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
 2+
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
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.
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
Rj  Rj'
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.
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”:
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
eN

 Vˆee  elect ( R, r ) = Eelect  elect ( R, r )
 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), 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).
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
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.
+= A(1)B(2) + A(2)B(1)
When the atoms are close, it’s not
possible to know whether it is
electron 1 that is on A or electron 2.
 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) → between the two nuclei: |+|2>0 
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”.
-= A(1)B(2) - A(2)B(1)
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
4.3.2 Homonuclear diatomic molecules
N
1s2 2s2

2px1 2py1 2pz1
Valence electrons
y

z
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.
x



4.3.3 Polyatomic molecules
O
x
1s2 2s2
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°.
2px2 2py1 2pz1
H 1s1
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”
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-
e-
4.4. Molecular orbital (MO) theory
4.4.1 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.
Example for += N(A + B):
 * d  N  A
2
2

d   B 2 d  2 AB d  1
 N 2 (1  1  2S )  1
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
2 2
H 
e  V
2me
e2  1 1 1 
   
V =40  rA rB R 
+= {2(1+S)}-1/2(A + B)
E  = E H 1s 
H=E
-= {2(1-S)}-1/2(A - B)
e2
 jk 
E  = E H 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.
4.4.2 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
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) (see Chap 3)
The two 2px give one x and one x*
The two 2py give one y and one y*
Bond order = 2
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
|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.
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:
M=½ (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. EA>0
when the addition of the electron releases energy, i.e. when it stabilizes the molecule.
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
4.4.4 Energy in the LCAO approach
(1)
 * H  d

E
 * d
 Numerator:  * H  d   c A A  cB B H c A A  cB B  d
 c A2  AHA d  cB2  BHB d  c AcB  AHB d  c AcB  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
Let’s find the coefficients cA and cB:
cA should minimize E:
cB should minimize E:
E
0
c A
E
0
c B


c A  A  E  cB   SE   0


c A   SE   cB  B  E  0
Secular
equations
See next
page
In order to have a solution, other than the simple solution cA= cB= 0, we must have:

   SE   0
  SE    E 
A
E
Secular determinant should be zero
B

A


 E  B  E    SE   0
2
The 2 roots give the energies of the bonding
and antibonding molecular orbitals formed
from the AOs
How to get the secular equations?
c A2 A  cB2 B  2c AcB  P
E

c A2  cB2  2c AcB S
R
 P  P' R  RP '
D  
R2
R

 

E
2c A A  2cB  R  2c A  2cB S P

C A
R2

 

P

R  2c A A  2cB   2c A  2cB S 
E
R
 
0
C A
R2

 

if 2c A A  2cB   2c A  2cB S E  0
P
R
i.e. c A  A  E  cB   SE   0
E


D. Homonuclear diatomic molecules
1) : =cAA + cBB with A= B → A= B= 
E 
1 S
 
E 
1 S
  E 2    SE 2  0
c A   E   cB   SE   0
c A   SE   cB   E   0
(1)
(2)

(1)
(2)
cA 
1
21  S 1/ 2
cB  c A
cA 
1
21  S 1/ 2
c B  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 
E 

1 S
  S
1 S
  S
Ebonding  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
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, 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 7
Ethene
ethene deshydrogenation
nickel
Ethyne
http://www.fhi-berlin.mpg.de/th/personal/hermann/pictures.html
4.5.2 Evolution of electronic structure with the size of the molecules
A. Butadiene and delocalization energy
  E 

0

  E 
0

  E 
0
0


0
0
 4th order polynomial  4 roots E
0

There is 1e- in each 2pz orbital of the four carbon atoms
  E   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
0

  E 
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