1. Molecular Orbitals

2. Atomic orbitals interact to form molecular orbitals
Electrons are placed in molecular orbitals
following the same rules as for atomic orbitals
In terms of approximate solutions to the Scrödinger equation
Molecular Orbitals are linear combinations of atomic orbitals (LCAO)
Y = caya + cbyb (for diatomic molecules)
Interactions depend on the symmetry properties
and the relative energies of the atomic orbitals

3. If the total energy of the electrons in the molecular orbitals
As the distance between atoms decreases
Atomic orbitals overlap
Bonding takes place if:
the orbital symmetry must be such that regions of the same sign overlap
the energy of the orbitals must be similar
the interatomic distance must be short enough but not too short
If the total energy of the electrons in the molecular orbitals
is less than in the atomic orbitals, the molecule is stable compared with the atoms

4. Y = N[caY(1sa) ± cbY (1sb)]
Combinations of two s orbitals (e.g. H2)
Antibonding
Bonding
More generally:
Y = N[caY(1sa) ± cbY (1sb)]
n A.O.’s
n M.O.’s

5. (total energy is raised)
Electrons in antibonding orbitals cause mutual repulsion between the atoms
(total energy is raised)
Electrons in bonding orbitals concentrate between the nuclei and hold the nuclei together
(total energy is lowered)

6. Both s (and s*) notation means symmetric/antisymmetric with respect to rotation
Not s

7. Combinations of two p orbitals (e.g. H2)
s (and s*) notation means no
change of sign upon rotation
p (and p*) notation means
change of sign upon C2 rotation

8. Combinations of two p orbitals

9. Combinations of two sets of p orbitals

10. Combinations of s and p orbitals

11. Combinations of d orbitals
No interaction – different symmetry
d means change of sign upon C4

12. Is there a net interaction?NO NO
YES

13. Relative energies of interacting orbitals must be similar
Strong interaction
Weak interaction

14. for diatomic molecules From H2 to Ne2
Molecular orbitals
for diatomic molecules
From H2 to Ne2
Electrons are placed
in molecular orbitals
following the same rules
as for atomic orbitals:
Fill from lowest to highest
Maximum spin multiplicity
Electrons have different quantum numbers including spin (+ ½, - ½)

16. symmetric/antisymmetric
O2 (2 x 8e)
1/2 (10 - 6) = 2
A double bond
Or counting only
valence electrons:
1/2 (8 - 4) = 2
Note subscripts
g and u
symmetric/antisymmetric
upon i

17. Place labels g or u in this diagram
s*u
p*g pu sg

18. s*u sg
g or u?
p*g pu
d*u dg

19. Orbital mixing Same symmetry and similar energies !
shouldn’t they interact?

20. When two MO’s of the same symmetry mix
s orbital mixing
When two MO’s of the same symmetry mix
the one with higher energy moves higher and the one with lower energy moves lower

21. for diatomic molecules
Molecular orbitals
for diatomic molecules
From H2 to Ne2
H2 sg2 (single bond)
He2 sg2 s*u2 (no bond)

22. E (Z*) DE s > DE p C2 pu2 pu2 (double bond)
Paramagnetic
due to mixing
C2 pu2 pu2 (double bond)
C22- pu2 pu2 sg2(triple bond)
O2 pu2 pu2 p*g1 p*g1 (double bond)
paramagnetic
O22- pu2 pu2 p*g2 p*g2 (single bond)
diamagnetic

23. Bond lengths in diatomic molecules
Filling bonding orbitals
Filling antibonding orbitals

24. Photoelectron Spectroscopy

25. O2 N2 sg (2p) p*u (2p) pu (2p) sg (2p) pu (2p) s*u (2s) s*u (2s)
Very involved in bonding
(vibrational fine structure)
s*u (2s)
(Energy required to remove electron, lower energy for higher orbitals)

26. Simple Molecular Orbital Theory
A molecular orbital, f, is expressed as a linear combination of atomic orbitals, holding two electrons.
The multi-electron wavefunction and the multi-electron Hamiltonian are
Where hi is the energy operator for electron i and involves only electron i

27. MO Theory - 2 Seek F such that
Divide by F(1,2,3…) recognizing that hi works only on electron i.
Since each term in the summation depends on the coordinates of a different electron then each term must equal a constant.

28. MO Theory - 3
Recall the expansion of a molecular orbital in terms of the atomic orbitals.
Multiply by uk and integrate.
Define
These integrals are fixed numerical values.
Substituting the expansion for f

29. MO Theory - 4
For k = 1 to AO
These are the secular equations. The number of such equations is equal to the number of atomic orbitals, AO.
There are AO equations with AO unknowns, the al.
For there to be a nontrivial (all al equal to zero) solution to the set of secular equations then the determinant below must equal zero

30. MO Theory 6
Drastic assumptions can now be made. We will use the simple Huckle approximations.
hi,i = a, if orbital i is on a carbon atom.
Si,i = 1, normalized atomic orbitals
hi,j = b, if atom i bonded to atom j, zero otherwise
Expand the secular determinant into a polynomial of degree AO in e. Obtain the allowed values of e by finding the roots of the polynomial. Choose one particular value of e, substitute into the secular equations and obtain the coefficients of the atomic orbitals within the molecular orbital.

31. Example The allyl pi system. The secular equations:
(a-e)a1 + b a a3 = 0
ba1 + (a-e) a2 + b a3 = 0
0 a1 +b a2 + (a-e) a3 = 0
Simplify by dividing every element by b and setting (a-e)/b = x

32. For x = -sqrt(2)
e = a + sqrt(2) b
For x = 0
normalized

33. Verify that
h f = e f

34. Perturbation Theory
The Hamiltonian is divided into two parts: H0 and H1
H0 is the Hamiltonian of for a known system for which we have the solutions: the energies, e0, and the wavefunctions, f0. H0f0 = e0f0
H1 is a change to the system and the Hamiltonian which renders approximation desirable. The change to the energies and the wavefunctions are expressed as a summation.

35. Corrections Energy Zero order (no correction): ei0
First Order correction:
Wave functions corrections to f0i
Zero order (no correction): f0i
First order correction:

36. Example Pi system only: Perturbed system: allyl system
Unperturbed system: ethylene + methyl radical

39. Mixes in anti-bonding
Mixes in bonding
Mixes in anti-bonding
Mixes in bonding

40. Projection Operator
Algorithm of creating an object forming a basis for an irreducible rep from an arbitrary function.
Where the projection operator results from using the symmetry operations multiplied by characters of the irreducible reps. j indicates the desired symmetry.
lj is the dimension of the irreducible rep.
Starting with the 1sA create a function of A1 sym
¼(E1sA + C21sA + sv1sA + sv’1sA) = ½(1sA + 1sB)