Transcript Molekylfysik - Leiden Univ

4. Molecular structure
The Schrödinger Equation for molecules

The Born-Oppenheimer approximation

4.1. Molecular orbital theory
4.1.1 The hydrogen molecule-ion
4.1.2 The structure of diatomic molecules
4.1.3 Heteronuclear diatomic molecules
4.1.4 Energy in the LCAO approach
4.2. Molecular orbitals for polyatomic systems
4.2.1 The Hückel approximation
4.2.2 The band theory of solids
 2+

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.
2

Tˆtot = 
2
Vˆtot  
1
4 0
n
M

i=1 j 1
2
M

j=1
j
2
2

2
mj
1
Z je
+
4 0
ri  R j
n

i=1 i>i
i

i=1 mi
n
2
e2
1
+
ri  ri '
4 0
M

j=1 j   j
2
Z j Z j e
R j  R j'
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.
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 16: the vibrational and rotational spectroscopies are 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”:
Tˆ Vˆ
e
eN

 Vˆee  elect(R, r) = Eelect  elect(R, r)
 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, we’ll only speak about
electronic wavefunction.
The electronic Schrödinger equation
Tˆ Vˆ
e

ˆ  elect(R, r) = E  elect(R, r)

V
eN
ee
elect
 The nuclear coordinate R appears as a parameter in the
expression of the electronic wave function.
R
 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), we can solve the electronic SE and evaluate the energy that
the molecule would have in this structure: the molecular
potential energy curve is obtained (see Figure).
D0
 The molecule is the most stable (minimum of energy) for
one specific position of the nuclei: the equilibrium position Re.
 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 ½ħ :
D0=De- ½ ħ
e-
4.2. Molecular orbital (MO) theory
4.2.1 The hydrogen molecule-ion:
rA
H-H+
A. Linear combination of atomic orbitals (LCAO)
A
rB
R
B
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. Let’s write the atomic orbitals on the two atoms by the letters A and B.
= N(A  B)
N is the normalization constant.


2
2
2

*

d


N
A
d


B


 d  2 AB d  1
 N (1  1  2S )  1
2
N
1
2(1  S )1/ 2
S   AB d is the overlap integral related to the overlap of the 2 AO
+
B. Bonding orbitals
The LCAO-MO: += {2(1+S)}-1/2(A + B), with A=1s, B=1s of H
Probability density: 2+= {2(1+S)}-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 AO’s 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)
 2-
-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
D. Energy of the states  and *
H-H+: One electron around 2 protons
e2  1 1 1 
   
V =4 0  rA rB R 
2 2
H 
e  V
2me
+= {2(1+S)}-1/2(A + B)
E = EH 1s 
H=E
-= {2(1-S)}-1/2(A - B)
e2
 j k 
E = EH 1s 


4 0 R  1  S 
erA
A
rB
R
=-
=+
B
rB
B   1Hs
1  a0

e
3
a0
A   1Hs
1  a0

e
a03
 
rA
 
 jk 


4 0 R  1  S 
e2
e2
j
4 0
e2
k
4 0

A2
d  0
rB
j= measure of the interaction between a nucleus and
the electron density centered on the other nucleus

AB
d  0
rA
k= measure of the interaction between a nucleus and
the excess probability in the internuclear region
S   AB d  0
S= measure of the overlap between 2 AO. S decreases
when R increases. Note: S=0 for 2 orthogonal AO.
4.2.2 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
He2: 4 electrons  ground-state configuration: 12 2*2
E
EE+
E+< E-
 He2 is not
stable and does not exist
Increase of electron density
B. Bond order
Bond order:
n= number of electrons in the bonding orbital
b=½(n-n*)
n*= number of electrons in the antibonding orbital
 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.
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 S=1
(paramagnetic)
The two 2px give one x and one x*
The two 2py give one y and one y*
Bond order = 2
Note: The  orbitals together give rise to an cylindrical
distribution of charge. Electrons can circulate around this torus
can create magnetic effect detected in NMR
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 AO’s, they appear as the resulting interference between s and p orbitals.
sp3 hybridization
sp3
h1= s + px + py + pz
h2= s - px + py - pz
+
h3= s - px - py + pz
h4= s + px - py - pz
Each hybrid orbital has the same energy and can be occupied by
one electron of the promoted atom  CH4 has 4 similar bonds.
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
120°
h3= s - (3/2)1/2 px - (1/2)1/2 py
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
H
H
H
H
 In ethene CH2=CH2, 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.
sp hybridization
In ethyne, HCCH: 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.
x
h1= s + pz
h2= s - pz
y
z
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.
H
Sketch of the transition state
species during a nucleophilic
substitution SN2
Nu
C
H H
X
CH3X + Nu-  CH3Nu + X-
4.2.3 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
|ci|2= 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): |cA|2= |cB|2
2) The ionic bond in A+B- : |cA|2= 0 and |cB|2=1
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.
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, we’ll 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.
(cA, cB)
=cAA + cBB
H=E 
E >Etrue
Aim: Try to find the (cA, cB) that minimize the calculated E
4.2.4 Energy in the LCAO approach
(1)
 * H  d

E
 * d
 Numerator:  * H  d   cA A  cB B H cA A  cB B  d
 cA2  AHA d  cB2  BHB d  c AcB  AHB d  cAcB  BHA d
 A   AHA d
 B   BHB d
   AHB d   BHA d
A is a Coulomb integral: it is related to the energy of the e- when it occupies A. ( < 0)
 is a Resonance integral: it is zero if the orbital don’t overlap. (at Re, <0)

2
2

*
H

d


c


c
 B  2c AcB 
A
A
B

2
 Denominator:  d   c A A  cB B  d
2
 c A2  A2 d  cB2  B 2 d  2c AcB  AB d
 c  c  2c AcB S
2
A
2
B
(1)
c A2 A  cB2 B  2c AcB 
E
c A2  cB2  2c AcB S
(1)
c A2 A  cB2 B  2c AcB 
E
c A2  cB2  2c AcB S
X  cA2 A  cB2 B  2cAcB   cA2 E  cB2 E  2cAcB SE  0
Let’s find the “zeros” or roots of the polynomial vs. cA and cB
X
0
c A
We want the cA minimizing E, we then impose:
X
0
cB
We want the cB minimizing E, we then impose: E  0
cB
2c A A  2cB   2c A E  2cB SE  0
2c A   2cB B  2cB E  2c A SE  0


E
0
c A
c A  A  E  cB   SE  0


c A   SE  cB  B  E  0
Secular
equations
In order to have a solution, other than the simple solution cA= cB= 0, we must have:

   SE  0
  E 
E
  SE

A
A

Secular determinant should be zero
B

 E  B  E    SE  0
2
The 2 roots give the energies of the bonding
and antibonding molecular orbitals formed
from the AOs
D. Two simple cases
1) Homonuclear diatomic molecules: =cAA + cBB with A= B  A= B= 
E 
1 S
 
E 
1 S
  E 2    SE2  0


c A   E  cB   SE  0
(1)
c A   SE  cB   E  0
(2)



(1)
(2)
cA 
1
21  S 1/ 2
cB  c A
cA 
1
21  S 1/ 2
cB  c A
bonding
antibonding
0

E 
 
1 S

antibonding= {2(1-S)}-1/2(A - B)
E 

1 S
bonding= {2(1+S)}-1/2(A + B)
0
E 
 
1 S

Eantibonding= - EEbonding = E+-
Eantibonding    E 
Ebonding
E 

1 S
  S
1 S
  S
 E   
1 S
Since: 0 < S < 1  Eantibonding > 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 (S=0), Eantibonding = Ebonding = 
 The resonance integral  is an indicator of the strength of covalent bonds
2) Heteronuclear diatomic molecules when we neglect the overlap integral: S=0

   SE  0
  E 
E     E  
E
  SE
A

S=0
A
B

E


B
2
A
2


B
E


0
A

  A B  0

   A   B  4 2
2
E
 A B
2


 E B  E   2  0
  A B



2
2


 4 2
A B
2
2
Limit case: if (A -B)>> :
A
B
E+A and + A: the 2 electrons are localized
on one atom, it’s the case of a 100% ionic bond.
 A B
E-
E+
The higher the difference (A -B), the more the
ionic character of the bond will be. The smaller
(A -B), the more covalent the bond will be.
2


  B  4 2
Note: for a complete resolution of the problem, we need to inject the values
2
2
A
of E in the secular equations and find the coefficients of the wavefunction.
Eb a  E  E 

A  B

2
 4 2
4.3. Molecular orbitals for polyatomic systems
4.3.1 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, CH2=CH2: =cAA + cBB, where A and B are the C2pz orbitals of each carbon atoms.
 Butadiene, CH2=CH-CH=CH2: =cAA + cBB+ccC + 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.
Following these methods and since A= B= , we obtain those secular determinants:
 Ethene, CH2=CH2:
  E    SE
0
  SE   E 
  E 
 AB  S AB E   AC  S AC E   AD  S AD E 
  S BA E 
  E 
 BC  S BC E   BD  S BD E 
 Butadiene, CH2=CH-CH=CH2: BA
0
 CA  SCA E   CB  SCB E 
  E 
 CD  SCD E 
 DA  S DA E   DB  S DB E   DC  S DC E 
  E 
Hückel approximation:
1) All overlap integrals Sij= 0 (ij).
2) All resonance integrals between non-neighbors, i,i+n=0 with n 2
3) All resonance integrals between neighbors are equal, i,i+1= i+1,i+2 =
 Severe approximation, but it allows us to calculate the general picture of the molecular
orbital energy levels.
B. Ethene and frontier orbitals
Within the Hückel approximation, the secular determinant becomes:
  E 


  E 
   E    2  0
2
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 17
Ethene
ethene deshydrogenation
nickel
Ethyne
http://www.fhi-berlin.mpg.de/th/personal/hermann/pictures.html
C. Butadiene and delocalization energy
  E 

0

  E 
0
0

  E 
0


0
0

  E 
0
 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
3 nodes
= E4
 The greater the number of internuclear nodes, the
higher the energy of the orbital
2 nodes
= E3
LUMO= 3*
= E2
1 node
HOMO= 1
0 node
= E1
Top view of the MOs
 Butadiene C4H6: total -electron binding energy, E is
E = 2E1+2E2= 4 + 4.48 with two -bonds
 Ethene C2H4:E = 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”
D. Benzene and aromatic stability
Each C has: - 3 electrons in (sp2) hybrid orbitals  3 bonds per C
Scheme of the different
orbital overlaps
- 1 electron in 2pz  one  bond per C
  E 
6*

0
0
0


0
0
0

0
0

0

  E 
0

  E 
0
0

  E 
0
0
0

  E 

0
0
0

E6 =  - 2
0

  E 
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
5*
E2,3 =  + 
2
1
3
E1 =  + 2
E = 2E1+4E2= 6 + 8 with three -bonds
 Three ethene molecules give: E = 6 + 6 for 3 separated
-bonds.
 The delocalization energy is 2 (-150kJ/mol)
 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*
2
5*
3
3
2
1
1
E. Is it possible to extend the conjugated chain?
What happens to the electronic structure?
We are used to the great impact scientific discoveries have on our ways of thinking. This year's Nobel Prize in Chemistry is no exception. What
we have been taught about plastic is that it is a good insulator - otherwise we should not use it as insulation in electric wires. But now the time
has come when we have to change our views. Plastic can indeed, under certain circumstances, be made to behave very like a metal - a
discovery for which Alan J. Heeger, Alan G. MacDiarmid and Hideki Shirakawa are to receive the Nobel Prize in Chemistry 2000.
How can plastic become conductive?
Plastics are polymers, molecules that form long chains, repeating themselves like pearls in a necklace. In becoming electrically conductive, a
polymer has to imitate a metal, that is, its electrons need to be free to move and not bound to the atoms. The first condition for this is that the
polymer consists of alternating single and double bonds, called conjugated double bonds. Polyacetylene is prepared through polymerization of
the hydrocarbon acetylene.
Polyacetylene
However, it is not enough to have conjugated double bonds. To become electrically conductive, the plastic has to be disturbed - either by
removing electrons from (oxidation), or inserting them into (reduction), the material. The process is known as doping.
4.3.2 The band theory of solids
 Solids are composed of a 3-dimensional array of atoms bound to each other via their valence
atomic orbitals. The resulting combinations of these atomic orbitals, involved in the bonds
forming this 3-D array, are orbitals spread all over the solid.
 Solids can be classified with respect to the behavior of
their electrical conductivity () vs. Temperature (T):
 A metallic conductor:  decreases as T is raised
 A semiconductor:  increases as T is raised
Note that an insulator appears as a semiconductor with
very low conductivity.
doped
Conductivity 
Note: S= -1
A. Formation o f bands
 Simple model for a solid: the one-dimensional solid, which consists of a single, infinitely long
line of atoms, each one having one s orbital available for forming molecular orbitals (MOs).
When the chain is extended:
 The range of energies covered by the MOs
is spread
 This range of energies is filled in with more
and more orbitals
 For a chain of N, the width of the range
of energies of the MOs is finite, while the
number of molecular orbitals is infinite: This is
called a band .
4
“s” band
 For a chain of N atoms, i.e. for N atomic orbitals 1s, the chain has N molecular orbitals. The
energy of these MOs can be found by solving the secular determinant:
  E 

0
0
....
0

0
....
0

....
0
....
0
....


  E 
0

  E 
0
0

  E 




0
0
0
0
0
 k 
Ek    2  cos

 N 1
k  1,2,...,N
....   E 
 Each molecular orbital is characterized by a
energy labeled with k.
 For N, the Ek-Ek+1 0. But the “s” band
still has finite width: EN -E1  4
 When the overlap of “p” atomic orbitals is
taken into account, a “p” band is formed. Their
may be a “band gap”: a range of energies to
which no orbitals correspond.
k=N: highest energy orbital  fully antibonding
k=1: lowest energy orbital  fully bonding
B. Influence of the temperature
 If each atom of the solid (composed of N atoms) contributes with one electron and since 2
electrons can be in one molecular orbital,
 then only half of the N molecular orbitals are filled at T=0. The HOMO is called the
Fermi level. The HOMO-LUMO energy difference is zero this is a metal.
 Since the energy between the HOMO and the LUMO is zero, then just a
little energy can promote an electron in an unoccupied level. Under
electrical potential difference, some electrons are therefore very mobile and
give rise to electrical conductivity.
 The excitation energy can be provided via an
increase of temperature. The population of the
orbitals is given by the Fermi-Dirac distribution:
1
P   E    / kT
e
1
 is the electron chemical potential, that is -EF for metals (T=0)
 When T increases, there are more electrons excited towards
empty orbitals…. However the conductivity decreases because the
vibration of the nuclei increases  more collisions between the
transported electrons and the nuclei  less efficient transport.
 When the HOMO-LUMO energy difference is non-zero 
there is an electronic gap. It is the case of semiconductors.
 If the band gap is small, thermal excitations can promote
electrons to unoccupied levels; consequently, those electrons
can participate to the electrical conductivity. That’s why the
conductivity of semiconductor increases with the temperature.
 The insulator are characterized by a huge band gap  the
electrons cannot reach the unoccupied levels: no conductivity.