Download Electronic structure and spectroscopy

Electronic structure and spectroscopy
Péter G. Szalay
Institute of Chemistry, Eötvös Loránd University, Budapest
2014. április 23.
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1.
Today’s concept of the atom
Shortly after Dalton’s atomic theory became widely accepted (matter consists of undividable particles called atoms) experiments by physicists indicated that the atoms indeed
have some structure.
1.1.
Discovery of electron
Joseph John Thomson (1856-1940)
His famous experiment with the cathode ray tube (1897):
In a cathode ray tube, independently of the material of the cathode, the same event
can be observed: light spots (flashes) appear on the screen, i.e. a particles leave the
cathode and fly against the plate. In an electric field, this particle deviates towards the
positive pole, i.e. it must have a negative charge. Its mass (measured by magnitude of
the deviation) is much smaller than the mass of the atom (in fact Thomson meassured
the ratio of mass and charge). He called it a „Corpuscle”, the name „electron” was later
introduced by G. Johnstone Stoney (1853-1928), who worked with electricity and called
the unit of electricity as electron.
Thomson was also a bit skeptical: "Could anything at first sight seem more impractical
than a body which is so small that its mass is an insignificant fraction of the mass of an
atom of hydrogen?" (Thomson)
He got the Nobel prize in 1906.
But how the atom should look like? There is a positive charge with large mass and
a tiny electron bearing the negative charge. It can be like a plum pudding; electrons are
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the plums, and the pudding is the distributed positive charge.
1.2.
Discovery of the nucleus
Ernest Rutherford (1871-1937)
Experiment: alpha particles originating from nuclear decay were used to bomb gold foil
(Geiger és Mardsen):
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The atom must have a nucleus which is much smaller than the atom and it is bearing
the positive charge.
„It is like to fire with a cannon on a sheet of paper and it would bounce back”
Rutherford’s atomic model (1911): an atom consists of a positively charged small nucleus
bearing almost all mass of it; the electrons are orbiting around it like the moon around
the Earth.
There are lots of unanswered question, though. For example, why the electron does not
fall into the nucleus? (This system is different from the Earth-Moon system, since it has
charges!!!)
Today we know that even the nucleus has a structure: it consists of protons and neutrons,
and even these can be divided into smaller elementary particles. For chemistry this is,
however, not relevant, chemistry in most part can be described by a nucleus „surrounded”
by the electrons. This is, on the other hand, already very relevant for chemical analysis!
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2.
Development of quantum mechanical view
The atomic theory allowed the development of modern chemistry, but lots of questions
remained unanswered, and in particular the WHY is not being explained:
• What is the binding force between atoms. It is not the charge since atoms are
neutral. Why can even two atoms of the same kind (like H-H) form a bond?
• Why atoms can form molecules only with certain rates?
• What is the reason of the periodic table of Mendeleev?
At the turning of the 19th and 20st century new experiments appeared which could
not be explained by the tools of the classical (Newtonian) mechanics. For the new theory
new concepts were needed:
• quantization: the energy can not have arbitrary value
• particle-wave dualism
⇒ development of QUANTUM MECHANICS
The new theory was developed along a long root (which in time was not that long at
all!). We will follow this route now and stop at the most important steps.
2.1.
Introduction: same basic terms related to light
In the strict sense, „light” is a narrow range of electromagnetic radiation, what we can
sense with our eyes. In physics very often the term „light” is used for the entire spectrum.
The electromagnetic radiation consists of oscillating magnetic and electric fileds wich are
perpedicular to each other and the direction of its propagation.
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Basic terms:
• ν: frequency of the oscillation [1/s]
• ν ∗ : wavenumber [1/m]
• λ: wave length [m]
• c: speed of light
• polarization: there is oscillation only in a plane
Important relations:
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Magspingerjesztés
Ranges of the electromagnetic radiation:
Ionizáció
1
λ
Molekularezgések
gerjesztése
ν∗ =
Molekulákforgásának
gerjesztése
c
ν
Elektrongerjesztés
Maggerjesztések
λ=
What is spectroscopy?
The matter can absorbe or emit light. The absorbed/emitted light can be divided into
its components, and these will be characteristic for the matter.
The light, thus, can be divided into its components, for example by a prism.
2.2.
2.2.1.
Observations leading to quantum mechanics
Black Body Radiation
Planck in 1900 came up with a new, unusual explanation: according to his theory,
the energy of the radiation is quantized, it can only be hν, 2hν, 3hν ..., thus it does not
change continuously. Here h is the so called Planck constant: h = 6.626 · 10−34 Js
(Planck himself did not like his own theory, since it required an assumption (postulate),
i.e. the existence of constant h; he wanted to derive this from the existing theory. He
was not successful with this; now we know it is not possible to derive since it follows from
a new theory. Thus, despite of his genius discovery, he could not participate in further
development of quantum mechanics. )
2.2.2.
Heat capacity
According to the Dulong-Petit rule, the heat capacity is given by cv,m ≈ 3R, i.e. it is
independent of temperature. This is valid at temperatures people could investigate until
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the 19th century, but then it turned out that at low temperatures it goes to zero:
Einstein explained this using Planck’s idea: the matter is also quantized, the oscillators
(vibrations) can not have any energy, like the oscillators causing the black body radiation.
(The final form of the theory with several oscillators was derived by Debye.)
2.2.3.
Photoelectric effect
Shining light on the metal plate can result in electric current in the circuit. However,
there is a threshold frequency, below this there is no current, irrespective of the intensity
of the light, i.e.
• below the threshold frequency, no electron leaves the metal plate
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• increasing the intensity of the light, the energy of the emitted electron does not
change, only their number grows.
According to the measurements, the following relation exists between the kinetic
energy of the electron (Te ) and the frequency of the light (ν):
Tel = hν − A
where A depends on the quality of the metal plate (called work function).
Explanation was given again by Einstein using the quantization introduced by Planck:
the light consist of tiny particles which can have energy of hν only. (Note that Planck
opposed the use of his „uncompleted” theory!!)
2.2.4.
The Compton effect
A photon collides with a resting electron, it looses energy. Therefore its frequency also
changes. The photon acted as a particle in this experiment!! Note a wave scattering on
an object would not change its wave length or frequency!!!
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2.2.5.
Scattering of electron beam
Experiment by Davisson and Germer (1927), as well as G.P. Thomson (1928). There are
interference circles on the photographic plate, just like in case of X-ray radiation → electron acted as a wave.
2.2.6.
The hydrogen atom
Hydrogen atom has four lines in his emission spectra in the visible range (experiment by
Ångsröm):
The position of the lines have been described by Balmer (so called Balmer formula):
1
1
1
= R 2− 2
λ
2
n
10
n = 3, 4, 5, 6
(1)
where R is the so called Rydberg constant, λ is the wave length.
The energy of the hydrogen atom must be quantized, too!!
Explanation by Bohr: in his atomic model, the electron must fullfil some „quantum”
relations:
• in case of orbits having certain radius, the electron do not dissipate energy; these
are the so called stationary states;
• if the electron jumps from one orbit to the other, it emits (or absorbes) energy in
form of electromagnetic field („light”).
• the possible values for the energy are:
E = −
1 e2
2n2 a0
n is real number
(2)
(e is the charge of the electron, a0 unit length (1 bohr)).
Gives the Balmer formula back. HOMEWORK: SHOW THAT THIS IS TRUE.
However, can not be applied for the He or any other atom!!!
2.2.7.
Summary
Event
New term
black body radiation
energy quantized (hν)
photoelectric effect
energy of the light is quantized
heat capacity at low temperature
matter is quantized
goes to zero
Compton effect
electromagnetic radiation
acts like a particle
scattering of electron
electron acts like a wave
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Discoverer
Planck (1900)
Einstein (1905)
Einstein (1905),
Debye
Compton (1923)
Davisson (1927),
G.P. Thomson (1928)
Remember:
• ν is the frequency of light
• λ is the wave length of light (λ = νc )
• c speed of light
• h = 6.626 10−34 Js a Planck constant
• h̄ =
h
2π
Important consequence of all these: particle-wave dualism (dual nature of the
matter)
The existing theories need to be revised completely! Although Bohr could „fix” this old
theory with quantum condition to describe the hydrogen atom, but the theory does not
work for other atoms!
New theory:
• Heisenberg (1925): Matrix mechanics
• Schrödinger (1926): Wave mechanics
It turned out later that the two theories are equivalent, they use only a slightly
different mathematics. Now we call this theory as (non-relativistic) quantum
mechanics.
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2.3.
2.3.1.
Basic concepts of quantum mechanics
Operators
What is an operator? It acts on a function and produces an other function:
Âf (x) = g(x)
There are special functions called eigenfunctions of an operator: if the operator acts
on its eigenfunction, it results in a constant times the same function:
Âf (x) = af (x)
where a is a constant, the eigenvalue of the operator.
Demonstration: Assume that
d2
 =
dx2
Then cos(x) is an eigenfunction of this operator since:
 cos(x) =
d2 cos(x)
= − cos(x)
dx2
its eigenvalue being −1.
2.3.2.
Schrödinger-equation
The stationary states of a system (e.g. atom, molecule) can be obtained by solving the
so called (time independent) Schrödinger-equation:
ĤΨ = EΨ
(3)
with:
• Ĥ being the Hamilton operator of the system;
• Ψ is the state function of the system;
• E is the energy of the system.
This is an eigenvalue equation, Ψ being the eigenfunction of Ĥ, E is the eigenvalue. This
has to be solved in order to obtain the states of, e.g. molecules.
According to Dirac (1929) the whole chemistry is included in this equation:
P. A. M. Dirac, "Quantum Mechanics of Many-Electron Systems", Proceedings of the Royal
Society of London, Series A, Vol. CXXIII (123), April 1929, pp 714.:
„The general theory of quantum mechanics is now almost complete ... The underlying physical
laws necessary for the mathematical theory of a large part of physics and the whole of chemistry
are thus completely known, and the difficulty is only that the exact application of these equations
leads to equations much too complicated to be soluble. It therefore becomes desireable that
approximate practical methods of applying quantum mechanics should be developed, which
can lead to an explanation of the main features of complex atomic systems without too much
computation."
My own interpretation (2013) :
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• to describe molecules one need quantum mechanics;
• we need to develop methods which can give more and more accurate solutions to
the Schrödinger equation
• we also need approximate methods which support chemical intuition without expensive calculations.
In practice: one has to approximate Ψ. Chemists are very good at this!
2.3.3.
The Hamilton operator
The Hamilton operator of the system (Ĥ) consists of the sum of the kinetic (T̂ ) and the
potential energy (V̂ ) operators.
Ĥ = T̂ + V̂
(4)
the form of T̂ is the same for all systems, while the potential energy represents the molecule
by including the interactions between the electrons and nuclei.
2.3.4.
State function
In quantum mechanics the state of the system is represented by the wave function (or
state function) which depends on the coordinates of the particles:
Ψ = Ψ(x, y, z) = Ψ(r)
(5)
Ψ = Ψ(x1 , y1 , z1 , x2 , y2 , z2 , ..., xn , yn , zn ) = Ψ(r1 , r2 , ..., rn )
(6)
or in case of n particles:
The wave function has no physical meaning, but its square, the so called probability
density can be given a probability interpretation:
Ψ∗ (x0 , y 0 , z 0 ) · Ψ(x0 , y 0 , z 0 )dx dy dz
(7)
is the probability of finding a particle at point (x0 , y 0 , z 0 ) (more precisely in the infinitesimal proximity).
Shorter notation: Ψ∗ Ψdv or |Ψ|2 dv
We have to chose the wave function normalized, otherwise the valószínűség of finding
the particle in the whole space would not be one:
Z Z Z
Ψ∗ · Ψ dx dy dz = 1
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(8)
2.3.5.
Other physical quantities
Other physical properties are also described by operators. Most important operators are:
• position operator: x̂ (x̂f (x) = xf (x))
∂
• momentum operator: p̂x = −ih̄ ∂x
• kinetic energy operator: T̂ =
1 2
p̂
2m
2
2
h̄ d
1
= − 2m
≡ − 2m
∆
dx2
• angular momentum operator: ˆl = (ˆlx , ˆly , ˆlz )
• ...
Operators of position (x̂) and momentum (p̂x ) do not commute, meaning that
[x̂, p̂x ] = x̂p̂x − p̂x x̂ = −ih̄
(9)
which means that the two operators can not be interchanged (the results depends on,
which operator acts first).
Consequence: the two quantities (coordinate and momentum) can be measured in the
same time, the product of their uncertainty (∆x and ∆px must be larger the a given value:
1
h̄
2
This is the famous Heisenberg uncertainty principle.
∆x · ∆px ≥
2.3.6.
(10)
„The particle in the box” model
This is a very instructive model system which shows nicely the new properties of quantum
objects:
Hamiltonian:
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V (x) = 0, 0 < x < L
V (x) = ∞, otherwise
Within the box of length L the Hamiltonian is equal to the kinetic energy:
Ĥ = T̂ +V (x),
| {z }
0
The particle can not leave the box, the probability finding it outside the box is zero,
therefore the wave function must also vanish there. The keep the wave function continuos,
it has to vanish at the walls, as well (boundary condition):
Ψ(0) = Ψ(L) = 0
(11)
Therefore the Schrödinger equation to solve reads:
T̂ Ψ(x) = EΨ(x)
(12)
After a short (and instructive) calculation one gets the following result:
h2
; n = 1, 2, ...
2
s 8mL
2
π
Ψ(x) =
sin n x
L
L
E = n2 ·
The form of the wave function:
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Notes:
• The energy is quantized, it grows quadratically with the quantum number n, it is
invers proportional to L2 and m.
If L → ∞, E2 − E1 ∼
L = ∞.
22 −12
L2
→ 0. This means, the quantization disappears with
The same is true for growing mass m → ∞.
• There is a zero point energy (ZPE)!
The energy is not 0 for the lowest level (ground state).
If, however, L → ∞, E0 → 0.
Why is ZPE there? This is an unknown term for classical mechanics!
It can be explained by the uncertainty principle: ∆x · ∆p ≥ 21 h̄.
Since here we have V̂ = 0, E ∼ p2 , i.e. the energy of the particle originates in his
momentum only. Assume that E = 0, than p = 0, therefore ∆x = ∞, which is a
contradiction since ∆x ≤ L, the particle must be in the box. We conclude that the
energy can never get zero, since in this case its uncertainty would also be zero which
is possible only for very large box where the uncertainty of the coordinate is large.
Or alternatively, one can also say: if L → 0 =⇒ ∆x → 0 =⇒ ∆p → ∞ =⇒ ∆E →
∞. This means the energy of all levels MUST BE larger and larger if the size of the
box gets smaller.
• Wave function: the larger n is the more nodes are on the wave function: ground
state has none, first excited state has one, etc. (Node: where the wave function
changes sign).
• How does the solution looks like in 3D?
π 2 h̄2
E =
2m
n2a n2b n2c
+ 2 + 2 ,
a2
b
c
!
where a, b, c are the three measures of the box and na , nb , nc = 1, 2, ... are the the
quantum numbers.
If a = b = L, then
na
1
2
1
nb
1
1
2
E
h2
8mL2
2
5
5
We have found degeneracy which is caused by symmetry of the system (two measures
are the same).
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2.4.
Quantum Mechanical description of hydrogen atom
Atomic units
Quantity
Ang. mom.
Mass
Charge
Permittivity
Length
Energy
Atomic unit
h̄
me
e
4πε0
a0 (bohr)
Eh (hartree)
SI
[J s]
[kg]
h [C] i
Conversion
h̄ = 1, 05459 · 10−34 Js
me = 9, 1094 · 10−31 kg
e = 1, 6022 · 10−19 C
C2
4πε0 = 1, 11265 · 10−10 Jm
C2
Jm
From these one can derive:
[m]
[J]
2
0 h̄
1 bohr = 4πε
= 0, 529177 · 10−10 m
me e2
2
1 hartree = 4πεe 0 a0 = 4, 359814 · 10−18 J
1 Eh ≈ 27, 21 eV
Eh ≈ 627 kcal/mol
The model of the hydrogen atom:
• an electron is „situated” around the nuclei which is not moving;
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• the interaction potential is given by the Coulomb interaction: V = − er
In quantum mechanics we have to solve the Schrödinger equation:
ĤΨi = Ei Ψi
with
• Ĥ is the Hamiltonian including the interactions within the system (kinetic and potential
energy): Ĥ = T̂ + V
• Ei is the total energy of the system
• Ψi (x, y, z) is the wave function describing the system, also called state function, here also
can be called the orbital of the electron.
Notes:
1. the energy is quantized;
2. i index denotes that there are several such states. The one with the lowest energy is called
the ground state, the others are the excited states.
About the solution: during the calculations it turns out that the states should not be labeled
with a simple index i, but rather with a triplet of numbers, the so called quantum numbers:
i → (n, l, m)
It also comes out from the calculation that quantum numbers can not have arbitrary values:
this is where the name is from! For the hydrogen atom the possible values of the quantum
numbers are:
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• n – principal quantum number : 1, 2, 3, . . ..
• l – angular momentum quantum number : 0, 1, 2, . . .(n − 1)
• m – magnetic quantum number : −l, − l + 1, . . ., 0, 1, . . ., l (2l+1 different values)
The quantum numbers are related to physical quantities:
• n: determines the energy: En = − 2n1 2 (Eh )
EXACTLY LIKE IN BOHR THEORY!!!
• l: determines the size of the angular momentum: |l| =
p
l(l + 1)(h̄)
• m: determines the z component of the angular momentum:
−l, −l + 1, ..., 0, 1, ...l
lz = m(h̄)
m =
What is the angular momentum?
Classical definition of the angular momentum:
L = r × p = mr × v
where m is the mass, v is the speed, p is the momentum, r is the position of the particle (see
figure).
Why is m called the magnetic quantum number?
m determines the z component of the angular momentum. Since the electron is moving
around the nuclei, and has a charge, it creates magnetic moment. There is a proportional
relation between angular momentum and magnetic moment:
µ = −µB l
µz = −m · µB
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where µB is a constant (called the Bohr-magneton).
How many different values m can have?
m = − l + 1, ..., 0, ..., l, i.e. 2l + 1 values.
Since the interaction with the magnetic filed will be proportional to the magnetic moment, its
magnitude depends on m. → in magnetic field the energy levels split up to 2l + 1 different values.
This is the so called Zeeman-effect.
l=0 → 1 energy level
l=1 → 3 energy levels
l=2 → 5 energy levels
etc.
Notation of the orbitals:
principal
quant. number (n)
1
ang. mom.
quant. number (l)
0
subshell
l
1s
magnetic
quant. number (m)
0
number of orbitals
on the subshell
1
2
0
1
2s
2p
0
-1,0,1
1
3
3
0
1
2
3s
3p
3d
0
-1,0,1
-2,-1,0,1,2
1
3
5
4
0
1
2
3
4s
4p
4d
4f
0
-1,0,1
-2,-1,0,1,2
-3,-2,-1,0,1,2,3
1
3
5
7
20
Representation of the orbitals: 1s and 2s orbitals
There is a node on the 2s orbital, where the value of the wave function gets zero.
21
Representation of the orbitals: 2p orbitals
22
Representation of the orbitals: d orbitals
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24
Representation of the orbitals: dotting – the frequency of the dots represent the value: more
points mean larger value of the wave function.
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Radial electron density:
probability of finding the electron at distance r from the nuclei (i.e. in a shell of the spere).
Radial density for orbitals 1s, 2s and 2p:
Radial density for orbitals 3s, 3p and 3d:
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The spin of the electron
We want to prove that in the ground state of hydrogen atom l=0: we put it into the magnetic
field. We assume one beam:
This is the so called Stern-Gerlach experiment.
The beam of ground state hydrogen atom splits into two beams. This contradicts the theory,
since we have expected 1, 3, 5,. . . beams!
Conclusion:
• Pauli (1925): a „fourth quantum number” is needed;
• Goudsmit and Uhlenbeck suggested the concept of spin, as the „internal angular momentum”
Classically: if the electron is not a pointwise particle, it can rotate around its axis, either to the
right or to the left.
In quantum mechanics: the electron as a particle has „intrinsic” angular momentum, which is its
own property, like its charge.
What do we know about it?
• it is like the angular momentum since there is magnetic moment associated with it;
• its projection can have two different values.
p
magnitude: s(s + 1)h̄
s quantum number
z component: ms h̄
ms quantum number, or spin quantum number
ms = −s, −s + 1, . . . , s − 1, s
⇒ s = 12 since in this case ms = − 12 , + 12
FOR ELECTRONS s = 21 always!!!!!
The fourth quantum number is: m s
Thus, the electron has spin.
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What is spin?
Where it does originate from?
Bad question, we would not ask: why the electron has a charge?
Properties of the electron:
charge: −1
spin: 1/2
Spin is the intrinsic momentum of the electron.
Spectroscopic application: Electron Spin Resonance (ESR):
Summarized:
The states of the hydrogen atom are quantized and are characterized by quantum numbers:
n = 1, 2, . . .
l = 0, 1, . . ., n − 1
m = − l, − l + 1, . . ., l
ms = − 12 , 21
The energy depends only on n: En = −
1 1
2 n2 (Eh ).
There is a many-fold degeneracy!
In magnetic field the energy splits up according to the magnetic quantum number m.
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3.
Electronic structure of atoms
3.1.
The Hamiltonian
Ĥ = T̂e + Ve−e + Ve−n
with
• T̂e kinetic energy of the nuclei;
• Ve−e repulsion of the electrons;
• Ve−n attraction of the electron and the nuclei.
3.2.
Wave function of the many electron system
Ψ = Ψ(r1 , r2 , r3 , ..., rn ) ≡ Ψ(1, 2, ..., n)
i.e. a function with 4n variables.
3.3.
The Schrödinger equation
ĤΨ(1, 2, ..., n) = EΨ(1, 2, ..., n)
(13)
Problem: the Hamiltonian can not be written as a sum of terms corresponding to individual
P
electrons ( i ), therefore the wave function is not a product:
• Schrödinger equation can not be solved exactly
• the solution is not intuitive
3.4.
Approximation of the wave function in a product form
Physical meaning:
• Independent particle approximation, or
• Independent Electron Model (IEM)
a) assume, there is no interaction between the electrons
This unphysical picture gives as an idea for the approximation.
Ĥ =
X
i
hi (i) ⇒ Ψ(1, 2, ..., n) = φ1 (1) · φ2 (2)... · φn (n)
|
{z
}
wave function
|
{z
product of spin orbitals
(14)
}
In this case the Schrödinger equation reduces to one-electron equations:
ĤΨ = EΨ ⇒ ĥ1 (1)φ1 (1) = ε1 φ1 (1)
(15)
ĥ2 (2)φ2 (2) = ε2 φ2 (2)
(16)
...
ĥn (n)φn (n) = εn φn (n)
29
(17)
One n-electron equation ⇒ system of n one-electron equations
P
Total energy in this case is a simple sum: E = i εi
What is ĥi ?
ĥi = t̂(i) + ve−n (i)
(18)
with t̂(i) is the kinetic energy of electron i, and ve−n (i) attraction between electron i and the
nuclei.
This resembles the Hamiltonian for the H-atom ⇒ eigenfunctions will be hydrogen-like!
Problem: electron-electron interaction is missing!!!
b) Hartree-method:
Consider the one-electron problem of the first electron, but let us complete ĥ1 with the
interaction to the other electrons:
f
ĥ1 → ĥef
= t̂(1) + ve−n (1) + v1ef f
1
(19)
where v1ef f is the interaction of electron 1 with all other electrons.
The energy and orbital of electron 1 can be obtained by solving the eigenvalue equation of
f
ĥef
1 :
f
ĥef
1 φ1 (1) = ε1 φ1 (1)
(20)
Similarly, for electron 2
f
= t̂(2) + ve−n (2) + v2ef f
ĥ2 → ĥef
2
f
ĥef
2 φ2 (2)
=
ε2 φ2 (2)
(21)
(22)
Finally for electron n:
f
ĥn → ĥef
= t̂(n) + ve−n (n) + vnef f
n
f
ĥef
n φn (n)
=
εn φn (n)
(23)
(24)
These equations are not independent since we have to know the orbitals of all other electrons
to get vief f . Therefore these equations can be solved iteratively.
P
Total energy: E 6= i εi , i.e. not a sum of the orbital energies since in this case we would
count electron-electron interaction twice.
3.5.
Pauli principle and the Slater determinant
An important principle of quantum mechanics: identical particles can not be distinguished.
Therefore the operator permuting two electrons (P̂12 ) can not change the wave function, or at
the most it can change its sign:
P̂12 Ψ(1, 2, ..., n) = ±Ψ(1, 2, ..., n)
(25)
Change of the sign is therefore eligible since only the square of the wave function has physical
meaning which does not change in this case, either.
30
According to one of the postulates of quantum mechanics (so called Pauli principle) the wave
function of the electrons must be anti-symmetric with respect to the interchange of two particles.
In case of two electrons:
P̂12 Ψ(1, 2, ..., n) = −Ψ(1, 2, ..., n)
(26)
The product wave function used in the Hartree method does not fulfill this requirement, it
isnot ant-symmetric. Therefore we have to use instead of product, a determinantal wave function.
Ψ(1, 2, ..., n) =
φ (1) φ (1) · · ·
2
1
1 φ1 (2) φ2 (2) · · ·
√ ..
..
..
n
.
.
.
φ1 (n) φ2 (n) · · ·
φn (n) φn (1)
φn (2)
..
.
(27)
This type of wave function is called the Slater determinant.
Remember the properties of determinants:
a) Interchanging two rows of a determinant, the determinant will change its sign. → interchanging two orbitals, the wave function will change sign;
b) If two columns of a determinant are equal, the value of the determinant is 0 → if two electrons
are on the same orbital, the wave function vanishes;
c) If we add a row of the determinant to an other one, the value of the determinant will not
change → any combination of the orbitals will give the same wave function.
Conclusions:
a) and b) Pauli principle is fulfilled automatically
c) the orbitals do not have physical meaning, only the space spaned by them!!!
Hartree-Fock method
The most accurate version of the Independent Particle Approximation: same as Hartree
method, but the wave function is determinant instead of product of orbitals.
The orbitals are obtained by the so called Hartree-Fock equations:
fˆ(i)φi (i) = εi φi (i)
i = 1, · · · , n
(28)
where the Fock operator reads:
fˆ(i) = t̂(i) + ve−n (i) + U HF
(29)
with U HF being an averaged electron-electron attraction. Notice that while in case of Hartree
f
method we had different hef
operator for the individual electrons, here all electrons share the
i
same operator, i.e. these are not distinguishable.
31
3.6.
3.6.1.
Electronic structure of atoms in the Independent Particle
Approximation
Energy, orbital, wave function
According to the discussion above, we should solve the Hartree-, or the Hartree-Fock equations
first. In both cases we will get orbital energies (εi ) and orbitals (φi ) therefore for a quantitative
discussion it does not matter which one we use. The form of the equation reads:
ĥ(i)φi = εi φi
ĥ(i) = t̂(i) + ve−n (i) + ve−e
(30)
(31)
where ve−e denotes the electron-electron repulsion and are given for both Hartree and HartreeFock methods above.
As a solution we get:
• φi orbitals
• εi orbital energies
Since ĥ is similar to the Hamiltonian of the hydrogen atom, the solutions will also be similar:
The angular part of the wave functions will be the SAME Therefore we can again classify the
orbitals as 1s, 2s, 2p0 , 2p1 , 2p−1 , etc.
The radial part: R(r) will differ, since the potential is different here than for the H atom: since
it is not a simple Coulomb-potencial, the degeneracy according to l quantum number will be
lifted, i.e. the orbital energies will depend not only on n but also on l (ε = εnl ).
Wave function: created from the occupied orbitals by a product (Hartree) or as a determinant
(Hartree-Fock); the occupied orbitals are selected according to the increasing value of the orbital
energy (so called Aufbau principle).
Some important terms:
• Shell: collection of the orbitals with the same n quantum number;
• Subshell: collection of orbitals with common n and l quantum numbers, which are degenerate according to the discussion above. Orbital 1s and 2s form subshells alone, while
2p0 , 2p1 és 2p−1 (or 2px , 2py , 2pz ) orbitals form the subshell 2p. Subshell 3d has five
components, 4f seven, etc.
• Configuration: defines the occupation of the subshells. Examples:
He: 1s2
C: 1s2 2s2 2p2
3.6.2.
Angular momentum of the atoms
one particle:
many particles:
ˆl2
L̂2
ˆlz
L̂z
ŝ2
Ŝ 2
ŝz
Ŝz
32
The angular momentum of the system is given by the sum of the individual angular momentum of the particles ( so called vector model or Sommerfeld model):
L̂ =
X
ˆl(i)
(32)
ŝ(i)
(33)
i
Ŝ =
X
i
It follows that the z component of L̂ is simply the sum of the z component of the individual
vectors:
ML =
X
m(i)
Ms =
i
X
ms (i)
(34)
i
The length of the vector is much more complicated: due to the quantizations and uncertainty
principle, we can get different results: For exemple for two particles:
L = (l1 + l2 ), (l1 + l2 − 1), · · · , |l1 − l2 |
S = (s1 + s2 ), s1 − s2
Since L and S are again angular momentum operators, the eigenvalues are given by similar
rules:
L̂2 → L(L + 1) [h̄2 ] L = 0, 1, 2, . . .
L̂Z
Ŝ 2
ŜZ
3.6.3.
(35)
→ ML [h̄] ML = −L, −L + 1, . . . , L
1
3
→ S(S + 1) [h̄2 ] S = 0, , 1, , 2, ...
2
2
→ MS [h̄] MS = −S, −S + 1, . . . , S
(36)
(37)
(38)
Classification and notation of the atomic states
The Hamiltonian commute with L̂2 , L̂z , Ŝ 2 and Ŝz operators ⇒ we can chose the eigenfunction
of the Hamiltonian such that these are eigenfunctions of the angular momentum operators at the
same time. This mens we can classify the atomic states by the corresponding quantum numbers
of the angular momentum operators:
ΨL,ML ,S,Ms
= |L, ML , S, Ms i
(39)
The latter notation is more popular.
In analogy to the hydrogen atom, the states can be classified according to the quantum numbers.
For this, L and S suffices, since the energy depends only from these two quantum numbers.
L=
notation:
degeneracy
S=
multiplicity (2S+1):
denomination:
0
S
1
0
1
singlet
1
P
3
1
2
2
doublet
2
D
5
1
3
triplet
3
F
7
3
2
4
quartet
4
G
9
2
···
···
5
H
11
···
···
···
···
In the full notation one takes the notation of the above table for the given L and writes the
multiplicity as superscript before it:
33
Examples:
L = 0, S = 0: 1 S read: singlet S
L = 2, S = 1: 3 D read: triplet D
Total degeneracy is (2S+1)(2L+1)-fold!!
3.6.4.
Construction of the atomic states
Since there is high-level degeneracy for the orbitals, most of the time we face open shell systems,
where the degenerate orbitals are not fully occupied. In this case one can construct several states
for the same configuration, i.e. configuration is not sufficient to represent the atomic state.
Example: carbon atom
1s2 2s2 2p2
2p is open subshell, since only two electrons are there for six possible places on the 2p subshell.
What are the possibilities to put the two electrons onto these orbitals?
spacial part: 2p0 , 2p1 , 2p−1
spin part: α, β
These gives all together six spin orbitals,
wich will be occupied by two electrons. The number
!
6
of the possibilities are given by
which results in 15 different determinants. This means
2
there will be 15 states in this case. Do the determinants form the states? With other words: are
these determinants eigenfunctions of L̂2 and Ŝ 2 ?
To see this, let us construct the states by summing the angular momenta: Since we do not
know angular momentum vectors completely (remember the uncertainty principle applying for
the components!), the summation of two angular momentum vectors will not be unique, either,
we get different possibilities:
l(1) = 1
1
s(1) =
2
l(2) = 1
1
s(2) =
2
⇓
L = l(1) + l(2), l(1) + l(2) − 1, . . . , |l(1) − l(2)| = 2, 1, 0
S = s(1) + s(2), s(1) + s(2) − 1, . . . , |s(1) − s(2)| = 1, 0
This is represented by the following figure (not exactly for our case): since we do not know the
direction of the angular momentum vectors, only the cone they are located on, there are different
results of the summation.
34
The possible states therefore:
1
3
S
S
1P
1
D
3P
3
D
Considering the degeneracy: 1 S gives one state, 1 P gives three states, 1 D gives five, 3 S gives
three states, 3 P gives nine states (three times three), 3 D gives fifteen states (three times five),
which is all together 36 states. But we can have only 15, as was shown above!
What is the problem? We also have to consider Pauli principle, which says that two electrons
can not be in the same state.
If we consider this, too, the following states will be allowed:
1
S
1
3P
D
These give exactly 15 states, so that everything is round now!
Summarized: carbon atom in the 2p2 configuration has three energy levels.
What is the order of these states?
Hund’s rule (from experiment):
• the state with the maximum multiplicity is the most stable (there is an interaction called
„exchange” which exists only between same spins);
• if multiplicities are the same, the state with larger L value is lower in energy;
In case of the carbon atom:
E3 P < E1 D < E1 S
3.6.5.
(40)
Spin-orbit interaction, total angular momentum
As has been discussed in case of the hydrogen atom, the orbital and spin angular momentums
interact. The formula for the interaction energy reads for several electrons as:
Ĥ → Ĥ +
X
i
35
ζ ˆl(i) · ŝ(i)
(41)
Consequence: L̂2 and Ŝ 2 do not commute with Ĥ anymore, thus L and S will not be suitable
to label the states („not good quantum numbers”). One can, however, define the total angular
momentum operator as:
Jˆ = L̂ + Ŝ
(42)
which
[Ĥ, Jˆ2 ] = 0
[Ĥ, Jˆz ] = 0
(43)
i.e. the eigenvalues of Jˆ2 and Jˆz are good quantum numbers. These eigenvalues again follow the
same pattern than in case of other angular-momentum-type operators we have already observed:
Jˆ2 → J(J + 1) [h̄2 ]
Jˆz → MJ [h̄]
(44)
(45)
The quantum numbers J and MJ of the total angular momentum operators follow the same
summation rule which was discussed above, i.e.
J
= L + S, L + S − 1, · · · , |L − S|
(46)
The energy depends on J only, therefore degenerate energy level might split!!
Notation: even though L and S are not good quantum numbers, we keep the notation but
we extend it with a subscript giving the value of J.
Example I.: carbon atom, 3 P state:
L = 1, S = 1 → J = 2, 1, 0
3
→
P
3
P2 , 3 P1 , 3 P0
The energy splits into three levels!
Example II.: C atom 1 D state:
L = 2, S = 0 → J = 2
1
D →
1
D2
There is no splitting of the energy here, J can have only one value. This should not be a surprise
since S = 0 means zero spin momentum, therefore no spin-orbit inetarction!!!
3.6.6.
Atom in external magnetic field
Considering the total angular momentum, the change of the energy in magnetic field reads:
∆E = MJ · µB · Bz
MJ
= −J, −J + 1, . . . , J
This means the levels will split into 2J + 1 sublevels!
36
(47)
(48)
3.6.7.
Summarized
Carbon atom in 2p2 configuration:
Other configuration for p shell:
p1 and p5
2
P
3
P, 1 D, 1 S
p3
4
S, 2 D, 2 P
p6 (closed shell)
1
S
p
2
and p
4
37
4.
Theory of molecular symmetry
4.1.
Symmetry operations
What is symmetry? In everyday life we think of it as a spacial regularity. Mathematically, an
object is symmetric, if it is does not change for certain transformations.
For example:
• A symmetric building will not change if we reflect it on the palne cutting it in the middle.
• Merry-go-round can be rotated such that one seat gets to an other, nothing changes.
• An egg can be rotated by any angle about its axis, we will not see any change.
Molecules also own similar properties. For example:
• Water molecule can be reflected on the plane cutting the oxygen atom and intersecting in
the middle of the two hydrogen atoms. Also, we can rotate it by 180 degrees around the
axis going through the oxygen atom.
• In case of ammonia, one can rotate it by 120 degrees around the axis going through the
nitrogen and middle of the triangle defined by the three hydrogen atoms. Also, there are
three planes for reflection including the oxygen and one of the hydrogens.
Let us gather all possible symmetry operations:
• Cn – rotation around an axis through an angel of 2π/n radian (gir)
• σ – reflection on a mirror plane (special cases: σv , σh , σd , see later)
• Sn – improper rotation (rotation-reflection): a combination of Cn rotation és σh reflection
(plane is orthogonal to the axis) as an individual operation (giroid)
38
• i – inversion, i.e. reflection on a point (i = S2 )
• E – identity operation; leaves the object unchanged, only for mathematical purposes.
4.2.
Point groups
Now we can return to the question: what is symmetry? Symmetry is the collection of symmetry
operations which do not change the object.
Chemical examples:
• water: C2 (z), σzx , σzy , E
39
• ammonia: C3 (z), 3 times σv , E
• benzene: C6 , 6 times C2 , σh (horizontal, perpendicular to main axis), 6 times σv , i, and
some more...
• formaldehyde: C2 (z), σzx , σzy , E
One can conclude that the operators not modifying the nuclear frame, characterize the molecules. These are characteristic for the symmetry, since, for example, the same operators are
there for water and formaldehyde. Such group of operators are called point group. (The name
come from the fact that these operators form a „group” in mathematical sense.) Point groups
are labeled by the so called Schönflies-symbols:1
• Cn : the group including the Cn operations (special case: C1 no symmetry at all);
• Cnv : the group including the Cn , as well as σv mirror plane;
• Cnh : the group includes the Cn , as well as σh mirror plane;
• Dn : the group includes Cn as well as n times C2 rotations which are perpendicular to the
main axis;
1
Arthur Moritz Schönflies (1853-1928) german mathematician
40
• Dnh : same as Dn but also includes mirror planes perpendicular to the main axis;
• Dnd : same as Dn but also includes mirror planes including the main axis;
• Sn : the group includes Sn operation;
• Td : group having tetrahedron symmetry;
• ...
• C∞v : this group includes a rotation through any angle (n = ∞);
• D∞h : this group includes a rotation through any angle; (n = ∞) and, in addition, mirror
plan perpendicular to this axis;
• O3+ : point group for spherical symmetry.
• ...
Chemical examples:
molecule
water
ammonia
benzene
formaldehyde
ethylene
acetilene
carbon monoxide
symmetry operation
C2 , σzx , σzy , E
C3 (z), 3 times σv , E
C6 , 6 times C2 , σh , 6 times σv , i, etc.
C2 (z), σzx , σzy , E
41
point group
C2v
C3v
D6h
C2v
D2h
D∞h
C∞v
4.3.
Representation, character table
We have seen that the operators of the molecule’s point group do not change the molecular
frame. It follows that the wave function of the molecule can not change either, or eventually it
can change sign. Again, the explanation of the minus sign can be given by the physical meaning
of the square of the wave function.
We can classify the wave function according to the sign caused by the symmetry operation.
But how many possibilities are there? Look at again water as an example!
Sorszám
1
2
3
4
E
1
1
1
1
C2
1
1
-1
-1
σzx
1
-1
1
-1
σzy
1
-1
-1
1
There are no other possibility, since certain combinations are excluded. Therefore we can
42
state that the wave function of water can belong to only four different classes with symmetry
properties according to the table above. These correspond to one of the rows of the above table.
These „classes” represent the point group, therefore are termed „representation”. In particular,
these are the elementary representations meaning that they can not be reduced further, therfore
these are called as irreducible representations (in short irreps).
The irreducible representations and their properties (signs resulting out of the symmetry
operations performed on them) are collected in the so called character tables. The character
table of the C2v point group is given by:
C2v E C2 σzx σzy
A1 1
1
1
1
A2 1
1
-1
-1
B1 1 -1
1
-1
B2 1 -1
-1
1
Notice that above we have constructed this table by putting in the possible signes.
In case of „very symmetric” systems the irreps are not one-dimensional, certain symmetry
operation can be described only by two or more functions. As example we mention ammonia
(C3v point group).
C3v
A1
A2
E
E
1
1
2
2C3
1
1
-1
3σv
1
-1
0
If a point group has many dimensional irreps, then the system will have degeneracy, the
manyfold of the degeneracy is the same as the dimension of the corresponding irrep.
In the above example:
• ammonia will have double degenerate energy levels;
• water does not have degeneracy;
• in atoms there are single-fold, triple-fold, quintuple-fold, ... degeneracies, since the O3+
point group has irreps of one, three, five, etc. dimension.
Some nomenclature:
Character: the elements of the character table are called „character”; these correspond to the
symmetry operation of their column and irrep of their row. In one dimensional case this is very
expressive: if the character is 1, the irrep is symmetric, if -1, the irrep is antisymmetric with
respect to the given operation.
Symboles denoting the irreps: these are given in the first row of the character table. If this letter
is A or B, the irrep is one dimensional, in case E it is two-dimensional, in case of F and T
three-dimensional. Exception: C∞v and D∞h groups, where the one-dimensional irrep is called
Σ, the two-dimensional is Π and ∆, etc.
43
Total symmetric representation: in this case character for all operations is 1, i.e. such function
is symmetric for all operations, and it does not change sign. Its symbole usually includes the
letter A1 , but in case of D∞h point group its is denoted by Σ+
g.
Dimension of the representation: given by the character of the identity operator (E).
44
45
5.
Quantum mechanics of chemical binding
According to quantum mechanics:
ĤΨ = EΨ
where:
• Ĥ is the Hamiltonian, includes the interactions:
Ĥ = T̂el (r) + V̂e−n (r, R) + V̂e−e (r) + V̂n−n (R) + T̂nucl (R)
|
{z
Ĥe (r,R)
}
|
{z
}
T̂n (R)
Ĥ(r, R) = Ĥe (r, R) + T̂n (R)
where
– r stands for the coordinates of the electrons
– R stands for the coordinates of the nuclei
– see also earlier notations.
The Born-Oppenheimer approximation
the electrons are much lighter than the nuclei ( mMel ≈ 1836)
⇓ equipartition principle
electrons are much faster
⇓
electrons follow the nuclei instantaneously (adiabatic approximation)
⇓
from the point of view of the electrons the nuclei are not moving
⇓
equation for the electronic part: Ĥe (r; R)Φ(r; R) = E(R)Φ(r; R)
equation for nuclei: (T̂n (R) + E(R)) χ(R) = ET OT χ(R)
Notes:
• in Born-Oppenheimer (BO) approximation the equations for nuclei and electrons have
separated;
• nuclei are not still (not motionless);
• the potential the nuclei feel is E(R), the electronic energy calculated at different bond
distances;
• E(R) potential energy surface is a consequence of the Born-Oppenheimer approximation;
• Usually very good approximation, but fails if the potential energy surfaces cross each other
(photochemical reactions).
46
From now on, we deal with the electronic part:
Ĥe (r; R)Φ(r; R) = E(R)Φ(r; R)
How can we deal with this? The same technique as for atoms:
• use the Independent Particle Approximation: obtain orbitals (φi ) and orbital energies (εi );
• use the Aufbau principle to occupy the orbitals and get the configurations;
• obtain the states.
5.1.
Simplest example: H2 +
This is a single electron problem, like hydrogen atom, and can be solved analytically (within the
Born-Oppenheimer approximation): we get the wave function of the system (orbital) and the
corresponding energies (orbital energies).
The two lowest energy orbitals:
The energy of these orbitals depend on the H-H distance:
47
Why one decreases, the other increases? Look at an other representation of the orbitals
(along the bond)
u1 is a binding orbital, u2 is anti-bonding (meaning it eases the bond).
Prinzipal question: what is chemical bond according to quantum mechanics?
Answer from the above figures:
• according to the energy curves: energy decreases if the two atoms get closer;
• form of the orbitals and the density: there is an increase in electron density between the
two atoms.
Quantum mechanics can not say more, but this is mathematically a perfect explanation. For
chemists there are of course other explanations based on approximate models.
How can we approximate the orbitals? Simplest is to start with atoms and approximate the
molecular orbital by linear combination:
48
By equations:
u1 = N1 · (1sA + 1sb )
u2 = N2 · (1sA − 1sb )
where 1sA and 1sB are the 1s orbital of the two atoms.
The energy of one of the combinations (plus) increases, that of the other one (minus) decreases
with respect to the energy of the atomic 1s orbital:
The single electron of the H2+ should be put on the orbital with lower energy.
49
5.2.
The simplest molecule: H2
The same MO-s can be constructed than in case of H2+ . Now we have to put two electrons to
the orbitals. One with α spin (ms =1/2), one with β spin (ms =-1/2):
5.3.
Other diatomic molecules
We just follow the same route: obtain MO-s from the atomic 1s, 2s, 2p orbitals.
He2 :
He atomic configuration: 1s2
Both binding and anti-binding orbitals are double occupied: no energy gain with respect to
the atoms.
Li2 :
Li atomic configuration: 1s2 2s1
We create MOs also from 2s orbitals:
50
One bond, energy gain.
Be2 :
Be atomic configuration : 1s2 2s2
No bond!!
51
N2 :
N atomic configuration 1s2 2s2 2p3
New MOs from the 2p orbitals
The orbital diagram:
Triple bond, since three bonding orbitals are double occupied.
52
O2
O atomic configuration 1s2 2s2 2p4
Az oxigénmolekula elektronszerkezete
2p
2p
2s
2s
1s
1s
O
O2
O
Double bond, but open shell!
(Remember C atom: we got three states there.)
1 −
1
Here also three states can be constructed: 3 Σ+
g , Σg and ∆g .
Lowest energy is the 3 Σ+
g : triplet, therefore paramagnetic property.
F2
F atomic configuration 1s2 2s2 2p5
Three bonding and two anti-bonding orbitals → single bond.
53
5.4.
Electronic structure of water
Water belongs to C2v point group, we have discussed the corresponding character table. Here it
is again:
C2v
A1
A2
B1
B2
E
1
1
1
1
C2
1
1
-1
-1
σv (yz)
1
-1
1
-1
σv (xz)
1
-1
-1
1
Remember the discussion on symmetry: the orbitals must belong to one of the four irreps.
The orbitals will be labeled according to the irreps.
Orbitals can be obtained from Independent Particle Approximation, using the following atomic functions as basis:
• H: 1sa , 1sb
• O: 1s, 2s, 2px , 2py , 2pz
One obtains the following orbitals (called canonical orbitals) and orbital energies:
1a1 orbital (-20.52 hartree)
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
Edge = 4.35 Space = 0.1000 Psi = 1
1a1 : 1s (core)
54
2a1 orbital ( -1.33 hartree)
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
Edge = 4.35 Space = 0.1000 Psi = 2
2a1 : 2s (–2pz )+ 1sa + 1sb bonding
1b1 orbital ( -0.67 hartree)
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
Edge = 4.35 Space = 0.1500 Psi = 3
1b1 : 2py + 1sa – 1sb bonding
3a1 orbital ( -0.56 hartree)
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
Edge = 4.35 Space = 0.2000 Psi = 4
3a1 : 2pz (+ 2s) non-bonding
55
1b2 orbital ( -0.52 hartree)
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
Edge = 4.35 Space = 0.2000 Psi = 5
1b2 : 2px non-bonding
4a1 orbital ( 0.33 hartree)
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
Edge = 4.35 Space = 0.2000 Psi = 6
4a1 : 2s + 2pz – 1sa – 1sb anti-bonding
2b1 orbital ( 0.49 hartree)
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
Edge = 4.35 Space = 0.2000 Psi = 7
2b1 : 2py – 1sa +1sb anti-bonding
Configuration: (1a1 )2 (2a1 )2 (1b1 )2 (3a1 )2 (1b2 )2
56
State: 1 A1 (orbitals are all occupied ⇒ total symmetric singlet)
Excited states:
Configuration: (1a1 )2 (2a1 )2 (1b1 )2 (3a1 )2 (1b2 )1 (4a1 )1
state: 3 B2 or 1 B2
Configuration: (1a1 )2 (2a1 )2 (1b1 )2 (3a1 )2 (1b2 )1 (2b1 )1
State: 3 A2 or 1 A2
Localized orbitals
The bonding orbitals extend to all three atoms: these are not the types of orbitals we really
like in chemistry. Chemical intuition requires different orbitals!!!!
Remember: in case of anti-symmetric wave function (Slater determinant) the orbitals have no
„physical meaning”: in a determinant we can form any linear combination of the occupied orbitals,
the total wave function does not change.
Let us form linear combination of the occupied orbitals!
2a1
1b1
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
defaults used
Edge = 4.35 Space = 0.2000 Psi = 1
Edge = 4.35 Space = 0.2000 Psi = 2
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
defaults used
Edge = 4.35 Space = 0.2000 Psi = 3
Edge = 4.35 Space = 0.2000 Psi = 4
2a1 – 1b1
2a1 +1b1
We obtain to bonding orbitals which correspond to chemical intuition!
57
Now consider the two non-bonding orbitals:
3a1
1b2
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
defaults used
Edge = 4.35 Space = 0.2000 Psi = 1
Edge = 4.35 Space = 0.2000 Psi = 2
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
M
OOOOLLLLDDDDEEEENNNN
M
M
M
M
M
MOOOLLLDDDEEENNN
defaults used
defaults used
Edge = 4.35 Space = 0.2000 Psi = 3
Edge = 4.35 Space = 0.2000 Psi = 4
3a1 + 1b2
3a1 –1b2
We obtained two non-bonding orbitals which correspond to chemical intuition (lone pairs).
Comparison of localized and canonic orbitals:
orbital energy
symmetry
bond of an atom pair
lone pair of an atom
58
cononic
yes
yes
no
no
localized
no
no
yes
yes
5.5.
The Hückel approximation
For larger molecules the description by the MO theory becomes complicated. In same cases it is
enough to consider only a subset of orbitals. For example, in case of conjugated molecules, we
only consider the 2p orbitals perpendicular to the molecular plane.
In the Hückel theory, in addition, also some further approximations are introduced, what we
do not detail here. We mention only the that the MOs of the π system are obtained by the linear
combination of the 2pz orbitals of the carbon atoms (one each), and the total energy is the sum
of the energy of the occupied orbitals.
5.5.1.
Ethylene
Orbitals:
Orbital energies:
ε1 = α + β
ε2 = α − β
Energy diagram:
Total energy:
E = 2α + 2β
59
(49)
5.5.2.
Butadiene
Point group: C2h
Four orbitals from the four 2pz orbitals:
2bg
2au
1bg
1au
Orbital energies
60
2bg : ε1 = α + 1.62β
2au : ε2 = α + 0.62β
1bg : ε3 = α − 0.62β
1au : ε4 = α − 1.62β
Energy diagram:
Configuration: (1au )2 (1bg )2
State: 1 Ag (total symmetric)
Total energy: Ebutadien = 4α + 4.48β
Delocalization energy: Ebutadien − 2Eetilen = (4α + 4.48β) − (4α + 4β) = 0.48β
61
5.5.3.
Benzene
Point group: D6h
Orbitals (six from six 2pz orbitals):
1b2g
1e2u (degenerate)
1e2u (degenerate)
62
1a2u
Orbital energies:
a2u :
ε1 = α + 2β
e1g :
ε2 = α + β
e2u :
ε3 = α − β
b2g :
ε4 = α − 2β
Orbitals e1g and e2u orbitals are degenerate due to their symmetry!
Energy diagram:
Configuration: (a2u )2 (e1g )4
State: 1 A1g
Energy: E = 6α + 8β
Delocalization energy: Ebenzol − 3Eetilen = (6α + 8β) − 6(α + β) = 2β
63
5.6.
Electronic structure of transition metal complexes
System:
• „transition metal”: atom or positively charged ion
→ open shell, can take additional electrons
• „ligands”: negative ion, or strong dipole, usually closed shell
→ donate electrons (non-bonding pair, π-electrons)
Two theories:
• Cristal field theory: only symmetry
• Ligand field theory: simple MO theory
Questions to answer:
• why are they stable?
• why is the typical color?
• why do they have typical ESR spectrum?
64
5.6.1.
Cristal field theory (Bethe, 1929)
Basic principle:
• the ligands (bound by electrostatic interaction) perturb the electronic structure of the
central atom (ion)
• electrons of the ligands are absolutely not considered
Denomination comes from the theory of crystals where the field of neighboring ions has similar
effect on the electronic structure of an ion considered.
atom
complex
pointgroup
O3+
lower symmetry
orbitals
degenerate d
(partial) break off of degeneracy
The theory is purely based on symmetry!!
Example: [T i(H2 O)6 ]3+
Pointgroup: Oh
Character table of the pointgroup Oh :
65
Lower symmetry, the five d functions form a reducible representation: feszít ki:
Γ(5 db d f v) = T2g + Eg
T2g : dz 2 , dx2 −y2
Eg : dxy , dxz , dyz
Energy levels:
Degree of the splitting:
• The theory does not say a word about this
• However: 6 · ∆t2g = 4 · ∆eg , i.e. the average energy does not change!
66
[T i(H2 O)6 ]3+ in more detail:
Ti: ...3d2 4s2
Ti3+ : ...3d1
configuration:
State:
d1
2D
atom
(t2g )1
2T
2g
(eg )1
2E
g
complex
ground state excited state
Energy difference between ground and excited states are small → violet color (20400 cm−1 )
How does this work for more electrons? Use the Aufbau-principle:
S
1/2
1
3/2
2
1
From the fourth electron, the occupation depends whether ∆ or the exchange interaction
(K) is larger:
• if ∆ > K, the electrons goes to the lower level (complex with small spin)
• if ∆ < K, the electron goes to the higher level (complex with large spin)
Strong cristal field: splitting is large enough so that the low spin case will be more stable.
Weak cristal field: the splitting is small and the high spin case will be more stable
Experiment: ESR spectroskopy (see later)
67
5.6.2.
Ligand field theory
Basic principle: MO theory
• the orbitals of the central atom interact with the orbitals of the ligands → bonding and
anti-bonding orbitals are formed
• symmetry is again important: which orbitals do mix?
Basis:
• atom (ion): 3d, 4s, 4p orbitals
• ligands (closed shell): s-type orbital per ligand („superminimal basis”)
(sometimes eventually also π orbitals)
Symmetrized basis:
according to the pointgroup of the complex, we split it into irreducible representations.
Exampel: Octahedral complex (Oh point group)
Basis:
• atom (ion): 3d, 4s, 4p orbitals →
Γ(3d) = T2g ⊕ Eg
Γ(4s) = A1g
Γ(4p) = T1u
• ligands:
Γ(λ1 , ...λ6 ) = A1g ⊕ Eg ⊕ T1u
Symmetry adapted orbitals of the ligands (6 water molecule):
68
MO diagram:
One has to put 1+12 electrons on these orbitals:
configuartion: (a1g )2 , (t1u )6 , (eg )4 , (t2g )1
The occupancy is the same as in cristal field theory, but
• energy of the t2g orbital does not change with respect to atomic orbital, while that of the
eg orbital grows
• stabilization of the complex is due to the stabilization of the orbitals of the ligands
69
Other example: tetrahedral complex (e.g.. [CoCl4 ]2− , [Cu(CN)4 ]3− ):
Character table of the pointgroup Td
Reductions:
Γ(3d) = E ⊕ T2
Γ(4s) = A1
Γ(4p) = T2
Γ(ligands) = A1 ⊕ T2
MO diagram:
70
6.
Structural chemistry
6.1.
6.1.1.
General considerations
What does structure determination do?
• obtains of chemical formula
• determines molecular geometry, configuration, dynamical structure (molecular forces, vibrations, etc.)
Grouping of the methods:
• spectroscopy: absorption or emission of electromagnetic radiation
• diffraction methods
• observation of ionized states (mass spectrometry and similar methods)
There are lots of development on this field:
• improvement of methods (e.g. use of Furier techniques)
• lasers
• new and „tricky” setups
6.1.2.
Common basis of spectroscopical methods
Event in all cases: interaction of matter and electromagnetic radiation
What can happen during the interaction?
Arrangement of the measurement
71
The spectrum: relativ or absolute intensity as a function of the wave number (λ), frequency (ν)
or wavenumber (ν ∗ ):
• frequency: ν, measured in [1/s], e.g. in Hz, MHz, but also in energy unit as eV
• wavenumber: ν ∗ measured in [1/m], e.g. in cm−1
• wave length: λ measured in [m], e.g nm, Å
Important relations:
λ=
c
ν
ν∗ =
1
λ
where c is the speed of light.
Lambert-Beer law:
I = I0 exp(−ε n l)
where:
I0 is the intensity of the incoming light
I is the intensity of the outcoming light
ε: molar absorption coefficient (also called as extinction coefficient)
n: concentration
l: thickness of the sample
72
This means that the intensity drops exponentially in the sampel. This law aplies only for small
intensity and small thickness.
Absorbance: A = ln( II0 ) = ε n l, which is proportional with the concentration
Transmittance: II0
Common physical background:
During the spectroscopical event there is a transition between energy levels: incresing the
energy of the molecule by absorption, and lowering the energy by emission.
It is seen that the the energy of the matter (molecule) changes exactly with the energy of
photon:
E2 − E1 = ∆E = hν
Factors determining the spectrum
a) energy levels
nothing to be said about this
b) selection rules
The spectrum can be described by the tools of quantum mechanics. It can be derived that
the probability of a transition between two energy levels is:
P1↔2 = |
Z
Ψ1 K̂ Ψ2 dv|2
where
Ψ1 , Ψ2 : are the wave functions of the two sates
73
K̂: operator of the interaction between the molecule and the electromagnetic field. Usually it
is approximated by the dipole moment.
The above integral is called transition integral.
Transition occurs, if the transition integral does not vanish: it is allowed. If it vanishes, we
say that the transition is forbidden.
When is the transition allowed? This can be derived by, e.g. consideration of molecular
symmetry.
c) population of the energy levels
The transition probability under b) is the same in both direction, i.e. absorption and
induced emission has the same probability!!
But: the population of the energy levels are different and given by the Boltzmann-distribution:
Ni = N0 exp ( −
∆Ei
)
kT
where
∆Ei : relative energy of energy level i (∆Ei = Ei − E0 )
k: Boltzmann-constant
T : temperature
N0 : population of the ground state
The observed absorption is determined by the population and the transition probability at the
same time.
6.1.3.
The lasers
LASER: Light Amplification of Stimulated Emission of Radiation
Usually, the population of the lower states are significantly larger ⇒ probability of absorption is
larger than that of stimulated emission.
However, by the so called population inversion (doing a long and intense irradiation) we can
get the higher population for the higher states ⇒ emission will be more likely.
Properties of the lasers:
• monochromatic: the radiation includes only a narrow range of the electromagnetic spectrum
• coherence: the waves are in a common phase
• directed (small radial distribution)
• large intensity
74
6.1.4.
Distinction of different spectroscopic methods
Principle: the different motions in molecules can be distinguished by a good approximation:
Ψspatial (r, R) ≈ Ψelectron,spatial (r) · Ψnuclei,spatial (R) ≈ Ψelectron,spatial (r) · Ψvibration · Ψrotation · Ψtranslation
|
{z
Bohr−Oppenheimer approximation
}
where r and R are the coordinates of electrons and nuclei, respectively.
If the above approximation for the wave function works, we can also approximate the energy
as:
Espatial = Eelectron + Evibration + Erotation + Etranslation
Additional degree of freedom is spin, which both electrons and nuclei have:
E = Espatial + Eelectron
spin
+ Enuclei
spin
The different contributions are of different size and therefore
⇒ they will be obtained with electromagnetic radition belonging to different region of the
electromagnetic spectrum
⇒ can be investigated by different methods of spectroscopy
75
6.2.
6.2.1.
Rotation of molecules – microwave spectroscopy
Physical background
Event:
The molecule as a rotating dipole interacts with the electromagnetic radiation. Therefore,
the molecules with dipol moment will give rotational spectrum.
Quantum mechanical description of rotation also results in quantized energy levels.
The difference in the energy levels corresponds to microwave (MW) region of the electromagnetic radiation → microwave spectroscopy
Experimental details:
source: klystron, the radiation is caused by vibrating electrons
sample: gase phase, the cell is often several meter long
information to be obtained: most accurate method for molecular geometry (bond length).
6.2.2.
Rotation of diatomic molecules
The rotating molecule possesses moment of inertia:
IB =
m1 r12
+
m2 r22
=
m1 m22
m21 m2
+
(m1 + m2 )2 (m1 + m2 )2
!
r2
m1 m2 (m1 + m2 ) 2
m1 · m2 2
r =
r ≡ mred r2
(m1 + m2 )2
m1 + m2
From the point of view of the mathematical treatment, it can be described as a single particle
with mass mred rotating on an orbit of radius r (c.f. with the electron in hydrogen atom; it is
simpler here since r is fixed).
Classically: there is only kinetic energy:
E = T = (1/2)Iω 2 ,
where ω is the angular velocity (c.f. translational motion, where E = (1/2)mv 2 ). Alternatively,
with angular momentum: (L = Iω):
E=T =
1 2
L
2I
The corresponding Hamiltonian:
Ĥ =
1 2
L̂
2I
76
i.e. it differs from L̂2 only by a constant → eigenvalues will be the same but multipied by the
constant. L̂2 eigenvalues (see hydrogen atom):
Λ = l(l + 1)h̄2 ,
l = 0, 1, 2, . . .
The energy of rotation of diatomic molecules differs from this only by a factor
number is traditionally denoted J.
EJrot =
1
2I .
The quantum
1
J(J + 1)h̄2 ≡ BJ(J + 1)hc J = 0, 1, 2, . . .
2I
with B being the rotational constant.
Similarly to the z component of the angular momentum, the z component is given by a
quantum number M :
M = −J, −J + 1, 0, 1, ..J
and the energy is independent of this → the rotational energy levels are (2J+1)-fold degenerate!
From the formula of the energy above one can conclude that with increasing J the energy levels
get further and further away!
There is no zero point energy!!!
Selection rules:
• µ 6= 0, only molecules with dipole moment have rotational spectrum;
• ∆J = ±1, transition only between neighboring states are possible.
A rotational spectrum of diatomic molecules consists of equidistante lines:
77
The distribution of the intensity follows maximum curve: this is because the population of the
energy levels, despite of the Boltzman distribution, grows first due to the increasing degeneracy
(2J + 1).
Informations from the rotational spectrum:
Distance of the lines: rotational constant (B) → moment of inertia (I) → bond distance (r),
or in generla, the molecular structure.
A „true” rotational spectrum:
6.2.3.
Rotation of molecules with more than two atoms
Basic concept:
The moment of inertia is in this case a tensor of dimension 3x3 (I).
The I tensor of the moment of inertia defines three main axes a, b, c and the corresponding
three main moments of inertia (Ia , Ib , Ic ). The spectrum can be described by this three numbers
(or the corresponding rotational constants Ba , Bb and Bc ).
In the main-axis system the Hamiltonian is given as the independent summ of the three
components:
Ĥ =
1 2
1 2
1 2
L̂ +
L̂ +
L̂
2Ia a 2Ib b 2Ic c
78
The molecules can be grouped according to the relative size of the main moments of inertia:
• spherical rotor: Ia = Ib = Ic ;
• symmetric rotor: Ia 6= Ib = Ic ;
• asymmetric rotor: Ia 6= Ib 6= Ic .
Examples:
• spherical rotor: CX4 , but these does not posses dipole, therefore nor rotational spectrum.
• symmetric rotor: favorite of the MW-spectroscopists, easy to evaluate; e.g. NH3 , CH3 X,
C6 H6 (For the latter, however, no dipol moment!!)
Summarized:
The rotational (microwave - MW) spectroscopy results in the most accurate molecular structure for small és symmetric molecules in the gase phase.
79
6.3.
6.3.1.
Vibration of molecules – IR and Raman spectroscopy
Physical background
Event:
The atoms building up the molecules are in constant vibration, their vibrational states are
quantized. We can observe transition between these quantum states:
• directly: in the infra red (IR) region, which frequency corresponds to the energy differences
between vibrational states.
• indirectly, when the frequency difference of the absorbed and emitted light corresponds to
the energy differences between vibrational states (Raman effect, inelastic scattering)
Experimental details of the IR measurement
Classical arrangement (as in the general case, see figure above):
• source: IR lamp
• sample: film or solution, eventually solid
• monochromator: optical grid or prism
• detector: thermo-cell
Sample rack and monochromator: salt of alkali metals (e.g. KI) since glas absorbes in IR.
Details of the Raman measurement
Effect: inelastic scattering; first excite to an electrically excited state, which will emit light
and lead back to a vibrationally excited level of the ground electronic state.
Generally, the left (Stokes) effect is used in Raman spectroscopy
Experimental instruments: similar to the VIS and UV measurements (see later)
80
6.3.2.
Vibrations of diatomic molecules
Classical description
Model: harmonic oscillator, which follows the Hook-law: F = −kx, where F is the force, x
is the displacement, k force constant of the spring.
= −k(r − re )
F
where re is the equilibrium distance between the two atoms, r is the actual distance (changes
during vibration).
The vibration of the diatomic molecule can be considered as a vibration of a particle of mass
mred around its equilibrium position; the frequency of the vibration is given by:
ν =
1
2π
s
k
mred
ṁ2
i.e. it depends on the reduced mass (mred = mm11+m
) and the force constant (k).
2
The energy of the system is continuous, it can have any value, even 0!
Quantum mechanical description in the harmonic approximation:
Hamiltonian:
Ĥ = T̂ + V̂
Z
Z
1 2
V =
kq (= − F = kq)
2
By solving the Schrödinger equation, we get the energy levels:
1
= hν v +
2
Ev
Thus the energy:
• quantized
• equidistant (Ev+1 − Ev = hν)
• there is zero point energy ( 12 hν)
Wave function:
81
v = 0, 1, 2, ...
i.e. similar to the wave functions of the „particle in the box model”. (On the figure: ω = 2πν.)
The figure also shows the distribution function (|Ψ|2 ), i.e. the probability of finding the
vibrating particle. These can be compared with the classical probability of the same energy (the
figure uses the letter m for the quantum number instead of v):
It is seen that
• in the classical case the probability at q = 0 is the smallest since the particle runs here
the fastest. In the quantum case, however, the largest probability is at q = 0. (This
contradiction can be resolved if we do not compare states with the same energy, but the
lowest energy state with the lowest energy state. In the classical case it has enery zero,
so it does not move, i.e. the particle is at q = 0 which corresponds to a maximum of the
quantized system.)
• from v = 2 the two cases hardly resemble each other, the maximum of the classical case is
still at the turning point.
• in case of v = 10 it is seen already that classical and quantum distributions are similar
(correspondence principle).
The form of the vibrational wave function has an important role in case of the fine structure
of the UV spectrum (see later).
Selection rules
82
The transition probability between levels v and v 0 in case of IR measurement:
P
v,v 0
2
Z
0
= Ψv µ̂Ψv dv where µ̂ is the dipole operator.
Skipping the details, this probability is not zero if
•
dµ
dq
6= 0, i.e. the dipole moement changes during vibration
• v 0 = v ± 1, i.e. the vibrational quantum number changes by 1.
This means that in the spectrum of diatomic molecule only one line will appear and ∆E = hν!
Please note that homo-nuclear diatomic molecules have no IR spectrum, since the dipole
moment remains zero during the vibration.
The transition probability between levels v and v 0 in case of Raman measurement:
Pv,v0
Z
= 2
Ψv α̂Ψv0 dv where α̂ is the operator of polarizability.
Skipping the details, this probability is not zero if
•
dα
dq
6= 0, i.e. the polarizability changes along the coordinate
• v 0 = v ± 1, i.e. the vibrational quantum number changes by 1.
This means that in the Raman spectrum of diatomic molecule only one line will appear and
∆E = hν!
Please note that homo-nuclear diatomic molecules have Raman spectrum, since the polarizability changes if we elongate the bond.
Illustration:
Vibrational frequency of hidrogene halogenides
HX
HF
HCl HBr
HI
−1
ν (cm ) 3958 2886 2559 2230
Structural information: ν → k, i.e. from the vibrational frequency we can get the force
constant, which is representative for the strength of the bond. Above the strongest is the bond
of the HF molecule.
Anharmonicity
For the real molecules the harmonic approximation is not quite valid, the potential differs
from the harmonic one:
83
This can be described the best by a so called Morse-potencial:
V (q) = De (1 − exp(−αq))2
where De is the dissociation energy (measured from the minimum of the curve), α is a constant.
Consequence:
• the energy levels are not equidistant – they get more dens with growing energy
• ∆v = ±1 will not be true strictly – in the spectrum so called overtones will aslo apear at
∼ 2ν, ∼ 3ν.
This effect is, however, very weak, the vibrational spectrum can approximated by the harmonic
picture in most cases: there will be only some weak lines beside the strong ones corresponding
to the harmonic approximation.
6.3.3.
Vibrations of molecules with more than two atoms
Internal coordinates
The number of internal degrees of freedom for a molecule with N atoms is 3N − 6 (in case of
linear molecules 3N −5). The use of internal coordinates is better than the Cartesian coordinates
(3N ), and also chemical intuition can be used. Main types:
• bond stretching
• bond bending
• torsion
• out-of-plane
84
Examples:
• water: three coordinates are necessary, r1 , r2 , α
• ammonia: 6 coordinates are necessary, there are different choices, e.g. r1 , r2 , r3 , α1 , α2 , α3 ,
but the description of inversion would require the out-of-plane coordinate.
• ethylene: 12 coordinates are necessary, one possible choice is: five stretching, four bending
(note that two of the possible six bendings are redundant), τ3126 torsion, two out-of-plane
(CH2 groups).
Normal coordinates
The theory gives a set of coordinates in which the 3N-6 (3N-5) degrees of freedom get independent, the Hamiltonian reads:
Ĥ =
X
ĥ(Qi )
i
Here Qi stands for the normal coordinates.
Normal coordinates are built from combination internal coordinates, therefore they extend
over the whole molecule.
The energy is the sum of the energy of the 3N-6 oscillators:
E(v1 , v2 , ..., vn ) =
n
X
i
1
hνi (vi + )
2
(50)
with νi being the frequency of the ith normal coordinate, with the corresponding vibrational
frequency of vi .
P
The ZPE is: EZP E = h2 i νi .
Selection rules:
Without derivations:
• ∆vi = ±1 és ∆vj6=i = 0, i.e. only one oscillator can be excited by one photon;
•
dµ
dQi
6= 0 (IR) or
dα
dQi
6= 0 (Raman).
85
Considering anharmonicity, beside the overtones, in the spectrum so called combination bends
will also appear, which are the results of simultaneous excitation of more than one mode.
Thus, the spectrum
• is dominated by 0 → 1 transition of symmetry-allowed normal modes. These are called
fundamental bands.
• Due to anharmonicity, also overtones and combination bends will apear.
Typical normal vibrations are shown on the following figure:
As an example here I show the IR and Raman spektra of toluene:
86
6.3.4.
Practical application of vibrational spectroscopy
• identification of the molecule
• assignation of typical groups
• quantitative characterization
Identification
Without theoretical characterization, the so called „finger-print” regions are compared with
databases. The fingerprint region is the low frequency (< 500 cm−1 ) part of the spectrum,
which mostly corresponds to deformation of the molecular backbone. These are very sensitive
to the molecular environment, thus very characteristic for individual molecules.
Assignation of typical groups
Although the normal vibrations extend over the whole molecule, these very often dominated
by some local internal coordinates. Such vibration will apear in the same region of the spectrum
even if the molecular environment is different. Mostly stretching vibration can be used for this
purpose. The following figure demonstrate this and shows the regions characteristic for different
functional groups (you should remember these!).
87
The vibrational spectroscopy can be used for identification of groups of atoms.
Qualitative characterization
From the vibrational frequencies, with mathematical tools, one can obtain the force constants,
which are characteristic for the strength of the bonds. One measurement is usually not enough
to obtain all force constants, therefore izotopomers are used, which give independent data. Here
theoretical methods have a very important role.
88
6.4.
Electron spectroscopy
Two different methods:
UV-VIS spectroscopy: M + hν → M ∗
PES, XPS (ESCA): M + hν → M + + e−
6.4.1.
UV-VIS – ultraviolet-visible spectroscopy
Experimental details:
visible
Regions: UV
VUV (Vacum UV)
400-800 nm
200-400 nm
< 200 nm
source:
The instrument is simple: optic:
detector:
wolfram lamp, hydrogen lamp, etc.
quartz
photocell
Physical background: transition between two electronic levels ⇒ „electron spectrum”
Theory: everything, what we have learned about electronic structure of molecules
Selection rules:
Transition probability between state i and j is given here also by the transition integral:
Z
Ψi µ̂ Ψj dv
where Ψ is the total wave function:
Ψ = Ψvib · Ψel = Ψvib · Ψspatial · Ψspin
therefore
Z
Z
Ψi µ̂ Ψj dv =
Ψvibr
i
Ψvibr
j
dv ·
Z
Ψspin
i
Ψspin
j
dv ·
Z
Ψspatial
µ̂ Ψspatial
dv
i
j
R
Ψvibr
Ψvibr
dv: called as the Frank-Condon factor (see later).
i
j
spin
spin
Ψj
dv: transition only between states of the same multiplicity is possible.
Ψi
R
Ψspatial
µ̂ Ψspatial
dv: determined by molecular symmetry
i
j
R
Example: benzene HOMO-LUMO excitation (HOMO: highest occupied MO, LUMO: Lovest
unoccupied MO)
89
Excitation with π electron configuration:
(a2u )2 (e1g )4 → (a2u )2 (e1g )3 (e2u )1
With states:
1A
1g → B2u , B1u , E1u (singlet or triplet)
The transition can be classified as:
spin allowed
spin forbidden
symmetry allowed
1E
1u
3E
1u
The spectrum:
excitation
1A
1E
1g →
1u
1B
1u
1B
2u
symemetry forbidden
1 B ,1 B
2u
1u
3 B ,3 B
2u
1u
wave number (Å)
1830
2050
2550
extinction coefficient
47000
7000
220
UV spectrum of benzene:
90
The triplet states can not be seen at all → spin prohibition is much stronger than symmetry!
Vibrational fine structure, Frank-Condon prinziple
The vibrational levels superimpose on the electronic transitions:
E = Eel + Evibr
These are called vibronic transitions.
Question: what are the relative intensities of the the vibronic transitions?
Transition integral:
Z
Ψvibr
i,v
Ψvibr
j,v 0
dv ·
Z
Ψspatial
µ̂ Ψspatial
dv
i
j
91
The second term is the same for all vibrational levels, therefore it is influenced only by the first
term: the individual transition probabilities are given by the overlap of the vibrational wave
functions in the ground and excited state.
Qualitatively: the motion of the nuclei are slow compered to the motion of electrons → the
so called vertical excitations will be the most probable, i.e. transition to vibrational levels where
the probability of the nuclear arrangement of the ground state is large.
In more detail for the diatomic molecules2 :
a) The potential energy surface does not change too much (e.g. excitation from lone pair)
See figure on the blackboard
In this case the vibrational wave function is almost the same in the two electronic state,
therefore v = 0 → v 0 = 0 transition will be the most probable.
b) The potential energy surface of the two electronic states differ considerably, also the equilibrium bond length differ:
See figure on the blackboard
In this case some other but the first vibrational wave function will have larger overlap, the
corresponding vibronic band will be the most intense, the lines left and right will have decreasing
intensity (so called progressions will appear. If the resolution of the spectrum is not large, we
can see only the envelop of the individual lines.
c) Dissociation
See figure on the blackboard
The spectrum becomes continuous above the dissociation energy, since the energy of the
fragments can take any value (translation).
2
The graphical representation of the vibrational wave function is seen in subsection 6.3.2, page 75.
92
Practical UV spectroscopy
Importance:
• quantitativ analysis
• structural chemistry: detection of chromophores
chromophores: are molecular fragments (group, bond, etc.) which can be excited easily.
According to the chromophores the following types of excitation can be distinguished:
1. π → π ∗ – Chromophore: double bond or π-system in the molecule
excitation from binding π orbital to antibonding π ∗ orbital
λ(Å)
ethylene
1620
butadiene
2170
hexatriene
2510
octatetraene
3040
so called batochrome shift: λ increases
Explanation? Yes, with the energy levels of the „particle in the box” they decrease with
2
02
the increasing size of the box (∆En−n0 ∼ n L−n
)
2
2. n → π ∗ – chromophore lone pair
3. σ → σ ∗ , n → σ ∗ , stb
4. „charge transfer” excitation
excitation from an orbital localized on one side of the molecule to a one which is localized
to an other side → there will be positive charge at one and negative charge on the other
end, therefore big dipole moment. Example:
C6 H6 · I2 → C6 H6 + · I2 −
Such transitions have large intensity because of the large change in dipole moment!
5. transition metal complexes
Relaxation of excited states
According to quantum mechanics, the excited states are only „quasi-stationary”, i.e. they can
not leave for ever (finite life time).
Possible routes:
• without radiation: with a complicated mechanism the energy transfers to other degrees of
freedom (e.g. vibration, rotation → the sample gets warm)
• with radiation
– fluorescence – immediately (nano-seconds time scale)
– phosphorescence – timely delayed (seconds or even minutes time scale)
Fluorescence
93
It can be seen on the figure:
• λout ≥ λin
• absorption shows the vibrations of the excited state
• emission shows the vibrations of the ground state
Phosphorescence
What is the reason for the delay?
94
Since triplet → singlet transition is forbidden, it can proceed only with low probability, i.e.
slow. The change in wave number is much larger than in case of fluorescence.
6.4.2.
PES – photoelectron spectroscopy and ESCA
Principle: electromagnetic radiation ejects an electron from the molecule:
M + hν → M + + e−
The kinetic energy of the electron (Telectron ) will be measured, and the ionization energy (IE)
can be calculated using the frequency of the radiation (ν).
hν = Eion − Emolecule +Telectron
|
{z
}
IE
In practice:
• UPS: UV light is used, the valence electrons are ejected → characteristic for the molecule
• XPS: X-ray is used: electrons of the core (1s) are ejected → will be characteristic for the
atoms of the molecule.
Photoelectron spectroscopy
Source: helium lamp, so called lines I and II (∼ 20-21 eV).
What will be seen in the spectrum?
• the molecule is usually in its ground state (both electronic and vibrational)
95
• several state of ion can be produced → the spectrum consist of several lines.
• different vibrational states of the ion can be reached → vibrational fine structure (as in
case of UV)
Example: photoelectron spectrum of the CO2 molecule:
We can see:
• there are three bands without vibrational structure – these are ionization from lone pair
orbitals
• one band has strong vibrational structure – ionization from a bonding orbital.
Explanation is very similar to that of the UV spectroscopy. The overlap of the vibrational
wave functions of the molecule and the ion will determine the intensity of the vibrational bands.
If we eject electron form a lone pair, the bonding of the molecule does not change much, vibration
will be the same. On the other hand, ejecting an electron form a bonding orbitals, results in a
big change in bindings and consequently in vibrational modes.
Information from the photoelectric spectrum:
• ionization energy
• vibrational modes of the ion
Application: mostly systems with lone pairs (the spectrum is simpler):
• transition metal complexes
• organic molecules with heterocyclic (aromatic) fragments
ESCA – Electronspectroscopy for chemical analysis
Excitation by X-ray radiation which ejects electrons from inner orbitals (mostly 1s)
Inner shells:
• are with good approximation identical to atomic orbitals (see e.g. the orbitals of water)
→ the ionization energy will be characteristic for the atom
• small dependence of the chemical environment → the ionization energy of the atoms in
different environment will be slightly different: „chemical shift”.
96
Example: ethylpropionate:
ESCA spectrum:
Notes to the spectrum:
• the bands corresponding to O and C atoms differ considerably (540 and 290 eV, respectively)
• the signal of the O atom splits into two: there are two types of O atoms in this molecule
• the signal of the C atom splits into three: there are three types of C atoms in this molecule
97
6.5.
Mágneses rezonancia spektroszkópiai módszerek (NMR, ESR)
Alapelv:
Mágneses momentum kölcsönhat a mágneses térrel:
∆EM
= µz · Bz
ahol µz a mágneses momentum, Bz a mágneses tér z komponensei.
Mitől függ a mágneses momentum? Mint láttuk, arányos az impulzusmomentummal, például
egy elektron esetén (l. H-atom):
µz = µ B · m
ahol µB az ún. Bohr magneton (konstans), m pedig az ˆlz operátor kvantumszáma.
Általánosan felírva:
µz = g · µB · Iz
ahol g a vizsgált rendszer ún. Lande-faktora, Iz a megfelelő impulzusmomentum z komponense,
melyre fennáll: Iz = −I, −I + 1, −I + 2, ..., I, ahol I a megfelelő momentum operátor sajátértéke.
g
I
pálya impulzusmomentum
1
0,1,2,...(l)
elektron spin
2 (2.00232)
1
2
magspin
(kísérletből határozható meg)
0, 12 ,1, 32 ,...
Így az energia:
∆EM
= Bz · g · µB · Iz
azaz az energia 2I + 1 értékre hasad fel!
Az energiaszintek között átmenet érhető el. A mágnesesség eredete alapján megkülönböztetünk:
• magspin: NMR (Nuclear Magnetic Resonance)
• elektronspin: ESR (Electron Spin Resonance)
spektroszkópia módszereket.
98
6.5.1.
Az NMR spektroszkópia fizikai alapjai
Történet: Bloch és munkatársai (Stanford), valamint Purcell és munkatársai (Harvard) 1946-ban
egymástól függetlenül publikálták a jelenséget, majd 1952 közös Nobel-díjat kaptak.
Bizonyos magoknak van nullától különböző spinje:
mag
I=
szintek száma (2I+1)
1H
2H
13 C
14 N
19 F
29 Si
31 P
1/2
2
1
3
1/2
2
1
3
1
3
1/2
2
1/2
2
Legegyszerűbb, egyben a leggyakrabban használt eset: I=1/2
µz = g · µB ·Iz = ±1/2γ 0
| {z }
≡γ 0
így az energia különbsége a két állapot között:
∆EM
= Bz · γ 0
A rendszer tehát ∆EM -nek megfelelő frekvenciájú fotont abszorbeál:
ν =
ahol γ =
∆EM
γ0
= Bz ·
≡ Bz γ
h
h
γ0
h
Vegyük észre, hogy a frekvencia függ a mágneses indukció (Bz ) nagyságától is. Azaz: az
átmenet két paraméter változtatásával érhető el, változtathatjuk az elektromágneses tér frekvenciáját és az indukciót tartjuk konstans értéken, illetve konstans mágneses tér mellett a frekvenciát
változtatjuk. Megvalósítható mágneses terek esetén a ν frekvencia a rádióhullámok tartományába
esik.
Gyakorlatban a B-t változtatják. A mérés elrendezését az alábbi ábra mutatja.
99
Cél: Minél nagyobb legyen a mágneses tér, hogy mérhető legyen az energiakülönbség. Az első
kommerciális készülékek 60 Mhz körül dolgoztak, a mai legmodernebb készülékek már az 1 Ghz-et
is túllépik.
Ezen az ábrán pedig az épületben lévő 500 MHz-es (a használt rádióhullám frekvenciája)
készüléket látjuk:
6.5.2.
Az NMR spektrum kvalitatív leírása
a) a kémiai eltolódás
Tekintsük a legegyszerűbb magot, a 1 H-t (proton NMR).
Hány átmenetet tapasztalunk az etilalkohol (CH3 –CH2 –OH ) molekula esetén?
Az eddig elmondottak alapján egyet, azonban a valóságban három sáv lesz. Mi ennek az
oka?
A molekulában ugyanis nem csupasz magok vannak, ezeket elektronok veszik körül. Az
elektronok árnyékolják a mágneses teret, tehát egy adott protonnál észlelhető mágneses tér a
külső mágneses téren kívül függni fog a mag környezetétől is.
100
Képletekkel: Legyen ν a méréshez használt frekvencia. Ehhez tartozik a fentiek szerint
B0 = ν/γ mágneses indukció, aminél rezonanciát észlelünk. Ennek kell lennie a magnál. Az
elektronok árnyékolnak, tehát ennél nagyobb teret kell a rendszerre kapcsolnunk:
B0 = B A (1 − σA )
ahol B A a külső tér, σA az árnyékolási tényező. Mivel ez utóbbi a molekula különböző környezetben lévő atomjaira különböző, rezonanciát különböző külső indukciónál (B A ) fogunk észlelni.
Tekintsünk egy referencia rendszert:
ν → B0 = B ref (1 − σref )
B ref természetesen különbözik B A -tól. Ami jellemző lesz az A protonra, az a mágneses indukciók
különbsége:
B A − B ref
Probléma:
különböző laboratóriumokban különböző frekvenciát használnak, ettől függeni fog az indukció
különbsége is, így a spektrumok nem lennének összehasonlíthatók. Ezért bevezetjük a kémiai
eltolódás (δA ) fogalmát:
δA =
σref − σA
B ref − B A
=
≈ σref − σA
B ref
1 − σA
A kémiai eltolódás dimenziótlan mennyiség, ppm-ben (part per million) mérjük.
Referenciaként a TMS (tetrametilszilán) használatos. Így a különböző laboratóriumokban
mért spektrumok összehasonlíthatók! (Azért TMS, mert sok egyforma proton van benne.)
Az így definiált kémiai eltolódás:
• különböző molekulák protonjai különböző helyen jelennek meg
• adott molekula különböző környezetben lévő protonjai más és más helyen jelennek meg
Az előbbi megjegyzéshez:
Jellemző kémiai eltolódások:
101
Kvalitatív észrevételek:
• nagy elektronegativitású szomszédos atom elszívja az elektront → a proton „csupaszabb”
lesz (nagyobb eltolódás)
• aromás gyűrű: általában erősebb (pozitívabb) eltolódás. Ennek a π rendszerben fellépő
köráramokat szokás okként megadni.
Az utóbbi megjegyzéshez:
A molekulán belüli eltérés az egyik legfontosabb az alkalmazás szempontjából, mert ez alapján
megadható, hogy hány féle proton van a molekulában, sőt az intenzitások alapján ezek relatív
száma is kiderül.
Példák:
Metanol:
CH3 -CH2 -OH (etilalkohol):
6.5.3.
A spin-spin csatolás
Finomszerkezet, pl. etanol nagyobb felbontásban:
102
(Hogy az OH sáv miért nem bomlik fel, azt majd később)
Magyarázat: a szomszédos, de különböző kémiai eltolódású magok között csatolás lép fel.
Nagysága attól függ, hogy a szomszédos ekvivalens magoknak mennyi az eredő spinje, azaz
P
i Iz (i).
Például tekintsük a CH2 csoportot! Itt két ekvivalens mag van, ezek együtt a következő
állapotokatPvehetik fel:
Iz statisztikus súly
αα
1
1
αβ
0
2
βα
0
ββ
-1
1
Azaz, három állapot lehetséges, 1,0,-1 értékkel, 1,2,1 valószínűséggel. Ezért a szomszédos
csoport sávja három sávra hasad, melyek intenzitása 1:2:1 arányú.
Tekintsük
P a CH3 csoportot:
Iz statisztikus súly
ααα 3/2
1
ααβ
αβα 1/2
3
βαα
αββ
ββα -1/2
3
βαβ
βββ -3/2
1
A CH3 csoport hatására a szomszédos csoport sávja tehát négy sávra hasad, 1:3:3:1 arányban.
Ez magyarázza az etanol fenti spektrumát: CH2 csoporthoz tartozó sáv négy sávra bomlik
a szomszédos CH3 csoport hatására; a CH3 csoport sávja a szomszédos CH2 csoport hatására
három sávra hasad fel.
Mikor látjuk a felhasadást? Ha nagyobb a felbontás, azaz a különböző csoportok sávjai lehető
legjobban eltávolodnak egymástól; ez akkor valósul meg, ha B elegendően nagy. Tehát: minél
nagyobb a B, azaz minél nagyon frekvenciával működik az NMR készülék, annál inkább lehet
látni a spin-spin felhasadást.
Példák:
C2 H 4 O
megoldás: acetaldehid
C3 H5 O2 Cl
103
megoldás: CH3 -CHCl-COOH
n-propil-alkohol és izopropil-alkohol. Melyik melyik spektruma?
6.5.4.
Gyakorlati NMR spektroszkópia
a) Szerkezeti képlet megállapítása
• hány féle proton (vagy C atom) van benne
• ezek relatív száma
• spin-spin csatolás alapján a csoportok relatív helyzete
b) Időfüggő tényező
Határozatlansági reláció alapján:
∆E · ∆t =
h̄
2
ahol ∆t a két szint kötött átmenet ideje (szint élettartama). Mivel az NMR-ben a legkisebb a
gerjesztési energia, ezért itt a legnagyobb az idő. Ha a valamely folyamat (pl. protoncsere) ennél
gyorsabban játszódik le, akkor a jel a folyamatra jellemző átlag lesz.
Okok:
104
• természetes
– intermolekuláris (pl. protoncsere)
etanolban az OH proton szignálja nem hasad fel, mert H-kötésben gyors protoncsere
(ha eléggé kiszárítjuk, felhasad)
– intramolekuláris (pl. konformációváltozás, belső rotáció)
dimetil-amid esetén a parciális kettőskötés miatt a rotáció gátolt. A gát magassága
megmérhető a spektrum hőmérsékletfüggéséből.
ciklohexán: az axiális és ekvatoriális protonok jele nem válik szét, mert gyors konformációváltozás, szobahőmérsékleten egyetlen éles sáv.
• mesterséges – telítjük a szintet, attól megszűnik a csatolás
(erre 13 C spektrum esetén van szükség, különben a protonokkal való csatolás borzalmasan
bonyolult spektrumot adna. (13 C-13 C csatolás nem lép fel, mert kicsi az előfordulása, így
egy molekulában nagy valószínűséggel csak egy ilyen mag van.)
6.5.5.
Az ESR spektroszkópia
Mágnesesség eredete: kompenzálatlan elektron spin → gyökök vizsgálata.
µB =
e
h̄
2mel
Az elektron tömeg három nagyságrenddel kisebb, mint a magé, ezért a felhasadás itt jóval nagyobb lesz → átmenet a mikrohullámú tartományba esik.
Legegyszerűbb: H-atom, ahol két szint van; l. korábban
Alkalmazás:
• átmeneti fémek komplexei: megkülönböztethetők a gyenge és erős kristályterek (l. korábban)
• gyökök vizsgálata: pl. gyökös polimerizáció követése
105