Molekylfysik

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4. Electronic structure of molecules
?
4.1 The Schrödinger Equation for molecules 4.2
The Born-Oppenheimer approximation 4.3
Valence-bond theory 4.3.1 The hydrogen
molecule 4.3.2 Polyatomic molecules 4.3.3
Hybridization 4.4. Molecular orbital
theory 4.4.1 The hydrogen molecule-ion 4.4.2
The structure of diatomic molecules 4.4.3
Heteronuclear diatomic molecules 4.4.4 Energy in
the LCAO approach 4.5 Molecular orbitals for
small conjugated molecules 4.5.1 The Hückel
approximation 4.5.2 Evolution of electronic
structure with the size of the molecules
?
?
?2
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4.1 The Schrödinger Equation for molecules
? All properties of a molecule ( M nuclei n
electrons) can be evaluated if we find the
wavefunction ?(x1, x2,..., xn) by solving the
Schrödinger equation (SE) H? E?. H Ttot
Vtot (TN Te ) (VeN Vee VNN) The total
kinetic operator of the molecule is composed of a
part for the nuclei TN and one for the electrons
Te. The total potential energy operator is the
sum of the electron/electron (Vee),
electron/nucleus (VeN) and nucleus/nucleus (VNN)
interactions. Rj are the nucleus
coordinates and ri the electronic coordinates.
Mj is the mass of the nucleus j and mi is the
electron mass. Zj is the number of protons in the
nucleus j and e is the charge of an electron. ?
Solving with the best approximation the SE is the
challenge of quantum chemistry (W. Kohn and J.
Pople Nobel Prize in Chemistry 1998). In this
chapter, we introduce the vocabulary and the
basic principles to understand the electronic
structure of molecules.
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4.2 The Born-Oppenheimer approximation
? The electrons are much lighter than the nuclei
(me/mH?1/1836) ? their motion is much faster than
the vibrational and rotational motions of the
nuclei within the molecule. ? A good
approximation is to neglect the coupling terms
between the motion of the electrons and the
nuclei this is the Born-Oppenheimer
approximation. The Schrödinger equation can then
be divided into two equations 1) One describes
the motion of the nuclei. The eigenvalues of this
nuclear part of the SE gives the discrete
energetic levels of the vibration and rotation of
the molecule. ? see Chap 6 the vibrational
spectroscopy is used to observed transition
between these energetic levels. 2) The other
one describes the motion of the electrons around
the nuclei whose positions are fixed. This
electronic part of the SE is the electronic
Schrödinger equation ? The knowledge of the
electronic wavefunction is necessary to
understand chemical bonding, electronic and
optical properties of the matter. In the rest of
the chapter, well only speak about electronic
wavefunction.
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The electronic Schrödinger equation
? The nuclear coordinate R appears as a parameter
in the expression of the electronic wave
function. ? An electronic wave function
?elect(R,r) and an energy Eelect are associated
to each structure of the molecule (set of nuclei
coordinates R). ? For each variation of bond
length in the molecule (each new R), the
electronic SE can be solved and the energy that
the molecule would have in this structure can be
estimated the molecular potential energy curve
is obtained (see Figure). ? The molecule is the
most stable (minimum of energy) for one specific
position of the nuclei the equilibrium position
Re.
R
? The zero energy corresponds to the dissociated
molecule. ? The depth of the minimum, De, gives
the bond dissociation energy, D0, considering the
fact that vibrational energy is never zero, but
½h? D0De- ½ h?
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4.3 Valence-bond theory
4.3.1 The hydrogen molecule
A(1) ?H1sA(r1) is the wavefunction of the
electron 1 on the 1s orbital of Hydrogen A. When
the atoms are close, its not possible to know
whether it is electron 1 that is on A or electron
2.
? A(1)B(2) A(2)B(1)
? The system is described by a superposition of
the wavefunctions for each possibility A(1)B(2)
and A(2)B(1). Two linear combinations are
possible
?- A(1)B(2) - A(2)B(1)
? A(1)B(2) A(2)B(1) ? between the two nuclei
?2gt0 ? creation of a ? bond described by a
bonding molecular orbital stabilization of the
2-H system. ?- A(1)B(2) - A(2)B(1) ? between
the two nuclei, ?- changes sign ? ?-2 0 ?
creation of a ? bond described by a antibonding
molecular orbital.
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What is the spin of the 2 electrons involved in
the ? bond?
? The total electronic wavefunction ?(x1,x2) of
H2 must be antisymmetric. The spatial orbital
forming the ? bond is symmetric ? the spin
function S(1,2) should be antisymmetric (see Chap
3). The only antisymmetric spinfunction for a
system of 2 e- corresponds to antiparallel
spins. ?(1,2) ?(1,2) S(1,2) 1/21/2 A(1)B(2)
A(2)B(1) ?(1)?(2) - ?(1)?(2) ? The spatial
orbital ?(1,2) gives an energy for the 2-H
system that is lower than the energy of 2
Hydrogen atoms far from each other bond
formation. ? The spinfunction corresponding to
this stable situation shows that the spins are
paired.
? The formation of a bond is achieved if the
electron spins are paired
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4.3.2 Homonuclear diatomic molecules
N 1s2 2s2 2px1 2py1 2pz1
?
Valence electrons
?
?
In N2 3 bonds are formed by combining the 3
different 2p orbitals of the 2 nitrogen atoms.
This is possible because of the symmetry and the
position of those 2p orbitals with respect to
each other. ? one ? bond and 2 ? bonds
(perpendicular to each other) are formed by spin
pairing.
?
?
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4.3.3 Polyatomic molecules
O 1s2 2s2 2px2 2py1 2pz1
x
H 1s1 H 1s1
y
z
H2O This simple model suggests that the formed
angle between the ?(O-H) bonds is 90, as well as
their position vs. the paired 2px electrons
whereas the actual bond angle is 104.5.
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A. Promotion
? C 1s2 2s2 2px1 2py1. With the valence bond
theory, we expect maximum 2 bonds. ? But,
tetravalent carbon atoms are well known e.g.,
CH4. This deficiency of the theory is
artificially overcome by allowing for
promotion. Promotion the excitation of an
electron to an orbital of higher energy. This is
not what happens physically during bond
formation, but it allows to feel the energetics.
Indeed, following this artificial excitation, the
atom is allowed to create bonds and
consequently, the energy is stabilized more than
the cost of the excitation energy. ? C 1s2 2s2
2px1 2py1 ? 1s2 2s1 2px1 2py1 2pz1. CH4
should be composed of 3? bonds due to the overlap
between the 2p of C and the 1s of H, and another
? bond coming from the 2s of C and the 1s of
H. ? But, it is known that CH4 has 4 similar
bonds. This problem is overcome by realizing that
the wavefunction of the promoted atom can be
described on different orthonormal basis sets 1)
either the orthogonal hydrogenoic atomic orbitals
(AO) 1s, 2s, 2px, 2py, 2pz. 2) or equivalently,
from another set of orthonormal functions the
hybrid orbitals
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B. Hybridization
An orthonormal set of hybrid orbitals is created
by applying a transformation on the orthonormal
hydrogenic orbitals. The sp3, sp2 or sp hybrid
orbitals are linear combinations of the AOs,
they appear as the resulting interference between
s and p orbitals. sp3 hybridization

-
Each hybrid orbital has the same energy and can
be occupied by one electron of the promoted atom
? CH4 has 4 similar bonds.
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sp2 hybridization
hi
? The sp2 hybrid orbitals lie in a plane and
points towards the corners of an equilateral
triangle. 2pz is not involved in the
hybridization, and its axis is perpendicular to
the plane of the triangle.
h1 s 21/2 py h2 s (3/2)1/2 px - (1/2)1/2 py
h3 s - (3/2)1/2 px - (1/2)1/2 py
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An hybrid orbital has pronounced directional
character because it has enhanced amplitude in
the internuclear region, coming from the
constructive resulting interference.
Consequently, the bond formation is accompanied
with a high stability gain.
2pz
? In ethene CH2CH2, the hybrid orbitals of each
C atom create the backbone of the molecule via 3?
bonds (2 C-H and 2 C-C). The remaining 2pz of the
2 C atoms create a ? bond preventing internal
rotation.
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sp hybridization
In ethyne, HCCH Formation of 2 ? bonds (with C
and H) using the 2 hybrid orbitals h1 and h2. The
remaining 2px and 2py can form two ? bonds
between the two carbon atoms.
h1 s pz h2 s - pz
Other possible hybridization ?
Note 'Frozen' transition states pentavalent
carbon et al Martin-JC Science.vol.221,
no.4610 5 Aug. 1983 p.509-14. Organic ligands
have been designed for the stabilization of
specific geometries of compounds of nonmetallic
elements. These ligands have made possible the
isolation, or direct observation, of large
numbers of trigonal bipyramidal
organo-nonmetallic species. Many of these species
are analogs of transition states for nucleophilic
displacement reactions and have been stabilized
by the ligands to such a degree that they have
become ground-state energy minima. Ideas derived
from research on these species have been applied
to carbon species to generate a molecule that is
an analog of the transition state for the
associative nucleophilic displacement reaction.
The molecule is a pentavalent carbon species that
has been observed by nuclear magnetic resonance
spectroscopy.
Sketch of the transition state species during a
nucleophilic substitution SN2
CH3X Nu- ? CH3Nu X-
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4.4. Molecular orbital (MO) theory
4.4.1 The hydrogen molecule-ion H-H
A. Linear combination of atomic orbitals (LCAO)
The electron can be found in an atomic orbital
(AO) belonging to atom A (i.e. 1s of H) and also
in an atomic orbital belonging to B (i.e. 1s of
H) ? The total wavefunction should be a
superposition of the 2 AO and it is called
molecular orbital LCAO-MO. Lets write the atomic
orbitals on the two atoms by the letters A and
B. ? N(A B) N is the normalization constant.

Example for ? N(A B)
is the overlap integral related to the overlap of
the 2 AO
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?
B. Bonding orbitals
The LCAO-MO ? 2(1S)-1/2(A B), with A1s,
B1s of H
Probability density ?2 2(1S)-1(A2 B2
2AB) A2 probability density to find the e- in
the atomic orbital A B2 probability density to
find the e- in the atomic orbital B 2AB the
overlap density represents an enhancement of the
probability density to find the e- in the
internuclear region the electron accumulates in
regions where AOs overlap and interfere
constructively which creates a bonding orbital
?1 with one electron.
?2
C. Antibonding orbitals
?-
?- 2(1-S)-1/2(A - B) Probability density
?2- 2(1-S)-1(A2 B2 - 2AB) -2AB reduction
of the probability density to find the e- in the
internuclear region there is a destructive
interference where the two AO overlap, which
creates an antibonding orbital ?. ? Nodal plane
between the 2 nuclei
?2-
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D. Energy of the states ? and ?
H-H One electron around 2 protons
? 2(1S)-1/2(A B)
?-
H ? E ?
?- 2(1-S)-1/2(A - B)
?
j measure of the interaction between a nucleus
and the electron density centered on the other
nucleus
k measure of the interaction between a nucleus
and the excess probability in the internuclear
region
S measure of the overlap between 2 AO. S
decreases when R increases. Note S0 for 2
orthogonal AO.
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4.4.2 Structure of diatomic molecules
Now, we use the molecular orbitals (? ? and ?
?-) found for the one-electron molecule H-H in
order to describe many-electron diatomic
molecules.
A. The hydrogen and helium molecules
H2 2 electrons ? ground-state configuration 1?2
E
Increase of electron density
He2 4 electrons ? ground-state configuration
1?2 2?2
E
E-
Elt E- ? He2 is not stable and does not exist
E
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B. Bond order
n number of electrons in the bonding orbital n
number of electrons in the antibonding orbital
Bond order b½(n-n) ? The greater the bond
order between atoms of a given pair of elements,
the shorter is the bond and the greater is the
bond strength.
C. Period 2 diatomic molecules
According to molecular orbital theory, ? orbitals
are built from all orbitals that have the
appropriate symmetry. In homonuclear diatomic
molecules of Period 2, that means that two 2s and
two 2pz orbitals should be used. From these four
orbitals, four molecular orbitals can be built
1?, 2?, 3?, 4?.
1?, 2?, 3?, 4?.
With N atomic orbitals ? the molecule will have N
molecular orbitals, which are combinations of the
N atomic orbitals.
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dioxygen O2 12 valence electrons
The two last e- occupy both the ?x and the ?y
in order to decrease their repulsion. The more
stable state for 2e- in different orbitals is a
triplet state. O2 has total spin S1
(paramagnetic) (see Chap 3)
The two 2px give one ?x and one ?x The two 2py
give one ?y and one ?y Bond order 2
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4.4.3 Heteronuclear diatomic molecules
? A diatomic molecule with different atoms can
lead to polar bond, a covalent bond in which the
electron pair is shared unequally by the 2 atoms.
A. Polar bonds
? 2 electrons in an molecular orbital composed of
one atomic orbital of each atom (A and B). ? cA
A cB B ci2 proportion of the atomic orbital
i in the bond
? The situation of covalent polar bonds is
between 2 limit cases 1) The nonpolar bond
(e.g. the homonuclear diatomic molecule) cA2
cB2 2) The ionic bond in AB- cA2 0 and
cB21
Example HF The H1s electron is at higher energy
than the F2p orbital. The bond formation is
accompanied with a significant partial negative
charge transfer from H to F.
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B. Electronegativity
? A measure of the power of an atom to attract
electron to itself when it is part of a
compound. There are different electronegativity
definitions, e.g. the Mulliken electronegativity
?M½ (IP EA) ? IP is the ionization potential
the minimum energy to remove an electron from the
ground state of the molecule (Chap 3, p14). ? EA
is the electron affinity energy released when
an electron is added to a molecule. EAgt0 when the
addition of the electron releases energy, i.e.
when it stabilizes the molecule.
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C. Variation principle
All the properties of a molecule can be found if
the wavefunction is known e.g., the electron
density distribution or the partial atomic
charge. The wavefunction can be found by solving
the Schrödinger equation. But the latter is
impossible to solve analytically!! We use the
variation principle to go around that problem and
find approximate wavefunction. The variation
principle is the basic principle to determine
wavefunction of complicated molecular systems.
The idea is to optimize the coefficient cA and cB
of the wavefunction, ?cAA cBB, such that the
system is the most stable, i.e. the energy is
minimal. ? Variation principle If an arbitrary
wavefunction is used to calculate the energy, the
value calculated is never less than the true
energy. ? In the end of the chapter, well try
to find the best coefficients ci to give to the
trial wavefunction ? in order to approach the
true energy. ? More sophisticated methods (i)
increase the set of atomic orbitals, called the
basis set from which the molecular orbital is
expressed (ii) improve the description of the
system by using more correct Hamiltonian H.
?cAA cBB
H?E ?
(cA, cB)
E gtEtrue
Aim Try to find the (cA, cB) that minimize the
calculated E
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4.4.4 Energy in the LCAO approach
(1)
? Numerator
?A is a Coulomb integral it is related to the
energy of the e- when it occupies A. (? lt 0) ? is
a Resonance integral it is zero if the orbital
dont overlap. (at Re, ?lt0)
(1)
?
? Denominator
(1)
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Lets find the coefficients cA and cB
cA should minimize E
Secular equations
cB should minimize E
See next page
In order to have a solution, other than the
simple solution cA cB 0, we must have
Secular determinant should be zero
The 2 roots give the energies of the bonding and
antibonding molecular orbitals formed from the AOs
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How to get the secular equations?
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D. Homonuclear diatomic molecules
1) ?cAA cBB with A B ? ?A ?B ?
(1)
(2)
?bonding
(1)
(2)
?antibonding
0
?
?
?antibonding 2(1-S)-1/2(A - B)
?bonding 2(1S)-1/2(A B)
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0
?
?Eantibonding? - E-
?Ebonding E-?
Since 0 lt S lt 1 ? ?Eantibonding gt ?Ebonding
Note 1 He2 has 4 electrons ? ground-state
configuration 1?2 2?2 ? He2 is not
stable! Note 2 If we neglect the overlap
integral (S0), ?Eantibonding ?Ebonding ? ?
The resonance integral ? is an indicator of the
strength of covalent bonds
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4.5. Molecular orbitals of small conjugated
molecules
4.5.1 The Hückel approximation
Here, we investigate conjugated molecules in
which there is an alternation of single and
double bonds along a chain of carbon atoms. In
the Hückel approach, the ? orbitals are treated
separately from the ? orbitals, the latter form a
rigid framework that determine the general shape
of the molecule. All C are considered similar ?
only one type of coulomb integral ? for the C2p
atomic orbitals involved in the ? molecular
orbitals spread over the molecule.
A. The secular determinant
The ? molecular orbitals are expressed as linear
combinations of C2pz atomic orbitals (LCAO),
which are perpendicular to the molecular
plane. ? Ethene, CH2CH2 ?cAA cBB, where A
and B are the C2pz orbitals of each carbon
atoms. ? Butadiene, CH2CH-CHCH2 ?cAA
cBBccC cDD The coefficients can be optimized
by the same procedure described before express
the total energy E as a function of the ci and
then minimize the E with respect to those
coefficients ci. Inject the energy solutions in
the secular equations and extract the
coefficients minimizing E.
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Following these methods and since ?A ?B ?, we
obtain those secular determinants
? Ethene, CH2CH2
? Butadiene, CH2CH-CHCH2
Hückel approximation 1) All overlap integrals
Sij 0 (i?j). 2) All resonance integrals between
non-neighbors, ?i,in0 with n? 2 3) All
resonance integrals between neighbors are equal,
?i,i1 ?i1,i2 ? ? Severe approximation, but
it allows us to calculate the general picture of
the molecular orbital energy levels.
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B. Ethene and frontier orbitals
Within the Hückel approximation, the secular
determinant becomes
E- ? - ? energy of the Lowest Unoccupied
Molecular Orbital (LUMO) E ? ? energy of the
Highest Occupied Molecular Orbital (HOMO)
LUMO 2?
2?
HOMO 1?
? HOMO and LUMO are the frontier orbitals of a
molecule. ? those are important orbitals because
they are largely responsible for many chemical
and optical properties of the molecule. Note The
energy needed to excite electronically the
molecule, from the ground state 1?2 to the first
excited state 1?1 2?1 is provided roughly by
2? (? is often around -0.8 eV) ? Chap 7
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Ethene
ethene deshydrogenation
nickel
Ethyne
http//www.fhi-berlin.mpg.de/th/personal/hermann/p
ictures.html
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4.5.2 Evolution of electronic structure with the
size of the molecules
A. Butadiene and delocalization energy
? 4th order polynomial ? 4 roots E
There is 1e- in each 2pz orbital of the four
carbon atoms ? 4 electrons to accommodate in the
4 ?-type molecular orbitals ? the ground state
configuration is 1?2 2?2 ? The greater the
number of internuclear nodes, the higher the
energy of the orbital ? Butadiene C4H6 total
?-electron binding energy, E? is E? 2E12E2 4?
4.48? with two ?-bonds ? Ethene C2H4E? 2?
2? with one ?-bond ? Two ethene molecules give
E? 4? 4? for two separated ?-bonds. ? The
energy of the butadiene molecule with two ?-bonds
lies lower by 0.48? (-36kJ/mol) than the sum of
two individual ?-bonds this extra-stabilization
of a conjugated system is called the
delocalization energy
3 nodes
E4
2 nodes
E3
LUMO 3?
E2
1 node
HOMO 1?
E1
0 node
Top view of the MOs
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D. Benzene and aromatic stability
Scheme of the different orbital overlaps
Each C has - 3 electrons in (sp2) hybrid
orbitals ? 3? bonds per C - 1 electron in
2pz ? one ? bond per C
6?
E6 ? - 2?
6 electrons 2pz to accommodate in the 6
?-molecular orbitals ? the ground state
configuration is 1?2 2?2 3?2.
E4,5 ? - ?
4?
? Benzene C6H6 total ?-electron binding energy,
E? is E? 2E14E2 6? 8? with three ?-bonds ?
Three ethene molecules give E? 6? 6? for 3
separated ?-bonds. ? The delocalization energy is
2? (-150kJ/mol)
5?
E2,3 ? ?
2?
3?
E1 ? 2?
1?
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? Benzene C6H6 is more stable than the
hexatriene. Both molecules has 3 ?-bonds, but the
cyclic structure of benzene stabilizes even more
the ?-electrons. The symmetry of benzene creates
two degenerated ?-bonds (2? and 3?). When they
are occupied, this is a more stable situation
than the 1?2 2?2 3?2 configuration for hexatriene
? aromatic stability
6?
4?
5?
3?
2?
3?
2?
1?
1?