1 electronic spectroscopy outlines -introduction -molecular term symbols -transitions between...
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1
Electronic Spectroscopy
Outlines
- Introduction
- Molecular Term Symbols
- Transitions between Electronic States of Diatomic Molecules
- Vibrational Fine Structure of Electronic Transitions in Diatomic
Molecules : Franck-Condon Principle
- UV-Visible Light Absorption in Polyatomic Molecules
- Transitions Between Ground and Excited States
- Jablonski diagram, Fluorescence, Phosphorescence
2
- Excitations between molecular electronic states (the
electron is excited from an initial low energy state to a
higher state by absorbing photon energy).
- Molecular electronic transitions are induced by UV/VIS
radiation (Eelec >> Evib >> Erot)
Introduction
3
• Describe the electronic states of molecules.• Molecular electronic configuration relied on the orbital and
spin angular momentums.
Molecular Term Symbols: diatomic molecules
2S + 1L
, (or 2S + 1 L g/u for homonuclear diatomic molecules indicate by inversion symmetry of MO’s functions)
“Spin multiplicity”(S : the total spin quanmtum number)
The “z-component” orbital momentum quanmtum number = |ML|
L 0 1 2 3 4L S P D F G
Principle z-axis
(similar to atomic term symbol)
4
1. Only unfilled subshells contribute to the term orbital and
spin angular momentum.
2. In the first or second row diatomic molecules, MOs are
either or type.
Determine molecular term symbol
H2 CO
MO s p d f g
ml 0 1 2 3 4
5
3. Calculate ML and MS
where mli, mls = z components of orbital and spin angular momentum for the i th electron in its molecular orbital.
n
isiS
n
iliL mMmM
11
and
For MO's with symmetry, ml = 0
For MO's with symmetry, ml = +1, -1
Note that there is no ml = 0 for MO's since the Pz
atomic orbital is associated with the MO
4. Determine L value from |ML| to assign symbol, and S
value from |MS| to calculate 2S+1
5. Generate term symbols2S + 1L
-L ML L, -S MS S
6
Molecular term symbol for Ground state of H2
1 = 0
msi mli
0 0 + 0 = 0
MO
ML = 0
MS = 0
2
1
2
1
|ML | = 0
|MS| = 0 S = 0 2S+1 = 1
L= 0 ; L
1STerm symbol
Singlet
Note that H2 is homonuclear diatomic molecules. Inversion symmetry of g MO’s wave functions is also considered. Using multiplication rules (g)(g) = g(g)(u) = (u)(g) = u(u)(u) = g
1S g
Ground-state Electronic configuration of H2 : (1g)2
7
Ground state of
* = 0 0
0
MO
ML = 0
MS = 1/2, -1/2
|ML | = 0
|MS | = 1/2 S = 1/2 2S+1 = 2
L = 0; L
1
1*
Fully occupied, not need to be considered
2
1
2
1
2S u
2STerm symbol
An electron fills in *u orbital
1*u
msi mli
1*g
Gr. state configuration (1g)2 (1*u)2
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0
1
0
0
-1
0
MO
2
1
2
1
1 ,1type- lm2 e- in * MO
+1 -1
1+1 = 2
1-1 = 0
1-1 = 0
1-1 = 0
1-1 = 0
-1+-1 = -2
msi mli LS (|MS|)
0
1
0
0
1
0
2
0
0
0
0
2
L
Term symbol
1 3
1 1 3 1
Three terms : 1, 3 , 1
Possible combination = 4C2 = 6
10
Gerade = symmetric
with inversion
Ungerade = antisymmetric with inversion
inversion
inversion
Parity :Gerade (g) and ungerade (u)
Gerade = symmetry with respect to a center of inversion
represents center of inversion
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From s orbs From pz orbs
From px orbs From py orbs
From s orbs From p orbs
From px orbs From py orbs
πu πgπg
σg σg σu
πu
Gerade and ungerade
Bonding MO Antibonding MO
σu
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(1g)2 (1*u)2 (2g)2 (2*u)2 (3g)2 (1u)2 (1u)2 (1*g)1 (1*g)1
Energy diagram and electronic configuration of O2
1g
1*u
2g
2*u
3g
1u
1*g
3*u
En
erg
y
Ground state electronic configuration of O2
1s
*1s
2s
*2s
2p
2px, 2py
*2px , *2py
*2p
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MO
+1 -1
Term symbol
1 g
3 g
1 g
1 g
3 g
1 g
Because 2 e- are in * MO so we use the notation g
*g
*g*g = *g
En
erg
y
3 g
1 g
1 g
3 g
Electrons of opposite spins separated in two orbitals1 g
Electrons of opposite spins paired in a single orbital1 g
Electrons of the same spins separated in two orbitals (lowest energy)
magnetic attraction between electrons of opposite spins
Hund’s rule
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3 g1 g
1 g
(1g)2 (1*u)2 (2g)2 (2*u)2 (3g)2 (1u)2 (1u)2 (1*g)1 (1*g)1
Electronic states and molecular term symbol of O2
lowest energy due to greatest spin multiplicity (Hund’s rules)
Excited states Ground state
Ground state configuration of O2
Triplet state O2
Singlet state O2
15
the +/- superscript applies only to states, and indicate whether the wavefunction is symmetric or antisymmetric with respect to reflection through a plane containing the two nuclei.
16
Assigning + and – sign to terms of diatomic molecules
+ and - refer to the change in sign of the molecular wave function on reflection in a plane that contains the molecular axis.
“+” refers to no change in sign of . “-” refers to does change sign.General rule for “+”:- All MOs are filled, - Unpaired electron in MOs
“+” superscript because MOs has plan of reflection. Therefore unchanges the sign.
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“-” superscript because MOs does not have plan of reflection. Therefore changes the sign.
+
- +
-
+
-
-
+
“+” superscript because MOs does not have plan of reflection. Therefore changes the sign.
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(1g)2 (1*u)2 (2g)2 (2*u)2 (3g)2 (1u)2 (1u)2 (1*g)1 (1*g)1
For the case of O2, electronic configuration is
(1*g)1 (1*g)1 Consider only
There are six possible combinations of wave functions for ml = 1 and for ms = 1
)2()1()(
)]2()1()2()1()[(
)2()1()(
)]2()1()2()1()[(
)]2()1()2()1([
)]2()1()2()1([
11116
11115
11114
11113
112
111
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Spatial (orbital) function
Symmetric orb
function
Spin function
|S| symbol
1 +1+1symmetry 12 - 12 2 0 1g
2 -1-1symmetry 12 - 12 2 0 1g
3 +1-1 + -1+1symmetry 12 - 12 0 0 1
+g
4 +1-1 - -1+1anti-symmetry
12 0 1 3-g
5 +1-1 - -1+1anti-symmetry
12 + 12 0 1 3-g
6 +1-1 - -1+1anti-symmetry
12 0 1 3-g
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Ex. Predict molecular term symbol of the ground state for Li2.
(1g)2 (1*u)2 (2g)2
the electronic configuration for Li2
= 0
1Sg or 1S+g
Two electron fills in *g whose the orbital is assigned as gerade. According to the multiplication rule (g) (g) = g
For the total orbital angular momentum:
mli = 0 + 0 = 0 =
For the total spin angular momentum
msi = 1/2 + -1/2 = 0 2S+1 = 1
Molecular term symbol of the ground state for Li2 is
each electron is in the same orbitals
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Electronic states of excited state configuration of O2
Possible combination (4C3 )= 4 4 = 16 states
(1g)2 (1*u)2 (2g)2 (2*u)2 (3g)2 (1u)1 (1u)2 (1*g)1 (1*g)2
Excited state configuration of O2
(1g)2 (1*u)2 (2g)2 (2*u)2 (3g)2 (1u)2 (1u)2 (1*g)1 (1*g)1
Ground state configuration of O2
1u 1*g
Ground state Excited state
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Description for potential energy curves of electronic states
Ground state of O2
the four lowest excited states of O2
- As the energy states become
higher in energy, the bond length
increases.
- Excited states have more anti-
bonding character (less bonding
order, less bonding energy and
longer bond).X refers to the ground state.A, B, … refers to the higher energy states with the same multiplicity as the ground state.a, b, … refers to the higher energy states with the different multiplicity as the ground state.
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Transitions Between Electronic States of Diatomic Molecules
Selection rules:
Note - The rule is applicable for the atomic number < 40.- All of these selection rules can be derived by calculating the transition dipole element.
Allowed transition
Examples
= 0, 1 S = 0 11 33 11 33g u 1g1u 1u1g
+ + or - - 1+g1+
u 3-u3-
g
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Ex. Which of the following electronic transitions are allowed or forbidden?
1+u1-
g1)
= 0, S = 0, u g, + - transitions forbidden1) 1+u1-
g
1+u1g2)
= 2, S = 0, u g transitions forbidden2) 1+u1g
3) 3u3g
= 1, S = 0, u g transitions allowed3) 3u3g
4) 1-u3-
g
4) 1-u3-
g = 0, S = 2, u g, - - transitions forbidden
5) 1g1 -g
5) 1g1 -g
= 1, S = 0, g g transitions forbidden
30
Ex. Using molecular term symbols of the ground and excited states of O2 to indicate which transition is allowed or forbidden.
1. X3-ga1g forbidden (S0)
2. X3-gb 1+
g forbidden (S0)
3. X3-g A3+
u forbidden (- +)
4. X3-g B3-
u allowed(the lowest allowed transition)
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The X3-g B3-
u transition is allowed
Absorption from the ground state into various vibrational levels of the B3-
u excited states is possible (vibronic coupling).
Note that the selection rule n = ±1 does not apply to vibrational transitions between different electronic states.
32Ex. The energy difference between the two lowest vibrational states in the electronic ground state (X3-
g) and in the first excited state (B3-
u) of O2 is 49000 cm-1. The observed spectral line of 52500 cm-1 corresponds to the transition of the ground vibrational state of X3-
g to the n vibrational state of B3-u . Find
the n value of this transition? Given the vibrational wavenumber for an allowed vibrational transition in B3-
u is 700 cm-1 (Ignore any rotational structure or anharmonicity.)
X3-g
B3-u
n=0
n=1n=2
n=3
n'=0n'=1
n'=2
n'=?
700 cm-1
49,0
00
cm-1
n'=3
rotationvibrationelectronictotal EEEE Ignore
52,5
00 c
m-1
)-() ( 0,00,10,1',1, EEEEE nvibelec
-10,00,1 cm 49000 EE
0,0',1 EE n
49000) ( 52500 0,1',1 EE n
-10,1',1 cm 5003 EE n
-10,1',1 cm 5003 '' vnEE n
cm 5003 )cm 700(' -1-1 n
5'nFor transition between different electronic states, the selection rule for vibrational transitions n = ±1 does not need to consider.
35
Vibrational Fine Structure of Electronic Transitions in Diatomic Molecules
• Vibrational and rotational quantum numbers can change during
electronic excitation.
• Born-Oppenheimer approximation can be used to determine
vibrational transition between electronic states.
Wave functions
nuclei (associated with vibration of the molecule)
electrons (associated with the motion of electrons)
Born-Oppenheimer approximation
Nuclei are much more massive than electrons
36
The total wave functions is a product of vibrational (nuclei) and
electronic parts within the Born-Oppenheimer approximation
where R1,…,Rm depends on position of the nuclei
r1,…rn depends on the position of electrons
0,...,,,...,ˆ,...,,,..., 1111* dRRrrRRrr mnimnf
fi
The spectral line of an electronic transition (initial final) has a
measurable intensity if the transition of electric dipole moment is not zero:
mnucleifixedm
fixedn
electronicmn RRRRrrRRrr ,...,,...,,,...,,...,,,..., 11111
n
jj
n
jj RZere
11
ˆthe dipole moment operator is given by
37
dRRrrRRrr mnimnffi ,...,,,...,ˆ,...,,,..., 1111
*
diffi ˆ*
Nemn RRrr ,...,,,..., 11
iNiei ,,
fNfef ,,
Ne ˆˆˆ
For simplification, we consider the short notation :
mnucleifixedm
fixedn
electronicmn RRRRrrRRrr ,...,,...,,,...,,...,,,..., 11111
For initial state:
For final state:
the dipole moment operator :
diNieNefNfefi
,,*
,*, )ˆˆ(
dd iNieNfNfeiNieefNfe ,,*
,*,,,
*,
*, ˆˆ
dddd iNNfNiefeieefeiNfN ,*
,,*,,
*,,
*, ˆˆ
38
,...,,..., 1
*
1,*
, dRRRRdS mvib
mvib
iNfN if
2*2 dS vibvibif
Represents the overlap between the vibrational wave functions in the initial and final states.
Franck-Condon factor
a measure of the expected intensity of an electronic transition
dd
dddd
ieefeiNfN
iNNfNiefeieefeiNfNfi
,*,,
*,
,*
,,*,,
*,,
*,
ˆ
ˆˆ
0 (orthogonal if)
dS ieefefi
,*, ˆ
39
the spectral line of the corresponding transition will be observed.
0,*
, dS iNfN
If the overlap between the vibrational wave functions in the initial and final states is not zero (S 0),
40
Franck-Condon principle
States that transitions between electronic states correspond to vertical lines on an energy versus inter-nuclear distance diagram.
• Electronic transitions occur at a much faster rate than the nuclei’ motion (The atoms do not move during the transition).
• The electronic transition occur at the initial state that have the ground (n=0) vibrational state (equilibrium bond distance).
Morse potential
41
Franck-Condon principle
1) Separation distance remains constant during electronic transitions 2) Later moves to new equilibrium position
Separation distance doesNOT change during transition
3) An electronic transition can go to any number of different vibrational levels in the excited electronic state depending on the energy.
No longer have selection rule for vibrations (n=1)
42
The Franck-Condon principle determine the n values in the excited state that give the most intense spectral lines.
nearly all of the molecules in the ground vibrational state
• The electronic transition will lift the highest populated molecules in the n=0 vibrational state, therefore the n excited state can be from the peak with the highest intensity.
43
Ex. Consider the refined structure of UV/VIS absorption spectrum for a diatomic molecule and sketch the energy diagram with the corresponding state of electronic transition
44
UV-Visible Light Absorption in Polyatomic Molecules
• Absorption lines in condense phases are usually board and obscure fine structure.
• Polyatomic molecules exhibit many rotational and vibrational transitions. Because their spectral lines overlap, absorption lines are board and featureless.
atom diatomic molecule polyatomic molecule
• This makes it difficult to extract information on the initial and final states involved in an electronic transition in polyatomic molecule.
45
• a chemical entity embedded within a molecule that absorbs radiation at the same wavelength in different molecules.
• Common chromophores: C=C, C=O, C=S, CN• Electronic excitation from HOMO to LUMO with the
configurations : nπ*, π π*, and σσ*
Chromophores
47
Types of transitions
1) Radiative Transitions: photons absorbed or emitted 1.1 Fluorescence: S = 0 singlet-singlet transition 1.2 Phosphorescence: S 0 singlet-triplet transition 2) Nonradiative Transitions: energy transfer between internal degrees of freedom of a molecule or to surroundings: 2.1 Internal Conversion: transition without a change in energy between states of the same multiplicity (S = 0, i.e. singlet-singlet transition) 2.2 Intersystem Crossing: transition without a change in energy between states of the different multiplicity (S 0, i.e. singlet-triplet)
Transitions Between Ground and Excited States
48
Jablonski diagram
IC
IC
IC ISC
ISC
Phosp
hore
scen
ce
Flu
ores
cenc
e
Fluorescence: singlet-singlet transition (S = 0)
Phosphorescence: singlet-triplet transition (S 0)
49
Singlet-Singlet Transitions: Absorption and Fluorescence
Fluorescence is a radiative transition from the lowest vibrational state of excited states back to the ground state.
The fluorescence process involves:
1) Absorption from the lowest vibrational level of the ground state to the various vibrational levels of the excited (singlet state)
2) Internal conversion of energy (non-radiative): molecules in the excited vibrational levels of the excited state collide with other molecules. (Note: Non-radiative transitions occur much more rapidly compared to radiative transitions from excited vibrational levels of the excited state)
50
The fluorescence process involves:
3) Once in the ground vibrational level of the excited state, the molecule undergoes radiative transition to any vibrational level in the ground state
51
Intersystem Crossing and Phosphorescence
Singlet-triplet transition (S 0).
Although intersystem crossing between singlet and triplet electronic states is forbidden, the transition probability is enhanced by two factors: - very similar molecular geometry in the excited singlet and triplet states, - a strong spin-orbit coupling
54
Ex. Consider the transition from one electronic state to another, their bond lengths being Re and R’e and their force constants unchanged. Calculate the Franck–Condon factor for the 0–0 transition and show that the transition is most intense when the bond lengths are equal.
dS 0'0)0,0(
We need to calculate S(0,0), the overlap integral of the two ground-state vibrational wavefunctions,
We use 22 '2/1
4/1'0
2/14/1
0 ; xx ee
2/ k
deeS xx 22 2/1
4/1'2/1
4/1
)0,0(
dee xx 22 2/1'2/1
2/1
ee RRxRRx '' ,
55
Singlet-Singlet Transitions:
Transitions Between Ground and Excited StatesFor transitions from ground singlet states to excited singlet or triplet states,three types of transitions are possible: Radiative Transitions: photons absorbed or emitted (Fluorescence) Nonradiative Transitions: energy transfer between internal degrees of freedom of a molecule or to surrondings. Intersystem Crossing (singlet-triplet). (Phosphorescence)
58
the Schrödinger equation can be solved using separation of variables.
)()(),( Y
0),(sin2),(),(
sinsin 222
2
EY
IYY
From the previous slide
2
22
2
),(),(sin
2sinsin
Y
YEI
Only Only
Rearranging the differential equation separating the θ-dependent terms from the -dependent terms:
For the J=0 → J=2 transition,
dzx
dxxdx
dzxzx
sin
1 ,sin ,cos ,
xzdzx
xzxdxzxdxx 44333 cos4
3
4
3
sin
1sin3sin3sincos3
Consider
Use the substitution method (similar to the previous one)
Replace x with and integrate from 0 to , we get:
0
3 sincos3 d
0
sincos d
0)1)1((4
3cos
4
3sincos3 44
04
0
3
d
0cos2
1
2
1
sin
1sinsincos 22
xzdz
xxzxdxx
Do the same for
0
sincos d
0)sincossincos3(4
5
0
320
dz
For the J=0 → J=2 transition,
0sincos30
3
d 0sincos0
d
From the previous derivation:
Therefore:
Thus:
020 z
Thus, the J=0 → J=2 transition is forbidden.
61
Home Work 2
1. Spectral line spacing of rotational microwave spectrum of OH radical is 37.8 cm-1 . Determine the OH bond length (in pm unit) and moment of inertia (in kg m2)
Spectrum (cm-1)
37.8 cm-1
mO = 15.994 amu
mH = 1.008 amu
molecule bond length (pm)
HF 91.7
HI 161
HCl 128
HBr 141
2. Use the bond length of diatomic molecules in Table to predict line spacing (in cm-1 unit) of rotational microwave spectrum.
62
Bond length (pm)
OH 37.80
ICl 0.11
ClF 1.03
AlH 12.60
3. Determine the bond length of these diatomic gases in Table and arrange them in order of increasing the bond length.
)cm( -1B
4. Using the information in the Table to calculate the ratio between the transition energy of rotation from J=0 to J =1 and vibration from n=0 to n=1 for H2
?10,
10,
nnvibration
JJrotation
E
EAtomic mass 1.008 amu
Bond length of H2 74.14 pm
Force constant of H2 575 N/m