8. UV-Vis Spectroscopies
Atomic Absorption
Molecular Absorption
Molecular Fluorescence
Raman Spectroscopy
Circular Dichroism
SELECTIVITY
Review of – will use to evaluate the UV-Vis methods
To date:
1. LOD (signal)
2. LOD (conc.)
3. Linearity
Signal
(mV, I, A)
LOD (signal)
Xblank + 3s
90%
10%
LOD (signal)
[conc analyte]
To date:
1.
2.
3.
4.
LOD (signal)
Linearity
LOD (conc.)
Selectivity
Selectivity is the response of the method to
The analyte relative to the response to an
interferent
Signal
(mV, I, A)
The best method
Would have no
Response to an interferent
[conc analyte]
[conc interferents]
Ag2S/PbS ion selective electrode
0
At pH values , ~5 the electrode is sensitive to
Protons (a measurable slope in a calibration
Curve)
-50
log[Pb2+]
-100
MORE selective!
mV
-150
Log[H+]
-200
-250
-300
-350
-10
-8
-6
-4
log [conc]
-2
0
0
-50
mV
-100
-150
pH 2
-200
pH 3
-250
pH 4
pH 5
-300
-9
-8
-7
-6
-5
-4
Log [Pb2+]
-3
-2
-1
0
Atomic Absorption
Big Advantage = Selectivity
Monitor absorption lines that are specific to lead
And not likely to overlap with other atoms
The use of atomic specific lines adds SELECTIVITY to this method
In this figure the
Darker the line the
More intense the
Absorption line
BUT: the use
of atomic
lines for
absorbance
Also
produces
A signifcant
instrumental
problem
Io is the area under the Io curve
I will be the area under the Io curve – the area inside the blue, natural Line width
1
0.9
0.8
Io
Bandwidth from a slit
0.7 nm
0.7
Intensity
0.6
0.5
0.4
0.3
0.2
Natural line width
Which will absorb
light
0.1
0
274
276
278
280
282
284
286
288
Wavelength
Will we be able to distinguish between Io and I?
290
292
294
 I
A   log 
 Io 
To make it simple consider the light passed as a rectangle
Io
x
0.7 nm region passed by slits
282.85
283.35
283.2
Amount removed
By 10-5 nm width of
Atomic absorption line
(Max absorption)
Amax
wavelength
 0.7 x  10  5 x 
  log
  0.0000006204 AU
0.7 x


Boy – that’s tough!
 0.001x .00001x 
Amax   log
  0.004


0.001x
If we could get the source light
Down to 0.0001
Band width we could maybe do it
Io
Natural linewidth
~10-5
282.9
283nm
0.001 nm or less
To do this implies that we need a line source
283.9
wavelength
Atomic Absorption Source: Hollow Cathode Lampe
Cup made
Of element, M
M go,*  M go  atomic,M
cathode
Ms  M
V
+

g , KE
ArKE
Ar
anode
What things might affect this process?
Why is the linewidth from atomic emission coming out of the lamp wider
Than the natural linewidth?
Natural Line Widths
Line broadens due
1. Uncertainty
t  1
2. Electric and magnetic fields
3. Doppler effect

0

Typical natural line widths are 10-5 nm

c
0   0i
0
4. Pressure (1 atm)
i
Single Atom
Pressure broadening, P
P
 P  0
P0
Multiple
Interacting
Atoms
Effect is to broaden
Line width by
2 orders of
Magnitude
T0
T
Is related to pressure, at 1 atm
It is ~1 pm (10-3 nm)
10-3 nm = 1 pm
Pressure broadened line widths are
(102 to 103 )(10-5 nm)= 10-3 to 10-2 nm
Atomic Absorption Source: Hollow Cathode Lamp
Set V to control total pressure in the hollow (low pressure) tube
If P = 0.01 atm then we can get the line width close to the natural linewidth!
Cup made
Of element, M
P
 P  0
P0
M go,*  M go  atomic,M
cathode
Ms  M
V
+
anode

g , KE
ArKE
Ar
T0
T
1
0.9
0.8
Relative Intensity
1
0.9
0.7
0.6
0.5
0.4
0.3
0.2
Relative Intensity
0.8
0.1
0
283.19
0.7
283.195
283.2
283.205
283.21
Wavelength
0.6
Pressure Broadened
linewidths
0.5
10-4
0.4
0.3
0.2
Natural Line width, ~10-5
0.1
0
283
283.05
283.1
283.15
283.2
Wavelength
283.25
283.3
283.35
283.4
 I
A   log 
 Io 
 0.0001x .00001x 
Amax   log
  0.04


0.0001x
Gap is
Light
absorbed
1
Wavelengths
that can be
absorbed by
atom
0.9
0.7
0.6
0.5
0.4
1
0.3
0.9
0.2
0.8
0.1
0
283.19
283.195
283.2
Wavelength
Source pressure
Broadened emitted light
283.205
Intensity after Absorption
Relative Intensity
0.8
0.7
0.6
283.21
0.5
0.4
0.3
0.2
0.1
0
283.17
283.18
283.19
283.2
Wavelength
283.21
283.22
283.23
We have another problem: Molecular absorption bands by
MgO CaO etc.
Atomic Spectroscopy by James Robinson
Ameasured
ANickle line
Amolecular oxide
A0
How can we make a measure of the background in order to
Correct the total measured absorbance?
To measure the background use a broad band source: The amount of light
Absorbed by the line will be so small that it does not contribute
 I
A   log 
 Io 
Amount of light removed by molecular oxide =30%
Io
x
0.7 nm region passed by slits
282.85
283.55
wavelength
283.2
Amount removed
By 10-5 nm width of
 0.7 x  10  5 x 
  0.0000006204 AU
Atomic absorption line Aatomic   log
0.7 x

(Max line absorption)
1x  0.7x 
Amolecularoxide   log 
 0.15490196
 1x 
Atotal  Aatomicline  Amolecularoxide  6.204x107  0.15490196  0.154902584

 0.15490196 
%Aatomicline  100
 0.0004%
0.154902584 
A is 99.96% due to molecular oxide
Atrue  Abroadband source  Aline source
Schematic of our PE AA
2 broadband sources, why?
Instrumental Requirements:
1.
True monochromatic source
2.
Monochromator selected “monochromatic source”
3.
Minimize molecular oxides (Drive CMO down)
4.
Get Pb into gas phase
5.
Get Pb into gas phase as an atom
Events required to transform sample
Molecular species
Ionic Species
M+*gas
MX*gas
Molecular Oxides
MO*gas
Line source
MOgas
MXaq
evaporate
MXgas
ionize
M+gas +e
Mogas
M*ogas
heat
“atomization”
M*ogas
Mogas
Thermally
electronically
excited atoms
Intensity of the absorption signal varies if the pathway to
and equilibrium state of the atomic gas is altered
Background grows with presence of alternative species
emission
The type of atomization will affect the
Prevalence of the different types of species
Types of atomization:
a.Flame
b.Electrothermal (graphite tube)
c.plasma
Envelope, high in oxygen due
To turbulent mixing
Acetylene/Air Flame
CO  O2  H 2  CO2  H 2O
Temp
Stable
Conc. of
H2, H2O,
CO, CO2
C2 H 2  O2  2CO  H 2
Primary combustion
Zone, lots of radicals
Formed, reducing
Maximum
Temperature
A flame changes temperature vertically and horizontally.
A flame has a larger oxygen content at the edges
http://ristretto.ecn.purdue.edu/projects/soot.html
Spectrochemical Methods of Analysis
Simulation of turbulent flame flow
J. D. Winefordner, chap. 1
axial
radial
Temp of air/Acetylene flame is ~2500K
Applied Spectroscopy 52, 2, 72A, 1998
A flame changes temperature vertically and horizontally.
A flame has a larger oxygen content at the edges
Cr
Mg
4
5cm
1750
3
1830
2
1863
1
1830
What do you observe?
Can you explain it?
Hint: Cr3+, Mg2+, Ag+
Absorbance
Ag
Height
Above
Entrance
To flame
Temperature
MOgas
MXaq
evaporate
MXgas
ionize
M+gas +e
Mogas
M*ogas
Absorbance
A flame changes temperature vertically and horizontally.
A flame has a larger oxygen content at the edges
Ag
Cr
Mg
1
2
3
4
5cm
Mg2+ is less reactive to oxygen
MgO
MgX
AgX
2
3
4
5cm
1863
1830
1750
1
1830
Ag+ is very unreactive to oxygen
Absorbance
CrO
Cr oxidizes to Cr3+ at low temp;
Cr3+ is highly reactive to oxygen
Ago
Mgo
Temperature
Height
Above
Entrance
To flame
What else happens in the flame that we can control?
1.Flow rate
2.Height at which we observe the populations
Two opposite trends in flow rate:
initially totally moles of atom increase
volume  moles 
moles  

min
 min volume 

but the increase also increases the amount
of water that consumes the available
energy on dehydrating
Flame Atomization: Evaporation
1.
2.
3.
4.
Aspirate
Breakup aerosol into droplets
Collect large size fraction to discard
Distribute Atom vapor along long cell path
(the “b” in Beer’s Law)
Solution to the problem of MO formed in flames
Graphite Furnace Atomization
1.
Dry drop onto surface
MX aq  MX s  other
2.
Ash (burn off organics)
MX s  other  MX s
3.
atomize
MX s  M go  other
Used to
Shield region
From oxygen
Advantages of GFAA
↓liquid volume sample
 Mog in a small volume
↓time involved
↓ MO due to
a) no O2 required for flame
b) Ar shielding of cell
Disadvantages of GFAA
Not isothermal across the tube deposition of sample at cooler ends of tube
a) decreases Mog
b) Also revaporizes on a second run = background
Pyrocoated tubes
Solve disadvantages
a) L’vov platform creates an isothermal environment
b) Block the porous surface: graphite treatment
Programmed temperature
What you want is to produce the population
To be sampled all at one time with no
Other species capable of producing an
Absorbance signal.
Avoid
molecular
Species
(avoid Cl-)
Avoid ionized species????
Molecular species
Ionic Species
M+*gas
MX*gas
Molecular Oxides
MO*gas
Line source
MOgas
MXaq
evaporate
MXgas
ionize
Need to remove water and volatilize
Make certain that MX can volatilize:
EDTA complexation!!
M+gas + e
Mogas
M*ogas
heat
M*ogas
Mogas
Thermally
electronically
excited atoms
emission
Reduce Ionization
M  M e
x
M e
K

M 
M    e
x2 p
 x 1  0
K
M   M 
M 
K
 2
 2
M 
Calculate the fraction ionized as a
Function of total metal (p) and K
Where K depends on T
K
Mass Balance
 RT ln K  G
 
p  M   M 
Fractions (alphas)
M   M 
x
M   M  p
M   1
x
M   M  1  M 
K
K




 2


1
xp
1
K
Atom
Cs
Rb
K
Na
Li
Ba
Sr
Ca
Mg
kJ/mole (1st ionization)
377
402
418
498
519
502
548
590
736
Atom
Cs
Rb
K
Na
Li
Ba
Sr
Ca
Mg
1.2
Cs
p=1e-6
1
kJ/mole (1st ionization)
377
402
418
498
519
502
548
590
736
p=1e-4
Fraction Ionized
0.8
Mg
0.6
0.4
What should you add
To the analyte soln to
Suppress ionization?
0.2
0
0
2000
4000
6000
Temperature, K
8000
10000
12000
Spectral line interferences:
Darker lines are
More probable
Transitions, higher
Molar absorptivity
They all originate
6p2 to 7s transitions
K, 404.4 and 404.7 nm
Mn, 403.1, 403.3, 403.5 nm
Why don’t we use these transitions?
Pb: 1s2 2s2p6 3s2p6d10 4s2p6d10f14 5s2p6d10 6s2p2
Outer valence electrons
Ionic Species
Molecular species
Molecular Oxides
M+*gas
MX*gas
MO*gas
Line source
MOgas
MXaq
evaporate
MXgas
ionize
M+gas + e
Mogas
M*ogas
heat
M*ogas
Mogas
Thermally
electronically
excited atoms
Monitor emission – generally requires
More energy to get all atoms excited
emission
RAISE the Temperature!
Graphite (Electrothermal)
Temp, oC
1200-3000
Flame
1700-1900
2000-2100
2100-2400
2550-2700
2600-2800
2700-2800
3050-3150
1.2
p=1e-6
1
p=1e-4
0.8
Fraction Ionized
Natural gas/air
H2/air
Acetylene/air
H2/O2
Acetylene/N2)
Natural gas/O2
Acetylene/O2
0.6
0.4
0.2
0
0
Inductively Coupled Plasma (ICP)
4000-6000
2000
4000
6000
8000
10000
Temperature, K
At this temperature you get enhanced selectivity because you
Can choose not only from atomic lines but ion lines
12000
K
http://www.mrl.ucsb.edu/mrl/centralfacilities/chemistry/icp.pdf
http://www.unilim.fr/theses/2006/sciences/2006limo0029/xml/ressources/image039.jpg
1.2
p=1e-6
1
p=1e-4
0.8
Fraction Ionized
When coupled to emission can simulataneously
Determine multiple elements, because every
Element has at least one line either as an atom or
As an ion that is free of spectral interferences!
0.6
0.4
0.2
0
0
2000
4000
6000
Temperature, K
8000
10000
12000
Line least likely to have
interferences
http://marine.rutgers.edu/LAICPMSintro/ICP-OES.html
Notice that
The various
Wavelengths
Of light are
Spread into
Two dimensions
Allows simultaneous
determination
http://www.sseau.unsw.edu.au/icp_instruments.htm
ICP AES
Not as
sensitive
As GFAA but
Has advantage
that can make
simultaneous
measurements
An atomic tangent (not UV-vis):
These electron transitions represent only a fraction of the possible events:
Inner shell electrons can also be probed in a way that has excellent selectivity
An atomic tangent (not UV-vis):
An atomic tangent (not UV-vis):
Inner Shell absorption edges for lead
What goes up must come down: inner shell fluorescence
An atomic tangent (not UV-vis):
http://www.learnxrf.com/
This site gives a reasonable written
Introduction to XRF, very few images
An atomic tangent (not UV-vis):
Field Portable X-ray Fluorescence
An atomic tangent (not UV-vis):
Source is usually a cobalt
source
http://www.niton.com/Portable-XRF-Technology/how-xrf-works.aspx
Molecular Spectroscopy
Where are the electrons that can be interrogated by molecular spectroscopy?
-in different orbitals
, , n
C
xx
O
x


n
POSSIBLE USEFUL TRANSITIONS
Criteria
1.  molar absorptivity
2. Peak absorbance between 150 and 600 nm
* antibonding
LUMO
Lowest Occupied Molecular Orbital
n*
n*
*
*
Energy
*, antibonding
n, nonbonding
, bonding
HOMO
Highest Occupied Molecular Orbital
Group Transition
C-C
*
O:
n*
n*
*
, bonding
(L/cm-mol)
100-3000
10-100
103 – 104
(nm)
<125
150-250
long wavelength
shorter wavelengths
Why is 100-800 nm the useful range?
Why
might
You
monitor
this
region at
the end
of a
chromato
graphic
column?
To observe lead using spectroscopies other than atomic absorption we have had to
Attach it to a signaling compound:
Calcein Blue
H3C
Chromophore – part of molecule sensitive to light
HO
O
O
+
N
H
O
O
O
-
“Selectivity” arm – complexes the metal ion and turns
On and off fluorescence
-
O
Dithizone
Chromophore
N
H
N
HS
N
N
O
N
N
HO
N
O
N
SH
H
Ethylenediaminetetraacetic acid
HO
N
Selective part
“Selectivity” arm
N
O
O
1.
2.
Which one(s) of the three molecules will
be most UV-Vis sensitive?
Why?
OH
OH
n*
Energy
* antibonding
n*
* antibonding
n, nonbonding
n, nonbonding
, bonding
, bonding
Large number of electrons on lead perturbs lone pair electron on
nitrogen making visible an n* transition
n* transition, short wavelength, weak molar absorptivity
n π* Transitions should occur at longer wavelengths and should have strong
solvent effects
Consider the effect of stabilizing/destabilizing non-bonding electrons on O or N
Π*
Π*
Π*
n
n
n
destabilizing those electrons without altering the energy of the π* orbital
decreases the energy gap and shifts the absorption wavelength ….?
H3C
“Parent” compound of CB
HO
O
O
HO
O
O
+
N
H
O
O
O
-
O
-
What happens to the absorption band for CB parent
Molecule with different solvents?
N
As the deprotonated oxygen is pushed into
A hydrogen bond by the proton from the nitrogen
Group the wavelength for absorbance
Shifts to longer wavelengths, see especially
The developing shoulder
O
N
O
Mizoguichi, Hiroki et al, Ber. Bunsenges Phys Chem 1997 101, 12, 1914-1920
* transitions are most probable and occur when
There are ……?
Calcein Blue
H3C
Chromophore – part of molecule sensitive to light
HO
O
O
+
N
H
O
O
O
-
“Selectivity” arm – complexes the metal ion and turns
On and off fluorescence
-
O
Dithizone
Chromophore
N
H
N
HS
N
N
N
SH
H
N
N
N
“Selectivity” arm
Effect of independent * chromophore is additive
Compound
CH3CH2CH2CH=CH2
CH2=CHCH2CH2CH=CH2
Molar Absorptivity
~10,000
~20,000
Additivity of the pi bonds is what gives intense color to the reagents we
Have worked with.
N
H
N
HS
N
N
N
SH
H
N
N
N
Making a commercial color test for lead???
Chromophore
H3C
N
H
O
O
H
N
CH3
N
H
Selectivity
O
O
P
HN
O
O
O
N
N
H
H
N
O
N
O
CH3
O
H
N
N
P
O
O
CH3
O
O
P
O
O
CH3
CH3
More selectivity
Solubility
Single bonds
N to pi star (Pb-EDTA)
Double bonds & Rings
D orbital splitting
Metal to Ligand charge transfer
Intervalence charge transfer
T2g d orbitals on the
Central metal lie
Somewhat out of
The path of the
Incoming octahedrally
Oriented ligands
Eg d orbitals on central metal
lie in the path
Of incoming octahedrally
Oriented ligands
Image: A Van der Ven and G. Ceder, p 47 in Lithium Batteries Science and
Technology, Nazri and Pistoia, eds., Kluwer, 2004
z
Images of electron
Density of d orbitals
x
y
http://vinobalan.tripod.com/sitebuildercontent/sitebuilderpictures/picture1.gif
z
z
x
y
dx2-y2
∆
z
Energy
z
x
y
dxy
No incoming ligands
y
dxz
z
x
y
z
x
dxy
y
z
z
x
x
y
dx2-y2
xy
dz2
dyz
dz2
ligands
z
z
x
y
x
y
x
y
dyz
dxz
z
Incoming Clsee little e
Cl-
Incoming Clbumps into d orbital e
xy
x
+
z
y
z
x
y
Clz
x
xz
+
y
Cl-
+
Cl-
x
+
y
yz
z
x
+
y
Cl-
1.
2.
3.
4.
5.
Initially all orbitals same energy
Incoming anion interacts with d orbitals
Energy levels change due to the interaction
3 orbitals move down in energy
2 move up
z
Cl-
z
Cl-
x
+
y
z
Cl-
xy
x
z
+
y
Cl-
y
+
x
+
y
z
T2g d orbitals on the
Central metal lie
Somewhat out of
The path of the
Incoming octahedrally
Oriented ligands
Eg d orbitals on central metal
lie in the path
Of incoming octahedrally
Oriented ligands
Image: A Van der Ven and G. Ceder, p 47 in Lithium Batteries Science and
Technology, Nazri and Pistoia, eds., Kluwer, 2004
Suppose it is :NH3 instead of ClWhat do you think will happen?
z
z
x
x
+
Clz2
+
.. N
y
y
z
D orbital energy in absence
of ligands
z
Greater Orbital Splitting with Nitrogen
charge dense lone pair (localized)
CN->NO2->en>NH3>NCS->H2O>F->OH->Cl->SCN->S2->Br->IStrong field
Weak field
eg
6Dq (2*6=+12)
10Dq
10
Dq
4Dq (3*4= -12)
t2g
CN->NO2->en>NH3>NCSStrong Field
OH->Cl->SCN->S2->Br->IWeak Field
http://www.chm.bris.ac.u
k/webprojects2003/roger
s/998/chemla2.gif
Note the spectral region
Single bonds
N to pi star (Pb-EDTA)
Double bonds & Rings
D orbital splitting
Metal to Ligand charge transfer
Intervalence charge transfer
Ligand to metal charge transfer

O   Pb 2  
 O   Pb 
Colors are different because the electrons on
Oxygen occupy different energy levels based
On the “flip” of the chain
Intervalence charge transfer
Pb4+
Pb2+

Pb 2   Pb 4  

 Pb  3  Pb 3
http://www.newport.com/Helios-Transient-Absorption-Spectroscopy-System/477063/1033/catalog.aspx?Section=Drawing#
2009 price: $80,000
Can you trace the optical beam in this instrument and name the component
Parts?
Might be useful for an exam
HP 8452 and 8453 spectrophotomers were used in the dithizone lab
Now known as Agilent 8452 and 8453 (2009)
8453
Deuterium and tungsten light sources
Why does it have two light sources?
We never discusse
Or measured
The wavelength
Accuracy.
(e.g. calibrating
The x axis of a
Spectra).
How would you do
It?
Transition
*
n*
n*
*
dd
Intervalence
LMCT
(L/cm-mol)
100-3000
10-100
103 – 104
(nm)
<125
150-250
long wavelength
shorter wavelengths
500
600-700
200/600-700
MOLECULAR LUMINESCENCE
Types of Molecular Luminescence
1.
Photoluminescence
a. Fluorescence
b. Phosphorescence
M  1  M *  M  2
2. Chemiluminescence
M  N  M *  P  M  2
Comments
inherent sensitivity
LOD ~103xmolecular absorption
X LOD  X blank  3s,blank
X blank  0 (light on exp eriment )
Linear range > molecular absorption
Selectivity > molecular absorption
WHY?
Hmm, at the time of this publication did we see any F?, why?
The probability to making an electronic
Transition to the nth vibrational state
Is similar both in excitation and in
Emission
This gives rise to “mirror” image
spectra
“Mirror image” spectra
Rhodamine Red
Factors Affecting the Yield of Fluorescence
Quantum Yield, , = ratio of photons emitted
To photons absorbed
+
NH2
NH3
>>
What might account for these results?
H
H
N
Resonance stabilization of aniline
H
H
N
H
+
H
H
H
+
N
N
H
+
H
N
H
-
CH
-
CH
1.
Resonance stability
Observations?
F
H
Relative 
10
10
Cl
7
Br
5
I
0
Heavy atom =  spin/orbit coupling
=  ki
= ↓
1.
2.
Resonance stability
Loss of electrons by spin flipping
+
Electron spin flip is related to magnetic field, B
Those who indulge in diatribes are most likely to be single
Spin of electron couples with orbital precession in spin/orbit interaction.
Interaction is greatest with highly charged nucleus
Greek
Dia
in opposition to
(as in diatribe:
a bitter, abusive attack)
Symbolism
↓
States
singlet
ki
Para
beside, for
(as in paramour:
lover)

triplet
Rate of intersystem crossing, or spin flipping, ki
Texas Red
Rhodamine Red
TAMRA
cy5
Oregon Green
FITC….(h)?
FAM=fluoroscein
What trends do you observe with
a) Temperature……b) Structure?
a) ↓as T 
1.  collisions
2. ↓viscosity
Rate of external conversion = kec
1.
2.
3.
Resonance stability
Loss of electrons by spin flipping
External collisions
b) Structures with more rotatable groups seem to have less fluorescence
Effect of structural rigidity
↓ rigidity = ↓
rigidity = 
=0.2
Fluorene
=1
=0.7
biphenyl
=0.2
FAM
=0
1.
2.
3.
4.
5.
Resonance stability
Loss of electrons by spin flipping
External collisions
Vibrational or internal Rotational which
moves system to a different, non fluorescing singlet state
Structural rigidity
=0.9
b) Molecules with more allowed internal rotations quench
with temperature faster
Rate of internal conversion = kic
1.
2.
3.
4.
Resonance stability
Loss of electrons by spin flipping
External collisions
Vibrational or internal Rotational
H3C
1.2
Fluorescence Intensity at max.
(A*)
1
O
CB
-
O
O
0.8
(A*)
(A*) quenched
0.6
by amino group
0.4
H3C
7HC
(T*)
0.2
O
-
O
O
N
0
0
2
4
6
8
pH
What do you observe?
10
12
14
O
O
O
-
O
-
H3C
O
O
O
-
+
N
H
O
O
O
-
-
O
Rate of electron transfer = ket
Ground state
Is occupied
:N
H3C
O
-
O
O
N
O
O
O
-
O
-
1.
2.
3.
4.
5.
6.
Resonance stability
Loss of electrons by spin flipping
External collisions
Vibrational or internal Rotational
Structural rigidity
Internal (also external) electron transfer
Quantitating Fluorescence
We observed:
1.
2.
3.
4.
5.
6.
Resonance stability
Loss of electrons by spin flipping
External collisions
Vibrational or internal Rotational
Structural rigidity
Electron Transfer quenching
QUANTUM YIELDS or EFFICIENCY, 

# molecules fluorescin g / s
# molecules excited / s

# molecules fluorescin g / s
# molecules decaying by all processes / s

rate fluorescin g
rate decaying
kf= fluorescence
ki = intersystem crossing (ST)
kic =internal conversion (S2S1)
kec = external conversion
(*ground state, no photon)
ko = other, dissociation
ket = electron transfer quenching
Jablonski diagram
Singlet (fluorescence)
kec
kic
ket
kf
ki
Triplet (phosphorescence)
Several orders
of magnitude
slower
Because the
electron
Has to flip on the
way down
Quantum Yield

k f *
k f *  k other ( dissociation) *  ki *  kic *  k ec *

kf
k f  k other ( dissociation )  k i  k ic  k ec
chemistry
environmental
k other ( dissociation)  for   250 nm
Why?
Label these as kf, kdissociation, ki, kic, kec
kic
kf
kf
kf
kd
kec
Fluorescence Intensity
F  K Po  P
F  K 
P
 10  bC  e  2.303bC
Po
P  Po e 2.303bC

F  KPo 1  e
 2 .303bC

2
3
4
x
x
x
ex  1 x 

 .....
2! 3! 4!
1  e x
1  e x
2
3


 x
 x


 1  1   x 

.....
2!
3!


2
3
x
x


 x

.....
2!

Point: strongly absorbing
Molecules may give
Non-linear fluorescence
response
3!
F  KPo 1  e  2.303bC

2
3
4
5


2
.
303

bC
2
.
303

bC
2
.
303

bC
2
.
303

bC








F  KPo 2.303bC 



.........
2!
3!
4!
5!


0.7
2
3
4
5


bC bC bC bC

F  KPo bC 



.........
2!
3!
4!
5!


Actual
Expected
0.5
bC
0.3
3
3!
F
0.1
-0.1 0
-0.3
0.2
0.4
0.6
bC
2
2!
-0.5
-0.7
Conc.
(Or eta, b)
0.8
bC
4!
1
4
2
3
4
5


bC bC bC bC

F  KPo bC 



.........
2!
3!
4!
5!


0.7
Sum 3
0.5
Sum first 2
0.3
F
0.1
-0.1 0
0.2
0.4
0.6
-0.3
-0.5
-0.7
Conc.
(Or eta, b)
0.8
1
0.7
true
expected
0.5
0.3
F
0.1
-0.1 0
0.2
0.4
0.6
-0.3
-0.5
-0.7
Conc.
0.8
1
Coupling Fluoresence to Analytical Measurements
1. signal off
2. signal on
Two examples of signal off
understanding fate of Pb in soils
Our Calcein Blue lab
Signal Off- Example 1
Brown stuff in soils = humic acid
http://www.rsc.org/ej/gt/2000/b001869o/
Rings which can
fluoresce
http://www.chem.neu.edu/web/faculty/davies.html
Lots of nice
Binding sites
Signal off – example 1
Here you can see the ability to use the two point recognition (selectivity) of F wel
Both excitation and emission can be used
HO
O
O
+
N
H
O
O
O
-
H3C
Excited
State
Proton
transfer
O
Signal off example 2
OH
O
-
O
O
O
O
-
Emission
Spectra, excitation at 320
Absorbance
spectra
+
N
H
-
O
1
0.4
0.9
0.35
0.8
0.3
480-490
Absorbance
0.7
0.25
0.6
0.5
0.2
0.4
0.15
0.3
0.1
0.2
320
0.05
0.1
0
200
250
300
350
400
450
500
550
0
600
wavelength, nm
H3C
HO
O
O
+
N
H
O
O
pH 6-8
O
-
-
O
Carboxyl groups only deprotonated
Stop fluorescence by interfering
With excited state proton transfer
Relative Fluorescence Intensity
H3C
Signal off example 2
Why might Pb quench the emission?
Lead quenches
emission
Emission spectra
Interferes with excited state proton transfer
1.
2.
Weaker Ca-O bond?
Effect of 2s lone electron pair on Pb?
Structures as determined from NMR
Coupling Fluoresence to Analytical Measurements
1. signal off
2. signal on
Three examples of signal on:
rigidity for metal ion analysis
Ca by Calcein blue to stopping internal electron transfer
Pb by protein fragment switching to stop internal electron transfer
Signal On, Example 1
Use chelation to achieve rigidity and decrease rate of external conversion
OH
O
N
O
-
OH
N
2+
Zn
O
-
N
benzoin
8-hydroxyquinoline
O
OH
O
OH
Flavanol
OH
HO
N
N
Alizarin Garnet R
Undergraduate Instrumental Analysis
By James W. Robinson, Eileen M. Skelly Frame
SO 3Na
Signal On, example 2: Ca2+ by Calcein Blue pH 11
1.2
H3C
-
O
O
N
O
O
pH 11
O
-
-
O
No F
Fluorescence Intensity at max.
O
(A*)
1
CB
0.8
Ca2+
(A*)
(A*) quenched
0.6
by amino group
0.4
7HC
(T*)
0.2
Ca2+
0
0
-
O
N
O
O Ca2+ OO
-
4
6
8
10
pH
H3C
O
2
O
F
Getting rid of N: by binding them to Ca2+ turns F back
on
12
14
Signal On: Example 3 Pb2+ by nucleic acid coil switch
Fluorophore, as an electron donor in
An electron transfer reaction
Fluorophore
(e donor)
Light off
Quencher
(e acceptor)
analyte
Light on
Acceptor
molecule
Donor
molecule
Signal On: Example 3 Pb2+ by nucleic acid coil switch
Fluorophore
(e donor)
Fluorophore donor
Quencher
(e acceptor)
analyte
Dabsyl
Quencher (acceptor)
Signal On: Example 3 Pb2+ by nucleic acid coil switch
Response Time
Signal On
Pb2+
Selectivity
Co2+
Li and Lu, JACS 2000 122 42 10467
Fluorescence
Iemitted
Wavelength 2
Io
Sample
Wavelength 1
Wavelength 1
I
Absorbance
180o
90o
320 nm
excitation
Excess
excitation
450 nm
emission
450 nm
excitation
355
nm
sample
450
nm
450
nm
450
nm
Describe this instrument
http://www.iss.com/resources/tech7/
1.
2.
3.
4.
What is the purpose of the double beam?
Why is the detector at right angles to the excitation?
Could we use other angles?
What is the purpose of the reference PMT?
This experiment used an excitation monochromator set at 355 nm
Lead quenches
emission
All emitted light observed
What is happening ?
We are now “foreshadowing” the next section……………..
Raman Spectroscopy
Uses UV-Vis spectrum to measure vibrational changes
Emission of Photons
Electromagnetic radiation is emitted when electrons relax from excited states.
A photon of the energy equivalent to the difference in electronic states
Is emitted
Ehi
Elo
e
hc
 E  h 

Lower energy
Longer 
Higher energy
Shorter 
“virtual state”
Why do you get this pattern?
E  E 0 cos2ex t 
m  E  E0 cos2ex t 
Excitation frequency
Induced dipole moment depends on polarizability of the bond
  
   0  r  req  
 r 


req
r
Polarizability changes with bond length changes
r  r   r
eq
m
cos2 v t 
Vibrational frequency
m  E  E0 cos2ex t 
  
   0  r  req  
 r 


r  r   r
eq
m
Combine
cos2 v t 

   
m    o  rm cos2v t     E 0 cos2 ex t 
 r  

  
m  E 0 cos2 ex t  o  rm cos2 v t   E 0 cos2 ex t 
 r 
cos x cos y 
cos x  y   cos x  y
2
  
m  E 0 cos2 ex t  o  rm cos2 v t   E 0 cos2 ex t 
 r 
cos x cos y 
m   o E 0 cos2 ex t  
cos x  y   cos x  y
2
E 0   
E
  
rm   cos2  ex  v v t   0 rm   cos2  ex  v v t 
2  r 
2  r 
Selection rules
REQUIRE a change
In polarizability

1
r
Cool thing: can measure
Vibrational spectroscopy
Using UV-Vis optical
1. Not H2O sensitive
optics

ex
 vv 
 ex
ex  vv 
A Case of Forensic Chemistry: Art and Forgeries
Lead Tin II, Paolo Veronese,
Allegory of Love
Lead Antimonate
Peter Rubens, The Dying Seneca
Lead Tin I
Forensic Art Chemistry
Two Sb octahedra
Linked via vertices
to a) eight pointed
polyhedra Of Pb &
b) Hexagonal
bipyramid
Lead Antimonate
Lead Tin I
Lead Tin II
Chains of
Sn octahedra
Joined by
Pyramidally
Coordinated Pb(II)
Circular dichroism
Consider a peptide bond
H2N
O
H2N
Beta Sheet
NH2
Polymerize here
H2N
H2N
O
NH
H2N
H bonding between N and O leads to three types
Of polymeric secondary structures
O
H2N
NH
O
NH2
NH
O
NH
NH2
O
NH2
Beta Sheet
Alpha Helix
Random Coil
R
R
O
R
NH
O
HN
R
Cis
Trans
Peptide bond absorption between 190-250 nm,
-*
Resonance stabilized
R
O
R
NH
Trans in polymeric folded state is preferred
n*
Very weak shoulder
NV1(-55o)
mn*
Magnetic dipole transition moment
O
NV2(60o)
R
N
R
H
Electric dipole transition moments
In an alpha helix these moments have pairwise interactions
With neighboring peptide bonds so that the sum of the
Moments is large enough to enhance the molar absorptivity
Of the n* shoulder
Notice also that the orientation of the dipole moments is not perpendicular
Leads to absorption of circularly polarized light
http://www.enzim.hu/~szia/cddemo/edemo0.htm
Animation of circular
polarization
Beta Sheet
Of poly-l-lysine
Fix this image for alpha helix
in your mind
For comparison below
Poly-l-lysine
http://www.ap-lab.com/circular_dichroism.htm#CD_secondary
Osteocalcin
Ca2+ or Pb2+ binding
At carboxylic acid groups
Will affect the coil region
www.earlbenjamin.com/images/Osteocalcin3.jpg
Alpha helix will be sensitive to polarized light measurements
http://www.flickr.com/photos/grrlscientist/1086037481/
Random coil
Alpha helix
0, 0.1, 0.5, 0.7, 1.0, 3.0, 5.0, 6.0 mM Ca2+
0, 10, 25, 31, 40, 48, 55 uM Pb2+
Note that the transitions occur at
Much lower concentrations of Pb2+
Data at 220 nm
Olis DSM 20 CD
Sample holders
(temp controlled with
2 detectorsWater bath)
Light
From
monochromator
Optional F detection (note 90o angle for detection)
Polarization beam splitter, modulated between R and L at 50 times
Per millisecond;
followed by circular polarization by photoelastic device
First polarize linearly
CaCO3 separates different polarities of light
Transmitted parallel light goes to zero
This simulation from molecular modeling shows how electrons at the surface of
A media are excited and re-radiate as individual radiators. The individual radiator
spread and form a wavefront (Hyugen’s Principle). For this simulation the electric field
stimulation was parallel to the surface and re-radiation results in scattering (reflection).
TM polarized light is parallel in our terminology
http://www.smart-systems.at/rd/rd_optoelectronic_VMM_convergence_de.html?printview=true
Then polarize circularly
Hinds PEM 90 – photoelastic modulators for polarization
Expansion slows the blue component
So it lags, out of phase
Unstressed plate passes both
Components in phase
Compression speeds the blue component
So it is now out of phase
http://www.hindsinstruments.com/PEM_Components/Technology/principlesOfOperation.aspx
Hinds PEM 90 – photoelastic modulators for polarization
When the phase separation is ¼ the wavelength of the light right circularly
Polarized light is obtained.
http://www.hindsinstruments.com/PEM_Components/Technology/principlesOfOperation.aspx
Animation of circular polarization at a ¼ wave retardation
http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_apr_09_2003.shtml
This site also has animations
http://cct.rncan.gc.ca/glossary/index_e.php?id=3059
END HERE