14: ELECTRONIC SPECTROSCOPY
TERM SYMBOLS FOR DIATOMIC MOLECULES
2S+1
|ML| g/u
|ML| = molecular angular momentum
(replaces atomic L)
2S+1 = spin multiplicity (same as atoms)
g/u = total inversion symmetry for homonuclear diatomic molecules
Must have inversion symmetry, so cannot specify with heteronuclear diatomic molecules
Similar to the term symbols used for atoms
Atomic term symbols based on looking at combination of electrons in partially filled orbitals.
Similarly, each electron in an MO has spin and orbital momentum associated with it
Molecular term symbols look at all the possible combinations of spin and z-axis orbital
momentum for electrons in the highest-occupied degenerate sub-shell MO’s.
E.g., all possible combinations of 2 electrons in O2 molecule occupying two degenerate
1πg MO’s.
Atoms
2S+1
Term Symbols
Orbital Momentum
- along z axis
- total
LJ
Molecules
|ML| g/u
2S+1
Choice of z axis is arbitrary for
spherical atom
ml = component along z axis for
individual electron
ML = Σ ml,i
z-axis defined as
internuclear axis, po
ml = component along z
axis for individual
electron
ML = Σ ml,i
|ML| = 1: Σ (s = sigma)
2: Π (p = pi)
3: ∆ (d = delta)
L = 1:
2:
3:
ML = L,
Electron Spin
- along z axis
- total
ms = ± ½
MS = Σ ms,i
L not fully applicable for
molecules due to lower
symmetry of molecules
(not spheres)
ms = ± ½
MS = Σ ms,i
Ms = S, S-1, ... –S
Spin multiplicity = 2S+1
Ms = S, S-1, ... –S
Spin multiplicity = 2S+1
Total Combined
Angular Momentum
J=L+S
J = L + S, L+S-1, … |L-S|
Inversion Symmetry
Gentry, 2013
S
P
D
L-1, ... 0 ..., –L
If inversion symmetry
exists:
gi *gj = g,
gi*uj = u, ui *uj =g
REMINDER ON HOW TO CONSTRUCT ATOMIC TERM SYMBOLS
EXAMPLE: One electron in a 2p orbital (2p1)
A) Map all the possible “microstate” combinations of sub-orbitals and spins.
For the 2p1 example:
2p orbitals can be 2px, 2pz, or 2py (ml = -1, 0, or 1)
For an electron in each sub-orbitals, the spin can be up (+ ½) or down (- ½)
Thus, there are a total of 6 possible microstates as shown below.
2p
ml= +1
ml= 0
ml= -1
B) For each microstate, add each electron’s mS to get total Ms for the atom
and similarly add each electron’s ml to get Ml
For the 2p1 example:
There is only one electron for each microstate.
Therefore summing over all electrons in the microstate is a trivial exercise
Ms=Σms
+½
+½
+½
-½
-½
-½
Ml= Σml
-1
0
+1
-1
0
+1
C) Back-calculate using allowed values of Ms to find S and similarly Ml and L.
Allowed ML values = L, L-1, ..., –L
Allowed Ms values = S, S-1, ..., –S
For the 2p1 example:
Ml = +1, 0, of -1, therefore only way to have these Ml values is if L = 1
Ms = + ½ or – ½ , therefore only way to have these Ms values is if S = + ½
Spin multiplicity = 2S+1 = 2
D) Find total angular momentum, J, by combining S and L values
J = L+S, L+S-1, … |L-S|
For the 2p1 example:
Maximum value of J = L+S = 1 + ½ = 3/2
Minimum value of J = |L-S| = 1 – ½ = ½
No room for additional values between J = 3/2 and 1/2 .
E) Form atomic term symbol and calculate spin-orbit coupling
2S+1
LJ
EL,S,J = ½ hcA {J(J+1) – L(L+1) – S(S+1)} A = spin-orbit coupling constant
For the 2p1 example:
2
State 1: P1/2 , E1, ½ , ½ = ½ hcA {3/4 – 2 – 3/4} = - hcA
2
State 2: P3/2 , E1, ½ , 3/2 = ½ hcA {15/4 – 2 – 3/4} = + ½ hcA
Ch 14: Electronic Spectroscopy
-2-
CONSTRUCTING DIATOMIC MOLECULAR TERM SYMBOLS
Similar Steps to Atomic Term Symbols
A) Construct all possible microstates
ml = angular momentum along z axis for individual electron in molecular orbital
Atomic orbitals have complete set of ml values along z axis for a given l value
ml = +l, +l-1, ..., -l
HOWEVER…
Molecular orbitals, on the other hand, do not use a complete set of ml for given MO
Symmetry of molecule means that not all atomic sub-orbitals contribute to a set of
degenerate molecular orbitals.
EXAMPLE: diatomic molecules
Atomic 2p orbitals (l=1) have ml =+1,0,-1 corresponding to 2px, 2pz, and 2py
But molecular orbitals either formed by mixing 2s and 2pz (=2po) to form σ MO
Or mix 2px and 2py (formed from 2p-1 and 2p+1) to form π MO.
The set of possible allowed microstates must reflect these symmetry considerations
B) Add up the individual ms and λ values to find MS and Λ
MS = Σ ms,i
ML = Σ ml i
C) Use allowed values of MS to find S
Allowed Ms values = –S, ..., +S
D) L is no longer a true quantum number due to lack of spherical symmetry in a molecule
(atoms on the other hand do have spherical symmetry)
However still use same methodology to identify L for term state
Allowed ML values ~ -L, ..., +L
Hence identify set of ML values associated with a given maximum |ML|
For maximum |ML| = 0,1,2,3 have
|ML|max = 0 ⇒ Σ (s= sigma)
|ML|max = 1 ⇒ Π (p = pi)
|ML|max = 2 ⇒ ∆ (d = delta)
E) For molecules having inversion symmetry, find total g or u symmetry
Multiply the symmetry element for each electron (1 and 2) in the MO
g1*g2 = g, g1*u2 = u, u1*u2=g
I.e. follows multiplication rules as if each g=+1 and u=‒1
F) Term symbol =
Ch 14: Electronic Spectroscopy
2 S +1
ML
g /u
-3-
Example 1: H2
MO configuration: 1σg2
H2
1σu
1s
1s
1σg
Spin:
All electrons paired, therefore # spin up equals # spin down; and MS=0
1σg (ml=0)
(1s atomic orbitals each have ml=0)
↑↓
0
MS = Σms
If MS = 0, then S=0
z-component Angular Momentum:
σ molecular orbital, therefore only ml = 0 atomic orbitals can contribute
|ML| = sum of individual ml = 0 (L=0 means s for atoms, so for molecules = Σ )
Inversion symmetry g/u
H2 has inversion symmetry, thus g/u notation is applicable
Electron #1 is in g MO, and electron #2 is in g MO, therefore g1 * g2 = g
Term symbol:
1
Σg
Example 2: H2‒
MO configuration: 1σg2 1σu1
1σg2 is fully occupied so does not need to be considered
1σu1 is partially occupied with a single unpaired electron
Spin:
1σg
← Can ignore fully occupied 1σg orbital
1σu (ml =0)
↑
↓
MS
½
-½
MS = S, S-1, ..., -S
Therefore S = ½ and spin multiplicity = 2S+1 = 2
z-component Angular Momentum:
Electron is in a σ orbital made up of 1s atomic orbitals, so ml =0 for that lone electron
Therefore |ML| = 0 (Σ)
Inversion symmetry g/u
H2- has inversion symmetry, therefore g/u is applicable
The one electron in an unfilled suborbital is in a u orbital, therefore total g/u
symmetry is u.
Term symbol:
2
Σu
Ch 14: Electronic Spectroscopy
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Example 3: O2
Ground state configuration: 1σg21σ∗u22σg22σ∗u23σg21πu41π∗g2
All sub-shells fully occupied except the antibonding HOMO, 1π∗g2
O2 has two degenerate 1π∗g orbitals, which are partially occupied with 2 e-‘s
Furthermore, π MO’s can only have AO contributions from 2px and 2py (ml = +1 or -1)
ml = +1
2p ml = 0
ml = -1
M s = Σ ms
|M L| = Σ ml
+1
0
0
-1
0
0
0
0
0
0
-2
+2
Scenario 1: 2 Electrons in Separate πg Orbitals
Angular Momentum: one electron in π orbital made of p-1 atomic orbitals ml = ‒1, and
other electron in π orbital made from in p+1 atomic orbitals (ml = +1)
... therefore |ML| = (+1) + (-1) = 0 (L=0 is s for atom so for molecule = Σ ).
Spin: Since the electrons are in separate MO’s, they can have spin in either the same or
opposite direction
↑↑
↑↓
↓↑
↓↓
MS
1
0
0
-1
MS = S, S-1, ..., -S
Therefore S = 1 (MS=‒1,0,+1) or S = 0 (MS=0)
g/u symmetry: the two electrons are both in g orbitals
g1 *g2 = g
Term Symbol: Electrons in separate 1πg orbitals can be in one of two term states
3
Σg (spins aligned to give triplet state)
... or 1Σg (spins in opposite direction for singlet)
Scenario 2: 2 Electrons in Same πg Orbital
Spin: Since electrons in same orbital, they must have opposite spin
↑↓
MS
0
Therefore S=0
Angular Momentum: If electrons in same πg orbital
ml 1 = ml 2 = ±1
... and therefore |ML| = 2 ( L=2 means d for atoms, so for molecules = ∆).
g/u symmetry: The two electrons are both in g orbitals
g1 *g2 = g
Term Symbol: 2 electrons in the same 1πg orbital gives
1
∆g
Ch 14: Electronic Spectroscopy
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SPIN-ORBIT COUPLING
Atomic term states lead to spin-orbit coupling
EL,S,J = ½ hcA {J(J+1) – L(L+1) – S(S+1)} A = spin-orbit coupling constant
Caused by magnetic interactions of orbital angular momentum and spin angular
momentum
Similarly, molecular term states give rise to splitting of otherwise degenerate sub-orbitals.
However, due to complexity of MO energies, there is no simplifying expression for spinorbit energy like there is with atomic term states
For O2 1π
πg2 sub-orbitals:
To first approximation the two π anti-bonding MO’s are degenerate
But with spin-orbit coupling from the two electrons, there is a splitting of energies for the
different term states
Term
State
3
Σg
1
∆g
1
Σg
ν~
Relative Energy
(cm-1)
0
7,882
13,121
Ro
(pm)
120.7
121.6
122.7
(cm-1)
1580
1509
1433
from Atkins, pg 483
Energy
O2
1π
πg2 Occupancy
with spin-orbit coupling
2σu
1πg
1
Σg
1∆
2p
2p
3
Σg
1πu
r
2σg
1σu
2s
2s
1σg
Ch 14: Electronic Spectroscopy
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g
SELECTION RULES FOR ELECTRONIC TRANSITIONS
∆|ML| = 0, ±1
Therefore Σ→Σ and Σ→Π are allowed transitions, but Σ→∆ is forbidden
∆S = 0
Change in spin multiplicity such as singlet→triplet means that an electron would have to
change the direction of its spin, and not just its orbital and energy level.
g → u,
Must change inversion symmetry between initial and final state... if inversion symmetry
exists.
Rule is a result of transition moment integral ∫ψ f µψ i dτ
µ acts like x, y, or z axis, therefore has u inversion symmetry
Only way integral can be non-zero is if overall integrand has even symmetry, g
Odd symmetry, u, would mean positive coordinates would cancel negative
coordinates
For g→u: ∫ u ⋅ u ⋅ g = ∫ g ≠ 0 but for g→g: ∫ g ⋅ u ⋅ g = ∫ u = 0
None of these selection rules are absolute, since all have built in assumptions.
But forbidden transitions are much weaker and take much longer to occur.
Ch 14: Electronic Spectroscopy
-7-
FRANCK-CONDON PRINCIPLE
Because nuclei are much more massive than electrons, transitions between electronic
states occur at a much faster rate than the nuclei can respond.
Background
Ground electronic states and excited electronic states all generally follow a Morse
potential dependence of energy vs. nuclear separation distance, r.
However, the equilibrium separation distance often differs from one electronic state to
the next depending on the atomic orbitals that contribute to the specific MO.
... and the width of the potential well also differs for the two states (giving rise to
different spring constants and vibrational frequencies)
Oxygen – Bond Lengths of Molecular States
Ground State
Excited State
X 3Σ g -
2σ
σ∗u
1π*g
A 3Σ u+
Bond Order
=2
Bond Length
= 120pm
2σ
σ ∗u
Bond Order
=1
Bond Length
= 150pm
1π*g
1πu
1πu
2σ
σg
2σ
σg
A molecule undergoing an electronic transition to an excited state will redistribute its nuclei
to reach its new excited-state equilibrium geometry.
But the Franck-Condon principle says that this re-positioning of nuclei is MUCH
SLOWER than the actual electronic transition between states.
Therefore the electronic transition is assumed to occur at the initial, ground-state
equilibrium position, with any re-positioning of the nuclei occurring at some later
time.
Franck-Condon Principle
1) Separation distance remains
constant during electronic transitions
2) Later moves to new equilibrium position
2
Separation distance does
NOT change during transition
ψf
ψf
1
X
ψi
ro,i ro,f
ro,i ro,f
Ch 14: Electronic Spectroscopy
-8-
ψi
Franck-Condon Simplification
Formal solution requires calculating the transition dipole for various vibronic transitions.
µ if = ∫ψ *f µψ i dτ
where ψi and ψf each are combinations of electronic and vibrational wave functions
BUT… Because of the fixed nuclear distance during transition, it is possible to construct a
simplified model for transitions to excited vibrational states.
An electronic ground-state molecule will likely sit in its ground-state vibrational state since
vibrational energies are usually much larger than available thermal energy.
The position of this ground state electron can be thought to be the equilibrium center
position.
When the molecule undergoes an electronic transition, it can transition up to any number of
different vibrational levels in the excited electronic state depending on the energy
of the light.
No longer have ∆v = ±1 selection rule for vibrations since going to a different Morse
potential well during an electronic transition.
Given that the nuclear separation distance does not change, the vibrational level in the
excited state with the highest probability of being found at that distance will
be the vibrational state whose potential-energy boundary is at that distance.
That is to say, when averaged over time, vibrations are most likely to be found at the
extremes of the vibration where the atoms slow down and then reverse direction,
rather than at the center of the vibration where the atoms have reached maximum
velocity.
Thus the most likely transition, and the transition having the highest absorbance, will be for
the excited vibrational state whose potential well is bounded at the same distance
as the ground state equilibrium separation.
For the schematic figure below
The molecule begins in the zero vibrational state for the ground electronic state, ψi.
It undergoes an electronic transition to the ψf excited electronic state
Since the nuclei do not have a chance to reposition themselves, the most probable
transition will be to the point where the excited potential well for ψf intersects
with the equilibrium distance for the ground state.
For the diagram below, this intersection occurs at the υ’=2 vibrational level in the
excited electronic state.
The absorption spectrum will reflect this maximum probability in terms to the maximum
absorbance measured when doing UV/Vis spectroscopy.
Transition Profiles
3
4
ψf
ψi
2
Absorbance ?
2
1
υ’=0
Absorption Spectrum
1
υ’=0
3
4
5
6
υ=0
ro,i ro,f
Ch 14: Electronic Spectroscopy
Absorption Energy ?
-9-