{align} | \psi_1 \rangle &= \dfrac{1}{\sqrt{6}} \left[ | 2p_{z1} \rangle+ | 2p_{z2} \rangle + | 2p_{z3} \rangle + | 2p_{z4} \rangle + | 2p_{z5} \rangle + | 2p_{z6} \rangle \right] \ | \psi_2 \rangle &= \dfrac{1}{\sqrt{4}} \left[ | 2p_{z2} \rangle + | 2p_{z3} \rangle - | 2p_{z4} \rangle - | 2p_{z5} \rangle \right] \ | \psi_3 \rangle &= \dfrac{1}{\sqrt{3}} \left[ | 2p_{z1} \rangle + \dfrac{1}{2}| 2p_{z2} \rangle - \dfrac{1}{2} | 2p_{z3} \rangle - | 2p_{z4} \rangle - \dfrac{1}{2} | 2p_{z5} \rangle + \dfrac{1}{2} | 2p_{z6} \rangle \right] \ | \psi_4 \rangle &= \dfrac{1}{\sqrt{4}} \left[ | 2p_{z2} \rangle - | 2p_{z3} \rangle + | 2p_{z4} \rangle - | 2p_{z5} \rangle \right] \ | \psi_5 \rangle &= \dfrac{1}{\sqrt{3}} \left[ | 2p_{z1} \rangle - \dfrac{1}{2}| 2p_{z2} \rangle - \dfrac{1}{2} | 2p_{z3} \rangle + | 2p_{z4} \rangle - \dfrac{1}{2} | 2p_{z5} \rangle - \dfrac{1}{2} | 2p_{z6} \rangle \right] \ | \psi_6 \rangle &= \dfrac{1}{\sqrt{6}} \left[ | 2p_{z1} \rangle- | 2p_{z2} \rangle + | 2p_{z3} \rangle - | 2p_{z4} \rangle + | 2p_{z5} \rangle - | 2p_{z6} \rangle \right] \end{align} \nonumber
Each of the carbons in benzene contributes one electron to the $\pi$-bonding framework (Figure 10.7.3 ). This means that all bonding molecular orbitals are fully occupied and benzene then has an electron configuration of $\pi_1^2 \pi_2^2 \pi_3^2$. With the three lowest molecular orbitals occupied, the total $\pi$-bonding energy is
$E_{tot} (benzene)=2(\alpha+2\beta)+4(\alpha+\beta)=6\alpha+8\beta\label{32}$
Since the energy of a localized double bond is $2(\alpha+\beta)$, as determined from the analysis of ethylene, the delocalization energy of benzene is
$\Delta E = E_{tot} (benzene) - 3 E_{tot} (ethylene) = (6\alpha+8\beta ) - 3 \times 2(\alpha+\beta) = 2\beta \nonumber$
The experimental thermochemical value is -152 kJ mol-1.
Aromaticity
In general, cyclic polyenes are only closed shell (i.e., each electron paired up) and extra stable for with (4n+2) $\pi$ electrons (n=0,1,2…). These special molecules have the highest delocalization energies and are said to be “aromatic”. For benzene this is $2\beta$ (Equation $\ref{32}$), which is the energy by which the delocalized $\pi$ electrons in benzene are more stable than those in three isolated double bonds.
Hückel's Rule
A stable, closed-shell conjugated cyclic structure is obtained for molecules with (4n+ 2) electrons with n=2, 6, 10, .... electrons.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 11.9 kJ mol-1. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 23.9 kJ mol-1 on complete hydrogenation, and 1,3,5-cyclohexatriene to release 35.9 kJ mol-1. These heats of hydrogenation $\Delta H_{hyd}$ reflect the relative thermodynamic stability of the compounds (Figure 10.7.4 ). In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 8.1 kJ mol-1, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 15 kJ mol-1 more stable than expected. This additonal stability is a characteristic of all aromatic compounds.
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These are homework exercises to accompany Chapter 10 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Q10.8
Show that the four $sp^3$ orbitals are orthonormal.
S10.8
This means showing that each pair of $|sp^3 \rangle$ hybrid orbitals meets the criteria: $\langle sp^3_i | sp^3_j \rangle = \delta_{ij}$
Designating the four sp3 orbitals as:
$| 1 \rangle = \dfrac{1}{2}( | s \rangle + | p_x \rangle + | p_y \rangle + | p_z \rangle$)
$| 2 \rangle = \dfrac{1}{2}( | s \rangle - | p_x \rangle - | p_y \rangle + | p_z \rangle$)
$| 3 \rangle = \dfrac{1}{2}( | s \rangle + | p_x \rangle - | p_y \rangle - | p_z \rangle$)
$| 4 \rangle = \dfrac{1}{2}( | s \rangle - | p_x \rangle + | p_y \rangle - | p_z \rangle$)
Normality:
$\langle 1 | 1 \rangle = \left(\dfrac{1}{2}( \langle s| + \langle p_x | + \langle p_y| + \langle p_z|)\right) \left(\dfrac{1}{2}( |s \rangle + | p_x \rangle + |p_y \rangle + | p_z \rangle) \right)$
$= \dfrac{1}{4} \left(\langle s|s \rangle + \langle s| p_x \rangle + \langle s| p_y \rangle + \langle s| p_z \rangle + \langle p_x| s \rangle + \langle p_x| p_x \rangle + \langle p_x| p_y \rangle + \langle p_x | p_z \rangle + \langle p_y| s \rangle + \langle p_y | p_x \rangle + \langle p_y | p_y \rangle + \langle p_y| p_z \rangle + \langle p_z| s \rangle + \langle p_z | p_x \rangle + \langle p_z | p_y \rangle + \langle p_z| p_z \rangle \right)$
$= \dfrac{1}{4}( 1+0+0+0+0+1+0+0+0+0+1+0+0+0+0+1) = 1$
$\langle 2 | 2 \rangle = \left(\dfrac{1}{2}( \langle s| - \langle p_x | - \langle p_y| + \langle p_z|)\right) \left(\dfrac{1}{2}( |s \rangle - | p_x \rangle - |p_y \rangle + | p_z \rangle)\right)$
$= \dfrac{1}{4} \left(\langle s|s \rangle - \langle s| p_x \rangle - \langle s| p_y \rangle + \langle s| p_z \rangle - \langle p_x| s \rangle + \langle p_x| p_x \rangle + \langle p_x| p_y \rangle - \langle p_x | p_z \rangle - \langle p_y| s \rangle + \langle p_y | p_x \rangle + \langle p_y | p_y \rangle - \langle p_y| p_z \rangle + \langle p_z| s \rangle - \langle p_z | p_x \rangle - \langle p_z | p_y \rangle + \langle p_z| p_z \rangle \right)$
$= \dfrac{1}{4}( 1+0+0+0+0+1+0+0+0+0+1+0+0+0+0+1) = 1$
$\langle 3 | 3 \rangle = \left(\dfrac{1}{2}( \langle s| + \langle p_x | - \langle p_y| - \langle p_z|) \right) \left(\dfrac{1}{2}( |s \rangle + | p_x \rangle - |p_y \rangle - | p_z \rangle) \right)$
$= \dfrac{1}{4} \left(\langle s|s \rangle + \langle s| p_x \rangle - \langle s| p_y \rangle - \langle s| p_z \rangle + \langle p_x| s \rangle + \langle p_x| p_x \rangle - \langle p_x| p_y \rangle - \langle p_x | p_z \rangle - \langle p_y| s \rangle - \langle p_y | p_x \rangle + \langle p_y | p_y \rangle + \langle p_y| p_z \rangle - \langle p_z| s \rangle - \langle p_z | p_x \rangle + \langle p_z | p_y \rangle + \langle p_z| p_z \rangle \right)$
$= \dfrac{1}{4}( 1+0+0+0+0+1+0+0+0+0+1+0+0+0+0+1) = 1$
$\langle 4 | 4 \rangle = \left(\dfrac{1}{2}( \langle s| - \langle p_x | + \langle p_y| - \langle p_z|)\right) \left(\dfrac{1}{2}( |s \rangle - | p_x \rangle + |p_y \rangle - | p_z \rangle)\right)$
$= \dfrac{1}{4}(\langle s|s \rangle - \langle s| p_x \rangle + \langle s| p_y \rangle - \langle s| p_z \rangle - \langle p_x| s \rangle + \langle p_x| p_x \rangle - \langle p_x| p_y \rangle + \langle p_x | p_z \rangle + \langle p_y| s \rangle - \langle p_y | p_x \rangle + \langle p_y | p_y \rangle - \langle p_y| p_z \rangle - \langle p_z| s \rangle + \langle p_z | p_x \rangle - \langle p_z | p_y \rangle + \langle p_z| p_z \rangle)$
$= \dfrac{1}{4}( 1+0+0+0+0+1+0+0+0+0+1+0+0+0+0+1) = 1$
Orthogonality:
For the longer wavefunctions, only surviving terms in the dot product are included:
$\langle 1 | 2 \rangle = \langle 2 | 1 \rangle = \left(\dfrac{1}{2}( \langle s| - \langle p_x | - \langle p_y| + \langle p_z|)\right) \left(\dfrac{1}{2}( |s \rangle + | p_x \rangle + |p_y \rangle + | p_z \rangle)\right)$
$= \dfrac{1}{4}(\langle s|s \rangle - \langle p_x| p_x \rangle - \langle p_y | p_y \rangle + \langle p_z| p_z \rangle) = \dfrac{1}{4}( 1-1-1+1) = 0$
$\langle 1 | 3 \rangle = \langle 3 | 1 \rangle = \left(\dfrac{1}{2}( \langle s| + \langle p_x | - \langle p_y| - \langle p_z|)\right) \left(\dfrac{1}{2}( |s \rangle + | p_x \rangle + |p_y \rangle + | p_z \rangle)\right)$
$= \dfrac{1}{4}(\langle s|s \rangle + \langle p_x| p_x \rangle - \langle p_y | p_y \rangle - \langle p_z| p_z \rangle) = \dfrac{1}{4}( 1+1-1-1) = 0$
$\langle 1 | 4 \rangle = \langle 4 | 1 \rangle = \left(\dfrac{1}{2}( \langle s| - \langle p_x | + \langle p_y| - \langle p_z|)\right) \left(\dfrac{1}{2}( |s \rangle + | p_x \rangle + |p_y \rangle + | p_z \rangle)\right)$
$= \dfrac{1}{4}(\langle s|s \rangle - \langle p_x| p_x \rangle + \langle p_y | p_y \rangle - \langle p_z| p_z \rangle) = \dfrac{1}{4}( 1-1+1-1) = 0$
$\langle 2 | 3 \rangle = \langle 3 | 2 \rangle = \left(\dfrac{1}{2}( \langle s| + \langle p_x | - \langle p_y| - \langle p_z|)\right) \left(\dfrac{1}{2}( |s \rangle - | p_x \rangle - |p_y \rangle + | p_z \rangle)\right)$
$= \dfrac{1}{4}(\langle s|s \rangle - \langle p_x| p_x \rangle + \langle p_y | p_y \rangle - \langle p_z| p_z \rangle) = \dfrac{1}{4}( 1-1+1-1) = 0$
$\langle 2 | 4 \rangle = \langle 4 | 2 \rangle = \left(\dfrac{1}{2}( \langle s| - \langle p_x | + \langle p_y| - \langle p_z|) \right) \left(\dfrac{1}{2}( |s \rangle - | p_x \rangle - |p_y \rangle + | p_z \rangle)\right)$
$= \dfrac{1}{4}(\langle s|s \rangle + \langle p_x| p_x \rangle - \langle p_y | p_y \rangle - \langle p_z| p_z \rangle) = \dfrac{1}{4}( 1+1-1-1) = 0$
$\langle 3 | 4 \rangle = \langle 4 | 3 \rangle = (\dfrac{1}{2}( \langle s| - \langle p_x | + \langle p_y| - \langle p_z|)) (\dfrac{1}{2}( |s \rangle + | p_x \rangle - |p_y \rangle - | p_z \rangle))$
$= \dfrac{1}{4}(\langle s|s \rangle - \langle p_x| p_x \rangle - \langle p_y | p_y \rangle + \langle p_z| p_z \rangle) = \dfrac{1}{4}( 1-1-1+1) = 0$
Q10.9
The $sp^{3}d^{2}$ hybrid orbitals are given by:
$\chi_1 (r)=\dfrac{1}{\sqrt{6}}\psi_{3s}(r)-\dfrac{1}{\sqrt{2}}\psi_{3p_x}(r)-\dfrac{1}{\sqrt{12}}\psi_{3d_{z^{2}}}+\dfrac{1}{\sqrt{4}}\psi_{3d_{x^{2}-y^{2}}}$
$\chi_2 (r)=\dfrac{1}{\sqrt{6}}\psi_{3s}(r)+\dfrac{1}{\sqrt{2}}\psi_{3p_x}(r)-\dfrac{1}{\sqrt{12}}\psi_{3d_{z^{2}}}+\dfrac{1}{\sqrt{4}}\psi_{3d_{x^{2}-y^{2}}}$
$\chi_3 (r)=\dfrac{1}{\sqrt{6}}\psi_{3s}(r)-\dfrac{1}{\sqrt{2}}\psi_{3p_y}(r)-\dfrac{1}{\sqrt{12}}\psi_{3d_{z^{2}}}-\dfrac{1}{\sqrt{4}}\psi_{3d_{x^{2}-y^{2}}}$
$\chi_4 (r)=\dfrac{1}{\sqrt{6}}\psi_{3s}(r)+\dfrac{1}{\sqrt{2}}\psi_{3p_y}(r)-\dfrac{1}{\sqrt{12}}\psi_{3d_{z^{2}}}-\dfrac{1}{\sqrt{4}}\psi_{3d_{x^{2}-y^{2}}}$
$\chi_5 (r)=\dfrac{1}{\sqrt{6}}\psi_{3s}(r)-\dfrac{1}{\sqrt{2}}\psi_{3p_z}(r)+\dfrac{1}{\sqrt{12}}\psi_{3d_{z^{2}}}$
$\chi_6 (r)=\dfrac{1}{\sqrt{6}}\psi_{3s}(r)+\dfrac{1}{\sqrt{2}}\psi_{3p_z}(r)+\dfrac{1}{\sqrt{12}}\psi_{3d_{z^{2}}}$
Determine the angles of $SF_{6}$ using the vector approach (dot product formula).
S10.9
The $s$ orbitals are spherical and therefor do not contribute to the directional vectors for this problem. The $d_{z^2}$ orbital only has $z$ directionality and the $d_{x^{2}-y^{2}}$ orbital has equal parts $x$ and $y$ directionality. The equations can be rewritten:
$\psi_{1}=\left( -\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{4}} \right) \textbf{i} + \dfrac{1}{\sqrt{4}}\textbf{j} - \dfrac{1}{\sqrt{12}} \textbf{k} \hspace{1cm} \psi_{2}=\left( \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{4}} \right) \textbf{i} + \dfrac{1}{\sqrt{4}}\textbf{j} - \dfrac{1}{\sqrt{12}} \textbf{k}$
$\psi_{3}=-\dfrac{1}{\sqrt{4}}\textbf{i} + \left( -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{4}} \right) \textbf{j} - \dfrac{1}{\sqrt{12}} \textbf{k} \hspace{1cm} \psi_{4}= -\dfrac{1}{\sqrt{4}}\textbf{i} + \left( \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{4}} \right) \textbf{j} - \dfrac{1}{\sqrt{12}} \textbf{k}$
$\psi_{5}= \left( -\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{12}} \right) \textbf{k}\hspace{1cm} \psi_{6}= \left( \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{12}} \right) \textbf{k}$
For any two $\psi$ the angel can be calculated using:
$\left( \sqrt{A^{2}_{x}+A^{2}_{y}+A^{2}_{z}} \right) \left( \sqrt{B^{2}_{x}+B^{2}_{y}+B^{2}_{z}} \right) cos\theta = A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$
So, choosing $\psi_{5}$ and $\psi_{6}$ we get:
$\left( -\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{12}} \right) \left( \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{12}} \right) cos\theta = \left( -\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{12}} \right) \left( \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{12}} \right)$
$-\dfrac{5}{12} cos \theta = -\dfrac{5}{12}$
$cos\theta = 0$
$\theta = 90^{\circ}$
Q10.10
Given $\xi_1 = \dfrac{1}{\sqrt{4}}2s + \sqrt{\dfrac{3}{4}}2p_z$
$\xi_2 = \dfrac{1}{\sqrt{4}}2s + \sqrt{\dfrac{2}{3}}2p_x - \dfrac{1}{\sqrt{12}}2p_z$
what is the angle between $\xi_1$ and $\xi_2$?
What is the purpose of the square root constants before the orbitals?
S10.10
Since we know that
$(\sqrt{A_x^2 + A_y^2 + A_z^2})(\sqrt{B_x^2 + B_y^2 + B_z^2})\cos{\theta} = A_xB_x + A_yB_y + A_zB_z$
When we plug in our constants in the x, y, and z directions for the two equations we get
$(\sqrt{\dfrac{2}{3} + \dfrac{1}{12}})(\sqrt{\dfrac{3}{4}})\cos{\theta} = - \sqrt{\dfrac{1}{12}}\sqrt{\dfrac{3}{4}}$
isolating the theta term we get
$\theta = \arccos(- \sqrt{\dfrac{1}{12}}/\sqrt{\dfrac{12}{9}}$
yielding
$\theta=109.47^{\circ}$
The square root constants are the necessary normalization constants.
Q10.11
Given that $\psi_{1} = 0.71j + 0.55k$ and $\psi_{2} = -0.71j + 0.55k$ Find the bond angle between $\psi_1$ and $\psi_2$ using the vector approach.
S10.11
Using this equation below:
$\left(\sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}} \right) \left(\sqrt{B_{x}^{2} + B_{y}^{2} + B_{z}^{2}} \right) cos\theta = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}$
use to find the angle between the orbitals,
$(0.71^{2} + 0.55^{2})^{0.5} (0.71^{2} + 0.55^{2})^{0.5} cos\theta = -0.71^{2} + 0.55^{2}$
$cos\theta = -0.25$
$\theta = 104.5^{\circ}$
Q10.12
Assuming a water molecule sits in the yz-plane, show that two bonding hybrid atomic orbitals on the oxygen atom can be expressed as
$\psi_{1} = N[\gamma 2s + (sin[\theta]2p_{y} + cos[\theta]2p_{z}]$
and
$\psi_{2} = N[\gamma 2s - (sin[\theta]2p_{y} + cos[\theta]2p_{z}].$
Additionally, find $\gamma$ assuming the bond angles to be $104.5°$.
S10.12
Because the molecules are in the yz plane, the $p_{x}$ orbital can be neglected because it is completely orthogonal. The hybrid orbitals are a linear combination of the $2s,\:2p_{y},\text{ and }2p_{z}$ orbitals.
$\psi_{bonding} = N[\gamma2_{s} + c_{1}2p_{y} + c_{2}2p_{z}]$
$c_{1}$ and $c_2$ can be found by thinking about how the bonds are oriented. The two bonds both have $cos[\theta]$ character in the z-direction and have $\pm sin[\theta]$ in the y-direction. The two functions can therefore be written as
$\psi_{1} = N[\gamma 2s + (sin[\theta]2p_{y} + cos[\theta]2p_{z})$
and
$\psi_{2} = N[\gamma 2s - (sin[\theta]2p_{y} + cos[\theta]2p_{z})$
when $c_{1}$ and $c_{2}$ are replaced with the sin and cos functions above.
The two functions when integrated will equal zero due to orthogonality.
$\langle\psi_{1}|\psi_{2}\rangle = 0$
$N^{2}(\gamma^{2} + cos^{2}[\theta] - sin^{2}[\theta])$
$\gamma^{2} = sin^{2}[\theta] - cos^{2}[\theta]$
The bond angles in water are 104.5° which can be substituted into $\theta$.
$\gamma = 0.5$
Q10.13
The lone pair wave functions of H2O can be described as:
$\psi_{l1} = 0.54(2s) - 0.44(2pz) + 0.72(2px)$
$\psi_{l2} = 0.54(2s) - 0.44(2pz) - 0.72(2px)$
Confirm the orthonormality of these wave functions.
S10.13
The two wave functions must be normalized and orthogonal to each other.
$\int d\tau \psi_{l1}^{*} \psi_{l1} = 1$
$= (0.54)^2(2s) + (-0.44)^2(2pz) + (0.72)^2(2px) =1$
$\int d\tau \psi_{l2}^{*} \psi_{l2} = 1$
$= (0.54)^2(2s) + (-0.44)^2(2pz) + (-0.72)^2(2px) =1$
$\int d\tau \psi_{l1}^{*} \psi_{l2} = 0$
$= (0.54)^2(2s) + (-0.44)^2(2pz) - (0.72)^2(2px) =0$
The atomic orbitals within the linear combination of the lone pair wave functions are normalized and orthogonal, zeroing out cross products and leaving only the squares of the coefficients.
10.14
Molecular orbitals for a linear $XY_{2}$ molecule can be represented as
Draw a schematic representation for the $3\sigma_{g}, 4\sigma_{g}, 1\pi_{g}, 2\sigma_{g}$ orbitals
Which has the highest energy?
S10.14
$3\sigma_{g}$
NOTE:
$3\sigma_g$
$4\sigma_{g}$
$1\pi_{g}$
$2\sigma_{g}$
$4\sigma_{g}$ has the highest energy
The $4\sigma_g$ moleclar orbital has the highest energy; as expected since it more nodes.
Q10.18
Use the given Walsh diagram to predict the geometry of the following molecules:
1. $H_{2}O$
2. $H_{2}S$
3. $H_{2}Be$
Walsh Diagram of an HAH molecule.Public Domain
S10.18
The Walsh Diagram predicts the geometry of a molecule by assigning its valence electrons to the appropriate energy levels. In general, the lowest energy configuration is preferred.
1. $H_{2}O$ has 8 valence electrons, which corresponds to the 4th highest orbital on the diagram. The bent configuration (90º) is lower in energy in this case.
2. $H_{2}S$ has 8 valence electrons as well, because sulfur and oxygen are in the same periodic group. Therefore, the bent configuration will be favored.
3. $H_{2}Be$ has 4 valence electrons, which corresponds to the 2nd highest orbital on the diagram. The linear configuration (180º) is lower in energy in this case.
Q10.19
Use the Walsh diagram for the valence electrons of a XY2 molecule to predict whether the following molecules are linear or bent:
a. (CO2) b. (CO2+) c. (CO2-) d. (SO2-) b. (CF2+)
S10.19
Valence Electrons Geometry
2-16 linear
17-20 bent
21-24 linear
1. CO2 16 valence electrons Linear
2. CO2+ 15 valence electrons Linear
3. CO2- 17 valence electrons Bent
4. SO2- 19 valence electrons Bent
5. CF2+ 18 valence electrons Bent
Q10.20
Walsh correlation diagrams can be used to predict the shapes of polyatomic molecules that contain more than three atoms. In this and the following three problems we consider molecules that have the general formula XH3. We will restrict our discussion to XH3 molecules, where all the H—X—H bond angles are the same. If the molecule is planar, then the H—X—H bond angle is 120°. A nonplanar XH3 molecule, then, has an H—X—H bond angle that is less than 120°. Figure 10.26 shows the Walsh correlation diagram that describes how the energies of the molecular orbitals for an XH3 molecule change as a function of the H—X—H bond angle. Note that because XH3 is not linear, the labels used to describe the orbitals on the two sides of the correlation diagram do not have designations such as and . We see that the lowest-energy molecular orbital is insensitive to the H—X—H bond angle. Which atomic orbital(s) contribute to the lowest-energy molecular orbital? Explain why the energy of this molecular orbital is insensitive to changes in the H—X—H bond angle.
S10.20
The lowest energy molecular orbital is the 1s orbital, which is a core atomic orbital instead of a bonding atomic orbital.
Q10.21
Consider the BH2 where the general Walsh diagram for a XH2 is shown below. What is the geometric preference of the molecule in ground and excited state?
S10.21
The BH2 molecule has the same geometric shape as water, it is bent where the HOMO is ${\pi_u}$. The first excited state relies on the degree of bending, and the 2a1 is unoccupied, and the next 1b2 is the highest occupied where the preferred geometry is linear, so at the first excited state will be linear.
$BH_{2}$ is a linear molecule. It has 4 valance electrons which fill 2 of the lines in the Walsh diagram. This second line has lower energy towards linear conformation and $1 \sigma_{u}$.
Q10.24
Solve for $\psi_\pi$ corresponding to the energy $E = \alpha + \beta$ for ethene.
S10.24
The bonding Huckel molecular orbitals is
$\psi_\pi = c_1 2p_{z1} + c_2 2p_{z2}$
the relationship of the coefficients can be defined as the following from the secular determinate:
$c_1 (\alpha - E) + c_2 \beta = 0$
$c_1 \beta+ c_2 (\alpha - E) = 0$
Substituting $E = \alpha + \beta$ into these expressions and solving gives
$-c_1\beta+c_2\beta = 0$
$c_1 = c_2$
Then plugging back into the original equation gives
$\psi_\pi = c_1 (2p_{zA} + 2p_{zB} )$
Now we can solve for $c_1$ by normalizing the wavefunction
$c^2_1(1 + 2S + 1) = 1$
where S = 0 so solving yields
$c_1 = \frac{1}{\sqrt{2}}$.
The final wave function can be written as
$\psi_\pi = \frac{1}{\sqrt{2}} (2p_{z1} + 2p_{z2})$
Q10.25
Generalize the molecular orbital treatment of propene allyl cation. Find the energies and wave function of this molecule.
S10.25
The Huckel secular determinant for propene is $\left. \begin{vmatrix} \alpha - E & \beta & 0 \\ \beta & \alpha - E & \beta \ 0 & \beta & \alpha - E \end{vmatrix} \right. = 0$
making a substitution for $x = \frac{\alpha - E }{\beta}$ the secular determinant becomes $\left. \begin{vmatrix} x & 1 & 0 \ 1 & x & 1 \ 0 & 1 & x \end{vmatrix} \right. = 0$
Solving the determinant yields a cubic polynomial $x^3 - 2x = 0$ the roots of this polynomial are $x = 0, \pm \sqrt{2}$
Replacing x with the previous substitution made it is found that $E = \alpha \pm \sqrt{2} \beta$ and $E = 0$
To find the wave function of propene you must find the constants $\left. \begin{vmatrix} c_{1}(x) & c_{2}& 0 \ c_{1} & c_{2}(x) & c_{3} \ 0 & c_{2}& c_{3}(x) \end{vmatrix} \right. = 0$
solving the determinant yields $c_{1} = c_{3} = \frac{1}{2}$ $c_{2} = \frac{1}{\sqrt{2}}$
therefore the wave function is $\psi = \frac{1}{2} 1s + \frac{1}{\sqrt{2}}2s + \frac{1}{2} 2p_{z}$
Q10.26
Show that the six molecular orbitals for Benzene consturcted from the $2Pp_x$ atomic orbital on each of the six carbon atoms:
$\psi_{i}=\sum_{j=1}^{6} c_{ij}2p_{xj}$
leads to a secular determinant.
S10.26
The above equation for benzene is :
$\psi_{i} = c_{i1}2p_{z1}+c_{i2}2p_{z2}+c_{i3}2p_{z3}+c_{i4}2p_{z4}+c_{i5}2p_{z5}+c_{i6}2p_{z6}$
The secular determinant for the benzene is :
$\begin{vmatrix}H_{11}-ES_{11}&H_{12}-ES_{12}&H_{13}-ES_{13}&H_{14}-ES_{14}&H_{15}-ES_{15}&H_{16}-ES_{16}\H_{12}-ES_{12}&H_{22}-ES_{22}&H_{23}-ES_{23}&H_{24}-ES_{24}&H_{25}-ES_{25}&H_{26}-ES_{26}\H_{13}-ES_{13}&H_{23}-ES_{23}&H_{33}-ES_{33}&H_{34}-ES_{34}&H_{35}-ES_{35}&H_{36}-ES_{36}\H_{14}-ES_{14}&H_{24}-ES_{24}&H_{34}-ES_{34}&H_{44}-ES_{44}&H_{45}-ES_{45}&H_{46}-ES_{46}\H_{15}-ES_{15}&H_{25}-ES_{25}&H_{35}-ES_{35}&H_{45}-ES_{45}&H_{55}-ES_{55}&H_{56}-ES_{56}\H_{16}-ES_{16}&H_{26}-ES_{26}&H_{36}-ES_{36}&H_{46}-ES_{46}&H_{56}-ES_{56}&H_{66}-ES_{66} \end{vmatrix}=0$
$H_{11}=H_{22}=H_{33}=H_{44}=H_{55}=H_{66}=\alpha$
$H_{12}=H_{23}=H_{34}=H_{45}=H_{56}=\beta$ The $H_{ij}=H_{ji}$ is a hermitian operator when $i$ and $j$ are neighbors
$H_{ij}=0$ when $i$ and $j$ are not neighbors
$S_{11}=S_{22}=S_{33}=S_{44}=S_{55}=S_{66}= 1$
$S_{ij}=0$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/10%3A_Bonding_in_Polyatomic_Molecules/10.E%3A_Bonding_in_Polyatomic_Molecules_%28Exercises%29.txt
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Computational chemistry is the field of chemistry that uses mathematical approximations and computer programs to solve problems of chemical interest. Quantum chemistry is a subfield that addresses the equations and approximations derived from the postulates of quantum mechanics; specifically involving solving the Schrödinger equation for molecular systems. Quantum chemistry is typically separated into ab initio, which uses methods that do not include any empirical parameters or experimental data and semi-empirical which do.
• 11.1: Overview of Quantum Calculations
The variational principle says an approximate energy is an upper bound to the exact energy, so the lowest energy that we calculate is the most accurate. This limiting energy is the lowest that can be obtained with a single determinant wavefunction . This limit is called the Hartree-Fock limit, the energy is the Hartree-Fock energy, the molecular orbitals producing this limit are called Hartree-Fock orbitals, and the determinant is the Hartree-Fock wavefunction.
• 11.2: Gaussian Basis Sets
A minimal basis set of STOs for a molecule includes only those STOs that would be occupied by electrons in the atoms forming the molecule. A larger basis set, however, improves the accuracy of the calculations by providing more variable parameters to produce a better approximate wavefunction, but at the expense of increased computational time. STOs have the following radial part (the spherical harmonic functions are used to describe the angular part) .
• 11.3: Extended Basis Sets
Today, there are hundreds of basis sets composed of Gaussian Type Orbitals (GTOs). The smallest of these are called minimal basis sets, and they are typically composed of the minimum number of basis functions required to represent all of the electrons on each atom. The largest of these can contain literally dozens to hundreds of basis functions on each atom.
• 11.4: Orbital Polarization Terms in Basis Sets
The use of a minimal basis set with fixed zeta parameters severely limits how much the electronic charge can be changed from the atomic charge distribution to describe molecules and chemical bonds. Expanding the basis set with more flexible functions can obtain more accurate results. Such functions are called polarization functions because they allow for charge polarization away form the atomic distribution to occur.
• 11.E: Computational Quantum Chemistry (Exercises)
These are homework exercises to accompany Chapter 11 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
11: Computational Quantum Chemistry
Multielectron Electronic Wavefunctions
We could symbolically write an approximate two-particle wavefunction as $\psi (r_1, r_2)$. This could be, for example, a two-electron wavefunction for helium. To exchange the two particles, we simply substitute the coordinates of particle 1 ($r_l$) for the coordinates of particle 2 ($r_2$) and vice versa, to get the new wavefunction $\psi (r_1, r_2)$. This new wavefunction must have the property that
$|\psi (r_1, r_2)|^2 = \psi (r_2, r_1)^*\psi (r_2, r_1) = \psi (r_1, r_2)^* \psi (r_1, r_2) \label{9-38}$
Equation $\ref{9-38}$ will be true only if the wavefunctions before and after permutation are related by a factor of $e^{i\varphi}$,
$\psi (r_1, r_2) = e^{i\varphi} \psi (r_1, r_2) \nonumber$
so that
$\left ( e^{-i\varphi} \psi (r_1, r_2) ^*\right ) \left ( e^{i\varphi} \psi (r_1, r_2) ^*\right ) = \psi (r_1 , r_2 ) ^* \psi (r_1 , r_2) \label{9-40}$
If we exchange or permute two identical particles twice, we are (by definition) back to the original situation. If each permutation changes the wavefunction by $e^{i \varphi}$, the double permutation must change the wavefunction by $e^{i\varphi} e^{i\varphi}$. Since we then are back to the original state, the effect of the double permutation must equal 1; i.e.,
$e^{i\varphi} e^{i\varphi} = e^{i 2\varphi} = 1 \nonumber$
which is true only if $\varphi = 0$ or an integer multiple of π. The requirement that a double permutation reproduce the original situation limits the acceptable values for $e^{i\varphi}$ to either +1 (when $\varphi = 0$) or -1 (when $\varphi = \pi$). Both possibilities are found in nature, but the behavior of elections is that the wavefunction be antisymmetric with respect to permutation $(e^{i\varphi} = -1)$. A wavefunction that is antisymmetric with respect to electron interchange is one whose output changes sign when the electron coordinates are interchanged, as shown below.
$\psi (r_2 , r_1) = e^{i\varphi} \psi (r_1, r_2) = - \psi (r_1, r_2) \nonumber$
Blindly following the first statement of the Pauli Exclusion Principle, that each electron in a multi-electron atom must be described by a different spin-orbital, we try constructing a simple product wavefunction for helium using two different spin-orbitals. Both have the 1s spatial component, but one has spin function $\alpha$ and the other has spin function $\beta$ so the product wavefunction matches the form of the ground state electron configuration for He, $1s^2$.
$\psi (\mathbf{r}_1, \mathbf{r}_2 ) = \varphi _{1s\alpha} (\mathbf{r}_1) \varphi _{1s\beta} ( \mathbf{r}_2) \label{8.6.1}$
After permutation of the electrons, this becomes
$\psi ( \mathbf{r}_2,\mathbf{r}_1 ) = \varphi _{1s\alpha} ( \mathbf{r}_2) \varphi _{1s\beta} (\mathbf{r}_1) \label{8.6.2}$
which is different from the starting function since $\varphi _{1s\alpha}$ and $\varphi _{1s\beta}$ are different spin-orbital functions. However, an antisymmetric function must produce the same function multiplied by (–1) after permutation, and that is not the case here. We must try something else.
To avoid getting a totally different function when we permute the electrons, we can make a linear combination of functions. A very simple way of taking a linear combination involves making a new function by simply adding or subtracting functions. The function that is created by subtracting the right-hand side of Equation $\ref{8.6.2}$ from the right-hand side of Equation $\ref{8.6.1}$ has the desired antisymmetric behavior. The constant on the right-hand side accounts for the fact that the total wavefunction must be normalized.
$\psi (\mathbf{r}_1, \mathbf{r}_2) = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(\mathbf{r}_1) \varphi _{1s\beta}( \mathbf{r}_2) - \varphi _{1s\alpha}( \mathbf{r}_2) \varphi _{1s\beta}(\mathbf{r}_1)] \nonumber$A linear combination that describes an appropriately antisymmetrized multi-electron wavefunction for any desired orbital configuration is easy to construct for a two-electron system. However, interesting chemical systems usually contain more than two electrons. For these multi-electron systems a relatively simple scheme for constructing an antisymmetric wavefunction from a product of one-electron functions is to write the wavefunction in the form of a determinant. John Slater introduced this idea so the determinant is called a Slater determinant.
The Slater determinant for the two-electron wavefunction for the ground state $H_2$ system (with the two electrons occupying the $\sigma_{1s}$ molecular orbital)
$\psi (\mathbf{r}_1, \mathbf{r}_2) = \dfrac {1}{\sqrt {2}} \begin {vmatrix} \sigma_{1s} (1) \alpha (1) & \sigma _{1s} (1) \beta (1) \ \sigma _{1s} (2) \alpha (2) & \sigma_{1s} (2) \beta (2) \end {vmatrix} \nonumber$
We can introduce a shorthand notation for the arbitrary spin-orbital
$\chi_{i\alpha}(\mathbf{r}) = \varphi_i \alpha \nonumber$
or
$\chi_{i\beta}(\mathbf{r}) = \varphi_i \beta \nonumber$
as determined by the $m_s$ quantum number. A shorthand notation for the determinant in Equation 8.6.4 is then
$\psi (\mathbf{r}_1 , \mathbf{r}_2) = 2^{-\frac {1}{2}} Det | \chi_{1s\alpha} (\mathbf{r}_1) \alpha \chi_{1s\beta} ( \mathbf{r}_2) \beta| \nonumber$
The determinant is written so the electron coordinate changes in going from one row to the next, and the spin orbital changes in going from one column to the next. The advantage of having this recipe is clear if you try to construct an antisymmetric wavefunction that describes the orbital configuration for uranium! Note that the normalization constant is
$(N!)^{-\dfrac {1}{2}}$
for a system of $N$ electrons.
The generalized Slater determinant for a multe-electrom atom with N electrons is then
$\psi(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N)=\dfrac{1}{\sqrt{N!}} \left| \begin{matrix} \chi_1(\mathbf{r}_1) \alpha & \chi_1(\mathbf{r}_1) \beta& \cdots & \chi_{N/2}(\mathbf{r}_1) \beta\ \chi_1(\mathbf{r}_2) \alpha & \chi_2(\mathbf{r}_2)\beta & \cdots & \chi_{N/2}(\mathbf{r}_2)\beta \ \vdots & \vdots & \ddots & \vdots \ \chi_1(\mathbf{r}_N) \alpha & \chi_2(\mathbf{r}_N)\beta & \cdots & \chi_{N/2}(\mathbf{r}_N) \beta \end{matrix} \right| \label{slater}$
In a modern ab initio electronic structure calculation on a closed shell molecule, the electronic Hamiltonian is used with a single determinant wavefunction. This wavefunction, $\Psi$, is constructed from molecular orbitals, $\psi$ that are written as linear combinations of contracted Gaussian basis functions, $\varphi$
$\varphi _j = \sum \limits _k c_{jk} \psi _k \label {10.69}$
The contracted Gaussian functions are composed from primitive Gaussian functions to match Slater-type orbitals. The exponential parameters in the STOs are optimized by calculations on small molecules using the nonlinear variational method and then those values are used with other molecules. The problem is to calculate the electronic energy from
$E = \dfrac {\int \Psi ^* \hat {H} \Psi d \tau }{\int \Psi ^* \Psi d \tau} \label {10.70}$
or in bra-ket notation
$E = \dfrac {\left \langle \Psi |\hat {H} | \Psi \right \rangle}{\left \langle \psi | \psi \right \rangle} \nonumber$
The the optimum coefficients $c_{jk}$ for each molecular orbital in Equation $\ref{10.69}$ by using the Self Consistent Field Method and the Linear Variational Method to minimize the energy as was described previously for atoms.
The variational principle says an approximate energy is an upper bound to the exact energy, so the lowest energy that we calculate is the most accurate. At some point, the improvements in the energy will be very slight. This limiting energy is the lowest that can be obtained with a single determinant wavefunction (e.g., Equation $\ref{slater}$). This limit is called the Hartree-Fock limit, the energy is the Hartree-Fock energy, the molecular orbitals producing this limit are called Hartree-Fock orbitals, and the determinant is the Hartree-Fock wavefunction.
Hartree-Fock Calculations
You may encounter the terms restricted and unrestricted Hartree-Fock. The above discussion pertains to a restricted HF calculation. In a restricted HF calculation, electrons with $\alpha$ spin are restricted or constrained to occupy the same spatial orbitals as electrons with $\beta$ spin. This constraint is removed in an unrestricted calculation. For example, the spin orbital for electron 1 could be $\psi _A (r_1) \alpha (1)$, and the spin orbital for electron 2 in a molecule could be $\psi _B (r_2) \beta (2)$, where both the spatial molecular orbital and the spin function differ for the two electrons. Such spin orbitals are called unrestricted. If both electrons are constrained to have the same spatial orbital, e.g. $\psi _A (r_1) \alpha (1)$ and $\psi _A (r_2) \beta (2)$, then the spin orbital is said to be restricted. While unrestricted spin orbitals can provide a better description of the electrons, twice as many spatial orbitals are needed, so the demands of the calculation are much higher. Using unrestricted orbitals is particular beneficial when a molecule contains an odd number of electrons because there are more electrons in one spin state than in the other.
Example 11.1.1 : Carbon Monoxide
It is well known that carbon monoxide is a poison that acts by binding to the iron in hemoglobin and preventing oxygen from binding. As a result, oxygen is not transported by the blood to cells. Which end of carbon monoxide, carbon or oxygen, do you think binds to iron by donating electrons? We all know that oxygen is more electron-rich than carbon (8 vs 6 electrons) and more electronegative. A reasonable answer to this question therefore is oxygen, but experimentally it is carbon that binds to iron.
A quantum mechanical calculation done by Winifred M. Huo, published in J. Chem. Phys. 43, 624 (1965), provides an explanation for this counter-intuitive result. The basis set used in the calculation consisted of 10 functions: the ls, 2s, 2px, 2py, and 2pz atomic orbitals of C and O. Ten molecular orbitals (mo’s) were defined as linear combinations of the ten atomic orbitals (Equation $\ref{10.69}$. The ground state wavefunction $\Psi$ is written as the Slater Determinant of the five lowest energy molecular orbitals $\psi _k$. Equation $\ref{10.70}$ gives the energy of the ground state, where the denominator accounts for the normalization requirement. The coefficients $C_{kj}$ in the linear combination are determined by the variational method to minimize the energy. The solution of this problem gives the following equations for the molecular orbitals. Only the largest terms have been retained here. These functions are listed and discussed in order of increasing energy.
• $1s \approx 0.94 1s_o$. The 1 says this is the first $\sigma$ orbital. The $\sigma$ says it is symmetric with respect to reflection in the plane of the molecule. The large coefficient, 0.94, means this is essentially the 1s atomic orbital of oxygen. The oxygen 1s orbital should have a lower energy than that of carbon because the positive charge on the oxygen nucleus is greater.
• $2s \approx 0.92 1s_c$. This orbital is essentially the 1s atomic orbital of carbon. Both the $1\sigma$ and $2 \sigma$ are “nonbonding” orbitals since they are localized on a particular atom and do not directly determine the charge density between atoms.
• $3s \approx (0.72 2s_o + 0.18 2p_{zo}) + (0.28 2s_c + 0.16 2p_{zc})$. This orbital is a “bonding” molecular orbital because the electrons are delocalized over C and O in a way that enhances the charge density between the atoms. The 3 means this is the third $\sigma$ orbital. This orbital also illustrates the concept of hybridization. One can say the 2s and 2p orbitals on each atom are hybridized and the molecular orbital is formed from these hybrids although the calculation just obtains the linear combination of the four orbitals directly without the à priori introduction of hybridization. In other words, hybridization just falls out of the calculation. The hybridization in this bonding LCAO increases the amplitude of the function in the region of space between the two atoms and decreases it in the region of space outside of the bonding region of the atoms.
• $4s \approx (0.37 2s_c + 0.1 2p_{zc}) + (0.54 2p_{zo} - 0.43 2s_{0})$. This molecular orbital also can be thought of as being a hybrid formed from atomic orbitals. The hybridization of oxygen atomic orbitals, because of the negative coefficient with 2sO, decreases the electron density between the nuclei and enhances electron density on the side of oxygen facing away from the carbon atom. If we follow how this function varies along the internuclear axis, we see that near carbon the function is positive whereas near oxygen it is negative or possibly small and positive. This change means there must be a node between the two nuclei or at the oxygen nucleus. Because of the node, the electron density between the two nuclei is low so the electrons in this orbital do not serve to shield the two positive nuclei from each other. This orbital therefore is called an “antibonding” molecular orbital and the electrons assigned to it are called antibonding electrons. This orbital is the antibonding partner to the $3 \sigma$ orbital.
• $1\pi \approx 0.32 2p_{xc} + 0.44 2p_{xo} \text {and} 2\pi \approx 0.32 2p_{yc} + 0.44 2p_{yo}$. These two orbitals are degenerate and correspond to bonding orbitals made up from the px and py atomic orbitals from each atom. These orbitals are degenerate because the x and y directions are equivalent in this molecule. $\pi$ tells us that these orbitals are antisymmetric with respect to reflection in a plane containing the nuclei.
• $5\sigma \approx 0.38 2_{sC} - 0.38 2_{pC} - 0.29 2p_{zO}$. This orbital is the sp hybrid of the carbon atomic orbitals. The negative coefficient for 2pC puts the largest amplitude on the side of carbon away from oxygen. There is no node between the atoms. We conclude this is a nonbonding orbital with the nonbonding electrons on carbon. This is not a “bonding” orbital because the electron density between the nuclei is lowered by hybridization. It also is not an antibonding orbital because there is no node between the nuclei. When carbon monoxide binds to Fe in hemoglobin, the bond is made between the C and the Fe. This bond involves the donation of the $5\sigma$ nonbonding electrons on C to empty d orbitals on Fe. Thus molecular orbital theory allows us to understand why the C end of the molecule is involved in this electron donation when we might naively expect O to be more electron-rich and capable of donating electrons to iron.
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A basis set in theoretical and computational chemistry is a set of functions (called basis functions) which are combined in linear combinations (generally as part of a quantum chemical calculation) to create molecular orbitals. For convenience these functions are typically atomic orbitals centered on atoms, but can theoretically be any function; plane waves are frequently used in materials calculations.
The Variational Method and Basis Sets
To describe the electronic states of molecules, we construct wavefunctions for the electronic states by using molecular orbitals. These wavefunctions are approximate solutions to the Schrödinger equation. A mathematical function for a molecular orbital is constructed, $\psi _i$, as a linear combination of other functions, $\varphi _j$, which are called basis functions because they provide the basis for representing the molecular orbital.
$\psi _i = \sum _j c_{ij} \varphi _j \label {10.8}$
The linear variational method is used to find values for parameters in the basis functions and for the constant coefficients in the linear combination that optimize these functions, i.e. make them as good as possible. The criterion for quality in the variational method is making the ground state energy of the molecule as low as possible. Here and in the rest of this chapter, the following notation is used: $\sigma$ is a general spin function (can be either $\alpha$ or $\beta$), $\varphi$ is the basis function (this usually represents an atomic orbital), $\psi$ is a molecular orbital, and $\Psi$ is the electronic state wavefunction (representing a single Slater determinant or linear combination of Slater determinants).
The ultimate goal is a mathematical description of electrons in molecules that enables chemists and other scientists to develop a deep understanding of chemical bonding and reactivity, to calculate properties of molecules, and to make predictions based on these calculations. For example, an active area of research in industry involves calculating changes in chemical properties of pharmaceutical drugs as a result of changes in chemical structure.
Selecting the ab initio model for a chemical system is almost always involves a trade-off between accuracy and computational cost. More accurate methods and larger basis sets make jobs run longer.
In modern computational chemistry, quantum chemical calculations are typically performed using a finite set of basis functions. In these cases, the wavefunctions of the system in question are represented as vectors, the components of which correspond to coefficients in a linear combination of the basis functions in the basis set used.
The molecular spin-orbitals that are used in the Slater determinant usually are expressed as a linear combination of some chosen functions, which are called basis functions. This set of functions is called the basis set. The fact that one function can be represented by a linear combination of other functions is a general property. All that is necessary is that the basis functions span-the-space, which means that the functions must form a complete set and must be describing the same thing. For example, spherical harmonics cannot be used to describe a hydrogen atom radial function because they do not involve the distance r, but they can be used to describe the angular properties of anything in three-dimensional space.
This span-the-space property of functions is just like the corresponding property of vectors. The unit vectors $(\overrightarrow {x}, \overrightarrow {y}, \overrightarrow {z})$ describe points in space and form a complete set since any position in space can be specified by a linear combination of these three unit vectors. These unit vectors also could be called basis vectors.
Exercise 11.2.1 : "Spanning the Space"
Explain why the unit vectors $(\overrightarrow {x}, \overrightarrow {y})$ do not form a complete set to describe your (three-dimensional) classroom.
Just as we discussed for atoms, parameters in the basis functions and the coefficients in the linear combination can be optimized in accord with the Variational Principle to produce a self-consistent field (SCF) for the electrons. This optimization means that the ground state energy calculated with the wavefunction is minimized with respect to variation of the parameters and coefficients defining the function. As a result, that ground state energy is larger than the exact energy, but is the best value that can be obtained with that wavefunction.
Slater Type Orbitals (STOs)
Intuitively one might select hydrogenic atomic orbitals as the basis set for molecular orbitals. After all, molecules are composed of atoms, and hydrogenic orbitals describe atoms exactly if the electron-electron interactions are neglected. At a better level of approximation, the nuclear charge that appears in these functions can be used as a variational parameter to account for the shielding effects due to the electron-electron interactions. Also, the use of atomic orbitals allows us to interpret molecular properties and charge distributions in terms of atomic properties and charges, which is very appealing since we picture molecules as composed of atoms. As described in the previous chapter, calculations with hydrogenic functions were not very efficient so other basis functions, Slater-type atomic orbitals (STOs), were invented.
A minimal basis set of STOs for a molecule includes only those STOs that would be occupied by electrons in the atoms forming the molecule. As with most variational method calculations, a larger basis set improves the accuracy of the calculations by providing more variable parameters to produce a better approximate wavefunction. However, this is at the expense of increased computational time (i.e., calculation "cost" or "expense"). STOs have the following radial part (the spherical harmonic functions are used to describe the angular part)
$R(r) = N r^{n − 1} e^{−\zeta r} \nonumber$
where
• $n$ is a natural number that plays the role of principal quantum number, $n = 1,\,2,\,...,$
• $N$ is a normalizing constant,
• $r$ is the distance of the electron from the atomic nucleus, and $\zeta$ is a constant related to the effective charge of the nucleus, the nuclear charge being partly shielded by electrons. Historically, the effective nuclear charge was estimated by Slater's rules.
Double-zeta basis Sets
One can use more than one STO to represent one atomic orbital, as shown in Equation $\ref{10.11}$, and rather than doing a nonlinear variational calculation to optimize each $\zeta$ value, use two STOs with different $\zeta$ variables. The linear variation calculation then will produce the coefficients ($C_1$ and $C_2$) for these two functions in the linear combination that best describes the charge distribution in the molecule (for the ground state). The function with the large zeta accounts for charge near the nucleus, while the function with the smaller zeta accounts for the charge distribution at larger values of the distance from the nucleus. This expanded basis set is called a double-zeta basis set.
$R_{2s} (r) = C_1re^{-\zeta _1r} + C_2 r e^{-\zeta _2 r} \label {10.11}$
The use of double zeta functions in basis sets is especially important because without them orbitals of the same type are constrained to be identical even though in the molecule they may be chemically inequivalent. For example, in acetylene the $p_z$ orbital along the internuclear axis is in a quite different chemical environment and is being used to account for quite different bonding than the $p_x$ and $p_y$ orbitals. With a double zeta basis set the $p_z$ orbital is not constrained to be the same size as the $p_x$ and $p_y$ orbitals.
Example 11.2.1
Explain why the $p_x$, $p_y$, and $p_z$ orbitals in a molecule might be constrained to be the same in a single-zeta basis set calculation, and how the use of a double-zeta basis set would allow the $p_x$, $p_y$, and $p_z$ orbitals to differ.
Gaussian Orbitals
Although any basis set that sufficiently spans the space of electron distribution could be used, the concept of Molecular Orbitals as Linear Combinations of Atomic Orbitals (LCAO) suggests a very natural set of basis functions: AO-type functions centered on each nuclei. One obvious choice are the exact hydrogen AO's, known as Slater-type orbitals (STO) -describing the radial component of the functions. However, the computation of the integrals is greatly simplified by using Gaussian-type orbitals (GTO) for basis functions.
While the STO basis set was an improvement over hydrogenic orbitals in terms of computational efficiency, representing the STOs with Gaussian functions produced further improvements that were needed to accurately describe molecules. A Gaussian basis function has the form shown in Equation $\ref{10.12}$. Note that in all the basis sets, only the radial part of the orbital changes, and the spherical harmonic functions are used in all of them to describe the angular part of the orbital.
$G_{nlm} (r, \theta , \psi ) = N_n \underbrace{r^{n-1} e^{-\alpha r^2}}_{\text{radial part}} \underbrace{Y^m_l (\theta, \psi)}_{\text{angular part}} \label{10.12}$
Unfortunately Gaussian functions do not match the shape of an atomic orbital very well. In particular, they are flat rather than steep near the atomic nucleus at $r = 0$, and they fall off more rapidly at large values of $r$ (Figure 11.2.1 ).
To compensate for this problem, each STO is replaced with a number of Gaussian functions with different values for the exponential parameter. These Gaussian functions form a primitive Gaussian basis set. Linear combinations of the primitive Gaussians are formed to approximate the radial part of an STO. This linear combination is not optimized further in the energy variational calculation, but rather is frozen and treated as a single function. The linear combination of primitive Gaussian functions is called a contracted Gaussian function. Although more functions and more integrals now are part of the calculation, the integrals involving Gaussian functions are quicker to compute than those involving exponentials, so there is a net gain in the efficiency of the calculation.
Gaussian basis sets are identified by abbreviations such as N-MPG*. N is the number of Gaussian primitives used for each inner-shell orbital. The hyphen indicates a split-basis set where the valence orbitals are double zeta. The M indicates the number of primitives that form the large zeta function (for the inner valence region), and P indicates the number that form the small zeta function (for the outer valence region). G identifies the set a being Gaussian. The addition of an asterisk to this notation means that a single set of Gaussian 3d polarization functions (discussed elsewhere) is included. A double asterisk means that a single set of Gaussian 2p functions is included for each hydrogen atom.
For example, 3G means each STO is represented by a linear combination of three primitive Gaussian functions. 6-31G means each inner shell (1s orbital) STO is a linear combination of 6 primitives and each valence shell STO is split into an inner and outer part (double zeta) using 3 and 1 primitive Gaussians, respectively (see Table 11.2.1 for other examples).
Basis set # functions Basis set # functions Basis set # functions
Table 11.2.1 : Different Gaussian Basis sets
STO-3G 5 6-31G 9 6-311G 13
3-21G 9 6-31G* 15 6-311G* 18*
4-31G 9 6-31+G* 19 6-311+G* 22*
Example 11.2.2
The 1s Slater-type orbital $S_1 (r) = \sqrt {4 \zeta _1 e^{-\zeta _1 r}}$ with $\zeta _1 = 1.24$ is represented as a sum of three primitive Gaussian functions,
$S_G (r) = \sum _{j=1}^3 C_j e^{-\alpha _j r^2} \nonumber$
This sum is the contracted Gaussian function for the STO.
1. Make plots of the STO and the contracted Gaussian function on the same graph so they can be compared easily. All distances should be in units of the Bohr radius. Use the following values for the coefficients, C, and the exponential parameters, $\alpha$.
index j $\alpha _j$ $C_j$
1 0.1688 0.4
2 0.6239 0.7
3 3.425 1.3
2. Change the values of the coefficients and exponential parameters to see if a better fit can be obtained.
3. Comment on the ability of a linear combination of Gaussian functions to accurately describe a STO.
Summary
When molecular calculations are performed, it is common to use a basis composed of a finite number of atomic orbitals (Equation $\ref{10.8}$), centered at each atomic nucleus within the molecule (linear combination of atomic orbitals ansatz). These atomic orbitals are well described with Slater-type orbitals (STOs), as STOs decay exponentially with distance from the nuclei, accurately describing the long-range overlap between atoms, and reach a maximum at zero, well describing the charge and spin at the nucleus. STOs are computationally difficult and it was later realized by Frank Boys that these Slater-type orbitals could in turn be approximated as linear combinations of Gaussian orbitals instead. Because it is easier to calculate overlap and other integrals with Gaussian basis functions, this led to huge computational savings
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Today, there are hundreds of basis sets composed of Gaussian Type Orbitals (GTOs). The smallest of these are called minimal basis sets, and they are typically composed of the minimum number of basis functions required to represent all of the electrons on each atom. The largest of these can contain literally dozens to hundreds of basis functions on each atom.
Minimum Basis sets
A minimum basis set is one in which a single basis function is used for each orbital in a Hartree-Fock calculation on the atom. However, for atoms such as lithium, basis functions of p type are added to the basis functions corresponding to the 1s and 2s orbitals of each atom. For example, each atom in the first row of the periodic system (Li - Ne) would have a basis set of five functions (two s functions and three p functions).
In a minimum basis set, a single basis function is used for each atomic orbital on each constituent atom in the system.
The most common minimal basis set is STO-nG, where n is an integer. This $n$ value represents the number GTOs used to approximate the Slater Type orbital (STO) for both core and valence orbitals. Minimal basis sets typically give rough results that are insufficient for research-quality publication, but are much cheaper (less calculations requires) than the larger basis sets discussed below. Commonly used minimal basis sets of this type are: STO-3G, STO-4G, and STO-6G.
Two is Often Better than One
Minimal basis sets are not flexible enough for accurate representation of, which requires the use multiple functions to represent each atomic orbital. The distribution of the electron density of valence electrons is better represented by the sum of two orbitals with different "effective charges". This is a double-$\zeta$ basis sets and includes split-valence set (inner and valence) and linear combination of two orbitals of same type, but with different effective charges (i.e., $\zeta$). This flexibility can be used to generate atomic orbital of adjustable sizes.
For example, the double-zeta basis set allows us to treat each orbital separately when we conduct the Hartree-Fock calculation.
$\phi_i = a_1 \phi_{2s}^{STO}(r, \zeta_1) + a_2\phi_{2s}^{STO}(r, \zeta_2) \label{11.2.1}$
The 2s atomic orbital approximated as a sum of two STOs. The two equations are the same except for the value of $\zeta$ which accounts for how large the orbital is. The constants $a_1$ and $a_2$ determines how much each STO contributes to the final atomic orbital, which will vary depending on the type of atom that the atomic orbit (i.e., hydrogen and lithium orbitals will have different $a_1$, $a_2$, $\zeta_1$, and $\zeta_2$ values).
Extended Basis Sets
The triple and quadruple-zeta basis sets work the same way, except use three and four STOs instead of two like in $\ref{11.2.1}$. The typical trade-off applies here as well, better accuracy, however with more expensive calculations. There are several different types of extended basis sets including: n split-valence, n polarized sets, n diffuse sets, and n correlation consistent sets. The notation of this sort of basis set (with a Gaussian basis) is
$N-MPG \nonumber$
for describing split-valence basis set. $N$ is the number of Gaussian functions describing inner-shell orbitals, while the hyphen denotes a split-valence set. $M$ and $P$ designate the number of Gaussian functions used to fit the two orbitals of the valence shell:
• M corresponds to number of Gaussian functions used to describe the smaller orbital
• P corresponds to number of Gaussian functions used to describe the larger orbital (e.g., 6-31G and 3-21G).
A minimal basis set is when one basis function for each atomic orbital in the atom, while a double-$\zeta$, has two two basis functions for each atomic orbital. Correspondingly, a triple and quadruple-$\zeta$ set had three and four basis functions for each atomic orbital, respectively. Higher order basis set have been constructed too, e.g., 5Z, 6Z,)..
There are hundreds of basis sets composed of Gaussian-type orbitals (Figure Figure 11.3.1 ). The smallest of these are called minimal basis sets, and they are typically composed of the minimum number of basis functions required to represent all of the electrons on each atom. The largest of these can contain dozens to hundreds of basis functions on each atom.
Figure 11.3.1 : Commonly used split-valence basis sets
3-21G 3-21G 3-21G 3-21G* - Polarized 3-21+G - Diffuse functions 3-21+G* - With polarization and diffuse functions
4-21G 4-31G 4-31G 4-31G 4-31G
6-21G 6-31G 6-31G* 6-31+G* 6-31G(3df, 3pd) 6-311G
6-311G 6-311G* 6-311+G*
11.04: Orbital Polarization Terms in Basis Sets
Polarization functions denoted in Pople’s sets by an asterisk. Two asterisks, indicate that polarization functions are also added to light atoms (hydrogen and helium). n Polarization functions have one additional node. For example, the only basis function located on a hydrogen atom in a minimal basis set would be a function approximating the 1 s atomic orbital. When polarization is added to this basis set, a p -function is also added to the basis set. The 6-31G** is synonymous to 6-31 G(d,p).
The use of a minimal basis set with fixed zeta parameters severely limits how much the electronic charge can be changed from the atomic charge distribution to describe molecules and chemical bonds. This limitation is removed if STOs with larger n values and different spherical harmonic functions, the $Y^m_l (\theta , \varphi )$ in the definition of STO’s are included. Adding such functions is another way to expand the basis set and obtain more accurate results. Such functions are called polarization functions because they allow for charge polarization away form the atomic distribution to occur.
The most common addition to minimal basis sets is probably the addition of polarization functions, denoted (in the names of basis sets developed by Pople) by an asterisk, *. Two asterisks, **, indicate that polarization functions are also added to light atoms (hydrogen and helium). These are auxiliary functions with one additional node. For example, the only basis function located on a hydrogen atom in a minimal basis set would be a function approximating the 1s atomic orbital. When polarization is added to this basis set, a p-function is also added to the basis set. This adds some additional needed flexibility within the basis set, effectively allowing molecular orbitals involving the hydrogen atoms to be more asymmetric about the hydrogen nucleus.
This is an important result when considering accurate representations of bonding between atoms, because the very presence of the bonded atom makes the energetic environment of the electrons spherically asymmetric. Similarly, d-type functions can be added to a basis set with valence p orbitals, and f-functions to a basis set with d-type orbitals, and so on. Another, more precise notation indicates exactly which and how many functions are added to the basis set, such as (d, p).
Diffuse Functions
Another common addition to basis sets is the addition of diffuse functions, denoted in Pople-type sets by a plus sign, +, and in Dunning-type sets by "aug" (from "augmented"). Two plus signs indicate that diffuse functions are also added to light atoms (hydrogen and helium). These are very shallow Gaussian basis functions, which more accurately represent the "tail" portion of the atomic orbitals, which are distant from the atomic nuclei. These additional basis functions can be important when considering anions and other large, "soft" molecular systems.
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These are homework exercises to accompany Chapter 11 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
What is meant by the expression ab initio calculation?
List all the terms in a complete molecular Hamiltonian.
Why are calculations on closed-shell systems more easily done than on open-shell systems?
How is it possible to reduce a multi-electron Hamiltonian operator to a single-electron Fock operator?
Why is the calculation with the Fock operator called a self-consistent field calculation?
What is the physical meaning of a SCF one-electron energy?
Why is the nonlinear variational method not used in every case to optimize basis functions, and what usually is done instead?
Why is it faster for a computer to use the variational principle to determine the coefficients in a linear combination of functions than to determine the parameters in the functions?
Identify the characteristics of hydrogenic, Slater, and Gaussian basis sets.
What is meant by the Hartree-Fock wavefunction and energy?
What is neglected that makes the Hartree-Fock energy necessarily greater than the exact energy?
What is meant by correlation energy?
What purpose is served by including configuration interaction in a calculation?
Q11.1
Prove that a three dimensional Gaussian function centered at $r_1$ = $x_1$i + $y_1$j + $z_1$k is a product of three one-dimensional Gaussian functions centered on $x_1$, $y_1$, $z_1$.
S11.1
$e^{-a(r-r_0)^2}$ = $e^{-a[(x-x_1)i+(y-y_1)j+(z-z_1)k]^2}$
= $e^{-a[(x-x_1)^2+(y-y_1)^2+(z-z_1)^2]}$
= $e^{-a(x-x_1)^2}$ $e^{-a(y-y_1)^2}$ $e^{-a(z-z_1)^2}$
Q11.2
Show that
$\int_{0}^{\infty} e^{-(x-x_0)^2} dx = \int_{0}^{\infty} e^{-x^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$
S11.2
The equalities are all equivalent since in the first integral $x_0$ is a constant and the second and third are even.
Q11.3
The Gaussian Integral $I_{0} = \int^\infty_{-\infty} e^{-4x^2}dx$
Convert the integration variables from Cartesian coordinates to polar coordinates and show that $I_{0} = \dfrac{\sqrt{\pi}}{2}$
S11.3
We first write
$I^2_{0} = \left( \int^\infty_{-\infty} e^{-4x^2}dx \right)^2 = \int^\infty_{-\infty} e^{-4x^2}dx \int^\infty_{-\infty} e^{-4y^2}dy$
the product of two integrals can be expressed as a double integral
$I^2_{0} = \int^\infty_{-\infty} \int^\infty_{-\infty} e^{-4(x^2 + y^2)}dxdy$
In polar coordinates $x^2 + y^2 = r^2$ and $dxdy = rdrd\theta$. The limits of integration in polar coordinates corresponding to the limits in Cartesian coordinates are 0 $\le r < \infty$ and $0 \le \theta \le 2\pi$.
The double integral becomes
$I^2_{0} = \int^\infty_{0} \int^{2\pi}_{0} e^{-4r^2}rdrd\theta = 2\pi \int^{\infty}_{0} e^{-4r^2}rdr .$
The integration over $\theta$ gives a factor of $2\pi$.The integral over r can be done using a U substitution, $u = 4r^2$ and $du = 8rdr$.
Therefore
$\int^{\infty}_{0} e^{-4r^2}rdr = \dfrac{1}{8} \int^{\infty}_{0} e^{-u^2}du = \dfrac{1}{8}$
meaning that $I^2 = 2\pi \times \dfrac{1}{8}$, so $I_{0} = \dfrac{\sqrt{\pi}}{2}$.
Q11.4
Show that the integral
$I_{2n}=\int_{-\infty}^{\infty} x^{2n}e^{-ax^2}dx$
can be obtained from $I_{o}$
$I_{o}=\int_{-\infty}^{\infty} e^{-ax^2}dx$
by differentiating $n$ times with respect to $a$ when $I_{o}$ is
$\ I_{0}=\frac{1}{2}\big(\frac{\pi}{a}\big)^{\frac{1}{2}}$
gives in a general form:
$I_{2n}=\frac{1\cdot3\cdot5\cdots\big(2n-1\big)}{2^{n+1}a^{n}}\big(\frac{a}{2}\big)^{\frac{1}{2}}$
S11.4
The first step is to take the derivative of $I_{o}$ about 3 times with respect to $a$:
$I_{o}=\int_{-\infty}^{\infty} e^{-ax^2}dx$
$\frac{dI_{o}}{da}=\int_{-\infty}^{\infty} -x^{2}e^{-ax^2}dx=-\int_{-\infty}^{\infty} x^{2}e^{-ax^2}dx$
$\frac{d^2I_{o}}{da^2}=\int_{-\infty}^{\infty} -x^{2\cdot2}e^{-ax^2}dx=\int_{-\infty}^{\infty} x^{4}e^{-ax^2}dx$
solve the integrals for the first $I_{2n}$ starting with $I_{o}$,
$I_{o}=\int_{-\infty}^{\infty} e^{-ax^2}dx = \frac{1}{2} \big(\frac{\pi}{a}\big)^{\frac{1}{2}}$
$I_{2}=\int_{-\infty}^{\infty} x^{2}e^{-ax^2}dx=\frac{dI_{o}}{da}=\frac{1}{4a}\big(\frac{\pi}{a}\big)^{\frac{1}{2}}$
$I_{4}=\int_{-\infty}^{\infty} x^{4}e^{-ax^2}dx=\frac{d^2I_{o}}{da^2}=-\frac{dI_{2}}{da}=\frac{3}{8a}\big(\frac{\pi}{a}\big)^{\frac{1}{2}}$
$I_{6}=\int_{-\infty}^{\infty} x^{6}e^{-ax^2}dx=-\frac{d^3I_{o}}{da^3}=-\frac{dI_{4}}{da}=\frac{3\cdot5}{16a}\big(\frac{\pi}{a}\big)^{\frac{1}{2}}$
in general
$I_{2n}=\frac{1\cdot3\cdot5\cdots\big(2n-1\big)}{2^{n+1}a^{n}}\big(\frac{a}{2}\big)^{\frac{1}{2}}$
(note: if you look at the equations sheet provided in the Gaussian integrals sections the limits of integration from -$\infty$ to +$\infty$ and from 0 to +$\infty$ give the same end result with a minor difference in the exponent for the two in the denominator)
Q11.17
Using the Figure below, specify the coordinates of the atoms that comprise the molecule methane. Determine a set of Cartesian coordinates of the atoms in the molecule. The HCH bond angle is 110.0° and the C-H bond length is 109.1 pm.
S11.17
This figure can represent methane if and only if the central atom is carbon and the 4 atoms at the vertices are hydrogen atoms. We then must assign the origin of our coordinate system to be at the carbon atom. Considering the length of the edge of this cube is 2a, then the bond length from the vertices (hydrogen atoms) to the center (carbon atom) is ${\sqrt{3}}$ times the length of one edge of the cube, so
$\dfrac{109.1\; pm}{ \sqrt{3}} = 63 \; pm = a$
Diagonal of cube length 2a would be $2a\sqrt(3)$ - but we need half that.
Q11.18
Determine a rough set of Cartesian coordinates of the atoms in the molecule $SiH_{3}F$ given the bond angle of $H-Si-H$ is $109.5^{\circ}$ and the $Si-H$ and $Si-F$ bond lengths are $146.0$ and $159.5 \hspace{.1cm} pm$, respectively. (Hint: locate the origin at the Silicon.)
S11.18
For a simpler case of $SiH_{4}$, The four hydrogen would be equally far from the central atom (origin). The coordinates can be calculated as $(a,a,a)$ ,$(-a,-a,a)$,$(a,-a,-a)$, and $(-a,-a,-a)$. The value of $a$ can be determined by $a = \dfrac{l}{\sqrt{3}}$ where $l$ is the bond length. For hydrogen:
$a = \dfrac{146 \hspace{.1cm} pm}{\sqrt{3}} = 84.29 \hspace{.1cm} pm$
For florine:
$a = \dfrac{159.5 \hspace{.1cm} pm}{\sqrt{3}} = 92.09 \hspace{.1cm} pm$
One set of solutions is:
x/pm y/pm z/pm
C 0 0 0
H 84.29 84.29 84.29
H -84.29 -84.29 84.29
H 84.29 -84.29 -84.29
F -92.09 -92.09 -92.09
Q11.19
Molecule Frequency [cm-1] Re [pm]
H2 4647 73.2
CO 2438 111.4
HCl 2886 130
Given the above table of calculated vibrational frequencies and bond lengths, calculate the vibrational force constant of each of the molecules. Do you expect that the calculated values are higher or lower than the experimental values? Are bond length calculations or vibrational-frequency calculations more accurate? Why?
S11.19
The relationship between wave number and harmonic force constant can be expressed as
$\tilde{\nu} = (2c\pi)^{-1} \sqrt{\dfrac{k}{\mu}}$
which can be rewritten as
$k=4\mu(\tilde{\nu}c\pi)^2.$
The reduced masses can be found to be
$\mu_{H_2} = 8.38\times10^{-28}$kg, $\mu_{CO} = 1.14\times10^{-26}$kg, and $\mu_{HCl} = 1.626\times10^{-27}$kg.
Now we can find our force constants by plugging in the given values.
• $k_{H_2} = 642 N/m$
• $k_{CO} = 2404 N/m$
• $k_{HCl} = 481 N/m$
We should expect that the values we found are higher than what is experimentally measured, as other forces are unaccounted for. Bond length calculations are more accurate because it requires a smaller basis set to calculate accurately.
Q11.20
Normalize the following Gaussian function:
$\phi(r) = xe^{-\alpha r^{2}}$
S11.20
We write $\phi (r)$ in spherical coordinates and then apply the normalization condition of the normalized function $A\phi (r)$
The normalization condition is
$\int A^{2}x^{2}e^{-2\alpha r^{2}} d\tau = 1$
or in bra-ket notation
$\langle \phi(r) | \phi(r) \rangle = 1$.
where A is the normalization constant. In spherical coordinates,
$1 = \int A^{2}r^{2} \sin ^{2}\theta\cos^{2}\phi e^{-2\alpha r^{2}} d\tau$
$= A^{2} \int_{0}^{\infty} r^{4}e^{-2\alpha r^{2}} dr \int_{0}^{\pi} \sin^{3}\theta d\theta \int_{0}^{2\pi} \cos^{2}\phi d\phi$
$\dfrac{1}{A^{2}} = \dfrac{3}{8(2\alpha)^{2}} \Big(\dfrac{\pi}{2\alpha}\Big)^{1/2} \Big(\dfrac{4}{3}\Big) (\pi)$
$= \dfrac{\pi^{3/2}}{2^{7/2}\alpha^{5/2}}$
Therefore, the nomalization constant will the inverse of this result:
$A = \Big(\dfrac{128\alpha^{5}}{\pi^{3}}\Big)^{1/4}$
Q11.21
Which hydrogen atomic orbital corresponds to the following normalized Gaussian orbital?
$G(x, y, z; \alpha) = \Big(\dfrac{128\alpha^{5}}{\pi^{3}}\Big)^{0.25}ye^{-\alpha r^2}$
How many radial and angular nodes does the above function have? Is this result what you would expect for the corresponding hydrogen function?
S11.21
The typical form is:
$G_{nlm}(r, \theta, \phi) = N_{n}r^{n-1}e^{-\alpha r^{2}}Y_{l}^{m}(\theta, \phi)$.
From this, we can see the function in the question shows n = 2 and l = 1. Because n = 2, there is 1 node and l = 1 tells us that there is 1 angular node. Therefore, there are no radial nodes.This is consistent with the $2p_{y}$ orbital in a hydrogenic function.
Q11-22
Slater type orbitals have the form, $\chi_{nlm} = R_{n}(r)Y_{lm}(\theta,\phi)$ where the second term is the spherical harmonic given by
$R_{n}(r)= \frac{(2\alpha)^{(n+1/2)}}{\sqrt{(2n)!}} r^{(n-1)}e^{(-\alpha r)}$
Define the 1s-slater type orbital.
S11-22
For n=1, the slater-type orbital is
$\phi_{nlm}= \frac{(2\zeta)^{(n+1/2)}}{\sqrt{(2n)!}} r^{(n-1)}e^{(-\zeta r)}Y_{lm}(\theta,\phi)$
$\phi_{1s}(r,\zeta) = S_{100}(r,\zeta) = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta r}$
11.23
Consider the normalized functions
$G_{1}(x, y, z; \alpha) = (\frac{2048\alpha^7}{9\pi^3})^(\frac{1}{4})x^2e^(-\alpha r^2)$
$G_{2}(x, y, z; \alpha) = (\frac{2048\alpha^7}{9\pi^3})^(\frac{1}{4})y^2e^(-\alpha r^2)$
$G_{3}(x, y, z; \alpha) = (\frac{2048\alpha^7}{9\pi^3})^(\frac{1}{4})z^2e^(-\alpha r^2)$
$G_{4}(x, y, z; \alpha) = (\frac{2048\alpha^7}{9\pi^3})^(\frac{1}{4})(x^2-y^2)e^(-\alpha r^2)$
Which hydrogen atomic orbital corresponds to the linear combination
$G_{3}(x, y, z; \alpha) +G_{1}(x, y, z;\alpha)?$
S11.23
$G_{3}(x, y, z; \alpha) +G_{1}(x, y, z;\alpha) = (\frac{2048\alpha^7}{9\pi^3})^(\frac{1}{4})(z^2+x^2)e^(-\alpha r^2)$
Corresponds to the $3d_{z^2+x^2}$ hydrogen atomic orbital.
This is a good tricky question because usually people would think that $H_2$ only has two energy levels, but really there are more, just not occupied. Once you excite/add a good amount of energy, it could change to different orbitals.
( The math is right but Hydrogen has only five 3d orbitals and they are $3d_{xy}$, $3d_{xz}$, $3d_{yz}$, $3d_{y^2}$, and $3d_{(x^2-y^2)}$ so the $3d_{z^2+x^2}$ is not consistent . -RM)
Q11.27
Scientists are trying to theoretically predict the dipole moment of a CO molecule using the STO-3G and 6-31G* basis sets. When compared to their experimental data, the 6-31G* basis set provided a more accurate calculation than did the STO-3G basis set. Why is this?
S11.27
To calculate the dipole moment of a molecule, one needs an accurate description of the electron densities and molecular orbitals. This description becomes more accurate when a larger basis set is used, which is why the 6-31G* basis set gave more accurate calculations than did the STO-3G basis set.
Q11.28
The orbital energies calculated for formaldehyde using STO-3G an 3-21G basis sets are given below.
Orbital energy/Eh energy/Eh
1a1 -20.3217 -20.4856
2a1 -11.1250 -11.2866
3a1 -1.3373 -1.4117
4a1 -0.8079 -0.8661
1b2 -0.6329 -0.6924
5a1 -0.5455 -0.6345
1b1 -0.4431 -0.5234
2b2 -0.3545 -0.4330
2b1 0.2819 0.1486
6a1 0.6291 0.2718
3b2 0.7346 0.3653
7a1 0.9126 0.4512
Determine the ground-state electronic configuration of water. The photoelectron spectrum of water is shown below.
Assign the bands. Which calculated set of energies shows the best agreement with the photoelectron spectrum? Predict the ionization energy and electron affinity of water for each calculated set of energy levels. How do these compare with the experimental values?
S11.28
There are 8 electrons in water (2 from water and 6 from oxygen). This gives use the ground-state electronic configuration of
1a122a12 1b22 3a12 1b12
The band at approx. 15 eV corresponds to the 1b12 electrons, the bands at 15.5 eV correspond to 1b22 3a12 electrons. 18.5 ev = 1a122a12.
IE = -E2h2 = 0.6924*15 ev = 10.386 ev
EA = -E2h1 = 0.5234*18.5 ev = 9.6829 ev
Q11.29
The units of dipole moment given by Gaussian 94 are called debyes (D), after the Dutch-American chemist, Peter Debye, who was awarded the Nobel Prize for chemistry in 1936 for his work on dipole moments. One dehye is equal to 10-18 esu•cm where esu (electrostatic units) is a non-SI unit for electric charge. Given that a 9v battery is 3.0 x 10-2 esu, show that the conversion factor between debyes and C • m (coulomb • meters) is 1 D = 5.34 x 10^-38 C•m.
S11.29
$1 D = 1*10^{-18} esu•cm\,\left( \frac{1.6022*10^{-19} \, C}{4.803*10^{-10}\, esu} *\dfrac{1 m}{100 cm} \right) = 3.3407*10^{-30} \, C•m$
Q11.30
Determine the dipole moment SnCl2 by using the geometry and charges:
$z= {e}\sum\, {X_i}{r_i}$
S11.30
${e}((0.41sin(52.57)i+.041cos(52.75)j)94.7*10^{-12}m+(0.41sin(52.57)i+0.41cos(52.75)j)94.7x10^{-12}m)$
${=2.3D}$
Notes:
The equation I found for dipole moment is:
$\vec{\mu} = \sum_i q_i \, \vec{r}_i$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/11%3A_Computational_Quantum_Chemistry/11.E%3A_Computational_Quantum_Chemistry_%28Exercises%29.txt
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Symmetry can help resolve many chemistry problems and usually the first step is to determine the symmetry. If we know how to determine the symmetry of small molecules, we can determine symmetry of other targets which we are interested in. Usually, it is not only the symmetry of molecule but also the symmetries of some local atoms, molecular orbitals, rotations and vibrations of bonds, etc. that are important. For example, if the symmetries of molecular orbital wave functions are known, we can find out information about the binding. Also, by the selection rules that are associated with symmetries, we can explain whether the transition is forbidden or not and also we can predict and interpret the bands we can observe in Infrared or Raman spectrum. The qualitative properties of molecular orbitals can be obtained using symmetry from group theory (whereas their precise energetics and ordering have to be determined by a quantum chemical method). Group Theory is a branch of the mathematical field of algebra. In quantum chemistry, group theory can applied to ab initio or semi-empirical calculations to significantly reduce the computational cost. Symmetry operations and symmetry elements are two basic and important concepts in group theory. When we perform an operation to a molecule, if we cannot tell any difference before and after we do the operation, we call this operation a symmetry operation. This means that the molecule seems unchanged before and after a symmetry operation. As Cotton defines it in his book, when we do a symmetry operation to a molecule, every points of the molecule will be in an equivalent position.
12: Group Theory - The Exploitation of Symmetry
To a fully understand the math behind group theory one needs to take a look at the theory portion of the Group Theory topic or refer to one of the reference text listed at the bottom of the page. Never the less as Chemist the object in question we are examining is usually a molecule. Though we live in the 21st century and much is known about the physical aspects that give rise to molecular and atomic properties. The number of high level calculations that need to be performed can be both time consuming and tedious. To most experimentalist this task is takes away time and is usually not the integral part of their work. When one thinks of group theory applications one doesn't necessarily associated it with everyday life or a simple toy like a Rubik's cube. A Rubik's cube is an a cube that has a $3 \times 3$ array of different colored tiles on each of its six surfaces, for a total of 54 tiles. Since the cube exist in 3D space, the three axis are $x$, $y$, $z$. Since the rubik's cube only allows rotation which are called operations, there are three such operations around each of the $x$, $y$, $z$ axis.
Of course the ultimate challenge of a Rubik's cube is to place all six colors on each of the six faces. By performing a series of such operations on the Rubik's cube one can arrive at a solution (A link of a person solving a Rubik's cube1 in 10.4s with operations performed noted, the operations performed will not translate to chemistry applications but it is a good example of how symmetry operations arrive at a solution). The operations shown in the Rubik's cube case are inherent to the make up of the cube, i.e., the only operations allowed are the rotations along the x, y, z axis. Therefore the Rubik's cube only has x,y,z rotation operations. Similarly the operations that are specific to a molecule are dependent on its symmetry.
Using group theory, we can exploit the symmetry of molecules to give us a rich amount of information on the molecular orbitals, rotations, and vibrations of bonds, to name a few. Making symmetry arguments, we can skip complicated quantum calculations to gain qualitatively accurate information.
In section 10.7, we used Hückel theory to explore the $\pi$ bonding network of benzene by constructing linear combinations of $2p_x$ atomic orbitals on the carbon atoms. In doing so, the roots of the secular equations were found via solving the $6 \times 6$ secular determinant.
$\left|\begin{array}{cccccc}x&1&0&0&0&1\1&x&1&0&0&0\0&1&x&1&0&0\0&0&1&x&1&0\0&0&0&1&x&1\1&0&0&0&1&x\end{array}\right|=0\label{31}$
Since the secular determinant is a $6 \times 6$ matrix, there are six solutions or values of $x$ that can be determined after expanding the determinant into the resulting (6th-order) polynomial.
$x^6-6x^4 + 9x^2 -4 =0 \label{poly1}$
Secular determinants are formulated in terms of a specific basis set; i.e., a set of functions that describe the wavefunctions. For the determinnat in Equation $\ref{31}$, that basis set is the the $\{|2p_z \rangle \}$ orbitals on the carbons. However, any basis set can be used to represent the determinant (long as it span the same space). For example, the following linear combination of $\{|2p_z \rangle \}$ orbitals could also be used:
\begin{align} & | \phi_1 \rangle = \dfrac{1}{\sqrt{6}} \left[ | 2p_{z1} \rangle+ | 2p_{z2} \rangle + | 2p_{z3} \rangle + | 2p_{z4} \rangle + | 2p_{z5} \rangle + | 2p_{z6} \rangle \right] \nonumber \ & | \phi_2 \rangle = \dfrac{1}{\sqrt{4}} \left[ | 2p_{z2} \rangle + | 2p_{z3} \rangle - | 2p_{z4} \rangle - | 2p_{z5} \rangle \right] \nonumber \ & | \phi_3 \rangle = \dfrac{1}{\sqrt{3}} \left[ | 2p_{z1} \rangle + \dfrac{1}{2}| 2p_{z2} \rangle - \dfrac{1}{2} | 2p_{z3} \rangle - | 2p_{z4} \rangle - \dfrac{1}{2} | 2p_{z5} \rangle + \dfrac{1}{2} | 2p_{z6} \rangle \right] \nonumber \ & | \phi_4 \rangle = \dfrac{1}{\sqrt{4}} \left[ | 2p_{z2} \rangle - | 2p_{z3} \rangle + | 2p_{z4} \rangle - | 2p_{z5} \rangle \right] \nonumber \ & | \phi_5 \rangle = \dfrac{1}{\sqrt{3}} \left[ | 2p_{z1} \rangle - \dfrac{1}{2}| 2p_{z2} \rangle - \dfrac{1}{2} | 2p_{z3} \rangle + | 2p_{z4} \rangle - \dfrac{1}{2} | 2p_{z5} \rangle - \dfrac{1}{2} | 2p_{z6} \rangle \right] \nonumber \ & | \phi_6 \rangle = \dfrac{1}{\sqrt{6}} \left[ | 2p_{z1} \rangle- | 2p_{z2} \rangle + | 2p_{z3} \rangle - | 2p_{z4} \rangle + | 2p_{z5} \rangle - | 2p_{z6} \rangle \right] \nonumber \end{align}
In this new basis set $\{\phi \rangle \}$, the secular determinant Equation $\ref{31}$ is represented as
$\left|\begin{array}{cccccc} x+2&0&0&0&0&0 \0&x-2&0&0&0&0 \0&0&x+1& \dfrac{x+1}{2}&0&0 \0&0& \dfrac{x+1}{2} &x+1&0&0 \0&0&0&0&x-1& \dfrac{x-1}{2} \0&0&0&0& \dfrac{x-1}{2} &x-1\end{array}\right|=0\label{32}$
This is the determinant into a bock diagonal form; which can be expanded into a product of smaller determinants to give the polynomial
$\dfrac{9}{16} ( x +2)(x-2)(x+1)^2(x-1)^2=0 \nonumber$
The roots to this equation are $\pm2$, $\pm1$ and $\pm 1$. This is not surprising since these are the same roots obtained from expanding the determinant in the original basis set (Equation $\ref{poly1}$). You may remember that the selection of a specific basis set to represent a function does not change the fundamental nature of the function (e.g., a parabola in 2D space is the same curve if represented in terms of Cartesian coordinates ($x$ and $y$) or polar coordinates ($\theta$ and $r$), which both span 2-D space).
As you recall, Hückel theory (irrespective of the basis set ) was used to simplify the general secular determinant (e.g., for benzene)
$\left|\begin{array}{cccccc} H_{11} - ES_{11} & H_{12} - ES_{12} & H_{13} - ES_{13} & H_{14} - ES_{14} & H_{15} - ES_{15} & H_{16} - ES_{16} \ H_{21} - ES_{21} & H_{22} - ES_{22} & H_{23} - ES_{23} & H_{24} - ES_{24} & H_{25} - ES_{25} & H_{26} - ES_{26} \ H_{31} - ES_{31} & H_{32} - ES_{32} & H_{33} - ES_{33} & H_{34} - ES_{34} & H_{35} - ES_{35} & H_{36} - ES_{36} \ H_{41} - ES_{41} & H_{42} - ES_{42} & H_{43} - ES_{43} & H_{44} - ES_{44} & H_{45} - ES_{45} & H_{46} - ES_{46} \ H_{51} - ES_{51} & H_{52} - ES_{52} & H_{53} - ES_{53} & H_{54} - ES_{54} & H_{55} - ES_{55} & H_{56} - ES_{56} \ H_{61} - ES_{61} & H_{62} - ES_{62} & H_{63} - ES_{63} & H_{64} - ES_{64} & H_{65} - ES_{65} & H_{66} - ES_{6} \end{array}\right|=0\label{33}$
where $H_{ij}$ are the Hamiltonian matrix elements
$H_{ij} = \langle \phi_i | \hat{H} | \phi_j \rangle = \int \phi _{i}H\phi _{j}\mathrm {d} v \nonumber$
and $S_{ij}$ are the overlap integrals.
$S_{ij}= \langle \phi_i | \phi_j \rangle = \int \phi _{i}\phi _{j}\mathrm {d} v \nonumber$
In general, this involves solving 36 Hamiltonian matrix elements ($H_{ij}$) and 36 overlap integrals ($S_{ij}$), which can be a daunting task to do by hand without the assumptions of Hückel theory to help out. As with the application of Hückel theory, which was used to set most of these integrals to zero, solving for the energies from Equation $\ref{33}$ can be simplified by using the intrinsic symmetry of the benzene system to demonstrate (rigorously) that many of these integrals are zero. This is the subject of group theory. Group theory is used to exploit the symmetry of molecules to quickly gain insights into their properties, such as vibrations and molecular orbitals.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/12%3A_Group_Theory_-_The_Exploitation_of_Symmetry/12.01%3A_The_Exploitation_of_Symmetry.txt
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Symmetry Elements
A symmetry operation is an action that leaves an object looking the same after it has been carried out. For example, if we take a molecule of water and rotate it by 180° about an axis passing through the central O atom (between the two H atoms) it will look the same as before. Each symmetry operation has a corresponding symmetry element, which is the axis, plane (2-dimensional), line (1-dimensional) or point (0-dimensional) with respect to which the symmetry operation is carried out:
The symmetry element consists of all the points that stay in the same place when the symmetry operation is performed. In a rotation, the line of points that stay in the same place constitute a symmetry axis; in a reflection the points that remain unchanged make up a plane of symmetry. The symmetry of a molecule or ion can be described in terms of the complete collection of symmetry operations it possesses. The symmetry elements that a molecule (and any other 3-D object) may possess are discussed below.
Symmetry Operations
The symmetry of a molecule or ion can be described in terms of the complete collection of symmetry operations it possesses. There are five types of operations:
1. Identity, /(E/)
2. Rotation, /(C_n/)
3. Reflection, /(\sigma\)
4. Inversion, $i/) 5. Improper Rotation, \(S_n$
Symmetry Operations
A symmetry operation is a permutation of atoms such that the molecule is transformed into a state indistinguishable from the starting state.
Identity Symmetry, $E$
The identity operator, $E$, consists of doing nothing, and the corresponding symmetry element is the entire molecule. Every molecule possesses at least this operation. For example, the $\ce{CHFClBr}$ molecule in Figure 12.2.1 . The identify symmetry operation is not indicated since all molecule exhibit this symmetry.
$n$-fold Axis of Rotation, $C_n$
The rotation operation (sometimes called proper rotation), $C_n$, rotates an object about an axis by $2\pi/n$ radians or $360^{\circ}/n$. Rotation by $C_n$ leaves the molecule unchanged. The $H_2O$ molecule has a $C_2$ axis (Figure 12.2.2 ). Molecules can have more than one $C_n$ axis, in which case the one with the highest value of $n$ is called the principal axis. In some high symmetry systems, there may be more than one principal axis. Note that by convention, rotations are counterclockwise about the axis. $C_n$ rotations are indicated via vectors with labels as indicated below.
We always want to express rotations in their simplest equivalent fractions of $m/n$:
$C_4^2=C_2$
$C_6^4=C_3^2$
$C_8^6=C_4^3$
Rotating an object $n$ times brings the object back to the original object and is equivalent to the identity operation, $E$:
$C_n^n = E \nonumber$
Linear molecules have a very high rotational symmetry: $C_\infty$. Examples include diatomics such as $\text{CO}$, $\text{NO}$, and $\text{CO}_2$.
Reflection, $\sigma$
Reflection, $\sigma$, defines the bilateral symmetry about a plane (mirror plane / reflection plane). Reflection in the plane leaves the molecule looking the same. In a molecule that also has an axis of symmetry, a mirror plane that includes the axis is called a vertical mirror plane and is labeled $\sigma_v$, while one perpendicular to the axis is called a horizontal mirror plane and is labeled $\sigma_h$. A vertical mirror plane that bisects the angle between two $C_2$ axes is called a dihedral mirror plane, $\sigma_d$. If no principal axis exist, $\sigma_h$ is defined as the plane of the molecule. $\sigma$ symmetry is indicated as a plane on molecules; since they often bisect atoms, which should be clearly indicated.
For any mirror plane, performing two successive reflections about the same plane brings objects back to their original configuration:
$\sigma\sigma=\sigma^2=E \nonumber$
Inversion, $i$
Inversion, $i$, through the center of symmetry leaves the molecule unchanged. Inversion consists of passing each point through the center of inversion and out to the same distance on the other side of the molecule. If inversion symmetry exists, a line drawn from any atom through the center will connect with an equivalent atom at an equivalent distance from the center. Examples of molecules with centers of inversion is shown in Figure 12.2.4 . Centers of inversion are indicated via a point, which may or may not overlap with an atoms. The inversion center is always located at the central point of the molecule and there can only be one inversion center in any system. The centers of inversion in the examples below do not overlap with atoms.
Performing inversion twice in succession brings every point back to its original position:
$ii=i^2=E \nonumber$
Molecules with no inversion symmetry are said to be centrosymmetric.
$n$-fold Axis of Improper Rotation, $S_n$
Improper rotations, $S_n$, are also called rotation-reflections. The rotation-reflection operation consists of rotating by $C_n$ about an axis, followed by reflecting in a plane perpendicular to the same axis. Improper rotation symmetry is indicated with both an axis and a plan as demonstrated in the examples in Figure 12.2.5 .
Note
$S_1$ is the same as reflection and $S_2$ is the same as inversion.
The lowest-order improper rotation that is not a simpler operation is $S_3$. The pattern of successive operations depends on if $n$ is even or odd. The general relationships for $S_n$ operations are:
• If $n$ is even, $S_n^n=E$
• Molecule returns to original orientation configuration after each full rotation
• There are an even number of rotation operators and reflection operators
• If $n$ is odd, $S_n^n=\sigma$ and $S_n^{2n}=E$
• First full rotation:
• Molecular does not return to original configuration after first rotation
• There are an odd number of rotation operators and reflection operators
• Second full rotation:
• Molecule returns to original orientation configuration
• There are an even number of rotation operators and reflection operators
• When $m$ is even, there is always a corresponding proper rotation ($C_n$):
• $S_n^m=C_n^m$ when $m<n$
• $S_n^m=C_n^{m-n}$ when $m>n$ (2nd rotation)
• If $S_n$ with even $n$ exists, $C_{n/2}$ exists
• If $S_n$ with odd $n$ exists, then both $C_n$ and $\sigma$ perpendicular to $C_n$ exist
Summary of Symmetry Operations
The identity $E$ and rotations $C_n$ are symmetry operations that could actually be carried out on a molecule. For this reason they are called proper symmetry operations. Reflections, inversions and improper rotations can only be imagined (it is not actually possible to turn a molecule into its mirror image or to invert it without some fairly drastic rearrangement of chemical bonds) and as such, are termed improper symmetry operations. These five symmetry operations are tabulated in Table 12.2.1 .
Table 12.2.1 : The five principal symmetry elements and their operators for 3D space
Symbol Elements Description Symbol Operator Symbol
$E$ identity $\hat{E}$ no change
$C_n$ $n$-fold axis of rotation $\hat{C}_n$ Rotation by $360°/n$ leaves the molecule unchanged
$\sigma$ plane of symmetry $\hat{\sigma}$ Reflection in the plane leaves the molecule unchanged
$i$ center of symmetry. $\hat{i}$ Inversion through the center of symmetry leaves the molecule unchanged.
$S_n$ $n$-fold improper rotation $\hat{S}_n$ The rotary reflection operation consists of rotating through an angle $360°/n$ about the axis, followed by reflecting in a plane perpendicular to the axis.
Defining the Coordinate System
Axis Definitions
Conventionally, when imposing a set of Cartesian axes on a molecule (as we will need to do later on in the course), the $z$ axis lies along the principal axis of the molecule, the $x$ axis lies in the plane of the molecule (or in a plane containing the largest number of atoms if the molecule is non-planar), and the $y$ axis makes up a right handed axis system.
Generally, the following conventions are observed to define the coordinate system:
1. The origin of the coordinate system is located at the central atom or the center of the molecule.
2. The $z$-axis is collinear with the highest-order rotational axis (principal axis).
3. For planar molecules, if the z-axis is perpendicular to the molecular plane:
• The $x$-axis lies in the plane of the molecule and passes through the greatest number of atoms.
If the $z$-axis lies in the plane of the molecule:
• The $x$-axis stands perpendicular to the plane
4. For non-planar molecules, once the z-axis has been defined, the x-axis is the usually chosen so that the xz plane contains as many atoms as possible.
Molecular Point Groups
It is only possible for certain combinations of symmetry elements to be present in a molecule (or any other object). As a result, we may group together molecules that possess the same symmetry elements and classify molecules according to their symmetry. These groups of symmetry elements are called point groups (due to the fact that there is at least one point in space that remains unchanged no matter which symmetry operation from the group is applied). There are two systems of notation for labeling symmetry groups, called the Schönflies and Hermann-Mauguin (or International) systems. Schönflies notation is used by chemists and spectroscopists, while Crystallographers prefer Hermann-Mauguin notation.The symmetry of individual molecules is usually described using the Schönflies notation, which is used below. The common point groups can be categorized into the following:
• Nonrotational groups
• Single-axis rotationa groups
• Dihedral groups
• Cubic groups
Shared Names
Some of the point groups share their names with symmetry operations, so be careful you do not mix up the two. It is usually clear from the context which one is being referred to.
Nonrotational groups
Nonrotational groups represent the lowest symmetry groups. They include:
$C_1$ - Contains only the identity (a $C_1$ rotation is a rotation by 360° and is the same as the identity operation, $E$). Molecules that belongs to the $C_1$ group have no symmetry and are therefore asymmetric.
• Example: CHDFCl
$C_i$ - Contains the identity $E$ and a center of inversion $i$.
• Example: C2H2F2Cl2
$C_S$ - Contains the identity $E$ and a plane of reflection $\sigma$.
Single-axis rotation groups
Single-axis rotation groups are examples of cyclic groups. In cyclic groups, all of the operators commute (Abelian). In their multiplication tables, elements appear along right-to-left diagonals. Knowing this pattern makes it easy to construct multiplication tables! Single-axis rotational groups include:
$C_n$ - Contains the identity and an $n$-fold axis of rotation.
$C_{nv}$ - Contains the identity, an $n$-fold axis of rotation, and $n$ vertical mirror planes $\sigma_v$.
$C_{nh}$ - Contains the identity, an $n$-fold axis of rotation, and a horizontal reflection plane $\sigma_h$ (note that in $C_{2h}$ this combination of symmetry elements automatically implies a center of inversion).
$S_n$ - Contains the identity and one $S_n$ axis. Note that molecules only belong to $S_n$ if they have not already been classified in terms of one of the preceding point groups (e.g. $S_2$ is the same as $C_i$, and a molecule with this symmetry would already have been classified).
Dihedral groups
The dihedral groups have $n$ two-fold axes perpendicular to the principal $n$-fold axis. These $C_2$ axis are called dihedral axes. Dihedral groups groups include:
$D_n$ - Contains the identity, an $n$-fold axis of rotation, and $n$ 2-fold rotations about axes perpendicular to the principal axis.
$D_{nd}$ - Contains the same symmetry elements as $D_n$ with the addition of $n$ dihedral mirror planes.
$D_{nh}$ - Contains the same symmetry elements as $D_n$ with the addition of a horizontal mirror plane.
• There are $n$-fold vertical mirror places ($\sigma_v$)
• Includes $n$-fold improper axis when $n>2$
• They are centrosymmetric when $n$ is even
$D_{\infty h}$ - Same as $D_{nh}$, except the principal rotational axis is an infinite-fold $C_\infty$.
• This is the point group for all linear centrosymmetric linear molecules (e.g. H2, CO2)
Cubic groups
The following groups are the cubic groups, which contain more than one principal axis. The cubic groups are associated with polyhedra that are geometrically related to the cube. All are characterized by the presence of multiple, intersecting, high-order rotational axes. They separate into the tetrahedral groups ($T_d$, $T_h$ and $T$) and the octahedral groups ($O$ and $O_h$).
$T_d$ - Contains all the symmetry elements of a regular tetrahedron, including the identity, 4 $C_3$ axes, 3 $C_2$ axes, 6 dihedral mirror planes, and 3 $S_4$ axes e.g. $\ce{CH_4}$.
• Example: Methane, $\ce{C_4}$
$T$ - Same as for $T_d$ but no planes of reflection.
$T_h$ - Same as for $T$ but contains a center of inversion.
$O_h$ - The group of the regular octahedron.
• Example: Sulfur hexafluoride, $\ce{SF_6}$
$O$ - Same as for $O_h$, but with no planes of reflection.
$I_h$ - Icoshedral, a geometric shape with 20 faces.
• Example: Buckminsterfullerene, ($\ce{C_{60}}$)
Summary of groups
The final group is the full rotation group $R_3$, which consists of an infinite number of $C_n$ axes with all possible values of $n$ and describes the symmetry of a sphere. Atoms (but no molecules) belong to $R_3$, and the group has important applications in atomic quantum mechanics. However, we won’t be treating it any further here.
Table 12.2.2 : Common Point Groups for Molecules
Nonaxial groups C1 Cs Ci - - - - - -
Cn groups C2 C3 C4 C5 C6 C7 C8 - -
Dn groups D2 D3 D4 D5 D6 D7 D8 - -
Cnv groups C2v C3v C4v C5v C6v C7v C8v - -
Cnh groups C2h C3h C4h C5h C6h - - - -
Dnh groups D2h D3h D4h D5h D6h D7h D8h - -
Dnd groups D2d D3d D4d D5d D6d D7d D8d - -
Sn groups S2 - S4 - S6 S8 S10 S12
Cubic groups T Th Td O Oh I Ih - -
Linear groups Cv Dh - - - - - - -
Once you become more familiar with the symmetry elements and point groups described above, you will find it quite straightforward to classify a molecule in terms of its point group. In the meantime, the flowchart shown below provides a step-by-step approach to the problem.
1Though the Hermann-Mauguin system can be used to label point groups, it is usually used in the discussion of crystal symmetry. In crystals, in addition to the symmetry elements described above, translational symmetry elements are very important. Translational symmetry operations leave no point unchanged, with the consequence that crystal symmetry is described in terms of space groups rather than point groups.
Symmetry and Physical Properties
Carrying out a symmetry operation on a molecule must not change any of its physical properties. It turns out that this has some interesting consequences, allowing us to predict whether or not a molecule may be chiral or polar on the basis of its point group.
For a molecule to have a permanent dipole moment, it must have an asymmetric charge distribution. The point group of the molecule not only determines whether the molecule may have a dipole moment, but also in which direction(s) it may point. If a molecule has a $C_n$ axis with $n > 1$, it cannot have a dipole moment perpendicular to the axis of rotation (for example, a $C_2$ rotation would interchange the ends of such a dipole moment and reverse the polarity, which is not allowed – rotations with higher values of $n$ would also change the direction in which the dipole points). Any dipole must lie parallel to a $C_n$ axis.
Also, if the point group of the molecule contains any symmetry operation that would interchange the two ends of the molecule, such as a $\sigma_h$ mirror plane or a $C_2$ rotation perpendicular to the principal axis, then there cannot be a dipole moment along the axis. The only groups compatible with a dipole moment are $C_n$, $C_{nv}$ and $C_s$. In molecules belonging to $C_n$ or $C_{nv}$ the dipole must lie along the axis of rotation.
One example of symmetry in chemistry that you will already have come across is found in the isomeric pairs of molecules called enantiomers. Enantiomers are non-superimposable mirror images of each other, and one consequence of this symmetrical relationship is that they rotate the plane of polarized light passing through them in opposite directions. Such molecules are said to be chiral,2 meaning that they cannot be superimposed on their mirror image. Formally, the symmetry element that precludes a molecule from being chiral is a rotation-reflection axis $S_n$. Such an axis is often implied by other symmetry elements present in a group.
For example, a point group that has $C_n$ and $\sigma_h$ as elements will also have $S_n$. Similarly, a center of inversion is equivalent to $S_2$. As a rule of thumb, a molecule definitely cannot have be chiral if it has a center of inversion or a mirror plane of any type ($\sigma_h$, $\sigma_v$ or $\sigma_d$), but if these symmetry elements are absent the molecule should be checked carefully for an $S_n$ axis before it is assumed to be chiral.
Chirality
The word chiral has its origins in the Greek word for hand ($\chi$$\epsilon$$\rho$$\iota$, pronounced ‘cheri’ with a soft ch as in ‘loch’). A pair of hands is also a pair of non-superimposable mirror images, and you will often hear chirality referred to as ‘handedness’ for this reason.
Summary
All molecules can be described in terms of their symmetry or lack thereof, which may contain symmetry elements (point, line, plane). Identity, rotation, reflection, and inversion are symmetry operations (movement of the molecules such that after the movement, all the atoms of the molecules is coincidental with equivalent atom of the molecule in original).
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Properties of Groups
Now that we have explored some of the properties of symmetry operations and elements and their behavior within point groups, we are ready to introduce the formal mathematical definition of a group. A mathematical group is defined as a set of elements ($g_1$, $g_2$, $g_3$...) together with a rule for forming combinations $g_j$. The number of elements $h$ is called the order of the group. For our purposes, the elements are the symmetry operations of a molecule and the rule for combining them is the sequential application of symmetry operations investigated in the previous section. The elements of the group and the rule for combining them must satisfy the following criteria:
1. Identity
2. Closure
3. Associativity
4. Reciprocality
These criteria are explained below.
Identity
The group must include the identity, $E$. $E$ commutes with any other elements of the group, $g_i$, such that:
$E g_i= g_i E = g_i \label{7.1}$
This requirement explains the need to define the symmetry operation of identity.
Closure
The elements must satisfy the group property of closure, meaning that the combination of any pair of elements is also an element of the group.
Closure is a mathematical definition. In mathematics, a group has closure under an operation if performance of that operation on members of the group always produces a member of the same group:
If $A$ and $B$ are elements of the group $G$, and if $AB=g_i$, then $g_i$ is also in the group $G$
Reciprocality
To satisfy reciprocality, each element $g_i$ must have an inverse $g_i^{-1}$, which is also an element of the group, such that:
$g_i g_i^{-1} = g_i^{-1}g_i = E \label{7.2}$
Some symmetry operations are their own inverses:
• $C_2 C_2=E$
• $\sigma \sigma=E$
• $ii=E$
• $EE=E$
The inverse of each of these operations effectively 'undoes’ the effect of the symmetry operation. Most other operations are not the inverse of themselves. For example, in $C_{3v}$ the inverse of $C_3$ is $C_3^{-1}$.
Associativity
The associative law of combination states that all combinations of elements of a group must be associative:
$(g_i g_j )(g_k) = g_i(g_jg_k) \label{7.3}$
The above definition does not require the elements to commute, which would require:
$g_i g_k =g_k g_i \label{7.4}$
Group Multiplication
As we discovered in the $C_{3v}$ example above, in many groups the outcome of consecutive application of two symmetry operations depends on the order in which the operations are applied.
Commuting is not a Requirement of Group Elements
Groups for which the elements do not commute are called non-Abelian groups; those for which they elements do commute are Abelian.
Group theory is an important area in mathematics, and luckily for chemists the mathematicians have already done most of the work for us. Along with the formal definition of a group comes a comprehensive mathematical framework that allows us to carry out a rigorous treatment of symmetry in molecular systems and learn about its consequences.
Many problems involving operators or operations (such as those found in quantum mechanics or group theory) may be reformulated in terms of matrices. Any of you who have come across transformation matrices before will know that symmetry operations such as rotations and reflections may be represented by matrices. It turns out that the set of matrices representing the symmetry operations in a group obey all the conditions laid out above in the mathematical definition of a group, and using matrix representations of symmetry operations simplifies carrying out calculations in group theory. Before we learn how to use matrices in group theory, it will probably be helpful to review some basic definitions and properties of matrices.
Now we will investigate what happens when we apply two symmetry operations in sequence. As an example, consider the $NH_3$ molecule, which belongs to the $C_{3v}$ point group. Consider what happens if we apply a $C_3$rotation followed by a $\sigma_v$ reflection. We write this combined operation $\sigma_v$$C_3$ (when written, symmetry operations operate on the thing directly to their right, just as operators do in quantum mechanics – we therefore have to work backwards from right to left from the notation to get the correct order in which the operators are applied). As we shall soon see, the order in which the operations are applied is important.
The combined operation $\sigma_v$$C_3$ is equivalent to $\sigma_v''$, which is also a symmetry operation of the $C_{3v}$ point group. Now let’s see what happens if we apply the operators in the reverse order i.e. $C_3$$\sigma_v$ ($\sigma_v$ followed by $C_3$).
Again, the combined operation $C_3$$\sigma_v$ is equivalent to another operation of the point group, this time $\sigma_v'$.
There are two important points that are illustrated by this example:
1. The order in which two operations are applied is important. For two symmetry operations $A$ and $B$, $AB$ is not necessarily the same as $BA$, i.e. symmetry operations do not in general commute. In some groups the symmetry elements do commute; such groups are said to be Abelian.
2. If two operations from the same point group are applied in sequence, the result will be equivalent to another operation from the point group. Symmetry operations that are related to each other by other symmetry operations of the group are said to belong to the same class. In $NH_3$, the three mirror planes $\sigma_v$, $\sigma_v'$ and $\sigma_v''$ belong to the same class (related to each other through a $C_3$ rotation), as do the rotations $C_3^+$ and $C_3^-$ (anticlockwise and clockwise rotations about the principal axis, related to each other by a vertical mirror plane
The effects of applying two symmetry operations in sequence within a given point group are summarized in group multiplication tables. As an example, the complete group multiplication table for $C_{3v}$ using the symmetry operations as defined in the figures above is shown below. The operations written along the first row of the table are carried out first, followed by those written in the first column (note that the table would change if we chose to name $\sigma_v$, $\sigma_v'$ and $\sigma_v''$ in some different order).
$\begin{array}{l|llllll} C_{3v} & E & C_3^+ & C_3^- & \sigma_v & \sigma_v' & \sigma_v'' \ \hline E & E & C_3^+ & C_3^- & \sigma_v & \sigma_v' & \sigma_v'' \ C_3^+ & C_3^+ & C_3^- & E & \sigma_v' & \sigma_v'' & \sigma_v \ C_3^- & C_3^- & E & C_3^+ & \sigma_v'' & \sigma_v & \sigma_v' \ \sigma_v & \sigma_v & \sigma_v'' & \sigma_v' & E & C_3^- & C_3^+ \ \sigma_v' & \sigma_v' & \sigma_v & \sigma_v'' & C_3^+ & E & C_3^- \ \sigma_v'' & \sigma_v'' & \sigma_v' & \sigma_v & C_3^- & C_3^+ & E \end{array} \label{5.1}$
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Matrices can be used to map one set of coordinates or functions onto another set. Matrices used for this purpose are called transformation matricesThe symmetry operations in a group may be represented by a set of transformation matrices $\Gamma$$(g)$, one for each symmetry element $g$. Each individual matrix is called a representative of the corresponding symmetry operation, and the complete set of matrices is called a matrix representation of the group. The matrix representatives act on some chosen basis set of functions, and the actual matrices making up a given representation will depend on the basis that has been chosen. The representation is then said to span the chosen basis. The basis we will use are unit vectors pointing in the $x$, $y$, and $z$ directions.
The transformation matrix for any operation in a group has a form that is unique from the matrices of the other members of the same group; however, the character of the transformation matrix for a given operation is the same as that for any other operation in the same class. Each symmetry operation below will operate on an arbitrary vector, $\bf{u}$:
$\bf{u} = \begin{pmatrix} x \ y \ z \end{pmatrix}$
The Identity Operation, $E$
The first rule is that the group must include the identity operation $E$ (the ‘do nothing’ operation). The matrix representative of the identity operation is simply the identity matrix and leaves the vector unchanged:
$E\bf{u} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x \ y \ z \end{pmatrix} \label{9.1}$
Every matrix representation includes the appropriate identity matrix.
The Reflection Operation, $\sigma$
The reflection operation reflects the vector $\bf{u}$ over a plane. This can be the $xy$, $xz$, or $yz$ plane. The matrix is similar to the identity matrix, with the exception that there is a sign change for the appropriate element. The reflect matrix in the $xy$ plane is:
$\sigma (xy) \bf{u} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x \ y \ -z \end{pmatrix} \label{9.2}$
Notice that the element for the dimension being reflected is the on that is negative. In the above case, since $z$ is being reflected over the $xy$ plane, the $z$ element in the matrix is negative. If we were to reflect over the $xz$ plane instead, the $y$ element would be the one that is negative:
$\sigma (xz) \bf{u} = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x \ -y \ z \end{pmatrix}$
The $n$-fold Rotation Operation, $C_n$
The $C_n$ operator rotates the molecule about an axis. The counterclockwise rotation of vector $\bf{u}$ about the $z$ axis is:
$C_n\bf{u} = \begin{pmatrix} \cos{\frac{2\pi}{n}} & -\sin{\frac{2\pi}{n}} & 0 \ \sin{\frac{2\pi}{n}} & \cos{\frac{2\pi}{n}} & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x' \ y' \ z \end{pmatrix}$
For clockwise rotation, the sign on the $\sin{\theta}$ terms are reversed. This matrix simplifies dramatically for the $C_2$ rotation:
$C_2\bf{u} = \begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -x \ -y \ z \end{pmatrix}$
Rotation matrices operating about the $x$, $y$ and $z$ axes are given by:
$R_{x}(\theta) = \begin{pmatrix} 1 & 0 & 0 \ 0 & \cos\theta & -\sin\theta \ 0 & \sin\theta & \cos\theta \end{pmatrix} \label{9.6a}$
$R_{y}(\theta) = \begin{pmatrix} \cos\theta & 0 & -\sin\theta \ 0 & 1 & 0 \ \sin\theta & 0 & \cos\theta \end{pmatrix} \label{9.6b}$
$R_{z}(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \ \sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{pmatrix} \label{9.6c}$
The Inversion operation, $I$
The inversion operation inverts every point:
$I\bf{u} = \begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -x \ -y \ -z \end{pmatrix}$
$C_{2v}$ Point Group
Now that we have seen the matrix form of our operators, we can see how the multiplication of each operator leads to another operations in the group. The $C_{2v}$ multiplication table is:
$C_{2v}$ $E$ $C_2$ $\sigma_v$ $\sigma_v'$
$E$ $E$ $C_2$ $\sigma_v$ $\sigma_v'$
$C_2$ $C_2$ $E$ $\sigma_v'$ $\sigma_v$
$\sigma_v$ $\sigma_v$ $\sigma_v'$ $E$ $C_2$
$\sigma_v'$ $\sigma_v'$ $\sigma_v$ $C_2$ $E$
For multiplication tables, the standard order of operation is the row elements (top) first, following by the column element (side). Every row or column includes every operation once and is different from any other row or column.
Let's look at $C_2 \sigma_v$ multiplication, where $\sigma_v$ is a reflection across the $xz$ plane:
$C_2 \sigma_v = \begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \sigma_v'$
where $\sigma_v'$ is the reflection across the $yz$ plane.
Matrix Representations and Groups
Before proceeding any further, we must check that a matrix representation of a group obeys all of the rules set out in the formal mathematical definition of a group.
1. Identity.The first rule is that the group must include the identity operation $E$ (the ‘do nothing’ operation). We showed above that the matrix representative of the identity operation is simply the identity matrix. As a consequence, every matrix representation includes the appropriate identity matrix.
2. Closure. The second rule is that the combination of any pair of elements must also be an element of the group (the group property). If we multiply together any two matrix representatives, we should get a new matrix which is a representative of another symmetry operation of the group. In fact, matrix representatives multiply together to give new representatives in exactly the same way as symmetry operations combine according to the group multiplication table. For example, in the $C_{2v}$ point group, we showed that the combined symmetry operation $C_2$$\sigma_v$ is equivalent to $\sigma_v'$. In a matrix representation of the group, if the matrix representatives of $C_2$ and $\sigma_v$ are multiplied together, the result will be the representative of $\sigma_v'$.
3. Associativity. The third rule states that the rule of combination of symmetry elements in a group must be associative. This is automatically satisfied by the rules of matrix multiplication.
4. Reciprocality. The final rule states that every operation must have an inverse, which is also a member of the group. The combined effect of carrying out an operation and its inverse is the same as the identity operation. It is fairly easy to show that matrix representatives satisfy this criterion. For example, the inverse of a reflection is another reflection, identical to the first. In matrix terms we would therefore expect that a reflection matrix was its own inverse, and that two identical reflection matrices multiplied together would give the identity matrix. This turns out to be true, and can be verified using any of the reflection matrices in the examples above. The inverse of a rotation matrix is another rotation matrix corresponding to a rotation of the opposite sense to the first.
Example 12.4.1 : Matrix Representation of the $C_{2v}$ Point Group (the allyl radical)
We often use sets of atomic orbitals as basis functions for matrix representations. In this example, we’ll take as our basis a $p$ orbital on each carbon atom $\begin{pmatrix} p_1, p_2, p_3 \end{pmatrix}$.
Note that the $p$ orbitals are perpendicular to the plane of the carbon atoms (this may seem obvious, but if you’re visualizing the basis incorrectly it will shortly cause you a not inconsiderable amount of confusion). The symmetry operations in the $C_{2v}$ point group, and their effect on the three $p$ orbitals, are as follows:
$\begin{array}{ll} E & \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \rightarrow \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \ C_2 & \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \rightarrow \begin{pmatrix} -p_3 \ -p_2 \ -p_1 \end{pmatrix} \ \sigma_v & \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \rightarrow \begin{pmatrix} -p_1 \ -p_2 \ -p_3 \end{pmatrix} \ \sigma_v' & \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \rightarrow \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \end{array} \nonumber$
The matrices that carry out the transformation are
$\begin{array}{ll} \Gamma(E) & \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} = \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \ \Gamma(C_2) & \begin{pmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} = \begin{pmatrix} -p_3 \ -p_2 \ -p_1 \end{pmatrix} \ \Gamma(\sigma_v) & \begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} = \begin{pmatrix} -p_1 \ -p_2 \ -p_3 \end{pmatrix} \ \Gamma(\sigma_v') & \begin{pmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} = \begin{pmatrix} p_1 \ p_2 \ p_3 \end{pmatrix} \end{array} \nonumber$
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In addition to operators, we can define properties of molecules using a matrix representation. Before making the matrix, we need to carefully choose a basis set that defines the information we want to extract For example, let's say we want to know the symmetry of the valence $s$ orbitals in ammonia, $\sf NH_3$, which is in the $C_{3v}$ point group. We will select a basis $\begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix}$ that consists of the valence s orbitals on the nitrogen and the three hydrogen atoms. We need to consider what happens to this basis when it is acted on by each of the symmetry operations in the $C_{3v}$ point group, and determine the matrices that would be required to produce the same effect. The basis set and the symmetry operations in the $C_{3v}$ point group are summarized in the figure below.
The effects of the symmetry operations on our chosen basis are as follows:
$\begin{array}{ll} E & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \rightarrow \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \ C_3^+ & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \rightarrow \begin{pmatrix} s_N, s_2, s_3, s_1 \end{pmatrix} \ C_3^- & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \rightarrow \begin{pmatrix} s_N, s_3, s_1, s_2 \end{pmatrix} \ \sigma_v & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \rightarrow \begin{pmatrix} s_N, s_1, s_3, s_2 \end{pmatrix} \ \sigma_v' & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \rightarrow \begin{pmatrix} s_N, s_2, s_1, s_3 \end{pmatrix} \ \sigma_v'' & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \rightarrow \begin{pmatrix} s_N, s_3, s_2, s_1 \end{pmatrix} \end{array} \label{10.1}$
By inspection, the matrices that carry out the same transformations are:
$\begin{array}{ll} \Gamma(E) & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \ \Gamma(C_3^+) & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} s_N, s_2, s_3, s_1 \end{pmatrix} \ \Gamma(C_3^-) & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 0 & 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} s_N, s_3, s_1, s_2 \end{pmatrix} \ \Gamma(\sigma_v) & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} s_N, s_1, s_3, s_2 \end{pmatrix} \ \Gamma(\sigma_v') & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} s_N, s_2, s_1, s_3 \end{pmatrix} \ \Gamma(\sigma_v'') & \begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} s_N, s_3, s_2, s_1 \end{pmatrix} \end{array} \label{10.2}$
These six matrices therefore form a reducible representation for the $C_{3v}$ point group in the $\begin{pmatrix} s_N, s_1, s_2, s_3 \end{pmatrix}$ basis as we will see in the next chapter that these matrices reduce down to the irreducible representations found in the character tables. These reducible representations multiply together according to the group multiplication table and satisfy all the requirements for a mathematical group.
We choose different basis sets to extract different properties of molecules. For example, we could include representations of the valence $p$ orbitals in N in our basis set to obtain the structure and symmetry of the molecular orbitals for ammonia. To understand understand the molecular motions of ammonia (translates, rotates, and vibrates), we could place a $x$, $y$, and $z$ unit vectors on each atom to represent the their motion, and then construct our matrices.
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Now that we’ve learned how to create a matrix representation of a point group within a given basis, we will move on to look at some of the properties that make these representations so powerful in the treatment of molecular symmetry.
Similarity Transforms
Suppose we have a basis set $\begin{pmatrix} x_1, x_2, x_3, ... x_n \end{pmatrix}$, where $x_i$ represents so position in $x$ of each atom, and we have determined the matrix reducible representatives for the basis in a given point group. There is nothing particularly special about the basis set we have chosen, and we could equally well have used any set of linear combinations of the original functions (provided the combinations were linearly independent). Consider an additional basis set $\begin{pmatrix} x_1', x_2', x_3', ... x_n' \end{pmatrix}$, in which each basis function $x_i'$ is a linear combination of our original basis $\begin{pmatrix} x_1, x_2, x_3, ... x_n \end{pmatrix}$:
$x_j' = \Sigma_i x_ic_{ji} = x_1c_{j1} + x_2c_{j2} + ... \tag{11.1}$
The $c_{ji}$ appearing in the sum are coefficients; $c_{ji}$ is the coefficient multiplying the original basis function $x_i$ in the new linear combination basis function $x_j'$. We could also represent this transformation in terms of a matrix equation $\textbf{x'} = C\textbf{x}$:
$\begin{pmatrix} x_1' \ x_2' \ ... \ x_n' \end{pmatrix} = \begin{pmatrix} c_{11} & c_{12} & ... & c_{1n} \ c_{21} & c_{22} & ... & c_{2n} \ ... & ... & ... & ... \ c_{n1} & c_{n2} & ... & c_{nn} \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ ... \ x_n \end{pmatrix} \tag{11.2}$
where $C$ is the transformation matrix. The matrix representatives for the two basis sets will certainly be different, but we would expect them to be related to each other in some way. As we shall show shortly, they are in fact related by a similarity transform. Now we look at what happens when we apply a symmetry operation $g$ to our two basis sets. If $\Gamma(g)$ and $\Gamma'(g)$ are matrix representatives of the symmetry operation in the $\textbf{x}$ and $\textbf{x'}$ bases, then we have:
$\begin{array}{rcll} g\textbf{x'} & = & \textbf{x'}\Gamma'(g) \ g\textbf{x}C & = & \textbf{x}C\Gamma'(g) & \text{since} \: \textbf{x'} = \textbf{x}C \ g\textbf{x} & = & \textbf{x}C\Gamma'(g)C^{-1} & \text{multiplying on the right by} \: C^{-1} \text{and using} \: CC^{-1} = I \ & = & \textbf{x}\Gamma(g) \end{array} \tag{11.3}$
We can therefore identify the similarity transform relating $\Gamma(g)$, the matrix representative in our original basis, to $\Gamma'(g)$, the representative in the transformed basis. The transform depends only on the matrix of coefficients used to transform the basis functions:
$\Gamma(g) = C\Gamma'(g)C^{-1} \tag{11.4}$
Also:
$\Gamma'(g) = C^{-1}\Gamma(g)C \tag{11.5}$
Characters of Representations
The trace of a matrix representative $\Gamma(g)$ is usually referred to as the character of the representation under the symmetry operation $g$. We will soon come to see that the characters of a matrix representation are often more useful than the matrix representatives themselves. Characters have several important properties.
1. The character of a symmetry operation is invariant under a similarity transform
2. Symmetry operations belonging to the same class have the same character in a given representation. Note that the character for a given class may be different in different representations, and that more than one class may have the same character.
Proofs of the above two statements are given in the Appendix.
Character Tables
A character table summarizes the behavior of all of the possible irreducible representations of a group under each of the symmetry operations of the group. The character table for $C_{3v}$ is shown below. All operations in the character table are contained in the first row of the character table, in this case $E$, $C_3$, & $\sigma_v$, these are all of the operations that can be preformed on the molecule that return the original structure. The first column contains the three irreducible representations $A_1$, $A_2$ & $E$. The character of the irreducible representation denotes what the operation does. A value of 1 represents no change, -1 opposite change and 0 is a combination of 1 & -1 (0’s are found in degenerate molecules. The final two columns Rotation and Translation represented by $R_x$,$R_y$, $R_z$ & $x$, $y$, $z$ respectively. Each $R_x$, $R_y$, $R_z$ & $x$, $y$, $z$ term is the irreducible symmetry of a rotation or translation operation. Like wise the final column the orbital symmetries relates the orbital wave function to a irreducible representation.
$\begin{array}{lllll} \hline C_{3v} & E & 2C_3 & 3\sigma_v & h=6 \ \hline A_1 & 1 & 1 & 1 & z, z^2, x^2+y^2 \ A_2 & 1 & 1 & -1 & R_z \ E & 2 & -1 & 0 & \begin{pmatrix} x, y \end{pmatrix}, \begin{pmatrix} xy, x^2+y^2 \end{pmatrix}, \begin{pmatrix} xz, yz \end{pmatrix}, \begin{pmatrix} R_x, R_y \end{pmatrix} \ \hline \end{array} \label{14.1}$
The various sections of the table are as follows:
1. The first element in the table gives the name of the point group, usually in both Schoenflies ($C_{3v}$) and Hermann-Mauguin ($3m$) notation.
2. Along the first row are the symmetry operations of the group, $E$, $2C_3$ and $3\sigma_v$, followed by the order of the group. Because operations in the same class have the same character, symmetry operations are grouped into classes in the character table and not listed separately.
3. In the first column are the irreducible representations of the group. In $C_{3v}$ the irreducible representations are $A_1$, $A_2$ and $E$ (the representation we considered above spans $2A_1$ + $E$).
4. The characters of the irreducible representations under each symmetry operation are given in the bulk of the table.
5. The final column of the table lists a number of functions that transform as the various irreducible representations of the group. These are the Cartesian axes $\begin{pmatrix} x, y, z \end{pmatrix}$, the Cartesian products $\begin{pmatrix} z^2, x^2 + y^2, xy, yz \end{pmatrix}$, and the rotations $\begin{pmatrix} R_x, R_y, R_z \end{pmatrix}$.
The functions listed in the final column of the table are important in many chemical applications of group theory, particularly in spectroscopy. For example, by looking at the transformation properties of $x$, $y$ and $z$ (sometimes given in character tables as $T_x$, $T_y$, $T_z$) we can discover the symmetry of translations along the $x$, $y$, and $z$ axes. Similarly, $R_x$, $R_y$ and $R_z$ represent rotations about the three Cartesian axes. As we shall see later, the transformation properties of $x$, $y$, and $z$ can also be used to determine whether or not a molecule can absorb a photon of $x$-, $y$-, or $z$-polarized light and undergo a spectroscopic transition. The Cartesian products play a similar role in determining selection rules for Raman transitions, which involve two photons.
Character tables for common point groups are given in Appendix B.
A simple way to determine the characters of a representation
In many applications of group theory, we only need to know the characters of the representative matrices, rather than the matrices themselves. Luckily, when each basis function transforms as a 1D irreducible representation (which is true in many cases of interest) there is a simple shortcut to determining the characters without having to construct the entire matrix representation. All we have to do is to look at the way the individual basis functions transform under each symmetry operation. For a given operation, step through the basis functions as follows:
1. Add $1$ to the character if the basis function is unchanged by the symmetry operation (i.e. the basis function is mapped onto itself);
2. Add $-1$ to the character if the basis function changes sign under the symmetry operation (i.e the basis function is mapped onto minus itself);
3. Add $0$ to the character if the basis function moves when the symmetry operation is applied (i.e the basis function is mapped onto something different from itself).
Try this for the $s$ orbital basis we have been using for the $C_{3v}$ group. You should find you get the same characters as we obtained from the traces of the matrix representatives.
We can also work out the characters fairly easily when two basis functions transform together as a 2D irreducible representation. For example, in the $C_{3v}$ point group $x$ and $y$ axes transform together as $E$. If we carry out a rotation about $z$ by an angle $\theta$, our $x$ and $y$ axes are transformed onto new axes $x'$ and $y'$. However, the new axes can each be written as a linear combination of our original $x$ and $y$ axes. Using the rotation matrices introduced in Section 9, we see that:
$\begin{array}{ccc}x' & = & \cos\theta \: x + \sin\theta \: y \ y' & = & -\sin\theta \: x + \cos\theta \: y \end{array} \label{14.2}$
For one-dimensional irreducible representations we asked if a basis function/axis was mapped onto itself, minus itself, or something different. For two-dimensional irreducible representations we need to ask how much of the ‘old’ axis is contained in the new one. From the above we see that the $x'$ axis contains a contribution $\cos{\theta}$ from the $x$ axis, and the $y'$ axis contains a contribution $\cos{\theta}$ from the $y$ axis. The characters of the $x$ and $y$ axes under a rotation through $\theta$ are therefore $\cos{\theta}$, and the overall character of the $E$ irreducible representation is therefore $\cos{\theta}+\cos{\theta}=2\cos{\theta}$. For a $C_3$ rotation through 120 degrees, the character of the $E$ irreducible representation is therefore $2\cos{120}$° $=-1$.
In general, when an axis is rotated by an angle $\theta$ by a symmetry operation, its contribution to the character for that operation is $\cos{\theta}$.
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Symbols of irreducible representations
The two one-dimensional irreducible representations spanned by $s_N$ and $s_1'$ are seen to be identical. This means that $s_N$ and $s_1'$ have the ‘same symmetry’, transforming in the same way under all of the symmetry operations of the point group and forming bases for the same matrix representation. As such, they are said to belong to the same symmetry species. There are a limited number of ways in which an arbitrary function can transform under the symmetry operations of a group, giving rise to a limited number of symmetry species. Any function that forms a basis for a matrix representation of a group must transform as one of the symmetry species of the group. The irreducible representations of a point group are labeled according to their symmetry species as follows:
$A$ Nondegenerate ($d_i=1$) representation that is symmetric (character $1$) with respect to rotation about the principal axis in finite-order ($h\ne\infty$) groups.
$B$ Nondegenerate ($d_i=1$) representation that is antisymmetric (character $-1$) with respect to rotation about the principal axis in finite-order ($h\neq\infty$) groups.
$E$ Doubly-degenerate ($d_i=2$) representation in finite-order ($h\neq\infty$) groups.
$T$ Triply-degenerate ($d_i=3$) representation in finite-order ($h\neq\infty$) groups.
$\Sigma$ Nondegenerate ($d_i=1$) representation in infinite-order ($h=\infty$) groups.
$\Pi$, $\Delta$, $\Phi$ Doubly-degenerate ($d_i=2$) representation in infinite-order ($h=\infty$) groups.
1. In groups containing a center of inversion, $g$ and $u$ labels (from the German gerade and ungerade, meaning symmetric and antisymmetric) denote the character of the irreducible representation under inversion ($+1$ for $g$, $-1$ for $u$)
2. In groups with a horizontal mirror plane but no center of inversion, the irreducible representations are given prime and double prime labels to denote whether they are symmetric (character $+1$ or antisymmetric (character $-1$) under reflection in the plane.
3. If further distinction between irreducible representations is required, subscripts $1$ and $2$ are used to denote the character with respect to a $C_2$ rotation perpendicular to the principal axis, or with respect to a vertical reflection if there are no $C_2$ rotations.
The 1D irreducible representation in the $C_{3v}$ point group is symmetric (has character $+1$) under all the symmetry operations of the group. It therefore belongs to the irreducible representation $A_1$. The 2D irreducible representation has character $2$ under the identity operation, $-1$ under rotation, and $0$ under reflection, and belongs to the irreducible representation $E$.
Sometimes there is confusion over the relationship between a function $f$ and its irreducible representation, but it is quite important that you understand the connection. There are several different ways of stating the relationship. For example, the following statements all mean the same thing:
• "$f$ has $A_2$ symmetry"
• "$f$ transforms as $A_2$"
• "$f$ has the same symmetry
Irreducible representations with complex characters
In many cases (see Appendix B), the characters for rotations $C_n$ and improper rotations $S_n$ are complex numbers, usually expressed in terms of the quantity $\epsilon$ = exp(2$\pi$i/n). It is fairly straightforward to reconcile this with the fact that in chemistry we are generally using group theory to investigate physical problems in which all quantities are real. It turns out that whenever our basis spans an irreducible representation whose characters are complex, it will also span a second irreducible representation whose characters are the complex conjugates of the first irreducible representation i.e. complex irreducible representations occur in pairs. According to the strict mathematics of group theory, each irreducible representation in the pair should be considered as a separate representation. However, when applying such irreducible representations in physical problems, we add the characters for the two irreducible representations together to get a single irreducible representation whose characters are real.
As an example, the ‘correct’ character table for the group $C_3$ takes the form:
$\begin{array}{l|l} C_3 & E \: \: \: \: \: \: \: \: C_3 \: \: \: \: \: \: \: \: C_3^2 \ \hline A & 1 \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: 1 \ \hline E & \begin{Bmatrix} 1 & \epsilon & \epsilon* \ 1 & \epsilon* & \epsilon \end{Bmatrix} \end{array} \label{14.3}$
Where $\epsilon$ = exp(2$\pi$i/3). However, as chemists we would usually combine the two parts of the $E$ irreducible representation to give:
$\begin{array}{l|lll} C_3 & E & C_3 & C_3^2 \ \hline A & 1 & 1 & 1 \ E & 2 & -1 & 1 \end{array} \label{14.4}$
Groups and subgroups have well-defined relationships as they descend or ascend in symmetry
Molecules can undergo structural changes through conformations or chemical reactions. Remember that the order of the subgroup must be an integer divisor from the order of the group. If the basic geometry of the molecule is preserved, the structure after the change may be a subgroup of the structure before the change, or vice versa. If the basic geometry does change, there may not be a relationship between the groups before and after. When point groups are related as group and subgroup, their irreducible representations are also related. A property that transforms as one representation in a group will transform as its correlated representation in a subgroup. Correlation diagrams show the relationships between subgroups and groups. Often, to or more bases of separate representations of a group yield the same set of $\chi(R)$ values for those operations that are carried over into the subgroup. In many cases, degenerate representations of a group ($E$ or $T$) may become two or three distinguishable bases in a subgroup.
Reduce representations of infinite groups by approximating them as finite groups
We cannot use the tabular method for infinite-order groups since we cannot divide an infinite quantity by $h$. We will use group-subgroup relations to use the tabular method:
• Set up the reducible representation in any convenient subgroup
• For $C_{\infty v}$, use $C_{2v}$
• For $D_{\infty h}$, use $D_{2h}$
• Correlate the component irreducible representations with the species for the infinite-order group.
While somewhat limiting, this method is fairly effective.
The direct product of two irreducible representations give either a reducible or irreducible representation of the same group
The last column in the character table shows the direct product between any two linear vectors. Direct products can also be taken between any number of irreducible representations:
$\Gamma_a \Gamma_b \Gamma_c=\Gamma_{abc}$
The characters of the direct product representation $\Gamma_{abc}$ for each operator R of the group are given by:
$\chi_a (R) \chi_b (R) \chi_c (R)=\chi_{abc}(R)$
The resulting representation may be reducible or irreducible. The dimension of the product, $D_p$, is the product of the dimensions of all the component representations:
$d_p=\prod_i{d_i}$
These properties will become useful later, so we will reference them as needed. Briefly:
1. If all the combined irreducible representations are nondegenerate, then the product will be a nondegenerate representation too.
2. The product of a nondegenerate representation and a degenerate representation is a degenerate representation.
3. The direct product of any representation with the totally symmetric representation is the representation itself.
4. The direct product of degenerate representations is a reducible representation.
5. The direct product of an irreducible representation with itself is or contains the totally symmetry representation.
6. Only the direct product of a representation with itself is or contains the totally symmetric representation.
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As we continue with this course, we will discover that there are many times when we would like to know whether a particular integral is necessarily zero, or whether there is a chance that it may be non-zero. We can often use group theory to differentiate these two cases.
You will have already used symmetry properties of functions to determine whether or not a one-dimensional integral is zero. For example, cos(x) is an ‘even’ function (symmetric with respect to reflection through the origin), and it follows from this that
$\int^{\infty}_{-\infty} \cos(x) dx = 0 \nonumber$
In general integral between these limits for any other even function will be also be zero.
In the general case we may have an integral of more than one dimension. The key to determining whether a general integral is necessarily zero lies in the fact that because an integral is just a number, it must be invariant to any symmetry operation. For example, bonding in a diatomic (see next section) depends on the presence of a non-zero overlap between atomic orbitals on adjacent atoms, which may be quantified by an overlap integral. You would not expect the bonding in a molecule to change if you rotated the molecule through some angle $\theta$, so the integral must be invariant to rotation, and indeed to any other symmetry operation.
In group theoretical terms, for an integral to be non-zero, the integrand must transform as the totally symmetric irreducible representation in the appropriate point group. In practice, the integrand may not transform as a single irreducible representation, but it must include the totally symmetric irreducible representation. These ideas should become more clear in the next section.
It should be noted that even when the irreducible representations spanned by the integrand do include the totally symmetric irreducible representation, it is still possible for the integral to be zero. All group theory allows us to do is identify integrals that are necessarily zero based on the symmetry (or lack thereof) of the integrand.
Secular Equations
As we have seen already, any set of linear equations may be rewritten as a matrix equation $A\textbf{x}$ = $\textbf{b}$. Linear equations are classified as simultaneous linear equations or homogeneous linear equations, depending on whether the vector $\textbf{b}$ is non-zero or zero. For a set of simultaneous linear equations (non-zero $\textbf{b}$) it is fairly apparent that if a unique solution exists, it can be found by multiplying both sides by the inverse matrix $A^{-1}$ (since $A^{-1}A$ on the left hand side is equal to the identity matrix, which has no effect on the vector $\textbf{x}$)
$\begin{array}{rcl} A\textbf{x} & = & \textbf{b} \ A^{-1}A\textbf{x} & = & A^{-1}\textbf{b} \ \textbf{x} & = & A^{-1}\textbf{b} \end{array} \label{21.1}$
In practice, there are easier matrix methods for solving simultaneous equations than finding the inverse matrix, but these need not concern us here. We previously argued that in order for a matrix to have an inverse, it must have a non-zero determinant. Since $A^{-1}$ must exist in order for a set of simultaneous linear equations to have a solution, this means that the determinant of the matrix $A$ must be non-zero for the equations to be solvable.
For a matrix to have an inverse, it must have a non-zero determinant.
The reverse is true for homogeneous linear equations. In this case the set of equations only has a solution if the determinant of $A$ is equal to zero. The secular equations we want to solve are homogeneous equations, and we will use this property of the determinant to determine the molecular orbital energies. An important property of homogeneous equations is that if a vector $\textbf{x}$ is a solution, so is any multiple of $\textbf{x}$, meaning that the solutions (the molecular orbitals) can be normalized without causing any problems.
Solving for the orbital energies and expansion coefficients
Recall the secular equations for the $A_1$ orbitals of $NH_3$ derived in the previous section
$\begin{array}{rcl} c_1(H_{11} - ES_{11}) + c_2(H_{12} - ES_{12}) & = & 0 \ c_1(H_{12} - ES_{12}) + c_2(H_{22} - ES_{22}) & = & 0 \end{array} \label{21.2}$
where $c_1$ and $c_2$ are the coefficients in the linear combination of the SALCs $\phi_1$ = $s_N$ and $\phi_2$ = $\dfrac{1}{\sqrt{3}}(s_1 + s_2 + s_3)$ used to construct the molecular orbital. Writing this set of homogeneous linear equations in matrix form gives
$\begin{pmatrix} H_{11} - ES_{11} & H_{12} - ES_{12} \ H_{12} - ES_{12} & H_{22} - ES_{22} \end{pmatrix} \begin{pmatrix} c_1 \ c_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \label{21.3}$
In order for the equations to have a solution, the determinant of the matrix must be equal to zero. Writing out the determinant will give us a polynomial equation in $E$ that we can solve to obtain the orbital energies in terms of the Hamiltonian matrix elements $H_{ij}$ and overlap integrals $S_{ij}$. The number of energies obtained by ‘solving the secular determinant’ in this way is equal to the order of the matrix, in this case two.
The secular determinant for Equation (21.3) is (noting that $S_{11}$ = $S_{22} = 1$ since the SALCs are normalized)
$(H_{11} - E)(H_{22} - E) - (H_{12} - ES_{12})^2 = 0 \label{21.4}$
Expanding and collecting terms in $E$ gives
$E^2(1-S_{12}^2) + E(2H_{12}S_{12} - H_{11} - H_{22}) + (H_{11}H_{22} - H_{12}^2) = 0 \label{21.5}$
which can be solved using the quadratic formula to give the energies of the two molecular orbitals.
$E_\pm = \dfrac{-(2H_{12}S_{12} - H_{11} - H_{22}) \pm \sqrt{(2H_{12}S_{12} - H_{11} - H_{22})^2 - 4(1-S_{12}^2)(H_{11}H_{22} - H_{12}^2)}}{2(1-S_{12}^2)} \label{21.6}$
To obtain numerical values for the energies, we need to evaluate the integrals $H_{11}$, $H_{22}$, $H_{12}$, and $S_{12}$. This would be quite a challenge to do analytically, but luckily there are a number of computer programs that can be used to calculate the integrals. One such program gives the following values.
$\begin{array}{rcl} H_{11} & = & -26.0000 \: eV \ H_{22} & = & -22.2216 \: eV \ H_{12} & = & -29.7670 \: eV \ S_{12} & = & \: 0.8167 \: eV \end{array} \label{21.7}$
When we substitute these into our equation for the energy levels, we get:
$\begin{array}{rcl} E_+ & = & \: 29.8336 \: eV \ E_- & = & -31.0063 \: eV \end{array} \label{21.8}$
We now have the orbital energies and the next step is to find the orbital coefficients. The coefficients for an orbital of energy $E$ are found by substituting the energy into the secular equations and solving for the coefficients $c_i$. Since the two secular equations are not linearly independent (i.e. they are effectively only one equation), when we solve them to find the coefficients what we will end up with is the relative values of the coefficients. This is true in general: in a system with $N$ coefficients, solving the secular equations will allow all $N$ of the coefficients $c_i$ to be obtained in terms of, say, $c_1$. The absolute values of the coefficients are found by normalizing the wavefunction.
Since the secular equations for the orbitals of energy $E_+$ and $E_-$ are not linearly independent, we can choose to solve either one of them to find the orbital coefficients. We will choose the first.
$(H_{11} - E_{\pm})c_1 + (H_{12} - E_{\pm}S_{12})c_2 = 0 \label{21.9}$
For the orbital with energy $E_-$ = -31.0063 eV, substituting numerical values into this equation gives
$\begin{array}{rcl} 5.0063 c_1 - 4.4442 c_2 & = & 0 \ c_2 & = & 1.1265 c_1 \end{array} \label{21.10}$
The molecular orbital is therefore
$\Psi = c_1(\phi_1 + 1.1265\phi_2) \label{21.11}$
Normalizing to find the constant $c_1$ (by requiring $<\Psi|\Psi>$ = 1) gives
$\begin{array}{rcll} \Psi_1 & = & 0.4933\phi_1 + 0.5557\phi_2 & \ & = & 0.4933s_N + 0.3208(s_1 + s_2 + s_3) & (\text{substituting the SALCs for} \: \phi_1 \: \text{and} \: \phi_2) \end{array} \label{21.12}$
For the second orbital, with energy $E_+$ = 29.8336 eV, the secular equation is
$\begin{array}{rcl} -55.8336c_1 - 54.1321c_2 & = & 0 \ c_2 & = & -1.0314c_1 \end{array} \label{21.13}$
giving
$\begin{array}{rcll} \Psi_2 & = & c_1(\phi_1 - 1.0314\phi_2) & \ & = & 1.6242\phi_1 - 1.6752\phi_2 & \text{(after normalization)} \ & = & 1.6242s_N -0.9672(s_1 + s_2 + s_3) \end{array} \label{21.14}$
These two $A_1$ molecular orbitals $\Psi_1$ and $\Psi_2$, one bonding and one antibonding, are shown below.
The remaining two SALCs arising from the $s$ orbitals of $NH_3$ ($\phi_3 = \dfrac{1}{\sqrt{6}}\begin{pmatrix} 2s_1 - s_2 - s_3 \end{pmatrix}$ and $\phi_4 = \dfrac{1}{\sqrt{2}}\begin{pmatrix} s_2 - s_3 \end{pmatrix}$), form an orthogonal pair of molecular orbitals of $E$ symmetry. We can show this by solving the secular determinant to find the orbital energies. The secular equations in this case are:
$\begin{array}{rcl} c_1(H_{33} - ES_{33}) + c_2(H_{34} -ES_{34}) & = & 0 \ c_1(H_{34} -ES_{34}) + c_2(H_{44} - ES_{44}) & = & 0 \end{array} \label{21.15}$
Solving the secular determinant gives
$E_\pm = \dfrac{-(2H_{34}S_{34} - H_{33} - H_{44}) \pm \sqrt{(2H_{34}S_{34} - H_{33} - H_{44})^2 - 4(1-S_{34}^2)(H_{33}H_{44} - H_{34}^2)}}{2(1-S_{34}^2)} \label{21.16}$
The integrals required are
$\begin{array}{rcl} H_{33} & = & -9.2892 \: eV \ H_{44} & = & -9.2892 \: eV \ H_{34} & = & 0 \ S_{34} & = & 0 \end{array} \label{21.17}$
Using the fact that $H_{34}$ = $S_{34} = 0$, the expression for the energies reduces to
$E_\pm = \dfrac{(H_{33} + H_{44}) \pm (H_{33} - H_{44})}{2} \label{21.18}$
giving $E_+$ = $H_{33}$ = -9.2892 eV and $E_-$ = $H_{44}$ = -9.2892 eV. Each SALC therefore forms a molecular orbital by itself, and the two orbitals have the same energy; the two SALCs form an orthogonal pair of degenerate orbitals. These two molecular orbitals of $E$ symmetry are shown below.
Summary of the steps involved in constructing molecular orbitals
1. Choose a basis set of functions $f_i$ consisting of the valence atomic orbitals on each atom in the system, or some chosen subset of these orbitals.
2. With the help of the appropriate character table, determine which irreducible representations are spanned by the basis set using Equation (15.20) to determine the number of times $a_k$ that the $k^{th}$ irreducible representation appears in the representation. $a_k = \dfrac{1}{h}\sum_C n_C \chi(g) \chi_k(g) \label{22.1}$
3. Construct the SALCs $\phi_i$ that transform as each irreducible representation using Equation 16.1 $\phi_i = \sum_g \chi_k(g) g f_i \label{22.2}$
4. Write down expressions for the molecular orbitals by taking linear combinations of all the irreducible representations of the same symmetry species.
5. Write down the secular equations for the system.
6. Solve the secular determinant to obtain the energies of the molecular orbitals.
7. Substitute each energy in turn back into the secular equations and solve to obtain the coefficients appearing in your molecular orbital expressions in step 4.
8. Normalize the orbitals.
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Determining the Symmetries of Molecular Motions
We mentioned above that the procedure for determining the normal vibrational modes of a polyatomic molecule is very similar to that used in previous sections to construct molecular orbitals. In fact, virtually the only difference between these two applications of group theory is the choice of basis set.
As we have already established, the motions of a molecule may be described in terms of the motions of each atom along the $x$, $y$ and $z$ axis. Consequently, it probably won’t come as too much of a surprise to discover that a very useful basis for describing molecular motions comprises a set of $\begin{pmatrix} x, y, z \end{pmatrix}$ axes centered on each atom. This basis is usually known as the $\textit{3N}$ Cartesian basis since there are $3N$ Cartesian axes, $3$ axes for each of the $N$ atoms in the molecule. In other words, each atom as 3 degrees of freedom. Degrees of freedom are the number of independent ways a molecule can move. An atom has 3 degrees of freedom as it can move in $x$, $y$, and $z$, but cannot rotate or vibrate. Note that each molecule will have a different $3N$ Cartesian basis, just as every molecule has a different atomic orbital basis.
Our first task in investigating motions of a particular molecule is to determine the characters of the matrix representatives for the $3N$ Cartesian basis under each of the symmetry operations in the molecular point group. We will use the $H_2O$ molecule, which has $C_{2v}$ symmetry, as an example.
$H_2O$ has three atoms, so the $3N$ Cartesian basis will have $9$ elements. The basis vectors are shown in the diagram below.
One way of determining the characters would be to construct all of the matrix representatives and take their traces. While you are more than welcome to try this approach if you want some practice at constructing matrix representatives, there is an easier way. Recall that we can also determine the character of a matrix representative under a particular symmetry operation by stepping through the basis functions and applying the following rules:
1. Add $1$ to the character if the basis function is unchanged by the symmetry operation;
2. Add $-1$ to the character if the basis function changes sign under the symmetry operation;
3. Add $0$ to the character if the basis function moves when the symmetry operation is applied.
For $H_2O$, this gives us the following characters for the $3N$ Cartesian basis (check that you can obtain this result using the rules above and the basis vectors as drawn in the figure):
$\begin{array}{lcccc} \text{Operation:} & E & C_2 & \sigma_v(xz) & \sigma_v'(yz) \ \chi_{3N}: & 9 & -1 & 3 & 1 \end{array} \tag{24.1}$
There is an even quicker way to work out the characters of the $3N$ Cartesian basis if you have a character table in front of you. The character for the Cartesian basis is simply the sum of the characters for the $x$, $y$, and $z$ (or $T_x$, $T_y$, and $T_z$) functions listed in the character table. To get the character for the $\textit{3N}$ Cartesian basis, simply multiply this by the number of atoms in the molecule that are unshifted by the symmetry operation.
The $C_{2v}$ character table is shown below.
$\begin{array}{l|cccc|l} C_{2v} & E & C_2 & \sigma_v & \sigma_v' & h = 4 \ \hline A_1 & 1 & 1 & 1 & 1 & z, x^2, y^2, z^2 \ A_2 & 1 & 1 & -1 & -1 & xy, R_z \ B_1 & 1 & -1 & 1 & -1 & x, xz, R_y \ B_2 & 1 & -1 & -1 & 1 & y, yz, R_x \ \hline \end{array} \tag{24.2}$
$x$ transforms as $B_1$, $y$ as $B_2$, and $z$ as $A_1$, so the characters for the Cartesian basis are
$\begin{array}{lcccc} \text{Operation:} & E & C_2 & \sigma_v(xz) & \sigma_v'(yz) \ \chi_{3N}: & 3 & -1 & 1 & 1 \end{array} \tag{24.3}$
We multiply each of these by the number of unshifted atoms ($3$ for the identity operation, $1$ for $C_2$, $3$ for $\sigma_v$ and $1$ for $\sigma_v'$) to obtain the characters for the $3N$ Cartesian basis.
$\begin{array}{lcccc} \chi_{3N}: & 9 & -1 & 3 & 1 \end{array} \tag{24.4}$
Reassuringly, we obtain the same characters as we did previously. Which of the three methods you use to get to this point is up to you.
We now have the characters for the molecular motions (described by the $3N$ Cartesian basis) under each symmetry operation. At this point, we want to separate these characters into contributions from translation, rotation, and vibration. This turns out to be a very straightforward task. We can read the characters for the translational and rotational modes directly from the character table, and we obtain the characters for the vibrations simply by subtracting these from the $3N$ Cartesian characters we’ve just determined. The characters for the translations are the same as those for $\chi_{Cart}$. We find the characters for the rotations by adding together the characters for $R_x$, $R_y$, and $R_z$ from the character table (or just $R_x$ and $R_y$ if the molecule is linear). For $H_2O$, we have:
$\begin{array}{lcccc} \text{Operation:} & E & C_2 & \sigma_v(xz) & \sigma_v'(yz) \ \chi_{3N}: & 9 & -1 & 3 & 1 \ \chi_{\text{Trans}}: & 3 & -1 & 1 & 1 \ \chi_{\text{Rot}}: & 3 & -1 & -1 & -1 \ \chi_{\text{Vib}} = \chi_{3N} - \chi_{\text{Trans}} - \chi_{\text{Rot}}: & 3 & 1 & 3 & 1 \end{array} \tag{24.5}$
The characters in the final row are the sums of the characters for all of the molecular vibrations. We can find out the symmetries of the individual vibrations by using the reduction equation (Equation (15.20)) to determine the contribution from each irreducible representation.
In many cases you won’t even need to use the equation, and can work out which irreducible representations are contributing just by inspection of the character table. In the present case, the only combination of irreducible representations that can give the required values for $\chi_{\text{Vib}}$ is $2A_1 + B_1$. As an exercise, you should make sure you are also able to obtain this result using the reduction equation.
So far this may all seem a little abstract, and you probably want to know is what the vibrations of $H_2O$ actually look like. For a molecule with only three atoms, it is fairly easy to identify the possible vibrational modes and to assign them to the appropriate irreducible representation.
For a larger molecule, the problem may become much more complex, and in that case we can generate the SALCs of the $3N$ Cartesian basis, which will tell us the atomic displacements associated with each vibrational mode. We will do this now for $H_2O$.
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As we saw in the previous section, the complete motion (translations, rotations, and vibrations) of ammonia (NH3) can be represented by the following reducible representation:
$C_{3v}$ $E$ $2C_3$ $3\sigma_v$
$\Gamma$ $12$ $0$ $2$
We want to relate the reducible form representation to the irreducible representation. To do this, we use the tabular method. First, create a new table:
$C_{3v}$ $E$ $2C_3$ $3\sigma_v$
$\Gamma$ $12$ $0$ $2$
$A_1$
$A_2$
$E$
We will need the $C_{3v}$ character table for ammonia:
$C_{3v}$ $C_{3}$ $C_{s}$
$A_1$ $A$ $A'$
$A_2$ $A$ $A"$
$E$ $E$ $A'+A"$
Fill in each number in our table by using the following equation:
$g_c \chi_i X_r$
$g_e$ Number of operations (order) in the class
$\chi_i$ Character of the irreducible representation from the character table
$\chi_r$ Character of the reducible representation from $\Gamma$
For example, the top-left value would be:
$1\times 1\times 12=12$
Where:
• $1$ is the number of operations in the $E$ class
• $1$ is the character of the irreducible representation
• 9 is the character of the reducible representation
The table becomes:
$C_{3v}$ $E$ $2C_3$ $3\sigma_v$
$\Gamma$ $12$ $0$ $2$
$A_1$ $12$ $0$ $6$
$A_2$ $12$ $0$ $-6$
$E$ $24$ $0$ $0$
Sum up each row:
$C_{3v}$ $E$ $2C_3$ $3\sigma_v$ $\sum$
$\Gamma$ $12$ $0$ $2$
$A_1$ $12$ $0$ $6$ $18$
$A_2$ $12$ $0$ $-6$ $6$
$E$ $24$ $0$ $0$ $24$
Now divide the summed values by the order of the group to obtain the number of times the irreducible representation appears ($n_i$). Ammonia has order $h=6$:
$C_{3v}$ $E$ $2C_3$ $3\sigma_v$ $\sum$ $n_i=\frac{\sum}{h}$
$\Gamma$ $12$ $0$ $2$
$A_1$ $12$ $0$ $6$ $18$ $3$
$A_2$ $12$ $0$ $-6$ $6$ $1$
$E$ $24$ $0$ $0$ $24$ $4$
The reducible representation can be broken down to its irreducible forms:
$\Gamma=3 A_1+A_2+4 E$
Now that we have the irreducible representations for the motion of ammonia, we can determine which are associated with rotations, vibrations, and translations. To start, we turn to the $C_{3v}$ character table:
$C_{3v}$ E 2C3 v
A1 1 1 1 z x2+y2, z2
A2 1 1 -1 Rz
E 2 -1 0 (Rx, Ry), (x,y) (xz, yz) (x2-y2, xy)
The first column to the right of the characters includes to the terms $x$, $y$, $z$, $R_x$, $R_y$, and $R_z$. The rows with $x$, $y$, and $z$ represent the irreducible representations for the translational modes in those directions:
$\Gamma_\text{trans} = A_1 + E$
The rows with $R_x$, $R_y$, and $R_z$ represent the irreducible representations for the rotational modes about those axes:
$\Gamma_\text{rot} = A_2 + E$
We can subtract the translational and rotational irreducible representations from our \Gamma to get the irreducible representations for the normal vibrational modes:
$\Gamma_\text{vib} = \Gamma - \Gamma_\text{trans} -\Gamma_\text{rot}$
Doing this, we obtains:
$\Gamma_\text{vib} = 2 A_1 + 2 E$
12.12: Normal Modes of Vibrations Describe how Molecules Vibrate
Normal modes of vibration
All molecules vibrate. The simplest vibration is the one that takes place between two atoms in a diatomic molecule. Vibrational motion in diatomic molecules is often discussed within the context of the simple harmonic oscillator in quantum mechanics. A diatomic molecule has only a single bond that can vibrate; we say it has a single vibrational mode. As you may expect, the vibrational motions of polyatomic molecules are much more complicated than those of a diatomic. First, there are more bonds that can vibrate; and secondly, in addition to stretching vibrations, the only type of vibration possible in a diatomic, we can also have bending and torsional vibrational modes. Since changing one bond length in a polyatomic will often affect the length of nearby bonds, we cannot consider the vibrational motion of each bond in isolation; instead we talk of normal modes of vibration involving the concerted motion of groups of bonds. As a simple example, the normal modes of a linear triatomic molecule are shown below.
Once we know the symmetry of a molecule at its equilibrium structure, group theory allows us to predict the vibrational motions it will undergo using exactly the same tools we used above to investigate molecular orbitals. Each vibrational mode transforms as one of the irreducible representations of the molecule’s point group. Before moving on to an example, we will quickly review how to determine the number of vibrational modes in a molecule.
Molecular degrees of freedom – determining the number of normal vibrational modes
An atom can undergo only translational motion, and therefore has three degrees of freedom corresponding to motion along the $x$, $y$, and $z$ Cartesian axes. Translational motion in any arbitrary direction can always be expressed in terms of components along these three axes. When atoms combine to form molecules, each atom still has three degrees of freedom, so the molecule as a whole has $3N$ degrees of freedom, where $N$ is the number of atoms in the molecule. However, the fact that each atom in a molecule is bonded to one or more neighboring atoms severely hinders its translational motion, and also ties its motion to that of the atoms to which it is attached. For these reasons, while it is entirely possible to describe molecular motions in terms of the translational motions of individual atoms (we will come back to this in the next section), we are often more interested in the motions of the molecule as a whole. These may be divided into three types: translational; rotational and vibrational.
Just as for an individual atom, the molecule as a whole has three degrees of translational freedom, leaving $3N - 3$ degrees of freedom in rotation and vibration.
The number of rotational degrees of freedom depends on the structure of the molecule. In general, there are three possible rotational degrees of freedom, corresponding to rotation about the $x$, $y$, and $z$ Cartesian axes. A non-linear polyatomic molecule does indeed have three rotational degrees of freedom, leaving $3N - 6$ degrees of freedom in vibration (i.e $3N - 6$ vibrational modes). In a linear molecule, the situation is a little different. It is generally accepted that to be classified as a true rotation, a motion must change the position of one or more of the atoms. If we define the $z$ axis as the molecular axis, we see that spinning the molecule about the axis does not move any of the atoms from their original position, so this motion is not truly a rotation. Consequently, a linear molecule has only two degrees of rotational freedom, corresponding to rotations about the $x$ and $y$ axis. This type of molecule has $3N - 5$ degrees of freedom left for vibration, or $3N - 5$ vibrational modes.
In summary:
• A linear molecule has $3N - 5$ vibrational modes
• A non-linear molecule has $3N - 6$ vibrational modes.
Symmetry
Let’s work through an example: Ammonia ($NH_3$) with a $C_{3v}$ symmetry. Consequently, all of the properties contained in the $C_{3v}$ character table above are pertinent to the ammonia molecule.
The principle axis is the axis that the highest order rotation can be preformed. In this case the z-axis pass through the lone pairs (pink sphere), which contains a $C_3$ axis. The ?’s or mirror planes ($\sigma_v$ parallel to z-axis & $\sigma_h$ perpendicular to the z-axis). In ammonia there is no $\sigma_h$ only three $\sigma_v$’s. The combination of $C_3$ & $\sigma_v$ leads to $C_{3v}$ point group, which leads to the C3v character table.
The number of transitions is dictated by 3N-6 for non-linear molecules, so in the case of Ammonia, there will be $3(4)-6=6$ vibrational modes. This can be confirmed by working through the vibrations of the molecule. This work is shown in the table below.
$C_{3v}$ $E$ $2C_3$ $3\sigma_v$
$\Gamma_{xyz}$ 3 0 1
Unmoved Atom 4 1 1
$\Gamma_\text{total}$ 12 0 1
$\Gamma_\text{translational}$ 3 0 1
$\Gamma_\text{rotational}$ 3 0 -1
$\Gamma_\text{vibrational}$ 6 0 1
$2A_1 + 2E$
The irreducible representations for the vibrations are $2A_1) and \(2E$ (where $E$ is doubly degenerate, meaning two vibration modes each), which total 6 vibrations. This calculation was done by using the character table to find out the rotation and translation values and what atoms move during each operation. Using the character table we can characterize the $A_1$ vibration as IR active along the z-axis and raman active as well. The $E$ vibration is IR active along both the x & y axis and is Raman active as well. From the character table the IR symmetries correspond to the x, y & z translations. Where the Raman active vibrations correspond to the symmetries of the d-orbitals.
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Vibrational transitions in molecules
The potential energy surface (electronic state), often approximated as a Morse potential, describes the energy of the eigenstate as a function of the interatomic distance. When an electron is excited form one eigenstate to another within the electronic state there is a change in interatomic distance, this result in a vibration occurring.
Vibrational transitions oftne occur when a molecule absorbs or emits infrared light, though, as we will see later, there are other processes that can cause vibrational transitions. During the vibrational transition, the electrons remains in the same electronic state, but that vibrational state changes from one eigenstate to another. In the case of the Morse diagram above, the vibrational eigenstates are denoted as $\nu$. As you can see from the diagram the eigenstate is a function of energy versus interatomic distance. To predict whether a vibrational transition will occur, or for that matter a transition of any kind, we use the transition moment integral:
$\int \Psi_i*\mu \Psi_f d\tau=\langle \Psi_i | \mu| \Psi_f \rangle$
The transition moment integral is written here in standard integral format, but this is equivalent to Bra & Ket format which is standard in most chemistry quantum mechanical text (The $\langle \Psi_i |$ is the Bra portion, $| \Psi_f \rangle$ is the Ket portion). The transition moment operator $\mu$ is the operator the couples the initial state $\Psi_i$ to the final state $\Psi_f$, which is derived from the time independent Schrödinger equation. However using group theory we can ignore the detailed mathematical methods. We can use the $\Gamma_{ir}$ of the vibrational energy levels and the symmetry of the transition moment operator to find out if the transition is allowed by selection rules. The selection rules for vibrations or any transition is that is allowed, for it to by allowed by group theory the answer must contain the totally symmetric $\Gamma_{ir}$, which is always the first $\Gamma_{ir}$ in the character table for the molecule in question.
Infrared Spectroscopy
Infrared Spectroscopy (IR) measures the vibrations that occur within a single electronic state, such as the one shown above. Because the transition occurs within a single electronic state there is a variation in interatomic distance. The dipole moment is dictate by the equation.
$\vec{\mu} = \alpha\vec{E}$
Where $\vec{\mu}$ is the magnitude of dipole moment; $\alpha$ is the polarizability constant (actually a tensor) & $E$ is the magnitude of the electric field which can be described as the electronegitivity.3 Therefore when a vibration occurs within a single electronic state there is a change in the dipole moment, which is the definition of an active IR transition.
$\left ( \frac{\mathrm{d\mu} }{\mathrm{d} q} \right )_{eq} \neq 0$
In terms of group theory a change in the dipole is a change from one vibrational state to another, as shown by the equation above. A picture of the vibrational states with respect ot the rotational states and electronic states is given below. In IR spectroscopy the transition occurs only from on vibrational state to another all within the same electronic state, shown below as B.
Light polarized along the $x$, $y$, and $z$ axes of the molecule may be used to excite vibrations with the same symmetry as the $x$, $y$ and $z$ functions listed in the character table.
For example, in the $C_{2v}$ point group, $x$-polarized light (light polarized in the $x$ direction) may be used to excite vibrations of $B_1$ symmetry, $y$-polarized light to excite vibrations of $B_2$ symmetry, and $z$-polarized light to excite vibrations of $A_1$ symmetry. In $H_2O$, we would use $z$-polarized light to excite the symmetric stretch and bending modes, and $x$-polarized light to excite the asymmetric stretch. Shining $y$-polarized light onto a molecule of $H_2O$ would not excite any vibrational motion.
Raman Scattering
If there are vibrational modes in the molecule that may not be accessed using a single photon, it may still be possible to excite them using a two-photon process known as Raman scattering. An energy level diagram for Raman scattering is shown below.
The first photon excites the molecule to some high-lying intermediate state, known as a virtual state. Virtual states are not true stationary states of the molecule (i.e. they are not eigenfunctions of the molecular Hamiltonian), but they can be thought of as stationary states of the ‘photon + molecule’ system. These types of states are extremely short lived, and will quickly emit a photon to return the system to a stable molecular state, which may be different from the original state. Since there are two photons (one absorbed and one emitted) involved in Raman scattering, which may have different polarizations, the transition dipole for a Raman transition transforms as one of the Cartesian products $x^2$, $y^2$, $z^2$, $xy$, $xz$, $yz$ listed in the character tables.
Vibrational modes that transform as one of the Cartesian products may be excited by a Raman transition, in much the same way as modes that transform as $x$, $y$, or $z$ may be excited by a one-photon vibrational transition.
In $H_2O$, all of the vibrational modes are accessible by ordinary one-photon vibrational transitions. However, they may also be accessed by Raman transitions. The Cartesian products transform as follows in the $C_{2v}$ point group.
$\begin{array}{clcl} A_1 & x^2, y^2, z^2 & B_1 & xz \ A_2 & sy & B_2 & yz \end{array} \tag{27.5}$
The symmetric stretch and the bending vibration of water, both of $A_1$ symmetry, may therefore be excited by any Raman scattering process involving two photons of the same polarization ($x$-, $y$- or $z$-polarized). The asymmetric stretch, which has $B_1$ symmetry, may be excited in a Raman process in which one photon is $x$-polarized and the other $z$-polarized.
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The construction of linear combinations of the basis of atomic movements allows the vibrations belonging to irreducible representations to be investigated. The wavefunction of these symmetry equivalent orbitals is referred to as Symmetry Adapted Linear Combinations, or SALCs. SALCs (Symmetry Adapted Linear Combinations) are the linear combinations of basis sets composed of the stretching vectors of the molecule. The SALCs of a molecule can help determine binding schemes and symmetries. The procedure used to determine the SALCs of a molecule is also used to determine the LCAO of a molecule. The LCAO, Linear Combination of Atomic Orbitals, uses the basis set of atomic orbitals instead of stretching vectors. The LCAO of a molecule provides a detailed description of the molecular orbitals, including the number of nodes and relative energy levels.
Symmetry adapted linear combinations are the sum over all the basis functions:
$\phi_{i} =\displaystyle\sum_{j} c_{ij} b_{j} \label{1}$
$ϕ_i$ is the ith SALC function, bj is the jth basis function, and $c_{ij}$ is a coefficient which controls how much of $b_j$ appears in $ϕ_i$. In method two, the projection operator is used to obtain the coefficients consistent with each irreducible representation.1
The SALCs of a molecule may be constructed in two ways. The first method uses a basis set composed of the irreducible representation of the stretching modes of the molecule. On the other hand, the second method uses a projection operator on each stretching vector. When determining the irreducible representations of the stretching modes, the reducible representations for all the vibrational modes must first be determined. Basis vectors are assigned characters and are treated as individual objects. A
Background
In order to understand and construct SALCs, a background in group theory is required. The identification of the point group of the molecule is essential for understanding how the application of operations affects the molecule. This allows for the determination of the nature of the stretching modes. As a review, let’s first determine the stretching modes of water together. Water has the point group C2v. Table 1 is the character table for the C2v point group.
Table $1$: C2v Character Table
C2v E C2 σv(xz) σv'(yz)
A1 1 1 1 1 z x2, y2, z2
A2 1 1 −1 −1 Rz xy
B1 1 −1 1 −1 x, Ry xz
B2 1 −1 −1 1 y, Rx yz
The first step in determining stretching modes of a molecule is to add the characters contained in the x, y, and z rows to obtain the total reducible representation of the xyz coordinates, ΓXYZ. ΓXYZ can also be found by applying the symmetry operations to the three vectors (x, y, and z) of the coordinate system of the molecule. The next step involves the investigation of the atoms that remain unchanged when an operation is applied, ΓUMA. This step refers to the unmoved atoms (UMA). Multiplying ΓXYZ and ΓUMA gives the reducible representation for the molecule referred to as ΓTOTAL. The ΓTOTAL is the reducible representation for all the modes of the molecule (vibrational, rotational, and translational) and can also be determined by applying the symmetry operations to each coordinate vector (x, y, and z) on each atom.
Table $2$: C2v Reducible Representation for H2O
C2v E C2 σv(xz) σv'(yz)
ΓXYZ 3 -1 1 1
ΓUMA 3 1 1 3
ΓTOTAL 9 -1 1 3
ΓTOTAL is then reduced to later give the stretching modes that are unique to the molecule. First, the reduction formula is applied to decompose the reducible representation:
$a_{i}=\dfrac{1}{h}\displaystyle\sum_{R}(X^{R}X_{i}^{R}C^{R}) \label{2}$
Here, ai is the number of times the irreducible representation will appear in the initial reducible representation. The order of the point group is represented by h; R is an operation of the group; XR is a character of the operation R in the reducible representation; XiR is a character of the operation R in the irreducible representation, and CR is the number of members in the class to which R belongs. Applying this formula and subtracting the representations obtained from the basis functions x, y, z, Rx, Ry, and Rz (for the translations and rotations of the molecule) gives the irreducible representation that corresponds to the vibrational states of the molecule:
ΓVibration = 2a1 + b2
A simple check can be performed to determine that the right number of modes was obtained. For linear molecules (3N-5) gives the correct number of normal modes. For molecules with any other shape otherwise known as non-linear molecules, the formula is (3N-6). N represents the number of atoms in a molecule. Let’s double check the above water example:
$(3N-6) N=3 \label{3}$
$[3(3)-6] = 3$
Water should have three vibrational modes. When the irreducible representation was obtained, it was seen that water has two a1 modes and a b2mode for a total of three.
When double checking that you have the correct number of normal modes for other molecules, remember that the irreducible representation E is doubly degenerate and counts as two normal modes. T is triply degenerate and counts for three normal modes, etc.
Constructing SALCs
Method 1
There are multiple ways of constructing the SALCs of a molecule. The first method uses the known symmetries of the stretching modes of the molecule. To investigate this method, the construction of the SALCs of water is examined. Water has three vibrational states, 2a1+b2. Two of these vibrations are stretching modes. One is symmetric with the symmetry A1, and the other is antisymmetric with the symmetry B2. While looking for the SALC of a molecule, one uses vectors represented by bj as the basis set. The vectors demonstrate the irreducible representations of molecular vibrations.
The SALCs of water can be composed by creating a linear combination of the stretching vectors.
$\phi(A_{1})=b_{1}+b_{2}$
and
$\phi(B_{1})=b_{1}-b_{2} \label{4}$
Normalization
The final step in constructing the SALCs of water is to normalize expressions. To normalize the SALC, multiply the entire expression by the normalization constant that is the inverse of the square root of the sum of the squares of the coefficients within the expression.
$\phi_{i}=N\displaystyle\sum_{j}c_{ij}b_{j} \label{5}$
$N=\dfrac{1}{\sqrt{\displaystyle\sum_{j=1}^{n}c_{ij}^{2}}} \label{6}$
$\phi(A_{1})=\dfrac{1}{\sqrt{2}}(b_{1}+b_{2})$
and
$\phi(B_{1})=\dfrac{1}{\sqrt{2}}(b_{1}-b_{2})$
Normalizing the SALCs ensures that the magnitude of the SALC is unity, and therefore the dot product of any SALC with itself will equal one.
Method 2
The other method for constructing SALCs is the projection operator method. The SALC of a molecule can be constructed in the same manner as the LCAO, Linear Combination of Atomic Orbitals, however the basis set differs. While looking for the SALCs of a molecule, one uses vectors represented by bj, on the other hand, while looking for the LCAO of a molecule, one uses atomic orbitals as the basis set. The vectors demonstrate the possible vibration of the molecule. While constructing SALCs, the basis vectors can be treated as individual vectors.
Example $1$: Water
Let’s take a look at how to construct the SALC for water. The first step in constructing the SALC is to label all vectors in the basis set. Below are the bond vectors of water that will be used as the basis set for the SALCs of the molecule.
Next, the basis vector, v, is transformed by Tj, the jth symmetry operation of the molecule’s point group. As the vector of the basis set is transformed, record the vector that takes its place. Water is a member of the point group C2V. The Symmetry elements of the C2V point group are E, C2, σv, σv’.
The ith SALC function, ϕiis shown below using the vector v=b1.
Once the transformations have been determined, the SALC can be constructed by taking the sum of the products of each character of a representation within the point group and the corresponding transformation. The SALCs functions are the collective transformations of the basis sets represented by ϕiwhere Xi(j) is the character of the ith irreducible representation and the jth symmetry operation.
$\phi_{i}=\displaystyle\sum_{j}X_{i}(j)T_{j}\nu \label{7}$
Table $3$: Projection Operator method for C2v
The final step in constructing the SALCs of water is to normalize expressions.
Table $4$: Normalized SALCs of H2O
There are two SALCs for the water molecule, ϕ1(A1) and ϕ1(B2). This demonstrates that water has two stretching modes, one is a totally symmetric stretch with the symmetry, A1, and the other is an antisymmetric stretch with the symmetry B2.
Interpreting SALCs
Both methods of construction result in the same SALCs. Only irreducible representations corresponding to the symmetries of the stretching modes of the molecule will produce a SALC that is non-zero. Method 1 only utilized the known symmetries of the vibrational modes. All irreducible representations of the point group were used, but the representations that were not vibrational modes resulted in SALCs equal to zero.
Therefore, with the SALCs of a molecule given, all the symmetries of the stretching modes are identified. This allows for a clearer understanding of the spectroscopy of the molecule. Even though vibrational modes can be observed in both infrared and Raman spectroscopy, the SALCs of a molecule cannot identify the magnitude or frequency of the peak in the spectra. The normalized SALCs can, however, help to determine the relative magnitude of the stretching vectors. The magnitude can be determined by the equation below.
$a \cdot b= |a||b|cos \theta \label{8}$
The resulting A1 and B1 symmetries for the above water example are each active in both Raman and IR spectroscopies, according to the C2v character table. If the vibrational mode allows for a change in the dipole moment, the mode can be observed through infrared spectroscopy. If the vibrational mode allows for a change in the polarization of the molecule, the mode can be observed through Raman spectroscopy. Both stretching and bending modes are seen in the spectra, however only stretching modes are expressed in the SALCs.
Example $2$: Difluorobenzene
The SALCs of a molecule can also provide insight to the geometry of a molecule. For example, SALCs can aid in determining the differences between para-difluorobenzene and ortho-difluorobenzene. The SALCs for these two molecules are given below.
$\phi (A_{g})= \dfrac{1}{2} (b_{1} + b_{2} + b_{3} +b_{4})$
$\phi (B_{1g})= \dfrac{1}{2} (b_{1} - b_{2} + b_{3} -b_{4})$
$\phi (B_{2u})= \dfrac{1}{2} (b_{1} - b_{2} - b_{3} +b_{4})$
$\phi (B_{3u})= \dfrac{1}{2} (b_{1} + b_{2} - b_{3} -b_{4})$
$\phi _{1} (A_{1})= \dfrac{1}{\sqrt{2}} (b_{1} + b_{2} + b_{3} +b_{4})$
$\phi _{2} (A_{1})= \dfrac{1}{\sqrt{2}} (b_{1} - b_{2} + b_{3} -b_{4})$
$\phi _{1} (B_{1})= \dfrac{1}{\sqrt{2}} (b_{1} - b_{2} - b_{3} +b_{4})$
$\phi _{2} (B_{1})= \dfrac{1}{\sqrt{2}} (b_{1} + b_{2} - b_{3} -b_{4})$
From the SALCs, it is seen that para-difluorobenzene has four stretching modes and ortho-difluorobenzene has only two. Therefore, it is no surprise that the vibrational spectroscopy of the para-difluorobenzene shows more peaks than the ortho-difluorobenzene.
Applications
The SALCs of a molecule can be used to understand the stretching modes and binding schemes of a molecule. More information can also be interpreted when applying the projection operator used in SALCs on the atomic orbitals of the molecule. This results in the determination of the linear combination of atomic orbitals (LCAO), which gives information on the molecular orbitals of the molecule. The molecular orbitals (MO) of a molecule are often constructed as LCAOs. Each MO is a solution to the Schrödinger equation and is an eigenfunction of the Hamiltonian operator. The LCAOs can be determined in the same manner as the SALCs of a molecule, with the use of a projection operator. The difference is that the basis set is no longer stretching vectors, but instead the atomic orbitals of the molecule. Hydrogen only has s orbitals, but oxygen has s and p orbitals, where the px, py, and pz all transform differently and therefore must be treated differently.
Once the LCAOs of the molecule have been determined, the expressions can be interpreted into images of the orbitals bonding. If two orbitals are of the same sign in the expression, the electrons in the orbitals are in phase with each other and are bonding. If two orbitals are of the opposite sign in the expression, the electrons in the orbitals are out of phase with each other and are antibonding. The image below shows the atomic orbitals' phases (or signs) as red or blue lobes.
Any separation between two antibonding atomic orbitals is a planar node. As the number of nodes increases, so does the level of antibonding. This allows for the LCAO to place the molecular orbitals in order of increasing energy, which can be used in constructing the molecular orbital (MO) diagram of the molecule. The irreducible representation used to construct the LCAO is used to describe the MOs. The LCAOs for water are shown below with red dotted lines showing the nodes. Notice the nodes for the px orbitals are in a different plane than the s orbitals of the hydrogens, so these are degenerate and nonbonding.
Information from the LCAO of water can also be used to analyze and anticipate the adsorption of water onto various surfaces. Evarestov and Bandura used this technique to identify the water adsorption on Y-doped BaZrO3 and TiO2 (Rutile) respectively.2,3
2
Applying a combination of Methods 1 and 2, the SALCs for CBr2H2 can be determined. The point group of this molecule is C2v, making it similar to the determination of SALCs for water.
However, the central carbon contains more than one type of attached atom; therefore, the stretching analysis must be performed in pieces. First, the C-H stretches are examined, followed by the C-Br stretches:
Table 5: Irreducible Representations for C-H and C-Br stretches in CBr2H2.
C2V E C2 σV σV’ Irreducible Representation
ΓC-H 2 0 0 2 ΓC-H = A1 + B2
ΓC-Br 2 0 2 0 ΓC-Br = A1 + B1
Applying the projection operator method to C-Br and C-H stretches individually, the SALCs are obtained in the same fashion as before.
Table 6: SALCs for CBr2H2.
ΓC-Br E C2 σV σV’ SUM
A1Tj(b1) b1 b2 b1 b2 2(b1 + b2)
B1Tj(b1) b1 -b2 b1 -b2 2(b1 - b2)
Table 7: SALCs for CBr2H2.
ΓC-H E C2 σV σV’ SUM
A1Tj(a1) a1 a2 a2 a1 2(a1 + a2)
B2Tj(a1) a1 -a2 -a2 a1 2(a1 - a2)
The results are normalized and the following SALCs are obtained for the C2v molecule CBr2H2:
$\phi CBr(A_{1})=\frac{1}{\sqrt{2}}(b_{1}+b_{2})$
$\phi CBr(B_{1}) =\frac{1}{\sqrt{2}} (b_{1}-b_{2})$
$\phi CH(A_{1}) =\frac{1}{\sqrt{2}} (a_{1}+a_{2})$
$\phi CH(B_{2}) =\frac{1}{\sqrt{2}} (a_{1}-a_{2})$
4
To obtain the SALCs for PtCl4, the same general method is applied.
However, even though the point group of the molecule is D4h, the cyclic subgroup C4 may be used (this is a more simplified character table used for spherically symmetrical molecules). Some manipulation is required in order to use this cyclic subgroup and will be discussed. Below is the C4 cyclic character table.
Table $8$: C4 cyclic character table.
C4 E C41 C42 C43
A 1 1 1 1
B 1 -1 1 -1
E1 E2 1 1 i -i -1 -1 -i i
Notice, there are two rows for E, each singly degenerate. To solve for the characters of E, one must take the sum and difference of the two rows. Then, a reduction can be applied to obtain the easiest possible characters by dividing each row by a common factor (removing the common factor is not necessary, but it does simplify the problem as well as remove any imaginary terms):
Sum = [ (1+1) (i-i) (-1-1) (-i+i) ] = (2 0 -2 0) ÷ 2 = E1 (1 0 -1 0)
Difference = [ (1-1) (i+i) (-1+1) (-i-i) ] = (0 2i 0 -2i) ÷ 2i = E2 (0 1 0 -1)
Using the above cyclic group, and the newly obtained characters for E, the projection operator can be applied using Method 2 for the construction of SALCs.
Table $9$: SALCs for PtCl4 using Method 2.
C4 E C41 C42 C43 SUM
ATj(b1) b1 b2 b3 b4 b1 + b2 + b3 + b4
BTj(b1) b1 -b2 b3 -b4 b1 - b2 + b3 - b4
E1Tj(b1) b1 0 -b3 0 b1 - b3
E2Tj(b1) 0 b2 0 -b4 b2 - b4
Normalizing the sum as mentioned in Method 1, the following SALCs are obtained for the D4h molecule PtCl4:
$\phi (A) =\frac{1}{2} (b_{1}+b_{2}+b_{3}+b_{4})$
$\phi (B)=\frac{1}{2} (b_{1}-b_{2}+b_{3}-b_{4})$
$\phi (E^{1})=\frac{1}{\sqrt{2}} (b_{1}-b_{3})$
$\phi (E^{2})=\frac{1}{\sqrt{2}} (b_{2}-b_{4})$
3
Applying a combination of Methods 1 and 2, the SALCs for the C-H stretches of PF2H3 can be determined. The point group of this molecule is Cs. The central carbon contains more than one type of attached hydrogen; therefore, the stretching analysis must be performed in pieces. First, the C-HA stretches are examined, followed by the C-HB stretches:
Table $10$: Irreducible Representations for C-Ha and C-Hb stretches in PF2H3.
Cs E σh Irreducible Representation
ΓC-Ha 2 0 ΓC-Ha = A’ + A”
ΓC-Hb 1 1 ΓC-Hb = A’
Applying the projection operator method to C-HA and C-HB stretches individually, the SALCs are obtained in the same fashion as before.
Table $11$: SALCs for PF2H3 using Method 2.
ΓC-HB E σh SUM
A’ Tj(b1) b1 b2 b1 + b2
A” Tj(b1) b1 -b2 b1 - b2
A’ Tj(a1) a1 a1 a1 + a1
Table $12$: SALCs for PF2H3 using Method 2.
ΓC-HA E C2 σV σV’ SUM
A1Tj(a1) a1 a2 a2 a1 2(a1+ a2)
B2Tj(a1) a1 -a2 -a2 a1 2(a1- a2)
The results are normalized and the following SALCs are obtained for the Cs molecule PF2H3:
$\phi_{1} A'=\frac{1}{\sqrt{2}} (b_{1} +b_{2} )+a_{1} = \frac{1}{sqrt{3}} (b_{1} +b_{2} +a_{1})$
$\phi_{2} A'=\frac{1}{\sqrt{2}} (b_{1} +b_{2} )-a_{1} = \frac{1}{sqrt{3}} (b_{1} +b_{2} -a_{1})$
$\phi A''=\frac{1}{\sqrt{2}} (b_{1} -b_{2} )$
Problems
1. Construct the SALCs for C-H stretches of ortho-difluorobenzene.
2. Construct the SALCs for ammonia.
3. Draw the nodes for the MOs of BeH2 (determined by used of LCAOs) and rank the MOs in order of increasing energy.
Solutions to Practice Problems
1. A1Tj(b1)= \frac{1}{\sqrt{2}} (b_{1} +b_{2})\)
B2Tj(b1)= \frac{1}{\sqrt{2}} (b_{1} -b_{2})\)
A1Tj(b3)= \frac{1}{\sqrt{2}} (b_{3} +b_{4})\)
B2Tj(b3)= \frac{1}{\sqrt{2}} (b_{3} -b_{4})\)
Then add and subtract (for in phase and out of phase) the individual linear combinations found by the projection operator to give the SALCs.
ϕ1(A1) = \frac{1}{2} (b_{1} +b_{2} +b_{3} +b_{4} )\)
ϕ2(A1) = \frac{1}{2} (b_{1} +b_{2} -b_{3} -b_{4} )\)
ϕ1(B2) = \frac{1}{2} (b_{1} -b_{2} +b_{3} -b_{4} )\)
ϕ2(B2) = \frac{1}{2} (b_{1} -b_{2} -b_{3} +b_{4} )\).
2. ATj(b1)= \frac{1}{\sqrt{3}} (b_{1} +b_{2} +b_{3} )\)
E1Tj(b1)= \frac{1}{\sqrt{6}} (2b_{1} -b_{2} -b_{3} )\)
E2Tj(b1)= \frac{1}{\sqrt{2}} (b_{2} -b_{3} )\)
3.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/12%3A_Group_Theory_-_The_Exploitation_of_Symmetry/12.14%3A_Symmetry_Adapted_Linear_Combinations_are_the_Sum_over_all_Basis_functions.txt
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Bonding in Diatomics
You will already be familiar with the idea of constructing molecular orbitals from linear combinations of atomic orbitals from previous courses covering bonding in diatomic molecules. By considering the symmetries of $s$ and $p$ orbitals on two atoms, we can form bonding and antibonding combinations labeled as having either $\sigma$ or $\pi$ symmetry depending on whether they resemble $s$ or $p$ orbitals when viewed along the bond axis (see diagram below). In all of the cases shown, only atomic orbitals that have the same symmetry when viewed along the bond axis $z$ can form a chemical bond e.g. two $s$ orbitals, two $p_z$ orbitals , or an $s$ and a $p_z$ can form a bond, but a $p_z$ and a $p_x$ or an $s$ and a $p_x$ or a $p_y$ cannot. It turns out that the rule that determines whether or not two atomic orbitals can bond is that they must belong to the same symmetry species within the point group of the molecule.
We can prove this mathematically for two atomic orbitals $\phi_i$ and $\phi_j$ by looking at the overlap integral between the two orbitals.
$S_{ij} = \langle \phi_i|\phi_j \rangle = \int \phi_i^* \phi_j d\tau \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text(18.1)$
In order for bonding to be possible, this integral must be non-zero. The product of the two functions $\phi_1$ and $\phi_2$ transforms as the direct product of their symmetry species i.e. $\Gamma_{12}$ = $\Gamma_1 \otimes \Gamma_2$. As explained above, for the overlap integral to be non-zero, $\Gamma_{12}$ must contain the totally symmetric irreducible representation ($A_{1g}$ for a homonuclear diatomic, which belongs to the point group $D_{\infty h}$). As it happens, this is only possible if $\phi_1$ and$\phi_2$ belong to the same irreducible representation. These ideas are summarized for a diatomic in the table below.
$\begin{array}{lllll} \hline \text{First Atomic Orbital} & \text{Second Atomic Orbital} & \Gamma_1 \otimes \Gamma_2 & \text{Overlap Integral} & \text{Bonding?} \ \hline s \: (A_{1g}) & s \: (A_{1g}) & A_{1g} & \text{Non-zero} & \text{Yes} \ s \: (A_{1g}) & p_x \: (E_{1u}) & E_{1u} & \text{Zero} & \text{No} \ s \: (A_{1g}) & p_z \: (A_{1u}) & A_{1u} & \text{Zero} & \text{No} \ p_x \: (E_{1u}) & p_x \: (E_{1u}) & A_{1g} + A_{2g} + E_{2g} & \text{Non-zero} & \text{Yes} \ p_X \: (E_{1u}) & p_z \: (A_{1u}) & E_{1g} & \text{Zero} & \text{No} \ p_z \: (A_{1u}) & p_z \: (A_{1u}) & A_{1g} & \text{Non-zero} & \text{Yes} \end{array} \tag{18.2}$
Bonding in Polyatomics- Constructing Molecular Orbitals from SALCs
In the previous section we showed how to use symmetry to determine whether two atomic orbitals can form a chemical bond. How do we carry out the same procedure for a polyatomic molecule, in which many atomic orbitals may combine to form a bond? Any SALCs of the same symmetry could potentially form a bond, so all we need to do to construct a molecular orbital is take a linear combination of all the SALCs of the same symmetry species. The general procedure is:
1. Use a basis set consisting of valence atomic orbitals on each atom in the system.
2. Determine which irreducible representations are spanned by the basis set and construct the SALCs that transform as each irreducible representation.
3. Take linear combinations of irreducible representations of the same symmetry species to form the molecular orbitals. E.g. in our $NH_3$ example we could form a molecular orbital of $A_1$ symmetry from the two SALCs that transform as $A_1$,
$\begin{array}{rcl} \Psi(A_1) & = & c_1 \phi_1 + c_2 \phi_2 \ & = & c_1 s_N + c_2 \dfrac{1}{\sqrt{3}}(s_1 + s_2 + s_3) \end{array} \tag{19.1}$
Unfortunately, this is as far as group theory can take us. It can give us the functional form of the molecular orbitals but it cannot determine the coefficients $c_1$ and $c_2$. To go further and obtain the expansion coefficients and orbital energies, we must turn to quantum mechanics. The material we are about to cover will be repeated in greater detail in later courses on quantum mechanics and valence, but they are included here to provide you with a complete reference on how to construct molecular orbitals and determine their energies.
Summary of the Steps Involved in Constructing Molecular Orbitals
1. Choose a basis set of functions $f_i$ consisting of the valence atomic orbitals on each atom in the system, or some chosen subset of these orbitals.
2. With the help of the appropriate character table, determine which irreducible representations are spanned by the basis set using Equation (15.20) to determine the number of times $a_k$ that the $k^{th}$ irreducible representation appears in the representation. $a_k = \dfrac{1}{h}\sum_C n_C \chi(g) \chi_k(g) \label{22.1}$
3. Construct the SALCs $\phi_i$ that transform as each irreducible representation using Equation 16.1 $\phi_i = \sum_g \chi_k(g) g f_i \label{22.2}$
4. Write down expressions for the molecular orbitals by taking linear combinations of all the irreducible representations of the same symmetry species.
5. Write down the secular equations for the system.
6. Solve the secular determinant to obtain the energies of the molecular orbitals.
7. Substitute each energy in turn back into the secular equations and solve to obtain the coefficients appearing in your molecular orbital expressions in step 4.
8. Normalize the orbitals.
A more complicated bonding example
As another example, we will use group theory to construct the molecular orbitals of $H_2O$ (point group $C_{2v}$) using a basis set consisting of all the valence orbitals. The valence orbitals are a $1s$ orbital on each hydrogen, which we will label $s_H$ and $s_H'$, and a $2s$ and three $2p$ orbitals on the oxygen, which we will label $s_O$, $p_x$, $p_y$, $p_z$ giving a complete basis $\begin{pmatrix} s_H, s_H', s_O, p_x, p_y, p_z \end{pmatrix}$.
The first thing to do is to determine how each orbital transforms under the symmetry operations of the $C_{2v}$ point group ($E$, $C_2$, $\sigma_v$ and $\sigma_v'$), construct a matrix representation and determine the characters of each operation. The symmetry operations and axis system we will be using are shown below.
The orbitals transform in the following way
$\begin{array}{lrcl} E & \begin{pmatrix} s_H, s_H', s_O, p_x, p_y, p_z \end{pmatrix} & \rightarrow & \begin{pmatrix} s_H, s_H', s_O, p_x, p_y, p_z \end{pmatrix} \ C_2 & \begin{pmatrix} s_H, s_H', s_O, p_x, p_y, p_z \end{pmatrix} & \rightarrow & \begin{pmatrix} s_H', s_H, s_O, -p_x, -p_y, p_z \end{pmatrix} \ \sigma_v(xz) & \begin{pmatrix} s_H, s_H', s_O, p_x, p_y, p_z \end{pmatrix} & \rightarrow & \begin{pmatrix} s_H, s_H', s_O, p_x, -p_y, p_z \end{pmatrix} \ \sigma_v'(yz) & \begin{pmatrix} s_H, s_H', s_O, p_x, p_y, p_z \end{pmatrix} & \rightarrow & \begin{pmatrix} s_H', s_H, s_O, -p_x, p_y, p_z \end{pmatrix} \end{array} \label{23.1}$
A short aside on constructing matrix representatives
After a little practice, you will probably be able to write matrix representatives straight away just by looking at the effect of the symmetry operations on the basis. However, if you are struggling a little the following procedure might help.
Remember that the matrix representatives are just the matrices we would have to multiply the left hand side of the above equations by to give the right hand side. In most cases they are very easy to work out. Probably the most straightforward way to think about it is that each column of the matrix shows where one of the original basis functions ends up. For example, the first column transforms the basis function $s_H$ to its new position. The first column of the matrix can be found by taking the result on the right hand side of the above expressions, replacing every function that isn’t $s_H$ with a zero, putting the coefficient of $s_H$ ($1$ or $-1$ in this example) in the position at which it occurs, and taking the transpose to give a column vector.
Rotation
Consider the representative for the $C_2$ operation. The original basis $\begin{pmatrix} s_H, s_H', s_O, p_x, p_y, p_z \end{pmatrix}$ transforms into $\begin{pmatrix} s_H', s_H, s_O, -p_x, -p_y, p_z \end{pmatrix}$. The first column of the matrix therefore transforms $s_H$ into $s_H'$. Taking the result and replacing all the other functions with zeroes gives $\begin{pmatrix} 0, s_H, 0, 0, 0, 0 \end{pmatrix}$. The coefficient of $s_H$ is $1$, so the first column of the $C_2$ matrix representative is
$\begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \label{23.2}$
Matrix representation, characters and SALCs
The matrix representatives and their characters are
$\begin{array}{cccc} E & C_2 & \sigma_v & \sigma_v' \ \scriptsize{\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}} & \scriptstyle{\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}} & \scriptstyle{\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 &1 & 0 & 0 \ 0 & 0 & 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}} & \scriptstyle{\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}} \ \chi(E) = 6 & \chi(C_2) = 0 & \chi(\sigma_v) = 4 & \chi(\sigma_v') = 2 \end{array} \label{23.3}$
Now we are ready to work out which irreducible representations are spanned by the basis we have chosen. The character table for $C_{2v}$ is:
$\begin{array}{l|cccc|l} C_{2v} & E & C_2 & \sigma_v & \sigma_v' & h = 4 \ \hline A_1 & 1 & 1 & 1 & 1 & z, x^2, y^2, z^2 \ A_2 & 1 & 1 & -1 & -1 & xy, R_z \ B_1 & 1 & -1 & 1 & -1 & x, xz, R_y \ B_2 & 1 & -1 & -1 & 1 & y, yz, R_x \ \hline \end{array}$
As before, we use Equation (15.20) to find out the number of times each irreducible representation appears.
$a_k = \dfrac{1}{h}\sum_C n_C \chi(g) \chi_k(g) \label{23.4}$
We have
$\begin{array}{rcll} a(A_1) & = & \dfrac{1}{4}(1 \times 6 \times 1 + 1 \times 0 \times 1 + 1\times 4\times 1 + 1\times 2\times 1) & = 3 \ a(A_2) & = & \dfrac{1}{4}(1\times 6\times 1 + 1\times 0\times 1 + 1\times 4\times -1 + 1\times 2\times -1) & = 0 \ a(B_1) & = & \dfrac{1}{4}(1\times 6\times 1 + 1\times 0\times -1 + 1\times 4\times 1 + 1\times 2\times -1) & = 2 \ a(B_2) & = & \dfrac{1}{4}(1\times 6\times 1 + 1\times 0\times -1 + 1\times 4\times -1 + 1\times 2\times 1) & = 1 \end{array} \label{23.5}$
so the basis spans $3A_1 + 2B_1 + B_2$. Now we use the projection operators applied to each basis function $f_i$ in turn to determine the SALCs $\phi_i = \Sigma_g \chi_k(g) g f_i$
The SALCs of $A_1$ symmetry are:
$\begin{array}{rclll} \phi(s_H) & = & s_H + s_H' + s_H + s_H' & = & 2(s_H + s_H') \ \phi(s_H') & = & s_H' + s_H + s_H' + s_H & = & 2(s_H + s_H') \ \phi(s_O) & = & s_O + s_O + s_O + s_O & = & 4s_O \ \phi(p_x) & = & p_x - p_x + p_x - p_x & = & 0 \ \phi(p_y) & = & p_y - p_y + p_y - p_y & = & 0 \ \phi(p_z) & = & p_z + p_z + p_z + p_z & = & 4p_z \end{array} \label{23.6}$
The SALCs of $B_1$ symmetry are:
$\begin{array}{rclll} \phi(s_H) & = & s_H - s_H' + s_H - s_H' & = & 2(s_H - s_H') \ \phi(s_H') & = & s_H' - s_H + s_H' - s_H & = & 2(s_H' - s_H) \ \phi(s_O) & = & s_O - s_O + s_O - s_O & = & 0 \ \phi(p_x) & = & p_x + p_x + p_x + p_x & = & 4p_x \ \phi(p_y) & = & p_y + p_y - p_y - p_y & = & 0 \ \phi(p_z) & = & p_z - p_z + p_z - p_z & = & 0 \end{array} \label{23.7}$
The SALCs of $B_2$ symmetry are:
$\begin{array}{rclll} \phi(s_H) & = & s_H - s_H' -s_H = s_H' & = & 0 \ \phi(s_H') & = & s_H' - s_H - s_H' + s_H & = & 0 \ \phi(s_O) & = & s_O - s_O - s_O + s_O & = & 0 \ \phi(p_x) & = & p_x + p_x - p_x - p_x & = & 0 \ \phi(p_y) & = & p_y + p_y + p_y + p_y & = & 4p_y \ \phi(p_z) & = & p_z - p_z - p_z + p_z & = & 0 \end{array} \label{23.8}$
After normalization, our SALCs are therefore:
A1 symmetry
$\begin{array}{rcl} \phi_1 & = & \dfrac{1}{\sqrt{2}}(s_H + s_H') \ \phi_2 & = & s_O \ \phi_3 & = & p_z \end{array} \label{23.9}$
B1 symmetry
$\begin{array}{rcl} \phi_4 & = & \dfrac{1}{\sqrt{2}}(s_H - s_H') \ \phi_5 & = & p_x \end{array} \label{23.10}$
B2 symmetry
$\begin{array}{rcl} \phi_6 & = & p_y \end{array} \label{23.11}$
Note that we only take one of the first two SALCs generated by the $B_1$ projection operator since one is a simple multiple of the other (i.e. they are not linearly independent). We can therefore construct three molecular orbitals of $A_1$ symmetry, with the general form
$\begin{array}{rcll} \Psi(A_1) & = & c_1 \phi_1 + c_2 \phi_2 + c_3 \phi_3 & \ & = & c_1'(s_H + s_H') + c_2 s_O + c_3 p_z & \text{where} \: c_1' = \dfrac{c_1}{\sqrt{2}} \end{array} \label{23.12}$
two molecular orbitals of $B_1$ symmetry, of the form
$\begin{array}{rcl} \Psi(B_1) & = & c_4 \phi_4 + c_5 \phi_5 \ & = & c_4'(s_H - s_H') + c_5 p_z \end{array} \label{23.13}$
and one molecular orbital of $B_2$ symmetry
$\begin{array}{rcl} \Psi(B_2) & = & \phi_6 \ & = & p_y \end{array} \label{23.14}$
To work out the coefficients $c_1$ - $c_5$ and determine the orbital energies, we would have to solve the secular equations for each set of orbitals in turn. We are not dealing with a conjugated $p$ system, so in this case Hückel theory cannot be used and the various $H_{ij}$ and $S_{ij}$ integrals would have to be calculated numerically and substituted into the secular equations. This involves a lot of tedious algebra, which we will leave out for the moment. The LCAO orbitals determined above are an approximation of the true molecular orbitals of water, which are shown on the right. As we have shown using group theory, the $A_1$ molecular orbitals involve the oxygen $2s$ and $2p_z$ atomic orbitals and the sum $s_H + s_H'$ of the hydrogen $1s$ orbitals. The $B_1$ molecular orbitals involve the oxygen $2p_x$ orbital and the difference $s_H -s_H'$ of the two hydrogen $1s$ orbitals, and the $B_2$ molecular orbital is essentially an oxygen $2p_y$ atomic orbital.
Electronic transitions in molecules
When an electron is excited from one electronic state to another, this is what is called an electronic transition. A clear example of this is part C in the energy level diagram shown above. Just as in a vibrational transition the selection rules for electronic transitions are dictated by the transition moment integral. However we now must consider both the electronic state symmetries and the vibration state symmetries since the electron will still be coupled between two vibrational states that are between two electronic states. This gives us this modified transition moment integral:
Where you can see that the symmetry of the initial electronic state & vibrational state are in the Bra and the final electronic and vibrational states are in the Ket. Though this appears to be a modified version of the transition moment integral, the same equation holds true for a vibrational transition. The only difference would be the electronic state would be the same in both the initial and final states. Which the dot product of yields the totally symmetric representation, making the electronic state irrelevant for purely vibrational spectroscopy.
Raman
In Resonance Raman spectroscopy transition that occurs is the excitation from one electronic state to another and the selection rules are dictated by the transition moment integral discussed in the electronic spectroscopy segment. However mechanically Raman does produce a vibration like IR, but the selection rules for Raman state there must be a change in the polarization, that is the volume occupied by the molecule must change. But as far as group theory to determine whether or not a transition is allowed one can use the transition moment integral presented in the electronic transition portion. Where one enters the starting electronic state symmetry and vibrational symmetry and final electronic state symmetry and vibrational state, perform the direct product with the different M's or polarizing operators For more information about this topic please explore the Raman spectroscopy portion of the Chemwiki
Fluorescence
For the purposes of Group Theory Raman and Fluorescence are indistinguishable. They can be treated as the same process and in reality they are quantum mechanically but differ only in how Raman photons scatter versus those of fluorescence.
Phosphorescence
Phosphorescence is the same as fluorescence except upon excitation to a singlet state there is an interconversion step that converts the initial singlet state to a triplet state upon relaxation. This process is longer than fluorescence and can last microseconds to several minutes. However despite the singlet to triplet conversion the transition moment integral still holds true and the symmetry of ground state and final state still need to contain the totally symmetric representation.
Symmetry
Assume that we have a molecule in some initial state $\Psi_i$. We want to determine which final states $\Psi_f$ can be accessed by absorption of a photon. Recall that for an integral to be non-zero, the representation for the integrand must contain the totally symmetric irreducible representation. The integral we want to evaluate is
$\hat{\boldsymbol{\mu}}_{fi} = \int \Psi_f^* \hat{\boldsymbol{\mu}} \Psi_i d\tau \tag{27.4}$
so we need to determine the symmetry of the function $\Psi_f^* \hat{\boldsymbol{\mu}} \Psi_i$. As we learned in Section $18$, the product of two functions transforms as the direct product of their symmetry species, so all we need to do to see if a transition between two chosen states is allowed is work out the symmetry species of $\Psi_f$, $\hat{\boldsymbol{\mu}}$ and $\Psi_i$ , take their direct product, and see if it contains the totally symmetric irreducible representation for the point group of interest. Equivalently (as explained in Section $18$), we can take the direct product of the irreducible representations for $\hat{\boldsymbol{\mu}}$ and $\Psi_i$ and see if it contains the irreducible representation for $\Psi_f$. This is best illustrated using a couple of examples.
Earlier in the course, we learned how to determine the symmetry molecular orbitals. The symmetry of an electronic state is found by identifying any unpaired electrons and taking the direct product of the irreducible representations of the molecular orbitals in which they are located. The ground state of a closed-shell molecule, in which all electrons are paired, always belongs to the totally symmetric irreducible representation$^7$. As an example, the electronic ground state of $NH_3$, which belongs to the $C_{3v}$ point group, has $A_1$ symmetry. To find out which electronic states may be accessed by absorption of a photon, we need to determine the irreducible representations for the electric dipole operator $\hat{\boldsymbol{\mu}}$. Light that is linearly polarized along the $x$, $y$, and $z$ axes transforms in the same way as the functions $x$, $y$, and $z$ in the character table$^8$. From the $C_{3v}$ character table, we see that $x$- and $y$-polarized light transforms as $E$, while $z$-polarized light transforms as $A_1$. Therefore:
1. For $x$- or $y$-polarized light, $\Gamma_\hat{\boldsymbol{\mu}} \otimes \Gamma_{\Psi 1}$ transforms as $E \otimes A_1 = E$. This means that absorption of $x$- or $y$-polarized light by ground-state $NH_3$ (see figure below left) will excite the molecule to a state of $E$ symmetry.
2. For $z$-polarized light, $\Gamma_\hat{\boldsymbol{\mu}} \otimes \Gamma_{\Psi 1 }$ transforms as $A_1 \otimes A_1 = A_1$. Absorption of $z$-polarized light by ground state $NH_3$ (see figure below right) will excite the molecule to a state of $A_1$ symmetry.
Of course, the photons must also have the appropriate energy, in addition to having the correct polarization to induce a transition.
We can carry out the same analysis for $H_2O$, which belongs to the $C_{2v}$ point group. We showed previously that $H_2O$ has three molecular orbitals of $A_1$ symmetry, two of $B_1$ symmetry, and one of $B_2$ symmetry, with the ground state having $A_1$ symmetry. In the $C_{2v}$ point group, $x$-polarized light has $B_1$ symmetry, and can therefore be used to excite electronic states of this symmetry; $y$-polarized light has $B_2$ symmetry, and may be used to access the $B_2$ excited state; and $z$-polarized light has $A_1$ symmetry, and may be used to access higher lying $A_1$ states. Consider our previous molecular orbital diagram for $H_2O$.
The electronic ground state has two electrons in a $B_2$ orbital, giving a state of $A_1$ symmetry ($B_2 \otimes B_2 = A_1$). The first excited electronic state has the configuration $(1B_2)^1(3A_1)^1$ and its symmetry is $B_2 \otimes A_1 = B_2$. It may be accessed from the ground state by a $y$-polarized photon. The second excited state is accessed from the ground state by exciting an electron to the $2B_1$ orbital. It has the configuration $(1B_2)^1(2B_1)^1$, its symmetry is $B_2 \otimes B_1 = A_2$. Since neither $x$-, $y$- or $z$-polarized light transforms as $A_2$, this state may not be excited from the ground state by absorption of a single photon.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/12%3A_Group_Theory_-_The_Exploitation_of_Symmetry/12.15%3A_Molecular_Orbitals_can_be_Constructed_on_the_Basis_of_Symmetry.txt
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These are homework exercises to accompany Chapter 12 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Q12.1
Normalize the following equation:
${ψ(x)}={Z}{x}{e^{-kx^2}}$
S12.1
$\int_{-\infty}^{\infty} {ψ(x)^*}{ψ(x)} dx$
$1=Z^2\dfrac{1}{4}\surd{\dfrac{\pi}{2}}\alpha^\dfrac{-3}{2}$
$Z^2=4\surd{\dfrac{2}{\pi}}(\dfrac{m\omega}{2\hbar})^{3/2}$
notes:
When I integrated $(xe^{-kx^2})^{2}$ the answer had an $erf$ term in it. I think that the normalization of this specific function is more complex than was intended
Q12.3
List the symmetry elements for the bent molecule $H_2O$.
S12.3
Identity element $E$, two reflection planes $\sigma_{xz}$ and $\sigma_{yz}$, one 2-fold rotation axis $C_2$, and it belongs to the point group $C_2v$.
Q12.4
Verify that an ethene molecule has the symmetry elements given in Table 12.2.
S12.4
The point group of ethene is $D_{2h}$. The identity of element is given. There are three $C_2$ axes and three vertical axes.
Q12.5
Verify that a water molecule has the symmetry elements given in Table 12.2.
S12.5
The point group of water is $C_{2v}$. A water molecule contains the symmetry elements $E, C_2$, and $2\sigma_v$. Water contains a two- fold $C_2$ axis through the oxygen molecule located directly on the Z axis. Water also contains two vertical planes of symmetry. The first mirror plane cuts vertically through all three molecules, H-O-H. The second mirror plane cuts through the water molecule perpendicular to the other vertical plane. The $C_2$ axis lies along the intersection of the two $\sigma$ planes.
Q12.6
What is the point group of tetrachlropalladate [PdCl4]2- and show the symmetry elements.
S12.6
The symmetry elements for tetrachlropalladate are $E,i,C_{4},4C_{2},S_{4},\sigma_{h},2\sigma_{v},2\sigma_{d}$.
(left): proper rotations ($C_2$ and $C_4$), (center) improper $S_4$ rotation, (right) relection planes ($\sigma_{h}$ $\sigma_d$ and $\sigma_h$)
Q12.31
Considering the allyl anion, $CH_2CHCH_2$- , which belongs to the $C_{2v}$ point group, calculate the Huckel secular determinant using $|\psi_1 \rangle$, $|\psi_2 \rangle$, and $|\psi_3 \rangle$ ( $2_{pz}$ on each carbon atom). Then find the reducible representation for the allyl anion using $|\psi_j \rangle$ as the basis.
Show that the reducible representation $\Gamma$ = $A_2 + 2B_1$. What does this say about the expected secular determinant? Now, use the generating operator (Equation 13.2) to derive three symmetry orbitals for the allyl anion. Normalize them and calculate the Huckel secular determinant equation and solve for the $\pi$ electron energies.
S12.31
Applying the Huckel theory to the allyl anion yields the secular determinant given as
$\begin{vmatrix}\alpha-E&\beta&0\\beta&\alpha-E&\beta\0&\beta&\alpha-E\end{vmatrix} = 0$
Dividing the matrix by $\beta$ and using the variable $x = \dfrac{\alpha - E}{\beta}$, we can solve a determinant of the form:
$\begin{vmatrix}x&1&0\1&x&1\0&1&x\end{vmatrix} = 0$
Expanding this determinant gives the equation
$x^3-2x = 0$
Solving this equation gives $x = 0, \pm \sqrt{2}$.
The reducible representation can be found by looking at the four operators in the $C_{2v}$ point group, which are $E, C_2, \sigma_v, \sigma'_v$. The operator $E$ leaves all three orbitals unchanged (reducible representation of 3). The $C_{2v}$ operator inverts just one of the orbitals (a reducible representation of -1). The $\sigma_v$ operator leaves just one of the orbitals unchanged but does not invert any (reducible representation of 1). Lastly, the $\sigma'_v$ operator inverts all three orbitals (reducible representation of -3). Thus, the reducible representation of the $C_{2v}$ point group is
$\Gamma = 3 \hspace{1pc} -1 \hspace{1pc} 1 \hspace{1pc} -3$
Using equation 12.23, we find the irreducible representations to be
$a_{A_1} = \dfrac{1}{4} ( 3 -1 +1 -3) = 0$
$a_{A_2} = \dfrac{1}{4} ( 3 -1 -1 +3) = 1$
$a_{B_1} = \dfrac{1}{4} ( 3 +1 +1 +3) = 2$
$a_{B_2} = \dfrac{1}{4} ( 3 +1 -1 -3) = 0$
We therefore yield the reducible representation $\Gamma = A_2 + 2B_1$. This result shows us that the secular determinant can be written in either a 1 x 1 or 2 x 2 block diagonal form corresponding to the $A_2$ or $B_1$ representation, respectively. The three symmetry orbitals are found by
$P_{A_2} \psi_1 = \dfrac{1}{4}(\psi_1 - \psi_3 - \psi_3 + \psi_1) \propto \psi_1 - \psi_3$
$P_{B_1} \psi_1 = \dfrac{1}{4}(\psi_1 + \psi_3 + \psi_3 + \psi_1) \propto \psi_1 + \psi_3$
$P_{B_1} \psi_2 = \dfrac{1}{4}(\psi_2 + \psi_2 + \psi_2 + \psi_2) = \psi_2$
using generating operators for $A_2$ and $B_1$.
The three normalized symmetry orbitals are
$\Phi_1 = \dfrac{1}{\sqrt{2}}(\psi_1 - \psi_3)$
$\Phi_2 = \psi_2$
$\Phi_3 = \dfrac{1}{\sqrt{2}}(\psi_1 + \psi_3)$
Thus, these three orbitals give the symmetry elements below.
$H_{11} = \dfrac{1}{2} (2\alpha) = \alpha$
$H_{22} = \alpha$
$H_{33} = \dfrac{1}{2} (2\alpha) = \alpha$
$H_{12} = \dfrac{1}{2} (\beta - \beta) = 0$
$H_{13} = \dfrac{1}{2} (\alpha - \alpha) = 0$
$H_{23} = \dfrac{1}{\sqrt{2}} (2\beta) = \sqrt{2}\beta$
$S_{11} = S_{22} = S_{33} = 1$
$S_{12} = S_{13} = S_{23} = 0$
This gives the secular determinant
$\begin{vmatrix}\alpha-E&0&0\0&\alpha-E&\sqrt{2}\beta\0&\sqrt{2}\beta&\alpha-E\end{vmatrix} = 0$
Dividing the matrix by $\beta$ and using the variable $x = \dfrac{\alpha - E}{\beta}$, we can solve a determinant of the form:
$\begin{vmatrix}x&0&0\0&x&\sqrt{2}\0&\sqrt{2}&x\end{vmatrix} = 0$
which gives roots $x = 0, \pm \sqrt{2}$. Using the substitution that $x = \dfrac{\alpha - E}{\beta}$, we get the energies to be
$E_1 = \alpha - \sqrt{2}\beta$
$E_2 = \alpha$
$E_3 = \alpha + \sqrt{2}\beta$
Q12.32
How will the secular determinant for $SF_{6}$ look if we use group theory to generate symmetry orbitals?
S12.32
The reducible representation for an octahedral is
$O_{h}\ \ E\ \ 8C_{3}\ \ 6C_{2}\ \ 6C_{4}\ \ 3C_{2}\ \ i\ \ 6S_{4} \ \ 8S_{6} \ \ 3 \sigma _{h} \ \ 6 \sigma _{d} \ \Gamma \hspace{.5cm} 6\hspace{.5cm} 0 \hspace{.8cm} 0 \hspace{.8cm} 2 \hspace{.8cm} 2 \hspace{.5cm} 0 \hspace{.8cm} 0 \hspace{.8cm} 0\hspace{.6cm} 4 \hspace{.8cm} 2$
Use this equation:
$a_{i} = \dfrac{1}{h} \sum \chi ( \hat{R}) \chi _{i} ( \hat{R})$
to get:
$a_{E} = \dfrac{1}{10} \big( 6+2+2+4+2 \big)$
However, the reducible representation is better represented in a table format:
$O_{h}$ $E$ $8C_{3}$ $6C_{2}$ $6C_{4}$ $3C_{2}$ $i$ $6S_{4}$ $8S_{6}$ $3\sigma_{h}$ $6\sigma_{d}$
$\Gamma$ $6$ $0$ $0$ $2$ $2$ $0$ $0$ $0$ $4$ $2$
Q12.33
Apply the Great Orthogonality Theorem,
$\sum_{\hat{R}}\Gamma_i(\hat{R})_{nm}\Gamma_{ij}(\hat{R})_{n'm'} = \dfrac{h}{d_i}\delta_{ij}\delta_{mm'}\delta_{nn'}$,
to $C_{3v}$ point group given in which
$\Gamma_E = [ E_1 E_2 E_3 E_4 E_5 E_6 ]$
where
$E_1= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$
$E_2= \begin{bmatrix} -1/2 & -\sqrt{3}/2 \ -\sqrt{3}/2 & -1/2 \end{bmatrix}$
$E_3= \begin{bmatrix} -1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & -1/2 \end{bmatrix}$
$E_4= \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}$
$E_5= \begin{bmatrix} -1/2 & \sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{bmatrix}$
$E_6= \begin{bmatrix} -1/2 & -\sqrt{3}/2 \ -\sqrt{3}/2 & 1/2 \end{bmatrix}$
($h$ is the number of elements of $\Gamma_i$ and $d_i$ is the length of the diagonal of the matrix element of $\Gamma_i$)
S12.33
If we assume that $i=j=E_i$ and that $m=m',n=n'$, the general equation looks like
$\sum_{\hat{R}}[\Gamma_E(\hat{R})_{nm}]^2$
we must to pick the same element of each matrix, square it, and add them all together. All of them should equal $/frac{h}{l}=3$.
$\sum_{\hat{R}}[\Gamma_E(\hat{R})_{11}]^2 = 1+1/4+1/4+1+1/4+1/4=3$
$\sum_{\hat{R}}[\Gamma_E(\hat{R})_{12}]^2 = 0+3/4+3/4+0+3/4+3/4=3$
$\sum_{\hat{R}}[\Gamma_E(\hat{R})_{21}]^2 = 0+3/4+3/4+0+3/4+3/4=3$
$\sum_{\hat{R}}[\Gamma_E(\hat{R})_{22}]^2 = 1+1/4+1/4+1+1/4+1/4=3$
for unequal cases,($m\neq m'$ and $n\neq n'$) we can use the products of the elements and they should sum to zero.
$\sum_{\hat{R}}\Gamma_E(\hat{R})_{11}\Gamma_{E}(\hat{R})_{12} = 0+\sqrt{3}/4-\sqrt{3}/4+0-\sqrt{3}/4+\sqrt{3}/4=0$
$\sum_{\hat{R}}\Gamma_E(\hat{R})_{11}\Gamma_{E}(\hat{R})_{21} = 0-\sqrt{3}/4+\sqrt{3}/4+0-\sqrt{3}/4+\sqrt{3}/4=0$
$\sum_{\hat{R}}\Gamma_E(\hat{R})_{12}\Gamma_{E}(\hat{R})_{21} = 0-3/4-3/4+0+3/4+3/4=0$
$\sum_{\hat{R}}\Gamma_E(\hat{R})_{12}\Gamma_{E}(\hat{R})_{22} = 1+1/4+1/4-1-1/4-1/4=0$
$\sum_{\hat{R}}\Gamma_E(\hat{R})_{21}\Gamma_{E}(\hat{R})_{22} = 0-\sqrt{3}/4+\sqrt{3}/4+0+\sqrt{3}/4-\sqrt{3}/4=0$
Q12.34
Using the Great Orthogonality Theorem, let i = j, m = n, and m' = n' and sum over n and n' to show that
$\sum_{\hat{R}} [\chi_{j} (\hat{R}]^{2} = h$.
S12.34
Recall that $\chi_{j}(\hat{R})$ is defined as the character of the jth irreducible representation of $\hat{R}$, which in terms of matrix elements, is given by
$\chi_{i}(\hat{R}) = \sum_{m}\Gamma_{i}(\hat{R})_{mm}$
We now use the great orthogonality theorem to find the summed equation:
$\sum_{\hat{R}}\Gamma_{i}(\hat{R})_{mn}\Gamma_{j}(\hat{R})_{m'n'} = \dfrac{h}{l_i}\delta_{ij}\delta_{mm'}\delta_{nn'}$
Let i = j, m = n, and m' = n'. Then
$\sum_{\hat{R}}\Gamma_{i}(\hat{R})_{nn}\Gamma_{i}(\hat{R})_{n'n'} = \dfrac{h}{l_i}\delta_{nn'}$
$\sum_{\hat{R}}\sum_{n}\Gamma_{i}(\hat{R})_{nn}\sum_{n'}\Gamma_{i}(\hat{R})_{n'n'} = \dfrac{h}{l_i}\delta_{nn'}$
$\sum_{\hat{R}} [ \chi_{i}(\hat{R})]^{2} = \dfrac{h}{l_i} = h$
Q12.35
Determine the character table for $C_i$ which has the symmetry elements E and i.
S12.35
Because there are two symmetry elements, there are two rows to the character table also to have a 2x2. The first row is completely symmetric to both operations while the second is antisymmetric with respect to the inversion center. Therefore, the character table is as shown below.
Q12-36
The $C_i$ point group character table is given by
Ci E i
Ag +1 +1
Au +1 -1
Show that the basis for this point group are the even and odd functions over an interval (-a,a). Evaluate the integrals of this basis set using group theory in order to establish symmetry principles.
S12-36
Applying the inversion operator to a function,
$i f_{even} = f_{even}$
$i f_{odd} = -f_{odd}$
This demonstrates that f(even) belongs to Ag and f(odd) belongs to Au. As a result these functions are a basis of the $C_i$ point group.
$S_{ij}=\int{\phi_{i}}^*\phi_{j} d\tau$
$RS_{ij}=\int{R\phi_{i}}^*R\phi_{j} d\tau = S_{ij}=\int{\phi_{i}}^*\phi_{j} d\tau$ with the value of Sij unchanged by the symmetry operation of the point group.
$S_{ij} = \int_{-a}^{a} f_{even}(x)f_{even}(x) dx$
$iS_{ij} = \int_{-a}^{a} if_{even}(x)if_{even}(x) dx = \int_{-a}^{a} f_{even}(x)f_{even}(x) dx = 1$
$S_{ij} = \int_{-a}^{a} f_{odd}(x)f_{odd}(x) dx$
$iS_{ij} = \int_{-a}^{a} if_{odd}(x)if_{odd}(x) dx = \int_{-a}^{a} -f_{odd}(x)-f_{odd}(x) dx = 1$
$S_{ij} = \int_{-a}^{a} f_{even}(x)f_{odd}(x) dx$
$iS_{ij} = \int_{-a}^{a} if_{even}(x)if_{odd}(x) dx = -\int_{-a}^{a} f_{even}(x)f_{odd}(x) dx = 0$
12.37
Derive the symmetry orbitals for the pi- orbitals of butadiene by applying the generating operator
${P}_{j} = \dfrac{d_{j}}{h} \sum_{R} \chi_{j} ({R}){R}$
to the atomic 2pz orbital on each carbon atom. Identify the irreducible representation to which each resulting symmetry orbital belong. Derive the Huckel secular determinant.
S12.37
Butadiene belongs to the C2h point-group. Denote the 2pz orbital on $C_{i}$ by $\psi_{i}$
${P} \psi_{1} = \dfrac{1}{4}\sum_{R} \chi ({R}){R}$
$=\dfrac{1}{4}[(1){E}\psi_{1}+(1){C}_{2}\psi_{1} +(1){i}\psi_{1}+(1){\sigma}\psi_{1}]$
$=\dfrac{1}{4}(\psi_{1}+\psi_{4}-\psi_{4}+\psi_{2}) = 0$
Similarly,
${P} \psi_{2} = \dfrac{1}{4}(\psi_{2}+\psi_{3}-\psi_{3}-\psi_{2}) = 0$
$P\psi_2=\dfrac{1}{4}(\psi_2+\psi_3-\psi_3-\psi_2)=0$
Using Psi 1 and 2 ( things get very confusing after this line, especially the part of the equation "=0 $\alpha \psi_{1}- \psi_{4}$ ? -RM)
${P}\psi_{1} = \dfrac{1}{4}(\psi_{1}-\psi_{4}-\psi_{4}+\psi_{1}) = 0 \alpha \psi_{1}- \psi_{4}$
${P}\psi_{2} = \dfrac{1}{4}(\psi_{2}-\psi_{3}-\psi_{3}+\psi_{2}) = 0 \alpha \psi_{2}- \psi_{3}$
${P}\psi_{3} = \dfrac{1}{4}(\psi_{1}+\psi_{4}+\psi_{4}+\psi_{1}) = 0 \alpha \psi_{1}+\psi_{4}$
${P}\psi_{4} = \dfrac{1}{4}(\psi_{2}+\psi_{3}+\psi_{3}+\psi_{2}) = 0 \alpha \psi_{2}+\psi_{3}$
${P}\psi_{1} = \^{P}\psi_{2} = 0$
The process isnt very clear as to how you got to the solution...perhaps explain a little better how the math works.
Q12.41
An arbitrary tetrahedral molecule ( $AB_{4}$ ) belonging to the Td point group has the reducible representation: $\Gamma$ = 4 1 0 0 2. Show that:
1. the symmetry elements of the point group give this representation, and
2. it can be reduced as $\Gamma$ = $A_{1} + T_{2}$.
Finally, prove that an sp3 orbital with Td symmetry can be formed.
S12.41
a.) Applying the symmetry elements, we see that:
• $\hat{E}$ leaves all 4 bonds unmoved
• $\hat{C_{3}}$ leaves 1 bond unmoved
• $\hat{C_{2}}$ leaves 0 bonds unmoved
• $\hat{S_{4}}$ leaves 0 bonds unmoved
• $\hat{\sigma_{d}}$ leaves 2 bonds unmoved
The result is the reducible representation $\Gamma$ = 4 1 0 0 2.
b.) Rewriting the symmetry elements in terms of the irreducible representations, we see that:
• $\alpha_{A_{1}} = \dfrac{1}{24} (4+8+0+0+12) = 1$
• $\alpha_{A_{2}} = \dfrac{1}{24} (4+8+0+0-12) = 0$
• $\alpha_{E} = \dfrac{1}{24} (8-8+0+0+0) = 0$
• $\alpha_{T_{1}} = \dfrac{1}{24} (12+0+0+0-12) = 0$
• $\alpha_{T_{2}} = \dfrac{1}{24} (12+0+0+0+12) = 1$
Using $\alpha$ as a coefficient and taking the sum of these 5 equations, we can rewrite the reducible representation as $\Gamma$ = $A_{1} + T_{2}$.
c.) The 2 p orbitals all have $T_{2}$ symmetry for a $T_{d}$ molecule, so they can combine to form a hybrid $T_{2}$ orbital. All s orbitals are totally symmetric due to their spherical shape, making them $A_{1}$. Summing the 3 p orbitals and an s orbital will give a hybrid orbital of the desired $A_{1} + T_{2}$ symmetry.
-Interesting question. I like the explanation on how an Sp3 orbital with $T_2$ symmetry can be formed
Q12.42
Consider an octahedral molecule XY6 whose point group is Oh. Prove the irreducible representation of Oh is $\Gamma$ = A1g + Eg + T1u.
S12.43
$\ a_{A_{1g}} = \dfrac{1}{48}(6+0+0+12+6+0+0+0+12+12)=1 \ a_{A_{2g}} = \dfrac{1}{48}(6+0+0-12+6+0+0+0+12-12)=0 \ a_{E_{g}} = \dfrac{1}{48}(12+0+0+0+12+0+0+0+24+0)=1 \ a_{T_{1g}} = \dfrac{1}{48}(18+0+0+12-6+0+0+0-12-12)=0 \ a_{T_{2g}} = \dfrac{1}{48}(18+0+0-12-6+0+0+0-12+12)=0 \ a_{A_{1u}} = \dfrac{1}{48}(6+0+0+12+6+0+0+0-12-12)=0 \ a_{A_{2u}} = \dfrac{1}{48}(6+0+0-12+6+0+0+0-12+12)=0 \ a_{E_{u}} = \dfrac{1}{48}(12+0+0+0+12+0+0+0-24+0)=0 \ a_{T_{1u}} = \dfrac{1}{48}(18+0+0+12-6+0+0+0+12+12)=1 \ a_{T_{2u}} =\dfrac{1}{48}(18+0+0-12-6+0+0+0+12-12)=0$
= $\Gamma$ = A1g + Eg + T1u.
Q12.43
Consider an octahedral molecule XY6 whose point group is Oh. Prove the irreducible representation of Oh is $\Gamma$ = A1g + Eg + T1u.
S12.43
$\ a_{A_{1g}} = \dfrac{1}{48}(6+0+0+12+6+0+0+0+12+12)=1 \ a_{A_{2g}} = \dfrac{1}{48}(6+0+0-12+6+0+0+0+12-12)=0 \ a_{E_{g}} = \dfrac{1}{48}(12+0+0+0+12+0+0+0+24+0)=1 \ a_{T_{1g}} = \dfrac{1}{48}(18+0+0+12-6+0+0+0-12-12)=0 \ a_{T_{2g}} = \dfrac{1}{48}(18+0+0-12-6+0+0+0-12+12)=0 \ a_{A_{1u}} = \dfrac{1}{48}(6+0+0+12+6+0+0+0-12-12)=0 \ a_{A_{2u}} = \dfrac{1}{48}(6+0+0-12+6+0+0+0-12+12)=1 \ a_{E_{u}} = \dfrac{1}{48}(12+0+0+0+12+0+0+0-24+0)=0 \ a_{T_{1u}} = \dfrac{1}{48}(18+0+0+12-6+0+0+0+12+12)=1 \ a_{T_{2u}} =\dfrac{1}{48}(18+0+0-12-6+0+0+0+12-12)=0$
Therefore, the irreducible representation becomes $\Gamma = A_1 + A_2 + E$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/12%3A_Group_Theory_-_The_Exploitation_of_Symmetry/12.E%3A_Group_Theory_-_The_Exploitation_of_Symmetry_%28Exercises%29.txt
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Nonaxial Groups
These groups are characterized by a lack of a proper rotation axis.
$C_1$ E
A 1
$C_s$ E σh
A' 1 1 x, y, Rz x2, y2, z2, xy
A" 1 -1 z, Rx, Ry yz, xz
$C_i$ E i
Ag 1 1 Rx, Ry, Rz x2, y2, z2, xy, yz, zx
Au 1 -1 x,y,z
Cyclic $C_n$ Groups
These groups are characterized by an n-fold proper rotation axis $C_n$.
C2 E C2
A 1 1 z, Rz x2, y2, z2, xy
B 1 -1 x, y, Rx, Ry yz,xz
C3 E C3 C32 ε=exp(2π/3)
A 1 1 1 z, Rz x2+y2, z2
E $\left\{\begin{matrix} 1\ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\left. \begin{matrix} \epsilon^* \ \epsilon\; \end{matrix} \right\}$ (x,y),(Rx,Ry) (x2-y2, xy), (xz, yz)
C4 E C4 C2 C43
A 1 1 1 1 z, Rz x2+y2, z2
B 1 -1 1 -1 x2-y2, xy
E $\left\{ \begin{matrix}1 \ 1 \end{matrix} \right.$ $\begin{matrix} \;i \ -i\end{matrix}$ $\begin{matrix}-1 \ -1 \end{matrix}$ $\left. \begin{matrix}-i \ \;i \end{matrix} \right\}$ (x,y), (Rx,Ry) (xz, yz)
C5 E C5 C52 C53 C54 ε=exp(i2π/5)
A 1 1 1 1 1 Z, Rz x2+y2, z2
E1 $\left\{ \begin{matrix}\sf 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^2\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix}$ $\left. \begin{matrix} \epsilon^* \ \epsilon\; \end{matrix} \right\}$ (x, y), (Rx,Ry) (xz, yz)
E2 $\left\{ \begin{matrix}\sf 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon^2\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\left. \begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix} \right\}$ (x2-y2, xy)
C6 E C6 C3 C2 C32 C65 ε=exp(i2π/6)
A 1 1 1 1 1 1 z, Rz x2+y2, z2
B 1 -1 1 -1 1 -1
E1 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} -\epsilon^* \ -\epsilon\;\end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\left. \begin{matrix} \epsilon^* \ \epsilon\; \end{matrix} \right\}$ (Rx,Ry), (x,y) (xz, yz)
E2 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} -\epsilon^*\; \ -\epsilon\; \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} -\epsilon^*\; \ -\epsilon\; \end{matrix}$ $\left. \begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix} \right\}$ (x2-y2, xy)
C7 E C7 C72 C73 C74 C75 C76 ε=exp(i2π/7)
A 1 1 1 1 1 1 1 z, Rz x2+y2, z2
E1 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^2\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^3\; \ \epsilon^{*3} \end{matrix}$ $\begin{matrix} \epsilon^{*3}\; \ \epsilon^3\; \end{matrix}$ $\begin{matrix} \epsilon^{*2}\; \ \epsilon^2\; \end{matrix}$ $\left. \begin{matrix} \epsilon^{*}\ \epsilon\; \end{matrix} \right\}$ (Rx,Ry), (x,y) (xz, yz)
E2 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon^2\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^{*3}\; \ \epsilon^3\; \end{matrix}$ $\begin{matrix} \epsilon^{*}\ \epsilon\; \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^3\; \ \epsilon^{*3} \end{matrix}$ $\left. \begin{matrix} \epsilon^{*2}\; \ \epsilon^2\; \end{matrix} \right\}$ (x2-y2, xy)
E3 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon^3\; \ \epsilon^{*3} \end{matrix}$ $\begin{matrix} \epsilon^{*}\ \epsilon\; \end{matrix}$ $\begin{matrix} \epsilon^2\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^{*2}\; \ \epsilon^2\; \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\left. \begin{matrix} \epsilon^{*3}\; \ \epsilon^3\; \end{matrix} \right\}$
C8 E C8 C4 C83 C2 C85 C43 C87 ε=exp(i2π/8)
A 1 1 1 1 1 1 1 1 z, Rz x2+y2, z2
B 1 -1 1 -1 1 -1 1 -1
E1 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \epsilon\; \\epsilon^*\end{matrix}$ $\begin{matrix} \;i \ -i \end{matrix}$ $\begin{matrix} -\epsilon^*\ -\epsilon\;\end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} -i \ \;i \end{matrix}$ $\left. \begin{matrix} \epsilon^{*}\ \epsilon\; \end{matrix} \right\}$ (Rx,Ry), (x,y) (xz, yz)
E2 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} \;i \ -i \end{matrix}$ $\begin{matrix} -1 \-1 \end{matrix}$ $\begin{matrix} -i \ \;i \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \;i \ -i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\left. \begin{matrix} -i \ \;i \end{matrix} \right\}$ (x2-y2, xy)
E3 $\left\{ \begin{matrix} 1 \ 1 \end{matrix} \right.$ $\begin{matrix} -\epsilon\; \ -\epsilon^*\end{matrix}$ $\begin{matrix} \;i\ -i\; \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} -1 \-1 \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} -i \ \;i \end{matrix}$ $\left. \begin{matrix} -\epsilon^*\ -\epsilon\;\end{matrix} \right\}$
Pyramidal $C_{nv}$ Groups
These groups are characterized by an n-fold proper rotation axis $C_n$ and n mirror planes $σ_v$ which contain $C_n$
$C_{2v}$ E C2 σ(xz) σ(yz)
A1 1 1 1 1 z x2, y2, z2
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
$C_{3v}$ E 2C3 v
A1 1 1 1 z x2+y2, z2
A2 1 1 -1 Rz
E 2 -1 0 (Rx, Ry), (x,y) (xz, yz) (x2-y2, xy)
$C_{4v}$ E 2C4 C2 v d
A1 1 1 1 1 1 z x2+y2, z2
A2 1 1 1 -1 -1 Rz
B1 1 -1 1 1 -1 x2-y2
B2 1 -1 1 -1 1 xy
E 2 0 -2 0 0 (Rx, Ry), (x,y) (xz, yz)
$C_{5v}$ E 2C5 2C52 v
A1 1 1 1 1 z x2+y2, z2
A2 1 1 1 -1 Rz
E1 2 $2\cos 72^\circ$ $2\cos 144^\circ$ 0 (Rx, Ry), (x,y) (xz, yz)
E2 2 $2\cos{144^\circ}$ $2\cos 72^\circ$ 0 (x2-y2, xy)
$C_{6v}$ E 2C6 2C3 C2 v d
A1 1 1 1 1 1 1 z x2+y2, z2
A2 1 1 1 1 -1 -1 Rz
B1 1 -1 1 -1 1 -1
B2 1 -1 1 -1 -1 1
E1 2 1 -1 -2 0 0 (Rx, Ry), (x,y) (xz, yz)
E2 2 -1 -1 2 0 0 (x2-y2, xy)
C∞v E 2C ... ∞σv
A1 1 1 ... 1 z x2+y2, z2
A2 1 1 ... -1 Rz
E1 2 $2\cos{\phi}$ ... 0 (x,y), (Rx, Ry) (xz, yz)
E2 2 $2\cos{2\phi}$ ... 0 (x2-y2, xy)
E3 2 $2\cos{3\phi}$ ... 0
... ... ... ... ...
Reflection $C_{nh}$ Groups
These groups are characterized by an n-fold proper rotation axis $C_n$ and a mirror plane $\sigma_h$ normal to $C_n$.
$C_{2h}$ E C2 i σh
Ag 1 1 1 1 Rz x2, y2, z2
Bg 1 -1 1 -1 Rx, Ry xz, yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
$C_{3h}$ E C3 C32 σh S3 S35 ε=exp(i2π/3)
A' 1 1 1 1 1 1 Rz x2+y2, z2
E' $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\left.\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}\right\}$ (x,y) (x2-y2, xy)
A" 1 1 1 -1 -1 -1 z
E" $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\left.\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}\right\}$ (Rx, Ry) (xz, yz)
$C_{4h}$ E C4 C2 C43 i S43 σh S4
Ag 1 1 1 1 1 1 1 1 Rz x2+y2, z2
Bg 1 -1 1 -1 1 -1 1 -1 x2-y2, xy
Eg $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \; i \ -i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -i \ \; i \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \; i \ -i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\left.\begin{matrix} -i \ \; i \end{matrix}\right\}$ (Rx, Ry) (xz, yz)
Au 1 1 1 1 -1 -1 -1 -1 z
Bu 1 -1 1 -1 -1 1 -1 1
Eu $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \; i \ -i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -i \ \; i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -i \ \; i \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\left.\begin{matrix} \; i \ -i \end{matrix}\right\}$ (x,y)
$C_{5h}$ E C5 C52 C53 C54 σh S5 S57 S53 S59 ε=exp(i2π/5)
A' 1 1 1 1 1 1 1 1 1 1 Rz x2+y2, z2
E1' $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^{2}\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^{2}\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix}$ $\left.\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}\right\}$ (x, y)
E2' $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon^{2}\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \epsilon^{2}\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\left.\begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix}\right\}$ (x2-y2, xy)
A" 1 1 1 1 1 -1 -1 -1 -1 -1 z
E1" $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^{2}\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} -\epsilon^{2}\; \ -\epsilon^{*2} \end{matrix}$ $\begin{matrix} -\epsilon^{*2} \ -\epsilon^{2}\; \end{matrix}$ $\left.\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}\right\}$ (Rx, Ry) (xz, yz)
E2" $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon^{2}\; \ \epsilon^{*2} \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^{*2} \ \epsilon^2\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon^{2}\; \ -\epsilon^{*2} \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\left.\begin{matrix} -\epsilon^{*2} \ -\epsilon^{2}\; \end{matrix}\right\}$
$C_{6h}$ E C6 C3 C2 C32 C65 i S35 S65 σh S6 S3 ε=exp(i2π/6)
Ag 1 1 1 1 1 1 1 1 1 1 1 1 Rz x2+y2, z2
Bg 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1
E1g $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\left.\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}\right\}$ (Rx, Ry) (xz, yz)
E2g $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} -\epsilon^* \ -\epsilon\; \end{matrix}$ $\left.\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}\right\}$ (x2-y2, xy)
Au 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 z
Bu 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1
E1u $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\left.\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}\right\}$ (x, y)
E2u $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\left.\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}\right\}$
Dihedral $D_n$ Groups
$D_2$ E C2(z) C2(y) C2(x)
A 1 1 1 1 x2, y2, z2
B1 1 1 -1 -1 z, Rz xy
B2 1 -1 1 -1 y, Ry zx
B3 1 -1 -1 1 x, Rx yz
$D_3$ E 2C3 3C2
A1 1 1 1 x2+y2, z2
A2 1 1 -1 z, Rz
E 2 -1 0 (Rx, Ry), (x,y) (x2-y2, xy) (xz, yz)
$D_4$ E 2C4 C2(C42) 2C2' 2C2"
A1 1 1 1 1 1 x2+y2, z2
A2 1 1 1 -1 -1 z, Rz
B1 1 -1 1 1 -1 x2-y2
B2 1 -1 1 -1 1 xy
E 2 0 -2 0 0 (Rx, Ry), (x,y) (xz, yz)
$D_5$ E 2C5 2C52 5C2
A1 1 1 1 1 x2+y2, z2
A2 1 1 1 -1 z, Rz
E1 2 $2\cos{72^\circ}$ $2\cos{144^\circ}$ (Rx, Ry), (x,y) (xz, yz)
E2 2 $2\cos{144^\circ}$ $2\cos{72^\circ}$ (x2-y2, xy)
$D_6$ E 2C6 2C3 C2 2C2' 3C2"
A1 1 1 1 1 1 1 x2+y2, z2
A2 1 1 1 1 -1 -1 z, Rz
B1 1 -1 1 -1 1 -1
B2 1 -1 1 -1 -1 1
E1 2 1 -1 -2 0 0 (Rx, Ry), (x,y) (xz, yz)
E2 2 -1 -1 2 0 0 (x2-y2, xy)
Prismatic $D_{nh}$ Groups
These groups are characterized by
1. an n-fold proper rotation axis $C_n$
2. n 2-fold proper rotation axes $C_2$ normal to $C_n$
3. a mirror plane $\sigma_h$ normal to $C_n$ and containing the $C_2$ axes.
$D_{2h}$ E C2(z) C2(y) C2(x) i σ(xy) σ(xz) σ(yz)
Ag 1 1 1 1 1 1 1 1 x2, y2, z2
B1g 1 1 -1 -1 1 1 -1 -1 Rz xy
B2g 1 -1 1 -1 1 -1 1 -1 Ry xz
B3g 1 -1 -1 1 1 -1 -1 1 Rx yz
Au 1 1 1 1 -1 -1 -1 -1
B1u 1 1 -1 -1 -1 -1 1 1 z
B2u 1 -1 1 -1 -1 1 -1 1 y
B3u 1 -1 -1 1 -1 1 1 -1 x
$D_{3h}$ E 2C3 3C2 σh 2S3 v
A1' 1 1 1 1 1 1 x2+y2, z2
A2' 1 1 -1 1 1 -1 Rz
E' 2 -1 0 2 -1 0 (x,y) (x2-y2, xy)
A1" 1 1 1 -1 -1 -1
A2" 1 1 -1 -1 -1 1 z
E" 2 -1 0 -2 1 0 (Rx, Ry) (xz, yz)
$D_{4h}$ E 2C4 C2 2C2' 2C2" i 2S4 σh v d
A1g 1 1 1 1 1 1 1 1 1 1 x2+y2, z2
A2g 1 1 1 -1 -1 1 1 1 -1 -1 Rz
B1g 1 -1 1 1 -1 1 -1 1 1 -1 x2-y2
B2g 1 -1 1 -1 1 1 -1 1 -1 1 xy
Eg 2 0 -2 0 0 2 0 -2 0 0 (Rx, Ry) (xz, yz)
A1u 1 1 1 1 1 -1 -1 -1 -1 -1
A2u 1 1 1 -1 -1 -1 -1 -1 1 1 z
B1u 1 -1 1 1 -1 -1 1 -1 -1 1
B2u 1 -1 1 -1 1 -1 1 -1 1 -1
Eu 2 0 -2 0 0 -2 0 2 0 0 (x,y)
$D_{5h}$ E 2C5 2C52 5C2 σh 2S5 2S53 v
A1' 1 1 1 1 1 1 1 1 x2+y2, z2
A2' 1 1 1 -1 1 1 1 -1 Rz
E1' 2 $2\cos{72^\circ}$ $2\cos{144^\circ}$ 0 2 $2\cos{72^\circ}$ $2\cos{144^\circ}$ (x,y)
E2' 2 $2\cos{144^\circ}$ $2\cos{72^\circ}$ 0 2 $2\cos{144^\circ}$ $2\cos{72^\circ}$ (x2-y2, xy)
A1" 1 1 1 1 -1 -1 -1 -1
A2" 1 1 1 -1 -1 -1 -1 1 z
E1" 2 $2\cos{72^\circ}$ $2\cos{144^\circ}$ 0 -2 $-2\cos{72^\circ}$ $-2\cos{144^\circ}$ 0 (Rx, Ry) (xz, yz)
E2" 2 $2\cos{144^\circ}$ $2\cos{72^\circ}$ 0 -2 $-2\cos{144^\circ}$ $-2\cos{72^\circ}$ 0
D6h E 2C6 2C3 C2 3C2' 3C2" i 2S3 2S6 σh d v
A1g 1 1 1 1 1 1 1 1 1 1 1 1 x2+y2, z2
A2g 1 1 1 1 -1 -1 1 1 1 1 -1 -1 Rz
B1g 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1
B2g 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1
E1g 2 1 -1 -2 0 0 2 1 -1 -2 0 0 (Rx, Ry) (xz, yz)
E2g 2 -1 -1 2 0 0 2 -1 -1 2 0 0 (x2-y2, xy)
A1u 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1
A2u 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 z
B1u 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1
B2u 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1
E1u 2 1 -1 -2 0 0 -2 -1 1 2 0 0 (x,y)
E2u 2 -1 -1 2 0 0 -2 1 1 -2 0 0
D8h E 2C8 2C38 2C4 C2 4C2' 4C2" i 2S8 2S38 2S4 σh d v
A1g 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x2+y2, z2
A2g 1 1 1 1 1 -1 -1 1 1 1 1 1 -1 -1 Rz
B1g 1 -1 -1 1 1 1 -1 1 -1 -1 1 1 1 -1
B2g 1 -1 -1 1 1 -1 1 1 -1 -1 1 1 -1 1
E1g 2 -$\sqrt{2}$ $\sqrt{2}$ 0 -2 0 0 2 -$\sqrt{2}$ $\sqrt{2}$ 0 -2 0 0 (Rx, Ry) (xz, yz)
E2g 2 0 0 -2 2 0 0 2 0 0 -2 2 0 0 (x2-y2, xy)
E3g 2 $\sqrt{2}$ -$\sqrt{2}$ 0 -2 0 0 2 $\sqrt{2}$ -$\sqrt{2}$ 0 -2 0 0
A1u 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1
A2u 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 1 1 z
B1u 1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 1
B2u 1 -1 -1 1 1 -1 1 -1 1 1 -1 -1 1 -1
E1u 2 -$\sqrt{2}$ $\sqrt{2}$ 0 -2 0 0 -2 $\sqrt{2}$ -$\sqrt{2}$ 0 2 0 0 (x,y)
E2u 2 0 0 -2 2 0 0 -2 0 0 2 -2 0 0
E1u 2 $\sqrt{2}$ -$\sqrt{2}$ 0 -2 0 0 -2 -$\sqrt{2}$ $\sqrt{2}$ 0 2 0 0
D∞h E 2C ... ∞σv i 2S ... ∞ C2
Sg+ 1 1 ... 1 1 1 ... 1 x2+y2, z2
Sg- 1 1 ... -1 1 1 ... -1 Rz
πg 2 $2\cos{\phi}$ ... 0 2 $-2\cos{\phi}$ ... 0 (Rx, Ry) (xz, yz)
Dg 2 $2\cos{2\phi}$ ... 0 2 $2\cos{2\phi}$ ... 0 (x2-y2, xy)
... ... ... ... ...... ... ... ... ...
Su+ 1 1 ... 1 -1 -1 ... -1 z
Su- 1 1 ... -1 -1 -1 ... 1
πu 2 $2\cos{\phi}$ ... 0 -2 $2\cos{\phi}$ ... 0 (x, y)
Du 2 $2\cos{2\phi}$ ... 0 -2 $-2\cos{2\phi}$ ... 0
... ... ... ... ... ... ... ... ...
Antiprismatic $D_{nd}$ Groups
These groups are characterized by
1. an n-fold proper rotation axis Cn
2. n 2-fold proper rotation axes C2 normal to Cn
3. n mirror planes σd which contain Cn.
D2d E 2S4 C2 2C2' d
A1 1 1 1 1 1 x2+y2, z2
A2 1 1 1 -1 -1 Rz
B1 1 -1 1 1 -1 x2-y2
B2 1 -1 1 -1 1 z xy
E 2 0 -2 0 0 (x, y), (Rx, Ry) (xz, yz)
D3d E 2C3 3C2 i 2S6 d
A1g 1 1 1 1 1 1 x2+y2, z2
A2g 1 1 -1 1 1 -1 Rz
Eg 2 -1 0 2 -1 0 (Rx, Ry) (x2-y2, xy),(xz, yz)
A1u 1 1 1 -1 -1 -1
A2u 1 1 -1 -1 -1 1 z
Eu 2 -1 0 -2 1 0 (x, y)
D4d E 2S8 2C4 2S83 C2 4C2' d
A1 1 1 1 1 1 1 1 x2+y2, z2
A2 1 1 1 1 1 -1 -1 Rz
B1 1 -1 1 -1 1 1 -1
B2 1 -1 1 -1 1 -1 1 z
E1 2 $\sqrt{2}$ 0 $-\sqrt{2}$ -2 0 0 (x, y)
E2 2 0 -2 0 2 0 0 (x2-y2, xy)
E3 2 $-\sqrt{2}$ 0 $\sqrt{2}$ -2 0 0 (Rx, Ry) (xz, yz)
D5d E 2C5 2C52 5C2 i 2S103 2S10 d
A1g 1 1 1 1 1 1 1 1 x2+y2, z2
A2g 1 1 1 -1 1 1 1 -1 Rz
E1g 2 $2\cos 72^\circ$ $2\cos 144^\circ$ 0 2 $2\cos 72^\circ$ $2\cos 144^\circ$ 0 (Rx, Ry) (xz, yz)
E2g 2 $2\cos 144^\circ$ $2\cos 72^\circ$ 0 2 $2\cos 144^\circ$ $2\cos 72^\circ$ 0 (x2-y2, xy)
A1u 1 1 1 1 -1 -1 -1 -1
A2u 1 1 1 -1 -1 1 -1 1 z
E1u 2 $2\cos 72^\circ$ $2\cos 144^\circ$ 0 -2 $-2\cos 72^\circ$ $-2\cos 144^\circ$ 0 (x, y)
E2u 2 $2\cos 144^\circ$ $2\cos 72^\circ$ 0 -2 $-2\cos 144^\circ$ $-2\cos 72^\circ$ 0
D6d E 2S12 2C6 2S4 2C3 2S125 C2 6C2' d
A1 1 1 1 1 1 1 1 1 1 x2+y2, z2
A2 1 1 1 1 1 1 1 -1 -1 Rz
B1 1 -1 1 -1 1 -1 1 1 -1
B2 1 -1 1 -1 1 -1 1 -1 1 z
E1 2 $\sqrt{3}$ 1 0 -1 $-\sqrt{3}$ -2 0 0 (x, y)
E2 2 1 -1 -2 -1 1 2 0 0 (x2-y2, xy)
E3 2 0 -2 0 2 0 -2 0 0
E4 2 -1 -1 2 -1 -1 2 0 0
E5 2 $-\sqrt{3}$ 1 0 -1 $\sqrt{3}$ -2 0 0 (Rx, Ry) (xz, yz)
Improper Rotation $S_n$ Groups
These groups are characterized by an n-fold improper rotation axis $S_n$, where $n$ is necessarily even
$S_4$ E S4 C2 S43
A 1 1 1 1 Rz x2+y2, z2
B 1 -1 1 -1 z x2-y2, xy
E $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \; i \ -i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\left.\begin{matrix} -i \ \; i \end{matrix}\right\}$ (x, y); (Rx, Ry) (xz, yz)
S6 E C3 C32 i S65 S6
Ag 1 1 1 1 1 1 Rz x2+y2, z2
Eg $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\left.\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}\right\}$ (Rx, Ry) (x2-y2, xy), (xz, yz)
Au 1 1 1 -1 -1 -1 z
Eu $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\left.\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}\right\}$ (x, y)
S8 E S8 C4 S83 C2 S85 C43 S87 ε=exp(i2π/8)
A 1 1 1 1 1 1 1 1 Rz x2+y2, z2
B 1 -1 1 -1 1 -1 1 -1 z
E1 $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \; i \ -i \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} -i \ \; i \end{matrix}$ $\left.\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}\right\}$ (Rx, Ry), (x, y)
E2 $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \; i \ -i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -i \ \; i \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \; i \ -i \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\left.\begin{matrix} -i \ \; i \end{matrix}\right\}$ (x2-y2, xy)
E3 $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\begin{matrix} -i \ \; i \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} \; i \ -i \end{matrix}$ $\left.\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}\right\}$ (xz, yz)
Cubic Groups
These polyhedral groups are characterized by not having a $C_5$ proper rotation axis.
$T$ E 4C3 4C32 3C2
A 1 1 1 1 x2+y2+z2
E $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\left.\begin{matrix} 1 \ 1 \end{matrix}\right\}$ (2z2-x2-y2, x2-y2)
T 3 0 0 (Rx, Ry, Rz), (x, y, z) (xz, yz, xy)
Th E 4C3 4C32 3C2 i 4S6 4S65 h ε=exp(i2π/3)
Ag 1 1 1 1 1 1 1 1 x2+y2+z2
Eg $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\left.\begin{matrix} 1 \ 1 \end{matrix}\right\}$ (2z2-x2-y2, x2-y2)
Tg 3 0 0 -1 1 0 0 -1 (Rx, Ry, Rz) (xz, yz, xy)
Au 1 1 1 1 -1 -1 -1 -1
Eu $\left\{\begin{matrix} 1 \ 1 \end{matrix}\right.$ $\begin{matrix} \epsilon\; \ \epsilon^* \end{matrix}$ $\begin{matrix} \epsilon^* \ \epsilon\; \end{matrix}$ $\begin{matrix} 1 \ 1 \end{matrix}$ $\begin{matrix} -1 \ -1 \end{matrix}$ $\begin{matrix} -\epsilon\; \ -\epsilon^* \end{matrix}$ $\begin{matrix} -\epsilon^{*} \ -\epsilon\; \end{matrix}$ $\left.\begin{matrix} -1 \ -1 \end{matrix}\right\}$
Tu 3 0 0 -1 -1 0 0 1 (x, y, z)
Td E 8C3 3C2 6S4 d
A1 1 1 1 1 1 x2+y2+z2
A2 1 1 1 -1 -1
E 2 -1 2 0 0 (2z2-x2-y2, x2-y2)
T1 3 0 -1 1 -1 (Rx, Ry, Rz)
T2 3 0 -1 -1 1 (x, y, z) (xz, yz, xy)
O E 8C3 3C2 6C4 6C2
A1 1 1 1 1 1 x2+y2+z2
A2 1 1 1 -1 -1
E 2 -1 2 0 0 (2z2-x2-y2, x2-y2)
T1 3 0 -1 1 -1 (Rx, Ry, Rz), (x, y, z)
T2 3 0 -1 -1 1 (xz, yz, xy)
Oh E 8C2 6C2 6C4 3C2(C42) i 6S4 8S6 h d
A1g 1 1 1 1 1 1 1 1 1 1 x2+y2+z2
A2g 1 1 -1 -1 1 1 -1 1 1 -1
Eg 2 -1 0 0 2 2 0 -1 2 0 (2z2-x2-y2, x2-y2)
T1g 3 0 -1 1 -1 3 1 0 -1 -1 (Rx, Ry, Rz)
T2g 3 0 1 -1 -1 3 -1 0 -1 1 (xz, yz, xy)
A1u 1 1 1 1 1 -1 -1 -1 -1 -1
A2u 1 1 -1 -1 1 -1 1 -1 -1 1
Eu 2 -1 0 0 2 -2 0 1 -2 0
T1u 3 0 -1 1 -1 -3 -1 0 1 1 (x, y, z)
T2u 3 0 1 -1 -1 -3 1 0 1 -1
12.T: Correlation Tables
$C_{2v}$ $C_{2}$ $C_{s}\sigma(xz)$ $C_{s}\sigma(yz)$
$A_1$ $A$ $A'$ $A'$
$A_2$ $A$ $A"$ $A"$
$B_1$ $B$ $A'$ $A"$
$B_2$ $B$ $A"$ $A'$
$C_{3v}$ $C_{3}$ $C_{s}$
$A_1$ $A$ $A'$
$A_2$ $A$ $A"$
$E$ $E$ $A'+A"$
$C_{4v}$ $C_{4}$ $C_{2v}\;\sigma_v$ $C_{2v}\;\sigma_d$ $C_{2}$ $C_{s}\;\sigma_v$ $C_{s}\;\sigma_d$
$A_1$ $A$ $A_1$ $A_1$ $A$ $A'$ $A'$
$A_2$ $A$ $A_2$ $A_2$ $A$ $A"$ $A"$
$B_1$ $B$ $A_1$ $A_2$ $A$ $A'$ $A"$
$B_2$ $B$ $A_2$ $A_1'$ $A$ $A"$ $A'$
$E$ $E$ $B_1+B_2$ $B_1+B_2$ $2B$ $A'+A"$ $A'+A"$
$C_{5v}$ $C_{5}$ $C_{s}$
$A_1$ $A$ $A'$
$A_2$ $A$ $A"$
$E_1$ $\{E_1\}$ $A'+A"$
$E_2$ $\{E_2\}$ $A'+A"$
$C_{6v}$ $C_{6}$ $C_{3v}\;\sigma_v$ $C_{3v}\;\sigma_d$ $C_{2v}\;\sigma_v\rightarrow \sigma(xz)$ $C_{3}$ $C_{2}$ $C_{s}\;\sigma_v$ $C_{s}\;\sigma_d$
$A_1$ $A$ $A_1$ $A_1$ $A_1$ $A$ $A$ $A'$ $A'$
$A_2$ $A$ $A_2$ $A_2$ $A_2$ $A$ $A$ $A"$ $A"$
$B_1$ $B$ $A_1$ $A_2$ $B_1$ $A$ $B$ $A'$ $A"$
$B_2$ $B$ $A_2$ $A_1'$ $B_2$ $A'$ $B$ $A"$ $A'$
$E_1$ $\{E_1\}$ $E$ $E$ $B_1+B_2$ $\{E\}$ $2B$ $A'+A"$ $A'+A"$
$E_2$ $\{E_2\}$ $E$ $E$ $A_1+A_2$ $\{E\}$ $2A$ $A'+A"$ $A'+A"$
$C_{\infty v}$ $C_{2v}$
$\Sigma^+$ $A_1$
$\Sigma^-$ $A_2$
$\Pi$ $B_1 + B_2$
$\Delta$ $A_1 + A_2$
$D_{\infty h}$ $D_{2h}$
$\Sigma^+_g$ $A_g$
$\Sigma^-_g$ $B_{1g}$
$\Pi_g$ $B_{2g} + B_{3g}$
$\Delta_g$ $A_g + B_{1g}$
$\Sigma^+_u$ $B_{1u}$
$\Sigma^-_u$ $A_{u}$
$\Pi_u$ $B_{2u} + B_{3u}$
$\Delta_u$ $A_u + B_{1u}$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/12%3A_Group_Theory_-_The_Exploitation_of_Symmetry/12.T%3A_Character_Tables.txt
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Spectroscopy generally is defined as the area of science concerned with the absorption, emission, and scattering of electromagnetic radiation by atoms and molecules, which may be in the gas, liquid, or solid phase. Visible electromagnetic radiation is called light, although the terms light, radiation, and electromagnetic radiation can be used interchangeably. Spectroscopy played a key role in the development of quantum mechanics and is essential to understanding molecular properties and the results of spectroscopic experiments. It is used as a “stepping stone” to take us to the concepts of quantum mechanics and the quantum mechanical description of molecular properties in order to make the discussion more concrete and less abstract and mathematical.
• 13.1: The Electromagnetic Spectrum
Electromagnetic radiation—light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another are explained best by describing light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle.
• 13.2: Rotations Accompany Vibrational Transitions
Below, will learn how the rotational transitions of molecules can accompany the vibrational transitions. It is important to know how each peak correlates to the molecular processes of molecules. Rovibrational spectra can be analyzed to determine average bond length.
• 13.3: Unequal Spacings in Vibration-Rotation Spectra
As molecules are excited to higher rotational energies they spin at a faster rate. The faster rate of spin increases the centrifugal force pushing outward on the molecules resulting in a longer average bond length. Looking back, B and l are inversely related. Therefore the addition of centrifugal distortion at higher rotational levels decreases the spacing between rotational levels.
• 13.4: Unequal Spacings in Pure Rotational Spectra
Vibrational energy which is a consequence of the oscillations/ vibrations of the nuclei along inter nuclear axis, is possible only when the distance between the nuclei is not fixed/ rigid; that means the separation between the two nuclei is flexible/ elastic (non-rigid rotator). Consequently, centrifugal force, when the molecule is rotating, tends to fly the reduced mass μ away from the axis of rotation. To keep the mass rotating about the axis, there must be some restoring force to counter bal
• 13.5: Vibrational Overtones
Combination bands, overtones, and Fermi resonances are used to help explain and assign peaks in vibrational spectra that do not correspond with known fundamental vibrations. Combination bands and overtones generally have lower intensities than the fundamentals. Hot bands will also be briefly addressed.
• 13.6: Electronic Spectra Contain Electronic, Vibrational, and Rotational Information
Molecules can also undergo changes in electronic transitions during microwave and infrared absorptions. The energy level differences are usually high enough that it falls into the visible to UV range; in fact, most emissions in this range can be attributed to electronic transitions.
• 13.7: The Franck-Condon Principle
The Franck-Condon Principle describes the intensities of vibronic transitions, or the absorption or emission of a photon. It states that when a molecule is undergoing an electronic transition, such as ionization, the nuclear configuration of the molecule experiences no significant change. This is due in fact that nuclei are much more massive than electrons and the electronic transition takes place faster than the nuclei can respond. When the nucleus realigns itself with with the new electronic c
• 13.8: Rotational Spectra of Polyatomic Molecules
To consider the rotational energy of molecules, it is useful to divided molecules into five categories: Diatomic, linear, symmetric tops, spherical tops, and asymmetric tops. The principle moments of inertial of polyatomic molecules: Rotation of the molecule can take places about any axis passing through the center of mass. There are two unique axes that are at 90º of each other, and about which the moment of inertia is a minimum or a maximum.
• 13.9: Normal Modes in Polyatomic Molecules
Normal modes are used to describe the different vibrational motions in molecules. Each mode can be characterized by a different type of motion and each mode has a certain symmetry associated with it. Group theory is a useful tool in order to determine what symmetries the normal modes contain and predict if these modes are IR and/or Raman active. Consequently, IR and Raman spectroscopy is often used for vibrational spectra.
• 13.10: Irreducible Representation of Point Groups
Each of these coordinates belongs to an irreducible representation of the point the molecule under investigation. Vibrational wavefunctions associated with vibrational energy levels share this property as well. The normal coordinates and the vibration wavefunction can be categorized further according to the point group they belong to. From the character table predictions can be made for which symmetries can exist.
• 13.11: Time-Dependent Perturbation Theory
Time-dependent perturbation theory, developed by Paul Dirac, studies the effect of a time-dependent perturbation V(t) applied to a time-independent Hamiltonian. Since the perturbed Hamiltonian is time-dependent, so are its energy levels and eigenstates. Thus, the goals of time-dependent perturbation theory are slightly different from time-independent perturbation theory.
• 13.12: The Selection Rule for the Rigid Rotor
A selection rule describes how the probability of transitioning from one level to another cannot be zero. This presents a selection rule for rigid rotors that transitions are forbidden for Δl=0.
• 13.13: The Harmonic Oscillator Selection Rule
Transitions with Δv= ±1, ±2, ... are all allowed for anharmonic potential, but the intensity of the peaks become weaker as Δv increases. v=0 to v=1 transition is normally called the fundamental vibration, while those with larger Δv are called overtones. Δv=0 transition is allowed between the lower and upper electronic states with energy E1 and E2 are involved, i.e. (E1, v''=n) →→ (E2, v'=n), where the double prime and prime indicate the lower and upper quantum state.
• 13.14: Group Theory Determines Infrared Activity
Group theory makes it easy to predict which normal modes will be IR and/or Raman active. If the symmetry label of a normal mode corresponds to x, y, or z, then the fundamental transition for this normal mode will be IR active. If the symmetry label of a normal mode corresponds to products of x, y, or z (such as \(x^2\) or yz) then the fundamental transition for this normal mode will be Raman active.
• 13.E: Molecular Spectroscopy (Exercises)
These are exercises for Chapter 13 of the McQuarrie and Simon Textmap for Physical Chemistry.
Thumbnail: White light is dispersed by a prism into the colors of the visible spectrum. (CC BY-SA 3.0; D-Kuru).
13: Molecular Spectroscopy
An important aspect of studying Physical Chemistry is to be able to recognize the interaction of molecules to the surroundings. Molecular Spectroscopy provides a clear image of how diatomic and polyatomic molecules interact by looking at the Frequency, Wavelength, Wave number, Energy, and molecular process. We will also be able to see the absorption properties of molecules in various regions from the electromagnetic spectrum.
Electromagnetic Radiation
Electromagnetic radiation—light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another are explained best by describing light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle. The exact nature of electromagnetic radiation remains unclear, as it has since the development of quantum mechanics in the first quarter of the 20th century. Nevertheless, the dual models of wave and particle behavior provide a useful description for electromagnetic radiation.
Electromagnetic radiation consists of oscillating electric and magnetic fields that propagate through space along a linear path and with a constant velocity. In a vacuum electromagnetic radiation travels at the speed of light, $c$, which is $2.997 92 \times 10^8\, m/s$. When electromagnetic radiation moves through a medium other than a vacuum its velocity, $v$, is less than the speed of light in a vacuum. The difference between $v$ and $c$ is sufficiently small (<0.1%) that the speed of light to three significant figures, $3.00 \times 10^8\, m/s$, is accurate enough for most purposes.
The oscillations in the electric and magnetic fields are perpendicular to each other, and to the direction of the wave’s propagation. Figure 13.1.1 shows an example of plane-polarized electromagnetic radiation, consisting of a single oscillating electric field and a single oscillating magnetic field.
An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation.2 For example, the amplitude of the oscillating electric field at any point along the propagating wave is
$A_\ce{t} = A_\ce{e}\sin(2πνt + \phi) \nonumber$
where $A_t$ is the magnitude of the electric field at time $t$, $A_e$ is the electric field’s maximum amplitude, $\nu$ is the wave’s frequency—the number of oscillations in the electric field per unit time—and $\phi$ is a phase angle, which accounts for the fact that $A_t$ need not have a value of zero at $t = 0$. The identical equation for the magnetic field is
$A_\ce{t} =A_\ce{m}\sin(2πνt + \phi) \nonumber$
where $A_m$ is the magnetic field’s maximum amplitude.
The frequency and wavelength of electromagnetic radiation vary over many orders of magnitude. For convenience, we divide electromagnetic radiation into different regions—the electromagnetic spectrum—based on the type of atomic or molecular transition that gives rise to the absorption or emission of photons (Figure 13.1.2 ). The boundaries between the regions of the electromagnetic spectrum are not rigid, and overlap between spectral regions is possible.
Other Units
Other properties also are useful for characterizing the wave behavior of electromagnetic radiation. The wavelength, λ, is defined as the distance between successive maxima (Figure 13.1.1 ). For ultraviolet and visible electromagnetic radiation the wavelength is usually expressed in nanometers (1 nm = 10–9 m), and for infrared radiation it is given in microns (1 μm = 10–6 m). The relationship between wavelength and frequency is
$λ = \dfrac{c}{ν} \nonumber$
Another unit useful unit is the wavenumber, $\tilde{ν}$, which is the reciprocal of wavelength
$\tilde{ν} = \dfrac{1}{λ} \nonumber$
Wavenumbers are frequently used to characterize infrared radiation, with the units given in cm–1.
Example 13.1.1
In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observing a continuous spectrum with numerous dark lines. Fraunhofer labeled the most prominent of the dark lines with letters. In 1859, Gustav Kirchhoff showed that the D line in the sun’s spectrum was due to the absorption of solar radiation by sodium atoms. The wavelength of the sodium D line is 589 nm. What are the frequency and the wavenumber for this line?
Solution
The frequency and wavenumber of the sodium D line are
$ν = \dfrac{c}{λ} = \mathrm{\dfrac{3.00×10^8\: m/s}{589×10^{−9}\: m} = 5.09×10^{14}\: s^{−1}} \nonumber$
$\tilde{ν} = \dfrac{1}{λ} = \mathrm{\dfrac{1}{589×10^{−9}\: m} × \dfrac{1\: m}{100\: cm} = 1.70×10^4\: cm^{−1}} \nonumber$
Exercise 13.1.1
Another historically important series of spectral lines is the Balmer series of emission lines form hydrogen. One of the lines has a wavelength of 656.3 nm. What are the frequency and the wavenumber for this line?
Above, we defined several characteristic properties of electromagnetic radiation, including its energy, velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. A spectroscopic measurement is possible only if the photon’s interaction with the sample leads to a change in one or more of these characteristic properties. We can divide spectroscopy into two broad classes of techniques. In one class of techniques there is a transfer of energy between the photon and the sample. Table 13.1.1 provides a list of several representative examples.
Table 13.1.1 : Examples of Spectroscopic Techniques Involving an Exchange of Energy Between a Photon and the Sample
Type of Energy Transfer Region of Electromagnetic Spectrum Spectroscopic Technique
absorption γ-ray Mossbauer spectroscopy
X-ray X-ray absorption spectroscopy
UV/Vis
UV/Vis spectroscopy
atomic absorption spectroscopy
IR
infrared spectroscopy
raman spectroscopy
Microwave microwave spectroscopy
Radio wave
electron spin resonance spectroscopy
nuclear magnetic resonance spectroscopy
emission (thermal excitation) UV/Vis atomic emission spectroscopy
photoluminescence X-ray X-ray fluorescence
UV/Vis
fluorescence spectroscopy
phosphorescence spectroscopy
atomic fluorescence spectroscopy
chemiluminescence UV/Vis chemiluminescence spectroscopy
Electromagnetic spectrum provides clearly information of molecules if they are rotational transitions, vibrational transitions, or electronic transitions. A molecule or a set of molecules can be read by the absorption of microwave radiation which provides transitions between rotational energy levels. In addition, if the molecules absorbs infrared radiation provides the transitions between vibrational levels follows by transitions between rotational energy levels. Finally, when molecules absorbs visible and ultraviolet radiation gives transitions between electronic energy levels follows by simultaneous transitions between vibrational and rotational levels.
When given the energy level of the molecules along with wavelength, we can easily figure the frequency of the molecules where they fall in the electromagnetic spectrum regions:
$\Delta E=E_u-E_l=h \nu \nonumber$
The above equation describes the energy change between upper state and lower state of energy.
• Frequency falls between 109 - 1011 which is in the microwave range correlates to the rotation of polyatomic molecules.
• Frequency falls between 1011 - 1013 which is in the far infrared range correlates to the rotation of small molecules.
• Frequency falls between 1013 - 1014 which is in the infrared range correlates to the vibrations of flexible bonds.
• Frequency falls between 1014 - 1016 which is in the visible and ultraviolet range correlates to the electronic transitions.
The powerful technique of figuring out the the frequency of the molecules can help us determine the bond length, temperature, probability distribution as you will learn later on from the degree of freedoms and how the process is undergo in specific a reaction.
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Each of the modes of vibration of diatomic molecules in the gas phase also contains closely-spaced (1-10 cm-1 difference) energy states attributable to rotational transitions that accompany the vibrational transitions. A molecule’s rotation can be affected by its vibrational transition because there is a change in bond length, so these rotational transitions are expected to occur. Since vibrational energy states are on the order of 1000 cm-1, the rotational energy states can be superimposed upon the vibrational energy states.
Selection Rules
Rotational and Vibration transitions (also known as rigid rotor and harmonic oscillator) of molecules help us identify how molecules interact with each other, their bond length as mentioned in previous section. In order to know each transitions, we have to consider other terms like wavenumber, force constant, quantum number, etc. There are rotational energy levels associated with all vibrational levels. From this, vibrational transitions can couple with rotational transitions to give rovibrational spectra. Rovibrational spectra can be analyzed to determine average bond length.
We treat the molecule's vibrations as those of a harmonic oscillator (ignoring anharmonicity). The energy of a vibration is quantized in discrete levels and given by
$E_v=h\nu \left(v+\dfrac{1}{2} \right) \nonumber$
Where v is the vibrational quantum number and can have integer values 0, 1, 2..., and $\nu$ is the frequency of the vibration given by:
$\nu=\dfrac{1}{2\pi}\left(\dfrac{k}{\mu}\right)^\dfrac{1}{2} \nonumber$
where $k$ is the force constant and $\mu$ is the reduced mass of a diatomic molecule with atom masses $m_1$ and $m_2$, given by
$\mu=\dfrac{{m}_1{m}_2}{{m}_1+{m}_2} \label{reduced mass}$
We treat the molecule's rotations as those of a rigid rotor (ignoring centrifugal distortion from non-rigid rotor aspects). The energy of a rotation is also quantized in discrete levels given by
$E_r=\dfrac{h^2}{8\pi^2I} J(J+1) \nonumber$
In which $I$ is the moment of inertia, given by
${I}=\mu{r}^2 \nonumber$
where $\mu$ is the reduced mass (Equation \ref{reduced mass}) and $r$ is the equilibrium bond length.
Experimentally, frequencies or wavenumbers are measured rather than energies, and dividing by $h$ or $hc$ gives more commonly seen term symbols, $F(J)$ using the rotational quantum number $J$ and the rotational constant $B$ in either frequency
$F(J)=\dfrac{E_r}{h}=\dfrac{h}{8\pi^2I} J(J+1)=BJ(J+1) \nonumber$
or wavenumbers
$\tilde{F}(J)=\dfrac{E_r}{hc}=\dfrac{h}{8\pi^2cI} J(J+1)=\tilde{B}J(J+1) \nonumber$
It is important to note in which units one is working since the rotational constant is always represented as $B$, whether in frequency or wavenumbers.
• Vibrational Transition Selection Rules: At room temperature, typically only the lowest energy vibrational state v= 0 is populated, so typically v0 = 0 and ∆v = +1. The full selection rule is technically that ∆v = ±1, however here we assume energy can only go upwards because of the lack of population in the upper vibrational states.
• Rotational Transition Selection Rules: At room temperature, states with J≠0 can be populated since they represent the fine structure of vibrational states and have smaller energy differences than successive vibrational levels. Additionally, ∆J = ±1 since a photon contains one quantum of angular momentum and we abide by the principle of conservation of energy. This is also the selection rule for rotational transitions.
These two selection rules mean that the transition ∆J = 0 (i.e. J" = 0 and J' = 0, but $\nu_0 \neq 0$ is forbidden and the pure vibrational transition is not observed in most cases. The rotational selection rule gives rise to an R-branch (when ∆J = +1) and a P-branch (when ∆J = -1). Each line of the branch is labeled R(J) or P(J), where J represents the value of the lower state Figure 13.2.1 ).
R-branch
When ∆J = +1, i.e. the rotational quantum number in the ground state is one more than the rotational quantum number in the excited state – R branch (in French, riche or rich). To find the energy of a line of the R-branch:
\begin{align} \Delta{E} &=h\nu_0 +hB \left [J(J+1)-J^\prime (J^\prime{+1}) \right] \[4pt] &=h\nu_0 +hB \left[(J+1)(J+2)-J(J+1)\right] \[4pt] &= h\nu_0 +2hB(J+1) \end{align} \nonumber
P-branch
When ∆J = -1, i.e. the rotational quantum number in the ground state is one less than the rotational quantum number in the excited state – P branch (in French, pauvre or poor). To find the energy of a line of the P-branch:
\begin{align} \Delta{E} &=h\nu_0 +hB \left [J(J+1)-J^\prime(J^\prime+1) \right] \[4pt] &= h\nu_0 +hB \left [J(J-1)-J(J+1) \right] \[4pt] &= h\nu_0 -2hBJ \end{align} \nonumber
Q-branch
When ∆J = 0, i.e. the rotational quantum number in the ground state is the same as the rotational quantum number in the excited state – Q branch (simple, the letter between P and R). To find the energy of a line of the Q-branch:
\begin{align} \Delta{E} &=h\nu_0 +hB[J(J+1)-J^\prime(J^\prime+1)] \[4pt] &=h\nu_0 \end{align} \nonumber
The Q-branch can be observed in polyatomic molecules and diatomic molecules with electronic angular momentum in the ground electronic state, e.g. nitric oxide, NO. Most diatomics, such as O2, have a small moment of inertia and thus very small angular momentum and yield no Q-branch.
As seen in Figure 13.2.2 , the lines of the P-branch (represented by purple arrows) and R-branch (represented by red arrows) are separated by specific multiples of $B$ (i.e, $2B$), thus the bond length can be deduced without the need for pure rotational spectroscopy.
The total nuclear energy of the combined rotation-vibration terms, $S(v, J)$, can be written as the sum of the vibrational energy and the rotational energy
$S(v,J)=G(v)+F(J) \nonumber$
where $G(v)$ represents the energy of the harmonic oscillator, ignoring anharmonic components and $S(J)$ represents the energy of a rigid rotor, ignoring centrifugal distortion.
From this, we can derive
$S(v,J)=\nu_0 \left(v+\dfrac{1}{2}\right) +BJ(J+1) \nonumber$
The spectrum we expect, based on the conditions described above, consists of lines equidistant in energy from one another, separated by a value of $2B$. The relative intensity of the lines is a function of the rotational populations of the ground states, i.e. the intensity is proportional to the number of molecules that have made the transition. The overall intensity of the lines depends on the vibrational transition dipole moment.
In Figure 13.2.2 , between $P(1)$ and $R(0)$ lies the zero gap, where the the first lines of both the P- and R-branch are separated by $4B$, assuming that the rotational constant B is equal for both energy levels. The zero gap is also where we would expect the Q-branch, depicted as the dotted line, if it is allowed.
Advanced Concept: Occupations (Peak Intensities)
The relative intensity of the P- and R-branch lines depends on the thermal distribution of electrons; more specifically, they depend on the population of the lower J state. If we represent the population of the Jth upper level as NJ and the population of the lower state as N0, we can find the population of the upper state relative to the lower state using the Boltzmann distribution:
$\dfrac{N_J}{N_0}={(2J+1)e}^\left(-\dfrac{E_r}{kT}\right) \nonumber$
(2J+1) gives the degeneracy of the Jth upper level arising from the allowed values of MJ (+J to –J). As J increases, the degeneracy factor increases and the exponential factor decreases until at high J, the exponential factor wins out and NJ/N0 approaches zero at a certain level, Jmax. Thus, when
$\dfrac{d}{dJ} \left( \dfrac{N_J}{N_0} \right)=0 \nonumber$
by differentiation, we obtain
$J_{max}=\left(\dfrac{kT}{2hB}\right)^\dfrac{1}{2}-\dfrac{1}{2} \nonumber$
This is the reason that rovibrational spectral lines increase in energy to a maximum as J increases, then decrease to zero as J continues to increase, as seen in Figure 13.2.2 .
From this relationship, we can also deduce that in heavier molecules, $B$ will decrease because the moment of inertia will increase, and the decrease in the exponential factor is less pronounced. This results in the population distribution shifting to higher values of J. Similarly, as temperature increases, the population distribution will shift towards higher values of $J$.
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We find that real rovibrational spectra do not exhibit the equal spacing expectations from the treatment in the previous section and look more like the idealized spectrum in Figure 13.3.1 . As energy increases, the R-branch lines become increasingly similar in energy (i.e., the lines move closer together) and as energy decreases, the P-branch lines become increasingly dissimilar in energy (i.e. the lines move farther apart). This is attributable to two phenomena: rotational-vibrational coupling and centrifugal distortion.
Experimental data shows that the P branch has closer spaces between the lines as the frequencies decrease while R branch has further spaces between the lines as the frequencies increase. On the spectrum, R region is on the left and P region is on the right separated by a large gap in between. There are other graphs show differently, but the regions can be labeled using their spacing characteristic discussed below.
Vibration-Rotation Interaction
Consider the rotational constant for a rigid rotator $\tilde{B}=h/8\pi^2c\mu R_e^2$ where $\tilde{B}$ is the rotational constant, and $R_e^2$ is the bond length. It is clear from the equation that $\tilde{B}$ depends on $R_e^2$ in a inverse manner, so $\tilde{B}$ will decrease as $R_e^2$ increase, and vice versa.
Vibrational state and the bond length also possess a relationship, and that is bond length, $R_e^2$, will increase as the vibrational state, $v$ increases. High vibrational states have large vibrational amplitudes (the amplitude can be visualized as the distance from one side to the other on the harmonic-oscillator parabola, and higher states has greater distance). On a potential graph, imagine a diatomic molecule with one of the atom fixed at the origin, the bond is along the x-axis, and the molecule vibrate toward and away from the origin along x-axis. The range at how much the bond can stretch or squeeze depend on the vibrational amplitude. So, at high vibrational state (thus high amplitude), the bond can squeeze in and stretch out more (just a little more) on the regular harmonic-oscillator potential, but it can stretch out greatly on an anharmonic oscillator. That's why bond length increase with vibrational state.
Note
The two relationship established above
• $\tilde{B} \propto R_e^2$ (inversely proportional)
and
• $R_e^2 \propto v$ (directly proportional)
Therefore,
• $\tilde{B} \propto v$ (inversely proportional)
The rotational constant dependent on the vibrational state is denoted as $\tilde{B}_v$, and the dependence of $\tilde{B}$ on $v$ is the vibrational-rotational interaction.
Let's calculate the frequencies of R and P Branches with vibrational dependent rotational constants. The energies of the rigid rotator-harmonic oscillator is
$\tilde{E}_{v, J} = G(v) + F(J) = \tilde{\nu} (v+\dfrac{1}{2}) + \tilde{B} J (J+1) \nonumber$
or in term of $\tilde{B}_v$
$\tilde{E}_{v, J} = \tilde{\nu} \left(v+\dfrac{1}{2} \right) + \tilde{B}_v J (J+1) \nonumber$
According to the vibration-rotation interaction, $\tilde{B}_1 < \tilde{B}_0$ because $R_e^2 (v=1) > R_e^2 (v=0)$. Hence, the frequencies of rovibrational transitions from $v = 0 \rightarrow 1$ can be calculated as the following:
For R branch, allowed $J = 0, 1, 2, ...$
$\tilde{\nu}_R(\Delta J = +1) = \tilde{E}_{1, J+1} - \tilde{E}_{0, J} \nonumber$
$= \dfrac{3}{2} \tilde{\nu} + \tilde{B}_1 (J+1)(J+2) - \dfrac{1}{2} \tilde{\nu} - \tilde{B}_0 J(J+1) \nonumber$
$= \tilde{\nu}+2\tilde{B}_1+(3\tilde{B}_1-\tilde{B}_0)J+(\tilde{B}_1-\tilde{B}_0)J^2 \nonumber$
For P branch, allowed $J = 1, 2, 3, ...$
$\tilde{\nu}_P(\Delta J = -1) = \tilde{E}_{1, J-1} - \tilde{E}_{0, J} \nonumber$
$= \tilde{\nu}-(\tilde{B}_1+\tilde{B}_0)J+(\tilde{B}_1-\tilde{B}_0)J^2 \nonumber$
Decrease in Spacing of Lines in the R Branch with Increasing J
A closer look at
$\tilde{\nu}_R = \tilde{\nu}+2\tilde{B}_1+(3\tilde{B}_1-\tilde{B}_0)J+(\tilde{B}_1-\tilde{B}_0)J^2 \nonumber$
shows that the last term in the parentheses $(\tilde{B}_1-\tilde{B}_0)$ will be always negative because $\tilde{B}_1 < \tilde{B}_0$, and it also multiplies with $J^2$, so the square term $(\tilde{B}_1-\tilde{B}_0)J^2$ will give a larger negative value for increasing $J$. As a result, $\tilde{\nu}_R=\tilde{\nu} +$(smaller and smaller value) as J increase. Because the rotational frequencies keep getting smaller, the spacing between lines in R branch decreases as J increases.
Increasing in Spacing of Lines in the P Branch with Decreasing J
Using the same analysis as above,
$\tilde{\nu}_P = \tilde{\nu}-(\tilde{B}_1+\tilde{B}_0)J+(\tilde{B}_1-\tilde{B}_0)J^2 \nonumber$
shows that the last term in the parentheses $(\tilde{B}_1-\tilde{B}_0)$ will be always negative because $\tilde{B}_1 < \tilde{B}_0$, and it also multiplies with $J^2$, so the square term $(\tilde{B}_1-\tilde{B}_0)J^2$ will give a larger negative value even for decreasing $J$. As a result, $\tilde{\nu}_R=\tilde{\nu} -$(smaller and smaller value) as J decreases. Because the rotational frequencies keep getting larger, the spacing between lines in P branch increase as J decreases.
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Vibrational energy which is a consequence of the oscillations/ vibrations of the nuclei along inter nuclear axis, is possible only when the distance between the nuclei is not fixed/ rigid; that means the separation between the two nuclei is flexible/ elastic (non-rigid rotator). Consequently, centrifugal force, when the molecule is rotating, tends to fly the reduced mass $\mu$ away from the axis of rotation. To keep the mass rotating about the axis, there must be some restoring force to counter balance the centrifugal force. The work done to supply this force is stored as potential energy. Therefore, unlike the case of rigid rotator, total energy of rotation in a molecule comprises of kinetic and the potential energy corresponding to centrifugal force of rotation.
Let $r_e$ is the distance between the nuclei when the separation is taken to be rigid and $r_c$ under the action of the centrifugal force. According to Hook’s law, restoring force is proportional to change in internuclear distance $(r_c - r_e)\,$, or $= k(r_c - r_e)\,$ which in turn, will be equal to the centrifugal force.
Centrifugal force,
$F_c = \mu r_c \omega^2 = L^2/\mu r_c{^3} \label{16}$
Equating the restoring force to the centrifugal force, one gets
$k(r_c - r_e) = \dfrac{L^2}{\mu r_c{^3}}$
$r_c - r_e = \dfrac{L^2}{k\mu r_c{^3}} \cong \dfrac{L^2}{k\mu r_e{^3}} \label{17}$
Total energy, on adding kinetic energy, as expressed in Equation $\ref{17}$ to the potential energy,
$\dfrac{1}{2}k (r_c - r_e)^2\,$
is given by
$E_r = \dfrac{L^2}{ 2\mu r_c{^2}} + \dfrac{1}{2}k\, (r_c - r_e)^2 \label{18}$
Using $(r_c - r_e) = \Delta r\,$ and eliminating $r_c$ from equation $\ref{18}$, one gets
\begin{align} E_r &= \dfrac{L^2}{2\mu\,\, r_e{^2}(1 + \Delta r/r_e)^2} + \dfrac{1}{2} k \Delta r^2 \[5pt] &= \dfrac{ L^2(1 -2 \Delta r/r_e) } { 2\mu\,\, r_e{^2}} + \dfrac{1}{2} k \Delta r^2 \end{align}
using
$(1 + \Delta r/r_e)^{-2} \approx (1 -2 \Delta r/r_e)$
\begin{align} E_r &= \dfrac{L^2}{ 2\mu\,r_e{^2}} - \dfrac{L^4}{ k\mu^2\,r_e{^6}} + \dfrac{1}{2}k \dfrac{ L^4}{ k^{2\mu 2}r_e{^6}} \[5pt] & = \dfrac{L^2}{ 2\mu\,r_e{^2}} - \dfrac{1}{2} \dfrac{ L^4}{ k\mu^2\,r_e{^6}} \label{19} \[5pt] & = \dfrac{\hbar^2\,\,J(J + 1)}{ 2 \mu\,r_e{^2}} - \dfrac{\hbar^4\,\,J^2(J + 1)^2}{2k\mu^2\, r_e{^6}} \label{20} \end{align}
Use of Equation $\ref{17}$ and of the relation regarding angular momentum of a rotor
$L = \hbar\,\,\sqrt {J(J + 1)}$
has been made to obtain relation in Equation $\ref{20}$, which may be expressed in $cm^{-1}$ as
$\color{red} F(J) = \underbrace{\tilde{B} J(J + 1)}_{\text{rigid rotator term}} - \underbrace{\tilde{D} J^2(J + 1)^2 cm^{-1}}_{\text{centrifugal stretching}} \label{21}$
where
$\color{red} \tilde{B} = \dfrac{ \hbar}{4\pi\,\mu\,r_e{^2}c} \,\,(\text{in cm}^{-1})$
and
$\color{red} \tilde{D} = \dfrac{\hbar^3}{4\pi k\,\mu^2\,r_e{^6}\,c} \,\,(\text{in cm}^{-1}) \label{22}$
First term in the Equation $\ref{21}$ is same as for the rigid rotator; second term is the consequence of the centrifugal stretching. Recall Equation $\ref{17}$ wherein $k$ is the spring constant that, as we will see in the following section, plays the same role as in the vibrational motion. In other words, centrifugal stretching constant $\tilde{D}$ is not only measures the influence of centrifugal force, but also hints upon the interaction between the rotational and vibrational motions.
Since $\tilde{D}$ is positive, it is clear from Equation $\ref{21}$ that the energy levels for the non-rigid rotator are slightly lower on energy scale than those of rigid rotator for the corresponding $J$ values; the magnitude of decrease in energy of the non-rigid rotator states increases with $J$ as shown in the Figure $2$.
Consequently, on applying the selection rule $\Delta J =\pm 1$, rotational spectrum of a non-rigid rotator consists of a series of lines (red lines) wherein separation, unlike the case of spectral series (broken red lines) of a rigid rotator, between the consecutive rotational lines decreases with increase in $J$ , as shown in the Figure $2$. It may be noted that value of $D$ is very small compared to $B$ with the result that the influence of $D$ is significant only for very large $J$ values.
For example, for $HCl$ the values are $B \sim 10.4\,\,\,cm^{-1}$ and $D \sim 0.0004\,cm^{-1}$. Usually, $D$ is ignored in the calculations. Figure $2$ exaggerates the decrease in energy to visualize its effect. Nevertheless, non-rigid rotator is the model that describes the rotational motion more accurately and hence explains the spectral experimental observations not only in the microwave region but also the rotation-vibration spectra and the rotational structure of the electronic bands discussed in the later sections.
Contributors and Attributions
• 202.141.40.218/wiki/index.php...r_Spectroscopy
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Although the harmonic oscillator proves useful at lower energy levels, like $v=1$, it fails at higher numbers of $v$, failing not only to properly model atomic bonds and dissociations, but also unable to match spectra showing additional lines than is accounted for in the harmonic oscillator model.
Anharmonicity
Until this point, we have been using the harmonic oscillator to describe the internuclear potential energy of the vibrational motion. Fundamental vibrational frequencies of a molecule corresponds to transition from $\Delta v= \pm 1$. While this is a decent approximation, bonds do not behave like they do in the Harmonic Oscillator approximation (Figure 13.5.1 ). For exaple, unlike the parabola given in the Harmonic Oscillator approximation, atoms that are too far apart will dissociate.
As you can see in Figure 13.5.1 , the harmonic oscillator potential (in green) well only roughly fits over the more accurate anharmonic oscillator well (in blue). The solid line accounts for dissociation at large R values, which the dotted lines does not even remotely cover. However, this is just one important difference between the harmonic and anharmonic (real) oscillators.
The real potential energy can be expanded in the Taylor series.
$V(R) = V(R_e) + \dfrac{1}{2!}\left(\dfrac{d^2V}{dR^2}\right)_{R=R_e} (R-R_e)^2 + \dfrac{1}{3!}\left(\dfrac{d^3V}{dR^3}\right)_{R=R_e} (R-R_e)^3 + \dfrac{1}{4!}\left(\dfrac{d^4V}{dR^4}\right)_{R=R_e} (R-R_e)^4 + ... \label{taylor}$
This expansion was discussed in detail previously. The first term in the expansion is ignored since the derivative of the potential at $R_e$ is zero (i.e., at the bottom of the well). The Harmonic Oscillator approximation only uses the next term, the quadratic term, in the series
$V_{HO}(R) \approx V(R_e) + \dfrac{1}{2!}\left(\dfrac{d^2V}{dR^2}\right)_{R=R_e} (R-R_e)^2 \nonumber$
or in terms of a spring constant (and ignore the absolute energy term) and defining $r$ to equal the displacement from equilibrium ($r=R-R_e$), then we get the "standard" harmonic oscillator potential:
$V_{HO}(R) = \dfrac {1}{2} kr^2 \nonumber$
Alternatively, the expansion in Equation $\ref{taylor}$ can be shortened to the cubic term
$V(x) = \dfrac {1}{2} kr^2 + \dfrac {1}{6} \gamma r^3 \label{cubic}$
where
• $V(x_0) = 0$, and $r = R - R_0$.
• $k$ is the harmonic force constant, and
• $\gamma$ is the first (i.e., cubic) anharmonic term
It is important to note that this approximation is only good for $R$ near $R_0$.
The harmonic oscillator approximation and gives by the following energies:
$E_{v} = \tilde{\nu} \left (v + \dfrac{1}{2} \right) \nonumber$
When cubic terms in the expansion (Equation $\ref{cubic}$) is included, then Schrödinger equation solved, using perturbation theory, gives:
$E_{v} = \tilde{\nu} \left (v + \dfrac{1}{2} \right) - \tilde{\chi_e} \tilde{\nu} \left (v + \dfrac{1}{2} \right)^2 \nonumber$
where $\tilde{\chi_e}$ is the anharmonicity constant. It is much smaller than 1, which makes sense because the terms in the Taylor series approach zero. This is why, although $G(n)$ technically includes all of the Taylor series, we only concern ourselves with the first and second terms. The rest are so small and barely add to the total and thus can be ignored. To get a more accurate approximation, more terms can be included, but otherwise, can be ignored. Almost all diatomics have experimentally determined $\frac {d^2 V}{d x^2}$ for their lowest energy states. $\ce{H2}$, $\ce{Li2}$, $\ce{O2}$, $\ce{N2}$, and $\ce{F2}$ have had terms up to $n < 10$ determined of Equation $\ref{taylor}$.
Overtones
The Harmonic Oscillator approximation predicts that there will be only one line the spectrum of a diatomic molecule, and while experimental data shows there is in fact one dominant line--the fundamental--there are also other, weaker lines. How can we account for these extra lines?
Any resonant frequency above the fundamental frequency is referred to as an overtone. In the IR spectrum, overtone bands are multiples of the fundamental absorption frequency. As you can recall, the energy levels in the Harmonic Oscillator approximation are evenly spaced apart. Energy is proportional to the frequency absorbed, which in turn is proportional to the wavenumber, the first overtone that appears in the spectrum will be twice the wavenumber of the fundamental. That is, first overtone $v = 1 \rightarrow 2$ is (approximately) twice the energy of the fundamental, $v = 0 \rightarrow 1$.
The levels are not equally spaced, like in the harmonic oscillator, but decrease as $v$ increases, until it ultimately converges, is implied by Figure 13.5.4 . Also as a result of anharmonicity, the $\Delta v= \pm 1$ selection rule is no longer valid and $v$ can be any number. This leads to the observation of higher order transitions, or overtones, which result from the transition of the ground state to higher energy levels.
Anharmonic Oscillator Selection Rules
For the anharmonic oscillator, the selection rule is $\Delta V= \text{any number}$. That is, there are no selection rules (for state to state transitions)
Overtones occur when a vibrational mode is excited from $v=0$ to $v=2$ (the first overtone) or $v=0$ to $v=3$ (the second overtone). The fundamental transitions, $v=\pm 1$, are the most commonly occurring, and the probability of overtones rapid decreases as $\Delta v > \pm 1$ gets bigger. Based on the harmonic oscillator approximation, the energy of the overtone transition will be approximately $v$ times the fundamental associated with that particular transition. The anharmonic oscillator calculations show that the overtones are usually less than a multiple of the fundamental frequency. Overtones are generally not detected in larger molecules.
This is demonstrated with the vibrations of the diatomic $\ce{HCl}$ in the gas phase:
Transition obs [cm-1] obs Harmonic [cm-1] obs Anharmonic [cm-1]
Table 13.5.1 : HCl vibrational spectrum.
$0 \rightarrow 1$ (fundamental) 2885.9 2885.9 2,885.3
$0 \rightarrow 2$ (first overtone) 5668.0 5771.8 5,665.0
$0 \rightarrow 3$ (second overtone) 8347.0 8657.7 8,339.0
$0 \rightarrow 4$ (third overtone) 10 923.1 11 543.6 10,907.4
$0 \rightarrow 5$ (fourth overtone) 13 396.5 14 429.5 13,370
We can see from Table 13.5.1 that the anharmonic frequencies correspond much better with the observed frequencies, especially as the vibrational levels increase. Because the energy levels and overtones are closer together in the anharmonic model, they are also more easily reached. This means that there is a higher chance of that level possibly being occupied, meaning it can show up as additional, albeit weaker intensity lines (the weaker intensity indicates a smaller probability of the transition occuring).
Exercise 13.5.1
$\ce{HCl}$ has a fundamental band at 2885.9 cm−1 and an overtone at 5668.1 cm−1 Calculate $\tilde{\nu}$ and $\tilde{\chi_e}$.
Exercise 13.5.2
Write out the Taylor series, and comment on the trend in the increasing terms. Using a test number $x$, please add terms 3, 4, and 5, then compare this to term 2. How do they compare? We have seen that the anharmonic terms increase the accuracy of our oscillator approximation. Why don't we care so much about terms past the second?
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Molecules can also undergo changes in electronic transitions during microwave and infrared absorptions. The energy level differences are usually high enough that it falls into the visible to UV range; in fact, most emissions in this range can be attributed to electronic transitions.
Electron Transitions are not Purely Electronic
We have thus far studied rovibrational transitions--that is, transitions involving both the vibrational and rotational states. Similarly, electronic transitions tend to accompany both rotational and vibrational transitions. These are often portrayed as an electronic potential energy cure with the vibrational level drawn on each curve. Additionally, each vibrational level has a set of rotational levels associated with it.
Recall that in the Born-Oppenheimer approximation, nuclear kinetic energies can be ignored (e.g., fixed) to solve for electronic wavefunctions and energies, which are much faster than rotation or vibration. As such, it is important to note that unlike rovibrational transitions, electronic transitions aren't dependent on rotational or transitional terms and are assumed to be separate. Therefore, when using an anharmonic oscillator-nonrigid rotator approximation (and excluding translation energy), the total energy of a diatomic is:
$\tilde{E}_{total} = \tilde{\nu}_{el} + G(v) + F(J) \label{Eqa1}$
where $\tilde{\nu}_{el}$ is the electronic transition energy change in wavenumbers, $G(n)$ is the vibrational energy with energy level $v$ (assuming anharmonic oscillator), and $F(J)$ is the rotational energy, assuming a nonrigid rotor. Equation $\ref{Eqa1}$ can be expanded accordingly:
$\tilde{E}_{total} = \underbrace{\tilde{\nu}_{el}}_{\text{electronic}} + \underbrace{\tilde{\nu}_e \left (v + \dfrac{1}{2} \right) - \tilde{\chi}_e \tilde{\nu}_e \left (v + \dfrac{1}{2} \right)^2}_{\text{vibrational}} + \underbrace{\tilde{B} J(J + 1) - \tilde{D} J^2(J + 1)^2}_{\text{rotational}} \label{Eqa2}$
Notice that both the vibration constant ($\tilde{\nu}_e$) and anharmonic constant ( $\tilde{\chi}_e$) are electronic state dependent (and hence the rotational constants would be too, but are ignored here). Since rotational energies tend to be so small compared to electronic, their effects are minimal and are typically ignored when we do calculations and are referred to as vibronic transitions.
The eigenstate-to-eigenstate transitions (e.g., $1 \rightarrow 2$) possible are numerous and have absorption lines at
$\tilde{\nu}_{obs} = \tilde{E}_{2} - \tilde{E}_{1} \label{Eqa21}$
and for simplification, we refer to constants associated with these states as $| ' \rangle$ and $| '' \rangle$, respectively. So Equation $\ref{Eqa21}$ is
$\tilde{\nu}_{obs} = \tilde{E''(v'')} - \tilde{E'(v')} \nonumber$
Also important to note that typically vibronic transitions are usually the result of the vibrational $v'=0$ vibratonal state. Within this assumption and excluding the rotational contributions (due to their low energies), Equation $\ref{Eqa2}$ can be used with Equation $\ref{Eqa21}$ to get
$\tilde{\nu}_{obs} = \tilde{T}_{el} + \left( \dfrac{1}{2} \tilde{\nu}'_e - \dfrac{1}{4} \tilde{\chi}'_e \tilde{\nu}_e' \right) - \left( \dfrac{1}{2} \tilde{\nu}''_e - \dfrac{1}{4} \tilde{\chi}''_e \tilde{\nu}_e'' \right) + \tilde{\nu}'_e v'' - \tilde{\chi}'_e \tilde{\nu}_e' v''(v''+1) \label{Eqa3}$
A common transition of importance is the $\tilde{\nu}_{00}$, which is the $0 \rightarrow 0$ transition and include no vibrational change. For this case, equation $\ref{Eqa3}$ is then
$\tilde{\nu}_{00} = \tilde{T}_{el} + \left( \dfrac{1}{2} \tilde{\nu}'_e - \dfrac{1}{4} \tilde{\chi}'_e \tilde{\nu}_e' \right) - \left( \dfrac{1}{2} \tilde{\nu}''_e - \dfrac{1}{4} \tilde{\chi}''_e \tilde{\nu}_e'' \right) \nonumber$
This is the lowest energy possible to observe in an electronic transition although it may be of low intensity as discussed in the following section.
Iodine
The absorption spectrum of iodine yields information about the excited state well rather than the ground state well (notice that equation $\ref{Eqa3}$ depends primarily on excited state parameters). In this experiment you will characterize the excited state well by extracting values for the following excited state parameters.
Recall that as $v'$ increases, the vibrational energy spacing decreases. At the upper edge of the well, the vibrational energy spacing decreases to 0, which means that the energies form a continuum rather than being quantized. It is at this limit that bond dissociation occurs. The energy required to dissociate the bond is actually $D_o'$ rather than $D_e'$ because the molecule cannot have less than the zero point energy.
The vibrational-electronic spectrum of I2 in the region from 500-650 nm displays a large number of well-defined bands which, for the most part, correspond to v'<-- 0 transitions connecting the v" = 0 vibrational level of the ground electronic state (denoted as X1Σ+) to many different vibrational levels v' of the excited B3Π electronic state. Under the conditions of this experiment (i.e., low resolution), the rotational lines within each band are not resolved. However, the peaks may be identified as R-branch band heads (1). For a molecule as heavy as I2, the position of each band head is within a few tenths of one cm-1 of the band origin (2), and for the purposes of this experiment, the distinction between the two may be ignored. The general features of the absorption spectrum are shown below:
Each small bump, or peak, such as the (26,0) band labelled on the spectrum, corresponds to a transition between two vibrational levels and is called a band. Each band is comprised of several hundred lines, each of which involves different upper and lower rotational quantum numbers; as mentioned, these lines are not resolved in the present experiment. The region of maximum absorption in each band is caused by many of these lines falling together; it is called the band head. The set of all of these bands is referred to as the visible band system of I2.
If the sample is hot, then excited vibrational levels of the ground state may be populated, and these also will absorb light. The hot bands arising from absorption from v"=1 and v"=2 are shown very approximately on the absorption spectrum above.
At a point called the convergence limit, the spacing between bands decreases to zero. Beyond this convergence limit, the spectrum is continuous because the excited state of the I2 molecule is not bound. One of the purposes of this experiment is to identify this convergence limit accurately.
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The Franck-Condon Principle describes the intensities of vibronic transitions, or the absorption or emission of a photon. It states that when a molecule is undergoing an electronic transition, such as ionization, the nuclear configuration of the molecule experiences no significant change. This is due in fact that nuclei are much more massive than electrons and the electronic transition takes place faster than the nuclei can respond. When the nucleus realigns itself with the new electronic configuration, the theory states that it must undergo a vibration.
If we picture the vertical transition from ground to excited electronic state as occurring from a vibrational wavefunction that gives a probability distribution of finding the nuclei in a give region of space we can determine the probability of a given vibrational level from the overlap integral $S_{v’,v}$ which gives the overlap of the vibrational wavefunction in the ground and excited state. The $v’$ quantum numbers refer to the ground state and the $v$ quantum numbers refer to the excited state. The transition probability can be separated into electronic and nuclear parts using the Condon approximation.
In Figure 13.7.1 , the nuclear axis shows a consequence of the internuclear separation and the vibronic transition is indicated by the blue and green vertical arrows. This figure demonstrates three things:
1. An absorption leads to a higher energy state,
2. fluorescence leads to a lower energy state, and
3. the shift in nuclear coordinates between the ground and excited state is indicative of a new equilibrium position for nuclear interaction potential. The fact that the fluorescence arrow is shorter than the absorption indicates that it has less energy, or that its wavelength is longer.
The Classical Condon approximation
Condon approximation is the assumption that the electronic transition occurs on a time scale short compared to nuclear motion so that the transition probability can be calculated at a fixed nuclear position.
This change in vibration is maintained during a state termed the rapid electronic excitation. The resulting Coulombic forces produce an equilibrium as shown in the figure for the nuclei termed a turning point. The turning point can be mapped by drawing a vertical line from the minimum of the lower curve to the intersection of the higher electronic state. This procedure is termed a vertical transition and was discussed before in the context of photoelectron spectroscopy (another electronic spectroscoy)..
The Franck-Condon Principle explains the relative intensities of vibronic transitions by relating the probablity of a vibrational transition to the overlap of the vibrational wavefunctions. It states that the probability of a vibrational transition occurring is weighted by the Franck-Condon overlap integral:
$P_{i \rightarrow f} = | \langle \psi^*_{final} | \boldsymbol{\mu} | \psi_{initial} \rangle | ^2 = | \int \psi^*_{final} \boldsymbol{\mu} \psi_{initial} d\tau | ^2 \nonumber$
Within the Franck-Condon approximation, the nuclei are considered "fixed" during electronic transitions. Thus, electronic transitions can be considered vertical transitions on electronic potential energy curves (vierical transitions in Figure 13.7.1 ).
The Quantum Franck-Condon Principle
The Franck-Condon Principle has both a Classical and Quantum application. Classically, the Franck–Condon principle is the approximation that an electronic transition is most likely to occur without changes in the positions of the nuclei in the molecular entity and its environment. The resulting state is called a Franck–Condon state, and the transition involved, a vertical transition. The quantum mechanical formulation of this principle is that the intensity of a vibronic transition is proportional to the square of the overlap integral between the vibrational wavefunctions of the two states that are involved in the transition.
The Franck-Condon principle is based on the Born-Oppenheimer approximation, which allows separation of the electronic $q$ and nuclear $Q$ wavefunctions given the total wavefunction.
$| \psi_{total}(Q,q) = | \psi_{nuc}(Q) \rangle \psi_{el} (Q;q) \rangle \nonumber$
Since the transition operator, $\hat{\mu}(q)$, is dependent only on the electronic component, the nuclear components can be separated from the transition moment integral that dictates the probability of the transition occuring:
\begin{align} \langle \psi^{*}_{total, f} | \hat{\mu} | \psi_{total, i} \rangle &= \langle \psi^*_{nuc, f} | \langle \psi^{*}_{el, f} | {\boldsymbol{\mu}} | \psi_{el, i} \rangle | \psi_{nuc, i} \rangle \[4pt] &= \underbrace{ \langle \psi^{*}_{nuc, f} | \psi_{nuc, i} \rangle}_{\text{nuclear overlap}} \langle \psi^{*}_{el, f} | {\boldsymbol{\mu}} | \psi_{el, i} \rangle \end{align} \nonumber
If the nuclear overlap integral is zero for this transition, then the transition will not be observed, irrespective of the magnitude of the electronic factor.
$S_{00}$ Transition Evaluated within Harmonic Oscillator Model
The nuclear overlap for the zero-zero transition $S_{00}$ can be calculated quite simply using the definition of the Gaussian form of the harmonic oscillator wavefunctions.
The zero-point wavefunction in the ground electronic state is
$| \psi(R) \rangle = \big| \left(\dfrac{\alpha}{\pi} \right)^{1/4} e ^{-\alpha(R-R_e)^2/2} \big\rangle \nonumber$
The zero-point wavefunction in the excited electronic state is
$| \psi(R) \rangle = \big | \left(\dfrac{\alpha}{\pi} \right)^{1/4} e ^{-\alpha(R-Q_e)^2/2} \big\rangle \nonumber$
where
• $\alpha = \dfrac{\sqrt{mk}}{\hbar}$
• $R_e$ is the equilibrium bond length in the ground electronic state
• $Q_e$ is the equilibrium bond length in the excited electronic state
The nuclear overlap integral is
$S_{00}= \langle \psi^*_{nuc, f} | \psi_{nuc, i} \rangle = \sqrt{\dfrac{\alpha}{\pi}} \int_{\infty}^{\infty} e ^{-\alpha(R-R_e)^2/2} e ^{-\alpha(R-Q_e)^2/2} dR \label{FC1}$
The exponent in Equation $\ref{FC1}$ can be expanded as
$S_{00}= \sqrt{\dfrac{\alpha}{\pi}} \int_{\infty}^{\infty} e ^{-\alpha(2R^2-RR_e - 2RQ_e+ R^2_e + Q_e^2)/2} dR \label{FC2}$
and we use
$(R_e + Q_e)^2 = R_e^2 + Q_e^2 + 2R_eQ_e \nonumber$
and
$(R_e - Q_e)^2 = R_e^2 + Q_e^2 - 2R_eQ_e \nonumber$
to substitute and complete the square inside the integral. We can express
$R_e^2 + Q_e^2 = \dfrac{1}{2}[(R_e + Q_e)^2 + (R_e - Q_e)^2]. \nonumber$
Thus, the integral in Equation $\ref{FC2}$ is
$S_{00}= \sqrt{\dfrac{\alpha}{\pi}} e^{-\alpha(R_e -Q_e)^2/4} \int_{-\infty}^{\infty} e ^{-\alpha\{R- 1/2(R_e+Q_e)\}^2} dR \nonumber$
The integral is a Gaussian integral. You can show that if we let $z = \sqrt{\alpha}\{R-1/2(R_e + Q_e)\}$ then $dz = \sqrt{\alpha} dR$ and the integral becomes
$S_{00}= \sqrt{\dfrac{\alpha}{\pi}} e^{-\alpha(R_e -Q_e)^2/4} \dfrac{1}{\sqrt{\alpha}} \int_{-\infty}^{\infty} e^{z^2} dz \label{FC3}$
this integral has been solved already, from a table of integrals, Equation $\ref{FC3}$ becomes
$S_{00} = e^{-\alpha(R_e-Q_e)^2/4} \nonumber$
We would follow the same procedure to calculate that overlap of the zeroth level vibration in the ground to the first excited vibrational level of the excited state: $S_{01}$.
$S_{01}$ Transition Evaluated within Harmonic Oscillator Model
To calculate the overlap of zeroth ground state level ($v=0$) with the first excited state level ($v'=1$) we use the Hermite polynomial $H_1(x) =2x$ for describing the excited state wavefunction (see here for a review on harmonic oscillator wavefunctions). Here $x = \sqrt{\alpha}(R - Q_e)$.
$S_{01}= \langle \psi^{*}_{nuc, f} | \psi_{nuc, i} \rangle \label{FC01}$
with the zero-point wavefunction in the ground electronic state is
$| \psi(R) \rangle = \big| \left(\dfrac{\alpha}{\pi} \right)^{1/4} e ^{-\alpha(R-R_e)^2/2} \big\rangle \nonumber$
The first excited-state wavefunction in the excited electronic state is
$| \psi(R) \rangle = \big | \left(\dfrac{\alpha}{\pi} \right)^{1/4} \sqrt{\alpha}2 (R-Q_e) e^{-\alpha(R-Q_e)^2/2} \big\rangle \nonumber$
The overlap of zeroth ground state level with the first excited state level (Equation $\ref{FC01}$) is then
$S_{01} = \dfrac{1}{\sqrt{2}} \sqrt{\dfrac{\alpha}{\pi}} \int_{-\infty}^{\infty} e^{-\alpha(R-R_e)^2/2} \sqrt{\alpha}2 (R-Q_e) e^{-\alpha(R-Q_e)^2/2} \nonumber$
and
$S_{01} = \sqrt{\dfrac{2 \alpha^2}{\pi}} e^{-\alpha(R_e-Q_e)^2/4} \int_{-\infty}^{\infty} (R-Q_e) e^{-\alpha \{R- 1/2 (R_e+Q_e)^2\}} \nonumber$
The same substitutions can be made as above so that the integral can be written as (not shown and to be demonstrated in a homework exercises) and the final result is
$S_{01} = \sqrt{\dfrac{\alpha^2}{2}} (R_e-Q_e) e^{-\alpha(R_e-Q_e)^2/4} \nonumber$
We could continue and calculate that overlap of the zeroth level in the ground state with all the higher light vibrational levels: $S_{02}$, $S_{03}$, etc. Each term corresponds to a transition with a different energy since the vibrational levels have different energies. The absorption band then has the appearance of a progression (a Franck-Condon progression) of transitions between different levels each with its own probability.
Franck-Condon Progressions
To understand the significance of the above formula for the FC factor, let us examine a ground and excited state potential energy surface at $T = 0$ Kelvin. Shown below are two states separated by 8,000 cm-1 in energy. This is energy separation between the bottoms of their potential wells, but also between the respective zero-point energy levels. Let us assume that the wavenumber of the vibrational mode is 1,000 cm-1 and that the bond length is increased due to the fact that an electron is removed from a bonding orbital and placed in an anti-bonding orbital upon electronic excitation.
According to the above model for the Franck-Condon factor we would generate a "stick" spectrum (Figure 13.7.3 ) where each vibrational transition is infinitely narrow and transition can only occur when $E = h\nu$ exactly. For example, the potential energy surfaces were given for S = 1 and the transition probability at each level is given by the sticks (black) in the figure below.
The dotted Gaussians that surround each stick give a more realistic picture of what the absorption spectrum should look like. In this first place each energy level (stick) will be given some width by the fact that the state has a finite lifetime. Such broadening is called homogeneous broadening since it affects all of the molecules in the ensemble in a similar fashion. There is also broadening due to small differences in the environment of each molecule. This type of broadening is called inhomogeneous broadening. Regardless of origin the model above was created using a Gaussian broadening
The nuclear displacement between the ground and excited state determines the shape of the absorption spectrum. Let us examine both a smaller and a large excited state displacement. If $S = ½$ and the potential energy surfaces in this case are:
For this case the "stick" spectrum has the appearance in Figure 13.7.5
Note that the zero-zero or $S_{0,0}$ vibrational transition is much large in the case where the displacement is small.
As a general rule of thumb the $S$ constant gives the ratio of the intensity of the $v = 2$ transition to the $v = 1$ transition. In this case since $S = 0.5$, the $v=2$ transition is 0.5 the intensity of $v=1$ transition.
As an example of a larger displacement the disposition of the potential energy surfaces for S = 2 is shown below.
The larger displacement results in decreased overlap of the ground state level with the v = 0 level of the excited state. The maximum intensity will be achieved in higher vibrational levels as shown in the stick spectrum.
The absorption spectra plotted below all have the same integrated intensity, however their shapes are altered because of the differing extent of displacement of the excited state potential energy surface.
So the nature of the relative vibronic band intensities can tell us whether there is a displacement of the equilibrium nuclear coordinate that accompanied a transition. When will there be an increase in bond length (i.e., $Q_e > R_e$)? This occurs when an electron is promoted from a bonding molecular orbital to a non-bonding or anti-bonding molecular orbitals (i.e., when the bond order is less in the excited state than the ground state).
• Non-bonding molecular orbital $\rightarrow$ bonding molecular orbital
• Anti-bonding molecular orbital $\rightarrow$ bonding molecular orbital
• Anti-bonding molecular orbital $\rightarrow$ non-bonding molecular orbital
In short, when the bond order is lower in the excited state than in the ground state, then $Q_e > R_e$; an increase in bondlength will occur when this happens.
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Quick Review of Diatomic Rotation
As discussed previously, the Schrödinger equation for the angular motion of a rigid (i.e., having fixed bond length $R$) diatomic molecule is
$\dfrac{\hbar^2}{2 μ} \left[ \dfrac{1}{R^2 \sin θ} \dfrac{∂}{∂θ} \left(\sin θ \dfrac{∂}{∂θ} \right) + \dfrac{1}{R^2 \sin^2 θ} \dfrac{∂^2}{∂φ^2} \right] |ψ \rangle = E | ψ \rangle \nonumber$
or
$\dfrac{L^2}{2 μ R^2 } | ψ \rangle = E | ψ\rangle \nonumber$
The Hamiltonian in this problem contains only the kinetic energy of rotation; no potential energy is present because the molecule is undergoing unhindered "free rotation". The angles $θ$ and $φ$ describe the orientation of the diatomic molecule's axis relative to a laboratory-fixed coordinate system, and $μ$ is the reduced mass of the diatomic molecule
$μ=\dfrac{m_1m_2}{m_1+m_2} \nonumber$
The eigenvalues corresponding to each eigenfunction are straightforward to find because $H_{rot}$ is proportional to the $L^2$ operator whose eigenvalues have already been determined. The resultant rotational energies are given as:
$E_J= \dfrac{\hbar^2J(J+1)}{2μR^2} = B J(J+1) \label{Ediatomic}$
and are independent of $M$. $B$ is the rotational constant. Thus each energy level is labeled by $J$ and is $2J+1$-fold degenerate (because $M$ ranges from $-J$ to $J$). The rotational energy in Equation $\ref{Ediatomic}$ can be expressed in terms of the moment of inertia $I$
$I =\sum_i m_i R_i^2 \label{Idiatomic}$
where $m_i$ is the mass of the $i^{th}$ atom and $R$ is its distance from the center of mass of the molecule. This moment of inertia replaces $μR^2$ in the denominator of Equation $\ref{Ediatomic}$:
$E_J= \dfrac{\hbar^2J(J+1)}{2I} = B J(J+1) \label{Ediatomic2}$
Rotation of Polyatomic Molecules
In contrast to diatomic molecules (Equation \ref{Idiatomic}), the rotational motions of polyatomic molecules in three dimensions are characterized by multiple moments of inertia. typically reflected in an $3 \times 3$ inertia tensor. It is common in rigid body mechanics to express in these moments of inertia in lab-based Cartesian coordinates via a notation that explicitly identifies the $x$, $y$, and $z$ axes such as $I_{xx}$ and $I_{xy}$, for the components of the inertia tensor.
$I =\begin{bmatrix}I_{xx}&I_{xy}&I_{xz}\I_{yx}&I_{yy}&I_{yz}\I_{zx}&I_{zy}&I_{zz}\end{bmatrix} \label{inertiamatrix}$
The components of this tensor can be assembled into a matrix given by
$I_{xx}=\sum _{k=1}^{N}m_{k}(y_{k}^{2}+z_{k}^{2}) \nonumber$
$I_{yy}=\sum _{k=1}^{N}m_{k}(x_{k}^{2}+z_{k}^{2}) \nonumber$
$I_{zz}=\sum _{k=1}^{N}m_{k}(x_{k}^{2}+y_{k}^{2}) \nonumber$
$I_{yx}=I_{xy}=-\sum _{k=1}^{N}m_{k}x_{k}y_{k} \nonumber$
$I_{zx}=I_{xz}=-\sum _{k=1}^{N}m_{k}x_{k}z_{k} \nonumber$
$I_{zy}=I_{yz}=-\sum _{k=1}^{N}m_{k}y_{k}z_{k}. \nonumber$
The rotational motions of polyatomic molecules are characterized by moments of inertia that are defined in a molecule based coordinates with axes that are labeled $a$, $b$, and $c$. Measured in the body frame the inertia matrix (Equation $\ref{inertiamatrix}$) is a constant real symmetric matrix, which can be decomposed into a diagonal matrix, given by
$I =\left(\begin{array}{ccc}I_{a}&0&0\0&I_{b}&0\0&0&I_{c}\end{array}\right) \nonumber$
These labels are assigned so that $I_c$ is the largest principal moment of inertia with an order of the three moments set as
$I_a<I_b<I_c \nonumber$
The rotational kinetic energy operator for a rigid non-linear polyatomic molecule is then expressed as
$H_{rot} = \dfrac{J_a^2}{2I_a} + \dfrac{J_b^2}{2I_b} + \dfrac{J_c^2}{2I_c} \label{genKE}$
The components of the quantum mechanical angular momentum operators along the three principal axes are:
\begin{align} J_a &= -i\hbar \cos χ \left[\cot θ \dfrac{∂}{∂χ} - (\sin θ )^{-1} \dfrac{∂}{∂φ} \right] - -i\hbar \sin χ \dfrac{∂}{∂θ} \[4pt] J_b &= i\hbar \sin χ \left[\cot θ \dfrac{∂}{∂χ} - (\sin θ )^{-1} \dfrac{∂}{∂φ} \right] - -i\hbar \cos χ \dfrac{∂}{∂θ} \[4pt] J_c &= - \dfrac{ih ∂}{∂χ} \end{align} \nonumber
The angles $θ$, $φ$, and $χ$ are the Euler angles needed to specify the orientation of the rigid molecule relative to a laboratory-fixed coordinate system. The corresponding square of the total angular momentum operator $J^2$ can be obtained as
\begin{align} J^2 &= J_a^2 + J_b^ 2 + J_c^2 \[4pt] & = - \dfrac{∂^2}{∂θ^2} - \cot θ \dfrac{∂}{∂θ} - \left(\dfrac{1}{\sin θ} \right) \left( \dfrac{∂^2}{∂φ^2} + \dfrac{∂^2}{∂χ^2} - 2 \cos θ \dfrac{∂^2}{∂φ∂χ} \right) \end{align} \nonumber
and the component along the lab-fixed $Z$ axis is
$J_Z = - ih \dfrac{∂}{∂φ}. \nonumber$
Spherical Tops
When the three principal moment of inertia values are identical, the molecule is termed a spherical top. In this case, the total rotational energy Equation $\ref{genKE}$ can be expressed in terms of the total angular momentum operator $J^2$
$H_{rot} = \dfrac{J^2}{2I} \nonumber$
As a result, the eigenfunctions of $H_{rot}$ are those of $J^2$ (and $J_a$ as well as $J_Z$ both of which commute with $J_2$ and with one another; $J_Z$ is the component of $J$ along the lab-fixed Z-axis and commutes with $J_a$ because
$J_Z = - ih \dfrac{∂}{∂φ} \nonumber$
and
$J_a = - ih \dfrac{∂}{∂χ} \nonumber$
act on different angles. The energies associated with such eigenfunctions are
$E(J,K,M) = \dfrac{\hbar^2 J(J+1)}{2I^2} \nonumber$
for all K (i.e., J a quantum numbers) ranging from -J to J in unit steps and for all M (i.e., J Z quantum numbers) ranging from -J to J. Each energy level is therefore $(2J + 1)^2$ degenarate because there are $2J + 1$ possible K values and $2J + 1$ possible M values for each J. The eigenfunctions of $J^2$, $J_Z$ and $J_a$, $|J,M,K>$ are given in terms of the set of rotation matrices $D_{J,M,K}$ :
$|J,M,K \rangle = \sqrt{ \dfrac{2J + 1}{8 π^2}} D^* _{J,M,K} ( θ , φ , χ ) \nonumber$
which obey
$J^2 |J,M,K \rangle = \hbar^2 J(J+1) | J,M,K \rangle \nonumber$
$J_a |J,M,K \rangle = \hbar K | J,M,K \rangle \nonumber$
$J_Z |J,M,K \rangle = \hbar M | J,M,K \rangle \nonumber$
Symmetric Tops
Symmetrical tops are molecules with two rotational axes that have the same inertia and one unique rotational axis with a different inertia. Symmetrical tops can be divided into two categories based on the relationship between the inertia of the unique axis and the inertia of the two axes with equivalent inertia. If the unique rotational axis has a greater inertia than the degenerate axes the molecule is called an oblate symmetrical top (Figure 13.8.1 ). If the unique rotational axis has a lower inertia than the degenerate axes the molecule is called a prolate symmetrical top. For simplification think of these two categories as either frisbees for oblate tops or footballs for prolate tops.
Again, the rotational kinetic energy, which is the full rotational Hamiltonian, can be written in terms of the total rotational angular momentum operator $J^2$ and the component of angular momentum along the axis with the unique principal moment of inertia.
For prolate tops, Equation $\ref{genKE}$ becomes
$H_{rot} = \dfrac{J^2}{2I} + J_a^2 \left( \dfrac{1}{2I_a} - \dfrac{1}{2I} \right) \nonumber$
For oblate tops, Equation $\ref{genKE}$ becomes
$H_{rot} = \dfrac{J^2}{2I} + J_c^2 \left( \dfrac{1}{2I_c} - \dfrac{1}{2I} \right) \nonumber$
As a result, the eigenfunctions of $H_{rot}$ are those of $J^2$ and $J_a$ or $J_c$ (and of $J_Z$), and the corresponding energy levels.
The energies for prolate tops are
$E(J,K,M) = \dfrac{h^2 J(J+1)}{2I^2} + h^2 K^2 \left( \dfrac{1}{2I_a} - \dfrac{1}{2I} \right) \nonumber$
and the energies for oblate tops are
$E(J,K,M) = \dfrac{h^2 J(J+1)}{2I 2} + h^2 K^2 \left( \dfrac{1}{2I_c} - \dfrac{1}{2I} \right) \nonumber$
again for K and M (i.e., $J_a$ or $J_c$ and $J_Z$ quantum numbers, respectively) ranging from $-J$ to $J$ in unit steps. Since the energy now depends on K, these levels are only $2J + 1$ degenerate due to the $2J + 1$ different $M$ values that arise for each $J$ value. The eigenfunctions $|J, M,K>$ are the same rotation matrix functions as arise for the spherical-top case.
Asymmetric Tops
The rotational eigenfunctions and energy levels of a molecule for which all three principal moments of inertia are distinct (a asymmetric top) can not easily be expressed in terms of the angular momentum eigenstates and the $J$, $M$, and $K$ quantum numbers. However, given the three principal moments of inertia $I_a$, $I_b$, and $I_c$, a matrix representation of each of the three contributions to the general rotational Hamiltonian in Equation $\ref{genKE}$ can be formed within a basis set of the $\{|J, M, K \rangle\}$ rotation matrix functions. This matrix will not be diagonal because the $|J, M, K \rangle$ functions are not eigenfunctions of the asymmetric top $H_{rot}$. However, the matrix can be formed in this basis and subsequently brought to diagonal form by finding its eigenvectors {C n, J,M,K } and its eigenvalues $\{E_n\}$. The vector coefficients express the asymmetric top eigenstates as
$\psi_n ( θ , φ , χ ) = \sum_{J, M, K} C_{n, J,M,K} |J, M, K \rangle \nonumber$
Because the total angular momentum $J^2$ still commutes with $H_{rot}$, each such eigenstate will contain only one J-value, and hence $Ψ_n$ can also be labeled by a $J$ quantum number:
$\psi _{n,J} ( θ , φ , χ ) = \sum_{M, K} C_{n, J,M,K} |J, M, K \rangle \nonumber$
To form the only non-zero matrix elements of $H_{rot}$ within the $|J, M, K\rangle$ basis, one can use the following properties of the rotation-matrix functions:
$\langle j, \rangle = \langle j, \rangle = 1/2 <j, \rangle = h 2 [ J(J+1) - K 2 ], \nonumber$
$\langle j, \rangle = h^2 K^2 \nonumber$
$\langle j \rangle = - \langle j \rangle = h^2 [J(J+1) - K(K ± 1)] 1/2 [J(J+1) -(K ± 1)(K ± 2)] 1/2 \langle j \rangle = 0 \nonumber$
Each of the elements of $J_c^2$, $J_a^2$, and $J_b^2$ must, of course, be multiplied, respectively, by $1/2I_c$, $1/2I_a$, and $1/2I_b$ and summed together to form the matrix representation of $H_{rot}$. The diagonalization of this matrix then provides the asymmetric top energies and wavefunctions.
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Normal modes are used to describe the different vibrational motions in molecules. Each mode can be characterized by a different type of motion and each mode has a certain symmetry associated with it. Group theory is a useful tool in order to determine what symmetries the normal modes contain and predict if these modes are IR and/or Raman active. Consequently, IR and Raman spectroscopy is often used for vibrational spectra.
Degrees of Freedom
In general, a normal mode is an independent motion of atoms in a molecule that occurs without causing movement to any of the other modes. Normal modes, as implied by their name, are orthogonal to each other. In order to discuss the quantum-mechanical equations that govern molecular vibrations it is convenient to convert Cartesian coordinates into so called normal coordinates. Vibrations in polyatomic molecules are represented by these normal coordinates.
A molecule can have three types of degrees of freedom and a total of 3N degrees of freedom, where N equals the number of atoms in the molecule. These degrees of freedom can be broken down into three categories.
• Translational: These are the simplest of the degrees of freedom. These entail the movement of the entire molecule’s center of mass. This movement can be completely described by three orthogonal vectors and thus contains 3 degrees of freedom.
• Rotational: These are rotations around the center of mass of the molecule and like the translational movement they can be completely described by three orthogonal vectors. This again means that this category contains only 3 degrees of freedom. However, in the case of a linear molecule only two degrees of freedom are present due to the rotation along the bonds in the molecule having a negligible inertia.
• Vibrational: These are any other types of movement not assigned to rotational or translational movement and thus there are 3N – 6 degrees of vibrational freedom for a nonlinear molecule and 3N – 5 for a linear molecule. These vibrations include bending, stretching, wagging and many other aptly named internal movements of a molecule. These various vibrations arise due to the numerous combinations of different stretches, contractions, and bends that can occur between the bonds of atoms in the molecule.
Total Degrees of Freedom Translational degrees of freedom Rotational degrees of freedom Vibrational degrees of freedom
Table 13.9.1 : Overview of degrees of freedom
Nonlinear Molecules 3N 3 3 3N -6
Linear Molecules 3N 3 2 3N - 5
Each of these degrees of freedom is able to store energy. However, In the case of rotational and vibrational degrees of freedom, energy can only be stored in discrete amounts. This is due to the quantized break down of energy levels in a molecule described by quantum mechanics. In the case of rotations the energy stored is dependent on the rotational inertia of the gas along with the corresponding quantum number describing the energy level.
Example 13.9.1 : Ethane vs. Carbon Dioxide
Ethane, $C_2H_6$ has eight atoms ($N=8$) and is a nonlinear molecule so of the $3N=24$ degrees of freedom, three are translational and three are rotational. The remaining 18 degrees of freedom are internal (vibrational). This is consistent with:
$3N -6 =3(8)-6=18 \nonumber$
Carbon Dioxide, $CO_2$ has three atoms ($N=3$ and is a linear molecule so of the $3N=9$ degrees of freedom, three are translational and two are rotational. The remaining 4 degrees of freedom are vibrational. This is consistent with:
$3N - 5 = 3(3)-5 = 4 \nonumber$
The normal modes of vibration are: asymmetric, symmetric, wagging, twisting, scissoring, and rocking for polyatomic molecules.
Symmetricical Stretching Asymmetrical Stretching Wagging
Twisting Scissoring Rocking
Figure 13.9.1 : Six types of Vibrational Modes. Taken from publisher en.Wikipedia.org/wiki/Infrared_spectroscopy with permission from copyright holder.
Normal Modes
If there is no external field present, the energy of a molecule does not depend on its orientation in space (its translational degrees of freedom) nor its center of mass (its rotational degrees of freedom). The potential energy of the molecule is therefore made up of its vibrational degrees of freedom only of $3N-6$ (or $3N-5$ for linear molecules). The difference in potential energy is given by:
\begin{align} \Delta V &= V(q_1,q_2,q_3,...,q_n) - V(0,0,0,...,0) \label{1} \[4pt] &= \dfrac{1}{2} \sum_{i=1}^{N_{vib}} \sum_{j=1}^{N_{vib}} \left(\dfrac{\partial^2 V}{\partial q_i\partial q_j} \right) q_iq_j \label{2} \[4pt] &= \dfrac{1}{2}\sum_{i=1}^{N_{vib}} \sum_{j=1}^{N_{vib}} f_{ij} q_iq_j \label{3} \end{align}
where $q$ represents the equilibrium displacement and $N_{vib}$ the number of vibrational degrees of freedom.
For simplicity, the anharmonic terms are neglected in this equation (i.e., higher order terms are ignore). A theorem of classical mechanics states that the cross terms can be eliminated from the above equation (the details of the theorem are very complex and will not be discussed in detail). By using matrix algebra a new set of coordinates {Qj} can be found such that
$\Delta{V} = \dfrac{1}{2} \sum_{j=1}^{N_{vib}}{F_jQ_j^2} \label{4}$
Note that there are no cross terms in this new expression. These new coordinates are called normal coordinates or normal modes. With these new normal coordinates in hand, the Hamiltonian operator for vibrations can be written as follows:
$\hat{H}_{vib} = -\sum_{j=1}^{N_{vib}} \dfrac{\hbar^2}{2\mu_i} \dfrac{d^2}{dQ_j^2} + \dfrac{1}{2} \sum_{j=1}^{N_{vib}}F_jQ_j^2 \label{5}$
The total wavefunction is a product of the individual wavefunctions and the energy is the sum of independent energies. This leads to:
$\hat{H}_{vib} = \sum_{j=1}^{N_{vib}} \hat{H}_{vib,j} = \sum_{j=1}^{N_{vib}} \left( \dfrac{-\hbar^2}{2 \mu_j}\dfrac{d^2}{dQ_i^2} + \dfrac{1}{2}\sum_{j=1}^{N_{vib}} F_jQ_j^2 \right) \label{6}$
and the wavefunction is then
\begin{align*} \psi_{vib} &= Q_1,Q_2, Q_3 ..., Q_{vib} \[4pt] &= \psi_{vib,1}(Q_1) \psi_{vib,2}(Q_2) \psi_{vib,3}(Q_3) , ..., \psi_{vib,N_{vib}}(Q_{N_{vib}}) \end{align*} \nonumber
and the total vibrational energy of the molecule is
$E_{vib} = \sum_{j=1}^{N_{vin}} h\nu_j \left (v_j + \dfrac{1}{2}\right) \label{8}$
where $v_j= 0,1,2,3...$.
The consequence of the result stated in the above equations is that each vibrational mode can be treated as a harmonic oscillator approximation. There are $N_{vib}$ harmonic oscillators corresponding to the total number of vibrational modes present in the molecule.
Pictorial description of normal coordinates using CO
The normal coordinate $q$ is used to follow the path of a normal mode of vibration. As shown in Figure 13.9.2 the displacement of the $\ce{C}$ atom, denoted by $Δr_o(\ce{C})$, and the displacement of the $\ce{O}$ atom, denoted by $Δr_o(\ce{O})$, occur at the same frequency. The displacement of atoms is measured from the equilibrium distance in ground vibrational state, $r_o$.
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There exists an important fact about normal coordinates. Each of these coordinates belongs to an irreducible representation of the point the molecule under investigation. Vibrational wavefunctions associated with vibrational energy levels share this property as well. The normal coordinates and the vibration wavefunction can be categorized further according to the point group they belong to. From the character table predictions can be made for which symmetries can exist. The irreducible representation offers insight into the IR and/or Raman activity of the molecule in question.
Symmetry of normal modes
It is important to realize that every normal mode has a certain type of symmetry associated with it. Identifying the point group of the molecule is therefore an important step. With this in mind it is not surprising that every normal mode forms a basis set for an irreducible representation of the point group the molecule belongs to. For a molecule such as water, having a structure of $\ce{XY2}$, three normal coordinates can be determined. The two stretching modes are equivalent in symmetry and energy. The figure below shows the three normal modes for the water molecule:
By convention, with nonlinear molecules, the symmetric stretch is denoted v1 whereas the asymmetric stretch is denoted v2. Bending motions are v3. With linear molecules, the bending motion is $\nu_2$ whereas asymmetric stretch is v3.
The water molecule has C2v symmetry and its symmetry elements are E, C2, σ(xz) and σ(yz). To determine the symmetries of the three vibrations and how they each transform, symmetry operations will be performed. As an example, performing C2 operations using the two normal mode v2 and v3 gives the following transformation:
Once all the symmetry operations have been performed in a systematic manner for each modes the symmetry can be assigned to the normal mode using the character table for C2v:
C2v E C2 σ (xz) σ (yz)
Table 13.10.2 : Character table for the C2v point group
ν1 1 1 1 1 = a1
ν2 1 1 1 1 = a1
ν3 1 -1 -1 1 = b2
Water has three normal modes that can be grouped together as the reducible representation
$Γ_{vib}= 2a_1 + b_2. \nonumber$
Determination of normal modes becomes quite complex as the number of atoms in the molecule increases. Nowadays, computer programs that simulate molecular vibrations can be used to perform these calculations. The example of [PtCl4]2- shows the increasing complexity (Figure 13.10.2 ). The molecule has five atoms and therefore 15 degrees of freedom, 9 of these are vibrational degrees of freedom. The nine normal modes are exemplified below along with the irreducible representation the normal mode belongs to (D4h point group).
13.11: Time-Dependent Perturbation Theory
In time-independent perturbation theory the perturbation Hamiltonian is static (i.e., possesses no time dependence). Time-independent perturbation theory was presented by Erwin Schrödinger in a 1926 paper,shortly after he produced his theories in wave mechanics. Time-dependent perturbation theory, developed by Paul Dirac, studies the effect of a time-dependent perturbation V(t) applied to a time-independent Hamiltonian $H_0$. Since the perturbed Hamiltonian is time-dependent, so are its energy levels and eigenstates. Thus, the goals of time-dependent perturbation theory are slightly different from time-independent perturbation theory, where one may be interested in the following quantities:
• The time-dependent expectation value of some observable A, for a given initial state.
• The time-dependent amplitudes of those quantum states that are energy eigenkets (eigenvectors) in the unperturbed system.
The first quantity is important because it gives rise to the classical result of a measurement performed on a macroscopic number of copies of the perturbed system. The second quantity looks at the time-dependent probability of occupation for each eigenstate. This is particularly useful in laser physics, where one is interested in the populations of different atomic states in a gas when a time-dependent electric field is applied. We will briefly examine the method behind Dirac's formulation of time-dependent perturbation theory. Choose an energy basis ${\displaystyle {|n\rangle }}$$| n \rangle$ for the unperturbed system. (We drop the (0) superscripts for the eigenstates, because it is not useful to speak of energy levels and eigenstates for the perturbed system.)
If the unperturbed system is in eigenstate $|j \rangle$ at time $t = 0$, its state at subsequent times varies only by a phase (this is the Schrödinger picture, where state vectors evolve in time and operators are constant)
$|j(t)\rangle =e^{-iE_{j}t/\hbar }|j\rangle \nonumber$
Now, introduce a time-dependent perturbing Hamiltonian $H_1(t)$. The Hamiltonian of the perturbed system is
$H=H_{0}+H_1(t) \nonumber$
Let $|\psi (t)\rangle$ denote the quantum state of the perturbed system at time $t$ and obeys the time-dependent Schrödinger equation,
$H|\psi (t)\rangle =i\hbar {\dfrac {\partial }{\partial t}}|\psi (t)\rangle \nonumber$
The quantum state at each instant can be expressed as a linear combination of the complete eigenbasis of $| n \rangle$:
$|\psi (t)\rangle =\sum _{n}c_{n}(t)e^{-iE_{n}t/\hbar }|n\rangle \nonumber$
where the $c_n(t)$ coefficients are to be determined complex functions of t which we will refer to as amplitudes
We have explicitly extracted the exponential phase factors $\exp(-iE_{n}t/\hbar)$ on the right hand side. This is only a matter of convention, and may be done without loss of generality. The reason we go to this trouble is that when the system starts in the state ${\displaystyle |j\rangle }$$|j\rangle$ and no perturbation is present, the amplitudes have the convenient property that, for all t, $c_j(t) = 1$ and $c_n(t) = 0$ if $n \neq j$.
The square of the absolute amplitude $c_n(t)$ is the probability that the system is in state $n$ at time $t$, since
$|\psi (t)\rangle =\sum _{n}c_{n}(t)e^{-iE_{n}t/\hbar }|n\rangle \nonumber$
Plugging into the Schrödinger equation and using the fact that $\partial/ \partial t$ acts by a chain rule, one obtains
$\sum _{n}\left(i\hbar {\dfrac {\partial c_{n}}{\partial t}}-c_{n}(t)V(t)\right)e^{-iE_{n}t/\hbar }|n\rangle =0~. \nonumber$
By resolving the identity in front of V, this can be reduced to a set of partial differential equations for the amplitudes,
${\dfrac {\partial c_{n}}{\partial t}}={\dfrac {-i}{\hbar }}\sum _{k}\langle n|H_1(t)|k\rangle \,c_{k}(t)\,e^{-i(E_{k}-E_{n})t/\hbar }~. \nonumber$
The matrix elements of $H_1$ play a similar role as in time-independent perturbation theory, being proportional to the rate at which amplitudes are shifted between states. Note, however, that the direction of the shift is modified by the exponential phase factor. Over times much longer than the energy difference $E_k − E_n$, the phase winds around 0 several times. If the time-dependence of $H_1$ is sufficiently slow, this may cause the state amplitudes to oscillate (e.g., such oscillations are useful for managing radiative transitions in a laser).
Two-Level System
Consider the two level system (i.e. $n=1,2$)
$| \psi \rangle = \sum_ {n=1,2} c_n(t) | n \rangle_o \nonumber$
Solution of time-dependent perturbation for two level system:
$i \hbar \dfrac{\partial c_1(t)}{\partial t} = c_1(t) H_{11}(t) + c_2 e^{-i \omega_o t} H_{12} (t) \nonumber$
$i \hbar \dfrac{\partial c_2(t)}{\partial t} = c_2(t) H_{22}(t) + c_1 e^{+i \omega_o t} H_{21} (t) \nonumber$
where the matrix elements of the permutation (in terms of the eigenstates of $H(0)$) are
$\langle m | H_1(t) | n \rangle = H_{mn}(t) \nonumber$
Assume initial state is $n=1$, and $H_{11}=H_{22}=0$
$| \psi (t=0) \rangle = |1 \rangle \nonumber$
Probability of particle at $n=2$ at time $t$ after the perturbation is turned on (i.e., incident light):
$c_2(t) = \dfrac{-i}{\hbar} \int_o^t e^{i \omega_o t} dt' H_{21}(t') \label{EQ1}$
where
$H_1(t) = \cos (\omega t) V(r) \nonumber$
$V(r)$ is an amplitude of polarization vector, which we can ignore for now.
If we assume incident frequency of incident light $\omega$ is comparable to the natural frequency of oscillation from $\omega _o$
$\omega \approx \omega_o \nonumber$
then Equation $\ref{EQ1}$ can be simplified to
$c_2(t) = - \dfrac{2i}{\hbar} \dfrac{\sin (\omega-\omega_o)t /2}{ (\omega-\omega_o )t} e^{i (\omega_0-\omega)t/2} H_{21} \nonumber$
Transition Probability
Assume initial state is $n=1$, and probability of transition from $n=1$ state to $n=2$ state is:
$P_{12}(t) = | c_2(t) |^2 =\dfrac{4}{\hbar^2} | \dfrac{\sin (\omega-\omega_o)t /2}{ (\omega-\omega_o )t} |^2 | H_{21} |^2 \nonumber$
What does this mean? Strangely, it means that the probability of makinga transition is actually oscillating sinusoidally (squared)! If you want to cause a transition, should turn off perturbation after time $\pi / |\omega- \omega_o|$or some odd multiple, when the system is in upper state with maximum probability.
$P_{12}(t)$ is peaked at $\omega- \omega_o = 0$. The height of $|H_{12}t/2 \hbar |^2$ and width of $4\pi/t$ gets higher and narrower as time goes on. Recall this is perturbative treatment, however, and $P_{12}(t)$ cannot get bigger than 1, so perturbation theory breaks down eventually.
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A selection rule describes how the probability of transitioning from one level to another cannot be zero. It has two sub-pieces: a gross selection rule and a specific selection rule. A gross selection rule illustrates characteristic requirements for atoms or molecules to display a spectrum of a given kind, such as an IR spectroscopy or a microwave spectroscopy. Once the atom or molecules follow the gross selection rule, the specific selection rule must be applied to the atom or molecules to determine whether a certain transition in quantum number may happen or not.
Selection rules specify the possible transitions among quantum levels due to absorption or emission of electromagnetic radiation. Incident electromagnetic radiation presents an oscillating electric field $E_0\cos(\omega t)$ that interacts with a transition dipole. The dipole operator is $\mu = e \cdot r$ where $r$ is a vector pointing in a direction of space.
A dipole moment of a given state is
$\mu_z=\int\Psi_1 \,^{*}\mu_z\Psi_1\,d\tau \nonumber$
A transition dipole moment is a transient dipolar polarization created by an interaction of electromagnetic radiation with a molecule
$(\mu_z)_{12}=\int\Psi_1 \,^{*}\mu_z\Psi_2\,d\tau \nonumber$
In an experiment we present an electric field along the z axis (in the laboratory frame) and we may consider specifically the interaction between the transition dipole along the x, y, or z axis of the molecule with this radiation. If $\mu_z$ is zero then a transition is forbidden. The selection rule is a statement of when $\mu_z$ is non-zero.
The selection rule is a statement of when $\mu_z$ is non-zero.
Rotational transitions
We can use the definition of the transition moment and the spherical harmonics to derive selection rules for a rigid rotator. Once again we assume that radiation is along the z axis.
$(\mu_z)_{J,M,{J}',{M}'}=\int_{0}^{2\pi } \int_{0}^{\pi }Y_{J'}^{M'}(\theta,\phi )\mu_zY_{J}^{M}(\theta,\phi)\sin\theta\,d\phi,d\theta \nonumber$
Notice that m must be non-zero in order for the transition moment to be non-zero. This proves that a molecule must have a permanent dipole moment in order to have a rotational spectrum. The spherical harmonics can be written as
$Y_{J}^{M}(\theta,\phi)=N_{\,JM}P_{J}^{|M|}(\cos\theta)e^{iM\phi} \nonumber$
where $N_{JM}$ is a normalization constant. Using the standard substitution of $x = \cos q$ we can express the rotational transition moment as
$(\mu_z)_{J,M,{J}',{M}'}=\mu\,N_{\,JM}N_{\,J'M'}\int_{0}^{2 \pi }e^{I(M-M')\phi}\,d\phi\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx \nonumber$
The integral over f is zero unless M = M' so $\Delta M =$ 0 is part of the rigid rotator selection rule. Integration over $\phi$ for $M = M'$ gives $2\pi$ so we have
$(\mu_z)_{J,M,{J}',{M}'}=2\pi \mu\,N_{\,JM}N_{\,J'M'}\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx \nonumber$
We can evaluate this integral using the identity
$(2J+1)x\,P_{J}^{|M]}(x)=(J-|M|+1)P_{J+1}^{|M|}(x)+(J-|M|)P_{J-1}^{|M|}(x) \nonumber$
Substituting into the integral one obtains an integral which will vanish unless $J' = J + 1$ or $J' = J - 1$.
$\int_{-1}^{1}P_{J'}^{|M'|}(x)\Biggr(\frac{(J-|M|+1)}{(2J+1)}P_{J+1}^{|M|}(x)+\frac{(J-|M|)}{(2J+1)}P_{J-1}^{|M|}(x)\Biggr)dx \nonumber$
This leads to the selection rule $\Delta J = \pm 1$ for absorptive rotational transitions. Keep in mind the physical interpretation of the quantum numbers $J$ and $M$ as the total angular momentum and z-component of angular momentum, respectively. As stated above in the section on electronic transitions, these selection rules also apply to the orbital angular momentum ($\Delta{l} = \pm 1$, $\Delta{m} = 0$).
13.13: The Harmonic Oscillator Selection Rule
Selection rules are a very important concept in spectroscopy and physical chemistry, as they directly complement the concept that systems on the atomic level are quantized. Quantization tells us how electrons, neutrons, and protons are set up, and selection rules tell us how, energetically speaking, they may move about. For example, the principal quantum number, n, determines the energy state of particles in the Particle in a Box model. With respect to real systems, the quantum number n describes electronic transitions, such as the excitation of an electron. There are other transitions, such as the rotational transition, which applies to the rigid rotator model and occurs where there are spherical harmonics. Vibrational transitions occur where there is a system that can be visualized as two masses connected by a spring, such as a diatomic molecule.
Selection rules can be very useful in spectroscopy for obtaining information about an unknown substance; any given substance has properties and behaviors that operate within the selection rules and determine the wavelengths of electromagnetic radiation (light) that cause transitions. By tracking these properties, chemists can use techniques like IR spectroscopy to determine the chemical structure of a molecule. For example, a selection rule might govern how a quantum number changes during a transition. That quantum number appears in equations that determine the energy transition based on the mass and other structural properties of a molecule. Thus, if one knows the selection rules and the energy transition, he or she can determine other properties about the molecule, such as its mass or its bond strength.
Example: Evaluating vibrational transitions using a transition moment integral.
Vibrational systems are described by the harmonic oscillator wavefunctions, which look like
$\Psi_v(x)=N_vH_v(\alpha^{1/2}x)e^{-\alpha{x^2}/2} \nonumber$
A transition dipole moment can be written to describe a transition along the z-axis.
$\mu_z = \int\limits_{-\infty}^{\infty}N_vN_{v+1}H_{v+1}(\alpha^{1/2}x)e^{-\alpha{x^2}/2}H_{v+1}\mu_{z}(\alpha^{1/2}x)e^{-\alpha{x^2}/2}dx \nonumber$
$\mu$ is the dipole moment, so we pull it out of the integral.
$\mu_z = \mu_{z}\int\limits_{-\infty}^{\infty}N_vN_{v+1}H_{v+1}(\alpha^{1/2}x)e^{-\alpha{x^2}/2}H_{v+1}(\alpha^{1/2}x)e^{-\alpha{x^2}/2}dx \nonumber$
Using the idea that the transition moment $\mu_{z}$ changes on x, we express it as a derivative.
$\mu_z = \frac{d\mu}{dx}\int\limits_{-\infty}^{\infty}N_vN_{v+1}H_{v+1}(\alpha^{1/2}x)e^{-\alpha{x^2}/2}H_{v+1}(\alpha^{1/2}x)e^{-\alpha{x^2}/2}dx \nonumber$
To evaluate the integral, we take advantage of a special Hermite relationship: $\sqrt{\alpha}xH_v(\sqrt{\alpha}x)=vH_{v-1}(\sqrt{\alpha}x)+\frac{1}{2}H_{v+1}(\sqrt{\alpha}x)$
$\mu_z = \frac{N_vN_{v+1}}{\sqrt{\alpha}}(\frac{d\mu}{dx})\int\limits_{-\infty}^{\infty}H_{v+1}(\sqrt{\alpha}x)e^(-\alpha{x^2}/2)[vH_{v-1}(\sqrt{\alpha}x)+\frac{1}{2}H_{v+1}(\sqrt{\alpha}x)]dx \nonumber$
By inspection, it is clear that the integral only allows transitions of $\Delta{v}=\pm1$, because for any other value, the integral will be zero.
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Determining if a Normal Modes is IR or Raman Active
A transition from $v \rightarrow v'$ is IR active if the transition moment integral contains the totally symmetric irreducible representation of the point group the molecule belongs to. The transition moment integral is derived from the one-dimensional harmonic oscillator. Using the definition of electric dipole moment $\mu$, the integral is:
$M(v \rightarrow v') = \langle \text{final wavefunction} | \vec{\mu} | \text{initial wavefunction } \rangle \nonumber$
or in terms of vibrational wavefunctions for a specific normal mode $| \phi (v) \rangle$
$M(v \rightarrow v') = \langle \phi(v' \neq 0) | \vec{\mu} | \phi (v=0) \rangle \label{M1}$
assuming the transition from the $v=0$ wavefunction to the $v' \neq 0$ wavefunction.
Now, consider the case that $\vec{μ}$, is a constant and therefore independent of the vibration (i.e., the electric dipole moment does not change during the vibration). This it could be taken outside the integral in Equation $\ref{M1}$ becomes
$M(v \rightarrow v') = \vec{\mu} \langle \phi(v' \neq 0) | \phi (v=0) \rangle \label{M2}$
Since $|\phi(v=0) \rangle$ and $|\phi(v =\neq0) \rangle$ are mutually orthogonal to each other, the integral in Equation $\ref{M1}$ will equal zero and the transition will not be allowed (i.e., it is forbidden). For the $M$ to be nonzero, $\vec{μ}$ must change during a vibration. This selection rule explains why homonuclear diatomic molecules do not produce an IR spectrum. There is no change in dipole moment resulting in a transition moment integral of zero and a transition that is forbidden.
For a transition to be Raman active, the same rules apply. The transition moment integral must contain the totally symmetric irreducible representation of the point group. The integral contains the polarizability tensor $\alpha$ (usually represented by a square matrix):
$M(v \rightarrow v') = \langle \phi(v' \neq 0) | \alpha | \phi (v=0) \rangle \label{M3}$
Following a similar argument as above, $\alpha$ must be nonzero for the transition to be allowed and exhibits Raman scattering.
Character Table
For a molecule to be IR active the dipole moment has to change during the vibration. For a molecule to be Raman active the polarizability of the molecule has to change during the vibration. The reducible representation Γvib can also be found by determining the reducible representation of the 3N degrees of freedom of H2O, Γtot. By applying Group Theory it is straightforward to find Γx,y,z as well as UMA (number of unmoved atoms). Again, using water as an example with C2v symmetry where 3N = 9, Γtot can be determined:
C2v E C2 σ (xz) σ (yz)
Τx,y,z 3 -1 1 1
UMA 3 1 1 3
Γtot 9 -1 1 3 =3a1 + a2 + 2b1 + 3b2
Note that Γtot contains nine degrees of freedom consistent with 3N = 9.
Γtot contains Γtranslational, Γrotational as well as Γvibrational. Γtrans can be obtained by finding the irreducible representations corresponding to x,y and z in the right side of the character table, Γrot by finding the ones corresponding to Rx, Ry and Rz. Γvib can be obtained by Γtot - Γtrans - Γrot.
$Γ_{vib} (H_2O) = (3a_1 + a_2 + 2b_1+ 3b_2) - (a_1 + b_1 + b_2) - (a_2 + b_1 + b_2) = 2a_1 + b_2 \nonumber$
In order to determine which modes are IR active, a simple check of the irreducible representation that corresponds to x,y and z and a cross check with the reducible representation Γvib is necessary. If they contain the same irreducible representation, the mode is IR active.
For H2O, z transforms as a1, x as b1 and y as b2. The modes a1 and b2 are IR active since Γvib contains 2a1 + b2.
In order to determine which modes are Raman active, the irreducible representation that corresponds to z2, x2-y2, xy, xz and yz is used and again cross checked with Γvib. For H2O, z2 and x2-y2 transform as a1, xy as a2, xz as b1 and yz as b2.The modes a1 and b2 are also Raman active since Γvib contains both these modes.
The IR spectrum of H2O does indeed have three bands as predicted by Group Theory. The two symmetric stretches v1 and v2 occur at 3756 and 3657 cm-1 whereas the bending v3 motion occurs at 1595 cm-1.
In order to determine which normal modes are stretching vibrations and which one are bending vibrations, a stretching analysis can be performed. Then the stretching vibrations can be deducted from the total vibrations in order to obtain the bending vibrations. A double-headed arrow is drawn between the atom as depicted below:
Then a determination of how the arrows transform under each symmetry operation in C2v symmetry will yield the following results:
C2v E C2 σ (xz) σ (yz)
Γstretch 2 0 0 2 = a1 + b2
$Γ_{bend} = Γ_{vib} - Γ_{stretch} = 2a_1 + b_2 -a_1 - b_2 = a_1 \nonumber$
H2O has two stretching vibrations as well as one bending vibration. This concept can be expanded to complex molecules such as PtCl4-. Four double headed arrows can be drawn between the atoms of the molecule and determine how these transform in D4h symmetry. Once the irreducible representation for Γstretch has been worked out, Γbend can be determined by Γbend = Γvib - Γstretch.
Most molecules are in their zero point energy at room temperature. Therefore, most transitions do originate from the v=0 state. Some molecules do have a significant population of the v=1 state at room temperature and transitions from this thermally excited state are called hot bands.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/13%3A_Molecular_Spectroscopy/13.14%3A_Group_Theory_Determines_Infrared_Activity.txt
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These are homework exercises to accompany Chapter 13 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Q13.1
1. Calculate the energy difference for radiation wavenumber of, $\nu=1.00cm^{-1}$ including the type of molecular process that absorbs of this radiation corresponding to.
2. Given the wavelength of 3.4 x 10-4, which molecular process does it fall into?
3. What are other advantages of knowing the spectroscopy of molecules?
4. Do you think the spectroscopy and molecular processes of the molecules have relationship to temperature? Explain.
5. What is other form that you can express in above equation?
Q13.20
The vibrational term of a diatomic molecule is
$G(v) = (v + \dfrac{1}{2}) \nu_e - (v + \dfrac{1}{2})^2 \chi_e \nu_e$
Show that the spacing between adjacent levels is given by
$\Delta G = G(v + 1) - G(v) = \nu_e - 2 \chi_e \nu_e (v+1)$
Show that the maximum vibrational quantum number, $v_{max}$ is given by
$v_{max} = \dfrac{1}{2\chi_e} - 1$
Use this result to show that the dissociation energy of the diatomic molecule can be written as
$D_e = \dfrac{\nu_e - \nu_e \chi_e^2}{4\chi_e} \approx \dfrac{\nu_e}{4\chi_e}$
Explain how the constants $\nu_e$ and $\chi_e$ can be evaluated from a plot of $\Delta G$ versus $v+1$, called a Birge-Sponer plot. After finding these constants, determine the dissociation energy of the molecule. Use the following experimental data of $H_2$ to determine the dissociation energy of the molecule.
v G(v) / cm-1
0 4161.12
1 8087.11
2 11782.35
3 15250.36
4 18497.92
5 21505.65
6 24287.83
7 26830.97
8 29123.93
9 31150.19
10 32886.65
11 34301.83
12 35351.01
13 35972.97
Explain why your plot is not linear for high values of $v$. How does this dissocation energy differ with the experimental value of $38269.48 cm^{-1}$.
S13.20
$\Delta G = G(v + 1) - G(v) = \left(v + \dfrac{3}{2}\right) \nu_e - \left(v + \dfrac{3}{2}\right)^2 \chi_e \nu_e - \left(v + \dfrac{1}{2}\right) \nu_e + \left(v + \dfrac{1}{2}\right)^2 \chi_e \nu_e = \nu_e - \chi_e \nu_e (2v+2) = \nu_e - 2 \chi_e \nu_e (v+1)$
In the limit that $\Delta G \rightarrow 0 , v \rightarrow v_{max}$, solving for $v_{max}$ gives
$0 = \nu_e - 2 \chi_e \nu_e (v_{max}+1)$
$2\chi_e( v_{max} +1) = 1$
$v_{max} = \dfrac{1}{2\chi_e} -1$
The molecule dissociates in the limit $\Delta G \rightarrow 0$, so the dissociation energy is given by
$D_e = G(v_{max}) = (\dfrac{1}{2\chi_e} - \dfrac{1}{2}) \nu_e - (\dfrac{1}{2\chi_e} - \dfrac{1}{2})^2 \chi_e \nu_e = \dfrac{\nu_e}{2\chi_e}(1-\chi_e) - \dfrac{\nu_e}{4\chi_e}(1-\chi_e)^2 = \dfrac{\nu_e}{4\chi_e}(2 - 2\chi_e - 1 + 2\chi_e - \chi_e^2) = \dfrac{\nu_e}{4\chi_e}(1 - \chi_e^2) \approx \dfrac{\nu_e}{4\chi_e}$
The final step can be made if we assume $\chi_e$ to be very small.
Looking at the equation for $\Delta G$, we conclude that a plot of $\Delta G$ versus $v+1$ will have an intercept of $\nu_e$ and a slope of $-2\chi_e\nu_e$. The experimental data points for $H_2$ can be plotted as such:
If we use a best linear fit, we determine that the intercept is $\nu_e = 4164.4 cm^{-1}$ and the slope is $-2\chi_e \nu_e = -232.01 cm^{-1}$. Thus, $\chi_e = 0.0279$. Then, the dissociation energy is
$D_e = \dfrac{\nu_e}{4\chi_e} = 37400 cm^{-1}$
The plot is not linear for large values of $v$ because the potential curve is not well described by the anharmonic potential energy function.
Q13.21
An analysis of the vibrational spectrum of the ground-state homonuclear diatomic molecule $Na_{2}$ gives $\tilde{\nu}_{e} = 159.125 \hspace{.1cm}cm^{-1}$ and $\tilde{\nu}_{e}\tilde{x}_{e} = 0.7255 \hspace{.1cm}cm^{-1}$. Suggest an experimental method that can be used to determine there spectroscopic parameters. Use $\nu_{max} = \dfrac{1}{2\tilde{x}_{e}}-1$ to determine the number of bound vibrational levels for ground state of $Na_{2}$.
S13.21
First solve for $\tilde{x}$:
$\tilde{x}=\dfrac{0.7255 \hspace{.1cm}cm^{−1}}{159.125\hspace{.1cm}cm^{−1}} = 4.5593 \times 10^{-3} \hspace{.1cm}cm^{−1}$
Then plug it in to the equation:
$\nu_{max} = \dfrac{1}{2 \times 4.5593 \times 10^{-3} \hspace{.1cm}cm^{−1}}-1 = 109.67\hspace{.1cm}cm^{−1}$
There are $109$ bound vibrational levels for the ground state of $Ne_{2}$.
Q13.22
A Morse potential is a decent representation of internuclear potential and is modeled by
$U(q)=D_e(1-e^{-/betaq})^2$
where $q=R-R_e$. Prove the force constant for the Morse Potential is evaluated
$k = 2D_e\beta^2$
If $D_e= 6.23\times 10^{-19} J\cdot Molecule^{-1}$ and $\beta=1.37 \times 10^{10} m^{-1}$ for a given compound, calculate the force constant for that compound.
S13.22
Using a Maclaurin series expansion of the Morse potential, the formula can be rewritten
$U(q)=D_e(1-e^{-\beta q})^2 = D_e(1-(1-\beta q+\dfrac{\beta^2q^2}{2}O(q)^3))^2 = D_e(\beta^2q^2+O(q)^3)$
We know that
$U(q)=\dfrac{kq^2}{2}$
assuming that $O(q)^3$ approaches zero, we get
$\frac{kq^2}{2}= D_e(\beta^2q^2+O(q)^3) \implies k=2D_e\beta^2$
Plugging in our given values, we get
$l=2(6.23\times 10^{-19} J\cdot Molecule^{-1})(1.37 \times 10^{10})^2 = 234 N\cdot m^{-1}$
Q13.23
Given that $D_{e} = 7.33\times10^{-19} J\cdot molecule^{-1}, \tilde{v}_{e} = 1580.0cm^{-1}, and R_{e} = 121 pm$, find $k$ and $\beta$ for $^{16}O_{2}$ .
S13.23
Assuming a a harmonic oscillator model, we can use the equation:
$\tilde{v}_{e} = \dfrac{1}{2\pi c}\Big(\dfrac{k}{\mu}\Big)^{1/2}$
We can find $k$ from the parameters given in the problem. Solving for $k$ gives
$k = (2\pi c\tilde{v}_{e})^{2}\mu$
$= [2\pi(3\times10^{10} cm\cdot s^{-1})(1580.0 cm^{-1})]^{2} (7.9975 amu)(1.661x10^{-27} kg\cdot amu^{-1})$
$= 1176.3 N\cdot m^{-1}$
Using the equation:
$\beta = \Big(\dfrac{k}{2D_{e}}\Big)^{1/2}$
We can solve for $\beta$ using the information above
$\beta = \Bigg[\dfrac{1176.3N\cdot m^{-1}}{2(7.33\times10^{-19}J)}\Bigg]^{1/2}$
$= 2.84\times10^{10}m^{-1}$
Q13.24
The fundamental line in the IR spectrum of $^{12}C^{16}O$ $cm^{-1}$, and the first overtone occurs at 4260.0 $cm^{-1}$. Find $\tilde{v}_{e}$ and $\tilde{x}_{e}\tilde{v}_{e}$ for $^{12}C^{16}O$.
S13.24
The equations for the fundamental frequency and the overtone are
$\tilde{v} = \tilde{v}_{e} - 2\tilde{x}_{e}\tilde{v}_{e}$
and
$\tilde{v} = 2\tilde{v}_{e} - 6\tilde{x}_{e}\tilde{v}_{e},$
respectively.
We can set the fundamental frequency equal to $2143.0$ $cm^{-1}$ and the first overtone equal to $4260.0$ $cm^{-1}$ which is given from the problem statement. Then, we want to solve for $\tilde{x}_{e}\tilde{v}_{e}$.
Multiply the fundamental frequency by 3 and subtract the overtone.
$\tilde{v}_{e} = [3(2143.0) - 4260.0]cm^{-1} = 2169.0 cm^{-1}$
Then multiply the fundamental frequency by 2 and subtract from the overtone.
$\tilde{x}_{e}\tilde{v}_{e} = 13.0cm^{-1}$
Q13.24
Which of the following molecules exhibit a microwave rotational absorption spectrum: BF2,SO2,C2F2,NO3-
S13.24
SO2
notes: microwave rotational absorption correlates to the rotation of polyatomic molecules. Would $BF_{2}$ and $NO_{3}^{-}$ also fall in this range?
Q13-25 Slightly Incorrect
Calculate the fundamental and the first two overtones of $H^{35}Cl$ given
$\tilde{v_{e}} = 2990.946 cm^{-1}$
$\tilde{x_{e}}\tilde{v_{e}} = 52.819 cm^{-1}$
S13-25
Use equation, $\tilde{v_{obs}} = \tilde{v_{e}}\nu -\tilde{x_{e}}\tilde{v_{e}}\nu(\nu+1)$ with v =1,2...
The fundamental frequency is given by v=1 and the first two overtones are given by v=2 and v=3.
Fundamental: $\tilde{v_{obs}} = \tilde{v_{e}} -2\tilde{x_{e}}\tilde{v_{e}} = 2990.946cm^{-1} - 2(52.819cm^{-1}) = 2885.308cm^{-1}$
First overtone:
$\tilde{v_{obs}} = 2\tilde{v_{e}} -6\tilde{x_{e}}\tilde{v_{e}} = 2990.946cm^{-1} - 6(52.819cm^{-1}) = 5664.978cm^{-1}$
Second overtone:
$\tilde{v_{obs}} = 3\tilde{v_{e}} -12\tilde{x_{e}}\tilde{v_{e}} = 3(2990.946cm^{-1}) - 12(52.819cm^{-1}) = 8339.01cm^{-1}$
Q13.26
Plot the $\frac{\nu_{obs}}{\nu}$ versus $(\nu +1)$ for the anharmonic oscillator approximation and use it to determine the values of $\nu_{e}$ and $x_{e}\nu_{e}$ for $H^35Cl$. Use the information in Table13.4.
S13.26
${\nu}_{obs} = \nu_{obs}\nu - x_{e}\nu_{e}\nu(\nu+1)$
$\frac{\nu_{obs}}{\nu} = \nu_{e} - {x}_{e}\nu_{e}(\nu+1)$
The slope is $-x_{e}\nu_{e}$ and the intercept is $\nu_{e}$
The problem is nice and straight forward. It would be great for an explanation of why the slope is negative for an insight of the graph
Q13.31
Phosphorus nitride ($PN$) is a heteronuclear diatomic molecule that is a similar to $N_2$ with a triple bond, although is substantially weaker because $P$ is a larger atomic radius than $N$. Using the following data, calculate $\tilde{\nu_{e}}'$ and $\tilde{\chi_{e}}'$ $\tilde{\nu_{e}}'$ for $PN$.
IR Transition $\tilde{\nu_{abs}}$ [$cm^{-1}$]
$0 \rightarrow 0$ 39,699.10
$0 \rightarrow 1$ 40,786.80
$0 \rightarrow 2$ 41,858.90
S13.31
We will use the following equation:
$\tilde{\nu_{abs}} = \tilde{\nu_{0,0}}' + \tilde{\nu_{e}}' \tilde{\nu}' - \tilde{\chi_{e}}' \tilde{\nu_{e}}' \tilde{\nu}' (\nu' + 1)$
Plugging in the appropriate values of $\nu'$, we get:
According to your table $\tilde{\nu_{0,0}}'$ should be 39,699.10 not 29,699.10.
• $\nu' = 0$: $\tilde{\nu_{0,0}}' = 29,699.10 \; cm^{-1} \tag{1}$
• $\nu' = 1$: $\tilde{\nu_{0,0}}' + \tilde{\nu_{e}}' - 2 \tilde{\chi_{e}}' \tilde{\nu_{e}}' = 40,786.80 \; cm^{-1} \tag{2}$
• $\nu' = 2$: $\tilde{\nu_{0,0}}' + 2 \tilde{\nu_{e}}' - 6 \tilde{\chi_{e}}' \tilde{\nu_{e}}' = 41,858.90\; cm^{-1} \tag{3}$
Solving as a system of equations, we subtract:
(2) - (1) = (4), (3) - (1) = (5), and finally (5) - 2x(4) = (6)
Equation (6) can be solved for $\tilde{\chi_{e}}' \tilde{\nu_{e}}'$, and then plugged back in to equation (4) or equation (5) to solve for $\tilde{\nu_{e}}'$. The answer is:
$\tilde{\chi_{e}}' \tilde{\nu_{e}}' = 1103.3\; cm^{-1}$
and
$\tilde{\nu_{e}}' = 7.80 \; cm^{-1}$
from my calculations $\tilde{\nu_{e}}' = 1103.3$ and $\tilde{\chi_{e}}' \tilde{\nu_{e}}' = 7.80$ it looks like you had them switched
Q13.32
The frequencies of the first few vibronic transitions to an excited state of NaCl* are as follows:
Vibronic Transitions 0 $\rightarrow$ 0 0 $\rightarrow$ 1 0 $\rightarrow$ 2
$\tilde{v}_{obs}/cm^-1$ 12376 89706 15219
*not accurate values for NaCl
Use these data to calculate the values of $\tilde{v}_e$ and $\tilde{v}_e\tilde{x}_e$ for the excited state of NaCl
S13.32
$A = \tilde{v}_e', B = \tilde{v}_e'\tilde{x}_e'$
$12376 = \tilde{v}_0,0$
$89706 = \tilde{v}_0,0+\tilde{v}_e'-2\tilde{v}_e'\tilde{x}_e' - 12376 = \tilde{v}_0,0 = 77330 = +\tilde{v}_e'-2\tilde{v}_e'\tilde{x}_e'$ (1)
$15219 = \tilde{v}_0,0+2\tilde{v}_e'-6\tilde{v}_e'\tilde{x}_e' - 12376 = \tilde{v}_0,0 = 2843 = 2\tilde{v}_e'-6\tilde{v}_e'\tilde{x}_e'$ (2)
77330 + 2B = A (1a)
2843 = 2A - 6B (2a)
77330+2B = A
2843 - 154660 = -2B
B = 75908.5 cm-1 = $\tilde{v}_e'$
A = 229147 cm-1 = $\tilde{v}_e'\tilde{x}_e'$
Q13.33
Determine the number of translational, rotational, and vibrational degrees of freedom in
1. Xe
2. HCl
3. CS2
4. hemoglobin containing 9272 molecules
The total number of degrees of freedom is 3N, where N is the number of atoms in the molecule. All molecules have three translational degrees of freedom. A nonlinear molecule has three rotational degrees of freedom and a linear molecule has two rotational degrees of freedom. A linear molecule has $3N-5$ vibrational degrees of freedom and a nonlinear molecule has 3N — 6 vibrational degrees of freedom.
S13.33
The number of translational degrees of freedom is 3, the number of rotational degrees of freedom is 2 for linear molecules and 3 for non-linear molecules. Therefore, the remaining degrees of freedom, which is equal to 3N-5 for linear molecules or 3N-6 for non-linear molecules, is equal to the number of vibrational degrees of freedom.
Translation Rotation Vibration Total
Xe 3 0 0 3
HCl 3 2 1 6
CS2 3 2 4 9
hemoglobin 3 3 27810 27816
Q13.36
Classify each of the following molecules as either a prolate or an oblate symmetric top: $XeF_4$, $ClCH_3$, $NH_3$, and $C_6$$H_6$.
S13.36
• $XeF_4$ :oblate
• $ClCH_3$:prolate
• $NH_3$:prolate
• $C_6$$H_6$:oblate
Q13.37
Solve for the components of the moment of inertia of a trigonal planar molecule if all the masses are m, the bond lengths are unit length, and all bond angles are $120^\circ$.
S13.37
$I_{xx} = \sum m_jy^2_j = m(1)^2 + 2m (sin^2 30^\circ) = \frac{3}{2}m$
$I_{yy} = \sum m_jx^2_j = m(0)^2 + 2m (cos^2 30^\circ) = \frac{3}{2}m$
$I_{zz} = \sum m_jx^2_j + \sum m_jy^2_j = 3m$
Q13.38
This problem illustrates how the principal moments of inertia can be obtained as an eigenvalue problem. Consider a molecule where all the masses are unit masses and the long and short bond lengths are 2 and 1, respectively.
Show that
$I_{xx} = 2\cos^2\theta + 8\sin^2\theta$I_{yy} = 8\cos^2\theta + 2\sin^2\theta$I_{xy} = -6\cos\theta\sin\theta$
Solve the secular determinant equation for $\lambda$
$\left. \begin{vmatrix} I_{xx} - \lambda & I_{xy} \ I_{xy} & I_{yy} - \lambda \end{vmatrix} \right. = 0$
and compare the result with the values of $I_{xx}$ and $I_{yy}$ that you would obtain if you align the "molecule" and coordinate system such that $\theta = 180^{\circ}$. What does this comparison tell you? What are the values of $I_{xx}$ and $I_{yy}$ if $\theta = 90$?
S13.38
We use trigonometric functions to find the x and y components
$I_{xx} = \sum m_{i}y^2_{i} = 2(1)(2\sin\theta)^2 + 2(1)\left[\sin\left(\dfrac{\pi}{2}\right) - \theta\right]^2 = 8\sin^2\theta + 2\cos^2\theta$
$I_{yy} = \sum m_{i}x^2_{i} = 2(1)(2\cos\theta)^2 + 2(1)\left[\cos\left(\dfrac{\pi}{2}\right) - \theta\right]^2 = 8\cos^2\theta + 2\sin^2\theta$
$I_{xy} = -\sum m_{i}x_{i}y_{i} = -(2\cos\theta)(2\sin\theta) - 2\left[\cos\left(\pi\right) + \theta\right]\left[\sin\left(\pi\right) + \theta\right] - \left[\sin\left(\dfrac{\pi}{2}\right) - \theta\right]\left[\cos\left(\dfrac{\pi}{2}\right) - \theta\right] - \left[\cos\left(\dfrac{\pi}{2}\right) + \theta\right]\left[\sin\left(\dfrac{\pi}{2}\right) + \theta\right]$
= $-6\cos\theta\sin\theta$
The secular determinantal equation becomes
$\left. \begin{vmatrix} 8\sin^2\theta + 2\cos^2\theta - \lambda & -6\cos\theta\sin\theta \ -6\cos\theta\sin\theta & 8\cos^2\theta + 2\sin^2\theta - \lambda \end{vmatrix} \right. = 0$
Expanding the determinant gives
$0 = \lambda^2 - \lambda(10)(\sin\theta^2 + \cos\theta^2) + 16(\sin\theta^2 + \cos\theta^2)^2$
$0 = \lambda^2 - 10\lambda + 16$
$\lambda = 5 \pm 3$
If $\theta = 180^\circ$ then $I_{xx} = 2, I_{yy} = 8, and I_{xy} = 0$. If $\theta = 90^\circ$ then $I_{xx} = 8, I_{yy} = 2, and I_{xy} = 0$. This tells us that the coordinate system does not affect the values of the principal momentum of inertia.
Q13.39
Sketch the energy level diagram of NH3 ( a prolate symmetric top molecule ) and XeF4 ( an oblate symmetric top ). How do they differ?
S13.39
The energies increase as the J levels increases for the prolate symmetry and the energies decrease as J increases for the oblate symmetry
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/13%3A_Molecular_Spectroscopy/13.E%3A_Molecular_Spectroscopy_%28Exercises%29.txt
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Nuclear magnetic resonance (NMR) is a physical phenomenon in which nuclei in a magnetic field absorb and re-emit electromagnetic radiation. This energy is at a specific resonance frequency which depends on the strength of the magnetic field and the magnetic properties of the isotope of the atoms. Many scientific techniques exploit NMR phenomena to study molecular physics, crystals, and non-crystalline materials through nuclear magnetic resonance spectroscopy. NMR is also routinely used in advanced medical imaging techniques, such as in magnetic resonance imaging (MRI).
14: Nuclear Magnetic Resonance Spectroscopy
The electron, as well as certain other fundamental particles, possesses an intrinsic angular momentum or spin, in addition to its orbital angular momentum. These two types of angular momentum are analogous to the daily and annual motions, respectively, of the Earth around the Sun. To distinguish the spin angular momentum from the orbital, we designate the quantum numbers as s and $m_s$, in place of $\ell$ and m. For the electron, the quantum number s always has the value $\dfrac{1}{2}$, while $m_s$ can have one of two values,$\pm \dfrac{1}{2}$. The electron is said to be an elementary particle of spin $\dfrac{1}{2}$. The proton and neutron also have spin $\dfrac{1}{2}$ and belong to the classification of particles called fermions, which are governed by the Pauli exclusion principle. Other particles, including the photon, have integer values of spin and are classified as bosons. These do not obey the Pauli principle, so that an arbitrary number can occupy the same quantum state. A complete theory of spin requires relativistic quantum mechanics. For our purposes, it is sufficient to recognize the two possible internal states of the electron, which can be called spin up' and spin down.' These are designated, respectively, by $\alpha$ and $\beta$ as factors in the electron wavefunction. Spins play an essential role in determining the possible electronic states of atoms and molecules.
14.02: Magnetic Moments Interact with Magnetic Fields
In the mid 1920's the German physicist Werner Heisenberg showed that if we try to locate an electron within a region $Δx$; e.g. by scattering light from it, some momentum is transferred to the electron, and it is not possible to determine exactly how much momentum is transferred, even in principle. Heisenberg showed that consequently there is a relationship between the uncertainty in position $Δx$ and the uncertainty in momentum $Δp$.
$\Delta p \Delta x \ge \frac {\hbar}{2} \label {5-22}$
You can see from Equation $\ref{5-22}$ that as $Δp$ approaches 0, $Δx$ must approach ∞, which is the case of the free particle discussed previously.
This uncertainty principle, which also is discussed in Chapter 4, is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space. Activity 1 at the end of this chapter illustrates that a reduction in the spatial extent of a wavefunction to reduce the uncertainty in the position of a particle increases the uncertainty in the momentum of the particle. This illustration is based on the ideas described in the next section.
Exercise $1$
Compare the minimum uncertainty in the positions of a baseball (mass = 140 gm) and an electron, each with a speed of 91.3 miles per hour, which is characteristic of a reasonable fastball, if the standard deviation in the measurement of the speed is 0.1 mile per hour. Also compare the wavelengths associated with these two particles. Identify the insights that you gain from these comparisons.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/14%3A_Nuclear_Magnetic_Resonance_Spectroscopy/14.01%3A_Nuclei_Have_Intrinsic_Spin_Angular_Momenta.txt
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In the mid 1920's the German physicist Werner Heisenberg showed that if we try to locate an electron within a region $Δx$; e.g. by scattering light from it, some momentum is transferred to the electron, and it is not possible to determine exactly how much momentum is transferred, even in principle. Heisenberg showed that consequently there is a relationship between the uncertainty in position $Δx$ and the uncertainty in momentum $Δp$.
$\Delta p \Delta x \ge \frac {\hbar}{2} \label {5-22}$
You can see from Equation $\ref{5-22}$ that as $Δp$ approaches 0, $Δx$ must approach ∞, which is the case of the free particle discussed previously.
This uncertainty principle, which also is discussed in Chapter 4, is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space. Activity 1 at the end of this chapter illustrates that a reduction in the spatial extent of a wavefunction to reduce the uncertainty in the position of a particle increases the uncertainty in the momentum of the particle. This illustration is based on the ideas described in the next section.
Exercise $1$
Compare the minimum uncertainty in the positions of a baseball (mass = 140 gm) and an electron, each with a speed of 91.3 miles per hour, which is characteristic of a reasonable fastball, if the standard deviation in the measurement of the speed is 0.1 mile per hour. Also compare the wavelengths associated with these two particles. Identify the insights that you gain from these comparisons.
14.04: The Magnetic Field Acting upon Nuclei in Molecules Is Shielded
An isolated nucleus in an external magnetic field experiences the external magnetic field. However, a nucleus in a molecule, when the molecule is placed in an external magnetic field, does not experience the external magnetic field, due to interference by the magnetic field generated by the surrounding electrons. The magnetic field experienced by the nucleus is either slightly lower than the external magnetic field or slightly higher. If the magnetic field experienced by the nucleus is lower than the external magnetic field, the nucleus is said to be shielded; if it is higher, the nucleus is said to be deshielded.
• \(B_x\) = the external magnetic field
• \(B_y\) = the magnetic field experienced by the nucleus
If \(B_y < B_x\), the nucleus is shielded. If \(B_y > B_x\), the nucleus is deshielded
14.05: Chemical Shifts Depend upon the Chemical Environment of the Nucleus
The chemical shift in NMR is extremely important, as it gives vital information about the local structure surrounding the nucleus of interest. For a majority of scientists, the chemical shift is used exlusivley to determine structure, especially in organic systems. Additional information may be gained by examining the anisotropy of the chemical shift. This section will be devoted to looking at chemical shift from a mathematical standpoint including a full treatment of the chemical shift tensor and the relation to the NMR lineshape.
The Chemical Shift
The local magnetic field is the field felt by a particular nucleus, where the applied field $B$ induces currents in the electrons surrounding the nucleus give rise to a shielding. The shielding constant is $\sigma$. The local magnetic field is reduced by shielding by a factor $1 - σ$.
$B_{loc} = B + δB = (1 – σ)B \nonumber$
The chemical shift is the difference between the resonance frequency of a nucleus and that of a standard.
The Larmor frequency of a shielded nucleus is:
$ν_L = \dfrac{γB_{loc} }{2π} \nonumber$
Chemical shifts are reported on the $\delta$-scale.
$δ = \dfrac{ ν – ν_0}{ν_0} \times 10^6 \nonumber$
The resonance frequency of the standard is $ν_0$.
The shielding constant is the sum of three contributions.
$σ = σ(local) + σ(molecule) + σ(solvent) \nonumber$
The local contribution is due to electrons on the atom that contains the nucleus. The molecular contribution is from the rest of the molecule. The solvent contribution is from surrounding solvent molecules.
The Local Contribution
The local contribution is a sum of both diamagnetic $σ_d$ and paramagnetic $σ_p$ parts. The diamagnetic part arises from circulation of the electrons in response to $B$. The Lamb formula gives the magnitude of $σ_d$,
$σ_d = \dfrac{e^2 µ_0}{3m_e } \int _o^∞ ρ(r)r \,dr \nonumber$
where $ρ$ is the electron probability density $|Ψ^2|$. $σ_d$ is inversely proportional to the Bohr radius. The magnetic moment of a current loop is proportional to $a_o^2$ and the magnetic field generated at the nucleus is proportional $\frac{1}{a_o^3}$.
The Molecular Contribution
The applied magnetic field generates currents in neighboring groups proportional to the magnetic susceptibility $\chi$ of a group. The induced magnetic moment gives rise to a magnetic field that is inversely proportional to the cube of the distance from the nucleus.
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5.5A: The source of spin-spin coupling
The 1H-NMR spectra that we have seen so far (of methyl acetate and para-xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1H-NMR spectra of most organic molecules contain proton signals that are 'split' into two or more sub-peaks. Rather than being a complication, however, this splitting behavior actually provides us with more information about our sample molecule.
Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, we show enlargements of individual signals so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet. The Hb signal at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet.
The source of signal splitting is a phenomenon called spin-spin coupling, a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2 trichloromethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of Hb will be aligned with B0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B0. The Beff ‘felt’ by Ha is a slightly weaker if Hb is aligned against B0, or slightly stronger if Hb is aligned with B0. In other words, in half of the molecules Ha is shielded by Hb (thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb(and the NMR signal shifted slightly downfield). What would otherwise be a single Ha peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram, as shown below.
Now, let's think about the Hbsignal. The magnetic environment experienced by Hb is influenced by the fields of both neighboring Ha protons, which we will call Ha1 and Ha2. There are four possibilities here, each of which is equally probable. First, the magnetic fields of both Ha1 and Ha2 could be aligned with B0, which would deshield Hb, shifting its NMR signal slightly downfield. Second, both the Ha1 and Ha2 magnetic fields could be aligned opposed to B0, which would shield Hb, shifting its resonance signal slightly upfield. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1opposed to B0 and Ha2 with B0. In each of the last two cases, the shielding effect of one Ha proton would cancel the deshielding effect of the other, and the chemical shift of Hb would be unchanged.
So in the end, the signal for Hb is a triplet, with the middle peak twice as large as the two outer peaks because there are two ways that Ha1 and Ha2 can cancel each other out.
Now, consider the spectrum for ethyl acetate:
We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (Ha) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. The signal at 1.055 ppm for the Hc hydrogens is split into a triplet by the two Hb hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane.
The Hbhydrogens give rise to a quartet signal at 3.915 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three Hc hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns.
Example 5.6
1. Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the Hc hydrogens, why the Hb signal in ethyl acetate is split into a quartet.
2. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the Hb quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.)
Solution
By now, you probably have recognized the pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. Thus the two Hb hydrogens in ethyl acetate split the Hc signal into a triplet, and the three Hc hydrogens split the Hb signal into a quartet. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two `neighbors`. When we begin to determine structures of unknown compounds using 1H-NMR spectral data, it will become more apparent how this kind of information can be used.
Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, Ha1 in 1,1,2-trichloroethane is not split by Ha2, and vice-versa.
Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the Ha hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away, too far for coupling to occur.
Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more details about how we quantify coupling interactions is provided in section 5.5B). Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules.
Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules.
Example 5.7
1. How many proton signals would you expect to see in the 1H-NMR spectrum of triclosan (a common antimicrobial agent found in detergents)? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Solution
Example 5.8
Predict the splitting pattern for the 1H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin).
Solution
5.5B: Coupling constants
Chemists quantify the spin-spin coupling effect using something called the coupling constant, which is abbreviated with the capital letter J. The coupling constant is simply the difference, expressed in Hz, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two subpeaks are separated by 6.1 Hz, and thus we write 3Ja-b = 6.1 Hz.
The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between Ha and Hb. Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet. This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength.
When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! The coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Coupling constants between proton sets on neighboring sp3-hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2-hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement.
For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens.
Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values.
5.5C: Complex coupling
In all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to just one neighboring set of hydrogens. When a set of hydrogens is coupled to two or more sets of nonequivalent neighbors, the result is a phenomenon called complex coupling. A good illustration is provided by the 1H-NMR spectrum of methyl acrylate:
First, let's first consider the Hc signal, which is centered at 6.21 ppm. Here is a closer look:
With this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? Hc is coupled to both Ha and Hb , but with two different coupling constants. Once again, a splitting diagram can help us to understand what we are seeing. Ha is trans to Hc across the double bond, and splits the Hc signal into a doublet with a coupling constant of 3Jac = 17.4 Hz. In addition, each of these Hc doublet sub-peaks is split again by Hb (geminal coupling) into two more doublets, each with a much smaller coupling constant of 2Jbc = 1.5 Hz.
The result of this `double splitting` is a pattern referred to as a doublet of doublets, abbreviated `dd`.
The signal for Ha at 5.95 ppm is also a doublet of doublets, with coupling constants 3Jac= 17.4 Hz and 3Jab = 10.5 Hz.
The signal for Hb at 5.64 ppm is split into a doublet by Ha, a cis coupling with 3Jab = 10.4 Hz. Each of the resulting sub-peaks is split again by Hc, with the same geminal coupling constant 2Jbc = 1.5 Hz that we saw previously when we looked at the Hc signal. The overall result is again a doublet of doublets, this time with the two `sub-doublets` spaced slightly closer due to the smaller coupling constant for the cis interaction. Here is a blow-up of the actual Hbsignal:
Example 5.9
Construct a splitting diagram for the Hb signal in the 1H-NMR spectrum of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).
Solution
When constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).
When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `n + 1 rule` of non-complex splitting. In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for Hb to be split into a triplet by Ha, and again into doublets by Hc, resulting in a 'triplet of doublets'.
Ha and Hc are not equivalent (their chemical shifts are different), but it turns out that 3Jab is very close to 3Jbc. If we perform a splitting diagram analysis for Hb, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the n + 1 rule.
For similar reasons, the Hc peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined Hb and Hd protons. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.
Example 5.10
What splitting pattern would you expect for the signal coresponding to Hb in the molecule below? Assume that Jab ~ Jbc. Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.
Solution
In many cases, it is difficult to fully analyze a complex splitting pattern. In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for Hb to be a doublet, Hd a triplet, and Hc a triplet.
In practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult. In this case, we would refer to the aromatic part of the spectrum as a multiplet.
When we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments. Quite simply, the stronger our magnet is, the more resolution we get in our spectrum. In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency 'window' in which we can observe proton signals is 1200 Hz wide. In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz - five times wider. In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/14%3A_Nuclear_Magnetic_Resonance_Spectroscopy/14.06%3A_Spin-Spin_Coupling_Can_Lead_to_Multiplets_in_NMR_Spectra.txt
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When doing NMR spectroscopy, it is an observed fact that equivalent hydrogens do not split one another. Why don't equivalent hydrogens split each other's signals? For example, why is the NMR spectrum for ethane a singlet instead of a quartet or even a dodecuplet (due to the hydrogens on the same carbon)? What is so special about hydrogens being equivalent to one another that causes no splitting to be observed?
Finding the states between which transitions occur
The Hamiltonian for two coupled spins is
$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + \dfrac{2\pi J_{12}}{\hbar}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \label{1}$
where $\omega_1$ and $\omega_2$ are the Larmor frequencies[2] of the two nuclei and $J_{12}$ is the coupling constant (in Hz) between the two nuclei. (The factor of $2\pi/\hbar$ is simply there to bring it into energy units.) $\hat{I}_{\!i}$ is the operator for the spin angular momentum of nucleus $i$ and $\hat{I}_{\!iz}$ is the operator for its projection along the $z$-axis.
However, in NMR, it is customary to work in frequency units instead of energy units. Since $E = h\nu$, we simply need to divide through by $h$. Bearing in mind that $\omega = 2\pi\nu$ and $h = 2\pi\hbar$, we get:
$\hat{H}_\text{freq} = \dfrac{\nu_1}{\hbar}\hat{I}_{\!1z} + \dfrac{\nu_2}{\hbar}\hat{I}_{\!2z} + \dfrac{J_{12}}{\hbar^2}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \label{2}$
On top of that, to make the math easier, it is also quite common to set $\hbar = 1$. Therefore, we have
$\hat{H}_\text{freq} = \nu_1\hat{I}_{\!1z} + \nu_2\hat{I}_{\!2z} + J_{12}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \label{3}$
We will deal with two spin-$1/2$ nuclei here, where $|\alpha\rangle$ and $|\beta\rangle$ represent the "up" and "down" spins (recall we set $\hbar = 1$ so it doesn't appear in the eigenvalues):
\begin{align*} \hat{I}_{\!iz}|\alpha_i\rangle &= \dfrac{1}{2}|\alpha_i\rangle & \hat{I}_{\!iz}|\beta_i\rangle &= -\dfrac{1}{2}|\beta_i\rangle & (i = 1,2) \label{4} \end{align*}
Furthermore since we are dealing with equivalent nuclei we can simply set $\nu_1 = \nu_2 = \nu$ and drop the subscript in $J_{12}$ just to make it a bit cleaner:
$\hat{H} = \nu(\hat{I}_{\!1z} + \hat{I}_{\!2z}) + J(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \label{5}$
Now, we need to find the eigenstates and eigenvalues of $\hat{H}$. To do so, we will adopt the basis set of product functions $(|\alpha_1\alpha_2\rangle, |\alpha_1\beta_2\rangle, |\beta_1\alpha_2\rangle, |\beta_1\beta_2\rangle)$. From equation $(4)$ we have
\begin{align*} \hat{I}_{\!1z}|\alpha_1\alpha_2\rangle &= \dfrac{1}{2}|\alpha_1\alpha_2\rangle & \hat{I}_{\!2z}|\alpha_1\alpha_2\rangle &= \dfrac{1}{2}|\alpha_1\alpha_2\rangle \label{6} \ \hat{I}_{\!1z}|\alpha_1\beta_2\rangle &= \dfrac{1}{2}|\alpha_1\beta_2\rangle & \hat{I}_{\!2z}|\alpha_1\beta_2\rangle &= -\dfrac{1}{2}|\alpha_1\beta_2\rangle \label{7} \ \hat{I}_{\!1z}|\beta_1\alpha_2\rangle &= -\dfrac{1}{2}|\beta_1\alpha_2\rangle & \hat{I}_{\!2z}|\beta_1\alpha_2\rangle &= \dfrac{1}{2}|\beta_1\alpha_2\rangle \label{8} \ \hat{I}_{\!1z}|\beta_1\beta_2\rangle &= -\dfrac{1}{2}|\beta_1\beta_2\rangle & \hat{I}_{\!2z}|\beta_1\beta_2\rangle &= -\dfrac{1}{2}|\beta_1\beta_2\rangle \label{9} \end{align*}
The action of the scalar product $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ is more complicated. We need to introduce the shift operators (or ladder operators).
\begin{align*} \hat{I}_{\!i+} &= \hat{I}_{\!ix} + \mathrm{i}\hat{I}_{\!iy} & \hat{I}_{\!i-} &= \hat{I}_{\!ix} - \mathrm{i}\hat{I}_{\!iy} \label{10} \end{align*}
from which we can obtain
\begin{align*} \hat{I}_{\!ix} &= \dfrac{\hat{I}_{\!i+} + \hat{I}_{\!i-}}{2} & \hat{I}_{\!iy} &= \dfrac{\hat{I}_{\!i+} - \hat{I}_{\!i-}}{2\mathrm{i}} \label{11} \end{align*}
So, finally, we can write
\begin{align*} \hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}} &= \hat{I}_{\!1x}\hat{I}_{\!2x} + \hat{I}_{\!1y}\hat{I}_{\!2y} + \hat{I}_{\!1z}\hat{I}_{\!2z} \label{12} \ &= \dfrac{\hat{I}_{\!1+}\hat{I}_{\!2-} + \hat{I}_{\!1-}\hat{I}_{\!2+}}{2} + \hat{I}_{\!1z}\hat{I}_{\!2z} \label{13} \end{align*}
where in going from $(12)$ to $(13)$ one simply substitutes in $(11)$ and does a fair bit of algebraic manipulation. The action of the shift operators are
\begin{align*} \hat{I}_{\!i+}|\alpha_i\rangle &= 0 & \hat{I}_{\!i+}|\beta_i\rangle &= |\alpha_i\rangle \tag{14} \ \hat{I}_{\!i-}|\alpha_i\rangle &= |\beta_i\rangle & \hat{I}_{\!i-}|\beta_i\rangle &= 0 \tag{14} \ \end{align*} \nonumber
This allows you to work out the effect of $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ on our basis states. The actual math is left to the reader and I will simply quote the results:
\begin{align*} \hat{H}|\alpha_1\alpha_2\rangle &= \left(\nu + \dfrac{J}{4}\right)|\alpha_1\alpha_2\rangle \label{15} \ \hat{H}|\alpha_1\beta_2\rangle &= -\dfrac{J}{4}|\alpha_1\beta_2\rangle + \dfrac{J}{2}|\beta_1\alpha_2\rangle \label{16} \ \hat{H}|\beta_1\alpha_2\rangle &= \dfrac{J}{2}|\alpha_1\beta_2\rangle - \dfrac{J}{4}|\beta_1\alpha_2\rangle \label{17} \ \hat{H}|\beta_1\beta_2\rangle &= \left(-\nu + \dfrac{J}{4}\right)|\beta_1\beta_2\rangle \label{18} \ \end{align*}
Therefore in this basis the Hamiltonian matrix is
$\mathbf{H} = \begin{pmatrix} \nu + J/4 & 0 & 0 & 0 \ 0 & -J/4 & J/2 & 0 \ 0 & J/2 & -J/4 & 0 \ 0 & 0 & 0 & -\nu + J/4 \end{pmatrix} \label{19}$
Finding the eigenvectors and eigenvalues of this matrix is again left to the reader (it is not a difficult task) and they are (eigenvalues denoted E_i) \begin{align*} |1\rangle &= |\alpha_1\alpha_2\rangle & E_1 &= \nu + \dfrac{J}{4} \label{20} \ |2\rangle &= \dfrac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle + |\beta_1\alpha_2\rangle) & E_2 &= \dfrac{J}{4} \label{21} \ |3\rangle &= \dfrac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle - |\beta_1\alpha_2\rangle) & E_3 &= -\dfrac{3J}{4} \label{22} \ |4\rangle &= |\beta_1\beta_2\rangle & E_4 &= -\nu + \dfrac{J}{4} \label{23} \ \end{align*} The form of the eigenstates should be familiar: they are simply the triplet and singlet states of two spin-1/2 particles. These states arise from the coupling of two sources of angular momenta, $I_1$ and $I_2$, to form one overall angular momentum denoted $I$.
$\vec{I} = \vec{I}_{\!1} + \vec{I}_{\!2} \label{24}$
The allowed values of $I$ are determined by the Clebsch-Gordan series:
$I = I_1 + I_2, I_1 + I_2 - 1, \cdots, |I_1 - I_2| \label{25}$
Since $I_1 = I_2 = 1/2$, $I$ can take the values $1$ and $0$. The values of $M_I$, the projection of the total angular momentum along the $z-axis, are as usual $M_I = I, I-1, \cdots, -I \label{26}$ so the states with \(I = 1$ ("triplet") have $M_I = 1, 0, -1$ and the state with $I = 0$ ("singlet") has $M_I = 0$. One can use more quantum mechanics to work out which state is associated with which quantum numbers, but I will not do it here. They are:
$\begin{array}{ccc} \hline \text{State} & I & M_I \ \hline |1\rangle & 1 & 1 \ |2\rangle & 1 & 0 \ |3\rangle & 0 & 0 \ |4\rangle & 1 & -1 \ \hline \end{array} \nonumber$
Selection rules
We have four different states, which leads to ${4\choose 2} = 6$ different possible transitions. However, not all of these transitions are allowed.
The intensity of the transition is proportional to the square of the matrix element $\langle \psi_\mathrm{f} | \hat{H'} | \psi_\mathrm{i} \rangle$ (the so-called "transition dipole moment"), where $\hat{H'}$ is the Hamiltonian for the process that induces the transition. In the case of NMR transitions, the transition arises due to a magnetic field aligned along the $x-axis.[3] The corresponding Hamiltonian is therefore $\hat{H'} = \omega'(\hat{I}_{\!1x} + \hat{I}_{\!2x}) = \omega'\hat{I}_{\!x} \label{27}$ Exactly what \(\omega'$ represents is not important here because we are only really concerned about whether the transition dipole moment is zero or not.[4] Making use of the relations established in equations \ref{10} and \ref{11},[5] one can find that the selection rules are
$\Delta I = 0; \Delta M_I = \pm 1 \label{28}$
which means that the allowed transitions are $|4\rangle \leftrightarrow |2\rangle$ and $|2\rangle \leftrightarrow |1\rangle$. Transitions to and from the singlet state $|3\rangle$ are forbidden. The energies of the transitions are
\begin{align*} E_{4\leftrightarrow2} &= \dfrac{J}{4} - \left(-\nu + \dfrac{J}{4}\right) = \nu \label{29} \ E_{2\leftrightarrow1} &= \left(\nu + \dfrac{J}{4}\right) - \dfrac{J}{4} = \nu \label{30} \ \end{align*}
i.e. the two transitions are degenerate and only one line in the spectrum at frequency $\nu$ is observed. This is exactly what is depicted in the diagrams posted in the other answers.
the two transitions are degenerate and only one line in the spectrum at frequency $\nu$ is observed.
Notes and references
[1] I am assuming the reader has some knowledge of the quantum mechanical treatment of angular momentum, which is a topic that is treated thoroughly in most quantum mechanics textbooks. See, for example, chapter 4 of Atkins's Molecular Quantum Mechanics (5th ed.).
[2] The Larmor frequency is given by $\omega = -\gamma B_0$, where $\gamma$ is the magnetogyric ratio of the nucleus in question and $B_0$ is the strength of the external magnetic field. It represents the frequency with which a magnetic moment precesses about a magnetic field. See any textbook on magnetism for further details.
[3] I am glossing over some details here. The so-called magnetic field in the $x-axis is a component of the radiofrequency pulse applied in the \(xy$-plane. If you are interested please consult a textbook on the vector model of NMR. In particular I recommend Keeler's Understanding NMR Spectroscopy (2nd ed.).
[4] It is related to the strength of the magnetic field in the $x$-axis, $B_1$, by $w' = |\gamma|B_1$. The usual symbol is $\omega_1$, but I chose not to use this here to avoid potential confusion. Again, please consult a textbook on the vector model of NMR if you wish to find out more.
[5] A full proof can be found in J. Chem. Educ. 1982, 59 (10), 819. There is also some discussion of the selection rules in Gunther's NMR Spectroscopy (3rd ed.), p 156 onwards.
Note
The first important point to note is that magnetically equivalent nuclei do in fact couple to each other, however no splitting is observed in the spectrum. The second point is that chemically equivalent, but magnetically non-equivalent, nuclei couple to each other, and this coupling is observable in the NMR spectrum.
Spin coupling comes from a magnetic interaction between nuclear spins transmitted through the bonding electrons. Signals observed in the NMR spectrum are a transition between energy levels of allowed spin states. When two nuclei 'couple', the energy levels are stabilised or destabilsed slightly based on the relative orientations of the nuclear moments, so that (for a doublet) one transition is now δ+J/2, and the other transition is δ-J/2. These two transitions constitute the two lines of the doublet signal. When two equivalent nuclei couple, transitions between energy levels do not change because the interactions between the nuclear moments are the same, as are all other contributing factors such as the Fermi contact. As long as the transitions remain the same, all possible transitions will be equivalent.
The diagram below hopefully explains it a little clearer. The middle energy levels are for two non-coupled spins. Spin A (shown in red) has two possible transitions, both of which are equivalent. When the two spins are coupled, the energy levels are stabilised/destabilised as shown on the right. The transitions for spin A are now no longer equivalent, and they will appear as two lines (doublet). On the left, the equivalent nuclei have an overall change in energy level, but the transition between the levels remains the same, hence the observed line is still a singlet.
When there are two distinct nuclei (The AX case) and there is spin-spin interaction ($J$ coupling) between them (in addition to chemical shift) then fine structure can be observed in the nmr spectrum. The first diagram shows the energy levels and how they interact. Note how the $J$ coupling moves the levels up and down in addition to chemical shift.
Now there are selection rules that allow the radio frequency radiation to couple different levels together and so produce a spectrum. The rule is that the $m_z$ quantum number (variously called magnetic, or azimuthal or projection quantum number) has to change by $+1$ or $-1$. This means that in the first diagram only levels in which one alpha changes to beta or vice versa, as shown by the vertical arrows, are allowed transitions. (The selection rule occurs because the photon (even if at radio frequency) has one unit of angular momentum and total angular momentum is conserved.)
In the equivalent nuclei case (called A2) the interaction between spins is still present but because of the magnetically identical nuclei the spin states are not either symmetrical or anti symmetrical to exchange of nuclei and a linear combination has to be made. This is shown on the left of the figure below. The reason that the splitting in energy levels is not observed is that selection rules make the transitions unobservable
Advanced: Liouville–von Neumann Equation
An alternative answer using the product operator formalism:
The fundamental equation describing NMR quantum mechanically (neglecting relaxation) is the Liouville–von Neumann Equation (in frequency units, i.e. setting $\hbar = 1$):
$\dfrac{\mathrm d}{\mathrm dt}\hat{\rho}=-\mathrm i[\hat{H},\hat{\rho}] \nonumber$
For a 2-spin system with identical chemical shifts and a coupling
$\hat{H}=\Omega\hat{I}_{\!1z}+\Omega\hat{I}_{\!2z}+J(\vec{\hat{I}}_{\!1}\cdot\vec{\hat{I}}_{\!2}) \nonumber$
After a 90 degree pulse on both nuclei the density matrix $\hat{\rho}$ is of the form $\hat{I}_{\!1\chi} + \hat{I}_{\!2\chi}$, where $\chi = x$ or $y$. Now it is a bit lengthy but easy to show that
$[\hat{I}_{\!1\chi}+\hat{I}_{\!2\chi},J(\vec{\hat{I}}_{\!1}\cdot\vec{\hat{I}}_{\!2})]=0, \quad \text{with } \chi=x,y,z \nonumber$
meaning that the coupling Hamiltonian does not influence the signal after a 90 degree pulse, as $[\hat{H}_\text{coupling},\hat{\rho}] = 0$.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/14%3A_Nuclear_Magnetic_Resonance_Spectroscopy/14.07%3A_Spin-Spin_Coupling_Between_Chemically_Equivalent_Protons_is_Not_Observed.txt
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The (n+1) Rule, an empirical rule used to predict the multiplicity and, in conjunction with Pascal’s triangle, splitting pattern of peaks in 1H and 13C NMR spectra, states that if a given nucleus is coupled (see spin coupling) to n number of nuclei that are equivalent (see equivalent ligands), the multiplicity of the peak is n+1. eg. 1:
The three hydrogen nuclei in 1, $H_a$, $H_b$, and $H_c$, are equivalent. Thus, 1H NMR spectrum of 1 $H_a$s only one peak. $H_a$, $H_b$, and $H_c$ are coupled to no hydrogen nuclei. Thus, for $H_a$, $H_b$, and $H_c$, n=0; (n+1) = (0+1) = 1. The multiplicity of the peak of $H_a$, $H_b$, and $H_c$ is one. The peak $H_a$s one line; it is a singlet. eg. 2:
There are two sets of equivalent hydrogen nuclei in 2:
• Set 1: $H_a$
• Set 2: $H_b$, $H_c$
Thus, the 1H NMR spectrum of 2 $H_a$s two peaks, one due to $H_a$ and the other to $H_b$ and $H_c$.
The peak of $H_a$: There are two vicinal hydrogens to $H_a$: $H_b$ and $H_c$. $H_b$ and $H_c$ are equivalent to each other but not to $H_a$. Thus, for $H_a$, n=2; (n+1) = (2+1) = 3. The multiplicity of the peak of $H_a$ is three. The peak $H_a$s three lines; from the Pascal’s triangle, it is a triplet.
The peak of $H_b$ and $H_c$: There is only one vicinal hydrogen to $H_b$ and $H_c$: $H_a$. $H_a$ is not equivalent to $H_b$ and $H_c$. Thus, for $H_b$ and $H_c$, n=1; (n+1) = (1+1) = 2. The multiplicity of the peak of $H_b$ and $H_c$ is two. The peak $H_a$s two lines, from the Pascal’s triangle, it is a doublet.
To determine the multiplicity of a peak of a nucleus coupled to more than one set of equivalent nuclei, apply the (n+1) Rule independently to each other.
eg:
There are three set of equivalent hydrogen nuclei in 3:
• Set 1: $H_a$
• Set 2: $H_b$
• Set 3: $H_c$
peak of $H_a$:
multiplicity of the peak of $H_a = 2 \times 2 = 4$. To determine the splitting pattern of the peak of $H_a$, use the Pascal’s triangle, based on the observation that, for alkenyl hydrogens, $J_{cis} > J_{gem}$.
The peak of $H_a$ is a doublet of a doublet.
peak of $H_b$:
multiplicity of the peak of $H_b = 2 \times 2 = 4$. To determine the splitting pattern of the peak of $H_b$, use the Pascal’s triangle, based on the observation that, for alkenyl hydrogens, $J_{trans} > J_{gem}$.
The peak of $H_b$ is a doublet of a doublet.
peak of $H_c$:
multiplicity of the peak of $H_c = 2 \times 2 = 4$. To determine the splitting pattern of the peak of $H_c$, use the Pascal’s triangle based on the observation that, for alkenyl hydrogens, $J_{trans} > J_{cis}$.
The peak of $H_c$ is a doublet of a doublet.
14.09: Second-Order Spectra Can Be Calculated Exactly Using the Variational Method
under development
14.E: Nuclear Magnetic Resonance Spectroscopy (Exercises)
These are homework exercises to accompany Chapter 14 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Q14.1
Write the equation for a magnetic dipole in the angular momentum form starting from $\mu =\frac{q(\textbf{r} \space x\space \textbf{v})}{2}$.
S14.1
$\mu$ can be expressed in terms of angular momentum by using the fact $\textbf{L}=\textbf{r}$ x $\textbf{p}$ and that $\textbf{p}=\textit{m}\textbf{v}$.
By substituting $\textbf{p}=\textit{m}\textbf{v}$ in to $\textbf{L}$ gives :
$\textbf{L}=\textbf{r}$ x $\textit{m}\textbf{v}$
by dividing out the $\text{m}$ becuse it is a scalar $\mu$ becomes:
$\mu =\frac{q}{2\textit{m}}\text{L}$.
Q14.3
Show that the frequency from $\nu_{1\rightarrow3}$ given in Table 14.6 reduces to Equation 14.66 when $J\ll \nu_{0}(\sigma_{1} - \sigma_{2})$
S14.35
$\nu_{1\rightarrow3} = \dfrac{ \nu_{0}}{2} (2 - \sigma_{1} - \sigma_{2}) - \frac{J}{2} + \frac{1}{2}\left[\nu^2_{0}(\sigma_{1} - \sigma_{2})^2 + J^2 \right]^\frac{1}{2} \nonumber$
$=\dfrac{ \nu_{0}}{2} (2 - \sigma_{1} - \sigma_{2}) - \frac{J}{2} + \dfrac{\nu_{0}(\sigma_{1} - \sigma_{2})}{2} \left[1 + \frac{J^2}{\nu^2_{0}(\sigma_{1} - \sigma_{2})^2} \right]^\frac{1}{2} \nonumber$
Since $J\ll \nu_{0}(\sigma_{1} - \sigma_{2})$, we can use a Taylor expansion and keeping only the terms that are linear in J gives
$\dfrac{ \nu_{0}}{2} (2 - \sigma_{1} - \sigma_{2}) - \frac{J}{2} + \dfrac{\nu_{0}(\sigma_{1} - \sigma_{2})}{2} - O(J^2)\nonumber$
$= \nu_{0}(1-\sigma_{2}) - \frac{J}{2}\nonumber$
Q14.18
Nuclear spin operators $I_x , I_y, I_z$ all obey the commutation relations, that
$[I_x, I_y] = ihI_z$
$[I_y, I_z] = ihI_x$
$[I_z, I_x] = ihI_y$
Show that
$I_z I_+ = I_+ I_z + h I_+ \hspace{1in} and \hspace{1in} I_z I_- = I_- I_z - h I_-$
S14.18
We know that the commutation relations are as follows:
$I_x I_y - I_y I_x = ihI_z$
$I_y I_z - I_z I_y = ihI_x$
$I_z I_x - I_x I_z = ihI_y$
Therefore,
$I_z I_+ = I_z(I_x + i I_y) = ihI_y +I_x I_z + i I_z I_y = ih I_y + I_x I_z + i(I_y I_z - ihI_x) = I_x I_z + i I_y I_z + hI_x +ihI_y = I_+ I_z + h I_+$
and
$I_z I_- = I_z(I_x - i I_y) = ihI_y +I_x I_z - i I_z I_y = ih I_y + I_x I_z - i(I_y I_z - ihI_x) = I_x I_z - i I_y I_z - hI_x +ihI_y = I_- I_z - h I_-$
Q14.19
Given:
$\hat{I}_{+}\hat{I}_{-} = \hat{I}_{x}^{2}+ i \hat{I}_{x} \hat{I}_{y} -i \hat{I}_{x} \hat{I}_{y}+ \hat{I}_{y}^{2}\nonumber$
and
$\hat{I}^{2} = \hat{I}^{2}_{x}+\hat{I}_{y}^{2}+\hat{I}_{z}^{2}\nonumber$
Show:
$\hat{I}_{+}\hat{I}_{-} = \hat{I}^{2}-\hat{I}_{z}^{2}+\hbar\hat{I}_{z}\nonumber$
and
$\hat{I}_{-}\hat{I}_{+} = \hat{I}^{2}-\hat{I}_{z}^{2}-\hbar\hat{I}_{z}\nonumber$
S14.19
$\hat{I}^{2} -\hat{I}_{z}^{2} = \hat{I}_{x}^{2}+\hat{I}_{y}^{2}$, so
$\hat{I}_{+}\hat{I}_{-} = \hat{I}^{2}-\hat{I}_{z}^{2} + i \hat{I}_{y}\hat{I}_{x} - i \hat{I}_{x}\hat{I}_{y}\nonumber$
$=\hat{I}^{2} - \hat{I}_{z}^{2} -i ( i \hbar \hat{I}_{z}) \nonumber$
$=\hat{I}^{2} - \hat{I}_{z}^{2} +\hbar \hat{I}_{z} \nonumber$
and
$\hat{I}_{-}\hat{I}_{+} = \hat{I}^{2}_{x}+\hat{I}_{y}^{2} + i \hat{I}_{x}\hat{I}_{y} - i \hat{I}_{y}\hat{I}_{x}\nonumber$
$=\hat{I}^{2} - \hat{I}_{z}^{2} +i ( i \hbar \hat{I}_{z}) \nonumber$
$=\hat{I}^{2} - \hat{I}_{z}^{2} -\hbar \hat{I}_{z} \nonumber$
Q14.20
Using
$\hat{I}_z\hat{I}_+ = \hat{I}_+\hat{I}_z+\hbar\hat{I}_+\nonumber$
$\hat{I}_z\beta = -\frac{\hbar}{2}\beta\nonumber$
$\hat{I}_+\beta = \hbar\alpha \nonumber$
and $c=\hbar$,
derive $\hat{I}_x\alpha$, $\hat{I}_y\alpha$, $\hat{I}_x\beta$, and $\hat{I}_y\beta$ in terms of $\alpha,\beta,\hbar$.
S14.20
We know $\hat{I}_+=\hat{I}_x+i\hat{I}_y \text{ and } \hat{I}_-= \hat{I}_x -i \hat{I}_y\nonumber$
So we can show that
$\hat{I}_+\alpha=\hat{I}_x\alpha+i\hat{I}_y\alpha=0 \text{ and } \hat{I}_-\alpha= \hat{I}_x\alpha -i \hat{I}_y\alpha = \hbar\beta\nonumber$
by adding the equations we can show that $\hat{I}_x\alpha=\frac{\hbar}{2}\beta \text{ and } \hat{I}_y\alpha=\frac{i\hbar}{2}\beta\nonumber$
if we apply the same methodology as above but with $\beta)\ we can see that $\hat{I}_x\beta=\frac{\hbar}{2}\alpha \text{ and } \hat{I}_y\beta=\frac{i\hbar}{2}\alpha\nonumber$ Q14.21 This problem shows that the proportionality constant \(c$ in
$\hat{I}_{+}\beta = c\alpha \: or \: \hat{I}_{-}\alpha = c\beta \nonumber$
is equal to $\hbar$. Start with
$\int\alpha^{*}\alpha d\tau = 1 = \dfrac{1}{c^{2}}\int (\hat{I}_{+}\beta)^{*}(\hat{I}_{+}\beta) d\tau \nonumber$
Let $\hat{I}_{+} = \hat{I}_{x} + i\hat{I}_{y}$ in the second factor in the above integral and use the fact that $\hat{I}_{x}$ and $\hat{I}_{y}$ are Hermitian to get
$\int (\hat{I}_{x}\hat{I}_{+}\beta)^{*}\beta d\tau + i \int (\hat{I}_{y}\hat{I}_{+}\beta)^{*}\beta d\tau = c^{2} \nonumber$
Now take the complex conjugate of both sides to get
$\int \beta^{*}\hat{I}_{x}\hat{I}_{+}\beta d\tau - i \int \beta^{*}\hat{I}_{y}\hat{I}_{+}\beta d\tau = c^{2} = \int \beta^{*}\hat{I}_{-}\hat{I}_{+}\beta d\tau \nonumber$
Using the given equation:
$\hat{I}_{-}\hat{I}_{+} = \hat{I}^{2} - \hat{I}_{z}^{2} - \hbar\hat{I}_{z} \nonumber$
Show that:
$c^{2} = \int \beta^{*}\hat{I}_{-}\hat{I}_{+}\beta d\tau = \int \beta^{*}(\hat{I}^{2} - \hat{I}_{z}^{2} - \hbar\hat{I}_{z})\beta d\tau = \int \beta^{*}\Big(\dfrac{3}{4}\hbar^{2} - \dfrac{1}{4}\hbar^{2} + \dfrac{\hbar^{2}}{2}\Big) \beta d\tau = \hbar^{2} \nonumber$
or that $c = \hbar$
S14.21
Recall that for a Hermitian operator $\hat{A}$,
$\int f^{*}(x)\hat{A}g(x)dx = \int g(x)\hat{A}^{*}f^{*}(x)dx \nonumber$
Begin with the expression
$\int\alpha^{*}\alpha d\tau = 1 = \dfrac{1}{c^{2}}\int (\hat{I}_{+}\beta)^{*}(\hat{I}_{+}\beta) d\tau \nonumber$
Solving for $c^{2}$ gives
$c^{2} = \int (\hat{I}_{+}\beta)^{*}(\hat{I}_{+}\beta)d\tau = \int (\hat{I}_{+}\beta)^{*}(\hat{I}_{x}\beta+ i\hat{I}_{y}\beta)d\tau = \int (\hat{I}_{+}\beta)^{*}( \hat{I}_{+}\beta d\tau + i \int (\hat{I}_{+}\beta)^{*}\hat{I}_{y}\beta d\tau \nonumber$
We can use the fact that $\hat{I}_{x}$ and $\hat{I}_{y}$ are Hermitian to write this as
$c^{2} = \int \beta\hat{I}_{+}^{*}(\hat{I}_{+}\beta)^{*}d\tau + i \int \beta\hat{I}_{y}^{*}(\hat{I}_{+}\beta)^{*}d\tau \nonumber$
Take the complex conjugate of both sides of the last equation to find
$c^{2} = \int \beta^{*}\hat{I}_{x}\hat{I}_{+}\beta d\tau - i \int \beta^{*}\hat{I}_{y}\hat{I}_{+}\beta d\tau = \int \beta^{*}(\hat{I}_{x}-i\hat{I}_{y})\hat{I}_{+}\beta d\tau = \int \beta^{*}\hat{I}_{-}\hat{I}_{+}\beta d\tau \nonumber$
Substituting $\hat{I}_{-}\hat{I}_{+} = \hat{I}^{2} - \hat{I}_{z}^{2} - \hbar\hat{I}_{z}$, we obtain
$c^{2} = \int \beta^{*}\hat{I}_{-}\hat{I}_{+}\beta d\tau = \int \beta^{*}(\hat{I}^{2} - \hat{I}_{z}^{2} - \hbar\hat{I}_{z})\beta d\tau = \int \beta^{*}\Big(\dfrac{3 \hbar^{2} }{4} - \dfrac{1 \hbar^{2}}{4} + \dfrac{\hbar^{2}}{2}\Big) \beta d\tau = \hbar^{2} \nonumber$
where we have used the given equation to evaluate the various terms involving $\hat{I}^{2}$ and $\hat{I}_{z}$.
Taking the squareroot of both sides of the final equation proves that $c = \hbar$.
Q14.22
Show that $H_{y,11} = \dfrac{hJ_{12}}{\bar{h}^{2}}\int\int d\tau_{1}\tau_{2} \alpha$*$(1)\alpha$*$(2)\hat{I}_{y1}\hat{I}_{y2}\alpha(1)\alpha(2) = 0$.
S14.22
$H_{y,11} = \dfrac{hJ_{12}}{\bar{h}^{2}}\int\int d\tau_{1}\tau_{2} \alpha\text{*}(1)\alpha\text{*}(2)\hat{I}_{y1}\hat{I}_{y2}\alpha(1)\alpha(2)\nonumber$
$H_{y,11} = \dfrac{hJ_{12}}{\bar{h}^{2}}\int\int d\tau_{1}\tau_{2} \alpha\text{*}(1)\alpha\text{*}(2)[\hat{I}_{y1}\alpha(1)][\hat{I}_{y2}\alpha(2)] \nonumber$
$H_{y,11} = \dfrac{hJ_{12}}{\bar{h}^{2}}\int\int d\tau_{1}\tau_{2} \alpha\text{*}(1)\alpha\text{*}(2)\Big[\dfrac{i\bar{h}}{2}\beta(1)\Big]\Big[\dfrac{i\bar{h}}{2}\beta(2)\Big]\nonumber$
$= -\dfrac{hJ_{12}}{4}\int\int d\tau_{1}\tau_{2} \alpha\text{*}(1)\alpha\text{*}(2)\beta(1)\beta(2) = 0\nonumber$
Orthogonality of spin functions is used so that the equation equates to zero.
Q14-23 Nice work! Correct
The energy levels of a two-spin system can be calculated using first-order perturbation theory. Show this for the first energy level.
S14-23
$E_{j}=E_{j}^{(0)} + \int{d\tau_{1}d\tau_{2}\psi_{j}^{*}H^{(1)}\psi_{j}}$
$H^{(0)}\psi_{j} = E_{j}^{(0)}\psi_{j}$
$E_{j}= E_{j}^{(0)} + H_{xjj}^{(1)} + H_{yjj}^{(1)} + H_{zjj}^{(1)}$
The unperturbed first-order energy is calculated using $I_{zj}\alpha(j) = \frac{\hbar}{2}\alpha(j)$
$H^{(0)}\psi_{1} = H^{(0)}\alpha(1)\alpha(2)$
$=-\gamma B_{0}(1-\sigma_{1})I_{z1}\alpha(1)\alpha(2) - \gamma B_{0}(1-\sigma_{2})I_{z2}\alpha(1)\alpha(2)$
$E_{1}^{(0)}\alpha(1)\alpha(2) = E_{1}^{(0)}\psi_{1}$
$E_{1}^{(0)} = -\hbar\gamma B_{0}(1-\frac{\sigma_{1}+\sigma_{2}}{2})$
The first-order correction is defined as $H_{ii}=\frac{hJ_{12}}{\hbar^{2}}\int d\tau_{1}d\tau_{2}\psi_{i}^{*}I_{1}I_{2}\psi_{i}$
with the x and y terms in I1 and I2 not contributing to the first-order energy.
Considering the unperturbed wave function for a two-spin system $\psi_{1}=\alpha(1)\alpha(2)$
$I_{z1}I_{z2}\alpha(1)\alpha(2)=\frac{\hbar^{2}}{4}\alpha(1)\alpha(2)$
Furthermore, the perturbation to the first-order energy becomes $H_{z,11}= \frac{hJ_{12}}{\hbar^{2}}\int d\tau_{1} d\tau_{2} \alpha^{*}(1)\alpha^{*}(2) I_{z1}I_{z2}\alpha(1)\alpha(2) = \frac{hJ_{12}}{4}$
As a result the first order energy energy can be represented by
$E_{1}= -h\nu_{0}(1-\frac{\sigma_{1}+\sigma_{2}}{2})+ \frac{hJ_{12}}{4}$.
Q14.24
Derive the frequencies associated with the allowed transitions between nuclear spin up to n = 4.
S14.24
( its good that you started from the beginning to derive the frequencies good job--RM)
$E = h\nu$
$E_{1=>2} = E_{2} - E_{1}$
$= \frac{h\nu}{2}(\sigma_{2}-\sigma_{1})-\frac{hJ_{12}}{4}+h\nu_{0}(1-\frac{\sigma_{1}+\sigma_{2}}{2})-\frac{hJ_{12}}{4}$
$=h\nu_{0}(1-\sigma_{1})-\frac{hJ_{12}}{2}$
$\nu_{1=>2}=\nu_{0}(1-\sigma_{1})-\frac{hJ_{12}}{2}$
$E_{1=>3} = E_{3} - E_{1}$
$= \frac{h\nu}{2}(\sigma_{1}-\sigma_{2})-\frac{hJ_{12}}{4}+h\nu_{0}(1-\frac{\sigma_{1}+\sigma_{2}}{2})-\frac{hJ_{12}}{4}$
$=h\nu_{0}(1-\sigma_{3})-\frac{hJ_{12}}{2}$
$\nu_{1=>3}=\nu_{0}(1-\sigma_{2})-\frac{hJ_{12}}{2}$
$E_{2=>4} = E_{4} - E_{2}$
$= h\nu(\sigma_{1}+\sigma_{2})-\frac{hJ_{12}}{4}-\frac{h\nu_{0}}{2}(\sigma_{2}-\sigma_{1})+\frac{hJ_{12}}{4}$
$=h\nu_{0}(1-\sigma_{2})+\frac{hJ_{12}}{2}$
$\nu_{2=>4}=\nu_{0}(1-\sigma_{2})+\frac{hJ_{12}}{2}$
$E_{3=>4} = E_{4} - E_{3}$
$= h\nu(1-\frac{\sigma_{1}+\sigma_{2}}{2})+\frac{hJ_{12}}{4}-\frac{h\nu_{0}}{2}(\sigma_{1}-\sigma_{2})+\frac{hJ_{12}}{4}$
$=h\nu_{0}(1-\sigma_{1})+\frac{hJ_{12}}{2}$
$\nu_{2=>4}=\nu_{0}(1-\sigma_{1})+\frac{hJ_{12}}{2}$
Q14.28
Using the Hamiltonian for an $H_{2}$ molecule with 2 nonequivalent hydrogen atoms, prove that $H_{13} = \iint d\tau_{1} d\tau_{2} \alpha^{*}(1) \alpha^{*}(2) \hat{H} \beta(1) \alpha(2) = 0$.
S14.28
The necessary Hamiltonian is:
$\hat{H} = -\gamma B_{0} (1 - \sigma_{1} ) \hat{I_{1}} - \gamma B_{0} (1 - \sigma_{2} ) \hat{I_{2}} + \dfrac{h J_{12}}{\hbar^{2}} \hat{I_{1}} \cdot \hat{I_{2}}. \nonumber$
We can then plug in the Hamiltonian and write the second half of the equation as:
$\hat{H} \beta(1) \beta(2) = -\gamma B_{0} (1 - \sigma_{1} ) (-\dfrac{\hbar}{2} ) \beta(1) \alpha(2) -\gamma \beta_{0} (1 - \sigma_{2} ) (\dfrac{\hbar}{2} ) \beta(1) \beta(2) + \dfrac{h J_{23}}{\hbar^{2}} [\dfrac{\hbar^{2}}{4} \alpha(1) \beta(2) + \dfrac{h^{2}}{4} \alpha(1) \beta(2) - \dfrac{\hbar^{2}}{4} \beta(1) \alpha(2) ] \nonumber$
$= \dfrac{\hbar}{2} \gamma B_{0} \beta(1) \alpha(2) [(1 - \sigma_{1} ) - (1 - \sigma_{2} )] + \dfrac{h J_{12}}{4} [2\alpha(1) \beta(2) - \beta(1) \alpha(2) ] \nonumber$
Doing some algebra, we get that:
$H_{13} = \iint d\tau_{1} d\tau_{2} \alpha^{*}(1) \alpha^{*}(2) \left[\dfrac{\hbar}{2} \gamma B_{0} \beta(1) \alpha(2) [(1 - \sigma_{1} ) - (1 - \sigma_{2} )] + \dfrac{h J_{12}}{4} [2\alpha(1) \beta(2) - \beta(1) \alpha(2) ]\right] \nonumber$
Because $\alpha$ and $\beta$ are orthonormal, the integral goes to zero. Therefore, we have proved that $H_{13} = 0$.
Q14.29
Prove that
$H_{44} = \iint d\tau_(1)d\tau_(2)\beta^*(1)\beta^*(2)\hat{H}\beta(1)\beta(2)\nonumber$
$= -\dfrac{1}{2} h\nu_0 (1-\sigma_1) -\dfrac{1}{2} h\nu_0 (1-\sigma_2)+\dfrac{hJ_12}{4}\nonumber$
with
$\hat{H} = -\gamma B_{0} (1 - \sigma_{1} ) \hat{I_{1}} - \gamma B_{0} (1 - \sigma_{2} ) \hat{I_{2}} + \dfrac{h J_{12}}{\hbar^{2}} \hat{I_{1}} \cdot \hat{I_{2}}. \nonumber$
S14.29
$\hat{H} \beta(1) \beta(2) = -\gamma B_{0} (1 - \sigma_{1} ) (\dfrac{\hbar}{2} ) \beta(1) \beta(2) -\gamma \beta_{0} (1 - \sigma_{2} ) (\dfrac{\hbar}{2} ) \beta(1) \beta(2) + \dfrac{h J_{12}}{\hbar^{2}} [\dfrac{\hbar^{2}}{4} \beta(1) \beta(2) - \dfrac{h^{2}}{4} \beta(1) \beta(2) + \dfrac{\hbar^{2}}{4} \alpha(1) \alpha(2) ] \nonumber$
$= \bigg[ \dfrac{-\hbar \gamma B_{0} }{2} (1 - \sigma_{1} ) - \dfrac{-\hbar \gamma B_{0} }{2} (1 - \sigma_{2} ) + \dfrac{h J_{12}}{4}] [2\alpha(1) \beta(2) - \beta(1) \alpha(2) \bigg] \beta(1) \beta(2)\nonumber$
Calculated with matlab,
$H_{44} = \iint d\tau_{1} d\tau_{2} \bigg[ \dfrac{-\hbar \gamma B_{0} }{2} (1 - \sigma_{1} ) - \dfrac{-\hbar \gamma B_{0} }{2} (1 - \sigma_{2} ) + \dfrac{h J_{12}}{4}] [2\alpha(1) \beta(2) - \beta(1) \alpha(2) \bigg] \beta(1) \beta(2) \nonumber$
$= \dfrac{1}{2}h\nu_0(1-sigma_1)-\dfrac{1}{2}h\nu_0(1-\sigma_2)+h\dfrac{J_{12}}{4}\nonumber$
Q14.30
Using Equation 14.58, prove that
$H_{44} = -\dfrac{1}{2} hv_0 (1-\sigma_1) + \dfrac{1}{2} h_v0 (1-\sigma_2) + \dfrac{hJ_{12}}{4}$
S14.30
To find this matrix element, you must complete the integral given by $\langle \psi_4| H | \psi_4 \rangle$. By using the relationships in Table 14.4 to evaluate parts of the integral, Equation 14.45 and Equation 14.58, algebra, and calculus, this integral can be solved to give the final answer above.
Q14.31
Given the following matrix, expand the determinants to solve for the energies:
$\begin{pmatrix}\alpha-E&\beta&0&0&0&0 \ \beta&\alpha-E&\beta&0&0&0\ 0 &\beta&\alpha-E&\beta&0&0 \ 0&0&\beta&\alpha-E&\beta&0\ 0&0&0&\beta&\alpha-E&\beta\ 0&0&0&0&\beta&\alpha-E&\end{pmatrix}\nonumber$
S14.31
$x_{i} = \frac{\alpha E}{\beta}\nonumber$
$E_{j}=\alpha-\beta x_{j}\nonumber$
$x_{j}=-2cos(\frac{j\pi}{n_{c}+1})\nonumber$
$E_{j}=\alpha+2\beta*cos(\frac{j\pi}{n_{c}}+1)\nonumber$
$E_{1}=\alpha+2\beta*cos(\frac{\pi}{6+1})=\alpha+1.8\beta\nonumber$
$E_{2}=\alpha+2\beta*cos(\frac{2\pi}{6+1})=\alpha+1.24\beta\nonumber$
$E_{3}=\alpha+2\beta*cos(\frac{3\pi}{6+1})=\alpha+0.44\beta\nonumber$
$E_{4}=\alpha+2\beta*cos(\frac{4\pi}{6+1})=\alpha-0.44\beta\nonumber$
$E_{5}=\alpha+2\beta*cos(\frac{5\pi}{6+1})=a-1.25\beta\nonumber$
$E_{6}=\alpha+2\beta*cos(\frac{6\pi}{6+1})=\alpha-1.8\beta\nonumber$
Q14.33
Show that a two-spin system with $J=0$ consists of only two peaks with frequencies $v_0$$(1-\sigma_1$) and $v_0$$(1-\sigma_2$).
S14.33
The resonance frequencies is given by
$v_1\to_2$ = $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) - $\frac{J}{2}$-$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+J^2]^{\frac{1}{2}}$
with J = 0, this leads to
$v_1\to_2$ = $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) - $\frac{0}{2}$-$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+0^2]^{\frac{1}{2}}$
= $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) -$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2]^{\frac{1}{2}}$
= $v_0$$(1-\sigma_1$)
Similarly, for the other transitions:
$v_1\to_3$ = $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) - $\frac{J}{2}$+$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+J^2]^{\frac{1}{2}}$
= $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) - $\frac{0}{2}$+$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+0^2]^{\frac{1}{2}}$
= $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) +$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2]^{\frac{1}{2}}$
= $v_0$$(1-\sigma_2$)
$v_2\to_4$ = $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) + $\frac{J}{2}$+$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+J^2]^{\frac{1}{2}}$
= $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) + $\frac{0}{2}$+$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+0^2]^{\frac{1}{2}}$
= $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) +$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2]^{\frac{1}{2}}$
= $v_0$$(1-\sigma_2$)
$v_3\to_4$ = $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) + $\frac{J}{2}$-$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+J^2]^{\frac{1}{2}}$
= $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) + $\frac{0}{2}$-$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2+0^2]^{\frac{1}{2}}$
= $\frac{v_0}{2}$ $(2-\sigma_1-\sigma_2$) -$\frac{1}{2}$ $[v_0^2(\sigma_1-\sigma_2)^2]^{\frac{1}{2}}$
= $v_0$$(1-\sigma_1$)
Q14.34
Show that
$\nu_{2\to4} = \frac{\nu_0}{2} (2-\sigma_1 - \sigma_2) + \frac{J}{2}) + \frac{1}{2}[\nu^2_0()\sigma_1 - \sigma_2)^2 + J^2]^{1/2} \nonumber$
for a general two-spin system.
S14.34
$E_4-E_2 = [h\nu_0 (1 - \frac{\sigma_1 + \sigma_2}{2}) + \frac{hJ}{4}] - [-\frac{hJ}{4} - \frac{h}{2} [\nu^2_0(\sigma_1 - \sigma_2)^2 + J^2]^{1/2}] \nonumber$
$= h\nu_0(1 - \frac{\sigma_1 + \sigma_2}{2}) + \frac{hJ}{2} - \frac{h}{2} [\nu^2_0 (\sigma_1 - \sigma_2)^2 +J^2]^{1/2}\nonumber$
$\frac{E_4-E_2}{2} = \frac{\nu_0}{2} (2-\sigma_1 - \sigma_2) + \frac{J}{2}) + \frac{1}{2}[\nu^2_0()\sigma_1 - \sigma_2)^2 + J^2]^{1/2} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/14%3A_Nuclear_Magnetic_Resonance_Spectroscopy/14.08%3A_The_n1_Rule_Applies_Only_to_First-Order_Spectra.txt
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The word 'laser' is an acronym for "light amplification by stimulated emission of radiation." The use of lasers in science and in society has rapidly expanded since their development in the early 1960s. Lasers provides chemists with a powerful and versatile tool for probing the nature of and dynamics of species and chemical reactions. This chapter will discuss the foundations of lasers and the interaction of their output toward understanding atomic and molecular properties. We will describe the generation of laser light from electronically excited atoms using the rate-equation model developed by Einstein. Modern laser designs and applications will then be discussed.
• 15.1: Electronically Excited Molecules can Relax by a Number of Processes
A particle in an excited electronic state will eventually relax back to its electronic ground state, but several relaxation pathways are often available. These pathways may involve a combination of radiative decay and nonradiative decay, including a change in spin state.
• 15.2: The Dynamics of Transitions can be Modeled by Rate Equations
Einstein proposed that electrons may transition between energy levels by means of absorption, spontaneous emission, and stimulated emission. In this section, we will describe the rates of these transitions, introducing the terms of spectral radiant energy density and the proportionality constants called Einstein coefficients.
• 15.3: A Two-Level System Cannot Achieve a Population Inversion
In this section, we will show that achieving population inversion in a two-level system is not very practical. Such a task would require a very strong pumping transition that would send any decaying atom back into its excited state. This would be similar to reversing the flow of water in a waterfall. It can be done but is very energy costly and inefficient. In a sense, the pumping transition would have to work against the lasing transition.
• 15.4: Population Inversion can be Achieved in a Three-Level System
The presence of a third energy level in a system allows for a population inversion to be created. Thus, a three-level system can act as a gain medium and can serve as a laser. The two possible lasing mechanisms for a three-level system will be described in this section.
• 15.5: What is Inside a Laser?
Laser light is produced by a gain medium inside the laser optical cavity. The gain medium is a collection of atoms or molecules in a gaseous, liquid, or solid form. For lasing to take place, the gain medium must be pumped into an excited state by an electric current or an intense light source, such as a flashlamp. To induce stimulated emission, the laser cavity must reflect emitted light into the gain medium, but also must allow a portion of the laser light to leave the optical cavity.
• 15.6: The Helium-Neon Laser
The He-Ne laser was the first continuous-wave (cw) laser invented. A few months after Maiman announced his invention of the pulsed ruby laser, Ali Javan and his associates W. R. Bennet and D. R. Herriott announced their creation of a cw He-Ne laser. This gas laser is a four-level laser that uses helium atoms to excite neon atoms. The atomic transitions in the neon produce the laser light. The most commonly used neon transition in these lasers produces red light at 632.8 nm.
• 15.7: Modern Applications of Laser Spectroscopy
Laser light offers valuable tools to researchers who wish to use the interaction of light with matter to interrogate atomic and molecular systems. Most laser light is characterized by its near monochromaticity (relative to light from other sources), directionality, and coherence. Those characteristics are used in modern laser spectroscopy.
• 15.E: Lasers, Laser Spectroscopy, and Photochemistry (Exercises)
Thumbnail: Six commercial lasers in operation, showing the range of different colored light beams that can be produced, from red to violet. From the top, the wavelengths of light are: 660 nm, 635 nm, 532 nm, 520 nm, 445 nm, and 405 nm. Manufactured by Q-line. (CC BY-SA 3.0 Unported; Sariling gawa via Wikipedia)
15: Lasers Laser Spectroscopy and Photochemistry
To understand how lasers work we must first describe how a molecule in an excited state can relax back to the ground state because the light emitted by a laser is influenced by these relaxation processes. The two radiative decay pathways for an excited molecule are fluorescence and phosphorescence. Fluorescence differs from phosphorescence in that the energy transition that is responsible for fluorescence does not involve a change in electron spin multiplicity. Thus fluorescence lifetimes are short (10-9 - 10-6 s). In phosphorescence, there is a change in electron spin multiplicity, resulting in a longer lifetime of the excited state (second to minutes). A description of electron spin and the differences between singlet and triplet states will clarify the difference between fluorescence and phosphorescence.
Spin Multiplicity - Singlet and Triplet Excited States
The Pauli Exclusion principle states that two electrons in an atom cannot have the same four quantum numbers ($n$, $l$, $m_l$, $m_s$). Therefore, because two electrons can occupy each orbital, these two electrons must have opposite spin states. These opposite spin states are called spin pairing. Because of this spin pairing, most molecules are diamagnetic, and are not attracted or repelled by an external magnetic or electric field. Molecules that contain unpaired electrons (free radicals) do have magnetic moments that are attracted to an external magnetic or electric field.
A molecule is said to be in a singlet state when all the electron spins are paired in the molecular electronic state, and the electronic energy levels do not split when the molecule is exposed to a magnetic field. A doublet state occurs when there is an unpaired electron that gives two possible orientations when exposed to a magnetic field and imparts different energies to the system. A singlet or a triplet state can form when one electron is excited to a higher energy level. In an excited singlet state, the promoted electron retains the spin orientation it had in the ground state (i.e., paired). In a triplet excited state, the promoted electron undergoes a change in spin, and so has the same spin orientation (parallel) to the unpaired electron that remains in the ground state orbital. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in Figure 15.1.1 . The terms singlet, doublet and triplet are derived using the equation for multiplicity, 2S+1, where S is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up (s = +1/2) or spin down (s = -1/2). If we were to calculate the multiplicity for the singlet ground state or the singlet excited state, the equation would be
$2(+1/2 + -1/2)+1 = 2(0)+1 = 1 \nonumber$
In a similar fashion, the spin multiplicity for the triplet excited state can be calculated as
$2(+1/2 + +1/2)+1 = 2(1)+1 =3 \nonumber$
which gives a triplet state as expected.
The difference between a molecule in the ground and the triplet excited state is that the molecule is diamagnetic in the ground state and paramagnetic in the triplet excited state. This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) less likely than the singlet-to-singlet transitions. For this reason, the lifetime of the triplet state is longer than the lifetime of the singlet state by a factor of roughly 104 seconds. The transition from ground to excited triplet state has a low probability of occurring, thus these absorption bands are less intense than singlet-singlet state absorption bands. However, an excited triplet state can be populated from an excited singlet state of certain molecules. Jablonski diagrams can be used to explain transitions such as this that occur in photoluminescence molecules.
Jablonski Diagrams
Aleksander Jablonski was a Polish academic who devoted his life to the study of molecular absorbance and emission of light. He developed a graphic representation that shows the possible consequences of applying photons from the visible spectrum of light to a particular molecule. These schematics are referred to as Jablonski diagrams.
A Jablonski diagram is an energy diagram, arranged with energy on a vertical axis. The energy levels can be quantitatively denoted, but most of these diagrams use energy levels schematically. The rest of the diagram is arranged into columns. Every column usually represents a specific spin multiplicity for a particular species. However, some diagrams divide energy levels within the same spin multiplicity into different columns. Within each column, horizontal lines represent eigenstates for that particular molecule. Bold horizontal lines are representations of the limits of electronic energy states. Within each electronic energy state are multiple vibrational energy states that may be coupled with the electronic state. Usually only a portion of these vibrational eigenstates are represented due to the massive number of possible vibrations in a molecule. Each of these vibrational energy states can be subdivided even further into rotational energy levels; however, typical Jablonski diagrams omit such intense levels of detail.
Through the use of straight and wavy lines, these figures show transitions between eigenstates that occur from the exposure of a molecule to a particular wavelength of light. Straight lines show the conversion between a photon of light and the energy of an electron. Wavy lines show nonradiative transitions of electrons. Within a Jablonski diagram several different pathways show how an electron may accept and then dissipate the energy from a photon. Thus, most diagrams start with arrows originating from the ground electronic state and finish with arrows returning to the ground electronic state.
The Jablonski diagram that drawn below is a partial energy diagram that represents the energy of a photoluminescent molecule in its different energy states. The lowest and darkest horizontal line represents the ground-state electronic energy of the molecule which is the singlet state labeled as $S_o$. At room temperature, a majority of the molecules are in this state. The thicker lines on the left labeled S1, S2 , and S3 represent the excited electronic states for the molecule in the singlet state. The thicker lines on the right labeled T1, T2, and T3 represent excited triplet states.
Deactivation Processes
A molecule that is excited can return to the ground state by several combinations of mechanical steps that will be described below and shown in Figure 15.1.2 .The deactivation process of fluorescence and phosphorescence involve an emission of a photon radiation as shown by the straight arrow in Figure 15.1.2 . The wiggly arrows in Figure 15.1.2 are deactivation processes without the use of radiation. The favored deactivation process is the route that is most rapid and spends the least time in the excited state. If the rate constant for relaxation is more favorable in the radiationless path, the fluorescence will be less intense or absent.
Relaxation and Fluorescence
Often, when an excited state species relaxes, giving off a photon, the wavelength of the photon is different from the one that initially led to excitation. When this happens, the photon is invariably red-shifted; its wavelength is longer than the initial one. This situation is called "fluorescence" (Figure $3$).
How can that be? Isn't energy quantized? How is the molecule suddenly losing some of the energy that the original photon brought with it? This energy discrepancy is related to the Franck-Condon principle from the previous page. When an electron is promoted to an electronic excited state, it often ends up in an excited vibrational state as well (Figure $4$). Vibrational energy, however, is not exchanged exclusively by means of photons. It can be gained or lost through molecular collisions and heat transfer.
Thus, an excited state molecule with an electron in a high-energy vibrational level within an excited electronic state might simply re-emit a photon of exactly the same wavelength as the one that was absorbed. But the excited electron is much more likely to relax into the lowest vibrational state within the excited electronic state, losing some of that initial excitation energy as heat. When the electron relaxes to this lower vibrational state, the energy gap between this excited state and the ground state is a little smaller. The photon that is emitted upon fluorescence will have lower energy and longer wavelength than a photon emitted from the original, higher vibrational level, excited state. See Figure $5$.
Figure $6$ shows the fluorescence transitions of a hypothetical diatomic molecule in which the equilibrium bond length of the ground state and the first singlet excited state are identical. In this molecule all absorptions involve transitions from the lowest vibrational level of the electronic ground state ($v^" = 0$) to various vibrational levels in the excited electronic state. Because vibrational relaxation occurs more rapidly than fluorescence, the fluorescence spectrum is composed of lines showing transitions from the lowest vibrational level of the excited state ($v^{'} = 0$) to various vibrational levels in the electronic ground state.
Just how does a molecule undergo vibrational relaxation? Vibrational energy is the energy used to lengthen or shorten bonds, or to widen or squeeze bond angles. Given a big enough molecule, some of this vibrational energy could be transferred into bond lengths and angles further away from the electronic transition. Otherwise, if the molecule is small, it may transfer some of its energy in collisions with other molecules. In molecules, as one molecule drops to a lower vibrational state, the other will hop up to a higher vibrational state with the energy it gains. In Figure $7$ below, the red molecule is in an electronic excited and vibrational state. In a collision, it transfers some of its vibrational energy to the blue molecule.
Note
There are many examples of energy being transferred this way in everyday life. In a game of pool, one billiard ball can transfer its energy to another, sending it toward the pocket. Barry Bonds can transfer a considerable amount of energy through his bat into a baseball, sending it out of the park, just as Serena Williams can send a whole lot of energy whizzing back at her sister.
Exercise $1$
How does the energy of an electronic absorption compare to other processes? To find out, you might consider the excitation of an entire mole of molecules, rather than a single molecule absorbing a single photon. Calculate the energy in kJ/mol for the following transitions.
1. absorbance at 180 nm (ultraviolet)
2. absorbance at 476 nm (blue)
3. absorbance at 645 nm (red)
Answer
One method to carry out this calculation: $\dfrac{1}{x \text{nm}} \times \dfrac{1 \, \text{nm}}{1x10^{-9} \, \text{m}} \times 6.626 x 10^{-34} \text{Js} \times \dfrac{3.000 x 10^8 \text{m}}{\text{s}} \times \dfrac{6.022 x 10^{23} \text{photons}}{ 1.000 \text{mole}} \times \dfrac{1 \text{kJ}}{1000 \text{J}}$
1. 665.0 kJ/mol
2. 251.5 kJ/mol
3. 185.6 kJ/mol
Exercise $2$
How does the energy of an excitation between vibrational states compare to that of an electronic excitation? Typically, infrared absorptions are reported in cm-1, which is simply what it looks like: the reciprocal of the wavelength in cm. Because wavelength and frequency are inversely related, wavenumbers are considered a frequency unit. Calculate the energy in kJ/mol for the following transitions.
1. absorbance at 3105 cm-1
2. absorbance at 1695 cm-1
3. absorbance at 963 cm-1
Answer
One method to carry out this calculation: $x \text{cm}^{-1} \times \dfrac{100 \, \text{cm}}{1 \, \text{m}} \times 6.626 x 10^{-34} \text{Js} \times \dfrac{3.000 x 10^8 \text{m}}{\text{s}} \times \dfrac{6.022 x 10^{23} \text{photons}}{ 1.000 \text{mole}} \times \dfrac{1 \text{kJ}}{1000 \text{J}}$
1. 37.14 kJ/mol
2. 20.27 kJ/mol
3. 11.52 kJ/mol
By comparing these answers with those from Exercise $1$, we can see that the energy difference between vibrational states is an order of magnitude less than the energy difference between electronic states.
If electrons can get to a lower energy state, and give off a little energy at a time, relaxing to lower and lower vibrational levels, do they need to give off a photon at all? Maybe they can relax all the way down to the ground state via vibrational relaxation. That is certainly the case. Given many vibrational energy levels, and an excited state that is low enough in energy so that some of its lower vibrational levels overlap with some of the higher vibrational levels of the ground state, an electron can relax from an excited electronic state back to the ground state without releasing a photon (Figure $8). This event is called a "radiationless transition", because it occurs without release of a photon. The electron simply slides over from a low vibrational state of the excited electronic state to a high vibrational state of the electronic ground state. If the electron simply keeps dropping a vibrational level at a time back to the ground state, the process is called "internal conversion". Internal conversion has an important consequence. Because the absorption of UV and visible light can result in energy transfer into vibrational states, much of the energy that is absorbed from these sources is converted into heat. That can be a good thing if you happen to be a marine iguana trying to warm up in the sun after a plunge in the icy Pacific. It can also be a tricky thing if you are a process chemist trying to scale up a photochemical reaction for commercial production of a pharmaceutical, because you have to make sure the system has adequate cooling available. Intersystem Crossing Intersystem crossing is a process that leads to the electron getting caught between the excited state and the ground state. Just as, little by little, vibrational relaxation can lead the electron back onto the ground state energy surface, it can also lead the electron into states that are intermediate in energy in which the spin multiplicity has changed. For example, suppose an organic molecule undergoes electronic excitation. Generally, organic molecules have no unpaired electrons. Their ground states are singlet states. According to one of the selection rules for electronic excitation, the excited state must also have no unpaired electrons. In other words, the spin on the electron that gets excited is the same after excitation as it was before excitation. However, that's not the lowest possible energy state for that electron. When we think about atomic orbital filling, there is a rule that governs the spin on the electrons in degenerate orbitals: in the lowest energy state, spin is maximized (Hund's rule). In other words, when we draw a picture of the valence electron configuration of nitrogen, we show nitrogen's three p electrons each in its own orbital, with their spins parallel. In Figure \(9$, the diagram on the left, with three unpaired electrons, all with parallel spins, shows a nitrogen in the quartet spin state. This is the most stable state. Having one of those spins point the other way would result in a different, higher energy spin state. The diagram on the right shows a different a different spin state, in which one pair of electrons in the p level is spin-paired, one up and one down, even though they are in different p orbitals. That would leave one electron without an opposite partner, so the nitrogen would be in a doublet spin state. The spin state on the left is lower in energy than the state on the right. That's just one of the rules of quantum mechanics (Hund's rule): maximize spin when degenerate orbitals are singly occupied.
A similar argument can be made for a molecule with the triplet state lower in energy than the singlet state, as shown in Figure $10$. Why didn't the electron get excited to the triplet state in the first place? That is a forbidden (unlikely) electronic transition. But sliding down vibrationally onto the triplet state from the singlet excited state is not, because it doesn't involve absorption or emission of a photon.
Intersystem crossing can have important consequences in reaction chemistry because it allows access to triplet states that are not normally available in many molecules. Because triplet states feature unpaired electrons, their reactivity is often typified by radical processes. That means an added suite of reactions can be accessed via this process.
Phosphorescence: A Radiationless Transition Followed by Emission
Intersystem crossing is one way a system can end up in a triplet excited state. Even though this state is lower in energy than a singlet excited state, it cannot be accessed directly via electronic excitation because that would violate the spin selection rule (\Delta S=0\). Once intersystem crossing has occurred, the process of reattaining the ground state slows down dramatically. The quick way back to ground state is to emit a photon. But because that would involve a change in spin state, it is not allowed. Realistically speaking, that means it takes a long time. By "a long time", we might mean a few seconds, several minutes, or possibly even hours. Eventually, the electron can drop back down, accompanied by the emission of a photon. This situation is called "phosphorescence" (Figure \(11).
Many plants and animals use phosphorescence as a means of signaling. Molecules that display phosphorescence are also often incorporated into toys and shirts so that they will glow in the dark.
External Conversion
Deactivation of the excited electronic state may also involve the interaction and energy transfer between the excited state and the solvent or solute in a process called external conversion. Low temperature and high viscosity lead to enhanced fluorescence because they reduce the number of collision between molecules, thus slowing down this type of deactivation process.
Absorption and Emission Rates
The table below compares the absorption and emission rates of fluorescence and phosphorescence. The rate of photon absorption is very rapid. Fluorescence emission occurs at a slower rate. Since the triplet to singlet (or reverse) is a forbidden transition, meaning it is less likely to occur than the singlet-to-singlet transition, the rate of triplet to singlet is typically slower. Therefore, phosphorescence emission requires more time than fluorescence.
Table 15.1.1 : Rates of Absorption and Emission comparison.
Process Radiative Process? Transition Timescale (sec)
Light Absorption (Excitation) yes S0 → Sn ca. 10-15 (instantaneous)
Internal Conversion no Sn → S1 10-14 to 10-11
Vibrational Relaxation no Sn* → Sn 10-12 to 10-10
Intersystem Crossing no S1 → T1 10-11 to 10-6
Fluorescence yes S1 → S0 10-9 to 10-6
Phosphorescence yes T1 → S0 10-3 to 100
Non-Radiative Decay to Ground State (Internal Conversion) no S1 → S0
T1 → S0
10-7 to 10-5
10-3 to 100
Contributors and Attributions
• Diana Wong (UCD)
• Tom Neils (Grand Rapids Community College)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/15%3A_Lasers_Laser_Spectroscopy_and_Photochemistry/15.01%3A_Electronically_Excited_Molecules_can_Relax_by_a_Number_of_Processes.txt
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The Interaction of Light with Atoms
In 1916, Albert Einstein proposed that there are three processes occurring in the formation of an atomic spectral line. The three processes are referred to as spontaneous emission, stimulated emission, and absorption (figure $1$).
For now, let's make our discussion as simple as possible, by assuming we are talking about the interaction of light with an atom that has only two energy levels. There is a collection of these atoms present, and they are in an assortment of states, some in the ground state, and some in "the" excited state. Our discussion of photon emission to this point has generally centered around spontaneous emission, the triggering of which we attributed to a perturbation in the hamiltonian that comes from (if nothing else) vacuum fluctuations.
Suppose some of these atoms spontaneously emit photons, and now the collection of atoms is not sitting in a total vacuum, but is in the presence of many photons as well. We now consider what effects these present photons can have on the atoms. The answer depends upon which state the atom is in when it encounters the photon. If the atom is in the ground state, then it can absorb the photon and go to the excited state. If the atom is in its excited state, it can't absorb the photon, but the time-varying EM fields of the photon present a nice perturbation to the electron's hamiltonian (at just the right frequency!) to induce the emission of a photon. This is called stimulated emission. The photon that induces the emission is unaffected by the interaction, and continues on its merry way. The emitted photon naturally has the same frequency as the passing photon (the energy gap assured this), but it also emerges moving in the same direction as, and in phase with the stimulating photon. This completes the acronym - Light Amplification by the Stimulated Emission of Radiation.
It turns out that if a photon happens by two of these atoms, the first in the ground state and the second in the excited state, the probability that it will be absorbed by the first is equal to the probability that it will stimulate emission in the other – neither of these results is preferred over the other. Now consider a collection of these atoms in a state of some fixed total energy. This total energy comes from the number of atoms in the ground state multiplied by the energy of that state, plus the number of atoms in the excited state multiplied by the number of atoms in that state. As we will discuss in a future chapter, the populations of these two states subject to the constraint of the total energy can be shown to be a function that depends upon the energy difference of the two states and the temperature of the collection of atoms. Specifically, if N1 is the number of atoms in the ground state, N2 is the number of atoms in the excited state, and the energy difference is ΔE, then the following relation holds:
$N_1 = N_2 \, e^{\dfrac{\Delta E}{k_BT}} \nonumber$
where T is the temperature and $k_B$ is Boltzmann's constant. Clearly N1 is greater than N2, but how much greater? We can do a very quick calculation of this using an approximation that at room temperature, the constant $k_BT$ is about 140 eV. Let's assume that the energy difference between two energy states of an atom is on the order of 1eV. Then the number of ground state atoms would exceed the number of excited state atoms by a factor of e40 = 2.4×1017!
Thus, a photon passing through this collection has an equal probability of being absorbed or stimulating emission if it happens by an atom prepared for one of those actions. But there are so many more atoms available to absorb the photon, that this has a much greater chance of happening. While waiting for a photon to stimulate emission, an atom in an excited state will impatiently emit a photon spontaneously, which will then be much more likely to be absorbed than stimulate emission, and things continue like this, with some atoms spontaneously sending out photons while others absorb them... no laser.
Einstein Coefficients
Einstein proposed a separate rate equation for each of the three processes described above. The rate of each process is proportional to the number of particles in either the ground state (absorption) or the excited state (emission). Each rate is also proportional to the spectral radiant energy density, $\rho_{\nu}$, which is a measure of the radiant energy density ($\rho$) per unit frequency of the light associated with the transition. Thus, $\rho_{\nu} = \dfrac{d\rho}{d\nu}$, and has units of $\dfrac{J·s}{m^3}$. With each process there is also an associated Einstein coefficient, which is a proportionality constant.
Photon absorption
Absorption is the process by which a photon is absorbed by the atom, causing an electron to jump from a lower energy level to a higher one (figure $\PageIndex{1a}$). The process is described by the Einstein coefficient $B_{12} \left(\dfrac{m^3}{ J· s^2}\right)$, which gives the probability per unit time per unit spectral radiance of the radiation field that an electron in state 1 with energy $E_1$ will absorb a photon with an energy $E_2 − E_1 = h\nu_{12}$ and jump to state 2 with energy $E_2$. If $N_1(t)$ is the number density of atoms in state 1, then the change in the number density of atoms in state 1 per unit time due to absorption will be
$-\dfrac {dN_1(t)}{dt}= B_{12}\rho_{\nu} (\nu_{12}) N_{1}(t)$
If only absorption occurred, the increase of the excited state population as a function of time is equal to the decrease of the ground state population as a function of time.
Spontaneous emission
Spontaneous emission is the process by which an electron "spontaneously" (i.e. without any outside influence) decays from a higher energy level to a lower one (figure $\PageIndex{1b}$). The process is described by the Einstein coefficient $A_{21} (s^{-1})$, which gives the probability per unit time that an electron in state 2 with energy $E_2$ will decay spontaneously to state 1 with energy $E_1$, emitting a photon with an energy $E_2 - E_1 = h\nu_{21}$. Due to the energy-time uncertainty principle, the transition actually produces photons within a narrow range of frequencies called the spectral linewidth. If $N_2(t)$ is the number density of atoms in state 2, then the change in the number density of atoms in state 2 per unit time due to spontaneous emission will be
$-\dfrac {dN_2(t)}{dt}= A_{21}N_{2}(t)$
The same process results in increasing of the population of the state 1:
$\dfrac {dN_1(t)}{dt}= A_{21}N_{2}(t)$
Stimulated emission
Stimulated emission (also known as induced emission) is the process by which an electron is induced to jump from a higher energy level to a lower one by the presence of electromagnetic radiation at (or near) the frequency of the transition (figure $\PageIndex{1c}$). From the thermodynamic viewpoint, this process must be regarded as negative absorption. The process is described by the Einstein coefficient $B_{21} \left(\dfrac{m^3}{ J· s^2}\right)$, which gives the probability per unit time per unit spectral radiance of the radiation field that an electron in state 2 with energy $E_2$will decay to state 1 with energy $E_1$, emitting a photon with an energy $E_2 - E_1 = h\nu_{21}$. The change in the number density of atoms in state 1 per unit time due to induced emission will be
$\dfrac {dN_1(t)}{dt}= B_{21}\rho_{\nu} (\nu_{21}) N_{2}(t)$
The Relationship between A21, B12, and B21
In 1900, Max Planck derived a formula for the energy density per unit bandwidth of a blackbody radiator by making the assumption that only discrete energies are allowed. His work agreed with known experimental data, and it is one of the fundamental ideas of quantum mechanics. More specifically, the spectral energy density per unit bandwidth, $\rho_{\nu}(\nu_{12})$ in units $\dfrac{J⋅s}{m^3}$, is given by
$\rho_{\nu}(\nu_{12}) = \dfrac{8\pi h}{c^3} \dfrac{\nu_{12}^3}{e^{(h\nu_{12})/(k_BT)} -1} \label{15.2.1}$
In a given sample of atoms exposed to radiation of the correct frequency all three processes will be occurring simultaneously. Thus the rate of change of the populations of the two states will be
$-\dfrac {dN_1(t)}{dt} = \dfrac {dN_2(t)}{dt}= B_{12}\rho_{\nu} (\nu_{12}) N_{1}(t) - A_{21}N_{2}(t) - B_{21}\rho_{\nu} (\nu_{21}) N_{2}(t) \label{15.2.2}$
If we assume that the two energy states are in thermal equilibrium, then $N_1$ and $N_2$ are constant, and thus
$-\dfrac {dN_1(t)}{dt} = \dfrac {dN_2(t)}{dt}= B_{12}\rho_{\nu} (\nu_{12}) N_{1}(t) - A_{21}N_{2}(t) - B_{21}\rho_{\nu} (\nu_{21}) N_{2}(t) =0 \label{15.2.3}$
Upon solving for $\rho_{\nu}(\nu_{12})$, we get
$\rho_{\nu}(\nu_{12}) = \dfrac{A_{21}}{(N_1(t)/N_2(t))B_{12} - B_{21}} \label{15.2.4}$
Recall that in this equation $N_1/N_2$ represents the electron density in the lower energy state divided by the electron density in the upper state at equilibrium. This quantity is a function of temperature. Assuming many allowed energy states, the number of occupied states decreases exponentially with temperature, as per Boltzmann statistics.
$N_2/N_1 =\dfrac{g_2}{g_1} e^{(-h\nu_{12})/(k_BT)} \label{15.2.5}$
The quantity $\dfrac{g_2}{g_1}$ represents the degeneracy level, which is the number of allowed electrons in the upper state over the number of allowed electrons in the lower state. In this expression, $g_1$ and $g2$ are unitless measures of the number of ways electrons can occupy an energy states. Upon rearrangement, equation $\ref{15.2.5}$ becomes
$N_1/N_2 =\dfrac{g_1}{g_2} e^{(h\nu_{12})/(k_BT)} \label{15.2.6}$
Equations $\ref{15.2.6}$ and $\ref{15.2.4}$ can be combined to get
$\rho_{\nu}(\nu_{12}) = \dfrac{A_{21}}{\left(\dfrac{g_1}{g_2} e^{(h\nu_{12})/(k_BT)}\right)B_{12} - B_{21}} \label{15.2.7}$
which can be rearranged to
$\rho_{\nu}(\nu_{12}) = \dfrac{\dfrac{A_{21}}{B_{21}}}{\left(\dfrac{g_1B_{12}}{g_2B_{21}} e^{(h\nu_{12})/(k_BT)}\right) - 1} \label{15.2.8}$
One expression for the energy density per unit bandwidth of this system is given by Equation $\ref{15.2.1}$. Equation $\ref{15.2.8}$ gives a second expression for the energy density per unit bandwidth, and it was found by considering the relative rates of absorption, spontaneous emission, and stimulated emission. These equations can be combined to relate the rates of the different processes.
$\dfrac{8\pi h}{c^3} \dfrac{\nu_{12}^3}{e^{(h\nu_{12})/(k_BT)} -1} = \dfrac{\dfrac{A_{21}}{B_{21}}}{\left(\dfrac{g_1B_{12}}{g_2B_{21}} e^{(h\nu_{12})/(k_BT)}\right) - 1} \label{15.2.9}$
The above equation is true for the conditions
$\dfrac{A_{21}}{B_{21}} = \dfrac{8\pi h \nu_{12}^3}{c^3} \label{15.2.10}$
and
$\dfrac{g_1B_{12}}{g_2B_{21}} = 1 \label{15.2.11}$
If we know one of the Einstein coefficients, we can quickly calculate the other two Einstein coefficients using equations $\ref{15.2.10}$ and $\ref{15.2.11}$.
These equations also provide further insight into the operation of lasers and other devices based on stimulated emission. The overall nonequilibrium upper state population rate is given by
$\dfrac {dN_2(t)}{dt}= - A_{21}N_{2} - B_{21}\rho_{\nu} (\nu_{21}) N_{2} + B_{21}\dfrac{g_2}{g_1}\rho_{\nu} (\nu_{12}) N_{1} \label{15.2.12}$
which can be simplified with some algebra.
$\dfrac {dN_2(t)}{dt}= - A_{21}N_{2} - B_{21}\rho_{\nu} (\nu_{12}) \left(N_2 - \dfrac{g_2}{g_1}N_1\right) \label{15.2.13}$
The term in parenthesis is the net upper state population. Optical amplification and lasing can only occur when the term in parenthesis is positive. The condition
$N_2 - \dfrac{g_2}{g_1}N_1 > 0\label{15.2.14}$
is called a population inversion. It only occurs when enough energy is being supplied to the system, by optical, electrical, or thermal means, so that there are more electrons in the upper energy level than the lower energy level. This situation cannot be achieved for a two-level system, as will be shown in the next section. However, in section 15.4, it will be shown that this situation can be achieved if the system has more than two levels.
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Lasing in Two-Level Systems
For the sake of our studies, let's first consider a laser medium whose atoms have only two energy states: a ground state and one excited state. In such an idealized atom the only possible transitions are excitation from the ground state to the excited state and de-excitation from the excited state back into ground state. Could such an atom be used to make a laser?
There are several important conditions that our laser must satisfy. First of all, the light that it produces must be coherent. That is to say, it must emit photons that are in-phase with one another. Secondly, it should emit monochromatic light, i.e. photons of the same frequency (or wavelength). Thirdly, it would be desirable if our laser's output were collimated, producing a sharply defined "pencil-like" beam of light (this is not crucial, but clearly a desirable condition). Lastly, it would also be desirable for our laser to be efficient, i.e. the higher the ratio of output energy - to - input energy, the better.
Let us begin by examining the requirements for our first condition for lasing, coherence. This condition is satisfied only when the lasing transition occurs through stimulated emission. As we have already seen, stimulated emission produces identical photons that are of equal energy and phase and travel in the same direction. But for stimulated emission to take place a "passer-by" photon whose energy is just equal to the de-excitation energy must approach the excited atom before it de-excites via spontaneous emission. Typically, a photon emitted by the spontaneous emission serves as the seed to trigger a collection of stimulated emissions. Still, if the lifetime of the excited state is too short, then there will not be enough excited atoms around to undergo stimulated emission. So, the first criteria that we need to satisfy is that the upper lasing state must have a relatively long lifetime, otherwise known as a meta-stable state, with typical lifetimes in the milliseconds range. In addition to the requirement of a long lifetime, we need to ensure that the likelihood of absorption of the "passer-by" photons is minimized. This likelihood is directly related to the ratio of the atoms in their ground state versus those in the excited state. The smaller this ratio, the more likely that the "passer-by" photon will cause a stimulated emission rather than get absorbed. So, to satisfy this requirement, we need to produce a population inversion: create more atoms in the excited state than those in the ground state.
Another way of stating the above set of conditions is that the rate of absorption must be less than the rate of stimulated emission
$\underbrace{B_{21}\rho_{\nu} (\nu_{21}) N_{2}}_{\text{rate of stimulated emission}} > \underbrace{B_{12}\rho_{\nu} (\nu_{12}) N_{1}}_{\text{rate of absorption}} \label{15.3.1}$
As shown in the previous section, $B_{21} = B_{12}$ and $\rho_{\nu}(\nu_{12}) = \rho_{\nu} (\nu_{21})$, thus Equation $\ref{15.3.1}$ becomes
$N_2 > N_1. \nonumber$
Hence, if $N_2 > N_1$ then there is a population inversion.
Achieving population inversion in a two-level atom is not very practical. Such a task would require a very strong pumping transition that would send any decaying atom back into its excited state. This would be similar to reversing the flow of water in a waterfall. It can be done but is very energy costly and inefficient. In a sense, the pumping transition would have to work against the lasing transition.
It is clear, from figure $1$, that in the two-level atom the pump is, in a way, the laser itself! Such a two-level laser would work only in jolts. That is to say, once the population inversion is achieved the laser would lase. But immediately it would end up with more atoms in the lower level. Such two-level lasers involve a more complicated process. We will see, in later material, examples of these in the context of excimer lasers, which are pulsed lasers. For continuous lasing action, we need to consider other possibilities, such as a three-level atom.
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Optical pumping will at most achieve only an equal population in a two-level system. This is because the probabilities for raising an electron to the upper level and inducing the decay of an electron to the lower level (stimulated emission) are exactly the same! In other words, when both levels are equally populated, the numbers of electrons "going up" and "down" will be the same, so you cannot achieve population inversion which is required for lasers. The solution is to involve a third, metastable level of intermediate energy. The pumping will occur between the two levels with the greatest energy difference, creating a highly populated upper energy level. Two lasing processes are possible once this situation has been obtained. Lasing occurs between the upper level and the intermediate level if the electrons in the upper energy level decay slowly to the intermediate, metastable level, and if the electrons in the intermediate level decay rapidly to the ground state. However, if the electrons in the upper energy level quickly decay into the intermediate level, and the transition from the intermediate level to the ground level is slow, then lasing occurs between the intermediate level and the ground state. In both cases, the lasing frequency has a frequency different than that of the pumping frequency between the ground level and the upper level, so the pumping is off-resonant to the laser transition and will not trigger stimulated emission.
Lasing in Three-Level Systems
The first laser that was demonstrated to operate was a three-level laser, known as Maiman's ruby laser.
Figure $1$ shows the outside and inside of a three-level laser. A pump causes an excitation from the ground state to the second excited state. This state is a rather short-lived state so that the atom quickly decays into the first excited level. (Atoms in the second excited state may also decay directly back to the ground state, but these atoms can be pumped back to the second excited state again.) The first excited state is a long-lived (i.e. metastable) state which allows the atom to "wait" for the "passer-by" photon while building up a large population of atoms in this state. The lasing transition, in this laser, is due to the decay of the atom from this first excited metastable state to the ground state. If the number of atoms in the ground state exceeds the number of atoms that are pumped into the excited state, then there is a high likelihood that the "lasing photon" will be absorbed and we will not get sustained laser light. The fact that the lower lasing transition is the ground state makes it rather difficult to achieve efficient population inversion. In a ruby laser, this task is accomplished by providing the ruby crystal with a very strong pulsating light source, called a flash lamp. The flash lamp produces a very strong pulse of light that is designed to excite the atoms from their ground state into a short-lived upper level. In this way, the ground state is depopulated and population inversion is achieved until a pulse of laser light is emitted. In the ruby laser, the flash lamp light lasts for about 1/1000 of a second (1 ms) and can be repeated about every second. The duration of the laser pulse is shorter than this, typically 0.1 ms. In some pulsed lasers, the pulse duration can be tailored using special methods to be much shorter than this, down to about 10 fs (where 1 fs = 10-15 s or one thousandth of a millionth of a millionth of a second). So, the output of a three-level laser is not continuous but consists of pulses of laser light.
Transition Rates in Three-level Systems
In a three-level system, there are nine (9) events that can occur (Figure $2$):
1. absorption from level 1 to level 3, as described by $\rho_{\nu}(\nu_{13})B_{13}N_1$
2. spontaneous emission from level 3 to level 1, as described by $A_{31}N_3$
3. stimulated emission from level 3 to level 1, as described by $\rho_{\nu}(\nu_{31})B_{31}N_3$
4. spontaneous emission from level 3 to level 2, as described by $A_{32}N_3$
5. stimulated emission from level 3 to level 2 (possible lasing), as described by $\rho_{\nu}(\nu_{32})B_{32}N_3$
6. absorption from level 2 to level 3, as described by $\rho_{\nu}(\nu_{23})B_{23}N_2$
7. absorption from level 1 to level 2, as described by $\rho_{\nu}(\nu_{12})B_{12}N_1$
8. spontaneous emission from level 2 to level 1, as described by $A_{21}N_2$
9. stimulated emission from level 2 to level 1 (possible lasing), as described by $\rho_{\nu}(\nu_{21})B_{21}N_2$
In most three-level lasers, neither the spontaneous emission from level 3 to level 1, nor the stimulated emission from level 3 to level 1 occur to any significant extent, and so these events are ignored in terms of the kinetics of the lasing process. It is also true that there is no excitation source to cause absorption from level 1 to level 2 other than the possible reabsorption of light emitted from a level 2 to level 1 transition, but this event is unlikely because most atoms in the ground state are pumped up to level 3. Because the pump will be used to drive the absorption from level 1 to level 3, this event is essentially under the control of the experimenter. This leaves a maximum of five events that we need to take into consideration when studying the lasing process.
A Ruby Laser
In a three-level system ruby laser such as that described above, the flash lamp will serve as the pump to repeatedly excite most of the atoms from the ground state to the second excited state. In this laser, lasing occurs due to the stimulated emission from level 2 to level 1 (Figure $3$.)
Because the lasing will occur from the intermediate, metastable energy level 2, we will focus on the five events that affect its population:
1. absorption from level 2 to level 3, as described by $\rho_{\nu}(\nu_{32})B_{32}N_2$
2. spontaneous emission from level 3 to level 2, as described by $A_{32}N_3$
3. stimulated emission from level 3 to level 2, as described by $\rho_{\nu}(\nu_{32})B_{32}N_3$
4. spontaneous emission from level 2 to level 1, as described by $A_{21}N_2$
5. stimulated emission from level 2 to level 1 (lasing), as described by $\rho_{\nu}(\nu_{21})B_{21}N_2$
In this system, the rate of spontaneous emission from level 3 to level 2 is much more rapid than the rate of absorption from level 2 to level 3, and also much more rapid than the rate of stimulated emission from level 3 to level 2. Thus, any electrons pumped into the upper level are most likely to transition between level 2 and level 3 by spontaneous emission. This means that $N_2$ is greater than $N_1$ and thus, population inversion between states 2 and 1 can be achieved. A system of atoms that are able to attain such a population inversion is called a gain medium, and can lase.
A Copper Vapor Laser
In this three-level system, the flash lamp will again, serve as the pump to repeatedly excite most of the atoms from the ground state to the second excited state. However, in this laser, lasing occurs due to the stimulated emission from level 3 to level 2 (Figure $4$).
Because the lasing will occur from the higher energy excited state, we focus on the four events that affect the population ratio between level 3 and level 2:
1. absorption from level 2 to level 3, as described by $\rho_{\nu}(\nu_{32})B_{32}N_2$
2. spontaneous emission from level 3 to level 2, as described by $A_{32}N_3$
3. stimulated emission from level 3 to level 2 (lasing), as described by $\rho_{\nu}(\nu_{32})B_{32}N_3$
4. spontaneous emission from level 2 to level 1, as described by $A_{21}N_2$
If we assume that the atoms in the total population will be distributed between the three energy levels, then
$N_1(t) + N_2(t) + N_3(t) = N_{total} \nonumber$
At equilibrium, the population of each level will remain constant, and so
$0 = \dfrac{dN_1(t)}{dt} = \dfrac{dN_2(t)}{dt} =\dfrac{dN_3(t)}{dt}. \nonumber$
Thus,
$\dfrac{N_2(t)}{dt} = -\rho_{\nu}(\nu_{32})B_{32}N_2 + A_{32}N_3 - A_{21}N_2 + \rho_{\nu}(\nu_{32})B_{32}N_3 = 0 \label{15.4.1}$
Rearranging Equation $\ref{15.4.1}$ gives
$N_2(A_{21} + \rho_{\nu}(\nu_{32})B_{32}) = N_3(A_{32} + \rho_{\nu}(\nu_{32})B_{32}) \label{15.4.2}$
which rearranges to
$\dfrac{N_3}{N_2} = \dfrac{A_{21} + \rho_{\nu}(\nu_{32})B_{32}}{A_{32} + \rho_{\nu}(\nu_{32})B_{32}} \label{15.4.3}$
From Equation $\ref{15.4.3}$ we can see that if $A_{21}$ is greater than $A_{32}$, then $N_3$ can be greater than $N_2$. If $N_3$ is greater than $N_2$, then population inversion between states 3 and 2 can be achieved. In other words, if atoms in state 2 decay to the ground state more rapidly than atoms in state 3 decay to state 2, there will be a population inversion between states 3 and 2, and the system can lase.
Four-Level Lasers
There are also lasers based on transitions between four energy levels. These lasers can be more efficiently pumped, because the lower level of the lasing transition is not the ground state. Only four-level lasers can provide a continuous lasing output. He-Ne and Nd:YAG lasers are common four-level lasers and will be described in section 15.6.
References
D.W. Coutts, in Encyclopedia of Modern Optics, 2005, page 460, https://doi.org/10.1016/B0-12-369395-0/00847-2
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Most lasers consist of three basic components:
1. the gain medium from which light of a specific frequency is emitted
2. the pump source which provides energy to the gain medium to create the light
3. the laser optical cavity in which the light is amplified by repeatedly passing through the gain medium
The Gain Medium
The gain medium, or active medium, is a collection of atoms or molecules that can undergo stimulated emission. The active medium can be in a gaseous, liquid, or solid form. It can be a pure substance or a solution. Because the emitted light of a laser is created by the transition between two quantized states of the gain medium, laser light is always monochromatic. Laser light is also always coherent, meaning that the light waves emitted by the laser are all in phase.
Solids
The first laser, Maiman's ruby laser, described in section 15.4, used a solid rod of synthetic ruby as the gain medium (Figure \(1\)).
Ruby is a gem variety of the mineral corundum, Al2O3, in which Cr3+ ions replace roughly 0.05% (by mass) of the Al3+ ions. The electronic structure of the Cr3+ impurities enables to ruby to act as a gain medium. However, synthetic ruby must be used because natural rubies have too many crystal defects. Other solid-state gain media are shown in Table \(1\).
Table \(1\): Solid-phase gain media
Solid-state Host Active Ion Wavelength (nm) Output Lifetime
Al2O3 Ti3+ 780 continuous and pulsed 10 fs - 5 ps
Al2O3 Cr3+ 694.3 pulsed 10 ps
Glass (SiO2) Nd3+ 1059 pulsed 1 ps
YAG Y3Al5O15 Nd3+ 1064.1 continuous and pulsed 10-150 ps
YLF Y3LixFy Nd3+ 1054.3 continuous and pulsed 10-100 ps
Gases
A basic He-Ne laser is shown in figure \(2\). The gain medium is the gas mixture through which the laser beam passes.
Other common gas-phase media include noble gases and molecules, but also metal cations, and metal atoms as shown in Table \(2\).
Table \(2\): Gas-phase gain media
Gaseous Gain Medium Wavelength (nm) Output Pulse Duration
N2 337 pulsed 1 ns
Cu 510 pulsed 30 ns
CO2, He, N2 tunable near 10,000 pulsed > 100 ns
He, Ne 3391, 1152, 632, 544 continuous continuous
He, Cd 441, 325 continuous continuous
Ar+ 488, 515 continuous continuous
K+ 647 continuous continuous.
Exciplex Lasers
Exciplex lasers involve the formation of an exciplex, a gaseous species that forms and exists only in an excited state. Because only the upper, excited state can exist, there is no population in the lower, dissociated state, thus a population inversion is obtained. The lasing transition occurs as the excited state exciplex emits its excitation energy and simultaneously falls apart. Two common exciplex lasers are XeCl* (308 nm) and KrF* ((249 nm). These lasers are often referred to as excimer laser, but the term "excimer" refers to an excited dimer, AA*, and so should not be used to describe an exciplex, AB*.
Liquid Solutions
Many dye lasers use an organic dye, usually dissolved in a solution, as the gain medium. One advantage of using a dye solution as the gain medium is that it can create a tunable laser, which allows for continuous tuning over a wide range of wavelengths. For example, rhodamine 6G is a dye that can be tuned from 635 nm to 560 nm. The dye solution enters the laser cavity by passing through a cell or by streaming through the air using a dye jet. (Figure \(3\)).
Some common laser dyes other than rhodamine 6G and the rhodamines, include various coumarins, stilbenes, and fluorescein. There are solid state dye lasers (SSDL), in which the dye is dispersed uniformly within a solid polymer matrix.
The Pump Source
For lasing to take place, the gain medium must be pumped into an excited state capable of undergoing stimulated emission. The energy required for excitation is often supplied by an electric current or an intense light source, such as a flashlamp or an excitation laser. Diode lasers, excimer lasers, and YAG lasers have all been used as pump sources.
Lasers are categorized as having either a continuous beam or a pulsed beam. In theory, any laser could be run as a pulsed beam laser if the pump source is set to deliver the excitation energy in regularly repeated bursts. Not all lasers can operate as continuous-wave (CW) lasers, though, because there are gain media in which it is not possible to maintain a continuous population inversion. The ruby laser is an example of a laser that cannot produce a continuous wave.
The Laser Optical Cavity
Figure \(4\) shows the main components of a generic laser optical (or resonator) cavity. The optical cavity is formed by a pair of mirrors that surround the gain medium and enable feedback of light into the medium. This feedback is critical to the operation of a laser because it is where the amplification of the signal occurs. The gain medium is excited by an external pump source, such as a flash lamp, electric current, or another laser. The output coupler is a partially reflective mirror that allows a portion of the laser radiation to leave the cavity but reflects a majority of the light back through the gain medium. The light trapped between the mirrors forms standing wave structures called modes. Although beyond the scope of this discussion, the reader interested in cavity modes can consult the “Laser Radiation Properties” module.
Figure \(5\) shows the laser cavity of a tunable dye laser. The laser is tunable because the pump source provides several excitation energies, which result in the creation of several possible energies for the lasing beam. The user chooses the desired lasing energy by means of a tuning element, which is often a movable diffraction grating.
Diode Lasers
Diode lasers do not lase in the same manner as the lasers described above. There is no laser optical cavity, and thus there is no stimulated emission. Diodes made of materials such as GaAs emit light when the recombination of an electron and a hole in a semiconductor releases energy in the form of photons. A population inversion is maintained by rapidly removing the electrons that are falling into the holes of the p-type semiconductor. A pseudo-optical cavity is formed because the semiconductor materials generally have a very high refractive index, causing the photons to be trapped in the crystal of the semiconductor due to the large and abrupt difference in the refractive indices at the surfaces of the crystal. GaAs emits infrared light, but it can be doped to create a material that emits at another wavelength. For example, GaAs0.6P0.4 emits red light.
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The He-Ne laser was the first continuous-wave (cw) laser invented. A few months after Maiman announced his invention of the pulsed ruby laser, Ali Javan and his associates W. R. Bennet and D. R. Herriott announced their creation of a cw He-Ne laser. This gas laser is a four-level laser that uses helium atoms to excite neon atoms. The atomic transitions in the neon produce the laser light. The most commonly used neon transition in these lasers produces red light at 632.8 nm. These lasers can also produce green and yellow light in the visible region, as well as several UV and IR wavelengths (Javan's first He-Ne operated in the IR at 1152.3 nm). By using highly reflective mirrors designed for one of these many possible lasing transitions, a given He-Ne's output is made to operate at a single wavelength.
He-Ne lasers are not sources of high-power laser light, typically producing a few to tens of mW (milli-Watt, or $10^{-3}$ W) of power. Probably one of the most important features of these lasers is that they are highly stable, both in terms of their wavelength (mode stability) and intensity of their output light (low jitter in power level). For these reasons, He-Ne lasers are often used to stabilize other lasers. They are also used in applications, such as holography, where mode stability is important. Until the mid-1990's, He-Ne lasers were the dominant type of lasers produced for low-power applications - from range finding to scanning to optical transmission, to laser pointers, etc. Recently, however, other types of lasers, most notably the semiconductor lasers, seem to have won the competition because of reduced costs.
The energy level diagram in figure $1$ shows the two excited states of a helium atom, the 2 3S and 2 1S, that get populated as a result of electromagnetic pumping. Both of these states are metastable and do not allow de-excitations via radiative transitions. Instead, the helium atoms give off their energy to neon atoms through collisional excitation. In this way, the 4s and 5s levels in neon get populated.
$He^* (2s^3 \, S_1) + Ne(g) \rightarrow He(g) + Ne^*(2p^54s) \nonumber$
$He^* (2s^1 \, S_0) + Ne(g) \rightarrow He(g) + Ne^*(2p^55s) \nonumber$
These are the two upper lasing levels, each for a separate set of lasing transitions. Each of these upper lasing levels contains 4 states (3P2, 3P1, 3P0, 1P1). Radiative decay from the 5s to the 4s levels is forbidden. So, ten states associated with the 4p and 3p levels serve as the lower lasing levels and rapidly decay into the metastable 3s level. In this way, population inversion is easily achieved in the He-Ne mixture. The 632.8 nm laser transition, for example, involves the 5s and 3p levels, as shown above. Table $1$ lists parameters of several of the Ne transitions.
Table $1$: Several Lasing Transitions for an Excited State Ne Atom
Transition $\lambda$ (nm) Einstein Coefficient A (s-1) Relative Intensity
5s1 P1 → 3p3 P1 640.1 0.60 x 106 100
5s1 P1 → 3p3 P0 635.2 0.70 x 106 100
5s1 P1 → 3p3 P2 632.8 6.56 x 106 300
5s1 P1 → 3p1 P1 629.4 1.35 x 106 100
5s1 P1 → 3p1 D2 611.8 1.28 x 106 100
In most He-Ne lasers the gas, a mixture of 5 parts helium to 1 part neon, is contained in a sealed glass tube with a narrow (2 to 3 mm diameter) bore as shown above in figure $2$. Typically the laser's optical cavity mirrors, the high reflector and the output coupler form the two sealing caps for the narrow bore tube. High voltage electrodes create a narrow electric discharge along the length of this tube, which then leads to the narrow beam of laser light.
Many He-Ne lasers also contain a ballast which serves to maintain the desired gas mixture. Since some of the atoms may get embedded in the glass and/or the electrodes as they accelerate within the discharge, in the absence of a ballast the lasing mixture would not last very long. To further prolong the useable life of the laser, some of these lasers also use "getters", often metals such as titanium, that absorb impurities in the gas. Figure $3$ below shows a He-Ne laser with a ballast.
A typical commercially available He-Ne produces about a few mW of 632.8 nm light with a beam width of a few millimeters at an overall efficiency of near 0.1%. This means that for every 1 Watt of input power from the power supply, 1 mW of laser light is produced. Still, because of their long operating lifetime of 20,000 hours or more and their relatively low manufacturing cost, He-Ne lasers are among the most popular gas lasers.
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A Quick Overview
Laser light offers valuable tools to researchers who wish to use the interaction of light with matter to interrogate atomic and molecular systems. Most laser light is characterized by its near monochromaticity (relative to light from other sources), directionality, and coherence [1]. Those characteristics are used in modern laser spectroscopy. The monochromaticity of laser light allows it to be used to probe specific energy changes in atoms and molecules. This ability to select specific wavelengths allows scientists to focus on chosen components in mixtures, including complex reaction mixtures. For example, Park et. al. tracked the dynamics of the reaction of ground state oxygen with ethyl radical by using a 355 nm laser beam to track one of the products of this reaction [2]. Parsons et. al. used near monochromatic laser light sources to achieve state-selective ionization for the study of the products of a photodissociation reaction of atmospheric relevance [3]. Laser monochromaticity can be used to quantify the amount of greenhouse gases in the atmosphere [4]. Even when the samples are not mixtures, monochromaticity allows researchers to gather detailed information about atomic and/or molecular structure. Applications of this type are extremely numerous, including detailed studies of atomic systems [5], semiconductor materials [6], single molecules [7], and biological molecules [8]. Lasers find research applications in art and archeology, where the property of monochromaticity allows specific energies to be probed and laser beam directionality and small spot size curtail destruction of samples [9,10].
Other laser applications focus on the ability of pulsed lasers to provide short pulses. Short-pulsed lasers offer myriad tools for exploring chemical kinetics at a range of timescales, including very short ones. Short-pulse lasers open the opportunity for time-resolved studies of molecular processes such as reaction processes [2,11] and biological processes [12 - 15], and the properties of excited states [16]. Some research applications take advantage of the ability of certain lasers to produce high intensity light. One such application is laser-induced break down spectroscopy (LIBS), in which the high intensity of the laser creates a plasma from solid or liquid samples; that plasma can subsequently be probed [17]. Other techniques that exploit high laser intensities are the nonlinear spectroscopies, where the high intensities of some lasers can produce behaviors in samples that lower-power sources cannot sufficiently stimulate [18,19]. Although most lasers are monochromatic and coherent, some researchers have modified laser light to obtain information from broadband, incoherent laser light [20 – 22], which allows them to probe the response of samples to light that more resembles sunlight (incoherent) than do traditional laser sources [22, 23]. Another possibility is to probe multiple wavelengths with broadband laser sources that are coherent [24].
Conclusion
This brief overview has only skimmed the surface of the vast field of laser spectroscopy. The examples given here are a very small, somewhat arbitrary, and significantly biased subset of the enormous library of published laser-based experiments. References have been chosen to give examples of the topics covered here and have not been extensively reviewed by the author.
Overall, lasers are especially useful as light sources when one or more of the following properties are desired for light in an experiment:
1. A high degree of monochromaticity
2. A well-known central wavelength for a light source
3. A tunable light source
4. Spatial coherence
5. Phase coherence
6. Directionality
7. High intensity
8. Tight focus
9. Short pulses
A particular laser type will not necessarily have all the properties in the list above; however, the menu of available laser types allows a researcher to choose a laser system or systems with the characteristics needed for a particular experiment. New applications of laser spectroscopy and new spectroscopic techniques employing laser light continue to be invented.
Contributor
Stephanie Schaertel (Grand Valley State University)
(Thank you to Tom Neils (Grand Rapids Community College) for editing the references)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/15%3A_Lasers_Laser_Spectroscopy_and_Photochemistry/15.07%3A_High-Resolution_Laser_Spectroscopy.txt
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These are homework exercises to accompany Chapter 15 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Q15.9
The Einstein coefficient of a ground and excited state with a degeneracy of $g_1$ and $g_2$ is given by
$A = \dfrac{16\pi^3v^3g_1}{3\epsilon_0hc^3g_2} |\mu|^2$
where $|\mu|$ is the transition dipole moment. Consider the 1s --> 2p absorption of H(g), which is observed at 121.8 nm. The radiative lifetime of the triply degenerate excited 2p state of H(g) is 1.6 x 10-9 s. Determine the value of the transition dipole moment of this transition.
S15.9
Using the expression given in the problem, we can isolate $|\mu|$ to be given as
$|\mu| = \sqrt{\dfrac{3A\epsilon_0hc^3g_2}{16\pi^3v^3g_1}}$
Since we know the wavelength and the radiative lifetime, we can find frequency (v) and amplitude (A) with the following equations, respectively.
$v = \dfrac{c}{\lambda} = 2.46 \times 10^{15} s^{-1}$
$A = \dfrac{1}{\tau} = 6.25 \times 10^8 s^{-1}$
where $\tau$ represents radiative lifetime.
The 2p orbital is threefold degenerate $g_2 = 3$ and the 1s orbital is signly degenerate $g_1 = 1$. Therefore, we can plug the corresponding numbers into the equation below in order to yield the transition dipole moment, which would be
$|\mu| = \sqrt{\dfrac{3A\epsilon_0hc^3g_2}{16\pi^3v^3g_1}}$
$|\mu| = 1.1 \times 10^{-29} C*m$
Q15.10
$A = \dfrac{16\pi^{3}\nu^{3}g_{1}}{3\varepsilon_{0}hc^{3}g_{2}}|\mu|^{2}$
$A_{21} = \dfrac{8h\pi\nu^{3}_{12}}{c^{3}}B_{21}$
Use the equations above to derive the quantum mechanical expression for the Einstein $B$ coefficient. Consider the $5s^{1}P_{1} \rightarrow 3p^{1}S_{0}$ transition of neon at $730.0\ nm$. Einstein $A$ coefficient is $0.48 \times 10^{6} s^{-1}$. Determine the values of the Einstein $B$ coefficient and the transition moment dipole for this transition.
S15.10
$\mu = \dfrac{h \nu }{4 \pi } n_{2} A_{21}$
$A = \dfrac{32 h \pi ^{3} \nu ^{7}}{c^{3}\varepsilon_{0}}n_{2}B_{21}$
$B_{21}=\dfrac{c^{3}\varepsilon_{0}}{n_{2}A 32 h \pi ^{3} \nu ^{7}}$
$730 nm = 7.3*10^{-7} m$
$\lambda$ = $7.3*10^{-7} m$ and thus $\nu$ = $c\lambda$ = $2.189*10^{16} s^{-1}$
$B_{21}=\dfrac{c^{3}\varepsilon_{0}}{n_{2}A 32 h \pi ^{3} \nu ^{7}}$
$B_{21} = \dfrac{c^{3}A}{8h \pi \nu^{3}}$
$B_{21} = \dfrac{(2.998*10^{8} m *s^{-1})^{3} *(0.48*10^{6} s^{-1})}{8h \pi (2.189*10^{16})^{3}}$
$B_{21} = 7.420*10^{13} kg^{-1} m$
Q15.11
Given $N_{total} = N_1(t) + N_2(t) +N_3(t)$ find the rate equation for each $N_i$.
S15.11
For each one we must look at the excitation energy, and the stimulated and spontaneous emissions energies.
For example, the excitation from 1 to 3, stimulated emission from 3 to 1 and the spontaneous emission from 3 to 1 and 2 to 1.
$\dfrac{dN_1}{dt} = -B_{31}\rho_v\nu_{31}N_1 + B_{31}\rho_v\nu_{31}N_3 + A_{31}N_3+A_21N_2$
the others follow as
$\dfrac{dN_2}{dt} = -B_{32}\rho_v\nu_{32}N_2 + B_{32}\rho_v\nu_{32}N_3 + A_{32}N_3-A_21N_2$
and
$\dfrac{dN_3}{dt} = -B_{31}\rho_v\nu_{31}N_1 +-B_{31}\rho_v\nu_{31}N_3 - A_{32}N_3-A_31N_3-B_{32}\rho_v\nu_{32}N_3+B_{32}\rho_v\nu_{32}N_2$
where we estimate the emission from level 3 to level 2.
Q15.12
Consider a nondegenerate 3-level system. Suppose that an incident light beam of energy $h\nu = E_{3} - E_{1}$ is turned on for a while and then turned off. Show that the subsequent decay of the $E_{3}$ is given by
$N_{3}(t) = N_{3}^{0}e^{-(A_{32}+A_{31})t}$
Where $N_{3}^{0}$ is the number of atoms in state 3 at the instant the light source is turned off. What will be the observed radiative lifetime of this excited state?
S15.12
After the light is turned off, no stimulated processes will occur and the rate equation of $N_{3}$ becomes:
$N_{3} = Ce^{-(A_{32}+A_{31})t}$
At $t$ = 0 (when the light is turned off), $N_3(t) = N_{3}^{0} = C$, so
$N_3(t) = N_{3}^{0}e^{-(A_{32}+A_{31})t}$
The observed radiative lifetime will be $(A_{32}+A_{31})^{-1}$. (The radiative lifetime is the reciprocal of the coefficient of the $t$ in the exponential term.)
Q15-14
An excited state of Lithium has the electron configuration, $1s^{2}2p^{1}$ Define the lowest energy term symbol.
S15-14
Construct a micro state table with spin (M(S)) across the top and angular momentum (M(L)) down the left side.
-1/2 1/2
-1 -1(-) -1(+)
0 0(-) 0(+)
1 1(-) 1(+)
The highest spin value is +1/2, while the highest angular momentum value is 1. As a result, J can range from 3/2 (L+S) to 1/2 (L-S). Because the p-orbital is less than half-full, the term symbol with the lowest J value will be lower in energy.
Term symbol: $^{2}P_{1/2}$
Q15.15
The ground state energy of the He atom is -2.904 hartrees. Use this value to determine the energy of He^+
S15.15
( good job however you could have shown more steps as it us unclear how you made the jumps from energy to $He^{+}$ and then that the energy was 0.904, it took me some time to fiddle with what you gave me as a solution to arrive to that number so more in between steps are need --RM)
$E=\dfrac{-Z^2}{2n^2}$
$He^+=-2E_{h}$
Therefore the energy of He+ is 0.904 hartrees above that of He.
There isn't too many steps in solving the problem. It would have been great to compare it to another method or the actual energy of an H+ hydrogen, such as, using the Rydberg formula to evaluate the energy....
Q15.19
A laser operating at 640 nm produces pulses at a rate of 85 MHz. Calculate the radiant power of each pulse if the pulses last 15 fs each and the average radiant power of the laser is 2.6 W. How many photons are produced by the laser per second?
S15.19
We can first calculate the amount of energy per pulse:
$\dfrac{2.6 W}{85 \dfrac{pulses}{MHz}} = \dfrac{2.6 \dfrac{J}{s}}{85 \cdot 10^{6} \dfrac{pulses}{s}} = 3.059 \cdot 10^{-8} \dfrac{J}{pulse}$
Using this, we can then calculate the radiant energy of each laser pulse:
$\dfrac{3.059 \cdot 10^{-8} \dfrac{J}{pulse}}{15 \cdot 10^{-15} s} = 2039.216 \dfrac{kW}{pulse}$
Finally, we can calculate the radiant energy of a single photon:
$E_{photon} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \cdot 10^{-34} J \cdot s) (2.988 \cdot 10^{8} \dfrac{m}{s})}{640 \cdot 10^{-9} m} = 3.094 \cdot 10^{-19} J$
Using this energy, we can calculate the the number of photons produced per second by the laser:
$\dfrac{E_{laser}}{E_{photon}} = \dfrac{2.6 \dfrac{J}{s}}{3.094 \cdot 10^{-19} \dfrac{J}{photon}} = 8.40 \cdot 10^{18} \dfrac{photons}{second}$
Q15.21
Which laser pulse contains more photons, a 945 ns, 9.45 mJ pulse at 945 nm or a 23 ns, 2.30 mJ at 230 nm? Does the speed of the pulse effect the number of photons?
S15.21
• ${\bf E}$ = energy per photon
• ${\bf ħ}$ = plancks constant = $6.626*10^{-34} J \cdot s$
• ${\bf c}$ = speed of light = $3.00*10^{8} \dfrac{m}{s}$
• ${\bf λ}$ = wavelength = 945 nm or 230 nm = $540*10^{-9} m \,or\, 230*10^{-9} m$
$E_{photon} = \dfrac{(6.626*10^{-34} J\cdot s) * (3.00*10^8\, \dfrac{m}{s})}{945*10^-9 \,m} = 2.104x10^{-19} \,J/photon$
$E_{pulse} = 9.45 \,mJ = 9.45*10^{-3}J$
$\ \#_{photon} = \dfrac{E_{pulse}}{E_{photon}} = \dfrac {9.45*10^{-3}J}{2.104*10^{-19} \,J/photon} = {\bf 4.493*10^{16}\,photons}$
$E_{photon} = \dfrac{(6.626*10^{-34} J\cdot s) * (3.00*10^8\, \dfrac{m}{s})}{230*10^-9 \,m} = 8.64*10^{-19} J$
$E_{pulse} = 2.30 \,mJ = 2.30*10^{-3}J$
$\#_{photon} = \dfrac{E_{pulse}}{E_{photon}} = \dfrac {2.30*10^{-3}J}{8.64*10^{-19} \,J/photon} = {\bf 2.66*10^{15}\,photons}$
First and foremost, the speed of the laser DOES NOT effect the number of photons! The 9.45 mJ at 945 nm laser pulse has more photons than the 2.30 mJ at 230 nm pulse.
The pulse duration does not affect the number of photons a laser pulse contains. Energies of both laser pulses are equivalent. Energy per pulse is propotional to the number of photons and inversely proportional to the wavelength of the emitted photons. Therefore, the 945 pulse contains more photons.
Q15.22
Given the following laser pulses: a $10 \times 10^{-9}s$, $1.6 \times 10^{6}J$ pulse at $76 \times 10^{-10}m$ or a $5 \times 10^{8}s$, $1.6 \times 10^{6}J$ pulse at $5.32 \times 10^{-11}$, which one has the most photons?
S15.22
For 10-ns, 1.60-mJ pulse at 760 nm:
$Q_p=\dfrac{hc}{\lambda}$
$=\dfrac{(6.626 \times 10^{-34})(3 \times 10^8)}{760 \times 10^{-9}}$
$=2.62 \times 10^{-19}$
Number of photon in 1.6 mJ pulse is:
$N_1=\dfrac{Q}{Q_p}=6.12 \times 10^{15}$
For 500-ms, 1.60-mJ pulse at 5320 nm:
$Q_p=3.75 \times 10^{-18}$
Number of photon in 1.6 mJ pulse is:
$N_2=\dfrac{Q}{Q_p}=0.4 \times 10^{15}$
Thus, 10-ns, 1.60-mJ pulse at 760 nm contain more photon.
Q15.24
A $CO_2$ laser operating at 9.0 $\mu m$ uses an electrical power of 5.20 kW. If this laser produces 100-ns pulses at a repetition rate of 20Hz and has an efficiency of 29%, how many photons are in each laser pulses?
S15.24
The pump energy per pulse is:
$\dfrac{5200 J*s^{-1}}{20 s^{-1}}$ = $260 J*pulse^{-1}$
Given that the laser is 29% effcient, the radiant per pulse is $(260J)(0.29)$ = $67.6J$.
The number of photons per pulse, $n$ is
$n$ = $\dfrac{E\lambda}{hc}$,
by plugging in the values into this equation, we have
$n$ = $\dfrac{(260J)(0.29)(9.0*10^{-6}m)}{(6.626*10^{-34} J*s)(2.998*10^{8} m*s^{-1})}$
= $3.42*10^{37}$ photons
My calculation reveals the final answer to be $3.42\times 10^{21}$ photons (Aaron Choi)/ I believe you accidentally calculated using the speed of light as $2.998\times 10^{-8} \dfrac{m}{s}$
Q15.25
Figure 15.10 displays the energy levels of the $CO_2$ laser. Given the following spectroscopic data for $CO_2(g)$, calculate the spacing between the $J' = 3\to2$ laser lines for the $001\to100$ vibrational transition.
$Fundamental frequency(J' = 0\to0) 100\to001 = 960.80cm^{-1}$
$\tilde{B}(001) = 0.3871cm^{-1} \tilde{B}(100) = 0.3902cm^{-1}$
The fundamental frequency is $960.80cm^{-1} S15.25 Using the following equation F(J) for J = 2 and J = 3 can calculated: $F(J) = \tilde{B}J(J + 1)$ $F_{001}(3) = (0.3871cm^{-1})(3)(3 +1) = 4.645cm^{-1}$ $F_{100}(2) = (0.3902cm^{-1})(2)(2 +1) = 2.341cm^{-1}$ Therefore the spacing is $960.80cm^{-1} + 4.645cm^{-1} - 2.341cm^{-1} = 963.10cm^{-1}$ Q15.26 The upper level of the \(H_2(g)$ laser is the lowest excited state of the molecule, the $B^1\sum_{u}^{+}$ state, and the lower level is the $X^1 \sum_{g}^{+}$ ground state.The lasting occurs between the $v^{'} = 6$ level of the excited state and the $v^{''} = 13$ level of the ground state. Use the following spectroscopic data to determine the wavelength of the laser light from $H_2(g)$ laser.
State $\tilde{T}_{e}/cm^{-1}$ $\tilde{\nu}_{e}/cm^{-1}$ $\tilde{\nu}_{e} \tilde{x}_{e}/cm^{-1}$
$B^1\sum_{u}^{+}$ 91,689.9 1356.9 19.93
$X^1 \sum_{g}^{+}$ 0 4401.2 121.34
A 3.0 ns pulse can be generated with a pulse radiant power of 200kW. Calculate the radiant energy of such a laser pulse. How many photons are there in this pulse?
S15.26
To calculate the energy of the upper and lower lasing levels the following equation is used:
$G(v) = \tilde{\nu}_{e}\left(v + \dfrac{1}{2}\right) - \tilde{x}_{e} \tilde{\nu}_{e}\left(v + \dfrac{1}{2}\right)^2$
$G^{''}(13) = (4401.2cm^{-1})(13.5) - (121.34cm^{-1})(13.5)^2 = 37,301.98 cm^{-1}$
$G^{''}(6) = (1356.9cm^{-1})(6.5) - (19.93cm^{-1})(6.5)^2 = 7,977.81 cm^{-1}$
the transition is as follows
$\tilde{\nu} = \tilde{T}_{e} + G^{''}(6) - G^{''}(13)$
$= 91,689.9 cm^{-1} + 7,977.81 cm^{-1} -37,301.98 cm^{-1} = 62,365.73cm^{-1}$
therefore
$\lambda = 160.34 nm$
The radiant energy of the laser pulse is found using dimensional analysis
$\left(200 \dfrac{kJ}{s}\right)\left(2 \times 10^{-9} s \right) = 1x 10^{-4}J$
the number of photons per pulse is determined using $E = nhv = 8.06 \times 10^{13}$ photons.
Q15.27
Determine the excited-state rotational quantum number for $X$ $\rightarrow$ $A$ adsorption bands of $H_{2}$(g). The transition is from the $v^{"}=0$of the $X$ state to the excited vibrational level of the $A$ state $v^{'}=3$. To accurately calculate the vibrational term $G(v)$ for the excited state $A$, a second-order anharmonic correction is made to account for the shape of the potential curve, while first-order corrections are sufficient for the ground level electron state.
$G(v)=\tilde{v}_{e}(v+\dfrac{1}{2})-\tilde{v}_{e}\tilde{x}_{e}(v+\dfrac{1}{2})^{2}+\tilde{v}_{e}\tilde{y}_{e}(v+\dfrac{1}{2})^{3}$
The need spectroscopic constants are tabulated below for the ground state $X$ and the excited state $A$ of $H_{2}$(g).
State $\tilde{T}_{e}/cm^{-1}$ $\tilde{v}_{e}/cm^{-1}$ $\tilde{v}_{e}\tilde{x}_{e}/cm^{-1}$ $\tilde{v}_{e}\tilde{y}_{e}/cm^{-1}$ $\tilde{B}_{e}/cm^{-1}$ $\tilde{\alpha}_{e}/cm^{-1}$
$A$ 120952 2321.4 62.8 0 29.9 1.24
$X$ 0 2291.7 62.4
These values are from the NIST website for the $H_{2}$
Determine the value of $\tilde{v}$ corresponding to the transition $X(v^{"}=0,j^{"}=0)$ $\rightarrow$ $A(v^{'}=3,j^{'}=0)$. Given that the ground state levels are $v^{"}=0,j^{"}=1$ of the $X$ state and that the rotational term for this level is $F(1)=1.057cm^{-1}$, determine the closest value of $J^{'}$, the rotational number of the $v^{"}=3$ level of the excited $A$ state.
S15.27
For the ground state, the $G(v)$ value is :
$G^{"}(0)=(\dfrac{1}{2})(2291 cm^{-1}) - (\dfrac{1}{2})^{2}62.4cm^{-1}$
$G^{"}(0)= 1129.9 cm^{-1}$
For the excited state $G(v)$ value is :
$G^{"}(3)=(2321.4cm^{-1})(3+\dfrac{1}{2})-(62.8 cm^{-1})(3+\dfrac{1}{2})^{2}$
$G^{"}(32)=7354 cm^{-1}$
$\tilde{T}_{e}$is the difference in the minima of the electronic potential energy curve in wave numbers, so the transition will have the energy
$\tilde{v}_{e}=\tilde{T}_{e} + G^{'}(32)-G^{"}(0)$
$\tilde{v}_{e}=120952cm^{-1}+7354 cm^{-1}-1129.9cm{-1}$
$\tilde{v}_{e}=127176.1 cm^{-1}$
$\tilde{B}_{v}$ can be found for $A$ state, $v=3$ with this equation:
$\tilde{B}_{32}=\tilde{B}_{e}-\tilde{\alpha}_{e}(v+\dfrac{1}{2})$
$\tilde{B}_{32}=29.9cm^{-1}-(1.24 cm^{-1})(3+\dfrac{1}{2})=25.56 cm{-1}$.
the observed lines are in between 35714.29 cm–1 and 3448.28 $cm_{-1}$ recalling that $\tilde{E}_{v,j}=G(v)-F(j)$;
$35717.29cm^{-1}= 127176.1 cm^{-1}-1.057 cm{-1} + F^{'}(j)$
$35717.29 cm^{-1}= 127175.043 cm^{-1}+ F^{'}(j)$
Therefore 91461$\approx F^{'}(j)$ so $j$ would be around 10 or higher.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/15%3A_Lasers_Laser_Spectroscopy_and_Photochemistry/15.E%3A_Lasers_Laser_Spectroscopy_and_Photochemistry_%28Exercises%29.txt
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• 16.1: All Dilute Gases Behave Ideally
Gases behave according to the ideal gas law when interactions between the gas molecules and the container as well as the size of the particles can be ignored. At low pressures and high temperatures since the gas occupies a large volume, the volume occupied by the constituents of the gas become even more insignificant in comparison. Thus real gases approach ideal behavior at low \(P\) and high \(T\).
• 16.2: van der Waals and Redlich-Kwong Equations of State
The van der Waals Equation of State is an equation relating the density of gases and liquids to the pressure, volume, and temperature conditions. The Redlich–Kwong equation of state is an empirical, algebraic equation that relates temperature, pressure, and volume of gases. It is generally more accurate than the van der Waals equation and the ideal gas equation at temperatures above the critical temperature.
• 16.3: A Cubic Equation of State
Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. The Van der Waals equation of state is the most well known of cubic equations of state, but many others have been developed.
• 16.4: The Law of Corresponding States
An additional assumption about real gases made by van der Waals was that all gases at corresponding states should behave similarly. The corresponding state that van der Waals choose to use is called the reduced state, which is based on the deviation of the conditions of a substance from its own critical conditions.
• 16.5: The Second Virial Coefficient
Because the perfect gas law is an imperfect description of a real gas, we can combine the perfect gas law and the compressibility factors of real gases to develop an Equation to describe the isotherms of a real gas. This Equation is known as the virial Equation of state, which expresses the deviation from ideality in terms of a power series in the density. The second virial coefficient describes the contribution of the pair-wise potential to the pressure of the gas.
• 16.6: The Repulsive Term in the Lennard-Jones Potential
Proposed by Sir John Edward Lennard-Jones, the Lennard-Jones potential describes the potential energy of interaction between two non-bonding atoms or molecules based on their distance of separation. The potential equation accounts for the difference between attractive forces (dipole-dipole, dipole-induced dipole, and London interactions) and repulsive forces.
• 16.7: Van der Waals Constants in Terms of Molecular Parameters
The van der Waals equation of state assumes the hard sphere model at close distances and weak intermolecular attractions at larger distances. These are represented by the van der Waals coefficients \(a\) for intermolecular attractions and \(b\) for finite volume of particles (hard sphere model).
• 16.E: The Properties of Gases (Exercises)
These are homework exercises to accompany Chapter 16 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.
Thumbnail: Motion of gas molecules. (CC BY-SA 3.0; Greg L via Wikipedia)
16: The Properties of Gases
Ideal gases have no interactions between the particles, hence, the particles do not exert forces on each other. However, particles do experience a force when they collide with the walls of the container. Let us assume that each collision with a wall is elastic. Let us assume that the gas is in a cubic box of length $a$ and that two of the walls are located at $x = 0$ and at $x = a$. Thus, a particle moving along the $x$ direction will eventually collide with one of these walls and will exert a force on the wall when it strikes it, which we will denote as $F_x$. Since every action has an equal and opposite reaction, the wall exerts a force $-F_x$ on the particle.
According to Newton’s second law, the force $-F_x$ on the particle in this direction gives rise to an acceleration via
$-F_x = ma_x = m\dfrac{\Delta v_x}{\Delta t} \label{2.1}$
Here, $t$ represents the time interval between collisions with the same wall of the box. In an elastic collision, all that happens to the velocity is that it changes sign. Thus, if $v_x$ is the velocity in the $x$ direction before the collision, then $-v_x$ is the velocity after, and $\Delta v_x = -v_x - v_x = -2v_x$, so that
$-F_x = -2m\dfrac{v_x}{\Delta t} \label{2.2}$
Since the particles have no forces acting upon them, except for when they collide iwht the wall container, the particles move at constant speed. Thus, a collision between a particle and, say, the wall at $x = 0$ will not change the particle’s speed. Before it strikes this wall again, it will proceed to the wall at $x = a$ first, bounce off that wall, and then return to the wall at $x = 0$. The total distance in the $x$ direction traversed is $2a$, and since the speed in the $x$ direction is always $v_x$, the interval $\Delta t = \dfrac{2a}{v_x}$. Consequently, the force is:
$-F_x = -\dfrac{mv_x^2}{a} \label{2.3}$
Thus, the force that the particle exerts on the wall is:
$F_x = \dfrac{mv_x^2}{a} \label{2.4}$
The mechanical definition of pressure is the average force over area:
$P = \dfrac{\langle F \rangle}{A} \label{2.5}$
where $\langle F \rangle$ is the average force exerted by all $N$ particles on a wall of the box of area $A$. Here $A = a^2$. If we use the wall at $x = 0$ we have been considering, then
$P = \dfrac{N \langle F_x \rangle}{a^2} \label{2.6}$
because we have $N$ particles hitting the wall. Hence:
$P = \dfrac{N m \langle v_x^2 \rangle}{a^3} \label{2.7}$
from our study of the Maxwell-Boltzmann distribution, we found that:
$\langle v_x ^2 \rangle = \dfrac{k_B T}{m} \label{2.8}$
Hence, since $a^3 = V$:
$\begin{split}P &= \dfrac{N k_B T}{V} \ &= \dfrac{n R T}{V} \label{2.9} \end{split}$
which is the ideal gas law. It relates the pressure, volume, and temperature of an ideal gas and is referred to as an equation of state. An equation of state is a relation between variables and describes the state of matter under a set of given physical conditions. One way to visualize any equation of state is to plot isotherms, which are graphs of $P$ vs. $V$ at fixed values of $T$ . For the ideal-gas equation of state, some of the isotherms are shown in the figure below (left panel): If we plot $P$ vs. $T$ at fixed volume (called the isochores), we obtain the plot in the right panel. What is important to note, here, is that an ideal gas can exist only as a gas. It is not possible for an ideal gas to condense into some kind of “ideal liquid”. In other words, a phase transition from gas to liquid can be modeled only if interparticle interactions are properly accounted for.
We can rearrange the ideal gas equation of state to put all the constants and variables on one side:
$\dfrac{P V}{n R T} = 1 = Z \label{2.11}$
$Z$ is called the compressibility of the gas. In an ideal gas, if we compress the gas by increasing $P$ , the volume decreases as well so as to keep $Z =1$. For a real gas, $Z$, therefore, gives us a measure of how much the gas deviates from ideal-gas behavior.
Figure 16.1.2 shows a plot of $Z$ vs. $P$ for several real gases and for an ideal gas. The plot shows that for sufficiently low pressures (hence, low densities), each gas approaches ideal-gas behavior, as expected.
Extensive and Intensive Properties in Thermodynamics
Before we discuss ensembles and how we construct them, we need to introduce an important distinction between different types of thermodynamic properties that are used to characterize the ensemble. This distinction is extensive and intensive properties.
Thermodynamics always divides the universe into a system and its surroundings with a boundary between them. The unity of all three of these is the thermodynamic universe.
Now suppose we allow the system to grow in such a way that both the number of particles and the volume grow with the ratio $N/V$ remaining constant. Any property that increases as we grow the system in this way is called an extensive property. Any property that remains the same as we grow the system in this way is called intensive.
The ideal gas equation of state can be written in an extensive form:
$\begin{split} P V &= n R T \ P \bar{V} &= R T \ P &= \rho R T \label{2.10} \end{split}$
where $\bar{V} = V/n$ is called the molar volume and $\rho = N/V$ is called density. Unlike $V$ , which increases as the number of moles increases (an extensive quantity, $\bar{V}$ and $\rho$ do not exhibit this dependence and, therefore, are called intensive. Compressibility can also be written in an intensive form:
$\begin{split} Z &= \dfrac{PV}{nRT} \ &= \dfrac{P\bar{V}}{RT} \ &= \dfrac{P}{\rho RT} \end{split} \nonumber$
Some examples of intensive and extensive properties include:
• Extensive: number of particles ($N$), moles ($n$), volume ($V$), mass ($m$), energy, free energy
• Intensive: pressure ($P$), density ($\rho$), molar heat capacity ($C$), temperature ($T$)
Contributors and Attributions
• Jerry LaRue (Chapman University)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/16%3A_The_Properties_of_Gases/16.01%3A_All_Dilute_Gases_Behave_Ideally.txt
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