Download Lecture 9 - ChemWeb (UCC)

Electronic Configurations:
What is the symmetry of the molecular
electronic state?
Electrons occupy the molecular orbitals, the
overall wavefunction of the state is a
product of electrons in orbitals.
Molecular orbitals of molecules transform
as one of the representations of the
molecular point group.
The symmetry of the state is the product of
the symmetry of occupied molecular
orbitals.
Most orbitals are fully occupied (2, 4 or 6
electrons), which results in symmetry Γ1 .
The symmetry of the state is due to the
unpaired electrons. Multiply the symmetries
of the unpaired electrons.
e.g. H2O
b2
ground state
configuration
a12 b22 a12 b12
Symmetry of ground
state wavefunction A1
b1
1st excited state
configuration
a12 b22 a12 b11 b21
a1
Symmetry of excited
state wavefunction
b1 x b2 = A2
b2
a1
Is the spectroscopic
transition A1  A2
allowed? (See Later!)
Formaldehyde:
pi *, b1
n , b2
ground state: … (n , b2) 2 : A1
1st excited state: …(n , b2) 1(π , b1) 1 : A2
What about molecules with degenerate
orbitals?
benzene π  π* transition in the UV.
benzene π m.o.s, D6h symmetry in energy
order:
π
π*
a2u < e1g
<<
e2u < b2g
ground state:
… a2u 2 e1g 4 : A1g
1st excited state: … a2u 2 e1g 3 e2u 1
1 unpaired electron in e1g and 1 in e2u
e1g x e2u = B2u + B1u + E1u
(check by multiplying to get reducible
representation and then reducing it)
States of these three symmetries are
observed in the absorption spectrum of the
molecule.
A1g  B2u at ~ 220 nm (uv) very weak
A1g  B1u at ~ 180 nm (vuv) weak
A1g  E1u at ~ 160 nm (vuv) very strong
[Spin adds an extra complexity to this
analysis.]
Spin Quantum Numbers
The unpaired electrons determine the total spin
quantum number S.
0 unpaired electrons S = 0
1 unpaired electrons S = ½
2 unpaired electrons S = 1 and S = 0
(for molecules very unusual to have more than 2
unpaired electrons)
Why 2 values for 2 unpaired electrons? The
spins of the electrons can either add ½ + ½ = 1,
or subtract ½ - ½ = 0. (Really the spin angular
momenta can add or cancel.)
To describe the molecular state need to give both
S and symmetry (Γ).
Molecular term symbol (2S+1) Γ
(2S+1) is the spin degeneracy – called
multiplicity.
S=0
(2S+1) = 1
singlet
S=½
(2S+1) = 2
doublet
S=1
(2S+1) = 3
triplet
Transition Intensity, Selection Rules and
Symmetry
Transition from molecular State 1 to State 2
E,
b
b
h  E  E
b
a
E ,
a
a
The intensity, Iab, of a transition from state a,
energy Ea, wavefunction a to a state b, Eb,
b can be shown to obey:
I ~     d
ab
a
2
b
where  is the dipole moment of the system
d is integration over all coordinates
All selection rules can be derived from this
equation.
For an electronic transition:

2Sa 1

ground state
 
excited state
As  only involves spatial coordinates and
cannot change the spin of the electrons, the spin
selection rule ΔS = 0 applies to the molecular
states. (Sa = Sb)
a
a
2Sb 1
b
b
I12 is a number (scalar) and had symmetry Γ1
I ~     d
ab
a
2
b
means that
Γ1 = Γa x Γ x Γb
is required for an allowed transition (Iab ≠ 0)
Γ are the representations of the coordinates
(x, y, z) because  is a vector.
For a closed shell ground state molecule, Γa = Γ1
and therefore Γb = Γ
Molecular states of H2 CO:
…. (π, b1) (n, b2) (π , b1)
1
2
*
…. (π, b1) (n, b2) (π , b1)
2
1
*
1
A1
3
A1
1
A2
3
A2
1
1
….(π, b1) 2 (n, b2) 2
For an allowed transition ΔS = 0, therefore
1
A1 1 Γ
C2v: Γ = A1 (z) , B1 (x) , B2 (y)
For an allowed transition: Γ 1 = Γa x Γ x Γb
for the n  π* transition: does Γ1 = A1 x Γ x A2?
no – forbidden.
for the π  π* transition: does Γ1 = A1 x Γ x A1?
yes – allowed.
1
A1