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Learning Objectives
• Manipulate probability, energy, and transition problems involving multidimensional wavefunctions
• Apply the Separation of Variables to solve a multi-dimensional Schrödinger equation
• Manipulate problems with multiple quantum numbers
• To be introduced to the concepts of non-singular nodes (e.g., nodal lines)
• To be introduced to the concept of degeneracy
Let us now consider the Schrödinger Equation for an electron confined to a two dimensional box, $0 < x < a$ and $0 < y < b$. That is to say, within this rectangle the electron wavefunction behaves as a free particle ($V(x,y) = 0$), but the walls are impenetrable so the wavefunction $\psi(x,y,t)=0$ at the walls.
Extending the (time-independent) Schrödinger equation for a one-dimensional system
$−\frac{\hbar^2}{2m}\frac{d^ 2\psi(x)}{d x^2}+V(x)\psi(x)=E\psi(x) \label{3.1.3}$
to a two-dimensional system is not difficult
$- \dfrac{\hbar^2}{2m} \left (\dfrac{\partial^2 \psi(x,y)}{\partial x^2} + \dfrac{\partial ^2 \psi(x,y)}{\partial y^2} \right) + V(x,y)\psi(x,y) = E \psi(x,y). \label{e1}$
Equation \ref{e1} can be simplified for the particle in a 2D box since we know that $V(x,y) =0$ within the box and $V(x,y) = \infty$ outside the box, ii.e.
$V(x,y)=\begin{cases} 0 & 0\leq x\leq a \; \text{and}\; 0\leq y\leq b\ \infty & x< 0 \; \text{and}\; x> a \ \infty & y< 0 \; \text{and}\; y> b \end{cases}\nonumber$
So Equation \ref{e1} becomes
$- \dfrac{\hbar^2}{2m} \left (\dfrac{\partial^2 \psi(x,y)}{\partial x^2} + \dfrac{\partial ^2 \psi(x,y)}{\partial y^2} \right) = E \psi(x,y). \label{e2}$
Since the Hamiltonian (i.e. left side of Equation \ref{e2}) is the sum of two terms with independent (separate) variables, we try a product wavefunction like in the Separation of Variables approach used to separate time-dependence from the spatial dependence previously. Within this approach we express the 2-D wavefunction as a product of two independent 1-D components
$\psi(x,y) = X(x)Y(y). \label{product}$
This ansatz separates Equation \ref{e2} into two independent one-dimensional Schrödinger equations
$- \dfrac{\hbar^2}{2m} \left (\dfrac{d^2 X(x)}{d x^2} \right) = \varepsilon_x X(x). \label{e3a}$
$- \dfrac{\hbar^2}{2m} \left (\dfrac{d ^2 Y(y)}{d y^2} \right) = \varepsilon_y Y(y). \label{e3b}$
where the total energy of the particle is the sum of the energies from each one-dimensional Schrödinger equation
$E=\varepsilon_x +\varepsilon_y \label{sum}$
The differential equations in Equations \ref{e3a} and \ref{e3b} are familiar as they were found for the particle in a 1-D box previously. They have the general solution
$X(x) = A_x \sin (k_x x) + B_x \cos(k_xx) \label{e4a}$
$Y(y) = A_y \sin (k_y y) + B_y \cos(k_y y) \label{e4b}$
Applying Boundary Conditions
The general solutions in Equations \ref{e4a} and \ref{e4b} can be simplified to address boundary conditions dictated by the potential, i.e., $\psi(0,y)=0$ and $\psi(x,0)=0$. Therefore, $B_x=0$ and $B_y=0$.
Thus, we can combine Equations \ref{e4a}, \ref{e4b}, and \ref{product} to construct the wavefunction $\psi (x,y)$ for a particle in a 2D box of the form
$\psi (x,y)=N \sin\left ( \sqrt{\dfrac{2m\varepsilon_x}{\hbar^2}}x \right ) \sin \left ( \sqrt{\dfrac{2m\varepsilon_y}{\hbar^2}}y \right ) \label{wave}$
We still need to satisfy the remaining boundary conditions $\psi (L,y)=0$ and $\psi (x,L)=0$
$N \sin\left ( \sqrt{\dfrac{2m\varepsilon_x}{\hbar^2}}L \right ) \sin \left ( \sqrt{\dfrac{2m\varepsilon_y}{\hbar^2}}y \right ) = 0 \label{cond1}$
and
$N\sin\left ( \sqrt{\dfrac{2m\varepsilon_x}{\hbar^2}}x \right ) \sin \left ( \sqrt{\dfrac{2m\varepsilon_y}{\hbar^2}}L \right ) = 0 \label{cond2}$
Equation \ref{cond1} can be satisfied if
$\sin\left ( \sqrt{\dfrac{2m\varepsilon_x}{\hbar^2}}L \right ) =0$
independent of the value of $y$, while Equation \ref{cond2} can be satisfied if
$\sin\left ( \sqrt{\dfrac{2m\varepsilon_y}{\hbar^2}}L \right ) =0$
independent of the value of $x$. These are the same conditions that we encountered for the one-dimensional box, hence we already know the $\sin$ function in each case can be zero in many places. In fact, these two conditions are satisfied if
$\sqrt{\dfrac{2m\varepsilon_x}{\hbar^2}}L = n_x \pi$
and
$\sqrt{\dfrac{2m\varepsilon_y}{\hbar^2}}L = n_y \pi$
which yield the allowed values of $\varepsilon_x$ and $\varepsilon_y$ as
$\varepsilon_{n_x}=\dfrac{\hbar^2 \pi^2}{2mL^2}n_{x}^{2}$
and
$\varepsilon_{n_y}=\dfrac{\hbar^2 \pi^2}{2mL^2}n_{y}^{2}$
We need two different integers $n_x$ and $n_y$ because the conditions are completely independent and can be satisfied by any two different (or similar) values of these integers. The allowed values of the total energy are now given by
$E_{n_x, n_y}=\dfrac{\hbar^2 \pi^2}{2mL^2}(n_{x}^{2}+n_{y}^{2}) \label{squareE}$
Note that the allowed energies now depend on two integers $n_x$ and $n_y$ rather than one. These arise from the two independent boundary conditions in the $x$ and $y$ directions. As in the one-dimensional box, the values of $n_x$ and $n_y$ are both restricted to the natural numbers $1,2,3,...$ Note, therefore, that the ground state energy $E_{1,1}$ is
$E_{1,1}=\dfrac{\hbar^2 \pi^2}{mL^2}$
is larger than for the one-dimensional box because of the contributions from kinetic energy in the $x$ and $y$ directions.
Once the conditions on $\varepsilon_{n_x}$ and $\varepsilon_{n_y}$ are substituted into Equation \ref{wave}, the wavefunctions become
$\psi_{n_x, n_y}(x,y)=N\sin\left ( \dfrac{n_x \pi x}{L} \right ) \sin \left ( \dfrac{n_y \pi y}{L} \right )$
The constant $N$ is now determined by the normalization condition
\begin{align*}\int_{0}^{L} \int_{0}^{L}|\psi_{n_x, n_y}(x,y)|^2 dx\,dy &= 1\ N^2 \int_{0}^{L}\sin^2\left ( \dfrac{n_x \pi x}{L} \right ) dx\int_{0}^{L}\sin^2 \left ( \dfrac{n_y \pi y}{L} \right )dy &= 1\ N^2\dfrac{L}{2}\cdot \dfrac{L}{2} &= 1\ N &= \dfrac{2}{L}\end{align*}
so that the complete normalized 2D wavefunction is
$\psi_{n_x, n_y}(x,y)=\dfrac{2}{L}\sin\left ( \dfrac{n_x \pi x}{L} \right ) \sin\left ( \dfrac{n_y \pi y}{L} \right ) \label{squareWF}$
The Separability of the Hamiltonian
Because the Halmiltonian in Equation \ref{e2} can be expressed as a simple sum of independent terms, the solutions can be expressed as simple products of the one-dimensional solution (Equation \ref{squareWF}) with energies that are expressed as the sum of one-dimensional energies (Equation \ref{squareE}). This observation extends to higher dimensional wavefunctions as demonstrated later.
The wavefunctions in Equation \ref{squareWF} are somewhat more difficult that 1-D analogs to visualize because they are two dimensional. Nevertheless, we can still visualize them, and Figure $1$ shows the following wavefunctions: $\psi_{1,1}(x,y)$, $\psi_{2,1}(x,y)$ and $\psi_{2,2}(x,y)$.
Figure $1$: Visualizing the first six wavefunctions for a particle in a two-dimensional square box ($L_x=L_y=L$). Use the slide bar to independently change either $n_x$ or $n_y$ quantum number and see the changing wavefunction.
It is important to note that the wavefunctions can be either positive or negative, even though the associated probability density $p_{n_x, n_y}(x,y)=|\psi_{n_x, n_y}(x,y)|^2$ is strictly positive.
Figure $2$: Visualizing the first six wavefunctions and associated probability densities for a particle in a two-dimensional square box ($L_x=L_y=L$). Use the slide bar to independently change either $n_x$ or $n_y$ quantum number and see the changing wavefunction.
Unlike in the one-dimensional analoge, where nodes in the wavefunction are points where $\psi_{n}(x)=0$, here entire lines can be nodal (called nodal lines). For example, in the state $\psi_{2,1}(x,y)$, there is a nodal line at $\psi_{2,1}(L/2,y)$. Along the entire line $x=L/2$, the wavefunction is $0$ independent of the value of $y$. The wavefunction $\psi_{2,2}(x,y)$ has two nodal lines when $x=L/2$ and when $y=L/2$. The sign of the wavefunction and its nodal structure will play central roles later when we consider chemical bonding.
The fact that the wavefunction $\psi_{n_x n_y}(x,y)$ is a product of one-dimensional wavefunctions:
$\psi_{n_x n_y}(x,y)=\psi_{n_x}(x)\psi_{n_y}(y)$
makes the calculation of probabilities rather easy. The probability that a measurement the particle's position will yield a value of $x \in [a,b]$ and $y \in [c,d]$ is
\begin{align}P(x \in [a,b] \ and \ y \in [c,d]) &= \int_{a}^{b}dx\int_{c}^{d}dy|\psi_{n_x n_y}(x,y)|^2 \nonumber \ &= \int_{a}^{b}dx\int_{c}^{d}dy\psi_{n_x}^{2}(x)\psi_{n_y}^{2}(y) \nonumber \ &= \left [ \int_{a}^{b}\psi_{n_x}^{2}(x)dx \right ] \left [ \int_{c}^{d}\psi_{n_y}^{2}(y)dy \right ] \nonumber \end{align} \label{2DProb}
Example $1$: Probability
For a particle in a two-dimensional square box of length $L$, if the particle is in the $\psi_{1,2}(x,y)$ state, what is the probability that a measurement of the particle's position will yield $x \in [0,L/2]$ and $y \in [0,L/2]$?
Solution
Substituting the limits of integration into Equation \ref{2DProb}, we have
\begin{align*}P(x \in [0,L/2] \ and \ y \in [0,L/2] &= \left [ \int_{0}^{L/2}\psi_{1}^{2}(x)dx \right ] \left [ \int_{0}^{L/2}\psi_{2}^{2}(y)dy \right ]\ &= \left [ \dfrac{2}{L}\int_{0}^{L/2}\sin^2 \left ( \dfrac{\pi x}{L} \right ) dx \right ] \left [ \dfrac{2}{L}\int_{0}^{L/2}\sin^2 \left ( \dfrac{2\pi y}{L} \right ) dy \right ]\ &= \left [ \dfrac{2}{L} \dfrac{L}{4} \right ] \left [ \dfrac{2}{L}\dfrac{L}{4} \right ]\ &= \dfrac{1}{4}\end{align*}
This can be graphically confirmed by exploring the probability densities in Figure $2$ for the $\psi_{1,2}(x,y)$ state.
Exercise $1$
For a particle in a two-dimensional square box of length $L$, if the particle is in the state $\psi_{2,3}(x,y)$, what is the probability that a measurement of the particle's position will yield $x \in [0,L/2]$ and $y \in [0,L/3]$?
Answer
$P= \dfrac{1}{6} \nonumber$
Confirmed this by exploring the probability densities in Figure $2$ for the $\psi_{2,3}(x,y)$ state.
Note that if the box were rectangular rather than square, then instead of having a length of $L$ on both sides, there would be two different lengths $L_x$ and $L_y$. The formulas for the energies and wavefunctions become only slightly more complicated:
$E_{n_x, n_y}=\dfrac{\hbar^2 \pi^2}{2m}\left ( \dfrac{n_{x}^{2}}{L_{x}^{2}}+\dfrac{n_{y}^{2}}{L_{y}^{2}} \right ) \label{energy}$
instead of Equation \ref{squareE} and
$\psi_{n_x, n_y}(x,y)=\dfrac{2}{\sqrt{L_x L_y}}\sin\left ( \dfrac{n_x \pi x}{L_x} \right ) \sin \left ( \dfrac{n_y \pi y}{L_y} \right )$
instead of Equation \ref{squareWF}.
Degeneracy
Two distinct wavefunctions are said to be degenerate if they correspond to the same energy. If the sides a, b of the rectangle are such that a/b is irrational (the general case), there will be no degeneracies. The most degenerate case is the square, $L_x = L_y$, for which clearly $E_{m,n} = E_{n,m}$. Degeneracies in quantum physics are most often associated with symmetries in this way. Figure $3$ shows the wavefunctions (3,2) and (2,3) for a rectangle. These are contour maps for the time-independent solution, with white being the highest point. These two wavefunctions do not correspond to the same energy, although they would, of course, for a square.
The energy of the particle in a 2-D square box (i.e., $L_x=L_y=L$) in the ground state is given by Equation $\ref{energy}$ with $n_x=1$ and $n_y=1$. This energy ($E_{11}$) is hence
$E_{1,1} = \dfrac{2 \hbar^2 \pi^2}{2mL^2}$
For the ground state of the particle in a 2D box, there is one wavefunction (and no other) with this specific energy; the ground state and the energy level are said to be non-degenerate. However, in the 2-D box potential, the energy of a state depends upon the sum of the squares of the two quantum numbers. The particle having a particular value of energy in the excited state may has several different stationary states or wavefunctions. If so, these states and energy eigenvalues are said to be degenerate.
For the first excited state, three combinations of the quantum numbers $(n_x,\, n_y)$ are $(2,\,1)$ and $(1,2)$. The sum of squares of the quantum numbers in each combination is same (equal to 5). Each waveunction has same energy:
$E_{2,1} =E_{1,2} = \dfrac{5 \hbar^2 \pi^2}{2mL^2}$
Corresponding to these combinations three different wavefunctions and two different states are possible. Hence, the first excited state is said to be doubly degenerate. The number of independent wavefunctions for the stationary states with a shared energy is called as the degree of degeneracy of the energy level. The value of energy levels with the corresponding combinations and sum of squares of the quantum numbers
$n^2 \,= \, n_x^2+n_y^2$
as well as the degree of degeneracy are depicted in Table $1$.
Table $1$: Degeneracy properties of the particle in a 2-D box with $L_x=L$ and $L_y=L$.
$n_x^2+n_y^2$ Combinations of Degeneracy ($n_x$, $n_y$) Total Energy ($E_{n_x,n_y}$) Degree of Degeneracy
2 (1, 1) $\dfrac{2 \hbar^2 \pi^2}{2mL^2}$ 1
5 (2, 1) (1, 2) $\dfrac{5 \hbar^2 \pi^2}{2mL^2}$ 2
8 (2, 2) $\dfrac{8 \hbar^2 \pi^2}{2mL^2}$ 1
10 (3, 1) (1, 3) $\dfrac{10 \hbar^2 \pi^2}{2mL^2}$ 2
13 (3, 2) (2, 3) $\dfrac{13 \hbar^2 \pi^2}{2mL^2}$ 2
18 (3,3) $\dfrac{18 \hbar^2 \pi^2}{2mL^2}$ 1
Note that all the states above would be non-degenerate if $L_x \neq L_y$. However, a degeneracy will be "accidentally" observed for states that fulfill this requirement:
$\dfrac{n_x^2}{L_x^2} = \dfrac{n_y^2}{L_y^2}.$
Example $2$: Transition Energy
An electron in a 2D infinite potential well needs to absorb electromagnetic wave with wavelength 4040 nm (IR radiation) to be excited from lowest excited state to next higher energy state. What is the length of the box if this potential well is a square ($L_x=L_y=L$)?
Solution
The energy of a particle in a 2-D well is given by Equation \ref{energy}:
$E_{n_x, n_y}=\dfrac{\hbar^2 \pi^2}{2m}\left ( \dfrac{n_{x}^{2}}{L_{x}^{2}}+\dfrac{n_{y}^{2}}{L_{y}^{2}} \right ) \nonumber$
and for a square well $L_{x}=L_{y}=L$ this simplifies to Equation \ref{squareE}:
$E_{n_x, n_y}=\dfrac{\hbar^2 \pi^2}{2mL^2}\left ( n_{x}^{2}+ n_{y}^{2} \right ) \nonumber$
When the electron absorbs an electromagnetic wave and it transitions from one state (e.g., $(n_x=1,n_y=2)$ or $(n_x=2,n_y=1)$ since are degenerate states as Table $1$ demonstrates) to a higher energetic state, e.g., $(n_x=2,n_y=2$).
The energy of the lowest excited state is $\dfrac{5 h^{2}}{8mL^2}$ and the energy of the next higher excited state is $\dfrac{8 h^{2}}{8mL^2}$, so we can equate the difference in energies of these state to the energy of the absorbed photon
\begin{align*} h\nu &= E_{2,1} - E_{1,1} \[4pt] &=\dfrac{5\hbar^2 \pi^2}{2mL^2} - \dfrac{2\hbar^2 \pi^2}{2mL^2} \[4pt] &= \dfrac{3\hbar^2 \pi^2}{2mL^2} \end{align*}
Converting frequency ($\nu$) to wavelength ($\lambda$):
$\nu = \dfrac{c}{\lambda} \nonumber$
results in an expresstion relating wavelength of absorbed light to box length $L$:
$\dfrac{h c}{\lambda}= \dfrac{3 \hbar^2 \pi^2}{2mL^2} \nonumber$
and solving for $L$
$L = \sqrt{ \dfrac{3\hbar^2 \pi^2}{2mhc} \lambda} \nonumber$
or
\begin{align*} L &= \sqrt{ \dfrac{3\hbar \pi}{4mc} \lambda} \[4pt] &= 1.91\, nm\end{align*}
Exercise $2$
An electron in a 2D infinite potential well needs to absorb electromagnetic wave with wavelength 4040 nm to be excited from $(n_x=2, n_y=2)$ state to the $(n_x=3, n_y=3)$ state.
1. What is the length of the box if this potential well is a square ($L_x=L_y=L$)?
2. How many wavefunction exist between these two states (do not count the start or ending states)
Answer a
$L=3.5\, nm$
Answer b
From the energies in Table $1$, four wavefunctions exist between $(n_x=2, n_y=2)$ state to the $(n_x=3, n_y=3)$ state. However, due to degeneracy, only two possible energies exist.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/05.5%3A_Particle_in_Boxes/Particle_in_a_2-Dimensional_Box.txt
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Particle on a sphere is one out of the two models that describe rotational motion. A single particle travels on the surface of the sphere. Unlike particle in a box, the particle on a sphere requires angular momentum, $J$.
Introduction
There is a vector that contains direction of the axis of rotation. The magnitude of angular momentum of the particle that travels around the sphere can be defined as:
$J = pr$
where
• $p$ is linear momentum is the result of the mass and velocity of object (p=mv)
• $r$ is the radius of the sphere
The faster a particle travels in a sphere the higher the angular momentum. In other words, if we increase the velocity of a particle we get an increase in angular momentum. Therefore this required stronger torque to bring the particle to stop. Particle of mass is not restrict to move anywhere on the surface of the sphere radius. The potential energy of the particle on a sphere is zero because the particle can travel anywhere on the surface of the sphere without a preference in location; the particle on the sphere is infinity. Furthermore, the wave-function needs to satisfy two cyclic boundary conditions which are passing over the poles and around the equator of the sphere surrounding the central point. Using the Schrödinger equation we are able to find the energy of the particle:
$E = l (l +1) (h/2?) 2(1/2I)$
with $l = 0, 1, 2, 3, …$
We also know that the energy of the rotation of the particle is related to the classical angular momentum:
$E = \dfrac{J^2}{2I}$
$I$ is the moment of inertia of the particle; heavy mass in a large radius path has a large $I$. Because energy is quantized we can assume that these two equations can be compared with each other. Therefore the magnitude of the angular momentum is also limited to the values:
$J = \sqrt{L (L+1)} \hbar /2$
$L$ is the orbital angular momentum quantum number.
Considering motions in three dimensions, J has three components $J_x$, $J_y$, and $J_z$, along x, y, and z – axis. The angular momentums of z-axis are quantized and have the values as $J_z = m_l (h/2)$ with $m_l =l, …, 1, 0, -1, …, - l$ Where $m_l$ is the magnetic quantum number. The value of ml is restricted because of two cyclic boundary conditions, such that for ml equal to l there are 2l +1
Refrences
• Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. New York, N.Y.: W. H. Freeman Company, 2006. (361-362).
• Stephen Berry, and Stuart A. Rice, John Ross. Physical Chemistry. John Wiley and Sons 1980, R. (118-119).
• An Nguyen
Particle on a Ring
Learning Objectives
• To be familiar with a quantum system with angular symmetry.
• To be introduced to quantum angular momentum
The case of a quantum particle confined a one-dimensional ring is similar to the particle in a 1D box. Consider a variant of the one-dimensional particle in a box problem in which the x-axis is bent into a ring of radius $R$. We can write the same Schrödinger equation
$\dfrac{-\hbar^2}{2m} \dfrac{d^2 \psi(x)}{dx^2} = E \psi(x) \label{1}$
There are no boundary conditions in this case since the x-axis closes upon itself. A more appropriate independent variable for this problem is the angular position on the ring given by, $\phi = x {/} R$. The Schrödinger equation would then read
$-\dfrac{\hbar^2}{2mR^2} \dfrac{d^2 \psi (\phi)} {d (\phi)^2} = E \psi (\phi) \label{2}$
The kinetic energy of a body rotating in the xy-plane can be expressed as
$E = \dfrac{L_z^2}{2I} \label{3}$
where $I = mR^2$ is the moment of inertia and $L_z$, the z-component of angular momentum. (Since $L = r \times p$, if r and p lie in the xy-plane, L points in the z-direction.) The structure of Equation \ref{2} suggests that this angular-momentum operator is given by
$\hat{L_z} = -{i} \hbar \dfrac{\partial}{\partial \phi} \label{4}$
This result will follow from a more general derivation elsewhere. The Schrödinger Equation (Equation \ref{2}) can now be written more compactly as
$\psi \prime \prime \ (\phi) + m^2 \psi (\phi) = 0 \label{5}$
where
$m^2 \equiv 2IE/ \hbar^2\label{6}$
Do not confuse the variable $m$ with the mass of the particle!
Possible solutions to Equation \ref{5} are
$\psi (\phi) = \text{const}\, e^{\pm{i}m\phi} \label{7}$
For this wavefunction to be physically acceptable, it must be single-valued. Since $\phi$ increased by any multiple of 2$\pi$ represents the same point on the ring, we must have
$\psi (\phi + 2\pi ) = \psi (\phi) \label{8}$
and therefore
$e^{{i}m (\phi + 2\pi)} = e^{{i}m \phi} \label{9}$
This requires that
$e^{2\pi {i}m} = 1 \label{10}$
which is true only if m is an integer:
$m = 0, \pm 1, \pm 2...$
Using Equation \ref{6}, this gives the quantized energy values
$E_m = \dfrac{\hbar^2}{2I} m^2$
In contrast to the particle in a box, the eigenfunctions corresponding to $+m$ and $-m$ (Equation \ref{7}) are linearly independent, so both must be accepted. Therefore all eigenvalues, except $E_0$, are two-fold (or doubly) degenerate. The eigenfunctions can all be written in the form const $e^{{i}m \phi}$, with $m$ allowed to take either positive and negative values (or 0), as in Equation \ref{10}. The normalized eigenfunctions are
${\psi _{m}} (\phi) = \dfrac{1}{\sqrt{2 \pi}} e^{im\phi}$
and can be verified to satisfy the normalization condition containing the complex conjugate
$\int\limits_{0}^{2\pi} {\psi_{m}^*} (\phi) {\psi _{m}} (\phi) d\phi = 1$
where we have noted that ${\psi_{m}^*} (\phi) = (2\pi)^{-1/2} e^{-{i}m\phi}$. The mutual orthogonality of the functions also follows easily, for
\begin{align} \int\limits_{0}^{2\pi} {\psi_{m^\prime}^*} {\psi _{m}} (\phi) d\phi &= \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} e^{{i}(m-m^\prime) \phi} d\phi \[4pt] &= \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} [\cos(m-m^\prime)\phi + {i} \sin(m-m^\prime)\phi] d\phi =0 \end{align}
for $m^\prime \neq m$.
The solutions of Equation \ref{2} are also eigenfunctions of the angular momentum operator (Equation \ref{4}), with
$\hat{L_z} \psi_{m} (\phi) = m\hbar \psi_{m} (\phi), m = 0, \pm 1, \pm 2...$
This is a instance of a fundamental result in quantum mechanics, that any measured component of orbital angular momentum is restricted to integral multiples of $\hbar$. The Bohr theory of the hydrogen atom can be derived from this principle alone.
Example $1$: Free Electron Model for Aromatic Molecules
The benzene molecule consists of a ring of six carbon atoms around which six delocalized pi-electrons can circulate. A variant of the FEM for rings predicts the ground-state electron configuration which we can write as $1\pi^{2} 2\pi^{4}$, as shown here:
The enhanced stability the benzene molecule can be attributed to the complete shells of $\pi$-electron orbitals, analogous to the way that noble gas electron configurations achieve their stability. Naphthalene, apart from the central C-C bond, can be modeled as a ring containing 10 electrons in the next closed-shell configuration$1\pi^{2} 2\pi^{4} 3\pi^{4}$. These molecules fulfill Hückel's "4N+2 rule" for aromatic stability. The molecules cyclobutadiene ${(1\pi^{2} 2\pi^{2})}$ and cyclooctatetraene${(1\pi^{2} 2\pi^{4} 3\pi^{2})}$ , even though they consist of rings with alternating single and double bonds, do not exhibit aromatic stability since they contain partially-filled orbitals.
The longest wavelength absorption in the benzene spectrum can be estimated according to this model as
$\dfrac{hc}{\lambda} = E_2 - E_1 = \dfrac{\hbar^2}{2mR^2} {(2^2 -1^2)}$
The ring radius R can be approximated by the C-C distance in benzene, 1.39 Å. We predict $\lambda \approx$ 210 nm, whereas the experimental absorption has $\lambda_{max} \approx$ 268 nm.
Angular Momentum
One type of rotational motion in quantum mechanics is a particle in a ring. An important aspect of this is the angular momentum J which includes a vector with a direction that shows axis of rotation1. The particle’s magnitude of angular momentum that is traveling along a circular path of radius $r$ is classified as $J=p \times r$ where $p$ is the linear momentum at any moment. When the particle of mass m travels in the horizontal radius $r$, the particle has purely kinetic energy since potential energy is constant and is set to zero everywhere. The energy with regards of angular momentum can be expressed as:
$E = \dfrac{J^2_z}{2mr^2} \label{1C}$
• $J$ is the angular momentum in z-axis and
• $mr^2$ is the particle's moment of inertia I on the z-axis.
A particle has a moment of inertia I when traveling along a circular path. I is defined by m (mass) multiplied by $r^2$ (radius squared). The heavier particle in the top picture has a large moment of inertia on the central point while the lighter particle in the lower picture has a smaller moment of inertia while traveling on the path of the same radius1. For the heavier particle, the I is large and therefore, the particle's energy can be expressed by:
$E = \dfrac{J^2_z}{2I} \label{2C}$
We then use the de Broglie equation to quantize the energy of rotation. This is done by expressing the angular momentum in wavelengths:
$J_z = pr = \dfrac{hr}{\lambda} \label{3C}$
where p is the linear momentum and h is Planck's constant (6.626 x 10-34 Js). It can also be written:
$\lambda = \dfrac{h}{p}$
With this equation, de Broglie postulated that there is a wave correlated with the electron via wavelength. He had done this to explain Bohr's model of the Hydrogen atom, in which the electron is only allowed permitted to orbit from the nucleus at certain distances.
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• Anharmonic Oscillator
Anharmonic oscillation is defined as the deviation of a system from harmonic oscillation, or an oscillator not oscillating in simple harmonic motion. A harmonic oscillator obeys Hooke's Law and is an idealized expression that assumes that a system displaced from equilibrium responds with a restoring force whose magnitude is proportional to the displacement. In nature, idealized situations break down and fails to describe linear equations of motion.
• Harmonic Oscillator
The harmonic oscillator is a model which has several important applications in both classical and quantum mechanics. It serves as a prototype in the mathematical treatment of such diverse phenomena as elasticity, acoustics, AC circuits, molecular and crystal vibrations, electromagnetic fields and optical properties of matter.
• Kinetic Isotope Effects
The kinetic isotope effect (KIE) is a phenomenon associated with isotopically substituted molecules exhibiting different reaction rates. Isotope effects such as KIEs are invaluable tools in both physical and biological sciences and are used to aid in the understanding of reaction kinetics, mechanisms, and solvent effects.
06. One Dimensional Harmonic Oscillator
Anharmonic oscillation is defined as the deviation of a system from harmonic oscillation, or an oscillator not oscillating in simple harmonic motion.
Introduction
A harmonic oscillator obeys Hooke's Law and is an idealized expression that assumes that a system displaced from equilibrium responds with a restoring force whose magnitude is proportional to the displacement. In nature, idealized situations break down and fails to describe linear equations of motion. Anharmonic oscillation is described as the restoring force is no longer proportional to the displacement. Two forms of nonlinearity are used to describe real-world situations:
1. elastic anharmonicity
2. damping anharmonicity
Anharmonic oscillators can be approximated to a harmonic oscillator and the anharmonicity can be calculated using perturbation theory.
Figure $1$ shows the ground state potential well and is calculated using the energy levels of a harmonic oscillator with the first anharmonic correction. $D_o$ is the dissociation energy, which is different from the well depth $D_e$. The vibrational energy levels of this plot are calculated using the harmonic oscillator model:
$E_v = \left(v + \dfrac{1}{2}\right) v_e - \left(v + \dfrac{1}{2}\right)^2 v_e x_e + \left(v + \dfrac{1}{2}\right)^3 v_e y_e + higher \; terms$
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.
The lines in the first figure represent overtones correspond to the transitions of the quantum number $v$ which terminate at the top line = $v_{max}$. Because this line is less confining than a parabola, the energy levels become less widely spaced at high excitation. These overtones are present because the selection rule is derived from the properties of harmonic oscillator wavefunctions, which are only approximately valid in the presence of anharmonicity.
Harmonic Oscillator
The harmonic oscillator is a model which has several important applications in both classical and quantum mechanics. It serves as a prototype in the mathematical treatment of such diverse phenomena as elasticity, acoustics, AC circuits, molecular and crystal vibrations, electromagnetic fields and optical properties of matter.
Classical Oscillator
A simple realization of the harmonic oscillator in classical mechanics is a particle which is acted upon by a restoring force proportional to its displacement from its equilibrium position. Considering motion in one dimension, this means
$F = −kx \label{1}$
Such a force might originate from a spring which obeys Hooke’s law, as shown in Figure $1$. According to Hooke’s law, which applies to real springs for sufficiently small displacements, the restoring force is proportional to the displacement—either stretching or compression—from the equilibrium position.
The force constant $k$ is a measure of the stiffness of the spring. The variable $x$ is chosen equal to zero at the equilibrium position, positive for stretching, negative for compression. The negative sign in Equation $\ref{1}$ reflects the fact that $F$ is a restoring force, always in the opposite sense to the displacement $x$.
Applying Newton’s second law to the force from Equation $\ref{1}$, we find $x$
$F = m \dfrac{d^2 x}{dx^2} = -kx \label{2}$
where $m$ is the mass of the body attached to the spring, which is itself assumed massless. This leads to a differential equation of familiar form, although with different variables:
$\ddot{x}(t)+ \omega^2x(t)= 0 \label{3}$
with
$\omega^2 \equiv \dfrac{k}{m}$
The dot notation (introduced by Newton himself) is used in place of primes when the independent variable is time. The general solution to Equation $\ref{3}$ is
$x(t) = A\sin ωt + B\cos ωt \label{4}$
which represents periodic motion with a sinusoidal time dependence. This is known as simple harmonic motion and the corresponding system is known as a harmonic oscillator. The oscillation occurs with a constant angular frequency
$\omega = \sqrt{\dfrac{k}{m}}\; \text{radians per second} \label{5}$
This is called the natural frequency of the oscillator. The corresponding circular (or angular) frequency in Hertz (cycles per second) is
$\nu = \dfrac{\omega}{2\pi } = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}\; \text{Hz} \label{6}$
The general relation between force and potential energy in a conservative system in one dimension is
$F =\dfrac{−dV}{dx} \label{7}$
Thus the potential energy of a harmonic oscillator is given by
$V(x) = \dfrac{1}{2}kx^2 \label{8}$
which has the shape of a parabola, as drawn in Figure $2$. A simple computation shows that the oscillator moves between positive and negative turning points $\pm x_{max}$ where the total energy $E$ equals the potential energy $\dfrac{1}{2} k x_{max}^{2}$ while the kinetic energy is momentarily zero. In contrast, when the oscillator moves past $x = 0$, the kinetic energy reaches its maximum value while the potential energy equals zero.
Harmonic Oscillator in Quantum Mechanics
Given the potential energy in Equation $\ref{8}$, we can write down the Schrödinger equation for the one-dimensional harmonic oscillator:
$-\dfrac{\hbar^{2}}{2m} \psi''(x) + \dfrac{1}{2}kx^2 \psi(x) = E \psi(x) \label{9}$
For the first time we encounter a differential equation with non-constant coefficients, which is a much greater challenge to solve. We can combine the constants in Equation $\ref{9}$ to two parameters
$\alpha^2 = \dfrac{mk}{\hbar^2}$
and
$\lambda = \dfrac{2mE}{\hbar^2\alpha} \label{10}$
and redefine the independent variable as
$\xi = \alpha^{1/2}x \label{11}$
This reduces the Schrödinger equation to
$\psi''(\xi) + (\lambda-\xi^2)\psi(\xi) = 0\label{12}$
The range of the variable $x$ (also $\xi$) must be taken from $−\infty$ to $+\infty$, there being no finite cutoff as in the case of the particle in a box. A useful first step is to determine the asymptotic solution to Equation $\ref{11}$, that is, the form of $\psi(\xi)$ as $\xi\rightarrow\pm\infty$. For sufficiently large values of $\lvert\xi\rvert$, $\xi^{2} \gg \lambda$ and the differential equation is approximated by
$\psi''(\xi) - \xi^2\psi(\xi) \approx 0 \label{13}$
This suggests the following manipulation:
$\left(\dfrac{d^2}{d\xi^2} - \xi^2 \right) \psi(\xi) \approx \left( \dfrac{d}{d\xi}-\xi \right) \left( \dfrac{d}{d\xi}+\xi \right) \psi(\xi) \approx 0 \label{14}$
The first-order differential equation
$\psi'(\xi) + \xi\psi(\xi)=0 \label{15}$
can be solved exactly to give
$\psi(\xi) = \text{const.}\, e^{-\xi^2/2} \label{16}$
Remarkably, this turns out to be an exact solution of the Schrödinger equation (Equation $\ref{12}$) with $\lambda=1$. Using Equation $\ref{10}$, this corresponds to an energy
$E=\dfrac{\lambda\hbar^2\alpha}{2m} = \dfrac{1}{2}\hbar\sqrt{\dfrac{k}{m}} = \dfrac{1}{2} \hbar\omega \label{17}$
where $\omega$ is the natural frequency of the oscillator according to classical mechanics. The function in Equation $\ref{16}$ has the form of a Gaussian, the bell-shaped curve so beloved in the social sciences. The function has no nodes, which leads us to conclude that this represents the ground state of the system.The ground state is usually designated with the quantum number $n = 0$ (the particle in a box is a exception, with $n = 1$ labeling the ground state). Reverting to the original variable $x$, we write
$\psi_{0}(x) = \text{const} e^{-\alpha x^2/2}$
with
$\alpha=(mk/\hbar^2)^{1/2} \label{18}$
With help of the well-known definite integral (Laplace 1778)
$\int^{\infty}_{-\infty} e^{- \alpha x^{2}} dx= \sqrt{\dfrac{\pi}{\alpha}} \label{19}$
we find the normalized eigenfunction
$\psi_{0}(x)=(\dfrac{\alpha}{\pi})^{1/4} e^{-\alpha x^{2}/2} \label{20}$
with the corresponding eigenvalue
$E_{0}=\dfrac{1}{2}\hbar\omega \label{21}$
Drawing from our experience with the particle in a box, we might surmise that the first excited state of the harmonic oscillator would be a function similar to Equation $\ref{20}$, but with a node at $x=0$, say,
$\psi_{1}(x)=const x e^{-\alpha x^{2}/2} \label{22}$
This is orthogonal to $\psi_0(x)$ by symmetry and is indeed an eigenfunction with the eigenvalue
$E_{1}=\dfrac{3}{2}\hbar\omega \label{23}$
Continuing the process, we try a function with two nodes
$\psi_{2}= const (x^{2}-a) e^{-\alpha x^{2}/2} \label{24}$
Using the integrals tabulated in the Supplement 5, on Gaussian Integrals, we determine that with $a=\dfrac{1}{2}$ makes $\psi_{2}(x)$ orthogonal to $\psi_{0}(x)$ and $\psi_{1}(x)$. We verify that this is another eigenfunction, corresponding to
$E_{2}=\dfrac{5}{2}\hbar\omega \label{25}$
The general result, which follows from a more advanced mathematical analysis, gives the following formula for the normalized eigenfunctions:
$\psi_{n}(x)=(\dfrac{\sqrt{\alpha}}{2^{n}n!\sqrt{\pi}})^{1/2} H_{n}(\sqrt{\alpha}x) e^{-\alpha x^{2}/2} \label{26}$
where $H_{n}(\xi)$ represents the Hermite polynomial of degree $n$. The first few Hermite polynomials are
$H_{0}(\xi)=1$
$H_{1}(\xi)=2\xi$
$H_{2}(\xi)=4\xi^{2}-2$
$H_{3}(\xi)=8\xi^{3}-12\xi \label{27}$
The four lowest harmonic-oscillator eigenfunctions are plotted in Figure $3$. Note the topological resemblance to the corresponding particle-in-a-box eigenfunctions.
The eigenvalues are given by the simple formula
$E_{n}=\left(n+\dfrac{1}{2}\right)\hbar\omega \label{28}$
These are drawn in Figure $2$, on the same scale as the potential energy. The ground-state energy $E_{0}=\dfrac{1}{2}\hbar\omega$ is greater than the classical value of zero, again a consequence of the uncertainty principle. This means that the oscillator is always oscillating.
It is remarkable that the difference between successive energy eigenvalues has a constant value
$\Delta E=E_{n+1}-E_{n}=\hbar\omega=h\nu \label{29}$
This is reminiscent of Planck’s formula for the energy of a photon. It comes as no surprise then that the quantum theory of radiation has the structure of an assembly of oscillators, with each oscillator representing a mode of electromagnetic waves of a specified frequency.
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The kinetic isotope effect (KIE) is a phenomenon associated with isotopically substituted molecules exhibiting different reaction rates. Isotope effects such as KIEs are invaluable tools in both physical and biological sciences and are used to aid in the understanding of reaction kinetics, mechanisms, and solvent effects.
Introduction
Research was first introduced on this topic over 50 years ago and has grown into an enormous field. The scientists behind much of the understanding and development of kinetic isotope effects were Jacob Bigeleisen and Maria Goeppert Mayer who published the first paper on isotope effects [J. Chem. Phys., 15, 261 (1947)]. Kinetic isotope effects specifically explore the change in rate of a reaction due to isotopic substitution.
An element is identified by its symbol, mass number, and atomic number. The atomic number is the number of protons in the nucleus while the mass number is the total number of protons and neutrons in the nucleus. Isotopes are two atoms of the same element that have the same number of protons but different numbers of neutrons. Isotopes are specified by the mass number.
As an example consider the two isotopes of chlorine, you can see that their mass numbers vary, with 35Cl being the most abundant isotope, while their atomic numbers remain the same at 17.
$^{35}Cl \;\text{and}\; ^{37}Cl$
The most common isotope used in light atom isotope effects is hydrogen ($^{1}H$) commonly replaced by its isotope deuterium ($^{2}H$). Note: Hydrogen also has a third isotope, tritium ($^{2}H$). Isotopes commonly used in heavy atom isotope effects include carbon ($^{12}C$, $^{13}C$, nitrogen ($^{14}N$, $^{15}N$), oxygen, sulfur, and bromine. Not all elements exhibit reasonably stable isotopes (i.e. Fluorine, $^{19}F$), but those that due serve as powerful tools in isotope effects.
Potential Energy Surfaces
Understanding potential energy surfaces is important in order to be able to understand why and how isotope effects occur as they do. The harmonic oscillator approximation is used to explain the vibrations of a diatomic molecule. The energies resulting from the quantum mechanic solution for the harmonic oscillator help to define the internuclear potential energy of a diatomic molecule and are
$E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{1}$
where
• n is a positive integer (n=1,2,3...),
• h is Planck's constant and
• $\nu$ is the frequency of vibration.
The Morse potential is an analytic expression that is used as an approximation to the intermolecular potential energy curves:
$V(l) = D_e{\left(1-e^{-\beta(l-l_o)}\right)}^2 \label{2}$
where
• $V(l)$ is the potential energy,
• $D_e$ is the dissociation energy of the molecule,
• $\beta$ is the measure of the curvature of the potential at its minimum,
• $l$ is displacement, and
• $l_o$ is the equilibrium bond length.
The $D_e$, $\beta$, and $l_o$ variables can be looked up in a textbook or CRC handbook.
Below is an example of a Morse potential curve with the zero point vibrational energies of two isotopic molecules (for example R-H and R-D where R is a group/atom that is much heavier than H or D). The y-axis is potential energy and the x axis is internuclear distance. In this figure EDo and EHo correspond to the zero point energies of deuterium and hydrogen. The zero point energy is the lowest possible energy of a system and equates to the ground state energy. Zero point energy is dependent upon the reduced mass of the molecule as will be shown in the next section. The heavier the molecule or atom, the lower the frequency of vibration and the smaller the zero point energy. Lighter molecules or atoms have a greater frequency of vibration and a higher zero point energy. We see this is the figure below where deuterium is heavier than hydrogen and therefore has the lower zero point energy.
This results in different bond dissociation energies for R-D and R-H. The bond dissociation energy for R-D (ED) is greater than the bond dissociation energy of R-H (EH). This difference in energy due to isotopic replacement results in differing rates of reaction, the effect that is measured in kinetic isotope effects. The reaction rate for the conversion of R-D is slower than the reaction rate for the conversion of R-H.
p>
It is important to note that isotope replacement does not change the electronic structure of the molecule or the potential energy surfaces of the reactions the molecule may undergo. Only the rate of the reaction is affected.
Activation Energies
The energy of the vibrational levels of a vibration (i.e., a bond) in a molecule is given by
$E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{3}$
where we assume that the molecule is in its ground state and we can compare zero-point vibrational energies,
$E_o = \left(\dfrac{1}{2}\right)hv \label{4}$
Using the harmonic oscillator approximation the fundamental vibrational frequency is
$\nu = \dfrac{1}{2 \pi} \sqrt{ \dfrac {k}{\mu} } \label{5}$
where
• $k$ is the force constant of the bond and
• $\mu$ is the reduced mass
$\mu = \dfrac{m_1m_2}{m_1+m_2} \label{6}$
The Arrhenius equation is used to determine reaction rates and activation energies and since we are interested in the change in rate of reactions with different isotopes, this equation is very important,
$k = Ae^{-\frac{E_a}{kT}} \label{7}$
where
• $k$ is the reaction rate,
• $E_a$ is the activation energy, and
• $A$ is the Arrhenius constant.
The Arrhenius equation can be used to compare the rates of a reaction with R-H and R-D,
$k_H = A_He^{-\frac{E_a^H}{kT}} \label{8}$
$k_D = A_De^{-\frac{E_a^D}{kT}} \label{9}$
where kH and kD are the rates of reaction associated with R-H and the isotope substituted R-D. We will then assume the Arrhenius constants are equal ($A_H=A_D$). The ratio of the rates of reaction gives an approximation for the isotope effect resulting in:
$\dfrac{k_H}{k_D} = e^{-\frac{E_a^H - E_a^D}{kT}} \label{10}$
By using the relationship that for both R-H and R-D
$E_o = \left(\dfrac{1}{2}\right)h\nu \label{11}$
a substitution can be made resulting in
$\dfrac{k_H}{k_D} = e^{\frac{h(\nu_H - \nu_D)}{2kT}} \label{12}$
The vibrational frequency (Equation 5) can then be substituted for R-H and R-D and the value of the expected isotope effect can be calculated.
$\dfrac{k_H}{k_D} = e^{\dfrac {h \left( \dfrac{k_{RH}}{\mu_{RH}} - \dfrac{k_{RD}}{\mu_{RD}} \right)}{4\pi kT}} \label{13}$
The same general procedure can be followed for any isotope substitution.
In summary, the greater the mass the more energy is needed to break bonds. A heavier isotope forms a stronger bond. The resulting molecule has less of a tendency to dissociate. The increase in energy needed to break the bond results in a slower reaction rate and the observed isotope effect.
Kinetic Isotope Effects
Kinetic Isotope Effects (KIEs) are used to determine reaction mechanisms by determining rate limiting steps and transition states and are commonly measured using NMR to detect isotope location or GC/MS to detect mass changes. In a KIE experiment an atom is replaced by its isotope and the change in rate of the reaction is observed. A very common isotope substitution is when hydrogen is replaced by deuterium. This is known as a deuterium effect and is expressed by the ratio kH/kD (as explained above). Normal KIEs for the deuterium effect are around 1 to 7 or 8. Large effects are seen because the percentage mass change between hydrogen and deuterium is great. Heavy atom isotope effects involve the substitution of carbon, oxygen, nitrogen, sulfur, and bromine, with effects that are much smaller and are usually between 1.02 and 1.10. The difference in KIE magnitude is directly related to the percentage change in mass. Large effects are seen when hydrogen is replaced with deuterium because the percentage mass change is very large (mass is being doubled) while smaller percent mass changes are present when an atom like sulfur is replaced with its isotope (increased by two mass units).
Primary KIEs
Primary kinetic isotope effects are rate changes due to isotopic substitution at a site of bond breaking in the rate determining step of a reaction.
Example
Consider the bromination of acetone: kinetic studies have been performed that show the rate of this reaction is independent of the concentration of bromine. To determine the rate determining step and mechanism of this reaction the substitution of a deuterium for a hydrogen can be made.
When hydrogen was replaced with deuterium in this reaction a $k_H \over k_D$ of 7 was found. Therefore the rate determining step is the tautomerization of acetone and involves the breaking of a C-H bond. Since the breaking of a C-H bond is involved, a substantial isotope effect is expected.
Heavy Atom Isotope Effects
A rule of thumb for heavy atom isotope effects is that the maximum isotopic rate ratio is proportional to the square root of the inverse ratio of isotopic masses.
• Expected: $\dfrac{k_{32}}{k_{34}} = \sqrt{\dfrac{34}{32}}=1.031$
• Experimental: $\dfrac{k_{32}}{k_{34}} = 1.072$
Secondary KIEs
Secondary kinetic isotope effects are rate changes due to isotopic substitutions at a site other than the bond breaking site in the rate determining step of the reaction. These come in three forms: $\alpha$, $\beta$, and $\gamma$ effects.
$\beta$ secondary isotope effects occur when the isotope is substituted at a position next to the bond being broken.
$\ce{(CH3)2CHBr + H2O ->[k_H] (CH3)2CHOH}$
$\ce{(CD3)2CHBr + H2O ->[k_D] (CD3)2CHOH}$
This is thought to be due to hyperconjugation in the transition state. Hyperconjugation involves a transfer of electron density from a sigma bond to an empty p orbital (for more on hyperconjugation see outside links).
Solvent Effects in Reactions
Reactions may be affected by the type of solvent used (for example H2O to D2O or ROH to ROD). There are three main ways solvents effect reactions:
1. The solvent can act as a reactant resulting in a primary isotope effect.
2. Rapid hydrogen exchange can occur between substrate molecules labeled with deuterium and hydrogen atoms in the solvent. Deuterium may change positions in the molecule resulting in a new molecule that is then reacted in the rate determining step of the reaction.
3. The nature of solvent and solute interactions may also change with differing solvents. This could change the energy of the transition state and result in a secondary isotope effects.
References
1. Baldwin, J.E., Gallagher, S.S., Leber, P.A., Raghavan, A.S., Shukla, R.; J. Org. Chem. 2004, 69, 7212-7219 (This is a great paper using kinetic isotope effects to determine a reaction mechanism. It will interest the organic chemistry oriented reader.)
2. Bigeleisen, J., Goeppert, M., J. Chem. Phys. 1947, 15, 261.
3. Chang, R.; Physical Chemistry for the Chemical and Biological Sciences; University Science Books: Sausalito, CA, 2000, pp 480-483.
4. Isaacs, N.; Physical Organic Chemistry; John Wiley & Sons Inc.: New York, NY; 1995, 2nd ed, pp 287-313.
5. March, J., Smith, M.B.; March’s Advanced Organic Chemistry; John Wiley & Sons, Inc.: Hoboken, NJ, 2007; 6th ed.
6. McMurry, J.; Organic Chemistry; Brooks & Cole: Belmont, CA; 2004, 6th ed.
7. McQuarrie, D.; Quantum Chemistry; University Science Books: Sausalito, CA, 2008, 2nd ed.
8. Rouhi, A.; C&EN. 1997, 38-42.
Problems
1. Describe the difference between primary and secondary kinetic isotope effects.
2. Estimate the kN-H/kN-D for a deuterium substitution on nitrogen given that vH=9.3x1013 Hz and the activation energy is equal to 5.31 kJ/mol.
3. Using the 'rule of thumb' for heavy isotope effects, calculate the expected effect for a bromine isotope substitution, 79Br and 81Br.
4. Explain some of the main ways kinetic isotope effects are used.
5. As discussed, the rate-limiting step in the bromination of acetone is the breaking of a carbon-hydrogen bond. Estimate kC-H/KC-D for this reaction at 285 K. (Given: vtildeC-H=3000 cm-1 and vtildeC-D=2100 cm-1)
Solutions
1. Primary isotope effects involve isotopic substitution at the bond being broken in a reaction, while secondary isotope effects involve isotopic substituion on bonds adjacent to the bond being broken.
2. 8.5
3. 1.0126
4. To determine reaction mechanisms, to determine rate limiting steps in reactions, to determine transition states in reactions.
5. 9.685
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• 6: Angular Momentum
Angular momentum is the rotational analog of linear momentum. It is an important quantity in classical physics because it is a conserved quantity. The extension of this concept to particles in the quantum world is straightforward.
• 7.5: Rigid Rotor
Microwave rotational spectroscopy uses microwave radiation to measure the energies of rotational transitions for molecules in the gas phase. It accomplishes this through the interaction of the electric dipole moment of the molecules with the electromagnetic field of the exciting microwave photon.
• Spherical Harmonics
Spherical Harmonics are a group of functions used in math and the physical sciences to solve problems in disciplines including geometry, partial differential equations, and group theory.
07. Angular Momentum
Particle in a Ring
Consider a variant of the one-dimensional particle in a box problem in which the x-axis is bent into a ring of radius R. We can write the same Schrödinger equation
$\dfrac{-\hbar^2}{2m} \dfrac{d^2 \psi(x)}{dx^2} = E \psi(x) \label{1}$
There are no boundary conditions in this case since the x-axis closes upon itself. A more appropriate independent variable for this problem is the angular position on the ring given by, $\phi = x {/} R$ . The Schrödinger equation would then read
$-\dfrac{\hbar^2}{2mR^2} \dfrac{d^2 \psi (\phi)} {d (\phi)^2} = E \psi (\phi) \label{2}$
The kinetic energy of a body rotating in the xy-plane can be expressed as
$E = \dfrac{L_z^2}{2I} \label{3}$
where
$I = mR^2$
is the moment of inertia and $L_z$, the z-component of angular momentum. (Since $L = r \times p$, if r and p lie in the xy-plane, L points in the z-direction.) The structure of Equation $\ref{2}$ suggests that this angular-momentum operator is given by
$\hat{L_z} = -{i} \hbar \dfrac{\partial}{\partial \phi} \label{4}$
This result will follow from a more general derivation in the following Section. The Schrödinger equation (Equation $\ref{2}$) can now be written more compactly as
$\psi \prime \prime \ (\phi) + m^2 \psi (\phi) = 0 \label{5}$
where
$m^2 \equiv 2IE/ \hbar^2\label{6}$
(Please do not confuse this variable m with the mass of the particle!) Possible solutions to (Equation $\ref{5}$) are
$\psi (\phi) = \text{const}\, e^{\pm{i}m\phi} \label{7}$
For this wavefunction to be physically acceptable, it must be single-valued. Since $\phi$ increased by any multiple of $2\pi$ represents the same point on the ring, we must have
$\psi (\phi + 2\pi ) = \psi (\phi) \label{8}$
and therefore
$e^{{i}m (\phi + 2\pi)} = e^{{i}m \phi} \label{9}$
This requires that
$e^{2\pi {i}m} = 1 \label{10}$
which is true only if $m$ is an integer:
$m = 0, \pm 1, \pm 2... \label{11}$
Using Equation $\ref{6}$, this gives the quantized energy values
$E_m = \dfrac{\hbar^2}{2I} m^2 \label{12}$
In contrast to the particle in a box, the eigenfunctions corresponding to $+m$ and $-m$ (Equation $\ref{7}$) are linearly independent, so both must be accepted. Therefore all eigenvalues, except $E_0$, are two-fold (or doubly) degenerate. The eigenfunctions can all be written in the form const $e^{{i}m \phi}$, with m allowed to take either positive and negative values (or 0), as in Equation $\ref{10}$. The normalized eigenfunctions are
${\psi _{m}} (\phi) = \dfrac{1}{\sqrt{2 \pi}} e^{im\phi} \label{13}$
and can be verified to satisfy the normalization condition containing the complex conjugate
$\int\limits_{0}^{2\pi} {\psi_{m}^*} (\phi) {\psi _{m}} (\phi) d\phi = 1$
where we have noted that ${\psi_{m}^*} (\phi) = (2\pi)^{-1/2} e^{-{i}m\phi}$. The mutual orthogonality of the functions (Equation $\ref{13}$) also follows easily, for
$\int\limits_{0}^{2\pi} {\psi_{m^\prime}^*} {\psi _{m}} (\phi) d\phi = \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} e^{{i}(m-m^\prime) \phi} d\phi$
$= \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} [cos(m-m^\prime)\phi + {i} sin(m-m^\prime)\phi] d\phi =0$
for $m^\prime \neq m$.
The solutions in Equation $\ref{12}$ are also eigenfunctions of the angular momentum operator (Equation $\ref{4}$), with
$\hat{L_z} \psi_{m} (\phi) = m\hbar \psi_{m} (\phi), m = 0, \pm 1, \pm 2...$
This is a instance of a fundamental result in quantum mechanics, that any measured component of orbital angular momentum is restricted to integral multiples of $\hbar$. The Bohr theory of the hydrogen atom, to be discussed in the next Chapter, can be derived from this principle alone.
Free Electron Model for Aromatic Molecules
The benzene molecule consists of a ring of six carbon atoms around which six delocalized pi-electrons can circulate. A variant of the FEM for rings predicts the ground-state electron configuration which we can write as $1\pi^{2} 2\pi^{4}$, as shown here:
The enhanced stability the benzene molecule can be attributed to the complete shells of $\pi$-electron orbitals, analogous to the way that noble gas electron configurations achieve their stability. Naphthalene, apart from the central C-C bond, can be modeled as a ring containing 10 electrons in the next closed-shell configuration$1\pi^{2} 2\pi^{4} 3\pi^{4}$. These molecules fulfill Hückel's "4N+2 rule" for aromatic stability. The molecules cyclobutadiene ${(1\pi^{2} 2\pi^{2})}$ and cyclooctatetraene${(1\pi^{2} 2\pi^{4} 3\pi^{2})}$, even though they consist of rings with alternating single and double bonds, do not exhibit aromatic stability since they contain partially-filled orbitals.
The longest wavelength absorption in the benzene spectrum can be estimated according to this model as
$\dfrac{hc}{\lambda} = E_2 - E_1 = \dfrac{\hbar^2}{2mR^2} {(2^2 -1^2)}$
The ring radius R can be approximated by the C-C distance in benzene, 1.39 Å. We predict $\lambda \approx$ 210 nm, whereas the experimental absorption has $\lambda_{max} \approx$ 268 nm.
Spherical Polar Coordinates
The motion of a free particle on the surface of a sphere will involve components of angular momentum in three-dimensional space. Spherical polar coordinates provide the most convenient description for this and related problems with spherical symmetry. The position of an arbitrary point r is described by three coordinates $r , \theta, \phi$ as shown in Figure $2$.
These are connected to Cartesian coordinates by the relations
$x = r \sin\theta \cos \phi$
$y = r \sin \theta \sin \phi$
$z = r \cos \theta$
The radial variable r represents the distance from r to the origin, or the length of the vector r:
$r= \sqrt{x^2 +y^2 +z^2}$
The coordinate $\theta$ is the angle between the vector r and the z-axis, similar to latitude in geography, but with $\theta= 0$ and $\theta = \pi$ corresponding to the North and South Poles, respectively. The angle $\phi$ describes the rotation of r about the z-axis, running from 0 to $2 \pi$, similar to geographic longitude. The volume element in spherical polar coordinates is given by
$d \tau = r^2 \sin \theta dr d \theta d\phi,$
$r \in \{0, \infty \} , \theta \in \{0, \pi\}, \phi \in \{0, 2\pi \}$
and represented graphically by the distorted cube in Figure $1$.
Figure $3$: Volume element in spherical polar coordinates. (CC BY; OpenStax).
We also require the Laplacian operator
$\nabla^{2} = \dfrac{1}{r^{2}} \dfrac{\partial}{\partial r} r^{2} \dfrac{\partial}{\partial r} + \dfrac{1}{r^2 sin \theta} \dfrac{\partial}{\partial \theta } sin \theta \dfrac{\partial}{\partial \theta } + \dfrac{1}{r^2 sin^2 \theta} \dfrac{\partial^2}{\partial \phi^2}$
A detailed derivation is given in Supplement 6.
Rotation in Three Dimensions
A particle of mass M, free to move on the surface of a sphere of radius R, can be located by the two angular variables $\theta, \phi$. The Schrödinger equation therefore has the form
$-\dfrac{\hbar^2}{2M} \nabla^{2} Y ({\theta , \phi}) = E Y ({\theta , \phi})$
with the wavefunction conventionally written as $Y ({\theta , \phi})$. These functions are known as spherical harmonics and have been used in applied mathematics long before quantum mechanics. Since $r = R$, a constant, the first term in the Laplacian does not contribute. The Schrödinger equation reduces to
$\left\{ \dfrac{1}{sin \theta} \dfrac{\partial}{\partial \theta } sin \theta \dfrac{\partial}{\partial \theta } + \dfrac{1}{sin^2 \theta} \dfrac{\partial^2}{\partial \phi^2} + \lambda \right\} Y ({ \theta , \phi}) = 0$
where
$\lambda = \dfrac{2MR^2E}{\hbar^2} = \dfrac{2IE}{\hbar^2}$
again introducing the moment of inertia $I = MR^2$. The variables $\theta$ and $\phi$ can be separated in Equation $\ref{22}$ after multiplying through by $sin^2 \theta$. If we write
$Y ({\theta , \phi }) = \Theta ({\theta }) \Phi ({\phi })$
and follow the procedure used for the three-dimensional box, we find that dependence on $\phi$ alone occurs in the term
$\dfrac{\Phi^{\prime \prime} ({\phi)} }{\Phi ({\phi}) } = const$
This is identical in form to Equation $\ref{5}$, with the constant equal to $-m^2$, and we can write down the analogous solutions
$\Phi_{m}({\phi}) = \sqrt{\dfrac{1}{2 \pi}} e^{im \phi}, m=0, \pm 1, \pm 2 ...$
Substituting Equation $\ref{24}$ into Equation $\ref{22}$ and cancelling the functions $\Phi ({\phi })$, we obtain an ordinary differential equation for $\Theta ({\theta })$
$\left \{ \dfrac{1}{sin \theta} \dfrac{d}{d \theta} sin \theta \dfrac{d}{d \theta} - \dfrac{m^2}{sin^2 \theta} + \lambda \right \} \Theta ({\theta}) = 0$
Consulting our friendly neighborhood mathematician, we learn that the single-valued, finite solutions to (Equation $\ref{27}$) are known as associated Legendre functions. The parameters $\lambda$ and $m$ are restricted to the values
$\lambda = \ell ({ \ell + 1}) , \ell = 0, 1, 2 ...$
while
$m = 0, \pm 1, \pm 2 ... \pm \ell ({2 \ell +1 values})$
Putting Equation $\ref{28}$ into Equation $\ref{23}$, the allowed energy levels for a particle on a sphere are found to be
$E_{\ell} = \dfrac{\hbar^2}{2I} \ell ({ \ell + 1})$
Since the energy is independent of the second quantum number m, the levels (Equation $\ref{30}$) are $({2 \ell+1})$-fold degenerate. The spherical harmonics constitute an orthonormal set satisfying the integral relations
$\int_0^{\pi} \int_0^{2 \pi} Y_{\ell^{\prime} m^{\prime}}^* ({ \theta , \phi }) Y_{\ell m} ({\theta , \phi}) sin \theta d \theta d \phi = \delta_{\ell \ell^{\prime}} \delta_{mm^{\prime}}$
The following table lists the spherical harmonics through $\ell$ = 2, which will be sufficient for our purposes.
$Spherical Harmonics Y_{\ell m} ({\theta , \phi})$
$Y_{00} = \left({\dfrac{1}{4 \pi}} \right)^{1/2}$
$Y_{10} = \left({\dfrac{3}{4 \pi}} \right)^{1/2} cos \theta$
$Y_{1 \pm 1} = \mp \left({\dfrac{3}{4 \pi}} \right)^{1/2} sin \theta e^{\pm i \phi}$
$Y_{20} = \left({\dfrac{5}{16 \pi}} \right)^{1/2} ({ 3 cos^2 \theta - 1})$
$Y_{2 \pm 1} = \mp \left({\dfrac{15}{8 \pi}} \right)^{1/2} cos \theta sin \theta e^{\pm i \phi}$
$Y_{2 \pm 2} = \left({\dfrac{15}{32 \pi}} \right)^{1/2} sin^2 \theta e^{\pm 2i \phi}$
A graphical representation of these functions is given in Figure $4$. Surfaces of constant absolute value are drawn, positive where green and negative where red.
Theory of Angular Momentum
Generalization of the energy-angular momentum relation in Equation $\ref{3}$ to three dimensions gives
$E = \dfrac{L^2}{2I}$
Thus from Equation $\ref{21}$-$\ref{23}$ we can identify the operator for the square of total angular momentum
$\hat{L^2} = -\hbar^2 \left\{ \dfrac{1}{sin \theta} \dfrac{\partial}{\partial \theta} sin \theta \dfrac{\partial}{\partial \theta} + \dfrac{1}{sin^2 \theta} \dfrac{\partial^{2}}{\partial \phi^{2}} \right\}$
By Equations $\ref{28}$ and $\ref{29}$, the functions $Y ({ \theta , \phi})$ are simultaneous eigenfunctions of $\hat{L^2}$ and $\hat{L}_z$ such that
$\hat{L^2} Y_{\ell m} ({\theta , \phi}) = \ell ({\ell + 1}) \hbar^2 Y_{\ell m} ({\theta , \phi})$
and
$\hat{L}_z Y_{\ell m} ({\theta, \phi}) = m \hbar Y_{\ell m} ({\theta , \phi})$
But the $Y_{\ell m} ({\theta , \phi})$ are not eigenfunctions of either $L_x$ and $L_y$ (unless $\ell$ = 0). Note that the magnitude of the total angular momentum $\sqrt{\ell ({\ell +1}) } \hbar$ is greater than its maximum observable component in any direction, namely $\ell \hbar$. The quantum-mechanical behavior of the angular momentum and its components can be represented by a vector model, illustrated in Figure 5. The angular momentum vector L, with magnitude $\sqrt{\ell ({\ell +1}) } \hbar$, can be pictured as precessing about the z-axis, with its z-component $L_z$ constant. The components $L_x$ and $L_y$ fluctuate in the course of precession, corresponding to the fact that the system is not in an eigenstate of either. There are 2$\ell$ + 1 different allowed values for $L_z$, with eigenvalues $m \hbar ({ m = 0, \pm 1, \pm 2 ... \pm \ell })$ equally spaced between $+ \ell \hbar$ and $- \ell \hbar$.
Figure $5$: Vector model for angular momentum, showing the case $\ell$= 2. (Public Domain; Maschen).
This discreteness in the allowed directions of the angular momentum vector is called space quantization. The existence of simultaneous eigenstates of $\hat{L^2}$ and any one component, conventionally $\hat{L}_z$, is consistent with the commutation relations derived in Chap. 4:
$\left[ \hat{L}_x , \hat{L}_y \right] = i \hbar \hat{L}_z et cyc$
and
$\left[ \hat{L^2} , \hat{L}_z \right] = 0$
Electron Spin
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.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/07._Angular_Momentum/6%3A_Angular_Momentum.txt
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Microwave rotational spectroscopy uses microwave radiation to measure the energies of rotational transitions for molecules in the gas phase. It accomplishes this through the interaction of the electric dipole moment of the molecules with the electromagnetic field of the exciting microwave photon.
Introduction
To probe the pure rotational transitions for molecules, scientists use microwave rotational spectroscopy. This spectroscopy utilizes photons in the microwave range to cause transitions between the quantum rotational energy levels of a gas molecule. The reason why the sample must be in the gas phase is due to intermolecular interactions hindering rotations in the liquid and solid phases of the molecule. For microwave spectroscopy, molecules can be broken down into 5 categories based on their shape and the inertia around their 3 orthogonal rotational axes. These 5 categories include diatomic molecules, linear molecules, spherical tops, symmetric tops and asymmetric tops.
Classical Mechanics
The Hamiltonian solution to the rigid rotor is
$H = T$
since,
$H = T + V$
Where $T$ is kinetic energy and $V$ is potential energy. Potential energy, $V$, is 0 because there is no resistance to the rotation (similar to a particle in a box model).
Since $H = T$, we can also say that:
${T = }\dfrac{1}{2}\sum{m_{i}v_{i}^2}$
However, we have to determine $v_i$ in terms of rotation since we are dealing with rotation. Since,
$\omega = \dfrac{v}{r}$
where $\omega$ = angular velocity, we can say that:
$v_{i} = \omega{X}r_{i}$
Thus we can rewrite the T equation as:
$T = \dfrac{1}{2}\sum{m_{i}v_{i}\left(\omega{X}r_{i}\right)}$
Since $\omega$ is a scalar constant, we can rewrite the T equation as:
$T = \dfrac{\omega}{2}\sum{m_{i}\left(v_{i}{X}r_{i}\right)} = \dfrac{\omega}{2}\sum{l_{i}} = \omega\dfrac{L}{2}$
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,
$L = I\omega$
where I is the moment of inertia of the rigid body relative to the axis of rotation. We can rewrite the T equation as,
$T = \omega\dfrac{{I}\omega}{2} = \dfrac{1}{2}{I}\omega^2$
Quantum Mechanics
The internal Hamiltonian, H, is:
$H = \dfrac{i^{2}\hbar^{2}}{2I}$
and the Schrödinger Equation for rigid rotor is:
$\dfrac{i^{2}\hbar^{2}}{2I}\psi = E\psi$
Thus, we get:
$E_n = \dfrac{J(J+1)h^2}{8\pi^2I}$
where $J$ is a rotational quantum number and $\hbar$ is the reduced Planck's constant. However, if we let:
$B = \dfrac {h}{8 \pi^2I}$
where $B$ is a rotational constant, then we can substitute it into the $E_n$ equation and get:
$E_{n} = J(J+1)Bh$
Considering the transition energy between two energy levels, the difference is a multiple of 2. That is, from $J = 0$ to $J = 1$, the $\Delta{E_{0 \rightarrow 1}}$ is 2Bh and from J = 1 to J = 2, the $\Delta{E}_{1 \rightarrow 2}$ is 4Bh.
Theory
When a gas molecule is irradiated with microwave radiation, a photon can be absorbed through the interaction of the photon’s electronic field with the electrons in the molecules. For the microwave region this energy absorption is in the range needed to cause transitions between rotational states of the molecule. However, only molecules with a permanent dipole that changes upon rotation can be investigated using microwave spectroscopy. This is due to the fact that their must be a charge difference across the molecule for the oscillating electric field of the photon to impart a torque upon the molecule around an axis that is perpendicular to this dipole and that passes through the molecules center of mass.
This interaction can be expressed by the transition dipole moment for the transition between two rotational states
$\text{Probability of Transition}=\int \psi_{rot}(F)\hat\mu \psi_{rot}(I)d\tau$
Where Ψrot(F) is the complex conjugate of the wave function for the final rotational state, Ψrot(I) is the wave function of the initial rotational state , and μ is the dipole moment operator with Cartesian coordinates of μx, μy, μz. For this integral to be nonzero the integrand must be an even function. This is due to the fact that any odd function integrated from negative infinity to positive infinity, or any other symmetric limits, is always zero.
In addition to the constraints imposed by the transition moment integral, transitions between rotational states are also limited by the nature of the photon itself. A photon contains one unit of angular momentum, so when it interacts with a molecule it can only impart one unit of angular momentum to the molecule. This leads to the selection rule that a transition can only occur between rotational energy levels that are only one quantum rotation level (J) away from another1.
$\Delta\textrm{J}=\pm 1$
The transition moment integral and the selection rule for rotational transitions tell if a transition from one rotational state to another is allowed. However, what these do not take into account is whether or not the state being transitioned from is actually populated, meaning that the molecule is in that energy state. This leads to the concept of the Boltzmann distribution of states. The Boltzmann distribution is a statistical distribution of energy states for an ensemble of molecules based on the temperature of the sample2.
$\dfrac{n_J}{n_0} = \dfrac{e^{(-E_{rot}(J)/RT)}}{\sum_{J=1}^{J=n} e^{(-E_{rot}(J)/RT)}}$
where Erot(J) is the molar energy of the J rotational energy state of the molecule,
• R is the gas constant,
• T is the temperature of the sample.
• n(J) is the number of molecules in the J rotational level, and
• n0 is the total number of molecules in the sample.
This distribution of energy states is the main contributing factor for the observed absorption intensity distributions seen in the microwave spectrum. This distribution makes it so that the absorption peaks that correspond to the transition from the energy state with the largest population based on the Boltzmann equation will have the largest absorption peak, with the peaks on either side steadily decreasing.
Degrees of Freedom
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 3 categories3.
• 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.
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.
Rotational Symmetries
To analyze molecules for rotational spectroscopy, we can break molecules down into 5 categories based on their shapes and their moments of inertia around their 3 orthogonal rotational axes:4
1. Diatomic Molecules
2. Linear Molecules
3. Spherical Tops
4. Symmetrical Tops
5. Asymmetrical Tops
Diatomic Molecules
The rotations of a diatomic molecule can be modeled as a rigid rotor. This rigid rotor model has two masses attached to each other with a fixed distance between the two masses.
It has an inertia (I) that is equal to the square of the fixed distance between the two masses multiplied by the reduced mass of the rigid rotor.
$\large I_e= \mu r_e^2$
$\large \mu=\dfrac{m_1 m_2} {m_1+m_2}$
Using quantum mechanical calculations it can be shown that the energy levels of the rigid rotator depend on the inertia of the rigid rotator and the quantum rotational number J2.
$E(J) = B_e J(J+1)$
$B_e = \dfrac{h}{8 \pi^2 cI_e}$
However, this rigid rotor model fails to take into account that bonds do not act like a rod with a fixed distance, but like a spring. This means that as the angular velocity of the molecule increases so does the distance between the atoms. This leads us to the nonrigid rotor model in which a centrifugal distortion term ($D_e$) is added to the energy equation to account for this stretching during rotation.
$E(J)(cm^{-1}) = B_e J(J+1) – D_e J^2(J+1)^2$
This means that for a diatomic molecule the transitional energy between two rotational states equals
$E=B_e[J'(J'+1)-J''(J''+1)]-D_e[J'^2(J'+1)^2-J''^2(J'+1)^2]\label{8}$
Where J’ is the quantum number of the final rotational energy state and J’’ is the quantum number of the initial rotational energy state. Using the selection rule of $\Delta{J}= \pm 1$ the spacing between peaks in the microwave absorption spectrum of a diatomic molecule will equal
$E_R =(2B_e-4D_e)+(2B_e-12D_e){J}''-4D_e J''^3$
Linear Molecules
Linear molecules behave in the same way as diatomic molecules when it comes to rotations. For this reason they can be modeled as a non-rigid rotor just like diatomic molecules. This means that linear molecule have the same equation for their rotational energy levels. The only difference is there are now more masses along the rotor. This means that the inertia is now the sum of the distance between each mass and the center of mass of the rotor multiplied by the square of the distance between them2.
$\large I_e=\sum_{j=1}^{n} m_j r_{ej}^2$
Where mj is the mass of the jth mass on the rotor and rej is the equilibrium distance between the jth mass and the center of mass of the rotor.
Spherical Tops
Spherical tops are molecules in which all three orthogonal rotations have equal inertia and they are highly symmetrical. This means that the molecule has no dipole and for this reason spherical tops do not give a microwave rotational spectrum.
Examples:
Symmetrical 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. 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.
Figure $3$: Symmetric Tops: (Left) Geometrical example of an oblate top and (right) a prolate top. Images used with permission from Wikipedia.com.
In the case of linear molecules there is one degenerate rotational axis which in turn has a single rotational constant. With symmetrical tops now there is one unique axis and two degenerate axes. This means an additional rotational constant is needed to describe the energy levels of a symmetrical top. In addition to the rotational constant an additional quantum number must be introduced to describe the rotational energy levels of the symmetric top. These two additions give us the following rotational energy levels of a prolate and oblate symmetric top
$E_{(J,K)}(cm^{-1})=B_e*J(J+1)+(A_e-B_e)K^2$
Where Be is the rotational constant of the unique axis, Ae is the rotational constant of the degenerate axes, $J$ is the total rotational angular momentum quantum number and K is the quantum number that represents the portion of the total angular momentum that lies along the unique rotational axis. This leads to the property that $K$ is always equal to or less than $J$. Thus we get the two selection rules for symmetric tops
$\Delta J = 0, \pm1$
$\Delta K=0$
when $K\neq 0$
$\Delta J = \pm1$
$\Delta K=0$
when $K=0$
However, like the rigid rotor approximation for linear molecules, we must also take into account the elasticity of the bonds in symmetric tops. Therefore, in a similar manner to the rigid rotor we add a centrifugal coupling term, but this time we have one for each quantum number and one for the coupling between the two.
$E_{(J,K)}(cm^{-1})=B_e J(J+1)-D_{eJ} J^2(J+1)^2+(A_e-B_e)*K^2 \label{13}$
$-D_{ek} K^4-D_{ejk} J(J +1)K^2 \label{14}$
Asymmetrical Tops
Asymmetrical tops have three orthogonal rotational axes that all have different moments of inertia and most molecules fall into this category. Unlike linear molecules and symmetric tops these types of molecules do not have a simplified energy equation to determine the energy levels of the rotations. These types of molecules do not follow a specific pattern and usually have very complex microwave spectra.
Additional Rotationally Sensitive Spectroscopies
In addition to microwave spectroscopy, IR spectroscopy can also be used to probe rotational transitions in a molecule. However, in the case of IR spectroscopy the rotational transitions are coupled to the vibrational transitions of the molecule. One other spectroscopy that can probe the rotational transitions in a molecule is Raman spectroscopy, which uses UV-visible light scattering to determine energy levels in a molecule. However, a very high sensitivity detector must be used to analyze rotational energy levels of a molecule.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/07._Angular_Momentum/7.5%3A_Rigid_Rotor.txt
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Spherical Harmonics are a group of functions used in math and the physical sciences to solve problems in disciplines including geometry, partial differential equations, and group theory. The general, normalized Spherical Harmonic is depicted below:
$Y_{l}^{m}(\theta,\phi) = \sqrt{ \dfrac{(2l + 1)(l - |m|)!}{4\pi (l + |m|)!} } P_{l}^{|m|}(\cos\theta)e^{im\phi}$
One of the most prevalent applications for these functions is in the description of angular quantum mechanical systems.
A Brief History
Utilized first by Laplace in 1782, these functions did not receive their name until nearly ninety years later by Lord Kelvin. Any harmonic is a function that satisfies Laplace's differential equation:
$\nabla^2 \psi = 0$
These harmonics are classified as spherical due to being the solution to the angular portion of Laplace's equation in the spherical coordinate system. Laplace's work involved the study of gravitational potentials and Kelvin used them in a collaboration with Peter Tait to write a textbook. In the 20th century, Erwin Schrödinger and Wolfgang Pauli both released papers in 1926 with details on how to solve the "simple" hydrogen atom system. Now, another ninety years later, the exact solutions to the hydrogen atom are still used to analyze multi-electron atoms and even entire molecules. Much of modern physical chemistry is based around framework that was established by these quantum mechanical treatments of nature.
The "Basic" Description
The 2px and 2pz (angular) probability distributions depicted on the left and graphed on the right using "desmos".
As Spherical Harmonics are unearthed by working with Laplace's equation in spherical coordinates, these functions are often products of trigonometric functions. These products are represented by the $P_{l}^{|m|}(\cos\theta)$ term, which is called a Legendre polynomial. The details of where these polynomials come from are largely unnecessary here, lest we say that it is the set of solutions to a second differential equation that forms from attempting to solve Laplace's equation. Unsurprisingly, that equation is called "Legendre's equation", and it features a transformation of $\cos\theta = x$. As the general function shows above, for the spherical harmonic where $l = m = 0$, the bracketed term turns into a simple constant. The exponential equals one and we say that:
$Y_{0}^{0}(\theta,\phi) = \sqrt{ \dfrac{1}{4\pi} }$
What is not shown in full is what happens to the Legendre polynomial attached to our bracketed expression. In the simple $l = m = 0$ case, it disappears.
It is no coincidence that this article discusses both quantum mechanics and two variables, $l$ and $m$. These are exactly the angular momentum quantum number and magnetic quantum number, respectively, that are mentioned in General Chemistry classes. If we consider spectroscopic notation, an angular momentum quantum number of zero suggests that we have an s orbital if all of $\psi(r,\theta,\phi)$ is present. This s orbital appears spherically symmetric on the boundary surface. In other words, the function looks like a ball. This is consistent with our constant-valued harmonic, for it would be constant-radius.
Extending these functions to larger values of $l$ leads to increasingly intricate Legendre polynomials and their associated $m$ values. The ${Y_{1}^{0}}^{*}Y_{1}^{0}$ and ${Y_{1}^{1}}^{*}Y_{1}^{1}$ functions are plotted above. Recall that these functions are multiplied by their complex conjugate to properly represent the Born Interpretation of "probability-density" ($\psi^{*}\psi)$. It is also important to note that these functions alone are not referred to as orbitals, for this would imply that both the radial and angular components of the wavefunction are used.
Example $1$
Identify the location(s) of all planar nodes of the following spherical harmonic:
$Y_{2}^{0}(\theta,\phi) = \sqrt{ \dfrac{5}{16\pi} }(3cos^2\theta - 1)$
Solution
Nodes are points at which our function equals zero, or in a more natural extension, they are locations in the probability-density where the electron will not be found (i.e. $\psi^{*}\psi = 0)$. As this specific function is real, we could square it to find our probability-density.
$Y_{2}^{0} = [Y_{2}^{0}]^2 = 0$
As the non-squared function will be computationally easier to work with, and will give us an equivalent answer, we do not bother to square the function. The constant in front can be divided out of the expression, leaving:
$3cos^2\theta - 1 = 0$
$\theta = cos^{-1}\bigg[\pm\dfrac{1}{\sqrt3}\bigg]$
$\theta = 54.7^o \& 125.3^o$
The Advanced Description
We have described these functions as a set of solutions to a differential equation but we can also look at Spherical Harmonics from the standpoint of operators and the field of linear algebra. For the curious reader, a more in depth treatment of Laplace's equation and the methods used to solve it in the spherical domain are presented in this section of the text. For a brief review, partial differential equations are often simplified using a separation of variables technique that turns one PDE into several ordinary differential equations (which is easier, promise). This allows us to say $\psi(r,\theta,\phi) = R_{nl}(r)Y_{l}^{m}(\theta,\phi)$, and to form a linear operator that can act on the Spherical Harmonics in an eigenvalue problem. The more important results from this analysis include (1) the recognition of an $\hat{L}^2$ operator and (2) the fact that the Spherical Harmonics act as an eigenbasis for the given vector space.
The $\hat{L}^2$ operator is the operator associated with the square of angular momentum. It is directly related to the Hamiltonian operator (with zero potential) in the same way that kinetic energy and angular momentum are connected in classical physics.
$\hat{H} = \dfrac{\hat{L}^2}{2I}$
for $I$ equal to the moment of inertia of the represented system. It is a linear operator (follows rules regarding additivity and homogeneity). More specifically, it is Hermitian. This means that when it is used in an eigenvalue problem, all eigenvalues will be real and the eigenfunctions will be orthogonal.
In Dirac notation, orthogonality means that the inner product of any two different eigenfunctions will equal zero:
$\langle \psi_{i} | \psi_{j} \rangle = 0$
When we consider the fact that these functions are also often normalized, we can write the classic relationship between eigenfunctions of a quantum mechanical operator using a piecewise function: the Kronecker delta.
$\langle \psi_{i} | \psi_{j} \rangle = \delta_{ij} \, for \, \delta_{ij} = \begin{cases} 0 & i \neq j \ 1 & i = j \end{cases}$
This relationship also applies to the spherical harmonic set of solutions, and so we can write an orthonormality relationship for each quantum number:
$\langle Y_{l}^{m} | Y_{k}^{n} \rangle = \delta_{lk}\delta_{mn}$
Example $2$: Symmetry
The parity operator is sometimes denoted by "P", but will be referred to as $\Pi$ here to not confuse it with the momentum operator. When this Hermitian operator is applied to a function, the signs of all variables within the function flip. This operator gives us a simple way to determine the symmetry of the function it acts on.
Recall that even functions appear as $f(x) = f(-x)$, and odd functions appear as $f(-x) = -f(x)$. Combining this with $\Pi$ gives the conditions:
• If $\Pi Y_{l}^{m}(\theta,\phi) = Y_{l}^{m}(-\theta,-\phi)$ then the harmonic is even.
• If $\Pi Y_{l}^{m}(\theta,\phi) = -Y_{l}^{m}(\theta,\phi)$ then the harmonic is odd.
Using the parity operator and properties of integration, determine $\langle Y_{l}^{m}| Y_{k}^{n} \rangle$ for any $l$ an even number and $k$ an odd number.
Solution
As this question is for any even and odd pairing, the task seems quite daunting, but analyzing the parity for a few simple cases will lead to a dramatic simplification of the problem.
Start with acting the parity operator on the simplest spherical harmonic, $l = m = 0$:
$\Pi Y_{0}^{0}(\theta,\phi) = \sqrt{\dfrac{1}{4\pi}} = Y_{0}^{0}(-\theta,-\phi)$
Now we can scale this up to the $Y_{2}^{0}(\theta,\phi)$ case given in example one:
$\Pi Y_{2}^{0}(\theta,\phi) = \sqrt{ \dfrac{5}{16\pi} }(3cos^2(-\theta) - 1)$
but cosine is an even function, so again, we see:
$Y_{2}^{0}(-\theta,-\phi) = Y_{2}^{0}(\theta,\phi)$
It appears that for every even, angular QM number, the spherical harmonic is even. As it turns out, every odd, angular QM number yields odd harmonics as well! If this is the case (verified after the next example), then we now have a simple task ahead of us.
Note: Odd functions with symmetric integrals must be zero.
$\langle Y_{l}^{m}| Y_{k}^{n} \rangle = \int_{-\inf}^{\inf} (EVEN)(ODD)d\tau$
An even function multiplied by an odd function is an odd function (like even and odd numbers when multiplying them together). As such, this integral will be zero always, no matter what specific $l$ and $k$ are used. As one can imagine, this is a powerful tool. The impact is lessened slightly when coming off the heels off the idea that Hermitian operators like $\hat{L}^2$ yield orthogonal eigenfunctions, but general parity of functions is useful!
Consider the question of wanting to know the expectation value of our colatitudinal coordinate $\theta$ for any given spherical harmonic with even-$l$.
$\langle \theta \rangle = \langle Y_{l}^{m} | \theta | Y_{l}^{m} \rangle$
$\langle \theta \rangle = \int_{-\inf}^{\inf} (EVEN)(ODD)(EVEN)d\tau$
Again, a complex sounding problem is reduced to a very straightforward analysis. Using integral properties, we see this is equal to zero, for any even-$l$.
A photo-set reminder of why an eigenvector (blue) is special. From https://en.Wikipedia.org/wiki/Eigenvalues_and_eigenvectors.
Lastly, the Spherical Harmonics form a complete set, and as such can act as a basis for the given (Hilbert) space. This means any spherical function can be written as a linear combination of these basis functions, (for the basis spans the space of continuous spherical functions by definition):
$f(\theta,\phi) = \sum_{l}\sum_{m} \alpha_{lm} Y_{l}^{m}(\theta,\phi)$
While any particular basis can act in this way, the fact that the Spherical Harmonics can do this shows a nice relationship between these functions and the Fourier Series, a basis set of sines and cosines. Spherical Harmonics are considered the higher-dimensional analogs of these Fourier combinations, and are incredibly useful in applications involving frequency domains. In the past few years, with the advancement of computer graphics and rendering, modeling of dynamic lighting systems have led to a new use for these functions.
Example $3$
In order to do any serious computations with a large sum of Spherical Harmonics, we need to be able to generate them via computer in real-time (most specifically for real-time graphics systems). This requires the use of either recurrence relations or generating functions. While at the very top of this page is the general formula for our functions, the Legendre polynomials are still as of yet undefined. The two major statements required for this example are listed:
$P_{l}(x) = \dfrac{1}{2^{l}l!} \dfrac{d^{l}}{dx^{l}}[(x^{2} - 1)^{l}]$
$P_{l}^{|m|}(x) = (1 - x^{2})^{\tiny\dfrac{|m|}{2}}\dfrac{d^{|m|}}{dx^{|m|}}P_{l}(x)$
Using these recurrence relations, write the spherical harmonic $Y_{1}^{1}(\theta,\phi)$.
Solution
To solve this problem, we can break up our process into four major parts. The first is determining our $P_{l}(x)$ function. As $l = 1$:
$P_{1}(x) = \dfrac{1}{2^{1}1!} \dfrac{d}{dx}[(x^{2} - 1)]$
$P_{1}(x) = \dfrac{1}{2}(2x)$
$P_{1}(x) = x$
Now that we have $P_{l}(x)$, we can plug this into our Legendre recurrence relation to find the associated Legendre function, with $m = 1$:
$P_{1}^{1}(x) = (1 - x^{2})^{\tiny\dfrac{1}{2}}\dfrac{d}{dx}x$
$P_{1}^{1}(x) = (1 - x^{2})^{\tiny\dfrac{1}{2}}$
At the halfway point, we can use our general definition of Spherical Harmonics with the newly determined Legendre function. With $m = l = 1$:
$Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{(2(1) + 1)(1 - 1)!}{4\pi (1 + |1|)!} } (1 - x^{2})^{\tiny\dfrac{1}{2}}e^{i\phi}$
$Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} } (1 - x^{2})^{\tiny\dfrac{1}{2}}e^{i\phi}$
The last step is converting our Cartesian function into the proper coordinate system or making the switch from x to $\cos\theta$.
$Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} } (1 - (\cos\theta)^{2})^{\tiny\dfrac{1}{2}}e^{i\phi}$
$Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} } (sin^{2}\theta)^{\tiny\dfrac{1}{2}}e^{i\phi}$
$Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} }sin\theta e^{i\phi}$
As a side note, there are a number of different relations one can use to generate Spherical Harmonics or Legendre polynomials. Often times, efficient computer algorithms have much longer polynomial terms than the short, derivative-based statements from the beginning of this problem.
As a final topic, we should take a closer look at the two recursive relations of Legendre polynomials together. As derivatives of even functions yield odd functions and vice versa, we note that for our first equation, an even $l$ value implies an even number of derivatives, and this will yield another even function. When we plug this into our second relation, we now have to deal with $|m|$ derivatives of our $P_{l}$ function. We are in luck though, as in the spherical harmonic functions there is a separate component entirely dependent upon the sign of $m$. As such, any changes in parity to the Legendre polynomial (to create the associated Legendre function) will be undone by the flip in sign of $m$ in the azimuthal component. Parity only depends on $l$!
This confirms our prediction from the second example that any Spherical Harmonic with even-$l$ is also even, and any odd-$l$ leads to odd $Y_{l}^{m}$.
Contributors and Attributions
• Alexander Staat
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/07._Angular_Momentum/Spherical_Harmonics.txt
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s orbital
p orbital
d orbital (Shape 1)
d orbital (Shape 2)
f orbital (Shape 1)
f orbital (Shape 2)
f orbital (Shape 3)
g orbital (Shape 1)
g orbital (Shape 2)
g orbital (Shape 3)
g orbital (Shape 4)
Atomic Orbitals
Once principle quantum number n equals 3 or greater, angular quantum number can equal 2. When angular quantum number l=2, it is considered the d-orbital. For the d-orbital, the magnetic quantum number ml can equal -2 to 2, taking the possible values -2, -1, 0, 1, or 2. This gives rise to five d orbitals, dxy, dyz, dxz, dx2-y2, and dz2. The magnetic quantum numbers do not correlate to a specific orbital, rather the orbitals are a linear combination of the different ml values, similar to that of the px and py orbitals. The general shape of the d-orbitals can be described as "daisy-like" or "four leaf clover" with the exception of the the dz2 orbital which looks like the donut with a lobe above and below. All the d-orbitals contain 2 angular nodes. In the case of dxy, dyz, dxz, and dx2-y2 they are planar angular nodes, easily seen as the axes which bisect the lobes of the orbitals. In dz2 they are conical angular nodes which divide the "donut" part of the orbital with the upper and lower lobes. The d-orbitals are important in the transition metals because they are typically what are used in bonding. Crystal Field Theory, more specifically Crystal Field Splitting, uses the d-orbitals and their degeneracy to describe spectroscopic properties of transition metal complexes.
f Atomic Orbitals
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The following are images of the f-orbitals. They exist for principal quantum number 4 and higher.
g Atomic Orbitals
The following are images of the different shapes possible for g-orbitals. They exist for principal quantum number 5 and higher.
p Atomic Orbitals
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The following is are images of p-orbitals. They exist for principal quantum numbers 2 and higher.
px py pz
s Atomic Orbitals
Each n orbital has only one s orbital and therefore two s orbital electrons. Since its angular momentum quantum number(l) is 0, its magnetic quantum number(ml) is also 0. If there is only one electron, the electron can exist in either spin up(ms=1/2) or with spin down(ms=-1/2) configuration; if there are two electrons, they must be one spin up and one spin down.
Basic Description
The shape of the s orbital is a sphere; s orbitals are spherically symmetric. The nodes of s orbital is n-1; the angular nodes is l, which is 0 for all s orbitals; the radial nodes is n-l-1, which is n-1 for all s orbitals. Therefore, s orbital only has radial nodes, which are spheres.
If n increases, s orbitals become larger, extending farther from the nucleus. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Advanced Description
Wavefunction is a mathematical expression that can be used to calculate any property of an atom. In general, wavefunctions depend on both time and position. For atoms, solutions to the Schrödinger equation correspond to arrangements of the electrons, which, if left alone, remain unchanged and are thus only functions of position. S orbitals only have angular wavefunctions, $Y^{m_J}_J (\theta , \varphi)$ = $\dfrac {1}{\sqrt {4 \pi}}$, because all s orbitals are l = 0 and therefore ml=0. S orbitals have different radial wavefunctions; for example, for n=1 to n=3, they are R1,0(r) = $2 \left (\dfrac {Z}{a_0} \right )^{3/2} e^{-\rho}$, R2,0(r) = $\dfrac {1}{2 \sqrt {2}}\left (\dfrac {Z}{a_0} \right )^{3/2} (2 - \rho) e^{-\rho/2}$, and R3,0(r) = $\dfrac {1}{81 \sqrt {3}}\left (\dfrac {Z}{a_0} \right )^{3/2} (27 - 18 \rho + 2\rho ^2) e^{-\rho/3}$. Their complete wave functions are $\psi_{100} = \dfrac {1}{\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} e^{-\rho}$, $\psi_{200} = \dfrac {1}{\sqrt {32\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (2-\rho)e^{\dfrac {-\rho}{2}}$, and $\psi_{300} = \dfrac {1}{81\sqrt {3\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (27-18\rho +2\rho^2)e^{-\rho/3}$ ($\rho = \dfrac {Zr}{a_0}$, where $a_0$ is the Bohr radius and r is the radial variable).
Wavefunctions for each atom have some properties that are exact, for example each wavefunction describes an electron in quantum state with a specific energy.
Radial Probability density is the probability density for the electron to be at a point located the distance r from the proton. Radial probability densities for three types of atomic orbitals are plotted below.
S orbitals' radial probability densities at 0 are completely different from p orbitas'.
Radial Probability distribution is the probability density for an electron to be found anywhere on the surface of a sphere located a distance $r$ from the proton. The area of a spherical surface is $4 \pi r^2$ (The graph is showed at basic description part).
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Orbitals/d_Atomic_Orbitals.txt
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This module was created to better your understanding of quarks and other sub-nucleon particles. Both of these topics fall into the study of particle physics, which is the study of elementary particles that are too small to study directly.
Contributors and Attributions
• Ben Havemann (UCD), Michael Norton (UCD)
Atomic Spectra
When atoms are excited they emit light of certain wavelengths which correspond to different colors. The emitted light can be observed as a series of colored lines with dark spaces in between; this series of colored lines is called a line or atomic spectra. Each element produces a unique set of spectral lines. Since no two elements emit the same spectral lines, elements can be identified by their line spectrum.
Contributors and Attributions
• Drew Cylinder
Connecting Electronic Configurations to th
Electron configuration can be described as how electrons are assembled within the orbitals shells and subshells of an atom. It is important to understand what an electron is in order to fully understand the electron configuration. An electron is a sub atomic particle that is associated with a negative charge. Electrons are found outside of the nucleus, as opposed to neutrons (particles with neutral charge,) and protons (particles with positive charge.) Furthermore, electrons are associated with energy, more specifically quantum energy, and exemplify wave-like and particle-like characteristics. The word configuration simply means the arrangement of something. Therefore electron configuration in straightforward language means the arrangement of electrons.
Introduction
In general when filling up the electron diagram, it is customary to fill the lowest energies first and work your way up to the higher energies. Principles and rules such as the Pauli exclusion principle, Hund’s rule, and the Aufbau process are used to determine how to properly configure electrons. The Pauli exclusion rule basically says that at most, 2 electrons are allowed to be in the same orbital. Hund’s rule explains that each orbital in the subshell must be occupied with one single electron first before two electrons can be in the same orbital. Lastly, the Aufbau process describes the process of adding electron configuration to each individualized element in the periodic table. Fully understanding the principles relating to electron configuration will promote a better understanding of how to design them and give us a better understanding of each element in the periodic table. How the periodic table was formed has an intimate correlation with electron configuration. After studying the relationship between electron configuration and the period table, it was pointed out by Niels Bohr that electron configurations are similar for elements within the same group in the periodic table. Groups occupy the vertical rows as opposed to a period which is the horizontal rows in the table of elements.
S, P, D, and F Blocks
• It is easy to see how similar electron configurations are in a group when written out. (Allow “n” to be the principal quantum number.) Lets first take a look at group 1 atoms. Group 1 atoms are the alkali metals. Let n=1. Notice the similar configuration within all the group 1 elements.
Group Element Configuration
1 H 1s1
1 Li [He]2s1
1 Na [Ne]3s1
1 K [Ar]4s1
1 Rb [Kr]5s1
1 Cs [Xe]6s1
1 Fr [Rn]7s1
Now consider group 16 elements. These elements also will also have similar electron configurations to each another because they are in the same group; these elements have 6 valence electrons.
Group Element Configuration
16 O [He]2s22p4
16 S [Ne]3s23p4
16 Se [Ar]3d104s2 4p4
16 Te [Kr]4d105s2 5p4
16 Po [Xe]4f14 5d106s2 6p4
Outside links
• 1. Question, True or False: Elements in the same period have similar electron configurations.
Answer: False. Elements in the same GROUP have similar electron configurations.
2. Question: What element has the electron configuration [Ar] 4s2 3d10 4p5?
Answer: Bromine
3. Question: What element has the electron configuration [Xe] 4f14 5d10 6s2 6p3?
Answer: Bismuth
4. Question: Demonstrate how elements in a group share similar characteristics by filling in the electron configurations for the Group 18 elements:
Group Element Configuration
18 He
18 Ne
18 Ar
18 Kr
18 Xe
18 Rn
Answer: 1s2, [He]2s22p6, [Ne]3s23p6, [Ar]3d104s24p6, [Kr]4d105s25p6, [Xe]4f145d106s26p6
5. Question: How many valence electrons are there in Iodine?
Answer: Iodine, z=53, group 17. This means there are seven valence electrons.
6. Question: What is the highest number of electrons a 4p subshell can hold?
Answer: 6! Each 3 p orbital can hold 2 electrons so if they are all filled, the answer is 6. You get this by multiplying the three orbitals by 2 electrons per orbital, so 3 multiplied by 2 equals 6.
Make up some practice problems for the future readers.
Contributors and Attributions
• Mariana Gerontides (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Case_Study%3A_Quarks_and_other_sub-Nucleon_Particles.txt
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Electron Spin or Spin Quantum Number is the fourth quantum number for electrons in atoms and molecules. Denoted as $m_s$, the electron spin is constituted by either upward ($m_s=+1/2$) or downward ($m_s=-1/2$) arrows.
Introduction
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 ms. 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. This spin can be denoted by an arrow pointing up, which is +1/2, or an arrow pointing down, which is -1/2.
The experiment mentioned above by Otto Stern and Walter Gerlach was done with silver which was put in an oven and vaporized. The result was that silver atoms formed a beam that passed through a magnetic field in which it split in two.
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. There is a high likelihood for either spin due to the large number of electrons, so when it went through the magnetic field it split into two beams.
Brief Explanation of Quantum Numbers
Note: In this module, capital "L" will be used instead of small case "l" for angular momentum quantum number.
A total of four quantum numbers were developed to better understand the movement and pathway of electrons in its designated orbital within an atom.
1. Principal quantum number (n): energy level n = 1, 2, 3, 4, ...
2. Orbital Angular Momentum Quantum Number (L): shape (of orbital) L = 0, 1, 2, 3, ...n-1
3. Magnetic Quantum Number (mL): orientation mL = interval of (-L, +L)
4. Electron Spin Quantum Number (ms): independent of other three quantum numbers because ms is always = –½ or +½
(For more information about the three quantum numbers above, see Quantum Number.)
The lines represent how many orientations each orbital has, (e.g. the s-orbital has one orientation, a p-orbital has three orientations, etc.) and each line can hold up to two electrons, represented by up and down arrows. An electron with an up arrow means it has an electron spin of +$\frac{1}{2}$, and an electron with a down arrow means it has an electron spin of -$\frac{1}{2}$.
Electron Spin
Significance: determines if an atom will or will not generate a magnetic field (For more information, scroll down to Magnetic Spin, Magnetism, and Magnetic Field Lines). Although the electron spin is limited to or –½, certain rules apply when assigning electrons of different spins to fill a subshell (orientations of an orbital) . For more information, scroll down to Assigning Spin Direction.
Magnetic Spin, Magnetism, and Magnetic Field Lines
An atom with unpaired electrons are termed as paramagnetic
• results in a net magnetic field because electrons within the orbital are not stabilized or balanced enough
• atoms are attracted to magnets
An atom with paired electrons are termed as diamagnetic
• results in no magnetic field because electrons are uniform and stabilized within the orbital
• atoms are not attracted to magnets
Applying concepts of magnetism with liquid nitrogen and liquid oxygen:
The magnetic spin of an electron follows in the direction of the magnetic field lines as shown below.
{{media("www.youtube.com/watch?v=uj0DFDfQajw\)
Assigning Spin Direction
An effective visual on how to assign spin directions can be represented by the orbital diagram (shown previously and below.) Restrictions apply when assigning spin directions to electrons, so the following Pauli Exclusion Principle and Hund's Rule help explain this.
When one is filling an orbital, such as the p orbital, you must fill all orbitals possible with one electron spin before assigning the opposite spin. For example, when filling the fluorine, which will have a total of two electrons in the s orbital and a total of five electrons in the p orbital, one will start with the s orbital which will contain two electrons. So, the first electron one assigns will be spin up and the next spin down. Moving on to the three p orbitals that one will start by assigning a spin up electron in each of the three orbitals. That takes up three of the five electrons, so with the remaining two electrons, one returns to the first and second p orbital and assigns the spin down electron. This means there will be one unpaired electron in fluorine so it will be paramagnetic.
Pauli Exclusion Principle
The Pauli exclusion principle declares that there can only be a maximum of two electrons for every one orientation, and the two electrons must be opposite in spin direction; meaning one electron has $m_s = +\frac{1}{2}$ and the other electron has $m_s = -\frac{1}{2}$.
Hund's Rule
Hund's Rule declares that the electrons in the orbital are filled up first by the +$\frac{1}{2}$ spin. Once all the orbitals are filled with unpaired +$\frac{1}{2}$ spins, the orbitals are then filled with the -$\frac{1}{2}$ spin. (See examples below, labeled electronic configuration.)
(For more information on Pauli Exclusion Principle and Hund's Rules, see Electronic Configuration.)
Identifying Spin Direction
1. Determine the number of electrons the atom has.
2. Draw the electron configuration for the atom. See Electronic Configurations for more information.
3. Distribute the electrons, using up and down arrows to represent the electron spin direction.
Example 1: Sulfur
Sulfur - S (16 electrons)
Electronic Configuration:
1s2 2s2 2p6 3s2 3p4 OR [Ne] 3s2 3p4
As shown in the following image, this is a demonstration of a -$\frac{1}{2}$ and a +$\frac{1}{2}$ Electron Spin.
Principal Quantum Number & (s, p, d, f) Orbitals
When given a principal quantum number, n, with either the s, p, d or f-orbital, identify all the possibilities of L, mL and ms.
Example 2
Given 5f, identify all the possibilities of the four quantum numbers.
Solution
In this problem, the principal quantum number is n = 5 (the subshell number placed in front of the orbital, the f-orbital in this case). Since we are looking at the f-obital, therefore L = 3. (Look under "Subshells" in the module Quantum Numbers for more information) Knowing L = 3, we can interpret that mL = 0, $\pm$ 1, $\pm$ 2, $\pm$ 3 since mL = -L,...,-1, 0, 1,...+L. As for ms, since it isn't specified in the problem as to whether it is -$\frac{1}{2}$ or +$\frac{1}{2}$, therefore for this problem, it could be both; meaning that the electron spin quantum number is $\pm$$\frac{1}{2}$.
Example 3
Given 6s and mL= +1, identify all the possibilities of the four quantum numbers.
Solution
The principal quantum number is n = 6. Looking at the s-orbital, we know that L = 0. Knowing that mL = -L,...,-1, 0, 1,...+L, therefore mL = +1 is not possible since in this problem, the interval of mL can only equal to 0 according to the angular momentum quantum number, L.
Example 4
Given 4d and ms = +$\frac{1}{2}$, identify all the possibilities of the four quantum numbers.
Solution
The principal quantum number is n = 4. Given that it is a d-orbital, we know that L = 2. Therefore, mL = 0, $\pm$ 1, $\pm$ 2 since mL = -L,...,-1, 0, 1,...+L. For ms, this problem specifically said ms = +$\frac{1}{2}$; meaning that the electron spin quantum number is +$\frac{1}{2}$.
Electron Spin: Where to begin?
First, draw a table labeled n, L, mLand ms, as shown below:
n L mL ms
Then, depending on what the question is asking for, fill in the boxes accordingly. Finally, determine the number of electrons for the given quantum number, n, with regards to L, mL and ms.
Example 5
How many electrons can have n = 5 and L = 1? 6
n L mL ms
5 1 -1 -$\frac{1}{2}$, +$\frac{1}{2}$
0
+1
This problem includes both -$\frac{1}{2}$, +$\frac{1}{2}$, therefore the answer is 6 electrons based on the mL.
Example 6
How many electrons can have n = 5 and ms = -$\frac{1}{2}$? 5
n L mL ms
5 4 $\pm$ 4 -$\frac{1}{2}$
3 $\pm$ 3
2 $\pm$ 2
1 $\pm$ 1
0 0
This problem only wants the Spin Quantum Number to be -$\frac{1}{2}$, the answer is 5 electrons based on the mL.
Example 7
How many electrons can have n = 3, L = 2 and mL = 3? zero
Solution
n L mL ms
3 2 NOT POSSIBLE
Since mL = -L...-1, 0, +1...+L (See Electronic Orbitals for more information), mL is not possible because L = 2, so it is impossible for mL to be equal to 3. So, there is zero electrons.
Example 8
How many electrons can have n = 3, mL = +2 and ms = +$\frac{1}{2}$? 1
Solution
n L mL ms
3 2 -2, +2 +$\frac{1}{2}$
1 $\pm$ 1
0 0
This problem only wants the Spin Quantum Number to be +$\frac{1}{2}$, the answer is 1 electrons based on the mL.
Practice Problems
1. Identify the spin direction (e.g. ms = -$\frac{1}{2}$ or +$\frac{1}{2}$ or $\pm$$\frac{1}{2}$ ) of the outermost electron in a Sodium (Na) atom.
2. Identify the spin direction of the outermost electron in a Chlorine (Cl) atom.
3. Identify the spin direction of the outermost electron in a Calcium (Ca) atom.
4. Given 5p and ms = +$\frac{1}{2}$, identify all the possibilities of the four quantum numbers.
5. Given 6f, identify all the possibilities of the four quantum numbers.
6. How many electrons can have n = 4 and L = 1?
7. How many electrons can have n = 4, L = 1, mL = -2 and ms = +$\frac{1}{2}$?
8. How many electrons can have n = 5, L = 3, mL = $\pm$ 2 and ms = +$\frac{1}{2}$?
9. How many electrons can have n = 5, L = 4, mL = +3 and ms = -$\frac{1}{2}$?
10. How many electrons can have n = 4, L = 2, mL = $\pm$1 and ms = -$\frac{1}{2}$?
11. How many electrons can have n = 3, L = 3, mL = +2?
Solutions: Check your work!
Problem (1): Sodium (Na) --> Electronic Configuration [Ne] 3s1
Spin direction for the valence electron or ms = +$\frac{1}{2}$
Sodium (Na) with a neutral charge of zero is paramagnetic, meaning that the electronic configuration for Na consists of one or more unpaired electrons.
Problem (2): Chlorine (Cl) --> Electronic Configuration [Ne] 3s2 3p5
Spin direction for the valence electron or ms = +$\frac{1}{2}$
Chlorine (Cl) with a neutral charge of zero is paramagnetic.
Problem (3): Calcium (Ca) --> Electronic Configuration [He] 4s2
Spin direction for the valence electron or ms = $\pm$$\frac{1}{2}$
Whereas for Calcium (Ca) with a neutral charge of zero, it is diamagnetic; meaning that ALL the electrons are paired as shown in the image above.
Problem (4): Given 5p and ms = -$\frac{1}{2}$, identify all the possibilities of the four quantum numbers.
The principal quantum number is n = 5. Given that it is a p-orbital, we know that L = 1. And based on L, mL = 0, $\pm$ 1 since mL = -L,...,-1, 0, 1,...+L. As for ms, this problem specifically says ms = -$\frac{1}{2}$, meaning that the spin direction is -$\frac{1}{2}$, pointing downwards ("down" spin).
Problem (5): Given 6f, identify all the possibilities of the four quantum numbers.
The principal quantum number is n = 6. Given that it is a f-orbital, we know that L = 3. Based on L, mL = 0, $\pm$ 1, $\pm$ 2, $\pm$ 3 since mL = -L,...,-1, 0, 1,...+L. As for ms, since it isn't specified in the problem as to whether it is -$\frac{1}{2}$ or +$\frac{1}{2}$, therefore for this problem, it could be both; meaning that the electron spin quantum number is $\pm$$\frac{1}{2}$.
Problem (6): How many electrons can have n = 4 and L = 1? 6
n L mL ms
4 1 $\pm$ 1 -$\frac{1}{2}$, + $\frac{1}{2}$
0
This problem includes both -$\frac{1}{2}$, +$\frac{1}{2}$, therefore the answer is 6 electrons based on the mL.
Problem (7): How many electrons can have n = 4, L = 1, mL = -2 and ms = +$\frac{1}{2}$? zero
n L mL ms
4 1 NOT POSSIBLE + $\frac{1}{2}$
Since mL = -L...-1, 0, +1...+L, mL is not possible because L = 1, so it is impossible for mL to be equal to 2 when mL MUST be with the interval of -L and +L. So, there is zero electron.
Problem (8): How many electrons can have n = 5, L = 3, mL = $\pm$ 2 and ms = +$\frac{1}{2}$? 2
n L mL ms
5 3 $\pm$ 2 + $\frac{1}{2}$
$\pm$ 1
0
This problem only wants the Spin Quantum Number to be +$\frac{1}{2}$ and mL = $\pm$ 2, therefore 2 electrons can have n = 5, L = 3, mL = $\pm$ 2 and ms = +$\frac{1}{2}$.
Problem (9): How many electrons can have n = 5, L = 4 and mL = +3? 2
n L mL ms
5 4 -3, +3 -$\frac{1}{2}$, + $\frac{1}{2}$
$\pm$ 2
$\pm$ 1
0
This problem includes both -$\frac{1}{2}$ and +$\frac{1}{2}$ and given that mL = +3, therefore the answer is 2 electrons.
Problem (10): How many electrons can have n = 4, L = 2 and mL = $\pm$ 1? 4
n L mL ms
4 2 $\pm$ 1 -$\frac{1}{2}$, + $\frac{1}{2}$
0
This problem includes both -$\frac{1}{2}$ and +$\frac{1}{2}$ and given that mL = $\pm$ 1, therefore the answer is 4 electrons.
Problem (11): How many electrons can have n = 3, L = 3, mL = +2 and ms = -$\frac{1}{2}$? zero
n L mL ms
3 3 (NOT POSSIBLE) $\pm$ 2 -$\frac{1}{2}$
$\pm$ 1
0
Since L = n - 1, there is zero electron, not possible because in this problem, n = L = 3.
Contributors and Attributions
• Liza Chu (UCD), Sharon Wei (UCD), Mandy Lam (UCD), Lara Cemo (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Electron_Spin.txt
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An atom is composed of a nucleus containing neutrons and protons with electrons dispersed throughout the remaining space. Electrons, however, are not simply floating within the atom; instead, they are fixed within electronic orbitals. Electronic orbitals are regions within the atom in which electrons have the highest probability of being found.
Quantum Numbers describing Electronic Orbitals
There are multiple orbitals within an atom. Each has its own specific energy level and properties. Because each orbital is different, they are assigned specific quantum numbers: 1s, 2s, 2p 3s, 3p,4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. The numbers, (n=1,2,3, etc.) are called principal quantum numbers and can only be positive numbers. The letters (s,p,d,f) represent the orbital angular momentum quantum number () and the orbital angular momentum quantum number may be 0 or a positive number, but can never be greater than n-1. Each letter is paired with a specific value:
s: subshell = 0
p: subshell = 1
d: subshell = 2
f: subshell = 3
An orbital is also described by its magnetic quantum number (m). The magnetic quantum number can range from to +. This number indicates how many orbitals there are and thus how many electrons can reside in each atom.
Orbitals that have the same or identical energy levels are referred to as degenerate. An example is the 2p orbital: 2px has the same energy level as 2py. This concept becomes more important when dealing with molecular orbitals. The Pauli exclusion principle states that no two electrons can have the same exact orbital configuration; in other words, the same quantum numbers. However, the electron can exist in spin up (ms = +1/2) or with spin down (ms = -1/2) configurations. This means that the s orbital can contain up to two electrons, the p orbital can contain up to six electrons, the d orbital can contain up to 10 electrons, and the f orbital can contain up to 14 electrons.
s subshell p subshell d subshell f subshell
Table 1: Breakdown and Properties of Subshells
ℓ = 0 ℓ = 1 ℓ = 2 ℓ = 3
m = 0 m= -1, 0, +1 m= -2, -1, 0, +1, +2 m= -3, -2, -1, 0, +1, +2, +3
One s orbital Three p orbitals Five d orbitals Seven f orbitals
2 s orbital electrons 6 p orbital electrons 10 d orbital electrons 14 f orbital electrons
Visualizing Electron Orbitals
As discussed in the previous section, the magnetic quantum number (ml) can range from –l to +l. The number of possible values is the number of lobes (orbitals) there are in the s, p, d, and f subshells. As shown in Table 1, the s subshell has one lobe, the p subshell has three lobes, the d subshell has five lobes, and the f subshell has seven lobes. Each of these lobes is labeled differently and is named depending on which plane the lobe is resting in. If the lobe lies along the x plane, then it is labeled with an x, as in 2px. If the lobe lies along the xy plane, then it is labeled with a xy such as dxy. Electrons are found within the lobes. The plane (or planes) that the orbitals do not fill are called nodes. These are regions in which there is a 0 probability density of finding electrons. For example, in the dyx orbital, there are nodes on planes xz and yz. This can be seen in Figure \(1\).
Radial and Angular Nodes
There are two types of nodes, angular and radial nodes. Angular nodes are typically flat plane (at fixed angles), like those in the diagram above. The quantum number determines the number of angular nodes in an orbital. Radial nodes are spheres (at fixed radius) that occurs as the principal quantum number increases. The total nodes of an orbital is the sum of angular and radial nodes and is given in terms of the \(n\) and \(l\) quantum number by the following equation:
\[ N = n-l -1\]
For example, determine the nodes in the 3pz orbital, given that n = 3 and = 1 (because it is a p orbital). The total number of nodes present in this orbital is equal to n-1. In this case, 3-1=2, so there are 2 total nodes. The quantum number determines the number of angular nodes; there is 1 angular node, specifically on the xy plane because this is a pz orbital. Because there is one node left, there must be one radial node. To sum up, the 3pz orbital has 2 nodes: 1 angular node and 1 radial node. This is demonstrated in Figure 2.
Another example is the 5dxy orbital. There are four nodes total (5-1=4) and there are two angular nodes (d orbital has a quantum number =2) on the xz and zy planes. This means there there must be two radial nodes. The number of radial and angular nodes can only be calculated if the principal quantum number, type of orbital (s,p,d,f), and the plane that the orbital is resting on (x,y,z, xy, etc.) are known.
Electron Configuration within an Orbital
We can think of an atom like a hotel. The nucleus is the lobby where the protons and neutrons are, and in the floors above, we find the rooms (orbitals) with the electrons. The principal quantum number is the floor number, the subshell type lets us know what type of room it is (s being a closet, p being a single room, d having two adjoining rooms, and f being a suit with three rooms) , the magnetic quantum number lets us know how many beds there are in the room, and two electrons can sleep in one bed (this is because each has a different spin; -1/2 and 1/2). For example, on the first floor we have the s orbital. The s orbital is a closet and has one bed in it so the first floor can hold a total of two electrons. The second floor has the room styles s and p. The s is a closet with one bed as we know and the p room is a single with three beds in it so the second floor can hold a total of 8 electrons.
Each orbital, as previously mentioned, has its own energy level associated to it. The lowest energy level electron orbitals are filled first and if there are more electrons after the lowest energy level is filled, they move to the next orbital. The order of the electron orbital energy levels, starting from least to greatest, is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
Since electrons all have the same charge, they stay as far away as possible because of repulsion. So, if there are open orbitals in the same energy level, the electrons will fill each orbital singly before filling the orbital with two electrons. For example, the 2p shell has three p orbitals. If there are more electrons after the 1s, and 2s orbitals have been filled, each p orbital will be filled with one electron first before two electrons try to reside in the same p orbital. This is known as Hund's rule.
The way electrons move from one orbital to the next is very similar to walking up a flight of stairs. When walking up stairs, you place one foot on the first stair and then another foot on the second stair. At any point in time, you can either stand with both feet on the first stair, or on the second stair but it is impossible to stand in between the two stairs. This is the way electrons move from one electron orbital to the next. Electrons can either jump to a higher energy level by absorbing, or gaining energy, or drop to a lower energy level by emitting, or losing energy. However, electrons will never be found in between two orbitals.
Problems
1. Which orbital would the electrons fill first? The 2s or 2p orbital?
2. How many d orbitals are there in the d subshell?
3. How many electrons can the p orbital hold?
4. Determine the number of angular and radial nodes of a 4f orbital.
5. What is the shape of an orbital with 4 radial nodes and 1 angular node in the xy plane?
Solutions
1. The 2s orbital would be filled before the 2p orbital because orbitals that are lower in energy are filled first. The 2s orbital is lower in energy than the 2p orbital.
2. There are 5 d orbitals in the d subshell.
3. A p orbital can hold 6 electrons.
4. Based off of the given information, n=4 and =3. Thus, there are 3 angular nodes present. The total number of nodes in this orbital is: 4-1=3, which means there are no radial nodes present.
5. 1 angular node means =1 which tells us that we have a p subshell, specifically the pz orbital because the angular node is on the xy plane. The total number of nodes in this orbital is: 4 radial nodes +1 angular node=5 nodes. To find n, solve the equation: nodes=n-1; in this case, 5=n-1, so n=6. This gives us a: 6pz orbital
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Electronic_Orbitals.txt
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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.
Introduction
Let's review some basic concepts. First, electrons repel against each other. Particles with the same charge repel each other, while oppositely charged particles attract each other. For example, a proton, which is positively charged, is attracted to electrons, which are negatively charged. However, if we put two electrons together or two protons together, they will repel one another. Since neutrons lack a charge, they will neither repel nor attract protons or electrons.
Figure 1: a) The two electrons are placed together and repel each other because of the same charge. b) The two protons are repelling each other for the same reason. c) When oppositely charged particles, an electron and a proton, are placed together, they attract each other.
Protons and neutrons are located in an atom's nucleus. Electrons float around the atom in energy levels. Energy levels consist of orbitals and sub-orbitals. The lower the energy level the electron is located at, the closer it is to nucleus. As we go down and to the right of the periodic table, the number of electrons that an element has increases. Since there are more electrons, the atom experiences greater repulsion and electrons will tend to stay as far away from each other as possible. Our main focus is what effects take place when more electrons surround the nucleus. To better understand the following concepts it is a good idea to first review quantum mechanics.
Shielding (Screening)
With more protons in the nucleus, the attractive force for electrons to the nucleus is stronger. Thus, the orbital energy becomes more negative (less energy). Orbital energy also depends on the type of l orbital an electron is located in. The lower the number of l, the closer it is to the nucleus. For example, l=0 is the s orbital. S orbitals are closer to the nucleus than the p orbitals (l=1) that are closer to the nucleus than the d orbitals (l=2) that are closer to the f orbitals (l=3).
More electrons create the shielding or screening effect. Shielding or screening means the electrons closer to the nucleus block the outer valence electrons from getting close to the nucleus. See figure 2. Imagine being in a crowded auditorium in a concert. The enthusiastic fans are going to surround the auditorium, trying to get as close to the celebrity on the stage as possible. They are going to prevent people in the back from seeing the celebrity or even the stage. This is the shielding or screening effect. The stage is the nucleus and the celebrity is the protons. The fans are the electrons. Electrons closest to the nucleus will try to be as close to the nucleus as possible. The outer/valence electrons that are farther away from the nucleus will be shielded by the inner electrons. Therefore, the inner electrons prevent the outer electrons from forming a strong attraction to the nucleus. The degree to which the electrons are being screened by inner electrons can be shown by ns<np<nd<nf where n is the energy level. The inner electrons will be attracted to the nucleus much more than the outer electrons. Thus, the attractive forces of the valence electrons to the nucleus are reduced due to the shielding effects. That is why it is easier to remove valence electrons than the inner electrons. It also reduces the nuclear charge of an atom.
Penetration
Penetration is the ability of an electron to get close to the nucleus. The penetration of ns > np > nd > nf. Thus, the closer the electron is to the nucleus, the higher the penetration. Electrons with higher penetration will shield outer electrons from the nucleus more effectively. The s orbital is closer to the nucleus than the p orbital. Thus, electrons in the s orbital have a higher penetration than electrons in the p orbital. That is why the s orbital electrons shield the electrons from the p orbitals. Electrons with higher penetration are closer to the nucleus than electrons with lower penetration. Electrons with lower penetration are being shielded from the nucleus more.
Radial Probability Distribution
Radial probability distribution is a type of probability to find where an electron is mostly likely going to be in an atom. The higher the penetration, the higher probability of finding an electron near the nucleus. As shown by the graphs, electrons of the s orbital are found closer to the nucleus than the p orbital electrons. Likewise, the lower the energy level an electron is located at, the higher chance it has of being found near the nucleus. The smaller the energy level (n) and the orbital angular momentum quantum number (l) of an electron is, the more likely it will be near the nucleus. As electrons get to higher and higher energy levels, the harder it is to locate it because the radius of the sphere is greater. Thus, the probability of locating an electron will be more difficult. Radial probability distribution can be found by multiplying 4πr², the area of a sphere with a radius of r and R²(r).
Radial Probability Distribution = 4πr² X R²(r)
By using the radial probability distribution equation, we can get a better understanding about an electron's behavior, as shown on Figures 3.1-3.3.
Figures 3.1, 3.2, and 3.3 show that the lower the energy level, the higher the probability of finding the electron close to the nucleus. Also, the lower momentum quantum number gets, the closer it is to the nucleus.
Contributors and Attributions
• Mixue (Michelle) Xie (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Multi-electron_Atoms.txt
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By the end of the 19th century, most problems with what has come to be known as classical physics had been solved. There were only a few questions that remained unanswered so physicists of the time thought that they were ready to say they knew all about the field of physics. However, as they tried to answer these questions, they came to realize that they were not generating answers but more questions that required different kinds of answers. It was then that physicists came to see that these unanswered questions would not mark the end of physics, but rather the beginning of a new field: quantum theory.
Introduction
While classical physics is more than enough to explain what occurs at a macroscopic level (for example, throwing a ball or pushing a car) a new set of rules and ideas is required to deal with things that occur at the subatomic level that that is where quantum theory comes in. Quantum theory is a field of physics that is required to understand phenomena at the molecular and atomic levels. Quantum theory is simply a new way of looking at the world. The rules as they apply to us don't apply to the tiny particles that quantum theory deals with. However, it was a breakthrough that led physicists to discover more about the world of physics and to understand our own world better, starting from the tiny particles of matter that are its building blocks.
Blackbody Radiation and Planck's Equation
One of the first ideas proposed to set quantum mechanics apart from classical physics was Max Planck’s idea that energy, like matter, was discontinuous. This revolutionary idea stemmed from blackbody radiation. A blackbody is an object that absorbs all the radiation falling on it. An object that absorbs all the radiation can also perfectly emit all radiation therefore a blackbody will radiate maximum energy when heated to a given temperature. Under classical physics, this energy emitted was predicted to be infinite. However, when it didn’t radiate energy indefinitely, scientists were faced with the problem of explaining the phenomenon. This led to Planck’s proposed idea that unlike classical physics, quantum theory limits energy to a set of specific values. Each of these values isn’t continuous but rather increases from one allowed value to another by a small, or quantum, jump. More specifically, a quantum is the difference between two allowed values in a set.
Based on the assumption that all atoms on the surface of the heated solid vibrate at the frequency, Planck developed a model that came to be known as Planck’s equation. Through experiments of frequencies and temperature, Planck was able to generate a constant, Planck’s constant
$h = 6.62607 \times 10^{-34} J\, s$
Using this constant he was able to restate his theory: energy was directly proportional to frequency. He wrote his equation as
$E=h\nu$
where $E$ is energy, $h$ is Planck’s constant, and v is frequency.
However, due to lack of solid proof, scientists, including Planck were skeptical about the new field of quantum theory. Since Planck’s hypothesis couldn’t be applied to anything other than blackbody radiation, it was unaccepted until much later when it was successfully applied to other phenomena.
The Photoelectric Effect
The basic idea behind the photoelectric effect is that under certain conditions when light is shined on a sample, electrons are ejected from that sample. Experimentation showed that the frequency of the light has to be above a certain threshold value for electrons to be emitted. After studying the photoelectric effect under several conditions, scientists made three observations.
1. A certain minimum frequency is required for electrons to be emitted.
2. Kinetic energy is directly proportional to frequency.
3. The number of electrons emitted from the surface was not dependent on intensity.
Scientists realized that frequency, not intensity, controlled whether or not electrons were emitted. Under classical wave theory electrons would be ejected at any frequency as long as it was intense enough, but this doesn’t happen. Since this dependency on frequency couldn’t be classical physics, scientists had to turn to quantum theory.
In order to explain this effect, Albert Einstein proposed that electromagnetic radiation also had particle-like qualities. Each of these particles of light was called a photon. Einstein suggested that each photon has an energy equal to hv, which is called a quantum of energy. This quantum of energy is the energy that is required of each electron in order to leave the metal surface.
The above mentioned threshold value for the frequency comes from the work function
$h\nu=1/2(m)(u^2)+w$
where w is the potential energy that is required to remove the electron from the surface and $½ mu^2$ is the kinetic energy of the electron once it has left the surface of the solid. The threshold frequency, v0, is the energy that is just sufficient to remove one electron and is denoted by
$\nu_0=w/h$
A light of smaller frequency cannot eject an electron no matter how long it falls on the metal surface.
The reason the photoelectric effect was so significant was that the relationship between radiation and a particle of matter caused scientists to understand that the wave theory of radiation wasn’t going to be enough to explain a lot of phenomena. This led to the development of a new way of thinking: wave-particle duality.
Ideas That Led to Quantum Theory
One important idea that is the basis of quantum theory is wave-particle duality, first shown through the photoelectric effect. In order to prove that the electron was a wave G.P. Thomson designed an experiment—the double-slit experiment. When a stream of electrons was directed at metal foil through one slit, a thin band was formed on the foil as expected. Likewise, when the electrons were directed through two slits, two bands would be expected to form. However, the experiment showed that an interference pattern, like would be expected of a wave, was formed instead. This experiment was one of several that gave rise to Quantum Mechanics
This is an example of another separate rule for quantum mechanics. In the macroscopic world of classical theory, a wave is a wave and a particle is a particle. One cannot and will not ever be the other. However, in the microscopic quantum world, this isn’t true. Electrons of atoms and photons of light aren’t necessarily particles or waves. In fact, physicists are having a hard time determining just what they even are since they have the properties of both waves and particles.
Another important idea in the field of quantum mechanics is the Heisenberg uncertainty principle. From a broad perspective, the uncertainty principle states that the position and momentum of a particle can never be precisely measured simultaneously. If one is known, the other cannot be determined accurately. This principle is a consequence of wave-particle duality and therefore leads physicists to embrace a modern description of atoms.
Outside Links
• Dr. Quantum Double Slit Experiment: www.youtube.com/watch?v=O55Xi...eature=related (this video explains the double slit experiment and wave-particle duality)
• The Photoelectric Effect: www.jce.divched.org/JCEDLib/W...peeffect5.html (this website has great visual representations of the photoelectric effect)
Problems
1. What is a quantum of energy?
2. Explain the significance of the photoelectric effect.
3. If an atom has a frequency of 5.357 x 1014 s-1, using Planck's equation, what is the energy of a single photon?
4. Using the answer from number 3, calculate the energy of a mole of photons.
5. Assume you have light at a wavelength of 640nm. Using the equation below as well as Planck's equation, calculate the energy of one photon of light at that wavelength. Remember that c= 3.00 x 108 m s-1 and $\lambda$ is wavelength in meters.
$c=\nu\lambda$
Answers
1. A quantum of energy is the energy difference between the two allowed values in a set. It is a tiny jump that moves from one value to another without ever reaching intermediate values.
2. The photoelectric effect was especially significant in the field of quantum theory because it essentially proved the necessity of this new field. Since it showed the frequency, not intensity, causes electrons to be expelled, it disproved some of the theories of classical physics that physicists of the time thought to be true, causing the need for a new field: quantum theory.
3. Since we know the frequency of the photon, we simply plug it into the formula.
$E=(6.626 x 10-34 J s)(5.357 x 1014 s-1)$
The resulting energy is 3.550 x 10-19 J.
4. In order to find the energy of a mole of atoms, we use the energy we found in question 3. We then use dimensional analysis to calculate the energy. Since there is 3.550 x 10-19 J of energy in 1 photon, and 6.022 x 1023 photons in a mole, the energy per mole would be 213,781 J/mol.
5. We know that the wavelength is in nanometers and speed of light is in meters. So we first convert wavelength to meters so they're in the same units. 1nm = 1.0 x 10-9 m. So in meters, the wavelength would be 6.40 x 10-7 m. Next, we calculate frequency by dividing c by λ.
$\nu=c/\lambda$
$\nu$=3.00 x 108 m s-1/6.40 x 10-7m
The resulting frequency is 4.688 x 1014 s-1. Now we can use Planck's equation to find the energy of a photon.
E=(6.626 x 10-34 J s)(4.688 x 1014 s-1)
The energy of a photon at 640nm is 3.106 x 10-19 J.
Contributors and Attributions
• Akshaya Ramanujam (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Quantum_Theory.txt
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In classical physics, studying the behavior of a physical system is often a simple task due to the fact that several physical qualities can be measured simultaneously. However, this possibility is absent in the quantum world. In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously.
The Nature of Measurement
In order to understand the conceptual background of the Heisenberg Uncertainty Principle it is important to understand how physical values are measured. In almost any measurement that is made, light is reflected off the object that is being measured and processed. The shorter the wavelength of light used, or the higher its frequency and energy, the more accurate the results. For example, when attempting to measure the speed of a tennis ball as it is dropped off of a ledge, photons(measurement of light) are shot off the tennis ball, reflected, and then processed by certain equipment. Because the tennis ball is so large compared to the photons, it is unaffected by the efforts of the observer to measure its physical quantities. However, if a photon is shot at an electron, the minuscule size of the electron and its unique wave-particle duality introduces consequences that can be ignored when taking measurements of macroscopic objects.
Heisenberg himself encountered such limitations as he attempted to measure the position of an electron with a microscope. As noted, the accuracy of any measurement is limited by the wavelength of light illuminating the electron. Therefore, in principle, one can determine the position as accurately as one wishes by using light of very high frequency, or short wave-lengths. However, the collision between such high energy photons of light with the extremely small electron causes the momentum of the electron to be disturbed.
Thus, increasing the energy of the light (and increasing the accuracy of the electron's position measurement), increases such a deviation in momentum. Conversely, if a photon has low energy the collision does not disturb the electron, yet the position cannot be accurately determined. Heisenberg concluded in his famous 1927 paper on the topic,
"At the instant of time when the position is determined, that is, at the instant when the photon is scattered by the electron, the electron undergoes a discontinuous change in momentum. This change is the greater the smaller the wavelength of the light employed, i.e., the more exact the determination of the position. At the instant at which the position of the electron is known, its momentum therefore can be known only up to magnitudes which correspond to that discontinuous change; thus, the more precisely the position is determined, the less precisely the momentum is known..." (Heisenberg, 1927, p. 174-5).
Heisenberg realized that since both light and particle energy are quantized, or can only exist in discrete energy units, there are limits as to how small, or insignificant, such an uncertainty can be. As proved later in this text, that bound ends up being expressed by Planck's Constant, h = 6.626*10-34 J*s.
It is important to mention that The Heisenberg Principle should not be confused with the observer effect. The observer effect is generally accepted to mean that the act of observing a system will influence that which is being observed. While this is important in understanding the Heisenberg Uncertainty Principle, the two are not interchangeable. The error in such thinking can be explained using the wave-particle duality of electromagnetic waves, an idea first proposed by Louis de Broglie. Wave-particle duality asserts that any energy exhibits both particle- and wave-like behavior. As a consequence, in quantum mechanics, a particle cannot have both a definite position and momentum. Thus, the limitations described by Heisenberg are a natural occurrence and have nothing to do with any limitations of the observational system.
Heisenberg’s Uncertainty Principle
It is mathematically possible to express the uncertainty that, Heisenberg concluded, always exists if one attempts to measure the momentum and position of particles.First, we must define the variable “x” as the position of the particle, and define “p” as the momentum of the particle. The momentum of a photon of light is known to simply be its frequency, expressed by the ratio h/λ, where h represents Planck’s constant and λ represents the wavelength of the photon. The position of a photon of light is simply its wavelength, \lambda\).. In order to represent finite change in quantities, the Greek uppercase letter delta, or Δ, is placed in front of the quantity. Therefore,
$\Delta{p}=\dfrac{h}{\lambda}$
$\Delta{x}= \lambda$
By substituting $\Delta{x}$ for $\lambda$ in the first equation, we derive
$\Delta{p}=\dfrac{h}{\Delta{x}}$
or,
$\Delta{p}\Delta{x}=h$
Note, we can derive the same formula by assuming the particle of interest is behaving as a particle, and not as a wave. Simply let Δp=mu, and Δx=h/mu (from De Broglie’s expression for the wavelength of a particle). Substituting in Δp for mu in the second equation leads to the very same equation derived above-ΔpΔx=h. This equation was refined by Heisenberg and his colleague Niels Bohr, and was eventually rewritten as
$\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi}$
What this equation reveals is that the more accurately a particle’s position is known, or the smaller Δx is, the less accurately the momentum of the particle Δp is known. Mathematically, this occurs because the smaller Δx becomes, the larger Δp must become in order to satisfy the inequality. However, the more accurately momentum is known the less accurately position is known.
Understanding the Uncertainty Principle through Wave Packets and the Slit Experiment
It is hard for most people to accept the uncertainty principle, because in classical physics the velocity and position of an object can be calculated with certainty and accuracy. However, in quantum mechanics, the wave-particle duality of electrons does not allow us to accurately calculate both the momentum and position because the wave is not in one exact location but is spread out over space. A "wave packet" can be used to demonstrate how either the momentum or position of a particle can be precisely calculated, but not both of them simultaneously. An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the "wave packet". The more waves that are combined in the "wave packet", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the "wave packet" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously.
Several scientists have debated the Uncertainty Principle, including Einstein. Einstein created a slit experiment to try and disprove the Uncertainty Principle. He had light passing through a slit, which causes an uncertainty of momentum because the light behaves like a particle and a wave as it passes through the slit. Therefore, the momentum is unknown, but the initial position of the particle is known. Here is a video that demonstrates particles of light passing through a slit and as the slit becomes smaller, the final possible array of directions of the particles becomes wider. As the position of the particle becomes more precise when the slit is narrowed, the direction, or therefore the momentum, of the particle becomes less known as seen by a wider horizontal distribution of the light.
The Importance of the Heisenberg Uncertainty Principle
Heisenberg’s Uncertainty Principle not only helped shape the new school of thought known today as quantum mechanics, but it also helped discredit older theories. Most importantly, the Heisenberg Uncertainty Principle made it obvious that there was a fundamental error in the Bohr model of the atom. Since the position and momentum of a particle cannot be known simultaneously, Bohr’s theory that the electron traveled in a circular path of a fixed radius orbiting the nucleus was obsolete. Furthermore, Heisenberg’s uncertainty principle, when combined with other revolutionary theories in quantum mechanics, helped shape wave mechanics and the current scientific understanding of the atom.
Problems
1. What aspect of the Bohr model of the atom does the Heisenberg Uncertainty Principle discredit?
2. What is the difference between the Heisenberg Uncertainty Principle and the Observer Effect?
3. A Hydrogen atom has a radius of 0.05nm with a position accuracy of 1.0%. What is the uncertainty in determining the velocity?
4. What is the uncertainty in the speed of a beam of electrons whose position is known with an uncertainty of 10 nm?
5. Using the Uncertainty Principle, find the radius of an atom (in nm) that has an electron with a position accuracy of 3.0% and a known velocity of $2\times 10^9\, m/s$.
Answers:
1.) The Heisenberg Uncertainty Principle discredits the aspect of the Bohr atom model that an electron is constrained to a one-dimensional orbit of a fixed radius around the nucleus.
2.) The Observer Effect means the act of observing a system will influence what is being observed, whereas the Heisenberg Uncertainty Principle has nothing to do with the observer or equipment used during observation. It simply states that a particle behaves both as a wave and a particle and therefore cannot have both a definite momentum and position.
3.) Uncertainty principle: ΔxΔp≥h/4Π
Can be written (x)(m)(v)(%)=h/4Π
(position)(mass of electron)(velocity)(percent accuracy)=(Plank's Constant)/4Π
(.05*10-9 m)(9.11*10-31 kg)(v)(.01)=(6.626*10-34 J*s)/4Π
v=1*108 m/s
4.) Uncertainty principle: ΔxΔp≥h/4Π
(10*10-9 m)(Δp)≥(6.626*10-34 J*s)/4Π
Δp≥5*10-26 (kg*m)/s
5.) Uncertainty principle: ΔxΔp≥h/4Π
Can be written (x)(m)(v)(%)=h/4Π
(position)(mass of electron)(velocity)(percent accuracy)=h/4Π
(radius)(9.11*10-31 kg)(2*109)(.03)=(6.626*10-34 J*s)/4Π
r=9*10-12 m
r=9*10-3 nm
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Uncertainty_Principle.txt
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In 1923, Louis de Broglie, a French physicist, proposed a hypothesis to explain the theory of the atomic structure.By using a series of substitution de Broglie hypothesizes particles to hold properties of waves. Within a few years, de Broglie's hypothesis was tested by scientists shooting electrons and rays of lights through slits. What scientists discovered was the electron stream acted the same was as light proving de Broglie correct.
Contributors and Attributions
• Duy Nguyen (UCD)
Wave-Particle Duality II
Quantum mechanics incorporates the idea of wave mechanics that demonstrates the idea of wave-particle duality.This notion suggests that matter can display simultaneously both particle and wave-like properties. This new approach came from Louis de Broglie who built upon Einstein's conception that light possessed particle-like properties in his attempt to explain the photoelectric effect.
Photoelectric Effect
Albert Einstein showed that the dependence on frequency could not be justified by the classical wave theory alone, so he provided a particle perspective. In 1905 he declared that photons (named by G.N. Lewis), were "particles of light" that had similar energy to that of Planck's equation. This equation states that the frequency and energy of a quantum of electromagnetic radiation are proportional. Einstein's idea was revolutionary because he brought a new perspective at looking at light not only as a wave, but as a particle.
Planck's equation: E=hv Planck's constant: h=6.626x10-34 Js
The photoelectric effect phenomenon that electrons are emitted when light strikes the surface of metals was discovered by Heinrich Hertz in 1888. This process holds true when the incident light has a higher frequency than a certain threshold value. The amount of electrons ejected is determined by the intensity of the incident light, however, the frequency of the light effects the kinetic energies of the emitted electrons.
In other words, the intensity can be described as the brightness from a source of light. So by increasing brightness, intensity increases, and so does energy released. The energy output will be greater and when this happens the amplitude of the light wave increases. But it doesn't matter how much energy is increased or how much you increase the amplitude when it comes to trying to emit electrons from a metallic surface. To do so, frequency must increase.
Increase brightness (maintains frequency and energy)-->Increases Intensity (increases#of photons)-->Increases # of electrons emitted
Increase frequency-->Increases kinetic energy of electrons
Einstein explanation was that light had the characteristic of a particle (photon) with the photon energy of E=hv. He concluded that if the threshold frequency of the metal was greater than the frequency of the photon, then the photon will have no effect when it bombards the metal surface. However, if the photon reached the threshold frequency it could cause one electron to be emitted. In order to emit more electrons, the light source must be brightened to increase intensity, which still maintains frequency of light and same energy, but increases the number of photons.
Threshold frequency: Vo= (eVo)/h = work function/Planck's constant
The photoelectric effect can occur even with the lowest frequency light called the threshold frequency.
Photelectrons are released when photon energy (hv) is greater than the work function. Energy in excess is released as kinetic energy in the ejected photoelectron and is proportional to frequency of light.
The diagram above illustrates an electron being striked by a photon of energy, which allows it to overcome the work function binding it to the metal surface. As a result, a photoelectron is emitted with kinetic energy.
By applying the law of conservation of energy we get the equation: hv =eVo+(1/2)mv2
To summarize, regardless of the intensity of light, no electrons will be emitted if the frequency of light is below the threshold frequency (Vo) of the metal surface.
1. If the frequency of incident light is above the threshold frequency,then kinetic energy of the emitted particle will increase linearly in respect to the magnitude of frequency.
2. If incident light frequency is greater than threshold frequency, the number of emitted electrons is determined by intensity. (IMPORTANT NOTE: Kinetic energy per electron doesn't change if intensity is changed, only when frequency is manipulated)
3. Although each metal has it's own unique threshold frequency, they all have similar patterns.
De Broglie Wavelength
In 1924, Louis de Broglie used Einstein's equation E=mc2 and incorporated it with Planck's equation. This brought together the relationship of the mass of a photon and speed of light with the photon energy.
Einstein's equation: E=mc2 m=mass of photon c=speed of light COMBINED WITH Planck's equation: E=hv
E=mc2 & E=hv --> mc2=hv --> mc=(hv)/c=p where p is momentum
Since wavelegth is speed of light/frequency as shown by λ=c/f momentum is equivalent to p=(hv)/c=h/λ
In order for this equation to apply for the purpose of a material particle (electron), de Broglie changed the equation to its equivalent
So p=h/λ changed to λ=h/p=h/(mv) where p is the product of mass of particle and velocity
λ=h/(mv) demonstrates that particles have wavelike properties as shown by the integration of wavength (λ), Planck's constant (h), and momentum of particle (p=mv). This equation is also known as the "de Broglie wavelength." He reasoned that if things that acted like waves had particle characteristics then it should work vice versa. Therefore, things that behave like particles should also have wave characteristics. He coined the term "matter waves" to describe waves that involved material particles, such as electrons.
Particle Properties
Single Slit Diffraction
Pattern is made up of emitted particles
Double Slit Diffraction
Pattern is similar to wave interference
Wave Properties
When waves travel in the same medium, interference can occur. Interference is the net effect caused by two different waves. In the example below, the superposition principle is in effect. The diagram shows that when two waves are traveling with the same frequency through the same medium, the individual waves pass one another without disturbance.
There are two types of interferences known as Constructive and Destructive Interference.
1. Constructive Interference - When waves are "in phase" this means that their crests and troughs coincide. The net result produces a combination of the two waves that produces high crests and deep troughs causing amplification. The waves overlap and create an even larger wave.
2. Destructive Interference - This occurs when waves cancel one another because the crest of one wave occurs at the trough of another and is considered "out of phase."
Uncertainty Principle
The wave-particle duality limits our understanding in determining the position and momentum of subatomic particles. In the 1920's, Werner Heisenberg and Niels Bohr tried different experiments to precisely measure the behavior of particles. However, they were unable to simultaneously measure position and momentum at the same time. This meant that even if they knew where a particle was, they couldn't find where it was going nor where it came from, after all the wave is spread out. They concluded that it was impossible to accurately measure and determine an electron's momentum and position simultaneously, hence the uncertainty principle is also referred to as the principle of indeterminacy.
Examples
1. From this diagram determine the number of crests and troughs. Remember crests are the high peaks and troughs form the deep ends. Therefore, there are 4 crests and 3 troughs.
2. If a radiation has a wavelength of 302.5nm, what is the energy of one photon?
To do this, we must look at Planck's equation and find the frequency of the radiation.
Frequency is v= c/λ and c=speed of light (2.998x108 m/s) --> v= (2.998x108 m/s)/302.5x10-9m = 9.911x1014s-1
Planck's equation E=hv = (6.626x10-34Js/photon)(9.911x1014s-1)=6.567x10-19J/photon
3. A photon has a frequency of 5.5x103Hz. Convert this to wavelength.
c=λv so λ=c/v= (2.998x108 m/s)/5.5x103Hz=5.5x104m Remember Hz is equivalent to s-1
5.5x104m (1nm/10-9m)= 5.5x1013nm Remember wavelengths are expressed in nanometers (nm)
4. What is the de Broglie wavelength of a 1.5g bouncy ball traveling at 20m/s?
λ=h/mv Convert g to kg so 1.5g(1kg/103g)=,0015kg Joule=Nm Newton=kgm/s2
λ=h/mv = 6.626x10-34Js/(.0015kgx20m/s)=2.209x10-32m Stoichiometry=Js/[kg(m/s)]=(Ns2/kg)=(kgms2/kgs2)=m
2.209x10-32m (1nm/10-9m)= 2.2x10-23nm
5. If a metal had a threshold frequency of 7.38x1013Hz, would it display the photelectric effect with infrared light?
The maximum frequency of infrared light is about 3x1014Hz. Since this is above the threshold frequency for the metal, it will display the photoelectric effect. However, if the threshold frequency was greater than the infrared light frequency, no photoelectric effect would occur.
Problems
1. How many crests and troughs are shown here?
2. The higher the frequency, the __________(smaller/greater) the wavelength.
3. Intensity of the incident light effects the amount of ______ emitted, whereas the frequency of incident light effects the _____ energy.
4. Using Planck's equation determine the frequency of radiation if it has an energy of 3.21x10-28 J/photon.
5. What is the energy (in joules) of a photon with a threshold frequency radiation of 8.67X1012Hz?
6. What is the de Broglie wavelenth of a 8.5g tennis bal traveling at 12.6m/s?
Outside Links
• Interference (Wave Propogation)
• Double-Slit Experiment
Contributors and Attributions
• Maryann Hernandez (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality.txt
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The picture of electrons "orbiting" the nucleus like planets around the sun remains an enduring one, not only in popular images of the atom but also in the minds of many of us who know better. The proposal, first made in 1913, that the centrifugal force of the revolving electron just exactly balances the attractive force of the nucleus (in analogy with the centrifugal force of the moon in its orbit exactly counteracting the pull of the Earth's gravity) is a nice picture, but is simply untenable.
One origin for this hypothesis that suggests this perspective is plausible is the similarity of the gravity and Coulombic interactions. The expression for the force of gravity between two masses (Newton's Law of gravity) is
$F_{gravity} \propto \dfrac{m_1m_2}{r^2}\label{1}$
with
• $m_1$ and $m_2$ representing the mass of object 1 and 2, respectively and
• $r$ representing the distance between the objects centers
The expression for the Coulomb force between two charged species is
$F_{Coulomb} \propto \dfrac{q_1q_2}{r^2}\label{2}$
with
• $q_1$ and $q_2$ representing the charge of object 1 and 2, respectively and
• $r$ representing the distance between the objects centers
However, an electron, unlike a planet or a satellite, is electrically charged, and it has been known since the mid-19th century that an electric charge that undergoes acceleration (changes velocity and direction) will emit electromagnetic radiation, losing energy in the process. A revolving electron would transform the atom into a miniature radio station, the energy output of which would be at the cost of the potential energy of the electron; according to classical mechanics, the electron would simply spiral into the nucleus and the atom would collapse.
Quantum theory to the Rescue!
By the 1920's, it became clear that a tiny object such as the electron cannot be treated as a classical particle having a definite position and velocity. The best we can do is specify the probability of its manifesting itself at any point in space. If you had a magic camera that could take a sequence of pictures of the electron in the 1s orbital of a hydrogen atom, and could combine the resulting dots in a single image, you would see something like this. Clearly, the electron is more likely to be found the closer we move toward the nucleus.
This is confirmed by this plot which shows the quantity of electron charge per unit volume of space at various distances from the nucleus. This is known as a probability density plot. The per unit volume of space part is very important here; as we consider radii closer to the nucleus, these volumes become very small, so the number of electrons per unit volume increases very rapidly. In this view, it appears as if the electron does fall into the nucleus!
According to classical mechanics, the electron would simply spiral into the nucleus and the atom would collapse. Quantum mechanics is a different story.
The Battle of the Infinities Saves the electron from its death spiral
As you know, the potential energy of an electron becomes more negative as it moves toward the attractive field of the nucleus; in fact, it approaches negative infinity. However, because the total energy remains constant (a hydrogen atom, sitting peacefully by itself, will neither lose nor acquire energy), the loss in potential energy is compensated for by an increase in the electron's kinetic energy (sometimes referred to in this context as "confinement" energy) which determines its momentum and its effective velocity.
So as the electron approaches the tiny volume of space occupied by the nucleus, its potential energy dives down toward minus-infinity, and its kinetic energy (momentum and velocity) shoots up toward positive-infinity. This "battle of the infinities" cannot be won by either side, so a compromise is reached in which theory tells us that the fall in potential energy is just twice the kinetic energy, and the electron dances at an average distance that corresponds to the Bohr radius.
There is still one thing wrong with this picture; according to the Heisenberg uncertainty principle (a better term would be "indeterminacy"), a particle as tiny as the electron cannot be regarded as having either a definite location or momentum. The Heisenberg principle says that either the location or the momentum of a quantum particle such as the electron can be known as precisely as desired, but as one of these quantities is specified more precisely, the value of the other becomes increasingly indeterminate. It is important to understand that this is not simply a matter of observational difficulty, but rather a fundamental property of nature.
What this means is that within the tiny confines of the atom, the electron cannot really be regarded as a "particle" having a definite energy and location, so it is somewhat misleading to talk about the electron "falling into" the nucleus.
Arthur Eddington, a famous physicist, once suggested, not entirely in jest, that a better description of the electron would be "wavicle"!
Probability Density vs. Radial probability
We can, however, talk about where the electron has the highest probability of manifesting itself— that is, where the maximum negative charge will be found.
This is just the curve labeled "probability density"; its steep climb as we approach the nucleus shows unambiguously that the electron is most likely to be found in the tiny volume element at the nucleus. But wait! Did we not just say that this does not happen? What we are forgetting here is that as we move out from the nucleus, the number of these small volume elements situated along any radius increases very rapidly with $r$, going up by a factor of $4πr^2$. So the probability of finding the electron somewhere on a given radius circle is found by multiplying the probability density by $4πr^2$. This yields the curve you have probably seen elsewhere, known as the radial probability, that is shown on the right side of the above diagram. The peak of the radial probability for principal quantum number $n = 1$ corresponds to the Bohr radius.
To sum up, the probability density and radial probability plots express two different things: the first shows the electron density at any single point in the atom, while the second, which is generally more useful to us, tells us the the relative electron density summed over all points on a circle of given radius.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Why_atoms_do_not_Collapse.txt
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Niels Bohr introduced the atomic Hydrogen model in 1913. He described it as a positively charged nucleus, comprised of protons and neutrons, surrounded by a negatively charged electron cloud. In the model, electrons orbit the nucleus in atomic shells. The atom is held together by electrostatic forces between the positive nucleus and negative surroundings.
Hydrogen Energy Levels
The Bohr model is used to describe the structure of hydrogen energy levels. The image below represents shell structure, where each shell is associated with principal quantum number n. The energy levels presented correspond with each shell. The amount of energy in each level is reported in eV, and the maxiumum energy is the ionization energy of 13.598eV.
Figure 1: Some of the orbital shells of a Hydrogen atom. The energy levels of the orbitals are shown to the right.
Hydrogen Spectrum
The movement of electrons between these energy levels produces a spectrum. The Balmer equation is used to describe the four different wavelengths of Hydrogen which are present in the visible light spectrum. These wavelengths are at 656, 486, 434, and 410nm. These correspond to the emission of photons as an electron in an excited state transitions down to energy level n=2. The Rydberg formula, below, generalizes the Balmer series for all energy level transitions. To get the Balmer lines, the Rydberg formula is used with an nf of 2.
Rydberg Formula
The Rydberg formula explains the different energies of transition that occur between energy levels. When an electron moves from a higher energy level to a lower one, a photon is emitted. The Hydrogen atom can emit different wavelengths of light depending on the initial and final energy levels of the transition. It emits a photon with energy equal to the difference of square of the final ($n_f$) and initial ($n_i$) energy levels.
$\text{Energy}=R\left(\dfrac{1}{n^2_f}-\dfrac{1}{n^2_i}\right) \label{1}$
The energy of a photon is equal to Planck’s constant, h=6.626*10-34m2kg/s, times the speed of light in a vacuum, divided by the wavelength of emission.
$E=\dfrac{hc}{\lambda} \label{2}$
Combining these two equations produces the Rydberg Formula.
$\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^2_f}-\dfrac{1}{n^2_i}\right) \label{3}$
The Rydberg Constant (R) = $10,973,731.6\; m^{-1}$ or $1.097 \times 10^7\; m^{-1}$.
Limitations of the Bohr Model
The Bohr Model was an important step in the development of atomic theory. However, it has several limitations.
• It is in violation of the Heisenberg Uncertainty Principle. The Bohr Model considers electrons to have both a known radius and orbit, which is impossible according to Heisenberg.
• The Bohr Model is very limited in terms of size. Poor spectral predictions are obtained when larger atoms are in question.
• It cannot predict the relative intensities of spectral lines.
• It does not explain the Zeeman Effect, when the spectral line is split into several components in the presence of a magnetic field.
• The Bohr Model does not account for the fact that accelerating electrons do not emit electromagnetic radiation.
Problems
1. An emission spectrum gives one of the lines in the Balmer series of the hydrogen atom at 410 nm. This wavelength results from a transition from an upper energy level to n=2. What is the principal quantum number of the upper level?
2. The Bohr model of the atom was able to explain the Balmer series because:
1. larger orbits required electrons to have more negative energy in order to match the angular momentum.
2. differences between the energy levels of the orbits matched the difference between energy levels of the line spectra.
3. electrons were allowed to exist only in allowed orbits and nowhere else.
4. none of the above
3. One reason the Bohr model of the atom failed was because it did not explain why
1. accelerating electrons do not emit electromagnetic radiation.
2. moving electrons have a greater mass.
3. electrons in the orbits of an atom have negative energies.
4. electrons in greater orbits of an atom have greater velocities.
Answers
1. (1/λ) = R*[ 1/(22) - 1/(n2) ] , R=1.097x107 m-1 , λ=410nm
(1/410nm) = (1.097x107 m-1) * [ 1/(22) - 1/(n2) ]
[ (1/4.10x10-7m) / (1.097x107 m-1) ] - [ (1/4) ] = [ -1/(n2) ]
-1/-0.02778 = n2
36 = n2 , n=6 --> The emission resulted from a transition from energy level 6 to energy level 2.
2. (B) differences between the energy levels of the orbits matched the difference between energy levels of the line spectra.
3. (A) accelerating electrons do not emit electromagnetic radiation.
Contributors and Attributions
• Michelle Faust
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Bohr%27s_Hydrogen_Atom.txt
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Atomic Spectra
When gaseous hydrogen in a glass tube is excited by a $5000$-volt electrical discharge, four lines are observed in the visible part of the emission spectrum: red at $656.3$ nm, blue-green at $486.1$ nm, blue violet at $434.1$ nm and violet at $410.2$ nm:
Other series of lines have been observed in the ultraviolet and infrared regions. Rydberg (1890) found that all the lines of the atomic hydrogen spectrum could be fitted to a single formula
$\dfrac{1}{\lambda} = \mathcal{R} \left( \dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}} \right), \quad n_1 = 1, \: 2, \: 3..., \: n_2 > n_1 \label{1}$
where $\mathcal{R}$, known as the Rydberg constant, has the value $109,677$ cm-1 for hydrogen. The reciprocal of wavelength, in units of cm-1, is in general use by spectroscopists. This unit is also designated wavenumbers, since it represents the number of wavelengths per cm. The Balmer series of spectral lines in the visible region, shown in Figure $1$, correspond to the values $n_1 = 2, \: n_2 = 3, \: 4, \: 5$ and $6$. The lines with $n_1 = 1$ in the ultraviolet make up the Lyman series. The line with $n_2 = 2$, designated the Lyman alpha, has the longest wavelength (lowest wavenumber) in this series, with $1/ \lambda = 82.258$ cm-1 or $\lambda = 121.57$ nm.
Other atomic species have line spectra, which can be used as a "fingerprint" to identify the element. However, no atom other than hydrogen has a simple relation analogous to Equation $\ref{1}$ for its spectral frequencies. Bohr in 1913 proposed that all atomic spectral lines arise from transitions between discrete energy levels, giving a photon such that
$\Delta E = h \nu = \dfrac{hc}{\lambda} \label{2}$
This is called the Bohr frequency condition. We now understand that the atomic transition energy $\Delta E$ is equal to the energy of a photon, as proposed earlier by Planck and Einstein.
The Bohr Atom
The nuclear model proposed by Rutherford in 1911 pictures the atom as a heavy, positively-charged nucleus, around which much lighter, negatively-charged electrons circulate, much like planets in the Solar system. This model is however completely untenable from the standpoint of classical electromagnetic theory, for an accelerating electron (circular motion represents an acceleration) should radiate away its energy. In fact, a hydrogen atom should exist for no longer than $5 \times 10^{-11}$ sec, time enough for the electron's death spiral into the nucleus. This is one of the worst quantitative predictions in the history of physics. It has been called the Hindenberg disaster on an atomic level. (Recall that the Hindenberg, a hydrogen-filled dirigible, crashed and burned in a famous disaster in 1937.)
Bohr sought to avoid an atomic catastrophe by proposing that certain orbits of the electron around the nucleus could be exempted from classical electrodynamics and remain stable. The Bohr model was quantitatively successful for the hydrogen atom, as we shall now show.
We recall that the attraction between two opposite charges, such as the electron and proton, is given by Coulomb's law
$F = \begin{cases} -\dfrac{e^{2}}{r^{2}} \quad \mathsf{(gaussian \: units)} \ -\dfrac{e^{2}}{4 \pi \epsilon_0 r^{2}} \quad \mathsf{(SI \: units)} \end{cases} \label{3}$
We prefer to use the Gaussian system in applications to atomic phenomena. Since the Coulomb attraction is a central force (dependent only on r), the potential energy is related by
$F = -\dfrac{dV(r)}{dr} \label{4}$
We find therefore, for the mutual potential energy of a proton and electron,
$V(r) = -\dfrac{e^2}{r} \label{5}$
Bohr considered an electron in a circular orbit of radius $r$ around the proton. To remain in this orbit, the electron must be experiencing a centripetal acceleration
$a = -\dfrac{v^{2}}{r} \label{6}$
where $v$ is the speed of the electron. Using Equations $\ref{4}$ and $\ref{6}$ in Newton's second law, we find
$\dfrac{e^{2}}{r^{2}} = \dfrac{mv^{2}}{r} \label{7}$
where $m$ is the mass of the electron. For simplicity, we assume that the proton mass is infinite (actually $m_p \approx 1836 m_e$) so that the proton's position remains fixed. We will later correct for this approximation by introducing reduced mass. The energy of the hydrogen atom is the sum of the kinetic and potential energies:
$E = T + V = \dfrac{1}{2} mv^{2} - \dfrac{e^{2}}{r} \label{8}$
Using Equation $\ref{7}$, we see that
$T = -\dfrac{1}{2} V\ \qquad \mathsf{and} \qquad E = \dfrac{1}{2} V = -T \label{9}$
This is the form of the virial theorem for a force law varying as $r^{-2}$. Note that the energy of a bound atom is negative, since it is lower than the energy of the separated electron and proton, which is taken to be zero.
For further progress, we need some restriction on the possible values of $r$ or $v$. This is where we can introduce the quantization of angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. Since $\mathbf{p}$ is perpendicular to $\mathbf{r}$, we can write simply
$L = rp = mvr \label{10}$
Using Equation $\ref{9}$, we find also that
$r = \dfrac{L^{2}}{me^{2}} \label{11}$
We introduce angular momentum quantization, writing
$L = n\hbar, \qquad n = 1, \: 2... \label{12}$
excluding $n = 0$, since the electron would then not be in a circular orbit. The allowed orbital radii are then given by
$r_n = n^{2} a_0 \label{13}$
where
$a_0 \equiv \dfrac{\hbar^{2}}{me^{2}} = 5.29 \times 10^{-11} \: \mathsf{m} = 0.529Å \label{14}$
which is known as the Bohr radius. The corresponding energy is
$E_n = -\dfrac{e^{2}}{2a_0n^{2}} = -\dfrac{me^{4}}{2\hbar^{2}n^{2}}, \qquad n = 1, \: 2... \label{15}$
Rydberg's formula (Equation $\ref{1}$) can now be deduced from the Bohr model. We have
$\dfrac{hc}{\lambda} = E_{n_2} - E_{n_1} = \dfrac{2\pi^{2}me^{4}}{h^{2}} \left( \dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}} \right) \label{16}$
and the Rydbeg constant can be identified as
$\mathcal{R} = \dfrac{2\pi^{2}me^{4}}{h^{3}c} \approx 109,737 \: \mathsf{cm}^{-1} \label{17}$
The slight discrepency with the experimental value for hydrogen $(109,677)$ is due to the finite proton mass. This will be corrected later.
The Bohr model can be readily extended to hydrogenlike ions, systems in which a single electron orbits a nucleus of arbitrary atomic number $Z$. Thus $Z = 1$ for hydrogen, $Z = 2$ for $\mathsf{He}^{+}$, $Z = 3$ for $\mathsf{Li}^{++}$, and so on. The Coulomb potential $\ref{5}$ generalizes to
$V(r) = -\dfrac{Ze^{2}}{r}, \label{18}$
the radius of the orbit (Equation $\ref{13}$) becomes
$r_n = \dfrac{n^{2}a_0}{Z} \label{19}$
and the energy Equation $\ref{15}$ becomes
$E_n = -\dfrac{Z^{2}e^{2}}{2a_0n^{2}} \label{20}$
De Broglie's proposal that electrons can have wavelike properties was actually inspired by the Bohr atomic model. Since
$L = rp = n\hbar = \dfrac{nh}{2\pi} \label{21}$
we find
$2\pi r = \dfrac{nh}{p} = n\lambda \label{22}$
Therefore, each allowed orbit traces out an integral number of de Broglie wavelengths.
Wilson (1915) and Sommerfeld (1916) generalized Bohr's formula for the allowed orbits to
$\oint p \, dr = nh, \qquad n =1, \: 2... \label{23}$
The Sommerfeld-Wilson quantum conditions Equation $\ref{23}$ reduce to Bohr's results for circular orbits, but allow, in addition, elliptical orbits along which the momentum $p$ is variable. According to Kepler's first law of planetary motion, the orbits of planets are ellipses with the Sun at one focus. Figure $2$ shows the generalization of the Bohr theory for hydrogen, including the elliptical orbits. The lowest energy state $n = 1$ is still a circular orbit. But $n = 2$ allows an elliptical orbit in addition to the circular one; $n = 3$ has three possible orbits, and so on. The energy still depends on $n$ alone, so that the elliptical orbits represent degenerate states. Atomic spectroscopy shows in fact that energy levels with $n > 1$ consist of multiple states, as implied by the splitting of atomic lines by an electric field (Stark effect) or a magnetic field (Zeeman effect). Some of these generalized orbits are drawn schematically in Figure $2$.
The Bohr model was an important first step in the historical development of quantum mechanics. It introduced the quantization of atomic energy levels and gave quantitative agreement with the atomic hydrogen spectrum. With the Sommerfeld-Wilson generalization, it accounted as well for the degeneracy of hydrogen energy levels. Although the Bohr model was able to sidestep the atomic "Hindenberg disaster," it cannot avoid what we might call the "Heisenberg disaster." By this we mean that the assumption of well-defined electronic orbits around a nucleus is completely contrary to the basic premises of quantum mechanics. Another flaw in the Bohr picture is that the angular momenta are all too large by one unit, for example, the ground state actually has zero orbital angular momentum (rather than $\hbar$).
The assumption of well-defined electronic orbits around a nucleus in the Bohr atom is completely contrary to the basic premises of quantum mechanics.
Quantum Mechanics of Hydrogenlike Atoms
In contrast to the particle in a box and the harmonic oscillator, the hydrogen atom is a real physical system that can be treated exactly by quantum mechanics. In addition to their inherent significance, these solutions suggest prototypes for atomic orbitals used in approximate treatments of complex atoms and molecules.
For an electron in the field of a nucleus of charge $+Ze$, the Schrӧdinger equation can be written
$\left\{ -\dfrac{\hbar^{2}}{2m} \nabla^{2} - \dfrac{Ze^{2}}{r} \right\} \psi(r) = E\psi(r) \label{24}$
It is convenient to introduce atomic units in which length is measured in bohrs:
$a_0 = \dfrac{\hbar^{2}}{me^{2}} = 5.29 \times 10^{-11} \: \mathsf{m} \equiv 1 \: \mathsf{bohr}$
and energy in hartrees:
$\dfrac{e^2}{a_0} = 4.358 \times 10^{-18} \: \mathsf{J} = 27.211 \: \mathsf{eV} \equiv 1 \: \mathsf{hartree}$
Electron volts $(\mathsf{eV})$ are a convenient unit for atomic energies. One $\mathsf{eV}$ is defined as the energy an electron gains when accelerated across a potential difference of $1 \: \mathsf{volt}$. The ground state of the hydrogen atom has an energy of $-1/2 \: \mathsf{hartree}$ or $-13.6 \: \mathsf{eV}$. Conversion to atomic units is equivalent to setting
$\hbar = e = m = 1$
in all formulas containing these constants. Rewriting the Schrӧdinger equation in atomic units, we have
$\left\{ -\dfrac{1}{2} \nabla^{2} - \dfrac{Z}{r} \right\} \psi(r) = E\psi(r) \label{25}$
Since the potential energy is spherically symmetrical (a function of $r$ alone), it is obviously advantageous to treat this problem in spherical polar coordinates $r, \: \theta, \: \phi$. Expressing the Laplacian operator in these coordinates [cf. Eq (6-20)],
$-\dfrac{1}{2} \left\{ \dfrac{1}{r^{2}} \dfrac{\partial}{\partial r} r^{2} \dfrac{\partial}{\partial r} + \dfrac{1}{r^{2}\sin\theta} \dfrac{\partial}{\partial \theta} \sin\theta \dfrac{\partial}{\partial \theta} + \dfrac{1}{r^{2}\sin^{2}\theta} \dfrac{\partial^{2}}{\partial\phi^{2}} \right\} \ \times \psi(r, \: \theta, \: \phi) - \dfrac{Z}{r} \psi(r, \: \theta, \: \phi) = E\psi(r, \: \theta, \: \phi) \label{26}$
Equation $\ref{26}$ shows that the second and third terms in the Laplacian represent the angular momentum operator $\hat{L}^{2}$. Clearly, Equation $\ref{26}$ will have separable solutions of the form
$\psi(r, \: \theta, \: \phi) = R(r)Y_{\ell m}(\theta, \: \phi) \label{27}$
Substituting Equation $\ref{27}$ into Equation $\ref{26}$ and using the angular momentum eigenvalue Equation Equation $\ref{6-34}$, we obtain an ordinary differential equation for the radial function $R(r)$:
$\left\{ -\dfrac{1}{2r^{2}} \dfrac{d}{dr} r^{2} \dfrac{d}{dr} + \dfrac{\ell(\ell + 1)}{2r^{2}} - \dfrac{Z}{r} \right\} R(r) = ER(r) \label{28}$
Note that in the domain of the variable $r$, the angular momentum contribution $\ell (\ell + 1) / 2r^{2}$ acts as an effective addition to the potential energy. It can be identified with centrifugal force, which pulls the electron outward, in opposition to the Coulomb attraction. Carrying out the successive differentiations in Equation $\ref{29}$ and simplifying, we obtain
$\dfrac{1}{2}R''(r) + \dfrac{1}{r}R'(r) + \left[\dfrac{Z}{r} - \dfrac{\ell(\ell + 1)}{2r^{2}} + E\right]R(r) = 0 \label{29}$
another second-order linear differential equation with non-constant coefficients. It is again useful to explore the asymptotic solutions to Equation $\ref{29}$, as $r \rightarrow \infty$. In the asymptotic approximation,
$R''(r) - 2r\lvert E \rvert R(r) \approx 0 \label{30}$
having noted that the energy $E$ is negative for bound states. Solutions to Equation $\ref{30}$ are
$R(r) \approx \mathsf{const} \, e^{\pm\sqrt{2\lvert E \rvert}r} \label{31}$
We reject the positive exponential on physical grounds, since $R(r) \rightarrow \infty$ as $r \rightarrow \infty$, in violation of the requirement that the wavefunction must be finite everywhere. Choosing the negative exponential and setting $E = -Z^{2}/2$ the ground state energy in the Bohr theory (in atomic units), we obtain
$R(r) \approx \mathsf{const} \, e^{-Zr} \label{32}$
It turns out, very fortunately, that this asymptotic approximation is also an exact solution of the Schrӧdinger equation (Equation $\ref{29}$) with $\ell = 0$, just what happened for the harmonic-oscillator problem in Chap. 5. The solutions to Equation $\ref{29}$, designated $R_{n\ell}(r)$, are labeled by $n$, known as the principal quantum number, as well as by the angular momentum $\ell$, which is a parameter in the radial equation. The solution in Equation $\ref{32}$ corresponds to $R_{10}(r)$. This should be normalized according to the condition
$\int_{0}^{\infty} [R_{10}(r)]^{2} \, r^{2} \, dr = 1 \label{33}$
A useful definite integral is
$\int_{0}^{\infty} r^{n} \, e^{-\alpha r} \, dr = \dfrac{n!}{\alpha^{n + 1}} \label{34}$
The normalized radial function is thereby given by
$R_{10}(r) = 2Z^{3/2} e^{-Zr} \label{35}$
Since this function is nodeless, we identify it with the ground state of the hydrogenlike atom. Multipyling Equation $\ref{35}$ by the spherical harmonic $Y_{00} = 1/ \sqrt{4\pi}$, we obtain the total wavefunction (Equation $\ref{27}$)
$\psi_{100}(r, \theta, \phi) = \left( \dfrac{Z^{3}}{\pi} \right)^{1/2} e^{-Zr} \label{36}$
This is conventionally designated as the 1s function $\psi_{1s}(r)$.
Integrals in spherical-polar coordinates over a spherically-symmetrical integrand (like the 1s orbital) can be significantly simplified. We can do the reduction
$\int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi} f(r) \, r^{2} \, \sin\theta \, dr \, d\theta \, d\phi = \int_{0}^{\infty} f(r) \, 4\pi r^{2} \, dr \label{37}$
since integration over $\theta$ and $\phi$ gives $4\pi$, the total solid angle of a sphere. The normalization of the 1s wavefunction can thus be written as
$\int_{0}^{\infty} [\psi_{1s}(r)]^{2} \, 4\pi r^{2} \, dr = 1 \label{38}$
Hydrogen Atom Ground State
There are a number of different ways of representing hydrogen-atom wavefunctions graphically. We will illustrate some of these for the 1s ground state. In atomic units,
$\psi_{1s}(r) = \dfrac{1}{\sqrt{\pi}}e^{-r} \label{39}$
is a decreasing exponential function of a single variable $r$, and is simply plotted in Figure 3.
Figure $3$ gives a somewhat more pictorial representation, a three-dimensional contour plot of $\psi_{1s}(r)$ as a function of $x$ and $y$ in the $x$, $y$-plane.
According to Born's interpretation of the wavefunction, the probability per unit volume of finding the electron at the point $(r, \: \theta, \: \phi)$ is equal to the square of the normalized wavefunction
$\rho_{1s}(r) = [\psi_{1s}(r)]^{2} = \dfrac{1}{\pi}e^{-2r} \label{40}$
This is represented in Figure 5 by a scatter plot describing a possible sequence of observations of the electron position. Although results of individual measurements are not predictable, a statistical pattern does emerge after a sufficiently large number of measurements.
The probability density is normalized such that
$\int_{0}^{\infty} \rho_{1s}(r) \, 4\pi r^{2} \, dr = 1 \label{41}$
In some ways $\rho (r)$ does not provide the best description of the electron distribution, since the region around $r = 0$, where the wavefunction has its largest values, is a relatively small fraction of the volume accessible to the electron. Larger radii $r$ represent larger physical regions since, in spherical polar coordinates, a value of $r$ is associated with a shell of volume $4\pi r^{2} \, dr$. A more significant measure is therefore the radial distribution function
$D_{1s}(r) = 4\pi r^{2} [\psi_{1s}(r)]^{2} \label{42}$
which represents the probability density within the entire shell of radius $r$, normalized such that
$\int_{0}^{\infty} D_{1s}(r) \, dr = 1 \label{43}$
The functions $\rho_{1s}(r)$ and $D_{1s}(r)$ are both shown in Figure $6$. Remarkably, the 1s RDF has its maximum at $r = a_0$, equal to the radius of the first Bohr orbit
Atomic Orbitals
The general solution for $R_{n\ell}(r)$ has a rather complicated form which we give without proof:
$R_{n\ell}(r) = N_{n\ell} \, \rho^{\ell} \, L_{n + \ell}^{2\ell + 1} \, (\rho) e^{-\rho /2} \qquad \rho \equiv \dfrac{2Zr}{n} \label{44}$
Here $L_{\beta}^{\alpha}$ is an associated Laguerre polynomial and $N_{n\ell}$, a normalizing constant. The angular momentum quantum number $\ell$ is by convention designated by a code: s for $\ell\ = 0$, p for $\ell\ = 1$, d for $\ell\ = 2$, f for $\ell\ = 3$, g for $\ell\ = 4$, and so on. The first four letters come from an old classification scheme for atomic spectral lines: sharp, principal, diffuse and fundamental. Although these designations have long since outlived their original significance, they remain in general use. The solutions of the hydrogenic Schrӧdinger equation
in spherical polar coordinates can now be written in full
$\psi_{n\ell m}(r, \: \theta, \: \phi) = R_{n\ell}(r)Y_{\ell m}(\theta, \: \phi) \ n = 1, \: 2... \qquad \ell = 0, \: 1... \: n - 1 \qquad m = 0, \: \pm 1, \: \pm 2... \: \pm \ell \label{45}$
where $Y_{\ell m}$ are the spherical harmonics tabulated in Chap. 6. Table 1 below enumerates all the hydrogenic functions we will actually need. These are called hydrogenic atomic orbitals, in anticipation of their later applications to the structure of atoms and molecules.
Table 1. Real hydrogenic functions in atomic units.
$\psi_{1s} = \dfrac{1}{\sqrt{\pi}} e^{-r}$
$\psi_{2s} = \dfrac{1}{2\sqrt{2\pi}} \left( 1 - \dfrac{r}{2} \right) e^{-r/2}$
$\psi_{2p_z} = \dfrac{1}{4\sqrt{2\pi}} z \, e^{-r/2}$
$\psi_{2p_x}, \: \psi_{2p_y} \qquad \mathsf{analogous}$
$\psi_{3s} = \dfrac{1}{81\sqrt{3\pi}} (27 - 18r + 2r^{2}) e^{-r/3}$
$\psi_{3p_z} = \dfrac{\sqrt{2}}{81\sqrt{\pi}} (6 - r) z \, e^{-r/3}$
$\psi_{3p_x}, \: \psi_{3p_y} \qquad \mathsf{analogous}$
$\psi_{3d_{z^{2}}} = \dfrac{1}{81\sqrt{6\pi}}(3z^{2} - r^{2}) e^{-r/3}$
$\psi_{3d_{zx}} = \dfrac{\sqrt{2}}{81\sqrt{\pi}}zx \, e^{-r/3}$
$\psi_{3d_{yz}}, \: \psi_{3d_{xy}} \qquad \mathsf{analogous}$
$\psi_{3d_{x^{2} - y^{2}}} = \dfrac{1}{81\sqrt{\pi}}(x^{2} - y^{2}) e^{-r/3}$
The energy levels for a hydrogenic system are given by
$E_n = -\dfrac{Z^{2}}{2n^{2}} \: \mathsf{hartrees} \label{46}$
and depends on the principal quantum number alone. Considering all the allowed values of $\ell$ and $m$, the level $E_n$ has a degeneracy of $n^{2}$. Figure 7 shows an energy level diagram for hydrogen $(Z = 1)$. For $E \geq 0$, the energy is a continuum, since the electron is in fact a free particle. The continuum represents states of an electron and proton in interaction, but not bound into a stable atom. Figure $7$ also shows some of the transitions which make up the Lyman series in the ultraviolet and the Balmer series in the visible region.
The $ns$ orbitals are all spherically symmetrical, being associated with a constant angular factor, the spherical harmonic $Y_{00} = 1/ \sqrt{4\pi}$. They have $n - 1$ radial nodesspherical shells on which the wavefunction equals zero. The 1s ground state is nodeless and the number of nodes increases with energy, in a pattern now familiar from our study of the particle-in-a-box and harmonic oscillator. The 2s orbital, with its radial node at $r = 2$ bohr, is also shown in Figure $3$.
p- and d-Orbitals
The lowest-energy solutions deviating from spherical symmetry are the 2p-orbitals. Using Equations $\ref{44}$, $\ref{45}$ and the $\ell = 1$ spherical harmonics, we find three degenerate eigenfunctions:
$\psi_{210}(r, \: \theta, \: \phi) = \dfrac{1}{4\sqrt{2\pi}}re^{-r/2} \cos\theta \label{47}$
and
$\psi_{21 \pm 1}(r, \: \theta, \: \phi) = \mp \dfrac{1}{4\sqrt{2\pi}}re^{-r/2} \sin\theta e^{\pm i \phi} \label{48}$
The function $\psi_{210}$ is real and contains the factor $r \cos\theta$, which is equal to the cartesian variable $z$. In chemical applications, this is designated as a 2pz orbital:
$\psi_{2p_z} = \dfrac{1}{4\sqrt{2\pi}}ze^{-r/2} \label{49}$
A contour plot is shown in Figure $8$. Note that this function is cylindrically-symmetrical about the $z$-axis with a node in the $x$, $y$-plane. The $\psi_{21 \pm 1}$ are complex functions and not as easy to represent graphically. Their angular dependence is that of the spherical harmonics $Y_{1 \pm 1}$, shown in Figure 6-4. As noted in Chap. 4, any linear combination of degenerate eigenfunctions is an equally-valid alternative eigenfunction. Making use of the Euler formulas for sine and cosine
$\cos\phi = \dfrac{e^{i\phi} + e^{-i\phi}}{2} \qquad \mathsf{and} \qquad \sin\phi = \dfrac{e^{i\phi} - e^{-i\phi}}{2} \label{50}$
and noting that the combinations $\sin\theta\cos\phi$ and $\sin\theta\sin\phi$ correspond to the cartesian variables $x$ and $y$, respectively, we can define the alternative 2p orbitals
$\psi_{2p_x} = \dfrac{1}{\sqrt{2}}(\psi_{21-1} - \psi_{211}) = \dfrac{1}{4\sqrt{2\pi}} xe^{-r/2} \label{51}$
and
$\psi_{2p_y} = -\dfrac{i}{\sqrt{2}}(\psi_{21-1} + \psi_{211}) = \dfrac{1}{4\sqrt{2\pi}} ye^{-r/2} \label{52}$
Clearly, these have the same shape as the 2pz-orbital, but are oriented along the $x$- and $y$-axes, respectively. The threefold degeneracy of the p-orbitals is very clearly shown by the geometric equivalence the functions 2px, 2py and 2pz, which is not obvious for the spherical harmonics. The functions listed in Table 1 are, in fact, the real forms for all atomic orbitals, which are more useful in chemical applications. All higher p-orbitals have analogous functional forms $x \, f(r)$, $y \, f(r)$ and $z \, f(r)$ and are likewise 3-fold degenerate.
The orbital $\psi_{320}$ is, like $\psi_{210}$, a real function. It is known in chemistry as the $d_{z^{2}}$-orbital and can be expressed as a cartesian factor times a function of $r$:
$\psi_{3d_{z^{2}}} = \psi_{320} = (3z^{2} - r^{2}) f(r) \label{53}$
A contour plot is shown in Figure $9$. This function is also cylindrically symmetric about the $z$-axis with two angular nodesthe conical surfaces with $3z^{2} - r^{2} = 0$. The remaining four 3d orbitals are complex functions containing the spherical harmonics $Y_{2 \pm 1}$ and $Y_{2 \pm 2}$ pictured in Figure 6-4. We can again construct real functions from linear combinations, the result being four geometrically equivalent "four-leaf clover" functions with two perpendicular planar nodes. These orbitals are designated $d_{x^{2} - y^{2}}, \: d_{xy}, \: d_{zx}$ and $d_{yz}$. Two of them are shown in Figure 9. The $d_{z^{2}}$ orbital has a different shape. However, it can be expressed in terms of two non-standard d-orbitals, $d_{z^{2} - x^{2}}$ and $d_{y^{2} - z^{2}}$. The latter functions, along with $d_{x^{2} - y^{2}}$ add to zero and thus constitute a linearly dependent set. Two combinations of these three functions can be chosen as independent eigenfunctions.
Summary
The atomic orbitals listed in Table 1 are illustrated in Figure $20$. Blue and red indicate, respectively, positive and negative regions of the wavefunctions (the radial nodes of the 2s and 3s orbitals are obscured). These pictures are intended as stylized representations of atomic orbitals and should not be interpreted as quantitatively accurate.
The electron charge distribution in an orbital $\psi_{n\ell m}(\mathbf{r})$ is given by
$\rho(\mathbf{r}) = \lvert \psi_{n\ell m}(\mathbf{r}) \rvert ^{2} \label{54}$
which for the s-orbitals is a function of $r$ alone. The radial distribution function can be defined, even for orbitals containing angular dependence, by
$D_{n\ell}(r) = r^{2} [R_{n\ell}(r)]^{2} \label{55}$
This represents the electron density in a shell of radius $r$, including all values of the angular variables $\theta$, $\phi$. Figure $11$ shows plots of the RDF for the first few hydrogen orbitals.
Contributors and Attributions
Seymour Blinder (Professor Emeritus of Chemistry and Physics at the University of Michigan, Ann Arbor)
• Integrated by Daniel SantaLucia (Chemistry student at Hope College, Holland MI)
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A radial node is a sphere (rather than an angular node which is a flat plane) that occurs when the radial wavefunction for an atomic orbital is equal to zero or changes sign.
Introduction
There are two types of nodes within an atom: angular and radial. Angular nodes are or will be discussed in another section; this section is dedicated to the latter. Radial nodes, as one could guess, are determined radially. Using the radial probability density function, places without electrons, or radial nodes, can be found. A quick comparison of the two types of nodes can be seen in the diagram above. Angular nodes are either x, y, and z planes where electrons aren’t present while radial nodes are sections of these axes that are closed off to electrons.
For atomic orbitals, the wavefunction can be separated into a radial part and an angular part so that it has the form
\[Ψ(r,θ,ϕ)=R(r)Y(θ,ϕ)\]
where \(R(r)\) is the radial component which depends only on the distance from the nucleus and Y(θ,ϕ) is the angular component. The radial nodes consist of spheres whereas the angular nodes consist of planes (or cones).
A radial node will occur where the radial wavefunction, \(R(r)\), equals zero. At a node the probability of finding an electron is zero; which means that we will never find an electron at a node.
Basic description
To solve for the number of radial nodes, the following simple equation can be used.
Radial Nodes = n - 1 - ℓ
The ‘n’ accounts for the total amount of nodes present. The ‘-1’ portion accounts for the node that exists at the ends. (A half of one node exists at one end and since there are two ends, there’s a total of one node located at the ends.) The azimuthal quantum number determines the shape of the orbital and how many angular nodes there are. The remaining number, which currently doesn’t have a symbol, is the amount of radial nodes which are present. Here’s a quick example:
Radial nodes occur as the principle quantum number (n) increases and the number of radial nodes depends on the principle quantum number (n) and the number of angular nodes (l). The total number of nodes is found using
Total Nodes=n-1
From knowing the total nodes we can find the number of radial nodes by using
Radial Nodes=n-l-1
which is just the total nodes minus the angular nodes.
Example \(1\):
first shell (n=1) number of nodes= n-1=0 so there aren't any nodes
second shell (n=2) number of nodes=n-1=1 total nodes
for 2s orbital l=0 so there are 0 angular nodes and 1 radial node
for 2p orbital l=1 so there is 1 angular node and 0 radial nodes
third shell (n=3) number of nodes=n-1=2 total nodes
for 3s orbital l=0 so there are 0 angular nodes and 2 radial nodes
for 3p orbital l=1 so there is 1 angular node and 1 radial node
for 3d orbital l=2 so there are 2 angular nodes and 0 radial nodes
Exercise \(1\)
Find the radial nodes in a 3p orbital.
Answer
For the 3p orbital, the ‘3’ means that ‘n’ = 3 and ‘p’ shows that ‘ℓ’ = 1. ‘ℓ’ also equals the number of angular nodes which means there is one angular node present. Using the equation for radial nodes, n - ℓ - 1 = 3 - 1 - 1 = 1. Thus there is one radial nodes. The following section will show how to determine radial nodes in a more complex way.
Advanced description
We can calculate how many nodes there will be based off the equation above, however we can also see this from the wavefunction. For example the wavefunction for the Hydrogen atom 3d orbital:
From the equation above we can see that the number of total nodes is n-1=2 and the number of angular nodes (l)=2 so the number of radial nodes is 0. From the wavefunction for the \(3dz^2\) orbital, we can see that (excluding r=0 and as r goes to infinity) the radial wavefunction will never equal to zero so there are 0 radial nodes for this orbital. For the angular wavefunction, we see there will be an angular node when \(3cos^2θ-1=0\); which corresponds to the 2 solutions θ=54.7º and 125.3º.
As stated above, we know that at a node the probability of finding an electron is zero. The diagram below shows that as n increases, the number of radial nodes increases.
From Figure 2 we can see that for the 1s orbital there are not any nodes (the curve for the 1s orbital doesn't equal zero probability other than at r=0 and as r goes to infinity). This is expected since n-l-1 for the 1s orbital is 1-0-1=0 radial nodes. For the 2s orbital, the curve has zero probability at 1 point (again other than r=0 and as r goes to infinity); which is consistent with the n-l-1 for the 2s orbital 2-0-1=1 radial node. For the 3s orbital, the curve has zero probability at 2 points; which is consistent with the n-l-1 for the 3s orbital 3-0-1=2 radial nodes.
But what about for molecular orbitals?
To separate the wavefunction into a radial part and an angular part, the system needs to be spherically symmetric. For an atom this is the case but a molecule can never be. Thus radial nodes do not exist for molecular orbitals.
Contributors and Attributions
• Rachel Schonwit
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When the wavefunctions that solve Schrödinger's equation $\psi(x,y,z)$ are expressed in spherical coordinates, then the wavefunctions can be separated into a radial and angular components
$\psi(r,\theta,\phi) =R(r) Y(\theta, \phi) \label{1.1}$
The $Y (\theta , \varphi )$ 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. Both $R(r)$ and $Y (\theta , \varphi )$ depend on all three quantum numbers $n, l, m_l$, although energy of each wavefunction ($E_n$) depends on only $n$.
In general, the radial wavefunction ($R(r)$) can be expressed as
$R(r) = Q(r) e^{-r/a} \label{1.2}$
where $Q(r)$ is a polynomial. $R(r)$ is typically normalized so that it can used to calculate electron probability just like proper wavefunction (but only with respect to the radial, $r$ dimension only). A more complete 3D description requires addressing the complete wavefunction in Equation $\ref{1.1}$.
Since the square of a particle's wave function at some location yields the probability that the particle is at that location, one might think that we can determine the probability that the electron is within some range of radii by integrating; however this is incorrect.
$P ( r_1 < r < r_2) \neq \int_{r_1}^{r_2} R(r)^2 dr = \int_{r_1}^{r_2} Q(r)^2 e^{-2r/a} dr \label{1.3}$
The trouble is that once we leave the familiar territory of Cartesian coordinates and venture into the Land of Spherical Coordinates, the volume element changes.
"Volumes" in One Dimension
In one dimension, things are dead simple: a single coordinate x describes everything. If we want to integrate, we must move a little extra distance in all dimensions ... which means a little extra distance dx. That's all there is.
The "differential volume element" is simply
$dV = dx$
Actually this is the differential length element, but the connection to volume is better described below.
"Volumes" in Two Dimensions
In two dimensions, thing are more interesting because there are several ways to identify a location in space. The Cartesian system uses variables x and y.
To integrate, we move a little extra distance in each dimension:
This sweeps out a little square box.
The differential volume element is the area of this box:
$dV = dx\, dy$
Actually, this is the differential area element, but the connection to volume is described below.
Now, we can also use polar coordinates r and θ to describe the location of a point in a plane.
If we increase each of these variables by a teeny bit to integrate ...
... we sweep out an area which is not square. Instead, it has a curve along the inside and outside edges.
The area of this curvy box, and thus the differential volume element in polar coordinates on a plane, is
$dV = dr\, r d\theta$
Actually, this is the differential area element.
Volumes in Three dimensions
I hope you'll believe me when I say that the differential volume element in three-dimensional Cartesian coordinates is
$dV = dx\,dx\,dz$
On the other hand, in three-dimensional spherical coordinates, we describe the position of an object with radius $r$ and the angles $θ$ and $φ$.
If we extend each of these three variables just a bit, we sweep out a funnel-shaped box.
In one direction, the box has straight lines of length dr. In the other directions, the edges are curved. We can compute the length of these tiny arcs using the arclength formulae in spherical coordinates, and end up with a differential volume element
$dV = (dr) (rd\theta)(r \sin \theta \,d \phi)$
$= r^2 \sin \theta \, dr\, d \theta \, d \phi$
So, back to the electron within the hydrogen atom. The probability the electron is at location $(r, θ, φ)$ within a tolerance of $(dr, dθ, dφ)$ is
$P ( r; \theta; \phi) = \int (\psi( r; \theta; \phi))^2 dV$
$P ( r; \theta; \phi) = \int (\psi( r; \theta; \phi))^2 r^2 \sin \theta \, dr\, d \theta \, d \phi \label{Int1}$
If we can separate the wavefunction $\ref{1.1}$ further into the product of three functions, each one dealing with only a single variable,
$\psi(r,\theta,\phi) =R(r) \Theta(\theta) \Phi(\phi)$
then the integral in Equation $\ref{Int1}$ becomes
$P ( r; \theta; \phi) = \int ( R(r) \Theta(\theta) \Phi(\phi) )^2 r^2 \sin \theta \, dr\, d \theta \, d \phi \label{Int2}$
$= \int (R(r)^2 r^2 dr) (\Theta(\theta)^2 \sin \theta d \theta ) ( \Phi(\phi)^2 d \phi) \label{Int3}$
$= \int R(r)^2 r^2 dr \int \Theta(\theta)^2 \sin \theta d \theta \int \Phi(\phi)^2 d \phi \label{Int4}$
In the simplest case, corresponding to the ground state $1s$, the wave function does not depend on either angle. That means the angular functions in Equation $\ref{Int4}$ are simply equal to 1; and that makes doing those integrals trivial.
$\int 1 \sin \theta d\theta \label{Int5}$
$\int 1 d\phi \label{Int6}$
Exercise 1
What are the limits and values of the integrals in Equations $\ref{Int5}$ and $\ref{Int6}$?
So the probability of finding the electron within some range of radii (Equation $\ref{1.3}$) becomes
$P ( r_1 < r < r_2) = 4\pi \int_{r_1}^{r_2} R(r)^2 r^2 dr \label{Int7}$
Recall that the radial function (Equation $\ref{1.2}$) can be separated into two compontents. For the electron in the 1s wavefunction, $Q$ is a constant and then
$R(r) = Q e^{-r/a}$
We can simplify this just a little bit by making a substitution: let
$u = \dfrac{2r}{a}.$
then
$du =\dfrac{2}{a} dr$
and Equation $\ref{Int7}$ can be rewritten as
$P ( r_1 < r < r_2) = 4\pi \left(Q^2 \dfrac{a^3}{8} \right) \int_{2r_1/a}^{2r_2/a} u^2 e ^{-u} du \label{Int8a}$
You can look up that integral in a table; the result isn't too ugly.
$P ( r_1 < r < r_2) = 4\pi \left(Q^2 \dfrac{a^3}{8} \right) \left[ -e^{-u}(u^2 + 2u +2) \right]_{2r_1/a}^{2r_2/a} \label{Int9a}$
Exercise $3$
1. If we compute the probability that the radial position of the electron must be between zero and infinity, what must we get?
2. Can you solve for the value of the coefficient $Q$ in our radial wave function (for the 1s orbital)?
Visualizing Radial Probabilities
So, where is the electron most likely to be? While the $R^2(r)$ plot (Figure 1; red curve) tracks the probability along a vector away from the nucleus, the $r^2r^2(r)$ plot (blue curve) tracks the probability of a finding the electron at a radius $r$. The peak of this curve is the MOST LIKELY radius for the electron.
You can find more examples of wave functions and probability distributions in textbooks and on web sites. Below are some examples made using Paul Fastad's hydrogen atom applet. In each example, the radial function is shown in the top-right panel of the figure below, labelled "Rnl(r)". The bottom-left panel provides an illustration of this function, looking down at the atom from above.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Where_is_the_electron_in_a_hydrogen_atom.txt
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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: The Helium Atom
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.
• 9: Atomic Structure and The Periodic Law
Quantum mechanics can account for the periodic structure of the elements, by any measure a major conceptual accomplishment for any theory. Although accurate computations become increasingly more challenging as the number of electrons increases, the general patterns of atomic behavior can be predicted with remarkable accuracy.
• Electronic Angular Wavefunction
The electronic angular wavefunction is one spatial component of the electronic Schrödinger wave equation, which describes the motion of an electron. It depends on angular variables, θ and ϕ , and describes the direction of the orbital that the electron may occupy. Some of its solutions are equal in energy and are therefore called degenerate.
• Electron Configuration
• H Atom Energy Levels
• Koopmans' Theorem
Koopmans' theorem states that the first ionization energy of a molecule is equal to the negative of the energy of the highest occupied molecular orbital (HOMO).
• Quantum Mechanical H Atom
• Quantum Numbers for Atoms
A total of four quantum numbers are used to describe completely the movement and trajectories of each electron within an atom. The combination of all quantum numbers of all electrons in an atom is described by a wave function that complies with the Schrödinger equation. Each electron in an atom has a unique set of quantum numbers; according to the Pauli Exclusion Principle, no two electrons can share the same combination of four quantum numbers.
• Radial and Angular Parts of Atomic Orbitals
10: Multi-electron Atoms
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.
The helium atom has two electrons bound to a nucleus with charge Z = 2. The successive removal of the two electrons can be diagrammed as
$\ce{He}\xrightarrow{\textit{I}_1}\ce{He}^++e^-\xrightarrow{\textit{I}_2}\ce{He}^{++}+2e^-\label{1}$
The first ionization energy I1, the minimum energy required to remove the first electron from helium, is experimentally 24.59 eV. The second ionization energy, I2, is 54.42 eV. The last result can be calculated exactly since He+ is a hydrogen-like ion. We have
$\textit{I}_2=-\textit{E}_{ 1\textit{s}}(\ce{He}^+)=\dfrac{Z^2}{2n^2}=2 \mbox{ hartrees}=54.42\mbox{ eV}\label{2}$
The energy of the three separated particles on the right side of Equation $\ref{1}$ is, by definition, zero. Therefore the ground-state energy of helium atom is given by $E_0=-(\textit{I}_1+\textit{I}_2)=-79.02\mbox{ eV}=-2.90372\mbox{ hartrees}$. We will attempt to reproduce this value, as close as possible, by theoretical analysis.
Schrödinger Equation and Variational Calculations
The Schrödinger equation for He atom, again using atomic units and assuming infinite nuclear mass, can be written
$\bigg\{-\dfrac{1}{2}\nabla^2_1-\dfrac{1}{2}\nabla^2_2-\dfrac{Z}{r_1}-\dfrac{Z}{r_2}+\dfrac{1}{r_{12}}\bigg\}\psi(\text{r}_1,\text{r}_2)=E\psi(\text{r}_1,\text{r}_2)\label{3}$
The five terms in the Hamiltonian represent, respectively, the kinetic energies of electrons 1 and 2, the nuclear attractions of electrons 1 and 2, and the repulsive interaction between the two electrons. It is this last contribution which prevents an exact solution of the Schrödinger equation and which accounts for much of the complication in the theory. In seeking an approximation to the ground state, we might first work out the solution in the absence of the 1/r12-term. In the Schrödinger equation thus simplified, we can separate the variables r1 and r2 to reduce the equation to two independent hydrogen-like problems. The ground state wavefunction (not normalized) for this hypothetical helium atom would be
$\psi(\text{r}_1,\text{r}_2)=\psi_{1s}(r_1)\psi_{1s}(r_2)=e^{-Z(r_1+r_2)}\label{4}$
and the energy would equal $2\times(-Z^2/2)=-4$ hartrees, compared to the experimental value of $-2.90$ hartrees. Neglect of electron repulsion evidently introduces a very large error.
A significantly improved result can be obtained with the functional form ( Equation $\ref{4}$), but with Z replaced by a adjustable parameter $\alpha$, thus:
$\tilde{\psi}(r_1,r_2)=e^{-\alpha(r_1+r_2)}\label{5}$
Using this function in the variational principle [cf. Eq (4.53)], we have
$\tilde{E}=\dfrac{\int\psi(r_1,r_2)\hat{H}\psi(r_1,r_2)d\tau_1\tau_2}{\int\psi(r_1,r_2)\psi(r_1,r_2)d\tau_1d\tau_2}\label{6}$
where $\hat{H}$ is the full Hamiltonian as in Equation $\ref{3}$, including the $1/r_{12}$-term. The expectation values of the five parts of the Hamiltonian work out to
$\left\langle-\dfrac{1}{2}\nabla^2_1\right\rangle=\left\langle-\dfrac{1}{2}\nabla^2_2\right\rangle=\dfrac{\alpha^2}{2}$
$\left\langle-\dfrac{Z}{r_1}\right\rangle=\left\langle-\dfrac{Z}{r_2}\right\rangle=-Z\alpha, \left\langle\dfrac{1}{r_{12}}\right\rangle=\dfrac{5}{8}\alpha\label{7}$
The sum of the integrals in Equation $\ref{7}$ gives the variational energy
$\tilde{E}(\alpha)=\alpha^2-2Z\alpha+\dfrac{5}{8}\alpha\label{8}$
This will be always be an upper bound for the true ground-state energy. We can optimize our result by finding the value of $\alpha$ which minimizes the energy (Equation $\ref{8}$). We find
$\dfrac{d\tilde{E}}{d\alpha}=2\alpha-2Z+\dfrac{5}{8}=0\label{9}$
giving the optimal value
$\alpha=Z-\dfrac{5}{16}\label{10}$
This can be given a physical interpretation, noting that the parameter $\alpha$ in the wavefunction (Equation $\ref{5}$) represents an effective nuclear charge. Each electron partially shields the other electron from the positively-charged nucleus by an amount equivalent to $5/8$ of an electron charge. Substituting Equation $\ref{10}$ into Equation $\ref{8}$, we obtain the optimized approximation to the energy
$\tilde{E}=-\left(Z-\dfrac{5}{16}\right)^2\label{11}$
For helium ($Z = 2$), this gives $-2.84765$ hartrees, an error of about $2\%$ $(E_0 = -2.90372)$. Note that the inequality $\tilde{E} > E_0$ applies in an algebraic sense.
In the late 1920's, it was considered important to determine whether the helium computation could be improved, as a test of the validity of quantum mechanics for many electron systems. The table below gives the results for a selection of variational computations on helium.
wavefunction parameters energy
$e^{-Z(r_1+r_2)}$ $Z=2$ $-2.75$
$e^{-\alpha(r_1+r_2)}$ $\alpha=1.6875$ $-2.84765$
$\psi(r_1)\psi(r_2)$ best $\psi(r)$ $-2.86168$
$e^{-\alpha(r_1+r_2)}(1+c r_{12})$ best $\alpha, \textrm{c}$ $-2.89112$
Hylleraas (1929) 10 parameters $-2.90363$
Pekeris (1959) 1078 parameters $-2.90372$
The third entry refers to the self-consistent field method, developed by Hartree. Even for the best possible choice of one-electron functions $\psi(r)$, there remains a considerable error. This is due to failure to include the variable $r_{12}$ in the wavefunction. The effect is known as electron correlation.
The fourth entry, containing a simple correction for correlation, gives a considerable improvement. Hylleraas (1929) extended this approach with a variational function of the form
$\psi(r_1, r_2, r_{12})=e^{-\alpha(r_1+r_2)} \times \textrm{polynomial in} r_1, r_2, r_{12}$
and obtained the nearly exact result with 10 optimized parameters. More recently, using modern computers, results in essentially perfect agreement with experiment have been obtained.
Spinorbitals and the Exclusion Principle
The simpler wavefunctions for helium atom in Equation $\ref{5}$, can be interpreted as representing two electrons in hydrogen-like 1s orbitals, designated as a 1s2 configuration. According to Pauli's exclusion principle, which states that no two electrons in an atom can have the same set of four quantum numbers, the two 1s electrons must have different spins, one spin-up or $\alpha$, the other spin-down or $\beta$. A product of an orbital with a spin function is called a spinorbital. For example, electron 1 might occupy a spinorbital which we designate
$\phi(1)=\psi_{1s}(1)\alpha(1) \textrm{or} \psi_{1s}(1)\beta(1)\label{12}$
Spinorbitals can be designated by a single subscript, for example, $\phi_a$ or $\phi_b$, where the subscript stands for a set of four quantum numbers. In a two electron system the occupied spinorbitals $\phi_a$ and $\phi_b$ must be different, meaning that at least one of their four quantum numbers must be unequal. A two-electron spinorbital function of the form
$\Psi (1, 2) = \dfrac{1}{2} \bigg( \phi_a(1)\phi_b(2) - \phi_b(1)\phi_a(2)\bigg)\label{13}$
automatically fulflls the Pauli principle since it vanishes if $a=b$. Moreover, this function associates each electron equally with each orbital, which is consistent with the indistinguishability of identical particles in quantum mechanics. The factor $1/\sqrt{2}$ normalizes the two-particle wavefunction, assuming that $\phi_a$ and $\phi_b$ are normalized and mutually orthogonal. The function (Equation $\ref{13}$) is antisymmetric with respect to interchange of electron labels, meaning that
$\Psi (2,1) = -\Psi (1, 2)\label{14}$
This antisymmetry property is an elegant way of expressing the Pauli principle.
We note, for future reference, that the function in Equation $\ref{13}$ can be expressed as a $2 \times 2$ determinant:
$\Psi (1, 2) = \dfrac{1}{\sqrt{2}}\begin{vmatrix}\phi_a(1) & \phi_b(1)\\phi_a(2) & \phi_b(2)\end{vmatrix}\label{15}$
For the 1s2 configuration of helium, the two orbital functions are the same and Equation $\ref{13}$ can be written
$\Psi (1, 2) = \psi_{1s}(1)\psi_{1s}(2) \times \dfrac{1}{\sqrt{2}}\bigg(\alpha(1)\beta(2) - \beta(1)\alpha(2)\bigg)\label{16}$
For two-electron systems (but not for three or more electrons), the wavefunction can be factored into an orbital function times a spin function. The two-electron spin function
$\sigma_{0,0}(1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\alpha(1)\beta(2) - \beta(1)\alpha(2)\bigg)\label{17}$
represents the two electron spins in opposing directions (antiparallel) with a total spin angular momentum of zero. The two subscripts are the quantum numbers S and MS for the total electron spin. Eqution $\ref{16}$ is called the singlet spin state since there is only a single orientation for a total spin quantum number of zero. It is also possible to have both spins in the same state, provided the orbitals are different. There are three possible states for two parallel spins:
$\sigma_{1,1}(1, 2) = \alpha(1)\alpha(2)$
$\sigma_{1,0}(1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\alpha(1)\beta(2) + \beta(2)\alpha(2)\bigg)$
$\sigma_{1,-1}(1, 2) = \beta(1)\beta(2)\label{18}$
These make up the triplet spin states, which have the three possible orientations of a total angular momentum of 1.
Excited States of Helium
The lowest excitated state of helium is represented by the electron configuration 1s 2s. The 1s 2p configuration has higher energy, even though the 2s and 2p orbitals in hydrogen are degenerate, because the 2s penetrates closer to the nucleus, where the potential energy is more negative. When electrons are in different orbitals, their spins can be either parallel or antiparallel. In order that the wavefunction satisfy the antisymmetry requirement (Equation $\ref{14}$), the two-electron orbital and spin functions must have opposite behavior under exchange of electron labels. There are four possible states from the 1s 2s configuration: a singlet state
$\Psi^+ (1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\psi_{1s}(1)\psi_{2s}(2) + \psi_{2s}(1)\psi_{1s}(2)\bigg) \sigma_{0, 0}(1, 2)\label{19}$
and three triplet states
$\Psi^-(1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\psi_{1s}(1)\psi_{2s}(2) - \psi_{2s}(1)\psi_{1s}(2)\bigg)\begin{cases} \sigma_{1,1}(1, 2)\ \sigma_{1,0}(1, 2) \ \sigma_{1,-1}(1, 2)\end{cases}\label{20}$
Using the Hamiltonian in Equation $\ref{3}$, we can compute the approximate energies
$E^{\pm}=\iint\Psi^{\pm}(1,2) \hat{H} \Psi^{\pm}(1,2)d\tau_1d\tau_2\label{21}$
After evaluating some fierce-looking integrals, this reduces to the form
$E^{\pm}=I(1s)+I(2s)+J(1s, 2s) \pm K(1s, 2s)\label{22}$
in terms of the one electron integrals
$I(a)=\int \psi_a(\textrm{r})\left\{-\dfrac{1}{2}\nabla^2-\dfrac{Z}{r}\right\} \psi_a(\textrm{r})d\tau\label{23}$
the Coulomb integrals
$J(a, b)=\iint\psi_a(\textrm{r}_1)^2\dfrac{1}{r_{12}}\psi_b(\textrm{r}_2)^2d\tau_1d\tau_2\label{24}$
and the exchange integrals
$K(a, b)=\iint\psi_a(\textrm{r}_1)\psi_b(\textrm{r}_1)\dfrac{1}{r_{12}}\psi_a(\textrm{r}_2)\psi_b(\textrm{r}_2)d\tau_1d\tau_2\label{25}$
The Coulomb integral represents the repulsive potential energy for two interacting charge distributions $\psi_a(\textbf{r}_1)^2$ and $\psi_b(\textbf{r}_2)^2$. The exchange integral, which has no classical analog, arises because of the exchange symmetry (or antisymmetry) requirement of the wavefunction. Both J and K can be shown to be positive quantities. Therefore the lower sign in (22) represents the state of lower energy, making the triplet state of the configuration 1s 2s lower in energy than the singlet state. This is an almost universal generalization and contributes to Hund's rule, to be discussed in the next Chapter.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/8%3A_The_Helium_Atom.txt
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Quantum mechanics can account for the periodic structure of the elements, by any measure a major conceptual accomplishment for any theory. Although accurate computations become increasingly more challenging as the number of electrons increases, the general patterns of atomic behavior can be predicted with remarkable accuracy.
Slater Determinants
According to the orbital approximation, which was introduced in the last Chapter, an N-electron atom contains N occupied spinorbitals, which can be designated $\phi$a, $\phi$b . . . $\phi$n. In accordance with the Pauli exclusion principle, no two of these spinorbitals can be identical. Also, every electron should be equally associated with every spinorbital. A very neat mathematical representation for these properties is a generalization of the two-electron wavefunction (8.13) or (8.15) called a Slater determinant
$\Psi (1,2 \ldots N ) = \frac{1}{\sqrt{N !}} \begin{vmatrix} \phi_a(1) & \phi_b(1) & \ldots & \phi_n(1) \ \phi_a(2) & \phi_b(2) & \ldots & \phi_n(2) \ & & \vdots & \ \phi_a(N) & \phi_b(N) & \ldots & \phi_n(N) \ \end{vmatrix} \label{1}$
Since interchanging any two rows (or columns) of a determinant multiplies it by $-1$, the antisymmetry property (8.15) is fulfilled, for every pair of electrons.
The Hamiltonian for an atom with N electrons around a nucleus of charge Z can be written
$\hat{H} = \sum_{i=1}^N \left\{-\frac{1}{2}\bigtriangledown^2_i - \frac{Z}{r_i} \right\} + \sum_{i<j}^N \frac{1}{ r_{ij}}\label{2}$
The sum over electron repulsions is written so that each pair {i,,j} is counted just once. The energy of the state represented by a Slater determinant (Equation $\ref{1}$) can be obtained after a lengthy derivation. We give just the final result
$\tilde{E} = \sum_{a} I_a+\frac{1}{2}\sum_{a,b} \left( J_{ab}-K_{ab} \right) \label{3}$
where the sums run over all occupied spinorbitals. The one-electron, Coulomb and exchange integrals have the same form as those defined for helium atom in Eqs (8.22-24). The only difference is that an exchange integral equals zero unless the spins of orbitals a and b are both $\alpha$ or both $\beta$. The factor 1/2 corrects for the double counting of pairs of spinorbitals in the second sum. The contributions with a = b can be omitted since Jaa = Kaa. This effectively removes the Coulomb interaction of an orbital with itself, which is spurious.
The Hartree-Fock or self-consistent field (SCF) method is a procedure for optimizing the orbital functions in the Slater determinant (1), so as to minimize the energy (Equation $\ref{3}$). SCF computations have been carried out for all the atoms of the periodic table, with predictions of total energies and ionization energies generally accurate in the $1-2\%$ range.
Aufbau Principles and Periodic Structure
Aufbau means "building-up." Aufbau principles determine the order in which atomic orbitals are filled as the atomic number is increased. For the hydrogen atom, the order of increasing orbital energy is given by 1s < 2s = 2p < 3s = 3p = 3d, etc. The dependence of energy on n alone leads to extensive degeneracy, which is however removed for orbitals in many-electron atoms. Thus 2s lies below 2p, as already observed in helium. Similarly, 3s, 3p and 3d increase energy in that order, and so on. The 4s is lowered sufficiently that it becomes comparable to 3d. The general ordering of atomic orbitals is summarized in the following scheme:
$1s < 2s < 2p < 3s < 3p < 4s \sim 3d < 4p < 5s \sim 4d\< 5p < 6s \sim 5d \sim 4f < 6p < 7s \sim 6d \sim 5f \label{4}$
and illustrated in Figure 1. This provides enough orbitals to fill the ground states of all the atoms in the periodic table. For orbitals designated as comparable in energy, e.g., 4s $\sim$ 3d, the actual order depends which other orbitals are occupied. The sequence of orbitals pictured above increases in the order $n+\frac{1}{2}$l, except that l = 4 (rather than 3) is used for an f-orbital.
The tabulation below shows the ground-state electron configuration and term symbol for selected elements in the first part of the periodic table. From the term symbol, one can read off the total orbital angular momentum L and the total spin angular momentum S. The code for the total orbital angular momentum mirrors the one-electron notation, but using upper-case letters, as follows:
L = 0 1 2 3 4
S P D F G
The total spin S is designated, somewhat indirectly, by the spin multiplicity 2S + 1 written as a superscript before the S, P, D. . . symbol. For example 1S (singlet S) ,1P (singlet P). . . mean S = 0; 2S (doublet S) ,2P (doublet P). . . mean S = 1/2; 3S (triplet S) ,3P (triplet P). . . mean S = 1, and so on. Please do not confuse the spin quantum number S with the orbital designation S.
Atom Z Electron Configuration Term Symbol
H 1 1s 2S1/2
He 2 1s2 1S0
Li 3 [He]2s 2S1/2
Be 4 [He]2s2 1S0
B 5 [He]2s22p 2P1/2
C 6 [He]2s22p2 3P0
N 7 [He]2s22p3 4S3/2
O 8 [He]2s22p4 3P2
F 9 [He]2s22p5 2P3/2
Ne 10 [He]2s22p6 1S0
Na 11 [Ne]3s 2S1/2
Cl 17 [Ne]3s23p5 2P3/2
Ar 18 [Ne]3s23p6 1S0
K 19 [Ar]4s 2S1/2
Ca 20 [Ar]4s2 1S0
Sc 21 [Ar]4s23d 2D3/2
Ti 22 [Ar]4s23d2 3F2
V 23 [Ar]4s23d3 4F3/2
Cr 24 [Ar]4s3d5 7S3
Mn 25 [Ar]4s23d5 6S5/2
Fe 26 [Ar]4s23d6 5D4
Co 27 [Ar]4s23d7 4F9/2
Ni 28 [Ar]4s23d8 3F4
Cu 29 [Ar]4s3d10 2S1/2
Zn 30 [Ar]4s23d10 1S0
Ga 31 [Ar]4s23d104p 2P1/2
Br 35 [Ar]4s23d104p5 2P3/2
Kr 36 [Ar]3d104s24p6 1S0
The vector sum of the orbital and spin angular momentum is designated
$\bf{J} = \bf{L} + \bf{S} \label{5}$
The possible values of the total angular momentum quantum number J runs in integer steps between |L - S| and L + S. The J value is appended as a subscript on the term symbol, eg, 1S0, 2P1/2, 2P3/2. The energy differences between J states is a result of spin-orbit interaction, a magnetic interaction between the circulating charges associated with orbital and spin angular momenta. For atoms of low atomic number, the spin-orbit coupling is a relatively small correction to the energy, but it can become increasingly significant for heavier atoms.
We will next consider in some detail the Aufbau of ground electronic states starting at the beginning of the periodic table. Hydrogen has one electron in an s-orbital so its total orbital angular momentum is also designated S. The single electron has s = 1/2, thus S = 1/2. The spin multiplicity 2S + 1 equals 2, thus the term symbol is written 2S. In helium, a second electron can occupy the 1s shell, provided it has the opposite spin. The total spin angular momentum is therefore zero, as is the total orbital angular momentum. The term symbol is 1S, as it will be for all other atoms with complete electron shells. In determining the total spin and orbital angular moments, we need consider only electrons outside of closed shells. Therefore lithium and beryllium are a reprise of hydrogen and helium. The angular momentum of boron comes from the single 2p electron, with l = 1 and s = 1/2, giving a 2P state.
To build the carbon atom, we add a second 2p electron. Since there are three degenerate 2p orbitals, the second electron can go into either the already-occupied 2p orbital or one of the unoccupied 2p orbitals. Clearly, two electrons in different 2p orbitals will have less repulsive energy than two electrons crowded into the same 2p orbital. In terms of the Coulomb integrals, we would expect, for example
$J(2px, 2py) < J(2px, 2px) \label{6}$
For nitrogen atom, with three 2p electrons, we expect, by the same line of reasoning, that the third electron will go into the remaining unoccupied 2p orbital. The half-filled 2p3 subshell has an interesting property. If the three occupied orbitals are 2px, 2py and 2pz, then their total electron density is given by
$\rho_{2p} = \psi^{2}_{2p_{x}} + \psi^{2}_{2p_{y}} + \psi^{2}_{2p_{z}} = \left(x^2 + y^2 + z^2\right) \times \text{function of r} = \text{function of r} \label{7}$
noting that $x^2 + y^2 + z^2 = r^2$. But spherical symmetry implies zero angular momentum, like an s-orbital. In fact, any half filled subshell, such as p3, d5, f7, will contribute zero angular momentum. The same is, of course true as well for filled subshells, such as p6, d10, f14. These are all S terms.
Another way to understand this vector cancelation of angular momentum is to consider the alternative representation of the degenerate 2p-orbitals: 2p-1; 2p0 and 2p1. Obviously, the z-components of angular momentum now add to zero, and since only this one component is observable, the total angular momentum must also be zero.
Returning to our unfinished consideration of carbon, the 2p2 subshell can be regarded, in concept, as a half-filled 2p3 subshell plus an electron "hole." The advantage of this picture is that the total orbital angular momentum must be equal to that of the hole, namely l = 1. This is shown below:
Thus the term symbol for the carbon ground state is P. It remains to determine the total spins of these subshells. Recall that exchange integrals Kab are non-zero only if the orbitals a and b have the same spin. Since exchange integrals enter the energy formula (3) with negative signs, the more nonvanishing K integrals, the lower the energy. This is achieved by having the maximum possible number of electrons with unpaired spins. We conclude that S = 1 for carbon and S = 3/2 for nitrogen, so that the complete term symbols are 3P and 4S, respectively.
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.
Resuming Aufbau of the periodic table, oxygen with four 2p electrons must have one of the 2p-orbitals doubly occupied. But the remaining two electrons will choose unoccupied orbitals with parallel spins. Thus oxygen has, like carbon, a 3P ground state. Fluorine can be regarded as a complete shell with an electron hole, thus a 2P ground state. Neon completes the 2s2p shells, thus term symbol 1S. The chemical stability and high ionization energy of all the noble-gas atoms can be attributed to their electronic structure of complete shells. The third row of the periodic table is filled in complete analogy with the second row. The similarity of the outermost electron shells accounts for the periodicity of chemical properties. Thus, the alkali metals Na and K belong in the same family as Li, the halogens Cl and Br are chemically similar to F, and so forth.
The transition elements, atomic numbers 21 to 30, present further challenges to our understanding of electronic structure. A complicating factor is that the energies of the 4s and 3d orbitals are very close, so that interactions among occupied orbitals often determines the electronic state. Ground-state electron configurations can be deduced from spectroscopic and chemical evidence, and confirmed by accurate self-consisent field computations. The 4s orbital is the first to be filled in K and Ca. Then come 3d electrons in Sc, Ti and V. A discontinuity occurs at Cr. The groundstate configuration is found to be 4s3d5, instead of the extrapolated 4s23d4. This can be attributed to the enhanced stability of a half-filled 3d5-shell. All six electrons in the valence shells have parallel spins, maximizing the number of stabilizing exchange integrals and giving the observed 6S term. An analogous discontinuity occurs for copper, in which the 4s subshell is again raided to complete the 3d10 subshell.
The order in which orbitals are filled is not necessarily consistent with the order in which electrons are removed. Thus, in all the positive ions of transition metals, the two 4s-electrons are removed first. The inadequacy of any simple generalizations about orbital energies is demonstrated by comparing the three ground-state electron configurations: Ni 4s23d8, Pd 5s04d10 and Pt 6s5d9.
The periodic structure of the elements is evident for many physical and chemical properties, including chemical valence, atomic radius, electronegativity, melting point, density, and hardness. The classic prototype for periodic behavior is the variation of the first ionization energy with atomic number, which is plotted in in Figure 2.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/9%3A_Atomic_Structure_and_The_Periodic_Law.txt
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The electron configuration of an atomic species (neutral or ionic) allows us to understand the shape and energy of its electrons. Many general rules are taken into consideration when assigning the "location" of the electron to its prospective energy state, however these assignments are arbitrary and it is always uncertain as to which electron is being described. Knowing the electron configuration of a species gives us a better understanding of its bonding ability, magnetism and other chemical properties.
Introduction
The electron configuration is the standard notation used to describe the electronic structure of an atom. Under the orbital approximation, we let each electron occupy an orbital, which can be solved by a single wavefunction. In doing so, we obtain three quantum numbers (n,l,ml), which are the same as the ones obtained from solving the Schrödinger's equation for Bohr's hydrogen atom. Hence, many of the rules that we use to describe the electron's address in the hydrogen atom can also be used in systems involving multiple electrons. When assigning electrons to orbitals, we must follow a set of three rules: the Aufbau Principle, the Pauli-Exclusion Principle, and Hund's Rule.
The wavefunction is the solution to the Schrödinger equation. By solving the Schrödinger equation for the hydrogen atom, we obtain three quantum numbers, namely the principal quantum number (n), the orbital angular momentum quantum number (l), and the magnetic quantum number (ml). There is a fourth quantum number, called the spin magnetic quantum number (ms), which is not obtained from solving the Schrödinger equation. Together, these four quantum numbers can be used to describe the location of an electron in Bohr's hydrogen atom. These numbers can be thought of as an electron's "address" in the atom.
Notation
To help describe the appropriate notation for electron configuration, it is best to do so through example. For this example, we will use the iodine atom. There are two ways in which electron configuration can be written:
I: 1s22s22p63s23p64s23d104p65s24d105p5
or
I: [Kr]5s24d105p5
In both of these types of notations, the order of the energy levels must be written by increased energy, showing the number of electrons in each subshell as an exponent. In the short notation, you place brackets around the preceding noble gas element followed by the valence shell electron configuration. The periodic table shows that kyrpton (Kr) is the previous noble gas listed before iodine. The noble gas configuration encompases the energy states lower than the valence shell electrons. Therefore, in this case [Kr]=1s22s22p63s23p64s23d104p6.
Quantum Numbers
Principal Quantum Number (n)
The principal quantum number n indicates the shell or energy level in which the electron is found. The value of n can be set between 1 to n, where n is the value of the outermost shell containing an electron. This quantum number can only be positive, non-zero, and integer values. That is, n=1,2,3,4,..
For example, an Iodine atom has its outmost electrons in the 5p orbital. Therefore, the principle quantum number for Iodine is 5.
Orbital Angular Momentum Quantum Number (l)
The orbital angular momentum quantum number, l, indicates the subshell of the electron. You can also tell the shape of the atomic orbital with this quantum number. An s subshell corresponds to l=0, a p subshell = 1, a d subshell = 2, a f subshell = 3, and so forth. This quantum number can only be positive and integer values, although it can take on a zero value. In general, for every value of n, there are n values of l. Furthermore, the value of l ranges from 0 to n-1. For example, if n=3, l=0,1,2.
So in regards to the example used above, the l values of Iodine for n = 5 are l = 0, 1, 2, 3, 4.
Magnetic Quantum Number (ml)
The magnetic quantum number, ml, represents the orbitals of a given subshell. For a given l, ml can range from -l to +l. A p subshell (l=1), for instance, can have three orbitals corresponding to ml = -1, 0, +1. In other words, it defines the px, py and pzorbitals of the p subshell. (However, the ml numbers don't necessarily correspond to a given orbital. The fact that there are three orbitals simply is indicative of the three orbitals of a p subshell.) In general, for a given l, there are 2l+1 possible values for ml; and in a n principal shell, there are n2 orbitals found in that energy level.
Continuing on from out example from above, the ml values of Iodine are ml = -4, -3, -2, -1, 0 1, 2, 3, 4. These arbitrarily correspond to the 5s, 5px, 5py, 5pz, 4dx2-y2, 4dz2, 4dxy, 4dxz, and 4dyz orbitals.
Spin Magnetic Quantum Number (ms)
The spin magnetic quantum number can only have a value of either +1/2 or -1/2. The value of 1/2 is the spin quantum number, s, which describes the electron's spin. Due to the spinning of the electron, it generates a magnetic field. In general, an electron with a ms=+1/2 is called an alpha electron, and one with a ms=-1/2 is called a beta electron. No two paired electrons can have the same spin value.
Out of these four quantum numbers, however, Bohr postulated that only the principal quantum number, n, determines the energy of the electron. Therefore, the 3s orbital (l=0) has the same energy as the 3p (l=1) and 3d (l=2) orbitals, regardless of a difference in l values. This postulate, however, holds true only for Bohr's hydrogen atom or other hydrogen-like atoms.
When dealing with multi-electron systems, we must consider the electron-electron interactions. Hence, the previously described postulate breaks down in that the energy of the electron is now determined by both the principal quantum number, n, and the orbital angular momentum quantum number, l. Although the Schrödinger equation for many-electron atoms is extremely difficult to solve mathematically, we can still describe their electronic structures via electron configurations.
General Rules of Electron Configuration
There are a set of general rules that are used to figure out the electron configuration of an atomic species: Aufbau Principle, Hund's Rule and the Pauli-Exclusion Principle. Before continuing, it's important to understand that each orbital can be occupied by two electrons of opposite spin (which will be further discussed later). The following table shows the possible number of electrons that can occupy each orbital in a given subshell.
subshell number of orbitals total number of possible electrons in each orbital
s 1 2
p 3 (px, py, pz) 6
d 5 (dx2-y2, dz2, dxy, dxz, dyz) 10
f 7 (fz3, fxz2, fxyz, fx(x2-3y2), fyz2, fz(x2-y2), fy(3x2-y2)
14
Using our example, iodine, again, we see on the periodic table that its atomic number is 53 (meaning it contains 53 electrons in its neutral state). Its complete electron configuration is 1s22s22p63s23p64s23d104p65s24d105p5. If you count up all of these electrons, you will see that it adds up to 53 electrons. Notice that each subshell can only contain the max amount of electrons as indicated in the table above.
Aufbau Principle
The word 'Aufbau' is German for 'building up'. The Aufbau Principle, also called the building-up principle, states that electron's occupy orbitals in order of increasing energy. The order of occupation is as follows:
1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
Another way to view this order of increasing energy is by using Madelung's Rule:
Figure 1. Madelung's Rule is a simple generalization which
dictates in what order electrons should be filled in the
orbitals,
however there are exceptions such as
copper and chromium.
This order of occupation roughly represents the increasing energy level of the orbitals. Hence, electrons occupy the orbitals in such a way that the energy is kept at a minimum. That is, the 7s, 5f, 6d, 7p subshells will not be filled with electrons unless the lower energy orbitals, 1s to 6p, are already fully occupied. Also, it is important to note that although the energy of the 3d orbital has been mathematically shown to be lower than that of the 4s orbital, electrons occupy the 4s orbital first before the 3d orbital. This observation can be ascribed to the fact that 3d electrons are more likely to be found closer to the nucleus; hence, they repel each other more strongly. Nonetheless, remembering the order of orbital energies, and hence assigning electrons to orbitals, can become rather easy when related to the periodic table.
To understand this principle, let's consider the bromine atom. Bromine (Z=35), which has 35 electrons, can be found in Period 4, Group VII of the periodic table. Since bromine has 7 valence electrons, the 4s orbital will be completely filled with 2 electrons, and the remaining five electrons will occupy the 4p orbital. Hence the full or expanded electronic configuration for bromine in accord with the Aufbau Principle is 1s22s22p63s23p64s23d104p5. If we add the exponents, we get a total of 35 electrons, confirming that our notation is correct.
Hund's Rule
Hund's Rule states that when electrons occupy degenerate orbitals (i.e. same n and l quantum numbers), they must first occupy the empty orbitals before double occupying them. Furthermore, the most stable configuration results when the spins are parallel (i.e. all alpha electrons or all beta electrons). Nitrogen, for example, has 3 electrons occupying the 2p orbital. According to Hund's Rule, they must first occupy each of the three degenerate p orbitals, namely the 2px orbital, 2py orbital, and the 2pz orbital, and with parallel spins (Figure 2). The configuration below is incorrect because the third electron occupies does not occupy the empty 2pz orbital. Instead, it occupies the half-filled 2px orbital. This, therefore, is a violation of Hund's Rule (Figure 2).
Figure 2. A visual representation of the Aufbau Principle and Hund's Rule. Note that the filling of electrons in each orbital
(px, py and pz) is arbitrary as long as the electrons are singly filled before having two electrons occupy the same orbital.
(a)This diagram represents the correct filling of electrons for the nitrogen atom. (b) This diagramrepresents the incorrect
filling of the electrons for the nitrogen atom.
Pauli-Exclusion Principle
Wolfgang Pauli postulated that each electron can be described with a unique set of four quantum numbers. Therefore, if two electrons occupy the same orbital, such as the 3s orbital, their spins must be paired. Although they have the same principal quantum number (n=3), the same orbital angular momentum quantum number (l=0), and the same magnetic quantum number (ml=0), they have different spin magnetic quantum numbers (ms=+1/2 and ms=-1/2).
Electronic Configurations of Cations and Anions
The way we designate electronic configurations for cations and anions is essentially similar to that for neutral atoms in their ground state. That is, we follow the three important rules: Aufbau Principle, Pauli-exclusion Principle, and Hund's Rule. The electronic configuration of cations is assigned by removing electrons first in the outermost p orbital, followed by the s orbital and finally the d orbitals (if any more electrons need to be removed). For instance, the ground state electronic configuration of calcium (Z=20) is 1s22s22p63s23p64s2. The calcium ion (Ca2+), however, has two electrons less. Hence, the electron configuration for Ca2+ is 1s22s22p63s23p6. Since we need to take away two electrons, we first remove electrons from the outermost shell (n=4). In this case, all the 4p subshells are empty; hence, we start by removing from the s orbital, which is the 4s orbital. The electron configuration for Ca2+ is the same as that for Argon, which has 18 electrons. Hence, we can say that both are isoelectronic.
The electronic configuration of anions is assigned by adding electrons according to Aufbau Principle. We add electrons to fill the outermost orbital that is occupied, and then add more electrons to the next higher orbital. The neutral atom chlorine (Z=17), for instance has 17 electrons. Therefore, its ground state electronic configuration can be written as 1s22s22p63s23p5. The chloride ion (Cl-), on the other hand, has an additional electron for a total of 18 electrons. Following Aufbau Principle, the electron occupies the partially filled 3p subshell first, making the 3p orbital completely filled. The electronic configuration for Cl- can, therefore, be designated as 1s22s22p63s23p6. Again, the electron configuration for the chloride ion is the same as that for Ca2+ and Argon. Hence, they are all isoelectronic to each other.
Problems
1. Which of the princples explained above tells us that electrons that are paired cannot have the same spin value?
2. Find the values of n, l, ml, and ms for the following:
a. Mg
b. Ga
c. Co
3. What is a possible combination for the quantum numbers of the 5d orbital? Give an example of an element which has the 5d orbital as it's most outer orbital.
4. Which of the following cannot exist (there may be more than one answer):
a. n = 4; l = 4; ml = -2; ms = +1/2
b. n = 3; l = 2; ml = 1; ms = 1
c. n = 4; l = 3; ml = 0; ms = +1/2
d. n = 1; l = 0; ml = 0; ms = +1/2
e. n = 0; l = 0; ml = 0; ms = +1/2
5. Write electron configurations for the following:
a. P
b. S2-
c. Zn3+
Answers
1. Pauli-exclusion Principle
2. a. n = 3; l = 0, 1, 2; ml = -2, -1, 0, 1, 2; ms can be either +1/2 or -1/2
b. n = 4; l = 0, 1, 2, 3; ml = -3, -2, -1, 0, 1, 2, 3; ms can be either +1/2 or -1/2
c. n = 3; l = 0, 1, 2; ml = -2, -1, 0, 1, 2, 3; ms can be either +1/2 or -1/2
3. n = 5; l = 3; ml = 0; ms = +1/2. Osmium (Os) is an example.
4. a. The value of l cannot be 4, because l ranges from (0 - n-1)
b. ms can only be +1/2 or -1/2
c. Okay
d. Okay
e. The value of n cannot be zero.
5. a. 1s22s22p63s23p3
b. 1s22s22p63s23p6
c. 1s22s22p63s23p64s23d7
Contributors and Attributions
• Lannah Lua, Andrew Iskandar (University of California Davis, Undergraduate) Mary Magsombol (University of California Davis)
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The electronic angular wavefunction is one spatial component of the electronic Schrödinger wave equation, which describes the motion of an electron. It depends on angular variables, $\theta$ and $\phi$, and describes the direction of the orbital that the electron may occupy. Some of its solutions are equal in energy and are therefore called degenerate.
Introduction
Electrons can be described as a particle or a wave. Because they exhibit wave behavior, there is a wavefunction that is a solution to the Schrödinger wave equation:
$\hat{H}\Psi(r,\phi,\theta,t)=E\Psi(r,\phi,\theta,t)$
This equation has eigenvalues, $E$, which are energy values that correspond to the different wavefunctions.
Spherical Coordinates
To solve the Shrödinger equation, spherical coordinates are used. Spherical coordinates are in terms of a radius $r$, as well as angles $\phi$, which is measured from the positive x axis in the xy plane and may be between 0 and $2\pi$, and $\theta$, which is measured from the positive z axis towards the xy plane and may be between 0 and $\pi$.
$x=rsin(\theta)cos(\phi)$
$y=rsin(\theta)sin(\phi)$
$z=rcos(\theta)$
Electronic Wavefunction
The electronic wavefunction, $\Psi(r,\phi ,\theta ,t)$, describes the wave behavior of an electron. Its value is purely mathematical and has no corresponding measurable physical quantity. However, the square modulus of the wavefunction, $\mid \Psi(r,\phi ,\theta ,t)\mid ^2$ gives the probability of locating the electron at a given set of values. To use separation of variables, the wavefunction can be expressed as
$\Psi(r,\phi, \theta ,t)=R(r)Y_{l}^{m}(\phi, \theta)$
$R(r)$ is the radial wavefunction and $Y_{l}^{m}(\phi, \theta)$ is the angular wavefunction. Separating the angular variables in $Y_{l}^{m}(\phi, \theta)$ gives
$Y_{l}^{m}(\phi, \theta)=\left[\dfrac{2l+1}{4\pi}\left(\dfrac{(l-\mid m \mid)!}{(l+\mid m \mid)!}\right)\right]^{\frac{1}{2}}P_l^{\mid m \mid}(cos(\theta))e^{im\phi}$
where $P_l^{\mid m \mid}(cos(\theta))$ is a Legendre polynomial and is only in terms of the variable $\theta$. The exponential function, which is only in terms of $\phi$, determines the phase of the orbital.
For the angular wavefunction, the square modulus gives the probability of finding the electron at a point in space on a ray described by $(\phi, \theta)$. The angular wavefunction describes the spherical harmonics of the electron's motion. Because orbitals are a cloud of the probability density of the electron, the square modulus of the angular wavefunction influences the direction and shape of the orbital.
Quantum Numbers and Orbitals
There are 3 quantum numbers defined by the Schrödinger wave equation. They are $n$, $l$, and $m_{l}$. Each combination of these quantum numbers describe an orbital. Values for $n$ come from from the radial wavefunction. $n$ may be 1, 2, 3... Because they evolved from the separation of variables performed to solve the wavefunction, solutions to the angular wavefunction are quantized by the values for $l$ and $m_{l}$. Acceptable values for $l$ are given by $l=n-1$. The corresponding values for $m_{l}$ are integers between $-l$ and $+l$.
Degeneracy and p, d and f Orbitals
Orbitals descrbed by the same $n$ and $l$ values but different $m_{l}$ values are degenerate, meaning that they are equal in energy but vary in their direction and, sometimes, shape. For $p$ orbitals, $l=1$, giving three $m_{l}$ values and thus, 3 degenerate states. They are $p_{x}$, $p_{y}$and $p_{z}$. $d$ orbitals have $l=2$, giving 5 degenerate states. These are $d_{xy}$, $d_{xz}$, $d_{yz}$, $d_{z^2}$, $d_{x^2-y^2}$. $f$ orbitals have $l=3$, giving a total of 7 degenerate states.
Problems
1. Which quantum numbers depend on the angular wavefunction?
2. Give the quantum numbers defined by the angular wavefunction for a $d_{z^2}$ orbital.
3. Given that the spherical representation of the $d_{x^2-y^2}$ orbital is $r^2 sin^{2}(\theta) cos(2\phi)$, show that this matches the label for the orbital, $x^2-y^2$. Hint: $cos(2x)=cos^2(x)-sin^2(x)$
Solutions:
1. $l$ and $m_{l}$
2. $l=2$ and $m_{l}=2,1,0,-1,-2$
3. $r^2 sin^{2}(\theta)cos(2\phi)=r^2 sin^{2}(\theta)(cos^2(\phi)-sin^2(\phi))$
$r^2 sin^{2}(\theta)cos^2(\phi)-r^2 sin^{2}(\theta)sin^2(\phi)$
Since $x=rsin(\theta)cos(\phi)$ and $y=rsin(\theta)sin(\phi)$,
$r^2 sin^{2}(\theta) cos(2\phi)=x^2-y^2$
• Bryn Ellison
H Atom Energy Levels
Any electron associated with an atom has a wavefunction that describes its position around the nucleus as well as an energy. This energy consists of two components: kinetic and potential energies. The kinetic energy is a consequence of the electron having mass and moving at a certain speed. The potential energy is a result of the electrostatic, or Coulombic, force that attracts oppositely charged particles (i.e. the electron and the proton nucleus). The hydrogen atom energy levels can be obtained through solving the Schrödinger Equation.
Introduction
Using the power series method, the solution of the hydrogen atom Schrödinger Equation leads to quantized energy levels
$E_n=-\dfrac{m_ee^4}{8\epsilon{_0}^2h^2n^2}=-\dfrac{m_ee^4}{32\pi{^2}\epsilon{_0^2}\hbar{^2}n^2}\;\;n=1,2,..$
This energy can be written in terms of the Bohr radius, $a_0$:
$E_n=-\dfrac{e^2}{8\pi{^2}\epsilon{_0}a_0n^2}\;\; n=1,2,...$
The energy is negative due to the attractive nature of the Coulombic interaction. This is alternatively visualized as an atom whose electron has been moved infinitely far away. The potential energy of the electron is defined as zero as there is no interaction at infinite distance. As the electron approaches the nucleus, the amount of potential energy decreases as the electron gets nearer to the nucleus.
One interesting feature of the energy levels are that they are not spaced evenly, but rather the spacing diminishes exponentially as $E_n\propto 1/{n^2}$. The limit of these energies as n approaches infinity is the dissociation energy, the point at which the electron escapes the Coulombic pull of the nucleus.
Another important aspect to note is that the energy is solely dependent on the principal quantum number n, hence the 3s, 3p and 3d orbitals all have the same energy. This is not the case for multi-electron atoms, where the s, p, or d orbitals in a given n level have different energies.
Problems
1. How much energy is required to excite the hydrogen electron from its ground state to the first excited state?
2. According to quantum mechanics, what is the dissociation energy of an H atom?
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Koopmans' theorem states that the first ionization energy of a molecule is equal to the negative of the energy of the highest occupied molecular orbital (HOMO).
Introduction
Koopmans' theorem uses the Hartree-Fock method for approximation of orbital energy εi which is derived from the wavefunction of the spin orbital and the kinetic and nuclear attraction energies. This theorem applies when an electron is removed from a molecular orbital in order to form a positive ion. It was originally only used for ionization energies in a closed-shell system, but has been generalized to be used to calculate energy changes when electrons are added to or removed from a system. Based on this generalization, it is possible to use the same method to approximate the electron affinity. In this case, the molecular orbital energy would be the one associated with the orbital to which the electron is being added. Koopmans' theorem is useful because the use of this approximation means that it is not necessary to calculate the two separate energies of the original molecule and its ion in order to find the ionization energy and electron affinity.
Basic Description
In order to understand Koopmans' theorem, we must first understand its background, which is based in Hartree-Fock method. Hartree-Fock method is used to approximate the wave function and energy of a multi-electron system. This method starts by expressing the wave function of the system as a Slater determinant of the wave function of each single-particle orbital. The Slater determinant for a system with N electons is
${\displaystyle \Psi (\mathbf {x} _{1},\mathbf {x} _{2},\ldots ,\mathbf {x} _{N})={\frac {1}{\sqrt {N!}}}\left|{\begin{matrix}\chi _{1}(\mathbf {x} _{1})&\chi _{2}(\mathbf {x} _{1})&\cdots &\chi _{N}(\mathbf {x} _{1})\\chi _{1}(\mathbf {x} _{2})&\chi _{2}(\mathbf {x} _{2})&\cdots &\chi _{N}(\mathbf {x} _{2})\\vdots &\vdots &\ddots &\vdots \\chi _{1}(\mathbf {x} _{N})&\chi _{2}(\mathbf {x} _{N})&\cdots &\chi _{N}(\mathbf {x} _{N})\end{matrix}}\right|\equiv \left|{\begin{matrix}\chi _{1}&\chi _{2}&\cdots &\chi _{N}\\end{matrix}}\right|,}$
By breaking up the electrons into individual wave functions, we can create single-particle Hartree-Fock equations, which can be used as an operator whose eigenvalue is the energy of a particle in a particular orbital.
$\hat {F} | \varphi _i \rangle = \epsilon _i| \varphi _i \rangle$
Where $\hat {F}$ is the Fock operator corresponding to the Hartree-Fock equation, $\epsilon _i$ is the energy of the particle in the orbital, and $\varphi _i$ is the wave function of the particle. Using Koopmans' theorem, the equation from above and the wave function of the HOMO, we can approximate the first order ionization energy of a molecule.
Advanced Description
The Hartree-Fock equation for a particle takes the form
$-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r}) + V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r}) + V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r}) - \sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} = \epsilon_{i}\psi_{i}(\mathbf{r})$
So, the Fock operator derived from this is
$\hat {F} = \hat {H} ^0 + \sum _{j=1}^N ( 2 \hat {J} _j - \hat {K} _j ) = -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {Ze^2}{4 \pi \epsilon _0 r} + \sum _{j=1}^N (2\hat {J}_j - \hat {K} _j )$
This is the operator that is applied to the wave function, and gives the eigenvalues that describe the energy of a given orbital. Notice that the first two terms in this equation are the same as the operator which corresponds to the energy of a hydrogen atom. The third term accounts for the potential energy applied by the other electrons in the molecule. This method of calculating the potential energy assumes that the electron only interacts with the average charge of the electron cloud.
Ionization energies and electron affinities are equal to the negative of the energy of the orbital which is added or removed, i.e.
$E_{N-1} - E_N = - \epsilon_k$ for ionization, and $E_N - E_{N+1} = - \epsilon_k .$
for electron affinity.
Example $1$: Calculating Ionization Energy
Koopmans' theorem also applies to the calculation of electron affinity. We will use Hartree-Fock equations to calculate the energy change when an electron is added to the N+1 orbital. The energy of the N-electron determinant with spin-orbitals $\phi _1$ through $\phi _N$ occupied is
$E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \sum_{i=1}^{N} [ J_{i,j} - K_{i,j} ] \nonumber$
or
$E_N = \sum_{i=1}^N \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N} [ J_{i,j} - K_{i,j} ].\nonumber$
And the energy of the N+1 electron determinant is
$E_{N+1} = \sum_{i=1}^{N+1} \langle \phi_i | T + V | \phi_i \rangle + \frac{1}{2} \sum_{i,j=1}^{N+1} [ J_{i,j} - K_{i,j} ].\nonumber = \epsilon_{N+1}. \nonumber$
Since we are adding the N+1 orbital the electron affinity is equal to the negative of the energy of the N+1 electron determinant.
$EA = - \epsilon_{N+1}. \nonumber$
However, we must remember that Koopmans' theorem is merely an approximation of ionization energies and electron affinities. The more accurate method would be to calculate the separate energies of the parent and daughter molecules and subtract them to calculate the difference. This is because Koopmans' theorem does not account for orbital relaxation; it uses the orbitals of only one of the molecules to describe both of the species, which is not necessarily true.
Contributors and Attributions
• Wikipedia
• Charlotte Swaney
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Penetration and shielding are two underlying principles in determining the physical and chemical properties of elements. We can predict basic properties of elements by using shielding and penetration characteristics to assess basic trends.
Introduction
Electrons are negatively charged and are pulled pretty close to each other by their attraction to the positive charge of a nucleus. The electrons are attracted to the nucleus at the same time as electrons repel each other. The balance between attractive and repulsive forces results in shielding. The orbital (n) and subshell (ml) define how close an electron can approach the nucleus. The ability of an electron to get close to the nucleus is penetration.
Coulomb's Law (an analogy with classical physics) can be used to describe the attraction and repulsion between atomic particles:
$F=k \dfrac{q_1q_1}{r^2} \label{1}$
The force that an electron feels is dependent on the distance from the nearest charge (i.e., an electron, usually with bigger atoms and on the outer shells) and the amount of charge. More distance between the charges will result in less force, and more charge will have more force of attraction or repulsion.
In the simplest case, every electron in an atom would feel the same amount of "pull" from the nucleus. For example, in Li, all three electrons might "feel" the +3 charge from the nucleus. However, this is not the case when observing atomic behavior. When considering the core electrons (or the electrons closest to the nucleus), the nuclear charge "felt" by the electrons (Effective Nuclear Charge ($Z_{eff}$)) is close to the actual nuclear charge. As you proceed from the core electrons to the outer valence electrons, $Z_{eff}$ falls significantly. This is because of shielding, or simply the electrons closest to the nucleus decrease the amount of nuclear charge affecting the outer electrons. Shielding is caused by the combination of partial neutralization of nuclear charge by core electrons, and by electron-electron repulsion.
The amount of charge felt by an electron depends on its distance from the nucleus. The closer an electron comes to the nucleus, or the more it penetrates, the stronger its attraction to the nucleus. Core electrons penetrate more and feel more of the nucleus than the other electrons.
$F_{electron-nucleus}=k \dfrac{Ze^2}{r^2} \label{2}$
with
• $Z$ is the charge of the nucleus (i.e., number of protons)
• $e$ is the charge of an electron or proton
• r is the radius, or distance between the proton and the electron
Penetration and shielding result in an Effective force ($F_{eff}$) that holds the outer electrons to the atom, akin to Equation $\ref{2}$, but with $Z_{eff}$ substituted for $Z$:
$F_{eff}=k \dfrac{Z_{eff}e^2}{r^2} \label{3}$
Orbital Penetration
Penetration describes the proximity to which an electron can approach to the nucleus. In a multi-electron system, electron penetration is defined by an electron's relative electron density (probability density) near the nucleus of an atom. Electrons in different orbitals have different wavefunctions and therefore different radial distributions and probabilities (defined by quantum numbers n and ml around the nucleus). In other words, penetration depends on the shell (n) and subshell (ml). For example, we see that since a 2s electron has more electron density near the nucleus than a 2p electron, it is penetrating the nucleus of the atom more than the 2p electron. The penetration power of an electron, in a multi-electron atom, is dependent on the values of both the shell and subshell.
Within the same shell value (n), the penetrating power of an electron follows this trend in subshells (ml):
s>p>d>f
And for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:
1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....
and the energy of an electron for each shell and subshell goes as follows...
1s<2s<2p<3s<3p<4s<3d<4p....
The electron probability density for s-orbitals is highest in the center of the orbital, or at the nucleus. If we imagine a dartboard that represents the circular shape of the s-orbital and if the darts landed in correlation to the probability to where and electron would be found, the greatest dart density would be at the 50 points region but most of the darts would be at the 30 point region. When considering the 1s-orbital, the spherical shell of 53 pm is represented by the 30 point ring.
Electrons which experience greater penetration experience stronger attraction to the nucleus, less shielding, and therefore experience a larger Effective Nuclear Charge ($Z_{eff}$), but shield other electrons more effectively.
Shielding
An atom (assuming its atomic number is greater than 2) has core electrons that are extremely attracted to the nucleus in the middle of the atom. However the number of protons in the nucleus are never equal to the number of core electrons (relatively) adjacent to the nucleus. The number of protons increase by one across the periodic table, but the number of core electrons change by periods. The first period has no core electrons, the second has 2, the third has 10, and etc. This number is not equal to the number of protons. So that means that the core electrons feel a stronger pull towards the nucleus than any other electron within the system. The valence electrons are farther out from the nucleus, so they experience a smaller force of attraction.
Shielding refers to the core electrons repelling the outer rings and thus lowering the 1:1 ratio. Hence, the nucleus has "less grip" on the outer electrons and are shielded from them. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity. For example, $Z_{eff}$ is calculated by subtracting the magnitude of shielding from the total nuclear charge. The value of $Z_{eff}$ will provide information on how much of a charge an electron actually experiences.
Because the order of electron penetration from greatest to least is s, p, d, f; the order of the amount of shielding done is also in the order s, p, d, f.
Since the 2s electron has more density near the nucleus of an atom than a 2p electron, it is said to shield the 2p electron from the full effective charge of the nucleus. Therefore the 2p electron feels a lesser effect of the positively charged nucleus of the atom due to the shielding ability of the electrons closer to the nucleus than itself, (i.e. 2s electron).
These electrons that are shielded from the full charge of the nucleus are said to experience an effective nuclear charge ($Z_{eff}$)of the nucleus, which is some degree less than the full nuclear charge an electron would feel in a hydrogen atom or hydrogenlike ions. The effective nuclear charge of an atom is given by the equation:
$Z_{eff}=Z-S \label{4}$
where.
• $Z$ is the atomic number (number of protons in nucleus) and
• $S$ is the shielding constant
We can see from this equation that the effective nuclear charge of an atom increases as the number of protons in an atom increases. Therefore as we go from left to right on the periodic table the effective nuclear charge of an atom increases in strength and holds the outer electrons closer and tighter to the nucleus. This phenomena can explain the decrease in atomic radii we see as we go across the periodic table as electrons are held closer to the nucleus due to increase in number of protons and increase in effective nuclear charge.
Example $1$: Fluorine, Neon, and Sodium
What is the effective attraction $Z_{eff}$ experienced by the valence electrons in the three isoelectronic species: the fluorine anion, the neutral neon atom, and sodium cation?
Solution
Each species has 10 electrons, and the number of nonvalence electrons is 2 (10 total electrons - 8 valence) but the effective nuclear charge varies because each has a different atomic number $A$.
The charge $Z$ of the nucleus of a fluorine atom is 9, but the valence electrons are screened appreciably by the core electrons (four electrons from the 1s and 2s orbitals) and partially by the 7 electrons in the 2p orbitals.
Diagram of a fluorine atom showing the extent of effective nuclear charge. (CC BY-SA- 3.0; Wikipedia).
• $Z_\mathrm{eff}(\mathrm{F}^-) = 9 - 2 = 7+$
• $Z_\mathrm{eff}(\mathrm{Ne}) = 10 - 2 = 8+$
• $Z_\mathrm{eff}(\mathrm{Na}^+) = 11 - 2 = 9+$
So the sodium cation has the greatest effective nuclear charge, and thus the smallest radius.
Radial Distribution Graphs
A radial distribution function graph describes the distribution of orbitals with the effects of shielding (Figure $2$). The small peak of the 2s orbital shows that the electrons in the 2s orbital are closest to the nucleus. Therefore, it is the electrons in the 2p orbital of Be that are being shielded from the nucleus, by the electrons in the 2s orbital.
The following is the radial distribution of the 1s and 2s orbitals. Notice the 1s orbital is shifted to the right, while the 2s orbital has a node.
Periodic Trends Due to Penetration and Shielding
• Effective Nuclear Charge ($Z_{eff}$): The effective nuclear charge increases from left to right and increases from top to bottom on the periodic table.
• Atomic Radius: The atomic radius decreases from left to right, and increases from top to bottom.
• Ionization Energies: The ionization energies increase from left to right, and decrease from top to bottom.
• Electronegativity: The electronegativity of the elements is highest near flourine. In general, it increases from left to right and decreases from top to bottom.
Questions
1. Which orbital is more effective in shielding? 1s or 2p?
2. True/False: The greater the penetration of an orbital, the greater the shielding capability of that orbital.
3. Find the $Z_{eff}$of
1. Mg
2. C
3. F
4. Ca
4. Which of these have the smallest electron affinity? B, C, N, O, or F.
5. Which atom has a stronger effective nuclear charge and why? (assuming S is the same in both cases) Li, or N
6. Why does the Hydrogen electron experiences the full charge of the nucleus without any shielding?
7. Which atom has a smaller radii? Be or F?
8. Which electron has higher energy level? 2s or 2p? and why?
9. Why do the orbitals of a hydrogen atom increase energy as follows: 1s<2s=2p<3s=3p=3d<4s=4p=4d<....
10. Which electrons shields better in an atom? 2s or 2p? 3p or 3d?
11. Why can we relate classical physics to quantum mechanics when it comes to subatomic activity?
12. What is penetration?
Solutions
1. 1s
2. T
1. a. 12-2 =10
2. b. 6-4=2
3. c. 9-7=2
4. d. 20-2=18
3. -
4. nitrogen atom has a stronger effective nuclear charge than lithium due to its greater number of protons in the nucleus holding the electrons tighter.
5. Hydrogen atom has only one electron total, therefore there are no other, lower energy (more penetrating), electrons available to help shield this electron from the nucleus.
6. Fluorine has a smaller radii than Beryllium due to its greater number of protons providing a greater effective nuclear charge on the outer electrons and therefore pulling them in tighter and providing a smaller atomic radii.
7. 2p has higher energy level because the negatively charged electron experiences less of an effective nuclear charge than the 2s electron.
8. because a Hydrogen atom has only one electron, that experiences no shielding from other electrons and therefore its energy level only depends on its distance away from the nucleus, which is dependent on it value of (n).
9. 2s shields the atom better than 2p because the s orbitals is much closer and surrounds the nucleus more than the p orbitals, which extend farther out. 3p shields better than 3d, because p orbitals are closer to the nucleus than the 3d orbitals.
10. Classical physics and quantum mechanics both can deal with subatomic activity such as electron interactions, orbital location, size, and shape, and distances to find forces of attractions.
11. Penetration is how well the outer electrons are shielded from the nucleus by the core electrons. The outer electrons therefore experience less of an attraction to the nucleus.
Contributors and Attributions
• Sidra Ayub (UCD), Alan Chu (UCD)
Wave Function of Multi-electron Atoms
Unlike hydrogenic atoms, the wavefunctions satisfying Schrödinger's equation for multi-electron atoms cannot be solved analytically. Instead, various techniques are used for giving approximate solutions to the wave functions.
First Approximation
The wavefunctions of multi-electron atoms can be considered, as a first approximation, to be built up of components, where the combined wavefunction for an atom with k electrons is of the form:
\(\Psi = \Psi(1) \; \Psi(2) \;...\; \Psi(k)\)
Here, \(\Psi(1)\), \(\Psi(2)\), up to \(\Psi(k)\) represent wave functions for the first, second, up to kth electrons. Each \(\Psi(i)\) is considered to be in the form of a wave function for the single electron of the hydrogenic atom subject to the Pauli Exclusion Principle and after making adjustments to account for shielding and penetration.
The Pauli Exclusion Principle allows at most two electrons in any one orbital. This is explained by postulating an additional quantum number for electron spin, ms , which can have values of +1/2 or -1/2. The Dirac theory of quantum mechanics, applied to electron orbitals, more naturally explains this spin magnetic quantum number because the theory goes beyond the assumptions of the Schrödinger equation by also accounting for the relativistic behavior of orbiting electrons.
Shielding occurs because other electrons that are closer to the nucleus shield an electron from the attractive force of the nucleus. Adjusting the hydrogenic orbitals can be done by reducing the value of Z to Zeff. The amount of the reduction depends on which inner orbitals are occupied and how much the orbital being calculated is able to penetrate the shielding. As the amount of shielding increases, Zeff becomes smaller, and the energy levels of the orbital increase. Conversely, as the amount of penetration increases, the shielding is reduced, Zeff becomes bigger, and the energy level of the orbital decreases.
For example, when the 1s orbital is fully occupied by two electrons, the third electron could occupy any of 2s or 2p orbitals, which would have the same energy level in a hydrogenic atom. However, as illustrated in the diagrams below, the 2s orbital concentrates more of its probability near the nucleus than the 2p orbital does. As a result, the 2s orbital penetrates the 1s orbital shielding more than the 2p orbital does. Hence, a third electron will occupy the 2s orbital where it has a slightly lower energy than in the 2p orbital.
The values of Zeff can be calculated by various techniques. Slater's rule is a relatively simple ad hoc method of estimating Zeff for values of n up to 4, that is for electrons in the s, p, d, or f orbitals.
Contributors and Attributions
• Thanh Hua (UCD)
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A hydrogen atom consists of a single proton orbited by a single electron. When compared to the electron, the proton has such a large mass that it may be considered stationary while the electron circles around it.
Introduction
It is known that this model is acceptable when the reduced mass of the system is used. Reduced mass is defined below for two masses 1 and 2.
$\mu = \dfrac{m_1m_2}{m_1+m_2}$
Using the reduced mass effectively converts the two-body problem (two moving and interacting bodies in space) into a one-body problem (a single electron moving about a fixed point). The basic Schdinger equation is
$\hat{H}\Psi=E\Psi$
where $\hat{H}$ is the Hamiltonian operator, E is the energy of the particle and $\Psi$ is the particle's wavefunction that describes its spatial probability. The Hamiltonian operator is the sum of the kinetic energy operator and potential energy operator. The kinetic energy operator is the same for all models but the potential energy changes and is the defining parameter.
Hydrogen Atom Schrödinger Equation
The hydrogen atom's electron wavefunctions can be described using a variation of the rigid rotor-harmonic oscillator (RRHO) model. Rather than following a parabolic potential energy surface, the electron associated with the hydrogen experiences an exponential coulombic interaction.
This RRHO Hamiltonian combines the kinetic energy elements of both previous models as well as an associated potential energy (as that in the harmonic oscillator scenario). The Schrödinger Equation for the RRHO model involves the Hamiltonian operator acting on a wavefunction that similarly reflects both the rigid rotor and harmonic oscillator models.
In order for the Hamiltonian to fit the model, all component Hamiltonian elements need only be added together.
$\hat{H}_{RRHO}=\hat{H}_{RR}(\theta{,}\phi{})+\hat{H}_{HO}(r)$
$\hat{H}_{RRHO}=\frac{1}{2mr_0^2}\hat{L}^2(\theta{,}\phi{})+\frac{-\hbar{^2}}{2\mu{}}\frac{d^2}{dr^2}+V(r)$
As shown below, the solution wavefunction will be a multiplicative combination of the two model solutions.
$\psi{_{RRHO}}=\psi{}_{RR}*\psi{}_{OH}$
$(\hat{H}_{RR}+\hat{H}_{HO})(\psi{_{RR}}*\psi{_{HO}})=\psi{_{HO}}\hat{H}_{RR}\psi{_{RR}}+\psi{_{RR}}\hat{H}_{HO}\psi{_{HO}}$
$(\hat{H}_{RR}+\hat{H}_{HO})(\psi{_{RR}}*\psi{_{HO}})=\psi{_{HO}}*E_{RR}\psi{_{RR}}+\psi{_{RR}}*E_{HO}\psi{_{HO}}$
$(] (\hat{H}_{RR}+\hat{H}_{HO})(\psi_{RR}*\psi_{HO})=E_{RR}(\psi{_{RR}}*\psi{_{HO}})+E_{HO}(\psi_{RR}*\psi_{HO})$
$\hat{H}_{RRHO}\psi{_{RRHO}}=E_{RR}\psi{_{RRHO}}+E_{HO}\psi{_{RRHO}}") }} Thus, \( E_{RRHO}=E_{RR}+E_{HO}$
After further refinement the Hamiltonian operator for the hydrogen atom is found to be
$\hat{H}=-\dfrac{\hbar{^2}}{2m_e}\bigtriangledown{^2}-\dfrac{e^2}{4\pi{}\epsilon{r}}$
where the Laplacian operator is defined as
$\bigtriangledown{^2}=\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}$
To solve the Schrödinger Equation for the hydrogen atom, it is simplest to perform the quantum mechanical calculations using spherical coordinates (based on the three variables r, $\theta$ and $\phi$). After appropriate adjustments are made to compensate for the change of variables, the Schrödinger equation becomes:
$-\hbar{^2}\dfrac{\partial{}}{\partial{r}}\left(r^2\dfrac{\partial{}\psi{}}{\partial{r}}\right)+\hat{L}^2\psi{}+2m_er^2[V(r)-E]\psi{(}r,\theta{,}\phi{)}=0$
where $V(r)=\dfrac{e^2}{4\pi{}\epsilon{_0}r}$ is the Coulombic (electrostatic) potential between the nucleus and electron and $\hat{L}^2$ is the angular momentum operator also found from the quantum mechanical rigid rotor model.
Additionally, the assumption must be made that the wavefunction is of a form such that it can be arranged as the product of two functions using different variables (separation of variables). It is found that the wavefunction can be described as
$\psi{(}r,\theta{,}\phi{)}=R_{nl}(r)Y_l^m(\theta{,}\phi{)}$
where $R_{nl}(r)$ is the radial function and is defined as $R_{nl}(r)=N_{nl}e^{r/na_0}r^lL_{n+l}^{2l+1}\left(\frac{2r}{na_0}\right)$
Y is a spherical harmonic function, identical to the set of solutions to the rigid rotor quantum mechanical model. The function $L_{n+l}^{2l+1}\left(\frac{2r}{na_0}\right)$ is an associated Laguerre polynomial, determined by the two quantum numbers $l")}} and n. However, simply keeping the notation of R(r) will suffice for the following mathematics. If this separated wavefunction is used in the Schrödinger Equation and the substitution \(\hat{L}^2Y_l^{m_l}(\theta{,}\phi{)}=\hbar{^2}l(l+1)Y_l^{m_l}(\theta{,}\phi{)})$ is carried out, it is found that all of the angular elements of the equation cancel out to yield an equation containing solely radial functions.
$-\hbar{^2}\dfrac{\partial{}}{\partial{r}}\left(r^2\dfrac{\partial{}}{\partial{r}}\left(R(r)Y_l^{m_l}(\theta{,}\phi{})\right)\right)+R(r)\hbar{^2}l(l+1)Y_l^{m_l}(\theta{},\phi{})+2m_er^2[V(r)-E]R(r)Y_l^{m_l}(\theta{},\phi{})$
$-\hbar{^2}\dfrac{d}{dr}\left(r^2\dfrac{\partial{R}}{\partial{r}}\right)Y_l^{m_l}(\theta{},\phi{})+\hbar{^2}l(l+1)R(r)Y_l^{m_l}(\theta{},\phi{})+2m_er^2[V(r)-E]R(r)Y_l^{m_l}(\theta{},\phi{})=0$
$-\hbar{^2}\dfrac{d}{dr}\left(r^2\dfrac{\partial{R}}{\partial{r}}\right)+\hbar{^2}l(l+1)R(r)+2m_er^2[V(r)-E]R(r)=0$
$\dfrac{-\hbar{^2}}{2m_er^2}\dfrac{d}{dr}\left(r^2\dfrac{dR}{dr}\right)+\left[\dfrac{\hbar{^2}l(l+1)}{2m_er^2}+V(r)-E\right]R(r)=0$
To solve this Schrödinger equation the power series method, the same method used to solve the harmonic oscillator, has to be used.
Outside Links
• Link to outside sources. Wikipedia entries should probably be referenced here.
Problems
1. Show that the energy is dependent only on the radial portion of the wavefunction.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Quantum_Mechanical_H_Atom.txt
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A total of four quantum numbers are used to describe completely the movement and trajectories of each electron within an atom. The combination of all quantum numbers of all electrons in an atom is described by a wave function that complies with the Schrödinger equation. Each electron in an atom has a unique set of quantum numbers; according to the Pauli Exclusion Principle, no two electrons can share the same combination of four quantum numbers. Quantum numbers are important because they can be used to determine the electron configuration of an atom and the probable location of the atom's electrons. Quantum numbers are also used to understand other characteristics of atoms, such as ionization energy and the atomic radius.
In atoms, there are a total of four quantum numbers: the principal quantum number (n), the orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms). The principal quantum number, $n$, describes the energy of an electron and the most probable distance of the electron from the nucleus. In other words, it refers to the size of the orbital and the energy level an electron is placed in. The number of subshells, or $l$, describes the shape of the orbital. It can also be used to determine the number of angular nodes. The magnetic quantum number, ml, describes the energy levels in a subshell, and ms refers to the spin on the electron, which can either be up or down.
The Principal Quantum Number ($n$)
The principal quantum number, $n$, designates the principal electron shell. Because n describes the most probable distance of the electrons from the nucleus, the larger the number n is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. n can be any positive integer starting at 1, as $n=1$ designates the first principal shell (the innermost shell). The first principal shell is also called the ground state, or lowest energy state. This explains why $n$ can not be 0 or any negative integer, because there exists no atoms with zero or a negative amount of energy levels/principal shells. When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where $n=2$. This is called absorption because the electron is "absorbing" photons, or energy. Known as emission, electrons can also "emit" energy as they jump to lower principle shells, where n decreases by whole numbers. As the energy of the electron increases, so does the principal quantum number, e.g., n = 3 indicates the third principal shell, n = 4 indicates the fourth principal shell, and so on.
$n=1,2,3,4…$
Example $1$
If n = 7, what is the principal electron shell?
Example $2$
If an electron jumped from energy level n = 5 to energy level n = 3, did absorption or emission of a photon occur?
Answer
Emission, because energy is lost by release of a photon.
The Orbital Angular Momentum Quantum Number ($l$)
The orbital angular momentum quantum number $l$ determines the shape of an orbital, and therefore the angular distribution. The number of angular nodes is equal to the value of the angular momentum quantum number $l$. (For more information about angular nodes, see Electronic Orbitals.) Each value of $l$ indicates a specific s, p, d, f subshell (each unique in shape.) The value of $l$ is dependent on the principal quantum number $n$. Unlike $n$, the value of $l$ can be zero. It can also be a positive integer, but it cannot be larger than one less than the principal quantum number ($n-1$):
$l=0, 1, 2, 3, 4…, (n-1)$
Example $3$
If $n = 7$, what are the possible values of $l$?
Answer
Since $l$ can be zero or a positive integer less than ($n-1$), it can have a value of 0, 1, 2, 3, 4, 5 or 6.
Example $4$
If $l = 4$, how many angular nodes does the atom have?
Answer
The number of angular nodes is equal to the value of l, so the number of nodes is also 4.
The Magnetic Quantum Number ($m_l$)
The magnetic quantum number $m_l$ determines the number of orbitals and their orientation within a subshell. Consequently, its value depends on the orbital angular momentum quantum number $l$. Given a certain $l$, $m_l$ is an interval ranging from $–l$ to $+l$, so it can be zero, a negative integer, or a positive integer.
$m_l= -l, (-l +1),( -l +2),…, -2, -1, 0, 1, 2, … (l – 1), (l – 2), +l$
Example $5$
Example: If $n=3$, and $l=2$, then what are the possible values of $m_l$?
Answer
Since $m_l$ must range from $–l$ to $+l$, then $m_l$ can be: -2, -1, 0, 1, or 2.
The Electron Spin Quantum Number ($m_s$)
Unlike $n$, $l$, and $m_l$, the electron spin quantum number $m_s$ does not depend on another quantum number. It designates the direction of the electron spin and may have a spin of +1/2, represented by↑, or –1/2, represented by ↓. This means that when $m_s$ is positive the electron has an upward spin, which can be referred to as "spin up." When it is negative, the electron has a downward spin, so it is "spin down." The significance of the electron spin quantum number is its determination of an atom's ability to generate a magnetic field or not. (Electron Spin.)
$m_s= \pm \dfrac{1}{2}$
Example $5$
List the possible combinations of all four quantum numbers when $n=2$, $l=1$, and $m_l=0$.
Answer
The fourth quantum number is independent of the first three, allowing the first three quantum numbers of two electrons to be the same. Since the spin can be +1/2 or =1/2, there are two combinations:
• $n=2$, $l=1$, $m_l =0$, $m_s=+1/2$
• $n=2$, $l=1$, $m_l=0$, $m_s=-1/2$
Example $6$
Can an electron with $m_s=1/2$ have a downward spin?
Answer
No, if the value of $m_s$ is positive, the electron is "spin up."
A Closer Look at Shells, Subshells, and Orbitals
Principal Shells
The value of the principal quantum number n is the level of the principal electronic shell (principal level). All orbitals that have the same n value are in the same principal level. For example, all orbitals on the second principal level have a principal quantum number of n=2. When the value of n is higher, the number of principal electronic shells is greater. This causes a greater distance between the farthest electron and the nucleus. As a result, the size of the atom and its atomic radius increases.
Because the atomic radius increases, the electrons are farther from the nucleus. Thus it is easier for the atom to expel an electron because the nucleus does not have as strong a pull on it, and the ionization energy decreases.
Example $7$
Which orbital has a higher ionization energy, one with $n=3$ or $n=2$?
Answer
The orbital with n=2, because the closer the electron is to the nucleus or the smaller the atomic radius, the more energy it takes to expel an electron.
Subshells
The number of values of the orbital angular number l can also be used to identify the number of subshells in a principal electron shell:
• When n = 1, l= 0 (l takes on one value and thus there can only be one subshell)
• When n = 2, l= 0, 1 (l takes on two values and thus there are two possible subshells)
• When n = 3, l= 0, 1, 2 (l takes on three values and thus there are three possible subshells)
After looking at the examples above, we see that the value of n is equal to the number of subshells in a principal electronic shell:
• Principal shell with n = 1 has one subshell
• Principal shell with n = 2 has two subshells
• Principal shell with n = 3 has three subshells
To identify what type of possible subshells n has, these subshells have been assigned letter names. The value of l determines the name of the subshell:
Name of Subshell Value of $l$
s subshell 0
p subshell 1
d subshell 2
f subshell 3
Therefore:
• Principal shell with n = 1 has one s subshell (l = 0)
• Principal shell with n = 2 has one s subshell and one p subshell (l = 0, 1)
• Principal shell with n = 3 has one s subshell, one p subshell, and one d subshell (l = 0, 1, 2)
We can designate a principal quantum number, n, and a certain subshell by combining the value of n and the name of the subshell (which can be found using l). For example, 3p refers to the third principal quantum number (n=3) and the p subshell (l=1).
Example $8$
What is the name of the orbital with quantum numbers n=4 and l=1?
Answer
Knowing that the principal quantum number n is 4 and using the table above, we can conclude that it is 4p.
Example $9$
What is the name of the oribital(s) with quantum number n=3?
Answer
3s, 3p, and 3d. Because n=3, the possible values of l = 0, 1, 2, which indicates the shapes of each subshell.
Orbitals
The number of orbitals in a subshell is equivalent to the number of values the magnetic quantum number ml takes on. A helpful equation to determine the number of orbitals in a subshell is 2l +1. This equation will not give you the value of ml, but the number of possible values that ml can take on in a particular orbital. For example, if l=1 and ml can have values -1, 0, or +1, the value of 2l+1 will be three and there will be three different orbitals. The names of the orbitals are named after the subshells they are found in:
s orbitals p orbitals d orbitals f orbitals
l 0 1 2 3
ml 0 -1, 0, +1 -2, -1, 0, +1, +2 -3, -2, -1, 0, +1, +2, +3
Number of orbitals in designated subshell 1 3 5 7
In the figure below, we see examples of two orbitals: the p orbital (blue) and the s orbital (red). The red s orbital is a 1s orbital. To picture a 2s orbital, imagine a layer similar to a cross section of a jawbreaker around the circle. The layers are depicting the atoms angular nodes. To picture a 3s orbital, imagine another layer around the circle, and so on and so on. The p orbital is similar to the shape of a dumbbell, with its orientation within a subshell depending on ml. The shape and orientation of an orbital depends on l and ml.
To visualize and organize the first three quantum numbers, we can think of them as constituents of a house. In the following image, the roof represents the principal quantum number n, each level represents a subshell l, and each room represents the different orbitals ml in each subshell. The s orbital, because the value of ml can only be 0, can only exist in one plane. The p orbital, however, has three possible values of ml and so it has three possible orientations of the orbitals, shown by Px, Py, and Pz. The pattern continues, with the d orbital containing 5 possible orbital orientations, and f has 7:
Another helpful visual in looking at the possible orbitals and subshells with a set of quantum numbers would be the electron orbital diagram. (For more electron orbital diagrams, see Electron Configurations.) The characteristics of each quantum number are depicted in different areas of this diagram.
Restrictions
• Pauli Exclusion Principle: In 1926, Wolfgang Pauli discovered that a set of quantum numbers is specific to a certain electron. That is, no two electrons can have the same values for n, l, ml, and ms. Although the first three quantum numbers identify a specific orbital and may have the same values, the fourth is significant and must have opposite spins.
• Hund's Rule: Orbitals may have identical energy levels when they are of the same principal shell. These orbitals are called degenerate, or "equal energy." According to Hund's Rule, electrons fill orbitals one at a time. This means that when drawing electron configurations using the model with the arrows, you must fill each shell with one electron each before starting to pair them up. Remember that the charge of an electron is negative and electrons repel each other. Electrons will try to create distance between it and other electrons by staying unpaired. This further explains why the spins of electrons in an orbital are opposite (i.e. +1/2 and -1/2).
• Heisenberg Uncertainty Principle: According to the Heisenberg Uncertainty Principle, we cannot precisely measure the momentum and position of an electron at the same time. As the momentum of the electron is more and more certain, the position of the electron becomes less certain, and vice versa. This helps explain integral quantum numbers and why n=2.5 cannot exist as a principal quantum number. There must be an integral number of wavelengths (n) in order for an electron to maintain a standing wave. If there were to be partial waves, the whole and partial waves would cancel each other out and the particle would not move. If the particle was at rest, then its position and momentum would be certain. Because this is not so, n must have an integral value. It is not that the principal quantum number can only be measured in integral numbers, it is because the crest of one wave will overlap with the trough of another, and the wave will cancel out.
Problems
1. Suppose that all you know about a certain electron is that its principal quantum number is 3. What are the possible values for the other four quantum numbers?
2. Is it possible to have an electron with these quantum numbers: $n=2$, $l=1$, $m_l=3$, $m_s=1/2$? Why or why not?
3. Is it possible to have two electrons with the same $n$, $l$, and $m_l$?
4. How many subshells are in principal quantum level $n=3$?
5. What type of orbital is designated by quantum numbers $n=4$, $l=3$, and $m_l =0$?
Solutions
• When $n=3$, $l=0$, $m_l = 0$, and $m_s=+1/2 \text{ or } -1/2$
• $l=1$, $m_l = -1, 0, or +1$, and $m_s=+1/2 \text{ or } -1/2$
• $l=2$, $m_l = -2, -1, 0, 1, \text{ or }+2$, and $m_s=+1/2 \text{ or } -1/2$
1. No, it is not possible. $m_l=3$ is not in the range of $-l$ to $+l$. The value should be be either -1, 0, or +1.
2. Yes, it is possible to have two electrons with the same $n$, $l$, and $m_l$. The spin of one electron must be +1/2 while the spin of the other electron must be -1/2.
3. There are three subshells in principal quantum level $n=3$.
4. Since $l=3$ refers to the f subshell, the type of orbital represented is 4f (combination of the principal quantum number n and the name of the subshell).
Contributors and Attributions
• Anastasiya Kamenko, Tamara Enriquez (UCD), Mandy Lam (UCD)
• Dr. Craig Fisher (Japan Fine Ceramics Center)
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The solutions to Schrödinger's equation for atomic orbitals can be expressed in terms of spherical coordinates: $r$, $\theta$, and $\phi$. For a point $(r, \theta, \phi)$, the variable $r$ represents the distance from the center of the nucleus, $\theta$ represents the angle to the positive z-axis, and $\phi$ represents the angle to the positive x-axis in the xy-plane.
Separation of Variables
Because the atomic orbitals are described with a time-independent potential V, Schrödinger’s equation can be solved using the technique of separation of variables, so that any wavefunction has the form:
$\Psi(r,\theta,\phi) = R(r) Y(\theta,\phi)$
where $R(r)$ is the radial wavefunction and $Y(\theta,\phi)$ is the angular wavefunction:
$Y(\theta,\phi) = \Theta(\theta) \; \; \Phi(\phi)$
Each set of quantum numbers, ($n$, $l$, $m_l$), describes a different wave function. The radial wave function is only dependent on $n$ and $l$, while the angular wavefunction is only dependent on $l$ and $m_l$. So a particular orbital solution can be written as:
$\Psi_{n,l,m_l}(r,\theta,\phi) = {R}_{n,l}(r) Y_{l,m_l}(\theta,\phi)$
Where
$n = 1, 2, 3, …$
$l = 0, 1, …, n-1$
$m_l = -l, … , -2, -1, 0, +1, +2, …, l$
Nodes
A wave function node occurs at points where the wave function is zero and changes signs. The electron has zero probability of being located at a node.
Because of the separation of variables for an electron orbital, the wave function will be zero when any one of its component functions is zero. When $R(r)$ is zero, the node consists of a sphere. When $\Theta(\theta)$ is zero, the node consists of a cone with the z-axis as its axis and apex at the origin. In the special case $\Theta(\pi/2)$ = 0, the cone is flattened to be the x-y plane. When $\Phi(\phi)$ is zero, the node consists of a plane through the z-axis.
Bonding and sign of wave function
The shape and extent of an orbital only depends on the square of the magnitude of the wave function. However, when considering how bonding between atoms might take place, the signs of the wave functions are important. As a general rule a bond is stronger, i.e. it has lower energy, when the orbitals of the shared electrons have their wavefunctions match positive to positive and negative to negative. Another way of expressing this is that the bond is stronger when the wave functions constructively interfere with each other. When the orbitals overlap so that the wave functions match positive to negative, the bond will be weaker or may not form at all.
Radial wavefunctions
The radial wavefunctions are of the general form:
$R(r) = N \; p(r) \; e^{-kr}$
Where
• $N$ is a positive normalizing constant
• $p(r)$ is a polynomial in $r$
• $k$ is a positive constant
The exponential factor is always positive, so the nodes and sign of $R(r)$ depends on the behavior of $p(r)$. Because the exponential factor has a negative sign in the exponent, $R(r)$ will approach 0 as $r$ goes to infinity.
$\Psi^2$ quantifies the probability of the electron being at a particular point. The probability distribution, $P(r)$ is the probability that the electron will be at any point that is $r$ distance from the nucleus. For any type of orbital, since $\Psi_{n,0,0}$ is separable into radial and angular components that are each appropriately normalized, and a sphere of radius r has area proportional to $r^2$, we have:
$P(r) = r^2R^2(r)$
Angular wavefunctions
The angular wave function $Y(\theta,\phi)$does much to give an orbital its distinctive shape. $Y(\theta,\phi)$ is typically normalized so the the integral of $Y^2(\theta,\phi)$ over the unit sphere is equal to one. In this case, $Y^2(\theta,\phi)$ serves as a probability function. The probability function can be interpreted as the probability that the electron will be found on the ray emitting from the origin that is at angles $(\theta,\phi)$ from the axes. The probability function can also be interpreted as the probability distribution of the electron being at position $(\theta,\phi)$ on a sphere of radius r, given that it is r distance from the nucleus.
The angular wave functions for a hydrogen atom, $Y_{l,m_l}(\theta,\phi)$ are also the wavefunction solutions to Schrödinger’s equation for a rigid rotor consisting of two bodies, for example a diatomic molecule.
Hydrogen Atom
The simplest case to consider is the hydrogen atom, with one positively charged proton in the nucleus and just one negatively charged electron orbiting around the nucleus. It is important to understand the orbitals of hydrogen, not only because hydrogen is an important element, but also because they serve as building blocks for understanding the orbitals of other atoms.
s Orbitals
The hydrogen s orbitals correspond to $l=0$ and only allow $m_l = 0$. In this case, the solution for the angular wavefunction, $Y_{0,0}(\theta,\phi)$ is a constant. As a result, the $\Psi_{n,0,0}(r,\theta,\phi)$ wavefunctions only depend on $r$ and the s orbitals are all spherical in shape.
Because $\Psi_{n,0,0}$ depends only on r, the probability distribution function of the electron:
$\Psi^2_{n,0,0}(r,\theta,\phi) = \dfrac{1}{4\pi}R^2_{n,0}(r)$
Graphs of the three functions, $R(r)$ in green, $R^2(r)$ in purple and $P(r)$ in orange are given below for n = 1, 2, and 3. The graph of the functions have been variously scaled along the vertical axis to allow an easy comparison of their shapes and where they are zero, positive and negative. The vertical scales for different functions, either within or between diagrams, are not necessarily the same.
In addition, a cross-section contour diagram is given for each of the three orbitals. These contour diagrams indicate the physical shape and size of the orbitals and where the probabilities are concentrated. An electron will be in the most-likely-10% (purple) regions 10% of the time, and it will be in the most-likely-50% regions (including the most-likely-10% regions, dark blue and purple) 50% of the time. Nodes are shown in orange in the contour diagrams. In all of these contour diagrams, the x-axis is horizontal, the z-axis is vertical, and the y-axis comes out of the diagram. The actual 3-dimensional orbital shape is obtained by rotating the 2-dimensional cross-section about the axis of symmetry, which is shown as a blue dashed line. The contour diagrams also indicate for regions that are separated by nodes, whether the wave function is positive (+) or negative (-) in that region. In order for the wave function to change sign, one must cross a node.
From these diagrams, we see that the 1s orbital does not have any nodes, the 2s orbital has one node, and the 3s orbital has 2 nodes.
Because for the s orbitals, $\Psi^2 = R^2(r)$, it is interesting to compare the $R^2(r)$ graphs and the $P(r)$ graphs. By comparing maximum values, in the 1s orbital, the $R^2(r)$ graph shows that the most likely place for the electron is at the nucleus, but the $P(r)$ graph shows that the most likely radius for the electron is at $a_0$, the Bohr radius. Similarly, for the other s orbitals, the one place the electron is most likely to be is at the nucleus, but the most likely radius for the electron to be at is outside the outermost node. Something that is not readily apparent from these diagrams is that the average radius for the 1s, 2s, and 3s orbitals is 1.5 a0, 6 a0, and 13.5 a0, forming ratios of 1:4:9. In other words, the average radius is proportional to $n^2$.
p orbitals
The hydrogen p orbitals correspond to l = 1 when n ? 2 and allow ml = -1, 0, or +1. The diagrams below describe the wave function for ml = 0. The angular wave function $Y_{1,0}(\theta,\phi) = cos\;\theta$ only depends on $\theta$. Below, the angular wavefunction shown with a node at $\theta = \pi/2$.
The radial wavefunctions and orbital contour diagrams for the p orbitals with n = 2 and 3 are:
As in the case of the s orbitals, the actual 3-dimensional p orbital shape is obtained by rotating the 2-dimensional cross-sections about the axis of symmetry, which is shown as a blue dashed line.
The p orbitals display their distinctive dumbbell shape. The angular wave function creates a nodal plane (the horizontal line in the cross-section diagram) in the x-y plane. In addition, the 3p radial wavefunction creates a spherical node (the circular node in the cross-section diagram) at r = 6 a0. For $m_l = 0$, the axis of symmetry is along the z axis.
The wavefunctions for ml = +1 and -1 can be represented in different ways. For ease of computation, they are often represented as real-valued functions. In this case, the orbitals have the same shape and size as $m_l = 0$, except that they are oriented in a different direction: the axis of symmetry is along the x axis with the nodal plane in the y-z plane or the axis of symmetry is along the y-axis with the nodal plane in the x-z plane. These correspond to wavefunctions that are the sum and the difference of the two ml = +1 and -1 wavefunctions
$\psi_{x}= \psi_{m_j=+1} + \psi_{m_j=-1} \notag$
$\psi_{y}= \psi_{m_j=+1} - \psi_{m_j=-1} \notag$
the $\psi_{z}$ wavefunction has a magnetic quantum number of ml =0, but the $\psi_{x}$ and $\psi_{y}$ are mixtures of the wavefunctions corresponding to ml = +1 and -1 and do not have unique magnetic quantum numbers.
d Orbitals
The hydrogen d orbitals correspond to l = 2 when n = 3 and allow ml = -2, -1, 0, +1, or +2. There are two basic shapes of d orbitals, depending on the form of the angular wave function.
The first shape of a d orbital corresponds to ml = 0. In this case, $Y_{2,0}(\theta,\phi)$ only depends on $\theta$. The graphs of the angular wavefunction, and for $n = 3$, the radial wave function and orbital contour diagram are as follows:
As in the case of the s and p orbitals, the actual 3-dimensional d orbital shape is obtained by rotating the 2-dimensional cross-section about the axis of symmetry, which is shown as a blue dashed line.
This first d orbital shape displays a dumbbell shape along the z axis, but it is surrounded in the middle by a doughnut (corresponding to the regions where the wavefunction is negative). The angular wave function creates nodes which are cones that open at about 54.7 degrees to the z-axis. At n=3, the radial wave function does not have any nodes.
The second d orbital shape is illustrated for ml = +1 and n = 3. In this case, $Y_{2,1}(\theta,\phi)$ depends on both $\theta$ and $\phi$, and can be shown as a surface curving over and under a rectangular domain. As a result, separate diagrams are shown for $Y_{2,1}(\theta,\phi)$ on the left and $Y^2_{2,1}(\theta,\phi)$ on the right.
Unlike previous orbital diagrams, this contour diagram indicates more than one axis of symmetry. Each axis of symmetry is at 45 degrees to the x- and z-axis. Each axis of symmetry only applies to the region surrounding it and bounded by nodes. Each of the four arms of the contour is rotated about its axis of symmetry to produce the 3-dimensional shape. However, the rotation is a non-standard rotation, producing only radial symmetry about the axis, not circular symmetry as was the case with other orbitals. This produces a double dumbell shape, with nodes in the x-y plane and the y-z plane.
Similar to the p orbitals, the wavefunctions for ml=+2, -1, and -2 can be represented as real-valued functions that have the same shape as for ml=+1, just oriented in different directions. In two cases, the shape is re-oriented so the the axes of symmetry are in the x-y plane or in the z-y plane. In both of those cases, the axes of symmetry are at 45 degrees to their respective coordinate axes, just as with ml=+1. For the third and final case, the orbital shape is re-oriented so the axes of symmetry are in the x-y plane, but also laying along the x and y axes.
It is often the case that the orbitals in the d subshell corresponding to the magnetic quantum numbers ml = ±1 and ml = ±2 are, as for the $\psi_{x}$ and $\psi_{y}$ orbitals, represented as sums and differences of the wavefunctions corresponding to ml = ±1 and ml = ±2. This, as for the p orbitals, better represents the spatial orientation of bonds formed with these orbitals.
The orbitals dxz and dyz are sums and differences of the two orbitals with ml = ±1 and lie in the xz and yz planes. ml = ±2 similarly corresponds to dxy and dx2 − y2; both lie in the xy plane. ml = 0 is the dz2 orbital, which is oriented along the z-axis.
Hydrogenic Orbitals
Hydrogenic atoms are atoms that only have one electron orbiting around the nucleus, even though the nucleus may have more than one proton and one or more neutrons. In this case, the electron has the same orbitals as the hydrogen atom, except that they are scaled by a factor of 1/Z. Z is the atomic number of the atom, the number of protons in the nucleus. The increased number of positively charged protons shrinks the size of the orbitals.
Thus, the same graphs for hydrogen above apply to hydrogenic atoms, except that instead of expressing the radius in units of a0, the radius is expressed in units of a0/Z. Correspondingly, the values have to be renormalized by a factor of (Z/a0)3/2. So a He+ atom has orbitals that are the same shape but half the size of the corresponding hydrogen orbitals and a Li2+ atom has orbitals that are the same shape but one third the size of the corresponding hydrogen orbitals.
• Thanh Hua
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Thumbnail: Vector cones" of total angular momentum J (purple), orbital L (blue), and spin S(green). The cones arise due to quantum uncertainty between measuring angular momentum components (see vector model of the atom. Image is used with permission (Public Domain; Maschen).
10: The Chemical Bond
The Hydrogen Molecule
This four-particle system, two nuclei plus two electrons, is described by the Hamiltonian
$\hat{H} = -\frac{1}{2} \nabla^2_1 -\frac{1}{2} \nabla^2_2 -\frac{1}{2M_A} \nabla^2_A -\frac{1}{2M_B} \nabla^2_B -\frac{1}{r_{1A}} -\frac{1}{r_{2B}} -\frac{1}{r_{2A}} -\frac{1}{r_{1B}} +\frac{1}{r_{12}} +\frac{1}{R} \label{1}$
in terms of the coordinates shown in Figure $1$. We note first that
the masses of the nuclei are much greater than those of the electrons,Mproton = 1836 atomic units, compared to melectron = 1 atomic unit. Therefore nuclear kinetic energies will be negligibly small compared to those of the electrons. In accordance with the Born-Oppenheimer approximation, we can first consider the electronic Schrödinger equation
$\hat{H}_{elec} \psi(r_1,r_2,R) = E_{elec}(R) \psi(r_1,r_2,R) \label{2}$
where
$\hat{H} = -\frac{1}{2} \nabla^2_1 -\frac{1}{2} \nabla^2_2 -\frac{1}{r_{1A}} -\frac{1}{r_{2B}} -\frac{1}{r_{2A}} -\frac{1}{r_{1B}} +\frac{1}{r_{12}} +\frac{1}{R} \label{3}$
The internuclear separation R occurs as a parameter in this equation so that the Schrödinger equation must, in concept, be solved for each value of the internuclear distance R. A typical result for the energy of a diatomic molecule as a function of R is shown in Figure $2$. For a bound state, the energy minimum occurs at for R = Re, known as the equilibrium internuclear distance. The depth of the potential well at Re is called the binding energy or dissociation energy De. For the H2 molecule, De = 4.746 eV and Re=1.400 bohr = 0.7406 Å. Note that as R → 0, E(R) → $\infty$, since the 1/R nuclear repulsion will become dominant.
The more massive nuclei move much more slowly than the electrons. From the viewpoint of the nuclei, the electrons adjust almost instantaneously to any changes in the internuclear distance. The electronic energy Eelec(R) therefore plays the role of a potential energy in the Schrödinger equation for nuclear motion
$\left\{ -\frac{1}{2M_A} \nabla^2_A -\frac{1}{2M_B} \nabla^2_B + V(R)\right\} \chi (r_A,r_B) = E \chi (r_A,r_B) \label{4}$
where
$V(R) = E_{elec}(R) \label{5}$
from solution of Equation $\ref{2}$. Solutions of Equation $\ref{4}$ determine the vibrational and rotational energies of the molecule. These will be considered elsewhere. For the present, we are interested in the obtaining electronic energy from Equation $\ref{2}$ and $\ref{3}$. We will thus drop the subscript "elec" on $\hat{H}$ and E(R) for the remainder this Chapter.
The first quantum-mechanical account of chemical bonding is due to Heitler and London in 1927, only one year after the Schrödinger equation was proposed. They reasoned that, since the hydrogen molecule H2 was formed from a combination of hydrogen atoms A and B, a first approximation to its electronic wavefunction might be
$\psi(r_1,r_2) = \psi_{1s} (r_{1A})\psi_{1s} (r_{2B}) \label{6}$
Using this function into the variational integral
$\tilde{E}(R) = \frac{\int{ \psi \hat{H} \psi d\tau}}{\int{\psi^2 d\tau}} \label{7}$
the value Re $\approx$ 1.7 bohr was obtained, indicating that the hydrogen atoms can indeed form a molecule. However, the calculated binding energy De $\approx$ 0.25 eV, is much too small to account for the strongly-bound H2 molecule. Heitler and London proposed that it was necessary to take into account the exchange of electrons, in which the electron labels in Equation $\ref{6}$ are reversed. The properly symmetrized function
$\psi(r_1, r_2) = \psi_{1s} (r_{1A})\psi_{1s} (r_{2B}) +\psi_{1s} (r_{1B})\psi_{1s} (r_{2A}) \label{8}$
gave a much more realistic binding energy value of 3.20 eV, with Re = 1.51 bohr. We have already used exchange symmetry (and antisymmetry) in our treatment of the excited states of helium. The variational function (Equation $\ref{8}$) was improved (Wang, 1928) by replacing the hydrogen 1s functions $e^{-r}$ by $e^{-\zeta r}$. The optimized value $\zeta$ = 1.166 gave a binding energy of 3.782 eV. The quantitative breakthrough was the computation of James and Coolidge (1933). Using a 13-parameter function of the form
$\psi(r_1, r_2) = e^{- \alpha ( \xi_{1}+\xi_{2})} \mbox{ x polynomial in} \{ \xi_{1}, \xi_{2}, \eta_{1}, \eta_{2}, \rho \} , \xi_{i} \equiv \frac{r_{iA} + r_{iB}}{R}, \eta_{i} \equiv \frac{r_{iA}+r_{iB}}{R}, \rho \equiv \frac{r_{12}}{R} \label{9}$
they obtained Re = 1.40 bohr, De = 4.720 eV. In a sense, this result provided a proof of the validity of quantum mechanics for molecules, in the same sense that Hylleraas' computation on helium was a proof for many-electron atoms.
The Valence Bond Theory
The basic idea of the Heitler-London model for the hydrogen molecule can be extended to chemical bonds between any two atoms. The orbital function (8) must be associated with the singlet spin function $\sigma_{0,0}(1,2)$ in order that the overall wavefunction be antisymmetric. This is a quantum-mechanical realization of the concept of an electron-pair bond, first proposed by G. N. Lewis in 1916. It is also now explained why the electron spins must be paired, i.e., antiparallel. It is also permissible to combine an antisymmetric orbital function with a triplet spin function but this will, in most cases, give a repulsive state, as shown by the red curve in Figure $2$.
According to valence-bond theory, unpaired orbitals in the valence shells of two adjoining atoms can combine to form a chemical bond if they overlap significantly and are symmetry compatible. A $\sigma$-bond is cylindrically symmetrical about the axis joining the atoms. Two s AO's, two pz AO's or an s and a pz can contribute to a $\sigma$-bond, as shown in Figure $3$. The z-axis is chosen along the internuclear axis. Two px or two py AO's can form a $\pi$-bond, which has a nodal plane containing the internuclear axis. Examples of symmetry-incompatible AO's would be an s with a px or a px with a py. In such cases the overlap integral would vanish because of cancelation of positive and negative contributions. Some possible combinations of AO's forming $\sigma$ and $\pi$ bonds are shown in Figure $3$.
Bonding in the HCl molecule can be attributed to a combination of a hydrogen 1s with an unpaired 3pz on chlorine. In Cl2, a sigma bond is formed between the 3pz AO's on each chlorine. As a first approximation, the other doubly-occupied AO's on chlorine-the inner shells and the valence-shell lone pairs-are left undisturbed.
The oxygen atom has two unpaired 2p-electrons, say 2px and 2py. Each of these can form a $\sigma$-bond with a hydrogen 1s to make a water molecule. It would appear from the geometry of the p-orbitals that the HOH bond angle would be 90°. It is actually around 104.5°. We will resolve this discrepency shortly. The nitrogen atom, with three unpaired 2p electrons can form three bonds. In NH3, each 2p-orbital forms a $\sigma$-bond with a hydrogen 1s. Again 90° HNH bond angles are predicted, compared with the experimental 107°. The diatomic nitrogen molecule has a triple bond between the two atoms, one $\sigma$ bond from combining 2pz AO's and two $\pi$ bonds from the combinations of 2px's and 2py's, respectively.
Hybrid Orbitals and Molecular Geometry
To understand the bonding of carbon atoms, we must introduce additional elaborations of valence-bond theory. We can write the valence shell configuration of carbon atom as 2s22px2py, signifying that two of the 2p orbitals are unpaired. It might appear that carbon would be divalent, and indeed the species CH2 (carbene or methylene radical) does have a transient existence. But the chemistry of carbon is dominated by tetravalence. Evidently it is a good investment for the atom to promote one of the 2s electrons to the unoccupied 2pz orbital. The gain in stability attained by formation of four bonds more than compensates for the small excitation energy. It can thus be understood why the methane molecule CH4 exists. The molecule has the shape of a regular tetrahedron, which is the result of hybridization, mixing of the s and three p orbitals to form four sp3 hybrid atomic orbitals. Hybrid orbitals can overlap more strongly with neighboring atoms, thus producing stronger bonds. The result is four C-H $\sigma$-bonds, identical except for orientation in space, with 109.5° H-C-H bond angles.
Other carbon compounds make use of two alternative hybridization schemes. The s AO can form hybrids with two of the p AO's to give three sp2 hybrid orbitals, with one p-orbital remaining unhybridized. This accounts for the structure of ethylene (ethene):
The C-H and C-C $\sigma$-bonds are all trigonal sp2 hybrids, with 120° bond angles. The two unhybridized p-orbitals form a $\pi$-bond, which gives the molecule its rigid planar structure. The two carbon atoms are connected by a double bond, consisting of one $\sigma$ and one $\pi$. The third canonical form of sp-hybridization occurs in C-C triple bonds, for example, acetylene (ethyne). Here, two of the p AO's in carbon remain unhybridized and can form two $\pi$-bonds, in addition to a $\sigma$-bond, with a neighboring carbon:
Acetylene H-C$\equiv$C-H is a linear molecule since sp-hybrids are oriented 180° apart.
The deviations of the bond angles in H2O and NH3 from 90° can be attributed to fractional hybridization. The angle H-O-H in water is 104.5° while H-N-H in ammonia is 107°. It is rationalized that the p-orbitals of the central atom acquire some s-character and increase their angles towards the tetrahedral value of 109.5°. Correspondingly, the lone pair orbitals must also become hybrids. Apparently, for both water and ammonia, a model based on tetrahedral orbitals on the central atoms would be closer to the actual behavior than the original selection of s- and p-orbitals. The hybridization is driven by repulsions between the electron densities of neighboring bonds.
Valence Shell Model
An elementary, but quite successful, model for determining the shapes of molecules is the valence shell electron repulsion theory (VSEPR), first proposed by Sidgewick and Powell and popularized by Gillespie. The local arrangement of atoms around each multivalent center in the molecule can be represented by AXn-kEk, where X is another atom and E is a lone pair of electrons. The geometry around the central atom is then determined by the arrangement of the n electron pairs (bonding plus nonbonding), which minimizes their mutual repulsion. The following geometric configurations satisfy this condition:
n shape
2 linear 5 trigonal bipyramid
3 trigonal planar 6 octahedral
4 tetrahedral 7 pentagonal bipyramid
The basic geometry will be distorted if the n surrounding pairs are not identical. The relative strength of repulsion between pairs follows the order E-E > E-X > X-X. In ammonia, for example, which is NH3E, the shape will be tetrahedral to a first approximation. But the lone pair E will repel the N-H bonds more than they repel one another. Thus the E-N-H angle will increase from the tetrahedral value of 109.5°, causing the H-N-H angles to decrease slightly. The observed value of 107° is quite well accounted for. In water, OH2E2, the opening of the E-O-E angle will likewise cause a closing of H-O-H, and again, 104.5° seems like a reasonable value.
Valence-bond theory is about 90% successful in explaining much of the descriptive chemistry of ground states. VB theory fails to account for the triplet ground state of O2 or for the bonding in electron-deffcient molecules such as diborane, B2H6. It is not very useful in consideration of excited states, hence for spectroscopy. Many of these deficiencies are remedied by molecular orbital theory, which we take up in the next Chapter.
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In many cases, the symmetry of a molecule provides a great deal of information about its quantum states, even without a detailed solution of the Schrödinger equation. A geometrical transformation which turns a molecule into an indistinguishable copy of itself is called a symmetry operation. A symmetry operation can consist of a rotation about an axis, a reflection in a plane, an inversion through a point, or some combination of these.
The Ammonia Molecule
We shall introduce the concepts of symmetry and group theory by considering a concrete example–the ammonia molecule NH3. In any symmetry operation on NH3, the nitrogen atom remains fixed but the hydrogen atoms can be permuted in 3!=6 different ways. The axis of the molecule is called a C3 axis, since the molecule can be rotated about it into 3 equivalent orientations, $120^\circ$ apart. More generally, a Cn axis has n equivalent orientations, separated by $2\pi/n$ radians. The axis of highest symmetry in a molecule is called the principal axis. Three mirror planes, designated $\sigma_1,\sigma_2,\sigma_3$, run through the principal axis in ammonia. These are designated as $\sigma_v$ or vertical planes of symmetry. Ammonia belongs to the symmetry group designated C3v, characterized by a three-fold axis with three vertical planes of symmetry.
Let us designate the orientation of the three hydrogen atoms in Figure $1$ as {1, 2, 3}, reading in clockwise order from the bottom. A counterclockwise rotation by 120$^\circ$, designated
by the operator C3, produces the orientation {2, 3, 1}. A second counterclockwise rotation, designated $C_3^2$, produces {3, 1, 2}. Note that two successive counterclockwise rotations by 120$^\circ$ is equivalent to one clockwise rotation by 120$^\circ$, so the last operation could also be designated $C_3^{-1}$. The three reflection operations $\sigma_1,\sigma_2,\sigma_3$, applied to the original configuration {1, 2, 3} produces {1, 3, 2}, {3, 2, 1} and {2, 1, 3}, respectively. Finally, we must include the identity operation, designated E, which leaves an orientation unchanged. The effects of the six possible operations of the symmetry group C3v can be summarized as follows:
$E\{1,2,3\}=\{1,2,3\} C_3\{1,2,3\}=\{2,3,1\}$
$C_3^2\{1,2,3\}=\{3,1,2\} \sigma_1\{1,2,3\}=\{1,3,2\}$
$\sigma_2\{1,2,3\}=\{3,2,1\} \sigma_3\{1,2,3\}=\{2,1,3\}$
We have thus accounted for all 6 possible permutations of the three hydrogen atoms.
The successive application of two symmetry operations is equivalent to some single symmetry operation. For example, applying C3, then $\sigma_1$ to our starting orientation, we have
$\sigma_1 C_3\{1,2,3\}=\sigma_1\{2,3,1\}=\{2,1,3\}$
But this is equivalent to the single operation $\sigma_3$. This can be represented as an algebraic relation among symmetry operators
$\sigma_1 C_3=\sigma_3$
Note that successive operations are applied in the order right to left when represented algebraically. For the same two operations in reversed order, we find
$C_3 \sigma_1 \{1,2,3\} = C_3 \{1,3,2\} = \{3,2,1\} = \sigma_2 \{1,2,3\}$
Thus symmetry operations do not, in general commute
$A B \not\equiv B A \label{1}$
although they may commute, for example, $C_3$ and $C_3^2$.
The algebra of the group $C_{3v}$ can be summarized by the following multiplication table.
$\begin{matrix} & 1^{st} & E & C_3 & C_3^2 & \sigma_1 &\sigma_2 &\sigma_3 \ 2^{nd} & & & & & & & \ E & &E &C_3 &C_3^2 &\sigma_1 &\sigma_2 &\sigma_3 \ C_3& &C_3 &C_3^2 &E &\sigma_3 &\sigma_1 &\sigma_2 \ C_3^2& & C_3^2 &E &C_3 &\sigma_2 &\sigma_3 &\sigma_1 \ \sigma_1& &\sigma_1 &\sigma_2 &\sigma_3 &E &C_3 &C_3^2 \ \sigma_2 & &\sigma_2 &\sigma_3 &\sigma_1 &C_3^2 &E &C_3 \ \sigma_3 & &\sigma_3 &\sigma_1 &\sigma_2 &C_3 &C_3^2 &E \end{matrix}$
Notice that each operation occurs once and only once in each row and each column.
Group Theory
In mathematics, a group is defined as a set of g elements $\mathcal{G} \equiv \{G_1,G_2...G_h\}$ together with a rule for combination of elements, which we usually refer to as a product. The elements must fulfill the following four conditions.
1. The product of any two elements of the group is another element of the group. That is $G_iG_j=G_k$ with $G_k\in\mathcal{G}$
2. Group multiplication obeys an associative law, $G_i(G_jG_k)=(G_iG_j)G_k\equiv G_iG_jG_k$
3. There exists an identity element E such that $EG_i=G_iE=G_i$ for all i.
4. Every element $G_i$ has a unique inverse $G_i^{-1}$, such that $G_iG_i^{-1}=G_i^{-1}G_i=E$ with $G_i^{-1}\in\mathcal{G}$.
The number of elements h is called the order of the group. Thus $C_{3v}$ is a group of order 6.
A set of quantities which obeys the group multiplication table is called a representation of the group. Because of the possible noncommutativity of group elements [cf. Eq (1)], simple numbers are not always adequate to represent groups; we must often use matrices. The group $C_{3v}$ has three irreducible representations, or IR’s, which cannot be broken down into simpler representations. A trivial, but nonetheless important, representation of any group is the totally symmetric representation, in which each group element is represented by 1. The multiplication table then simply reiterates that $1\times 1=1$. For $C_{3v}$ this is called the $A_1$ representation:
$A_1: E=1,C_3=1,C_3^2=1,\sigma_1=1,\sigma_2=1,\sigma_3=1 \label{2}$
A slightly less trivial representation is $A_2$:
$A_2: E=1,C_3=1,C_3^2=1,\sigma_1=-1,\sigma_2=-1,\sigma_3=-1 \label{3}$
Much more exciting is the E representation, which requires $2\times 2$ matrices:
$E= \begin{pmatrix} 1 &0 \0 &1 \end{pmatrix} \qquad C_3=\begin{pmatrix} -1/2 &-\sqrt{3}/2 \ \sqrt{3}/2 &-1/2 \end{pmatrix} \ C_3^2=\begin{pmatrix} -1/2 &\sqrt{3}/2 \ -\sqrt{3}/2 &-1/2 \end{pmatrix} \qquad \sigma_1=\begin{pmatrix} -1 &0 \0 &1 \end{pmatrix} \ \sigma_2=\begin{pmatrix} 1/2 &-\sqrt{3}/2 \ -\sqrt{3}/2 &-1/2 \end{pmatrix} \qquad \sigma_3=\begin{pmatrix} 1/2 &\sqrt{3}/2 \ \sqrt{3}/2 &-1/2 \end{pmatrix} \label{4}$
The operations $C_3$ and $C_3^2$ are said to belong to the same class since they perform the same geometric function, but for different orientations in space. Analogously, $\sigma_1, \sigma_2$ and $\sigma_3$ are obviously in the same class. E is in a class by itself. The class structure of the group is designated by $\{E,2C_3,3\sigma_v\}$. We state without proof that the number of irreducible representations of a group is equal to the number of classes. Another important theorem states that the sum of the squares of the dimensionalities of the irreducible representations of a group adds up to the order of the group. Thus, for $C_{3v}$, we find $1^2+1^2+2^2=6$.
The trace or character of a matrix is defined as the sum of the elements along the main diagonal:
$\chi(M)\equiv\sum_kM_{kk} \label{5}$
For many purposes, it suffices to know just the characters of a matrix representation of a group, rather than the complete matrices. For example, the characters for the E representation of $C_{3v}$ in Eq (4) are given by
$\chi(E)=2,\quad \chi(C_3)=-1, \quad \chi(C_3^2)=-1, \ \chi(\sigma_1)=0, \quad \chi(\sigma_2)=0, \quad \chi(\sigma_3)=0 \label{6}$
It is true in general that the characters for all operations in the same class are equal. Thus Eq (6) can be abbreviated to
$\chi(E)=2,\quad \chi(C_3)=-1, \quad \chi(\sigma_v)=0 \label{7}$
For one-dimensional representations, such as $A_1$ and $A_2$, the characters are equal to the matrices themselves, so Equations $\ref{2}$ and $\ref{3}$ can be read as a table of characters.
The essential information about a symmetry group is summarized in its character table. We display here the character table for $C_{3v}$
$\begin{matrix} C_{3v} &E &2C_3 &3\sigma_v & & \\hline A_1 &1 &1 &1 &z &z^2,x^2+y^2 \A_2 &1 &1 &-1 & & \E &2 &-1 &0 &(x,y) &(xy,x^2-y^2),(xz,yz) \end{matrix}$
The last two columns show how the cartesian coordinates x, y, z and their products transform under the operations of the group.
Group Theory and Quantum Mechanics
When a molecule has the symmetry of a group $\mathcal{G}$, this means that each member of the group commutes with the molecular Hamiltonian
$[\hat G_i,\hat H]=0 \quad i=1...h \label{8}$
where we now explicitly designate the group elements $G_i$ as operators on wavefunctions. As was shown in Chap. 4, commuting operators can have simultaneous eigenfunctions. A representation of the group of dimension d means that there must exist a set of d degenerate eigenfunctions of $\hat H$ that transform among themselves in accord with the corresponding matrix representation. For example, if the eigenvalue $E_n$ is d-fold degenerate, the commutation conditions (Equation $\ref{2}$) imply that, for $i=1...h$,
$\hat G_i \hat H \psi_{nk} = \hat H \hat G_i \psi_{nk}=E_n \hat G_i \psi_{nk} \; \text{for} \;k=1...d \label{9}$
Thus each $\hat G_i \psi_{nk}$ is also an eigenfunction of $\hat H$ with the same eigenvalue $E_n$, and must therefore be represented as a linear combination of the eigenfunctions $\psi_{nk}$. More precisely, the eigenfunctions transform among themselves according to
$\hat G_i \psi_{nk}=\sum_{m=1}^d D(G_i)_{km}\psi_{nm} \label{10}$
where $D(G_i)_{km}$ means the $\{k,m\}$ element of the matrix representing the operator $\hat G_i$.
The character of the identity operation E immediately shows the degeneracy of the eigenvalues of that symmetry. The $C_{3v}$ character table reveals that $NH_3$, and other molecules of the same symmetry, can have only nondegenerate and two-fold degenerate energy levels. The following notation for symmetry species was introduced by Mulliken:
1. One dimensional representations are designated either A or B. Those symmetric wrt rotation by $2\pi/n$ about the $C_n$ principal axis are labeled A, while those antisymmetric are labeled B.
2. Two dimensional representations are designated E; 3, 4 and 5 dimensional representations are designated T, F and G, respectively. These latter cases occur only in groups of high symmetry: cubic, octahedral and icosohedral.
3. In groups with a center of inversion, the subscripts g and u indicate even and odd parity, respectively.
4. Subscripts 1 and 2 indicate symmetry and antisymmetry, respectively, wrt a $C_2$ axis perpendicular to $C_n$, or to a $\sigma_v$ plane.
5. Primes and double primes indicate symmetry and antisymmetry to a $\sigma_h$ plane.
For individual orbitals, the lower case analogs of the symmetry designations are used. For example, MO’s in ammonia are classified $a_1,a_2$ or e.
For ammonia and other $C_{3v}$ molecules, there exist three species of eigenfunctions. Those belonging to the classification $A_1$ are transformed into themselves by all symmetry operations of the group. The 1s, 2s and $2p_z$ AO’s on nitrogen are in this category. The z-axis is taken as the 3-fold axis. There are no low-lying orbitals belonging to $A_2$. The nitrogen $2p_x$ and $2p_y$ AO’s form a two-dimensional representation of the group $C_{3v}$. That is to say, any of the six operations of the group transforms either one of these AO’s into a linear combination of the two, with coefficients given by the matrices (4). The three hydrogen 1s orbitals transform like a $3\times 3$ representation of the group. If we represent the hydrogens by a column vector {H1,H2,H3}, then the six group operations generate the following algebra
$\begin{matrix} E=\begin{pmatrix} 1 &0 &0 \0 &1 &0 \0 &0 &1 \end{pmatrix} & C_3=\begin{pmatrix} 0 &1 &0 \0 &0 &1 \1 &0 &0 \end{pmatrix} \ C_3^2=\begin{pmatrix} 0 &0 &1 \1 &0 &0 \0 &1 &0 \end{pmatrix} & \sigma_1=\begin{pmatrix} 1 &0 &0 \0 &0 &1 \0 &1 &0 \end{pmatrix} \ \sigma_2=\begin{pmatrix} 0&0 &1 \0 &1 &0 \1 &0 &0 \end{pmatrix} & \sigma_3=\begin{pmatrix} 0&1 &0 \1 &0 &0 \0 &0 &1 \end{pmatrix} \end{matrix} \label{11}$
Let us denote this representation by $\Gamma$. It can be shown that $\Gamma$ is a reducible representation, meaning that by some unitary transformation the $3 \times 3$ matrices can be factorized into blockdiagonal form with $2 \times 2$ plus $1 \times 1$ submatrices. The reducibility of $\Gamma$ can be deduced from the character table. The characters of the matrices (Equation $\ref{11}$) are
$\Gamma: \qquad \chi(E)=3, \quad \chi(C_3)=0, \quad \chi_(\sigma_v)=1 \label{12}$
The character of each of these permutation operations is equal to the number of H atoms left untouched: 3 for the identity, 1 for a reflection and 0 for a rotation. The characters of $\Gamma$ are seen to equal the sum of the characters of $A_1$ plus E. This reducibility relation is expressed by writing
$\Gamma=A_1\oplus E \label{13}$
The three H atom 1s functions can be combined into LCAO functions which transform according to the IR’s of the group. Clearly the sum
$\psi=\psi_{1s}(1)+\psi_{1s}(2)+\psi_{1s}(3) \label{14}$
transforms like $A_1$. The two remaining linear combinations which transform like E must be orthogonal to (Equation $\ref{14}$) and to one another. One possible choice is
$\psi'=\psi_{1s}(2)-\psi_{1s}(3), \quad \psi''=2\psi_{1s}(1)-\psi_{1s}(2)-\psi_{1s}(3) \label{15}$
Now, Equation $\ref{14}$ can be combined with the N 1s, 2s and $2p_z$ to form MO’s of $A_1$ symmetry, while Equation $\ref{15}$ can be combined with the N $2p_x$ and $2p_y$ to form MO’s of E symmetry. Note that no hybridization of AO’s is predetermined, it emerges automatically in the results of computation.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/11%3A_Molecules/11%3A_Molecular_Symmetry.txt
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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 $H_{2}^{+}$.
The Hydrogen Molecule-Ion
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. It also has been detected in outer space. The Schrödinger equation forH$H_{2}^{+}$ can be solved exactly within the Born-Oppenheimer approximation. For fixed internuclear distance R, this reduces to a problem of one electron in the field of two protons, designated A and B. We can write
$\left\{-\dfrac{1}{2}\nabla^2\dfrac{1}{r_A}-\dfrac{1}{r_B}+\dfrac{1}{R} \right\} \psi(r)=E\psi(r) \label{1}$
where rA and rB are the distances from the electron to protons A and B, respectively. This equation was solved by Burrau (1927), after separating the variables in prolate spheroidal coordinates. We will write down these coordinates but give only a pictorial account of the solutions. The three prolate spheroidal coordinates are designated $\xi$, $\eta$, $\phi$. the first two are defined by
$\xi=\dfrac{r_{A}+r_{B}}{R}$
and
$\eta=\dfrac{r_{A}-r_{B}}{R}\label{2}$
while $\phi$ is the angle of rotation about the internuclear axis. The surfaces of constant $\xi$ and $\eta$ are, respectively, confocal ellipsoids and hyperboloids of revolution with foci at A and B. The two-dimensional analog should be familiar from analytic geometry, an ellipse being the locus of points such that the sum of the distances to two foci is a constant. Analogously, a hyperbola is the locus whose difference is a constant. Figure $1$ shows several surfaces of constant $\xi$, $\eta$ and $\phi$. The ranges of the three coordinates are $\xi$ $\in$ $\{1,\infty\}$, $\eta$ $\in$ $\{-1,1\}$, $\phi$ $\in$ $\{0,2\pi\}$. The prolate-spheroidal coordinate system conforms to the natural symmetry of the $H_{2}^{+}$ problem in the same way that spherical polar coordinates were the appropriate choice for the hydrogen atom.
The first few solutions of the $H_2^+$ Schrödinger equation are sketched in Figure $2$, roughly in order of increasing energy. The $\phi$-dependence of the wavefunction is contained in a factor
$\Phi(\phi)=e^{i\lambda\phi},\ \ \ \ \ \lambda=0,\pm1,\pm2,\ldots\label{3}$
which is identical to the $\phi$-dependence in atomic orbitals. In fact, the quantum number $\lambda$ represents the component of orbital angular momentum along the internuclear axis, the only component which has a definite value in systems with axial (cylindrical) symmetry. The quantum number $\lambda$ determines the basic shape of a diatomic molecular orbital, in the same way that $\ell$ did for an atomic orbital. An analogous code is used $\sigma$ for $\lambda$ = 0, $\pi$ for $\lambda$ = $\pm$1, $\delta$ for $\lambda$ = $\pm$2, and so on. We are already familiar with $\sigma$- and $\pi$-orbitals from valence-bond theory. A second classification of the $H_{2}^{+}$ eigenfunctions pertains to their symmetry with respect to inversion through the center of the molecule, also known as parity. If $\psi$(-r) = +$\psi$(r), the function is classified gerade or even parity, and the orbital designation is given a subscript g, as in $\sigma_{g}$ or $\pi_{g}$. If $\psi$(-r) = -$\psi$(r), the function is classified as ungerade or odd parity, and we write instead $\sigma_{u}$ or $\pi_{u}$. Atomic orbitals can also be classified by inversion symmetry. However, all s and d atomic orbitals are g, while all p and f orbitals are u, so no further designation is necessary. The molecular orbitals of a given symmetry are numbered in order of increasing energy, for example, 1$\sigma_{g}$, 2$\sigma_{g}$, 3$\sigma_{g}$.
The lowest-energy orbital, as we have come to expect, is nodeless. It obviously must have cylindrical symmetry ($\lambda$ = 0) and inversion symmetry (g). It is designated 1$\sigma_{g}$ since it is the first orbital of this classification. The next higher orbital has a nodal plane, with $\eta$ = 0, perpendicular to the axis. This function still has cylindrical symmetry ($\sigma$) but now changes sign upon inversion (u). It is designated 1$\sigma_{u}$, as the first orbital of this type. The next higher orbital has an inner ellipsiodal node. It has the same symmetry as the lowest orbital and is designated 2$\sigma_{g}$. Next comes the 2$\sigma_{u}$ orbital, with both planar and ellipsoidal nodes. Two degenerate $\pi$-orbitals come next, each with a nodal plane containing the internuclear axis, with $\phi$=const. Their classification is 1$\pi_{u}$. The second 1$\pi_{u}$-orbital, not shown in Figure $2$, has the same shape rotated by 90°. The 3$\sigma_{g}$ orbital has two hyperbolic nodal surfaces, where $\eta$ = $\pm$const. The 1$\pi_{g}$, again doubly-degenerate, has two nodal planes, $\eta$ = 0 and $\phi$=const. Finally, the 3$\sigma_{u}$, the last orbital we consider, has three nodal surfaces where $\eta$=const.
An molecular orbital is classified as a bonding orbital if it promotes the bonding of the two atoms. Generally a bonding molecular orbital has a significant accumulation of electron charge in the region between the nuclei and thus reduces their mutual repulsion. The 1$\sigma_{g}$, 2$\sigma_{g}$, 1$\pi_{u}$ and 3$\sigma_{g}$ are evidently bonding orbitals. An molecular orbital which does not significantly contribute to nuclear shielding is classified as an antibonding orbital. The 1$\sigma_{u}$, 2$\sigma_{u}$, 1$\pi_{g}$ and 3$\sigma_{u}$ belong in this category. Often an antibonding molecular orbital is designated by $\sigma$* or $\pi$*.
The actual ground state of $H_{2}^{+}$ has the 1$\sigma_{g}$ orbital occupied. The equilibrium internuclear distance Re is 2.00 bohr and the binding energy De is 2.79 eV, which represents quite a strong chemical bond. The 1$\sigma_{u}$ is a repulsive state and a transition from the ground state results in dissociation of the molecule.
The LCAO Approximation
In Figure $3$, the 1$\sigma_{g}$ and 1$\sigma_{u}$ orbitals are plotted as functions of z, along the internuclear axis. Both functions have cusps, discontinuities in slope, at the positions of the two nuclei A and B. The 1s orbitals of hydrogen atoms have the same cusps. The shape of the 1$\sigma_{g}$ and 1$\sigma_{u}$ suggests that they can be approximated by a sum and difference, respectively, of hydrogen 1s orbitals, such that$\psi(1\sigma_{g,u})\approx\psi(1s_{A})\pm\psi(1s_{B})\label{4}$
This linear combination of atomic orbitals is the basis of the so-called LCAO approximation. The other orbitals pictured in Figure $2$ can likewise be approximated as follows:
$\psi(2\sigma_{g,u})\approx\psi(2s_{A})\pm\psi(2s_{B})$
$\psi(3\sigma_{g,u})\approx\psi(2p\sigma_{A})\pm\psi(2p\sigma_{B})$
$\psi(1\pi_{u,g})\approx\psi(2p\pi_{A})\pm\psi(2p\pi_{B})\label{5}$
The 2$p\sigma$ atomic orbital refers to 2pz, which has the axial symmetry of a $\sigma$-bond. Likewise 2$p\pi$ refers to 2px or 2py, which are positioned to form $\pi$-bonds. An alternative notation for diatomic molecular orbitals which specifies their atomic origin and bonding/antibonding character is the following:
1$\sigma_{g}$ 1$\sigma_{u}$ 2$\sigma_{g}$ 2$\sigma_{u}$ 3$\sigma_{g}$ 3$\sigma_{u}$ 1$\pi_{u}$ 1$\pi_{g}$
$\sigma$1s $\sigma$*1s $\sigma$2s $\sigma$*2s $\sigma$2p $\sigma$*2p $\pi$2p $\pi$*2p
Almost all applications of molecular-orbital theory are based on the LCAO approach, since the exact $H_{2}^{+}$ functions are far too complicated to work with.
The relationship between molecular orbitals and their constituent atomic orbitals can be represented in correlation diagrams, show in Figure $4$.
MO Theory of Homonuclear Diatomic Molecules
A sufficient number of orbitals is available for the Aufbau of the ground states of all homonuclear diatomic species from H2 to Ne2. Table 1 summarizes the results. The most likely order in which the molecular orbitals are filled is given by
$1\sigma_{g}<1\sigma_{u}<2\sigma_{g}<2\sigma_{u}<3\sigma_{g}\sim1\pi_{u}<1\pi_{g}<3\sigma_{u}$
The relative order of 3$\sigma_{g}$ and 1$\pi_{u}$ depends on which other molecular orbitals are occupied, much like the situation involving the 4s and 3d atomic orbitals. The results of photoelectron spectroscopy indicate that 1$\pi_{u}$ is lower up to and including N2, but 3$\sigma_{g}$ is lower thereafter.
The term symbol $\Sigma,\Pi,\Delta\ldots$, analogous to the atomic S, P, D. . . symbolizes the axial component of the total orbital angular momentum. When a $\pi$-shell is filled (4 electrons) or half-filled (2 electrons), the orbital angular momentum cancels to zero and we find a $\Sigma$ term. The spin multiplicity is completely analogous to the atomic case. The total parity is again designated by a subscript g or u. Since the many electron wavefunction is made up of products of individual MO's, the total parity is odd only if the molecule contains an odd number of u orbitals. Thus a $\sigma_{u}^{2}$ or a $\pi_{u}^{2}$ subshell transforms like g.
For $\Sigma$ terms, the superscript $\pm$ denotes the sign change of the wavefunction under a reflection in a plane containing the internuclear axis. This is equivalent to a sign change in the variable $\phi\rightarrow-\phi$. This symmetry is needed when we deal with spectroscopic selection rules. In a spin-paired $\pi_{u}^{2}$ subshell the triplet spin function is symmetric so that the orbital factor must be antisymmetric, of the form
$\dfrac{1}{\sqrt{2}} \biggl( \pi_{x}(1)\pi_{y}(2)-\pi_{y}(1)\pi_{x}(2) \biggr) \label{6}$
This will change sign under the reflection, since $x\rightarrow{x}$ but $y\rightarrow{-y}$. We need only remember that a $\pi_{u}^{2}$ subshell will give the term symbol $^{3}\Sigma_{g}^{-}$.
The net bonding effect of the occupied molecular orbitals is determined by the bond order, half the excess of the number bonding minus the number antibonding. This definition brings the molecular orbital results into correspondence with the Lewis (or valence-bond) concept of single, double and triple bonds. It is also possible in molecular orbital theory to have a bond order of 1/2, for example, in $H_{2}^{+}$ which is held together by a single bonding orbital. A bond order of zero generally indicates no stable chemical bond, although helium and neon atoms can still form clusters held together by much weaker van der Waals forces. Molecular-orbital theory successfully accounts for the transient stability of a $^{3}\Sigma_{u}^{+}$ excited state of He2, in which one of the antibonding electrons is promoted to an excited bonding orbital. This species has a lifetime of about 10-4 sec, until it emits a photon and falls back into the unstable ground state. Another successful prediction of molecular orbital theory concerns the relative binding energy of the positive ions N$_{2}^{+}$ and O$_{2}^{+}$, compared to the neutral molecules. Ionization weakens the N–N bond since a bonding electron is lost, but it strengthens the O–O bond since an antibonding electron is lost.
One of the earliest triumphs of molecular orbital theory was the prediction that the oxygen molecule is paramagnetic. Figure $5$ shows that liquid O2 is a magnetic substance, attracted to the region between the poles of a permanent magnet. The paramagnetism arises from the half-filled 1$\pi_{g}^{2}$ subshell. According to Hund's rules the two electrons singly occupy the two degenerate 1$\pi_{g}$ orbitals with their spins aligned parallel. The term symbol is $^{3}\Sigma_{g}^{-}$ and the molecule thus has a nonzero spin angular momentum and a net magnetic moment, which is attracted to an external magnetic field. Linus Pauling invented the paramagnetic oxygen analyzer, which is extensively used in medical technology.
Variational Computation of Molecular Orbitals
Thus far we have approached molecular orbital theory from a mainly descriptive point of view. To begin a more quantitative treatment, recall the LCAO approximation to the $H_{2}^{+}$ ground state, Equation $\ref{4}$, which can be written
$\psi=c_{A}\psi_{A}+c_{B}\psi_{B}\label{7}$
Using this as a trial function in the variational principle, we have
$E(c_{A},c_{B})=\dfrac{\int\psi\hat{H}\psi{d}\tau}{\int\psi^{2}d\tau}\label{8}$
where $\hat{H}$ is the Hamiltonian from Equation $\ref{1}$. In fact, these equations can be applied more generally to construct any molecular orbital, not just solutions for $H_{2}^{+}$. In the general case, $\hat{H}$ will represent an effective one-electron Hamiltonian determined by the molecular environment of a given orbital. The energy expression involves some complicated integrals, but can be simplified somewhat by expressing it in a standard form. Hamiltonian matrix elements are defined by
$H_{AA}=\int\psi_{A}\hat{H}\psi_{A}d\tau$H_{BB}=\int\psi_{B}\hat{H}\psi_{B}d\tau$H_{AB}=H_{BA}=\int\psi_{A}\hat{H}\psi_{B}d\tau\label{9}$
while the overlap integral is given by
$S_{AB}=\int\psi_{A}\psi_{B}d\tau\label{10}$
Presuming the functions $\psi_{A}$ and $\psi_{B}$ to be normalized, the variational energy (Equation $\ref{8}$) reduces to
$E(c_{A}c_{B})=\dfrac{c_{A}^{2}H_{AA}+2c_{A}c_{B}H_{AB}+c_{B}^{2}H_{BB}}{c_{A}^{2}+2c_{A}c_{B}S_{AB}+c_{B}^{2}}\label{11}$
To optimize the MO, we find the minimum of E wrt variation in cA and cB, as determined by the two conditions
$\dfrac{\partial{E}}{\partial{c_{A}}}=0,\ \ \ \ \ \dfrac{\partial{E}}{\partial{c_{B}}}=0\label{12}$
The result is a secular equation determining two values of the energy:
$\left|\begin{array}{ll}H_{AA}-E&H_{AB}-ES_{AB}\H_{AB}-ES_{AB}&H_{BB}-E\end{array}\right|=0\label{13}$
For the case of a homonuclear diatomic molecule, for example $H_{2}^{+}$, the two Hamiltonian matrix elements $H_{AA}$ and $H_{BB}$ are equal, say to $\alpha$. Setting $H_{AB}=\beta$ and $S_{AB}=S$, the secular equation reduces to
$\left|\begin{array}{ll}\alpha-E&\beta-ES\\beta-ES&\alpha-E\end{array}\right|=(\alpha-E)^{2}-(\beta-ES)^{2}=0\label{14}$
with the two roots
$E^{\pm}=\dfrac{\alpha\pm\beta}{1\pm{S}}\label{15}$
The calculated integrals $\alpha$ and $\beta$ are usually negative, thus for the bonding orbital
$E^{+}=\dfrac{\alpha+\beta}{1+S}\ \ \ \ \ (bonding)\label{16}$
while for the antibonding orbital
$E^{-}=\dfrac{\alpha-\beta}{1-S}\ \ \ \ \ (antibonding)\label{17}$
Note that $(E^{-}-\alpha)>(\alpha-E^{+})$, thus the energy increase associated with antibonding is slightly greater than the energy decrease for bonding. For historical reasons, $\alpha$ is called a Coulomb integral and $\beta$, a resonance integral.
Heteronuclear Molecules
The variational computation leading to Equation $\ref{13}$ can be applied as well to the heteronuclear case in which the orbitals $\psi_{A}$ and $\psi_{B}$ are not equivalent. The matrix elements $H_{AA}$ and $H_{BB}$ are approximately equal to the energies of the atomic orbitals $\psi_{A}$ and $\psi_{B}$, respectively, say $E_{A}$ and $E_{B}$ with $E_{A}>E_{B}$. It is generally true that $|E_{A}|,|E_{B}|\gg|H_{AB}|$. With these simplifications, secular equation can be written
$\left|\begin{array}{ll}E_{A}-E&H_{AB}-ES_{AB}\H_{AB}-ES_{AB}&E_{B}-E\end{array}\right|=(E_{A}-E)(E_{B}-E)-(H_{AB}-ES_{AB})^{2}=0\label{18}$
This can be rearranged to $E-E_{A}=\dfrac{(H_{AB}-ES_{AB})^{2}}{E-E_{B}}\label{19}$To estimate the root closest to EA, we can replace E by EA on the right hand side of the equation. This leads to$E^{-}\approx{E_{A}+\dfrac{(H_{AB}-E_{A}S_{AB})^{2}}{E_{A}-E_{B}}}\label{20}$and analogously for the other root,
$E^{+}\approx{E_{B}-\dfrac{(H_{AB}-E_{B}S_{AB})^{2}}{E_{A}-E_{B}}}\label{21}$
The following correlation diagram represents the relative energies of these atomic orbitals and MO's:
A simple analysis of Equations $\ref{18}$ implies that, in order for two atomic orbitals $\psi_{A}$ and $\psi_{B}$ to form effective molecular orbitals the following conditions must be met:
1. (The atomic orbitals must have compatible symmetry. For example, $\psi_{A}$ and $\psi_{B}$ can be either s or p$\sigma$ orbitals to form a $\sigma$-bond or both can be p$\pi$ (with the same orientation) to form a $\pi$-bond.
2. The charge clouds of $\psi_{A}$ and $\psi_{B}$ should overlap as much as possible. This was the rationale for hybridizing the s and p orbitals in carbon. A larger value of SAB implies a larger value for HAB.
3. The energies EA and EB must be of comparable magnitude. Otherwise, the denominator in (20) and (21) will be too large and the molecular orbitals will not differ significantly from the original AO's. A rough criterion is that EA and EB should be within about 0.2 hartree or 5 eV. For example, the chlorine 3p orbital has an energy of -13.0 eV, comfortably within range of the hydrogen 1s, with energy -13.6 eV. Thus these can interact to form a strong bonding (plus an antibonding) molecular orbital in HCl. The chlorine 3s has an energy of -24.5 eV, thus it could not form an effective bond with hydrogen even if it were available.
Hückel Molecular Orbital Theory
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 $\sigma$-bonds. This is, in fact, a more sophisticated version of a free-electron model. We again illustrate the model using butadiene CH2=CH-CH=CH2. From four p atomic orbitals with nodes in the plane of the carbon skeleton, one can construct four $\pi$ molecular orbitals by an extension of the LCAO approach:
$\psi=c_{1}\psi_{1}+c_{2}\psi_{2}+c_{3}\psi_{3}+c_{4}\psi_{4}\label{22}$
Applying the linear variational method, the energies of the molecular orbitals are the roots of the 4 x 4 secular equation
$\left|\begin{array}{lcc}H_{11}-E&H_{12}-ES_{12}&\ldots\ \ \ \H_{12}-ES_{12}&H_{22}-E&\ldots\ \ \ \\ \ \ \ \ \ \ \ldots&\ldots&\ldots\ \ \ \end{array}\right|=0\label{23}$
Four simplifying assumptions are now made
1. All overlap integrals Sij are set equal to zero. This is quite reasonable since the p-orbitals are directed perpendicular to the direction of their bonds.
2. All resonance integrals Hij between non-neighboring atoms are set equal to zero.
3. All resonance integrals Hij between neighboring atoms are set equal to $\beta$.
4. All coulomb integrals Hii are set equal to $\alpha$.
The secular equation thus reduces to
$\left|\begin{array}{cccc}\alpha-E&\beta&0&0\\beta&\alpha-E&\beta&0\0&\beta&\alpha-E&\beta\0&0&\beta&\alpha-E\end{array}\right|=0\label{24}$
Dividing by $\beta^{4}$ and defining
$x=\dfrac{\alpha-E}{\beta}\label{25}$
the equation 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 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 gives the 4th order polynomial equation
$x^{4}-3x^{2}+1=0\label{27}$
Noting that this is a quadratic equation in $x^{2}$, the roots are found to be $x^{2}=(3\pm\sqrt{5})/2$, so that $x=\pm0.618,\pm1.618$. This corresponds to the four MO energy levels
$E=\alpha\pm1.618\beta,\ \ \ \ \ \alpha\pm0.618\beta\label{28}$
Since $\alpha$ and $\beta$ are negative, the lowest molecular orbitals have
$E(1\pi)=\alpha+1.618\beta$
and
$E(2\pi)=\alpha+0.618\beta$
and the total $\pi$-electron energy of the $1\pi^{2}2\pi^{2}$ configuration equals
$E_{\pi}=2(\alpha+1.618\beta)+2(\alpha+0.618\beta)=4\alpha+4.472\beta\label{29}$
The simplest application of Hückel theory, to the ethylene molecule CH2=CH2 gives the secular equation
$\left|\begin{array}{cc}x&1\1&x\end{array}\right|=0\label{30}$
This is easily solved for the energies $E=\alpha\pm\beta$. The lowest orbital has $E(1\pi)=\alpha+\beta$ and the 1$\pi^{2}$ ground state has $E_{\pi}=2(\alpha+\beta)$. If butadiene had two localized double bonds, as in its dominant valence-bond structure, its $\pi$-electron energy would be given by $E_{\pi}=4(\alpha+\beta)$. Comparing this with the Hückel result (Equation $\ref{29}$), we see that the energy lies lower than the that of two double bonds by $0.48\beta$. The thermochemical value is approximately -17 kJmol-1. This stabilization of a conjugated system is known as the delocalization energy. It corresponds to the resonance-stabilization energy in valence-bond theory.
Aromatic systems provide the most significant applications of Hückel theory. For benzene, we find the secular equation
$\left|\begin{array}{cccccc}x&1&0&0&0&1\1&x&1&0&0&0\0&1&x&1&0&0\0&0&1&x&1&0\0&0&0&1&x&1\1&0&0&0&1&x\end{array}\right|=0\label{31}$
with the six roots $x=\pm2,\pm1,\pm1$. The energy levels are $E=\alpha\pm2\beta$ and two-fold degenerate $E=\alpha\pm\beta$. With the three lowest molecular orbitals occupied, we have
$E_{\pi}=2(\alpha+2\beta)+4(\alpha+\beta)=6\alpha+8\beta\label{32}$
Since the energy of three localized double bonds is $6\alpha+6\beta$, the delocalization energy equals $2\beta$. The thermochemical value is -152 kJmol-1.
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Arriving at solutions of many complex quantum mechanical systems have posed great challenges to the scientists of both the past and the present. Few quantum systems have been solved analytically and even fewer without the help of approximations. The goal of this overview is to review the fundamentals of solving one of these complicated systems, the diatomic molecule.
Introduction
In contrast to atoms, diatomics show additional degrees of freedom of nuclear motion, e.g. vibrations and rotations, along with the motion of electrons. Inclusion of this additional degree of freedom complicates the system by making the system a three (or more depending on the number of electrons) body system, the solution of which has only been solved with approximation techniques (e.g. mean field theory or ignoring weak couplings). More specifically, we will define such properties as the nuclear potential curves and electronic energy curves for diatomic and then address the connection between these curves under the validity of the Born-Oppenheimer approximation and adiabatic approximation. In addition, breakdowns of Born-Oppenheimer approximation will be discussed.
Neglecting higher order relativistic interactions, the total Hamiltonian for a diatomic system is
$\left[ -\sum_{\alpha = 1}^2\dfrac{\hbar^2}{2M_{\alpha}}\bigtriangledown_{\alpha}^{2} - \sum_{i - 1 }^n\dfrac{\hbar^2}{2m}\bigtriangledown_{i}^{2} + \dfrac{Z_{\alpha}Z_{\beta}e^2}{4\pi\epsilon_o \mid R_{\alpha} - R_{\beta}\mid} - \sum_{\alpha = 1}^2 \sum_{i = 1}^n \dfrac{Z_{\alpha}e^2}{4\pi\epsilon_o \mid R_{\alpha} - r_i \mid} + \sum_{i = 1}^{n} \sum_{j > 1}^n \dfrac{e^2}{4\pi\epsilon_o \mid r_i - r_j \mid} + \hat{H}_{SO} \right] \Psi (Q,q) = E\Psi (Q,q) \label{1}$
where the first and second summations represent the kinetic energy of the nuclei, $\hat{T}_N$, and electrons, $\hat{T}_E$, respectively. The next three terms express the potential energy of the system, $V_{NE}$: the nuclei-nuclei repulsion, the electron-nuclei attraction and the electron-electron repulsion respectively. The last term is the spin-orbit operator. Note that the system wavefunction naturally includes both nuclear and electronic positions and that the Hamiltonian is not separable into nuclear and electronic motion, i.e., the motions of the electrons and nuclei are coupled by the fourth term. The total Hamiltonian can then be expressed as:
$\; \; \hat{H} = \hat{T}_{N} + \hat{H}_{el} + \hat{H}_{SO} \label{2}$
where the nuclei-nuclei repulsion term as well as the nuclei-electronic attractive and electronic-electronic repulsive terms is included in the electronic Hamiltonian:
$\displaystyle \; \hat{H}_{el} = -\sum_{i - 1 }^n\dfrac{\hbar^2}{2m}\bigtriangledown_{i}^{2} + \dfrac{Z_{\alpha}Z_{\beta}e^2}{4\pi\epsilon_o \mid R_{\alpha} - R_{\beta}\mid} - \sum_{\alpha = 1}^{2} \sum_{i = 1}^{n}\dfrac{Z_{\alpha}e^2}{4\pi\epsilon_o \mid R_{\alpha} - r_i \mid} + \sum_{i = 1}^{n} \sum_{j > 1}^{n}\dfrac{e^2}{4\pi\epsilon_o\mid r_i - r_j \mid} \label{3}$
Should the Hamiltonian, indeed, be made separable, then the corresponding wavefunction can be expressed as the product of the eigenfunctions of the nuclear and electronic motion. To do this, the nuclear/electronic coupling summation in Equation $\ref{1}$ would have to be either ignored or the approach to solving the system be modified.
Nuclear-Electronic Motion
In order to solve this problem, we expand the total wavefunction for the system $\Psi (Q,q)$ an assumed complete and orthonormal basis of adiabatic electronic wavefunctions. The intuitive basis of modifying this expansion lies in the relative velocity of the electrons versus the nuclei. Electron velocities are approximately ~3000 km/sec whereas the nuclei velocities are three orders of magnitude slower at around 3 km/sec. Such a disparaging difference in velocities implies that the electrons can be assumed to respond “instantaneously” to the motion of the nuclei and hence the electronic motion can be described as the motion of the electrons within the field of stationary nuclei, i.e. nuclear positions are parameters for solving the electronic wavefunction, not variables as are the electronic positions. The term adiabatic is used to describe systems as such. Hence the molecular wavefunction can be expressed as:
$\; \; \Psi (Q,q) = \sum_{k}^{} a_k \mid x_k (Q) \rangle \mid \phi_{k} (q;Q)\rangle \label{4}$
where $\mid x_k \rangle$ is the nuclear wavefunction that describes the motion of the nuclei on the potential energy surface associated with the kth adiabatic electronic state $\mid \phi_k \rangle$. This expansion shows that nuclear motion occurs on all electronic states simultaneously. A common approximation is to assume that this expansion can be represented by one term in the summation, with only one ak coefficient being non-zero, and hence nuclear motion occurs on only one electronic state. This approximation is the Born-Oppenheimer approximation and when a wavefunction can be represented as such a product it is called a Born-Oppenheimer state. The adiabatic electronic eigenstates used in the previous expansion are calculated from solving the Schrödinger equation with the electronic Hamiltonian from Equation $\ref{3}$:
$\; \; \hat{H}_{el} \mid \phi_k(q;Q)\rangle = \epsilon_k(Q) \mid \phi_k (q;Q)\rangle \label{5}$
Note that the adiabatic electronic eigenstates, $\mid \phi_k \rangle$ depend parametrically on the values of the nuclear positions. Additionally, $\epsilon_k(Q)$, which represent the adiabatic electronic energy curve of the kth electronic eigenstate, varies with the values of the nuclear positions. By substituting this expanded wavefunction into the Schrödinger equation with total Hamiltonian (Equation $\ref{3}$), then multiplying by the kth adiabatic wavefunction and integrating over electronic coordinates we obtain an infinite set of coupled equations:
$\displaystyle \; \; a_k \left[\hat{T}_N + T^{'}_{kk} + T^{''}_{kk} + \epsilon_k(Q) + SO_{kk} - E \right] |\chi_k(Q)\rangle = -\sum_{k^{'}\not= k}^{} a_{k'} \left[T^{'}_{kk'} + T^{''}_{kk'} + SO_{kk'}\right] | \chi_k (Q)\rangle \label{6}$
where
$T^{''}_{kk'} = \dfrac{-\hbar^2}{2\mu_R} \langle \phi_k \mid \bigtriangledown^{2}_{R} \mid \phi_{k'} \rangle \label{7}$
$T^{''}_{kk'} = \dfrac{-\hbar^2}{\mu_R} \langle \phi_k \mid \bigtriangledown_R \mid \phi_{k'} \rangle \cdot \bigtriangledown_R\label{8}$
$SO_{kk'} = \langle \phi_k \mid \hat{H}_{SO} \mid \phi_{k'} \rangle \label{9}$
where the nuclear kinetic energy operator can be expanded in a center of mass term and a internal motion term, where $R$ is the internuclear distance and $\mu$ is the reduced mass:
$-\sum^{2}_{\alpha = 1} \dfrac{\hbar^2}{2M_{\alpha}} \bigtriangledown^2_{\alpha} = - \dfrac{\hbar^2}{2M} \bigtriangledown^2_{cm} - \dfrac{\hbar^2}{2\mu_{\alpha}} \bigtriangledown^2_R \label{10}$
The center of mass has been omitted in Equation $\ref{6}$ for clarity.
The Born-Oppenheimer Approximation and Adiabatic Approximation
The most common and easiest approach to solving Equation $\ref{6}$ is to assume that the wavefunction can be represented with one term in the expansion of Equation $\ref{4}$. This approximation thus assumes that the motion of the electrons do not depend on the motion of the nuclei, just the nuclear positions. Another interpretation of this approximation is that nuclear motion occurs on only one electronic energy curve and hence in only a single electronic state. Max Born and his student Robert Oppenheimer demonstrated that under many situations the coupling terms on the right side of Equation $\ref{6}$ can be ignored, thus simplifying the problem immensely by converting a large coupled system into two simpler uncoupled systems. Hence this approximation is commonly called the Born-Oppenheimer (BO) approximation. Often this approximation is referred to as the adiabatic approximation. The BO approximation will then be a further simplification of the problem (vida infra). Thus the total wavefunction can be expressed as only one term in Equation $\ref{4}$, then the right side of Equation $\ref{6}$ is zero since all other ak’s are zero. Thus we obtain the Schrödinger’s equation for the motion of the nuclei under the adiabtic approximation:
$\left[\hat{T}_N + V_R(Q) - E_{TOT}\right]\mid \chi_K(Q) \rangle = 0 \label{11}$
where is the effective nuclear potential energy operator that governs the motion of the nuclei and is given by:
$V_R(Q) = T^{'}_{kk} + T^{''}_{kk} + \epsilon_k(Q) + SO_{kk} \label{12}$
Note that nuclear potential energy is a function of both the nuclear position and nuclear momentum. The Born Oppenheimer approximation further simplifies the problem by neglecting the $T^{'}_{kk}$ and $T^{''}_{kk}$ terms above with the following rational. The term is usually small as the electronic eigenstates do not normally oscillate greatly with internuclear separation. Equally, when nuclear motion is slow and the adiabatic wavefunction doesn’t oscillate greatly, the $T^{'}_{kk}$ can be neglected. So in the absence of appreciable spin-orbit coupling, the effective nuclear potential energy curve is the adiabatic electronic energy curve. These additional terms in Equation $\ref{12}$ represent non-adiabatic corrections to the nuclear potential energy curves but the adiabatic approximation of nuclear motion still occurring in only one electronic energy state holds. In this BO situation, where the nuclear and electronic motions are separable, the system can be solved in two stages. The first stage involves solving Equation $\ref{5}$ for the adiabatic electronic state $\mid \phi_k \rangle$ and energy curve $\epsilon_{k}(Q)$that nuclear motion occurs on and then constructing a nuclear potential energy curve, $V_{r}(Q)$, via Equation $\ref{12}$. The next step involves solving the nuclear Schrödinger equation, Equation $\ref{11}$, with the constructed potential energy curve for the nuclear wavefunction, and then uses Equation $\ref{4}$ with only one expansion term to create a total wavefunction for the system. This occurs only within the adiabatic approximation where the wavefunction can be expressed as only one product of the adiabatic electronic and nuclear wavefunctions.
Breakdown of the Born-Oppenheimer Approximation
The essence of the Born-Oppenheimer approximation is that the nuclear motion occurs on only one electronic energy surface. When in the course of the nuclear motion, possibly in intramolecular vibrations or intermolecular reactions, the nuclear motion can’t be expressed as occurring in one electronic state throughout the motion, then the Born-Oppenheimer approximation breaks down. In other words, the right side of Equation $\ref{6}$ can’t be approximated as zero and thus the nuclear motion couples different electronic states. This is often called vibronic coupling. Any of the terms on the ride side of Equation $\ref{6}$ can be responsible for non-adiabatic coupling of electronic energy states. Certainly, for diatomics with large nuclei where spin-orbit coupling becomes appreciable, the $SO_{kk’}$ term can become a significant coupling factor. The $T^{''}_{kk'}$ term though is generally small for the same reason that the $T^{''}_{kk}$ was considered negligible in the previous section. The term mostly responsible for non-adiabatic transitions is the term $T^{'}_{kk'}$. This term is velocity dependent since it involves the $\bigtriangledown_{R}$ acting on the nuclear wavefunction $\mid \chi_{k} \rangle$. Hence when nuclear motion is larger, the probability of non-adiabatic transitions increases, and thus non-adiabatic transitions can be observed in many high-energy chemical reactions. Equally, when the nature of the electronic wavefunction changes significantly within a short distant, e.g. an avoided crossing, this term increases in magnitude. Considerable interest has been generated in studying the effects of non-adiabatic transitions within the recent past and their effects in activated barrier crossings.
Conclusion
There exists a connection between the adiabatic electronic energy curves constructed by solving Equation $\ref{5}$ and the nuclear potential energy curves, $V_{R}(Q)$, used to solve for nuclear motion. This connection can be a complicated one depending on the validity of the Born-Oppenheimer and adiabatic approximations used to solve the system. When the adiabatic approximation can be used then this relationship can be expressed in Equation $\ref{12}$. And under circumstances when the BO approximation is valid and the spin-orbit coupling can be ignored, the nuclear potential curve can then be further approximated by the electronic energy curve $\epsilon_{k}(Q)$. Naturally, further additions to the molecular Hamiltonian will introduce extra corrections to the nuclear potential energy curve. An example of such is the LM2M2* nuclear potential curve used for the helium dimer, in which relativistic retardation effects are introduced to account for the finite time interactions between the electrostatic Coulomb forces of the electrons in the rather large diatomic helium.
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A molecule of dihydrogen contains two atoms, in which the nuclei of both the atoms are spinning. Depending upon the direction of the spin of the nuclei, the hydrogens are of two types:
Ortho hydrogen molecules are those in which the spins of both the nuclei are in the same direction. Molecules of hydrogen in which the spins of both the nuclei are in the opposite direction are called para hydrogen.
Ordinary dihydrogen is an equilibrium mixture of ortho and para hydrogen.
$\text{ortho hydrogen} \ce{<=>} \text{para hydrogen} \nonumber$
The amount of ortho and para hydrogen varies with temperature as:
• At 0°K, hydrogen contains mainly para hydrogen which is more stable.
• At the temperature of liquefaction of air, the ratio of ortho and para hydrogen is 1 : 1.
• At the room temperature, the ratio of ortho to para hydrogen is 3 : 1.
• Even at very high temperatures, the ratio of ortho to para hydrogen can never be more than 3 : 1.
Thus, it has been possible to get pure para hydrogen by cooling ordinary hydrogen gas to a very low temperature (close to 20 K) but it is never possible to get a sample of hydrogen containing more than 75% of ortho hydrogen.
Overlap Integral
The Overlap Integral is a quantitative measure of the overlap of atomic orbitals situated on different atoms. An orbital overlap is the amount of orbitals of adjacent atoms that are in the same regions of space. The overlap of the atomic orbital of an atom A and the atomic orbital of an atom B is called their overlap integral. It is defined as $S_{AB} = \int_{}^{} \psi^{*}_{A} \psi_{B} dr$ extending over all space.
Basic description
To understand the overlap integral, consider the wave function $\psi_{+}$ to calculate the energy of $H^{+}_{2}$ as a function of internuclear separation, R.
The energy associated with $\psi_{+}$, E+ has the equation
$\hat{H} \psi_{+} (r; R) = E_{+} \psi_{+} (r; R)$
Multiply the left by $\psi^{*}_{+} (r;R)$ and then integrate over the allowed values of r to get E+
$E_{+} = \dfrac{\int_{}^{} dr \psi^{*}_{+} \hat{H} \psi_{+}}{\int_{}^{} dr \psi^{*}_{+} \psi_{+}}$
The wave function $\psi_{+}$ is given by $\psi_{+} = c(1s_{A} + 1s_{B})$. To normalize this wave function, $\int_{}^{} dr \psi^{*}_{+} \psi_{+} = 1$, so
$\int_{}^{} dr \psi^{*}_{+} \psi_{+} = \int_{}^{} dr (1s^{*}_{A} + 1s^{*}_{B})(1s_{A} + 1s_{B})$
$\int_{}^{} dr \psi^{*}_{+} \psi_{+} = \int_{}^{} dr (1s^{*}_{A}1s_{A} + 1s^{*}_{A}1s_{B} + 1s^{*}_{B}1s_{A} + 1s^{*}_{B}1s_{B})$
$\int_{}^{} dr \psi^{*}_{+} \psi_{+} = (\int_{}^{} dr 1s^{*}_{A}1s_{A} + \int_{}^{} dr 1s^{*}_{A}1s_{B} + \int_{}^{} dr 1s^{*}_{B}1s_{A} + \int_{}^{} dr 1s^{*}_{B}1s_{B})$
The first and fourth integrals are simply the normalization expressions of the hydrogen atomic orbitals, so
$\int_{}^{} dr 1s^{*}_{A}1s_{A} = \int_{}^{} dr 1s^{*}_{B}1s_{B} = 1$
The second and third integrals involve the product of the hydrogen atomic orbital focus on nucleus A and the hydrogen atomic orbital focus on nucleus B. This product is larger when the two atomic orbitals have a larger overlap. These two integrals equal each other because the hydrogen atomic 1s orbital is expressed by a real function, or $1s^{*} = 1s$, and S is the overlap integral, so
$\int_{}^{} dr 1s^{*}_{A}1s_{B} = \int_{}^{} dr 1s^{*}_{B}1s_{A} = S$
This then gives
$\int_{}^{} dr \psi^{*}_{+} \psi_{+} = 1 + S + S + 1$
$\int_{}^{} dr \psi^{*}_{+} \psi_{+} = 2 + 2S$
Thus, the denominator is
$\int_{}^{} dr (1s^{*}_{A} + 1s^{*}_{B})(1s_{A} + 1s_{B}) = 2 + 2S$
Example $1$: Determine the normalized wave function for $\psi_{+}$
The wave function $\psi_{+}$ is given by $\psi_{+} = c(1s_{A} + 1s_{B})$. To normalize this wave function, $\int_{}^{} dr \psi^{*}_{+} \psi_{+} = 1$, so
$1 = \int_{}^{} dr \psi^{*}_{+} \psi_{+}$
$1 = \int_{}^{} dr \space c(1s^{*}_{A} + 1s^{*}_{B}) c(1s_{A} + 1s_{B})$
$1 = c^{2} \int_{}^{} dr (1s^{*}_{A} + 1s^{*}_{B})(1s_{A} + 1s_{B})$
$1 = c^{2} \int_{}^{} dr (1s^{*}_{A}1s_{A} + 1s^{*}_{A}1s_{B} + 1s^{*}_{B}1s_{A} + 1s^{*}_{B}1s_{B})$
$1 = c^{2} (\int_{}^{} dr 1s^{*}_{A}1s_{A} + \int_{}^{} dr 1s^{*}_{A}1s_{B} + \int_{}^{} dr 1s^{*}_{B}1s_{A} + \int_{}^{} dr 1s^{*}_{B}1s_{B})$
Using the concepts of the Overlap Integral, $\int_{}^{} dr 1s^{*}_{A}1s_{A} = \int_{}^{} dr 1s^{*}_{B}1s_{B} = 1$ and $\int_{}^{} dr 1s^{*}_{A}1s_{B} = \int_{}^{} dr 1s^{*}_{B}1s_{A} = S$, so
$1 = c^{2} (1 + S + S + 1)$
$1 = c^{2} (2 + 2S)$
$1 = c^{2} 2(1 + S)$
$c^{2} = \dfrac{1}{2(1 + S)}$
$c = \sqrt{\dfrac{1}{2(1 + S)}} = \dfrac{1}{\sqrt{2(1 + S)}}$
Thus, the normalized wave function is
$\psi_{+} = c(1s_{A} + 1s_{B})$
$\psi_{+} = \dfrac{1}{\sqrt{2(1 + S)}} (1s_{A} + 1s_{B})$
Advanced description
If the wave functions do not overlap, then the overlap integral is zero. The integral can also be zero if the wave functions have positive and negative aspects that cancel out. If the overlap integral is zero, then the wave functions are called orthogonal. As the distance of the functions get closer to R = 0, then the overlap integral gets closer to the maximum overlap integral of S = 1.
Overall, the full equation for the overlap integral, which is hard to calculate but is explained here, is
$S(R)= \left \langle 1s_A | 1s_B \right \rangle = e^{-R/a_0} \left (1 + \dfrac{R}{a_0} + \dfrac{R^2}{3a_0^2} \right)$
Example $2$: The overlap integral at different separations
$S(R)= e^{-R/a_0} \left( 1 +\dfrac{R}{a_0} + \dfrac{R^2}{3a_0^2} \right)$
Calculating the different separations of Figure 1.2 above:
At $\dfrac{R}{a_0}$ = 0,
$S(0) = e^{-0} (1 + 0 + \dfrac{0^{2}}{3}) = e^{0}(1 + 0 + 0) = 1(1) = 1$
At $\dfrac{R}{a_0}$ = 1,
$S(1) = e^{-1} (1 + 1 + \dfrac{1^{2}}{3}) = e^{-1}(1 + 1 + \dfrac{1}{2}) = e^{-1}(\dfrac{5}{2}) = 0.9197$
At $\dfrac{R}{a_0}$ = 2,
$S(2) = e^{-2} (1 + 2 + \dfrac{2^{2}}{3}) = e^{-2}(1 + 2 + \dfrac{4}{2}) = e^{-2}(1 + 2 + 2) = e^{-2}(5) = 0.6767$
At $\dfrac{R}{a_0}$ = 4,
$S(4) = e^{-4} (1 + 4 + \dfrac{4^{2}}{3}) = e^{-4}(1 + 4 + \dfrac{16}{2}) = e^{-4}(1 + 2 + 8) = e^{-4}(11) = 0.2015$
As evident from these calculations, the distance of R/a0 = 0 or R = 0 gives the maximum overlap integral of S = 1. After that, the overlap integral S decreases exponentially with distance R.
Contributors and Attributions
• Marilyn McBryan
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Atoms in a molecule are held together by chemical bonds. When the atom is distorted, the bonds are stretched or compressed, in which increases the potential energy of its system. As the new geometry is formed, the molecule stays stationary. Therefore, the energy of the system is not caused by the kinetic energy, but depending on the position of the atoms (potential). Energy of a molecule is a function of the position of the nuclei. When nuclei moves, electron readjusts quickly. The relationship between this molecular energy and molecular geometry (position) is mapped out with potential energy surface.
Basic Description
We use Born-oppenheimer approximation to separate electronic and nuclear motion. This is useful in analyzing properties of structures composed of atoms. PES typically has the same dimensionality as the number of geometric degrees of freedom of the molecule (3N-6) where N is number of atoms and has to be greater than 2.
PES depends on Born-Oppenheimer approximation, which states that the nuclei in a molecule are essentially stationary compared to the electrons. This approximation is significant in computational chemistry as it simplifies the application of Schrödinger equation to molecules by focusing on the electronic energy and adding in the nuclear repulsion energy. Inherently a classical construct with respect to the nuclei.
We always want to reach minimum on the PES - as molecules want to be at the lowest possible potential energy. A minimum on the PES is defined by curvatures that are all positives. A saddle point is characterized by a negative curvature in one direction and positive curvature in all other directions.
At the minima and saddle point, the slope of the function is set to zero, where
$(\frac{df}{dx})_y = 0 , (\frac{df}{dy})_x = 0 . At the minima, the second derivatives are positive, due to positive curvature, that is \[ (\frac{d^2f}{dx^2})_y = > 0 , (\frac{d^2f}{dy^2})_x = > 0 . And at the saddle point, one of the second derivative is negative (negative curvature), and all other derivatives are positives, so that \[ (\frac{d^2f}{dx^2})_y = < 0 \[ , (\frac{d^2f}{dy^2})_x = > 0$
The model surface: $H+ H_2$
Three dimensional PES is used for triatomic systems by substituting the collision reaction of an atom with a diatomic molecule. The very first study done on this topic is $H_3$ atom. The reaction is
$H + H_2 \rightarrow H_2 + H$
In this case, the triatomic molecule is the H-H-H transition state. The potential energy surface at this point will have a barrier as it has a higher energy than the reactants or products. Quantum mechanical and quasiclassical trajectories calculations require the PES for the H3 system. The PES does not dependent on the masses of the atoms, thus, the H3 surface can be used for any isotopic variant of this reaction. As three nuclei (ABC) are involved, the PES depends on three coordinates. Therefore, one coordinate needs to be fixed in order to plot the PES as a function of the remaining two.
The potential energy surface of H + H2 can be seen in figure below
(Potential energy surface for the H3 (H+H2) system (Springer, 1998))
The three dimensional potential energy surfaces can be seen in this contour plot, where A-B-C transition state is marked by a red cross.
The barrier at the transition state is called the saddle point, a minimum along the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates.
12: Electron Scattering
More discussion on scattering in the Physics Library.
Thumbnail: Collimated homogeneous beam of monoenergetic particles, long wavepacket which is approximately a planewave, but strictly does not extend to infinity in all directions, is incident on a target and subsequently scattered into the detector subtending a solid angle. The detector is assumed to be far away from the scattering center. (Department of Physics Wiki @ Florida State University).
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Fine structure describes the splitting of the spectral lines of atoms due to electron spin and relativistic corrections to the non-relativistic Schrödinger equation. Hyperfine structure, with energy shifts typically orders of magnitudes smaller than those of a fine-structure shift, results from the interactions of the nucleus (or nuclei, in molecules) with internally generated electric and magnetic fields.
• Fine Structure
Spin–orbit coupling is an interaction of a particle's spin with its motion. This interaction leading to shifts in an electron's atomic energy levels, due to electromagnetic interaction between the electron's spin and the magnetic field generated by the electron's orbit around the nucleus. This is detectable as a splitting of spectral lines, which can be thought of as a Zeeman effect due to the internal field.
• Hyperfine Structure
The hyperfine structure is caused by interaction between magnetic field (from electron movement) and nuclear spin.
Thumbnail: Schematic illustration of fine and hyperfine structure in hydrogen atom. (Public Domain; Edudobay).
13: Fine and Hyperfine Structure
From classical electrodynamics, a rotating electrically charged body creates a magnetic dipole with magnetic poles of equal magnitude but opposite polarity. This analogy holds as an electron indeed behaves like a tiny bar magnet. In the classical model, the electron (Bohr atom model) is moving along a closed path. A revolving electron is equivalent to a current loop (Figure $\PageIndex{1a}$).
The current arising from the orbiting of the electron around the nucleus is
$I = \dfrac{e}{T}$
where $T$ is the orbital period or revolution. We can define the velocity then as
$v=\dfrac{2πr}{T}$
where $r$ is the radius of orbit. The "orbital" magnetic (dipole) moment is then
$\vec{\mu}_L = \dfrac{\vec{\mu}e}{T} (π r^2) \label{eq1}$
and the angular momentum associated with that orbital motion is
\begin{align} L &= m_e v_r \[4pt] &= \dfrac{m_e 2\pi r^2}{T} \label{eq2} \end{align}
and $m_e$ is the electron rest mass.
Next we combine Equation \ref{eq1} and \ref{eq2} to get how the orbital angular momentum induced a magnetic moment (in vector form)
$\color{red} \vec{\mu}_L = - \dfrac{e}{2m_e} \vec{L} \label{classical}$
For a more detailed derivation of Equation \ref{classical}, look here. We can glean two important feature from Equation \ref{classical}:
1. If the electron has zero orbital angular momentum, it will have a zero-amplitude orbital magnetic moment.
2. The orientation of the orbital angular moment is parallel to the angular momentum.
The Quantum Orbital Magnetic Dipole Moment
Equation \ref{classical} is applicable to classical systems and since $\vec{L}$ can have any values (amplitude and orientation), so can the magnetic moment. That is a different story in the quantum world. where the orbital angular momentum is quantized and hence so is $\mu_L$. The amplitude of is $\vec{L}$ represented by the $\hat{L}^2$ operator and the projection of $\vec{L}$ on the z-axis is represented by $\hat{L}_z$ operator.
$\hat{L^2} | \psi \rangle = \ell(\ell+1) \hbar^2 | \psi \rangle$
and
$\hat{L}_z | \psi \rangle = m_l \hbar | \psi \rangle$
So we obtain the eigenvalues $µ_{L_z}$ of z-component of the magnetic moment
$µ_{L_z} = −\dfrac{ e}{2m_e} \hbar m_l \label{eq5}$
It is usual to express the magnetic moment (Equation \ref{eq5}) in terms the Bohr magneton $\mu_B$
$µ_{L_z} = −µ_B \,m_l$
where $µ_B$ is the Bohr Magneton:
$µ_B = \dfrac{e\hbar}{2m_e}$
The Bohr magneton is a physical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital angular momentum and (and spin as discussed below) and is numerically
\begin{align} µ_B &= 9.273 \times 10^{-24}\, J/K \[4pt] &= 5.656 \times 10^{-5}\, eV/T \[4pt] &= 1.4 \times 10^{10}\, Hz/T \,(T \,= \,Tesla) \end{align}
Coupling Orbital Magnetic Dipole Moment to an External Magnetic Field
When a magnetic field is applied to the electron, then the energies of the eigenstates will depends on the magnitude $B$ of the applied magnetic field $\vec{B}$ via
\begin{align} E_B &= −µ_{L_z} \, B \[4pt] &= µ_B\, m_l\, B \end{align}
and the total energy of that state is
\begin{align} E &= E_0 + E_B \[4pt] &= E_0 + µ_B\, m_l\, B \end{align}
Hence, the energy levels having a angular momentum with quantum number $\ell$ are split into $2\ell + 1$ new levels (Figure $2$):
This effect is the "ordinary" or "normal" Zeeman Effect.
Coupling Spin Magnetic Dipole Moment to an External Magnetic Field
The hydrogen ground state ($m_l = 0$) is uninfluenced in Figure $1$. However, we know that a H atom is paramagnetic; this is because of electron spin. Now we consider the spin in classical mechanics as rotating around the axis electron. We also find here
$µ_{S_z} = −g_S\, µ_B \, m_s$
$E_B = g_S\, µ_B\, m_s\, B$
The so-called gyromagnetic factor $g_S$ is obtained from the relativistic Dirac equation and experiments quantify it at
$g_S = 2.00231930438(6)$
The value
$g_S\, µ_B ≈ 28\, GHz/T$
shows us which energy state is higher according to electron spin interaction with magnetic field. Since now it's possible to produce magnetic fields with the strength of a few Teslas, we expect to detect transitions in GHz region (microwaves) when applying such fields. That's why Electron-Spin-Resonance (ESR) Spectroscopy is involved with microwaves. When applying NMR method we have a deal with MHz region. We will discuss it in more detail later.
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Splittings from Internal Magnetic Field of Nuclear Spins
The hyperfine structure is caused by interaction between magnetic field (from electron movement) and nuclear spin. For instance, the hydrogen atom has one proton with spin $I = ½$ and corresponding magnetic moment
$\vec{\mu}_P = −g_I \dfrac{eh}{2m_p} \dfrac{\vec{I}}{\hbar} = \dfrac{µ_K\vec{I}}{hbar}$
Since a proton is heavier than an electron, its magneton $µ_K$
$µ_K = g_I \dfrac{eh}{2m_p}$
is lighter by a factor $\dfrac{m_e}{m_p}$.
Similar to fine structure coupling, we must connect $\vec{J}$ and $\vec{I}$ now having "new" total angular momentum:
$\vec{F} = \vec{J} + \vec{I}$
where
$\hat{F}^2 |\psi \rangle = F(F +1) h^ | \psi \rangle$
and
$\hat{F}_z | \psi \rangle= m_F \hbar | \psi \rangle$
where
$m_F = −F, −F+1, ... , F−1, F$
Squaring the equation for $\vec{F}$ and solving it relatively $\vec{J} \cdot \vec{I}$ one can obtain the hyperfine energy structure:
$E_{HFS} = \dfrac{cHFS}{2}\{F(F + 1) - J(J + 1) - I(I + 1)\}$
The lowest term of hydrogen 1²S½ is then split into two terms with F = 1 (spin electron and spin proton ) and F = 0 (spin electron and spin proton ¯ antiparallel). The transition between these two levels can only occur when the spin turns over. This, however, is a very rare phenomenon in an isolated hydrogen atom (it, on average, would happen once every 10 million years for a single molecule) however because of the exceedingly high number of hydrogen atoms in the universe it can be observed. Further, collisions can flip the spin. It lies near n» 1,4204 GHz (λ = 21 cm).
Photons have Angular Momenta Too
Since this transition changes the system angular momentum from $F = 1h$ into $F = 0h$ but from the fact that the total angular momentum should be the same follows photon must take the hydrogen atom angular momentum. And so a photon also possesses angular momentum. However, the photon spin is as follows
$S_{Photon} = 1\,h$
The photon spin isn't a good designation because the "photon spin" projection on the flight axis can take values m = +1 or m = -1 for right- and lefthand circular polarized light, correspondingly. And it can never be m=0, that would correspond to the longitudinal polarization which is never realized!
The proton spin has interesting consequences for optical spectra since we obtain the selection rule for transitions. We will have for the final total angular momentum $\vec{J_f}$ of the system that was originally at some initial state $\vec{J_i}$ according to the angular momentum conservation law:
$\vec{J}_i = \vec{J}_f + \vec{J}_{photon}$
Certainly $\vec{J}_{photon}$ shouldn't be necessarily equal to $\vec{1}$ because photon can also possess (together with spin) the orbital moment relative to an atom. The photon emission with unchanged orbital moment visually means that photon has left the boundary areas. Nevertheless, this process is very improbable so that quantum numbers $J_i$ and $J_f$ have the following relation according to the angular momentum conservation law:
$ΔJ = J_f − J_i = 0,\, ±1$
• $ΔJ = −1$
• $ΔM = 0 , ±1$
P-Branch
• ΔJ = 0
• ΔM = ±1
Q-Branch
• ΔJ = +1
• ΔM = 0 , ±1
R-Branch
If $J_f = 0$ and $J_i = 0$ then $ΔJ=0$ is forbidden according to the angular momentum conservation law. If there is small interaction between $\vec{L}$ and $\vec{S}$ then there is special selection rule for spin that states:
$ΔS = 0$
The selection rule for electron orbital moment has been given here without any derivation:
$Δl = ±1$
The possible transitions are shown on the illustration to the left (Balmer a-Line, n = 3 ↔ n = 2). Since H atoms move there are Doppler distribution of these optical lines, that can be resolved by using Doppler free spectroscopy which can help us to prove the theory.
Certainly we must also have angular momentum conservation for increasing rotation of a molecule. In the simplest case of a linear molecule for which electron contribution into angular momentum is small so that $ΔJ = ±1$, i.e. the rotation can be increased by $1h$ or decreased by $1h$ when photon is absorbed. It's said about P-Branch (the rotation is decreased by 1h) and R-Branch (the rotation is increased by 1h).
If total electron angular momentum is not equal to 0 or if we have a deal with a non-linear molecule then the kinetic impulse vector character according to the selection rule undergoes some changes, i.e. there is also a transition $\Delta J = 0$ (Q-Branch) possible (when \(J = 0 → J = 0 !).
In Raman-Spectroscopy two photons are used: the molecule undergoes a transition from the initial state i into virtual state and then there is "emission" from this virtual state into the final state f. Since the angular momentum of both photons should compensate or add up to each other (the angular momentum conservation law) it follows the selection rules for the Raman spectroscopy: ΔJ = 0, ±2. When ΔJ = 0 the angular momentums of both photons compensate each other (¯); and when ΔJ = ±2 they are added to each other ( or ¯¯). One can talk about S-Branch (ΔJ = +2) and O-Branch (ΔJ = -2) in this case.
14: Systems of Identical Particles
Thumbnail: Antisymmetric wavefunction for a (fermionic) 2-particle state in an infinite square well potential. (CC BY 3.0; TimothyRias).
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The Schrödinger equation is the basis for quantum mechanics. The solutions to the equation, known as wave functions, give complete quantum mechanical insight into the system under observation. The Hamiltonian operator, which is specific to the system's environment, acts upon the wavefunction to yield the wavefunction again, accompanied by the energy of the system (the eigenvalue). The time-dependant Schrödinger equation describes the evolution of a system’s wavefunction through time.
Thumbnail: A wave packet without dispersion (real or imaginary part). (Public Domain; Fffred~commonswiki).
15: Time-dependent Quantum Dynamics
Postulate V states how the time-dependent Schrödinger equation describes the evolution of a system’s wavefunction through time.
Contributors and Attributions
• Eric Gobrogge (Hope College), Alex Ketchum (Hope College)
WavePackets
De Broglie's explanation of the Bohr atom quantization rules, together with the accidental discovery of electron diffraction scattering by Davisson and Germer, make a very convincing case for the wave nature of the electron. Yet the electron certainly behaves like a particle sometimes. An electron has a definite mass and charge, it can move slowly, it can travel through a piece of apparatus from a gun to a screen. What, then, is the relationship between the wave and particle viewpoints? De Broglie himself always felt both were always present. He called the wave a pilot wave, and thought it guided the motion of the particle. Unfortunately, that viewpoint leads to contradictions. The standard modern interpretation is that the intensity of the wave (measured by the square of its amplitude) at any point gives the relative probability of finding the particle at that point. This interpretation, originally presented by Max Born in 1926, is parallel to the relation between the electromagnetic field and quanta -- the probability of finding a quantum (photon) at any point is proportional to the energy density of the field at that point, which is the square of the electric field vector plus the square of the magnetic field vector. The standard notation for the de Broglie wave function associated with the electron is $\psi (x,t)$. Thus, $|\psi (x,t)|^2\Delta x$ is the relative probability of finding the electron in a small interval of length $\Delta x$ near point $x$ at time $t$. (For the moment, we restrict the electron to move in one dimension for simplicity. The generalization is straightforward.)
Keeping the Wave and the Particle Together?
Suppose following de Broglie we write down the relation between the "particle properties" of the electron and its "wave properties":
$\dfrac{mv^2}{2} = E = h\nu \label{1a}$
$mv = p = \dfrac{h}{\lambda} \label{1b}$
It would seem that we can immediately figure out the speed of the wave, just using $\lambda \nu = v$, say. We find:
$\lambda \nu = \left(\dfrac{h}{mv} \right) \times \left(\dfrac{mv^2}{2h} \right) = \dfrac{v}{2} \label{2}$
So the speed of the wave seems to be only half the speed of the electron! How could they stay together? What's wrong with this calculation?
Localizing an Electron
To answer this question, it is necessary to think a little more carefully about the wave function corresponding to an electron traveling through a vacuum tube, say. The electron leaves the cathode, shoots through the vacuum, and impinges on an anode of a grid. At an intermediate point in this process, it is moving through the vacuum and the wave function must be nonzero over some volume, but zero in the places the electron has not possibly reached yet, and zero in the places it has definitely left.
However, if the electron has a precise energy, say fifty electron volts, it also has a precise momentum. This necessarily implies that the wave has a precise wavelength. But the only wave with a precise wavelength $\lambda$ has the form
$\psi(x,t)= A\sin(kx-\omega t) \label{3}$
where $k = 2\pi /\lambda$, and $\omega = 2\pi \nu$. The problem is that this plane sine wave extends to infinity in both spatial directions, so cannot represent a particle whose wave function is non zero in a limited region of space.
Therefore, to represent a localized particle, we must superpose waves having different wavelengths. The principle is best illustrated by superposing two waves with slightly different wavelengths, and using the trigonometric addition formula:
$\sin [(k-\Delta k)x - (\omega - \Delta \omega)t] + \sin [(k+\Delta k)x -(\omega + \Delta \Omega)t]=$
$\sin (kx - \omega t) \cos [(\Delta k)x - (\Delta \omega) t ] \label{4}$
This formula represents the phenomenon of beats between waves that are close in frequency. The first term, $\sin(kx-\omega t)$, oscillates at the average of the two frequencies. It is modulated by the slowly varying second term, which oscillates once over a spatial extent of order $\frac{\pi}{\Delta k}$. This is the distance over which waves initially in phase at the origin become completely out of phase. Of course, going a further distance of order $\frac{\pi }{\Delta k}$, the waves will become synchronized again.
That is, beating two close frequencies together breaks up the continuous wave into a series of packets, the beats. To describe a single electron moving through space, we need a single packet. This can be achieved by superposing waves having a continuous distribution of wavelengths, or wave numbers within of order $\Delta k$, say of $k$. In this case, the waves will be out of phase after a distance of order $\frac{\pi}{\Delta k}$, but since they have many different wavelengths, they will never get back in phase again.
The Uncertainty Principle
It should be evident from the above argument that to construct a wave packet representing an electron localized in a small region of space, the component waves must get out of phase rapidly. This means their wavelengths cannot be very close together. In fact, it is not difficult to give a semi-quantitative estimate of the spread in wavelength necessary, just from a consideration of the two beating waves.
A packet localized in a region of extent $\Delta x$ can be constructed of waves having k's spread over a range $\Delta k$, where $\Delta x ~ \dfrac{\pi}{\Delta k}$.
Now, $k = 2\pi /\lambda$, and $p = \dfrac{h}{\lambda}$, so $k = 2\pi \dfrac{p}{h}$.
Therefore the uncertainly of $k$ can be expanded
$\Delta k = 2\pi \dfrac{\Delta p}{h} \label{5a}$
and
$\Delta x ~ \dfrac{p}{\Delta k} \approx \dfrac{h}{\Delta p} \label{5b}$
(ignoring the factor of 2 for now).
Thus:
$\Delta x \Delta p \approx h \label{6}$
This is Heisenberg's Uncertainty Principle.
Reconciling the "conflicting" Wave/Particle Properties
Establishing that an electron moving through space must be represented by a wave packet also resolves the paradox that the velocity of the waves seems to be different from the velocity of the electron. The point is that the electron waves, like water waves but unlike electromagnetic waves, have differing phase and group velocities. To see this, consider again the beating of two waves of slightly different wavelengths in Equation $\ref{4}$.
The waves described by the term $\sin(kx - \omega t)$ have velocity $½v$, as previously derived. However, the envelope, the shape of the wave packet, has velocity $\frac{\Delta \omega}{\Delta k}$ rather than $\frac{\omega}{k}$. These velocities would be the same if $\omega$ were linear in $k$, as it is for ordinary electromagnetic waves. However, the $\omega - k$ relationship follows from the energy-momentum relationship for the (non-relativistic) electron:
$E = \dfrac{mv^2}{2} = \dfrac{p^2}{2m}. \label{7}$
So
$\dfrac{dE}{dp} = \dfrac{p}{m} = v. \label{8}$
Equations $\ref{7}$ and $\ref{8}$ address the particle nature, but considering the wave nature:
$E = h\nu = \dfrac{h\omega}{2pi}\label{9}$
and
$p = \dfrac{h}{\lambda} = \dfrac{hk}{2\pi}. \label{10}$
Therefore,
$\dfrac{\Delta \omega}{\Delta k} = \dfrac{dE}{dp} = v \label{11}$
So the packet travels at the speed we know the electron must travel at, even though the wave peaks within the packet travel at one-half the speed.
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Our most detailed knowledge of atomic and molecular structure has been obtained from spectroscopy-study of the emission, absorption and scattering of electromagnetic radiation accompanying transitions among atomic or molecular energy levels. Whereas atomic spectra involve only electronic transitions, the spectroscopy of molecules is more intricate because vibrational and rotational degrees of freedom come into play as well. Early observations of absorption or emission by molecules were characterized as band spectra-in contrast to the line spectra exhibited by atoms. It is now understood that these bands reflect closely-spaced vibrational and rotational energies augmenting the electronic states of a molecule. With improvements in spectroscopic techniques over the years, it has become possible to resolve individual vibrational and rotational transitions. This has provided a rich source of information on molecular geometry, energetics and dynamics. Molecular spectroscopy has also contributed significantly to analytical chemistry, environmental science, astrophysics, biophysics and biochemistry.
Reduced Mass
Consider a system of two particles of masses $m_1$ and $m_2$ interacting with a potential energy which depends only on the separation of the particles. The classical energy is given by $E=\frac{1}{2} m_1 \dot{\vec{r}}_1^2+\frac{1}{2} m_2 \dot{\vec{r}}_2^2+V (|\vec{r}_2-\vec{r}_1|) \label{1}$ the dots signifying derivative wrt time. Introduce two new variables, the particle separation $\vec{r}$ and the position of the center of mass $\vec{R}$:
$\vec{r}=\vec{r}_2-\vec{r}_1 \mbox{,}\hspace{20pt}\vec{R}=\dfrac{m_1 \vec{r}_1+m_2\vec{r}_2}{m}\label{2}$
where $m=m_1+m_2$. In terms of the new coordinates
$\vec{r}_1=\vec{R}+\frac{m_2}{m} \vec{r} \mbox{,}\hspace{20pt}\vec{r}_2=\vec{R}-\frac{m_1}{m} \vec{r}\label{3}$
and
$E=\dfrac{1}{2}m\dot{\vec{R}}^2+\dfrac{1}{2}\mu\dot{\vec{r}}^2+V(r)\label{4}$
where $r=|\vec{r}|$ and $\mu$ is called the $reduced\hspace{2pt}mass$
$\mu\equiv\dfrac{m_1 m_2}{m_1+m_2}\label{5}$
An alternative relation for reduced mass is
$\dfrac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}\label{6}$
reminiscent of the formula for resistance of a parallel circuit. Note that, if $m_2\rightarrow\infty$, then $\mu\rightarrow m_1$. The term containing $\dot{\vec{R}}$ represents the kinetic energy of a single hypothetical particle of mass $\mu$ located at the center of mass $\vec{R}$. The remaining terms represent the $relative$ motion of the two particles. It was the appearance of a $single$ particle of effective mass $\mu$ moving in the potential field $V(r)$.
$E_{rel}=\dfrac{1}{2} \mu \dot{\vec{r}^2}+V(r)= \dfrac{\vec{p}^2}{2\mu}+V(r)\label{7}$
We can thus write the Schrödinger equation for the relative motion
$\left\{-\dfrac{\hbar^2}{2 \mu} \bigtriangledown^2+V(r) \right\}\psi (\vec{r})= E \psi (\vec{r}) \label{8}$
When we treated the hydrogen atom, it was assumed that the nuclear mass was infinite. In that case we can set $\mu =m$, the mass of an electron. The Rydberg constant for infinite nuclear mass was calculated to be
$R_\infty = \dfrac{2\pi^2me^4}{h^3c}=109,737 \text{cm} ^{-1}\label{9}$
If instead, we use the reduced mass of the electron-proton system
$\mu = \dfrac{mM}{m+M} \approx \dfrac{1836}{1837}\, m \approx 0.999456 \, m \label{10}$.
This changes the Rydberg constant for hydrogen to
$R_{H}\approx 109,677 \, cm^{-1}\label{11}$
in perfect agreement with experiment.
In 1931, H. C. Urey evaporated four liters of hydrogen down to one milliliter and measured the spectrum of the residue. The result was a set of lines displaced slightly from the hydrogen spectrum. This amounted to the discovery of deuterium, or heavy hydrogen, for which Urey was awarded in 1934 Nobel Prize in Chemistry. Estimating the mass of the deuteron, 2H1, as twice that of the proton, gives
$R_{D}\approx 109,707 \, cm^{-1}\label{12}$
Another interesting example involving reduced mass concerns positronium, a short-lived combination of an electron and a positron-the electron's antiparticle. The electron and position mutually annihilate with a half-life of approximately 10-7 sec. and positroium decays into gamma rays. The reduced mass of positronium is
$\mu = \frac{m \times m}{m+m} = \frac{m}{2} \label{13}$
half the mass of the electron. Thus the ionization energy equals 6.80 eV, half that of hydrogen atom. Positron emission tomography (PET) provides a sensitive scanning technique for functioning living tissue, notably the brain. A compound containing a positron-emitting radionuclide, for example, 11C, 13N, 15O or 18F, is injected into the body. The emitted positrons attach to electrons to form short-lived positronium, and the annihilation radiation is monitored.
Vibration of Diatomic Molecules
A diatomic molecule with nuclear masses MA, MB has a reduced mass
$\mu =\frac{M_{A}M_{B}}{M_{A}+M_{B}}\label{14}$
Solution of the electronic Schrödinger equation gives the energy as a function of internuclear distance Eelec(R). This plays the role of a potential energy function for motion of the nuclei V(R), as sketched in Fig. 2. We can thus write the Schrödinger equation for vibration
$\begin{Bmatrix} -\frac{\hbar^2}{2\mu }\frac{d^2}{dR^2} +V(R)\end{Bmatrix}\chi (R)=E_{\chi }(R)\label{15}$
If the potential energy is expanded in a Taylor series about R = Re
$V(R)=V(R_{e})+(R-R_{e})V'(R_{e})+\frac{1}{2}(R-R_{e})^2V"(R_{e})+...\label{16}$
An approximation for this expansion has the form of a harmonic oscillator with
$V(R)\approx \frac{1}{2}k(R-R_{e})^2\label{17}$
The energy origin can be chosen so V (Re) = 0. At the minimum of the potential, V'(Re) = 0. The best fit to the parabola (17) is obtained with a force constant set equal to
$k\approx \frac{d^2V(R)}{dR^2}\mid _{R\, =\, R_{e}}\label{18}$
From the solution for the harmonic oscillator, we identify the ground state vibrational energy, with quantum number $\nu$ = 0
$E_{0}=\hbar\omega =\hbar\sqrt{\frac{k}{\mu }}\label{19}$
The actual dissociation energy from the ground vibrational state is then approximated by
$D_{0}\approx D_{e}-\frac{1}{2}\hbar\omega\label{20}$
In wavenumber units
$hcD_{0}\approx hcD_{e}-\frac{1}{2}\tilde{\nu }\: cm^{-1}\label{21}$
An improved treatment of molecular vibration must account for anharmonicity, deviation from a harmonic oscillator. Anharmonicity results in a finite number of vibrational energy levels and the possibility of dissociation of the molecule at sufficiently high energy. A very successful approximation for the energy of a diatomic molecule is the Morse potential:
$V(R)=hcD_{e}\begin{Bmatrix}1-e^{a(R-R_{e})}\end{Bmatrix}^2\; \; \; a=\begin{pmatrix}\frac{\mu \omega ^2}{2hcD_{e}}\end{pmatrix}^{\frac{1}{2}}\label{22}$
Note that V (Re) = 0 at the minimum of the potential well. The Schrödinger equation for a Morse oscillator can be solved to give the energy levels
$E_{\upsilon }=(\upsilon +\frac{1}{2})\hbar\omega-(\upsilon+\frac{1}{2} )^2\hbar\omega x_{e}\label{23}$
or, expressed in wavenumber units,
$hcE_{\upsilon }=(\upsilon +\frac{1}{2})\tilde{\nu }-(\upsilon +\frac{1}{2})^2x_{e}\tilde{\nu }\label{24}$
Higher vibrational energy levels are spaced closer together, just as in real molecules. Vibrational transitions of diatomic molecules occur in the infrared, roughly in the range of 50-12,000 cm-1. A molecule will absorb or emit radiation only if it has a non-zero dipole moment. Thus HCl is infrared active while H2 and Cl2 are not.
Vibration of Polyatomic Molecules
A molecule with N atoms has a total of 3N degrees of freedom for its nuclear motions, since each nucleus can be independently displaced in three perpendicular directions. Three of these degrees of freedom correspond to translational motion of the center of mass. For a nonlinear molecule, three more degrees of freedom determine the orientation of the molecule in space, and thus its rotational motion. This leaves 3N - 6 vibrational modes. For a linear molecule, there are just two rotational degrees of freedom, which leaves 3N - 5 vibrational modes. For example, the nonlinear molecule H2O has three vibrational modes while the linear molecule CO2 has four vibrational modes. The vibrations consist of coordinated motions of several atoms in such a way as to keep the center of mass stationary and nonrotating. These are called the normal modes. Each normal mode has a characteristic resonance frequency $\tilde{\nu _{i}}$, which is usually determined experimentally. To a reasonable approximation, each normal mode behaves as an independent harmonic oscillator of frequency $\tilde{\nu _{i}}$. The normal modes of H2O and CO2 are pictured below.
A normal mode will be infrared active only if it involves a change in the dipole moment. All three modes of H2O are active. The symmetric stretch of CO2 is inactive because the two C-O bonds, each of which is polar, exactly compensate. Note that the bending mode of CO2 is doubly degenerate. Bending of adjacent bonds in a molecule generally involves less energy than bond stretching, thus bending modes generally have lower wavenumbers than stretching modes.
Rotation of Diatomic Molecules
The rigid rotor model assumes that the internuclear distance R is a constant. This is not a bad approximation since the amplitude of vibration is generally of the order of 1% of R. The Schrödinger equation for nuclear motion then involves the three-dimensional angular momentum operator, written $\hat{J}$ rather than $\hat{L}$ when it refers to molecular rotation. The solutions to this equation are already known and we can write
$\frac{\hat{J}^2}{2\mu R^2}Y_{JM}(\theta ,\phi )=E_{J}Y_{JM}(\theta,\phi)\; \; \; J=0,1,2...\; \; \; M=0,\pm \, 1...\pm\,J\label{25}$
where YJM($\theta,\phi$) are spherical harmonics in terms of the quantum numbers J and M, rather than l and m. Since the eigenvalues of $\hat{J}^2$ are $J(J +1)\hbar^2$, the rotational energy levels are
$E_{J}=\frac{\hbar^2}{2I}J(J+1)\label{26}$
The moment of inertia is given by
$I=\mu R^2=M_{A}R^{2}_{A}+M_{B}R^{2}_{B}\label{27}$
where RA and RB are the distances from nuclei A and B, respectively, to the center of mass. In wavenumber units, the rotational energy is expressed
$hcE_{J}=BJ(J+1)cm^{-1}\label{28}$
where B is the rotational constant. The rotational energy-level diagram is shown in Fig.5. Each level is (2J + 1)-fold degenerate. Again, only polar molecules can absorb or emit radiation in the course of rotational transitions. The radiation is in the microwave or far infrared region. The selection rule for rotational transitions is $\Delta$J = $\pm$1.
Molecular Parameters from Spectroscopy
Following is a table of spectroscopic constants for the four hydrogen halides:
The force constant can be found from the vibrational constant. Equating the energy quantities $\hbar\omega=hc\tilde{\nu }$, we find
$\omega=2\pi c\tilde{\nu }=\sqrt{\frac{k}{\mu }}\label{29}$
Thus
$k=(2\pi c\tilde{\nu })^2\mu\label{30}$
with
$\mu=\frac{m_{A}m_{B}}{m_{A}+m_{B}}=\frac{M_{A}M_{B}}{M_{A}+M_{B}}u\label{31}$
where u = 1.66054 x 10-27 kg, the atomic mass unit. MA and MB are the conventional atomic weights of atoms A and B (on the scale 12C = 12). Putting in numerical factors
$k=58.9\, \times \, 10^{-6}(\tilde{\nu }/cm^{-1})^2\frac{M_{A}M_{B}}{M_{A}+M_{B}}N/m \label{32}$
This gives 958.6, 512.4, 408.4 and 311.4 N/m for HF, HCl, HBr and HI, respectively. These values do not take account of anharmonicity.
The internuclear distance R is determined by the rotational constant. By definition,
$hcB=\frac{\hbar^2}{2I}\label{33}$
Thus
$B=\frac{\hbar}{4\pi cI}\label{34}$
with
$I=\mu R^2=\frac{m_{A}m_{B}}{m_{A}+m_{B}}R^2=\frac{M_{A}M_{B}}{M_{A}+M_{B}}uR^2\; \; kg\: m^2\label{35}$
Solving for R,
$R=410.6 / \sqrt{\frac{M_{A}M_{B}}{M_{A}+M_{B}}(B/cm^{-1})}\: \: pm\label{36}$
For the hydrogen halides, HF, HCl, HBr, HI, we calculate R = 92.0, 127.9, 142.0, 161.5 pm, respectively.
Rotation of Nonlinear Molecules
A nonlinear molecule has three moments of inertia about three principal axes, designated Ia, Ib and Ic. The classical rotational energy can be written
$E=\dfrac{J^2_{a}}{2I_{a}}+\frac{J^2_{b}}{2I_{b}}+\dfrac{J^2_{c}}{2I_{c}}\label{37}$
where Ja, Jb, Jc are the components of angular momentum about the principal axes. For a spherical rotor, such as CH4 or SF6, the three moments of inertia are equal to the same value I. The energy simplifies to J2/2I and the quantum-mechanical Hamiltonian is given by
$\hat{H}=\frac{\hat{J}^2}{2I}\label{38}$
The eigenvalues are
$E_{J}=\frac{\hbar^2}{2I}J(J+1)\; \;\; J=0,1,2..\label{39}$
just as for a linear molecule. But the levels of a spherical rotor have degeneraciesof (2J + 1)2 rather than (2J + 1).
A symmetric rotor has two equal moments of inertia, say Ic = Ib $\neq$ Ia. The molecules NH3, CH3Cl and C6H6 are examples. The Hamiltonian takes the form
$\hat H = \frac{\hat J_a^2}{2I_a}+\frac{\hat J_b^2 + \hat J_c^2}{2I_b} = \frac{\hat J^2}{2I_b} + (\frac{1}{2I_a} - \frac{1}{2I_b}) \hat J_a^2 \label{40}$
Since it is possible to have simultaneous eigenstates of $\hat J^2$ and one of its components $\hat J_a$, the energies of a symmetric rotor have the form
$E_{JK}=\frac{J(J+1)}{2I_{b}}+(\frac{1}{2I_{a}}-\frac{1}{2I_{b}})K^2\; \; J=0,1,2...\; \;\; K=0\pm 1,\pm2,...\pm J\label{41}$
There is, in addition, the (2J + 1)-fold M degeneracy.
Electronic Excitations in Diatomic Molecules
The quantum states of molecules are composites of rotational, vibrational and electronic contributions. The energy spacing characteristic of these different degrees of freedom vary over many orders of magnitude, giving rise to very different spectroscopic techniques for studying rotational, vibrational and electronic transitions. Electronic excitations are typically of the order of several electron volts, 1 eV being equivalent to approximately 8000 cm-1 or 100 kJ mol-1. As we have seen, typical energy differences are of the order of 1000 cm-1 for vibration and 10 cm-1 for rotation. Fig. 6 gives a general idea of the relative magnitudes of these energy contributions. Each electronic state has a vibrational structure, characterized by vibrational quantum numbers v and each vibrational state has a rotational structure, characterized by rotational quantum numbers J and M.
Every electronic transition in a molecule is accompanied by changes in vibrational and rotational states. Generally, in the liquid state, individual vibrational transitions are not resolved, so that electronic spectra consist of broad bands comprising a large number of overlapping vibrational and rotational transitions. Spectroscopy on the gas phase, however, can often resolve individual vibrational and even rotational transitions.
When a molecule undergoes a transition to a different electronic state, the electrons rearrange themselves much more rapidly than the nuclei. To a very good approximation, the electronic state can be considered to occur instantaneously, while the nuclear configuration remains fixed. This is known as the Franck-Condon principle. It has the same physical origin as the Born-Oppenheimer approximation, namely the great disparity in the electron and nuclear masses. On a diagram showing the energies of the ground and excited states as functions of internuclear distance, Franck-Condon behavior is characterized by vertical transitions, in which R remains approximately constant as the molecule jumps from one potential curve to the other.
In a vibrational state $\upsilon$ = 0 the maximum of probability for the internuclear distance R is near the center of the potential well. For all higher values vibrational states, maxima of probability occur near the two turning points of the potential-where the total energy equals the potential energy. These correspond on the diagrams to the end points of the horizontal dashes inside the potential curve.
Contributors and Attributions
Seymour Blinder (Professor Emeritus of Chemistry and Physics at the University of Michigan, Ann Arbor)
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Density-functional theory is a set of theories in statistical mechanics that profit from the fact that the Helmholtz energy function of a system can be cast as a functional of the density. That is, the density (in its usual sense of particles per volume), which is a function of the position in inhomogeneous systems, uniquely defines the Helmholtz energy. By minimizing this Helmholtz energy one arrives at the true Helmholtz energy of the system and the equilibrium density function. The situation parallels the better known electronic density functional theory, in which the energy of a quantum system is shown to be a functional of the electronic density (see the theorems by Hohenberg, Kohn, Sham, and Mermin).
Starting from this fact, approximations are usually made in order to approach the true functional of a given system. An important division is made between local and weighed theories. In a local density theory in which the dependence is local, as exemplified by the (exact) Helmholtz energy of an ideal system:
$A_{id}=k_BT\int dr \rho(r) [\log \rho(r) -1 -U(r)],$
where $U(r)$ is an external potential. It is an easy exercise to show that Boltzmann's barometric law follows from minimization. An example of a weighed density theory would be the (also exact) excess Helmholtz energy for a system of 1-dimensional hard rods:
$A_{ex}=-k_BT\int dz \rho(z) \log [1-t(z)],$
where $t(z)=\int_{z-\sigma}^z dy \rho(y)$, precisely an average of the density over the length of the hard rods, $\sigma$. "Excess" means "over ideal", i.e., it is the total $A=A_{id}+A_{ex}$ that is to be minimized.
Variational Method
The Variational Method is a mathematical method that is used to approximately calculate the energy levels of difficult quantum systems. It can also be used to approximate the energies of a solvable system and then obtain the accuracy of the method by comparing the known and approximated energies. The Schrödinger equation can be solved exactly for our model systems including Particle in a Box (PIB), Harmonic Oscillator (HO), Rigid Rotor (RR), and the Hydrogen Atom. However, for systems that have more than one electron, the Schrödinger equation cannot be analytically solved and requires approximation like the variational method to be used.
Example $1$: Helium
The Schrödinger equation takes the form:
$\hat{H} \psi = E \psi \label{schro}$
with the Hamiltonian operator $\hat{H}$ representing the sum of kinetic energy ($T$) and potential energy ($V$):
$\hat {H} = T + V$
For the helium atom, the Hamiltonian can be expanded to the following:
$\hat{H} = -\dfrac{\hbar^2}{2m_e}\bigtriangledown_{el_{1}}^2 -\dfrac{\hbar^2}{2m_e}\bigtriangledown_{el_{2}}^2 - \dfrac {Ze^2}{4\pi\epsilon_0 r_1} - \dfrac {Ze^2}{4\pi\epsilon_0 r_2} + \dfrac {e^2}{4\pi \epsilon_0 r_{12}}\label{8}$
where
• $r_1$ and $r_2$ are distances of electron 1 and electron 2 from the nucleus
• $r_{12}$ is the distance between the two electrons ($r_{12})= | r_1 - r_2|$
• $Z$ is the charge of the nucleus (2 for helium)
Solving the Schrödinger equation for helium is impossible to solve because of the electron-electron repulsion term in the potential energy:
$\dfrac {e^2}{4\pi\epsilon_0 r_{12}}$
Because of this, approximation methods were developed to be able to estimate energies and wavefunctions for complex systems. The variational method is one such approxation and perturbation theory is another.
Because of a chemist's dependence on said approximation methods, it is very important to understand the accuracy of these methods. This can be done by applying the method to simple known systems.
The variational method is useful because of its claim that the energy calculated for the system is always more than the actual energy.
$E_{\phi} \ge E_o$
It does this by introducing a trial wavefunction and then calculating the energy based on it. If the trial wavefunction is chosen correctly, the variational method is quite accurate. If the trial wavefunction is poor, the energy calculated will not be very accurate, but it will always be larger than the true value. The greater than or equal symbol is used because if by chance the trial wavefunction that is guessed is the actual wavefunction that describes a system, then the trial energy is equal to the true energy.
Example $2$: Particle in a 1D Box
When trying to find the energy of a particle in a box, set the boundaries at x = 0 and x = L as shown in the diagram below.
The true solution of the Schrödinger equation is well known as:
$\psi _{n}(x)=\sqrt{\dfrac{2}{L}} sin \dfrac{n\pi x}{L}$
With the energy levels being:
$E_{n}=\dfrac{\hbar^2\pi^2}{2mL^2}\, n^2=\dfrac{h^2}{8mL^2}n^2\; \; \; \; n=1,2,,3...$
The following describes the variational method equation that is used to find the energy of the system. H is the Hamiltonian operator for the system.
How to choose a Trial Wavefunction?
• The trial function must have the characteristics that classify it as a wavefunction, ie. continuous, etc.
• The trial function need to have the same general shape as the true wavefunction.
• The trial function must have the same boundary conditions.
• To improve accuracy, the trial wavefunction can be represented as linear combinations of single trial wavefunctions.
Variational Theorem
$E_{trial} = \dfrac{\langle \phi_{trial}| \hat{H} | \phi_{trial} \rangle }{\langle \phi_{trial}| \phi_{trial} \rangle} \label{theorem}$
The denominator above is only necessary if the trial wavefunction needs to normalized.
How to know if a given trial wavefunction is normalized
• When given a trial wavefunction, sometimes the problem states if it is normalized or not.
• When forced to decide, if there is a constant in front of the function, that is usually assumed to be the normalization constant. If a constant is not present then there is no normalization constant and the denominator in Equation \ref{theorem} is necessary.
Picking a trial wavefunction for particle in a box
A trial function for the $n=1$ system is:
$\phi_{trial} = x(L-x)$
where this function is not normalized. Is this a good guess for the system? To find out we must apply the variational theorem to find the energy.
Example $3$: Find Trial Energy
$E_{trial} = \dfrac{\langle \phi_{trial}| \hat{H} | \phi_{trial} \rangle }{\langle \phi_{trial}| \phi_{trial} \rangle}$
For PIB we know our Hamiltonian is $\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}$
$E_{trial} = \dfrac{\langle \phi_{trial}| \dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2} | \phi_{trial} \rangle }{\langle \phi_{trial}| \phi_{trial} \rangle}$
By putting in our trial $\phi$, our trial energy becomes:
$E_{trial} = \dfrac{\langle Nx(L-x)| \dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2} |Nx(L-x)\rangle}{\langle Nx(L-x)|Nx(L-x)\rangle}$
Solve the numerator:
Then we calculate the numerator of $(1)$:
$\langle\varphi | H | \varphi\rangle = \int_{0}^{L}x(L−x) (- \dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}) x(L−x)dx = \- \dfrac{\hbar^2}{2m} \int_{0}^{L} (xL - x^2) (-2) dx = \dfrac{\hbar^2}{m} (L\dfrac{x^2}{2} - \dfrac{x^3}{3}) \Biggr\rvert_{0}^{L} = \dfrac{\hbar^2}{m} (\dfrac{L^3}{2} - \dfrac{L^3}{3}) = \dfrac{\hbar^2}{m} \dfrac{L^3(3-2)}{6} = \dfrac{\hbar^2 L^3}{6m}$
Find the denominator:
$\langle x(L-x)|x(L-x)\rangle = N^2$
$N^2 = \langle (xL-x^2)(xL-x^2)\rangle = langle x^2L^2-x^3L-x^3L+x^4\rangle = \int_{0}^{L} x^2L^2-2x^3L+x^4 dx = \dfrac{L^5}{3}-\dfrac{L^5}{2}+\dfrac{L^5}{5} = \dfrac{L^5}{30}$
$N = \sqrt{\dfrac{L^5}{30}}$
Now put them together
$\dfrac{\langle\varphi| H | \varphi\rangle}{\langle\varphi |\varphi\rangle} = \dfrac{30}{L^5} \dfrac{\hbar^2 L^3}{6m} = \dfrac{5\hbar^2}{mL^2}$
The conclusion is:
$E_1 \leq \dfrac{5\hbar^2}{mL^2}$
As seen in the diagram above, the trial wavefunction follows the shape of the true wavefunction and has the same boundary conditions, so it is a good guess for the system.
A Different Trial Wavefunction: Linear Combination of Wavefunctions
The accuracy of the variational method can be greatly enhanced by the use of a trial function with additional terms. We can try this out by repeating the earlier steps with the following wavefunction:
$\phi_{trial} = x(L-x)+Cx^2(L-x)$
The normalization constant was omitted because it is not necessary to find the energy. $C$ in this equation is a variational parameter. When inserting this equation into our variational theorem and then into a programming application we yield:
$E_{trial} = \dfrac{\int_{0}^{L} \phi_{trial}(\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\phi_{trial})dx}{\int_{0}^{L} \phi_{trial}\phi_{trial} dx} = \dfrac{\dfrac{1}{105}C^2+\dfrac{1}{15}C+\dfrac{1}{6}}{\dfrac{1}{630}C^2+\dfrac{1}{70}C+\dfrac{1}{30}}$
Because the trial energy is always larger than the actual energy, we can minimize the trial energy by taking the derivative with respect to C, setting it equal to zero and solving for C. The smaller energy when plugging in all found values of C is the closest to the true energy.
$\dfrac{d}{dC}(E_{trial}) = \dfrac{\dfrac{1}{105}C^2+\dfrac{1}{15}C+\dfrac{1}{6}}{\dfrac{1}{630}C^2+\dfrac{1}{70}C+\dfrac{1}{30}}-\dfrac{\dfrac{1}{105}C^2+\dfrac{1}{15}C+\dfrac{1}{6}}{(\dfrac{1}{630}C^2+\dfrac{1}{70}C+\dfrac{1}{30})^2}\dfrac{1}{315}C+\dfrac{1}{70}$
$0=3\dfrac{4C^2+14C-21}{(C^2+9C+21)^2}$
$C = -\dfrac{7}{4}+\dfrac{1}{4}\sqrt{133} ; -\dfrac{7}{4}-\dfrac{1}{4}\sqrt{133}$
Plugging in we get a smaller value when using the first term for $C$ and we get that
$E_{trial} = 4.9348748$ $\%error = 0.0015\%$
This error is much smaller than that of our first wavefunction, which shows that a linear combination of terms can be more accurate than one term by itself and describe the system much better. It is actually necessary to use this method of guessing for the wavefunction for excited states of a system. If we were to do the same for the first excited state of the particle in a box, then the percent error would go from 6.37% error to 0.059% error. This shows how crucial this method of linearly combining terms to form trial wavefunctions becomes with the excited states of systems. Without this method the excited states would not be nearly as accurate as needed.
The Variational Theorem states that the trial energy can be only greater or equal to the true energy (Equation \ref{theorem}). But when does the Variational Method give us the exact energy that we are looking for?
Example $4$: Achieving True Energy with the Variational Method
Let's use the Harmonic Oscillator as our system.
$\phi_{trial} = e^{-\alpha x^2}$
$V = \dfrac{1}{2} kx^2$
$T = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}$
First find the denominator:
$\langle \phi_{trial}|\phi_{trial}\rangle$
$\int_{-\infty}^{\infty} dx(e^{-\alpha x^2})*(e^{-\alpha x^2}) = \int_{-\infty}^{\infty}(e^{-2\alpha x^2})dx$
$\langle \phi_{trial}|\phi_{trial}\rangle = \sqrt{\dfrac{\pi}{2\alpha}}$
Then find the numerator:
$\langle\varphi | H | \varphi\rangle = \langle\varphi | T | \varphi\rangle +\langle\varphi | V | \varphi\rangle$
$= \dfrac{1}{2k}\dfrac{1}{4\alpha}\sqrt{\dfrac{\pi}{2\alpha}}\dfrac{\hbar^2 \alpha}{2m}$
$E_{\phi} = \dfrac{k}{8 \alpha} +\dfrac{\hbar^2 \alpha}{2m}$
Now because there is a variational constant, $\alpha$ we need to minimize it
$\dfrac{dE_{\phi}}{d\alpha} = -\dfrac{k}{8\alpha^2}+\dfrac{\hbar^2}{2m} = 0$
$\alpha = \dfrac{\sqrt{km}}{2\hbar}$
Now we plug this into the $E_{\phi}$ for $\alpha$ and we will find $E_{\phi min}$
$E_{\phi min} = \dfrac{\hbar}{4} \sqrt{\dfrac{k}{m}} + \dfrac{\hbar}{4} \sqrt{\dfrac{k}{m}}$
Where $w = \sqrt{\dfrac{k}{m}}$
$E_{\phi min} = \dfrac{1}{2} \hbar w$
Now we can plug in our minimum $\alpha$ into our $\phi_{trial}$ and we need to introduce our normalization constant.
$\phi_{\alpha min} = (\dfrac{2\alpha}{\pi})^{\dfrac{1}{4}} e^{-\dfrac{\sqrt{km}x^2}{2\hbar}}$
If you notice, this is the exact equation for the Harmonic Oscillator ground state. We were able to find this by initially guessing a good wave function, and varying and minimizing the variational constant.
For a more in depth step by step video on this example: Click here
Linear Variational Method
Another approximation method that is used to study molecules is the linear variational method. It includes having a trial wavefunction with a linear combination of $n$ linearly independent functions of f. More information can be found here.
Contributors and Attributions
• Ashleigh Falasco
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Historically, the quantum calculations for molecules were performed as LCAO MO, i.e. Linear Combination of Atomic Orbitals - Molecular Orbitals. This means that molecular orbitals are formed as a linear combination of atomic orbitals:
$|\psi_i \rangle = \sum_i^n c_{ij} | \phi _i \rangle$
where $|\psi_i \rangle$ is the i-th molecular orbital, $c_{ij}$ are the coefficients of linear combination, $\phi_i$ is the $j^{th}$ atomic orbital, and $n$ is the number of atomic orbitals.
Strictly speaking, Atomic Orbitals (AO) are solutions of the Hartree-Fock equations for the atom, i.e., a wavefunctions for a single electron in the atom. Anything else is not really an atomic orbital. Some things are similar though, and there is a lot of confusion in the terminology used. Later on, the term atomic orbital was replaced by "basis function" or "contraction," when appropriate. Early, the Slater Type Orbitals (STO's) were used as basis functions due to their similarity to atomic orbitals of the hydrogen atom. They are described by the function depending on spherical coordinates:
$| \phi_i (\zeta,n,l,m; r, \theta, \phi) \rangle = N r^{n-1}e^{\zeta r} Y_l^m (\theta,\phi)$
where $N$ is a normalization constant, $\zeta$ is called "exponent". The $r$, $\theta$, and $\phi$ are spherical coordinates, and $Y_l^m$ is the angular momentum part (function describing "shape"). The $n$, $l$, and $m$ are quantum numbers: principal, angular momentum, and magnetic; respectively.
Unfortunately, functions of this kind are not suitable for fast calculations of necessary two-electron integrals. That is why, the Gaussian Type Orbitals (GTOs) were introduced. You can approximate the shape of the STO function by summing up a number of GTOs with different exponents and coefficients. Even if you use 4 or 5 GTO's to represent STO, you will still calculate your integrals much faster than if original STOs are used. The GTO (called also Cartesian Gaussian) is expressed as:
$G_{l, m, n}(x, y,z) = N e^{-\alpha r^2}x^ly^mz^n$
• where $N$ is a normalization constant,
• $\alpha$ is called "exponent" and
• the $x$, $y$, and $z$ are Cartesian coordinates.
The $l$, $m$, and $n$ are not quantum numbers, but simply integral exponents at Cartesian coordinates:
$r^2 = x^2 + y^2 + z^2$
Calling Gaussians GTOs is probably a misnomer, since they are not really orbitals; they are simpler functions. In recent literature, they are frequently called Gaussian primitives. The main difference is that $r^{n-1}$, the preexponential factor, is dropped, the $r$ in the exponential function is squared, and angular momentum part is a simple function of Cartesian coordinates. The absence of $r^{n-1}$ factor restricts single Gaussian primitive to approximating only 1s, 2p, 3d, 4f ... orbitals (with no nodes). It was done for practical reasons, namely, for fast integral calculations. However, combinations of Gaussians are able to approximate correct nodal properties of atomic orbitals by taking them with different signs (see below). Following Gaussian functions are possible:
• $1s = N e^{-\alpha r^2}$
• $2p_x = N e^{-\alpha r^2}x$
• $2p_y = N e^{-\alpha r^2}y$
• $2p_z = N e^{-\alpha r^2}z$
• $3d_{xx} = N e^{-\alpha r^2}x^2$
• $3d_{xy} = N e^{-\alpha r^2}xy$
• $3d_{xz} = N e^{-\alpha r^2}xz$
• $3d_{yy} = N e^{-\alpha r^2}y^2$
• $3d_{yz} = N e^{-\alpha r^2}yz$
• $3d_{zz} = N e^{-\alpha r^2}z^2$
• $4d_{xxx} = N e^{-\alpha r^2}x^3$
• $4d_{xxy} = N e^{-\alpha r^2}x^2y$
• $4d_{xyz} = N e^{-\alpha r^2}zxyz$
etc.
Sometimes, the so-called scale factor, $f$, is used to scale all exponents in the related Gaussians. In this case, the Gaussian function is written as:
$G_{l,m,n,f}(x, y,z) = N e^{-\alpha r^2}x^ly^mz^n$
Be careful not to confuse it with "f" for the f-orbital. The sum of exponents at Cartesian coordinates,
$L = l+m+n$
is used analogously to the angular momentum quantum number for atoms, to mark functions as s-type (L=0), p-type (L=1), d-type (L=2), f-type (L=3), etc.
There is a problem with d-type and higher functions. There are only five linearly independent and orthogonal d orbitals, while there are 6 possible Cartesian Gaussians. If we use all six, we are also introducing a 3s type function since:
$3d_{xx} + 3d_{yy} + 3d_{zz} = N ( x^2 + y^2 + z^2) e^{-\alpha r^2} = N r^2e^{-\alpha r^2} = 3s$
More recently, this effect was studied for sulfur by (Sapio & Topiol, 1989).
Examination of f-type functions shows that there are 10 possible Cartesian Gaussians, which introduce , and type contamination. However, there are only 7 linearly independent f-type functions. This is a major headache since some programs remove these spurious functions and some do not. Of course, the results obtained with all possible Cartesian Gaussians will be different from those obtained with a reduced set.
Contractions
Gaussian primitives are usually obtained from quantum calculations on atoms (i.e. Hartree-Fock or Hartree-Fock plus some correlated calculations, e.g. CI). Typically, the exponents are varied until the lowest total energy of the atom is achieved (Clementi et al., 1990). In some cases, the exponents are optimized independently. In others, the exponents are related to each other by some equation, and parameters in this equation are optimized (e.g. even-tempered or "geometrical" and well-tempered basis sets). The primitives so derived describe isolated atoms and cannot accurately describe deformations of atomic orbitals brought by the presence of other atoms in the molecule. Basis sets for molecular calculations are therefore frequently augmented with other functions which will be discussed later.
For molecular calculations, these Gaussian primitives have to be contracted, i.e., certain linear combinations of them will be used as basis functions. The term contraction means "a linear combination of Gaussian primitives to be used as basis function." Such a basis function will have its coefficients and exponents fixed. The contractions are sometimes called Contracted Gaussian Type Orbitals (CGTO). To clear things up, a simple example from Szabo and Ostlund, 1989. The coefficients and exponents of Gaussian expansion which minimizes energy of the hydrogen atom were derived by Huzinaga, 1965. Four s-type Gaussians were used to represent 1s orbital of hydrogen as:
$N_i$ is a normalization constant for a given primitive. In the case of Gaussians of type s it is equal to $(2\alpha/\pi)^{3/4}}. These primitives may be grouped in 2 contractions. The first contraction contains only 1 primitive: $| \phi_1 \rangle = N_1 e^{-0.123317 r^2}$ Three primitives are present in the second contraction: \(N$ is a normalization constant for the whole contraction.
In this case, 4 primitives were contracted to 2 basis functions. It is frequently denoted as (4s) [2s] contraction (some use (4s)/[2s] notation). The coefficients in function are then fixed in subsequent molecular calculations.
The way in which contractions are derived is not easy to summarize. Moreover, it depends upon the intended use for the basis functions. It is a good idea to always read the original paper which describes the way in which contractions have been done. Some basis sets are good for geometry and energies, some are aimed at properties (e.g. polarizability), some are optimized only with Hartree-Fock in mind, and some are tailored for correlated calculations. Finally, some are good for anions and other for cations and neutral molecules. For some calculations, a good representation of the inner (core) orbitals is necessary (e.g. for properties required to analyze NMR spectrum), while other require best possible representation of valence electrons. WHY ARE CONTRACTIONS DONE
Obviously, the best results could be obtained if all coefficients in Gaussian expansion were allowed to vary during molecular calculations. Moreover, the computational effort (i.e. "CPU time") for calculating integrals in the Hartree-Fock procedure depends upon the 4th power in the number of Gaussian primitives. However, all subsequent steps depend upon the number of basis functions (i.e. contractions). Also, the storage required for integrals (when Direct SCF is not used) is proportional to the number of basis functions (not primitives!). Frequently the disk storage and not the CPU time is a limiting factor. The CPU time requirements are more acute when post-Hartree- Fock (e.g. correlated methods) are used, since the dependence upon the number of basis functions here is more steep than the 4th power.
There are two basic forms of contractions, namely "segmented" and "general". The segmented contractions are disjointed, i.e., given primitive appears only in one contraction. The example given above (4s) [2s] is a segmented contraction. Occasionally, one or two primitives may appear in more than one contraction, but this is an exception to the rule. The general contractions, on the contrary, allow each of the primitives to appear in each basis function (contraction). The segmented contractions are far more popular and will be described first. The reason for their popularity is not that they are better, but simply, that the most popular ab initio packages do not implement efficient integral calculations with general contractions. The computer code to perform integral calculations with general contractions is much more complex than that for the segmented case.
SEGMENTED CONTRACTIONS. TERMS AND NOTATION
The segmented basis sets are usually structured in such a way that the most diffuse primitives (primitives with the smallest exponent) are left uncontracted (i.e. one primitive per basis function). More compact primitives (i.e. those with larger exponents) are taken with their coefficients from atomic Hartree-Fock calculations and one or more contractions are formed. Then the contractions are renormalized. Sometimes different contractions share one or two functions (the most diffuse function(s) from the first contraction enter the next one).
Cartesian Gaussians are grouped in shells coresponding to the same value of angular momentum quantum number. Of course, these shells should not be confused with electron shells (i.e. electrons with the same principal quantum number: K n=1, L n=2, etc.). Quantum chemists must have run out of words on this one. And hence, we have s-shell, p-shell, d-shell, f-shell, g-shell, etc. The shell is a collection of Cartesian Gaussians that have the same L (see definition of Cartesian Gaussian above). Strictly speaking, the s-shell is a collection of s type Gaussians; p-shell is a collection of p-type Gaussians; d-shell is a collection of d-type Gaussians; and so on. Of course, combining primitives belonging to different shells within the same contraction does not make sense because primitives from different shells are orthogonal.
But even here there is a room for more confusion. Many basis sets use the same exponents for functions corresponding to the same principal quantum number, i.e., electronic shell. STO-3G is an example, as well as other basis sets from Pople's group. Atoms of the first and second row (i.e. Li - Ne, Na - Cl) have the same exponents for s- and p-type Gaussians formally associated with a given electron shell of the isolated atom. For the basis sets in which s- and p-type functions share the same exponents, the term SP-shell is used. Sometimes term L-shell is used by analogy to the 2nd electron shell. This approximation works very well in practice. Moreover, it is possible to write efficient code for calculating integrals for such cases. It is important to stress here that the distinction between inner orbitals and valence orbitals is kind of arbitrary and lingers from the past era of Slater orbitals. Contrac- tions consisting of primitives with large exponents are associated with inner atomic orbitals while more diffuse fuctions are allied with valence orbitals. Basis functions are not usually atomic orbitals, and in many cases, they do not even resemble orbitals of isolated atoms. In fact, examining coefficients of molecular orbitals frequently reveals that these "core" basis functions contribute substantially to the Highest Occupied Molecular Orbital (HOMO). It comes as a consequence of the fact that basis functions on a given center are usually not orthogonal to each other and "core" basis functions on different centers overlap to a great extent - situation not likely to occur with true atomic orbitals.
The early Gaussian contractions were obtained by a least square fit to Slater atomic orbitals. The number of contractions (not primitives!) used for representing a single Slater atomic orbital (i.e. zeta) was a measure of the goodness of the set. From this era we have terms like single zeta (SZ), double zeta (DZ), triple zeta (TZ), quadruple zeta (QZ), etc. In the minimal basis set (i.e. SZ) only one basis function (contraction) per Slater atomic orbital is used. DZ sets have two basis functions per orbital, etc. Since valence orbitals of atoms are more affected by forming a bond than the inner (core) orbitals, more basis functions are assigned frequently to describe valence orbitals. This prompted development of split-valence (SV) basis sets, i.e., basis sets in which more contractions are used to describe valence orbitals than core orbitals. That more basis functions are assigned to valence orbitals does not mean the valence orbitals incorporate more primitives. Frequently, the core orbitals are long contractions consisting of many primitive Gaussians to represent well the "cusp" of s type function at the position of the nucleus. The "zeta" terminology is often augmented with a number of polarization functions which will be described later. So, DZP means double-zeta plus polarization, TZP stands for triple-zeta plus polarization, etc. Sometimes the number of polarization functions is given, e.g. TZDP, TZ2P, TZ+2P stands for triple-zeta plus double polarization. Letter V denotes split valence basis sets, e.g., DZV represents basis set with only one contraction for inner orbitals, and two contractions for valence orbitals. The creativity here is enormous and spontaneous.
The minimal basis set is the smallest possible set, i.e., it contains only one function per occupied atomic orbital in the ground state. Actually, it always includes all orbitals from partially occupied subshells and valence p-type functions for elements from the first 2 groups of the periodic table. So for Li and Be atoms it has 2 s-type contractions and 1 p-type contraction. Minimal basis set for S atom has 3 s-type contractions and 2 p-type contractions. The most popular minimal basis sets are the STO-nG, where n denotes number of primitives in the contraction. These sets were obtained by least square fit of the combination of n Gaussian functions to a Slater type orbital of the same type with zeta = 1.0, For this set additional constraint is used, that exponents of corresponding Gaussian primitives are the same for basis functions describing orbitals with the same principal quantum number (e.g. the same primitives are used for 2s and 2p function). Then, these exponents are multipled by the square of zeta in the Slater orbital which described best the set of molecules. For details, see Szabo and Ostlund (1989) or original literature quoted on page 71 of Hehre et al. (1986). The STO-3G (i.e. 3 primitives per each function) is the most widely used set.
For other sets a more complicated notation needs to be used to specify the number of primitives and contractions explicitly. The parentheses () embrace the number of primitives that are given in the order of angular momentum quantum number. Square brackets [] are used to specify the number of resulting contractions. For example: (12s,9p,1d) means 12 primitives on s-shell, 9 primitives on p-shell, and 1 primitive on d-shell. This is sometimes abbreviated even further by skipping the shell symbols (12,9,1). The [5,4,1] means that s-shell has 5 contractions, p-shell has 4 contractions and d-shell has 1 contraction. To denote how contractions were performed, the following notation is frequently used: (12,9,1) [5,4,1] or (12,9,1)/[5,4,1] or (12s,9p,1d) [5s,4p,1d]. This means that 12 s-type primitives were contracted to form 5 s-type contractions, 9 p-primitives were contracted to 4 basis functions and 1 d-primitive was used as a basis function by itself. Note of caution here. The statement "9 p-primitives were contracted to 4 basis functions" actually means that 12 basis functions were created. Each p-type basis functions has 3 variants: , , and which differ in their Cartesian part (i.e., angular part). The same is true for d-, f-, and higher angular momentum functions.
The notation above does not say how many primitives are used in each contraction. The more elaborate notation explicitly lists the number of primitives in each contraction. For example: (63111,4311,1) means that there are 5 s-type contractions consisting of 6, 3, 1, 1 and 1 primitives, repectively. The p-shell consists of 4 basis functions with 4, 3, 1 and 1 primitives, and d-shell has 1 uncontracted primitive. Sometimes slashes are used instead of commas: (63111/4311/1). This is sometimes "abbreviated" to (633x1,432x1,1). There is also another notation to denote contractions as L(i/j/k/l...) for each shell corresponding to angular momentum quantum number equal to L. For example, the (63111,4311,1) basis set is represented as: s(6/3/1/1/1), p(4/3/1/1), and d(1). Of course, variants of this notation are also used. You can find this set written as: (6s,3s,1s,1s,1s/4p,3p,1p,1p/1d) or (6,6,1,1,1/4,3,1,1/1) or [6s,3s,1s,1s,1s/4p,3p,1p,1p/1d] (sic!). I did not study the combinatorics of this, but quantum chemists might have exhausted all combinations of digits, brackets and commas. However, if you ask 10 quantum chemists which notation is considered standard, you will get 20 different answers.
Sometimes the same primitive is incorporated in two contractions (i.e. is "doubled"); e.g., the popular Chandler-McLean (12,9) sulphur basis set (McLean and Chandler, 1980) is contracted as [6,5] with the scheme (631111,42111). If you count primitives contained in contractions for the s-shell, you get 13 primitives instead of 12. This means that one primitive is shared (i.e. doubled) between two contractions, 6- and 3-contraction in this case. It would make little sense to share a primitive between 6- and 1- or 3- and 1-contrac- tion since such contraction whould yield the basis set of the same quality as "undoubled" one. In some cases the smallest exponent from the first contraction is repeated in the next contraction as the largest one. In the above case, the basis set formaly represents a general contraction, but since only one function is doubled, it is used frequently in programs that do not support general contractions.
By convention, the primitives are listed as exponents and coefficients starting from the highest exponent. In tables of exponents and coefficients the numbers are frequently represented in an interesting way, with powers of 10 in parentheses, e.g. 457.3695 is denoted as 4.573696(+2) and 0.01403732 as 1.403732(-2). Of course, it is obvious if you know it. The typical basis set specification (Gordon, 1980, modified) is given below as an example:
Table: 66-31G basis set for silicon.
In the example above, corresponding exponents for s- and p-type contractions are equal but coefficients in s- and p-type contractions are different. Gaussian primitives are normalized here since coefficients for basis functions consisting of one primitive (last row) are exactly 1.0. The basis set above represents the following contraction (16s,10p) [4s,3p] or (6631,631).
To add to the confusion, the coefficients are sometimes listed either as original coefficients in atomic orbitals or are renormalized for the given contraction. In some cases coefficients are premultiplied by a normalization constant for a Gaussian primitive, but in most cases it is assumed that is already normalized (and this is the correct way!). You have to be prepared for surprises when entering explicit basis sets from the literature. Program manuals neglect basis sets description assuming it is common knowledge. When specifying structure of the basis sets for the entire molecule, slashes are used to separate information for different atoms (or rows, if basis sets for a given row have the same structure for all atoms). The information is given starting from the heaviest atoms. For example, the basis set for water would be given as (10s,5p,1d/5s,1p) [4s,2p,1d/2s,1p] in which case the contractions for oxygen atoms are (10,5p,1d) [4s,2p,1d] and for the hydrogen (5s,1p) [2s,1p].
Pople's basis sets
A different convention was adopted by Pople and coworkers. The basis set structure is given for the whole molecule, rather than particular a atom. This notation emphasizes also a split valence (SV) nature of these sets. Symbols like n-ijG or n-ijkG can be encoded as: n - number of primitives for the inner shells; ij or ijk - number of primitives for contractions in the valence shell. The ij notations describes sets of valence double zeta quality and ijk sets of valence triple zeta quality. Generally, in basis sets derived by Pople's group, the s and p contractions belonging to the same "electron shell" (i.e. corresponding formally to the same principal quantum number n) are folded into a sp-shell. In this case, number of s-type and p-type primitives is the same, and they have identical exponents. However, the coefficients for s- and p-type contractions are different.
Now, some examples. The 4-31G basis set for hydrogen (hydrogen has only valence electrons!) is a contraction (31) or (4s) [2s]; for first row atoms (8s,4p) [3s,2p] or (431,31); and for 2nd row atoms the contraction scheme is (12s,8p) [4s,3p] or (4431,431). For water molecule, these contractions could be encoded as (431,31/31). The 6-311G set represents the following contractions for water (6311,311)/(311) or (11s,5p/5s) [4s,3p/3s].
The Pople's basis sets can also be augmented with d type polarization functions on heavy atoms only (n-ijG* or n-ijkG*) or on all atoms, with p-functions on hydrogens (n-ijG** or n-ijkG**). In methane, the 4-31G* encodes following split (431,31,1)/(31) or (8s,4p,1d/4s) [3s,2p,1d/2s], while 6-311G** for HCN molecule would involve following contractions: (6311,311,1)/(311,1) or (11s,5p,1d/5s,1p) [4s,3p,1d/3s,1p]. Currently, the 6-311G keyword for second row atoms, as implemented in Gaussian90 program, does not actually correspond to the true 6-311G set. It is explicitly mentioned in Gaussian90 manual. For these atoms, 6-311G keyword defaults to MC basis sets (McLean and Chandler, 1980) of the type (12s,9p) [6,5] with contraction scheme (631111,42111). Note, that one of the s-type functions is doubled. The basis sets for P, S and Cl correspond actually to the "anion" basis sets in the original paper since "these were deemed to give better results for neutral molecules as well."
Sometimes, for atoms of the second row nm-ijG notation is used. For example, 66-31G means that there is:
• 1 function containing 6 primitives on the innermost s-shell;
• 1 set of functions belonging to the inner SP-shell (i.e. 2SP shell), each consisting of 6 Gaussian primitives (i.e. 1 s-type function and , , functions consisting of 6 primitives with the same exponents). Note though that coefficients in s and p type contractions are different,
• 2 sets of SP functions for valence SP shell (one set consisting of contractions with 3 primitives and the other with 1 primitive).
It is possible to write this as (16s,10p) [4s,3p] or in more details as (6631,631) contraction scheme or alternatively as s(6/6/3/1), p(6/3/1).
Polarization and diffuse functions
The original contractions derived from atomic Hartree-Fock calculations are frequently augmented with other functions. The most popular are the polarization and diffuse functions. The polarization functions are simply functions having higher values of L than those present in occupied atomic orbitals for the corresponding atom. At least for me, there is some ambiguity here, since for lithium, the p-type functions are not considered polarization functions, while for sulphur, the d-functions are considered polarization functions. In both cases these orbitals are not populated in the ground electronic state of the atom. The reason for including p-type functions in the Li and Be atoms, even in the minimal basis sets, is prac- tical, however. Without these functions, the results are extremely poor. The reason for not including d-type functions for sulphur should be the same as for other atoms, i.e., you can obtain reasonable results without them. I wish, I could believe that.
The exponents for polarization functions cannot be derived from Hartree-Fock calculations for the atom, since they are not populated. However, they can be estimated from correlated calculations involving atoms. In practice, however, these exponents are estimated "using well established rules of thumb or by explicit optimization" (Dunning, 1989).
The polarization functions are important for reproducing chemical bonding. They were frequently derived from optimizing exponents for a set of molecules. They should also be included in all correlated calculations. They are usually added as uncontracted Gaussians. It is important to remember that adding them is costly. Augmenting basis set with d type polarization functions adds 5 (or 6) basis function on each atom while adding f type functions adds 7 (or 10, if spurious combinations are not removed). This brings us to the problem of specifying the number of d, f, g, etc. polarization functions in a form of some compact notation. Unfortunately, there is no provision for this information in the notations described above. Pople's 6-31G* basis uses 6 d type functions as polarization functions, while 6-311G* uses 5 of them. The () notation is no better. If the paper does not say explicitly how many d or f functions are used, you are on your own. The only way to find out is to repeat the calcula- tions or contact the author. Many papers do not specify this important information. The Pople's group introduced yet another more general notation to encode type of polarization functions. The easiest way is to explain an example. The 6-31G** is synonymous to 6-31G(d,p); the 6-311G(3d2f,2p) represents 6-311G set augmented with 3 functions of type d and 2 functions of type f on heavy atoms, and 2 functions of type p on hydrogens or specifically (6311,311,111,11)/(311,11), i.e. (11s,4p,3d,2f/5s,2p) [4s3p3d2f/3s2p] contraction. The 6 d-type polarization function is added to 6-31G set, while only 5 to 6-311G. For both 6-31G and 6-311G set, f-type functions are added in groups of 7. Polarization functions are, as a rule, used uncontracted. More information can be found in the following papers: (Dunning, 1989), (Francl et al., 1982), (Gutowski et al., 1987), (Jankowski, 1985), (Krishnan et al., 1980).
The basis sets are also frequently augmented with the so-called diffuse functions. The name says it all. These Gaussians have very small exponents and decay slowly with distance from the nucleus. Diffuse Gaussians are usually of s and p type, however sometimes diffuse polarization functions are also used. Diffuse functions are necessary for correct description of anions and weak bonds (e.g. hydrogen bonds) and are frequently used for calculations of properties (e.g. dipole moments, polarizabilities, etc.). For the Pople's basis sets the following notaton is used: n-ij+G, or n-ijk+G when 1 diffuse s-type and p-type Gaussian with the same exponents are added to a standard basis set on heavy atoms. The n-ij++G, or n-ijk++G are obtained by adding 1 diffuse s-type and p-type Gaussian on heavy atoms and 1 diffuse s-type Gaussian on hydrogens. For example, the 6-31+G* represents (6311,311,1)/(31) or (11s,5p,1d/4s) [4s,3p,1d/2s]. The 6-311+G(2d1f,2p1d) stands for (63111,3111,11,1)/(311,11,1) split, or (12s,6p,2d,1f) [5s,2p,1d]. For more information about diffuse functions see, for example (Clark et al., 1983), (Del Bene, 1989), and (Frisch et al., 1984).
To calculate total number of primitives/basis functions in your molecule, you sum up the number of primitives/basis functions for each partaking atom. As an example, let us compute the number of functions for H2SO3 molecule assuming the use of Gaussian90 program. The 6-311++G(3df,2p) basis set is used as an example. In this case the reduced set of d and f Gaussians is used, i.e., 5 d-type functions and 7 f-type functions. It corresponds to the following contractions:
S: (6311111,421111,111,1)
(for sulphur, Gaussian90 defaults to McLean-Chandler basis set (631111,42111) for sulphur anion which is augmented with one diffuse s and one diffuse p function, and three d and one f polarization functions)
O: (63111,3111,111,1)
(this is 6-311G for oxygen augmented with one s- and one p-type diffuse function, and three d and one f polarization function)
H: (3111,11)
(this is 6-311G augmented with one diffuse s and two p-functions for polarization)
Number of basis functions:
S:
7 s-type functions, 6 3 p-type functions, 3 5 d-type functions and 1 7 f-type functions
O:
5 s-type functions, 4 3 p-type functions, 3 5 d-type functions and 1 7 f-type functions
H:
4 functions of type s and 2 3 functions of type p (there are 3 p function for each p type contraction, i.e. , , )
H2SO3 = (4 + 2 3) 2 + (7 + 6 3 + 3 5 + 1 7) + (5 + 4 3 + 3 5 + 1 7) 3 = 184
Total number of Gaussian primitives:
S:
1 (6+3+1+1+1+1+1) + 3 (4+2+1+1+1+1) + 5 (1+1+1) + 7 1 = 66
O:
1 (6+3+1+1+1) + 3 (3+1+1+1) + 5 (1+1+1) + 7 1 = 52
H:
1 (3+1+1+1) + 3 (1+1) = 12
H2SO3 = 2 12 + 66 + 3 52 = 246 primitives.
GENERAL CONTRACTIONS. TERMS AND NOTATION
Raffenetti (1973) introduced term "general contraction" for basis sets in which the same Gaussian primitives can appear in several basis functions. In general contraction scheme, the basis functions are formed as different linear combinations of the same primitives. This is clearly in contrast with the segmented scheme described above. Please do not confuse general contrac- tions with a term "general basis set" used in some program manuals to denote "user defined segmented basis sets".
General contractions have many advantages from the theoretical point of view. The most important is that they might be chosen to approximate true atomic orbitals which makes interpretation of coefficients in molecular orbitals meaningful. Also for correlated calculations their performance is praised (Almlof and Taylor, 1987; Almlof et al., 1988; Dunning, 1989) Secondly, they can be chosen in a more standard way than segmented contractions, either as true atomic orbitals obtained from Hartree- Fock calculations for the atom with uncontracted primitives as basis functions, or as Atomic Natural Orbitals (ANO). For description of ANO's consult papers by Almlof and coworkers or read appropriate chapter in Szabo and Ostlund, (1989). The only problem with general contractions is that only a few programs support them. The code for integral package is much more complicated in this case, since it has to work on a block of integrals at each time, to compute the contribution from the given primitive set only once. Of course, you can always enter general contractions as "user defined segmented basis sets," by repeating the same primitives over and over again in different contractions. This will cost you, however, immensely in computer time at the integral computation stage. Remember, the time required for calculating integrals is proportional to the 4th power in the number of Gaussian primitives, and most programs assume that primitives entering different contractions are different.
As an example, the general contractions of (8s4p) set of primitives for oxygen by Huzinaga et al., 1971 (taken from: Raffenetti, 1973).
In the table above, integer numbers in parantheses denote powers of 10 multiplying number in front of them.
The set above can be described as (8s,4p) [4s,3p] contraction. Clearly, the notation giving the number of primitives in each contraction as (abcd...) is not really useful here. It is especially true with newer sets implementing general contractions, where each primitive has all nonzero coefficients in practically every column.
EFFECTIVE CORE POTENTIALS (EFFECTIVE POTENTIALS)
It was known for a long time that core (inner) orbitals are in most cases not affected significantly by changes in chemical bonding. This prompted the development of Effective Core Potential (ECP) or Effective Potentials (EP) approaches, which allow treatment of inner shell electrons as if they were some averaged potential rather than actual particles. ECP's are not orbitals but modifications to a hamiltonian, and as such are very efficient computationally. Also, it is very easy to incorporate relativistic effects into ECP, while all-electron relativistic computations are very expensive. The relativistic effects are very important in describing heavier atoms, and luckily ECP's simplify calculations and at the same time make them more accurate with popular non-relativistic ab initio packages (provided that such packages have support for ECP's). The core potentials can only be specified for shells that are filled.
[ ANDREAS SAVIN correction: You state that core potentials can only be specified for filled shells. In fact, they can be also defined for open shell cores (see, for example, the pseudopotentials for the lanthanides, Dolg et al., Theor. Chim. Acta 75 (1989) 173 ). We would like to use this opportunity to point out that pseudopotentials and corresponding valence basis sets are available from us, not only for the lanthanides, but for all the elements up to Uranium. A comparison of our pseudopotentials with other ones shows their quality (see, e.g., Andrae et al., Theor. Chim. Acta 78 (1991) 247).]
For the rest of electrons (i.e. valence electrons), you have to provide basis functions. These are special basis sets optimized for the use with specific ECP's. These basis sets are usually listed in original papers together with corresponding ECP's. Some examples of papers describing ECP's: (Durand and Bartelat, 1975), (Hay and Wadt, 1985ab), (Hurley et al., 1986), (Pacios and Christensen, 1985), (Stevens ey al., 1984), (Wadt and Hay, 1985), (Walace et al., 1991). The ECP are tabulated in the literature as parameters of the following expansion:
where M is the number of terms in the expansion, is a coefficient for each term, r denotes distance from nucleus, is a power of r for the i-th term, and represents the exponent for the i-th term.
To specify ECP for a given atomic center, you need to include typically: the number of core electrons that are substituted by ECP, the largest angular momentum quantum number included in the potential (e.g., 1 for s only, 2 for s and p, 3 for s, p, and d; etc.), and number of terms in the "polynomial Gaussian expansion" shown above. For each term in this expansion you need to specify: coefficient ( ), power of r ( ) and exponent in the Gaussian function ( ). Also you need to enter basis set for valence electrons specific to this potential. As a result of applying the ECP's you drastically reduce number of needed basis functions, since only functions for valence electrons are required. In many cases, it would simply be impossible to perform some calculations on systems involving heavier elements without ECP's (try to calculate number of functions in TZ2P basis set for e.g. U, and you will know why).
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18. P.J. Hay, W.R. Wadt, 1985b, "Ab Initio effective core potentials for molecular calculations. Potentials for K to Au including outermost core orbitals," J. Chem. Phys. 82, 299-310.
19. W.J. Hehre, L. Radom, P.v.R Schleyer, J.A. Pople, 1986, "Ab Initio Molecular Orbital Theory," Willey & Sons, New York.
20. M.M. Hurley, L.F. Pacios, P.A. Christiansen, R.B. Ross, W.C. Ermler, 1986, "Ab initio relativistic effective potentials with spin-orbit operators. II. K through Kr," J. Chem. Phys. 84, 6840-6853.
21. S. Huzinaga, 1965, "Gaussian-type functions for polyatomic systems," J. Chem. Phys., 42, 1293-1302.
22. S. Huzinaga, D. McWilliams, B. Domsky, 1971, "Approximate Atomic Function," J. Chem. Phys. 54, 2283-2284.
23. K. Jankowski, R. Becherer, P. Scharf, H. Schiffer, R. Ahlrichs, 1985, "The impact of higher polarization basis functions on molecular ab initio results," J. Chem. Phys. 82, 1413-1419.
24. R. Krishnan, J.S. Binkley, R. Seeger, J.A. Pople, 1980, "Self-consistent orbital methods. XX. A basis set for correlated wave functions," J. Chem. Phys. 72, 650-654.
25. A.D. McLean, G.S. Chandler, 1980, "Contracted Gaussian basis sets for molecular calculations. I. Second row atoms, Z=11-18," J. Chem. Phys., 72, 5639-5648.
26. L.F. Pacios, P.A. Christiansen, 1985, "Ab initio relativistic effective potentials with spin-orbit operators. I. Li through Ar.
27. R. Poirier, R. Kari, I.G. Csizmadia, 1985, "Handbook of Gaussian Basis Sets," Elsevier Science, New York.
28. R.C. Raffenetti, 1973, "General contraction of Gaussian atomic orbitals: Core, valence, polarization and diffuse basis sets; Molecular integral evaluation," J. Chem. Phys, 58, 4452-4458.
29. M. Sabio, S. Topiol, 1989, " 3s- versus 1s-type gaussuan primitives: Modification of the 3-21G(*) basis set for sulphur atom. J. Comput. Chem., 10, 660-672.
30. W.J. Stevens, H. Bash, M. Krauss, 1984, J. Chem. Phys. 81, 6026-6033.
31. A. Szabo, N.S. Ostlund, 1989, "Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory". MacMillan Publishing Co., New York.
32. W.R. Wadt, P.J. Hay, 1985, "Ab Initio effective core potentials for molecular calculations. Potentials for main group elements Na to Bi," J. Chem. Phys. 82, 284-298.
33. N.M. Wallace, J.P. Blaudeau, R.M. Pitzer, 1991, "Optimized Gaussian Basis Sets for use with Relativistic Effective (core) Potentials: Li-Ar,"\ Int. J. Quantum Chem. (in press).
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Nuclear magnetic resonance (NMR) is a versatile and highly-sophisticated spectroscopic technique which has been applied to a growing number of diverse applications in science, technology and medicine. This chapter will consider, for the most part, magnetic resonance involving protons.
Magnetic Properties of Nuclei
In all our previous work, it has been sufficient to treat nuclei as structureless point particles characterized fully by their mass and electric charge. On a more fundamental level, as was discussed in Chap. 1, nuclei are actually composite particles made of nucleons (protons and neutrons) and the nucleons themselves are made of quarks. The additional properties of nuclei which will now become relevant are their spin angular momenta and magnetic moments. Recall that electrons possess an intrinsic or spin angular momentum s which can have just two possible projections along an arbitrary spacial direction, namely $\pm \frac{1}{2} \hbar$. Since $\hbar$ is the fundamental quantum unit of angular momentum, the electron is classified as a particle of spin one-half. The electron’s spin state is described by the quantum numbers $s=1$ and $m_s = \pm 1$. A circulating electric charge produces a magnetic moment $\vec{\mu}$ proportional to the angular momentum J. Thus
$\vec{\mu} = \gamma \vec{J} \label{1}$
where the constant of proportionality γ is known as the magnetogyric ratio. The z-component of $\vec{\mu}$ has the possible values
$\vec{\mu}_z = \gamma\hbar m_J \mbox{where} m_J = -J, -J+1,..., +J \label{2}$
determined by space quantization of the angular momentum J. The energy of a magnetic dipole in a magnetic field B is given by
$E= -\vec{\mu} \cdot \vec{B} = -\vec{\mu}_z B \label{3}$
where magnetic field defines the z-axis. The SI unit of magnetic field (more correctly, magnetic induction) is the tesla, designated T. Electromagnets used in NMR produce fields in excess of 10 T. Small iron magnets have fields around .01 T, while some magnets containing rare-earth elements such as NIB (niobium-iron-boron) reach 0.2 T. The Earth’s magnetic field is approximately 5×10−5T (0.5 gauss in alternative units), dependent on geographic location. At the other extreme, a neutron star, which is really a giant nucleus, has a field predicted to be of the order of 108 T. The energy relation (3) determines the most conveniently units for magnetic moment, namely joules per tesla, J T1.
For orbital motion of an electron, where the angular momentum is l, the magnetic moment is given by
$\vec{\mu}_z = -\frac{e{\hbar}}{2m} m_l = -\vec{\mu}_B m_l \label{4}$
where the minus sign reflects the negative electric charge. The Bohr magneton is defined by
$\vec{\mu}_B = -\frac{e{\hbar}}{2m} = 9.274\times10^{-24} JT^{-1} \label{5}$
The magnetic moment produced by electron spin is written
$\vec{\mu}_z = -g \vec{\mu}_B m_s \label{6}$
with introduction of the g-factor. Eq (4) implies g = 1 for orbital motion. For electron spin, however, g = 2 (more exactly, 2.0023). The factor 2 compensates for ms = 1 such that spin and l = 1 orbital magnetic moments 2 are both equal to one Bohr magneton.
Many nuclei possess spin angular momentum, analogous to that of the electron. The nuclear spin, designated I, has an integral or half-integral value: 0, 1 , 1, 3 , and so on. Table 1 lists some nuclei of importance in chemical applications of NMR. The proton and the neutron both are spin 2 particles, like the electron. Complex nuclei have angular momenta which are resultants of the spins of their component nucleons. The deuteron 2H, with I = 1, evidently has parallel proton and neutron spins. The 4He nucleus has I = 0, as do 12C, 16O, 20Ne, 28Si and 32S. These nuclei contain filled shells of protons and neutrons with the vector sum of the component angular momenta equal to zero, analogous to closed shells of electrons in atoms and molecules. In fact, all even-even nuclei have spins of zero. Nuclear magnetic moments are of the order of a nuclear magneton
$\vec{\mu}_N = \frac{e\hbar}{2M} = 5.051\times 10^{-27} JT^{-1} \label{7}$
where M is the mass of the proton. The nuclear magneton is smaller than the Bohr magneton by a factor m/M ≈ 1836.
Table 1: Some common nuclei in NMR spectroscopy
In analogy with Equations $\ref{2}$ and $\ref{6}$, nuclear moments are represented by
$\vec{\mu}_z = g_I \vec{\mu}_N m_I = \hbar \gamma_I m_I \label{8}$
where gI is the nuclear g-factor and γI, the magnetogyric ratio. Most nuclei have positive g-factors, as would be expected for a rotating positive electric charge. It was long puzzling that the neutron, although lacking electric charge, has a magnetic moment. It is now understood that the neutron is a composite of three charged quarks, udd. The negatively-charged d- quarks are predominantly in the outermost regions of the neutron, thereby producing a negative magnetic moment, like that of the electron. The g- factor for 17O, and other nuclei dominated by unpaired neutron spins, is consequently also negative.
Nuclear Magnetic Resonance
The energy of a nuclear moment in a magnetic field, according to Equation $\ref{3}$, is given by
$E_{m_I} = -\hbar \gamma_I m_I B \label{9}$
For a nucleus of spin $I$, the energy of a nucleus in a magnetic field is split into $2I+1$ Zeeman levels. A proton and other nuclei with spin $\frac{1}{2}$ particles will have just two possible levels:
$E_{\pm \frac{1}{2}} = \pm \dfrac{1}{2} \hbar \gamma B \label{10}$
with the α-spin state ( $m_I = -\frac{1}{2}$) lower in energy than the β-spin state ($m_I = +\frac{1}{2}$) by
$\Delta E = \hbar \gamma B \label{11}$
Fig. 1 shows the energy of a proton as a function of magnetic field. In zero field (B = 0), the two spin states are degenerate. In a field B, the energy splitting corresponds to a photon of energy $\Delta E = \hbar \omega = h \nu$ where
$\omega_L = \gamma_B\space \mbox{or}\space \nu_L = \gamma_B \label{12}$
known as the Larmor frequency of the nucleus. For the proton in a field of 1 T, $\nu_L$ = 42.576 MHz, as the proton spin orientation flips from $+\frac{1}{2}$ to $-\frac{1}{2}$. This transition is in the radiofrequency region of the electromagnetic spectrum. NMR spectroscopy consequently exploits the technology of radiowave engineering.
Figure 1. Energies of spin $\frac{1}{2}$ in magnetic field showing NMR transition at Larmor frequency $\nu_L$.
A transition cannot occur unless the values of the radiofrequency and the magnetic field accurately fulfill Eq (12). This is why the technique is categorized as a resonance phenomenon. If some resonance condition is not satisfied, no radiation can be absorbed or emitted by the nuclear spins. In the earlier techniques of NMR spectroscopy, it was found more convenient keep the radiofrequency fixed and sweep over values of the magnetic field B to detect resonances. These have been largely supplanted by modern pulse techniques, to be described later.
The transition probability for the upward transition (absorption) is equal to that for the downward transition (stimulated emission). (The contribution of spontaneous emission is neglible at radiofrequencies.) Thus if there were equal populations of nuclei in the α and β spin states, there would be zero net absorption by a macroscopic sample. The possibility of observable NMR absorption depends on the lower state having at least a slight excess in population. At thermal equilibrium, the ratio of populations follows a Boltzmann distribution
$\frac{N_{\beta}}{N_\alpha} = \frac{e^{\frac{-E_\beta}{kT}}}{e^{\frac{-E_\alpha}{kt}}} = e^{- \frac{\hbar \gamma B}{kT}} \label{13}$
Thus the relative population difference is given by
$\dfrac{\Delta N}{N_\alpha} = \frac{N_\alpha - N_{\beta}}{N_\alpha + N_{\beta}} \approx \frac{ \hbar \gamma B}{2kT} \label{14}$
Since nuclear Zeeman energies are so small, the populations of the α and β spin states differ very slightly. For protons in a 1 T field, N/N 3×106. Although the population excess in the lower level is only of the order of parts per million, NMR spectroscopy is capable of detecting these weak signals. Higher magnetic fields and lower temperatures are favorable conditions for enhanced NMR sensitivity.
The Chemical Shift
NMR has become such an invaluable technique for studying the structure of atoms and molecules because nuclei represent ideal noninvasive probes of their electronic environment. If all nuclei of a given species responded at their characteristic Larmor frequencies, NMR might then be useful for chemical analysis, but little else. The real value of NMR to chemistry comes from minute differences in resonance frequencies dependent on details of the electronic structure around a nucleus. The magnetic field induces orbital angular momentum in the electron cloud around a nucleus, thus, in effect, partially shielding the nucleus from the external field B. The actual or local value of the magnetic field at the position of a nucleus is expressed
$B_{loc} = (1- \sigma)B \label{15}$
where the fractional reduction of the field is denoted by σ, the shielding constant, typically of the order of parts per million. The actual resonance frequency of the nucleus in its local environment is then equal to
$\nu = (1- \sigma) \frac{\gamma B}{2 \pi} \label{16}$
A classic example of this effect is the proton NMR spectrum of ethanol CH3CH2OH, shown in Fig. 2. The three peaks, with intensity ratios 3:2:1 can be identified with the three chemically-distinct environments in which the protons find themselves: three methyl protons (CH3), two methylene protons (CH2) and one hydroxyl proton (OH).
Figure 2. Oscilloscope trace showing the first NMR spectrum of ethanol, taken at Stanford University in 1951. Courtesy Varian Associates, Inc.
The variation in resonance frequency due to the electronic environment of a nucleus is called the chemical shift. Chemical shifts on the delta scale are defined by
$\delta = {\frac{ \nu - \nu^{O}}{\nu^{O}}} \times 10^{6} \label{17}$
where νo represents the resonance frequency of a reference compound, usually tetramethylsilane Si(CH3)4, which is rich in highly-shielded chemically-equivalent protons, as well as being unreactive and soluble in many liquids. By definition δ = 0 for TMS and almost everything else is “downfield” with positive values of δ. Most compounds have delta values in the range of 0 to 12 (hydrogen halides have negative values, e.g. δ 13 for HI). The hydrogen atom has δ 13 while the bare proton would have δ 31. Conventionally, the δ-scale is plotted as increasing from right to left, in the opposite sense to the magnitude of the magnetic field. Nuclei with larger values of δ are said to be more deshielded, with the bare proton being the ultimate limit. Fig. 3 shows some typical values of δ for protons in some common organic compounds.
Figure 3. Ranges of proton chemical shifts for common functional groups. From P. Atkins, Physical Chemistry, (Freeman, New York, 2002).
Fig. 4 shows a high-resolution NMR spectrum of ethanol, including a δ-scale. The “fine structure” splittings of the three chemically-shifted components will be explained in the next Section. The chemical shift of a nucleus is very difficult to treat theoretically. However, certain empirical regularities, for example those represented in Fig. 3, provide clues about the chemical environment of the nucleus. We will not consider these in any detail except to remark that often increased deshielding of a nucleus (larger δ) can often be attributed to a more electronegative neighboring atom. For example the proton in the ethanol spectrum (Fig. 4) with δ 5 can be identified as the hydroxyl proton, since the oxygen atom can draw significant electron density from around the proton.
Figure 4. High-resolution NMR spectrum of ethanol showing δ scale of chemical shifts. The line at δ = 0 corresponds to the TMS trace added as a reference.
Neighboring groups can also contribute to the chemical shift of a given atom, particularly those with mobile π-electrons. For example, the ring current in a benzene ring acts as a secondary source of magnetic field. Depending on the location of a nucleus, this can contribute either shielding or deshielding of the external magnetic field, as shown in Fig. 5. The interaction of neighboring groups can be exploited to obtain structural information by using lanthanide shift reagents. Lanthanides (elements 58 through 71) contain 4f-electrons, which are not usually involved in chemical bonding and can give large paramagnetic contributions. Lanthanide complexes which bind to organic molecules can thereby spread out pro- ton resonances to simplify their analysis. A popular chelating complex is Eu(dpm)3, tris(dipivaloylmethanato)europium, where dpm is the group (CH3)3C–CO=CH–CO–C(CH3)3.
Figure 5. Magnetic field produced by ring current in benzene, shown as red loops. Where the arrows are parallel to the external field B, including protons directly attached to the ring, the effect is deshielding. However, any nuclei located within the return loops will experience a shielding effect.
Spin-Spin Coupling
Two of the resonances in the ethanol spectrum shown in Fig. 4 are split into closely-spaced multiplets—one triplet and one quartet. These are the result of spin-spin coupling between magnetic nuclei which are relatively close to one another, say within two or three bond separations. Identical nuclei in identical chemical environments are said to be equivalent. They have equal chemical shifts and do not exhibit spin-spin splitting. Nonequivalent magnetic nuclei, on the other hand, can interact and thereby affect one another’s NMR frequencies. A simple example is the HD molecule, in which the spin-1proton can interact with the spin-1 deuteron, even though the atoms are chemically equivalent. The proton’s energy is split into two levels by the external magnetic field, as shown in Fig. 1. The neighboring deuteron, itself a magnet, will also contribute to the local field at the pro- ton. The deuteron’s three possible orientations in the external field, with MI = −1,0,+1, with different contributions to the magnetic field at the proton, as shown in Fig. 6. The proton’s resonance is split into three evenly spaced, equally intense lines (a triplet), with a separation of 42.9 Hz. Corespondingly the deuteron’s resonance is split into a 42.9 Hz doublet by its interaction with the proton. These splittings are independent of the external field B, whereas chemical shifts are proportional to B. Fig. 6 represents the energy levels and NMR transitions for the proton in HD.
• Figure 6. Nuclear energy levels for proton in HD molecule. The two Zeeman levels of the proton when B > 0 are further split by interaction with the three possible spin orientations of the deuteron Md= 1, 0, +1. The proton NMR transition, represented by blue arrows, is split into a triplet with separation 42.9 Hz.
Nuclear-spin phenomena in the HD molecule can be compactly represented by a spin Hamiltonian
$\hat{H} = -\hbar \gamma_H M_H(1- \sigma_H) - \hbar \gamma_D M_D(1-\sigma_D)B + h J_{HD} I_H \cdot I_D \label{18}$
• The shielding constants σH and σD are, in this case, equal since the two nuclei are chemically identical. For sufficiently large magnetic fields B, the last term is effectively equal to hJHDMHMD. The spin-coupling constant J can be directly equated to the splitting expressed in Hz. We consider next the case of two equivalent protons, for example, the CH2 group of ethanol. Each proton can have two possible spin states with $M_I = \pm\frac{1}{2}$ , giving a total of four composite spin states. Just as in the case of 2 electron spins, these combine to give singlet and triplet nuclear-spin states with M = 0 and 1, respectively. Also, just as for electron spins, transitions between singlet and triplet states are forbidden. The triplet state allows NMR transitions with $\Delta M = \pm 1$ to give a single resonance frequency, while the singlet state is inactive. As a consequence, spin-spin splittings do not occur among identical nuclei. For example, the H2 molecule shows just a single NMR frequency. And the CH2 protons in ethanol do not show spin- spin interactions with one another. They can however cause a splitting of the neighboring CH3 protons. Fig. 7 (left side) shows the four possible spin states of two equivalent protons, such as those in the methylene group CH2, and the triplet with intensity ratios 1:2:1 which these produce in nearby protons. Also shown (right side) are the eight possible spin states for three equivalent protons, say those in a methyl group CH3, and the quartet with intensity ratios 1:3:3:1 which these produce. In general, n equivalent protons will give a splitting pattern of n + 1 lines in the ratio of binomial coefficients 1:n:n(n 1)/2 . . . The tertiary hydrogen in isobutane (CH3)3CH, marked with an asterisk, should be split into 10 lines by the 9 equivalent methyl protons.
• Figure 7. Splitting patterns from methylene and methyl protons.
The NMR spectrum of ethanol CH3CH2OH (Fig. 4) can now be interpreted. The CH3 protons are split into a 1:2:1 triplet by spin-spin interaction with the neighboring CH2. Conversely, the CH2 protons are split into a 1:3:3:1 quartet by interaction with the CH3. The OH (hydroxyl) proton evidently does not either cause or undergo spin-spin splitting. The explanation for this is hydrogen bonding, which involves rapid exchange of hydroxyl protons among neighboring molecules. If this rate of exchange is greater than or comparable to the NMR radiofrequency, then the splittings will be “washed out.” Only one line with a motion-averaged value of the chemical shift will be observed. NMR has consequently become a useful tool to study intramolecular motions.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Chapter_14.__Nuclear_Magnetic_Resonance.txt
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(2) radial density versus r.
1. Density versus radius r: In this case, the square of the wave function is plotted against r. These plots are sometimes misleading. For example, the 1s orbital plot looks like
```probability
|
|.
|
| .
|
| .
|
| .
| .
| .
| .
| .
|________________________________________._______ r
```
You may feel the probability of finding the electron is the highest in the nucleus, yet you have learned that the electron is most likely at a distance r = 53 pm from the center of the atom.
2. Radial density (RD) versus r: To really represent the probability of finding the electron at r at a given time, the radial distribution against r is often plotted. In this plot, instead of plotting square-of-the-wave-function, we modify square-of-the-wave-function by the volume associated with r, (4*pi*r2). This modification converts electron density to radial electron density.
The radial density plot of 1s orbital has a shape as shown below:
```probability
|
|
|
|
|
| . ' .
| . .
| . .
| . .
|________________________________________._______ r
```
At the center of the atom, the value of the wavefunction is large, but when r = 0, the volume element (4*pi*r2) is almost zero when r -> 0. Thus, the radial distribution rises as r increases, reaching a maximum at some value of r. For the H atom, the maximum of the radial distribution is at r = 53 pm.
Electromagnetic Waves
Electromagnetic waves are used to transmit long/short/FM wavelength radio waves, and TV/telephone/wireless signals or energies. They are also responsible for transmitting energy in the form of microwaves, infrared radiation (IR), visible light (VIS), ultraviolet light (UV), X-rays, and gamma rays. Each region of this spectrum plays an important part in our lives, and in the business involving communication technology. The list given above is in increasing frequency (or decreasing wavelength) order. Here again is the list of regions and the approximate wavelengths in them. For simplicity, we choose to give only the magnitudes of frequencies. That is we give log (frequency) (log(f)).
Region: Radio, FM, TV, microwave, IR, VIS, UV, X-rays, Gamma rays.
Wavelength: 600 m 20 m 1 mc 1 mm 0.1 mm 1e-9 m 1e-12 m 1e-15 m
log (f): 6 7 8 9 10 11 12 13 14 15 20 23
Electromagnetic radiations are usually treated as wave motions. The electronic and magnetic fields oscillate in directions perpendicular to each other and to the direction of motion of the wave.
The wavelength, the frequency, and the speed of light obey the following relationship:
$\mathrm{wavelength \times frequency = speed\: of\: light}$.
The speed of light is usually represented by c, the wavelength by the lower case Greek letter lambda, $\lambda$, and the frequency by lower case Greek letter nu $\nu$. In these symbols, the above formula is:
$\lambda \nu = c$
The electromagnetic radiation is the foundation for radar, which is used for guidance and remote sensing for the study of the planet Earth.
The Visible Spectrum
Wavelengths of the visible region of the spectrum range from 700 nm for red light to 400 nm for violet light.
red 700 nm
orange 630
yellow 550
green 500
blue 450
violet 400
There is no need to memorize these numbers, but knowing that the visible region has such a narrow range of 400-700 nm is handy at times when referring to certain light.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Gen_Chem_Quantum_Theory/Atomic_Orbitals.txt
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Learning Objectives
• Explain the rules for filling electrons in atomic orbitals -- Pauli exclusion principle and Hund's rule.
• Fill electrons in atomic orbitals--Aufbau process.
• Explain the arrangement of elements in terms of quantum numbers.
• Explain the systematic variation of element properties.
Mendeleev noticed the recurrence of properties of elements as the atomic weight increased, and he invented the Periodic Table of Elements, which is a useful tool for organizing and correlating chemical and physical properties of chemical elements. Today, the most popular Periodic Table form is shaped by results of quantum theory.
Quantum theory rationalized the existence of and arrangement of all elements in today's Periodic Table. It has also been applied to explain their chemical properties.
Energy Levels in Many-Electron Atoms
In order to fill the electrons in various atomic orbitals, we need to know how the energy levels vary as the nuclear charge increases. For hydrogen-like atoms, the approximate energy levels are as indicated below:
Energy levels of \(\ce{H}\)-like atoms
``` :::::: : ::: ::::: :::::::
4s4p4d4f - --- ----- -------
3s3p3d - --- -----
2s2p - ---
(a large gap)
1s -
```
The shielding effect and electron-electron interactions cause the energy levels of subshells such as 2s & 2p to be different from those of \(\ce{H}\)-like atoms. This is done by treating the electron shield cores as a proton but the core has an effective nuclear charge Z.
For the \(\ce{H}\)-like atoms, energy levels for 2s, 2p stay the same, but the separation between 2s and 2p energy levels increases as the atomic number (Z) increases. Similar situations happen for 3s, 3p, and 3d energy levels. The energy diagrams of \(\ce{H}\), \(\ce{Li}\) & \(\ce{K}\) are used to illustrate this point. The color diagram is from a Hyperion website discussing quantum numbers and structure of atoms.
Variation of energy levels for atomic orbitals of some elements
\(\ce{H}\)
_2s_ _ _2p
_ 1s
\(\ce{Li}\)
_ _ _ 2p
_ 2s
_ 1s
\(\ce{Be}\)
_ _ _ 2p
_ 2s
_ 1s
\(\ce{B}\)
_ _ _ 2p
_ 2s
_ 1s
\(\ce{C}\)
_ _ _ 2p
_ 2s
_ 1s
\(\ce{N}\)
_ _ _ 2p
_ 2s
_ 1s
\(\ce{O}\)
_ _ _ 2p
_ 2s
_ 1s
\(\ce{F}\)
_ _ _ 2p
_ 2s
_ 1s
Understanding how the energy levels vary is the key to the Aufbau process, because Electrons tend to occupy the lowest energy level available. But before we talk about the Aufbau process, we need to be aware of the Pauli exclusion principle and the Hund's rule.
The Pauli Exclusion Principle
The Pauli exclusion principle suggests that only two electrons with opposite spin can occupy an atomic orbital. Stated another way, no two electrons have the same 4 quantum numbers n, l, m, s. Pauli's exclusion principle can be stated in some other ways, but the idea is that energy states have limited room to accommodate electrons. A state accepts two electrons of different spins.
In applying this rule, you should realize that an atomic orbital is an energy state.
Hund's Rule
Hund's rule suggests that electrons prefer parallel spins in separate orbitals of subshells. This rule guides us in assigning electrons to different states in each sub-shell of the atomic orbitals. In other words, electrons fill each and all orbitals in the subshell before they pair up with opposite spins.
Pauli exclusion principle and Hund's rule guide us in the Aufbau process, which is figuring out the electron configurations for all elements.
The Aufbau Procedure
The Aufbau procedure (filling order of atomic orbitals) is used to work out the electron configurations of all atoms. However, modification should be made by applying Hund's rule to be discussed in the next section.
The Aufbau procedure is based on a rough energy levels diagram of many-electron atoms as shown below:
```7s |~ 5d |----- 5f |_______
Actanides
6s |_ 6p |~~~ 5d |===== 4f |=======
Lanthanides
5s |_ 5p |~~~ 4d |-----
Note the trend 5s 4d 5p
& develope a pattern
4s |_ 4p |~~~ 3d |-----
Transition elements
3s |_ 3p |---
2s |_ 2p |---
1s |-
```
The fine difference between energy levels cannot be adequately shown.
You've learned various techniques to work out the electronic configurations of elements. Here is yet another form of an energy level diagram:
``` 7p _ _ _
7s _ 5f - - - - - - - 6d ~ ~ ~ ~ ~
6p _ _ _
6s _ 4f - - - - - - - 5d ~ ~ ~ ~ ~
5s _ 4d - - - - - 5p ~ ~ ~
4s _ 3d - - - - - 4p ~ ~ ~
3s _ 3p - - -
2s _ 2p - - -
1s _
```
In order to master the technique of the Aufbau procedure, you should apply Hund's rule and the Pauli exclusion principle to work out the electronic configuration for closed shells of inert gases. After you have it worked out, you may compare your result with the following. Do not try to remember it; it is important to know how to work it out.
```Z= 2 10 18 36 54 86
1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6 6s24f145d106p6
He Ne Ar Kr Xe Rn
```
Block of Elements by Highest Occupied Atomic Orbitals
Block of elements by
last filled atomic orbitals
1s
2s
3s
4s
5s
6s
7s
4f - - - - - 4f
5f - - - - - 5f
3d - - - 3d
4d - - - 4d
5d - - - 5d
6d - - - 6d
2p - 2p
3p - 3p
4p - 4p
5p - 5p
6p - 6p
7p - 7p
The highest atomic orbitals occupied by electrons determine the properties of the elements. According to this scheme, the periodic table can be divided into s, p, d, and f blocks as seen in the table on the right.
This table shows the filling order of atomic orbitals as
1s
2s 2p
3s 3p
4s 3d 4p
5s 4d 5p
6s 4f 5d 6p
7s 5f 6d 7p
The s- and p-blocks of elements are called main group elements. The d-block elements are called transition elements The f-block elements are called the inner transition elements.
In an ordinary periodic table, the s, p, and d block elements are in the main body of the Periodic Table, whereas the f block elements are placed below the main body. If we placed them on the same period where they belong, the Periodic Table would be too long for the screen to accommodate. Thus, we keep the Periodic Table in the usual (long) form.
Special Electronic Configurations
When two electrons occupy the same orbital, they not only have different spins (Pauli exclusion principle), the pairing raises the energy slightly. On the other hand, a half filled subshell and a full filled subshell lower the energy, gaining some stability. Bearing this in mind, you will be able to understand why we have the following special electronic configurations.
\(\ce{Cr\: [Ar]}4s^1 3d^5\) <=All s and d subshells are half full
\(\ce{Cu\: [Ar]}4s^1 3d^{10}\) <=Prefers a filled d subshell, leaving s with 1
\(\ce{Nb\: [Kr]}5s^1 4d^4\) <=5s and 4d energy levels are close
\(\ce{Mo\: [Kr]}5s^1 4d^5\) Similar to \(\ce{Cr}\) above
\(\ce{Tc\: [Kr]}5s^2 4d^5\) (Not special, but think of Hund's rule)
\(\ce{Ru\: [Kr]}5s^1 4d^7\) <= Only 1 5s electron
\(\ce{Rh\: [Kr]}5s^1 4d^8\) <= In both
\(\ce{Pd\: [Kr]}5s^0 4d^{10}\) <= Note filled 4d and empty 5s
\(\ce{Ag\: [Kr]}5s^1 4d^{10}\) <= Partially filled 5s, but filled d
For CHEM120 students, you are not required to remember the special ones, but you should take notice of the electronic configurations of \(\ce{Cr}\) and \(\ce{Cu}\) to realize the gaining of stability due to half and full filled subshells.
Confidence Building Questions
1. For the \(\ce{H}\)-like atom, which subshell has the highest energy level?
4f, 3d, 2p, 1s
Hint: 4f
Skill:
Describe the energy levels of hydrogen atoms.
2. For an element with atomic number 20, which is the last or highest occupied subshell of atomic orbitals?
1s 2s 2p 3s 3p 3d 4s 5s
Hint: 4s
Discussion:
The element is Calcium, \(\ce{Ca}\).
3. How many electrons are required to fill all the following sub shells?
1s 2s 2p 3s 3p 4s 3d 4p
Hint: 2 + 8 + 8 + 18
Discussion:
The element is Krypton, \(\ce{Kr}\).
4. Which of the following two electronic configurations is more stable?
a
\(\ce{[Ar]}4s^1 3d^5\)
b \(\ce{[Ar]}4s^2 3d^4\)
Hint: a
Discussion:
Check the electronic configuration for \(\ce{Cu}\) and \(\ce{Cr}\).
5. Which of the following two electronic configurations is more stable?
a
\(\ce{[Ar]}4s^2 3d^9\)
b \(\ce{[Ar]}4s^1 3d^{10}\)
Hint: b
Discussion:
This is the electronic configuration for copper.
6. Choose the electronic configuration for palladium, \(\ce{Pd}\) (Z = 46).
a
\(\ce{[Kr]}5s^1 4d^7\)
b \(\ce{[Kr]}5s^1 4d^8\)
c \(\ce{[Kr]}5s^0 4d^{10}\)
d \(\ce{[Kr]}5s^1 4d^{10}\)
Hint: c
Discussion:
Excellent. This is a special case. One can argue that the filled 4d subshell has lower energy than 5s2 4d8.
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Learning Objectives
• Explain the spectrum from hydrogen gas.
• Describe Rydberg's theory for the hydrogen spectra.
• Interpret the hydrogen spectrum in terms of the energy states of electrons.
Hydrogen Spectra
Lasers emit radiation which is composed of a single wavelength. However, most common sources of emitted radiation (i.e. the sun, a lightbulb) produce radiation containing many different wavelengths. When the different wavelengths of radiation are separated from such a source a spectrum is produced. A rainbow represents the spectrum of wavelengths of light contained in the light emitted by the sun. Sun light passing through a prism (or raindrops) is separated into its component wavelengths and is made up of a continuous spectrum of wavelengths (from red to violet); there are no gaps.
Not all radiation sources emit a continuous spectrum of wavelengths of light. When high voltage is applied to a glass tube containing various gasses under low pressure different colored light is emitted: neon gas produces a red-orange glow and sodium gas produces a yellow glow. When such light is passed through a prism only a few wavelengths are present in the resulting spectra that appear as lines separated by dark areas, and thus are called line spectra
Hydrogen, the simplest but the most abundant element in the universe, is also the most studied element. During the early development of science, people had been investigating light emitted by heated tubes of hydrogen gas. When these lights passed prisms, they saw some lines in the visible region. When the spectrum emitted by hydrogen gas was passed through a prism and separated into its constituent wavelengths four lines appeared at characteristic wavelengths in the visible spectral range: 656 nm, 486 nm, 434 nm, and 410 nm
wl
These lines are shown here together with lines emitted by hot gases of $\ce{Hg}$ and $\ce{He}$. These lines are called the Balmer Series, because Balmer saw some regularity in their wavelength, and he has given a formula to show the regularity.
The Balmer Series
J.J. Balmer analyzed these lines and identified the following relationship:
$\lambda = 364.56\,\dfrac{n_2^2}{n^2 - 2^2}\textrm{ nm}$
for the regularity in terms of integers $n_2$. For some integers of $n_2 \ge 3$, you can confirm the $\lambda$ to be
Table 1: The Balmer Series of Hydrogen Emission Lines
$n_2$ 3 4 5 6 7 8 9 10
$\lambda$ 656 486 434 410 397 389 383 380
color red teal blue indigo violt not visible not visible not visible
As the $n_2$ increase, the lines are getting closer together. If you plot the lines according to their $\lambda$ on a linear scale, you will get the appearance of a spectrum as observed by experimentalists; these lines are called the Balmer series.
The Rydberg Formula
Rydberg inverted both sides of Balmer's formula and gave
$\dfrac{1}{\lambda}= R_{\textrm H}\left(\dfrac{1}{2^2}-\dfrac{1}{n^2} \right )$
This is known as the Rydberg formula, and R is known as the Rydberg constant; its numerical values depends on the units used
\begin{align} R_{\textrm H} &= \mathrm{0.010972\: nm^{-1}}\ &= \mathrm{10972\: mm^{-1}}\ &= \mathrm{109721\: cm^{-1}}\ &= \mathrm{10972130\: m^{-1}} \end{align}
This formula shows that if you plot $\widetilde{\nu}$ vs. $1/n^2$, you will get a straight line.
Other Series
The results given by Balmer and Rydberg for the spectrum in the visible region of the electromagnetic radiation start with n = 3, and the other integer is 2. Is there a series with the following formula?
$\dfrac{1}{\lambda} = R_{\textrm H} \left(\dfrac{1}{1^2} - \dfrac{1}{n^2} \right )$
If the law of nature is simple and regular, a series should exist, and the values for n and wavenumber (wn) should be:
Table 2: The Lyman Series of Hydrogen Emission Lines
$n_2$ 2 3 4 5 ...
$\lambda$ (nm) 121 102 97 94 ...
$\widetilde{\nu}$ (cm-1) 82,2291 97,530 102,864 105,332 ...
Do you know in what region of the electromagnetic radiation these lines are? Of course, these lines are in the UV region, and they are not visible, but they are detected by instruments; these lines form a Lyman series. The existences of the Lyman series and Balmer's series suggest the existence of more series, and a generalized formula is suggested.
$wn = R_{\textrm H} \left(\dfrac{1}{n_{\textrm f}^2} - \dfrac{1}{n_{\textrm i}^2} \right )$
Actually, the series with $n_2^2$ = 3, and $n_1^2$ = 4, 5, 6, 7, ... is called Pashen series.
Questions
1. Calculate the wave numbers of the lines with the longest wavelength in the Pashen series.
Hint: 5334 /cm
Discussion:
Calculate the wavelength for three lines in this series. What region are these lines?
2. Draw an energy level diagram to show the transition for the emission of the various series of lines by hydrogen.
Hint: See the discussion below:
Discussion:
A diagram showing the energy levels is shown here.
==== ..... very closely spaced lines
---- n = 3
---- n = 2
---- n = 1
Interpretation of the hydrogen spectrum led to the development of quantum mechanics.
3. Among all possible photons emitted by hydrogen atoms, what is the shortest wavelength possible?
Hint: The shortest wavelength = 1/R, where R is the Rydberg constant.
Discussion:
The photo with the highest energy from a hot hydrogen gas is in the Lyman series. The wavelength is about 91 nm. Confirm this by using the following equations.
$wn = R\left(\dfrac{1}{n_{\textrm f}^2} - \dfrac{1}{n_{\textrm i}^2}\right )$
Spectra of Elements shows the spectra of elements in the periodic table.
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Learning Objectives
• Give a very brief history on the development of chemistry as a science leading to the development of quantum theory.
• Explain these terms: electron, atomic nucleus, electromagnetic radiation, spectroscopy.
Light Spectra
Learning Objectives
• Explain the terms white light, line, emission, & absorption spectra
• Interpret the regularity of hydrogen spectra
Quantum Numbers and Atomic Orbitals
follow the rules:
``` for the principal q.n. n
azimuthal q.n. l = 0, 1, 2, ..., n-1
magnetic q.n. m = -l, -(l-1), ..., (l-1), l
```
Rules are algorithms, by which we generate possible quantum numbers. The lowest value of n is 1 (NOT zero). For n = 1, the only possible value for quantum number l is 0, and m = 0. Each set of quantum numbers is called a state. Thus, for n = 1, there is only one state (1,0,0). The states are represented by symbols, and special symbols have been used to represent the quantum number l as follows:
``` l = 0, 1, 2, 3, 4, ...
symbol = s, p, d, f, g, ...
```
Using symbols, the valid quantum states can be listed in the following manner:
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f 5g
6s 6p 6d 6f 6g 7h
7s 7p 7d 7f 7g 7h 8i
For hydrogen-like atoms, that is atoms or ions with one electron, the energy level is solely determined by the principle quantum number n, and the energy levels of the subshells np and nd etc. are the same as the ns. For these species, the energy levels have the reverse order as the list given earlier:
``` = 7s 7p--- 7d----- .... ] These are very close together!
= 6s 6p--- 6d----- .... ]
= 5s 5p--- 5d----- 5f------- 5g---------
- 4s 4p--- 4d----- 4f-------
- 3s 3p--- 3d-----
- 2s 2p---
- 1s
```
Confidence Building Questions
1. How many possible orbitals are there if n = 3?
Hint: There are nine (9) possible orbitals.
Discussion:
3s, 3 3p, 5 3d; total = 1 + 3 + 5 = ?
2. How many possible orbitals are there in the subshell [n=5, l=4]?
Hint: There are 9 such orbitals.
Discussion:
For 5g, l = 4; ml = -4, -3, -2, -1, 0, 1, 2, 3, 4. A total of 9.
3. How many electrons can be accommodated in the subshell 4f?
Hint: A total of 14 electrons.
Discussion:
Each of the 7 4f orbitals accommodates a pair of electrons. There are 14 elements in the lanthanides, the 4f block elements.
4. How many atomic orbitals are there for the subshell with [n = 3, l = 2]?
Hint: There are 5 3d orbitals.
Discussion:
The magnetic q.n. = -2, 1, 0, 1, 2. The number of orbitals associated with a given value of l is (2l + 1).
5. What is the symbol representing the set of orbitals in the subshell with [n = 3, l = 2]?
Hint: The symbol is 3d.
Discussion:
For symbols, consider the following:
```Symbol = s p d f g correspond to
l = 0 1 2 3 4
```
The d-block in each period of the periodic table has 10 elements.
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Electronegativity of some elements are given below for your reference:
``` B C N O F
2.0 2.6 3.0 3.4 4.0
Cs Si P S Cl Kr Xe
0.8 1.9 2.2 2.6 3.2 2.9 2.6
```
The higher the electronegativity, the more attraction the element has towards bonding electrons. Bonding between two elements with a large electronegativity difference tends to be ionic. It is interesting to note that inert gases \(\ce{Kr}\) and \(\ce{Xe}\) have elenegativities similar to those of nitrogen and carbon.
On the periodic table, there is a general trend. An element close to \(\ce{F}\) has a large electronegativity, whereas an element close to \(\ce{Cs}\) on the opposit corner of the periodic table from \(\ce{F}\) has the least electronegativity.
The Periodic Table as a Summary of Chemical Properties
The periodic table is a convenient way to correlate chemical properties. For example, from their position on the periodic table, we easily recognize them as metals, semimetals (metalloids), or nonmetals.
Most elements are metals (M); the rest are metalloids (o), nonmetals (-), and inert gases (i), on the top right hand of the long periodic table.
``` - i
Mo o----i
MM oo---i
MM MMMMMMMMMMMoo--i
MMMMMMMMMMMMMMMMMMMMMMMMMMMMoo-i
MMMMMMMMMMMMMMMMMMMMMMMMMMMMMooi
```
Look at some metals, and see if you can describe their characteristics. If not, please check some resource books to see if they give a description you like?
Metals are characterized by having a high melting point, high electric and heat conductivity, metallic bonding, being malleable and ductile, forming positive ions \(\ce{Cu^2+}\), \(\ce{Fe^3+}\), and forming alloys with one another. Check out a metal, and you'll be amazed how close it has come to your life.
Nonmetals have low melting points, form molecules, and atoms in their solids are covalently bonded. They are poor conductors of heat and electricity, and they form negative ions or molecular compounds. Here are some non-metals. Note, however, that diamond is an exceptionally good heat conductor, but it does not conduct electricity.
Noble gases: \(\ce{He}\), \(\ce{Ne}\), \(\ce{Ar}\), \(\ce{Kr}\), \(\ce{Xe}\), \(\ce{Rn}\)
Oxygen group: \(\ce{O}\), \(\ce{S}\), \(\ce{Se}\)
Nitrogen group: \(\ce{N}\), \(\ce{P}\), \(\ce{As}\)
Carbon group: \(\ce{C}\), \(\ce{Si}\)
Boron
Between metals and nonmetals lie the metalloids as indicated above.
The chemistry of these elements is best discussed in GROUPS such as alkali metals, carbon group, halogens, inert gases, etc.
Confidence Building Questions
1. All material exhibit diamagnetism, true or false?
Hint: True, but paramagnetism and ferromagnetism may overcome diamagnetism.
Skill:
Explain what diamagnetism is and how this type of material behaves in a magnetic field.
2. Atoms of chromium have five 3d electrons. What type of magnetic material is chromium?
Hint: Chromium is a paramagnetic element.
Discussion:
Paramagnetism is due to unpaired electrons in the material.
3. Which of the alkali elements has the highest ionization energy (IE): \(\ce{Li}\), \(\ce{Na}\), \(\ce{K}\), \(\ce{Rb}\), or \(\ce{Cs}\)?
Hint: \(\ce{Li}\)
Skill
Explain what ionization energy is, and describe the trend for a group of elements.
It takes the most energy to remove an electron from \(\ce{Li}\) in this group. Which of the inert gases has the highest IE: \(\ce{He}\), \(\ce{Ne}\), \(\ce{Ar}\), \(\ce{Kr}\), or \(\ce{Rn}\)?
4. Which of the following elements releases the most energy when acquiring an electron?
In other words, which has the largest absolute value of electron affinity?
Hint: \(\ce{Cl}\)
Discussion:
The electron affinities are given below:
``` Cs C O Cl F
-45.5 -122 -141 -349 -322 kJ/mol
```
5. \(\ce{Cs}\) \(\ce{C}\) \(\ce{O}\) \(\ce{Cl}\) \(\ce{F}\)
Hint: \(\ce{F}\)
Discussion:
You should know that \(\ce{F}\) is the most electronegative element, and \(\ce{Cs}\) is the least electronegative. Francium should be, but there is no natural occuring francium because its isotopes are all radioactive.
6. Should you compare \(\ce{O}\) and \(\ce{Cl}\), which one is more electronegative?
Hint: The Pauling scale electronegativities are: oxygen, 3.4; chlorine, 3.2.
Discussion:
Due to the large electronegativity and the small size of oxygen atoms, oxygen compounds such as water and ethanol form hydrogen bonds.
Trivialities of Chemical Elements
Everyone loves sports and that is the talk of the town. Knowing the winner is a great asset, because that is a great ice breaker in a conversation. Knowing it makes you a person fun to be with.
Chemical trivialities are also ice breakers and fun things to talk about. You attract attention for being knowledgeable about chemical elements. These trivialities are zests of life.
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We have discussed earlier that, within an atom, there are magnetic moments owing to the orbital and spin motion of the electrons. Magnetic interaction between these moments was discussed as spin- orbit interaction that caused each term of a given $l$ split into terms each with same $l$ but different $j(l \pm s$ value. In other words, there is a net magnetic moment, $\mu_j$ in an atom. When atom is subjected to external magnetic field, $H$ of few thousand gauss (1 Tesla = $10^{4}$ Gauss), there will be magnetic interaction between $H$ and $\mu_j$ resulting in further splitting of the spectral lines. Such splitting, in the presence of magnetic field of few thousand gauss, of the sodium $D$ lines was first observed by Pieter Zeeman, a Dutch physicist, in 1896; the effect was subsequently named after him – Zeeman Effect. The Beauty of these findings is that splitting was observed not only before Bohr’s theory of hydrogen atom (1913) but was also before the discovery of electron in 1897. The importance of the Zeeman Effect can be judged by its application to the fields of NMR, ESR, MRI, Mӧssbauer Spectroscopy etc., each is now a well established field of study in itself.
Zeeman Effect
Singlet series (i.e. spin is zero) of atoms was observed to split into three components when viewed in transverse direction with respect to the direction of the magnetic field and to two components in the longitudinal direction. In transverse view, the outer two components are equally placed (on energy scale) around the central component that has the same energy as that of the original line. Electric vector () of the outer components, designated as (also known as (senkrecht) components) is perpendicular to the magnetic field direction whereas electric field vector of the undisplaced central component, named as (also known as (parallel) component) is parallel to the field lines of force. In the longitudinal view, the outer components are circularly polarized in the opposite direction and the central undisplaced component is not observed because of the transverse nature of the electromagnetic radiation (Fig. 5).
Multiplet lines split into several components. We will see below that strength of the external magnetic field relative to the internal decides whether there are three components with equal spacing (Normal Zeeman Effect) or more than three components with unequal spacing (Anomalous Zeeman Effect). H.A. Lorentz successfully interpreted the normal Zeeman Effect by using the laws of classical physics. He assumed that the motion of electron in atom is harmonic. Since the electron is revolving (say with angular velocity ) around the nucleus in an orbit of radius , the restoring force is centrifugal force. Therefore,
$m_e \omega^2 r_o = k r_0 \label{35}$
Application of homogeneous magnetic field, H in a direction perpendicular to the plane of the orbit provides additional radial force, Hev. The motion of a linearly oscillating electron may be divided into two components; first, a linear oscillation along the direction of the field and second, two rotary oscillations (with opposite directions of rotation) in a plane perpendicular to H (Fig. 6). In case of clockwise rotary motion, the additional force (Hev) is directed outward (away from the nucleus) and for the anticlockwise additional force is directed inward as shown in the fig. The magnetic field does not act on the component parallel to it and therefore the
Fig. 6 Classical picture explaining Zeeman effect & the rotary oscillatory components in green colours, of linear oscillation.
Frequency remains unchanged, undisplaced line ( component). That is, there is additional radial force, $Hev$ on the rotary components that makes the motion complicated. The resulting motion turns out to be Larmor’s precession yielding different frequencies; the difference in frequencies same in magnitude but opposite in sign ( component). According to Larmor’s theorem frequencies of the outer components are (in mks system) is Larmor’s frequency.
Alternatively, one can interpret the Normal Zeeman Effect as follows:
To compensate the additional radial force due to the external magnetic field, centrifugal force has to change keeping the orbit radius constant. Therefore, if new angular velocity is$\omega'$
$m_e \omega'^2 r_o - m_e \omega^2 r_o = He \omega' r_o \label{36}$
or
$m_e ( \omega'^2 - \omega^2) = He\omega'$
of
$(\omega'^2 - \omega) (\omega'+ \omega) = \dfrac{He \omega'}{m_e} \label{37}$
If ; i.e. ; therefore, to first approximation ; equation $\ref{37}$ then reduces to
$\Delta \omega = \dfrac{eH}{2m_e}$
or
$\Delta \nu = \dfrac{eH}{4 \pi m_e} = \nu_L \label{39}$
where $\nu_L$ is the Larmor’s frequency.
It may be mentioned that there had been numerous studies to arrive at some general interpretation that might explain normal as well as anomalous Zeeman Effect but all of them resulted in the explanation of normal Zeeman triplet. A survey of all the attempts is given in ‘The historical development of quantum theory’, Jagdish Mehra, Helmut Rechenberg (Springer, 2001). In most of the earlier interpretations, either spin was not introduced or if it was, its gyro magnetic ratio (ratio of its magnetic dipole moment to its angular momentum) was not correctly understood. To explain the anomalous Zeeman Effect, one not only needs to introduce the concept of the spin of the electron & magnetic moment associated with its mechanical motion, but also its (spin magnetic moment) differentiation from the orbital magnetic moment and finally the interaction between the two in the presence of the external magnetic field.
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To determine the number of components (Zeeman lines) one has to find the magnitude of separations among Zeeman levels (split energy levels under Zeeman Effect); and to know the separations, knowledge of the total magnetic moment of the atom is a prime requirement. Let us take simple case of a single valence electron atom in which lone electron is moving around the core (stationary nucleus plus completely filled orbit(s)) whose net moments is zero. That means whatever magnetic moments an atom has is all exclusively due to the valence electron.
Whatever magnetic moments an atom has is all exclusively due to the valence electron.
We have already discussed that, because of the orbital motion,
---------------[I.45']
and for the spin motion,
---------------[I.49']
is Bohr Magneton and $g$ is Lande’ $g$ factor and has value 1 for pure orbital motion and 2 for pure spin motion. Since, $l$ and precess about their resultant and must also precess about . However, the total magnetic moment, because does not coincide with (Figure $1$).
As resultant mechanical moment, is invariant, all the moment vectors and precess around it (Figure $1$). Recalling that mechanical and magnetic moments (for orbital as well as for spin motion) lie along the same line but in opposite direction, the total magnetic moment of the atom is, likewise expected to be in line with its resultant mechanical moments though directed in opposite directions. To meet this requirement, let us resolve into components, one component parallel to and the second perpendicular to it. The perpendicular component will constantly be changing its direction; consequently its time average over a period is zero. Under this assumption, net effective magnetic moments of the atom, , is the component parallel to whose magnitude will be vector sum of the components of and along .
Therefore, //components
---------------[39]
From the figure (7) it is clear that leading to and similarly . Substitution of the values for reduce the relation for , to
---------------[40]
In quantum mechanics, the equation is expressed in terms of the operator and is expressed as
---------------[40']
These operators commute with each other. Using the properties of the commuting operators and the digitalization of their matrix, operators can be replaced by the corresponding eigenvalues. Therefore, the relation is rewritten as
---------------[41]
or
---------------[42]
$\vec{mu}_j = g \dfrac{\hbar}{2 m_e} j = g \dfrac{e \hbar}{2m_e} j \dfrac{\mu_B}{\hbar}j \label{42}$
where is Bohr Magneton
and
$g = \dfrac{1+[j(j+1) + s(2+1)- l(l+1)}{2j(j+1)} \label{43}$
In case of a pure orbital motion which means $s=0$; value of $g$ turns out to be 1. For pure spin motion, $l=0$; consequently value of $g$ is 2.
Equation $\ref{42}$ is similar to the ones we obtained earlier for orbital and spin motion of the electron (equations I.45' and I.49'). The precession of around the external field is the result of the torque acting on both $l$ and $s$. When the external field is not very strong, $j$ retains its significance and therefore precesses with a compromise angular velocity (as given by Larmor’s theorem)
$\omega_L = g \dfrac{eH}{2M_e} \label{44}$
Equation $\ref{44}$ is same as for the pure orbital motion except for the Landeʹ g-factor which, in any case, is 1 for the orbital motion.
Note that, in the above treatment, no where external magnetic field is considered. is the resultant magnetic moment in an atom arising out of the internal motions of the electron and hence refers to as ‘INNER PRECESSION’.
When the atom is subjected to external magnetic field, $H$, it will start precessing around the field direction; the precession is termed as ‘OUTER PRECESSION’. Although the atom precesses as a whole around $H$, yet the nature of the internal precessions (and hence the magnetic interactions) depend largely on the relative strength of the external magnetic field, $H$.
1. If field strength of $H$ is weak compared to that of internal magnetic moment, , spin –orbit coupling remains intact and therefore, continues to have physical meaning. That means is well defined and precess around when atom is subjected to it. coupling is intact means precessions of and as a rigid system around is more rapid compared to that of around ; and under this condition one can legitimately assume that perpendicular components of average out to zero over a period of rotation. We will see below that weak external field compared to internal leads to the anomalous Zeeman Effect.
2. If field strength of $H$ is fairly strong compared to internal magnetic moment, , spin –orbit coupling is weakened or even broken. Under this condition starts losing its physical significance thereby start precessing independently around . This means precessions of and about is more rapid compared to their precession about , a must requirement for Paschen-Back effect, which is discussed later.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Zeeman_Effect/2%3A_Anomalous_Zeeman_Effect_%28Vector_Model%29.txt
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The interaction energy of the magnetic moments of the atom, with the external field is given by
---------------[45]
With the help of relation , where , the equation (45) is simplified to
---------------[45']
With projection of on , i.e. component of along , relation (45') reduces to
(in mks units)
(in cgs units) with ---------------[46]
Finally, in terms of Bohr magneton,
---------------[47]
Or ---------------[47']
is Lorentz unit ---------------[48]
gives the change in energy for each of from the original line. Obviously, apart from the dependence on , the Landeʹg factor plays an important role to find the relative separations of the Zeeman levels. The value of the g-factor depends on , and ; consequently, value of and therefore the energy differences, is different for different terms i.e. different values. This results in the appearance of the several lines in accordance with the selection rule . Further, when spin is disregarded , value of turns out to be one; in such a case, the differences between the Zeeman levels again reduce to , that is the equation reduces to Lorentz’s classical formula.
Levels in Anomalous Zeeman levels split into equidistant levels for a given but with a magnitude times , the separation observed in the normal Zeeman Effect. In the absence of anomalous behaviour of the spin, is expected to be equal to would have been in line with resultant mechanical moment; consequently
A relation similar to the one obtained for orbital motion except that is replaced by . Magnetic interaction energy (in ) then is given by
or ---------------[49]
Separations between Zeeman lines (because ) would have again been same like in the case of normal Zeeman Effect.
Zeeman patterns for yellow sodium lines (first member of the Principal series) are derived as an
factor for are
Similarly, g- factor for and are and 2 respectively. Self explanatory figure (8) shows splitting of Zeeman levels, transitions in accordance with the selection rules , and finally pattern of expected anomalous Zeeman spectrum. Dotted horizontal line on the left side represents the centre of gravity of the levels, dotted vertical transitions indicate forbidden transitions and the heights of the expected spectral lines (shown at the bottom of the Fig. 8) indicate relative intensities.
Fig.8 Anomalous Zeeman pattern of Sodium lines.
Electromagnetic waves that have undergone Zeeman Effect get polarized; nature of polarization as mentioned in the beginning of the section 3 depends on the direction (with respect to the magnetic field) of the field of view. Polarization rules can be derived both from the classical theory as well as from the quantum mechanics. These are summarised below:
Transverse View (viewed perpendicular to the direction of magnetic field, )
1. Transitions with yield plane polarized light having its electric vector oscillating perpendicular to , termed as - components.
2. Transitions with yield plane polarized light having its electric vector oscillating parallel to , termed as - components.
Longitudinal View (viewed parallel to the direction of magnetic field, )
1. Transitions with yield circularly polarized light, termed as - components.
2. Transitions with are forbidden.
Problem: Discuss the Anomalous Zeeman spectra of the green and violet lines of mercury.
4: Intensity of Zeeman components
Intensity sum rules of the alkali spectra are equally applicable to provide relative intensities of the Zeeman components also. In this case the initial and the final states are Zeeman levels. This may be justified on the basis of the correspondence principle. Assuming that the magnetic field strength is such that the degeneracy is removed; the populations, at equilibrium, of the various Zeeman levels, are proportional to the respective statistical weights. Further, taking into account the relation between the Einstein emission and absorption coefficients (statement to be verified), i.e.
,
the sum rules may be stated as:
The arguments, on which the above sum rules are based, can be extended to more general case of transitions: . Calculations are somewhat tedious and therefore the resulting formulae for the intensity, , are given below.
are constants; their values need not be determined for the relative intensity considerations within a Zeeman pattern. Since final and initial states can be interchanged, transitions are included in these relations.
It is important to note that half of the circularly polarized light is seen in the transverse direction and the second half of the circularly polarized is observed in the longitudinal direction. On the other hand, total intensity of the linearly polarized light is seen in the transverse direction and zero intensity (no light) in the longitudinal direction. These considerations have been taken into account in the above equations.
A careful observation of the figure (8) showing Zeeman transitions and of the equation (47), reveal that the magnitude of separation between the consecutive Zeeman levels for a given values of and of the magnetic field, is same; the quantum, of course, depends on – factor. This is true for given values irrespective of the electronic state. This fact coupled with selection rule yields or –components and or –components) lead to a quick method to obtain the Zeeman pattern. The method is explained taking example of Zeeman pattern of sodium line.
Write values in a row. Below this are the separation factors for both the initial and final states in two rows such that equal values of lie directly one above the other (Fig. 9).
Fig. 9 Zeeman components
Vertical arrows indicate (p)-components and the diagonal arrows components . The differences between the separation factors (i.e. the difference between values of the start and the end of the arrows) with a least common denominators are:
Vertical differences ( (p)-components)
Diagonal differences ( (s)-components)
In short the shifts of the lines are .
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Zeeman_Effect/3%3A_Magnetic_Interaction_Energy_%28Anomalous_Zeeman_Effect%29.txt
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Spectrum of the field free hydrogen atom is simple, in the sense that $L_{\alpha}$ (first member of Lyman series) has doublet fine structure whereas $H_{\alpha}$ (first member of the Balmer series) has very close lying seven components. Though the fine structure is very small (less than half a wave number), it could be experimentally observed. Zeeman spectrum of hydrogen is not that simple and can be understood by taking $L_{\alpha}$ and as $H_{\alpha}$ lines as the examples.
The doublet fine structure of hydrogen Lyman alpha line arises as a result of two transitions:
$2^2P_{1/2} \rightarrow 1^2S_{1/2}$
and
$2^2P_{3/2} \rightarrow 1^2S_{1/2}$
In the presence of an external weak magnetic field, the Zeeman effect splits the $^1S_{1/2}$ and $^1P_{1/2}$ states into 2 levels each with $m_J = \pm 1/2$ and the $^2P_{3/2}$ state into four Zeeman levels $m_J = 3/2$, $m_J = 1/2$, $m_J = -1/2$, $m_J = -3/2$). The corresponding Lande g-factors for these three levels are:
• $g = 2$ for $^2S_{1/2}$ state (i.e., (J=1/2, l=0)
• $g = 2/3$ for $^2P_{1/2}$ state (i.e., (J=1/2, l=1)
• $g = 4/3$ for $^2P_{3/2}$ state (i.e., (J=3/2, l=1)
Note in particular that the size of the energy splitting $m_jg$ is different for the different states, because the $g$ values are different. On the left of the Figure $1$, the fine structure splitting is because of the spin-orbit coupling that occurs even in the absence of a magnetic field. Zeeman levels that show up only in the presence of external magnetic field are shown on the right side of the figure.
Under normal magnetic field, say 1 tesla, all the Zeeman components will be within less than half a wave number rendering the structures too close to observe all the components. Consequently, one may not confirm experimentally the value of with Zeeman spectra in . When the external magnetic field is stronger than the internal magnetic field, the orbital and spin magnetic moments become quantized independently in the direction of the magnetic field, . is then replaced by the quantum number . This reduces the anomalous Zeeman spectra to a normal Zeeman triplet. Therefore, Paschen-Back effect of the hydrogen atom is just a normal Zeeman effect.
has fine structure comprising of the seven components arising from the transitions and . When subjected to weak magnetic field, each of the seven fine structure components will exhibit anomalous Zeeman components placed symmetrically around the fine component. Again all the seven anomalous Zeeman patterns will lie within half a wave number thereby making the observation of anomalous Zeeman effect of line extremely difficult. With the increase of the field strength, Zeeman levels of all the fine components start overlapping until in a field strength of few tesla, the Paschen Back effect sets in. Under these conditions, the strong field Zeeman pattern turns, approximately out to a normal triplet. Paschen and Back observed such normal triplet when the atom was subjected to a field of greater than three teslas.
Quantum mechanical treatment also yields the same Zeeman splitting of the field free lines. The Hamiltonian in the presence of magnetic field is sum of the field free Hamiltonian $(H_o)$ and the magnetic interaction energy $(\Delta E)$:
$H = H_0 + \Delta E$
Average value of the has been derived to be . If this interaction energy is small compared to the internal spin-orbit interaction, which is usually the case with external magnetic field of 1 Tesla, it can be treated as perturbation. Application of first-order perturbation theory yields the shift equal to the magnitude of the perturbation, i.e. .
The theoretical value of the spin g-factor is 2.001041589 (T. Beier et al, Phys. Rev. Lett. 88, 011603 (2002), V. Shabaev et al, Phys. Rev. Lett. 88, 091801 (2002), V. Yerokhin et al, Phys. Rev. Lett. 89, 143001 (2002)), slightly greater than 2. The deviation has largely been attributed to first order and second order Quantum Electrodynamics corrections.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Zeeman_Effect/5%3A_Zeeman_Effect_in_Hydrogen_atom.txt
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In describing the anomalous Zeeman effect, it is assumed that external magnetic field is weak compared to the internal fields; consequently the interaction between \)J\) and $H$ is weak compared to the interaction of orbital and spin magnetic fields. If the strength of the external magnetic field is increased to the extent that it starts competing with the internal field, the interaction between $l$ and $s$ starts loosening or gets broken at sufficiently large value. Under this condition starts losing its significance and & interact independently with the external magnetic field resulting in the independent precessions of & about ; their precession much faster than the precession of residual interaction of & about . This is known as Paschen-Back Effect. It may be recalled that, as emphasized earlier, the separation between the spin components or between the Zeeman components is a measure of the corresponding precessional frequency.
------------[50]
Using equations for respectively and substituting the values of , the equation (50) reduces to
is Bohr magneton ---------------[51]
In terms of wave numbers, ---------------[51']
Although precess independently, yet each produces magnetic field on the electron causing some perturbation to other’s motion. In other words, a small interaction is inherent that one has to take into considerations despite it being small compared to the effect of the external magnetic field. The separation due to this residual interaction is of the same order of magnitude as the field free fine structure doublet separations. Therefore, using equation and that the is vector sum of , its contribution can be written as
---------------[52]
In field free interaction (sec 9.2, chapter I), and rotate together as a rigid system about thereby the angle between and is constant rendering easy evaluation of . In the present case angle between and is varying continuously because of the anomalous behavior of spin precess faster than . This necessitate the use of average value of that can be evaluated using trigonometry theorem, . Using this theorem along with the values of and from the figure (11), one finds
---------------[52']
Adding this to equation (51'), total energy shift becomes
---------------[53]
A general relation for the term values may, thus, be written as
---------------[54]
is the term value of the hypothetical center of gravity of the fine structure doublet. It may be reminded that is no longer well defined quantum number; therefore, Paschen Back effect is described in terms of quantum number . As an example, Paschen Back effect of sodium lines: quantum numbers for the states involved in the transitions exhibiting Paschen Back effect are given below.
Two highlighted states have same value of Paschen Back quantum number and will, therefore, constitute a single component of the state . States corresponding to these quantum numbers are shown in the figure 12. Selection rules allow only six out of possible ten transitions. If resolution of the detecting system is small enough to neglect the interaction, the Paschen Back will constitute three lines; each is a coinciding pair (a coincides with with and with ) since the values of for each of the components of a pair of lines are , 0 and respectively (fig 12). Dotted lines represent the forbidden transitions. Three pair of lines is obtained under the assumption that ls- interaction is zero.
The effect of the external magnetic field (weak to strong with respect to internal magnetic field) on the spectrum can be summarized as: as long as the external magnetic field is unable to perturb the inner precession, one gets anomalous Zeeman spectrum. On the other side, if the strength of the external field yields the magnetic resolution
more than the spin –orbit fine structure, one ends up with a normal Zeeman triplet, shown in the Figure 13 (transition of interaction for WEAK to STRONG field). In a way, these are the two extreme situations; what about the intermediate fields, i.e. during the transition from relatively weak to strong external field? Usually this transitional zone is referred to as the Paschen Back effect. We will see below that the Paschen Back spectrum is not as simple as discussed and shown in figure 12; it is somewhat complicated and separation between the various magnetic field components is a function of external magnetic field.
The situation can be addressed by using the classical laws of the conservation of angular momentum. The conservation law demands that , projection of on , that describes the internal magnetic field & is well defined under weak field, must be equal to the total angular momentum when the field is relatively strong. That is, conservation law says that . In addition, in keeping with the quantum mechanics, levels with same must not cross in the correlation between the two extreme conditions. Correlation between the magnetic levels of the transitions are shown in the fig 14.
In a relatively strong field, as discussed above, the extreme limit (spin is not coupled to orbital motion) of the Zeeman effect leads to normal Zeeman triplet. Quantum numbers and , since are independently & strongly coupled to , are well defined and, therefore, keep their physical significance. Conservation of angular momentum accounts for the selection rules: for the electric dipole radiation. Since the orientation of the spin does not alter during any radiation process (emission or absorption), the selection rules and for dipole radiation hold well. Together with these, selection rules and allow us to ignore spin. Or one may argue that on substituting in the Equation 51, that is magnetic energy shift , the shift turns out to be proportional to ; that is spin is completely dropped out. Consequently, only three lines corresponding to are observed. Implication of the selection rule is that the polarization remains unaltered throughout all the magnetic field strengths. It may be remarked that should , there will be residual spin-orbit coupling yielding group of several transitions around each of the three components.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Zeeman_Effect/6%3A_Paschen%E2%80%93Back_Effect.txt
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Electron spectroscopy is an analytical technique to study the electronic structure and its dynamics in atoms and molecules. In general an excitation source such as x-rays, electrons or synchrotron radiation will eject an electron from an inner-shell orbital of an atom.
• Circular Dichroism
Circular Dichroism, an absorption spectroscopy, uses circularly polarized light to investigate structural aspects of optically active chiral media. It is mostly used to study biological molecules, their structure, and interactions with metals and other molecules.
• Electronic Spectroscopy: Application
Electronic Absorption and Fluorescence spectroscopy are both analytical methods that center around the idea that when one perturbs a known or unknown solution with a spectrum of energetic photons, those photons that have the correct energy to interact with the molecules in solution will do so, and those molecules under observation will always interact with photons of energies characteristic to that molecule.
• Electronic Spectroscopy - Interpretation
Electronic Spectroscopy relies on the quantized nature of energy states. Given enough energy, an electron can be excited from its initial ground state or initial excited state (hot band) and briefly exist in a higher energy excited state. Electronic transitions involve exciting an electron from one principle quantum state to another. Without incentive, an electron will not transition to a higher level. Only by absorbing energy, can an electron be excited.
• Electronic Spectroscopy Basics
Explains the origin of UV-visible absorption spectra, how they are measured, and how they can be used in the analysis of organic compounds.
• Fluorescence and Phosphorescence
Fluorescence and phosphorescence are types of molecular luminescence methods. A molecule of analyte absorbs a photon and excites a species. The emission spectrum can provide qualitative and quantitative analysis. The term fluorescence and phosphorescence are usually referred as photoluminescence because both are alike in excitation brought by absorption of a photon.
• Jablonski diagram
A Jablonski diagram is basically 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.
• Metal to Ligand and Ligand to Metal Charge Transfer Bands
In the field of inorganic chemistry, color is commonly associated with d–d transitions. If this is the case, why is it that some transition metal complexes show intense color in solution, but possess no d electrons? In transition metal complexes a change in electron distribution between the metal and a ligand gives rise to charge transfer (CT) bands when performing Ultraviolet-visible spectroscopy experiments. A brief introduction to electron transfer reactions and Marcus-Hush theory is given.
• Radiative Decay
Spontaneous emission is the process in which a quantum mechanical system (such as an atom, molecule or subatomic particle) transitions from an excited energy state to a lower energy state (e.g., its ground state) and emits a quantum in the form of a photon.
• Selection Rules for Electronic Spectra of Transition Metal Complexes
The Selection Rules governing transitions between electronic energy levels.
• Spin-orbit Coupling
Spin-orbit coupling refers to the interaction of a particle's "spin" motion with its "orbital" motion.
• Two-photon absorption
Two-photon absorption is one of a variety of two-photon processes. In this specific process, two photons are absorbed by a sample simultaneously. Neither photon is at resonance with the available energy states of the system, however, the combined frequency of the photons is at resonance with an energy state.
Electronic Spectroscopy
Circular Dichroism, an absorption spectroscopy, uses circularly polarized light to investigate structural aspects of optically active chiral media. It is mostly used to study biological molecules, their structure, and interactions with metals and other molecules.
Introduction
Circular Dichroism (CD) is an absorption spectroscopy method based on the differential absorption of left and right circularly polarized light. Optically active chiral molecules will preferentially absorb one direction of the circularly polarized light. The difference in absorption of the left and right circularly polarized light can be measured and quantified. UV CD is used to determine aspects of protein secondary structure. Vibrational CD, IR CD, is used to study the structure of small organic molecules, proteins and DNA. UV/Vis CD investigates charge transfer transitions in metal-protein complexes.
Circular Polarization of Light
Electromagnetic radiation consists of oscillating electric and magnetic fields perpendicular to each other and the direction of propagation. Most light sources emit waves where these fields oscillate in all directions perpendicular to the propagation vector. Linear polarized light occurs when the electric field vector oscillates in only one plane. In circularly polarized light, the electric field vector rotates around the propagation axis maintaining a constant magnitude. When looked at down the axis of propagation the vector appears to trace a circle over the period of one wave frequency (one full rotation occurs in the distance equal to the wavelength). In linear polarized light the direction of the vector stays constant and the magnitude oscillates. In circularly polarized light the magnitude stays constant while the direction oscillates.
As the radiation propagates the electric field vector traces out a helix. The magnetic field vector is out of phase with the electric field vector by a quarter turn. When traced together the vectors form a double helix.
Light can be circularly polarized in two directions: left and right. If the vector rotates counterclockwise when the observer looks down the axis of propagation, the light is left circularly polarized (LCP). If it rotates clockwise, it is right circularly polarized (RCP). If LCP and RCP of the same amplitude, they are superimposed on one another and the resulting wave will be linearly polarized.
Interaction with Matter
As with linear polarized light, circularly polarized light can be absorbed by a medium. An optically active chiral compound will absorb the two directions of circularly polarized light by different amounts $\Delta A = A_l - A_r$
This can be extended to the Beer-Lambert Law. The molar absorpitivty of a medium will be different for LCP and RCP. The Beer-Lambert Law can be rewritten as $A = (\varepsilon_l -\varepsilon_r)cl$
The difference in molar absorptivity is also known as the molar circular dichroism $\Delta \varepsilon = \varepsilon_l -\varepsilon_r$
The molar circular dichroism is not only wavelength dependent but also depends on the absorbing molecules conformation, which can make it a function of concentration, temperature, and chemical environment.
Any absorption of light results in a change in amplitude of the incident wave; absorption changes the intensity of the light and intensity of the square of the amplitude. In a chiral medium the molar absorptivities of LCP and RCP light are different so they will be absorbed by the medium in different amounts. This differential absorption results in the LCP and RCP having different amplitudes which means the superimposed light is no longer linearly polarized. The resulting wave is elliptically polarized.
Molar Ellipticity
The CD spectrum is often reported in degrees of ellipticity, $\theta$, which is a measure of the ellipticity of the polarization given by:
$tan \theta = \frac{E_l-E_r}{E_l+E_r}$
where E is the magnitude of the electric field vector.
The change in polarization is usually small and the signal is often measured in radians where $\theta = \frac{2.303}{4}(A_l - A_r)$ and is a function of wavelength. $\theta$ can be converted to degrees by multiplying by $\frac{180}{\pi}$ which gives $\theta = 32.98 \Delta A$
The historical reported unit of CD experiments is molar ellipticity, $[\theta]$, which removes the dependence on concentration and path length
$[\theta] = 3298 \Delta \varepsilon$
where the 3298 converts from the units of molar absorptivity to the historical units of degrees$\cdot$ cm2$\cdot$dmol-1.
Applications
Instrumentation
Most commercial CD instruments are based on the modulation techniques introduced by Grosjean and Legrand. Light is linearly polarized and passed through a monochromator. The single wavelength light is then passed through a modulating device, usually a photoelastic modulator (PEM), which transforms the linear light to circular polarized light. The incident light on the sample switches between LCP and RCP light. As the incident light swtches direction of polarization the absorption changes and the differention molar absorptivity can be calculated.
Biological molecules
The most widely used application of CD spectroscopy is identifying structural aspects of proteins and DNA. The peptide bonds in proteins are optically active and the ellipticity they exhibit changes based on the local conformation of the molecule. Secondary structures of proteins can be analyzed using the far-UV (190-250 nm) region of light. The ordered $\alpha$-helices, $\beta$-sheets, $\beta$-turn, and random coil conformations all have characteristic spectra. These unique spectra form the basis for protein secondary structure analysis. It should be noted that in CD only the relative fractions of residues in each conformation can be determined but not specifically where each structural feature lies in the molecule. In reporting CD data for large biomolecules it is necessary to convert the data into a normalized value that is independent of molecular length. To do this the molar ellipticity is divided by the number of residues or monomer units in the molecule.
The real value in CD comes from the ability to show conformational changes in molecules. It can be used to determine how similar a wild type protein is to mutant or show the extent of denaturation with a change in temperature or chemical environment. It can also provide information about structural changes upon ligand binding. In order to interpret any of this information the spectrum of the native conformation must be determined.
Some information about the tertiary structure of proteins can be determined using near-UV spectroscopy. Absorptions between 250-300 nm are due to the dipole orientation and surrounding environment of the aromatic amino acids, phenylalanine, tyrosine, and tryptophan, and cysteine residues which can form disulfide bonds. Near-UV techniques can also be used to provide structural information about the binding of prosthetic groups in proteins.
Metal containing proteins can be studied by visible CD spectroscopy. Visible CD light excites the d-d transitions of metals in chiral environments. Free ions in solution will not absorb CD light so the pH dependence of the metal binding and the stoichiometry can be determined.
Vibrational CD (VCD) spectroscopy uses IR light to determine 3D structures of short peptides, nucleic acids, and carbohydrates. VCD has been used to show the shape and number of helices in A-, B-, and Z-DNA. VCD is still a relatively new technique and has the potential to be a very powerful tool. Resolving the spectra requires extensive ab initio calculations, as well as, high concentrations and must be performed in water, which may force the molecule into a nonnative conformation.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Circular_Dichroism.txt
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Electronic Absorption and Fluorescence spectroscopy are both analytical methods that center around the idea that when one perturbs a known or unknown solution with a spectrum of energetic photons, those photons that have the correct energy to interact with the molecules in solution will do so, and those molecules under observation will always interact with photons of energies characteristic to that molecule. These methods have tremendously helped scientists elucidate and characterize the physical properties of a variety of molecules by giving a means to probe the fundamental electronic structure of those molecules.
Introduction
The theory behind Electronic Absorption and Fluorescence was described in a previous text. If you are unfamiliar with electronic spectroscopy, browsing the theory might help paint a better picture of what will be discussed in this module.But even if you understand what a Jablonski diagram represents or the mathematical description behind the transition dipole moment operator, eventually you want to know how to apply the information you've learned to your world.
What this means to you is that you now have the ability that allow you to find the reason why carrots are orange or why plants are green. You could take leaves and carrots, chop them into fine pieces, extract and separate the molecules inside of them, and use Electronic Absorption or Fluorescence spectroscopy to figure out that the orange color in carrots is due to a family of molecules known as the carotenoids and that the green color in the leaves is due to a variety of different chlorophyll molecules. So excited by this knowledge, you could take the carotenoids that you now know are responsible for the color orange and dye cloth to make yourself a carrot costume!
But what if you wanted a different color? What if you wanted a lighter shade of orange? What if you wanted a very exact and very specific color? Before you can go explore the world (or lab) to find (or design) a molecule that has those properties you need to be able to understand the instruments that allow you to do the work so you can be certain you get the results you desire.
The remainder of this wikitext describes the design and components of a photospectrometer to measure the the Ultra Violet through Visable (UV-Vis) region of the electromagnetic spectrum. Basic instrumental design and theory behind a spectrofluorometer will also be discussed along with the errors and limits known to plague these instruments. By the end of this text, you should be fully able to design a a photospectrometer in order to characterize the electronic behavior of molecules in order to prove to yourself that carotenoids are responsible for the color orange (building of a carrot costume is optional)
Basic Design
As stated in the introduction, the purpose of a spectrophotometer is to characterize how molecules react with photons of varying wavelengths. This easiest way to do this is to design an instrument that can control what type of light is in contact with the sample and the amount of that light that makes it through. Shown below is a simple spectrophotometer that does just what was described. A light source produces light that can be separated and controlled by the monochrometer which then allows only the desired light to impinge upon the sample. The light that travels through the sample is detected and sent to a recorder. Keep this overall picture in mind when thinking about the components and what each of their parts are in whole experiment.
Sources
As shown above, the first piece of equipment needed is a source of photons (light). Since we are interested in the UV-Vis region of the electromagnetic spectrum, we need to assure that this source is able to produce photons with the correct wavelengths (180-780 nm) that are characteristic to the spectrums of interest so that the experiment can be completed. Typical sources come in two forms, either in the form of a lamp, or in the form of a laser. Lamps can provide a wide spectrum of light that can be used for almost any purpose at a very reasonable cost, while lasers provide very intense monochromatic light and tend can be used for fluorescence and other specialized experiments due to their high costs.
Tungsten Filament Lamps
A common lamp source that might be familiar is the tungsten filament lamp or sometimes referred to as the incandescent light bulb. This type of lamp produces a spectrum of light perfect for the visible spectrum (320 nm - 3.5 µm)1 by applying a voltage through a thin filament of tungsten until it heats up enough to produce black body radiation. The Tungsten-Halogen lamp is a significant improvement over the normal tungsten filament lamp due to its longer lifetime and the wider range of light it produces. This is achieved by incasing the filament with quartz instead of silica so the filament can be forced to higher temperatures (thus increasing the range of light produced) and by trapping a halogen gas (normally Iodine) that can react with sublimated tungsten atoms to form WI2 that can then redeposit the tungsten by coming into contact with the filament (thus increasing the lifespan).
Hydrogen/Deuterium Lamps
This type of lamp is used to obtain a spectrum of light in the Ultra Violet region (190-400 nm). The spectrum is obtained by trapping Hydrogen at low pressures inside of a glass tube and creating an electrical arc inside of the cell. This electrical arc creates high energy hydrogen gas atoms that dissociate into two hydrogen atoms along with producing photons. The wavelength of the photons created from the dissociation is dependent on the resulting kinetic energies of the two hydrogen atoms after the dissociation. Since the kinetic energies of the dissociated atoms can vary from zero to the original energy of the excited atom, the photon produced can also vary which leads to a spectrum of photons. Deuterium is often used in lieu of hydrogen because it can produce more a more intense spectrum and has a longer lifetime. Both types of this lamp are housed within a quartz shell since silica absorbs in the Ultraviolet range.
Lasers
Tunable dye and fixed wavelength lasers that produce photons within the UV-Vis spectrum are commonly used for fluorescence experiments where the intensity of the resulting fluorescence is directly proportional to the intensity of the source. Lasers are inherently focused which provides the benefit of being able to work with very small or dilute samples. The caveats of using a laser are that the spectrum produced is physically limited and that care must also be taken to avoid photodegredation of your samples by attenuating the power of the laser. A description of the construction and theory behind lasers won't be discussed here, but can be found through a variety of sources.1,2,3,4,5,6
Filters and Monochrometers
Unless a laser is used, a device is needed to restrict and isolate the wavelengths that our sources provide in order to control the experiment. If a isolated wavelength is needed or isn't needed, interference or absorption filters can be used. If a spectrum of wavelengths needs to be passed through the sample, monochrometers provide the ability to separate and control the light passed through the sample while preserving the available spectrum.
Filters
Intereference filters are designed to provide constructive or destructive interference of light by taking advantage of the refraction of light through different materials. As light passes from one medium to the other the direction and wavelength of light can be changed based on the index of refraction of both mediums involved and the angle of the incident and exiting light (For more look at Snell's Law). Due to this behavior, constructive and destructive interference can be controlled by varying the thickness (d) of a transparent dielectric material between two semi-reflective sheets and the angle the light is shined upon the surface. As light hits the first semi-reflective sheet, a portion is reflected, while the rest travels through the dielectric to be bent and reflected by the second semi-reflective sheet. If the conditions are correct, the reflected light and the initial incident light will be in phase and constructive interference occurs for only a particular wavelength.
Another common filter is the Absorption Filter. Absorption filters work on the premise of being able to filter light by absorbing all other wavelengths that aren't of interest. They are normally constructed from a colored glass that absorbs over a wide range. Although normally cheaper than interference filters, absorbance filters tend to be less precise at filtering for a selected wavelength and also have the added penalty of absorbing some of the selected light thus lowering the intensity.
Monochrometers
Monochrometers, as previously mentioned, are used to control and separate light so that a sample can be subjected to a span of wavelengths. Light entering a monochrometer is filtered by a thin slit. The filtered light is then focused by a mirror onto a dispersing element that separates the light into its different wavelengths. The separated light is then focused again and angled toward an exit slit which then filters against all wavelengths except the desired one. The wavelength selected can then easily be changed by rotating the dispersing element to support the transmission of the new desired wavelength. Though monochrometers nearly all have the same practical design, the difference normally is determined by the type of dispersing element.
One type of dispersing element is a prism which disperses light by refraction through two angled surfaces. In order to separate the light into different wavelengths, the prism needs to be made of material that has a change in the index of refraction with respect to wavelength so that each wavelength is bent at a different angle. The larger the difference in the index of refraction the better the separation is between wavelengths. For the UV-Vis range, a typical prism is cut from left handed quartz at a thirty degree angle and attached to another piece of quarts cut the same way except from right handed quartz to make a Conru Prism. The pitfall of prism based dispersing elements is the fact that the index of refraction for most materials varies nonlinearly when compared to wavelength which results in a smaller degree of separation at longer wavelengths.
More commonly used due to less expensive fabrication costs and its ability to separate light in a linear fashion are grating dispersing elements. Gratings are designed to diffract light which is a separation based on the angle of the incident light to the grating normal and the spacing between groves. When beams of light impinge upon the grating some beams travel farther than others and this causes an effect akin to a interference filter which allows for constructive and destructive interference and provides the means to reflect a specific wavelength based on the angle of incidence. The most common type of grating is the echellette grating in which the grooves are angled in order to provide maximum reflection of the incident light into a single order of reflected light.
Sample Cell
Care must be taken when considering a sample container to avoid unwanted absorption in the range of interest. While glass might be perfect for the visible range, it absorbs in the UV range. Quartz can be used for both but most likely would cost considerably more. Disposable Plastic cuvets created from polystyrene or polymethyl methacrylate are in common use today as they are cheap to purchase and eliminate the need for cleaning cuvettes in order to analyze multiple samples. The shape of the sample holder is also very important as unwanted scattering of light should be minimized and pathlength through the cell should remain constant . Square (1 cm) cuvettes with frosted sides have been designed to minimize scattering and provide a surface to hold the cuvette, thereby reducing smudges or smears of the optical window all the while keeping a fixed path length. Cuvettes for fluorescent experiments cannot allow the frosted sides due to the 90° angle of most instruments and have to be carefully cleaned as bodily oils have the ability to fluoresce. Placement of the sample within either a absorption or fluorescence device is imperative and is normally fixed by a sample holder to insure reproducible results.
Detectors
After the light has passed through the sample, we want to be able to detect and measure the resulting light. These types of detectors come in the form of transducers that are able to take energy from light and convert it into an electrical signal that can be recorded, and if necessary, amplified.Photomultiplier tubes are a common example of a transducer that is used in a variety of devices. The idea is that when a photon hits the top of the tube electrons are released, which are pulled toward the other end of the tube by an electric field. The way the electrons are multiplied is due to the fact that along the length of the tube there are several dynodes that have a slightly less negative potential than surface before it which causes the electrons traveling down the tube to hit each surface which then in turn produces more electrons until at the very end there is a large amount of electrons (~1,000,000) representing the one photon that started the cascade.
Another type of transducer is a Charge injection device (CID). This device works by having a p-type and n-type semiconductor next to each other in which the n-type material is separated from the anodes by a silica layer that acts as a capacitor. As a photon interacts with the n-type layer, an electron migrates to the p-type semiconductor and a positive charge is generated that migrates toward the silica capacitor. This process continues to happen until the potential is measured by means of comparing the more negative of the anodes to ground (Vi) . The charge is then transferred to the other anode and measured (Vf). The difference between Vf and Vi is directly proportional to the amount of photons that collided with the n-type layer. The positive charges are then repelled when the anodes switch momentarily to having a positive charge and then the cycle can be repeated. Thousands of these little detectors can be aligned and used to describe the light that interacts with it and can rival the performance of the photomultiplier tube.
Signal Processor
In the end of the experiment all of the data collected needs to be stored or written down. In the past printers have been used to record the transmittance as the experiment progresses. These days, computers are used to record, store, and even manipulate data along with controlling the devices within the spectrometer. This ability allowed by having a computer that can quickly integrate, compare, or annotate spectra dramatically decreases the learning curve needed to operate the instrument as well as reduces the labor associated with simple or even complex experiments. Instead of having to record the transmittance, convert it to absorbance and then subtract the background noise, the computer can accomplish all that for you in the matter of a few seconds .
Instrumental Designs
Now that we've covered the instrument and its components in some detail, there are some different modifications that can be made in order to tailor the device to the experiment it'll be performing. The device introduced in the beginning of this text is the basic single beam spectrometer. This device can measure the transmittance or absorbance of a particular analyte at a given wavelength provided by the source and monochrometer. Background subtraction in these machines needs to be done separately before the analyte is inserted and can be stored and subtracted by the signal processor attached. Although this is the simplest design, the cost of these types of instruments can vary greatly depending on the components that the machine is comprised of.
For fluorescence spectroscopy the equivalent to a single beam spectrometer has a few slight modifications in that the detector is perpendicular to the source, and that there is an additional monochrometer that can be used to vary the wavelength detected. In this fashion, the source light is unable to interfere with the fluorescence light being measured, and the fluorescent light produced can be separated and described as the resulting wavelengths of fluorescence as a function of incident light.
A double beam spectrometer takes a more analytical approach to the design. Because of fluctuations in the source intensity and inconsistencies in the transducer, the source light is split into two beams, one which travels through the sample, and another that is sent through a blank or standard solution. Both beams are then read by separate transducers and the difference between the two is recorded as the corrected transmittance. This allows for quick screening of analytes and negates the need for two separate scans to complete a background subtraction. The same idea can be applied to a fluorometer as well to obtain the same benefit
The latest instrument designs use a multichannel detector, such as an array of CID's, that allow for the spectrum of an analyte to be gathered in seconds due to the fact that the light transmitted through the sample can be split and a spectrum of wavelengths can be monitored simultaniously instead of individually. Also fiber optic cables can be used to transfer the light from the source to the sample or from the sample to the dispersion grating and negate the need to consider the slit when thinking about resolution.
Instrumental Noise
As with any instrument, the measurements we make are limited by the tools we use. In spectrophotometers, the only measurement they are designed to monitor is the transmittance of light through the sample. Thus any noise associated with the transmittance will correlate with errors in our analyses. The analysis of the noise in spectrophotometers has been done7 and is outlined by Skoog and others in "Principles of Analytical Chemistry"3. In the outline they summarize that the types of errors commonly experienced fall into three cases and are listed below.
In the first case, the standard deviation of the transmittance is a constant.
$S_T=k_1$
Errors associated with this situation are due to detectors with limited readout resolution and dark currant and amplifier noise. Limited readout resolution keeps the deviation constant because the measurements more precise than the readout can not be expressed or displayed by the machine. Dark Currant and Amplifier noise are only an issue when the lamp intensity and transducer sensitivity are low and random fluctuations in the currant become the dominant source of error.
In the second case, the deviation in transmittance varies with the equation.
$S_T=K_2 \sqrt{T^2+T}$
Errors associated with this equation are those related to shot noise or the transfer of electrons across any sort of junction or barrier like those found in the photomultiplier tube in the detector. Since the currant is dependant on this transfers, which happen randomly, the currant becomes a random distribution that centers around and average. This error becomes considerable when dealing with either really low or really high transmittances.
The last situation deals with errors that relate proportionally to the transmittance. Errors related to this type of noise have to do with source flickering along with cell positioning. Both of these errors can be easily corrected either by attaching a constant voltage source on the lamp or by placing a fixed cuvette holde
$S_T=k_3T$
Other big sources of error can be introduced from the entrance and exit slits not having the appropriate separation or if the experiment being conducted is outside or near the limit of the instrumentation. In older machines, a scan time that was faster than the recording printer could be a source of error as the printer would not be able to record the data as fast it was being given.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy%3A_Application.txt
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Electronic Spectroscopy relies on the quantized nature of energy states. Given enough energy, an electron can be excited from its initial ground state or initial excited state (hot band) and briefly exist in a higher energy excited state. Electronic transitions involve exciting an electron from one principle quantum state to another. Without incentive, an electron will not transition to a higher level. Only by absorbing energy, can an electron be excited. Once it is in the excited state, it will relax back to it's original more energetically stable state, and in the process, release energy as photons.
Introduction
Often, during electronic spectroscopy, the electron is excited first from an initial low energy state to a higher state by absorbing photon energy from the spectrophotometer. If the wavelength of the incident beam has enough energy to promote an electron to a higher level, then we can detect this in the absorbance spectrum. Once in the excited state, the electron has higher potential energy and will relax back to a lower state by emitting photon energy. This is called fluorescence and can be detected in the spectrum as well.
Embedded into the electronic states (n=1,2,3...) are vibrational levels (v=1,2,3...) and within these are rotational energy levels (j=1,2,3...). Often, during electronic transitions, the initial state may have the electron in a level that is excited for both vibration and rotation. In other words, n=0, v does not = 0 and r does not =0. This can be true for the ground state and the excited state. In addition, due to the Frank Condon Factor, which describes the overlap between vibrational states of two electronic states, there may be visible vibrational bands within the absorption bands. Therefore, vibrational fine structure that can be seen in the absorption spectrum gives some indication of the degree of Frank Condon overlap between electronic states.
When interpreting the absorbance and fluorescence spectra of a given molecule, compound, material, or an elemental material, understanding the possible electronic transitions is crucial. Assigning the peaks in the absorption spectrum can become easier when considering which transitions are allowed by symmetry, the Laporte Rules, electron spin, or vibronic coupling. Knowing the degree of allowedness, one can estimate the intensity of the transition, and the extinction coefficient associated with that transition. These guidelines are a few examples of the selection rules employed for interpreting the origin of spectral bands. Only a complete model of molecular energy diagrams for the species under investigation can make clear the possible electronic transitions. Every different compound will have unique energy spacing between electronic levels, and depending on the type of compound, one can categorize these spacings and find some commonality. For example, aromatic compounds pi to pi* and n to pi* transitions where as inorganic compounds can have similar transitions with Metal to Ligand Charge Transfer (MLCT) and Ligand to Metal Charge Transfer (LMCT) in addition to d-d transitions, which lead to the bright colors of transition metal complexes. Although surprises in science often lead to discovery, it is more fortuitous for the interpreter to predict the spectra rather than being baffled by the observation.
The following section will discuss the interpretation of electronic absorption spectra given the nature of the chemical species being studied. This includes an understanding of the molecular or elemental electronic state symmetries, Russell-Sanders states, spin multiplicities, and forbidden and allowed transitions of a given species.
As the light passes through the monochrometer of the spectrophotometer, it hits the sample with some wavelength and corresponding energy. The ratio of the initial intensity of this light and the final intensity after passing through the sample is measured and recorded as absorbance (Abs). When absorbance is measured at different wavelengths, an absorbance spectrum of Abs vs wavelength can be obtained. This spectra reveals the wavelengths of light that are absorbed by the chemical specie, and is specific for each different chemical. Many electronic transitions can be visible in the spectrum if the energy of the incident light matches or surpasses the quantum of energy separating the ground state and that particular excited state. An example of an absorbance spectrum is given below.
Temperature Effects
Here we can see the effect of temperature and also the effect of solvents on the clarity of the spectrum. We can see from these anonymous compounds that decreasing the temperature allows the vibrational fine structure to emerge. These vibrational bands embedded within the electronic bands represent the transitions from v=n to v'=n. Generally, the v=0 to v'=0 transition is the one with the lowest frequency. From there, increasing energy, the transitions can be from v=0 to v'=n, where n=1,2,3... With a higher temperature, the vibrational transitions become averaged in the spectrum due to the presence of vibrational hot bands and Fermi Resonance, and with this, the vibrational fine structure is lost at higher temperatures.
Solvent Effects
The effect that the solvent plays on the absorption spectrum is also very important. It is clear that polar solvents give rise to broad bands, non-polar solvents show more resolution, though, completely removing the solvent gives the best resolution. This is due to solvent-solute interaction. The solvent can interact with the solute in its ground state or excited state through intermolecular bonding. For example, a polar solvent like water has the ability of hydrogen bonding with the solute if the solute has a hydrogen bonding component, or simply through induced dipole-dipole interactions. The non-polar solvents can interact though polarizability via London interactions also causing a blurring of the vibronic manifold. This is due to the solvent's tendency to align its dipole moment with the dipole moment of the solute. Depending on the interaction, this can cause the ground state and the excited state of the solute to increase or decrease, thus changing the frequency of the absorbed photon. Due to this, there are many different transition energies that become average together in the spectra. This causes peak-broadening. The effects of peak broadening are most severe for polar solvent, less so for non-polar solvents, and absent when the solute is in vapor phase.
Group Theory and The Transition Moment Integral
When estimating the intensities of the absorption peaks, we use the molar absorptivity constant (epsilon). If the transition is "allowed" then the molar absorptivity constant from the Beer's Law Plot will be high. This means that the probability of transition is large. If the transition is not allowed, then there will be no intensity and no peak on the spectrum. Transitions can be "partially allowed" as well, and these bands appear with a lower intensity than the full allowed transitions. One way to decide whether a transition will be allowed or not is to use symmetry arguments with Group Theory.
If the symmetries of the ground and final state of a transition are correct, then the transition is symmetry allowed. We express this by modifying the transition moment integral from an integral of eigenstates to an orthogonally expressed direct product of the symmetries of the states.
$\int \psi_{2} \mu \psi_{1} d \mathcal{T} \longrightarrow \Gamma_{2} \otimes \Gamma \mu_{x y z} \otimes \Gamma_{1} \label{1}$
$\int \psi_{v_{2}} \psi_{e l_{2}} \mu \psi_{v_{1}} \psi_{e l_{1}} d \mathcal{T} \longrightarrow \Gamma_{\mathrm{v}_{2}} \otimes \Gamma_{\mathrm{el}_{2}} \otimes \Gamma \mu_{x y z} \otimes \Gamma_{\mathrm{v}_{1}} \otimes \Gamma_{\mathrm{el}_{1}} \label{2}$
The conversions of integration to direct products of symmetry as shown gives spectroscopists a short cut into deciding whether the transition will be allowed or forbidden. A transition will be forbidden if the direct products of the symmetries of the electronic states with the coupling operator is odd. More specifically, if the direct product does not contain the totally symmetric representation, then the transition is forbidden by symmetry arguments. If the product does contain the totally symmetric representation (A, A1, A1g...etc) then the transition is symmetry allowed.
Some transitions are forbidden by the Equation \ref{1} and one would not expect to be able to see the band that corresponds to the transition; however, a weak absorbance band is quite clear on the spectrum of many compounds. The transition may be forbidden via pure electronic symmetries; however, for an octahedral complex for example since it has a center of inversion, the transition is weakly allowed because of vibronic coupling. When the octahedra of a transition metal complex is completely symmetric (without vibrations), the transition cannot occur. However, when vibrations exist, they temporarily perturb the symmetry of the complex and allow the transition by equation (2). If the product of all of these representations contains the totally symmetric representation, then the transition will be allowed via vibronic coupling even if it forbidden electronically.
Hot Bands
Some transitions are forbidden by symmetry and do not appear in the absorption spectrum. If the symmetries are correct, then another state besides the ground state can be used to make the otherwise forbidden transition possible. This is accomplished by hot bands, meaning the electrons in the ground state are heated to a higher energy level that has a different symmetry. When the transition moment integral is solved with the new hot ground state, then the direct product of the symmetries may contain the totally symmetric representation. If we employ the old saying, "You can't get there from here!" then we would be referring to the transition from the ground state to the excited state. However, if we thermally excite the molecules from out of the ground state, then, "we can get there from here!"
Knowing whether a transition will be allowed by symmetry is an essential component to interpreting the spectrum. If the transition is allowed, then it should be visible with a large extinction coefficient. If it is forbidden, then it should only appear as a weak band if it is allowed by vibronic coupling. In addition to this, a transition can also be spin forbidden. The examples below of excited state symmetries, give an indication of what spin forbidden means:
These states are derived from the electron configuration of benzene. Once we have the molecular orbital energy diagram for benzene, we can assign symmetries to each orbital arrangement of the ground state. From here, we can excite an electron from the Highest Occupied Molecular Orbital (HOMO) to the Lowest Unoccupied Molecular Orbital (LUMO). This is the lowest energy transition. Other transitions include moving the electron above the LUMO to higher energy molecular orbitals. To solve for the identity of the symmetry of the excited state, one can take the direct product of the HOMO symmetry and the excited MO symmetry. This give a letter (A, B, E..) an the subscript (1u, 2u, 1g...). The superscript is the spin multiplicity, and from single electron transitions, the spin multiplicity is 2S+1 = M, where S = 1 with two unpaired electrons having the same spin and S=0 when the excited electron flips its spin so that the two electrons have opposite spin. This gives M=1 and M=3 for benzene above. From the results above, we have three transitions that are spin allowed and three that are spin forbidden.
Once we take the direct product of the symmetries and the coupling operator for each of these states given above, we find that only the A1g to E1u transition is allowed by symmetry. Therefore, we have information regarding spin and symmetry allowedness and we have an idea of what the spectra will look like:
When interpreting the spectrum, it is clear that some transitions are more probable than others. According to the symmetry of excited states, we can now order them from low energy to high energy based on the position of the peaks (E1u is the highest, then B1u, and B2u is lowest). The A1g to E1u transition is fully allowed and therefore the most intense peak. The A1g to B1u and A1gto B2u transitions are symmetry forbidden and thus have a lower probability which is evident from the lowered intensity of their bands. The singlet A1g to triplet B1u transition is both symmetry forbidden and spin forbidden and therefore has the lowest intensity. This transition is forbidden by spin arguments; however, a phenomenon known as spin-orbit coupling can allow this transition to be weakly allowed as well. If spin-orbit coupling exists, then the singlet state has the same total angular momentum as the triplet state so the two states can interact. A small amount of singlet character in the triplet state leads to a transition moment integral that is non-zero, so the transition is allowed.
Organic Molecule Spectra
From the example of benzene, we have investigated the characteristic pi to pi* transitions for aromatic compounds. Now we can move to other organic molecules, which involves n to pi* as well as pi to pi*. Two examples are given below:
The highest energy transition for both of these molecules has an intensity around 10,000 cm-1 and the second band has an intensity of approximately 100 cm-1. In the case of formaldehyde, the $n \rightarrow \pi^*$ transition is forbidden by symmetry where as the pi to pi* is allowed. The opposite is true for As(Ph)3 and the difference in molar absorptivity is evidence of this.
$n \rightarrow \pi^*$ transitions
These transitions involve moving an electron from a nonbonding electron pair to a antibonding $\pi^*$ orbital. They tend to have molar absorbtivities less than 2000 and undergo a blue shift with solvent interactions (a shift to higher energy and shorter wavelengths). This is because the lone pair interacts with the solvent, especially a polar one, such that the solvent aligns itself with the ground state. When the excited state emerges, the solvent molecules do not have time to rearrange in order to stabilize the excited state. This causes a lowering of energy of the ground state and not the excited state. Because of this, the energy of the transition increases, hence the "blue shift".
$\pi \rightarrow \pi^*$ transitions
These transitions involve moving an electron from a bonding $\pi$ orbital to an antibonding $\pi^*$ orbital. They tend to have molar absorptivities on the order of 10,000 and undergo a red shift with solvent interactions (a shift to lower energy and longer wavelengths). This could either be due to a raising of the ground state energy or lowering of the excited state energy. If the excited state is polar, then it will be solvent stabilized, thus lowering its energy and the energy of the transition.
Inorganic Molecule Spectra
Speaking of transition probabilities in organic molecules is a good introduction to interpreting the spectra of inorganic molecules. Three types of transitions are important to consider are Metal to Ligand Charge Transfer (MLCT), Ligand to Metal Charge Transfer (LMCT), and d-d transitions. To understand the differences of these transitions we must investigate where these transitions originate. To do this, we must define the difference between pi accepting and pi donating ligands:
d-d Transitions
From these two molecular orbital energy diagrams for transition metals, we see that the pi donor ligands lie lower in energy than the pi acceptor ligands. According to the spectral chemical series, one can determine whether a ligand will behave as a pi accepting or pi donating. When the ligand is more pi donating, its own orbitals are lower in energy than the t2g metal orbitals forcing the frontier orbitals to involve an antibonding pi* (for t2g) and an antibonding sigma* (for eg). This is in contrast to the pi accepting ligands which involve a bonding pi (t2g) and an antibonding sigma* (eg). Because of this, the d-d transition (denoted above by delta) for the pi acceptor ligand complex is larger than the pi donor ligand. In the spectra, we would see the d-d transitions of pi acceptor ligands to be of a higher frequency than the pi donor ligands. In general though, these transitions appear as weakly intense on the spectrum because they are Laporte forbidden. Due to vibronic coupling; however, they are weakly allowed and because of their relatively low energy of transition, they can emit visible light upon relaxation which is why many transition metal complexes are brightly colored. The molar extinction coefficients for these transition hover around 100.
LMCT Transitions
At an even higher energy are the LMCT which involve pi donor ligands around the metal. These transitions arise because of the low-lying energy of the ligand orbitals. Therefore, we can consider this as a transition from orbitals that are ligand in character to orbitals that are more metal in character, hence the name, Ligand to Metal Charge Transfer. The electron travels from a bonding pi or non-bonding pi orbital into a sigma* orbital. These transitions are very strong and appear very intensely in the absorbance spectrum. The molar extinction coefficients for these transitions are around 104. Examples of pi donor ligands are as follows: F-, Cl-, Br-, I-, H2O, OH-, RS-, S2-, NCS-, NCO-,...
MLCT Transitions
The somewhat less common MLCT has the same intensity and energy of the LMCT as they involve the transition of an electron from the t2g (pi) and the eg (sigma*) to the t1u (pi*/sigma*). These transitions arise from pi acceptor ligands and metals that are willing to donate electrons into the orbitals of Ligand character. This is the reason that they are less frequent since metals commonly accept electrons rather than donate them. All the same, both types of Charge Transfer bands are more intense than d-d bands since they are not Laporte Rule forbidden. Examples of pi accepting ligands are as follows: CO, NO, CN-, N2, bipy, phen, RNC, C5H5-, C=C double bonds, C=C triple bonds,...
From this spectra of an octahedral Chromium complex, we see that the d-d transitions are far weaker than the LMCT. Since Chlorine is a pi donor ligand in this example, we can label the CT band as LMCT since we know the electron is transitioning from a MO of ligand character to a MO of metal character. The Laporte forbidden (symmetry forbidden) d-d transitions are shown as less intense since they are only allowed via vibronic coupling.
In addition, the d-d transitions are lower in energy than the CT band because of the smaller energy gap between the t2g and eg in octahedral complexes (or eg to t2g in tetrahedral complexes) than the energy gap between the ground and excited states of the charge transfer band.
These transitions abide by the same selection rules that organic molecules follow: spin selection and symmetry arguments. The Tanabe and Sugano diagrams for transition metal complexes can be a guide for determining which transitions are seen in the spectrum. We will use the [CrCl(NH3)5]2+ ion as an example for determining the types of transitions that are spin allowed. To do this we look up the Tanabe and Sugano diagrams for Octahedral fields. Since Cr in the complex has three electrons, it is a d3 and so we find the diagram that corresponds to d3 metals:
Based on the TS diagram on the left, and the information we have already learned, can you predict which transition will be spin allowed and which ones will be forbidden? From the diagram we see that the ground state is a 4A2. This is because of the three unpaired electrons which make M=2S+1= 4. The A comes from the fact that there is only one combination of electrons possible. With a spin multiplicity of 4, by the spin selection rules, we can only expect intense transitions between the ground state 4A2 and 4T2, 4T1, and the other 4T1 excited state. The other transitions are spin forbidden. Therefore, we would expect to see three d-d transitions on the absorption spectra.
For us to visualize this, we can draw these transitions in order of increasing energy and then plot the spectrum as we would expect it for only the d-d transitions in a d3 octahedral complex:
From three spin allowed transitions, we would expect to see three d-d bands appear on the spectrum. In addition to these of course, the LMCT band will appear as well.
Fluorescence
Now that we have discussed the nature of absorption involving an electron absorbing photon energy to be excited to a higher energy level, now we can discuss what happens to that excited electron. Due to its higher potential energy, the electron will relax back to its initial ground state, and in the process, emit electromagnetic radiation. The energy gap between the excited state and the state to which the electron falls determines the wavelength of light that will be emitted. This process is called fluorescence. Generally, the wavelengths of fluorescence are longer than absorbance, can you explain why? Given the following diagram, one can see that vibrational relaxation occurs in the excited electronic state such that the electronic relaxation occurs from the ground vibrational state of the excited electronic state. This causes lower energy electronic relaxations than the previous energy of absorption.
Here we see that the absorption transitions by default involve a greater energy change than the emission transitions. Due to vibrational relaxation in the excited state, the electron tends to relax only from the v'=0 ground state vibrational level. This gives emission transitions of lower energy and consequently, longer wavelength than absorption. When obtaining fluorescence, we have to block out the transmitted light and only focus on the light being emitted from the sample, so the detector is usually 90 degrees from the incident light. Because of this emission spectra are generally obtained separately from the absorption spectra; however, they can be plotted on the same graph as shown.
Generally separated by ~10 nm, the fluorescence peak follows the absorption peak according to the spectrum. With that, we conclude our discussion of electronic spectroscopy interpretation. Refer to outside links and references for additional information.
Practice Problems!
1. From what we've discussed so far, if we change the solvent from non-polar to polar what effect will this have on the frequency of absorption if the ground state is non-polar and the excited state is polar? Will it increase or decrease?
2. What causes peak broadening in absorption spectra?
3. What are the little spikes in the more broad electronic transition bands? Draw potential energy wells to show their order and use the Frank Condon factor to describe your answer.
4. Why are fluorescence bands lower in energy than absorption bands?
5. If an electronic transition is symmetry forbidden and spin forbidden, list two ways of overcoming this to explain why the bands are still seen in the spectrum.
6. Define MLCT, LMCT, and d-d transitions and label the molar extinction coefficients associated with each.
7. How do the spectra of transition metal complexes differ with organic molecule?
8. What is a "blue shift" and a "red shift" and what solvent conditions would cause these to occur?
9. From the Tanabe Sugano diagram of a d2 metal complex, list all of the transitions that are spin allowed.
10. Define the coupling operator that sits between the excited state wave function and the ground state wave function in the transition moment integral!
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy_-_Interpretation.txt
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Explains the origin of UV-visible absorption spectra, how they are measured, and how they can be used in the analysis of organic compounds.
Electronic Spectroscopy Basics
This page describes a double beam UV-visible absorption spectrometer.
The overall design
If you pass white light through a colored substance, some of the light gets absorbed. A solution containing hydrated copper(II) ions, for example, looks pale blue because the solution absorbs light from the red end of the spectrum. The remaining wavelengths in the light combine in the eye and brain to give the appearance of cyan (pale blue).
Some colorless substances also absorb light - but in the ultra-violet region. Since we can't see UV light, we don't notice this absorption. Different substances absorb different wavelengths of light, and this can be used to help to identify the substance - the presence of particular metal ions, for example, or of particular functional groups in organic compounds.
The amount of absorption is also dependent on the concentration of the substance if it is in solution. Measurement of the amount of absorption can be used to find concentrations of very dilute solutions. An absorption spectrometer measures the way that the light absorbed by a compound varies across the UV and visible spectrum.
A simple double beam spectrometer
We'll start with the full diagram, and then explain exactly what is going on at each stage.
The light source
You need a light source which gives the entire visible spectrum plus the near ultra-violet so that you are covering the range from about 200 nm to about 800 nm. (This extends slightly into the near infra-red as well.)
You can't get this range of wavelengths from a single lamp, and so a combination of two is used - a deuterium lamp for the UV part of the spectrum, and a tungsten / halogen lamp for the visible part. The combined output of these two bulbs is focused on to a diffraction grating.
The diffraction grating and the slit
You are probably familiar with the way that a prism splits light into its component colors. A diffraction grating does the same job, but more efficiently.
The blue arrows show the way the various wavelengths of the light are sent off in different directions. The slit only allows light of a very narrow range of wavelengths through into the rest of the spectrometer. By gradually rotating the diffraction grating, you can allow light from the whole spectrum (a tiny part of the range at a time) through into the rest of the instrument.
The rotating disks
This is the clever bit! Each disk is made up of a number of different segments. Those in the machine we are describing have three different sections - other designs may have a different number.
The light coming from the diffraction grating and slit will hit the rotating disk and one of three things can happen.
1. If the original beam of light from the slit hits the mirrored section of the first rotating disk, it is bounced down along the green path. After the mirror, it passes through a reference cell (more about that later). Finally the light gets to the second disk which is rotating in such a way that it meets the transparent section. It goes straight through to the detector.
2. If the light meets the first disk at the black section, it is blocked - and for a very short while no light passes through the spectrometer. This just allows the computer to make allowance for any current generated by the detector in the absence of any light.
The sample and reference cells
These are small rectangular glass or quartz containers. They are often designed so that the light beam travels a distance of 1 cm through the contents. The sample cell contains a solution of the substance you are testing - usually very dilute. The solvent is chosen so that it doesn't absorb any significant amount of light in the wavelength range we are interested in (200 - 800 nm). The reference cell just contains the pure solvent.
The detector and computer
The detector converts the incoming light into a current. The higher the current, the greater the intensity of the light. For each wavelength of light passing through the spectrometer, the intensity of the light passing through the reference cell is measured. This is usually referred to as Io - that's I for Intensity.
The intensity of the light passing through the sample cell is also measured for that wavelength - given the symbol, I.
If I is less than Io, then obviously the sample has absorbed some of the light. A simple bit of math is then done in the computer to convert this into something called the absorbance of the sample - given the symbol, A.
For reasons which will become clearer when we do a bit of theory on another page, the relationship between A and the two intensities is given by:
$A=\log \left( \dfrac{I}{I_o} \right)$
On most of the diagrams you will come across, the absorbance ranges from 0 to 1, but it can go higher than that. An absorbance of 0 at some wavelength means that no light of that particular wavelength has been absorbed. The intensities of the sample and reference beam are both the same, so the ratio Io/I is 1. Log10 of 1 is zero.
An absorbance of 1 happens when 90% of the light at that wavelength has been absorbed - which means that the intensity is 10% of what it would otherwise be.
In that case, Io/I is 100/I0 (=10) and log10 of 10 is 1.
The absoprtion spectrum is a plot of absorbance against wavelength. The output might look like this:
This particular substance has what are known as absorbance peaks at 255 and 395 nm. How these arise and how they are interpreted are diskussed on another page.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy_Basics/A_Double_Beam_Absorption_Spectrometer.txt
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This page takes an introductory look at two areas of bonding theory needed for a proper understanding of how organic compounds absorb some of the UV or visible light that passes through them.
It looks simply at anti-bonding orbitals, and what is meant by conjugation in compounds and how it contributes to the delocalization of electrons.
Anti-bonding orbitals
Bonding and anti-bonding orbitals in a simple hydrogen molecule
I am assuming that you know how a simple covalent bond between two atoms forms. Half-filled atomic orbitals on each atom overlap in space to form a new orbital (a molecular orbital) containing both electrons. In the case of two hydrogen atoms, each has one electron in the 1s orbital. These come together to make a new orbital surrounding both of the hydrogen nuclei.
It is important to understand exactly what this molecular orbital means. The two electrons are most likely to be found in this region of space - and the most likely place to find them within this space is on the line between the two nuclei.
The molecule holds together because both nuclei are strongly attracted to this same pair of electrons. This most simple of bonds is called a sigma bond - a sigma bond is one where the electron pair is most likely to be found on the line between the two nuclei.
However . . . This is all a bit of a simplification! Molecular orbital theory demands that if you start with two atomic orbitals, you must end up with two molecular orbitals - and we seem to be only producing one. A second molecular orbital is formed, but in most cases (including the hydrogen molecule) it is left empty of electrons. It is described as an anti-bonding orbital. The anti-bonding orbital has a quite different shape and energy from the bonding orbital.
The next diagram shows the relative shapes and energies of the various atomic and molecular orbitals when two hydrogen atoms combine.
An anti-bonding orbital is always shown by the use of a star after its symbol.
Notice that when a bonding orbital forms, it is at a lower energy than the original atoms. Energy is released when the bonding orbital is formed, and the hydrogen molecule is more energetically stable than the original atoms. However, an anti-bonding orbital is less energetically stable than the original atoms.
A bonding orbital is stable because of the attractions between the nuclei and the electrons. In an anti-bonding orbital there are no equivalent attractions - instead you get repulsions. There is very little chance of finding the electrons between the two nuclei - and in fact half-way between the nuclei there is zero chance of finding them. There is nothing to stop the two nuclei from repelling each other apart.
So in the hydrogen case, both of the electrons go into the bonding orbital, because that produces the greatest stability - more stable than having separate atoms, and a lot more stable than having the electrons in the anti-bonding orbital.
Why doesn't helium form an He2 molecule?
You might reasonably say that helium can't form an He2 molecule because it doesn't have any unpaired electrons to share. Fine! But let's also look at it from the point of molecular orbital theory. The diagram for helium is just a small modification of the last one.
This time we have a total of 4 electrons in the original atomic orbitals. Two atomic orbitals have to form two molecular orbitals. That means that this time, we would have to use both the bonding and anti-bonding molecular orbitals to accommodate them.
But any gain in energetic stability due to the formation of the bonding orbitals would be countered by the loss of energetic stability because of the anti-bonding ones. There is no energetic advantage in He2 forming - and so it doesn't.
Anti-bonding orbitals in double bonds
You are probably familiar with a picture of the double bond in ethene shown as:
The pi bond shown in red is, of course, a normal bonding orbital. It was formed by sideways overlap between a half-filled p-orbital on each of the two carbon atoms. Remember that the two red shapes shown in the diagram are part of the same pi bonding orbital. But if you overlap two atomic orbitals, you must get two molecular orbitals according to molecular orbital theory. The second one is an anti-bonding pi orbital - and we never draw it under normal circumstances.
The anti-bonding pi orbital is (just like the anti-bonding sigma one) at a higher energy than the bonding orbital - and so isn't used to hold electrons. Both of the electrons in the pi bond are found in the pi bonding orbital.
Summarizing the relative energies of various kinds of orbitals
The next diagram gives a general impression of how the energies of various types of orbital relate to each other in the sort of compounds we will be looking at when we try to explain the absorption of light. It is not to scale. You will find that a new sort of orbital has crept into the diagram - labelled "n" (for non-bonding). The sort of non-bonding orbitals that we will be interested in contain lone pairs of electrons on, for example, oxygen, nitrogen and halogen atoms.
So . . . think of non-bonding orbitals as those containing lone pairs of electrons at the bonding level.
When light passes through a compound, some of the energy in the light kicks an electron from one of the bonding or non-bonding orbitals into one of the anti-bonding ones. The energy gaps between these levels determine the frequency (or wavelength) of the light absorbed, and those gaps will be different in different compounds. This is covered in detail on another page.
Conjugation
We are going to leave explaining what conjugation is for a while - it is necessary to look at some more bonding first.
The simple ethene double bond
To understand about conjugated double bonds, you first need to be sure that you understand simple double bonds. Ethene contains a simple double bond between two carbon atoms, but the two parts of this bond are different. Part of it is a simple sigma bond formed from end-to-end overlap between orbitals on each carbon atom, and part is caused by sideways overlap between a p-orbital on each carbon.
The important diagram is the one leading up to the formation of the pi bond - where the two p-orbitals are overlapping sideways:
. . . giving the familiar pi bond.
Conjugated double bonds in buta-1,3-diene
Bonding in buta-1,3-diene
Buta-1,3-diene has the structure:
Now picture the formation of the various molecular orbitals as if you were thinking about two ethene molecules joined together. You would have sigma bonds formed by the end-to-end overlap of various orbitals on the carbons and hydrogens. That would leave you with a p-orbital on each carbon atom.
Those p-orbitals will overlap sideways - all of them! A system of delocalised pi bonds is formed, similar to the benzene case that you are probably familiar with. The diagram shows one of those molecular orbitals.
To stress again - the diagram shows only one of the delocalised molecular orbitals. Remember that both of the red bits in the diagram are part of the same orbital. The interaction of the two double bonds with each other to produce a delocalised system of pi electrons over all four atoms is known as conjugation. Conjugation in this context literally means "joining together".
In reality, if you start by overlapping four atomic orbitals, you will end up with four molecular orbitals. The four electrons will go into the two lowest energy of these - two in each. That means that you get two pi bonding orbitals. We just draw one of these for simplicity - the other one has a different shape.
There are also two pi anti-bonding orbitals, but these are normally empty. For most purposes, we ignore these entirely - although not for this topic because energy from light can promote electrons from a pi bonding orbital into one of the anti-bonding orbitals (as you will see on the next page).
Recognizing conjugated double bonds in a molecule
You can recognize the presence of conjugated double bonds in a molecule containing more than one double bond because of the presence of alternating double and single bonds. The double bonds don't have to be always between carbon atoms. All of the following molecules contain conjugated double bonds, although in the last case, the conjugation doesn't extend over the whole molecule:
However, although the next molecule contains two double bonds, they aren't conjugated. They are separated by two single bonds.
The reason why it is important to have the double and single bonds alternating is that this is the only way you can get all the p-orbitals overlapping sideways. In the last case, you will get sideways overlap at each end of the molecule to get two individual pi bonds. But the extra single bond in the middle stops them from interacting with each other.
Benzene rings
You are almost certainly familiar with the delocalization which occurs in a benzene ring. If you think about benzene using the Kelulé structure, you have a perfect system of alternating single and double bonds around the molecule.
These conjugate to give the familiar delocalised pi system.
Once again, remember that this only shows one of the molecular orbitals formed. There will actually be three bonding pi orbitals and three anti-bonding ones - because they arise from combining a total of six atomic orbitals. The extra bonding orbitals aren't usually drawn.
Phenylamine and phenol
delocalization can also extend beyond pi bonds to include lone pairs on atoms like nitrogen or oxygen. Two simple examples of this are phenylamine (aniline) and phenol. Writing these using the Kekulé structure for benzene:
You can see the alternating double and single bonds around the benzene ring. This conjugation leads to the familiar delocalised electron system in benzene which we usually show as:
But the delocalization doesn't stop at the ring. It extends out to the nitrogen or oxygen atoms. In the phenylamine case, there is a lone pair on the nitrogen atom which can overlap with the ring electrons . . .
. . . leading to delocalization which takes in both ring and nitrogen.
Exactly the same thing happens with phenol. One of the oxygen lone pairs overlaps with the ring electrons. The other one is pointing in the wrong direction to get involved.
So . . . if you are trying to work out how far delocalization extends in a molecule, don't forget to look for atoms with lone pairs that might get involved in the delocalization.
Other groups to look out for
Look especially for benzene rings with groups attached which contain double bonds. Let's start with a couple of simple ones - phenylethene (styrene) and benzaldehyde.
In each case, you've got delocalization over the ring. Does it extend to the attached group? Do you have anything like alternating single and double bonds? Yes, you do. You have the double bond in the side group, then a single bond, then the ring delocalization. Looking at this in the phenylethene case, and imagining the arrangement of orbitals just before delocalization over the side group:
You can see that the double bond and ring electrons will overlap to form a delocalised system looking something like this:
The benzaldehyde case is very similar, except that this time instead of the CH2 group at the end there is an oxygen with two lone pairs. The delocalization is just the same.
Be careful, though! Remember that to get this extended delocalization, any double bond in the side chain must be able to conjugate with the ring electrons - the two bits must be close enough to join together. Molecules like those in the next diagram don't have the delocalization extending out into the side chain. The extra CH2 group prevents the necessary sideways overlap between the p orbitals of the double bond and the ring electrons.
The other side group which it would be useful to know about is the nitro group, NO2 - for example in nitrobenzene.
The bonding in the nitro group is surprisingly awkward to work out. It is often shown with a double bond between the nitrogen and one of the oxygens, and a co-ordinate (dative covalent) bond to the other.
This structure is actually misleading. Both nitrogen-oxygen bonds are identical and the group already has delocalization. This is often shown as:
The dotted half-circle suggests the delocalization. Think of it as being like the circle you draw in the middle of the benzene hexagon. This delocalization is just one single bond away from the ring delocalization. You get conjugation between the two, and the delocalization takes in the whole molecule.
A case to beware of
Another case we need to look at (because it occurs in a molecule we'll explore on the next page) is an SO3- group attached to a benzene ring.
It looks as if the double bonds are in just the right position relative to the ring for the delocalization to extend out over this group. However, I have been told on good authority that the delocalization doesn't extend from the side group into the ring. I can't, though, find any reference to this anywhere on the web or in the textbooks that I have available. The bonding in the sulfonate group isn't at all easy to describe in orbital terms. In fact, it is most easily explained in terms of co-ordinate (dative covalent) bonds, but introducing that now is just complicating things pointlessly.
Summary
When you are trying to work out how far delocalization extends in a molecule, look for:
• alternating double and single bonds - not just between carbon and carbon, but including C=O, C=N, N=N, N=O. Carbon-carbon triple bonds can also be involved in place of a carbon-carbon double bond.
• benzene rings.
• possible involvement of lone pairs on nitrogen or oxygen.
• NO2 groups.
And finally . . . why does all this matter? The wavelength of UV or visible light absorbed by organic compounds depends largely on the extent of delocalization in the molecules. That means that you may well have to look at an unfamiliar molecule and make a reasonable estimate of whether it is highly delocalised or not very delocalised. It will always be pretty clear-cut at this level, so don't worry too much about it!
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This page is a basic introduction to the electromagnetic spectrum sufficient for chemistry students interested in UV-visible absorption spectroscopy. If you are looking for any sort of explanations suitable for physics courses, then I'm afraid this isn't the right place for you.
Light as a wave form
Any wave is essentially just a way of shifting energy from one place to another - whether the fairly obvious transfer of energy in waves on the sea or in the much more difficult-to-imagine waves in light.
In waves on water, the energy is transferred by the movement of water molecules. But a particular water molecule doesn't travel all the way across the Atlantic - or even all the way across a pond. Depending on the depth of the water, water molecules follow a roughly circular path. As they move up to the top of the circle, the wave builds to a crest; as they move down again, you get a trough.
The energy is transferred by relatively small local movements in the environment. With water waves it is fairly easy to draw diagrams to show this happening with real molecules. With light it is more difficult.
The energy in light travels because of local fluctuating changes in electrical and magnetic fields - hence "electromagnetic" radiation.
Wavelength, frequency and the speed of light
If you draw a beam of light in the form of a wave (without worrying too much about what exactly is causing the wave!), the distance between two crests is called the wavelength of the light. (It could equally well be the distance between two troughs or any other two identical positions on the wave.)
You have to picture these wave crests as moving from left to right. If you counted the number of crests passing a particular point per second, you have the frequency of the light. It is measured in what used to be called "cycles per second", but is now called Hertz, Hz. Cycles per second and Hertz mean exactly the same thing.
Orange light, for example, has a frequency of about 5 x 1014 Hz (often quoted as 5 x 108 MHz - megahertz). That means that 5 x 1014 wave peaks pass a given point every second. Light has a constant speed through a given substance. For example, it always travels at a speed of approximately 3 x 108 meters per second in a vacuum. This is actually the speed that all electromagnetic radiation travels - not just visible light.
There is a simple relationship between the wavelength and frequency of a particular color of light and the speed of light:
. . . and you can rearrange this to work out the wavelength from a given frequency and vice versa:
$\lambda = \dfrac{c}{\nu}$
$\nu= \dfrac{c}{\lambda}$
These relationships mean that if you increase the frequency, you must decrease the wavelength.
Compare this diagram with the similar one above. . . and, of course, the opposite is true. If the wavelength is longer, the frequency is lower. It is really important that you feel comfortable with the relationship between frequency and wavelength. If you are given two figures for the wavelengths of two different colors of light, you need to have an immediate feel for which one has the higher frequency.
For example, if you were told that a particular color of red light had a wavelength of 650 nm, and a green had a wavelength of 540 nm, it is important for you to know which has the higher frequency. (It's the green - a shorter wavelength means a higher frequency. Don't go on until that feels right!)
The frequency of light and its energy
Each particular frequency of light has a particular energy associated with it, given by another simple equation:
You can see that the higher the frequency, the higher the energy of the light. Light which has wavelengths of around 380 - 435 nm is seen as a sequence of violet colours. Various red colours have wavelengths around 625 - 740 nm. Which has the highest energy?
The light with the highest energy will be the one with the highest frequency - that will be the one with the smallest wavelength. In other words, violet light at the 380 nm end of its range.
The Electromagnetic Spectrum
Visible light
The diagram shows an approximation to the spectrum of visible light.
The main colour regions of the spectrum are approximately:
colour region wavelength (nm)
violet 380 - 435
blue 435 - 500
cyan 500 - 520
green 520 - 565
yellow 565 - 590
orange 590 - 625
red 625 - 740
Don't assume that there is some clear cut-off point between all these colours. In reality, the colours just merge seamlessly into one another - much more seamlessly than in my diagram!
Placing the visible spectrum in the whole electromagnetic spectrum
The electromagnetic spectrum doesn't stop with the colors you can see. It is perfectly possible to have wavelengths shorter than violet light or longer than red light. On the spectrum further up the page, I have shown the ultra-violet and the infra-red, but this can be extended even further into x-rays and radio waves, amongst others. The diagram shows the approximate positions of some of these on the spectrum.
Once again, don't worry too much about the exact boundaries between the various sorts of electromagnetic radiation - because there are no boundaries. Just as with visible light, one sort of radiation merges into the next. Just be aware of the general pattern.
Also be aware that the energy associated with the various kinds of radiation increases as the frequency increases.
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The Beer-Lambert law relates the attenuation of light to the properties of the material through which the light is traveling. This page takes a brief look at the Beer-Lambert Law and explains the use of the terms absorbance and molar absorptivity relating to UV-visible absorption spectrometry.
The Absorbance of a Solution
For each wavelength of light passing through the spectrometer, the intensity of the light passing through the reference cell is measured. This is usually referred to as $I_o$ - that's $I$ for Intensity.
The intensity of the light passing through the sample cell is also measured for that wavelength - given the symbol, $I$. If $I$ is less than $I_o$, then the sample has absorbed some of the light (neglecting reflection of light off the cuvette surface). A simple bit of math is then done in the computer to convert this into something called the absorbance of the sample - given the symbol, $A$. The absorbance of a transition depends on two external assumptions.
1. The absorbance is directly proportional to the concentration ($c$) of the solution of the sample used in the experiment.
2. The absorbance is directly proportional to the length of the light path ($l$), which is equal to the width of the cuvette.
Assumption one relates the absorbance to concentration and can be expressed as
$A \propto c \label{1}$
The absorbance ($A$) is defined via the incident intensity $I_o$ and transmitted intensity $I$ by
$A=\log_{10} \left( \dfrac{I_o}{I} \right) \label{2}$
Assumption two can be expressed as
$A \propto l \label{3}$
Combining Equations $\ref{1}$ and $\ref{3}$:
$A \propto cl \label{4}$
This proportionality can be converted into an equality by including a proportionality constant ($\epsilon$).
$A = \epsilon c l \label{5}$
This formula is the common form of the Beer-Lambert Law, although it can be also written in terms of intensities:
$A=\log_{10} \left( \dfrac{I_o}{I} \right) = \epsilon l c \label{6}$
The constant $\epsilon$ is called molar absorptivity or molar extinction coefficient and is a measure of the probability of the electronic transition. On most of the diagrams you will come across, the absorbance ranges from 0 to 1, but it can go higher than that. An absorbance of 0 at some wavelength means that no light of that particular wavelength has been absorbed. The intensities of the sample and reference beam are both the same, so the ratio $I_o/I$ is 1 and the $\log_{10}$ of 1 is zero.
Example $1$
In a sample with an absorbance of 1 at a specific wavelength, what is the relative amount of light that was absorbed by the sample?
Solution
This question does not need Beer-Lambert Law (Equation $\ref{5}$) to solve, but only the definition of absorbance (Equation $\ref{2}$)
$A=\log_{10} \left( \dfrac{I_o}{I} \right)\nonumber$
The relative loss of intensity is
$\dfrac{I-I_o}{I_o} = 1- \dfrac{I}{I_o}\nonumber$
Equation $\ref{2}$ can be rearranged using the properties of logarithms to solved for the relative loss of intensity:
$10^A= \dfrac{I_o}{I}\nonumber$
$10^{-A}= \dfrac{I}{I_o}\nonumber$
$1-10^{-A}= 1- \dfrac{I}{I_o}\nonumber$
Substituting in $A=1$
$1- \dfrac{I}{I_o}= 1-10^{-1} = 1- \dfrac{1}{10} = 0.9\nonumber$
Hence 90% of the light at that wavelength has been absorbed and that the transmitted intensity is 10% of the incident intensity. To confirm, substituting these values into Equation $\ref{2}$ to get the absorbance back:
$\dfrac{I_o}{I} = \dfrac{100}{10} =10 \label{7a}$
and
$\log_{10} 10 = 1 \label{7b}$
The Beer-Lambert Law
You will find that various different symbols are given for some of the terms in the equation - particularly for the concentration and the solution length.
The Greek letter epsilon in these equations is called the molar absorptivity - or sometimes the molar absorption coefficient. The larger the molar absorptivity, the more probable the electronic transition. In uv spectroscopy, the concentration of the sample solution is measured in mol L-1 and the length of the light path in cm. Thus, given that absorbance is unitless, the units of molar absorptivity are L mol-1 cm-1. However, since the units of molar absorptivity is always the above, it is customarily reported without units.
Example $2$: Guanosine
Guanosine has a maximum absorbance of 275 nm. $\epsilon_{275} = 8400 M^{-1} cm^{-1}$ and the path length is 1 cm. Using a spectrophotometer, you find the that $A_{275}= 0.70$. What is the concentration of guanosine?
Solution
To solve this problem, you must use Beer's Law.
$A = \epsilon lc$
0.70 = (8400 M-1 cm-1)(1 cm)($c$)
Next, divide both side by [(8400 M-1 cm-1)(1 cm)]
$c$ = 8.33x10-5 mol/L
Example $3$
There is a substance in a solution (4 g/liter). The length of cuvette is 2 cm and only 50% of the certain light beam is transmitted. What is the extinction coefficient?
Solution
Using Beer-Lambert Law, we can compute the absorption coefficient. Thus,
$- \log \left(\dfrac{I_t}{I_o} \right) = - \log(\dfrac{0.5}{1.0}) = A = {8} \epsilon$
Then we obtain that
$\epsilon$ = 0.0376
Example $4$
In Example 3 above, what is the molar absorption coefficient if the molecular weight is 100?
Solution
It can simply obtained by multiplying the absorption coefficient by the molecular weight. Thus,
$\epsilon$ = 0.0376 x 100 = 3.76 L·mol-1·cm-1
The Importance of Concentration
The proportion of the light absorbed will depend on how many molecules it interacts with. Suppose you have got a strongly colored organic dye. If it is in a reasonably concentrated solution, it will have a very high absorbance because there are lots of molecules to interact with the light. However, in an incredibly dilute solution, it may be very difficult to see that it is colored at all. The absorbance is going to be very low. Suppose then that you wanted to compare this dye with a different compound. Unless you took care to make allowance for the concentration, you couldn't make any sensible comparisons about which one absorbed the most light.
Example $4$
In Example $3$ above, how much is the beam of light is transmitted when 8 g/liter ?
Solution
Since we know $\epsilon$, we can calculate the transmission using Beer-Lambert Law. Thus,
$log(1) - log(I_t) = 0 - log(I_t)$ = 0.0376 x 8 x 2 = 0.6016
$log(I_t)$ = -0.6016
Therefore, $I_t$ = 0.2503 = 25%
Example $5$
The absorption coefficient of a glycogen-iodine complex is 0.20 at light of 450 nm. What is the concentration when the transmission is 40 % in a cuvette of 2 cm?
Solution
It can also be solved using Beer-Lambert Law. Therefore,
$- \log(I_t) = - \log_{10}(0.4) = 0.20 \times c \times 2$
Then $c$ = 0.9948
The importance of the container shape
Suppose this time that you had a very dilute solution of the dye in a cube-shaped container so that the light traveled 1 cm through it. The absorbance is not likely to be very high. On the other hand, suppose you passed the light through a tube 100 cm long containing the same solution. More light would be absorbed because it interacts with more molecules. Again, if you want to draw sensible comparisons between solutions, you have to allow for the length of the solution the light is passing through. Both concentration and solution length are allowed for in the Beer-Lambert Law.
Molar Absorptivity
The Beer-Lambert law (Equation $\ref{5}$) can be rearranged to obtain an expression for $\epsilon$ (the molar absorptivity):
$\epsilon = \dfrac{A}{lc} \label{8}$
Remember that the absorbance of a solution will vary as the concentration or the size of the container varies. Molar absorptivity compensates for this by dividing by both the concentration and the length of the solution that the light passes through. Essentially, it works out a value for what the absorbance would be under a standard set of conditions - the light traveling 1 cm through a solution of 1 mol dm-3. That means that you can then make comparisons between one compound and another without having to worry about the concentration or solution length.
Values for molar absorptivity can vary hugely. For example, ethanal has two absorption peaks in its UV-visible spectrum - both in the ultra-violet. One of these corresponds to an electron being promoted from a lone pair on the oxygen into a pi anti-bonding orbital; the other from a $\pi$ bonding orbital into a $\pi$ anti-bonding orbital. Table 1 gives values for the molar absorptivity of a solution of ethanal in hexane. Notice that there are no units given for absorptivity. That's quite common since it assumes the length is in cm and the concentration is mol dm-3, the units are mol-1 dm3 cm-1.
Table $1$
electron jump wavelength of maximum absorption (nm) molar absorptivity
lone pair to $\pi$ anti-bonding orbital 290 15
$\pi$ bonding to $\pi$ anti-bonding orbital 180 10,000
The ethanal obviously absorbs much more strongly at 180 nm than it does at 290 nm. (Although, in fact, the 180 nm absorption peak is outside the range of most spectrometers.) You may come across diagrams of absorption spectra plotting absorptivity on the vertical axis rather than absorbance. However, if you look at the figures above and the scales that are going to be involved, you aren't really going to be able to spot the absorption at 290 nm. It will be a tiny little peak compared to the one at 180 nm. To get around this, you may also come across diagrams in which the vertical axis is plotted as log10(molar absorptivity).
If you take the logs of the two numbers in the table, 15 becomes 1.18, while 10,000 becomes 4. That makes it possible to plot both values easily, but produces strangely squashed-looking spectra!
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This page takes a brief look at how UV-visible absorption spectra can be used to help identify compounds and to measure the concentrations of colored solutions. It assumes that you know how these spectra arise, and know what is meant by terms such as absorbance, molar absorptivity and lambda-max. You also need to be familiar with the Beer-Lambert Law.
Using UV-absorption spectra to help identify organic compounds
If you have worked through the rest of this section, you will know that the wavelength of maximum absorption (lambda-max) depends on the presence of particular chromophores (light-absorbing groups) in a molecule.
For example, on another page you will have come across the fact that a simple carbon-carbon double bond (for example in ethene) has a maximum absorption at 171 nm. The two conjugated double bonds in buta-1,3-diene have a maximum absorption at a longer wavelength of 217 nm. We also talked about the two peaks in the spectrum of ethanal (containing a simple carbon-oxygen double bond) at 180 and 290 nm.
In carefully chosen simple cases (which is all you will get at this level), if you compared the peaks on a given UV-visible absorption spectrum with a list of known peaks, it would be fairly easy to pick out some structural features of an unknown molecule.
Lists of known peaks often include molar absorptivity values as well. That might help you to be even more sure. For example (again using the simple carbon-oxygen double bond), data shows that the peak at 290 nm has a molar absorptivity of only 15, compared with the one at 180 nm of 10000. If your spectrum showed a very large peak at 180 nm, and an extremely small one at 290 nm, that just adds to your certainty.
Using UV-absorption spectra to find concentrations
You should remember the Beer-Lambert Law:
The expression on the left of the equation is known as the absorbance of the solution and is measured by a spectrometer. The equation is sometimes written in terms of that absorbance.
$A = \epsilon \, l \, c$
The symbol epsilon is the molar absorptivity of the solution.
Finding concentration using the molar absorptivity
If you know the molar absorptivity of a solution at a particular wavelength, and you measure the absorbance of the solution at that wavelength, it is easy to calculate the concentration. The only other variable in the expression above is the length of the solution. That's easy to measure and, in fact, the cell containing the solution may well have been manufactured with a known length of 1 cm.
For example, let's suppose you have a solution in a cell of length 1 cm. You measure the absorbance of the solution at a particular wavelength using a spectrometer. The value is 1.92. You find a value for molar absorptivity in a table of 19400 for that wavelength. Substituting those values:
$A = \epsilon l c$
$1.92 = 19400 \times 1 \times c$
$c=\dfrac{1.92}{19400}$
$= 9.90 \times 10^{-5}\; mol/l$
Notice what a very low concentration can be measured provided you are working with a substance with a very high molar absorptivity. This method, of course, depends on you having access to an accurate value of molar absorptivity. It also assumes that the Beer-Lambert Law works over the whole concentration range (not true!). It is much better to measure the concentration by plotting a calibration curve.
Finding concentration by plotting a calibration curve
Doing it this way you don't have to rely on a value of molar absorptivity, the reliability of the Beer-Lambert Law, or even know the dimensions of the cell containing the solution.
What you do is make up a number of solutions of the compound you are investigating - each of accurately known concentration. Those concentrations should bracket the concentration you are trying to find - some less concentrated; some more concentrated. With colored solutions, this isn't a problem. You would just make up some solutions which are a bit lighter and some a bit darker in color.
For each solution, you measure the absorbance at the wavelength of strongest absorption - using the same container for each one. Then you plot a graph of that absorbance against concentration. This is a calibration curve.
According to the Beer-Lambert Law, absorbance is proportional to concentration, and so you would expect a straight line. That is true as long as the solutions are dilute, but the Law breaks down for solutions of higher concentration, and so you might get a curve under these circumstances. As long as you are working from values either side of the one you are trying to find, that isn't a problem.
Having drawn a best fit line, the calibration curve will probably look something like the next diagram. (I've drawn it as a straight line because it is easier for me to draw than a curve(!), and it's what you will probably get if you are working with really dilute solutions. But if it turns out to be a curve, so be it!)
Notice that no attempt has been made to force the line back through the origin. If the Beer-Lambert Law worked perfectly, it would pass through the origin, but you can't guarantee that it is working properly at the concentrations you are using.
Now all you have to do is to measure the absorbance of the solution with the unknown concentration at the same wavelength. If, for example, it had an absorbance of 0.600, you can just read the corresponding concentration from the graph as above.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
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This page explains what happens when organic compounds absorb UV or visible light, and why the wavelength of light absorbed varies from compound to compound.
What happens when light is absorbed by molecules?
When we were talking about the various sorts of orbitals present in organic compounds on the introductory page (see above), you will have come across this diagram showing their relative energies:
Remember that the diagram isn't intended to be to scale - it just shows the relative placing of the different orbitals. When light passes through the compound, energy from the light is used to promote an electron from a bonding or non-bonding orbital into one of the empty anti-bonding orbitals. The possible electron jumps that light might cause are:
In each possible case, an electron is excited from a full orbital into an empty anti-bonding orbital. Each jump takes energy from the light, and a big jump obviously needs more energy than a small one. Each wavelength of light has a particular energy associated with it. If that particular amount of energy is just right for making one of these energy jumps, then that wavelength will be absorbed - its energy will have been used in promoting an electron.
We need to work out what the relationship is between the energy gap and the wavelength absorbed. Does, for example, a bigger energy gap mean that light of a lower wavelength will be absorbed - or what? It is easier to start with the relationship between the frequency of light absorbed and its energy:
You can see that if you want a high energy jump, you will have to absorb light of a higher frequency. The greater the frequency, the greater the energy. That's easy - but unfortunately UV-visible absorption spectra are always given using wavelengths of light rather than frequency. That means that you need to know the relationship between wavelength and frequency.
You can see from this that the higher the frequency is, the lower the wavelength is. So, if you have a bigger energy jump, you will absorb light with a higher frequency - which is the same as saying that you will absorb light with a lower wavelength.
Important summary: The larger the energy jump, the lower the wavelength of the light absorbed.
Some jumps are more important than others for absorption spectrometry
An absorption spectrometer works in a range from about 200 nm (in the near ultra-violet) to about 800 nm (in the very near infra-red). Only a limited number of the possible electron jumps absorb light in that region. Look again at the possible jumps. This time, the important jumps are shown in black, and a less important one in grey. The grey dotted arrows show jumps which absorb light outside the region of the spectrum we are working in.
Remember that bigger jumps need more energy and so absorb light with a shorter wavelength. The jumps shown with grey dotted arrows absorb UV light of wavelength less that 200 nm. The important jumps are:
• from pi bonding orbitals to pi anti-bonding orbitals;
• from non-bonding orbitals to pi anti-bonding orbitals;
• from non-bonding orbitals to sigma anti-bonding orbitals.
That means that in order to absorb light in the region from 200 - 800 nm (which is where the spectra are measured), the molecule must contain either pi bonds or atoms with non-bonding orbitals. Remember that a non-bonding orbital is a lone pair on, say, oxygen, nitrogen or a halogen.
Groups in a molecule which absorb light are known as chromophores.
What does an absorption spectrum look like
The diagram below shows a simple UV-visible absorption spectrum for buta-1,3-diene - a molecule we will talk more about later. Absorbance (on the vertical axis) is just a measure of the amount of light absorbed. The higher the value, the more of a particular wavelength is being absorbed.
You will see that absorption peaks at a value of 217 nm. This is in the ultra-violet and so there would be no visible sign of any light being absorbed - buta-1,3-diene is colorless. You read the symbol on the graph as "lambda-max". In buta-1,3-diene, CH2=CH-CH=CH2, there are no non-bonding electrons. That means that the only electron jumps taking place (within the range that the spectrometer can measure) are from pi bonding to pi anti-bonding orbitals.
A chromophore producing two peaks
A chromophore such as the carbon-oxygen double bond in ethanal, for example, obviously has pi electrons as a part of the double bond, but also has lone pairs on the oxygen atom. That means that both of the important absorptions from the last energy diagram are possible. You can get an electron excited from a pi bonding to a pi anti-bonding orbital, or you can get one excited from an oxygen lone pair (a non-bonding orbital) into a pi anti-bonding orbital.
The non-bonding orbital has a higher energy than a pi bonding orbital. That means that the jump from an oxygen lone pair into a pi anti-bonding orbital needs less energy. That means it absorbs light of a lower frequency and therefore a higher wavelength. Ethanal can therefore absorb light of two different wavelengths:
• the pi bonding to pi anti-bonding absorption peaks at 180 nm;
• the non-bonding to pi anti-bonding absorption peaks at 290 nm.
Both of these absorptions are in the ultra-violet, but most spectrometers won't pick up the one at 180 nm because they work in the range from 200 - 800 nm.
The importance of conjugation and delocalisation
Consider these three molecules:
Ethene contains a simple isolated carbon-carbon double bond, but the other two have conjugated double bonds. In these cases, there is delocalization of the pi bonding orbitals over the whole molecule. Now look at the wavelengths of the light which each of these molecules absorbs.
molecule wavelength of maximum absorption (nm)
ethene 171
buta-1,3-diene 217
hexa-1,3,5-triene 258
All of the molecules give similar UV-visible absorption spectra - the only difference being that the absorptions move to longer and longer wavelengths as the amount of delocalization in the molecule increases.
Why is this? You can actually work out what must be happening.
• The maximum absorption is moving to longer wavelengths as the amount of delocalization increases.
• Therefore maximum absorption is moving to shorter frequencies as the amount of delocalization increases.
• Therefore absorption needs less energy as the amount of delocalization increases.
• Therefore there must be less energy gap between the bonding and anti-bonding orbitals as the amount of delocalization increases.
. . . and that's what is happening.
Compare ethene with buta-1,3-diene. In ethene, there is one pi bonding orbital and one pi anti-bonding orbital. In buta-1,3-diene, there are two pi bonding orbitals and two pi anti-bonding orbitals. This is all discussed in detail on the introductory page that you should have read.
The highest occupied molecular orbital is often referred to as the HOMO - in these cases, it is a pi bonding orbital. The lowest unoccupied molecular orbital (the LUMO) is a pi anti-bonding orbital. Notice that the gap between these has fallen. It takes less energy to excite an electron in the buta-1,3-diene case than with ethene.
In the hexa-1,3,5-triene case, it is less still.
If you extend this to compounds with really massive delocalisation, the wavelength absorbed will eventually be high enough to be in the visible region of the spectrum, and the compound will then be seen as colored. A good example of this is the orange plant pigment, beta-carotene - present in carrots, for example.
Why is beta-carotene orange?
Beta-carotene has the sort of delocalization that we've just been looking at, but on a much greater scale with 11 carbon-carbon double bonds conjugated together. The diagram shows the structure of beta-carotene with the alternating double and single bonds shown in red.
The more delocalization there is, the smaller the gap between the highest energy pi bonding orbital and the lowest energy pi anti-bonding orbital. To promote an electron therefore takes less energy in beta-carotene than in the cases we've looked at so far - because the gap between the levels is less.
Remember that less energy means a lower frequency of light gets absorbed - and that's equivalent to a longer wavelength. Beta-carotene absorbs throughout the ultra-violet region into the violet - but particularly strongly in the visible region between about 400 and 500 nm with a peak about 470 nm. If you have read the page in this section about electromagnetic radiation, you might remember that the wavelengths associated with the various colors are approximately:
color region wavelength (nm)
violet 380 - 435
blue 435 - 500
cyan 500 - 520
green 520 - 565
yellow 565 - 590
orange 590 - 625
red 625 - 740
So if the absorption is strongest in the violet to cyan region, what color will you actually see? It is tempting to think that you can work it out from the colors that are left - and in this particular case, you wouldn't be far wrong. Unfortunately, it isn't as simple as that!
Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors.
Complementary colors
If you arrange some colors in a circle, you get a "color wheel". The diagram shows one possible version of this. An internet search will throw up many different versions!
colors directly opposite each other on the color wheel are said to be complementary colors. Blue and yellow are complementary colors; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colors of light will give you white light.
What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. In the beta-carotene case, the situation is more confused because you are absorbing such a range of wavelengths. However, if you think of the peak absorption running from the blue into the cyan, it would be reasonable to think of the color you would see as being opposite that where yellow runs into red - in other words, orange.
Phenolphthalein
You have probably used phenolphthalein as an acid-base indicator, and will know that it is colorless in acidic conditions and magenta (bright pink) in an alkaline solution. How is this color change related to changes in the molecule? The structures of the two differently colored forms are:
Both of these absorb light in the ultra-violet, but the one on the right also absorbs in the visible with a peak at 553 nm. The molecule in acid solution is colorless because our eyes can't detect the fact that some light is being absorbed in the ultra-violet. However, our eyes do detect the absorption at 553 nm produced by the form in alkaline solution.
553 nm is in the green region of the spectrum. If you look back at the color wheel, you will find that the complementary color of green is magenta - and that's the color you see.
So why does the color change as the structure changes? What we have is a shift to absorption at a higher wavelength in alkaline solution. As we've already seen, a shift to higher wavelength is associated with a greater degree of delocalisation.
Here is a modified diagram of the structure of the form in acidic solution - the colorless form. The extent of the delocalization is shown in red.
Notice that there is delocalization over each of the three rings - extending out over the carbon-oxygen double bond, and to the various oxygen atoms because of their lone pairs.
But the delocalization doesn't extend over the whole molecule. The carbon atom in the centre with its four single bonds prevents the three delocalized regions interacting with each other.
Now compare that with the magenta form:
The rearrangement now lets the delocalization extend over the entire ion. This greater delocalization lowers the energy gap between the highest occupied molecular orbital and the lowest unoccupied pi anti-bonding orbital. It needs less energy to make the jump and so a longer wavelength of light is absorbed.
Increasing the amount of delocalization shifts the absorption peak to a higher wavelength.
Methyl orange
You will know that methyl orange is yellow in alkaline solutions and red in acidic ones. The structure in alkaline solution is:
In acid solution, a hydrogen ion is (perhaps unexpectedly) picked up on one of the nitrogens in the nitrogen-nitrogen double bond.
This now gets a lot more complicated! The positive charge on the nitrogen is delocalized (spread around over the structure) - especially out towards the right-hand end of the molecule as we've written it. The normally drawn structure for the red form of methyl orange is . . .
But this can be seriously misleading as regards the amount of delocalization in the structure for reasons discussed below (after the red warning box) if you are interested.
Which is the more delocalized structure?
Let's work backwards from the absorption spectra to see if that helps. The yellow form has an absorption peak at about 440 nm. That's in the blue region of the spectrum, and the complementary color of blue is yellow. That's exactly what you would expect. The red form has an absorption peak at about 520 nm. That's at the edge of the cyan region of the spectrum, and the complementary color of cyan is red. Again, there's nothing unexpected here.
Notice that the change from the yellow form to the red form has produced an increase in the wavelength absorbed. An increase in wavelength suggests an increase in delocalisation. That means that there must be more delocalization in the red form than in the yellow one. Here again is the structure of the yellow form:
delocalization will extend over most of the structure - out as far as the lone pair on the right-hand nitrogen atom.
If you use the normally written structure for the red form, the delocalization seems to be broken in the middle - the pattern of alternating single and double bonds seems to be lost.
But that is to misunderstand what this last structure represents.
Canonical forms
If you draw the two possible Kekulé structures for benzene, you will know that the real structure of benzene isn't like either of them. The real structure is somewhere between the two - all the bonds are identical and somewhere between single and double in character. That's because of the delocalization in benzene.
The two structures are known as canonical forms, and they can each be thought of as adding some knowledge to the real structure. For example, the bond drawn at the top right of the molecule is neither truly single or double, but somewhere in between. Similarly with all the other bonds.
The two structures we've previously drawn for the red form of methyl orange are also canonical forms - two out of lots of forms that could be drawn for this structure. We could represent the delocalized structure by:
These two forms can be thought of as the result of electron movements in the structure, and curly arrows are often used to show how one structure can lead to the other.
In reality, the electrons haven't shifted fully either one way or the other. Just as in the benzene case, the actual structure lies somewhere in between these.
You must also realize that drawing canonical forms has no effect on the underlying geometry of the structure. Bond types or lengths or angles don't change in the real structure.
For example, the lone pairs on the nitrogen atoms shown in the last diagram are both involved with the delocalisation. For this to happen all the bonds around these nitrogens must be in the same plane, with the lone pair sticking up so that it can overlap sideways with orbitals on the next-door atoms. The fact that in each of the two canonical forms one of these nitrogens is shown as if it had an ammonia-like arrangement of the bonds is potentially misleading - and makes it look as if the delocalization is broken.
The problem is that there is no easy way of representing a complex delocalized structure in simple structural diagrams. It is bad enough with benzene - with something as complicated as methyl orange any method just leads to possible confusion if you aren't used to working with canonical forms.
It gets even more complicated! If you were doing this properly there would be a host of other canonical forms with different arrangements of double and single bonds and with the positive charge located at various places around the rings and on the other nitrogen atom.
The real structure can't be represented properly by any one of this multitude of canonical forms, but each gives a hint of how the delocalization works.
If we take the two forms we have written as perhaps the two most important ones, it suggests that there is delocalization of the electrons over the whole structure, but that electron density is a bit low around the two nitrogens carrying the positive charge on one canonical form or the other.
Why is the red form more delocalized
Finally, we get around to an attempt at an explanation as to why the delocalization is greater in the red form of methyl orange in acid solution than in the yellow one in alkaline solution. The answer may lie in the fact that the lone pair on the nitrogen at the right-hand end of the structure as we've drawn it is more fully involved in the delocalization in the red form. The canonical form with the positive charge on that nitrogen suggests a significant movement of that lone pair towards the rest of the molecule.
Doesn't the same thing happen to the lone pair on the same nitrogen in the yellow form of methyl orange? Not to the same extent.
Any canonical form that you draw in which that happens produces another negatively charged atom somewhere in the rest of the structure. Separating negative and positive charges like this is energetically unfavourable. In the red form, we aren't producing a new separation of charge - just shifting a positive charge around the structure.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy_Basics/What_Causes_Molecules_to_Absorb_UV_and_Visible_.txt
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Fluorescence and phosphorescence are types of molecular luminescence methods. A molecule of analyte absorbs a photon and excites a species. The emission spectrum can provide qualitative and quantitative analysis. The term fluorescence and phosphorescence are usually referred as photoluminescence because both are alike in excitation brought by absorption of a photon. Fluorescence differs from phosphorescence in that the electronic energy transition that is responsible for fluorescence does not change in electron spin, which results in short-live electrons (<10-5 s) in the excited state of fluorescence. In phosphorescence, there is a change in electron spin, which results in a longer lifetime of the excited state (second to minutes). Fluorescence and phosphorescence occurs at longer wavelength than the excitation radiation.
Introduction
Fluorescence can occur in gaseous, liquid, and solid chemical systems. The simple kind of fluorescence is by dilute atomic vapors. A fluorescence example would be if a 3s electron of a vaporized sodium atom is excited to the 3p state by absorption of a radiation at wavelength 589.6 and 589.0 nm. After 10-8 s, the electron returns to ground state and on its return it emits radiation of the two wavelengths in all directions. This type of fluorescence in which the absorbed radiation is remitted without a change in frequency is known as resonance fluorescence. Resonance fluorescence can also occur in molecular species. Molecular fluorescence band centers at wavelengths longer than resonance lines. The shift toward longer wavelength is referred to as the Stokes Shift.
Singlet and Triplet Excited State
Understanding the difference between fluorescence and phosphorescence requires the knowledge of electron spin and the differences between singlet and triplet states. The Pauli Exclusion principle states that two electrons in an atom cannot have the same four quantum numbers ($n$, $l$, $m_l$, $m_s$) and only two electrons can occupy each orbital where they must have opposite spin states. These opposite spin states are called spin pairing. Because of this spin pairing, most molecules do not exhibit a magnetic field and are diamagnetic. In diamagnetic molecules, electrons are not attracted or repelled by the static electric field. Free radicals are paramagnetic because they contain unpaired electrons have magnetic moments that are attracted to the magnetic field.
Singlet state is defined 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 into a magnetic field. A doublet state occurs when there is an unpaired electron that gives two possible orientations when exposed in a magnetic field and imparts different energy to the system. A singlet or a triplet can form when one electron is excited to a higher energy level. In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired). In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in Figure $1$. Singlet, doublet and triplet is 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 calculated the S for the excited singlet state, the equation would be 2(+1/2 + -1/2)+1 = 2(0)+1 = 1, therefore making the center orbital in the figure a singlet state. If the spin multiplicity for the excited triplet state was calculated, we obtain 2(+1/2 + +1/2)+1 = 2(1)+1 =3, which gives a triplet state as expected.
The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in electronic state. For this reason, the lifetime of the triplet state is longer the singlet state by approximately 104 seconds fold difference.The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram.
Jablonski Diagrams
The Jablonski diagram that drawn below is a partial energy diagram that represents the energy of 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, majority of the molecules in a solution are in this state.
The upper lines represent the energy state of the three excited electronic states: S1and S2 represent the electronic singlet state (left) and T1 represents the first electronic triplet state (right). The upper darkest line represents the ground vibrational state of the three excited electronic state.The energy of the triplet state is lower than the energy of the corresponding singlet state.
There are numerous vibrational levels that can be associated with each electronic state as denoted by the thinner lines. Absorption transitions (blues lines in Figure $2$) can occur from the ground singlet electronic state (So) to various vibrational levels in the singlet excited vibrational states. It is unlikely that a transition from the ground singlet electronic state to the triplet electronic state because the electron spin is parallel to the spin in its ground state (Figure $1$). This transition leads to a change in multiplicity and thus has a low probability of occurring which is a forbidden transition. Molecules also go through vibration relaxation to lose any excess vibrational energy that remains when excited to the electronic states ($S_1$ and $S_2$) as demonstrated in wavy lines in Figure $2$. The knowledge of forbidden transition is used to explain and compare the peaks of absorption and emission.
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.
Process Transition Timescale (sec)
Table $1$: Rates of Absorption and Emission comparison.
Light Absorption (Excitation) S0 → Sn ca. 10-15 (instantaneous)
Internal Conversion Sn → S1 10-14 to 10-11
Vibrational Relaxation Sn* → Sn 10-12 to 10-10
Intersystem Crossing S1 → T1 10-11 to 10-6
Fluorescence S1 → S0 10-9 to 10-6
Phosphorescence T1 → S0 10-3 to 100
Non-Radiative Decay S1 → S0
T1 → S0
10-7 to 10-5
10-3 to 100
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 $2$.The deactivation process of fluorescence and phosphorescence involve an emission of a photon radiation as shown by the straight arrow in Figure $2$. The wiggly arrows in Figure $2$ are deactivation processes without the use of radiation. The favored deactivation process is the route that is most rapid and spends less time in the excited state.If the rate constant for fluorescence is more favorable in the radiationless path, the fluorescence will be less intense or absent.
• Vibrational Relaxation: A molecule maybe to promoted to several vibrational levels during the electronic excitation process.Collision of molecules with the excited species and solvent leads to rapid energy transfer and a slight increase in temperature of the solvent. Vibrational relaxation is so rapid that the lifetime of a vibrational excited molecule (<10-12) is less than the lifetime of the electronically excited state. For this reason, fluorescence from a solution always involves the transition of the lowest vibrational level of the excited state. Since the space of the emission lines are so close together, the transition of the vibrational relaxation can terminate in any vibrational level of the ground state.
• Internal Conversion: Internal conversion is an intermolecular process of molecule that passes to a lower electronic state without the emission of radiation.It is a crossover of two states with the same multiplicity meaning singlet-to-singlet or triplet-to-triplet states.The internal conversion is more efficient when two electronic energy levels are close enough that two vibrational energy levels can overlap as shown in between S1 and S2. Internal conversion can also occur between S0 and S1 from a loss of energy by fluorescence from a higher excited state, but it is less probable. The mechanism of internal conversion from S1 to S0 is poorly understood. For some molecules, the vibrational levels of the ground state overlaps with the first excited electronic state, which leads to fast deactivation.These usually occur with aliphatic compounds (compound that do not contain ring structure), which would account for the compound is seldom fluorescing. Deactivation by energy transfer of these molecules occurs so rapidly that the molecule does not have time to fluoresce.
• 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 leads to enhanced fluorescence because they reduce the number of collision between molecules, thus slowing down the deactivation process.
• Intersystem Crossing: Intersystem crossing is a process where there is a crossover between electronic states of different multiplicity as demonstrated in the singlet state to a triplet state (S1 to T1) on Figure $1$. The probability of intersystem crossing is enhanced if the vibration levels of the two states overlap. Intersystem crossing is most commonly observed with molecules that contain heavy atom such as iodine or bromine. The spin and orbital interaction increase and the spin become more favorable.Paramagnetic species also enhances intersystem crossing, which consequently decreases fluorescence.
• Phosphorescence: Deactivation of the electronic excited state is also involved in phosphorescence. After the molecule transitions through intersystem crossing to the triplet state, further deactivation occurs through internal or external fluorescence or phosphorescence. A triplet-to-singlet transition is more probable than a singlet-to-singlet internal crossing. In phosphorescence, the excited state lifetime is inversely proportional to the probability that the molecule will transition back to the ground state. Since the lifetime of the molecule in the triplet state is large (10-4 to 10 second or more), transition is less probable which suggest that it will persist for some time even after irradiation has stopped. Since the external and internal conversion compete so effectively with phosphorescence, the molecule has to be observed at lower temperature in highly viscous media to protect the triplet state.
Variables that affect Fluorescence
After discussing all the possible deactivation processes, variable that affect the emissions to occur. Molecular structure and its chemical environment influence whether a substance will fluoresce and the intensities of these emissions. The quantum yield or quantum efficiency is used to measure the probability that a molecule will fluoresce or phosphoresce. For fluorescence and phosphorescence is the ratio of the number of molecules that luminescent to the total number of excited molecules. For highly fluoresce molecules, the quantum efficiency approaches to one.Molecules that do not fluoresce have quantum efficiencies that approach to zero.
Fluorescence quantum yield ($\phi$) for a compound is determined by the relative rate constants (k) of various deactivation processes by which the lowest excited singlet state is deactivated to the ground state. The deactivation processes including fluorescence (kf), intersystem crossing ($k_i$), internal conversion (kic), predissociation (kpd), dissociation (kd), and external conversion (kec) allows one to qualitatively interpret the structural and environmental factors that influence the intensity of the fluorescence. They are related by the quantum yield equation given below:
$\dfrac{k_f}{k_f+k_i+k_{ec}+k_{ic}+k_{pd}+k_d}$
Using this equation as an example to explain fluorescence, a high fluorescence rate (kf) value and low values of the all the other relative rate constant terms (kf +ki+kec+kic+kpd+kd) will give a large $\phi$, which suggest that fluorescence is enhanced. The magnitudes of kf , kd, and kpd depend on the chemical structure, while the rest of the constants ki, kec, and kic are strongly influenced by the environment.
Fluorescence rarely results from absorption of ultraviolet radiation of wavelength shorter than 250 nm because radiation at this wavelength has sufficient energy to deactivate the electron in the excited state by predissociation or dissociation. The bond of some organic molecules would rupture at 140 kcal/mol, which corresponds to 200-nm of radiation. For this reason, $\sigma \rightarrow \sigma^{*}$ transition in fluorescence are rarely observed. Instead, emissions from the less energetic transition will occur which are either $\pi^{*} \rightarrow \pi$ or $\pi^{*} \rightarrow n$ transition.
Molecules that are excited electronically will return to the lowest excited state by rapid vibrational relaxation and internal conversion, which produces no radiation emission. Fluorescence arises from a transition from the lowest vibrational level of the first excited electronic state to one of the vibrational levels in the electronic ground state. In most fluorescent compounds, radiation is produced by a $\pi^{*} \rightarrow \pi$ or $\pi^{*} \rightarrow n$ transition depending on which requires the least energy for the transition to occur.
Fluorescence is most commonly found in compounds in which the lowest energy transition is $\pi \rightarrow \pi^{*}$ (excited singlet state) than $n \rightarrow \pi^{*}$ which suggest that the quantum efficiency is greater for $\pi \rightarrow \pi^{*}$ transitions. The reason for this is that the molar absorptivity, which measures the probability that a transition will occur, of the $\pi \rightarrow \pi^{*}$ transition is 100 to 1000 fold greater than $n \rightarrow \pi^{*}$ process. The lifetime of $\pi \rightarrow \pi^{*}$ (10-7 to 10-9 s) is shorter than the lifetime of $n \rightarrow \pi^{*}$ (10-5 to 10-7).
Phosphorescent quantum efficiency is the opposite of fluorescence in that it occurs in the $n \rightarrow \pi^{*}$ excited state which tends to be short lived and less suceptable to deactivation than the $\pi \rightarrow \pi^{*}$ triplet state. Intersystem crossing is also more probable for $\pi \rightarrow \pi^{*}$ excited state than for the $n \rightarrow \pi^{*}$state because the energy difference between the singlet and triplet state is large and spin-orbit coupling is less likely to occur.
Fluorescence and Structure
The most intense fluorescence is found in compounds containing aromatic group with low-energy $\pi \rightarrow \pi^{*}$ transitions. A few aliphatic, alicyclic carbonyl, and highly conjugated double-bond structures also exhibit fluorescence as well. Most unsubstituted aromatic hydrocarbons fluoresce in solution too. The quantum efficiency increases as the number of rings and the degree of condensation increases. Simple heterocycles such as the structures listed below do not exhibit fluorescence.
Pyridine Pyrrole Furan Thiophene
With nitrogen heterocyclics, the lowest energy transitions is involved in $n \rightarrow \pi^{*}$ system that rapidly converts to the triplet state and prevents fluorescence. Although simple heterocyclics do not fluoresce, fused-ring structures do. For instance, a fusion of a benzene ring to a hetercyclic structure results in an increase in molar absorptivity of the absorption band. The lifetime of the excited state in fused structure and fluorescence is observed. Examples of fluorescent compounds is shown below.
quinoline
Benzene ring substitution causes a shift in the absorption maxima of the wavelength and changes in fluorescence emission. The table below is used to demonstrate and visually show that as benzene is substituted with increasing methyl addition, the relative intensity of fluorescence increases.
Table 2. Relative intensity of fluorescence comparison with alkane substituted benzenes.
Compound
Structure
Wavelength of Fluorescence (nm)
Relative intensity of Fluorescence
Benzene
270-310
10
Toluene
270-320
17
Propyl Benzene
270-320
17
The relative intensity of fluorescence increases as oxygenated species increases in substitution. The values for such increase is demonstrated in the table below.
Table $3$: Relative intensity of fluorescence comparison with benzene with oxygenated substituted benzene
Compound
Structure
Wavelength of Fluorescence (nm)
Relative intensity of Fluorescence
Phenol
285-365
18
Phenolate ion
310-400
10
Anisole
285-345
20
Influence of a halogen substitution decreases fluorescence as the molar mass of the halogen increases. This is an example of the “heavy atom effect” which suggest that the probability of intersystem crossing increases as the size of the molecule increases. As demonstrated in the table below, as the molar mass of the substituted compound increases, the relative intensity of the fluorescence decreases.
Table $4$: Relative intensity fluorescence comparison with halogen substituted compounds
Compound
Structure
Wavelength of Fluorescence (nm)
Relative intensity of Fluorescence
Fluorobenzene
270-320
10
Chlorobenzene
275-345
7
Bromobenzene
290-380
5
In heavy atom substitution such as nitro derivatives or heavy halogen substitution such as iodobenzene, the compounds are subject to predissociation. These compounds have bonds that easily rupture that can then absorb excitation energy and go through internal conversion. Therefore, the relative intensity of fluorescence and fluorescent wavelength is not observed and this is demonstrated in the table below.
Table $5$: Relative fluorescent intensities of iodobenzene and nitro derivative compounds
Compound
Structure
Wavelength of Fluorescence (nm)
Relative intensity of Fluorescence
Iodobenzene
None
0
Anilinium ion
None
0
Nitrobenzene
None
0
Carboxylic acid or carbonyl group on aromatic ring generally inhibits fluorescence since the energy of the $n \rightarrow \pi^*$ transition is less than $\pi \rightarrow \pi^*$ transition. Therefore, the fluorescence yield from $n \rightarrow \pi^*$ transition is low.
Table $6$: Relative fluorescent intensity of benzoic acid
Compound
Structure
Wavelength of Fluorescence (nm)
Relative intensity of Fluorescence
Benzoic Acid
310-390
3
Effect of Structural Rigidity on Fluorescence
Fluorescence is particularly favored in molecules with rigid structures. The table below compares the quantum efficiencies of fluorine and biphenyl which are both similar in structure that there is a bond between the two benzene group. The difference is that fluorene is more rigid from the addition methylene bridging group. By looking at the table below, rigid fluorene has a higher quantum efficiency than unrigid biphenyl which indicates that fluorescence is favored in rigid molecules.
Table $7$: Quantum Efficiencies in Rigid vs. Nonrigid structures
Compound
Structure
Quantum Efficiency
Fluorene
1.0
Biphenyl
0.2
This concept of rigidity was used to explain the increase in fluorescence of organic chelating agent when the compound is complexed with a metal ion. The fluorescence intensity of 8-hydroxyquinoline is much less than its zinc complex.
vs
8-hydroxyquinoline 8-hydroxyquinoline with Zinc complexed
The explanation for lower quantum efficiency or lack of rigidity in caused by the enhanced internal conversion rate (kic) which increases the probability that there will be radiationless deactivation. Nonrigid molecules can also undergo low-frequency vibration which accounts for small energy loss.
Temperature and Solvent Effects
Quantum efficiency of Fluorescence decreases with increasing temperature. As the temperature increases, the frequency of the collision increases which increases the probability of deactivation by external conversion. Solvents with lower viscosity have higher possibility of deactivation by external conversion. Fluorescence of a molecule decreases when its solvent contains heavy atoms such as carbon tetrabromide and ethyl iodide, or when heavy atoms are substituted into the fluorescing compound. Orbital spin interaction result from an increase in the rate of triplet formation, which decreases the possibility of fluorescence. Heavy atoms are usually incorporated into solvent to enhance phosphorescence.
Effect of pH on Fluorescence
The fluorescence of aromatic compound with basic or acid substituent rings are usually pH dependent. The wavelength and emission intensity is different for protonated and unprotonated forms of the compound as illustrated in the table below:
Table $8$: Quantum efficiency comparison due to protonation
Compound
Structure
Wavelength of Fluorescence (nm)
Relative intensity of Fluorescence
aniline
310-405
20
Anilinium ion
None
0
The emission changes of this compound arises from different number of resonance structures associated with the acidic and basic forms of the molecule.The additional resonance forms provides a more stable first excited state, thus leading to fluorescence in the ultraviolet region.The resonance structures of basic aniline and acidic anilinium ion is shown below:
basic Aniline Fluorescence of certain compounds have been used a detection of end points in acid-base titrations.An example of this type of fluorescence seen in compound as a function of pH is the phenolic form of 1-naphthol-4-sulfonic acid.This compound is not detectable with the eye because it occurs in the ultraviolet region, but with an addition of a base, it becomes converted to a phenolate ion, the emission band shifts to the visible wavelength where it can be visually seen. Acid dissociation constant for excited molecules differs for the same species in the ground state.These changes in acid or base dissociation constant differ in four or five orders of magnitude.
Dissolved oxygen reduces the intensity of fluorescence in solution, which results from a photochemically induced oxidation of fluorescing species.Quenching takes place from the paramagnetic properties of molecular oxygen that promotes intersystem crossing and conversion of excited molecules to triplet state.Paramagnetic properties tend to quench fluorescence.
Effects of Concentration on Fluorescence Intensity
The power of fluorescence emission $F$ is proportional to the radiant power is proportional to the radiant power of the excitation beam that is absorbed by the system. The equation below best describes this relationship.
(1)
Since $\phi_f K”$ is constant in the system, it is represented at K’. The table below defines the variables in this equation.
Table 9: Definitions of all the variables defined in the Fluorescence Emission (F) in Equation 1.
Variable
Definition
F
Power of fluorescence emission
P0
Power of incident beam on solution
P
Power after transversing length b in medium
K”
Constant dependent on geometry and other factors
f
Quantum efficiency
Fluorescence emission ($F$) can be related to concentration ($c$) using Beer’s Law stating:
$F = \epsilon b c \label{2}$
where $\epsilon$ is the molar absorptivity of the molecule that is fluorescing. Rewriting Equation 2 gives:
$P =P_o 10^{-\epsilon b c} \label{3}$
Plugging this Equation $\ref{3}$ into Equation $\ref{1}$ and factoring out $P_0$ gives us this equation:
$F=K^{\prime} P_{0}\left(1-10^{-\varepsilon b c}\right)$
The MacLaurin series could be used to solved the exponential term.
$F=K^{\prime} P_{0}\left[2.303 \varepsilon b c-\frac{(2.303 \varepsilon b c)^{2}}{2 !}+\frac{(2.303 \varepsilon b c)^{3}}{3 !}+\frac{(2.303 \varepsilon b c)^{4}}{4 !}+\ldots \frac{(2.303 \varepsilon b c)^{n}}{n !}\right]_{1}$
Given that $(2.303 \epsilon b c = \text{Absorbance} <0.05$, all the subsequent terms after the first can be dropped since the maximum error is 0.13%. Using only the first term, Equation $\ref{5}$ can be rewritten as:
$F=K^{\prime} P_{0} 2.303 \varepsilon b c$
Equation $\ref{6}$ can be expanded to the equation below and simplified to compare the fluorescence emission F with concentration. If the equation below were to be plotted with F versus c, a linear relation would be observed.
$F=\phi_{f} K^{\prime \prime} P_{0} 2.303 \varepsilon b c$
If $c$ becomes so great that the absorbance > 0.05, the higher terms start to become taken into account and the linearity is lost. F then lies below the extrapolation of the straight-line plot. This excessive absorption is the primary absorption. Another cause of this negative downfall of linearity is the secondary absorption when the wavelength of emission overlaps the absorption band. This occurs when the emission transverse the solution and gets reabsorbed by other molecules by analyte or other species in the solution, which leads to a decrease in fluorescence.
Quenching Methods
Dynamic Quenching is a nonradiative energy transfer between the excited and the quenching agent species (Q).The requirements for a successful dynamic quenching are that the two collision species the concentration must be high so that there is a higher possibility of collision between the two species.Temperature and quenching agent viscosity play a role on the rate of dynamic quenching.Dynamic quenching reduces fluorescence quantum yield and the fluorescence lifetime.
Dissolved oxygen in a solution increases the intensity of the fluorescence by photochemically inducing oxidation of the fluorescing species.Quenching results from the paramagnetic properties of molecular oxygen that promotes intersystem crossing and converts the excited molecules to triplet state.Paramagnetic species and dissolved oxygen tend to quench fluorescence and quench the triplet state.
Static quenching occurs when the quencher and ground state fluorophore forms a dark complex.Fluorescence is usually observed from unbound fluorophore.Static quenching can be differentiated from dynamic quenching in that the lifetime is not affected in static quenching.In long range (Förster) quenching, energy transfer occurs without collision between molecules, but dipole-dipole coupling occurs between excited fluorophore and quencher.
Emission and Excitation Spectra
One of the ways to visually distinguish the difference between each photoluminescence is to compare the relative intensities of emission/excitation at each wavelength. An example of the three types of photoluminescence (absorption, fluorescence and phosphorescence) is shown for phenanthrene in the spectrum below.In the spectrum, the luminescent intensity is measure in a wavelength is fixed while the excitation wavelength is varied. The spectrum in red represents the excitation spectrum, which is identical to the absorption spectrum because in order for fluorescence emission to occur, radiation needs to be absorbed to create an excited state.The spectrum in blue represent fluorescence and green spectrum represents the phosphorescence.
Fluorescence and Phosphorescence occur at wavelengths that are longer than their absorption wavelengths.Phosphorescence bands are found at a longer wavelength than fluorescence band because the excited triplet state is lower in energy than the singlet state.The difference in wavelength could also be used to measure the energy difference between the singlet and triplet state of the molecule. The wavelength ($\lambda$) of a molecule is inversely related to the energy ($E$) by the equation below:
$E= \dfrac{hc}{\lambda}$
As the wavelength increases, the energy of the molecule decrease and vice versa.
Problems
1. Draw and label the Jablonski Diagram.
2. How do spin states differ in ground singlet state versus excite singlet state and triplet excited state?
3. Describe the rates of deactivation process.
4. What is quantum yield and how is it used to compare the fluorescence of different types of molecule?
5. What roles do solvent play in fluorescence?
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Fluorescence_and_Phosphorescence.txt
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Aleksander Jablonski was a Polish academic who devoted his life to the study of molecular absorbance and emission of light. He developed a written representation that generally shows a portion of the possible consequences of applying photons from the visible spectrum of light to a particular molecule. These schematics are referred to as Jablonski diagrams.
Introduction
A Jablonski diagram is basically 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 vibronic 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. As electronic energy states increase, the difference in energy becomes continually less, eventually becoming a continuum that can be approach with classical mechanics. Additionally, as the electronic energy levels get closer together, the overlap of vibronic energy levels increases.
Through the use of straight and curved 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. Curved lines show transitions of electrons without any interaction with light. Within a Jablonski diagram several different pathways show how an electron may accept and then dissipate the energy from a photon of a particular wavelength. Thus, most diagrams start with arrows going from the ground electronic state and finish with arrows going to the ground electronic state.
Absorbance
The first transition in most Jablonski diagrams is the absorbance of a photon of a particular energy by the molecule of interest. This is indicated by a straight arrow pointing up. Absorbance is the method by which an electron is excited from a lower energy level to a higher energy level. The energy of the photon is transferred to the particular electron. That electron then transitions to a different eigenstate corresponding to the amount of energy transferred. Only certain wavelengths of light are possible for absorbance, that is, wavelengths that have energies that correspond to the energy difference between two different eigenstates of the particular molecule. Absorbance is a very fast transition, on the order of 10-15 seconds. Most Jablonski diagrams, however, do not indicate a time scale for the phenomenon being indicated. This transition will usually occur from the lowest (ground) electronic state due to the statistical mechanical issue of most electrons occupying a low lying state at reasonable temperatures. There is a Boltzmann distribution of electrons within this low lying levels, based on the the energy available to the molecules. This energy available is a function of the Boltzmann's constant and the temperature of the system. These low lying electrons will transition to an excited electronic state as well as some excited vibrational state.
Vibrational Relaxation and Internal Conversion
Once an electron is excited, there are a multitude of ways that energy may be dissipated. The first is through vibrational relaxation, a non-radiative process. This is indicated on the Jablonski diagram as a curved arrow between vibrational levels. Vibrational relaxation is where the energy deposited by the photon into the electron is given away to other vibrational modes as kinetic energy. This kinetic energy may stay within the same molecule, or it may be transferred to other molecules around the excited molecule, largely depending on the phase of the probed sample. This process is also very fast, between 10-14 and 10-11 seconds. Since this is a very fast transition, it is extremely likely to occur immediately following absorbance. This relaxation occurs between vibrational levels, so generally electrons will not change from one electronic level to another through this method.
However, if vibrational energy levels strongly overlap electronic energy levels, a possibility exists that the excited electron can transition from a vibration level in one electronic state to another vibration level in a lower electronic state. This process is called internal conversion and mechanistically is identical to vibrational relaxation. It is also indicated as a curved line on a Jablonski diagram, between two vibrational levels in different electronic states. Internal Conversion occurs because of the overlap of vibrational and electronic energy states. As energies increase, the manifold of vibrational and electronic eigenstates becomes ever closer distributed. At energy levels greater than the first excited state, the manifold of vibrational energy levels strongly overlap with the electronic levels. This overlap gives a higher degree of probability that the electron can transition between vibrational levels that will lower the electronic state. Internal conversion occurs in the same time frame as vibrational relaxation, therefore, is a very likely way for molecules to dissipate energy from light perturbation. However, due to a lack of vibrational and electronic energy state overlap and a large energy difference between the ground state and first excited state, internal conversion is very slow for an electron to return to the ground state. This slow return to the ground state lets other transitive processes compete with internal conversion at the first electronically excited state. Both vibrational relaxation and internal conversion occur in most perturbations, yet are seldom the final transition.
Fluorescence
Another pathway for molecules to deal with energy received from photons is to emit a photon. This is termed fluorescence. It is indicated on a Jablonski diagram as a straight line going down on the energy axis between electronic states. Fluorescence is a slow process on the order of 10-9 to 10-7 seconds; therefore, it is not a very likely path for an electron to dissipate energy especially at electronic energy states higher than the first excited state. While this transition is slow, it is an allowed transition with the electron staying in the same multiplicity manifold. Fluorescence is most often observed between the first excited electron state and the ground state for any particular molecule because at higher energies it is more likely that energy will be dissipated through internal conversion and vibrational relaxation. At the first excited state, fluorescence can compete in regard to timescales with other non-radiative processes. The energy of the photon emitted in fluorescence is the same energy as the difference between the eigenstates of the transition; however, the energy of fluorescent photons is always less than that of the exciting photons. This difference is because energy is lost in internal conversion and vibrational relaxation, where it is transferred away from the electron. Due to the large number of vibrational levels that can be coupled into the transition between electronic states, measured emission is usually distributed over a range of wavelengths.
Intersystem Crossing
Yet another path a molecule may take in the dissipation of energy is called intersystem crossing. This where the electron changes spin multiplicity from an excited singlet state to an excited triplet state. It is indicated by a horizontal, curved arrow from one column to another. This is the slowest process in the Jablonski diagram, several orders of magnitude slower than fluorescence. This slow transition is a forbidden transition, that is, a transition that based strictly on electronic selection rules should not happen. However, by coupling vibrational factors into the selection rules, the transition become weakly allowed and able to compete with the time scale of fluorescence. Intersystem crossing leads to several interesting routes back to the ground electronic state. One direct transition is phosphorescence, where a radiative transition from an excited triplet state to a singlet ground state occurs.This is also a very slow, forbidden transition. Another possibility is delayed fluorescence, the transition back to the first excited singlet level, leading to the emitting transition to the ground electronic state.
Other non-emitting transitions from excited state to ground state exist and account for the majority of molecules not exhibiting fluorescence or phosphorescent behavior. One process is the energy transfer between molecules through molecular collisions (e.g., external conversion). Another path is through quenching, energy transfer between molecules through overlap in absorption and fluorescence spectra. These are non-emitting processes that will compete with fluorescence as the molecule relaxes back down to the ground electronic state. In a Jablonski diagram, each of these processes are indicated with a curved line going down to on the energy scale.
Time Scales
It is important to note that a Jablonski diagram shows what sorts of transitions that can possibly happen in a particular molecule. Each of these possibilities is dependent on the time scales of each transition. The faster the transition, the more likely it is to happen as determined by selection rules. Therefore, understanding the time scales each process can happen is imperative to understanding if the process may happen. Below is a table of average time scales for basic radiative and non-radiative processes.
Table 1: Average timescales for radiative and non-radiative processes
Transition Time Scale Radiative Process?
Internal Conversion 10-14 - 10-11 s no
Vibrational Relaxation 10-14 - 10-11 s no
Absorption 10-15 s yes
Phosphorescence 10-4 - 10-1 s yes
Intersystem Crossing 10-8 - 10-3 s no
Fluorescence 10-9 - 10-7 s yes
Each process outlined above can be combined into a single Jablonski diagram for a particular molecule to give a overall picture of possible results of perturbation of a molecule by light energy. Jablonski diagrams are used to easily visualize the complex inner workings of how electrons change eigenstates in different conditions. Through this simple model, specific quantum mechanical phenomena are easily communicated.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Jablonski_diagram.txt
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In the field of inorganic chemistry, color is commonly associated with d–d transitions. If this is the case, why is it that some transition metal complexes show intense color in solution, but possess no d electrons? In transition metal complexes a change in electron distribution between the metal and a ligand gives rise to charge transfer (CT) bands when performing Ultraviolet-visible spectroscopy experiments. For complete understanding, a brief introduction to electron transfer reactions and Marcus-Hush theory is necessary.
Outer Sphere Charge Transfer Reactions
Electron transfer reactions(charge transfer) fall into two categories: Inner-sphere mechanisms and Outer-sphere mechanisms. Inner-sphere mechanisms involve electron transfer occurring via a covalently bound bridging ligand (Figure $1$).
Outer-sphere mechanisms involve electron transfer occurring without a covalent linkage between reactants, e.g.,
$\ce{[ML6]^{2+} + [ML6]^{3+} -> [ML6]^{3+} + [ML6]^{2+}} \label{eq1}$
Here, we focus on outer sphere mechanisms.
In a self-exchange reaction the reactant and product side of a reaction are the same. No chemical reaction takes place and only an electron transfer is witnessed. This reductant-oxidant pair involved in the charge transfer is called the precursor complex. The Franck-Condon approximation states that a molecular electronic transition occurs much faster than a molecular vibration.
Consider the reaction described in Equation \ref{1}. This process has a Franck-Condon restriction: Electron transfer can only take place when the $\ce{M–L}$ bond distances in the $\ce{ML(II)}$ and $\ce{ML(III)}$ states are the same. This means that vibrationally excited states with equal bonds lengths must be formed in order to allow electron transfer to occur. This would mean that the $\ce{[ML6]^{2+}}$ bonds must be compressed and $\ce{[ML6]^{3+}}$ bonds must be elongated for the reaction to occur.
Self exchange rate constants vary, because the activation energy required to reach the vibrational states varies according to the system. The greater the changes in bond length required to reach the precursor complex, the slower the rate of charge transfer.1
A Brief Introduction to Marcus-Hush Theory
Marcus-Hush theory relates kinetic and thermodynamic data for two self-exchange reactions with data for the cross-reaction between the two self-exchange partners. This theory determines whether an outer sphere mechanism has taken place. This theory is illustrated in the following reactions
Self exchange 1: [ ML6 ] 2+ + [ML6] 3+ → [ ML6 ] 3+ + [ ML6 ] 2+ ∆GO = 0
Self exchange 2: [ ML6 ] 2+ + [ML6] 3+ → [ ML6 ] 3+ + [ ML6 ] 2+ ∆GO = 0
Cross Reaction: [ ML6 ] 2+ + [ ML6] 3+ → [ ML6 ] 3+ + [ ML6 ] 2+
The Gibbs free energy of activation $\Delta G^{\mp }$ is represented by the following equation:
$\Delta G^{\mp } = \Delta_{w} G^{\mp } + \Delta _{o}G^{\mp } + \Delta _{s}G^{\mp }+ RT \ln ( k’T / hZ) \nonumber$
• T = temperature in K
• R = molar gas constant
• k’ = Boltzmann constant
• h = Planck's constant
• Z = effective frequency collision in solution ~ 1011 dm3 mol-1 s-1
• wGŦ = the energy associated with bringing the reactants together, includes the work done to counter any repulsion
• 0GŦ = energy associated with bond distance changes
• s ∆GŦ= energy associated with the rearrangements taking place in the solvent spheres
• ln ( k’T / hZ) = accounts for the energy lost in the formation of the encounter complex
The rate constant for the self-exchange is calculated using the following reaction
$k = \kappa Z e^{-\Delta G^{\mp }/RT} \nonumber$
where $\kappa$ is the transmission coefficient ~1.
The Marcus-Hush equation is given by the following expression
$k_{12}=(k_{11}k_{22}K_{12}f_{12})^{1/2} \label{marcus}$
where:
$\log f_{12}=\frac{(\log K_{12})^{2}}{4\log(\frac{k_{11}k_{22}}{Z^{2}})} \nonumber$
• $Z$ is the collision frequency
• k11 and ∆GŦ11 correspond to self exchange 1
• k22 and ∆GŦ22 correspond to self exchange 2
• k12 and ∆GŦ12 correspond to the cross-reaction
• K12 = cross reaction equilibrium constant
• ∆GO12= standard Gibbs free energy of the reaction
The following equation is an approximate from of the Marcus-Hush equation (Equation \ref{marcus}):
$\log k_{12}\approx 0.5\log k_{11}+0.5\log \log \nonumber$
since $f\approx 1$ and $\log f \approx 0$.
How is the Marcus-Hush equation used to determine if an outer sphere mechanism is taking place?
• values of k11, k22, K12, and k12 are obtained experimentally
• k11 and k22 are theoretically values
• $K_{12}$ is obtained from $E_{cell}$
If an outer sphere mechanism is taking place the calculated values of $k_{12}$ will match or agree with the experimental values. If these values do not agree, this would indicate that another mechanism is taking place.1
The Laporte Selection Rule and Weak d–d Transitions
d- d transitions are forbidden by the Laporte selection rule.
• Laporte Selection Rule: $∆l = + 1$
• Laporte allowed transitions: a change in parity occurs i.e. s → p and p → d.
• Laporte forbidden transitions: the parity remains unchanged i.e. p → p and d → d.
d-d transitions result in weak absorption bands and most d-block metal complexes display low intensity colors in solution (exceptions d0 and d10 complexes). The low intensity colors indicate that there is a low probability of a d-d transition occurring.
Ultraviolet-visible (UV/Vis) spectroscopy is the study of the transitions involved in the rearrangements of valence electrons. In the field of inorganic chemistry, UV/Vis is usually associated with d – d transitions and colored transition metal complexes. The color of the transition metal complex solution is dependent on: the metal, the metal oxidation state, and the number of metal d-electrons. For example iron(II) complexes are green and iron(III) complexes are orange/brown.2
Charge Transfer Bands
If color is dependent on d-d transitions, why is it that some transition metal complexes are intensely colored in solution but possess no d electrons?
In transition metal complexes a change in electron distribution between the metal and a ligand give rise to charge transfer (CT) bands.1 CT absorptions in the UV/Vis region are intense (ε values of 50,000 L mole-1 cm-1 or greater) and selection rule allowed. The intensity of the color is due to the fact that there is a high probability of these transitions taking place. Selection rule forbidden d-d transitions result in weak absorptions. For example octahedral complexes give ε values of 20 L mol-1 cm-1 or less.2 A charge transfer transition can be regarded as an internal oxidation-reduction process. 2
Ligand to Metal and Metal to Ligand Charge Transfer Bands
Ligands possess $σ$, $σ^*$, $π$, $π^*$, and nonbonding ($n$) molecular orbitals. If the ligand molecular orbitals are full, charge transfer may occur from the ligand molecular orbitals to the empty or partially filled metal d-orbitals. The absorptions that arise from this process are called ligand-to-metal charge-transfer (LMCT) bands (Figure 3).2 LMCT transitions result in intense bands. Forbidden d-d transitions may also take place giving rise to weak absorptions. Ligand to metal charge transfer results in the reduction of the metal.
If the metal is in a low oxidation state (electron rich) and the ligand possesses low-lying empty orbitals (e.g., $\ce{CO}$ or $\ce{CN^{-}}$) then a metal-to-ligand charge transfer (MLCT) transition may occur. MLCT transitions are common for coordination compounds having π-acceptor ligands. Upon the absorption of light, electrons in the metal orbitals are excited to the ligand $π^*$ orbitals.2 Figure 4 illustrates the metal to ligand charge transfer in a d5 octahedral complex. MLCT transitions result in intense bands. Forbidden d–d transitions may also occur. This transition results in the oxidation of the metal.
Effect of Solvent Polarity on CT Spectra
*This effect only occurs if the species being studied is an ion pair*
The position of the CT band is reported as a transition energy and depends on the solvating ability of the solvent. A shift to lower wavelength (higher frequency) is observed when the solvent has high solvating ability.
Polar solvent molecules align their dipole moments maximally or perpendicularly with the ground state or excited state dipoles. If the ground state or excited state is polar an interaction will occur that will lower the energy of the ground state or excited state by solvation. The effect of solvent polarity on CT spectra is illustrated in the following example.
Example $1$
You are preparing a sample for a UV/Vis experiment and you decide to use a polar solvent. Is a shift in wavelength observed when:
a) Both the ground state and the excited state are neutral
When both the ground state and the excited state are neutral a shift in wavelength is not observed. No change occurs. Like dissolves like and a polar solvent won’t be able to align its dipole with a neutral ground and excited state.
b) The excited state is polar, but the ground state is neutral
If the excited state is polar, but the ground state is neutral the solvent will only interact with the excited state. It will align its dipole with the excited state and lower its energy by solvation. This interaction will lower the energy of the polar excited state. (increase wavelength, decrease frequency, decrease energy)
c) The ground state and excited state is polar
If the ground state is polar the polar solvent will align its dipole moment with the ground state. Maximum interaction will occur and the energy of the ground state will be lowered. (increased wavelength, lower frequency, and lower energy) The dipole moment of the excited state would be perpendicular to the dipole moment of the ground state, since the polar solvent dipole moment is aligned with the ground state. This interaction will raise the energy of the polar excited state. (decrease wavelength, increase frequency, increase energy)
d) The ground state is polar and the excited state is neutral
If the ground state is polar the polar solvent will align its dipole moment with the ground state. Maximum interaction will occur and the energy of the ground state will be lowered. (increased wavelength, lower frequency, and lower energy). If the excited state is neutral no change in energy will occur. Like dissolves like and a polar solvent won’t be able to align its dipole with a neutral excited state. Overall you would expect an increase in energy (Illustrated below), because the ground state is lower in energy (decrease wavelength, increase frequency, increase energy).4
How to Identify Charge Transfer Bands
CT absorptions are selection rule allowed and result in intense (ε values of 50,000 L mole-1 cm-1 or greater) bands in the UV/Vis region.2 Selection rule forbidden d-d transitions result in weak absorptions. For example octahedral complexes give ε values of 20 L mol-1 cm-1 or less.2 CT bands are easily identified because they:
• Are very intense, i.e. have a large extinction coefficient
• Are normally broad
• Display very strong absorptions that go above the absorption scale (dilute solutions must be used)
Example $2$: Ligand to Metal Charge Transfer
$\ce{KMnO4}$ dissolved in water exhibits intense CT Bands. The one LMCT band in the visible is observed around 530 nm.
The band at 528 nm gives rise to the deep purple color of the solution. An electron from a “oxygen lone pair” character orbital is transferred to a low lying $\ce{Mn}$ orbital.1
Example $3$: Metal to Ligand Charge Transfer
Tris(bipyridine)ruthenium(II) dichloride ($\ce{[Ru(bpy)3]Cl2}$) is a coordination compound that exhbits a CT band is observed (Figure $6$)
A d electron from the ruthenium atom is excited to a bipyridine anti-bonding orbital. The very broad absorption band is due to the excitation of the electron to various vibrationally excited states of the π* electronic state.6
Practice Problems
1. You perform a UV/Vis on a sample. The sample being studied has the ability to undergo a charge transfer transition. A charge transfer transitions is observed in the spectra. Why would this be an issue if you want to detect d-d transitions? How can you solve this problem?
2. What if both types of charge transfer are possible? For example a complex has both σ-donor and π-accepting orbitals? Why would this be an issue?
3. If the ligand has chromophore functional groups an intraligand band may be observed. Why would this cause a problem if you want to observe charge transfer bands? How would you identify the intraligand bands? State a scenario in which you wouldn’t be able to identify the intraligand bands.
Answers to Practice Problems
1. This is an issue when investigating weak d-d transitions, because if the molecule undergoes a charge transfer transitions it results in an intense CT band. This makes the d – d transitions close to impossible to detect if they occur in the same region as the charge transfer band. This problem is solved by performing the UV/Vis experiment on a more concentrated solution, resulting in minor peaks becoming more prominent.
2. Octahedral complexes such as Cr(CO)6, have both σ-donor and π-accepting orbitals. This means that they are able to undergo both types of charge transfer transitions. This makes it difficult to distinguish between LMCT and MLCT.
3. This would cause a problem because CT bands may overlap intraligand bands. Intraligand bands can be identified by comparing the complex spectrum to the spectrum of the free ligand. This may be difficult, since upon coordination to the metal, the ligand orbital energies may change, compared to the orbital energies of the free ligand. It would be very difficult to identify an intraligand band if the ligand doesn’t exist as a free ligand. If it doesn’t exist as a free ligand you wouldn’t be able to take a UV/Vis, and thus wouldn’t be able to use this spectrum in comparison to the complex spectrum.
Literature: Marcus Theory and Charge Transfer Bands
1. Marcus, R. A. "Chemical and Electrochemical Electron-Transfer Theory." Annual Review of Physical Chemistry 15.1 (1964): 155-96.
2. Eberson, Lennart. "Electron Transfer Reactions in Organic Chemistry. II* An Analysis of Alkyl Halide Reductions by Electron Transfer Reagents on the Basis of Marcus Theory." Acta Chemica Scandinavica 36 (1982): 533-43.
3. Chou, Mei, Carol Creutz, and Norman Sutin. "Rate Constants and Activation Parameters for Outer-sphere Electron-transfer Reactions and Comparisons with the Predictions of Marcus Theory." Journal of the American Chemical Society 99.17 (1977): 5615-623.
4. Marcus, R. A. "Relation between Charge Transfer Absorption and Fluorescence Spectra and the Inverted Region." Journal of Physical Chemistry 93 (1989): 3078-086.
Literature: Examples of Charge Transfer Bands
1. Electron Transfer Reactions of Fullerenes:
• Mittal, J. P. "Excited States and Electron Transfer Reactions of Fullerenes." Pure and Applied Chemistry 67.1 (1995): 103-10.
• Wang, Y. "Photophysical Properties of Fullerenes/ N,N-diethylanaline Charge Transfer Complexes." Journal of Physical Chemistry 96 (1992): 764-67.
• Vehmanen, Visa, Nicolai V. Tkachenko, Hiroshi Imahori, Shunichi Fukuzumi, and Helge Lemmetyinen. "Charge-transfer Emission of Compact Porphyrin–fullerene Dyad Analyzed by Marcus Theory of Electron-transfer." Spectrochimica Acta Part A 57 (2001): 2229-244.
2. Electron Transfer Reactions in RuBpy:
• Paris, J. P., and Warren W. Brandt. "Charge Transfer Luminescence of a Ruthenium(II) Chelate." Communications to the Editor 81 (1959): 5001-002.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Metal_to_Ligand_and_Ligand_to_Metal_Charge_Transfer_Bands.txt
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Spontaneous emission is the process in which a quantum mechanical system (such as an atom, molecule or subatomic particle) transitions from an excited energy state to a lower energy state (e.g., its ground state) and emits a quantum in the form of a photon.
Radiative Decay
Most chemiluminescence methods involve only a few chemical components to actually generate light. Luminol chemiluminescence (Nieman, 1989), which has been extensively investigated, and peroxyoxalate chemiluminescence (Given and Schowen, 1989; Orosz et al., 1996) are both used in bioanalytical methods and will be the subject of this primer on chemiluminescence. In each system, a "fuel" is chemically oxidized to produced an excited state product. In many luminol methods it is this excited product that emits the light for the signal. In peroxyoxalate chemiluminescence, the initial excited state product does not emit light at all and instead it reacts with another compound.
Introduction
Chemiluminescence takes its place among other spectroscopic techniques because of its inherent sensitivity and selectivity. It requires:
• no excitation source (as does fluorescence and phosphorescence)
• only a single light detector such as a photomultiplier tube
• no monochromator and often not even a filter
Maybe this list should be entitled "What chemiluminescent systems do not require." Although not as widely applicable as excitation spectroscopy, the detection limits for chemiluminescent methods can be 10 to 100 times lower than other luminescence techniques.
Most chemiluminescence methods involve only a few chemical components to actually generate light. Luminol chemiluminescence (Nieman, 1989), which has been extensively investigated, and peroxyoxalate chemiluminescence (Given and Schowen, 1989; Orosz et al., 1996) are both used in bioanalytical methods and will be the subject of this primer on chemiluminescence. In each system, a "fuel" is chemically oxidized to produced an excited state product. In many luminol methods it is this excited product that emits the light for the signal. In peroxyoxalate chemiluminescence, the initial excited state product does not emit light at all and instead it reacts with another compound, often a compound also viable as a fluorescent dye, and it is this fluorophore which becomes excited and emits light. That said, the oxalate reactions, to have practical applicability in, for instance HPLC, require a mixed solvent system (buffer/organic solvent) to assure solubility of the reagents, optimized pH, and allow compatibility with the analytes.
A general discussion of these two methods, their applicability as reported in some of the recent literature, and a discussion of the emission spectra of each--complete with movies that show short experiments with each--will be presented.
Peroxyoxalate Chemiluminescence Primer
One of the suggested reaction sequences in the reaction of peroxyoxalates, of which bis(2,4,6-trichlorophenyl)oxlate (TCPO) is the most prominent example, follows. It involves the fuel (TCPO) plus the oxidant (H2O2) reacting to produce a proposed intermediate, in this example shown as a dioxetane; although, this reaction probably produces many intermediates, and others, such as hydroperoxyoxalate, have been proposed (Milofsky and Birks, 1991; Choksi et al., 1990).
The intermediate, shown here as 1,2-dioxetanedione, excites a fluorophore. In the included movie demonstrating TCPO chemiluminescence, 9,10-diphenylanthracene acts as the fluorophore; its lambda max is 425 nm in the solvent used, tetrahydrofuran. Its reaction with the intermediate produces the excited state product which quickly emits light.
The process of transferring the energy of the initial reaction, the chemical reaction of hydrogen peroxide with TCPO, to light emission from the excited state fluorophore (fluorophore*) can be sidetracked along the way by loses in each step of the process: the initial oxidation to produce the intermediate, the reaction of the intermediate with a fluorophore, and the reaction of the excited fluorophore to produce light (Orosz et al., 1996).
The initial oxidation can yield the high energy intermediate or
• TCPO can be hydrolyzed instead or
• oxidation can occur that doesn't yield chemiluminescent products.
The high energy intermediate can react to excite the fluorophore or
• the intermediate can react with a quencher more easily oxidized than the fluorophore
• the intermediate and fluorophore can react without yielding excited fluorophore
• the intermediate can decompose or be further oxidized by residual H2O2.
Finally the excited fluorophore can loose energy by emission of light or
• the excited fluorophore can de-excited by production of heat instead of light.
In normal chromatographic (HPLC) procedures, these alternate mechanistic routes can be effected by solvent and buffers (Orosz, 1989; Jennings and Capomacchia, 1988); pH (de Jong et al., 1986); catalyst (Orlovic et al., 1989; Alverez et al., 1986); and type of fuel (Orlovic et al., 1989; Orosz, 1989), oxidant (Orlovic et al., 1989), and fluorophore concentration and identify. Possibly most important for chromatographers, eluent and reagent flows (Givens and Schowen, 1989; Kwakman and Brinkman, 1992), detector volume and geometry (de Jong et al., 1990; Grayeski and Weber, 1984), and mixing parameters (Kobayashi and Imai, 1980; Sugiura et al., 1993) can all effect this method's light production.
This, therefore, sets the stage for analytical methods whereby manipulating the appropriate parameter allows for the sensitive determination of hydrogen peroxide (Pontén et al., 1996; Stigbrand et al., 1994) or fluorophore content.
Recently, for example, Hamachi et al. (1999) determined the concentration of propentofylline in hypocampus extracts from rats by derivitizing the analyte to create a fluorophore which would chemiluminesce with another peroxyoxalate, TDPO [bis(2-(3,6,9-trioadecanyloxycarbonyl)-4-nitrophenyl)oxalate, and hydrogen peroxide following HPLC. Propentofylline is a reported inhibitor of dopamine released during low oxygenation events in the cerebellum. The derivatization of propentofylline was carried out in trifluoracetic acid/acetonitrile solution using DBD-H (a benzoaxadiazole). The detection limit for the analyte, 31 fg/injection, was about 200 times better than comparable HPLC-UV methods.
Emission Spectrum of Diphenylanthracene as Chemiluminescent Fluorophore
A solution of TCPO and 9,10-diphenylanthracene (DPA; Aldrich Chemicals Co., Milwaukee, WI USA) both in the 1 x 10-3 M concentration range dissolved in tetrahydrofuran (THF) were mixed with a dilute solution of H2O2 in THF (~0.3%) at ~25oC. The resulting emission spectrum was recorded on a fluorescence spectrometer (Hitachi F-4500; 1 cm quartz cell) in chemiluminescence mode (with no excitation source).
The slit and PMT voltage were adjusted to allow for the detection of a strong signal without overloading the detector. The components were mixed and the emission spectrum scanned immediately (1200 nm/min). As the Figure below shows, the emission was centered around 425 nm. This is, of course, similar to DPA's "normal" fluorescent emission.
Movie of TCPO + H2O2 + Diphenylanthracene Chemiluminescence Reaction
The movie included here involves that same solution, TCPO and 9,10-diphenylanthracene dissolved in THF. If you look closely you may be able to see the milky consistency of the slightly yellow, initial mixture--shown under fluorescent lights, before hydrogen peroxide was added. Without a mixed solvent system, the solubility of each of these components is relatively low and so the solution is basically saturated with each of these reagents (but still in the low millimolar concentration range).
In the dark, a solution of ~0.3% H2O2 in THF was added dropwise to approximately 8 mL of the fuel + fluorophore in THF (~25oC) in an open-topped vial. The reaction(s) immediately produces light from the excited fluorophore. The emission is relatively short lived but since H2O2 is apparently limiting, a second and third dropwise addition of the oxidant yields additional bursts of light. If you will look carefully at the end of the movie you will see a clear--yet still yellow--solution in which all precipitates have dissolved. Also apparent to the experimenter, but undetectable in the movie, was the formation of a gas produced by the reaction; this appeared as a bubbling that could be seen while the reaction was still producing light yet which stopped as the reaction reached completion, about 30 seconds after the last (excess) H2O2 addition. This kind of gas production has been used as evidence for the production of CO2 as a product from the 1,2-dioxetandione intermediate as detailed in the figure above. Further peroxide addition does not yield more bubbling so this is not simply H2O2 decomposition. The process of filming this reaction is described below.
Luminol Chemiluminescence
Luminol is also widely used as a chemiluminescent reagent, but unlike the peroxyoxalate systems does not require an organic/mixed solvent system. The chemiluminescent emitter is a "direct descendent" of the oxidation of luminol (or an isomer like isoluminol) by an oxidant in basic aqueous solution. Probably the most useful oxidant is also hydrogen peroxide similar to peroxyoxalate chemiluminescence; however, other oxidants have been used such as perborate, permanganate (Lu and Lu, 1992), hypochlorite (Cunningham et al., 1998), and iodine (Seitz, 1981). If the fuel is luminol, the emitting species is 3-aminophthalate (see below); however, luminol-derivatized analytes allow for determination of compounds that would not normally chemiluminescence in this system and presumably have slightly different emitters (Edwards et al., 1995; Kawasaki et al., 1985; Lippman, 1980; Nakazone et al., 1992; Pontén et al., 1996).
The presence of a catalyst is paramount to this chemiluminescent method as an analytical tool. Many metal cations catalyze the reaction of luminol, H2O2, and OH- in aqueous solution to increase light emission or at least to increase the speed of the oxidation to produce the emitter and therefore the onset/intensity of light production. [Some metals, however, repress chemiluminescence at different concentrations (Yuan and Shiller, 1999; see below.] This therefore can be the foundation of significantly different analytical determinations. For instance, this system can be used:
• to determine luminol itself by holding other variables constant
• to determine luminol-like derivatives similarly (Edwards et al., 1995; Kawasaki et al., 1985; Lippman, 1980; Nakazone et al., 1992; Pontén et al., 1996)
• to determine hydrogen peroxide or the progress of reactions that produce H2O2 (Yuan and Shiller, 1999; Tsukagoshi et al., 1998)
• to determine the concentrations of metal cations (Kyaw et al., 1998; O'Sullivan et al., 1995; Robards and Worsfold, 1992; TheingiKyaw et al. 1999)
• or to determine analytes that effect the concentration of metal catalysts.
This last is particularly powerful feature of this system because many compounds complex metallic cations and thereby make themselves "known." Amino acids (Koerner and Nieman, 1987), fructose and tagatose (Valeri et al., 1997), glycerol (Robards and Worsfold, 1992), thiols (Sano and Nakamura, 1998), and serum albumin (Tie et. al., 1995) among many others have been determined using luminol chemiluminescence.
Most recently, Yuan and Shiller (1999) report a subnanomolar detection limit for H2O2using luminol chemiluminescence. Their method, which was used to determine hydrogen peroxide content in sea water, was based on the cobalt(II) catalytic oxidation of luminol. While Co is the most sensitive luminol metal catalyst, it is also present in sea water at very low concentrations. The pH of the luminol solution used in this work was 10.15, and interferences from seven different metals were investigated. Interestingly some metals interfered positively and some negatively, and Fe(III) interfered positively at one concentration and negatively at another. Finally, very low concentrations of iron(II) showed a significant positive interference in determination of H2O2, but the authors used the relatively short half life of Fe(II) in marine water as a means of eliminating Fe(II) interference in the determination of hydrogen peroxide in their analysis by storing samples for over 1 hr before analysis.
Light emission from 3-APA
Approximately 15 ml of a solution containing luminol, copper catalyst, and pH controllers were placed in a glass vial at ~25oC (1 x 10-3 M luminol; 0.05 M sodium carbonate; 0.3 M sodium bicarbonate; 5 x 10-3 M ammonium carbonate; 1.5 x 10-3 M Cu(II) added as sulfate salt). An aqueous solution of approximately 0.25% H2O2 was added dropwise.
The emission spectrum was taken as before using a fluorescence spectrometer with the excitation source off. The light intensity-time decay data were taken immediately after mixing the reagents and for 60 seconds. The lambda max is at approximately 445 nm, slightly longer wavelength than the TCPO/DPA system described above. Online presentations of the light intensity-time decay aspects of the luminol reaction with hydrogen peroxide and differing concentrations of Cu(II) as catalyst are also available elsewhere (Iwata and Locker, 1998); however, with this reagent mixture the onset of emission was almost instantaneous and reached a maximum within a few seconds.
As the figure shows the light intensity decayed to approximately 50% of maximum at about 8 seconds. Iwata and Locker found that both the initial intensity and rate of decay in this kind of system was dependent on Cu(II) content. In the TCPO system described above, Orosz et al. (1996) reported that decay rate, rise constant, maximal light intensity, and quantum efficiency depended on hydrogen peroxide concentration. These authors present a comprehensive review of efforts to model the optimization of reagent flow rates and concentrations on HPLC detector responses with the TCPO reaction(s).
Movie of Luminol Chemiluminescence
The luminol reaction described above was carried out by placing approximately 15 mL of a solution containing the fuel (luminol), Cu2+ (1.5 x 10-3 M as the sulfate), and buffers detailed above in a open-topped glass vial (~25oC). The initial solution is visible at the movie's beginning as light blue in color under the laboratory's fluorescent light due to aqueous copper cations. In the dark, aqueous hydrogen peroxide (~0.3%) was added dropwise four times (small 1 or 2 mL squirts is probably a better description). The light emission is also, as before with TCPO, almost simultaneous upon mixing. The light produced appears white/blue and, as in the TCPO/DPA movie, since fuel is initially in excess, multiple injections of the limiting H2O2 reagent are necessary to take the reaction nearer to completion. Finally, after the fourth addition, the mixture was allowed to decay undisturbed and the light intensity drops off rather quickly (see the time decay data above). Approximately 80 seconds after the initial mixing began, the overhead fluorescent light were turned on and the final frame shows that solution. The light blue solution then appear green with a finely dispersed, black precipitate.
Table 1: Analytically Useful Chemiluminescent Emitters
Chemiluminescent
Emitter
Lambda Max
(Wavelength Region)
Reference
3-Aminophthalate
(from luminol)
425 nm White and Roswell. In Burr, Ed., Chemi- and Bioluminescence, Marcel Dekker, New York, 1985, p. 215.
CH3Se 750-825 nm Glinski et al., J.A.C.S.. 108, 531 (1986).
CN 383-388 nm Sutton et al., Anal. Chem.. 51, 1399 (1979).
HCF 475-750 nm Glinski et al., J. Photochem.. 37, 217 (1987).
HCHO 350-500 nm Finlayson et al., J.A.C.S..96, 5356 (1974).
HF 670-700 nm Glinski et al., J. Photochem.. 37, 217 (1987).
HSO 360-380 nm Toby, Chem. Rev.. 84, 277, (1984).
IF 450-800 nm Getty and Birks, Anal. Lett.. 12, 469, (1979)
N-methylacridone
(from lucigene)
420-500 nm Totter, Photochem. Photobiol. 22, 203 (1975).
Na 589 Gough et al., J. Chromatogr.. 137, 293 (1977).
NO2 1200 nm Greaves and Gavin, Chem. Physics..30, 348, (1959).
OH 306 nm Toby, Chem. Rev.. 84, 277, (1984).
Oxyluciferin
(from luciferin)
562 nm Seitz, Crc Crit. Rev. Anal. Chem.. 13, 1 (1981).
Ru(bpy)3 600 nm Rubinstein et al., Anal. Chem.. 55, 1580 (1983).
S2 275-425 nm Spurlin and Yeung, Anal. Chem..54, 318 (1982).
SF2 550-875 nm Glinski and Taylor, Chem. Phys. Lett..155, 511 (1989).
SO2 260-480 nm Kenner and Ogryzlo, In Burr, Ed., Chemi- and Bioluminescence, Marcel Dekker, New York, 1985, p. 139.
• Luminol
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Fluorescence, a type of luminescence, occurs in gas, liquid or solid chemical systems. Fluorescence is brought about by absorption of photons in the singlet ground state promoted to a singlet excited state. The spin of the electron is still paired with the ground state electron, unlike phosphorescence. As the excited molecule returns to ground state, it involves the emission of a photon of lower energy, which corresponds to a longer wavelength, than the absorbed photon.
Introduction
The energy loss is due to vibrational relaxation while in the excited state. Fluorescent bands center at wavelengths longer than the resonance line. This shift toward longer wavelengths is called a Stokes shift. Excited states are short-lived with a lifetime at about 10-8 seconds. Molecular structure and chemical environment affect whether or not a substance luminesces. When luminescence does occur, molecular structure and chemical environment determine the intensity of emission. Generally molecules that fluoresce are conjugated systems. Fluorescence occurs when an atom or molecules relaxes through vibrational relaxation to its ground state after being electrically excited. The specific frequencies of excitation and emission are dependent on the molecule or atom.
$S_0 + h\nu_{ex} = S_1$
where
• $h\nu$ is a photon energy with
• $h$ is Planck's constant and
• $\nu$ is the frequency of light,
• $S_0$ is the ground state of the fluorophore and
• $S_1$ is its first electronically excited state.
Figure 1 is a Jablonski energy diagram representing fluorescence. The purple arrow represents the absorption of light. The green arrow represents vibrational relaxation from singlet excited state, S2 to S1. This process is a non-radiative relaxation in which the excitation energy is dispersed as vibrations or heat to the solvent, and no photon is emitted. The yellow arrow represents fluorescence to the singlet ground state, So.
The fluorescence quantum yield ((\Phi\)) gives the efficiency of the fluorescence process. It is the ratio of photons emitted to photons absorbed.
$\Phi = \dfrac{\text{ # emitted photons }}{\text{ # absorbed photons }} \label{Eq0}$
If every photon absorbed results in a photon emitted. The maximum fluorescence quantum yield is 1.0, and compounds with quantum yields of 0.10 are still considered fluorescent. Another way to define the fluorescence quantum yield is by the excited state decay rates:
$\Phi = \dfrac{k_f}{\sum_i k_i} \label{Eq1}$
where $k_f$ is the rate of spontaneous emission of radiation and the denominator is the sum of all rates of excited state decay for each deactivation process (ie phosphorescence, intersystem crossing, internal conversion…). The fluorescence lifetime is the average time the molecule remains in its excited state before emitting a photon. Fluorescence typically follows first-order kinetics:
$[S_1] = [S_1]_o e^{- t/\tau} \label{Eq2}$
where
• $[S_1]$ is the concentration of excited state molecules at time $t$,
• $[S_1]_0$ is the initial concentration and $\tau$ is the decay rate.
Various radiative and non-radiative processes can de-populate the excited state so the total decay rate is the sum over all rates:
$\tau_{tot}= \tau_{rad} + \tau_{nrad} \label{Eq3}$
where $\tau_{tot}$ is the total decay rate, $\tau_{rad}$ the radiative decay rate and $\tau_{nrad}$ the non-radiative decay rate. If the rate of spontaneous emission or any of the other rates are fast, the lifetime is short. The average lifetime of fluorescent compounds that emit photons with energies from the UV to near infrared are within the range of 0.5 to 20 nanoseconds.
The fluorescence intensity, $I_F$ is proportional to the amount of light absorbed and the fluorescence quantum yield, $\Phi$
$I_f=kI_o \phi [1-(10^{-εbc})] \label{Eq4}$
where
• $k$ is a proportionality constant attributed to the instrument
• $I_o\( is the incident light intensity • \(\epsilon$ is the molar absorptivity,
• $b$ is the path length, and
• $c$ is the concentration of the substrate.
If dilute solutions are used so that less than 2% of the excitation energy is absorbed, then an approximation can be made so that
$10^{x} \approx 1+x + ...$
so Equation $\ref{Eq4}$ can be simplified to
$I_f=kI_o\Phi [εbc] \label{Eq4a}$
This relationship shows that fluorescence intensity is proportional to concentration.
Fluorescence rarely results from absorption of UV-radiation of wavelengths shorter than 250 nm because this type of radiation is sufficiently energetic to cause deactivation of the excited state by predissociation or dissociation. Most organic molecules have at least some bonds that can be ruptured by energies of this strength. Consequently, fluorescence due to $\sigma^* \rightarrow \sigma$ transitions is rarely observed. Instead such emission is confined to the less energetic $\pi^* \rightarrow \pi$ and $\pi^* \rightarrow n$ processes. Fluorescence commonly occurs from a transition from the lowest vibrational level of the first excited electronic state to the one of the vibrational levels of the electronic ground state. Quantum yield ($\Phi$) is greater for $\pi^* \rightarrow \pi$ transition because these excited states show short average lifetimes (larger $k_f$) and because deactivation processes that compete with fluorescence is not as likely to happen. Molar absorptivity of ππ* transitions is 100-1000 fold greater. The average lifetime is 10-7 to 10-9 seconds for ?, ?* states and 10-5 to 10-7 seconds for n, π* states.
Figure 2 is a schematic of a typical filter fluorometer that uses a source beam for fluorescence excitation and a pair of photomultiplier tubes as transducers. The source beam is split near the source into a reference beam and a sample beam. The reference beam is attenuated by the aperture disk so that its intensity is roughly the same as the fluorescence intensity. Both beams pass through the primary filter, with the reference beam being reflected to the reference photomultiplier tube. The sample beam is focused on the sample by a pair of lenses and causes fluorescence emission. The emitted radiation passes through a second filter and then is focused on the sample photomultiplier tube. The electrical outputs from the two transducers are then processed by an analog to digital converter to compute the ratio of the sample to reference intensities, which can then be used for qualitative and quantitative analysis. To obtain an emission spectrum, the excitation monochromator is fixed and the emission monochromator varies. To obtain an excitation spectrum, the excitation monochromator varies while the emission monochromator is fixed.
Fluorescence spectroscopy can be used to measure the concentration of a compound because the fluorescence intensity is linearly proportional to the concentration of the fluorescent molecule. Fluorescent molecules can also be used as tags. For example, fluorescence in situ hybridization (FISH) is a method of determining what genes are present in an organism's genome. Single stranded DNA encoding a gene of interest is covalently bonded to a fluorescent molecule and washed over the organism's chromosome, binding to its complementary sequence. The presence and placement of the gene in the organism then fluoresces when shined with ultraviolet light. Green fluorescence protein (GFP) is used in molecular biology to monitor the activity of proteins. The gene encoding GFP can be inserted next to a gene encoding a protein that will be studied. When the genes are expressed, the protein will be attached to GFP and can be identified in the cell by its fluorescence.
Phosphorescence
Unlike florescence, phosphorescence does not re-emit the light immediately. Instead, phosphorescence releases light very slowly in the dark due to its energy transition state. When light such as ultraviolet light is shined upon a glow in dark object, the object emits light, creating phosphorescence.
Introduction
There is a similarity between the phosphorescent and the fluorescent materials. They both contain substances with identical atoms. It is very important to understand the impurity state energy band, which is located between the conduction and valence energy bands. In a phosphorescence event, the absorbed energy usually goes through a high energy state which happens to be triplet state. The energy gets trapped in the triplet state because its physical situation forbids the transition to return to lower energy state, also as known from impurity to valence band. In order to change the energy of valence band, electrons must regain the energy they had lost during the impurity band transitional process. If the quantum yield of the phosphorescence is high enough, a great amount of light will be released and thus making the object glow in the dark.
Most compounds have the ground state of singlet S0. When it absorbs light, the electrons in the molecule may move to excited state of S1, S2, Sn and so on. There are also triplet states T1 and T2. The energy of the T1 state is typically below the S1 state, while T2 is between S2 and S1 state. The wavelength of the radiation can determine which state the electron will move to. It is possible for the electron to return from excited state back to the ground state. An example is phosphorescence, where the emitting of radiation demotes the electrons from the excited state of T1 to ground state S0. The molecule of phosphorescence has long life time, it loses energy easily, so it is hard to observe phosphorescence.
Applications
Materials that can produce phosphorescence often contain zinc sulfide, sodium fluorescein, rhodamine, or strontium. The majority of phosphorescence is often used in drugs in pharmaceutical field. Some common drugs that have phosphorescence property include Aspirin, benzoic acid, morphine, and dopamine. Phosphorescence is also used to analyze water, air and chemical pollutions.
Reference
1. Atkins Physical chemistry 8th . Peter Atkins. Julio De Paula. Ch. 13 p.567-569
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The Selection Rules governing transitions between electronic energy levels of transition metal complexes are:
1. ΔS = 0 The Spin Rule
2. Δl = +/- 1 The Orbital Rule (or Laporte)
The first rule says that allowed transitions must involve the promotion of electrons without a change in their spin. The second rule says that if the molecule has a center of symmetry, transitions within a given set of p or d orbitals (i.e. those which only involve a redistribution of electrons within a given subshell) are forbidden.
Relaxation of these rules can occur through:
• Spin-Orbit coupling: this gives rise to weak spin forbidden bands
• Vibronic coupling: an octahedral complex may have allowed vibrations where the molecule is asymmetric.
Absorption of light at that moment is then possible.
• Mixing: π-acceptor and π-donor ligands can mix with the d-orbitals so transitions are no longer purely d-d.
Transition Types
1. Charge transfer, either ligand to metal or metal to ligand. These are often extremely intense and are generally found in the UV but they may have a tail into the visible.
2. d-d, these can occur in both the UV and visible region but since they are forbidden transitions have small intensities.
Expected intensities of electronic transitions
Transition type Example Typical values of ε /m2mol-1
Spin forbidden,
Laporte forbidden
[Mn(H2O)6]2+ 0.1
Spin allowed (octahedral complex),
Laporte forbidden
[Ti(H2O)6]3+ 1 - 10
Spin allowed (tetrahedral complex),
Laporte partially allowed
by d-p mixing
[CoCl4]2- 50 - 150
Spin allowed,
Laporte allowed
e.g. charge transfer bands
[TiCl6]2- or MnO4- 1,000 - 106
Expected Values
The expected values should be compared to the following rough guide.
• For M2+ complexes, expect Δ = 7,500 - 12,500 cm-1 or λ = 800 - 1,350 nm.
• For M3+ complexes, expect Δ= 14,000 - 25,000 cm-1 or λ = 400 - 720 nm.
For a typical spin-allowed, but Laporte (orbitally) forbidden transition in an octahedral complex, expect ε < 10 m2mol-1. Extinction coefficients for tetrahedral complexes are expected to be around 50-100 times larger than for octrahedral complexes. B for first-row transition metal free ions is around 1,000 cm-1. Depending on the position of the ligand in the nephelauxetic series, this can be reduced to as low as 60% in the complex.
Selection Rules for Electronic Spectra of Transition Metal Complexes
The Laporte Rule is a selection rule in electron absorption spectroscopy that applies to centrosymmetric molecules. It says that transitions between states of the same symmetry with respect to inversion are forbidden. Using the mathematical concept of even and odd functions, the Laporte Rule can be derived and summarized as follows: Electronic transitions from waveunfunctions with g symmetry to wavefunctions with g symmetry are forbidden, as are transitions from wavefunctions with u symmetry to wavefunctions with u symmetry. Transitions from g to u and u to g, where the symmetry switches, may be (but are not necessarily) allowed.
Introduction
When an incident photon is absorbed by an electron, the electron is excited from a lower energy state to a higher energy , excited state. Although there are multiple higher energy excited states, not all are available to the electron, and only certain transitions are allowed. Electron absorption spectroscopy measures the intensity of transmitted photons as a function of wavelength. From this the energy of absorbed photons is determined, and transitions are assigned using selection rules. Selection rules are the guidelines for determining if a transition will be allowed or forbidden. This module will further discuss the orbital selection rule called the Laporte Rule, but will leave the spin selection rule to be discussed in another module. Specifically, the Laporte selection rule determines whether an electonic transtion is orbitally allowed or forbidden. The Laporte Rule applies only to centrosymmetric molecules, or those that contain a center of inversion. An electronic transition is forbidden by the Laporte Rule if the ground and excited states have the same symmetry with respect to an inversion center. Electronic transitions are described by the transition moment integral.
$\int_{- \infty}^{ \infty} \Psi_{el} \hat{ M} \Psi^{ex}_{el} d \tau = \left | \left \langle i \left | \hat{M} \right | f \right \rangle \right | \label{1}$
(Ψ is the wavefunction, M is the transition dipole moment operator, i refers to the initial state and f refers to the final state). The Laporte Rule tells us that if the integrand of the transition moment integral does not contain the totally symmetric representation, then the transition is forbidden.
Background
In electron absorption spectroscopy, absorption of a photon excites an electron from one energy state to another. Such transitions are only detectable using optical spectroscopy when there is a corresponding change in the dipole moment of the molecule; transitions are detectable when the transition moment dipole is nonzero.
$\mu = \int_{- \infty }^{\infty} \Psi \hat{ \mu} \Psi^{ex} \label{2}$
The transition moment dipole given above includes the wave functions of the final and initial states, which contain both electronic and nuclear wave functions.
$\Psi = \Psi_{nuc} \Psi_{el} \label{3}$
The Born–Oppenheimer approximation tells us that electronic transitions happen on a faster time scale than nuclear transitions so the two can be separated, and when dealing with electronic transitions we can effectively ignore any nuclear motion. Thus, in discussing electronic spectroscopy and the "allowedness" of certain electronic transitions, we consider only the electronic integral.
$\int_{- \infty}^{ \infty} \Psi_{el} H \Psi_{el}^{ex}\label{4}$
Where H is the time-dependent Hamiltonian. We can make this integral easier to work with by invoking Fermi’s Golden Rule, which allows us to replace H with M, the electric dipole moment operator , which is part of the time-dependent Hamiltonian.
$\hat{ M} = e \sum_{ i} \vec{ r_{ i}} \label{5}$
We arrive at the following transition moment integral to describe electronic transitions from a ground state to an excited state.
$\int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \left | \left \langle i \left | \hat{M} \right | f \right \rangle \right | \label{6}$
The transition moment integral describes the probability of a transition taking place. The integral must be non-zero in order for a transition to occur. In other words, the two electronic states must overlap in order for a transition to occur.
Derivation
By the Laporte Rule in a centrosymmetric molecule, if the integrand of the transition moment integral contains the totally symmetric representation, then the transition is Laporte allowed. If the transition does not contain the totally symmetric representation, then the transition is Laporte forbidden. The totally symmetric representation is the irreducible symmetry representation that is even with respect to all symmetry operations. On a character table, this representation is usually listed first and has the designation A, A1, or A1g.
To determine if a transition is allowed, the point group of the molecule is determined, and the corresponding character table gives the symmetry representation of each electronic wavefunction. The character table is also used to determine the electric dipole moment operator, which is contained in the transition moment integral and has three parts that transform with the Cartesian x, y, and z axes.
$\hat{ M} = \begin {vmatrix} \hat {M_{x}}\ \hat {M_{y}}\ \hat {M_{z}} \end{vmatrix}\label{7}$
For calculating the integrand of the transition moment integral, the symmetry representations corresponding to x, y, and z are used for the electric dipole moment operator. The integrand of the transition moment integral is then given by the following direct products:
$\Psi_{el} \otimes \begin {vmatrix} \hat {M_{x}}\ \hat {M_{y}}\ \hat {M_{z}} \end{vmatrix} \otimes \Psi_{el}^{ex} \label{8}$
The component of the electric dipole moment operator that gives the totally symmetric representation corresponds to the direction of polarization of the electronic transition.
For a point group of low symmetry like C2h, computing the integrand for all possible transitions is a relatively short task. However, when considering a point group with a high degree of symmetry, like Oh, computing the integrand for all possible transitions in a point group becomes a long and tedious task. If we examine these computations through the lens of mathematics, this problem can be simplified. The Laporte Rule applies to centrosymmetric molecules, those containing a center of inversion. All states have a symmetry with respect to the inversion center, and all representations for a given point group have either a g (gerade) or u (ungerade) designation with respect to the center of inversion. Designations of g and u refer to even and odd symmetries with respect to the inversion center. Odd and even with respect to orbital symmetry correspond to the mathematical definitions of odd and even functions. M is an odd function. For the derivation of the Laporte Rule, allow g states to be represented by the even function cos(x), u states to be represented by the odd function sin(x), and M to be represented by the odd function x.
A transition between two even states is given by the integral
$F(x) = \int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \int \cos{x} \times{ x} \times{ \cos{x}} dx \label{9}$
The integrand is given by
$G(x) = x \times{ \cos^2 {x}}$
Evaluate the integrand
$G( x = \frac{ \pi}{4}) = \frac{ \pi}{4} \times{ \cos^2 {( \frac{ \pi}{4})}} = \frac { \pi}{8}$
Evaluate the integrand
$G( x = \frac{- \pi}{4}) = \frac{- \pi}{4} \times{ \cos^2 {( \frac{- \pi}{4})}} = \frac {- \pi}{8}$
$G(x) = - G(-x)$
Thus, $G(x) = x \times{ \cos^2 {x}}$ is an odd function, and the transition between two even states is forbidden.
Similarly, a transition between two odd states is given by the integral
$F(x) = \int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \int \sin{x} \times{x} \times{ \sin{x}} dx \label{10}$
The integrand is given by
$G(x) = x \times{ \sin^2 {x}}$
Evaluate the integrand
$G(x = \frac{ \pi}{4}) = (\frac{ \pi}{4}) \times{ \sin^2 { (\frac{ \pi}{4})}} = \frac { \pi}{8}$
Evaluate the integrand
$G(x = \frac{- \pi}{4}) = (\frac{- \pi}{4}) \times{ \sin^2 {( \frac{- \pi}{4})}} = \frac {- \pi}{8}$
$G(x) = - G(-x)$
Thus,
$G(x) = x \times{ \sin^2 {x}}$
is an odd function, and the transition between two odd states is forbidden.
A transition between an odd and an even state is given by the integral
$F(x) = \int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \int \cos{x} \times{x} \times{ \sin{x}} dx \label{11}$
The integrand is given by
$G(x) = \cos{x} \times{x} \times{ \sin{x}}$
Evaluate the integrand
$G(x = \frac{ \pi}{4}) = \sin{( \frac{ \pi}{4})} \times{( \frac{ \pi}{4})} \times{ \cos {( \frac{ \pi}{4})}} = \frac{ \pi}{8}$
Evaluate the integrand
$G(x = \frac{- \pi}{4}) = \sin{( \frac{- \pi}{4})} \times{( \frac{- \pi}{4})} \times{ \cos {( \frac{- \pi}{4})}} = \frac{ \pi}{8}$
$G(x) = G(-x)$
Thus, $G(x) = x \times{ \sin{x}} \times{ \cos{x}}$ is an even function, and the transition between an even and an odd state is not immediately forbidden.
These calculations can be summarized in the following set of rules:
g x g = g Forbidden
u x u = g Forbidden
g x u = u
May be Allowed
u x g = u
May be Allowed
It is shown above that the Laporte Rule expressly states that transitions between two even states or two odd states are forbidden. It should be emphasized, however, that the Laporte Rule does not state that transitions between an even and an odd state are allowed. The integrand of the transition moment integral for the transition between an odd and an even state is even, but there are multiple states with even inversion symmetry that are not the totally symmetric representation. Only an integrand that contains the totally symmetric representation is allowed. To determine which of the even transitions are actually allowed, the cross product of the representations of the initial and final electronic states with the electric dipole moment operator must be computed. All resulting cross products will be even; only those which contain the totally symmetric representation will be allowed.
Example 1: Octohedral Group
Oh (octahedral) point group
$M_x, M_y, M_z = T_{1u}$
$A_{1g} \otimes{T_{1u}} \otimes{A_{1g}} = T_{1u}$ Equation (1)
$A_{2u} \otimes{T_{1u}} \otimes{A_{2u}} = T_{1u}$ Equation (2)
$A_{2g} \otimes{T_{1u}} \otimes{A_{1g}} = A_{1g}$
Equation (3)
$A_{2u} \otimes{T_{1u}} \otimes{T_{1g}} = A_{2g} + E_{g} + T_{1g} + T_{2g}$ Equation (4)
Equation (1) shows a forbidden even-even transition. Equation (2) shows a forbidden odd-odd transition. Equations (3) and (4) demonstrate the subtlety of the Laporte Rule. Both are even transitions between an odd and an even state. However, (3) contains the totally symmetric representation and is an allowed transition, while (4) does not contain the totally symmetric representation and is thereby forbidden.
Problems
1. For the C4h point group, which transitions are forbidden by the Laporte Rule? Which transitions are orbitally allowed?
2. For the C2h point group, what are the polarizations of the Laporte allowed transitions?
3. Using the Laporte Rule, are transitions between d orbitals in a molecule of octahedral (Oh) symmetry allowed? Are transitions between p orbitals allowed? Are transitions between p and d orbitals allowed?
4. [Cu(H2O)6]2+ and [Cu(NH3)4]2+ both appear blue in solution because of the presence of copper ions. However, the two solutions are not identical. How would the appearance of these solutions differ? If given an unlabeled sample of each, how could the two solutions be distinguished without collecting any spectra?
Answers
1. The C4h point group contains the representations Ag, Bg, Eg, Au, Bu, and Eu. Via the Laporte rule: g to g and u to u transitions are orbitally forbidden. For g to u and u to g transitions, the integrand of the transition moment integral must contain the totally symmetric representation, which is Ag in this case.
$A_g \otimes{ A_g}$ Laporte Forbidden
$A_g \otimes{ B_g}$ Laporte Forbidden
$A_g \otimes{ E_g}$ Laporte Forbidden
$A_g \otimes{ A_u}$
$A_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} A_{u} = \begin {vmatrix} A_{g}\ E_{u} \end{vmatrix}$
Laporte Allowed
$A_g \otimes{ B_u}$
$A_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} B_{u} = \begin {vmatrix} E_{g}\ B_{g} \end{vmatrix}$
Laporte Forbidden
$A_g \otimes{ E_u}$
$A_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} E_{u} = \begin {vmatrix} A_{g} + A_{u} + 2B_{g}\ E_{g} \end{vmatrix}$
Laporte Allowed
$B_g \otimes{ B_g}$ Laporte Forbidden
$B_g \otimes{ E_g}$ Laporte Forbidden
$B_g \otimes{ A_u}$
$B_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} A_{u} = \begin {vmatrix} E_{g}\ B_{g} \end{vmatrix}$
Laporte Forbidden
$B_g \otimes{ B_u}$
$B_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} B_{u} = \begin {vmatrix} E_{g}\ B_{g} \end{vmatrix}$
Laporte Forbidden
$B_g \otimes{ E_u}$
$B_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} E_{u} = \begin {vmatrix} A_{g} + A_{u} + 2B_{g}\ E_{g} \end{vmatrix}$
Laporte Allowed
$E_g \otimes{ E_g}$ Laporte Forbidden
$E_g \otimes{ A_u}$
$E_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} A_{u} = \begin {vmatrix} A_{g} + A_{u} + 2B_{g}\ E_{g} \end{vmatrix}$
Laporte Allowed
$E_g \otimes{ B_u}$
$E_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} B_{u} = \begin {vmatrix} B_{u} + B_{g} + 2A_{g}\ E_{g} \end{vmatrix}$
Laporte Allowed
$E_g \otimes{ E_u}$
$E_{g} \begin {vmatrix} E_{u}\ A_{u} \end{vmatrix} E_{u} = \begin {vmatrix} E_{g} + E_{u} + 2E_{g}\ A_{g} +A_{u} + 2B_{g} \end{vmatrix}$
Laporte Allowed
$A_u \otimes{ A_u}$
Laporte Forbidden
$A_u \otimes{ B_u}$ Laporte Forbidden
$A_u \otimes{ E_u}$ Laporte Forbidden
$B_u \otimes{ B_u}$ Laporte Forbidden
$B_u \otimes{ E_u}$ Laporte Forbidden
$E_u \otimes{ E_u}$ Laporte Forbidden
2. The C2h point group contains Ag, Bg, Au, and Burepresentations. By the Laporte rule g to g and u to u transitions are orbitally forbidden. For g to u and u to g transitions, the integrand of the transition moment integral must contain the totally symmetric representation, which is Ag in this case. The component of the transition moment operator that gives the totally symmetric representation dictates the polarization of the transition.
$A_{g} \begin {vmatrix} A_{u}\ B_{u}\ B_{u} \end{vmatrix} A_{g} = \begin {vmatrix} A_{u}\ B_{u}\ B_{u} \end{vmatrix}$
Laporte Forbidden
$A_{g} \begin {vmatrix} A_{u}\ B_{u}\ B_{u} \end{vmatrix} B_{g} = \begin {vmatrix} B_{u}\ A_{u}\ A_{u} \end{vmatrix}$ Laporte Forbidden
$A_{g} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} A_{u} = \begin {vmatrix} A_{g}\ B_{g}\ B_{g} \end{vmatrix}$ Laporte Allowed: x-polarized
$A_{g} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} B_{u} = \begin {vmatrix} B_{g}\ A_{g}\ A_{g} \end{vmatrix}$ Laporte Allowed: y-, z-polarized
$B_{g} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} B_{g} = \begin {vmatrix} A_{u}\ B_{u}\ B_{u} \end{vmatrix}$ Laporte Forbidden
$B_{g} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} A_{u} = \begin {vmatrix} B_{g}\ A_{g}\ A_{g} \end{vmatrix}$ Laporte Allowed: y-, z-polarized
$B_{g} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} B_{u} = \begin {vmatrix} B_{g}\ A_{g}\ A_{g} \end{vmatrix}$ Laporte Allowed: y-, z-polarized
$A_{u} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} A_{u} = \begin {vmatrix} A_{u}\ B_{u}\ B_{u} \end{vmatrix}$ Laporte Forbidden
$A_{u} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} B_{u} = \begin {vmatrix} B_{u}\ A_{u}\ A_{u} \end{vmatrix}$ Laporte Forbidden
$B_{u} \begin {vmatrix} A_{u}\ B_{u}\ B_{g} \end{vmatrix} B_{u} = \begin {vmatrix} A_{u}\ B_{u}\ B_{u} \end{vmatrix}$
Laporte Forbidden
3. Using the character table, we see that d-orbitals in octahedral have the following symmetries:
$z^{2}, x^{2} - y^{2}$ $E_{g}$
$xy, xz, yz$ $T_{2g}$
Possible transitions are $E_{g} \times{ E_{g}}$ $E_{g} \times{ T_{2g}}$ $T_{2g} \times{ T_{2g}}$ All possible transitions are g to g, and are thus forbidden by the Laporte Rule.
P-orbitals have the following symmetries:
$x, y, z$ $T_{1u}$
All possible transitions are u to u, and are thus forbidden by the Laporte Rule.
P-orbital to d-orbital transitions would all be u to g, and are not forbidden by the Laporte Rule.
4. [Cu(NH3)4]2+ is a tetrahedral complex and is therefore non-centrosymmetric. Since it is non-centrosymmetric, it is not Laporte forbidden. [Cu(H2O)6]2+ is an octahedral complex whose d-d transitions are Laporte forbidden. [Cu(NH3)4]2+ will be a darker shade of blue in solution because its d-d transitions are not forbidden, and [Cu(H2O)6]2+ will be a paler shade of blue in solution because its d-d transitions are Laporte forbidden. Coloration and absorption of [Cu(H2O)6]2+ in the ultraviolet-visible range is attributed to vibronic coupling.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Selection_Rules_for_Electronic_Spectra_of_Transition_Metal_Complexes/Derivatio.txt
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Spin-orbit coupling refers to the interaction of a particle's "spin" motion with its "orbital" motion.
The Spin-orbit coupling Hamiltonian
The magnitude of spin-orbit coupling splitting is measured spectroscopically as
\begin{align*} H_{so} &=\dfrac{1}{2} hcA \left( (l+s)(l+s+1)-l(l+1)-s(s+1)\right) \[4pt] &= \dfrac{1}{2} hcA \left(l^2 +s^2 + ls + sl + l +s -l^2 -l -s^2 -s) \right) \[4pt] &= hcA \textbf{l} \cdot \textbf{s} \end{align*}
The expression can be modified by realizing that $j = l + s$.
$H_{so}=\dfrac{1}{2} hcA \biggr(j(j+1)-l(l+1)-s(s+1)\biggr)$
where $A$ is the magnitude of the spin-orbit coupling in wave numbers. The magnitude of the spin orbit coupling can be calculated in terms of molecule parameters by the substitution
$hcA\,\widehat{L}\cdot\widehat{S}=\dfrac{Z\alpha^2}{2}\dfrac{1}{r^3}\widehat{L}\cdot\widehat{S}$
where $a$ is the fine structure constant ($a = 1/137.037$) and the carrots indicate that $L$ and $S$ are operators. The fine structure constant is a dimensionless constant, $a = \dfrac{e^2}{ác}$. $Z$ is an effective atomic number. The spin orbit coupling splitting can be calculated from
$E_{so}=\int \Psi^*H_{SO}\Psi\,d\tau=\dfrac{Z}{2(137)^2}\int\Psi^*\dfrac{\widehat{L}\cdot\widehat{S}}{r^3}\Psi\,d\tau$
This expression can be recast to give an spin-orbit coupling energy in terms of molecular parameters
$E_{so}=\dfrac{1}{2} \biggr(j(j+1)-l(l+1)-s(s+1)\biggr)=\dfrac{Z}{2(137)^2}\biggr\langle\dfrac{1}{r^3}\biggr\rangle$
where
$\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\int\Psi^*\biggr(\dfrac{1}{r^3}\biggr)\Psi\,d\tau$
We can evaluate this integral explicitly for a given atomic orbital.
For example for Y210 we have
$\Psi_{210}=\dfrac{1}{4\sqrt{2\pi}}\biggr(\dfrac{Z}{a_0}\biggr)^\dfrac{3}{2}\dfrac{Zr}{a_0}e^{-Zr/2a_0}\cos\theta$
so that the integral is
$\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\dfrac{1}{32\pi}\biggr (\dfrac{Z}{a_0}\biggr)^5\int_{0}^{2z}d\phi\int_{0}^{z}\cos^2\theta\sin\theta\,d\theta cos\theta\int_{0}^{\infty}r^2e^{Zr/a_0}\biggr(\dfrac{1}{r^3}\biggr)r^2\,dr$
which integrates to
$\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\dfrac{1}{32\pi}\biggr(\dfrac{Z}{a_0}\biggr)^52\pi\biggr(\dfrac{2}{3}\biggr)\biggr(\dfrac{a_0\,^2}{Z^2}\biggr)=\dfrac{1}{24}\biggr(\dfrac{Z}{a_0}\biggr)^3$
Or $Z^3/24$ in atomic units.
Therefore in atomic units we have
$\biggr\langle\dfrac{1}{r^3}\biggr\rangle=\dfrac{Z^3}{n^3l(l+1/2)(l+1)}$
Therefore, in general the spin-orbit splitting is given by
$E_{so}=\dfrac{Z^4}{2(137)^2n^3}\Biggr(\dfrac{j(j+1)-l(l+1)-s(s+1)}{2l(l+1/2)(l+1)}\Biggr)$
Note that the spin-orbit coupling increases as the fourth power of the effective nuclear charge Z, but only as the third power of the principal quantum number n. This indicates that spin orbit-coupling interactions are significantly larger for atoms that are further down a particular column of the periodic table.
Spin-orbit Coupling
In electronic spectroscopy, an atomic term symbol specifies a certain electronic state of an atom (usually a multi-electron one), by briefing the quantum numbers for the angular momenta of that atom. The form of an atomic term symbol implies Russell-Saunders coupling. Transitions between two different atomic states may be represented using their term symbols, to which certain rules apply.
History
At the beginning, the spectroscopic notation for term symbols was derived from an obsolete system of categorizing spectral lines. In 1885, Johann Balmer, a Swiss mathematician, discovered the Balmer formula for a series of hydrogen emission lines.
$\lambda=B \left(\dfrac{m^2}{m^2-4} \right)$
where
• $B$ is constant, and
• $m$ is an integer greater than 2.
Later it was extended by Johannes Rydberg and Walter Ritz.
Yet this principle could hardly explain the discovery of fine structure, the splitting of spectral lines. In spectroscopy, spectral lines of alkali metals used to be divided into categories: sharp, principal, diffuse and fundamental, based on their fine structures. These categories, or "term series," then became associated with atomic energy levels along with the birth of the old quantum theory. The initials of those categories were employed to mark the atomic orbitals with respect to their azimuthal quantum numbers. The sequence of "s, p, d, f, g, h, i, k..." is known as the spectroscopic notation for atomic orbitals.
By introducing spin as a nature of electrons, the fine structure of alkali spectra became further understood. The term "spin" was first used to describe the rotation of electrons. Later, although electrons have been proved unable to rotate, the word "spin" is reserved and used to describe the property of an electron that involves its intrinsic magnetism. LS coupling was first proposed by Henry Russell and Frederick Saunders in 1923. It perfectly explained the fine structures of hydrogen-like atomic spectra. The format of term symbols was developed in the Russell-Saunders coupling scheme.
Term Symbols
In the Russell-Saunders coupling scheme, term symbols are in the form of 2S+1LJ, where S represents the total spin angular momentum, L specifies the total orbital angular momentum, and J refers to the total angular momentum. In a term symbol, L is always an upper-case from the sequence "s, p, d, f, g, h, i, k...", wherein the first four letters stand for sharp, principal, diffuse and fundamental, and the rest follow in an alphabetical pattern. Note that the letter j is omitted.
Angular momenta of an electron
In today's physics, an electron in a spherically symmetric potential field can be described by four quantum numbers all together, which applies to hydrogen-like atoms only. Yet other atoms may undergo trivial approximations in order to fit in this description. Those quantum numbers each present a conserved property, such as the orbital angular momentum. They are sufficient to distinguish a particular electron within one atom. The term "angular momentum" describes the phenomenon that an electron distributes its position around the nucleus. Yet the underlying quantum mechanics is much more complicated than mere mechanical movement.
• The azimuthal angular momentum: The azimuthal angular momentum (or the orbital angular momentum when describing an electron in an atom) specifies the azimuthal component of the total angular momentum for a particular electron in an atom. The orbital quantum number, l, one of the four quantum numbers of an electron, has been used to represent the azimuthal angular momentum. The value of l is an integer ranging from 0 to n-1, while n is the principal quantum number of the electron. The limits of l came from the solutions of the Schrödinger Equation.
• The intrinsic angular momentum: The intrinsic angular momentum (or the spin) represents the intrinsic property of elementry particles, and the particles made of them. The inherent magnetic momentum of an electron may be explained by its intrinsic angular momentum. In LS-coupling, the spin of an electron can couple with its azimuthal angular momentum. The spin quantum number of an electron has a value of either ½ or -½, which reflects the nature of the electron.
Coupling of the electronic angular momenta
Coupling of angular momenta was first introduced to explain the fine structures of atomic spectra. As for LS coupling, S, L, J and MJ are the four "good" quantum numbers to describe electronic states in lighter atoms. For heavier atoms, jj coupling is more applicable, where J, MJ, ML and Ms are "good" quantum numbers.
LS coupling
LS coupling, also known as Russell-Saunders coupling, assumes that the interaction between an electron's intrinsic angular momentum s and its orbital angular momentum L is small enough to be considered as an perturbation to the electronic Hamiltonian. Such interactions can be derived in a classical way. Let's suppose that the electron goes around the nucleus in a circular orbit, as in Bohr model. Set the electron's velocity to be Ve. The electron experiences a magnetic field B due to the relative movement of the nucleus
$\mathbf{B}=\dfrac{1}{m_ec^2}(\mathbf{E} \times \mathbf{p})= \dfrac{Ze}{4 \pi \epsilon_0m_ec^2r^3} \mathbf{L} \label{2}$
while E is the electric field at the electron due to the nucleus, p the classical monumentum of the electron, and r the distance between the electron and the nucleus.
The electron's spin s brings a magnetic dipole moment μs
$\mathbf \mu_s= \dfrac{-g_{se} \mathbf{s}}{2m_e}( \hat{ \mathbf{L}} \cdot \hat{ \mathbf{s}}) \label{3}$
where gs is the gyromagnetic ratio of an electron. Since the potential energy of the coulumbic attraction between the electron and nucleus is6
$V (r) = \dfrac{-Ze^2}{4πε_0r} \label{4}$
the interaction between μs and B is
$\hat{H}_{so}=\dfrac{g_s}{2m_e^2c^2} \dfrac{Ze^2}{4 \pi \epsilon_{_0}r^2}( \mathbf{L} \cdot \mathbf{s})=\dfrac{g_s}{2m_e^2c^2} \dfrac{ \partial{V}}{ \partial{r}}( \mathbf{L} \cdot \mathbf{s}) \label{5}$
After a correction due to centripetal acceleration10, the interaction has a format of
$\hat{H}_{so} = \dfrac{1}{2 \mu^2c^2} \dfrac{ \partial{V}}{ \partial{r}}( \mathbf{L} \cdot \mathbf{s}) \label{6}$
Thus the coupling energy is
$\langle \psi_{nlm}| \hat{H}_{so}| \psi_{nlm} \rangle \label{7}$
In lighter atoms, the coupling energy is low enough be treated as a first-order perturbation to the total electronic Hamiltonian, hence LS coupling is applicable to them. For a single electron, the spin-orbit coupling angular momentum quantum number j has the following possible values
j = |l-s|, ..., l+s
if the total angular momentum J is defined as J = L + s. The azimuthal counterpart of j is mj, which can be a whole number in the range of [-j, j].
The first-order perturbation to the electronic energy can be deduced so6
$\dfrac{\hbar^2[j(j+1)-l(l+1)-s(s+1)]}{4\mu^2c^2}\int_0^\infty r^2dr \dfrac{1}{r} \dfrac{ \partial{V}}{\partial{r}}R_{nl}^2(r)$
Above is about the spin-orbit coupling of one electron. For many-electron atoms, the idea is similar. The coupling of angular momenta is
$\mathbf{J}=\mathbf{L}+\mathbf{S} \label{9}$
thereby the total angular quantum number
J = |L-S|, ..., L+S
where the total orbital quantum number
$L=\sum\limits_i l_i \label{10}$
and the total spin quantum number
$S=\sum \limits_{i}s_i \label{11}$
While J is still the total angular momentum, L and S are the total orbital angular momentum and the total spin, respectively. The magnetic momentum due to J is
$\mathbf \mu_J=- \dfrac{g_Je \mathbf{J}}{2m_e} \label{12}$
wherein the Landé g factor is
$g_J=1+\dfrac{J(J+1)+S(S+1)-L(L+1)} {2J(J+1)} \label{13}$
supposing the gyromagnetic ratio of an electron is 2.
jj coupling
For heavier atoms, the coupling between the total angular momenta of different electrons is more significant, causing the fine structures not to be "fine" any more. Therefore the coupling term can no more be considered as a perturbation to the electronic Hamiltonian, so that jj coupling is a better way to quantize the electron energy states and levels.
For each electron, the quantum number j = l + s. For the whole atom, the total angular momentum quantum number
$J= \sum \limits_i j_i \label{14}$
Term symbols for an Electron Configuration
Term symbols usually represent electronic states in the Russell-Saunders coupling scheme, where a typical atomic term symbol consists of the spin multiplicity, the symmetry label and the total angular momentum of the atom. They have the format of
$\large ^{2S+1}L_J$
such as 3D2, where S = 1, L = 2, and J = 2.
Here is a commonly used method to determine term symbols for an electron configuration. It requires a table of possibilities of different "micro states," which happened to be called "Slater's table".6 Each row of the table represents a total magnetic quantum number, while each column does a total spin. Using this table we can pick out the possible electronic states easily since all terms are concentric rectangles on the table.
The method of using a table to count possible "microstates" has been developed so long ago and honed by so many scientists and educators that it is hard to accredit a single person. Let's take the electronic configuration of d3 as an example. In the Slater's table, each cell contains the number of ways to assign the three electrons quantum numbers according to the MS and ML values. These assignments follow Pauli's exclusion law. The figure below shows an example to find out how many ways to assign quantum numbers to d3 electrons when ML = 3 and MS = -1/2.
MS
-3/2 -1/2 1/2 3/2
ML
5 0 1 1 0
4 0 2 2 0
3 1 4 4 1
2 1 6 6 1
1 2 8 8 2
0 2 8 8 2
-1 2 8 8 2
-2 1 6 6 1
-3 1 4 4 1
-4 0 2 2 0
-5 0 1 1 0
Now we start to subtract term symbols from this table. First there is a 2H state. And now it is subtracted from the table.
MS
-3/2 -1/2 1/2 3/2
ML
5 0 0 0 0
4 0 1 1 0
3 1 3 3 1
2 1 5 5 1
1 2 7 7 2
0 2 7 7 2
-1 2 7 7 2
-2 1 5 5 1
-3 1 3 3 1
-4 0 1 1 0
-5 0 0 0 0
And now is a 4F state. After being subtracted by 4F, the table becomes
MS
-3/2 -1/2 1/2 3/2
ML
5 0 0 0 0
4 0 1 1 0
3 0 2 2 0
2 0 4 4 0
1 1 6 6 1
0 1 6 6 1
-1 1 6 6 1
-2 0 4 4 0
-3 0 2 2 0
-4 0 1 1 0
-5 0 0 0 0
And now 2G.
MS
-3/2 -1/2 1/2 3/2
ML
5 0 0 0 0
4 0 0 0 0
3 0 1 1 0
2 0 3 3 0
1 1 5 5 1
0 1 5 5 1
-1 1 5 5 1
-2 0 3 3 0
-3 0 1 1 0
-4 0 0 0 0
-5 0 0 0 0
Now 2F.
MS
-3/2 -1/2 1/2 3/2
ML
5 0 0 0 0
4 0 0 0 0
3 0 0 0 0
2 0 2 2 0
1 1 4 4 1
0 1 4 4 1
-1 1 4 4 1
-2 0 2 2 0
-3 0 0 0 0
-4 0 0 0 0
-5 0 0 0 0
Here in the table are two 2D states.
MS
-3/2 -1/2 1/2 3/2
ML
5 0 0 0 0
4 0 0 0 0
3 0 0 0 0
2 0 0 0 0
1 1 2 2 1
0 1 2 2 1
-1 1 2 2 1
-2 0 0 0 0
-3 0 0 0 0
-4 0 0 0 0
-5 0 0 0 0
4P.
MS
-3/2 -1/2 1/2 3/2
ML
5 0 0 0 0
4 0 0 0 0
3 0 0 0 0
2 0 0 0 0
1 0 1 1 0
0 0 1 1 0
-1 0 1 1 0
-2 0 0 0 0
-3 0 0 0 0
-4 0 0 0 0
-5 0 0 0 0
And the final deducted state is 2P. So in total the possible states for a d3 configuration are 4F, 4P, 2H, 2G, 2F, 2D, 2D and 2P. Taken J into consideration, the possible states are:
$^4F_2\; ^4F_3 \; ^4F_4\; ^4P_0\; ^4P_1\; ^4P_2\; ^2H_{\frac{9}{2}} \; ^2H_{\frac{11}{2}} \; ^2G_{\frac{9}{2}} \; ^2G_{\frac{7}{2}} \; ^2F_{\frac{7}{2}}\; ^2F_{\frac{5}{2}} \; ^2D_{\frac{5}{2}} \; ^2D_{\frac{5}{2}} \; ^2D_{\frac{3}{2}} \; ^2D_{\frac{3}{2}}\; ^2P_{\frac{3}{2}} \; ^2P_{\frac{3}{2}}$
For lighter atoms before or among the first-row transition metals, this method works well.
Using group theory to determine term symbols
Another method is to use direct products in group theory to quickly work out possible term symbols for a certain electronic configuration. Basically, both electrons and holes are taken into consideration, which naturally results in the same term symbols for complementary configurations like p2 vs p4. Electrons are categorized by spin, therefore divided into two categories, α and β, as are holes: α stands for +1, and βstands for -1, or vice versa. Term symbols of different possible configurations within one category are given. The term symbols for the total electronic configuration are derived from direct products of term symbols for different categories of electrons. For the p3 configuration, for example, the possible combinations of different categories are eα3, eβ3, eα2eβ and eαeβ2. The first two combinations were assigned the partial term of S.6 As eα2 and eβ were given an P symbol, the combination of them gives their direct product
P × P = S + [P] + D.
The direct product for eα and eβ2 is also
P × P = S + [P] + D.
Considering the degeneracy, eventually the term symbols for p3 configuration are 4S, 2D and 2P.
There is a specially modified version of this method for atoms with 2 unpaired electrons. The only step gives the direct product of the symmetries of the two orbitals. The degeneracy is still determined by Pauli's exclusion.
Determining the ground state
In general, states with a greater degeneracy have a lower energy. For one configuration, the level with the largest S, which has the largest spin degeneracy, has the lowest energy. If two levels have the same S value, then the one with the larger L (and also the larger orbital degeneracy) have the lower energy. If the electrons in the subshell are fewer than half-filled, the ground state should have the smallest value of J, otherwise the ground state has the greatest value of J.
Electronic transitions
Electrons of an atom may undergo certain transitions which may have strong or weak intensities. There are rules about which transitions should be strong and which should be weak. Usually an electronic transition is excited by heat or radiation. Electronic states can be interpreted by solutions of Schrödinger's equation. Those solutions have certain symmetries, which are a factor of whether transitions will be allowed or not. The transition may be triggered by an electric dipole momentum, a magnetic dipole momentum, and so on. These triggers are transition operators. The most common and usually most intense transitions occur in an electric dipolar field, so the selection rules are
1. ΔL = 0, ±1 except L = 0 ‡ L' = 0
2. ΔS = 0
3. ΔJ = 0; ±1 except J = 0 ‡ J' = 0
where a double dagger means not combinable. For jj coupling, only the third rule applies with an addition rule: Δj = 0; ±1.
Problems
1. Potassium has the electronic configuration of [Ar]4s1, what are the possible term symbols of a neutral K atom?
2. What is the ground state of Cr2+? Specify the value of J.
3. Why does not the spin selection rule apply for electronic transitions under jj coupling?
4. Why do different term symbols appear as rectangles on the Slater's table?
5. Tanabe-Sugano diagrams tell us the order of energy levels in a complex, usually the order of d electron levels. Look up for the possible term symbols for the d2 configuration, and determine how many fundamentals (transition from the ground state to an excited state) may occur when Δ = 0 i.e. minor field due to ligands.
Answers
1. 2S.
2. 5D.
3. Under jj coupling the spin quantum number is not a "good" quantum number any more, which means it cannot describe and differentiate electronic states properly.
4. A certain term symbol represents a certain value of L as well as of S, which implies limited possible values of ML and Ms. Following Pauli's exclusion rule, there must be only one possible way to assign the configuration a certain value of ML and Ms.
5. No appropriate answer.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Spin-orbit_Coupling/Atomic_Term_Symbols.txt
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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
The Russell Saunders Coupling Scheme
Review of Quantum Numbers
Electrons in an atom reside in shells characterized by a particular value of n, the Principal Quantum Number. Within each shell an electron can occupy an orbital which is further characterized by an Orbital Quantum Number, \(l\), where \(l\) can take all values in the range:
\[l = 0, 1, 2, 3, ... , (n-1),\]
traditionally termed s, p, d, f, etc. orbitals.
Each orbital has a characteristic shape reflecting the motion of the electron in that particular orbital, this motion being characterized by an angular momentum that reflects the angular velocity of the electron moving in its orbital. A quantum mechanics approach to determining the energy of electrons in an element or ion is based on the results obtained by solving the Schrödinger Wave Equation for the H-atom. The various solutions for the different energy states are characterized by the three quantum numbers, n, l and ml.
• ml is a subset of l, where the allowable values are: ml = l, l-1, l-2, ..... 1, 0, -1, ....... , -(l-2), -(l-1), -l.
• There are thus (2l +1) values of ml for each l value, i.e. one s orbital (l = 0), three p orbitals (l = 1), five d orbitals (l = 2), etc.
• There is a fourth quantum number, ms, that identifies the orientation of the spin of one electron relative to those of other electrons in the system. A single electron in free space has a fundamental property associated with it called spin, arising from the spinning of an asymmetrical charge distribution about its own axis. Like an electron moving in its orbital around a nucleus, the electron spinning about its axis has associated with its motion a well defined angular momentum. The value of ms is either + ½ or - ½.
In summary then, each electron in an orbital is characterized by four quantum numbers (Table 1).
Table 1: Quantum Numbers
symbol description range of values
n Principal Quantum Number - largely governs size of orbital and its energy 1,2,3 etc
l Azimuthal/Orbital Quantum Number - largely determines shape of subshell
0 for s orbital, 1 for p orbital etc
(0 ≤ l ≤ n-1) for n = 3 then l = 0, 1, 2 (s, p, d)
ml Magnetic Quantum Number - orientation of subshell's shape for example px with py and pz lml ≥ -l for l = 2, then ml = 2, 1, 0, -1, -2
ms Spin Quantum Number either + ½ or - ½ for single electron
Russell Saunders coupling
The ways in which the angular momenta associated with the orbital and spin motions in many-electron-atoms can be combined together are many and varied. In spite of this seeming complexity, the results are frequently readily determined for simple atom systems and are used to characterize the electronic states of atoms. The interactions that can occur are of three types.
• spin-spin coupling
• orbit-orbit coupling
• spin-orbit coupling
There are two principal coupling schemes used:
• Russell-Saunders (or L - S) coupling
• and jj coupling.
In the Russell Saunders scheme (named after Henry Norris Russell, 1877-1957 a Princeton Astronomer and Frederick Albert Saunders, 1875-1963 a Harvard Physicist and published in Astrophysics Journal, 61, 38, 1925) it is assumed that:
spin-spin coupling > orbit-orbit coupling > spin-orbit coupling.
This is found to give a good approximation for first row transition series where spin-orbit (J) coupling can generally be ignored, however for elements with atomic number greater than thirty, spin-orbit coupling becomes more significant and the j-j coupling scheme is used.
Spin-Spin Coupling
S - the resultant spin quantum number for a system of electrons. The overall spin S arises from adding the individual ms together and is as a result of coupling of spin quantum numbers for the separate electrons.
Orbit-Orbit Coupling
L - the total orbital angular momentum quantum number defines the energy state for a system of electrons. These states or term letters are represented as follows:
Total Orbital Momentum
L 0 1 2 3 4 5
S P D F G H
Spin-Orbit Coupling
Coupling occurs between the resultant spin and orbital momenta of an electron which gives rise to J the total angular momentum quantum number. Multiplicity occurs when several levels are close together and is given by the formula (2S+1). The Russell Saunders term symbol that results from these considerations is given by:
\[ \large ^{(2S+1)}L\]
Configuration
S= + ½, hence (2S+1) = 2
L=2 and the Ground Term is written as 2D
The Russell Saunders term symbols for the other free ion configurations are given in the Table below.
Terms for 3dn free ion configurations
Configuration # of quantum states # of energy levels Ground Term Excited Terms
d1,d9 10 1 2D -
d2,d8 45 5 3F 3P, 1G,1D,1S
d3,d7 120 8 4F 4P, 2H, 2G, 2F, 2 x 2D, 2P
d4,d6 210 16 5D 3H, 3G, 2 x 3F, 3D, 2 x 3P, 1I, 2 x 1G, 1F, 2 x 1D, 2 x 1S
d5 252 16 6S 4G, 4F, 4D, 4P, 2I, 2H, 2 x 2G, 2 x 2F, 3 x 2D, 2P, 2S
Note that dn gives the same terms as d10-n
Hund's Rules
The Ground Terms are deduced by using Hund's Rules. The two rules are:
1. The Ground Term will have the maximum multiplicity
2. If there is more than 1 Term with maximum multipicity, then the Ground Term will have the largest value of L.
A simple graphical method for determining just the ground term alone for the free-ions uses a "fill in the boxes" arrangement.
dn 2 1 0 -1 -2 L S Ground Term
d1 2 1/2
2D
d2 3 1
3F
d3 3 3/2
4F
d4 2 2
5D
d5 0 5/2
6S
d6 2 2
5D
d7 3 3/2
4F
d8 3 1
3F
d9 2 1/2
2D
To calculate S, simply sum the unpaired electrons using a value of ½ for each. To calculate L, use the labels for each column to determine the value of L for that box, then add all the individual box values together.
Configuration
For a d7 configuration, then:
• in the +2 box are 2 electrons, so L for that box is 2*2= 4
• in the +1 box are 2 electrons, so L for that box is 1*2= 2
• in the 0 box is 1 electron, L is 0
• in the -1 box is 1 electron, L is -1*1= -1
• in the -2 box is 1 electron, L is -2*1= -2
Total value of L is therefore +4 +2 +0 -1 -2 or L=3.
Note that for 5 electrons with 1 electron in each box then the total value of L is 0. This is why L for a d1 configuration is the same as for a d6.
The other thing to note is the idea of the "hole" approach. A d1 configuration can be treated as similar to a d9 configuration. In the first case there is 1 electron and in the latter there is an absence of an electron i.e., a hole.
The overall result shown in the Table above is that:
• 4 configurations (d1, d4, d6, d9) give rise to D ground terms,
• 4 configurations (d2, d3, d7, d8) give rise to F ground terms
• and the d5 configuration gives an S ground term.
The Crystal Field Splitting of Russell-Saunders terms
The effect of a crystal field on the different orbitals (s, p, d, etc.) will result in splitting into subsets of different energies, depending on whether they are in an octahedral or tetrahedral environment. The magnitude of the d orbital splitting is generally represented as a fraction of Δoct or 10Dq.
The ground term energies for free ions are also affected by the influence of a crystal field and an analogy is made between orbitals and ground terms that are related due to the angular parts of their electron distribution. The effect of a crystal field on different orbitals in an octahedral field environment will cause the d orbitals to split to give t2g and eg subsets and the D ground term states into T2g and Eg, (where upper case is used to denote states and lower case orbitals). f orbitals are split to give subsets known as t1g, t2g and a2g. By analogy, the F ground term when split by a crystal field will give states known as T1g, T2g, and A2g.
Note that it is important to recognize that the F ground term here refers to states arising from d orbitals and not f orbitals and depending on whether it is in an octahedral or tetrahedral environment the lowest term can be either A2g or T1g.
The Crystal Field Splitting of Russell-Saunders terms
in high spin octahedral crystal fields.
Russell-Saunders Terms Crystal Field Components
S (1)
A1g
P (3)
T1g
D (5)
Eg , T2g
F (7)
A2g , T1g , T2g
G (9)
A1g , Eg , T1g , T2g
H (11)
Eg , 2 x T1g , T2g
I (13)
A1g , A2g , Eg , T1g , 2 x T2g
Note that, for simplicity, spin multiplicities are not included in the table since they remain the same for each term. The table above shows that the Mulliken symmetry labels, developed for atomic and molecular orbitals, have been applied to these states but for this purpose they are written in CAPITAL LETTERS.
Mulliken Symbols
Mulliken Symbol
for atomic and molecular orbitals
Explanation
a Non-degenerate orbital; symmetric to principal Cn
b Non-degenerate orbital; unsymmetric to principal Cn
e Doubly degenerate orbital
t Triply degenerate orbital
(subscript) g Symmetric with respect to center of inversion
(subscript) u Unsymmetric with respect to center of inversion
(subscript) 1 Symmetric with respect to C2 perp. to principal Cn
(subscript) 2 Unsymmetric with respect to C2 perp. to principal Cn
(superscript) ' Symmetric with respect to sh
(superscript) " Unsymmetric with respect to sh
For splitting in a tetrahedral crystal field the components are similar, except that the symmetry label g (gerade) is absent. The ground term for first-row transition metal ions is either D, F or S which in high spin octahedral fields gives rise to A, E or T states. This means that the states are either non-degenerate, doubly degenerate or triply degenerate.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Spin-orbit_Coupling/Molecular_Term_Symbols.txt
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Two-photon absorption is one of a variety of two-photon processes. In this specific process, two photons are absorbed by a sample simultaneously. Neither photon is at resonance with the available energy states of the system, however, the combined frequency of the photons is at resonance with an energy state.
Introduction
Two-photon absorption may seem very similar to non-resonance Raman, and was in fact predicted due to this phenomena. In both cases, a non-resonant photon is used for excitation. However, in the absorption case a secondary non-resonant photon is used for excitation as well, while in Raman a second non-resonant photon is emitted. For Raman this results in occupation of an energy state at the difference of the frequencies of the absorbed and emitted photon. However, in two-photon absorption this results in the occupation of an energy state at the sum of the frequencies of the absorbed photons. The basic process is illustrated below in Figure 1.
Two-photon absorption is not a feature of a specific type of spectroscopy. It can be utilized in any type of spectroscopy including IR, NMR, XAS, and UV-VIS. However, it will result in a change of selection rules for many of these spectroscopies. This means that the peak shapes should remain the same, but the intensities should be significantly smaller due to the result of the second order perturbation.
History
In her 1931 dissertation, Maria Goeppert-Mayer postulated the existence of two-photon absorption for the first time. She says that she based the derivation of this phenomena off of Kramer and Heisenberg's derivation of the probability of two-photon emission which uses Dirac's dispersion theory for calculation.
The invention of the laser had a great impact on the field of two-photon spectroscopy because of the necessity of a high-intensity electromagnetic field to induce transitions. Although, lasers were not formally invented until 1969, Bell Labs was testing masers in 1958, which were only capable of short pulses of intense electromagnetic radiation. In 1961 Kaiser and Garret reported the first two-photon absorption of a compound. They used the new laser technology to excite CaF$_{2}$Eu$^{2+}$ with both red and blue light to induce a two photon transition.
General Perturbation Method
The following constants, variables, and operators are used in this section:
$\hbar$ Reduced Planck's constant = 1.058x10-34 Js
e Elementary electric charge = 1.602x10-19 Coulombs
t time
$\omega$ angular frequency
$\hat{H}_{0}$ $\frac{\hat{p}^{2}}{2m}$
$\hat{p}$ $-i\hbar\nabla$
First we must take a look at how the atomic system interacts with any perturbation
$\hat{H} = \hat{H}_{0} + \hat{H}_{1}\hspace{1cm}\label{1}$
If we are only going to look at the effects of the electric dipole then:
$\hat{H} = \hat{H}_{0} + \hat{H}_{\mu_{e}}\hspace{1cm}\label{2}$
Now we must consider how our states will be evolving in time because the Electric Dipole operator is time-dependent. According to the time-dependent Schrödinger equation:
$\Psi_{\textrm{i}}(t) = \textrm{exp}\left (-\frac{i\hat{H}(t-t_{0})}{\hbar}\right )\Psi_{\textrm{i}}(t_{0})\hspace{1cm} \label{3}$
We have described our ground state as it is moving in time so now we can look at our final state which we can only describe as being at a set energy value:
$\hat{H}_{0}\Psi_{f} = \hbar\omega_{f}\Psi_{f}\hspace{1cm} \label{4}$
Now that we have described both the final and the initial states.
We can find the probability of such a transition by finding the magnitude of the projection of the intial state upon the final state, otherwise known as the magnitude of the inner product.
$|\left <\Psi_{f}| \Psi_{i}(t) \right >|^{2} =$
$\left | \left <\Psi_{f} \left | \textrm{exp}\left (-\frac{i\hat{H}(t-t_{0})}{\hbar}\right )\right | \Psi_{\textrm{i}}(t_{0}) \right > \right | ^{2}\hspace{1cm} \label{5}$
Now that we've found the general formula for the probability of our transitions in a time-dependent system, we must apply the perturbation from earlier. To do this we use the following identity:
$\textrm{exp} \left (\frac{i\hat{H}_{0}t}{\hbar}\right )\hat{H}_{\mu_{e}}\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) = i\hbar\frac{\textrm{d}}{\textrm{d}t}\left [\textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right )\right ] \hspace{1cm}\label{6}$
The issue now is that the electric dipole Hamiltonian is not in the exponential form, this prevents us from simply solving for the total Hamiltonian exponential. To solve this we integrate with respect to time, which nullifies the derivative.
$\int^{t}_{t_{0}} \textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\hat{H}_{\mu_{e}}\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) \textrm{d}t_{1}$
$= i\hbar\left [\textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) -\textrm{exp}\left (\frac{i\hat{H}_{0}t_{0}}{\hbar}\right )\textrm{exp}\left (-\frac{i\hat{H}t_{0}}{\hbar}\right )\right ]\hspace{1cm} \label{7}$
This equation can be simplified and solved for using the approximation that the initial time was well before the incident of interaction with the perturbation. This allows us to assume that the electric dipole Hamiltonian would yield 0 at the initial time. This allows for the following iterative equation.
$\textrm{exp}\left (\frac{-i\hat{H}t}{\hbar} \right ) = \textrm{exp} \left ( -\frac{i\hat{H}_{0}t}{\hbar}\right ) \times \left [ 1 - \frac{i}{\hbar}\int^{t}_{-\infty} \textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\hat{H}_{\mu_{e}}\textrm{exp}(\epsilon t_{1})\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) \textrm{d}t_{1} \right ]\hspace{1cm} \label{8}$
Now for the purposes of two-photon absorption we will need to find the 2nd order perturbation. Thus we must find the second iteration of this equation. Where epsilon is a very small number, we come to the following equation for both the first and second order contribution:
$\left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \frac{1}{\hbar}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left ( \left< \Psi_{f} \left | \hat{H}_{\mu_{e}} \right | \Psi_{i} \right > + \frac{1}{\hbar}\sum_{l}\frac{ \left< \Psi_{f} \left | \hat{H}_{\mu_{e}} \right | \Psi_{l} \right > \left< \Psi_{l} \left | \hat{H}_{\mu_{e}} \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right )\hspace{1cm} \label{9}$
Electric Dipole Contribution
The following constants, variables and operators are introduced in this section:
$\xi$ Amplitude of Electric field
$\hat{\mu}$ Dipole operator = $e\sum_{i=1}^{N}q_{i}\hat{r}_{i}$
For the perturbation method described in the previous section the use of a single iteration accurately describes the transitions between a two levels. This describes the results of Fermi's Golden Rule. However, in order to describe these two-photon processes which transgress through more than one transition, we must look at the second iteration within the perturbation.
$\hat{H}_{\mu_{e}} = e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \hspace{1cm} \label{10}$
Where
$\hat{E_{n}} = \xi_{n}\textrm{cos}(\omega_{n} t)\hspace{1cm} \label{11}$
Thus
$\hat{H}_{\mu_{e}} = e\hat{\mu}(\xi_{1}\textrm{cos}(\omega_{1} t) + \xi_{2}\textrm{cos}(\omega_{2} t))\hspace{1cm} \label{12}$
Substituting this into the second iteration gives us:
$\left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > =$
$\frac{1}{\hbar}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left ( \left< \Psi_{f} \left |e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \right | \Psi_{i} \right > + \frac{1}{\hbar}\sum_{l}\frac{ \left< \Psi_{f} \left | e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right )\hspace{1cm} \label{13}$
Looking at this equation we see that the first term is that corresponding to single photon transitions. However, because we are looking for two-photon transitions we can ignore this term and focus on the second term. This second term has a summation to distinguish between the different permutations of arranging the perturbations and achieve the same final state but it includes the interaction of two photons from the same light source which we do not need to consider at the moment. If we eliminate the terms of the sum responsible for these guys we should find the expression for the two permutations of two-photon transitions from our two light sources. Now let us implement these changes.
$\left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > =$
$\frac{1}{\hbar^{2}}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left (\frac{ \left< \Psi_{f} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{i} \right > +\left< \Psi_{f} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right ) \hspace{1cm} \label{14}$
This equation describes the electric dipole contribution towards two-photon transitions.
Selection Rules
The following constants, variables and operators are introduced in this section:
$l$ Angular Momentum Quantum Number
$m_{l}$ Magnetic Quantum Number
$Y$ Spherical Harmonic
$\hat{Z}$
Position operator = $z$
$\hat{P}_{z}$ Momentum Operator in the Z direction $-i\hbar\frac{\textrm{d}}{\textrm{d}z}$
In order to take a thorough look at the selection rules for the two-photon system induced by the electric field contribution we must look qualitatively at the results of the equation for the probability of a transition. The Hamiltonian presented in the previous section for the electric dipole contribution is the result of a gauge transformation of the original electric dipole contribution. For the purposes of finding the electric dipole selection rules we shall revert back to the original form.
$\hat{H}_{\mu_{e}}(t) = \frac{qE}{m\omega}\hat{P}_{z}\textrm{sin }\omega t \hspace{1cm}\label{15}$
We will consider the effect of the sinusoid part of this equation later in this section. For now let us track the selection rules due only to the operator $\hat{P}_{z}$.
$\left [ \hat{Z}, \hat{H}_{0} \right ] = i\hbar\frac{\partial \hat{H}_{0}}{\partial \hat{P}_{z}} = i\hbar\frac{\hat{P}_{z}}{m} \hspace{1cm} \label{16}$
This can be shown to be true by simply working out the commutator and leads to the following statement
$\left < \Psi_{f} \left | \left [\hat{Z},\hat{H}_{0} \right ] \right | \Psi_{i} \right > = \left < \Psi_{f} \left | \hat{Z}\hat{H}_{0} -\hat{H}_{0}\hat{Z} \right | \Psi_{i} \right > \hspace{1cm} \label{17}$
$= -(E_{f} - E_{i})\left < \Psi_{f} \left | \hat{Z} \right | \Psi_{i} \right > = \frac{i\hbar}{m}\left < \Psi_{f} \left | \hat{P}_{z} \right | \Psi_{i} \right > \hspace{1cm} \label{18}$
This shows the result of an electric dipole induced transition for a two level system. However, we need to slightly change things in order to incorporate the three levels that will exist as a result of the two-photon transition as well as the fact that we will have two different perturbations. Thus the Hamiltonian for the two-photon system will be:
$\hat{H}_{\mu_{e}}(t) = \frac{qE_{1}}{m\omega_{1}}\hat{P}_{z}\textrm{sin }\omega_{1} t + \frac{qE_{2}}{m\omega_{2}}\hat{P}_{z}\textrm{sin }\omega_{2} t \hspace{1cm}\label{19}$
Similar to before we can ignore the sinusoid part for later in this section. If we do this we instead come to the result:
$(E_{f}-E_{l})(E_{l}-E_{i})\left < \Psi_{f} \left | \hat{Z} \right | \Psi_{l} \right > \left < \Psi_{l} \left | \hat{Z} \right | \Psi_{i} \right > \hspace{1cm}\label{20}$
Whereas the momentum operator is difficult to find the selection rules for, the position operator is fairly straight forward. This is because the variable $z$, which is the result of the operator $\hat{Z}$, is equal to the following wavefunction.
$z = \sqrt{\frac{4\pi}{3}}rY^{0}_{1}(\theta) \hspace{1cm}\label{21}$
This works under the assumption that the light is polarized in the $z$ direction but we can also polarize the light in multiple directions as there are two photons. If this is the case we need to expand to define both $x$ and $y$ in terms of an angular function. We'll only examine the x case here because it gives the same results.
$x = \sqrt{\frac{2\pi}{3}}r\left ( Y^{-1}_{1} - Y^{1}_{1} \right ) \hspace{1cm} \label{22}$
While for our case we have two position operators acting on different kets, however, that has no effect on the ket because multiplication is commutative. This means the angular part of the inner product will result in the integral for the double $z$ and double $x$ case respectively. Because all of the results retain the same $l$ values that will not be addressed until after we first look at the $m$ values.
$\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) Y^{0}_{1}(\theta)Y^{m_{l}}_{l_{l}}(\theta,\phi)Y^{m_{l}\ast}_{l_{l}}(\theta,\phi)Y^{0}_{1}(\theta)Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{23}$
$\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) \left ( Y^{-1}_{1} - Y^{1}_{1} \right )Y^{m_{l}}_{l_{l}}(\theta,\phi)Y^{m_{l}\ast}_{l_{l}}(\theta,\phi)\left ( Y^{-1}_{1} - Y^{1}_{1} \right )Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{24}$
The intermediate or virtual state angular functions have no effect since they are complex conjugates of each other, their $l$ and $m$ values will annihilate each other. Thus the integral we need to be concerned with is:
$\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) Y^{0}_{1}(\theta)Y^{0}_{1}(\theta)Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{25})$
$\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) \left ( Y^{-1}_{1} - Y^{1}_{1} \right )\left ( Y^{-1}_{1} - Y^{1}_{1} \right )Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{26}$
Because the basis to the selection rule is that only a sum of $m$ values that results in 0 will result in an even function to integrate over, we can now determine a range for $m$ values. For purely $Z$ polarized two-photon absorption the sum of the $m$ values without the initial and final states is still 0, thus the $\Delta m$ must remain 0. However, for the inclusion of $Z$ and $X$ or $Y$. the sum without the initial and final states will be $\pm 1$. Thus in order to retain a total sum of 0 the $\Delta m$ must be $\pm 1$. Finally for the pure $X$, pure $Y$, and $XY$ cases the sum from the operators can results in $\pm 2$ or 0. Therefore, the $\Delta m=\pm 2, 0$ to negate the operators.
Now to talk about the $l$ values. The selection rule that governs them is that we need the sum of all the $l$ values to be even. Complex conjugate $l$ values are negative, the same holds true for the $m$ values. Thus because each operator is has an $l = 1$ and we now have two operators thus we can have the following possible changes to result in even sums.
$\Delta l = 0, \pm 2$
Now the rationality behind the change in the selection rules will be discussed. Each photon has a unit of momentum with it. During a single photon excitation that momentum must be transferred due to conservation of momentum. Thus if it destructively interferes with the electron it can lower the total momentum by one unit and if it constructively interferes with the electron is can increase the total momentum by one unit. However, for the two photon case, we now have to consider not only the ways in which the photon and electron will interfere but also how the photons will interfere with each other. This results in a doubly constructive interference state with an increase of 2 units, one constructive and one destructive interference state that results in no change and a doubly destructive interference state that results in a decrease of 2 units.
Inversion Selection Rules
If we take a look at the equation presented at the end of the electric dipole section, it is a simple analysis to derive the basic selection rules for electric dipole induced two-photon transitions. However, we do not need to consider the constant values to determine selection rules as they only determine the intensity of the signal, not the existence of the peak. Where C is a constant:
$\left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > =$
$C \left ( \left< \Psi_{f} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{i} \right > +\left< \Psi_{f} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{i} \right > \right ) \hspace{1cm} \label{27}$
Although for one-photon transitions the dipole moment is of considerable importance. We notice that because both terms contain two dipole moments, regardless of whether they are odd or even, their combination will still become even. Furthermore, we can see that because the wave part of the electric field is an even function, that this too will preserve parity of the transition in both terms, even if it were an odd function as each term contains two of such functions. Thus the result is that as long as the final and initial state conserve parity (inversion symmetry or antisymmetry is maintained) the transition will be allowed. This is different from one-photon processes where parity conservation is forbidden and makes two-photon spectroscopy especially useful.
Magnetic Dipole Contribution
The following constants, variables, and operators are introduced in this section:
$\hat{L_{x}}$ Angular momentum Operator $i\hbar\left (\textrm{sin }\phi\frac{\partial}{\partial\theta} + \textrm{cot }\theta\textrm{cos }\phi\frac{\partial}{\partial \phi}\right )$
$\hat{S}_{x}$ Spin momentum Operator $\frac{\hbar}{2}\sigma_{x}$
$\sigma_{x}$ Pauli Matrix
B Strength of Magnetic Field
The magnetic dipole contribution is determined by the Magnetic dipole Hamiltonian:
$\hat{H}_{DM} = -\frac{q}{2m}(\hat{L_{x}} + 2\hat{S_{x}})B\textrm{cos}(\omega t) \hspace{1cm} \label{28}$
where B is the strength of the magnetic field. So for the two photon case we have:
$\hat{H}_{DM} = -\frac{q}{2m}\left [(\hat{L_{x}} + 2\hat{S_{x}})B_{1}\textrm{cos}(\omega_{1} t) +(\hat{L_{x}} + 2\hat{S_{x}})B_{2}\textrm{cos}(\omega_{2} t)\right ]\hspace{1cm} \label{29}$
Be aware that the photons do not need to be from the same polarization as they are here. One could be polarized in the X direction while the other is polarized in the Z direction. We will cover these situations in the derivation of the selection rules.
If we similarly substitute this Hamiltonian into the second iteration for perturbation we get:
$\left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \frac{1}{\hbar}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \Bigg( \left< \Psi_{f} \left |-\frac{q}{2m}(\hat{L_{x}} + 2\hat{S_{x}})\left [B_{1}\textrm{cos}(\omega_{1} t) +B_{2}\textrm{cos}(\omega_{2} t)\right ] \right | \Psi_{i} \right > +$
$\small{\sum_{l}\frac{ \left< \Psi_{f} \left |-\frac{q}{2m}(\hat{L}_{x} + 2\hat{S}_{x})\left [B_{1}\textrm{cos}(\omega_{1} t) +B_{2}\textrm{cos}(\omega_{2} t)\right ] \right | \Psi_{l} \right> \left<\Psi_{l} \left |-\frac{q}{2m}(\hat{L}_{x} + 2\hat{S}_{x})\left [B_{1}\textrm{cos}(\omega_{1} t) +B_{2}\textrm{cos}(\omega_{2} t)\right ] \right | \Psi_{i} \right >}{\hbar(\omega_{i} - \omega_{l})} \Bigg)} \hspace{1cm}\label{30}$
We can ignore the first term of this equation as it corresponds to the magnetic dipole contribution of the single photon transitions. From the summation we can ignore the squared terms as well because they represent the two photon transitions due to the beam interfering with itself. While that does exist and may be relevant for some spectroscopic work, here we are looking at the two-photon absorption from two light sources.
$\left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > =$
$\small{\frac{q^{2}\textrm{cos }(\omega_{1}t)\textrm{cos }(\omega_{2} t)}{4m^{2}\hbar^{2}}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left ( \frac{ \left< \Psi_{f} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{l} \right > \left< \Psi_{l} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{i} \right > +\left< \Psi_{f} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{l} \right > \left< \Psi_{l} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right )} \hspace{1cm} \label{31}$
Selection Rules
Just as before, our selection rules stem from the magnetic dipole Hamiltonian:
$\hat{H}_{DM} = -\frac{q}{2m}(\hat{L_{x}} + 2\hat{S_{x}})B\textrm{cos}(\omega t) \hspace{1cm} \label{32}$
Here, as before, the sinusoid part is even and thus will have no effect on the selection rules. We must look at the operators to determine the selection rules.
Because neither $\hat{L}_{x}$ or $\hat{S}_{x}$ have any effect on the value of $l$ the selection rule for $\Delta l$ is 0. However both $\hat{L}_{x}$ and $\hat{S}_{x}$ can change their corresponding $m_{l}$ and $m_{s}$ values by one unit. For $m_{s}$ this corresponds to changing the spin of the electron. If the magnetic field is perpendicular to the spin of the system, then the $\Delta m_{s}$ can also have a value of 0. If we extrapolate this to a two photon system then the same $m_{s}$ values should still apply given that even if it is changed by 1 by the first photon, it can change back by -1 or remain in the $+\frac{1}{2}$ state as those are the only two options. Therefore, we should expect the same selection rules for $\Delta m_{s} = 0, \pm1$ in a two-photon system.
Modern Research Interests
Two-photon absorption has traditionally been a large part of the spectroscopy field. However, it was only within the past decade that biomedical applications have returned it to the spotlight. Two-photon absorption, and potentially all nonlinear absorptions, allow for unprecedented depth in medical imaging technology. Traditional single photon methods result in an enormous amount of scattering from the biological tissue samples. However, these nonlinear optics allow for the assignment of the scattered photons to their origins.
Much of the research in this area is focused on maximizing the depth and clarity of the signal for this potentially non-intrusive medical procedure. Several parameters have been looked at including excitation wavelength, beam size, pulse width, and pulse frequency. As of now the largest depths achievable are around 1 mm.
Problems
1. Derive the expression for the quadrupole contribution to the probability of a two-photon transition.
2. Find the third iteration of the Hamiltonian from the Perturbation adjustment using the iterative equation.
3. Use this third iteration to find the quantum mechanical selection rules for the electric dipole contribution to a three-photon process.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Two-photon_absorption.txt
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• Electromagnetic Radiation
• Introduction to Spectroscopy
• Lineshape Functions
Four different mathematical descriptions of the lineshape for an absorptive transition are discussed below. These transitions may involve electronic, rotational, or vibrational (i.e. visible, microwave or infrared radiation) eigenstates.
• Selection Rules
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.
• Selection rules and transition moment integral
In chemistry and physics, selection rules define the transition probability from one eigenstate to another eigenstate. In this topic, we are going to discuss the transition moment, which is the key to understanding the intrinsic transition probabilities. Selection rules have been divided into the electronic selection rules, vibrational selection rules (including Franck-Condon principle and vibronic coupling), and rotational selection rules.
• The Power of the Fourier Transform for Spectroscopists
Fourier transform is a mathematical technique that can be used to transform a function from one real variable to another. It is a unique powerful tool for spectroscopists because a variety of spectroscopic studies are dealing with electromagnetic waves covering a wide range of frequency.
• Time-resolved vs. Frequency Resolved
Fundamentals of Spectroscopy
As you read the print off this computer screen now, you are reading pages of fluctuating energy and magnetic fields. Light, electricity, and magnetism are all different forms of electromagnetic radiation.
Introduction
Electromagnetic radiation is a form of energy that is produced by oscillating electric and magnetic disturbance, or by the movement of electrically charged particles traveling through a vacuum or matter. The electric and magnetic fields come at right angles to each other and combined wave moves perpendicular to both magnetic and electric oscillating fields thus the disturbance. Electron radiation is released as photons, which are bundles of light energy that travel at the speed of light as quantized harmonic waves. This energy is then grouped into categories based on its wavelength into the electromagnetic spectrum. These electric and magnetic waves travel perpendicular to each other and have certain characteristics, including amplitude, wavelength, and frequency.
General Properties of all electromagnetic radiation:
1. Electromagnetic radiation can travel through empty space. Most other types of waves must travel through some sort of substance. For example, sound waves need either a gas, solid, or liquid to pass through in order to be heard.
2. The speed of light is always a constant. (Speed of light : 2.99792458 x 108 m s-1)
3. Wavelengths are measured between the distances of either crests or troughs. It is usually characterized by the Greek symbol $\lambda$.
Waves and their Characteristics
Fig. 1 & 2: Electromagnetic Waves
Fig. 3: An EM Wave
Amplitude
Amplitude is the distance from the maximum vertical displacement of the wave to the middle of the wave. This measures the magnitude of oscillation of a particular wave. In short, the amplitude is basically the height of the wave. Larger amplitude means higher energy and lower amplitude means lower energy. Amplitude is important because it tells you the intensity or brightness of a wave in comparison with other waves.
Wavelength
Wavelength ($\lambda$) is the distance of one full cycle of the oscillation. Longer wavelength waves such as radio waves carry low energy; this is why we can listen to the radio without any harmful consequences. Shorter wavelength waves such as x-rays carry higher energy that can be hazardous to our health. Consequently lead aprons are worn to protect our bodies from harmful radiation when we undergo x-rays. This wavelength frequently relationship is characterized by:
$c = \lambda\nu$
where
• c is the speed of light,
• $\lambda$ is wavelength, and
• $\nu$ is frequency.
Shorter wavelength means greater frequency, and greater frequency means higher energy. Wavelengths are important in that they tell one what type of wave one is dealing with.
Fig. 4: Different Wavelengths and Frequencies
Remember, Wavelength tells you the type of light and Amplitude tells you about the intensity of the light
Frequency
Frequency is defined as the number of cycles per second, and is expressed as sec-1 or Hertz (Hz). Frequency is directly proportional to energy and can be express as:
$E = h\nu$
where
• E is energy,
• h is Planck's constant, (h= 6.62607 x 10-34 J), and
• $\nu$ is frequency.
Period
Period (T) is the amount of time a wave takes to travel one wavelength; it is measured in seconds (s).
Velocity
The velocity of wave in general is expressed as:
$velocity = \lambda\nu$
For Electromagnetic wave, the velocity in vacuum is $2.99 \times 10^8\;m/s$ or $186,282$ miles/second.
Electromagnetic spectrum
Figure 24.5.1: Electromagnetic spectrum with light highlighted. from Wikipedia.
As a wave’s wavelength increases, the frequency decreases, and as wave’s wavelength decreases, the frequency increases. When electromagnetic energy is released as the energy level increases, the wavelength decreases and frequency decreases. Thus, electromagnetic radiation is then grouped into categories based on its wavelength or frequency into the electromagnetic spectrum. The different types of electromagnetic radiation shown in the electromagnetic spectrum consists of radio waves, microwaves, infrared waves, visible light, ultraviolet radiation, X-rays, and gamma rays. The part of the electromagnetic spectrum that we are able to see is the visible light spectrum.
Fig. 6: Electromagnetic Spectrum with Radiation Types
Radiation Types
Radio Waves are approximately 103 m in wavelength. As the name implies, radio waves are transmitted by radio broadcasts, TV broadcasts, and even cell phones. Radio waves have the lowest energy levels. Radio waves are used in remote sensing, where hydrogen gas in space releases radio energy with a low frequency and is collected as radio waves. They are also used in radar systems, where they release radio energy and collect the bounced energy back. Especially useful in weather, radar systems are used to can illustrate maps of the surface of the Earth and predict weather patterns since radio energy easily breaks through the atmosphere. ;
Microwaves can be used to broadcast information through space, as well as warm food. They are also used in remote sensing in which microwaves are released and bounced back to collect information on their reflections.
Microwaves can be measured in centimeters. They are good for transmitting information because the energy can go through substances such as clouds and light rain. Short microwaves are sometimes used in Doppler radars to predict weather forecasts.
Infrared radiation can be released as heat or thermal energy. It can also be bounced back, which is called near infrared because of its similarities with visible light energy. Infrared Radiation is most commonly used in remote sensing as infrared sensors collect thermal energy, providing us with weather conditions.
This picture represents a snap shot in mid-infrared light.
Visible Light is the only part of the electromagnetic spectrum that humans can see with an unaided eye. This part of the spectrum includes a range of different colors that all represent a particular wavelength. Rainbows are formed in this way; light passes through matter in which it is absorbed or reflected based on its wavelength. Thus, some colors are reflected more than other, leading to the creation of a rainbow.
Color Region
Wavelength (nm)
Violet 380-435
Blue 435-500
Cyan 500-520
Green 520-565
Yellow 565-590
Orange 590-625
Red 625-740
Fig. 7: The color regions of the Visible Spectrum
Fig. 8: Dispersion of Light Through A Prism
Ultraviolet, Radiation, X-Rays, and Gamma Rays are all related to events occurring in space. UV radiation is most commonly known because of its severe effects on the skin from the sun, leading to cancer. X-rays are used to produce medical images of the body. Gamma Rays can used in chemotherapy in order to rid of tumors in a body since it has such a high energy level. The shortest waves, Gamma rays, are approximately 10-12 m in wavelength. Out this huge spectrum, the human eyes can only detect waves from 390 nm to 780 nm.
Equations of Waves
The mathematical description of a wave is:
$y = A\sin(kx - \omega{t})$
where A is the amplitude, k is the wave number, x is the displacement on the x-axis.
$k = \dfrac{2\pi}{\lambda}$
where $\lambda$ is the wavelength. Angular frequency described as:
$\omega = 2\pi \nu = \dfrac{2\pi}{T}$
where $\nu$ is frequency and period (T) is the amount of time for the wave to travel one wavelength.
Interference
An important property of waves is the ability to combine with other waves. There are two type of interference: constructive and destructive. Constructive interference occurs when two or more waves are in phase and and their displacements add to produce a higher amplitude. On the contrary, destructive interference occurs when two or more waves are out of phase and their displacements negate each other to produce lower amplitude.
Figure 9 & 10: Constructive and Destructive Interference
Interference can be demonstrated effectively through the double slit experiment. This experiment consists of a light source pointing toward a plate with one slit and a second plate with two slits. As the light travels through the slits, we notice bands of alternating intensity on the wall behind the second plate. The banding in the middle is the most intense because the two waves are perfectly in phase at that point and thus constructively interfere. The dark bands are caused by out of phase waves which result in destructive interference. This is why you observe nodes on figure 4. In a similar way, if electrons are used instead of light, electrons will be represented both as waves and particles.
Fig. 11 & 12: Double-Slit Interference Experiment
Wave-Particle Duality
Electromagnetic radiation can either acts as a wave or a particle, a photon. As a wave, it is represented by velocity, wavelength, and frequency. Light is an EM wave since the speed of EM waves is the same as the speed of light. As a particle, EM is represented as a photon, which transports energy. When a photon is absorbed, the electron can be moved up or down an energy level. When it moves up, it absorbs energy, when it moves down, energy is released. Thus, since each atom has its own distinct set of energy levels, each element emits and absorbs different frequencies. Photons with higher energies produce shorter wavelengths and photons with lower energies produce longer wavelengths.
Fig. 13: Photon Before and After Emission
Ionizing and Non-Ionizing Radiation
Electromagnetic Radiation is also categorized into two groups based, ionizing and non-ionizing, on the severity of the radiation. Ionizing radiation holds a great amount of energy to remove electrons and cause the matter to become ionized. Thus, higher frequency waves such as the X-rays and gamma-rays have ionizing radiation. However, lower frequency waves such as radio waves, do not have ionizing radiation and are grouped as non-ionizing.
Electromagnetic Radiation and Temperature
Electromagnetic radiation released is related to the temperature of the body. Stephan-Boltzmann Law says that if this body is a black body, one which perfectly absorbs and emits radiation, the radiation released is equal to the temperature raised to the fourth power. Therefore, as temperature increases, the amount of radiation released increases greatly. Objects that release radiation very well also absorb radiation at certain wavelengths very well. This is explained by the Kirchhoff’s Law. Wavelengths are also related to temperature. As the temperature increases, the wavelength of maximum emission decreases.
Problems
1. What is the wavelength of a wave with a frequency of 4.28 Hz?
2. What is the frequency of a wave with a wavelength of 200 cm?
3. What is the frequency of a wave with a wavelength of 500 pm?
4. What is the wavelength of a wave with a frequency of 2.998 × 105 Hz?
5. A radio transmits a frequency of 100 Hz. What is the wavelength of this wave?
Answers:
1. 700m
2. 1.5 × 108 Hz
3. 4.0 × 1017 Hz
4. 100m
5. 2.998 × 106 m
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Fundamentals_of_Spectroscopy/Electromagnetic_Radiation.txt
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Realizing that light may be considered to have both wave-like and particle-like characteristics, it is useful to consider that a given frequency or wavelength of light is associated with a "light quanta" of energy we now call a photon. As noted in the following equations, frequency and energy change proportionally, but wavelength has an inverse relationship to these quantities.
$\nu = \dfrac{c}{\lambda}$
and
$\Delta{E} = h\nu$
with
• $\nu$ is the frequency of light
• $\lambda$ is the wavelength of ligt
• $c$ is the speed of light ($3 \times 10^{8}\; m/sec$)
• $\Delta{E}$ is the transition energy (difference of energies between the initial and final states)
• $h$ is Planck's constant ($h = 6.626069 \times 10^{−34}\;J \,s$)
To "see" a molecule, we must use light having a wavelength smaller than the molecule itself (roughly 1 to 15 angstroms). Such radiation is found in the X-ray region of the spectrum, and the field of X-ray crystallography yields remarkably detailed pictures of molecular structures amenable to examination. The chief limiting factor here is the need for high quality crystals of the compound being studied. The methods of X-ray crystallography are too complex to be described here; nevertheless, as automatic instrumentation and data handling techniques improve, it will undoubtedly prove to be the procedure of choice for structure determination. The spectroscopic techniques described below do not provide a three-dimensional picture of a molecule, but instead yield information about certain characteristic features. A brief summary of this information follows:
• Ultraviolet-Visible Spectroscopy: Absorption of this relatively high-energy light causes electronic excitation. The easily accessible part of this region (wavelengths of 200 to 800 nm) shows absorption only if conjugated $\pi$ electron systems are present.
• Infrared Spectroscopy: Absorption of this lower energy radiation causes vibrational and rotational excitation of groups of atoms. within the molecule. Because of their characteristic absorptions, identification of functional groups is easily accomplished.
• Nuclear Magnetic Resonance (NMR) Spectroscopy: Absorption in the low-energy radio-frequency part of the spectrum causes excitation of nuclear spin states. NMR spectrometers are tuned to certain nuclei (e.g. 1H, 13C, 19F & 31P). For a given type of nucleus, high-resolution spectroscopy distinguishes and counts atoms in different locations in the molecule.
Lineshape Functions
Four different mathematical descriptions of the lineshape for an absorptive transition are discussed below. These transitions may involve electronic, rotational, or vibrational (i.e. visible, microwave or infrared radiation) eigenstates.
δ-function Lineshape
In an ideal gas phase experiment where there is no Doppler broadening or lifetime broadening the energy at which transition occurs is unique. This implies that the line is infinitely narrow. How can we describe an infinitely narrow line. It turns out that this is done by defining the delta function δ(x-x0). The delta function has the useful property that the integral over all space (i.e. from minus infinity to plus infinity along the x axis) gives a single value x0. We write this as
$\int_{-\infty}^{\infty} \delta{(x-x_o)}dx = x_o$
This integral is perhaps the hardest of all to grasp since according to its definition there is no finite extent along the x axis. The function defined is a line at x0 that is infinitely narrow. But notice that if we define the variable as an energy e then the delta function returns a unique energy e0.
$\int_{-\infty}^{\infty} \delta{(\epsilon-\epsilon_o)}d\epsilon = \epsilon_o$
This corresponds to our requirement for the idea line shape function of a transition with no broadening. Note that we observe a frequency (or wavenumber in an absorption spectrum). Therefore, we can also write the delta function in terms of the frequency ω.
$\int_{-\infty}^{\infty} \delta{(\omega-\omega_o)}d\omega = \omega_o$
Lorentzian Lineshape
Of course, real transitions always involve some line broadening. If nothing else there must be lifetime broadening since absorption of radiation always produces an excited state with a finite lifetime. Recall that the uncertainty principle states that there is a relationship between the lifetime and the energy width of the state. Usually, we describe the kinetic decay of an excited state using an exponential function. This emerges directly from first order kinetics. Looking ahead this means that if we define the rate of disappearance of the excited state E as
$\dfrac{\partial [E]}{\partial t} = -\dfrac{[E]}{T_1}$
Then we can solve this equation to find the $E(t) = E_0e^{-t/T_1}$ where $T_1$ is the excited state life time and $E_0$ is the initial concentration of the excited state.
The uncertainty principle states that the conjugate energy width can be obtained from the Fourier transform of the life time function.
$L(\omega)=\int_{0}^{\infty} e^{-t/T_1} e^{-i\omega t} dt$
This Fourier transform is easily solved since the integral is nothing more than an exponential integral. The solution for this integral is
$L(\omega) = \dfrac{1}{\dfrac{1}{T_1}-i\omega}$
$L(\omega) = \dfrac{1/T_1}{(1/T_1)^2+ \omega^2} + \dfrac{i\omega}{(1/T_1)^2 + \omega^2}$
which is complex. We can calculate real and imaginary parts by multiplying both numerator and denominator by $1/T_1 + i\omega$. This gives
$\Lambda(\omega) = \dfrac{1/T_1}{\pi((1/T_1)^2+\omega^2)}$
The real part of this integral is a Lorentzian line shape function. This is the line shape that will be observed for transitions that have no inhomogeneous broadening. Examples include, NMR spectra, Lamb dip spectra in the gas phase, etc. The normalized Lorentzian function is (i.e. the real part of the above function $L(\omega)$).
We can define the energy width G as being $1/T_1$, which corresponds to a Lorentzian linewidth. Using this definition and generalizing the function so that it can be used to describe the line shape function centered about any arbitrary frequency $\omega_o$, we have
$\Lambda(\omega) = \dfrac{\Gamma}{\pi(\Gamma^2+(\omega-\omega_0)^2)}$
Note that this is a normalized function so that the integral from $-\infty$ to $\infty$ is equal to one.
Gaussian Lineshape
A Gaussian function is also a useful lineshape function. Any source of inhomogeneous broadening such as the Doppler shift or site differences of molecules in crystals or solution can be described as a Gaussian lineshape.
$G(\omega)=\dfrac{1}{\sqrt{\pi} \Gamma}e^{-(\omega-\omega_o)^2/\Gamma^2}$
In considering the appearance of spectra we observe that Franck-Condon progressions often have the appearance of a progression of different levels with Gaussian broadening for molecules in solution. The progression of levels arises from the discrete transitions among vibrational levels of the molecule and the Gaussian function is superposed on this discrete "stick" spectrum to represent the broadening due to lifetime and solvent origins. Since the inhomogeneous broadening is often significantly larger than the lifetime broadening, the Gaussian lineshape often dominates in solution.
The Gaussian distribution in frequencies can be thought of as corresponding to a Gaussian distribution in relaxation times of the excited state. The Gaussian distribution of relaxation times is usually assumed to arise from an inhomogeneous distribution of the molecules in the sample. The molecules experience different environments due to solvation or matrix effects and therefore they all have slightly different properties. Since the Fourier transform of a Gaussian is also a Gaussian, the distribution of the relaxation times gives rise to a distribution of energies (and therefore a distribution of frequencies in the observed spectrum).
Voight Lineshape
The Voight lineshape is just a convolution of a Gaussian and a Lorentzian and can be interpreted as an inhomogeneous distributions of Gaussian lineshapes.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Fundamentals_of_Spectroscopy/Introduction_to_Spectroscopy.txt
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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
$\mu_z=\int\psi_1 \,^{*}\mu_z\psi_1\,d\tau$
A transition dipole moment is a transient dipolar polarization created by an interaction of electromagnetic radiation with a molecule
$(\mu_z)_{12}=\int\psi_1 \,^{*}\mu_z\psi_2\,d\tau$
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.
We can consider selection rules for electronic, rotational, and vibrational transitions.
Electronic transitions
We consider a hydrogen atom. In order to observe emission of radiation from two states $mu_z$ must be non-zero. That is
$(\mu_z)_{12}=\int\psi_1^{\,*}\,e\cdot z\;\psi_2\,d\tau\neq0$
For example, is the transition from $\psi_{1s}$ to $\psi_{2s}$ allowed?
$(\mu_z)_{12}=\int\psi_{1s}\,^{\,*}\,e\cdot z\;\psi_{2s}\,d\tau$
Using the fact that z = r cosq in spherical polar coordinates we have
$(\mu_z)_{12}=e\iiint\,e^{-r/a_0}r\cos \theta \biggr(2-\frac{r}{a_0}\biggr)e^{-r/a_0}r^2\sin\theta drd\theta\,d\phi$
We can consider each of the three integrals separately.
$\int_{0}^{\infty}e^{-r/a_0}r\biggr(2-\frac{r}{a_0}\biggr)e^{-r/a_0}r^2dr\int_{0}^{\pi}\cos\theta\sin\theta\,d\theta\int_{0}^{2\pi }d\phi$
If any one of these is non-zero the transition is not allowed. We can see specifically that we should consider the q integral. We make the substitution $x = \cos q, dx = -\sin\; q\; dq$ and the integral becomes
$-\int_{1}^{-1}x dx=-\frac{x^2}{2}\Biggr\rvert_{1}^{-1}=0$
which is zero. The result is an even function evaluated over odd limits. In a similar fashion we can show that transitions along the x or y axes are not allowed either. This presents a selection rule that transitions are forbidden for $\Delta{l} = 0$. For electronic transitions the selection rules turn out to be $\Delta{l} = \pm 1$ and $\Delta{m} = 0$. These result from the integrals over spherical harmonics which are the same for rigid rotator wavefunctions. We will prove the selection rules for rotational transitions keeping in mind that they are also valid for electronic transitions.
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.
$(\mu_z)_{J,M,{J}',{M}'}=\int_{0}^{2\pi } \int_{0}^{\pi }Y_{J'}^{M'}(\theta,\phi )\mu_zY_{J}^{M}(\theta,\phi)\sin\theta\,d\phi,d\theta$
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
$Y_{J}^{M}(\theta,\phi)=N_{\,JM}P_{J}^{|M|}(\cos\theta)e^{iM\phi}$
where $N_{JM}$ is a normalization constant. Using the standard substitution of $x = \cos q$ we can express the rotational transition moment as
$(\mu_z)_{J,M,{J}',{M}'}=\mu\,N_{\,JM}N_{\,J'M'}\int_{0}^{2 \pi }e^{I(M-M')\phi}\,d\phi\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx$
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
$(\mu_z)_{J,M,{J}',{M}'}=2\pi \mu\,N_{\,JM}N_{\,J'M'}\int_{-1}^{1}P_{J'}^{|M'|}(x)P_{J}^{|M|}(x)dx$
We can evaluate this integral using the identity
$(2J+1)x\,P_{J}^{|M]}(x)=(J-|M|+1)P_{J+1}^{|M|}(x)+(J-|M|)P_{J-1}^{|M|}(x)$
Substituting into the integral one obtains an integral which will vanish unless $J' = J + 1$ or $J' = J - 1$.
$\int_{-1}^{1}P_{J'}^{|M'|}(x)\Biggr(\frac{(J-|M|+1)}{(2J+1)}P_{J+1}^{|M|}(x)+\frac{(J-|M|)}{(2J+1)}P_{J-1}^{|M|}(x)\Biggr)dx$
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 ($\Delta{l} = \pm 1$, $\Delta{m} = 0$).
Vibrational transitions
The harmonic oscillator wavefunctions are
$\psi_{\,v}(q)=N_{\,v}H_{\,v}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}$
where $H_v(a1/2q)$ is a Hermite polynomial and a = (km/á2)1/2.
The transition dipole moment for electromagnetic radiation polarized along the z axis is
$(\mu_z)_{v,v'}=\int_{-\infty}^{\infty}N_{\,v}N_{\,v'}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}H\mu_z(\alpha^{1/2}q)e^{-\alpha\,q^2/2}dq$
Note that we continue to use the general coordinate q although this can be z if the dipole moment of the molecule is aligned along the z axis. The transition moment can be expanded about the equilibrium nuclear separation.
$\mu_z(q)=\mu_0+\biggr({\frac{\partial\mu }{\partial q}}\biggr)q+.....$
where m0 is the dipole moment at the equilibrium bond length and q is the displacement from that equilibrium state. From the first two terms in the expansion we have for the first term
$(\mu_z)_{v,v'}=\mu_0\int_{-\infty}^{\infty}N_{\,v}N_{\,v'}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}H_v(\alpha^{1/2}q)e^{-\alpha\,q^2/2}dq$
This term is zero unless v = v’ and in that case there is no transition since the quantum number has not changed.
$(\mu_z)_{v,v'}=\biggr({\frac{\partial\mu }{\partial q}}\biggr)\int_{-\infty}^{\infty}N_{\,v}N_{\,v'}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}H_v(\alpha^{1/2}q)e^{-\alpha\,q^2/2}dq$
This integral can be evaluated using the Hermite polynomial identity known as a recursion relation
$xH_v(x)=vH_{v-1}(x)+\frac{1}{2}H_{v+1}(x)$
where x = Öaq. If we now substitute the recursion relation into the integral we find
$(\mu_z)_{v,v'}=\frac{N_{\,v}N_{\,v'}}{\sqrt\alpha}\biggr({\frac{\partial\mu }{\partial q}}\biggr)$
$\int_{-\infty}^{\infty}H_{\,v'}(\alpha^{1/2}q)e^{-\alpha\,q^2/2}\biggr(vH_{v-1}(\alpha^{1/2}q)+\frac{1}{2}H_{v+1}(\alpha^{1/2}q)\biggr)dq$
which will be non-zero if v’ = v – 1 or v’ = v + 1. Thus, we see the origin of the vibrational transition selection rule that v = ± 1. We also see that vibrational transitions will only occur if the dipole moment changes as a function nuclear motion.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Fundamentals_of_Spectroscopy/Selection_Rules.txt
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In chemistry and physics, selection rules define the transition probability from one eigenstate to another eigenstate. In this topic, we are going to discuss the transition moment, which is the key to understanding the intrinsic transition probabilities. Selection rules have been divided into the electronic selection rules, vibrational selection rules (including Franck-Condon principle and vibronic coupling), and rotational selection rules.
Introduction
The transition probability is defined as the probability of particular spectroscopic transition to take place. When an atom or molecule absorbs a photon, the probability of an atom or molecule to transit from one energy level to another depends on two things: the nature of initial and final state wavefunctions and how strongly photons interact with an eigenstate. Transition strengths are used to describe transition probability. Selection rules are utilized to determine whether a transition is allowed or not. Electronic dipole transitions are by far the most important for the topics covered in this module.
Transition Moment
In an atom or molecule, an electromagnetic wave (for example, visible light) can induce an oscillating electric or magnetic moment. If the frequency of the induced electric or magnetic moment is the same as the energy difference between one eigenstate Ψ1 and another eigenstate Ψ2, the interaction between an atom or molecule and the electromagnetic field is resonant (which means these two have the same frequency). Typically, the amplitude of this (electric or magnetic) moment is called the transition moment. In quantum mechanics, the transition probability of one molecule from one eigenstate Ψ1 to another eigenstate Ψ2 is given by |$\vec{M}_{21}$|2, and $\vec{M}_{21}$ is called the transition dipole moment, or transition moment, from Ψ1 to Ψ2. In mathematical form it can be written as
$\vec{M}_{21}=\int \Psi_2\vec{\mu}\Psi_1d\tau$
The Ψ1 and Ψ2 are two different eigenstates in one molecule, $\vec{M}_{21}$ is the electric dipole moment operator. If we have a system with n molecules and each has charge Qn, and the dipole moment operator is can be written as
$\vec{\mu}=\sum_{n}Q_n\vec{x}_n$
the $\vec{x}_{n}$ is the position vector operator.
Transition Moment Integral
Based on the Born-Oppenheimer approximation, the fast electronic motion can be separated from the much slower motion of the nuclei. As a result, the total wavefunction can be separated into electronic, vibrational, and rotational parts:
$\Psi (r,R) = \psi _e (r,R_e )\psi _v (R)\psi _r (R)$
The Born-Oppenheimer approximation assumes that the electronic wavefunction, $\psi _e$, is approximated in all electronic coordinates at the equilibrium nuclear coordinates (Re). Since mass of electrons is much smaller than nuclear mass, the rotational wavefunction, $\psi _r$, only depends on nuclear coordinates. The rotational wavefunction could provide important information for rotational selection rules, but we will not consider the rotational wavefuntion any further for simplicity because most of the spectra are not rotationally resolved. With the rotational part removed, the transition moment integral can be expressed as
$\ M = \iint {\psi _e '(r,R_e )}\cdot \psi _v '(R)(\mu _e + \mu _n )\psi _e ''(r,R_e ) \cdot \psi _v ''(R)drdR$
where the prime and double prime represent the upper and lower states respectively. Both the nuclear and electronic parts contribute to the dipole moment operator. The above equation can be integrated by two parts, with $\mu _n$ and $\mu _e$ respectively. A product of two integral is obtained:
$\ M = \int {\psi _e '(r,R_e )} \cdot \mu _e \cdot \psi _e ''(r,R_e )dr\int {\psi _v '(R)} \cdot \psi _v ''(R)dR + \int {\psi _v '(R)} \cdot \mu _n \cdot \psi _v ''(R)dR\int {\psi _e '(r,R_e )} \psi _e ''(r,R_e )dr$
Because different electronic wavefunctions must be orthogonal to each other, hence $\int {\psi _e '(r,R_e )} \psi _e ''(r,R_e )dr$ is zero, the second part of the integral should be zero.
The transition moment integral can be simplified as
$\ M = \int {\psi _e '(r,R_e )} \cdot \mu _e \cdot \psi _e ''(r,R_e )dr\int {\psi _v '(R)} \cdot \psi _v ''(R)dR$
The above equation is of great importance because the first integral defines the electronic selection rules, while the second integral is the basis of vibrational selection rules.
Electronic Selection Rules
In atoms
Atoms are described by the primary quantum number n, angular momentum quantum number L, spin quantum number S, and total angular momentum quantum number J. Based on Russell-Saunders approximation of electron coupling, the atomic term symbol can be represented as 2S+1LJ.
1. The total spin cannot change, ΔS=0;
2. The change in total orbital angular momentum can be ΔL=0, $\pm$1, but L=0 $\leftrightarrow$ L=0 transition is not allowed;
3. The change in the total angular momentum can be ΔJ=0, $\pm$1, but J=0 $\leftrightarrow$ J=0 transition is not allowed;
4. The initial and final wavefunctions must change in parity. Parity is related to the orbital angular momentum summation over all elections Σli, which can be even or odd; only even $\leftrightarrow$ odd transitions are allowed.
In molecules
The electronic-state configurations for molecules can be described by the primary quantum number n, the angular momentum quantum number Λ, the spin quantum number S, which remains a good quantum number, the quantum number Σ (S, S-1, ..., -S), and the projection of the total angular momentum quantum number onto the molecular symmetry axis Ω, which can be derived as Ω=Λ+Σ. The term symbol for the electronic states can be represented as
$^{2S+1} \Lambda_{\Omega,(g/u)}^{(+/-)}$
Group theory makes great contributions to the prediction of the electronic selection rules for many molecules. An example is used to illustrate the possibility of electronic transitions via group theory.
1. The total spin cannot change, ΔS=0; the rule ΔΣ=0 holds for multiplets;
If the spin-orbit coupling is not large, the electronic spin wavefunction can be separated from the electronic wavefunctions. Since the electron spin is a magnetic effect, electronic dipole transitions will not alter the electron spin. As a result, the spin multiplicity should not change during the electronic dipole transition.
2. The total orbital angular momentum change should be ΔΛ=0, $\pm$1;
For heteronuclear diatomic molecules with $\ C_{\infty v}$ symmetry, a Σ+ $\leftrightarrow$ $\Pi$ transition is allowed according to ΔΛ selection rule. In order to prove the allowance of this transition, the direct product of $\Sigma ^ + \otimes \Pi$ yields $\Pi$ irreducible representation from the direct product table. Based on the $\ C_{\infty v}$ character table below, the operator in the x and y direction have doubly degenerate $\Pi$ symmetry. Therefore, the transition between Σ+ $\leftrightarrow$ $\Pi$ must be allowed since the multiplication of any irreducible representation with itself will provide the totally symmetric representation. The electronic transition moment integral can be nonzero. We can use the same kind of argument to illustrate that $\Sigma ^ - \leftrightarrow \Phi$ transition is forbidden.
$C_{\infty v}$ character table
$C_{\infty v}$ E $2C_{\infty ^\Phi}$ ... $\infty \sigma_v$
Σ+ 1 1 ... 1 z (x2+y2), z2
Σ- 1 1 ... -1 Rz
$\Pi$ 2 $2\cos (\Phi )$ ... 0 (x, y), (Rx, Ry) xz, yz
$\Delta$ 2 $2\cos (2 \Phi )$ ... 0 (x2-y2), xy
$\Gamma$ 2 $2\cos (3 \Phi )$ ... 0
... ... ... ... ...
3. Parity conditions are related to the symmetry of the molecular wavefunction reflecting against its symmetry axis. For homonuclear molecules, the g $\leftrightarrow$ u transition is allowed. For heteronuclear molecules, + $\leftrightarrow$ + and - $\leftrightarrow$ - transitions apply;
• For hetero diatomic molecules with $\ C_{\infty v}$ symmetry, we can use group theory to reveal that $\Sigma ^ + \leftrightarrow \Sigma ^ +$ and $\Sigma ^ - \leftrightarrow \Sigma ^ -$ transitions are allowed, while $\Sigma ^ + \leftrightarrow \Sigma ^ -$ transitions are forbidden. The direct product of either $\Sigma ^ + \otimes \Sigma ^ +$ or $\Sigma ^ - \otimes \Sigma ^ -$ yields the same irreducible representation Σ+ based on the direct product table. The z component of the dipole operator has Σ+ symmetry. The transition is allowed because the electronic transition moment integral can generate the totally symmetric irreducible representation Σ+. We can use the same method to prove that $\Sigma ^ + \leftrightarrow \Sigma ^ -$ transitions are forbidden.
• Similarly, for a molecule with an inversion center, a subscript g or u is used to reveal the molecular symmetry with respect to the inversion operation, i. The x, y, and z components of the transition dipole moment operator have u inversion symmetry in molecule with inversion center (g and u are short for gerade and ungerade in German, meaning even and odd). The electronic transition moment integral need to yield totally symmetric irreducible representation Ag for allowed transitions. Therefore, only g $\leftrightarrow$ u transition is allowed.
Vibrational Selection rules
1. Transitions with Δv=$\pm$1, $\pm$2, ... are all allowed for anharmonic potential, but the intensity of the peaks become weaker as Δv increases.
2. v=0 to v=1 transition is normally called the fundamental vibration, while those with larger Δv are called overtones.
3. Δv=0 transition is allowed between the lower and upper electronic states with energy E1 and E2 are involved, i.e. (E1, v''=n) $\to$ (E2, v'=n), where the double prime and prime indicate the lower and upper quantum state.
The geometry of vibrational wavefunctions plays an important role in vibrational selection rules. For diatomic molecules, the vibrational wavefunction is symmetric with respect to all the electronic states. Therefore, the Franck-Condon integral is always totally symmetric for diatomic molecules. The vibrational selection rule does not exist for diatomic molecules.
For polyatomic molecules, the nonlinear molecules possess 3N-6 normal vibrational modes, while linear molecules possess 3N-5 vibrational modes. Based on the harmonic oscillator model, the product of 3N-6 normal mode wavefunctions contribute to the total vibrational wavefunction, i.e.
$\psi _{vib} = \prod\limits_{3N - 6} {\psi _1 \psi _2 } \psi _3 ...\psi _{3N - 6}$
where each normal mode is represented by the wavefunction $\psi_i$. Comparing to the Franck-Condon factor for diatomic molecules with single vibrational overlap integral, a product of 3N-6 (3N-5 for linear molecules) overlap integrals needs to be evaluated. Based on the symmetry of each normal vibrational mode, polyatomic vibrational wavefunctions can be totally symmetric or non-totally symmetric. If a normal mode is totally symmetric, the vibrational wavefunction is totally symmetric with respect to all the vibrational quantum number v. If a normal mode is non-totally symmetric, the vibrational wavefunction alternates between symmetric and non-symmetric wavefunctions as v alternates between even and odd number.
If a particular normal mode in both the upper and lower electronic state is totally symmetric, the vibrational wavefunction for the upper and lower electronic state will be symmetric, resulting in the totally symmetric integrand in the Franck-Condon integral. If the vibrational wavefunction of either the lower or upper electronic state is non-totally symmetric, the Franck-Condon integrand will be non-totally symmetric.
We will use CO2 as an example to specify the vibrational selection rule. CO2 has four vibrational modes as a linear molecule. The vibrational normal modes are illustrated in the figure below:
The vibrational wavefunction for the totally symmetric C-O stretch, v1, is totally symmetric with respect to all the vibrational quantum numbers. However, the vibrational wavefunctions for the doubly degenerate bending modes, v2, and the antisymmetric C-O stretch, v3, are non-totally symmetric. Therefore, the vibrational wavefunctions are totally symmetric for even vibrational quantum numbers (v=0, 2, 4...), while the wavefunctions remain non-totally symmetric for v odd (v=1, 3, 5...).
Therefore, any value of Δv1 is possible between the upper and lower electronic state for mode v1. On the other hand, modes v2 and v3 include non-totally symmetric vibrational wavefunctions, so the vibrational quantum number can only change evenly, such as Δv=$\pm$ 2, $\pm$ 4, etc..
Franck-Condon Principle
Franck-Condon principle was proposed by German physicist James Franck (1882-1964) and U.S. physicist Edward U. Condon (1902-1974) in 1926. This principle states that when an electronic transition takes place, the time scale of this transition is so fast compared to nucleus motion that we can consider the nucleus to be static, and the vibrational transition from one vibrational state to another state is more likely to happen if these states have a large overlap. It successfully explains the reason why certain peaks in a spectrum are strong while others are weak (or even not observed) in absorption spectroscopy.
$\ M = \int {\psi _e '(r,R_e )} \cdot \mu _e \cdot \psi _e ''(r,R_e )dr\int {\psi _v '(R)} \cdot \psi _v ''(R)dR$
The second integral in the above equation is the vibrational overlap integral between one eigenstate and another eigenstate. In addition, the square of this integral is called the Franck-Condon factor:
$\displaystyle \ Franck-Condon \; Factor = \mid \int {\psi _v '}\psi _v ''dR\mid^2$
It governs the vibrational transition contribution to the transition probability and shows that in order to have a large vibrational contribution, the vibrational ground state and excited state must have a strong overlap.
The above figure shows the Franck-Condon principle energy diagram, since electronic transition time scale is small compared to nuclear motion, the vibrational transitions are favored when the vibrational transition have the smallest change of nuclear coordinate, which is a vertical transition in the figure above. The electronic eigenstates favors the vibrational transition v'=0 in the ground electronic state to v"=2 in the excited electronic state, while peak intensity of v'=0 to v"=0 transition is expected to be low because the overlap between the v'=0 wavefunction and v''=0 wavefunction is very low.
The figure above is the photoelectron spectrum of the ionization of hydrogen molecule (H2), it is also a beautiful example to formulate the Franck-Condon principle. It shows the appropriate energy curves and the vibrational energy levels, and with the help of the Franck-Condon principle, the transition between ground vibrational state v0" and excited vibrational state v2' is expected to be the most intense peak in the spectrum.
Vibronic Coupling
Why can some electronic-forbidden transitions be observed as weak bands in spectrum? It can be explained by the interaction between the electronic and vibrational transitions. The word "vibronic" is the combination of the words "vibrational" and "electronic." Because the energy required for one electronic state to another electronic state (electronic transition, usually in the UV-Vis region) is larger than one vibrational state to another vibrational state (vibrational transition, usually in the IR region), sometimes energy (a photon) can excite a molecule to an excited electronic and vibrational state. The bottom figure shows the pure electronic transition (no vibronic coupling) and the electronic transition couples with the vibrational transition (vibronic coupling).
We now can go back to the original question: Why can some electronic-forbidden transitions be observed as weak bands in spectrum? For example, the d-d transitions in the octahedral transition metal complexes are Laporte forbidden (same symmetry, parity forbidden), but they can be observed in the spectrum and this phenomenon can be explained by vibronic coupling.
Now we consider the Fe(OH2)62+ complex which has low spin d6 as ground state (see figure above). Let's examine the one-electron excitation from t2g molecular orbital to eg molecular orbital. The octahedral complex has Oh symmetry and therefore the ground state has A1g symmetry from the character table. The symmetry of the excited state, which is the direct product of all singly occupied molecular orbitals (eg and t2g in this case):
$\displaystyle \Gamma_{t_{2g}}\otimes\Gamma_{e_g}= T_{1g}+T_{2g}$
The transition moment integral for the electronic transition can be written as
$\vec{M}=\int \psi^{'*}\vec{\mu}\psi d\tau$
where ψ is the electronic ground state and ψ' is the electronic excited state. The condition for the electronic transition to be allowed is to make the transition moment integral nonzero. The $\hat{\mu}$ is the transition moment operator, which is the symmetry of the $\hat{x}$, $\hat{y}$, $\hat{z}$ operators from the character table. For octahedral symmetry, these operators are degenerate and have T1u symmetry. Therefore, we can see that both 1T1g1A1g and 1T2g1A1g are electronically forbidden by the direct product (do not contain the totally symmetric A1g):
$\displaystyle T_{1g}\times\ T_{1u}\times\ A_{1g}= A_{1u}+E_{u}+T_{1u}+T_{2u}$
$\displaystyle T_{2g}\times\ T_{1u}\times\ A_{1g}= A_{2u}+E_{u}+T_{1u}+T_{2u}$
For octahedral complex, there are 15 vibrational normal modes. From the Oh character table we can get these irreducible representations:
$\displaystyle \Gamma_{vib}= a_{1g}+e_{g}+2t_{1u}+t_{2g}+t_{2u}$
When we let the vibrational transition to couple with the electronic transition, the transition moment integral has the form:
$\displaystyle \vec{M}=\int \psi_{e'}^{*} \psi_{v'}^{*} \vec{\mu} \psi_e \psi_v d\tau$
For the vibronic coupling to be allowed, the transition moment integral has to be nonzero. Use these vibrational symmetries of the octahedral complex to couple with the electronic transition:
$\displaystyle A_{1g}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= A_{1u}+E_{u}+T_{1u}+T_{2u}$
$\displaystyle E_{g}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= E_{g}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1u}+A_{2u}+2E_{u}+2T_{1u}+2T_{2u}$
$\displaystyle T_{1u}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= T_{1u}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1g}+ \ldots$
$\displaystyle T_{2g}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= T_{2g}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1u}+A_{2u}+2E_{u}+3T_{1u}+4T_{2u}$
$\displaystyle T_{2u}\times \{T_{1g}\times\ T_{1u}\times\ A_{1g}\}= T_{2u}\times \{A_{1u}+E_{u}+T_{1u}+T_{2u}\}=A_{1g}+ \ldots$
Since the t1u and t2u representation can generate the totally symmetric representation in the integrand. Therefore, we can see that t1u and t2u can couple with the electronic transition to form the allowed vibronic transition. Therefore, the d-d transition band for Fe(OH2)62+ complex can be observed through vibronic coupling.
Rotational Selection rules
1. Transitions with ΔJ=$\pm$1 are allowed;
Photons do not have any mass, but they have angular momentum. The conservation of angular momentum is the fundamental criteria for spectroscopic transitions. As a result, the total angular momentum has to be conserved after a molecule absorbs or emits a photon. The rotational selection rule relies on the fact that photon has one unit of quantized angular momentum. During the photon emission and absorption process, the angular moment J cannot change by more than one unit.
Let's consider a single photon transition process for a diatomic molecule. The rotational selection rule requires that transitions with ΔJ=$\pm$1 are allowed. Transitions with ΔJ=1 are defined as R branch transitions, while those with ΔJ=-1 are defined as P branch transitions. Rotational transitions are conventional labeled as P or R with the rotational quantum number J of the lower electronic state in the parentheses. For example, R(2) specifies the rotational transition from J=2 in the lower electronic state to J=3 in the upper electronic state.
2. ΔJ=0 transitions are allowed when two different electronic or vibrational states are involved: (X'', J''=m) $\to$ (X', J'=m).
The Q branch transitions will only take place when there is a net orbital angular momentum in one of the electronic states. Therefore, Q branch does not exist for $\ {}^1\Sigma \leftrightarrow {}^1\Sigma$ electronic transitions because $\Sigma$ electronic state does not possess any net orbital angular momentum. On the other hand, the Q branch will exist if one of the electronic states has angular momentum. In this situation, the angular momentum of the photon will cancel out with the angular momentum of the electronic state, so the transition will take place without any change in the rotational state.
The schematic of P, Q, and R branch transitions are shown below:
With regard to closed-shell non-linear polyatomic molecules, the selection rules are more complicated than diatomic case. The rotational quantum number J remains a good quantum number as the total angular momentum if we don't consider the nuclear spin. Under the effect of single photon transition, the change of J is still limited to a maximum of $\pm$ 1 based on the conservation of angular momentum. However, the possibility of Q branch is greatly enhanced irrelevant to the symmetry of the lower and upper electronic states. The rotational quantum number K is introduced along the inertial axis. For polyatomic molecules with symmetric top geometry, the transition moment is polarized along inertial axis. The selection rule becomes ΔK=0.
Problems
1. What are the differences between Born-Oppenheimer approximation and Franck-Condon principle?
2. In evaluating the transition moment integral, why there must be a totally symmetric irreducible representation to make the transition allowed?
3. Because of the d-d transition, transition metal complexes are known for its various color. However, Mn2+ does not have intense color (pale pink), why?
4. Assume a Pd2+ square planar complex (D4h symmetry), is the following electronic transition allowed? If not, can it gain intensity by vibronic coupling?
1. What's the vibrational selection rule for polyatomic molecules?
2. Under what condition can we see the Q branch transition?
Answers to Problems
1. The Born Oppenheimer approximation states that the fast electronic motion can be separated from the much slower motion of the nuclei, so the total wavefunction can be separated into the electronic, vibrational and rotational part. The Franck-Condon principle states during an electronic transition, the timescale for an electron to move from one orbital to another is so short that the nuclear position does not change during the transition process. The Franck Condon factor is the square of the overlap integral, which defines the vibrational contribution to the transition probability.
2. n the character table, only A1 or A1g are totally symmetric irreducible representations. When you are integrating an odd function over space, the integrand will be 0. Only the integration of an even function over space will yield a non-zero value, representing the possibility of the transition.
3. The ground state Mn2+ has a high spin d5 electronic configuration, which has one electron in each of the eg and t2g orbitals, and according to the electronic transition selection rule, the total spin cannot change:
then we will find out it is impossible for a high spin d5 octahedral complex since it violates Pauli exclusion principle. Therefore, the d-d transition for this is Laporte forbidden and also spin forbidden, resulting in a pale color.
4. The symmetry of the excited state will have the symmetry:
$\Gamma_{a_{1g}}\otimes\Gamma_{b_{1g}}= B_{1g}$
In the point group of D4h, the dipole moment operator transforms as eu(x,y) and a2u(z). Then by evaluating the transition moment integral for the electronic transition, we can easily find out that there is no totally symmetric irreducible representation in the integrand.Therefore, the electronic transition is not allowed:
$B_{1g}\times\dbinom{E_u}{A_{2u}}\times\ A_{1g}= \dbinom{E_u}{B_{2u}}$
Next step we examine the vibrational irreducible representation a D4h complex has:
$\Gamma_{vib}= a_{1g}+b_{1g}+b_{2g}+a_{2u}+b_{2u}+2e_{u}$
we can see that eu and b2u can potentially serve as the promoting mode for vibronic transition, that is, generating the totally symmetric A1g in the integrand. Therefore, this electronic transition for D4h complex is forbidden, however, the transition can be allowed through vibronic coupling.
5. If the vibrational mode is totally symmetric, any change in the vibrational quantum number v is possible between the lower electronic state and upper electronic state. If the vibrational mode is non-totally symmetric, the vibrational wavefunction for the lower and upper electronic states are non-totally symmetric. The vibrational quantum number can only change evenly, such as Δv=$\pm$ 2, $\pm$ 4, etc..
6.The Q branch transitions will only take place when there is a net orbital angular momentum in one of the electronic states.
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Fourier transform is a mathematical technique that can be used to transform a function from one real variable to another. It is a unique powerful tool for spectroscopists because a variety of spectroscopic studies are dealing with electromagnetic waves covering a wide range of frequency. In Fourier transform term $\ e^{ - 2\pi ixy} \$, when x represents frequency, the corresponding y is time. This provides an alternate way to process signal in time domain instead of the conventional frequency domain. To realize this idea, Fourier transform from time domain to frequency domain is the essential process that enable us to translate raw data to readable spectra. Recent prosperity of Fourier transform in spectroscopy should also attribute to the development of efficient Fast Fourier Transform algorithm.
Introduction
The nature of trigonometric function enables Fourier transform to convert a function from the domain of one variable to another and reconstruct it later on. This is a robust mathematical tool to process data in different domains under different circumstances. Taking this principal idea and applying it in spectroscopy showed many impressive results in the early stage, which in other ways are very difficult to resolve. These benefits triggered a wide exploration of Fourier transform based methodology in a variety of spectroscopic techniques. At the same time, Fourier transform spectroscopic instruments are developed with great efforts by physicists and engineers. All these factors give rise to the wide use of Fourier transform spectroscopy.
In the following topics, the relevant mathematical background, the implementation of Fourier transform in spectroscopy and a brief overview of various Fourier transform Spectrometers will be addressed in sequence.
Fourier Series
The motivation of Fourier transform arises from Fourier series, which was proposed by French mathematician and physicist Joseph Fourier when he tried to analyze the flow and the distribution of energy in solid bodies at the turn of the 19th century. He claimed that the temperature distribution could be described as an infinite series of sines and cosines of the form shown in equation (1):
$f(x)= \dfrac{a_{0}}{2} + \displaystyle \sum_{n=1}^{\infty} \left(a_{n}\cos\dfrac{n\pi x}{L}+b_{n} \sin \frac{n \pi x}{L}\right) \tag{1}$
It turns out that this combination of sines and cosines series can be used to express any periodical function. As $n$ increases, the series will approach to the original function more closely.
If we use Euler's Identity (equation 2), as well as the exponential representations of sine (equation 3} and cosine (equation 4) in our Fourier Series, we will find a natural redefinition of our coefficients $a_n$ and $b_n$ into a single complex coefficient C_n (equation 5).
$e^{i\theta}=\cos{\theta}+ i\sin{\theta} \tag{2}$
$\cos{\theta}=\dfrac{1}{2} (e^{i\theta} + e^{-i\theta}) \tag{3}$
$\sin{\theta}=\dfrac{1}{2i} (e^{i\theta} - e^{-i\theta}) \tag{4}$
$C_n =\dfrac{1}{2}(C_n-ib_n) \tag{5}$
Using a bit of clever mathematics (complex conjugation, properties of odd function, rearranging summation set) we can represent our original Fourier series in terms of a complex exponential, shown as equation 6.
$f(x) = \sum_{n=-\infty}^{\infty} C_n e^{i\theta} \tag{6a}$
$\theta = \dfrac{2\pi n x}{L} \tag{6b}$
Writing the Fourier series in this exponential form helps to simplify many formulas and expressions involved in the transformation.
Fourier Transform
Then we can consider an extreme case, when L in equation 1, the summation becomes an integral as shown in equation (4)
$f(x) = \sum\limits_{-\infty} ^\infty c_n e^{n\frac{i\pi x}{L}} = \int_{ -\infty}^\infty c_n e^{n\frac{i\pi x}{L}} dn = \int_{-\infty}^\infty c_n e^{\frac{i\pi xn}{L}} dn \tag{7}$
This naturally gives the Fourier transform pair of f(x) and F(y). The relationships are shown below :
$F(y) = \int_{ -\infty }^{ +\infty } {f(x)e^{ - i2\pi yx} dx} \tag{8a}$
$f(x) = \int_{ -\infty }^{ +\infty } {F(y)e^{ i2\pi xy} dx} \tag{8b}$
Another important formula widely used to benefit In other cases, it is used to simplify the integral in the Fourier transform based on the symmetry of the function. But so far, all these are just about mathematics. Its story with spectroscopy should start from the mathematical description of electromagnetic waves.
Mathematical description of electromagnetic waves
Maxwell–Faraday equation and Ampère's circuital law give us electromagnetic wave equations to describe the characteristics of an electromagnetic wave.[1] Using the linearity of Maxwell's equations in a vacuum, the solutions of the equation can be decomposed into a superposition of sinusoids as shown below[3]:
$E(r,t) = \overrightarrow {E_0 } \cos (2\pi ft - \overrightarrow k \cdot r + \phi _0 ) \tag{9a}$
$B(r,t) = \overrightarrow {B_0 } \cos (2\pi ft - \overrightarrow k \cdot r + \phi _0 ) \tag{9b}$
Where t is time, f is the frequency, k=(kx,ky,kz) is the wave vector and $\phi$ is the phase angle.
This indicates that electromagnetic wave can be written as the sum of trigonometric functions with specific frequencies. Scientists already discovered the fact that frequency and time is a classic Fourier transform pair in Fourier transform relationship. All the Fourier transform pairs are connected by the Fourier transform term $e^{ - i2\pi yx}$. Regarding this case, we can use the term to transform between two variables in this pair, namely time and frequency. In this way, we can measure the properties of the electromagnetic wave in both conventional frequency domain and somehow more robust time domain.
Applying Fourier Transform-Fourier Transform Spectroscopy
Fourier transform are widely involved in spectroscopy in all research areas that require high accuracy, sensitivity, and resolution. All these spectroscopic techniques using Fourier transform are considered Fourier transform spectroscopy. By definition, Fourier transform spectroscopy is a spectroscopic technique where interferograms are collected by measurements of the coherence of an electromagnetic radiation source in the time-domain or space-domain, and translated into frequency domain through Fourier transform.
Interferometer-What it is used for and how it works?
How to introduce a time-domain or space-domain variable in the spectrometer is the primary question that needed to be addressed when we consider constructing a Fourier transform spectrometer. In the experimental set-up, a Michelson interferometer is commonly used to solve this problem. Different from the classical Michelson interferometer with two fixed mirrors (Figure 1.a), the interferometer used in Fourier transform spectrometer has a moving mirror at one arm (Figure 1.b).
a. b.
Figure 1. Scheme for Michelson interferometer [components: coherent light source; half-silvered beam-splitting mirror; two highly polished reflective mirrors; detector] (a) Stationary version [two fixed mirrors] (b) Movable version [One movable mirror and one movable mirror]
As shown in Figure 1.b, when a parallel beam of coherent light hits a half-silvered mirror, it is divided into two beams of equal intensities by partial reflection and transmission. After being reflected back, the two beams meet at the half-silvered mirror and recombine to produce an interference pattern, which is later detected by the detector. Manipulating the difference between these two paths of light is the core of Michelson interferometer. If these two paths differ by a whole number of wavelengths, the resulting constructive interference will give a strong signal at the detector. If they differ by a whole number and a half of wavelengths, destructive interference will cancel the intensity of the signal.
Measuring Interferograms
With a Fourier transform spectrometer equipped with an interferometer, we can easily vary the parameter in time domain or spatial domain by changing the position of the movable mirror.
But how data are collected by a Fourier transform spectrometer? A quick comparison between a conventional spectrometer and a Fourier transform Spectrometer may help to find the answer.
• Conventional spectrometer:
Monochromator is commonly used. It can block off all other wavelengths except for a certain wavelength of interest. Then measuring the intensity of a monochromic light with that particular wavelength becomes practical. To collect the full spectrum over a wide wavelength range, monochromator needs to vary the wavelength setting every time.
• Fourier transform spectrometer:
Rather than allowing only one wavelength to pass through the sample at a time, an interferometer can let through a beam with the whole wavelength range at once, and measure the intensity of the total beam at that optical path difference. Then by changing the position of the moving mirror, a different optical path difference is modified and the detector can measure another intensity of the total beam as the second data point. If the beam is modified for each new data point by scanning the moving mirror along the axis of the moving arm, a series of intensity versus each optical path length difference are collected.
So instead of obtaining a scan spectrum directly, raw data recorded by the detector in a Fourier transform spectrometer is less intuitive to reveal the property of the sample. The raw data is actually the intensity of the interfering wave versus the optical path difference (also called Interferogram). The spectrum of the sample is actually encoded into this interferogram.
Extracting the spectrum from raw data
Based on the previous discussion, it is predictable that, without further translation, the raw data collected on a Fourier transform spectrometer will be quite difficult to read. A Fourier transform needs to be performed to decode interferogram and extract actual spectrum I($\overline v$) from it. The following shows how to conduct a Fourier transform to decode:
The intensity collected by the detector is a function of the path length differences in the interferometer p and wavenumber $\overline v$[3]:
$I(p,\overline v ) = I(\overline v )[1 + \cos (2\pi \overline v p)] \tag{9}$
Thus, the total intensity measured at a certain optical path length difference (for each data point at a certain optical pathlength difference p) is:
$I(p) = \int_0^\infty {I(p,\overline v ) = I(\overline v )[1 + \cos (2\pi \overline v p)]} \cdot d\overline v \tag{10}$
It shows that they have a cosine Fourier transform relationship. So by computing an inverse Fourier transform, we can resolve the desired spectrum in terms of the measured raw data I(p) (10):
$I(\overline v ) = 4\int_0^\infty {[I(p) - \frac{1} {2}I(p = 0)]} \cos (2\pi \overline v p) \cdot dp \tag{11}$
An example to illustrate the raw data and the resolved spectrum is also shown in Figure 2.
Figure 2. Fourier transform between interferogram and actual spectrum[4]
The Fast Fourier Transform (FFT)
Fast Fourier Transform (FFT) is a very efficient algorithm to compute Fourier transform. It applies to Discrete Fourier Transform (DFT) and its inverse transform. DFT is a method that decomposes a sequence of signals into a series of components with different frequency or time intervals. This operation is useful in many fields, but in most cases computing it directly from definition is too slow to be practical. Fast Fourier Transform algorithm can help to reduce DFT computation time by several orders of magnitude without losing the accuracy of the result. This benefit becomes more significant when the number of the components is very large. FFT is considered a huge improvement to make many DFT-based algorithms practical. In Fourier transform spectrometer, signals are often collected by a series of optical or digital channels at the detector. Then FFT is of great importance to quickly achieve the following signal processing and data extraction based on DFT method.
Combining all these steps together, we can take a look at how the data from the sample are processed. The diagram is shown in Figure 3.
Figure 3. Data processing in FTIR
(Figure from ThermoNiolet at mmrc.caltech.edu/FTIR/FTIRintro.pdf)
Different operating modes in Fourier Transform Spectrometer
• Continuous/Scanning FTS
Continuous Fourier transform spectroscopy refers to the scanning form of FTS, in which by step moving one mirror, the whole range of optical path difference is measured. This is the most widely used mode in FTS, like most absorption spectra and emission spectra obtained by FTS.
• Pulsed FTS
In some Fourier transform spectrometers, depending on the feature of the involved spectroscopic technique and purpose of measurement, a pulsed Fourier transform technique may be applied instead of the scanning mode.
Pulsed FTS is different from conventional continuous FTS. It is not based on the transmittance technique, which is widely used in the absorption spectra, like FTIR. Instead, in pulse FTS, the idea is that the sample is first exposed to an energizing event, and this pulse induces a periodic response. The frequency of this response relative to the field strength is determined by the properties of the sample. Using Fourier transform to resolve the frequency will tell the information about the targeted analyte.
Pulse FTS is a relatively new improvement of FTS. Some examples are from pulse-Fourier Transform-Nuclear Magnetic Resonnance (FT-NMR), pulse-Fourier Transform-Electron Paramagnetic Resonnance (FFT-EPR) and Fourier Transform-Mass Spectrometry (FT-MS). Please refer to the following topics for more details about how they work.
• Stationary FTS
In addition to the continuous/scanning mode of FTS, a number of stationary Fourier transform spectrometers are also available to meet special needs.
The principle of the interferometer and the analysis of its output signal is similar to the typical scanning FTS. But the signal is collected at certain optical path length differences rather than scanning over the whole range of the path difference.
An overview of various FTS techniques
Fourier transform spectroscopy can be applied to a variety of regions of spectroscopy and it continues to grow in application and utilization including optical spectroscopy, infrared spectroscopy (IR), nuclear magnetic resonance, electron paramagnetic resonance spectroscopy, mass spectrometry, and magnetic resonance spectroscopic imaging (MRSI). Among them, Fourier Transform Infrared Spectroscopy (FTIR) has been most intensively developed, which uses scanning Fourier transform to measure the mid-IR absorption spectra. .
Nuclear Magnetic Resonance (NMR) and Electron Spin Resonance Spectroscopy (EPR) are two magnetic techniques that use pulse Fourier transform mode. A Radio Frequency Pulse (RF Pulse) in a strong ambient magnetic field background is used as the energizing event. This RF Pulse directs the magnetic particles at an angle to the ambient strong magnetic field, causing gyration of the particle. Then the resulting gyrating spin induces a periodic current in the detector coil. This periodic current is recorded as the signal. Each gyrating spin has a characteristic frequency relative to the strength of the ambient magnetic field, which is also governed by the properties of the sample.
Fourier Transform Mass Spectrometry (MS) is also operated at pulse Fourier transform mode. Different from NMR and EPR, the injection of the charged sample into the strong electromagnetic field of a cyclotron acts as the energizing event in MS. The injected charged particles travel in circles under the strong electromagnetic field. The circular pathway will thus induce a current in a fixed coil at one point in their circle. Each traveling particle exhibits a characteristic cyclotron frequency relative to the field strength, which is determined by the masses in the sample.
Outside Links
• Please visit the free source for a consice introductory liquid pulse FT-NMR textbook[5] : www.analytik.ethz.ch/praktika...nmr/ft-nmr.pdf
• Fourier Series en.Wikipedia.org/wiki/Fourier_series
• Electromagnetic wave equation en.Wikipedia.org/wiki/Electro..._wave_equation
• Fourier Transform en.Wikipedia.org/wiki/Fourier_transform
• Fourier Transform Spectroscopy en.Wikipedia.org/wiki/Fourier...m_spectroscopy
• Fast Fourier Transform en.Wikipedia.org/wiki/Fast_Fourier_transform
• Discrete Fourier Transform en.Wikipedia.org/wiki/Discret...rier_transform
Problems
1. Search literatures to find the advantages and the limitations of Fourier transform spectroscopic techniques.
2. Perform a Fourier transform to show how to extract spectrum, equation (11), from the raw data in equation (10).
3. Compare interferometer with monochromator (at least two aspects).
4. What are the important components to make Fourier transform spectrometer practical? What are they used for?
5. Based on the information introduced in this module, design any one of the Fourier transform spectrometers mentioned in the context.
Solutions
1. Advantages
Firstly, Fourier transform spectrometers have a multiplex advantage (Fellgett advantage) over dispersive spectral detection techniques for signal, but a multiplex disadvantage for noise; Moreover, measurement of a single spectrum is faster(in the FTIR technique) because the information at all frequencies is collected simultaneously. This allows multiple samples to be collected and averaged together also resulting in an improvement in sensitivity; In addition, FT spectrometers are cheaper than conventional spectrometers because building of interferometers is easier than the fabrication of a monochromator (in the FTIR technique). So most commercial IR spectrometers are built based on FTIR techniques.
Limitations: practical frequency regions limited (FT UV-vis is not quite practical)
2. See the following figure for the solution:
3. Interferometer vs. Monochromator
Interometer: a. Collect signal in time or spatial domain; b. Measure all frequencies in the incident beam at one time; c. Determined by the interferometer, raw data from FT spectrometer is an interogram, which needs to be Fourier Transform back to get spectrum.
Monochromator: a. Collect signal in frequency domain; b. Scan each wavelength and measure the intensity for each single wavelength at a time; c. Determined by the feasure of monochromator, spectrum can be directly collected from the spectrometer;
4. Interferometer and Fast Fourier Transform Data Analyzer
Interferometer: to generate continuous optical path length difference and enable the idea to collect data in the time or spatial domain;
Fast Fourier Transform Data Analyzer: to quickly transform the raw data (interferogram) to spectrum by using fast Fourier transform algorithm;
5. FTIR instrumentation: Figure 4
Figure 4. A FTIR Spectrometer Layout (Figure from ThermoNiolet at mmrc.caltech.edu/FTIR/FTIRintro.pdf)
FT-NMR instrumentation:
A modern high resolution liquid FT-NMR instrumentation is shown in Figure 5:
Figure 5. A schematic diagram of liquid FT-NMR[5]
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Spectroscopic measurements are typically taken in one of two domains: frequency or time. These measurements are given the terms frequency-resolved or time-resolved. Frequency-resolved measurements are the most familiar forms of spectroscopy. UV/Visible, IR, Raman, and X-ray spectroscopy are typically done in the frequency domain. This type of spectroscopy acquires data across a range of frequencies (or wavelengths). The data acquired is typically in the form of an light intensity which can in turn be interpreted as absorbance, transmittance, reflectance, or photon scattering depending on the instrument and technique being used.
Introduction
The less familiar time-resolved spectroscopy includes Ultrafast laser spectroscopy and florescence. In this form of spectroscopy data is acquired over a range of time. This data can be at a single wavelength or at multiple wavelengths, depending on the specific technique. Some spectroscopic techniques, such as Ultrafast laser spectroscopy, FT-NMR and FT-IR, span both frequency and time domains. In the case of Ultrafast laser spectroscopy, useful data is acquired in both the time and frequency domains. FT-NMR and FT-IR acquire data in the time domain. That data is then converted into a signal in the frequency domain using a process called Fourier Transform. This is covered in more detail below.
Frequency Resolution
Frequency is defined as inverse time. The unit is typically given in inverse seconds(s-1) or hertz(hz). Frequency is used to represent the number of cycles occurring in a given time period. These cycles could be any repetitive process including: the periodic motion of a harmonic oscillator, the sinusoidal propagation of electromagnetic radiation, or the rotation of a rigid rotator. The most relevant for spectroscopy is the propagation or electromagnetic radiation, or light. This is often represented in many different forms that, though not technically frequency, are related to frequency, and therefore fall into the frequency domain. These include representing frequency as wavelength (nm), wavenumber (cm-1), and photon energy (eV). These are all connected by a few simple equations given below. (c=speed of light, E=energy, v=frequency, w=wavenumber, h=Planck's constant,lambda=wavelength)
$\lambda = \dfrac{c}{\nu}$
$E = h\nu$
$w = \dfrac{1}{\lambda}$
The frequency domain is the most familiar domain in spectroscopy. UV/visible, infrared, photoelectron, microwave, and X-ray spectroscopy all have applications in the frequency domain. The results of these spectroscopic techniques are typically given in some form of intensity versus wavelength. The most familiar is likely the steady state ultraviolet/visible absorption spectrum(an example is shown below).
Frequency Resolved Visible Absorption Spectrum
The energy of absorbed light corresponds to the energy of transition between two eigen states of the system. In the case of visible spectroscopy these states are electronic states.
Time Resolution
Spectra can also be acquired in the time domain. Rather than acquiring spectra by averaging data over a relatively long time range, data is acquired over discrete time intervals or, in some cases, continuously. This may be done over one or many wavelengths. Time resolved spectroscopy observes the change in eigenstates with respect to time. In order for data from time-resolved spectroscopy to be useful, the spectroscopy must be suited to the time scale of the process of interest. Below is a table of the approximate time scales and spectral ranges of physical processes that may be interesting.
Process Time Applicable Spectral Range
Singlet Electronic Excited State Lifetime femto-nanoseconds Visible
Triplet Electronic Excited State Lifetime nanoseconds-minutes Visible
Molecular Vibration Excited State Lifetime pico-milliseconds Infrared
Nuclear Rotation pico-microseconds Radio
Molecular Reaction Kinetics eons-picoseconds Varies
If a spectroscopy with suitable spectral region and time resolution is available, time resolved spectroscopy can be used to study kinetics, reactions, and lifetimes. Common applications of time-resolved spectroscopy include ultrafast laser spectroscopy and time-resolved florescence.
Fourier Transform
The process of Fourier Transform is a mathematical process used to move from one set of coordinates to another. The most spectroscopically relevant fourier transform is from the time domain to the frequency domain. In this case, a signal originally measured in the time domain can be converted into a signal in the frequency domain. This is done via the mathematical process shown below.
$\displaystyle F(v) = \int_{-\infty}^{\infty} f(t)e^{-2i\pi vt}dt$
$\displaystyle f(t) = \int_{-\infty}^{\infty} F(v)e^{-2i\pi vt}dv$
A mathematical relation known as Euler's Formula is an important identity when using Fourier Transform, particularly with sine and cosine functions. This is shown below.
$e^{\pm ix} = cos(x) \pm isin(x)$
A simple graphic representation of Fourier Transform is shown below.
Original Signal in Time Domain
Fourier Transformed: Signal in Frequency Domain
Fourier Transformed Again: Signal in Time Domain (The Components of the Original Signal)
Sum of the Components(Equivalent to the Original Signal)
An example fourier transform with sine is given in the links section.
Common Application of Fourier Transform
Some fields of spectroscopy use measurements taken in the time domain to gain information about the frequency domain. These spectroscopies include Nuclear Magnetic Resonance, Fourier Transform Ion Cyclotron Resonance Mass Spectroscopy(FT-ICR MS) and Fourier Transform Infrared spectroscopy(FT-IR). Unlike the above graphic representation of Fourier Transform, these systems yield transforms that are impossible to compute by hand. Computer algorithms must be used.
The initial signal typically forms a beat pattern. This can be in the form of an interferogram or a free induction decay(FID), in the cases of FT-IR and NMR, respectively. Graphic representations of an interferogram and FID are shown below alongside IR and NMR spectra.
FT-IR Spectra
Interferogram(X-axis: Time|Y-Axis:Amplitude) Infrared Spectrum(X-axis: Wave Number (cm-1)|Y-axis: Percent Transmission)
FT-NMR Spectra
Free Induction Decay(X-axis: Time|Y-axis: Amplitude) NMR Spectrum(X-axis: Chemical Shift (ppm)|Y-axis: Amplitude)
Spectroscopy in Both Time and Frequency Domains
Some spectroscopies yield data in both time and frequency domains. The most prominent of these techniques is time-resolved laser spectroscopy. By measuring complete spectra at discrete time intervals, spectral evolution with respect to time can be monitored. This technique is unique in it's ability to collect data in multiple domains with femtosecond time resolution. This allows electronic states to be monitored both as they evolve over time and statically at any particular time along the time scale of the instrument. Below is an example of a typical ultrafast spectrum, notice both wavelength and time domains. This three-dimensional spectrum can be deconstructed to yield either time or frequency dependent results in two-dimensions. This is also shown below.
Time Resolved Visible Absorbance Spectrum
(X-axis:Wavelength(nm)|Y-axis:Time(ps)|Z-axis:(ABS)[Shown as color])
The time resolved spectrum shown above is plotted as a contour plot showing the Z-axis as a color gradient from red(low signal) to blue(high signal). This contour plot contains a multitude of information, but is by itself not terribly useful. It is most easily analyzed by taking crossections at a single wavelength or time. Each is shown below.
Frequency Domain Signal (A crossection of the Time Resolved Spectrum constant time)
The above spectrum is a crossection of the complete time resolved spectrum. This particular crossection is the frequency domain signal at 56.75 picoseconds after the sample has been excited by a laser pulse.
Time Domain Signal (A crossection of the Time Resolved Spectrum at a single Wavelength)
The above spectrum is a crossection of the complete time resolved spectrum. This crossection is the time resolved signal at 618 nanometers. This crossection contains kinetic data on the eigen state that absorbs at 618nm.
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Electron Paramagnetic Resonance (EPR) is a remarkably useful form of spectroscopy used to study molecules or atoms with an unpaired electron. It is less widely used than NMR because stable molecules often do not have unpaired electrons (i.e., paramagnetic). However, EPR can be used analytically to observe labeled species in situ either biologically or in chemical reactions.
• ENDOR - Theory
Electron-nuclear double resonance spectroscopy (ENDOR) is a powerful advanced EPR technique that probes the environment surrounding paramagnetic centers. It is of great use to further examine paramagnetic samples which give complicated spectra via the standard EPR method due to electronic-nuclear interactions manifested as the hyperfine interaction.
• EPR: Introduction
Though less used than Nuclear Magnetic Resonance (NMR), Electron Paramagnetic Resonance (EPR) is a remarkably useful form of spectroscopy used to study molecules or atoms with an unpaired electron. It is less widely used than NMR because stable molecules often do not have unpaired electrons. However, EPR can be used analytically to observe labeled species in situ either biologically or in a chemical reaction.
• EPR - Interpretation
Electron paramagnetic resonance spectroscopy (EPR), also called electron spin resonance (ESR), is a technique used to study chemical species with unpaired electrons. EPR spectroscopy plays an important role in the understanding of organic and inorganic radicals, transition metal complexes, and some biomolecules.
• EPR - Parallel Mode Operation
This module presents the theory that describes how EPR transitions can be induced in integer high spin systems by the application of a modulating magnetic field parallel to the bond axis (z-axis), as well as some of the applications of this technique to various molecular systems.
• EPR - Theory
Electron Paramagnetic Resonance (EPR), also called Electron Spin Resonance (ESR), is a branch of magnetic resonance spectroscopy which utilizes microwave radiation to probe species with unpaired electrons, such as radicals, radical cations, and triplets in the presence of an externally applied static magnetic field.
• Hyperfine Splitting
This splitting occurs due to hyperfine coupling (the EPR analogy to NMR’s J coupling) and further splits the fine structure (occurring from spin-orbit interaction and relativistic effects) of the spectra of atoms with unpaired electrons. Although hyperfine splitting applies to multiple spectroscopy techniques such as NMR, this splitting is essential and most relevant in the utilization of EPR.
Electron Paramagnetic Resonance
Electron-nuclear double resonance spectroscopy (ENDOR) is a powerful advanced EPR technique that probes the environment surrounding paramagnetic centers. It is of great use to further examine paramagnetic samples which give complicated spectra via the standard EPR method due to electronic-nuclear interactions manifested as the hyperfine interaction.
Introduction
The primary interest of this advanced EPR method lies in elucidation of the hyperfine interactions or coupling (hfc) of an unpaired electron with surrounding nuclei. In a standard CW-EPR experiment, one can often detect these hyperfine splittings, however accurate calculation of their values is generally a challenge. This is due to complexity of the environment of the spin system, which can result in splitting of the EPR transition due to coupling of the unpaired electron with several neighboring nuclei, resulting in overlapping signals that cannot always be distinguished. Inhomogeneous broadening of the EPR lineshapes due to a distribution in some property of the system, such as the orientation of the spins in a powder or frozen solution, also broadens each peak, thereby severely limiting assignment of hfc parameters and identification of these nuclei. ENDOR is able to greatly simplify these CW-EPR spectra by reducing the number of peaks, displaying only 2N peaks as opposed to 2N peaks for N-sets of equivalent nuclei, and allowing for selectivity of which couplings are observed through optimization of the ENDOR parameters. For a comprehensive discussion of EPR theory and relevant information to comprehensive of the content of this module please refer to the EPR: Theory module.
The Hyperfine Interaction
The permanent magnetic dipole of nuclei such as hydrogen, with I = ½, can interact with that of an unpaired electron if they are located closely in space. This hyperfine interaction is independent of the strength of the static magnetic field that is applied in a magnetic resonance experiment and leads to a splitting of the electronic spin states by an energy difference of half of the hyperfine coupling, $A$. The energy splitting can be described by the equation:
$E_{hfc}=A\ m_I\ m_S \label{1}$
where $A$ is the hyperfine coupling constant (hfc). This hfc constant can contain both isotropic and anisotropic contributions, the latter of which is discussed in the dipolar coupling section of the module, but depends on both the electronic and nuclear g values, the electronic and nuclear magnetic moments, and the distance between them. These magnetic moments with the B0 field along the z-axis are:
$\mu_{ez}=\gamma_e \hat{S_z} \hbar=-g\beta_e \hat{S_z} \label{2}$
$\mu_{nz}= \gamma_{n} \hat{I_z} \hbar=+g_n \beta_n \hat{I_z}\label{3}$
for the electron and nucleus respectively. The isotropic hyperfine constant, that in the absence of a dipolar coupling contribution, as well as the Hamiltonain describing isotropic hfc are given by the equations below.
$A_{iso}=\dfrac{2\mu_0}{3}g\beta_e \beta_n | \Psi(0) |^2 \label{4}$
$\hat{\mathcal{H}} =\dfrac{2\mu_0}{3}g\beta_e\beta_n | \Psi(0) |^2 \hat{S_z}\hat{I_z} \label{5}$
The hfc constant is generally expressed in terms of units of frequency as:
$\dfrac{A_{iso}}{h}\label{6}$
and can also be expressed in magnetic field units with the equation:
$a_{iso}=\dfrac{A_{iso}}{g_e\beta_e} \label{7}$
It is generally this parameter that is sought to be determined in via ENDOR experiments.
Fundamentals of the technique
The key to the ENDOR technique lies in the use of both microwave, MW, and radiofrequency, RF, (B1) applied fields, resulting in a hybrid EPR and NMR experiment. Like all EPR experiments, the sample is placed in a homogenous magnetic field, B0, and secondary fields are applied to induce transitions between electronic spin states, however ENDOR also induces nuclear spin transitions to extract hfc parameters. The important requirement of the technique is the saturation of a specific EPR transition, resulting in an almost equal population in each of the two spin states, for example ms = ± ½. Then an RF frequency signal is applied to induce and also partly saturate NMR transitions, which results in a desaturation of the corresponding EPR transition. The selection rules for ENDOR are the same as that for NMR, $\Delta M_S=0, \ \Delta M_I=\pm1$. The saturation requirement for ENDOR can be shown below by the following two equations:
$\gamma_e^2 {B_1}^2 T_{1e} T_{2e} \geq 1,\ \gamma_n^2 {B_1}^2 T_{1n} T_{2n} \geq 1\label{8}$
which show that the product of the gyromagnetic ratios, the saturating MW and RF field intensities, and the characteristic relaxation times of the system must be unity or larger.
From here we will consider the simple case of the interaction between one electron, S = ½, and one proton, I = ½, for which the energy levels are shown below. The first splitting arises from the electron zeeman splitting discussed below, which removes the degeneracy of the electron ms = ± ½ spin states. The next splitting arises from the hyperfine coupling of the electron and nuclear spins, split by half of the hfc magnitude. The final splittings show the ENDOR/NMR transitions which comprise the spectrum. The wide arrows indicate the EPR transitions which are saturated.
Population Polarization
The diagram below shows the changes in populations that occur during an ENDOR experiment, as it is these changes that lead to the ENDOR effect and that are detected. Prior to the application of MW frequency photons, or a preparation pulse, the system exists with the majority of the population in the ms = - ½ spin manifold, as determined by the Boltzmann population distribution and show in (a) of the figure below. With the application of a MW field, or preparation MW pulse, the population is moved from the ms = - ½ to ms = + ½ manifold as expected by the inversion sequence, seen in (b) of the figure. Finally, after the mixing period and induction of on-resonance NMR transitions, population transfer occurs between the upper mI levels as seen in (c.) of the figure below. This is then detected as a change in the EPR signal with the CW technique, or with an ESE, electron spin echo, sequence using a pulsed ENDOR strategy. If the RF $\pi$ pulse is off resonance, then this final population polarization does not occur.
Continuous Wave (CW) ENDOR
The first ENDOR technique invented was continuous wave ENDOR, in which the radio frequency source is modulated while the B0 applied field is held constant unlike a standard EPR experiment. The first derivative spectra are thus still obtained via this techniques. The EPR transitions seen above are saturatured, and an applied RF field desaturates these transitions, which are observed in the CW ENDOR spectrum as changes in the EPR signal intensity as a function of the frequency of the RF field applied. This is the most simple ENDOR experiment to discuss, however it has become a less commonly used method for acquisition of ENDOR spectra in recent years.
Pulsed ENDOR
Pulsed techniques offering several advantages over continuous wave ENDOR. CW ENDOR is the most sensitive of the techniques, however pulsed ENDOR provides higher resolution as well as allowing observation of weakly coupled nuclei. It also offers insight into the relaxation times of the system, is less susceptible to artifacts, and does not require balancing of relaxation times with RF power like the CW technique. Therefore, pulsed-ENDOR has been the most common technique in the past several decades, using pulse sequences predominantly developed by Mims and Davies. For an ENDOR experiment, either Mims or Davies, the homogenous applied magnetic field B0 is held constant, as opposed to the normal scanning of this field in a standard CW-EPR experiment. The field is selected where the EPR transition occurs by performing an ESE field sweep prior to beginning the experiment, thus fixing both the B0 and MW-B1 fields. This leaves the RF spectrum to be swept during the mixing period, much like in a standard NMR experiment. This is done by repeating the pulse sequence millions of times and using a different RF frequency during the mixing period for each sequence, which is done by randomly sampling each of the RF frequencies in the range. Electron spin echo, or ESE, is a technique used for signal detection in most pulsed advanced EPR methods
Davies ENDOR
In Davies ENDOR, a preparation $\pi$, 180o, pulse is used in order to invert the magnetization of the spins in the applied static B0 field. This essentially creates a hole in the EPR spectrum, whose width and depth depend on the length of the pulse applied, with long pulse producing narrow holes. During the mixing period a $\pi$ RF pulse is applied and if the RF frequency is resonant with an NMR transition, magnetization will be transferred to the other ms spin manifold, otherwise no mixing will occur to fill in the hole that the inversion pulse creates. During the detection period, the z-component of the magnetization is measured using a two pulse echo sequence, and one detects essentially the EPR signal that is restored during the mixing period. This technique is most suited to nuclei with large hfc values.
Mims ENDOR
The Mims ENDOR technique is based on a stimulated electron spin echo (ESE) sequence, using a two $\frac{\pi}{2}$, 90o, preparation pulse sequence to invert the electron spin population, and a final $\frac{\pi}{2}$ pulse after the mixing period to stimulate the ESE for signal detection. Between the preparation pulses and the final pulse, a radio frequency ppulse is used to invert the nuclear spin population, resulting in polarization transfer between the nuclear and electronic transitions in the so called “mixing period.”
This results in the actual ENDOR transitions, changes in the intensity of the EPR transition, that are detected by use of the ESE MW-pulse sequence. The echo intensity is subsequently measured as a function of the RF frequency to give the characteristic ENDOR spectrum shown below.
The hfc can be determined experimentally:
$\begin{vmatrix}{{\nu_n}_1}\mp{{\nu_n}_2}\end{vmatrix}=h^{-1}\begin{vmatrix}A\end{vmatrix} \label{9}$
where the plus sign is for the low field limit and the minus sign for the high field limit. In the low field limit, the nuclear Larmor frequency, or frequency of the precession of the nuclear spin in the presence of a magnetic field, is less than half of the hyperfine coupling, A. In the high field limit, the nuclear Larmor frequency is greater than half of the hyperfine coupling term, resulting in different assignments of the hfc parameter A according to the above equation.
The two techniques are very similar to each other in their implementation, however they are relatively complementary in their results and usefulness. The Mims ENDOR technique is most suited to weakly coupled nuclei, with small hyperfine coupling constants, A, generally less than 2 MHz and as low as 0.1 MHz. This makes the Davies ENDOR pulse sequence most useful for those nuclei with relatively large hyperfine couplings, therefore making the assignments of strongly and weakly coupled nuclei greatly simplified, as well as the corresponding spectra associated with each of the pulse sequences.
Experimental Parameters
The parameters to be specified in any given ENDOR experiment include the static B0 field, the microwave frequency, the RF power and range, the pulse lengths, the delay time ($\tau$), and the repetition rate. The delay time, or time between data collection at each RF frequency, is determined by the timescale on which equilibrium magnetization is restored. This is determined by a couple relaxation parameters, most importantly the spin lattice relaxation time (T1) and the transverse relaxation time (T2). T1 determines how quickly the magnitization realigns with the z-axis (B0) field, and T2 determines how quickly the magnetization in the x-y plane (transverse) disappears.
In a Davies ENDOR experiment, the microwave frequency bandwidth is the most important parameter to be set. The preparation inversion pulse burns a hole in the EPR spectrum, i.e. saturates an EPR transition, which is then filled in through population transfer by the RF inversion pulse. It is this parameter that needs to be optimized in order to get the best spectral intensity and resolution. In Mims ENDOR, the $\tau$ value is the most important parameter to set, and the microwave frequency bandwidth is less important, which is the opposite case for Davies. However, a hole is still burned in the EPR spectrum, now with the use of two $\frac{\pi}{2}$ pulses in the preparation period, but it is modulated according to the following equation.
$1+\cos{2\pi\frac{(B_0-B)}{g\beta}\tau} \label{10}$
Application of the RF inversion pulse during the mixing period shifts the magnetization away from B0 and changes this magnetization pattern. Mims ENDOR intensity then depends on the differences in the magnetization pattern following the preparation pulse, and those following the mixing period. Therefore Mims intensity is larger for larger differences in sinusoidal magnetization.
Spin Hamiltonian
Although the spin Hamiltonian applies to all EPR techniques, it is imperative to highlight the term of most importance for ENDOR spectroscopy which is the hyperfine coupling term describing the interaction of an unpaired electron with nuclei in its vicinity. The Hamiltonian for a system consisting of two spins, an electron with S = ½ and proton with I = ½ is shown below.
$\hat{H} = \hat{H}_{EZ}+\hat{H}_{NZ}+\hat{H}_{HF}+\hat{H}_{Q} \label{11}$
$\hat{H} = \mu_B {\bf{B}}^T \bf{g} \hat{\bf{S}} - \mu_n g_n {\bf{B}}^T\hat{\bf{I}} +\hat{\bf{S}}^T {\bf{A}} \hat{\bf{I}} +\hat{\bf{I}}^T {\bf{Q}} \hat{\bf{I}} \label{12}$
The terms shown are the electron Zeeman term, nuclear Zeeman term, hyperfine coupling term, and the nuclear quadrapole interaction term (which does not apply to nuclei with I < 1). Both the electron and nuclear Zeeman terms arise from the splitting of the spin states which are degenerate in the absence of an external magnetic field. The splitting of the spin up/down energy levels are linearly dependent on the strength of the applied B0 field according to these terms.
The third term describing the hyperfine interaction of the electron spin with nuclear magnetic moment is the most important term of the ENDOR technique. The hyperfine coupling magnitude and the identification of the nuclei from which the hfc arises is generally what ENDOR is used to elucidate, especially for the cases mentioned in the intro such as couplings to several nuclei or inhomogeneous broadening of a CW-EPR spectrum. The hyperfine coupling term $A$ has two components arising from both an isotropic term, $a_{iso}$, and in the case of a dipolar coupling interaction, an anisotropic term, $a_{dip}$. The final term is the nuclear quadrapole interaction that arises for nuclei with I > 1, and this term is inherently anisotropic.
$\hat{\mathcal{H}} = \mu_B \bf{B}}^T{\bf{g} \hat{\bf{S}} - \mu_n g_n \bf{B}}^T\hat{\bf{I}+ \hat{\bf{S}}^T \bf{A}}\hat{\bf{I}}+\hat{\bf{I}}^T \bf{Q} \hat{\bf{I}}+\hat{\bf{S}}^T \bf{D}\hat{\bf{S}}+\hat{\bf{S}_1}^T \bf{J}\hat{\bf{S}_2}} \label{13}$
The full spin Hamiltonian is shown above which has terms that do not arise from the one electron, S = ½, and one proton, I = ½ being described, which includes a zero field splitting term as well as an electron-electron interaction term, both of which arise from the presence of more than one electron spin.
Anisotropy and Dipolar Coupling
The nuclear quadrupole interaction is intrinsically an anisotropic parameter of a system with I > ½, but the most important parameters of interest to the ENDOR technique that may contain anisotropy are hyperfine coupling, and less so the electron g value. The electronic g value anisotropy arises from the effect of orbital angular momentum on electron spin energy levels, which can provide information on the orbitals containing unpaired electrons. Even more important to ENDOR spectroscopy is hyperfine anisotropy, which is manifested in the dipolar coupling component of the hyperfine interaction. Hfc has two components, the isotropic contribution Aiso, and the anisotropic dipolar coupling contribution, denoted by Adip or T. Aiso is solely a result of unpaired electron spin in s-type orbitals, as they are the only type with an electron density at the nucleus that is finite.
$A_{iso}=\frac{2}{3}\mu_0\mu_e\mu_n\begin{vmatrix}\Psi(0)\end{vmatrix}^2$
or
$A_{iso}=\frac{2\mu_0}{3}(g_e\beta_eg_n\beta_n)\langle\rho_x\rangle \label{14}$
This unpaired spin density at the nucleus can arise from contact with sp-type ligand orbitals as well as by the unpaired spin density in other orbitals causing polarization of core s electrons, and is classified as direct or local contact. The anisotropic component, arising from dipolar coupling, can be local or nonlocal, and is only seen in frozen solutions or powders by EPR. Indirect dipolar coupling is a through space interaction between an electron point-dipole and nuclear-point dipole, such as that between a paramagnetic metal and a hydrogen nucleus bound to the metal through an oxygen atom. The value of the dipolar coupling can be determined by the equation
$A_{dip}=\frac{-\mu_0}{4\pi}(g_e\beta_eg_n\beta_n)\frac{3\cos^2\theta-1}{R^3} \label{15}$
Where R is the distance from the electron point-dipole to nuclear point-dipole and theta is the angle between the bond connecting the two particles. Direct dipolar coupling would arise from the metal being directly bound to the hydrogen atom, not through a donor atom. These anisotropic effects of the hyperfine coupling can make the ENDOR spectrum distorted from the above sketches showing two symmetric peaks separated by a clean baseline.
The above X-band spectra simulated using the EasySpin toolbox for MATLAB, developed by Dr. Stefan Stoll, demonstrates this fact. An anisotropic hyperfine coupling tensor was used to simulate the above left Mims ENDOR spectrum of a single proton and single electron, and an isotropic g tensor was used for each to highlight the effect of an anisotropic hyperfine tensor. The spectrum no longer contains two symmetric peaks like the idealized spectra shown in the Mims ENDOR spectrum on the right, that was simulated with an isotropic hyperfine term, due to the anisotropy of the hyperfine term from the dipolar coupling contribution discussed above. This anisotropy can complicate the interpretation of ENDOR spectra, however it can also provide a plethora of information on the symmetry and environment of a paramagnetic center.
Problems
1. For a system with a single unpaired electron interacting with 4 identical nuclei all with I = 1/2 how many lines would be observed in the CW-EPR spectrum and the ENDOR spectrum?
2. What effect does an increase in the static B0 magnetic field have in a one electron one nucleus system (S = 1/2, I = 1/2) on the separation of the two peaks and the nuclear Larmor frequency?
3. Why is it that an ENDOR experiment is so much more sensitive than an ordinary NMR experiment?
4. For the Hydrogen atom, with S = 1/2 and I = 1/2, which electronic orbitals will give rise to a hyperfine interaction term in the spin Hamiltonian?
Solutions
1. The number of CW-EPR peaks for the one electron (S = 1/2) and 4 nuclei (I = 1/2) system is 24 = 16 peaks observed in this spectrum. ENDOR reduces the number of peaks in half, hence 2*4 = 8 peaks are seen in this spectrum.
2. An increase in the static applied magnetic field has no effect on the separation of the 2 ENDOR peaks, as the value of the hyperfine coupling remains constant regardless of the strength of the applied field. However, since the nuclear Larmor frequency (as well as electron Larmor frequency) is linearly dependent on the strength of the applied field, the spectrum would be shifted to a higher frequency as the strength of the field is increased as $g_n\beta_nB$.
3. This is mainly due to differences in population distribution in both types of experiments. ENDOR is more sensitive due to larger population distributions between spin up and down becuase the transitions are inherently larger energy, the rate of absoprtion of radiation is higher in EPR/ENDOR, and in an ENDOR experiment the electron magnetic field adds with the external field leading to a large net field at the nucleus and thus a larger population difference between spin states.
4. For all atoms, in order for isotropic hyperfine splitting to occur there must be a non-zero probability that the electron may reside at the nucleus. It is apparent that only s-type orbitals give rise to a non-zero probability at the nucleus, as p, d, f, ..., orbitals all contain a node at the nucleus (for $\begin{vmatrix}\Psi(0)\end{vmatrix}^2$). Therefore, the 1s1 ground state electronic configuration of Hydrogen will give rise to hyperfine splitting, as well as all excited states with the electron in an s orbital, e.g. 2s1, 3s1, 4s1, etc..
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Electron_Paramagnetic_Resonance/ENDOR_-_Theory.txt
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This article is about the experimental application of Electron Paramagnetic Resonance spectroscopy (EPR). For theoretical background on EPR, see alternate article EPR: Theory. The most basic application of Electron Paramagnetic Resonance spectroscopy requires the use of a microwave radiation source, an electromagnet, a resonator, and a detector.
Background
The experimental procedure is to place a paramagnetic sample in the resonator, which is designed to resonate at a specific microwave frequency. The resonator is positioned between two electromagnets, which vary in magnetic field strength depending on how much current is run through them. Electromagnetic radiation in the microwave frequency range is produced using a klystron and channeled into the resonator. While the microwave radiation is kept at a constant frequency, the strength of the applied magnetic field will be swept. Resonance is detected by a decrease in the amount of microwave radiation that is being reflected out of the resonator. Reflected microwaves are channeled specifically into the detector by a special component called a circulator. Resonance will be detected at the field strength that corresponds to the energy splitting of the spin states of the unpaired electron.
Figure 1: Basic experimental setup of an EPR spectrometer.
These are the most basic elements in an experimental setup for the observation of electron paramagnetic resonance. This article first elaborates on each of these basic elements of EPR Instrumentation and introduces other advanced elements necessary for collecting more sophisticated spectra, which are used to characterize an unpaired electron and its the environment. The experimental applications of this spectroscopy are then briefly reviewed.
Instrumentation
History of EPR
EPR was first detected in 1945 by Yevgeny Zaboisky at Kazan State University in the Soviet Union. The instrumentation required for EPR was greatly benefited by the development of RADAR during WWII, which required the use of reliable and tunable microwave sources. After the war, the necessary components for an EPR spectrometer were cheap and available. Initially, the field was advanced with the production of home-built EPR spectrometers. In the 1980s, commercially built EPR spectrometers became available. The German company Bruker leads the field in the production of commercially available EPR spectrometers, as well as NMR spectrometers and MRI detectors. Today, EPR spectroscopy is done at a wide range of frequencies and fields.
Microwave radiation source
A microwave radiation source is necessary for any EPR spectrometer, because this the energy that corresponds with a detectable splitting of the electronic spin states. Microwave radiation is typically produced and amplified by a klystron, which is capable of tuning waves to a precise frequency, amplitude, and phase. The microwaves are then channeled into the resonant cavity by use of either waveguides or coaxial cables. Waveguides are the most common method of microwave propagation. Waveguides are essentially open air, brass rectangular channels with dimensions that correspond to the wavelength of the radiation that is to be propagated. However, for low power, low frequency microwave radiation, like that often used in EPR, special coaxial cables have been found to be just as effective.
In order to observe resonance, the frequency of the microwaves must correspond to the splitting of the spin states of the electron, which is determined by the strength of the magnetic field. Thus, the strength of the magnetic field required is dependent on the frequency of radiation used, and vise versa. In addition, the dimensions of the waveguides and the resonator are specific to the microwave frequency. In this light, an experimental setup requires a corresponding combination of microwave energy, magnetic field strength, and resonator/waveguide dimensions. Experimental setups are classified by the frequency of microwave energy utilized, by the table below:
Table 1: Letter designations of various microwave frequency ranges.
Letter Designation Frequency Range
L Band 1-2 GHz
S Band 2-4 GHz
C Band 4-8 GHz
X Band 8-12 GHz
Ku Band 12-18 GHz
K Band 18-26.5 GHz
Ka Band 26.5-40 GHz
Q Band 30-50 GHz
U Band 40-60 GHz
V Band 50-75 GHz
E Band 60-90 GHz
W Band 75-110 GHz
F Band 90-140 GHz
D Band 110-170 GHz
This naming system is a vestigial organ of WW2 RADAR use. The letters assigned to a given range are random and were used as code to prevent the axis powers from knowing which frequency was being used to detect incoming bombers. The first observation of electron paramagnetic resonance was seen at S-band. The majority of EPR spectrometers operate at X-band, a mid-range frequency, due to the fact that equipment from this range was most easily available after WW2. In addition, this is a suitable range to observe many EPR transitions and does not require and exceptionally high magnetic field strength. However, in the past few decades, spectrometers operating at higher frequency ranges have become widely used, thus requiring greater applied field strengths. Different energy level experimental setups allow the observation of different transitions that are hidden at one applied field strength, but clearly visible at another. This is because at a higher magnetic field strength, the spin states will be split further, spreading apart transitions that are close in energy.
Resonator
A resonant cavity is the most common form of resonator used in EPR spectroscopy. The resonant cavity of an EPR spectrometer is designed with dimensions corresponding to the specific wavelength of microwave radiation that is to be used. For example, at X-band frequencies (~9.5 GHz), the wavelength of the electromagnetic wave is ~3 cm. In this case, the resonant cavity has dimensions of ~3 cm. This allows the microwaves to resonate in the cavity. In a resonant cavity, the level of microwave energy is 1000s of times greater than that in the waveguide. The sample is located in the cavity at a location that allows maximum absorption of magnetic energy from the microwaves, and minimum absorption of electric energy. This is because it is the magnetic field component of the microwave energy that excites and EPR transition, while the electric field component will be absorbed in a non-resonant manner. Typically, the sample is placed in a central location in the cavity, where it will absorb the greatest level of energy from the magnetic field of the microwaves.
Figure 2: An X-band resonant cavity with waveguide attached. Note the dimensions of ~3 cm.
To fine-tune the range and level of resonating microwaves, different procedures are used. The most common technique is by controlling the amount of microwaves allowed to enter the cavity with an iris screw. The iris is the hole by which the microwaves enter the cavity and a screw is used to precisely vary the size of this whole. "Tuning" the microwaves to resonate in the cavity must be done prior to each experiment. This is done by first finding the frequency that results in the least amount of reflection out of the cavity, and then fine-tuning the cavity to the optimum range or resonance desired for the experiment. The frequency and coupling required will change depending on sample and its position (this changes the dimensions of the cavity), and hence, must be done whenever either of these parameters is changed.
An optional characteristic of a resonance cavity is the ability to operate in either "parallel mode," or "perpendicular mode." In this case, the dimensions of the cavity will vary just slightly, so that a slightly different frequency of radiation will resonate parallel to the magnetic field as compared with that which resonates parallel to the magnetic field. Hence, the sample will absorb energy from the electromagnetic radiation either parallel or perpendicular to its induced magnetic dipole moment. The purpose of this component is not to change the frequency of radiation, but to change the direction that the electromagnetic radiation approaches the sample. This changes the number and nature of transitions that are allowed by quantum mechanical rules. These different transitions produce spectra that can provide different structural insights into the electronic nature of the molecule. Typically, a spectrometer operates in perpendicular mode.
Figure 3: the resonator of a high field EPR spectrometer.
This is a custom built resonator designed to resonate microwaves at a frequency of 130 GHz. This corresponds to a wavelength of ~2mm. Thus, the size of the cavity in this resonator has those dimensions. The probe that this resonator is attached to has coaxial cables to propagate radio frequency waves for ENDOR experiments (see optional section below), as well as a fiber optic cable to illuminate light-sensitive samples and observe changes in their spectra. When in the spectrometer, then resonant cavity sits directly on top of a waveguide, which propagates the microwaves into the cavity. This resonator has specially designed components that allow one to change the dimensions and orientation of the cavity, in order to tune the cavity to resonate.
Detection Methods
Various detection methods exist for EPR spectroscopy. The majority of EPR spectrometers are reflection spectrometers, meaning they measure the amount of radiation that is reflected back out of the resonator. Changes in the level of reflected microwave energy at various field strengths allow the observation of spectroscopic transitions. A special channeling device called a "circulator" is used to insure that radiation from the microwave source is sent only into the cavity and that reflected radiation is sent only to the detector. A diode is used to measure the power of reflected energy. To achieve an accurate quantitative measurement of absorbed microwave energy, the diode must sometimes be "biased," or supplied with a steady level of current which insures that the diode is reading in the correct linear range.
CW spectrometers (those which have been described up to this point) make use of a "field modulator," which creates an oscillating magnetic field around the sample, in addition to the applied magnetic field created by the electromagnet. This serves to fix the range of electromagnetic energy that reaches the detector, increasing resolution. Because the absorption of microwave energy is measured relative to this the modulating field, CW EPR spectra show up as a derivative line shape relative to actual energy absorption.
Electromagnet
An electromagnet is used in an EPR experiment because it allows one to vary the strength of the applied magnetic field precisely, quickly, and consistently. As mentioned previously, in any experimental setup, the strength of the applied magnetic field must correspond to the frequency of the electromagnetic radiation being applied to a sample. For example, if the experiment is being performed at X-band (~9.5 GHz), the field strength should be swept from a range of 0-4000 gauss (or 0-0.4 Tesla) in order to see all transitions. At higher frequencies, the field strength must be higher to see the same frequencies, because the degenerate spin states are be split more.
In some experiments, the magnetic field is kept constant for continuous measurement of a single transition. In spin labeling experiments, where the paramagnetic species are bound to specific positions on a molecule (usually, a protein), the field strength of the electromagnet only needs to be swept a couple hundred Gauss. This small sweep range makes EP spectrometers specially designed for spin labeling experiments much smaller than a traditional EPR spectrometer. Spin-label specific EPR spectrometers can sit on a lab bench, while a typical EPR spectrometer will take up a good 25 square feet of lab space including computer and power source.
As the field is swept, a significant amount of current is run through the electromagnetic coils. This creates a large amount of heat. For this reason, an EPR spectrometer must have some kind of cooling apparatus to keep it from getting hot. This is usually accomplished by continuously flowing water around the electromagnet.
Figure 4: Ka-band EPR spectrometer.
The two copper-colored discs in the Ka-band spectrometer pictured in figure 4 are the electromagnet. The sample cavity sits directly in the center of these two discs, as the center of the magnetic field. This is the conventional design for the applied magnetic field of an EPR spectrometer. The pictured magnet can be swept from 0-1.5 Tesla, allowing observation of EPR transitions excited with 31 GHz microwave energy.
Figure 5: D-band spectrometer.
Figure 5 pictures an alternate experimental setup for the applied magnetic field of an EPR spectrometer. This instrument utilizes a superconducting electromagnet that is kept at a constant temperature of ~4 Kelvin by a closed-loop liquid helium pump. This setup contains 2 magnetic coils, and can sweep an applied field of 0-8 Tesla. Powerful superconducting magnets like this are necessary to split paramagnetic spin states when high frequency microwaves are to be used to excite the transitions. This instrument operates at D-band frequencies of 130 GHz.
Optional Components
Though optional, the following components are commonly used in EPR experimental setups:
1. A cryogenic cooling device is used to cool samples down to ultra-cold temperatures. This leads to a greater population difference in the split spin states, increasing the signal level. Many transitions can only be seen at ultra-cold temperatures, especially in biological samples. A temperature as low as 4 Kelvin is often necessary to gain a resolved spectrum. Liquid helium must be used to maintain these ultra-cold temperatures. Two types of cryogenic cooling setups are common: A) A reservoir dewar, which maintain a constant level of liquid helium and/or liquid nitrogen surrounding the sample, or B) A continuously flowing liquid helium pump, which continuously pumps liquid helium into the space surrounding the sample cavity. The latter is the setup commonly utilized on commercially available spectrometers. Naturally, these setups require the use of high-powered vacuum pumps to evacuate the areas insulating the lines and reservoirs that transfer and contain the cryogen.
2. Most EPR experimental setups do not only operate in the "continuous wave" (CW) format described above, during which the sample is continuously bathed in microwave energy. Instead, advanced EPR experiments utilize pulsed microwave energy to selectively excite different transitions of a paramagnetic sample. These experiments have the same basic components as previously described, but require different amplifying and detecting equipment specifically designed for pulse creation and detection. Experiments that utilize pulse microwaves include Electron Spin Echo Envelope Modulation (ESEEM) and Hyperfine Sublevel Correlation Spectroscopy (HYSCORE).
3. Many EPR experimental setups contain a radio frequency amplifier, which is used to excite transitions between the spin states of magnetic moments of nuclei surrounding the paramagnetic nuclei. These advanced EPR experiments are called Electron Nuclear Double Resonance (ENDOR) experiments, and provide valuable information about nuclei in the environment surrounding an unpaired electron.
CalEPR Center
The CalEPR center on campus contains 5 EPR experimental setups, located in the basement of the Chemistry building. These were assembled and are maintained and used by the lab Dr. David R. Britt. The range of frequencies and the overall experimental capability makes the CalEPR center a truly world-class EPR lab. Of the 5 spectrometers, 3 are home-built by Dr. Britt and his graduate students, and 2 are commercially built by Bruker.
Figure 6: X-band CW spectrometer.
Figure 6 pictures the ECS-106, a Bruker instrument. This is an X-band instrument configured only for CW EPR at ~9.5 GHz. The resonant cavity of this instrument is capable of performing in parallel or perpendicular mode. The sample cavity of this instrument is cooled with continuous flowing liquid helium, pumped straight from the helium dewar.
The power supply is sitting to the left of the instrument, which can be identified by the large circular electromagnets. To the right of the instrument is the console, which contains the necessary electrical components. Above the magnet is the cryostat and bridge, which contains important components like the microwave amplifier, attenuators, and the detector.
Figure 7: Home-built pulsed X-band spectrometer.
Figure 7 pictures the oldest instrument in the lab, a home-built spectrometer capable of a wide frequency range (8-18 GHz). It is configured for pulsed EPR spectroscopy, including ESEEM and ENDOR. The sample of cavity of this spectrometer is cooled with a reservoir dewar which must be filled with liquid helium and/or nitrogen. This instrument is currently not in use because it is going through some necessary upgrades to increase its speed (as of March 2009).
Figure 8: X-band and Q-band CW and pulse spectrometer.
Figure 8 pictures the other commercially built spectrometer, the E560 ELEXYS EPR system built by Bruker. This spectrometer is capable of doing CW and pulsed EPR at both X-band (~9.5 GHz) and Q-band frequencies (~35 GHz). The sample cavity of this instrument is cooled with continuous flowing liquid helium, pumped straight from the helium dewar. The line to the helium dewar flowing into the sample cavity can be seen in this picture.
Figure 9: Home-built pulse Ka-band spectrometer.
Figure 9 pictures the home-built Ka-Band spectrometer, which performs experiments at ~31 GHz using a 100 W traveling wave tube amplifier. This spectrometer is configured for pulsed EPR spectroscopy, and has primarily been used for ESEEM studies. The sample of cavity of this spectrometer is cooled with a reservoir dewar which must be filled with liquid helium and/or nitrogen. This spectrometer was built by current Britt lab group member Michelle Dicus and former Britt lab group member Gregory Yeagle.
Figure 10: Home-built pulse D-band spectrometer.
Figure 10 pictures the home-built pulsed D-band spectrometer, which performs high-field pulsed EPR experiments at ~130 GHz. It utilizes a superconducting electromagnet to produce fields of up to 8 Tesla. The magnet is constantly kept at ~4 Kelvin by a closed helium loop pump system. This spectrometer is configured for ESEEM and ENDOR studies.This spectrometer was built and is maintained by current Britt lab group member Alex Gunn.
Experimental Applications
EPR spectroscopy can be applied to any sample that contains a paramagnetic electron. This includes a range of samples from basic organic and inorganic compounds to complex biomolecules such as proteins. Spectra are used to identify molecules within a sample and more importantly, to characterize the environment of the unpaired electron. EPR spectroscopy has advantages in that it is very sensitive (1000x more sensitive than NMR) and in that it has very good specificity (it looks only at the region that contains the unpaired electron). These characteristics make EPR especially useful for the study of metallic cofactor-containing proteins, which perform the most important enzymatic reactions on the planet (photosynthesis, nitrogenase, etc.). How substrates bind to the metallic cofactors in these enzymes can be elucidated best using EPR.
CW EPR
Continuous Wave EPR (CW) is an experiment during which a sample is continuously illuminated with microwave radiation at a fixed frequency and the strength of the applied magnetic field is swept, observing changes in microwave absorption. CW EPR is the most basic experiment performed. The instrumentation section of this article primarily describes the instrumental setup utilized to obtain CW spectra. Figure 11 displays a sample CW spectrum taken at X-band. This is a sample of isolated photosystem II, a complex of proteins involved in photosynthesis. Note that the X-axis displays magnetic field strength in Gauss. Also note that the spectrum displays the derivative of absorption of microwaves. The Y-axis displays the derivative of units of absorption in arbitrary units.
Figure 11: Sample CW EPR spectrum
Each peak of this spectrum represents a transition between the spin states of the unpaired electron at a specific energy. In a system as complex as PSII, each transition corresponds to an entirely different molecule, be it a metallic cofactor or a biomolecular cofactor such as chlorophyll. This spectrum is interpreted in light of theory which describes how the unpaired electron's spin states are affected by its environment, most importantly, coupling to the surrounding nuclei. A thorough interpretation of this spectrum would require comparisons with spectra of similar samples. Such an analysis can be used to determine precisely the orientation of the surrounding nuclei and the unpaired spin density on each.
Pulsed EPR
Pulse EPR varies from CW in that microwaves are applied to the sample as a series of nanoseconds-long pulses. The magnetic field strength is kept constant, and the sample is pulsed with microwave energy. The strategic pattern of pulse length and spacing allows the creation of a spin echo. The parameters of the pulse pattern can be changed and utilized in such a way that the hidden features of a transition observed at a specific field strength in a CW EPR spectrum can be observed. For example, figure 12 presents an ESEEM (electron spin echo envelope modulation) spectrum taken on the same PSII sample see in figure 11. The ESEEM spectrum is taken at the field strength corresponding to the largest (in amplitude) transition in figure 11 (~3450 G).
Figure 12: This sample ESEEM spectrum demonstrates the advanced capabilities of pulse EPR.
ESEEM spectra provide direct information about the frequencies that make up the less energetic transitions of an EPR transition. While CW EPR more bluntly shows the transition between one spin state to another. Pulse EPR spectra have the capability of showing the transitions that occur within a spin state, which are caused by the interaction of the unpaired electron with the surrounding nuclei. The peaks seen in figure 12 correspond to specific quantitative parameters that describe the coupling of the electron and the nucleus, thus corresponding to the distance between the nuclei and the electron. In short, pulse EPR allows the observation of transitions within the transitions seen in CS spectra.
ENDOR
Electron nuclear double resonance spectroscopy (ENDOR) further advances pulsed EPR techniques by including pulses in the radiofrequency region to the pulse patterns utilized to produce spin echoes. The result is the ability to measure the resonance energies of the specific nuclei that are coupled to the unpaired electron. Most importantly, it allows one to observe how these resonance energies are perturbed with respect to the coupling of the nuclear and electronic magnetic moments. The direct measurement of these coupling parameters allows understanding of physical relationship of the electron with its surrounding nuclei.
Spin-Spin Labeling Experiments
An entirely different application of EPR spectroscopy is done with spin-spin labeling experiments. In these, a protein is tagged with paramagnetic groups (usually NO2) that can be easily detected by EPR. The distance between parts of the molecule which contain the paramagnetic groups can observed by the interaction of their respective transitions in spectra. The most valuable information is gained from these experiments by altering the molecule and observing the whether the paramagnetic groups interact with each other. For example, this technique has been utilized to determine whether membrane-bound ion channels open or close in response to a stimulus.
Outside links
• Britt lab homepage contains more information about CalEPR center and EPR research at UCD.
• The National High Magnetic Field Laboratory in Tallahassee conducts EPR research at even higher frequencies and fields.
• Wikipedia on EPR spectroscopy contains useful background information.
• Bruker's EPR Product Lines contain information about commercially available EPR spectrometers.
Problems
1. Approximately what size waveguide would be appropriate if you were operating in mid-range S-band frequencies?
2. Would S-band frequencies require higher magnetic field strength or lower? Based on magnetic field ranges described in this article, what would expected the magnetic field range of an S-band spectrometer to be?
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Though less used than Nuclear Magnetic Resonance (NMR), Electron Paramagnetic Resonance (EPR) is a remarkably useful form of spectroscopy used to study molecules or atoms with an unpaired electron. It is less widely used than NMR because stable molecules often do not have unpaired electrons. However, EPR can be used analytically to observe labeled species in situ either biologically or in a chemical reaction.
Introduction
Electron Paramagnetic Resonance (EPR), also known as Electron Spin Resonance (ESR). The sample is held in a very strong magnetic field, while electromagnetic (EM) radiation is applied monochromatically (Figure 1).
Figure 1(3)-monochromatic electromagnetic beam
This portion of EPR is analogous to simple spectroscopy, where absorbance by the sample of a single or range of wavelengths of EM radiation is monitored by the end user ie absorbance. The unpaired electrons can either occupy +1/2 or -1/2 ms value (Figure 2). From here either the magnetic field "B0" is varied or the incident light is varied. Today most researchers adjust the EM radiation in the microwave region, the theory is the find the exact point where the electrons can jump from the less energetic ms=-1/2 to ms=+1/2. More electrons occupy the lower ms value (see Boltzmann Distribution).
Overall, there is an absorption of energy. This absorbance value, when paired with the associated wavelength can be used in the equation to generate a graph of showing how absorption relates to frequency or magnetic field.
$\Delta E=h\nu=g_e \beta_B B_0$
where ge equals to 2.0023193 for a free electron; $\beta_B$ is the Bohr magneton and is equal to 9.2740 * 10 -24 J T -1; and B0 indicates the external magnetic field.
Theory
Like NMR, EPR can be used to observe the geometry of a molecule through its magnetic moment and the difference in electron and nucleus mass. EPR has mainly been used for the detection and study of free radical species, either in testing or anylytical experimentation. "Spin labeling" species of chemicals can be a powerfull technique for both quantification and investigation of otherwise invisible factors.
The EPR spectrum of a free electron, there will be only one line (one peak) observed. But for the EPR spetrum of hydrogen, there will be two lines (2 peaks) observed due to the fact that there is interaction between the nucleus and the unpaired electron. This is also called hyperfine splitting. The distance between two lines (two peaks) are called hyperfine splitting constant (A).
By using (2NI+1), we can calculate the components or number of hyperfine lines of a multiplet of a EPR transtion, where N indicates number of spin, I indicates number of equivalent nuclei. For example, for nitroxide radicals, the nuclear spin of 14N is 1, N=1, I=1, we have 2 x 1 + 1 = 3, which means that for a spin 1 nucleus splits the EPR transition into a triplet.
To absorb microwave, there must be unpaired electrons in the system. no EPR signal will be observed if the system contains only paired electrons since there will be no resonant absorption of microwave energy. Molecules such as NO, NO2, O2 do have unpaired electrons in groud states. EPR can be also performed on proteins with paramagnetic ions such as Mn2+, Fe3+ and Cu2+. Additionally, molecules containing stable nitroxide radicals such as 2,2,6,6-tetramethyl-1-piperidinyloxyl (TEMPO, Figure 3) and di-tert-butyl nitroxide radical.
Examples of EPR spectra:
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Electron paramagnetic resonance spectroscopy (EPR), also called electron spin resonance (ESR), is a technique used to study chemical species with unpaired electrons. EPR spectroscopy plays an important role in the understanding of organic and inorganic radicals, transition metal complexes, and some biomolecules. For theoretical background on EPR, please refer to EPR:Theory.
Introduction
Like most spectroscopic techniques, EPR spectrometers measure the absorption of electromagnetic radiation. A simple absorption spectra will appear similar to the one on the top of Figure 1. However, a phase-sensitive detector is used in EPR spectrometers which converts the normal absorption signal to its first derivative. Then the absorption signal is presented as its first derivative in the spectrum, which is similar to the one on the bottom of Figure 1. Thus, the magnetic field is on the x-axis of EPR spectrum; dχ″/dB, the derivative of the imaginary part of the molecular magnetic susceptibility with respect to the external static magnetic field in arbitrary units is on the y-axis. In the EPR spectrum, where the spectrum passes through zero corresponds to the absorption peak of absorption spectrum. People can use this to determine the center of the signal. On the x-axis, sometimes people use the unit “gauss” (G), instead of tesla (T). One tesla is equal to 10000 gauss.
Proportionality factor (g-factor)
As a result of the Zeeman Effect, the state energy difference of an electron with s=1/2 in magnetic field is
$\Delta E = gβB \label{1}$
where β is the constant, Bohr magneton. Since the energy absorbed by the electron should be exactly the same with the state energy difference ΔE, ΔE=hv ( h is Planck’s constant), the Equation $\ref{1}$ can be expressed as
$h \nu=gβB \label{2}$
People can control the microwave frequency v and the magnetic field $B$. The other factor, $g$, is a constant of proportionality, whose value is the property of the electron in a certain environment. After plugging in the values of $h$ and $β$ in Equation $\ref{2}$, g value can be given through Equation $\ref{3}$:
$g = 71.4484v \text{(in GHz)/B (in mT)} \label{3}$
A free electron in vacuum has a $g$ value ($g_e= 2.00232$. For instance, at the magnetic field of 331.85 mT, a free electron absorbs the microwave with an X-band frequency of 9.300 GHz. However, when the electron is in a certain environment, for example, a transition metal-ion complex, the second magnetic field produced by the nuclei, ΔB, will also influence the electron. At this kind of circumstance, Equation $\ref{2}$ becomes
$h\nu = gβ(B_e+ \Delta B) \label{4}$
since we only know the spectrometer value of B, the Equation $\ref{4}$ is written as:
$h\nu = (g_e+ \Delta g)βB ] \label{5}$
From the relationship shown above, we know that there are infinite pairs of v and B that fit this relationship. The magnetic field for resonance is not a unique “fingerprint” for the identification of a compound because spectra can be acquired at different microwave frequencies. Then what is the fingerprint of a molecule? It is Δg. This value contains the chemical information that lies in the interaction between the electron and the electronic structure of the molecule, one can simply take the value of g = ge+ Δg as a fingerprint of the molecule. For organic radicals, the g value is very close to ge with values ranging from 1.99-2.01. For example, the g value for •CH3 is 2.0026. For transition metal complexes, the g value varies a lot because of the spin-orbit coupling and zero-field splitting. Usually it ranges from 1.4-3.0, depending on the geometry of the complex. For instance, the g value of Cu(acac)2 is 2.13. To determine the g value, we use the center of the signal. By using Equation $\ref{3}$, we can calculate the g factor of the absorption in the spectrum. The value of g factor is not only related to the electronic environment, but also related to anisotropy. About this part, please refer to EPR:Theory, Parallel Mode EPR: Theory and ENDOR:Theory. An example from UC Davis is shown below[1] (Britt group, Published in J.A.C.S. ):
Hyperfine Interactions
Another very important factor in EPR is hyperfine interactions. Besides the applied magnetic field B0, the compound contains the unpaired electrons are sensitive to their local “micro” environment. Additional information can be obtained from the so-called hyperfine interaction. The nuclei of the atoms in a molecule or complex usually have their own fine magnetic moments. Such magnetic moments occurrence can produce a local magnetic field intense enough to affect the electron. Such interaction between the electron and the nuclei produced local magnetic field is called the hyperfine interaction. Then the energy level of the electron can be expressed as:
$E = gm_BB_0M_S + aM_sm_I \label{6}$
with $a$ is the hyperfine coupling constant, $m_I$ is the nuclear spin quantum number.
Hyperfine interactions can be used to provide a wealth of information about the sample such as the number and identity of atoms in a molecule or compound, as well as their distance from the unpaired electron.
Table 1. Bio transition metal nuclear spins and EPR hyperfine patterns[3]
The rules for determining which nuclei will interact are the same as for NMR. For isotopes which have even atomic and even mass numbers, the ground state nuclear spin quantum number, I, is zero, and these isotopes have no EPR (or NMR) spectra. For isotopes with odd atomic numbers and even mass numbers, the values of I are integers. For example the spin of 2H is 1. For isotopes with odd mass numbers, the values of I are fractions. For example the spin of 1H is 1/2 and the spin of 23Na is 7/2. Here are more examples from biological systems:
Table 2. Bio ligand atom nuclear spins and their EPR hyperfine patterns[3]
The number of lines from the hyperfine interaction can be determined by the formula: 2NI + 1. N is the number of equivalent nuclei and I is the spin. For example, an unpaired electron on a V4+ experiences I=7/2 from the vanadium nucleus. We can see 8 lines from the EPR spectrum. When coupling to a single nucleus, each line has the same intensity. When coupling to more than one nucleus, the relative intensity of each line is determined by the number of interacting nuclei. For the most common I=1/2 nuclei, the intensity of each line follows Pascal's triangle, which is shown below:
For example, for •CH3, the radical’s signal is split to 2NI+1= 2*3*1/2+1=4 lines, the ratio of each line’s intensity is 1:3:3:1. The spectrum looks like this:
If an electron couples to several sets of nuclei, first we apply the coupling rule to the nearest nuclei, then we split each of those lines by the coupling them to the next nearest nuclei, and so on. For the methoxymethyl radical, H2C(OCH3), there are (2*2*1/2+1)*(2*3*1/2+1)=12 lines in the spectrum, the spectrum looks like this:
For I=1, the relative intensities follow this triangle:
The EPR spectra have very different line shapes and characteristics depending on many factors, such as the interactions in the spin Hamiltonian, physical phase of samples, dynamic properties of molecules. To gain the information on structure and dynamics from experimental data, spectral simulations are heavily relied. People use simulation to study the dependencies of spectral features on the magnetic parameters, to predict the information we may get from experiments, or to extract accurate parameter from experimental spectra.
EasySpin Simulations
Many methods were developed to simulate the EPR spectra. Dr. Stefan Stoll wrote EasySpin, a computational EPR package for spectral simulation. EasySpin is based on Matlab, which is a numerical computing environment and fourth-generation programming language. EasySpin is a powerful tool in EPR spectral simulation. It can simulate spectra under many different conditions. Some functions are shown below:
Spectral simulations and fitting functions:
• garlic: cw EPR (isotropic and fast motion)
• chili: cw EPR (slow motion)
• pepper: cw EPR (solid state)
• salt: ENDOR (solid state)
• saffron: pulse EPR/ENDOR (solid state)
• esfit: least-squares fitting
To learn more, please visit EasySpin: http://www.easyspin.org/.
References
1. Dicus, M.M; Conlan,A.; Nechushtai,R.; Jennings,P.A.;Paddock,M.L.; Britt,R.D.; Stoll S. J. AM. CHEM. SOC. 2010, 132, 2037–2049
2. Hagen,W.R. 2009. Biomolecular EPR Spectroscopy. Boca Raton: CRC Press.
3. Hagen,W.R. Dalton Trans., 2006, 4415–4434
4. Stoll,S., Schweiger,A. Journal of Magnetic Resonance 178 (2006) 42–55
Example $1$
If there is one unpaired electron in Cu2+ (I=3/2) and the copper ion is coordinated by one nitrogen atom (I=1) and one OH- (I=1/2), how many lines can be expected in the EPR spectrum?
Solution
$2 \times 1 \times 3/2 +1)(2 \times 1\times 1 +1)(2\times 1 \times 1/2+1)=24 \nonumber$
Example $2$
For a radical, the magnetic field is 3810 G, the frequency of the microwave is 9600 MHz. What is the value of its g-factor?
Solution
$g = \dfrac{ (71.4484) (9600 \,\text{x} \,10^{-3})}{3810\, \text{x}\, 10^{-1}}=1.800 \nonumber$
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This module presents the theory that describes how EPR transitions can be induced in integer high spin systems by the application of a modulating magnetic field parallel to the bond axis (z-axis), as well as some of the applications of this technique to various molecular systems.
Introduction
Standard mode Electron Paramagnetic Resonance (EPR), in which the modulating magnetic field is perpendicular to the applied field, is capable of detecting transitions between eigenstates in systems with fractional spins (i.e.; S=1/2, 3/2, ...), so called 'Kramer Systems'. This provides a sensitive experimental technique for detecting the electronic environment of unpaired electrons in various molecular systems. There are, however, systems with an integer spin value (ie; S=2), called 'Non-Kramer Systems'. In order to probe the electronic environment of such systems a method in which the modulating field is parallel to the applied field is employed. This method is called Parallel Mode EPR.
Spin Hamiltonian
To describe the theory of parallel mode EPR we must first define the system that we wish to characterize. If we have a ligand field that has axial symmetry, the collection of 2S+1 states of the system will be split, in the case of integral spin systems, into S doublets and a singlet. If we consider a system with S=2, the five states will be split into two doublets and a singlet. Application of a magnetic field gives the following spin Hamiltonian for the system.
$\hat{H}=D[S_z^2-\frac{1}{3}S(S+1)]+E(S_x^2-S_y^2)+\beta\mathbf{B}\cdot\mathbf{g}\cdot\mathbf{S}$
$=D[S_z^2-\frac{1}{3}S(S+1)]+E(S_x^2-S_y^2)+g_{\parallel}\beta BS_z\cos{\theta}+g_{\perp}\beta BS_z\sin{\theta}$
where the D is the axial field splitting term and E is the rhombic splitting term, these terms are a measure of the energy gap between the various states when no magnetic field is applied. These values are a reflection of the symmetry of the system. B is the magnetic field vector, $\beta$ is the Bohr magneton, Sz is the z projection of the spin, and $\theta$ is the angle of the applied magnetic field with respect to the symmetry axis of the system in the zx-plane.
Eigenfunctions and Eigenstates
If we have the condition that
$g\beta B\ll D$
and we neglect the rhombic field splitting term in the spin Hamiltonian, then the states, and their respective energy values, are given by the following, to first order in perturbation theory.
$\mid \pm2\rangle\Rightarrow E_{\pm2}=2D\pm 2g_{\parallel}\beta B\cos{\theta}$
$\mid \pm1\rangle\Rightarrow E_{\pm1}=-D\pm g_{\parallel}\beta B\cos{\theta}$
$\mid 0\rangle \rightarrow E_0=-2D$
Analysis of the energy matrix of these states shows that there are no non-zero elements for the terms that represent transitions in the doublet states. Therefore these states have no allowed transitions. We will employ a mathematical trick to describe these states in a way that allows us to ascertain the EPR transitions that are allowed for the given system. We will adopt a new set of basis functions
$\mid 2^s\rangle$, $\mid 2^a\rangle$, $\mid1^s\rangle$, and $\mid1^a\rangle$
which are symmetric and antisymmetric linear combinations, respectively, of the form
$\mid2^s\rangle=\dfrac{1}{\sqrt{2}}(\mid+2\rangle+\mid-2\rangle)$
$\mid2^a\rangle=\dfrac{1}{\sqrt{2}}(\mid+2\rangle-\mid-2\rangle)$
Since these basis functions are linear combinations of basis functions that are eigenfunctions of the spin Hamiltonian, it follows that these linear combinations are also eigenfunctions of the spin Hamiltonian. When this change of basis is adopted, and we account for the rhombic field splitting term in the spin Hamiltonian we obtain the following energies, correct to second order in perturbation theory.
$E_{\pm2}=2D+\dfrac{(g_{\perp}\beta B\sin{\theta})^2}{E_2-E_1}+\frac{1}{2}\Delta_2\pm[(2g_{\parallel}\beta B\cos{\theta})^2+\left(\dfrac{\Delta_2}{2}\right)^2]^{1/2}$
$E_{\pm1}=-D+\dfrac{(g_{\perp}\beta B\sin{\theta})^2}{E_1-E_2}+\dfrac{3(g_{\perp}\beta B\sin{\theta})^2}{2(E_1-E_0)}\pm\left[(g_{\parallel}\beta B \cos{\theta})^2+\left(3E+\dfrac{3(g_{\perp}\beta B\sin{\theta})^2}{2(E_1-E_0)}\right)^2\right]^{1/2}$
$E_0=-2D+\dfrac{12E^2}{E_0-E_2}+\dfrac{3(g_{\perp}\beta B\sin{\theta})^2}{E_0-E_1}$
If we have no rhombic splitting term (E=0), there are no transitions in the $\mid\pm2\rangle$ doublet that are allowed. However, for the $\mid\pm1\rangle$ doublet, the states are actually a linear combination of the $\mid1\rangle$, $\mid0\rangle$, and $\mid-1\rangle$ states and, due to this admixture, for some value of $\theta$ there is a weakly allowed transition. This is due to the second order Zeeman effects. This transition becomes forbidden in the case where $\theta=0$, ie; when the magnetic field is oriented along the z-axis.
If E is not equal to zero then the $\mid\pm2\rangle$ states are also linear combinations. If $\theta=0$, we have energy levels for the $\mid\pm2\rangle$ states that are given by the following
$E_{\pm2}=\pm\dfrac{1}{2}[(4g_{\parallel}\beta B\cos{\theta})^2+\Delta_2^2]^{1/2}$
where $\Delta_2=\dfrac{12E^2}{E_2-E_o}$. The states of this doublet can then be written in the form
$\mid+2^{\prime}\rangle=\cos{\alpha\mid+\rangle}+\sin{\alpha\mid-\rangle}$
$\mid-2^{\prime}\rangle=\cos{\alpha\mid+\rangle}-\sin{\alpha\mid-\rangle}$
These new states, along with their respective energies and transitions, are shown schematically in Figure (1).
Transitions and the Resonance Condition
The states of the doublet described above have a matrix element that couples them together that is given by
$\langle2+^{\prime}\mid 4g_{\parallel}\beta S_z\mid-^{\prime}\rangle=2g_{\parallel}\beta\sin{2\alpha}$
This shows that there are allowed transitions in the $\mid\pm2\rangle$ manifold as long as the modulating magnetic field is oriented with the z axis. The resonance condition for these transitions becomes
$\Delta E=h\nu=[(4g_{\parallel}\beta B\cos{\theta})^2+\Delta^2]^{1/2}$
and the probability of a transition occurring is given by
$\mid\mu_z\mid^2=4g_{\parallel}^2\beta^2\dfrac{\Delta^2}{(h\nu)^2}$
We have shown that, by taking linear combinations of the $\mid\pm1\rangle$ and $\mid\pm2\rangle$ states that show no transitions, we can construct a description of the system that accounts for the transitions observed in the parallel mode EPR spectra of integer spin systems.
Applications
The parallel mode electron paramagnetic resonance technique, in which the modulating magnetic field is parallel to the applied field, allows for the detection of transitions between eigenstates for systems with integer spin. This technique has been applied to a variety of systems to ascertain the nature of the spin states, some of which are described below.
Biophysical Applications
Both heme and non-heme iron proteins have been shown to posses high spin ferrous ions. The parallel mode EPR technique has been applied to several of these systems to verify their integer spin state. These proteins include the mononuclear ferrous sites of myoglobin and transferrin, as well as the polynuclear ferrous sites of methane monoxygenase, ferredoxin II, and aconitase. It has also been used to study the nature of the Mn(III)-Salen system and it's catalysis of the epoxidation of cis-beta-methylstyrene.
Inorganic Applications
Advances have been made in the field of synthetic inorganic chemistry such that there are now examples in the literature of synthetic high spin non-heme iron(IV) systems. This systems are predicted to have a high degree of reactivity and are being probed for their catalytic properties. The parallel mode EPR technique has been applied to these systems to extract the true nature of their spin state and to measure the g values of their absorptions. An example of a parallel mode EPR spectrum of a high spin (S=2) Fe(IV)-oxo compound, measured by Hendrich and coworkers at Pennsylvania State University, is shown in figure 2.
Parallel mode EPR has also been used to describe the spin state of other high spin inorganic complexes, including an S=6 chromium system synthesized by Piligko and coworkers at the University of Manchester.
Outside Links
1. CalEPR facility website, www.brittepr.ucdavis.edu.
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Electron Paramagnetic Resonance (EPR), also called Electron Spin Resonance (ESR), is a branch of magnetic resonance spectroscopy which utilizes microwave radiation to probe species with unpaired electrons, such as radicals, radical cations, and triplets in the presence of an externally applied static magnetic field. In many ways, the physical properties for the basic EPR theory and methods are analogous to Nuclear Magnetic Resonance (NMR). The most obvious difference is that the direct probing of electron spin properties in EPR is opposed to nuclear spins in NMR. Although limited to substances with unpaired electron spins, EPR spectroscopy has a variety of applications, from studying the kinetics and mechanisms of highly reactive radical intermediates to obtaining information about the interactions between paramagnetic metal clusters in biological enzymes. EPR can even be used to study the materials with conducting electrons in the semiconductor industry.
Historical Development of EPR
In 1896, the line splitting in optical spectra in a static magnetic field was first found by a Dutch physicist Zeeman. In 1920s, Stern and Gerlach sent a beam of silver atoms through an inhomogeneous magnetic field and the beam splits into two distinct parts, indicating the intrinsic angular momentum of electrons and atoms. Then Uhlenbeck and Goudsmit proposed that the electrons have an angular momentum. In 1938, Isidor Rabi measured the magnetic resonance absorption of lithium chloride molecules, which means he could measure different resonances to get more detailed information about molecular structure. After World War II, microwave instrumentation’s widespread availability sped up the development of electron paramagnetic resonance (EPR). The first observation of a magnetic resonance signal was detected by a Soviet physicist Zavoisky in several salts, including hydrous copper chloride, copper sulfate and manganese sulfate in 1944. Later the Oxford group proposed the basic theory of magnetic resonance. Contributed by many researchers, such as Cummerow & Halliday and Bagguley & Griffiths, EPR was extensively studied. Between 1960 and 1980, continuous wave (CW) EPR was developed and pulsed EPR was mainly studied in Bell laboratories. EPR was usually applied for organic free radicals. In the 1980s, the first commercial pulsed EPR spectrometer appeared in the market and was then extensively used for biological, medical field, active oxygen and so on. Nowadays, EPR has become a versatile and standard research tool.
Comparison between EPR and NMR
EPR is fundamentally similar to the more widely familiar method of NMR spectroscopy, with several important distinctions. While both spectroscopies deal with the interaction of electromagnetic radiation with magnetic moments of particles, there are many differences between the two spectroscopies:
1. EPR focuses on the interactions between an external magnetic field and the unpaired electrons of whatever system it is localized to, as opposed to the nuclei of individual atoms.
2. The electromagnetic radiation used in NMR typically is confined to the radio frequency range between 300 and 1000 MHz, whereas EPR is typically performed using microwaves in the 3 - 400 GHz range.
3. In EPR, the frequency is typically held constant, while the magnetic field strength is varied. This is the reverse of how NMR experiments are typically performed, where the magnetic field is held constant while the radio frequency is varied.
4. Due to the short relaxation times of electron spins in comparison to nuclei, EPR experiments must often be performed at very low temperatures, often below 10 K, and sometimes as low as 2 K. This typically requires the use of liquid helium as a coolant.
5. EPR spectroscopy is inherently roughly 1,000 times more sensitive than NMR spectroscopy due to the higher frequency of electromagnetic radiation used in EPR in comparison to NMR.
It should be noted that advanced pulsed EPR methods are used to directly investigate specific couplings between paramagnetic spin systems and specific magnetic nuclei. The most widely application is Electron Nuclear Double Resonance (ENDOR). In this method of EPR spectroscopy, both microwave and radio frequencies are used to perturb the spins of electrons and nuclei simultaneously in order to determine very specific couplings that are not attainable through traditional continuous wave methods.
Origin of the EPR Signal
An electron is a negatively charged particle with certain mass, it mainly has two kinds of movements. The first one is spinning around the nucleus, which brings orbital magnetic moment. The other is "spinning" around its own axis, which brings spin magnetic moment. Magnetic moment of the molecule is primarily contributed by unpaired electron's spin magnetic moment.
$M_{S}=\sqrt{S(S+1)} \dfrac{h}{2\pi } \label{1}$
• $M_S$ is the total spin angular moment,
• $S$ is the spin quantum number and
• $h$ is Planck’s constant.
In the z direction, the component of the total spin angular moment can only assume two values:
$M_{S_{Z}}=m_{S}\cdot \dfrac{h}{2\pi } \label{2}$
The term ms have (2S + 1) different values: +S, (S − 1), (S − 2),.....-S. For single unpaired electron, only two possible values for ms are +1/2 and −1/2.
The magnetic moment, μe is directly proportional to the spin angular momentum and one may therefore write
$\mu _{e}=-g_{e}\mu _{B}M_{S} \label{3}$
The appearance of negative sign due to the fact that the magnetic mom entum of electron is collinear, but antiparallel to the spin itself. The term (geμB) is the magnetogyric ratio. The Bohr magneton, μB, is the magnetic moment for one unit of quantum mechanical angular momentum:
$\mu_{B}=\dfrac{eh}{4\pi m_{e}} \label{4}$
where e is the electron charge, me is the electron mass, the factor ge is known as the free electron g-factor with a value of 2.002 319 304 386 (one of the most accurately known physical constant). This magnetic moment interacts with the applied magnetic field. The interaction between the magnetic moment (μ) and the field (B) is described by
$E=-\mu \cdot B\label{5}$
For single unpaired electron, there will be two possible energy states, this effect is called Zeeman splitting.
$E_{+\dfrac{1}{2}}=\dfrac{1}{2}g\mu _{B}B \label{6}$
$E_{-\dfrac{1}{2}}=-\dfrac{1}{2}g\mu _{B} B \label{7}$
In the absence of external magnetic field,
$E_{+1/2}=E{-1/2}=0$
However, in the presence of external magnetic field (Figure 1), the difference between the two energy states can be written as
$\Delta E=hv=g\mu _{B}B \label{8}$
With the intensity of the applied magnetic field increasing, the energy difference between the energy levels widens until it matches with the microwave radiation, and results in absorption of photons. This is the fundamental basis for EPR spectroscopy. EPR spectrometers typically vary the magnetic field and hold the microwave frequency. EPR spectrometers are available in several frequency ranges, and X band is currently the most commonly used.
Table 1: Different Microwave bands for EPR Spectroscopy
Microwave band Frequency/GHz Wavelength/cm B (electron)/Tesla
S 3.0 10.0 0.107
X 9.5 3.15 0.339
K 23 1.30 0.82
Q 35 0.86 1.25
W 95 0.315 3.3
Energy Level Structure and the g-factor
EPR is often used to investigate systems in which electrons have both orbital and spin angular momentum, which necessitates the use of a scaling factor to account for the coupling between the two momenta. This factor is the g-factor, and it is roughly equivalent in utility how chemical shift is used in NMR. The g factor is associated with the quantum number J, the total angular momentum, where $J=L+S$.
$g_J = \dfrac{J(J+1)(g_L+g_s)+(L(L+1)-S(S+1))(g_L-g_s)}{2J(J+1)} \label{9}$
Here, $g_L$ is the orbital g value and gs is the spin g value. For most spin systems with angular and spin magnetic momenta, it can be approximated that gL is exactly 1 and gs is exactly 2. T his equation reduces to what is called the Landé formula:
$g_J = \dfrac{3}{2}- \dfrac{L(L+1)-S(S+1))}{2J(J+1)} \label{10}$
And the resultant electronic magnetic dipole is:
$\mu_{J} = -g_{J}\mu_{B}J \label{11}$
In practice, these approximations do not always hold true, as there are many systems in which J-coupling does occur, especially in transition metal clusters where the unpaired spin is highly delocalized over several nuclei. But for the purposes of a elementary examination of EPR theory it is useful for the understanding of how the g factor is derived. In general this is simply referred to as the g-factor or the Landé g-factor.
The g-factor for a free electron with zero angular momentum still has a small quantum mechanical corrective $g$ value, with g=2.0023193. In addition to considering the total magnetic dipole moment of a paramagnetic species, the g-value takes into account the local environment of the spin system. The existence of local magnetic fields produced by other paramagnetic species, electric quadrupoles, magnetic nuclei, ligand fields (especially in the case of transition metals) all can change the effective magnetic field that the electron experiences such that
$B_{eff} = B_{0}+B_{local} \label{12}$
These local fields can either:
1. be induced by the applied field, and hence have magnitude dependence on $B_0$ o r are
2. permanent and independent of $B_0$ o ther than in orientation.
In the case of the first type, it is easiest to consider the effective field experienced by the electron as a function of the applied field, thus we can write:
$B_{eff} = B_{0}(1-\sigma)\label{13}$
where $\sigma$ is the shielding factor that results in decreasing or increasing the effective field. The g-factor must then be replaced by a variable g factor geff such that:
$B_{eff}=B_{0}\cdot (\dfrac{g}{g_{eff}}) \label{14}$
Many organic radicals and radical ions have unpaired electrons with $L$ near zero, and the total angular momentum quantum number J becomes approximately S. As result, the g-values of these species are typically close to 2. In stark contrast, unpaired spins in transition metal ions or complexes typically have larger values of L and S, and their g values diverge from 2 accordingly.
After all of this, the energy levels that correspond to the spins in an applied magnetic field can now be written as:
$E_{m_{s}}=m_{s}g_{e}\mu _{B}B_{0} \label{15}$
And thus the energy difference associated with a transition is given as:
$\Delta E_{m_{s}}=\Delta m_{s}g_{e}\mu _{B}B_{} \label{16}$
Typically, EPR is performed perpendicular mode, where the magnetic field component of the microwave radiation is oriented perpendicular to the magnetic field created by the magnet. Here, the selection rule for allowed EPR transitions is $\Delta m_{s} = \pm 1$, so the energy of the transition is simply:
(17) $\Delta E_{m_{s}}=g_{e}\mu _{B}B_{} \label{17}$
There is a method called Parallel Mode EPR in which the microwaves are applied parallel to the magnetic field, changing the selection rule to $\Delta m_{s} = \pm 2$. This is more fully explained in the Parallel Mode EPR: Theory module.
Sensitivity
At the thermal equilibrium and external applied magnetic field, the spin population is split between the two Zeeman levels (Figure 1) according to the Maxwell–Boltzmann law. Absorption can occur as long as the number of particles in the lower state is greater than the number of particles in the upper state. At equilibrium, the ratio predicted by the Boltzmann distribution:
$\dfrac{N_{upper}}{N_{lower}}=e^{\dfrac{-\Delta E}{k_BT}}=e^{-\dfrac{g\mu B}{k_BT}} \label{18}$
with kB is the Boltzmann constant.
At regular temperatures and magnetic fields, the exponent is very small and the exponential can be accurately approximated by the expansion,
$e^{–x} \approx 1 – x$
Thus
$\dfrac{N_{upper}}{N_{lower}}=1-\dfrac{g\mu B}{k_BT} \label{19}$
At 298 K in a field of about 3000 G the distribution shows that Nupper /Nlower=0.9986, which means the difference between is Nupperand Nlower is very small. The populations of the two Zeeman levels are nearly the same, but the slight excess in the lower level gives rise to a net absorption.
$N_{lower}-N_{upper}=N_{lower}\left [ 1-\left ( 1-\dfrac{g\mu B}{k_BT} \right ) \right ]=\dfrac{Ng\mu B}{2k_BT} \label{20}$
This expression tells us that EPR sensitivity (net absorption) increases as temperature decreases and magnetic field strength increases, and magnetic field is proportional to microwave frequency. Theoretically speaking, the sensitivity of spectrometer with K-band or Q-band or W-band shoulder be greater than spectrometer with X-band. However, since the K-, Q- or W-band waveguides are smaller, samples are necessarily smaller, thus canceling the advantage of a more favorable Boltzmann factor.
Spin Operators and Hamiltonians
Any system which has discrete energy levels and is described by defined quantum numbers can be represented by an eigenvalue equation, such that if we define an operator ($\hat{\Lambda }$ ) that is appropriate to the property being observed, the eigenfunction equation is:
$\hat{\Lambda }{\psi }_k=\ {\lambda }_k{\psi }_k \label{21}$
Here λk is an eigenvalue of a state “k” for which the eigenfunction is ψk. EPR is most concerned with the quantization of spin angular momentum, therefore, the operator must be defined is a spin operator that operated on a function that describes a spin state. In the case of a system with a total electron spin of S = ½, the two states are described by the quantum numbers Ms = +1/2 and Ms = -1/2, which measure the components Ms of angular momentum along the z-direction of the magnetic field. In most systems, it is convenient to treat the direction of the magnetic field as the z-direction, and thus the spin operator is denoted Ŝz, where Ŝ is the angular momentum operator. So, omitting the k index, the z-component of the angular momentum operator can be written as:
$\hat{S_z}\phi_e=M_s\phi_e \label{22}$
where ms is the eigenvalue of the operator Sz, and ϕe(Ms) is the corresponding eigenfunction. Adopting the α- notation for spin states, where α(e) = ϕe(Ms=+1/2) and β(e) = ϕe(Ms=-1/2), this expression can be written:
$\hat{S_z}\alpha \left(e\right)=+\dfrac{1}{2}\alpha \left(e\right) \label{23}$
$\hat{S_z}\beta \left(e\right)=-\dfrac{1}{2}\beta \left(e\right) \label{24}$
In a similar fashion, the eigenfunctions for the nuclear spin operator for a nucleus with spin = ½ can be written:
$\hat{I_z}\alpha \left(n\right)=+\dfrac{1}{2}\alpha \left(n\right) \label{25}$
$\hat{I_z}\beta \left(n\right)=-\dfrac{1}{2}\beta \left(n\right) \label{26}$
Written in the convenient Dirac notation, these expressions become:
$\left.\hat{S_z}\right|\left.\alpha \left(e\right)\right\rangle =+\left.\left.\dfrac{1}{2}\right|\alpha \left(e\right)\right\rangle \label{27}$
$\left.\hat{S_z}\right|\left.\beta \left(e\right)\right\rangle =+\left.\left.\dfrac{1}{2}\right|\beta \left(e\right)\right\rangle \label{28}$
and
$\left.\hat{I_z}\right|\left.\alpha \left(n\right)\right\rangle =+\left.\left.\dfrac{1}{2}\right|\alpha \left(n\right)\right\rangle \label{29}$
$\left.\hat{I_z}\right|\left.\beta \left(n\right)\right\rangle =+\left.\left.\dfrac{1}{2}\right|\beta \left(n\right)\right\rangle \label{30}$
Using the time-independent Schrödinger equation, we can define the energies associated with the systems described by these equations as such:
$\left.\hat{{{\mathcal H}}_e}\right|\left.\phi_{ek}\right\rangle =\left.E_{ek}\right|\left.\phi_{ek}\right\rangle \label{31}$
$\left.\hat{{{\mathcal H}}_n}\right|\left.\phi_{nk}\right\rangle =\left.E_{nk}\right|\left.\phi_{nk}\right\rangle \label{32}$
So that
$\left.\hat{{{\mathcal H}}_e}\right|\left.\alpha (e)\right\rangle =\left.E_{\alpha (e)}\right|\left.\alpha (e)\right\rangle \label{33}$
$\left.\hat{{{\mathcal H}}_e}\right|\left.\beta (e)\right\rangle =\left.E_{\alpha (e)}\right|\left.\beta (e)\right\rangle \label{34}$
$\left.\hat{{{\mathcal H}}_n}\right|\left.\alpha (n)\right\rangle =\left.E_{\alpha (n)}\right|\left.\alpha (n)\right\rangle \label{35}$
$\left.\hat{{{\mathcal H}}_n}\right|\left.\beta (n)\right\rangle =\left.E_{\alpha (n)}\right|\left.\beta (n)\right\rangle \label{36}$
Here Ĥ is the Hamiltonian operator and represents the operator for the total energy, and commutes with both I and S operators.
Electron/Nuclear Zeeman Interactions using Operators
Using the Hamiltonians derived in the last section, we can develop hamiltonians for the perturbed case in which an external magnetic field is introduced. For the simple case of the hydrogen atom with S=1/2 and I=1/2, interaction with a strong magnetic field oriented along the z-direction will be considered. Using the operator form, the Hamiltonian takes the form:
$\ {\hat{H}}= -B\hat{\mu }_z \label{37}$
Here, the electron magnetic moment operator μez is proportional to the electron spin operator. Likewise, the nuclear magnetic moment operator μnzis proportional to the nuclear spin operator Iz. Therefore,
$\ {\hat{\mu }}_{ez}=\ {\gamma }_e{\hat{S}}_zh=\ -g{\mu}_B{\hat{S}}_z \label{38}$
$\ {\hat{\mu }}_{nz}=\ {\gamma }_n{\hat{I}}_zh=\ +g_n{\mu }_n{\hat{I}}_z \label{39}$
Now the electron and nuclear spin Hamiltonians can be defined as:
$\ {\hat{\mathcal H}}_e=\ g_e{\mu}_B\hat{S_z} \label{40}$
$\ {\hat{\mathcal H}}_n=\ -g_n{\mu }_n\hat{I_z} \label{41}$
We now have a quantum mechanical framework for the energies of electronic and nuclear spin states that will be further useful in developing a description of the interactions between the magnetic moments of the two classes of particles.
Nuclear Hyperfine Structure
According to the figure 1, we should observe one spectra line in a paramagnetic molecule, but in reality, we usually observe more than one split line. The reason for that is hyperfine interactions, which results from interaction of the magnetic moment of the unpaired electron and the magnetic nuclei. The hyperfine patterns are highly valuable when it comes to determine the spatial structure of paramagnetic species and identify the paramagnetic species. As a result, nuclear spins act as probes which are sensitive to the magnetitude and direction of the field due to the unpaired electron.
In general, there are two kinds of hyperfine interactions between unpaired electron and the nucleus. The first is the interaction of two dipoles. We refer it as the anisotropic or dipolar hyperfine interaction, which is the interaction between electron spin magnetic moment and the nuclei magnetic moment, and it depends on the shape of electronic orbital and the average distance of electron and nucleus. This interaction can help us to determine the possible position of a paramagnetic species in a solid lattice.
The second interaction is known as the Fermi contact interaction, and only takes the electrons in s orbital into consideration, since p, d and f orbitals have nodal planes passing through the nucleus. We refer to this type of interaction as isotropic, which depends on the presence of a finite unpaired electron spin density at the position of the nucleus, not on the orientation of the paramagnetic species in the magnetic field.
$A=-\dfrac{8}{3}\pi \left \langle \mu _{n}\cdot \mu _{e} \right \rangle\cdot \left | \psi \left ( 0 \right ) \right |^{2} \label{42}$
A is the isotropic hyperfine coupling constant and is related to the unpaired spin density, μn is the nuclear magnetic moment, μe is the electron magnetic moment and Ψ(0) is the electron wavefunction at the nucleus. The Fermi contact interaction happens in s orbital when electron density is not zero. Thus nuclear hyperfine spectra not only includes the interaction of nuclei and their positions in the molecule but also the extent to which part or all of the molecule is free to reorientate itself according to the direction of the applied magnetic field.
Isotropic Hyperfine Interaction
In the case of one unpaired electron, the spin hamiltonian can be written as below for the isotropc part of nuclear hyperfine interaction.
$H=H_{EZ}-H_{NZ}-H_{HFS} \label{43}$
EZ means electron Zeeman, NZ means nuclear Zeeman and HFS represents hyperfine interaction. The equation can also be written as
$\hat{H}=g\mu _{B}HS_{Z}-g_{N}\mu _{N}\cdot BI_{Z}+h\cdot S\cdot aI \label{44}$
The term aS*I is introduced by Fermi contact interaction. I is the nucleus spin, H is the external field. Since μBis much larger than μN, the equation can take the form as:
$\hat{H}=g\mu _{B}HS_{Z}+h\cdot S\cdot aI \label{45}$
When one unpaired electron interacts with one nucleus, the number of EPR lines is 2I+1. When one unpaired electron interacts with N equivalent nuclei, the number of EPR lines is 2NI+1. When one electron interacts with nonequivalent nucleis (N1, N2.....), the numer of EPR lines is
$\prod_{i=1}^{k}(2N_{i}I_{i}+1) \label{46}$
In the case of DPPH, I=1 and two nitrogen nucleui are equivalent. 2NI+1=5, we can get five lines: 1:2:3:2:1.
The table below shows the relative intensities of the lines according to unpaired electrons interacting with multiple equivalent nuclei.
Number of Equivalent Nuclei
Relative Intensities
1
1:1
2
1:2:1
3
1:3:3:1
4
1:4:6:4:1
5
1:5:10:10:5:1
6
1:6:15:20:15:6:1
We can observe that increasing number of nucleuses leads to the complexity of the spectrum, and spectral density depends on the number of nuclei as equation shown below:
$Spectral density_{EPR}=\dfrac{\prod_{i=1}^{k}2N_{i}I_{i}+1}{\sum_{i=1}^{k}2\left | a_{i} \right |N_{i}I_{i}} \label{47}$
$a$ is the isotropic hyperfine coupling constant.
The g Anisotropy
From the below equation, we can calculate g in this way:
$\Delta E=hv=g\mu _{B}B \label{48}$
If the energy gap is not zero, g factor can be remembered as:
$g\approx \dfrac{1}{14}\dfrac{\nu \left [ GHZ \right ]}{B\left [ T \right ]} \label{49}$
The g factor is not necessarily isotropic and needs to be treated as a tensor g. For a free electron, g factor is close to 2. If electrons are in the atom, g factor is no longer 2, spin orbit coupling will shift g factor from 2. If the atom are placed at an electrostatic field of other atoms, the orbital energy level will also shift, and the g factor becomes anisotropic. The anisotropies lead to line broadening in isotropic ESR spectra. The Electron-Zeeman interaction depends on the absolute orientation of the molecule with respect to the external magnetic field. Anisotropic is very important for free electrons in non-symmetric orbitals (p,d).
In a more complex spin system, Hamiltonian is required to interpret as below:
$\hat{H}_{s}=\mu _{B} \vec{B} \cdot g\cdot \hat{S}+\sum _{i} \vec{I}_{i}\cdot A_{i}\cdot \vec{S} \label{50}$
g and Ai are 3*3 matrices representing the anisotropic Zeeman and nuclear hyperfine interactions, thus it is more accurate to describe g-factor as a tensor like:
$g=\sqrt{g_{x}^{2}\cdot sin^{2}\alpha \cdot cos^{2}\beta +g_{y}^{2}\cdot sin^{2}\alpha \cdot cos^{2}\beta +g_{z}^{2}cos^{2}\alpha } \label{51}$
Alpha and beta is the angle between magnetic field with respect to principle axis of g tensor. If gx=gy, it can be expressed as:
$g=\sqrt{g_{x}^{2}\cdot sin^{2}\alpha +g_{z}^{2}cos^{2}\alpha } \label{52}$
Thus, we can identify the g tensor by measuring the angular dependence in the above equation.
Spin Relaxation Mechanisms
The excess population of lower state over upper state for a single spin system is very small as we can calculate from the following example. With the temperature of 298K in a magnetic field of 3000G, Nupper /Nlower=0.9986, which means the populations of the two energy levels are almost equal, yet the slight excess in the lower level leads to energy absorption. In order to maintain a population excess in the lower level, the electrons from the upper level give up the hν energy to return to the lower level to satisfy the Maxwell–Boltzmann law. The process of this energy releasing is called spin relaxation process, of which there are two types, known as spin–lattice relaxation and spin–spin relaxation.
Spin-lattice relaxation
this implies interaction between the species with unpaired electrons, known as "spin system" and the surrounding molecules, known as "lattice". The energy is dissipated within the lattice as vibrational, rotational or translational energy. The spin lattice relaxation is characterized by a relaxation time T1e, which is the time for the spin system to lose 1/eth of its excess energy. Rapid dissipation of energy (short T1e) is essential if the population difference of the spin states is to be maintained. Slow spin-lattice relaxation, which is of frequent occurence in systems containing free radicals, especially at low temperatures, can cause saturation of the spin system. This means that the population difference of the upper and lower spin states approaches zero, and EPR signal ceases.
Spin-spin relaxation
Spin-spin relaxation or Cross relaxation, by which energy exchange happens between electrons in a higher energy spin state and nearby electrons or magnetic nuclei in a lower energy state, without transfering to the lattice. The spin–spin relaxation can be characterized by spin-spin relaxation time T2e.
When both spin–spin and spin–lattice relaxations contribute to the EPR signal, the resonance line width (ΔB) can be written as
$\Delta B\propto \dfrac{1}{T_{1e}}+\dfrac{1}{T_{2e}} \label{53}$
From the equation, we can tell that when T1e > T2e, ΔB depends primarily on spin–spin interactions. Decreasing the spin-spin distance, which is the spin concentration, T1e will become very short, approximately below roughly 10−7 sec, thus the spin-lattice relaxation will have a larger influence on the linewidth than spin-spin ralaxation. In some cases, the EPR lines are broadened beyond detection. When a spin system is weakly coupled to the lattice, the system tends to have a long T1e and electrons do not have time to return to the ground state, as a result the population difference of the two levels tends to approach zero and the intensity of the EPR signal decreases. This effect, known as saturation, can be avoided by exposing the sample to low intensity microwave radiation. Systems with shorter T1e are more difficult to saturate.
Parting Thoughts
Although EPR is limited to investigation of compounds and materials with unpaired electrons, it is undoubtedly the most direct and useful spectroscopic method for probing the properties of these specific systems. Another advantage is that sample preparation is simple and EPR does not cause destruction or activation in the sample. By probing the fundamental splitting of energy levels of spins with regard to their orientation in an external magnetic field, interactions between paramagnetic spin systems and their local environments can be detected. EPR spectra is highly sensitive to the local electonic structure, oxidation state and the proximity of magnetic nuclei to the system in question.
Problems
1. Why are microwaves necessary to study the electron spin resonances?
2. What is the g-factor? What does it mean if the value is 2?
Answer
1. EPR is targeted to unpaired electrons. Single unpaired electron behaves like a small magnetic bar when placed in a large maganetic field. It will orient itself parallel to the large magnetic field. At a particular magnetic field inensity, the microwave irradiation will induce unpaired electrons to orient against the large magnetic field. This effect will cause Zeeman splitting, when the energy difference between the lower and the higher energy level matches the microwave frequency. There will be absorption of energy, thus producing an EPR resonance, and detected by the spectrometer.
1. As we have discussed before, g-factor can be calculated from measuring the magnetic field and frequency. g-factor is an effective Zeeman factor. For a free electron, isotropic ge = 2.0023193043617. It is predicted by quantum electrodynamics. This free electron has spin angular momentum but no orbital angular momentum. For electrons in an atom, g factor will shift from ge due to spin orbit coupling. thus g factor is the characteristic of different electronic structures.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Electron_Paramagnetic_Resonance/EPR_-_Theory.txt
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This splitting occurs due to hyperfine coupling (the EPR analogy to NMR’s J coupling) and further splits the fine structure (occurring from spin-orbit interaction and relativistic effects) of the spectra of atoms with unpaired electrons. Although hyperfine splitting applies to multiple spectroscopy techniques such as NMR, this splitting is essential and most relevant in the utilization of electron paramagnetic resonance (EPR) spectroscopy.
Introduction
Hyperfine Splitting is utilized in EPR spectroscopy to provide information about a molecule, most often radicals. The number and identity of nuclei can be determined, as well as the distance of a nucleus from the unpaired electron in the molecule. Hyperfine coupling is caused by the interaction between the magnetic moments arising from the spins of both the nucleus and electrons in atoms. As shown in Figure $1$, in a single electron system the electron with its own magnetic moment moves within the magnetic dipole field of the nucleus.
This spin interaction in turn causes splitting of the fine structure of spectral lines into smaller components called hyperfine structure. Hyperfine structure is approximately 1000 times smaller than fine structure. Figure $2$ shows a comparison of fine structure with hyperfine structure splitting for hydrogen, though this is not to scale.
The total angular momentum of the atom is represented by F with regards to hyperfine structure. This is found simply through the relation F=J+I where I is the ground state quantum number and J refers to the energy levels of the system.
Results of Nuclear-Electron Interactions
These hyperfine interactions between dipoles are especially relevant in EPR. The spectra of EPR are derived from a change in the spin state of an electron. Without the additional energy levels arising from the interaction of the nuclear and electron magnetic moments, only one line would be observed for single electron spin systems. This process is known as hyperfine splitting (hyperfine coupling) and may be thought of as a Zeeman effect occurring due to the magnetic dipole moment of the nucleus inducing a magnetic field.
The coupling patterns due to hyperfine splitting are identical to that of NMR. The number of peaks resulting from hyperfine splitting of radicals may be predicted by the following equations where Mi is the number of equivalent nuclei:
• $\text{# of peaks}=M_{i}I+1$ for atoms having one equivalent nuclei
• $\text{# of peaks}=(2M_{1}I_{1}+1)(2M_{2}I_{2}+1)....$ for atoms with multiple equivalent nuclei
For example, in the case of a methyl radical 4 lines would be observed in the EPR spectra. A methyl radical has 3 equivalent protons interacting with the unpaired electron, each with I=1/2 as their nuclear state yielding 4 peaks.
The relative intensities of certain radicals can also be predicted. When I = 1/2 as in the case for 1H, 19F, and 31P, then the intensity of the lines produced follow Pascal's triangle. Using the methyl radical example, the 4 peaks would have relative intensities of 1:3:3:1. The following figures2 show the different splitting that results from interaction between equivalent versus nonequivalent protons.
It is important to note that the spacing between peaks is 'a', the hyperfine coupling constant. This constant is equivalent for both protons in the equivalent system but unequal for the unequivalent protons.
The Hyperfine Coupling Constant
The hyperfine coupling constant ($a$) is directly related to the distance between peaks in a spectrum and its magnitude indicates the extent of delocalization of the unpaired electron over the molecule. This constant may also be calculated. The following equation shows the total energy related to electron transitions in EPR.
$\Delta E=g_e \mu _e M_s B + \sum_i g_{N_i} \mu_{N_i}M_{I_i}(1-\sigma _{i})+\sum_{i}a_i M_s M_{I_i}$
The first two terms correspond to the Zeeman energy of the electron and the nucleus of the system, respectively. The third term is the hyperfine coupling between the electron and nucleus where $a_i$ is the hyperfine coupling constant. Figure $5$ shows splitting between energy levels and their dependence on magnetic field strength. In this figure, there are two resonances where frequency equals energy level splitting at magnetic field strengths of $B_1$ and $B_2$.
These parameters are essential in the derivation of the hyperfine coupling constant. By manipulating the total energy equation (those interested in the entire derivation, refer to the first outside link), the following two relations may be derived.
$B_1= \dfrac {h\nu -a/2} {g\mu_e}$
$B_2= \dfrac{h\nu +a/2}{g\mu_e}$
From this, the hyperfine coupling constant ($a$) may be derived where $g$ is the g-factor.
\begin{align} \Delta B &=B_{2}-B_{1} \[4pt] &=\dfrac {h\nu +a/2} {g\mu_{e}}- \dfrac {h\nu -a/2}{g\mu_{e}} \end{align}
so solving for hyperfine coupling constant results in the following relationship:
$a= g\mu_e \Delta B$
Isotropic and Anisotropic Interactions
Electron-nuclei interactions have several mechanisms, the most prevalent being Fermi contact interaction and dipole interaction. Dipole interactions occur between the magnetic moments of the nucleus and electron as an electron moves around a nucleus. However, as an electron approaches a nucleus, it has a magnetic moment associated with it. As this magnetic moment moves very close to the nucleus, the magnetic field associated with that nucleus is no longer entirely dipolar. The resulting interaction of these magnetic moments while the electron and nucleus are in contact is radically different from the dipolar interaction of the electron when it is outside the nucleus. This non-dipolar interaction of a nucleus and electron spin in contact is the Fermi contact interaction. A comparison of this is shown in Figure $6$. The sum of these interactions is the overall hyperfine coupling of the system.
Fermi contact interactions predominate with isotropic interactions, meaning sample orientation to the magnetic field does not affect the interaction. Due to the fact that this interaction only occurs when the electron is inside the nucleus, only electrons in the s orbital exhibit this kind of interaction. All other orbitals (p,d,f) contain a node at the nucleus and can never have an electron at that node. The hyperfine coupling constant in isotropic interactions is denoted 'a'.
Dipole interactions predominate with anisotropic interactions, meaning sample orientation does change the interaction. These interactions depend on the distance between the electron and nuclei as well as the orbital shape. The typical scheme is shown in Figure $7$.
Dipole interactions can allow for positioning paramagnetic species in solid lattices. The hyperfine coupling constant in isotropic interactions is denoted 'B'.
Superhyperfine Splitting
Further splitting may occur by the unpaired electron if the electron is subject to the influence of multiple sets of equivalent nuclei. This splitting is on the order of 2nI+1 and is known as superhyperfine splitting. As hyperfine structure splits fine structure into smaller components, superhyperfine structure further splits hyperfine structure. As a result, these interactions are extremely small but are useful as they can be used as direct evidence for covalency. The more covalent character a molecule exhibits, the more apparent its hyperfine splitting.
For example, in a CH2OH radical, an EPR spectrum would show a triplet of doublets. The triplet would arise from the three protons, but superhyperfine splitting would cause these to split futher into doublets. This is due to the unpaired electron moving to the different nuclei but spending a different length of time on each equivalent proton. In the methanol radical example, the electron lingers the most on the CH2 protons but does move occasionally to the OH proton.
Problems
1. What is the number of peaks of a benzene radical in EPR due to hyperfine coupling and what are their relative intensities?
2. What is the number of peaks for CH2(OCH3), a methoxymethyl radical in EPR due to hyperfine coupling?
3. Why are s orbitals only considered in Fermi contact interactions?
Answers:
1. 7 lines, 1:6:15:20:15:6:1 (benzene has 6 equivalent protons so the number of peaks is M+1 = (6+1) = 7 peaks, intensities come from pascal's triangle)
2. 12 lines (has two different equivalent nuclei, one with two protons and one with 3 so (M1+1)(M2+1) = (2+1)(3+1) = 12 peaks)
3. p, d, and f orbitals have nodes at the nucleus and do not exhibit Fermi contact interactions
Contributors and Attributions
• Stephanie Gray
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• 2D NMR: Indirect Detection
The reverse detection technique became increasingly popular due to the software improvements of the spectrometers from one hand and to the increase of the sensitivity of the detection of a X nucleus by the indirect way. In fact the reverse detection solves the problem of the low concentrations. Moreover, it allows to reach the NMR parameters such as chemical shifts, coupling constants and relaxation time spin lattice for nuclei impossible to study by the direct detection.
• 2D NMR Background
These atoms and some others behave, basically, as small magnets with a so called nuclear magnetic momentum quantified by a quantic value: the spin.
• 2D NMR Basics
The 2D NMR experiment belongs as well to the Fourier transform spectroscopy than to the impulsion one and relies on a sequence of three time intervals: preparation, evolution and detection (3). In some experiment another time interval is added before the detection: the mixing time.
• 2D NMR Experiments
An overview of 2D NMR fundamentals is given.
• 2D NMR Introduction
Some general principles and techniques used in two-dimensional NMR are discussed. Applications covered are mostly concerned with protein NMR, but additional 2D techniques and applications can be found in the references section.
• Heteronuclear Correlations
The heteronuclear correlation of chemical shifts by scalar coupling derives from the existence of heteronuclear scalar coupling all owing a magnetization transfer from the more sensitive nucleus (1H) toward the less sensitive nucleus (13C).
• Homonuclear Correlations
This sequence allows to get the coupling constant along the Fl axis and the chemical shift of each uncoupled proton along the F2 axis (Spectrum 1 and 2). Of course, the couplings involving two different nuclei (31P, 19P coupled to 1H) are always found in the F2 dimension. One can thus determine the coupling between a proton within a multiplet of a coupled spectrum and phosphorus.
2D NMR
The advantages of the indirect detection
The reverse detection technique became increasingly popular due to the software improvements of the spectrometers from one hand and to the increase of the sensitivity of the detection of a X nucleus by the indirect way. In fact the reverse detection solves the problem of the low concentrations. Moreover, it allows to reach the NMR parameters such as chemical shifts, coupling constants and relaxation time spin lattice for nuclei impossible to study by the direct detection (Fig. 28).
For example it is not always obvious to get NMR spectra for nuclei such as 57Fe, 183W, 187Os and sometimes even 15N in direct observation. Consequently, the reverse detection becomes the only possible way either for 1H-15N correlation in the case of peptides or proteins or also for 1H-X correlation in the case of organometallic compounds. In each of these examples the reverse detection allowed to determine the chemical shift of the heteronucleus together with the coupling constants J(X-H).
The arrows figure the magnetization transfer from the proton towards the 13C and then from the 13C towards the proton.
The available pulse sequences
The HSQC (Heteronuclear Single Quantum Coherence) is specially used for the structural determination of small proteins. The HMBC (Heteronuclear Multiple Bond Coherence) is optimized for long range couplings and works better than a COLOC.
Bibliography
1. La détection inverse en RMN ; M Boudonneu Analusis N°1, vol.18, 1990
2. La RMN concepts et méthodes; D. Canet inter éditions 1991
3. La spectroscope de RMN, H. Gunther MASSON 1994.
4. NMR in protein studies. M. P. Williamson. Chem. in Britain April 1991(335-337).
5. RMN 2D BRUKER 1993 Poly. 1, 2 et 3.
6. Resonance Magnétique Nucléaire; Evolution instrumentale et analyse de structure; J.P. Girault le technoscope de biofutur n°38 sept 90
7. Sensitivity enhancement by signal averaging in pulsed by Fourier transform NMR spectroscopy; D. L.Rabenteck; J. of Chem. Ed. Vol. 61, n°70, Oct 1984
8. Sequence Cosy Poly. BRUKER 1993.
9. Topic in chemical instrumentation part 3 and 4. 2 D Methods W. King and K. R. Williams J. of Chem. Educ. vol.67n°4.
10. Topics in chemical instrumentation (Definition) W. King and K. R. Williams
11. Two Dimensional NMR spectroscopy; Applications for chemists and biochemists; W. R. Crousmum, R. M. K. Carlson. Vol 9.
12. Two dimensional NMR Aspect 2000 - 3000 BRUKER
13. G. E. Reeman et Moms, J. ChemSoc, Chem.Comm, 684 (1978).
14. H. Kessler, W. Bernel and C. Giresinger, J. AM. Chem. Soc. 107, 1083 (1985).
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These atoms and some others behave, basically, as small magnets with a so called nuclear magnetic momentum quantified by a quantic value: the spin.
The spin ½ nuclei seen by the quantum mechanic
The example of the proton: The atomic physic teaches us that some nuclei have a kinetic momentum due to their spin state: P and also a magnetic momentum µ.1 which is defined by the following relation:
$\mu=\gamma P$
where is the gyro magnetic ratio, thus µ/P characterizes a given nucleus. The quantum mechanic describes an atom with wave functions solution of the Schrödinger equation [2]. For a proton, nucleus with spin quantum number I = 1/2 or mI = +/- ½ the own wave functions are the following: for for mI = -1/2.
mI corresponds to the magnetic quantum number which characterizes the stationary state of the nucleus, which depends on the spin numbers I of the nucleus. The total number of the possible own stationary states of the nucleus is then equal to 2I+1. Thus the proton may exist from the point of view of its spin momentum in only two stationary states.
Without any magnetic field the and states have the same energy they are, so called, degenerated. It is only in a static homogenous magnetic field of value Bo and following the interaction between Bo and that this degeneracy is suppressed (Figure $1$). The separation of energy then produced is proportional to the intensity of the field Bo and creates the necessary condition for the existence of spectroscopic transition and constitutes the basis of the spectroscopy by nuclear magnetic resonance.
The proton will occupy the state (i.e., the weakest energy one) with a higher probability. The greater the Bo field amplitude, the more the difference of energy (or ΔE) between the two spin states increases. We describe the difference of occupancy between the basic state and the excited one with the Boltzmann law:
$\left(\frac{N \beta}{N \alpha}=\exp \cdot\left(\frac{-\Delta E}{k T}\right)\right)$
Where ΔE is the difference of energy between the two states, k the Boltzmann constant, T the absolute temperature in °K, N the number of spins at the high energy level and N the number of spins at the low energy level.
At the equilibrium only a small excess of nuclei are at the fundamental state. For a field of 1.4 T at room temperature, ΔE=0.02.J.mol-1. According to the theory the population excess determines the probability of excitation. However, this population excess does not reach more than 0.001%
The Nuclear Magnetic Resonance phenomena
According to the Bohr equation, , we need for inducing a transition towards the high level energy, an energy quantum with a value of . This is a resonance condition, with which is the resonance frequency of the nucleus; ħ the Planck constant divided by and the gyro magnetic ratio. The resonance frequency of the nucleus, which is also called the Larmor frequency, is then proportional to the field Bo value following the equation .
Its range is in the high frequencies:
in a field of 1.4T at the level of a frequency of =60 MHz.
This corresponds to an ultra short radio wavelength of
Schematic representation of the coherence
When the sample is in a field Bo we create a coherence at the thermal equilibrium (Figure $2$). We create therefore also a magnetization along the Z axis [3].
This figure represents the spins precession movement which is induced by the field Bo. For example, we may consider the proton as a magnetic dipole whose component µz is either parallel or anti parallel to the z axis. The angular speed of the precession moment is
The rotating coordinates or rotating frame
It is convenient to figure only the magnetic momentum of the excess of the nuclei in their fundamental states (Figure $1$).
M is the macroscopic magnetization as a result of the individual nuclear moment µ. The xy plane does not contain any magnetization component. In addition to the coordinate system K(x,y,z), another system K'(x,y,z) rotates around the z axis with the angular speed thus . In this figure, is a virtual field Bf opposite to Bo, which is generated by the only relative rotation from one versus the other of the two coordinate systems K and K'. This means that the µ vector is static in the system K’ which is called rotating frame, it happens when has the same sign and the same value than 0. (3)
Requirements to get a resonance
The excitation of the nucleus is generated by a high frequency impulsion B1 at a high power (50 watts) for a short time (10 to 50 µs). The field Bo is high and constant, the field B1 is assumed to be weak, perpendicular to B0 and rotates in the plane XOY with an angular speed 1 and in the same sense than µ [3].
Practically, a small coil put perpendicular in the field Bo is run by a current (Figure $4$). This one causes an oscillatory induction along the axis of the coil. We may represent this induction Bx with two magnetic vectors rotating in opposite sense B(l) and B(r), one of which has the same rotation sense than the precession. The second vector is practically without any influence on the phenomenon.
The emitter coil on the x axis generates a weak electromagnetic field B1 linearly polarized with a frequency and with an amplitude 2B1.
The Fourier Transform
We may compare the NMR spectroscopy and the optic one. In the optical spectroscopy, the prism allows the monochromatic analysis of a polychromatic source interacting with the mater. In the NMR spectroscopy, the impulsion B1 yield a broad frequency spectrum and the Fourier transform allows a "monochromatic" analysis of the interaction beam/nucleus. The technique is to treat the signal M(t) according to the following relationship (2):
$F(v)=\int_{0}^{\infty} f(t) \exp (-2 i \pi v t) d t$
Since the pulse sequence last generally less than a second, we may achieve a great number of accumulations of data before to proceed to the Fourier transform. The signal to noise ratio thus improves (this ratio is proportional to where S = number of scans).
Bibliography
1. La détection inverse en RMN ; M Boudonneu Analusis N°1, vol.18, 1990
2. La RMN concepts et méthodes; D. Canet inter éditions 1991
3. La spectroscope de RMN, H. Gunther MASSON 1994.
4. NMR in protein studies. M. P. Williamson. Chem. in Britain April 1991(335-337).
5. RMN 2D BRUKER 1993 Poly. 1, 2 et 3.
6. Resonance Magnétique Nucléaire; Evolution instrumentale et analyse de structure; J.P. Girault le technoscope de biofutur n°38 sept 90
7. Sensitivity enhancement by signal averaging in pulsed by Fourier transform NMR spectroscopy; D. L.Rabenteck; J. of Chem. Ed. Vol. 61, n°70, Oct 1984
8. Sequence Cosy Poly. BRUKER 1993.
9. Topic in chemical instrumentation part 3 and 4. 2 D Methods W. King and K. R. Williams J. of Chem. Educ. vol.67n°4.
10. Topics in chemical instrumentation (Definition) W. King and K. R. Williams
11. Two Dimensional NMR spectroscopy; Applications for chemists and biochemists; W. R. Crousmum, R. M. K. Carlson. Vol 9.
12. Two dimensional NMR Aspect 2000 - 3000 BRUKER
13. G. E. Reeman et Moms, J. ChemSoc, Chem.Comm, 684 (1978).
14. H. Kessler, W. Bernel and C. Giresinger, J. AM. Chem. Soc. 107, 1083 (1985).
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Durations
The 2D NMR experiment belongs as well to the Fourier transform spectroscopy than to the impulsion one and relies on a sequence of three time intervals: preparation, evolution and detection (3). In some experiment another time interval is added before the detection: the mixing time ( Figure $1$).
The preparation time
Upon the preparation time the spin system under study is firstly prepared, for example it is submitted either to a decoupling experiment or just to a transverse magnetization by the means of a 90° impulsion. It allows the excited nuclei to get back their equilibrium state between two successively executed pulse sequence (5)
The evolution time t1
During the evolution time $t_1$, the spin system is evolving under the effect of different factors, each coherence evolves at its own characteristic frequency as a function of the chemical shift and of the scalar coupling of the corresponding nucleus.
The Mixing time
It is made of a pulse sequence which achieves coherence transfers in such a way that different frequencies can be correlated.
The detection time
The acquisition of the modulated signal takes place during the detection period. The sequence we just described does not constitute by itself a 2D NMR experiment.
JEENER
The idea of Jeener consists in the stepwise increasing of the evolution time tl. This will allow to get an NMR signal under the aspect of a sampling of free precession signals of the s(t2) type. These FID will differ from each other only by the tl period duration written under a matricial form s(tl ,t2) (5). The tl delay is the time between the first and the second pulse ( Figure $2$).
The first Fourier transform as a function of tl gives us an interferogram of the form ( Figure $3$).
A second Fourier transform, versus the second variable t2, gives an NMR spectrum with two frequencies dimensions F1 and F2 (Figure $3$). The result of this two fold Fourier transform does not get two spectra and but only one spectrum as a function of two independent frequencies, exhibiting a peak with the coordinates . Thus, an alimentation evolving with the frequency in the time course t1 has been converted in another coherence evolving with the frequency during the period t2.
This double Fourier transform in both dimension yields thus a matrix . ( Figure $3$ ).
2D NMR Experiments
The experiment
When $\omega_1 \neq \omega_o$: There is no interaction between the field $B_1$ and µ (they are not in accord) and the precession is kept around the axis Oz.
When $\omega_1 \approx \omega_o$: $B_1$ and $\mu$ are in accordance. Thus the couple applied by $B_1$ modifies the angle $\alpha$ of µ with Oz, and then causes transitions between the magnetic sublevels.
There is resonance. The value of the z magnetization decreases and it appears, in the $y$ direction, the transverse magnetization $M_y$.
The Rotating Coordinate System
Impulsion angle and position of the $\vec{M}$ vector (Figure $1$) is given by the following relation , with $B_1$ which represents the amplitude or the power of the impulsion and tp the width or duration of the impulsion. It is possible to make these two values vary together in such a way that particularly interesting rotation angles appear. One of these is the angle $\alpha = 90^o$.
In this case all the magnetization is orientated in the plane xy and the signal reaches its maximum of intensity. (7)
An other angle allows the inversion of the M vector by the application of an impulsion or = 180°. In this experiment, the vector M is orientated in –z. Practically, one uses the angle = 45° in order to get a good compromise between measurement time/quality of the response.
The Free Induction Decay Signal
The signal given by the receiver coil is known under the name of interferogram: Free Induction Decay (FID). In opposition to the continue wavelength signal, the time dependent signal we monitor in this case is an emission signal, because the radiofrequency field B1 is turned off upon the signal acquisition. The experiment gives us practically a variable field which is linear at high frequency following the y axis. In fact it is an oscillator or an emitter with the Larmor frequency of the nuclei under study (7) .
The time signal S(t) generated in the receiver coil by the xy component of M weakens under the relaxation process.
The notion of phase cycle
The phase cycles or phase programs allow to:
• choose the relevant signals and to neglect these ones which do not contain information and which, eventually, are susceptible to hide some other useful signals.
• discriminate the sign of the frequencies in the fl dimension.
• compensate the inhomogeneity from one or several impulsion in the sequence,
• to achieve an optimal quadratic detection in the f2 axis. (5)
Any impulsion in a sequence has its own phase cycle which may be more or less complex. The ideal number of scans must be a multiple of the phase number of the longer cycle. For a better understanding of this method, a standard nomenclature has been defined (Fig. 7).
Example $1$: The Cyclops Cycle
= x,y,-x,-y. 0,1,2,3. This cycle involves the accumulation of a multiple of 4 numbers between each free precession.
The first impulsion B1 occurs around the x and the emission will occur in the y axis, the third impulsion along the - x and the reception in -y, etc.
There are other variations of this cycle like the following:
$\Phi = x,-x, y,-y$
or
$\Phi = 0,2,1,3.$
or still:
$\Phi = x,x,-x,-x,y,y,-y,-y$
or
$\Phi 0,0,2,2,1,1,3,3.$
This last one implies a multiple of eight transient cycles in the accumulation.
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Some general principles and techniques used in two-dimensional NMR are discussed. Applications covered are mostly concerned with protein NMR, but additional 2D techniques and applications can be found in the references section.
Introduction
A two dimensional variation of NMR was first proposed by Jean Jeener in 1971; since then, scientists such as Richard Ernst have applied the concept to develop the many techniques of 2D NMR. Although traditional, one-dimensional NMR is sufficient to observe distinct peaks for the various funtional groups of small molecules, for larger, more complex molecules, many overlapping resonances can make interpretation of an NMR spectrum difficult. Two-dimensional NMR, however, allows one to circumvent this challenge by adding additional experimental variables and thus introducing a second dimension to the resulting spectrum, providing data that is easier to interpret and often more informative.
Basics of 2D NMR
Experimental Set-up
In traditional 1D Fourier transform NMR, a sample under a magnetic field is hit with a series of RF pulses, as seen in the pulse sequence below, and the Fourier transform of the outgoing signal results in a 1D spectra as a function chemical shift.
Figure 1: Pulse sequence of a typical FT-NMR experiment
A 2D NMR experiment, however, adds an additional dimension to the spectra by varying the length of time ($\tau$)) the system is allowed to evolve following the first pulse. The result is an outgoing signal f ($\tau$, t2), which, when Fourier transformed, gives a 2D spectrum of F ($\omega$1, $\omega$2).
The use of two-dimensional NMR allows the researcher to better resolve signals which would normally overlap in 1D NMR. Depending on the size of your molecule, different variations or combinations of 2D and multidimensional NMR experiments are utilized.
The Spin Hamiltonian
The spin of a given nuclei during any NMR experiment is governmed by the spin Hamiltonian. If long-range spin interactions are ignored, the spin Hamilitonian for a one-spin system is given the equation
$\hat{H}=\hat{H}^0+\hat{H}_{RF}$
The magnetic field along the z-axis, shielding, and J-coupling with nearby nuclei are all constant and are accounted for in H0. HRF is the induced magnetic field resulting from an RF pulse. For a system where two spins are coupled, the H0 is
$\hat{H}=\omega_{1}^0\hat{I}_{1z}+\omega_{2}^0\hat{I}_{2z}+{2}\pi{J}_{12}\hat{I}_{1}\hat{I}_{2}$
Where $\omega$ is the Larmor frequency, I is the net magnetization vector of the given nucleus or nuclei, and J is the observed J coupling between nuclei. $\omega$ is directly related to the chemical shift ($\delta$) by the equation
$\omega_{j}^0=-\gamma_{j}{B}^0(1+\delta_{j}^{iso})$
Where $\gamma$ is the gyromagnetic ratio of the given isotope. If nuclei 1 and 2 are of the same element and isotope, the system is referred to as homonuclear. If they are different, it is a heteronuclear spin system.
Correlation Spectroscopy (COSY)
The most basic form of 2D NMR is the 2D COSY (pulse sequence shown below) experiment, a homonuclear experiment with a pulse sequence similar to the procedure dicussed above. It consists of a 90o RF pusle followed by an evolution time and an additional 90o pulse. The resulting oscillating magnetization (symbolized by decaying the sinusoidal curve) is then acquired during t2.
Figure 2: Pulse sequence of a 2D COSY experiment
The analysis of the acquired spectrum is discussed below, making it useful for determining the coupling between nuclei that are connected through one to three bond lengths. However, in macromolecules such as proteins, coupling through bonds alone is not sufficient to obtain substantial structural information. For this reason, the Nuclear Overhauser Effect (NOE) is often used in protein NMR to obtain information on the distance between nuclei through space rather than through bonds.
Nuclear Overhauser Effect (NOE)
Thus far, only the coupling of nuclei through bonds has been considered. In bond coupling, the magnetization of nuclei affect those closely bound to them through the electrons that make up those bonds; however, coupling directly between nuclei that are in close spatial proximity to each other also occurs. This is called the Nuclear Overhauser Effect, and it arises when the spin relaxation of nuclei A is felt by nearby nuclei B, stimulating a corresponding change in magnetization in B. In a typical NMR spectrum, the interference of electrons makes this coupling undetectable. However, a sample can be decoupled to “neutralize” the bond coupling through electrons, allowing the space coupling of the NOE to be detected. This is called NOESY (NOE correlation spectroscopy) and is another type of homonuclear NMR.
The pulse sequence for a NOESY NMR experiment is depicted below.
Figure 3: Pulse sequence of NOESY experiment
Like COSY, the first step is a 90o pulse followed by a variable evolution time. Unlike COSY, however, pulse two actually consists of two 90 degree pulses separated by a short delay. The first pulse converts the bulk magnetization from the transverse plane to the z-plane, eliminating the effect of electron-aided bond coupling. Then, during the $\tau$m, there is cross relaxation between spatially adjacent nuclei. Finally, the last 90 degree pulse converts the space coupling of nuclei into an observable transverse magnetization, which can be detected during t2.
2D NMR Spectra
As discussed earlier, by performing multiple one dimensional experiments at varying lengths ($\tau$) of the evolution period and performing a Fourier transformation on the signal which converts f ($\tau$, t2) to F($\omega$1, $\omega$2), a two dimensional spectrum can be formed into a 3D contour map.
A more useful representation of 2D data is called a correlation map. The correlation map of the steroid progesterone is shown below.
Figure 4: 2D COSY spectrum of progesterone
In this representation, the x- and y-axes correspond to the frequencies resulting from the Fourier transforms, and the intensity of shade at each frequency coordinate indicates the peak intensity. Two types of peaks are observed in a homonuclear correlation map—diagonal peaks and cross peaks. Diagonal peaks are found along the diagonal of the map where the x- and y-axes have equal frequency values and simply correspond to the absorptions from a one-dimensional NMR experiment. Because heteronuclear NMR does not involve the same isotope, diagonal peaks are not observed. Cross peaks, on the other hand, give information on the coupling of two nuclei and are seen in both homo- and heteronuclear spectra.
Applications in Protein NMR
As previously mentioned, the major advantage of 2D NMR over 1D NMR is the ability to distinguish between the overlapping signals that exist in larger molecules. Heteronuclear two-dimensional NMR is especially important in biological chemistry in the elucidation of the three-dimensional structure of proteins.
Heteronuclear Single Quantum Coherence (HSQC)
A protein is make-up of a series of amino acid monomers. Although there are 19 different amino acids each with a distinct side chain, the protein backbone is an invariable pattern of NH-C-CO as shown in Figure 5.
When synthesized under the right conditions, a heavy atom protein can be produced which constains NMR active nuclei; however, a 15N nucleus has a very low gyromagnetic ratio. According to the Hamiltonian operators discussed above, it will give a very weak signal in traditional 2D NMR. Fortunately, the nucleus can be detected indirectly by transferring polarization through a 1H nucleus. This method is used in HSQC NMR.
In protein NMR, each HSQC experiment has three steps:
1. An INEPT (insensitive nuclei enhanced by polarization transfer) transfers the polarization of a 1H nuclei to the neighboring 15N (see figure below)
2. The polarization is transferred back to the 1H nuclei
3. Signal from the 1H nuclei is recorded
The pulse sequence for a typical HSQC experiment is detailed below.
Figure 6: Pulse sequence of a HSQC experiment involving 1H and 15N nuclei
A 90o1H RF pulse creates a transverse polarization in 1H nuclei. Following the pulse, the nuclei are allowed to evolve for a 1/(4J) time period, which is the longitudinal relaxation time. Next, a 180o 1H and 15N pulse are used at the same time. During the subsequent relaxation time, the 1H nuclei develop a polarization that is antiphase to 15N. Finally, a 90o 1H and 15N pulse, again simultaneous, enacts the INEPT transfer of antiphase magnetization from the 1H nucleus to the 15N nucleus. Following the INEPT transfer, the 15N nuclei are allowed to evolve during $\tau$before a reverse INEPT transfer moves the 15N polarization back to 1H and a 15N decoupled signal is recorded.
An example HSCQ spectrum from ubiquitin is shown below.
Figure 7: 1H15N HSQC spectrum of ubiquitin
Notice the greater clarity of spectra of the HSQC vs the COSY experiment. This is a strong advantage of heteronuclear NMR. In this diagram, each peak corresponds to a cross peak, showing coupling between sets of 1H and 15N nuclei. Each peak represents the 15N—1H of a unique amino acid along the backbone of the amino acid.
The 2D HSQC experiment which relates 1H and 15N is just the start on the long, complicated road of protein structural characterization using 2D and multidimensional NMR techniques. The next experiment is typically a 3D NMR technique in which coupling with 13C is also including in the spectra—which will give information on which NH peaks are associated with which type of amino acid residue—followed by a NOE experiment—which gives spatial distances between nuclei. These multidimensional techniques are outside the scope of this module; however, the principles which were applied in the construction and analysis of the 2D spectra can be carried over to 3D and 4D NMR. The only difference, naturally, is the additional dimensions. By the time multiple experiments have been carried out, information regarding proximity of nuclei, the types of amino acid associated with each nuclei, and secondary structure has been amassed, and by carefully piecing all the data together, a 3D structure of a given protein can be constructed.
Other Applications of 2D NMR
2D NMR has many more applications beyond protein NMR, including characterization of pharmaceuticals, temperature dependence of carbohydrate conformations, and metabolomics, to just name a few. For more information on these applications and the 2D NMR techniques that are used in them, please see the “Further Reading” section.
Problems
1. What is the effect of the 90° pulse on the bulk magnetization of a sample?
2. If a 90° pulse for a give sample is 4 fs long, how long is a 180° pulse on the same sample?
3. The magnetic effect of which type of particle must be removed from an NMR experiment in order to observe an NOE?
Solutions: 1) a 90o pulse moves the bulk magnetization into the transverse plane 2) 8 fs 3) electrons
Further Reading
1. Ludwig, C., Viant, Mark R., Two-dimensional J-resolved NMR spectroscopy: review of a key methodology in the metabolomics toobox. Phytochemical Analysis, 2010. 21(1): p. 22-32. DOI: 10.1002/pca.1186
2. Brown, S.P., Applications of high-resolution 1H solid-state NMR. Solida State Nuclear Magnetic Resonance, 2012. 41: p. 1-27. DOI: 10.1016/j.ssnmr.2011.11.006
3. Shrot, Y.F., Lucio, Ghost-peak suppression in ultrafast two-dimensional NMR. Journal of Magnetic Resonance, 2003. 164(2): p. 351-357. DOI: 10.1016/S1090-7807(03)00177-0
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The heteronuclear correlation of chemical shifts by scalar coupling derives from the existence of heteronuclear scalar coupling all owing a magnetization transfer from the more sensitive nucleus (1H) toward the less sensitive nucleus (13C). The model experiment is the correlation δ(13C)/ δ(1H) which uses the large direct coupling 1J(13C-1H) comprised between more 100 Hz and 200Hz and which allows to correlate the signals from a proton to that of a 13C to which it is bound (5).
The XHCORR sequence
This sequence allows to correlate the signals of the 1H and of the 13C bound to each other (13). Spectrum 11 (XHCORR), shows the correlations for the dihydrofuran (Fig. 31) between bound carbons and protons. The chemical shifts of the protons of these molecules are in the table 2, those of the carbon are in the table 3.
Table 2: DBF protons chemical shift.
Table 3: DBF carbons chemical shift
Spectrum: XHCORR of the DBF
Within the XHCORR pulse sequence (Fig. 25), transverse magnetization is caused by a impulsion which is evolving during the t1 period. The impulsion (13C), located in the middle of this period refocuses the heteronuclear couplings.
The optimization of the and delays allows the selection of the long range heteronuclear couplings, this means that instead of seeing the correlation between 13C and protons directly bound , we favor the appearance of the correlations spots between 13C and non bound protons (HNB).
For example, for a coupling constant J=10 Hz then =50 ms and =33 ms.
The COLOC sequence (COrrelation via LOng range Coupling)
The sequence which is used (Fig. 26) allows generally a good correlation between the quaternary carbons and the neighbouring protons. On the spectrum 12 (COLOC), the correlation between the quaternary carbon (13C = δ = 156. 2 ppm) and its vicinal proton (δ = 7.7 ppm) is named Cq. However, it does not give good results for long range correlations involving protonated carbons (14). This observation led to write a pulse sequence named XC ORFE (X nucleus proton CORrelation with Fixed Evolution time). This one is very closed to the COLOC sequence.
Spectrum 12: COLOC of DBF
Heteronuclear 2D J-resolved Pulse Sequence
In this experiment we observe the coupling J between the X nuclei (13C, 15N, 31P ...etc ) and the protons which are bound to them (Fig. 27). A coupled normal 1D spectrum XH may be very difficult to explain. This is due to the intercrossing of the multiplets (12). For simplifying the spectra, we will spread the coupling data in the Fl dimension and the informations arising from the chemical shifts in the F2 dimension.
The results are given for the saccharine (Fig. 29). The F2 carbon (108 ppm) is a quaternary one because we may observe only one spot devoid of coupling. The G1 carbon (95 ppm) is a CH because there are two correlation spots. Moreover, the coupling constant between this 13C and this 1H is of 80 Hz approximately (Spectrum 13 and 14).
Spectrum 13: Heteronuclear 2D J-resolved
The Heteronuclear correlation of chemical shifts in inverse detection. What is the reverse detection? The reverse detection or indirect detection affords the NMR parameters (chemical shifts, t1, ...) of a weakly sensitive X nucleus by observing it through a much more sensitive nucleus (generally the proton) (1.5).
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Homonuclear 2D J-resolved
This sequence allows to get the coupling constant along the Fl axis and the chemical shift of each uncoupled proton along the F2 axis (Spectrum 1 and 2). Of course, the couplings involving two different nuclei (3IP, 19P coupled to 1H) are always found in the F2 dimension. One can thus determine the coupling between a proton within a multiplet of a coupled spectrum and phosphorus (8). For the 2D J-resolved sequence (Fig. 12), the pulse sequence used is a spin echo sequence. In our example we studied the saccharose molecule (Fig. 29). As in any "echo" experiment, the evolution period tl is divided into two equivalent parts. After the first impulsion of (90°)x, all the magnetization vectors begin to precess as a function of their chemical shifts and the coupling constants. The impulsion of (180°) on the y axis allows for each refocused spin to reassociate the vectors with their chemical shifts.
Spectrum 1: 1D proton saccharin.
Spectrum 2: J-resolved 2D.
The doublet which is located at 4.22 ppm, shows that the f3 proton has a neighboring proton. The coupling constant is directly seen on the F2 axis. The triplet f4 at 4.05 ppm shows that this one has two protons which are closed to him ... (and so on).
The COSY 90 pulse sequence (Correlated SpectroscopY)
The example under study is the saccharine (Fig. 29). The aim of this sequence is to correlate signals of identical nuclei with a scalar coupling. For an isolated nucleus, it is made of two impulsions on both side of the evolution period t1 (Fig. 13).
The preparation begins with a relaxation delay allowing the magnetization to come back to its equilibrium, then the period t1 occurs with a stepwise increase and finally the t2 period is the acquisition time (8).
Following the double Fourier transform, we get a spectrum in chemical shift on the two axis. Moreover, the cross peaks indicate the nuclei which are coupled within each other and allow then a direct attribution of the neighboring protons.
We can understand the obtention of the diagonal and crossing signals of the COSY spectrum by the merely means of quantitative consideration. (8)
In fact, if we consider a singlet signal without scalar spin-spin coupling, the preparation allows to set the maximal amount of nuclei at their right position and the first impulsion generates a transverse magnetization M(A) which rotates around the Y axis (Fig. 14).
Fig. 14: After the first impulsion
During the evolution period, the aimantation M(A) precesses with the frequency of the corresponding nucleus resonance in the rotating frame; there is thus a sharing of the spins between the x axis and the y axis (Fig. 15).
The components Mx and My of the transverse magnetization at the end of the evolution period in the Ox and Oy axis of the rotating frame are then:
Where M0 is the aimantation contribution before the execution of the first impulsion and T2* the apparent transverse relaxation time of the corresponding nucleus.
The second impulsion allows to rotate the y component which stays in the xy plane, in the negative z part direction (-z). These last one will not be detected (Fig. 16). However, the x part, still in the xy plane will be taken into account.
The magnetization which is going to be detected is of the form: , where as the magnetization along the –z axis followed the formula: .
The intensity of the signal to be recorded depends thus of the time dependent position of the M(A) vector at the very time of the end of the evolution period. This intensity is determined by the Larmor frequency of the A nucleus. The amplitude is modulated in t1 versus this frequency and the Fourier transform yields in both dimensions the frequency and then a diagonal signal. In the case of a spin-spin coupling, the second impulsion does not only induce a change in the transverse A magnetization, but also changes of populations for other transitions in the spin system under study. Therefore, an exchange of magnetization occurs between all the nuclei which are coupled between each other. Their signals, in a set of experiments t1, are in the same way modulated versus the neighboring nuclei.
This ends up with cross peaks of the 2D spectrum in and and allows us to get a spectrum in which the homonuclear scalar coupling constants are figure out by spots outside of the diagonal.
For example, we might see the coupling between the f3 and f4 protons at the crossing of 4.05 and 4.18 ppm (Spectrum 3).
Spectrum 3: COSY 90° Both impulsions are at 90°.
The COSY 45 pulse sequence
The example is given for the saccharose (Fig. 29). The second impulsion is of 45°. The drawback of this second pulse is to diminish the signal to noise ratio (Fig 17). However it allows a more easy attribution of the signals of the spectrum in the cases of strong AB type coupling. In fact, the correlations spots near the diagonal are no more susceptible to be hidden by the auto correlation peaks (Spectrum 4).
Spectrum 4: COSY 45°. We observe a strong diminution of the autocorrelation peaks.
The long range coupling: COSY
This experiment allows the detection of the correlation peaks caused by weak coupling constant (less than one Hz) and which are not detectable with the COSY 9O° pulse sequence (8).
For the "long range" COSY sequence, we artificially increase the evolution and detection period by introducing a fixed delay on both side of the mixing time (Fig. 18).
The chosen example is the saccharine molecule (see also Fig 29).
The spectrum 5 is a 90° COSY pulse sequence: The scalar coupling between neighboring protons G1 and G2 are clearly seen. The spectrum 6 is a L.R. COSY pulse sequence.:The long range couplings appear between the G1 proton and the G3 and G4 protons.
Spectrum 5: 90° COSY
Spectrum 6: LR COSY
The relayed COSY or one homonuclear step relayed COSY
A COSY spectrum may sometimes be inefficient to clearly identify all the signals, particularly if per chance two nuclei are equivalent. A typical example is that of peptides (Fig 19) where protons born by carbon of different amino acids have closed chemical shifts; the COSY spectrum gives informations of the correlations between proton and protons in one hand and between protons and NH protons on the other hand. Unfortunately, due to the overlapping of the protons signals, the attribution for the protons signals can not be deduced from the NH protons (or the other way around). To solve this problem, we should be able to correlate the NH proton signals to those of the protons; such a correlation may be obtained either by a long range COSY if the 4J coupling is not too weak or, better, by a relayed COSY experiment which allows the correlation of the signal from two nuclei at the only requirement that both of them are coupled to a third one (8).
This experiment allows the differentiation of the belonging of the NH versus the or .
It is widely used for the sugar structure determination. The pulse sequence of the experiment is derived from the COSY sequence in which the mixing pulse is followed by a fixed time and by a 90° pulse (Fig. 20).
A 180° pulse, achieved in the middle of the new mixing period is also needed for the elimination of the evolution of the magnetization (refocalization) due to the chemical shifts during this additional delay.
Applied to the AMX type system, this pulse sequence achieves a first coherence transfer from A to M as that one observed in the COSY experiment. During the mixing time, the magnetization evolve in such a wav that ideally the different components of the magnetization of the M nucleus are in anti phase versus the JMX coupling. The third impulsion at 90° achieves then the transfer from the M nucleus toward the X nucleus. The phase cycles of the impulsions and of the receiver is derived from that of the COSY experiment. However, it needs an additional base cycle for the elimination of the artifacts due to transfers of the NOESY type. This consists to inverse the phase of the second impulsion without modifying the reception phase. The result is a global cycle with 8 components:
= 0, = 0 0 1 1 2 2 3 3 , = 0 2 1 3, = 0 0 2 2
For optimizing the mixing time, we have to define the delay between each impulsion. This one is deduced from the following formula where Jmax stands for the value of the biggest coupling constant:
COSY experiment with a double quantum filter (DQF COSY)
The COSY sequence with double quantum filter affords some advantages compared to the conventional cosy (2). The diagonal spots and the correlation spots are not phase shifted to 90° and they have the same anti phase structure (Fig. 21). It is thus possible to phase as a whole all the peaks of the spectrum in positive or in negative absorption allowing thus an easier identification of the peaks whose correlations are the closest to the diagonal. These diagonal peaks correspond to singlet signals and are thus eliminated. Unfortunately the sensitivity is divided by 2.
The molecule which is used as an example is the saccharine (Fig. 29). We may notice on the spectrum 7 (DQF COSY), a strong decrease of the singlet of the proton F1 which is located at 3.68 ppm compared to the spectrum 3.
Spectrum 7: DQF COSY.
The NOESY sequence (Nuclear Over Hauser Enhancement Spectroscopy)
It introduces the crossed relaxation swhich governs the Over Hauser effect (2). The NOESY spectroscopy is rather limited to the determination of the notion of crossed homonuclear relaxation. The pulse sequence scheme is given Figure 22.
The correlation spots (peaks outside of the diagonal) indicate either a dipolar interaction, thus a given spatial vicinity (2 to 5 A°) between the spins under study, or a chemical shift. This experiment allows the determination of large molecules such as the proteins.
The spectrum 8 is a COSY 45° of the phenanthro-benzofurane (PBF) (Fig.30). The spectrum 9 is the corresponding NOESY spectrum We observe an additional correlation spot between the proton which is at 8.8 ppm and the proton which has a chemical shift of 8.6 ppm. This spot is termed A, it means a dipolar interaction between two protons H1 and H2.
Spectrum 8: COSY spectrum of PBF
Spectrum 9: NOESY spectrum of PBF
Phase Sensitive COSY and NOESY
The TPPI technique (Time Proportional Phase incremantation) allows detection in a quadratic sequence. We may therefore solve the problem of the discrimination of the sign of the frequencies upon the detection of a free precession signal in an experiment of impulsional NMR There are other methods in the phase mode like the STATES technique where the detection is in simultaneous quadratic. It is easier to achieve a phased NOESY rather than a classical NOESY, because this technique is much more sensitive. Moreover, the correlation spots meaning a spatial vicinity are in the negative portion of the spectrum versus the correlation spots due to the spin-spin coupling. (5)
TOCSY and ROESY
These experiments are characterized by a mixing time during which the magnetization of a spin A, submitted to a radio wave frequency field B1 is transferred to a spin linked to A either by a dipolar coupling or by a scalar coupling. During the mixing time the magnetization is locked according to one direction of the space defined by the direction of the field B1 and the spin frequency shifting versus the main component. The block in the spin system is commonly named spin-lock (2).
TOCSY or HOHAHA
This method allows to demonstrate scalar couplings. The pulse sequence scheme is given Fig. 23. The first 90°x impulsion rotates the magnetization vectors in the xOy plane, in which they freely evolve during the time t1, under the influence of their chemical shift and the scalar coupling constants. We apply then a spin lock on the x axis, that means a low power impulsion or set of impulsions whose duration represents the mixing time tm. During this time, the spin coherence of the system is exchanged. By increasing the mixing time we increase the number of the visible transitions. That is to say that within the same experiment, we get the informations of a COSY, of a LR COSY and of a relayed COSY. The pulse sequence widely used is base upon the MLEV-17 sequence which contains a set of 17 pulses. (5).
The spectrum 10 shows the result of three different mixing time for the saccharose (Fig. 29). These are given in milliseconds (20, 60, 100 ms). For each mixing time, we see the appearance of new correlation spots. There is thus an increase of the number of relays with the spin lock time. These ones correspond then to long range correlations.
Spectrum 10: 20 ms, 60 ms, 100 ms.
ROESY and CAMELSPIN
The aim of the dipolar correlation experiments in two dimensions is to take advantage of the vicinity in space of some nuclei. The result of this kind of experiment is a 2D map in which the signals outside of the diagonal arise from the Over Hauser enhancement effect between two space coupled nuclei (2) (Fig. 24).
In this case, we apply a spin-lock onto the y axis. The correlation peaks arising from the ROE effect are on the opposite sign compared to the diagonal one their intensity is different from 0. However, the correlation peaks coming from a chemical exchange are of the same sign than the crossing one. The ROESY experiment allows by this way the separation of the contributions coming from the exchange and those coming from the dipolar interactions. The ROESY sequence is thus complementary of the NOESY one and it is more often used for the structural determination of the small molecules.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/2D_NMR/Homonuclear_Correlations.txt
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• Carbon-13 NMR
This page describes what a C-13 NMR spectrum is and how it tells you useful things about the carbon atoms in organic molecules.
• Diffusion Ordered Spectroscopy (DOSY)
Diffusion Ordered SpectroscopY (DOSY) utilizes magnetic field gradients to investigate diffusion processes occurring in solid and liquid samples.
• Magnetic Resonance Imaging
Magnetic resonance imaging is a widely used noninvasive medical imaging technique to visualize the inner part of human body. It applied the basic principles of nuclear magnetic resonance (NMR) spectroscopy, which provides both chemical and physical information of molecules.
• NMR - Interpretation
NMR interpretation plays a pivotal role in molecular identifications. As interpreting NMR spectra, the structure of an unknown compound, as well as known structures, can be assigned by several factors such as chemical shift, spin multiplicity, coupling constants, and integration.
• NMR Hardware
• NMR Spectroscopy in Lab: Complications
• Solid State Experiments
The solid state NMR community has developed hundreds of techniques to investigate different parts of the NMR Hamiltonian. This section gives a brief, by no means comprehensive of a few techniques that are helpful in probing select interactions.
• Solid State NMR Experimental Setup
This section is devoted to giving a practical guide for solid state NMR. Please follow all safety guidelines at your University. All experiments are performed on a 500MHz magnet using a Bruker AQS spectrometer running TopSpin 2.1. Please note that your spectrometer may be different and/or run different software in which some/all of the commands listed below will be different.
NMR: Experimental
This page describes what a C-13 NMR spectrum is and how it tells you useful things about the carbon atoms in organic molecules.
Carbon-13 Nuclei as Little Magnets
About 1% of all carbon atoms are the C-13 isotope; the rest (apart from tiny amounts of the radioactive C-14) is C-12. C-13 NMR relies on the magnetic properties of the C-13 nuclei. Carbon-13 nuclei fall into a class known as "spin ½" nuclei for reasons which do not really need to concern us at the introductory level this page is aimed at (UK A level and its equivalents). The effect of this is that a C-13 nucleus can behave as a little magnet. C-12 nuclei do not have this property.
If you have a compass needle, it normally lines up with the Earth's magnetic field with the north-seeking end pointing north. Provided it is not sealed in some sort of container, you could twist the needle around with your fingers so that it pointed south - lining it up opposed to the Earth's magnetic field. It is very unstable opposed to the Earth's field, and as soon as you let it go again, it will flip back to its more stable state.
Because a C-13 nucleus behaves like a little magnet, it means that it can also be aligned with an external magnetic field or opposed to it. Again, the alignment where it is opposed to the field is less stable (at a higher energy). It is possible to make it flip from the more stable alignment to the less stable one by supplying exactly the right amount of energy.
The energy needed to make this flip depends on the strength of the external magnetic field used, but is usually in the range of energies found in radio waves - at frequencies of about 25 - 100 MHz. If you have also looked at proton-NMR, the frequency is about a quarter of that used to flip a hydrogen nucleus for a given magnetic field strength.
It's possible to detect this interaction between the radio waves of just the right frequency and the carbon-13 nucleus as it flips from one orientation to the other as a peak on a graph. This flipping of the carbon-13 nucleus from one magnetic alignment to the other by the radio waves is known as the resonance condition.
The Importance of the Carbon's Environment
What we've said so far would apply to an isolated carbon-13 nucleus, but real carbon atoms in real bonds have other things around them - especially electrons. The effect of the electrons is to cut down the size of the external magnetic field felt by the carbon-13 nucleus.
Suppose you were using a radio frequency of 25 MHz, and you adjusted the size of the magnetic field so that an isolated carbon-13 atom was in the resonance condition. If you replaced the isolated carbon with the more realistic case of it being surrounded by bonding electrons, it wouldn't be feeling the full effect of the external field any more and so would stop resonating (flipping from one magnetic alignment to the other). The resonance condition depends on having exactly the right combination of external magnetic field and radio frequency.
How would you bring it back into the resonance condition again? You would have to increase the external magnetic field slightly to compensate for the shielding effect of the electrons. Now suppose that you attached the carbon to something more electronegative. The electrons in the bond would be further away from the carbon nucleus, and so would have less of a lowering effect on the magnetic field around the carbon nucleus.
Electronegativity is a measure of the ability of an atom to attract a bonding pair of electrons. If you are not happy about electronegativity, you could follow this link at some point in the future, but it probably is not worth doing it now!
The external magnetic field needed to bring the carbon into resonance will be smaller if it is attached to a more electronegative element, because the C-13 nucleus feels more of the field. Even small differences in the electronegativities of the attached atoms will make a difference to the magnetic field needed to achieve resonance.
Summary
For a given radio frequency (say, 25 MHz) each carbon-13 atom will need a slightly different magnetic field applied to it to bring it into the resonance condition depending on what exactly it is attached to - in other words the magnetic field needed is a useful guide to the carbon atom's environment in the molecule.
The C-13 NMR Spectrum for Ethanol
This is a simple example of a C-13 NMR spectrum. do not worry about the scale for now - we'll look at that in a minute.
The NMR spectra on this page have been produced from graphs taken from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.
It is possible that small errors may have been introduced during the process of converting them for use on this site, but these won't affect the argument in any way.
There are two peaks because there are two different environments for the carbons. The carbon in the CH3 group is attached to 3 hydrogens and a carbon. The carbon in the CH2 group is attached to 2 hydrogens, a carbon and an oxygen. The two lines are in different places in the NMR spectrum because they need different external magnetic fields to bring them in to resonance at a particular radio frequency.
The C-13 NMR Spectrum for a more complicated compound
This is the C-13 NMR spectrum for 1-methylethyl propanoate (also known as isopropyl propanoate or isopropyl propionate).
This time there are 5 lines in the spectrum. That means that there must be 5 different environments for the carbon atoms in the compound. Is that reasonable from the structure?
Well - if you count the carbon atoms, there are 6 of them. So why only 5 lines? In this case, two of the carbons are in exactly the same environment. They are attached to exactly the same things. Look at the two CH3 groups on the right-hand side of the molecule.
You might reasonably ask why the carbon in the CH3 on the left is not also in the same environment. Just like the ones on the right, the carbon is attached to 3 hydrogens and another carbon. But the similarity is not exact - you have to chase the similarity along the rest of the molecule as well to be sure.
The carbon in the left-hand CH3 group is attached to a carbon atom which in turn is attached to a carbon with two oxygens on it - and so on down the molecule. That's not exactly the same environment as the carbons in the right-hand CH3 groups. They are attached to a carbon which is attached to a single oxygen - and so on down the molecule. We'll look at this spectrum again in detail on the next page - and look at some more similar examples as well. This all gets easier the more examples you look at.
For now, all you need to realize is that each line in a C-13 NMR spectrum recognizes a carbon atom in one particular environment in the compound. If two (or more) carbon atoms in a compound have exactly the same environment, they will be represented by a single line.
If you are fairly wide-awake, you might wonder why all this works, since only about 1% of carbon atoms are C-13. These are the only ones picked up by this form of NMR. If you had a single molecule of ethanol, then the chances are only about 1 in 50 of there being one C-13 atom in it, and only about 1 in 10,000 of both being C-13.
But you have got to remember that you will be working with a sample containing huge numbers of molecules. The instrument can pick up the magnetic effect of the C-13 nuclei in the carbon of the CH3 group and the carbon of the CH2 group even if they are in separate molecules. There's no need for them to be in the same one.
The Need for a Standard: TMS
Before we can explain what the horizontal scale means, we need to explain the fact that it has a zero point - at the right-hand end of the scale. The zero is where you would find a peak due to the carbon-13 atoms in tetramethylsilane (TMS). Everything else is compared with this.
You will find that some NMR spectra show the peak due to TMS (at zero), and others leave it out. Essentially, if you have to analyse a spectrum which has a peak at zero, you can ignore it because that's the TMS peak. TMS is chosen as the standard for several reasons. The most important are:
• It has 4 carbon atoms all of which are in exactly the same environment. They are joined to exactly the same things in exactly the same way. That produces a single peak, but it's also a strong peak (because there are lots of carbon atoms all doing the same thing).
• The electrons in the C-Si bonds are closer to the carbons in this compound than in almost any other one. That means that these carbon nuclei are the most shielded from the external magnetic field, and so you would have to increase the magnetic field by the greatest amount to bring the carbons back into resonance.
• The net effect of this is that TMS produces a peak on the spectrum at the extreme right-hand side. Almost everything else produces peaks to the left of it.
The Chemical Shift
The horizontal scale is shown as $\delta$ (ppm). $\delta$ is called the chemical shift and is measured in parts per million - ppm. A peak at a chemical shift of, say, 60 means that the carbon atoms which caused that peak need a magnetic field 60 millionths less than the field needed by TMS to produce resonance. A peak at a chemical shift of 60 is said to be downfield of TMS. The further to the left a peak is, the more downfield it is.
If you are familiar with proton-NMR, you will notice that the chemical shifts for C-13 NMR are much bigger than for proton-NMR. In C-13 NMR, they range up to about 200 ppm. In proton-NMR they only go up to about 12 ppm. You do not need to worry about the reasons for this at this level.
Solvents for NMR Spectroscopy
NMR spectra are usually measured using solutions of the substance being investigated. A commonly used solvent is CDCl3. This is a trichloromethane (chloroform) molecule in which the hydrogen has been replaced by its isotope, deuterium. CDCl3 is also commonly used as the solvent in proton-NMR because it does not have any ordinary hydrogen nuclei (protons) which would give a line in a proton-NMR spectrum. It does, of course, have a carbon atom - so why doesn't it give a potentially confusing line in a C-13 NMR spectrum? In fact it does give a line, but the line has an easily recognisable chemical shift and so can be removed from the final spectrum. All of the spectra from the SDBS have this line removed to avoid any confusion.
Carbon-13 NMR
This page takes an introductory look at how you can get useful information from a C-13 NMR spectrum.
Introduction
Taking a close look at three 13C NMR spectra below. The 13C NMR spectrum for ethanol
The NMR spectra on this page have been produced from graphs taken from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.
Remember that each peak identifies a carbon atom in a different environment within the molecule. In this case there are two peaks because there are two different environments for the carbons. The carbon in the CH3 group is attached to 3 hydrogens and a carbon. The carbon in the CH2group is attached to 2 hydrogens, a carbon and an oxygen. So which peak is which?
You might remember from the introductory page that the external magnetic field experienced by the carbon nuclei is affected by the electronegativity of the atoms attached to them. The effect of this is that the chemical shift of the carbon increases if you attach an atom like oxygen to it. That means that the peak at about 60 (the larger chemical shift) is due to the CH2 group because it has a more electronegative atom attached.
In principle, you should be able to work out the fact that the carbon attached to the oxygen will have the larger chemical shift. In practice, you always work from tables of chemical shift values for different groups (see below).
What if you needed to work it out? The electronegative oxygen pulls electrons away from the carbon nucleus leaving it more exposed to any external magnetic field. That means that you will need a smaller external magnetic field to bring the nucleus into the resonance condition than if it was attached to less electronegative things. The smaller the magnetic field needed, the higher the chemical shift.
A table of typical chemical shifts in C-13 NMR spectra
carbon environment chemical shift (ppm)
C=O (in ketones) 205 - 220
C=O (in aldehydes) 190 - 200
C=O (in acids and esters) 170 - 185
C in aromatic rings 125 - 150
C=C (in alkenes) 115 - 140
RCH2OH 50 - 65
RCH2Cl 40 - 45
RCH2NH2 37 - 45
R3CH 25 - 35
CH3CO- 20 - 30
R2CH2 16 - 25
RCH3 10 - 15
In the table, the "R" groups will not necessarily be simple alkyl groups. In each case there will be a carbon atom attached to the one shown in red, but there may well be other things substituted into the "R" group.
If a substituent is very close to the carbon in question, and very electronegative, that might affect the values given in the table slightly. For example, ethanol has a peak at about 60 because of the CH2OH group. No problem! It also has a peak due to the RCH3 group. The "R" group this time is CH2OH. The electron pulling effect of the oxygen atom increases the chemical shift slightly from the one shown in the table to a value of about 18. A simplification of the table:
carbon environment chemical shift (ppm)
C-C 0 - 50
C-O 50 - 100
C=C 100 - 150
C=O 150 - 200
This may, of course, change and other syllabuses might want something similar. The only way to find out is to check your syllabus, and recent question papers to see whether you are given tables of chemical shifts or not.
The 13C NMR spectrum for but-3-en-2-one. This is also known as 3-buten-2-one (among many other things!)
Here is the structure for the compound:
You can pick out all the peaks in this compound using the simplified table above.
• The peak at just under 200 ppm is due to a carbon-oxygen double bond. The two peaks at 137 ppm and 129 ppm are due to the carbons at either end of the carbon-carbon double bond. And the peak at 26 is the methyl group which, of course, is joined to the rest of the molecule by a carbon-carbon single bond. If you want to use the more accurate table, you have to put a bit more thought into it - and, in particular, worry about the values which do not always exactly match those in the table!
• The carbon-oxygen double bond in the peak for the ketone group has a slightly lower value than the table suggests for a ketone. There is an interaction between the carbon-oxygen and carbon-carbon double bonds in the molecule which affects the value slightly. This isn't something which we need to look at in detail for the purposes of this topic.
• You must be prepared to find small discrepancies of this sort in more complicated molecules - but do not worry about this for exam purposes at this level. Your examiners should give you shift values which exactly match the compound you are given.
• The two peaks for the carbons in the carbon-carbon double bond are exactly where they would be expected to be. Notice that they aren't in exactly the same environment, and so do not have the same shift values. The one closer to the carbon-oxygen double bond has the larger value.
• And the methyl group on the end has exactly the sort of value you would expect for one attached to C=O. The table gives a range of 20 - 30, and that's where it is.
One final important thing to notice. There are four carbons in the molecule and four peaks because they are all in different environments. But they aren't all the same height. In C-13 NMR, you cannot draw any simple conclusions from the heights of the various peaks.
1-methylethyl propanoate is also known as isopropyl propanoate or isopropyl propionate.
Here is the structure for 1-methylethyl propanoate:
Two simple peaks
There are two very simple peaks in the spectrum which could be identified easily from the second table above.
• The peak at 174 is due to a carbon in a carbon-oxygen double bond. (Looking at the more detailed table, this peak is due to the carbon in a carbon-oxygen double bond in an acid or ester.)
• The peak at 67 is due to a different carbon singly bonded to an oxygen. Those two peaks are therefore due to:
If you look back at the more detailed table of chemical shifts, you will find that a carbon singly bonded to an oxygen has a range of 50 - 65. 67 is, of course, a little bit higher than that.
As before, you must expect these small differences. No table can account for all the fine differences in environment of a carbon in a molecule. Different tables will quote slightly different ranges. At this level, you can just ignore that problem!
Before we go on to look at the other peaks, notice the heights of these two peaks we've been talking about. They are both due to a single carbon atom in the molecule, and yet they have different heights. Again, you can't read any reliable information directly from peak heights in these spectra.
The three right-hand peaks
From the simplified table, all you can say is that these are due to carbons attached to other carbon atoms by single bonds. But because there are three peaks, the carbons must be in three different environments.
The easiest peak to sort out is the one at 28. If you look back at the table, that could well be a carbon attached to a carbon-oxygen double bond. The table quotes the group as \(\ce{CH_3CO-}\), but replacing one of the hydrogens by a simple CH3 group will not make much difference to the shift value.
The right-hand peak is also fairly easy. This is the left-hand methyl group in the molecule. It is attached to an admittedly complicated R group (the rest of the molecule). It is the bottom value given in the detailed table.
The tall peak at 22 must be due to the two methyl groups at the right-hand end of the molecule - because that's all that's left. These combine to give a single peak because they are both in exactly the same environment.
If you are looking at the detailed table, you need to think very carefully which of the environments you should be looking at. Without thinking, it is tempting to go for the R2CH2 with peaks in the 16 - 25 region. But you would be wrong! The carbons we are interested in are the ones in the methyl group, not in the R groups. These carbons are again in the environment: RCH3. The R is the rest of the molecule. The table says that these should have peaks in the range 10 - 15, but our peak is a bit higher. This is because of the presence of the nearby oxygen atom. Its electronegativity is pulling electrons away from the methyl groups - and, as we've seen above, this tends to increase the chemical shift slightly.
Working out Structures from C-13 NMR Spectra
So far, we have just been trying to see the relationship between carbons in particular environments in a molecule and the spectrum produced. We've had all the information necessary. Now let's make it a little more difficult - but we'll work from much easier examples! In each example, try to work it out for yourself before you read the explanation. How could you tell from just a quick look at a C-13 NMR spectrum (and without worrying about chemical shifts) whether you had propanone or propanal (assuming those were the only options)?
Because these are isomers, each has the same number of carbon atoms, but there is a difference between the environments of the carbons which will make a big impact on the spectra.
In propanone, the two carbons in the methyl groups are in exactly the same environment, and so will produce only a single peak. That means that the propanone spectrum will have only 2 peaks - one for the methyl groups and one for the carbon in the C=O group. However, in propanal, all the carbons are in completely different environments, and the spectrum will have three peaks.
Example \(3\): \(C_4H_{10}O\)
There are four alcohols with the molecular formula \(C_4H_{10}O\).
Which one produced the C-13 NMR spectrum below?
You can do this perfectly well without referring to chemical shift tables at all.
In the spectrum there are a total of three peaks - that means that there are only three different environments for the carbons, despite there being four carbon atoms.
In A and B, there are four totally different environments. Both of these would produce four peaks.
In D, there are only two different environments - all the methyl groups are exactly equivalent. D would only produce two peaks.
That leaves C. Two of the methyl groups are in exactly the same environment - attached to the rest of the molecule in exactly the same way. They would only produce one peak. With the other two carbon atoms, that would make a total of three. The alcohol is C.
Example \(4\):
This follows on from Example \(3\), and also involves an isomer of \(C_4H_{10}O\) but which isn't an alcohol. Its C-13 NMR spectrum is below. Work out what its structure is.
Because we do not know what sort of structure we are looking at, this time it would be a good idea to look at the shift values. The approximations are perfectly good, and we will work from this table:
carbon environment chemical shift (ppm)
C-C 0 - 50
C-O 50 - 100
C=C 100 - 150
C=O 150 - 200
There is a peak for carbon(s) in a carbon-oxygen single bond and one for carbon(s) in a carbon-carbon single bond. That would be consistent with C-C-O in the structure.
It is not an alcohol (you are told that in the question), and so there must be another carbon on the right-hand side of the oxygen in the structure in the last paragraph. The molecular formula is C4H10O, and there are only two peaks. The only solution to that is to have two identical ethyl groups either side of the oxygen. The compound is ethoxyethane (diethyl ether), CH3CH2OCH2CH3.
Example \(5\)
Using the simplified table of chemical shifts above, work out the structure of the compound with the following C-13 NMR spectrum. Its molecular formula is \(C_4H_6O_2\).
Let's sort out what we've got.
• There are four peaks and four carbons. No two carbons are in exactly the same environment.
• The peak at just over 50 must be a carbon attached to an oxygen by a single bond.
• The two peaks around 130 must be the two carbons at either end of a carbon-carbon double bond.
• The peak at just less than 170 is the carbon in a carbon-oxygen double bond.
Putting this together is a matter of playing around with the structures until you have come up with something reasonable. But you can't be sure that you have got the right structure using this simplified table. In this particular case, the spectrum was for the compound:
If you refer back to the more accurate table of chemical shifts towards the top of the page, you will get some better confirmation of this. The relatively low value of the carbon-oxygen double bond peak suggests an ester or acid rather than an aldehyde or ketone.
It can't be an acid because there has to be a carbon attached to an oxygen by a single bond somewhere - apart from the one in the -COOH group. We've already accounted for that carbon atom from the peak at about 170. If it was an acid, you would already have used up both oxygen atoms in the structure in the -COOH group. Without this information, though, you could probably come up with reasonable alternative structures. If you were working from the simplified table in an exam, your examiners would have to allow any valid alternatives.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Experimental/Carbon-13_NMR/Interpr.txt
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Diffusion Ordered SpectroscopY (DOSY) utilizes magnetic field gradients to investigate diffusion processes occurring in solid and liquid samples.
Theory
Spin Diffusion
In the classic formation of a spin-echo (i.e. 90o-$\tau$-180o-2$\tau$) the intensity of the echo will damp according to T2 relaxation processes. However, in the presence of inhomogeneities in the magnetic field, if a spin drifts to a location in which the magnetic field is different, the Larmor frequency changes and the spin will not refocus with the remaining spins. We can then re-write the bloch equation to account for diffusion processes
$\frac{dM}{dt}=\gamma MB-\frac{M_x+M_y}{T_2}-\frac{M_z-M_0}{T_1}+D\nabla ^2M$
Typically, the changes in the Larmor frequency have a negligible effect. However, application of a Gradient, G, will induce a large change in larmor frequencies experienced at places in the sample.
Magnetic Field Gradient Echo
The principle behind the DOSY experiment is the formation of a magnetic field gradient echo. This is a spin-echo in which a magnetic field gradient is applied during the spin evolution. Thus spins that diffuse during this time will not refocus in the spin echo, as the application of the gradient will impart their own magnetic field.
Pulse Program
The sequence consists of a conventional Hahn echo (spin echo) sequence in which two gradient pulses are applied at equal timings after the 90 and 180 degree pulses. The gradients are opposite in magnitude. The basic principle relies on the diffusion of spins in the sample. Initially the 90 degree pulse tips the magnetization into the x-y plane, where spins begin to precess with their characteristic Larmor frequencies. The application of a gradient sometime after the 90 degree pulse, encodes a spatial component to the spin. That is the gradient is not uniform over the sample and therefore, the processional frequencies will change. Next the application of the 180 flips the magnetization to refocus the spins. However, the spins, due to the application of the gradient will not refocus. The application of a gradient at the same time (but with opposite direction) will refocus the spins at a total time of 2 $\tau$. If a spin diffuses to a different place in the sample, the refocusing will not occur, leading the dampening of the echo intensity.
Below is the basic pulse sequence for a pulsed field gradient experiment.
Processing
Processing the data is fairly straight forward. Apply a Fourier Transformation along the F2 dimension. The F1 dimension, in which either the time or the gradient was incremented remains in the time domain. The echo intensity of a given peak is then described by:
$\ln \mu (g_a, t_c)-\ln \mu(0, t_c)=-C_n \gamma ^2 D \delta ^2 g_a ^2 t_D$
where $\mu((g_a, t_c)$ is the echo amplitude after the gradient application, $\ln m(0, t_c)$ is the echo amplitude with no gradient applied, Cn is a constant that depends on the particular pulse sequence used, $\gamma$ is the gyromagnetic ratio, D is the diffusion coefficient, $\delta$ is the width of the applied gradient, and tD is the diffusion time.
Magnetic Resonance Im
Magnetic resonance imaging is a widely used noninvasive medical imaging technique to visualize the inner part of human body. It applied the basic principles of nuclear magnetic resonance (NMR) spectroscopy, which provides both chemical and physical information of molecules.
Introduction
Since a large fraction of composition of the human body is water and fat, the abundance of hydrogen atoms in fat and water makes the human body approximately 63% hydrogen atoms. MRI uses a powerful magnetic field to align the nuclear magnetization of hydrogen atoms in water in the body. When Radio frequency (RF) fields are added to systematically alter the alignment of this magnetization, the hydrogen nuclei produce a rotating magnetic field detectable by the scanner. Magnetic resonance imaging primarily images the NMR signals provided by the hydrogen nuclei inside the body. Those signals can be detected by gradient magnetic fields and build up enough information to construct an image of the body, especially important since it can display the different properties between normal tissues and tumor. The ability of depicting inner structure of substances also makes MRI of great use in physics, chemistry and many other areas.
Compared to PET and CT, MRI has a rather good resolution and is harmless to patient, since it doesn’t require radioactive injections. This young and still growing technique should have a promising future. In 2003, there were approximately 10,000 MRI units worldwide, and approximately 75 million MRI scans per year performed.[2] As the field of MRI continues to grow, so do the opportunities in MRI.
History of NMR and MRI
Back in the 1930s, Isidor Rabi conducted research on the nature of the force binding protons to atomic nuclei and eventually discovered that the spin orientation of nuclei in a magnetic field changes when radio frequency is added. This was the first human investigation on the interaction of atomic nuclei, magnetic field and radio frequency. He was thus awarded the Nobel Prize for Physics in the year of 1944 for this great work.
A decade later, in 1946, Purcel of Harvard University and F. Block of Stanford University independently discovered that magnetic nuclei in a magnetic field absorb and re-emit electromagnetic radiation. Their discovery of nuclear magnetic resonance and further expansion on this technique won them the Nobel Prize for Physics in 1952.
Immediate after the discovery of nuclear magnetic resonance phenomena, people started to apply this technique on investigation on nuclei structure and properties, such as the measurement of nuclear magnetic moment, as well as other practical researches. Chemists developed NMR spectroscopy according to the influences of molecular structure on the surround magnetic field of Hydrogen. NMR spectroscopy becomes a prevailing technique for molecular structure determination and it develops from one dimensional 1H spectroscopy to 13 carbon spectroscopy, two dimensional spectroscopy and other advanced methods. In the 1990s, scientist even applied this technique on protein, which made the accurate determination of molecular structure of liquid phase protein possible. Nowadays, nuclear magnetic resonance technique is widely used in various areas in chemistry, biology and medical research, from molecular structure and composition analysis to medical diagnose.
In 1971, R. Damadian found out that both the NMR longitudinal relaxation time (T1) and transverse relaxation time (T2) in tumor are longer than normal tissue, which means tumor and normal tissue can be distinguished by NMR in vivo. This work provided a magnificent chance of bringing NMR technique into medical research. Magnetic resonance imaging was first demonstrated on small test tube samples by Paul Lauterbur in 1973 and the first cross-sectional image of a living mouse was published in January 1974. In 1975 Richard Ernst proposed magnetic resonance imaging using phase and frequency encoding, and the Fourier Transform, which is the basis of current MRI techniques. While Richard Ernst was rewarded for his achievements in pulsed Fourier Transform NMR and MRI with the Nobel Prize in Chemistry in 1991, Paul C. Lauterbur of the University of Illinois and Sir Peter Mansfield of the University of Nottingham were awarded the Nobel Prize in Medicine for their discoveries concerning magnetic resonance imaging in 2003. To date, MRI is widely used for tumor detection and diagnose.
Basic Principles
Spin physics
The hydrogen atom possesses a nuclear spin of 1/2. When placed in an external magnetic field,the spin vector of hydrogen atoms aligns itself with the magnetic field and the single energy state will split into two energy states. A particle in the lower energy state can absorb a photon, whose energy exactly matches the energy gap between the two states, and transit to the upper state. The energy of the photon can be described as:
Where h stands for the Planck’s constant and ν is called the resonance frequency and the Larmor frequency in MRI. It depends on the gyromagnetic ratio, γof the particleas:
For hydrogen, γ= 42.58 MHz / T.
Since the frequency of the photon lies in the radio frequency (RF) rangein NMR experiment, ν is between 60 and 800 MHz for hydrogen nuclei. In clinical MRI, ν is typically between 15 and 80 MHz for hydrogen imaging.For example, at the most commonly used field strength of 1.5 Tesla,ν is63.87MHz.
T1 processes
In the equilibrium state, the net magnetization vector lies along the direction of the external magnetic field Bo and is called the equilibrium magnetization Mo. If we refer MZ as the longitudinal magnetizationand MX and MY as the transverse magnetization, MZ equals M0 while both MX and MY equal zero at the equilibrium state. However, If we apply the nuclear spin system to a radio frequency equal to the energy gap between the two spin states, the net magnetization can be changed and eventually saturate the spin system, so that MZ reaches 0. The so called spin-lattice relaxation time (T1, also named as the longitudinal relaxation time) describes how MZ returns to its equilibrium value. The equation depicts this process is shown below:
Mz = Mo (1 - e-t/T1)
Obviously, T1 is the time to reduce the difference between the longitudinal magnetization (MZ) and its equilibrium value by a factor of e.
Fig.1 T1 process (a) after a 90o pulse (b) after an 180o pulse
If the net magnetization is saturated to the –Z axis (by an 180o pulse) and we record its return to the equilibrium state, the behavior can be described as:
Mz = Mo (1 - 2e-t/T1)
T2 processes
When the net magnetization was changed to the XY plane, after absorbing Energy from the radio frequency, it starts to dephase as different spin packets rotates at its own Larmor frequency due to experiencing slightly different magnetic field. The two main factors for this difference in magnetic field are molecular interactions and variations in B0. The former is considered as a pure T2 molecular effect while the latter is said to lead to an inhomogeneous T2 effect. The combination of these two factors is the realresult in the decay of transverse magnetizationandthe combined time constant is called T2 star. The relationship is showed in the following equation:
1/T2* = 1/T2 + 1/T2inhomo.
To describes the return to equilibrium of the transverse magnetization, MXY, the time constant T2 is defined as in the following equation.
MXY =MXYo e-t/T2
Fig.2 T2 process
T2 is called the spin-spin relaxation time or the transverse relaxation time and is always equal to or less than T1. The net magnetization in the XY plane goes to zero and then the longitudinal magnetization grows in until we have Mo along Z.
Rotating frame, Bloch equations and pulse magnetic field
If we define a rotating frame of reference which rotates about the Z axis at the Larmor frequency, it will be much easy to describe the behavior of the net magnetization vector in this frame than in the laboratory frame. In this case, the magnetization vector rotating at the Larmor frequency in the laboratory frame will be stationary in our rotating frame.
The Bloch equations are a set of coupled differential equations, used to describe the behavior of a magnetization vector under any conditions. When properly integrated, the Bloch equations will yield the X', Y', and Z components of magnetization as a function of time.
To create a magnetic field along the X axis, we can put a coil of wire around X axis and pass a current through it. An alternating current can create an oscillating magnetic field, but when we see it in the rotating frame, it just provides a constant magnetic field along the X’ axis. This works just as we move the coil about the rotating frame coordinate system at the Larmor frequency. So the magnetic field generated by the coil passing an alternating current at the Larmor frequency is called the B1 magnetic field. We can generate this pulsed magnetic field but turning the current on and off. 90o pulse and 180o are the mostly used pulses.
Magnetic field gradient
Magnetic field gradient makes it possible for different regions of spin to be exposed to a different magnetic field so that we are able to image their positions. In the following sections, we will use Gx, Gy, and Gz for a magnetic field gradient in the x, y and z directions. The strength of the magnetic field increases along the axis.
A magnetic field gradient in the z direction is highly important for slice selection, which is the selection of spins in a plane through the object. When a 90o pulse is applied, together with our Gz, only spins in the certain slice will be rotated to the XY plane. This is mainly because only spins in this slice matches the frequency provided by the RF, as showed in the following figure.
Fig.3 slice selection magnetic field gradient [2]
The phase encoding gradient is a gradient to impart a specific phase angle to a transverse magnetization vector. After slice selection, spins in that slice are rotated to the XY plane and if we consider them in the same chemical shift, they should have the same Larmor frequency. When a gradient in the magnetic field along the X axis is applied, the vectors will process about the direction of the applied magnetic field at the following frequency:
ν = γ ( Bo + x Gx) = νo + γ x Gx
So when we turn onthe phase encoding gradient, each transverse magnetization vector has its own unique Larmor frequencyand the description of phase encoding is the same as frequency encoding.
In summary, the slice selection gradient is perpendicular to the slice planewhile the phase encoding gradient is along one of the sides of the image planeand the frequency encoding gradient the other.
Pulse sequences
A pulse sequenceis defined as a set of RF pulses applied to the object for a specific form of NMR signal.The 90-FID (free induction decay) sequence, spin-echo sequence and inversion recovery sequence are the most used ones.
In the 90-FID pulse sequence, net magnetization is rotated down into the X'Y' plane with a 90o pulseand the magnitude of the vector decays with time.
The spin-echo sequence also uses a 90o pulse at first so that the magnetization rotates down into the X'Y' planeand the transverse magnetization begins to dephase. However, an 180o pulse is then appliedshortly after the 90o pulse and it rotates the magnetization by 180o about the X' axis. This creates a partially magnetization rephrase and produces the called echosignal.
Fig.4 (a) 90-FID sequence, (b) spin-echo sequence, (c) inversion recovery sequence
In the inversion recovery pulse sequence, an 180o pulse is first appliedto rotate the net magnetization down to the -Z axis.The magnetization undergoes T1 process and returns toward its equilibrium position along the +Z axis. A 90o pulse is applied before it reaches the equilibrium, so that the longitudinal magnetization is rotated into the XY plane. The dephasing process then gives free induction decay.
Basic Imaging Techniques
Now we will talk about basic imaging techniques using various sequences. A good way to show how each imaging method works is to use timing diagrams. A timing diagram normally includes the radio frequency, magnetic field gradients, and signal as a function of time. For example, the simplest FT imaging sequence contains a 90o slice selective pulse, a slice selection gradient, a phase encoding gradient, a frequency encoding gradient, and a signal. The slice selection gradientand the slice selection RF pulseare the first things to be turned on.When they are complete, a phase encoding gradient is applied to the object. A frequency encoding gradient is turned on afterwards and a signal is recorded, which is in the form of a free induction decay. To collect all the data needed for an image, the sequence of pulses is usually repeated 128 or 256 times. We define the time between the repetitions of the sequence as the repetition time, TR. Each time the sequence is repeated, the magnitude of the phase encoding gradient is changed in equal steps between the maximum amplitude of the gradient and the minimum value. In this case, we can record 128 or 256 different free induction decaysfor resolving 128 or 256 locations in the phase encoding direction. Those recorded signals described above can be Fourier transformed to create an image of the location of spins.
A widely used parameter to quantify image quality is the signal-to-noise ratio (SNR). In an image, it is the ratio of the average signal for the tissue to the standard deviation of the noise in the background. The SNR of an MRI image has the following dependent factors: (1) the number of signal producing water molecules in the image voxel (2) the size of the signal each molecule produces, (3) the quality of the signal detection, and (4) the amount of spurious noise and the statistics of noise averaging.
Gradient-Echo Imaging
The gradient-echo sequence is intrinsically more sensitive to magnetic field inhomogeneities because of the use of the refocusing gradient or wind-up gradient. Together with a slice selection gradient, the slice selective RF is turned on to creates a rotation angle of 90o. The phase encoding gradient Gφ is applied right after. Here, a negative dephasing frequency encoding gradient -Gf is on at the same time to cause the spins to be in phase at the center of the acquisition period. When the frequency encoding gradient Gf is applied, an echo is produced, because the gradient refocuses the dephasing which occurred from the dephasing gradient. This echo is named as a gradient echo, which is different from the echo created by an 180o pulse.
We define echo time (TE) as the time period from the start point of the RF pulse to the maximum of the signal. An equation to describe the signal intensity versus time and spin density can be written as:
S = k ρ (1-exp(-TR/T1)) exp(-TE/T2*)
Fig.5 Gradient-Echo Imaging timing diagram
Spin-Echo Imaging
In spin-echo imaging, a spin-echo sequence is used and it displays the transverse relaxation time dependence to the signal. This is especially useful for some tissues which have similar T1 values but different T2. So as showed in the timing diagram in figure 5, a slice selective 90o pulse, together with the slice selection gradient, is applied to the object. After a time period of TE/2, an 180o slice selective pulse is turned on in conjunction with the slice selection gradient. Between the two pulses, just like in the previous Gradient-echo imaging method, the phase encoding gradient Gφ and frequency encoding gradient Gf are applied at the same time. This time, Gf is necessary because it dephases the spins so that they will rephase by the center of the echo. Finally, the frequency encoding gradient applied after the 180o pulse, during the time that echo is collected. The signal, which is the collected echo, is as follows:
S = k ρ (1-exp(-TR/T1)) exp(-TE/T2)
Fig.6 Spin-Echo Imaging timing diagram
Inversion Recovery Imaging
An inversion recovery sequence is used in this imaging technique. It uses a spin-echo sequence to detect the signal so the RF pulses are 180-90-180. An inversion recovery sequence which uses a gradient-echo signal detection is similar, with the exception that a gradient-echo sequence is substituted for the spin-echo part of the sequence.
S = k ρ (1-2exp(-TI/T1)+exp(-TR/T1))
Fig.7 Inversion Recovery Imaging timing diagram
Multi-slice Imaging
Multi-slice imaging makes a volume of anatomy to be imaged in the shortest time possible. The time to obtain an image is equal to the product of the TR value and the number of phase encoding steps. While TR takes time in seconds, the large steps of phase encoding last much longer. So when we take a careful look at the timing diagram of the gradient-echo sequence introduced above, we can easily find out that the most of the sequence time is unused. To make full use of the scanning time, other slices can be excited as show in figure 8. However, other RF frequencies of the 90o should be used in order to prevent the interactions between those slices.
Fig.8 Multi-slice Imaging timing diagram
T1, T2, and ρ Images
Sometimes, properties of the spins in a tissue, such as the spin-lattice relaxation time (T1), spin-spin relaxation time (T2), and the spin density (ρ) are searched for. There are several methods of calculating T1, T2, and ρ values. A T1 image can be created from a series of images using the same pulse sequence with varying TR. The signal for a given pixel can be plotted for each TR value and the best fit line from the spin-echo equation drawn through the data to find T1, as show in figure 10.
Similarly, to produce a T2 image, a series of images using a spin-echo pulse sequence with varying TE is recorded. The signal for a given pixel can be plotted for each TE value and the best fit line from the spin-echo equation drawn through the data to find T2, as shown in figure 12. The spin-echo equation is mentioned in the previous Pulse Sequences section. Once T1 and T2 are obtained, it’s easy to acquire spin density by using the spin echo signal equation and any spin echo signal.
Fig.9 a series of imaging recorded with various TR under spin-echo sequence[2]
Fig.10 signal – TR graph
Fig. 11 a series of imaging recorded with various TE under spin-echo sequence[2]
Fig. 12 signal – TE graph
Chemical contrast agent
Although MRI has a good resolution and is very sensitive compared with other imaging modalities, such as positron emission tomography, chemical contrast agents (CA), substances which are introduced into the body to change the contrast between the tissues, play a vital role in the imaging procedure. Contrast agents may be injected intravenously to enhance the appearance of blood vessels, tumors or inflammation or directly injected into a joint in the case of arthrograms. Normally, they can be divided into two main categories, namely T1 contrast agents and T2 contrast agents and they shorten the according relaxation time, respectively. The most commonly used compounds for contrast enhancement are gadolinium-based or iron-based. The former is paramagnetic and belongs to T1 contrast agents, as it changes the contrast by creating time varying magnetic fields which promote spin-lattice and spin-spin relaxation of the water molecules. The latter is superparamagnetic and comes to the second group, as it changes T2 of the water molecules around by distorting the Bo magnetic field. Several well-known contrast agents are Magnevist (Gd-DPTA), Dotarem (Gd-DOTA) and superparamagnetic iron oxide (SPIO). Free Gadolinium ions in the human body may arose a certain kind of disease called nephrogenic systemic fibrosis(NSF). That’s why Gd needs to be connected with organic polymers with great care for safety reason.
Applications
In clinical practice, MRI is used to distinguish pathologic tissue, such as a brain tumor, from normal tissue.
Hardware
Fig. 13 Hardware overview[2]
Figure 13 is an overview of the magnetic resonance imaging scanning system. Computer plays the most important role in this procedure as it controls all the components. The magnet produces the Bo field which is critical for the imaging process, along with those gradient coils for gradients in Bo in the X, Y, and Z directions. The RF coil generates the B1 magnetic field for rotating the spins by 90o, 180o, or any other value selected by the pulse sequence and also functions as the signal detector from the spins within the body. The RF amplifier increases the pulses power from mW to kW while the gradient amplifier increases the power of the gradient pulses to a sufficient level to drive the gradient coils. Since outer RF pulses may cause an additional impact on imaging, this scan room is enclosure with RF shields.
Advantages and disadvantages
Despite of the rapid development and its improvement in the past several decades, MRI still has its shortcomings. The scan time of MRI for cardinal tissues lasts more than 30min, which is much longer than a CT scan and the expense of an MRI system is still high. Patients need to be stationary during the scan to prevent distortion of images. Besides, patients with pacemakers cannot have MRIs and claustrophobic patients cannot usually make it through a MRI. The machine makes a tremendous amount of noise during a scan due to the rising electrical current in the wires of the gradient.
Even with these disadvantages, we cannot ignore the tremendous contribution MRI has made in clinical practice in the past years. Because variations in T1 and T2 values are so much greater than variations in tissue density, MRI provides better soft-tissue contrast than plain radiography or computed tomography (CT).[3] In addition, it is easy to take a MRI slice in any direction, without changing the position of the patient. Compared to PET and CT, MRI is also harmless to the human body. The high resolution of imaging makes MRI an effective imaging modality and its opportunity in the future should be even more promising.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Experimental/Diffusion_Ordered_Spe.txt
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Nuclear Magnetic Resonance (NMR) interpretation plays a pivotal role in molecular identifications. As interpreting NMR spectra, the structure of an unknown compound, as well as known structures, can be assigned by several factors such as chemical shift, spin multiplicity, coupling constants, and integration. This Module focuses on the most important 1H and 13C NMR spectra to find out structure even though there are various kinds of NMR spectra such as 14N, 19F, and 31P. NMR spectrum shows that x- axis is chemical shift in ppm. It also contains integral areas, splitting pattern, and coupling constant.
Strategy for Solving Structure
Here is the general strategy for solving structure with NMR:
1. Molecular formula is determined by chemical analysis such as elementary analysis
2. Double-bond equivalent (also known as Degree of Unsaturation) is calculated by a simple equation to estimate the number of the multiple bonds and rings. It assumes that oxygen (O) and sulfur (S) are ignored and halogen (Cl, Br) and nitrogen is replaced by CH. The resulting empirical formula is CaHb
1. Structure fragmentation is determined by chemical shift, spin multiplicity, integral (peak area), and coupling constants ($^1J$, $^2J$)
2. Molecular skeleton is built up using 2-dimensional NMR spectroscopy.
3. Relative configuration is predicted by coupling constant (3J).
1H NMR
Chemical Shift
Chemical shift is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilane (TMS, $\ce{(CH3)4Si}$) is generally used as an internal standard to determine chemical shift of compounds: δTMS=0 ppm. In other words, frequencies for chemicals are measured for a 1H or 13C nucleus of a sample from the 1H or 13C resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane. 1H chemical shift play a role in identifying many functional groups. Figure $1$. indicates important example to figure out the functional groups.
Chemical equivalence
Protons with Chemical equivalence has the same chemical shift due to symmetry within molecule ($CH_3COCH_3$) or fast rotation around single bond (-CH3; methyl groups).
Spin-Spin Splitting
Spin-Spin splitting means that an absorbing peak is split by more than one “neighbor” proton. Splitting signals are separated to J Hz, where is called the coupling constant. The spitting is a very essential part to obtain exact information about the number of the neighboring protons. The maximum of distance for splitting is three bonds. Chemical equivalent protons do not result in spin-spin splitting. When a proton splits, the proton’s chemical shift is determined in the center of the splitting lines.
Spin Multiplicity (Splitting pattern)
Spin Multiplicity plays a role in determining the number of neighboring protons. Here is a multiplicity rules: In case of $A_mB_n$ system, the multiplicity rule is that Nuclei of $B$ element produce a splitting the $A$ signal into $nB+1$ lines. The general formula which applies to all nuclei is $2_nI+1$, where $I$ is the spin quantum number of the coupled element. The relative intensities of the each lines are given by the coefficients of the Pascal’s triangle (Figure $2$).
First-order splitting pattern
The chemical shift difference in Hertz between coupled protons in Hertz is much larger than the $J$ coupling constant:
$\dfrac{\Delta \nu }{J} \ge 8$
Where $\Delta \nu$ is the difference of chemical shift. In other word, the proton is only coupled to other protons that are far away in chemical shift. The spectrum is called first-order spectrum. The splitting pattern depends on the magnetic field. The second-order splitting at the lower field can be resolved into first-order splitting pattern at the high field. The first-order splitting pattern is allowed to multiplicity rule (N+1) and Pascal’s triangle to determine splitting pattern and intensity distribution.
Example $1$
The note is that structure system is A3M2X2. Ha and Hx has the triplet pattern by Hm because of N+1 rule. The signal of Hm is split into six peaks by Hx and Ha (Figure3) The First order pattern easily is predicted due to separation with equal splitting pattern.
High-order splitting pattern
High-order splitting pattern takes place when chemical shift difference in Hertz is much less or the same that order of magnitude as the j coupling.
$\frac{\Delta v}{J} \leq 10$
The second order pattern is observed as leaning of a classical pattern: the inner peaks are taller and the outer peaks are shorter in case of AB system (Figure $4$). This is called the roof effect.
Here is other system as an example: A2B2 (Figure $5$). The two triplet incline toward each other. Outer lines of the triplet are less than 1 in relative area and the inner lines are more than 1. The center lines have relative area 2.
Coupling constant (J Value)
Coupling constant is the strength of the spin-spin splitting interaction and the distance between the split lines. The value of distance is equal or different depending on the coupled nuclei. The coupling constants reflect the bonding environments of the coupled nuclei. Coupling constant is classified by the number of bonds:
Geminal proton-proton coupling (2JHH)
Germinal coupling generates through two bonds (Figure $6$). Two proton having geminal coupling are not chemically equivalent. This coupling ranges from -20 to 40 Hz. 2JHHdepends on hybridization of carbon atom and the bond angle and the substituent such as electronegative atoms. When S-character is increased, Geminal coupling constant is increased: 2Jsp1>2Jsp2>2Jsp3 The bond angle(HCH) gives rise to change 2JHH value and depend on the strain of the ring in the cyclic systems. Geminal coupling constant determines ring size. When bond angle is decreased, ring size is decreased so that geminal coupling constant is more positive. If a atom is replace to an electronegative atom, Geminal coupling constant move to positive value.
Vicinal proton-proton coupling (3JHH)
Vicinal coupling occurs though three bonds (Figure $7$.). The Vicinal coupling is the most useful information of dihedral angle, leading to stereochemistry and conformation of molecules. Vicinal coupling constant always has the positive value and is affected by the dihedral angle (?;HCCH), the valence angle (?; HCC), the bond length of carbon-carbon, and the effects of electronegative atoms. Vicinal coupling constant depending on the dihedral angle (Figure $8$) is given by the Karplus equation.
$^3 J=7.0-0.5 \cos \phi+4.5 \cos ^{2} \phi$
When ? is the 90o, vicinal coupling constant is zero. In addition, vicinal coupling constant ranges from 8 to 10 Hz at the and ?=180o, where ?=0o and ?=180o means that the coupled protons have cis and trans configuration, respectively.
The valence angle(?;Figure $8$) also causes change of 3JHH value. Valence angle is related with ring size. Typically, when the valence angle decreases, the coupling constant reduces. The distance between the carbons atoms gives influences to vicinal coupling constant
The coupling constant increases with the decrease of bond length. Electronegative atoms affect vicinal coupling constants so that electronegative atoms decrease the vicinal coupling constants.
Integral
Integral is referred to integrated peak area of 1H signals. The intensity is directly proportionally to the number of hydrogen.
13C NMR
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C- 13 C spin-spin splitting rarely exit between adjacent carbons because 13C is naturally lower abundant (1.1%)
• 13C-1H Spin coupling: 13C-1H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (1JCH), -CH, -CH2, and CH3 have respectively doublet, triplet, quartets for the 13C resonances in the spectrum. However, 13C-1H Spin coupling has an disadvantage for 13C spectrum interpretation. 13C-1H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of 1H.
• Decoupling: Decoupling is the process of removing 13C-1H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling 13C spectra shows only one peak(singlet) for each unique carbon in the molecule(Figure $10$.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
• Distortionless enhancement by polarization transfer (DEPT): DEPT is used for distinguishing between a CH3 group, a CH2 group, and a CH group. The proton pulse is set at 45o, 90o, or 135o in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Figure $11$. is an example about DEPT spectrum.
2-dimensional NMR spectroscopy (COSY)
COSY stands for COrrelation SpectroscopY. COSY spectrum is more useful information about what is being correlated.
1H-1H COSY (COrrelation SpectroscopY)
1H-1H COSY is used for clearly indicate correlation with coupled protons. A point of entry into a COSY spectrum is one of the keys to predict information from it successfully. Relation of Coupling protons is determined by cross peaks(correlation peaks) and in the COSY spectrum. In other words, Diagonal peaks by lines ar e coupled to each other. Figure $12$ indicates that there are correlation peaks between proton H1 and H2 as well as between H2 and H4. This means the H2 coupled to H1 and H4.
1H-13C COSY (HETCOR)
1H-13C COSY is the heteronuclear correlation spectroscopy. The HETCOR spectrum is correlated 13C nuclei with directly attached protons. 1H-13C coupling is one bond. The cross peaks mean correlation between a proton and a carbon (Figure $13$). If a line does not have cross peak, this means that this carbon atoms has no attached proton (e.g. a quaternary carbon atom)
Outside Links
• NMRShiftDB: a Free web database for NMR data : nmrshiftdb.chemie.uni-mainz.de/nmrshiftdb
• NMR database from ACD/LAbs : www.acdlabs.com/products/spec_lab/exp_spectra/spec_libraries/aldrich.html
• NMR database from John Crerar Library : http://crerar.typepad.com/crerar_lib...h_ir_nmr_.html
Problems
Draw the 1H NMR spectrum for 2-Hydroxypropane in CDCl3. Assume sufficient resolution to provide a first-order spectrum and ignore vicinal proton-proton coupling(3JHH)
Solution
1) the structure of 2-hydoroxyporpane is drawn
Figure out which protons are chemically equivalent, i.e., two methyl (-CH3) groups are chemical equivalent.
4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by Hb(#of proton=1). Hb has the septet pattern by Ha (#of proton=6). Hc has one peak.(Note that Hc has doublet pattern by Hb due to vicinal proton-proton coupling.)
• You Jin Seo
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The theory sections in this wiki have been devoted to understanding how NMR works and we have made several attempts to investigate atomic nuclei using Magnetic fields, coils, and RF fields. This section is devoted to developing and NMR spectrometer from the theory side then introducing modifications to this simple spectrometer that makes NMR spectroscopy much easier. It is the goal of this page to provide enough information about NMR hardware that one (given the proper materials) could build an NMR spectrometer in their garage!
Basic NMR Spectrometer
Shown below is the most basic NMR spectrometer you can make. For an NMR spectrometer to be operational we first need a magnet to split the degenerate nuclear energy levels of the sample. We also need a coil to apply an oscillating RF field which will be used to drive transitions between the energy levels as well as detect the signals. Therfore we will also need and RF oscillator and a oscilloscope to generate and detect the RF waves.
Magnet
The magnet is arguably the most important part of the NMR instrumentation. The magnetic field drives the splitting of the energy levels and allows for a change in the distribution of energy levels of the degenerate ground state. As we have seen previously, the larger the magnetic field, the larger the spin difference between the upper and lower levels allowing for a stronger signal to be observed. In addition to the splitting of energy levels, several NMR interactions are dependent on the magnetic field. Both the dipolar and CSA interactions are multiplied by an increase in magnetic field, while the Quarupolar coupling interaction is decreased with increasing magnetic field.
Insert picture
Both superconducting and electromagnets magnets are comprised of loops of wires with a current passing through, which results in the formation of a magnetic field. The "north pole" of the magnetic field (using the right hand rule) is generated in the direction perpendicular to the direction of the current flow.
show equation
Superconducting Magnets
The magnet is typically composed of an alloy that is super conducting at low temperature. While the theory of super conductivity is beyond the scope of this page, several unique properties of super conductors are important. This alloy is made into a wire which is wrapped into a coil and has a current passed through it. This generates the magnetic field! The wires are kept cooled with liquid helium at 4K, which significantly reduces the resistance of the wire to nearly zero. Therefore the current passing through the wire onyl needs to happen once and the spectrometer does not need to be plugged in. The liquid helium Dewar is insulated by a vacuum chamber and then a liquid nitrogen dwar which itself is insulated by another vacuum chamber.
Probe
The probe is a specialized piece of hardware that contians the coils and sample and therefore is responsible for tuning to the correct frequency, applying RF pulses, receiving the signal from the nuclei. More specialized probes can also spin the sample, vary the temperature, and change the angle of the sample and coil.
Lets first optimize the efficiency of the coil so that the maximum power is transferred to the sample from RF source and from the sample to the receiver. Since both the power transfer from the RF source and the sample to the receiver is done in the coil we can rite that the power is the
$P=VI$
were P is power, V is the voltage, and I is the amperage. Now an EMF is generated in the coil when the magnetization processes. ANd EMF is considered to be a pure voltage source and is measure in volts. However the coil has a finite resistance to it which is related to the voltage by
$V=IR$
so the total voltage measurable voltage is then
$V_{tot}=V_{EMF}-IR$
Now measuring a voltage is fairly straightforward and will not be discussed. However The concept of impedance needs to be addressed. Impedence is another type of resistance that stems from a phase lag between the current and the voltage. Specifically, the current lags behind the voltage. As implied by the term "phase" the impedance is a complex quantity, with the rela part being the resistance and the imaginary part being called the reactance. Reactance is the opposition of a circuit element to a change in the current or voltage.
We can then show the Impedance, Z, is then
$Z=\abs{Z}e^{i \theta}=R+iX$
where $\theta$ is the thase shift in polar coordinates, and X is the reactance.
Then
$V=IZ$
Now our total voltage can be reordered to show
$V_{tot}=V_{EMF}-IZ$
Lets first start with our coil. The coil is an inductor has an induction of L given by the coil dimensions as
$L=\frac{r^2n^2}{9r+10l}$
where r is the radius of the coil, n is the number of turns, and l is the length of the coil. In our coil we will have an oscillating magnetic field which generates both an oscillating current and voltage which can be described as
$V(t)=V_0 cos \omega t$
$I(t)=\frac{V_0}{\omega L} sin \omega (t)$
we can then immediately see that the voltage and current are 90 degrees out of phase leading to a completely imaginary impedence
$Z_L=i \omega L$
There are several points to this. First, we now see that the current is frequency dependent, that is at high frequencies the current is very small, but at low frequencies the current is maximized. This has ramification for both application of power to the coil and detection as high frequency nuclei will then be harder to detect! Remember that we are trying to measure a voltage and that any decrease in current will directly decrease the voltage and therefore the signal that we receive! The inductor also has a finite resistance, allowing it to behave as a resistor as well. For a oscillating voltage the current is then
$I(t)=\frac{V_0}{R}cos \omega t$
assuming the time dependent voltage is of the form
$V(t)=V_0cos \omega t$
as we can see there is no phase shift so the impedance of the resistor is then
$Z_R=R$
and the current is directly related to the voltage. We have now reached quite a predicament in the high frequency range. That is the power scales down significantly when we go to high frequency. Therefore we need some way to "reverse" this effect. One way to do this is to use a capacitor. The current flowing across a capacitor is given by
$-C \omega V_0 sin \omega t$
and we can see that this scales directly with frequency. Therefore capacitors allow high frequencies to pass through them and block low frequencies. Since they are 90 degrees out of phase with the voltage, they exhibit the same reactance that inductors do
Detector
A simple description of the NMR detector is given in the Detectors section. The function of the NMR detector is to detect the nuclear transitions occurring in the experiment. As the magnetization is precessing, it generates an electromotive force (EMF). The induced EMF is detected through the coil and sent to the receiver.
$EMF=\frac{d \phi}{dt}$
where $d\phi$ is the magnetic flux induced from the magnetization vector processing.
Let's assume our magnetization vector has length m. The time dependence of m is then
$m(t)=|m|[sin\psi cos(\omega_0 t +\xi_0)e_x+sin\psi sin(\omega_0 t +\xi_0)e_y +cos\psi e_z$
for a spherical basis set. The resultant magnetization at point r from the precession is
$B(r,t)=\frac{m_0}{4\pi r^3} [3(m(t) \cdot e_r)e_r-m(t)]$
the magnetic flux through the coil, distance r away from magnetic vector, is given by a surface integral
$\phi(t)=\int\int B(r,t)\cdot n r drd\theta$
Solving this integral results in
$\phi(t)=-\frac{m_0}{4\pi}\int_0^{r_{coil}}\frac{dr}{r^2}\int_0^{2\pi} d\theta m(t) \cdot n$
which can be simplified if the coil is positioned such that it is in the x-z or y-z plane (n=ey or ex). (check math).
Interestingly,using only one coil, we are only able to detect the the magnetization is precessing, not which direction it is precessing. We circumvent this issue by employing quadrature detection.
Quadrature Detection
To find the direction of the magnetization precession, the NMR signal is split into 2 parts and applying a phase to each part.
Relation to rotating frame
Problems with imperfect phases
reference coherence pathways
describe how its done
Shims
The shims are used to obtain a homogenous field around the sample. They are named after the thin metal pieces NMR technicians would insert into the magnetic field in the to manually make a homogenous magnetic field. Nowadays, the shims are coils of wire which pass current to generate magnetic fields in certain directions. These shims are not just about the z, y and z directions, the vary across a variety of planes to ensure a very homogenous field is achieved.
If we want our magnetic field to be homogenous, then the space we want to be homogenous must be absent of any charges or current. Therefore, a homogenous magnetic field obeys the Laplace Equation, more commonly written as
$\left(\frac{d^2}{d^2x}+\frac{d^2}{d^2y}+\frac{d^2}{d^2z}\right)B_0=0$
Those familiar quantum mechanics may recognize this from the Schroedinger equation in the forms
$\nabla^2\psi$
which represents the potential energy in three dimensions. Solving this second order differential is indeed challenging. As with any field, it can be represented by a linear of any complete orthogonal basis set. A particularly useful basis set is the spherical harmonics basis set. B) under this basis set is described as
$B_0=\sum_{n=0}^\infty \sum_{m=0}^{n} C_{nm} (\frac{r}{a})^n P_{nm}(\cos\theta)\cos[m(\phi-\psi_{nm})]$
where $r$ is a distance of some point, $a$ is the bore radius of the magnet, and $P_{nm}$ is the Legendre polynomial.
Zonal Harmonics
When m =0 the variation of B0 with (\phi\) goes away and the equation is reduced to
$B_0=\sum_{n=0}^\infty C_{n0} (\frac{r}{a})^n P_{nm}(cos\theta)$
These inhomogeneities can be cancelled by application of the z shims.
Tesseral Harmonics
For the remainder of the m values, the off-z axis shims must be used.
We must then apply magnetic fields in all directions in order to cancel the inhomegenties in B0.Since modern shims are coils of wire, the current must be adjusted to create the proper field to cancel to inhomogeneity. Commonly the shims are listed in cartesian coordinated.
Conversion equation
Show the Chart of conversion
Shim Mixing
Unfortunately, the shims are not independent, that is adjusting one shim changes the magnetic fields of the previously set shims.
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There will be cases in which you already know what the structure might be. In these cases:
• One of the most complicated problems to deal with is the analysis of a mixture. This situation is not uncommon when students run reactions in lab and analyse the data.
• Remember that integration ratios are really only meaningful within a single compound. If your NMR sample contains some benzene (C6H6) and some acetone (CH3COCH3), and there is a peak at 7.15 that integrates to 1 proton and a peak at 2.10 ppm integrating to 6 protons, it might mean there are 6 protons in acetone and 1 in benzene, but you can tell that isn't true by looking at the structure. There must be six times as many acetone molecules as benzene molecules in the sample.
There are six protons in the benzene, and they should all show up near 7 ppm. There are six protons in acetone, and they should all show up near 2 ppm. Assuming that small integral of 1H for the benzene is really supposed to be 6H, then the large integral of 6H for the acetone must also represent six times as many hydrogens, too. It would be 36 H. There are only six hydrogens in acetone, so it must represent six times as many acetone molecules as there are benzenes.
However, comparing the ratio of two integrals for two different compounds can give you the ratio of the two compounds in solution, just as we could determine the ratio of benzene to acetone in the mixture described above.
We will look at two examples of sample mixtures that could arise in lab. Results like these are pretty common events in the labIn the first example, a student tried to carry out the following reaction, a borohydride reduction of an aldehyde. The borohydride should give a hydride anion to the C=O carbon; washing with water should then supply a proton to the oxygen, giving an alcohol.
Her reaction produced the following spectrum.
(simulated data)
From this data, she produced the table below.
Notice how she calculated that ratio. She found a peak in molecule 1, the aldehyde, that she was pretty sure corresponded to the aldehydic hydrogen, the H attached to the C=O; in other words, the CH=O. She found another peak from molecule 2, the alcohol, that she was pretty sure represented the two hydrogens on the carbon attached to oxygen, the CH2-O.
The integrals for those two peaks are equal. They are both 2H in her table. However, she notes that within each molecule, the first integral really represents 1H and the second represents 2H. That means there must be twice as many of molecule 1 as there are molecule 2. That way, there would be 2 x CH=O, and its integral would be the same as the 1 x CH2-O in the other molecule.
One way to approach this kind of problem is to:
• So there is twice as much aldehyde as alcohol in the mixture. In terms of these two compounds alone, she has 33% alcohol and 66% aldehyde. That's ( 1/(1+2) ) x100% for the alcohol, and ( 2/(1+2) ) x100% for the aldehyde. That calculation just represents the amount of individual component divided by the total of the components she wants to compare.
There are a number of things to take note of here.
• Another student carried out a similar reaction, shown below. He also finished the reaction by washing with water, but because methanol is soluble in water, he had to extract his product out of the water. He chose to use dichloromethane for that purpose.
From this data, he constructed the following table.
There are some things to learn about this table, too.
• This student might not get a very good grade; the sample does not even show up in the spectrum, so he lost it somewhere. But his analysis is also poor, so he will really get a terrible grade.
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A pulse sequence is a succinct visual representation of the pulses and delays used in a certain NMR experiment. Depending on the experiment there may be hundreds of pulses! This page is dedicated to understanding what a pulse sequence is and how to understand the pictorial representations, as well as terms commonly used to describe parts of pulse sequences.
How to Read and Understand a Pulse Sequence
The most basic pulse sequence is a single pulse followed by detection of the signal. This is shown in figure 1. The pulse is represented by the large black bar and the acquisition or detection period is denoted by the squiggly line which represents the free induction decay (FID) of thr signal generated. The arrow below the sequence is the direction in which time is increasing. In all pulse sequences, time increases from left to right.
Pulses
A pulse is a collection of oscillating waves with a broad range of frequencies (~1/4*pulse length) used to rotated the bulk magnetization. Pulses are described using two parameters, the angle and the phase. Typically, the angle and the phase are shown above the pulse. The angle can vary from 0°-180° While the phase can take on values from 0°-360°. Commonly the phase of the pulse is referred to the axis along which the $B_1$ field is applied. For instance, shown below are the effects of a 90 pulse with a phase along x axis. This rotates the magnetization down along the y axis (right hand rule). While this example is fairly straightforward, pulse phases can become incredibly complex. More information about pulse phases will be covered in the phase cycling section. Pulses may be refereed to by just their angle ("a nintey pulse"), their angle in radians (a $\frac{\pi}{2}$ "pi over two pulse") or their angle and phase "ninety x", (90°)x.
Acquisition
Now we move further to the right, increasing in time and see a squiggly line. This line represents the signal, which shows decay, as expected from T2 and T2* relaxation.
Multiple Pulse Sequences
Most pulse sequences have more than one pulse, which add a degree of complexity in to interpretations of the pulse sequence. Using multiple pulses have many advantageous including separation of NMR interactions, measuring relaxation timescales, and signal enhancement. Two additional features are present in multiple pulse experiments, delays and loops.
Delays
Delays are times when pulsing and detection are not occurring and are very important in pulse sequences, as separation of NMR interactions require precise timings. Additionally, creating proper delays can preserve select information as well as enhance signals. Delays are often referred to as "times". Below is a list of common delays:
• Relaxation Delay: Amount of time needed to wait before the next experiment can begin. Typically this is on the order 3-5 T1.
• Rotor Synchronized Delay (pulses): These delays are set so pulses remain centered in time with an integer multiple of the rotor period. Delays will then be:
$Delay=n*\tau_r-\frac{1}{2}(P_i+P_f)$
where $n$ is an integer number, $\tau_r$ is the spinning speed, Pi is the pulse length of the pulse before the delay and Pf is the pulse length of the pulse after the delay.
• Evolution (time) Delay: This is time during which the magnetization is precessing to satisfy the an equation relating to the separation of a variable.
• Storage (time) Delay: This is the time at which the bulk magnetization is stored along the external magnetic field to allow for sample manipulation, such a hopping or flipping.
Loops
Loops are repeating parts of the pulse sequence. They are typically employed in sequences to boost the S/N.
Echoes
Echoes are appear in pulse sequences as 2 FIDs placed in opposite directions. Figure 3 shows what an echo looks like during the CPMG loop.
Pulse Sequences
In NMR simply throwing a string of pulses with random phases is a recipe for disaster. Phase cycling is a way to choose the proper coherence pathway in order to acquire the NMR signal which has the interactions you are trying to acquire. Additionally, phase cycling eliminates artifacts that can appear in the NMR spectrum which are not real NMR signals.
NMR Artifact
Two problems are commonly encountered during detection of the NMR signal. This first occurs when the detector should be detecting zero but actually records a non-zero value. This is known as a receiver baseline error, commonly known as a zero glitch. The other type of artifact occurs from errors in the quadrature detection of the NMR signal. NMR spectra can contain both of these and can severely complicate spectra. Fortunately, we can eliminate these by employing phase cycling.
Zero Glitch
If the detector is turned on and the voltage is non-zero, then a spike will form in the exact center of the spectrum. A zero glitch is easy to check for by changing the carrier frequency. If the spike moves to the center of the spectrum, you have a zero glitch.
Quadrature Ghost (Image)
From our discussion on the NMR receiver, quadrature detection is used. In this, the frequency is split and a reference phase is superimposed on one of the carrier frequencies. Successful recombination of the frequencies is of the form
$S=\cos+i\sin$
and the resulting Fourier Transform is a peak at a certain frequency. However, when the phase is not set correctly, or the efficiency of the hardware changes, the signal now has a different phase (not 90°). This results in a signal at the exact opposite frequency.
Coherence Transfer Pathway
Conferences are developed by the application of pulses of a phase $\phi$. As our discussion from the Bloch equations, three different conferences can develop p=-1,0,+1 with the application of the pulse, depending on the efficiency of the coherence transfer. We also saw that the relaxation processes decay the coherence. Additionally, the coherence will not change without the application of a pulse. We can then construct a coherence transfer diagram which illustrates the coherence path as a function of the pulse sequence.
As you can see from above, there are multiple coherences at the end of the pulse sequence. In fact, for a spin=1/2 system with n pulses, 3n coherence pathways evolve! We must then select the proper coherence to detect! Luckily, most receivers are equipped to only detect on the -1 coherence. This limits detection to only a fraction of the total coherences that evolve. Therefore, we must develop a way to selectively obtain the coherences with the interactions we want to probe.
Nested Phase Cycling
Phase cycling is a way to selectively extract a single coherence that has evolved with a selected pathway. The only way to change a coherence is to apply a pulse by phase $\phi$. Therefore, we must know the coherence pathway we are interested in and we need to set the phases of the pulses and receiver accordingly. Luckily it is fairly easy to calculate which phases are needed. We know that the coherence must be =-1 at the end of the pulse sequence for the signal to be detected by the receiver. The master equation is listed below
$\phi_{rec}=-\sum_{m=1}^{M_T}\Delta p \phi_m$
where $\phi_R$ is the receiver phase, $\Delta p$ is the desired change in coherence and $\phi_m$ is the phase of the pulse m.
If we only ever used one pulse, then only three coherences would evolve. However, with multiple pulses, multiple signals will be at the -1 coherence during detection. We can average these by changing the phase $\phi_m$. The phase should be incremented, at most by
$\Delta \phi_m=\frac{2\pi}{n_m}$
where
$n_m=\Delta p_m^{(max)}-\Delta p_m^{(min)} +1$
Let's consider the pulse sequence in figure 1. Let's also assume we do not have a perfect detector, and some of the magnetization from the +1 and 0 coherences can be detected.
Our desired pathway is 0 to -1. $\Delta p_1^{(max)} =1$ and $\Delta p_1^{(min)}=-1$ Then n1=3. Therefore we need to increment $\phi$ by steps of 120^$\circ$ to create the phase cycle. Our desired $\Delta p = -1-0=-1$. Thus
$\phi_{rec}=\phi_1$
Our full phase cycle is for both $\phi_{rec}$ and $\phi_1$ = 0, 120, 240. Therefore 3 scans are needed to complete the phase cycle. Experiments should have a multiple of 3 to make sure a proper spectrum is acquired. Let's take a look at a more complex pulse sequence, the Hahn echo sequence. The pulse sequences with the desired pathway is outline below.
We must once again calculate the the master equation, however we now have to consider the second pulse. Lets first break the sequence up into 2 steps, the first step will be for $\phi_1$ and the second for $\phi_2$. For $\phi_1$:
$\Delta p_1=1-0=1$
$n_m==1-(-1)+1=3$
$\Delta \phi_1=\frac{2\pi}{n_m}=120^\circ$
For $\phi_2$:
$\Delta p_2=-1-1=-2$
$n_m==1-(-1)+1=3$
$\Delta \phi_1=\frac{2\pi}{n_m}=120^\circ$
Then
$\phi_{rec}=-(\phi_1-2\phi_2)$
and the phase cycle would be
$\begin{matrix}\phi_1 & = & 0^\circ & 120^\circ & 240^\circ & 0^\circ & 120^\circ & 240^\circ & 0^\circ & 120^\circ & 240^\circ \ \phi_2 & =& 0^\circ & 0^\circ & 0^\circ & 120^\circ & 120^\circ & 120^\circ & 240^\circ & 240^\circ & 240^\circ \ \phi_{rec}&=&0^\circ& 240^\circ &120^\circ &240^\circ &120^\circ &0^\circ &120^\circ & 0^\circ & 240^\circ & \end{matrix}$
As you can see from above, this is a nested phase cycle, in which the three steps of $\phi_1$ are incremented successively, while $\phi_2$ is incremented only after completion of a full phase cycle is completed for $\phi_1$ while this creates a robust phase cycle, there are several shortcuts which enable the phase cycles to be shortened. For instance, if no quadrature ghosts are detected, meaning only to -1 coherence is observed, then only one pulse needs to be phase cycled. This gives
$\begin{matrix}\phi_1 & = & 0^\circ & 120^\circ & 240^\circ \ \phi_{rec}&=&0^\circ& 240^\circ &120^\circ & \end{matrix}$
or
$\begin{matrix}\phi_2 & = & 0^\circ & 120^\circ & 240^\circ \ \phi_{rec}&=&0^\circ& 240^\circ &120^\circ & \end{matrix}$
Note that the phases for $\phi_1$ and $\phi_2$ are obtained when the opposing $\phi$=0
Phase Cycling with Coherences>1
Until now, we have considered a spin=1/2 nucleus with coherences = -1, 0 ,+1, which we readily obtained from the spherical representation of the Bloch equations. However, the Bloch equations are only valid for a single uncouple spin, which is not the case for most systems. Therefore, access to higher coherences are possible for coupled spin=1/2 nuclei as well as quadrupolar nuclei.
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The solid state NMR community has developed hundreds of techniques to investigate different parts of the NMR Hamiltonian. This section gives a brief, by no means comprehensive of a few techniques that are helpful in probing select interactions. Additional resources useful for understanding 2D NMR are available.
Magic Angle Spinning (MAS) NMR
For a spin 1/2 nucleus, the CSA and the dipolar coupling significantly broaden the solids lineshape, even in crystalline compounds. By inspecting the Hamiltonians of each interaction, the $3cos(\theta)-1$ term appears. If this term was set to zero the dipolar coupling would go to zero. The CSA would only be dependent on $\phi$ and $\eta$. Rapid rotation about this angle would average all $\phi$ leading to a purely isotropic lineshapes. This is the logic behind (MAS) NMR. In this experiment, the sample is physically rotated around the magic angle at very high spinning speeds, on the order of kHz. The magic angle can be solved for numerically by,
$3 \cos\theta-1=0$
$\theta=54.74 = \text{Magic Angle}$
The chemical shift can be related to the individual orientation of the crystal in the rotor to the PAS using Euler angles, $\alpha, \beta,\gamma$. This conversion gives a chemical shift of
$-\omega_0{\sigma_{iso}+[A_1 \cos(\omega_r t+\gamma)+B_1 \sin(\omega_r t+\gamma)]+[A_2\cos(2\omega_r t+2\gamma)+B_2\sin(\omega_r t+2\gamma]}$
where
\begin{align} A_1 &=\frac{2}{3} \sqrt{2} \sin \beta cos\beta [\cos^2\alpha (\sigma_{xx}^{PAS}- \sigma_{zz}^{PAF}) +\sin^2\alpha (\sigma_{yy}^{PAS}- \sigma_{zz}^{PAF})] \[4pt] B_1 &=\frac{2}{3} \sqrt{2} \sin \alpha \cos\alpha \sin\beta (\sigma_{xx}^{PAS} -\sigma_{zz}^{PAF}) \[4pt] A_2 &=\frac{1}{3} ((\cos^2 \beta \cos^2\alpha -\sin^2\alpha) (\sigma_{xx}^{PAS} -\sigma_{zz}^{PAF}) +(\cos^2\beta \sin^2\alpha- \cos^2\alpha) (\sigma_{yy}^{PAS} -\sigma_{zz}^{PAF})) \[4pt] B_2 &=-\frac{2}{3} \sin\alpha \cos\alpha \cos\beta (\sigma_{xx}^{PAS}- \sigma_{zz}^{PAF}) \end{align}
This expression is useful when considering incomplete averaging of the CSA by MAS. Incomplete averaging of the CSA occurs when the spinning frequency is less than the size of the CSA. The first term of this equation is the isotropic component and the remaining terms are all dependent on the spinning frequency $\omega_r$. When the CSA averaging is incomplete, these terms describe rotation echoes observed in the FID. Rotational echoes form as the crystallite moves with the rotor. Due to the rotation, the orientation of the crystallite changes which changes the shielding around the nucleus. Once the sample returns to its original starting position, the next rotation will have the crystallite sampling the same orientation for the same time, imposing a separate FID on the FID. The additional FIDs, denoted by the red arrows, are known as rotational echoes and are denoted as sharp spike in the time dimension, shown in the figure below. Rotational echoes are easy to identify as they appear in the FID as sharp spikes spaced every $\frac{1}{\omega_r}$. When the sample is spinning faster than the anisotropy, the shielding lags behind the spinning speed and the crystallite shielding will not experience the same set of values over the same time period prevention rotational echo formation.
Fourier transforming this results in characteristic spinning sidebands which appear at integer multiples of the spinning speed. Spinning sideband patterns, shown above, provide information about the CSA and are therefore helpful to have when performing structural studies. In this figure, $\delta=200ppm$ and $\eta=0.5$.However, if a compound has multiple sites all with large CSA values, the spectrum becomes complicated and the isotropic peaks cannot be identified a priori. To identify the isotropic positions, a Herzfeld Berger analysis of the sideband patters may be performed. In a Herzfeld Berger analysis, spectra are collected at different spinning speeds. The more spinning speeds at which the spectrum is collected, the easier the assignment of the peaks. The sidebands appear at integer multiples of the rotor speed, while the isotropic peaks remain constant. Thus to find the peaks you only need to locate those peaks that do not move when the spinning speed is changed. The relative intensities of the isotropic peaks will not change, while the sideband patterns will change intensities depending on the spinning speed. However, an isotropic peak may be overlapping with a sideband, giving artificial intensity to that peak.
As alluded to earlier, the CSA parameters may be derived from the sideband patterns. When doing this one must consider a few things:
1. The dipolar coupling. If the spinning speed is not large enough to average the dipolar coupling between nuclei this can give additional sidebands which when fit lead to inaccurate determination of CSA
2. At least 4 sidebands in order to have enough degrees of freedom to accurately fit the chemical shift tensor. The chemical shift tensor is 3x3 matrix. If the matrix is diagonalized, there are 6 remaining values; 3 isotropic and 3 anisotropic. The anisotropy may be fit using 3 values in which case you need 3+1 degrees of freedom to fit accurately.
Below is a typical pulse sequence used for MAS experiments. It is sometimes called pulse and acquire, as only a $\frac{\pi}{2}$ pulse is used and the spectrum is immediately acquired. The pulse doesn't need to be $\frac{\pi}{2}$. In fact, normally a pulse with a smaller tip angle is used in order to acquire more scans in a shorter time. The optimum tip angle at which the signal is maximized at the shortest time is called the Ernst angle.
Hahn Echo
The Hahn echo sequence is named after Erwin Hahn, who in the 1950's discovered that by applying a pulse with one phase and then another pulse with a different phase, the signal would refocus and form what is known as an echo. There are many different types of echoes including solids echoes, Solomon echoes, and Hahn echoes. The Hahn echo is particularly important for spin =1/2 nuclei. By applying a $\frac{\pi}{2}$ pulse waiting a set delay and then pulsing with a $\pi$ pulse results in an echo forming. The delay between the $\frac{\pi}{2}$ and the $\pi$ pulses will be the time at which the echo reaches its maximum value, as shown in the pulse sequence below. In this experiment the delay is an integer multiple of the rotor period, or rotor-synchronized. This synchronization is important in order to retain any information about the CSA. Obviously if the rotor is no longer rotor synchronized the CSA will refocus at a different time yielding different (fake) interactions. The Hahn echo is often used to shift the signal out in time to negate any pulse ringdown effects that may add artificial broadening to the spectrum.
To understand how an echo forms an analogy to runners on a track (shown below) is used. Initially, all of the runners (circles) are at the starting line. Each runner (different colored circle) represents one spin in the ensemble. After the race starts, the runners begin to race around the track, separating based on speed (arrows). This is equivalent to T2 relaxation in the NMR system. However, as theses runners begin to lose their coherence a whistle is blown and the runners all turn around and run back toward the starting line at EXACTLY the same speed as before. When they reach the starting line they are all in phase again. This is the echo that is seen in the NMR experiment. Thus the Hahn echo can be used to measure T2 processes.
Solids Echo
In quadrupolar nuclei, the refocusing of the spins into echoes becomes challenging as there are multiple energy levels and certain transitions are forbidden $\Delta m \neq \pm 1$. Consequently, treating the quadrupolar nuclei with the Hahn echo sequence results in incomplete refocusing (Some of the runners do not hear the whistle and turn around). This leads to different interactions which need to be discussed more thoroughly. However, if instead a $\frac{\pi}{2}$ pulse is used instead of the $\pi$ pulse, the central transitions will experience the spin flip and reform a solids (quadrupolar) echo. The main advantage of this type of echo is that it includes no broadening to do satellite transitions and is therefore more quantitative than a Hahn echo for quadrupolar nuclei.
Multiple Quantum MAS (MQMAS)
MQMAS is a high resolution 2D NMR technique that allows for an isotropic spectrum of a quadrupolar nucleus to be detected. Frydman and Harwood showed that by manipulating spin coherences using pulses, the second and fourth rank Legendre Polynomials can be averaged over the entire experiment. Separating the experiment into 2 time domains t1 and t2 and setting the Legendre polynomials to zero yields
$C_4(I,m_1)t_1+C_4(I, m_2)t_2=0$.
Experimentally, t1 can be chosen and the resulting echo will form at t2. However, m2 is fixed as the final transition needs to be 1/2 so the signal can be detected. Solving for t2 yields:
$t_2=kt_1$
$k=\frac{C_4(I,m_1)}{C_4(I,\frac{1}{2})}$
In order to complete this experiment, the MQ transition must first be excited. This can be accomplished using a very short $\frac{\pi}{12}$ pulse. After waiting t1, the spins are then reconverted into the central transition so they can be detected. The resultant echo consists of spins which have satisfied the averaging of the fourth rank polynomials and therefore only contain isotropic information. Next, t1 is varied and the resultant echo is measured. The signal then evolves on a diagonal (k) which is then sheared an FT in both dimensions giving a v1' correlated with the isotropic and anisotropic components.
in this figure
$\nu_1=\frac{34\nu_{iso}-60\nu_0}{9}$
and
$\nu_2=\nu_{iso}+3\nu_0-21\nu_4(\theta,\phi)$
Two Dimensional Phase Adjusted Spinning Sidebands (2D PASS)
The 2D PASS experiment is an experiment which separates the CSA from the isotropic chemical shift. The pulse sequence consists of a $\frac{\pi}{2}$ pulse followed by a train of 5 $\pi$ pulses which are spaced at timings given by the PASS equations. The PASS equations are complex and will not be discussed further. Timings for the pulses may be found in the paper published by Antzutkin et al. in 1995.
The goal of this experiment is to separate the CSA into a separate dimension from the isotropic chemical shift. The basic principle is that the CSA may be separated out by changing the timings between the pi pulses. The change in the timings allows for the isotropic part of the chemical shift to evolve as a function of the timings and the CSA is directly acquired in each 1D slice. The two are then related to one another by a shearing transformation. The data acquisition and evolution of the CSA and isotropic chemical shift are shown in the figure below.
Dynamic Angle Spinning (DAS)
Unlike MQMAS which manipulated the spin of the nuclei, DAS manipulates the orientation of the spins in the rotor. As can be seen in the quadrupolar equations, no one angle can allow for complete averaging the of the quadrupolar interactions. However, solving these equations individually leads to sets of angles (left as an exercise for the reader).
Several techniques have been developed in order to exploit the different angles.
Double Rotation: This technique places a smaller rotor oriented at a different orientation inside a rotor which spins at the magic angle. The rotor inside also spins, which averages 2 of the above interactions. This is an extremely challenging experiment and the inner rotor is only able to spin at a max of 3 kHz.
Magic Angle Flipping: In this technique, the rotor is physically reoriented by a stepper motor during the experiment. Commonly, the experiment is performed at the magic angle and then is reoriented to 90 degrees to reintroduce the anisotropy. In this way the anisotropy can be correlated to the isotropic chemical shift, much like the 2D PASS experiment. However, when using a quadrupolar nucleus, the sample can be reoriented to another "magic" angle which will average the remainder of the quadrupolar interaction. In order for a successful experiment $\tau$, the time it takes to switch the rotor must be shorter than T1 or no signal will be observed.
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Performing a MAS Experiment
This section is devoted to giving a practical guide for solid state NMR. Please follow all safety guidelines at your University. All experiments are performed on a 500MHz magnet using a Bruker AQS spectrometer running TopSpin 2.1. Please note that your spectrometer may be different and/or run different software in which some/all of the commands listed below will be different.
The Probe
The probe has several different parts which are summarized below. The probe must house the sample, spin the sample, tune to the correct frequency, and allow for temperature dependence studies.
1. Probe Casing: The probe case shields the delicate components inside the probe. It is typically made out of aluminum and attaches to the base of the probe via a series of screws.
2. Variable Temperature (VT) line and Bearing Airline: The VT line is a glass rod that connects to the VT apparatus coming out the the liquid nitrogen dewar (see section x.x). It transfers the cold nitrogen gas to the sample
3. Drive Airline
4. X Channel
5. 1H Channel
6. X Channel Tuning Rod
7. X Channel Matching Rod
8. Magic Angle Adjustment Rod
9. Tachometer
10. Stator (Coil contained inside)
11. Variable capacitors for matching and tuning
Packing The Sample
It becomes very important in SSNMR to pack your sample very carefully. The sample is rotating several thousands of times per second and therefore must not wobble or suddenly stop. If the sample stops rotating in the stador, it will crash and this normally breaks the stater results in a several thousand dollar repair. In order to prevent this here are some steps to pack the sample.
1. Crush the sample thoroughly. Use an mortar and pestle. crush your sample into a fine powder by constantly grinding the sample for about 10 minutes. Not only does this help with the packing efficiency allowing for more sample to be added, but it also homogenizes the crystallite orientations.
2. Place the rotor in the packing tool.
3. Add a small amount of sample to the rotor. Next, using the packing stick, press the sample down into the bottom of the rotor. I normally make about 10-15 quick downward pushes to allow the sample to find its closest packing arrangement, then I press it hard into the bottom using the tool about 5 times.
4. Repeat Step 3 until the rotor is almost full
5. When your about to the top, you need to leave a little room to put in the cap. This is primarily the part when people tend to mess up packing. There can be no loose powder at the top. Therefore you must carefully push on the cap without shaking any loose powder at the top. If you're not interested in 13C or 1H NMR placing some Parafilm on top of the sample can prevent this from happening.
6. Place the cap on the rotor. Be careful not to put it on at an angle and the it fully closes. If there is a space between the rotor and the cap, you have too much sample and you need to unpack some of it.
7. Mark the bottom of the rotor with a black sharpie so the tachometer can read the spinning speed.
Spinning The Sample
To spin the sample, you must first insert the sample into the stador. Most probes are equipped with transfer lines that use compressed air to insert and eject samples. Once the sample is inserted, you may begin spinning the sample. Most spectrometers have a automatic program that will allow you to spin the sample by simply selecting the spinning speed you want. However, you may need to do it manually, wither on the console or by opening the air valves. When doing this it is important to have the bearing air at least 150 mbar higher than the drive pressue. This keeps the sample elevated and from crashing into the sides of the rotor. When stopping the rotor, you must slowly decrease the spinning speed or else you risk crashing the rotor.
Setting Up The Spectrometer
Once the sample is spinning, it is now important to make sure all of the hardware is set. The first thing to do is select the probe that you are using by typing "edhead" this brings up a list of available probes. Select the one you have already inseted your sample into. After selecting this, the screen will show the current hardware set up, along with the nucleus you will be probing. If this is not correct, select the nucleus you are interested in. You will also be able to see the frequency at which this nucleus is resonant at.
Next, make sure the band pass filter is selected for the RF range you are interested in. If you do not have the correct filter, then all the signal will be locked and your experiment will not collect any data.
Next read in the shim file. The shims are used to homogenize the field. type rsh to read in the shim profile for your probe.
Tuning The Probe
The probe must be tuned to the correct frequency range in order to probe the transitions. Additionally, the RF circuit must match impedance with the spectrometer, which is known as matching. To tune the probe here are the steps:
1. Type "wobb" on the command line.
2. Rotate the tuning rod clockwise if the tuning line needs increase in frequency and counter clockwise to decrease the frequency.
3. Use the matching rod to lower the line as low as it goes.
4. Once optimized, type "stop"
Note: you may need to increase sweep width on the wobble curve to see the line. You may change this under acquisition parameters.
Running the Experiment
Running the experiment is actually fairly easy, however there are several parameters which need to be addressed.
• Experiment: You can choose which experiment to run. Press the "..." next to the spectrometer and select "zg" for a simple pulse and acquire experiment.
• TD: This is the number of points your FID will consist of. Each data point will be acquired after a delay of DW
• NS: The is the number of scans which is to be completed for the experiment. It must be an integer number of the phase cycle. For MAS this is 8.
• SW: This is the sweep width in Hz of your spectrum. This must be large enough for your experimental spectrum to be seen. It is equal to 1/DW
• AQ: This is the acquisition time which is automatically set by TD*DW. Typically 1ms is long enough
• DW: The is the dwell time. The dwell time is the amount of time waited between individual points being sampled. A typical value for this is 5us.
• DS: This is the dummy scans which the experiment is performed but no data is acquired. The dummy scans are first completed then NS scans are completed.
• RG: This is the receiver gain. The receiver gain amplifies the signal. This is sample dependent. If the FID looks clipped i.e. looks like the tops of the peaks are being cut off in the time dimension, then the RG needs to be lowered.
• P1: This is the pulse length. You can calibrate this using "popt" command. The popt command will bring you to a separate window in which you define the parameter you would like to optimize. Type p1. Next, type in the minimum pulse length you want to sample and the interval you want to increment with. Press start and the maximum peak is the 90 degree pulse.
• PL1: The power level at which the pulse, P1, is delivered. This is in decibels so smaller numbers are high power.
• SFO1: This is the carrier frequency which is set by the nucleus of interest.
• D1: This is the amount of tiume between each scan. It should be approximately 5*T1 for good spectra to be acquired.
After there parameters are set, type "zg" to begin the experiment. It will complete NS scans and the time to complete it will be approximately NS*D1.
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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.
Background t
This page describes what a C-13 NMR spectrum is and how it tells you useful things about the carbon atoms in organic molecules.
The background to C-13 NMR spectroscopy
Nuclear magnetic resonance is concerned with the magnetic properties of certain nuclei. On this page we are focussing on the magnetic behavior of carbon-13 nuclei.
Carbon-13 nuclei as little magnets
About 1% of all carbon atoms are the C-13 isotope; the rest (apart from tiny amounts of the radioactive C-14) is C-12. C-13 NMR relies on the magnetic properties of the C-13 nuclei. Carbon-13 nuclei fall into a class known as "spin ½" nuclei for reasons which don't really need to concern us at the introductory level this page is aimed at (UK A level and its equivalents). The effect of this is that a C-13 nucleus can behave as a little magnet. C-12 nuclei don't have this property.
If you have a compass needle, it normally lines up with the Earth's magnetic field with the north-seeking end pointing north. Provided it isn't sealed in some sort of container, you could twist the needle around with your fingers so that it pointed south - lining it up opposed to the Earth's magnetic field. It is very unstable opposed to the Earth's field, and as soon as you let it go again, it will flip back to its more stable state.
Because a C-13 nucleus behaves like a little magnet, it means that it can also be aligned with an external magnetic field or opposed to it.
Again, the alignment where it is opposed to the field is less stable (at a higher energy). It is possible to make it flip from the more stable alignment to the less stable one by supplying exactly the right amount of energy.
The energy needed to make this flip depends on the strength of the external magnetic field used, but is usually in the range of energies found in radio waves - at frequencies of about 25 - 100 MHz. (BBC Radio 4 is found between 92 - 95 MHz!) If you have also looked at proton-NMR, the frequency is about a quarter of that used to flip a hydrogen nucleus for a given magnetic field strength.
It's possible to detect this interaction between the radio waves of just the right frequency and the carbon-13 nucleus as it flips from one orientation to the other as a peak on a graph. This flipping of the carbon-13 nucleus from one magnetic alignment to the other by the radio waves is known as the resonance condition.
The importance of the carbon's environment
What we've said so far would apply to an isolated carbon-13 nucleus, but real carbon atoms in real bonds have other things around them - especially electrons. The effect of the electrons is to cut down the size of the external magnetic field felt by the carbon-13 nucleus.
Suppose you were using a radio frequency of 25 MHz, and you adjusted the size of the magnetic field so that an isolated carbon-13 atom was in the resonance condition.
If you replaced the isolated carbon with the more realistic case of it being surrounded by bonding electrons, it wouldn't be feeling the full effect of the external field any more and so would stop resonating (flipping from one magnetic alignment to the other). The resonance condition depends on having exactly the right combination of external magnetic field and radio frequency.
How would you bring it back into the resonance condition again? You would have to increase the external magnetic field slightly to compensate for the shielding effect of the electrons.
Now suppose that you attached the carbon to something more electronegative. The electrons in the bond would be further away from the carbon nucleus, and so would have less of a lowering effect on the magnetic field around the carbon nucleus.
The external magnetic field needed to bring the carbon into resonance will be smaller if it is attached to a more electronegative element, because the C-13 nucleus feels more of the field. Even small differences in the electronegativities of the attached atoms will make a difference to the magnetic field needed to achieve resonance.
Summary
For a given radio frequency (say, 25 MHz) each carbon-13 atom will need a slightly different magnetic field applied to it to bring it into the resonance condition depending on what exactly it is attached to - in other words the magnetic field needed is a useful guide to the carbon atom's environment in the molecule.
Features of a C-13 NMR spectrum
The C-13 NMR spectrum for ethanol
This is a simple example of a C-13 NMR spectrum. Don't worry about the scale for now - we'll look at that in a minute.
There are two peaks because there are two different environments for the carbons. The carbon in the CH3 group is attached to 3 hydrogens and a carbon. The carbon in the CH2 group is attached to 2 hydrogens, a carbon and an oxygen. The two lines are in different places in the NMR spectrum because they need different external magnetic fields to bring them in to resonance at a particular radio frequency.
The C-13 NMR spectrum for a more complicated compound
This is the C-13 NMR spectrum for 1-methylethyl propanoate (also known as isopropyl propanoate or isopropyl propionate).
This time there are 5 lines in the spectrum. That means that there must be 5 different environments for the carbon atoms in the compound. Is that reasonable from the structure?
Well - if you count the carbon atoms, there are 6 of them. So why only 5 lines? In this case, two of the carbons are in exactly the same environment. They are attached to exactly the same things. Look at the two CH3 groups on the right-hand side of the molecule.
You might reasonably ask why the carbon in the CH3 on the left isn't also in the same environment. Just like the ones on the right, the carbon is attached to 3 hydrogens and another carbon. But the similarity isn't exact - you have to chase the similarity along the rest of the molecule as well to be sure.
The carbon in the left-hand CH3 group is attached to a carbon atom which in turn is attached to a carbon with two oxygens on it - and so on down the molecule.
That's not exactly the same environment as the carbons in the right-hand CH3 groups. They are attached to a carbon which is attached to a single oxygen - and so on down the molecule.
We'll look at this spectrum again in detail on the next page - and look at some more similar examples as well. This all gets easier the more examples you look at.
For now, all you need to realise is that each line in a C-13 NMR spectrum recognises a carbon atom in one particular environment in the compound. If two (or more) carbon atoms in a compound have exactly the same environment, they will be represented by a single line.
The need for a standard for comparison - TMS
Before we can explain what the horizontal scale means, we need to explain the fact that it has a zero point - at the right-hand end of the scale. The zero is where you would find a peak due to the carbon-13 atoms in tetramethylsilane - usually called TMS. Everything else is compared with this.
You will find that some NMR spectra show the peak due to TMS (at zero), and others leave it out. Essentially, if you have to analyse a spectrum which has a peak at zero, you can ignore it because that's the TMS peak.
TMS is chosen as the standard for several reasons. The most important are:
• It has 4 carbon atoms all of which are in exactly the same environment. They are joined to exactly the same things in exactly the same way. That produces a single peak, but it's also a strong peak (because there are lots of carbon atoms all doing the same thing).
• The net effect of this is that TMS produces a peak on the spectrum at the extreme right-hand side. Almost everything else produces peaks to the left of it.
The chemical shift
The horizontal scale is shown as (ppm). is called the chemical shift and is measured in parts per million - ppm. A peak at a chemical shift of, say, 60 means that the carbon atoms which caused that peak need a magnetic field 60 millionths less than the field needed by TMS to produce resonance. A peak at a chemical shift of 60 is said to be downfield of TMS. The further to the left a peak is, the more downfield it is.
Solvents for NMR spectroscopy
NMR spectra are usually measured using solutions of the substance being investigated. A commonly used solvent is CDCl3. This is a trichloromethane (chloroform) molecule in which the hydrogen has been replaced by its isotope, deuterium.
CDCl3 is also commonly used as the solvent in proton-NMR because it doesn't have any ordinary hydrogen nuclei (protons) which would give a line in a proton-NMR spectrum. It does, of course, have a carbon atom - so why doesn't it give a potentially confusing line in a C-13 NMR spectrum? In fact it does give a line, but the line has an easily recognizable chemical shift and so can be removed from the final spectrum. All of the spectra from the SDBS have this line removed to avoid any confusion.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
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There are a many ways we can use NMR spectroscopy to analyse compounds. One common application is in determination of an unknown structure. Given the MS, IR, 13C and 1H NMR spectra, what might be the structure of an unknown sample?
It is often easiest to start with the IR spectrum.
• identify at least three peaks in the IR spectrum. Which peaks seem to tell you the most information about this compound?
• don't think with your head; think with your hands. Write down ideas on the spectrum.
• if you are working on a formal proof of structure, on a class test or a lab report, you may be required to enter your data in a table correlating wavenumber with peak assignment:
cm-1
asst
For example, a student might obtain the following IR spectrum.
From that information, she constructs the following table. She might even write this table, by hand, directly on her spectrum. She makes useful notes on the edges, and might even include some guesses, which she later crosses out, but does not erase. She is assisted in this task by consulting an IR table, that suggests what some of these peaks might mean.
Remember:
• make special note of what atoms are present in the compound: C, H, N, O...
• also note your initial ideas about specific functional groups that may be present.
• if you are unsure of an assignment, put a question mark beside it to signal this uncertainty.
• some data may need to be discarded later if it is not consistent with other data.
Look at the 13C spectrum.
• How many different carbons are there, based on the number of peaks in the spectrum? This is the first step in estimating the molecular formula.
• Do you have reason to believe there is symmetry in the structure? In the entire compound or just part of it? Adjust the number of carbons you think you are dealing with.
• As in IR spectroscopy, begin assigning peaks, either on the spectrum or, if required, in a table:
ppm
asst
For example, a student might obtain the following 13C NMR spectrum:
From that information, she puts together the following table:
Remember:
• you will be able to assign all peaks in the NMR spectrum, not just a few like in IR.
You can get more information on the formula from the 1H NMR spectrum and the mass spectrum.
• In the 1H NMR spectrum, what does the sum of the integrations suggest about the number of hydrogens?
• At this point you may have a formula, CxHyOz. What would be the mass of a compound with this formula?
• Compare this mass to the mass spectrum. Does it match?
• If not, consider whether a common atom (such as oxygen) is missing from your formula, or if there might be symmetry that you have missed.
• Does the mass spectrum suggest the presence of a nitrogen (odd molecular weight) or chlorine or bromine (isotope pattern)?
Once you have the molecular formula, the number of possible structures is automatically limited. The number of units of unsaturation can help you narrow down the possibilities.
• The ratio of C : H in a saturated, acyclic hydrocarbon is n : 2n+2.
• Each pair of H missing from the formula corresponds to a multiple bond or a ring.
• The presence of oxygen (which is divalent) does not alter the C : H ratio.
• The presence of halogen (which is monovalent) means there is one H missing.
• The presence of a nitrogen (which is trivalent) means there is an extra H in the formula.
• In other words, C4H8O has one ring or double bond just like C4H8, and so do C4H7Br and C4H9N.
As in 13C NMR, you should be able to assign all peaks in the 1H NMR spectrum. You may be able to do so by making notes on the spectrum. If you think you know the structure, you may be able to draw it and note which peak belongs with which proton.
A formal proof of structure might require a table of assignments.
ppm
int
mult
partial structure
assignment
• This table demonstrates your ability to read the spectrum. Can you decide what ratio of protons is suggested by the integral line? Can you decide whether a peak is a quartet?
• The partial structure column should explain the shift, integration and multiplicity for the peak in that row. It should not show any other information from elsewhere in the structure. This restriction forces you to demonstrate a thorough understanding of the data in a way that "getting the right answer" does not.
• The partial structure column is best filled in with drawings, not words. The drawing is a partial structure.
• Because the partial structure will show the protons absorbing at the shift in that row as well the neighbouring protons, you need to distinguish between them in your picture. Most people circle or underline or make bold the protons that show up at the shift given in that row.
• When finished with the partial structure column, you should be able to link the partial structures together to make an entire structure in the assignment column.
An example of a spectrum and its accompanying data table is given below. Here is the spectrum:
Here is a data table:
Things to note:
• This student has used two integration columns instead of just one.
• The first column shows the integral measured from the spectrum. She probably used a ruler.
• The second column, which she called int(n), contains a convenient ratio taken from the raw data. This ratio is easier to use in her assignments.
• Also note that the peak at 9.7 ppm does not have a very good integral. There is either a "phasing" or a "level & tilt" problem here that can be corrected using the NMR software, but this is sometimes difficult to do. If she had taken an automatic printout of this integral measurement, she would have gotten a strange number; in this case, it would be about -5, because the end of the integral line is lower than the start. It clearly isn't a negative number of hydrogens, though. She has instead measured the vertical rise in the integral and recorded that; it isn't perfect, but is a fair estimate in this case.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Structural_Assignment/Determine_St.txt
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This page describes how you interpret simple high resolution nuclear magnetic resonance (NMR) spectra. It assumes that you have already read the background page on NMR so that you understand what an NMR spectrum looks like and the use of the term "chemical shift". It also assumes that you know how to interpret simple low resolution spectra.
The difference between high and low resolution spectra
What a low resolution NMR spectrum tells you
• The number of peaks tells you the number of different environments the hydrogen atoms are in.
• The ratio of the areas under the peaks tells you the ratio of the numbers of hydrogen atoms in each of these environments.
• The chemical shifts give you important information about the sort of environment the hydrogen atoms are in.
High resolution NMR spectra
In a high resolution spectrum, you find that many of what looked like single peaks in the low resolution spectrum are split into clusters of peaks.
1 peak a singlet
2 peaks in the cluster a doublet
3 peaks in the cluster a triplet
4 peaks in the cluster a quartet
You can get exactly the same information from a high resolution spectrum as from a low resolution one - you simply treat each cluster of peaks as if it were a single one in a low resolution spectrum. But in addition, the amount of splitting of the peaks gives you important extra information.
Interpreting a high resolution spectrum
The n+1 rule
The amount of splitting tells you about the number of hydrogens attached to the carbon atom or atoms next door to the one you are currently interested in. The number of sub-peaks in a cluster is one more than the number of hydrogens attached to the next door carbon(s). So - on the assumption that there is only one carbon atom with hydrogens on next door to the carbon we're interested in.
singlet next door to carbon with no hydrogens attached
doublet next door to a CH group
triplet next door to a CH2 group
quartet next door to a CH3 group
Using the n+1 rule
What information can you get from this NMR spectrum?
Assume that you know that the compound above has the molecular formula C4H8O2.
Treating this as a low resolution spectrum to start with, there are three clusters of peaks and so three different environments for the hydrogens. The hydrogens in those three environments are in the ratio 2:3:3. Since there are 8 hydrogens altogether, this represents a CH2 group and two CH3 groups. What about the splitting?
• The CH2 group at about 4.1 ppm is a quartet. That tells you that it is next door to a carbon with three hydrogens attached - a CH3 group.
• The CH3 group at about 1.3 ppm is a triplet. That must be next door to a CH2 group. This combination of these two clusters of peaks - one a quartet and the other a triplet - is typical of an ethyl group, CH3CH2. It is very common.
• Finally, the CH3 group at about 2.0 ppm is a singlet. That means that the carbon next door doesn't have any hydrogens attached.
So what is this compound? You would also use chemical shift data to help to identify the environment each group was in, and eventually you would come up with:
Alcohols
Where is the -O-H peak? This is very confusing! Different sources quote totally different chemical shifts for the hydrogen atom in the -OH group in alcohols - often inconsistently. For example:
• The Nuffield Data Book quotes 2.0 - 4.0, but the Nuffield text book shows a peak at about 5.4.
• The OCR Data Sheet for use in their exams quotes 3.5 - 5.5.
• A reliable degree level organic chemistry text book quotes1.0 - 5.0, but then shows an NMR spectrum for ethanol with a peak at about 6.1.
• The SDBS database (used throughout this site) gives the -OH peak in ethanol at about 2.6.
The problem seems to be that the position of the -OH peak varies dramatically depending on the conditions - for example, what solvent is used, the concentration, and the purity of the alcohol - especially on whether or not it is totally dry.
A clever way of picking out the -OH peak
If you measure an NMR spectrum for an alcohol like ethanol, and then add a few drops of deuterium oxide, D2O, to the solution, allow it to settle and then re-measure the spectrum, the -OH peak disappears! By comparing the two spectra, you can tell immediately which peak was due to the -OH group.
The reason for the loss of the peak lies in the interaction between the deuterium oxide and the alcohol. All alcohols, such as ethanol, are very, very slightly acidic. The hydrogen on the -OH group transfers to one of the lone pairs on the oxygen of the water molecule. The fact that here we've got "heavy water" makes no difference to that.
The negative ion formed is most likely to bump into a simple deuterium oxide molecule to regenerate the alcohol - except that now the -OH group has turned into an -OD group.
Deuterium atoms don't produce peaks in the same region of an NMR spectrum as ordinary hydrogen atoms, and so the peak disappears.
You might wonder what happens to the positive ion in the first equation and the OD- in the second one. These get lost into the normal equilibrium which exists wherever you have water molecules - heavy or otherwise.
The lack of splitting with -OH groups
Unless the alcohol is absolutely free of any water, the hydrogen on the -OH group and any hydrogens on the next door carbon don't interact to produce any splitting. The -OH peak is a singlet and you don't have to worry about its effect on the next door hydrogens.
The left-hand cluster of peaks is due to the CH2 group. It is a quartet because of the 3 hydrogens on the next door CH3 group. You can ignore the effect of the -OH hydrogen. Similarly, the -OH peak in the middle of the spectrum is a singlet. It hasn't turned into a triplet because of the influence of the CH2 group.
Equivalent hydrogen atoms
Hydrogen atoms attached to the same carbon atom are said to be equivalent. Equivalent hydrogen atoms have no effect on each other - so that one hydrogen atom in a CH2 group doesn't cause any splitting in the spectrum of the other one.
But hydrogen atoms on neighboring carbon atoms can also be equivalent if they are in exactly the same environment. For example:
These four hydrogens are all exactly equivalent. You would get a single peak with no splitting at all. You only have to change the molecule very slightly for this no longer to be true.
Because the molecule now contains different atoms at each end, the hydrogens are no longer all in the same environment. This compound would give two separate peaks on a low resolution NMR spectrum. The high resolution spectrum would show that both peaks subdivided into triplets - because each is next door to a differently placed CH2 group.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Structural_Assignment/High_Resolut.txt
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