158
Chapter 4 Ultraviolet and visible Absorption Spectroscopy
Chapter 4
Ultraviolet and visible Absorption
Spectroscopy
Properties of Electromagnetic Radiation
Electromagnetic Radiation
– energy radiated in the form of a WAVE
caused by an electric field interacting with
a magnetic field
– result of the acceleration of a charged
particle
– does not require a material medium and
can travel through a vacuum
Electromagnetic Radiation
Electromagnetic Radiation
vi = n li
where vi => velocity
n => frequency
li => wavelength
Electromagnetic Spectrum
Type of Transition Wavelengt
h Range
Frequency
Range (Hz)
Type of
Radiation
nuclear <1 pm 1020-1024 gamma-
rays
inner electron 1 nm-1 pm 1017-1020 X-rays
outer electron 400 nm-1
nm 1015-1017 ultraviolet
outer electron 750 nm-
400 nm 4-7.5x1014 visible
outer electron
molecular vibrations
2.5 µm-750
nm 1x1014-4x1014
near-
infrared
molecular vibrations 25 µm-2.5
µm 1013-1014 infrared
molecular rotations,
electron spin flips*
1 mm-25
µm 3x1011-1013
microwav
es
nuclear spin flips* >1 mm <3x1011 radio
waves
Electromagnetic Spectrum
Interaction of EMR with Matter
Jablonski diagram:
Selection Rules
• The electron must be promoted without
a change in its orientation. s = 0
• When s ≠ 0 transition is forbidden. It
may occur with very low probability
• Some other from quantum mechanics
Etotal (molecule) = Eelectronic + Evibrational +
Erotational + Enuclear
Absorption of Light
Uv & Vis
IR
Microwave
*
Molecular and Atomic Absorption
* Less extent
Collisions between molecules
lead to broadening of
absorption bands
Types of Transitions
Three types of transitions
1. , , and n electrons
2. d & f electrons
3. charge transfer electrons
Electronic transition in Formaldehyde
Spectroscopy Nomenclature
* vacuum UV
Effect of Structure on
lmax Cl < lmax Br < lmax I
n * transitions occur at longer wavelengths
is a function of
1. Cross sectional area of absorbing species ( )
2. Transition probability (P)
= 9X1019 P ( = 10-15 cm2 ( = about 105 for the
average organic molecule
Transition Multiplicity
Consider two electrons paired in an orbital, and their possible transitions to an empty orbital.
ground state excited singlet excited triplet
• the ground state has all electrons in the lowest energyorbital • organic compounds almost always have paired spins, thus their ground state is almost always a singlet • singlet-triplet transitions are optically forbidden - light cannot both promote an electron to a new orbital and change its spin • in an organic compound most absorption spectra are due to singlet-singlet electronic transitions
2 1
ii
S s
M S
Electronic Transitions in Ethylene
C C
H
H
H
H
*
*
pz pzsp2 sp2
• Attention will be restricted to electrons involved with carbon-carbon bonding • The two sp2 electrons form the -bond, while the two pz electrons form the -bond • Absorption of a photon will promote one of the bonding electrons into an anti- bonding orbital, preserving electron spin • The wavelength of absorbed light will follow Planck's Law, E = hc/l
• The transition energies are: * > *, * > *
• The * transitions are of most interest since they give us information about the conjugated double bond structure of a molecule
* Transitions in Butadiene
C C
H
H
H
C
H
C
H
H
pz
pz pz pz
1
2*
2
1*
• Each carbon atom has one electron in a pz-orbital • The four pz electrons create two bonding -orbitals and two anti-bonding *-orbitals • 2 2
* absorption is in the deep UV, it has an energy similar to that in ethylene • The longest wavelength absorption is due to the 1 1
* transition • intermediate wavelength absorption is due to 2 1
* and 1 2* transitions
• The long wavelength transition has an energy that decreases with the number of double bonds
Rotational Broadening
0 1000 20000
0.5
1
energy (cm-1)
pop
ula
tio
n (
rela
tiv
e)
298 K• Boltzmann's constant is 0.694 cm-1
• kT = ~200 cm-1 at room temperature • the spacing of molecular rotational levels is a few tenths of reciprocal centimeters • Thermal energy populates many rotational levels giving molecules an internal source of energy • rotational energy available within a molecule can add to that of a photon, making a range of optical energy that can satisfy Planck's Law, E = hn + Erot
• the graph shows the energy of thermally populated rotational levels; the distribution has a width of ~700 cm-1
• an electronic transition will be broadened by this width
500 nm transition will be 17 nm wide (491 - 508 nm) 400 nm transition will be 11 nm wide (394 - 405 nm) 300 nm transition will be 7 nm wide (296 - 303 nm)
Vibronic Transitions
0 2 4 60
5000
10000
15000
20000
25000
30000
35000
internuclear distance (Angstroms)
(a)
(b)
• A simultaneous change in vibrational and electronic quantum numbers is called a vibronic transition • if the inter-nuclear distances are not affected when the electron changes orbitals AND the transition is symmetry allowed, the spectrum will have a single peak and, no or very weak, vibronic bands • If one or more vibrations have different equilibrium inter-nuclear coordinates, a vibronic sequence will appear in the spectrum - this is shown in (a) for a single vibrational mode (more than one can be affected) • If the electronic transition is symmetry forbidden, vibronic bands will appear for those vibrations that deform the molecule into a shape which has an allowed transition (b) • symmetry allowed: = 103 - 105 M-1 cm-1
symmetry forbidden: 102 M-1 cm-1
State Diagrams and Absorption Spectra
absorption spectrum state diagram
5000(2 m)
10000(1 m)
15000(667 nm)
20000(500 nm)
25000(400 nm)
30000(333 nm)
E (cm-1)
S0 v = 0
v = 1
S1* v = 0
12
1
3
S2* v = 0
2
3
0-0transition
infraredabsorption
symmetry forbidden1 1
*
1 2*
with achange in
internucleardistance
lmax
0 cm-1
( nm)
long wavelengthabsorption band
ground state
excited singletstates
Chromophores
• They are groups with one element of
unsaturation (unsaturated linkages or
groups) and cause coloring to the
molecules when they are attached to a
non-absorbing hydrocarbon chain
Effect of Multichromophores on Absorption
• More chromophores in the same molecule cause bathochromic effect
(Red shift: shift to longer wavelength)
and hyperchromic effect (increase in intensity)
• Hypsochromic effect: Blue shift: shift to shorter wavelengths
• Hypochromic effect: decrease in intensity
• In the conjugated chromophores * electrons are delocalized over larger number of atoms causing a decrease in the energy of to * transitionsand an incrase in due to an increase in probability for transition
•Aromatic Hydrocabons
They absorb at three bands: 260, 200
and 180 nm
•Policyclic aromatic (Naphthalene):
exhibit regular shift towards longer
wavelength (Red shift)
•Azo Compounds with the linkage –N=N-
show low intensity bands in the near Uv
and Vis due to n to * transitions
•Azobenzenes absorb at about 445 nm the –
N=N- may be conjugated with the ring
system.
UV absorption spectra of benzene, naphthalene, and anthracene
Auxochromes
• They are groups that do not confer
color but increase the coloring power
of a chromophore.
• They are functional groups that have
non-bonded valence electrons and
show no absorption at l > 220 nm; they
absorb in the far UV
• -OH and -NH2 groups cause a red shift
Steric Effect
•Extended conjugation of orbitals requires
coplanarity of the atoms involved in the -
cloud delocalization for maximum resonance
interaction
•Large bulki groups cause a perturbation of
the coplanarity of the system .
•Thus lmax is usually shifted towards shorter
and also decreases
Linear Polyenes
14.2 : 2
n name lmax (nm) max (M-1 cm-1)
1 ethylene 163 ?
2 butadiene 217 21,000
3 hexatriene 268 35,000
4 octatetraene 304 ?
5 decapentaene 328 120,000
CH CHn
• As the number of double bonds increases, the long wavelength absorption shifts to higher values (called a red-shift) • The molar absorptivity increases as the molecular orbital size increases • To anticipate the spectrum, use the number of conjugated double bonds, i.e. CH2=CH-CH2-CH=CH2 has a spectrum closer to ethylene than butadiene.
Linear Fused Aromatics
structure name lmax (nm) max (M-1 cm-1)
benzene 255 220
naphthalene 315 320
anthracene 357 10,000
tetracene 471 10,000
• As the number of fused rings increases, the long wavelength absorption shifts to higher values • The long wavelength transition is forbidden in benzene and naphthalene, but allowed in anthracene and tetracene • To anticipate the spectrum use the number of conjugated double bonds, i.e. diphenylmethane has a spectrum that resembles toluene
Linear Fused Aromatics
Non-Linear Fused Aromatics
structure name 0-0 band (nm) 0-0 (M-1 cm-1)
3,4-benzo
phenanthrene 370 170
chrysene 360 800
pyrene 370 120
perylene 437 3,700
• The 0-0 band appears at lower wavelengths than would be predicted by the number of fused rings (379 for anthracene and 479 for tetracene) • The first three have band positions similar to anthracene and molar absorptivities similar to naphthalene • Perylene has properties between anthracene and tetracene
Non-Linear Fused Aromatics
Linear Polyphenyls
n name 0-0 band (nm) lmax(nm) max (M-1 cm-1)
1 benzene 264 255 220
2 biphenyl 288 248 1,600
3 p-terphenyl 320 276 3,300
4 p-quaterphenyl 340 294 4,000
n
• As the number of conjugated rings increases, the 0-0 band shifts to higher wavelengths • The increase in wavelength is not as fast as the polyenes or linear aromatics because of the bond between the rings is wisted • The spectrum is featureless because thermally induced oscillation about the twist angle adds width to the vibronic bands • The molar absorptivity increases because the number of double bonds is increasing
Linear Polyphenyls
Alkyl Substituents
name 0-0 band (nm) max (M-1 cm-1)
benzene 264 220
toluene 269 290
ethylbenzene 269 260
propylbenzene 268 230
anthracene 377 9,800
9-methylanthracene 387 10,200
• Alkyl substituents shift the 0-0 band of the parent aromatic a few nanometers to the red, and increase the molar absorptivity a small amount • Multiple substituents will increase the shift by smaller increments • The vibronic pattern in the spectrum will change because of the new vibrations
Alkyl Substituents
Substituents with Lone-Pairs of Electrons
14.3 : 4
N
H
H
•When atoms with lone-pairs of electrons are attached to aromatic compounds they can effectively increase the size of the ring system • An increase in the size of the ring system shifts the parent spectrum to the red • Lone-pairs often break the symmetry of a molecule, converting a forbidden transition into a moderately allowed transition • When a transition is made more allowed, there is an increase in the molar absorptivity • When aromatic compounds with hydroxyl or amine substituents are dissolved in hydrogen bonding solvents, the absorption bands become broad and vibronic structure is decreased or lost
Halogen Substituents
name 0-0 band (nm) max (M-1 cm-1)
benzene 264 220
fluorobenzene 266 1,100
toluene 269 290
p-chlorotoluene 278 570
anthracene 377 9,800
9,10-dichloroanthracene 402 14,300
• Halogen substituents shift the 0-0 band to the red • The larger the halogen the larger the shift • Halogens can break symmetry to make a transition more allowed • Multiple substituents will increase the red shift
Halogen Substituents
Hydroxide and Amine Substituents
name 0-0 band (nm) max (M-1 cm-1)
benzene 264 220
hydroxybenzene (phenol) 278 2,400
aminobenzene (aniline) ~313 1,450
naphthalene 315 320
1-hydroxynaphthalene 325 5,400
2-hydroxynaphthalene 328 2,700
• Hydroxide and amine substituents shift the 0-0 band to the red • Both substituents create broad bands when the compound is dissolved in a hydrogen bonding solvent • Both substituents can break symmetry to make a transition more allowed • Molar absorptivities are in the range of 1,000 - 6,000
Hydroxyl and Amine Substituents
cyclohexane cyclohexane ethanol
ethanol methanol ethanol
“Ultraviolet absorption spectra for
1,2,4,5-tetrazine (a.) in the vapor phase,
(b.) in hexane solution, and (c.) in
aqueous solution.”
Effect of polar solvents on transitions
Polar solvents stabilize both non-bonding electrons in
The ground state and * elctrons in the excited state
Absorption
Incident
beam Transmitted
beam
Ground state
Excited state
Absorption
Absorption
along radiation
beam
Loss of energy as
radiation, heat, etc ... Sample
•Empirical relationship between transmitted
intensity and number of absorbers.
•A. Beer, 1852. See H. G. Pfeiffer and H. A.
Liebhafsky, J. Chem. Ed. 1951, 28, 123-
125,“The origins of Beer’s law.”
•The incident radiation is monochromatic.
•The absorbing units (atoms, molecules, ions)
act independently of one another.
•The absorption is limited to a volume of uniform
cross section.
Beer’s law (Beer-Lambert Law)
Beer’s law (The Beer-Lambert Law)
• Exponential attenuation with
– concentration
– sample thickness (optical path length)
• Assumes
– sample is non-turbid (Non-scattering)
• Scattering effects:
– create losses out of the side of the sample
» apparent absorption is greater than actual
absorption
– optical path-length is now no longer simply the
length of the cuvette
– lead to requirement for model of light
propagation (diffusion theory, etc)
0
Absorbing medium
b
P Po
dx
x
bc =
bc
bc =
Beer’s Law Example
Example
known
Example
Solvents for UV-Visible Regions
Analysis of Mixtures of Absorbing Substances
Absorption spectrum of a two-component mixture
Solution of Binary Mixture
Schematic representation of the absorption spectra
of solutions containing
(1) c1 moles per liter of substance 1
(2) c2 moles per liter of substance 2
(3) c1 moles per liter of substance 1 and
c2 moles per liter of substance 2
Solution of Binary Mixture
Wavelength 1
Am,l1 = a1,l1*b*c1 + a2,l1*b*c2
Am,l1 = (a1,l1*c1 + a2,l1*c2)*b
Wavelength 2
Am,l2 = a1,l2*b*c1 + a2,l2*b*c2
Am,l2 = (a1,l2*c1 + a2,l2*c2)*b
Solution of Binary Mixture
let A1 = Am,l1 A2 = Am,l2
D1 = a1,l1 D2 = a1,l2
1 = a2,l1 2 = a2,l2
then A1 = (D1*c1 + 1*c2)*b
A2 = (D2*c1 + 2*c2)*b
solve for c2
A2/b = (D2*c1 + 2*c2)
A2/b - D2*c1 = 2*c2
2*c2 = A2/b - D2*c1
c2 = (A2/(b* 2) - (D2*c1)/ 2
then
A1 = (D1*c1 + 1((A2/(b* 2)-(D2*c1)/ 2))*b
A1/b = (D1*c1 + 1((A2/(b* 2)(D2*c1)/ 2))
A1/b = (c1(D1 - D2*(1/ 2))+(1/ 2)*(A2/b))
A1/b - (A2/b)*(1/ 2) = c1(D1-D2*(1/ 2))
thus
(A1/b - (A2/b)*(1/ 2)) c1 = ------------------------------- (D1 - D2*(1/ 2))
and
C2 = (A2/(2*b) - (D2*c1)/ 2
Example
Example
Deviation from Beer’s Law
•It is a deviation from direct proportionality
between A and C.
•It is either +ve (upward curvature) or –Ve
(downwod curvature) deviation
•Sources are: Real, Instrumental , or
Chemical factors
Real Deviations
• high concentration - particles too
close
• Dependence of absorptivity on
refractive index of solution
Real Factors
•Derivation did not take into consideration the
changes in Refractive index of the solution
due to concentration changes.
Refractive index increase as
concentration increases
•Consequently the proportionality constant is
not but n / (n2+2)2 where n is the refractive
index of the medium
Instrumental Factors
1. Polychromatic Radiation
Assume a radiation consisting of two wavelengths:
l and l’ and Beer’s Law applies at each
•Alterations in power supply voltage, light source or
detector response. Others include:
At l the absorbance will be given as follows
Ac = b c when = ’
• Departure increases by increasing the difference
between l Values of the polychromatic radiation
due to the increase in the difference between
and ’
• The steeper the absorption (A Vs l) curve
the greater the error
Overlap of Sample with Optical Beam
A
BC
• As long as the optical beam is narrower than the sample cell width (cases A and B in the figure) measured absorption is constant. • When the optical beam is larger than the sample cell width (case C) light misses the sample and cannot be absorbed. • The equation describing transmission when light misses the cell is given below. In the example, the true absorption is 0.30, the incorrect absorption is 0.22
( )( )0
( ) ( )0 0
0.5 0.5 0.10.6
1.0 0.9 0.1
outsideinside
true errorinside outside
I IT T T
I I
•This situation can occur when trying to make absorption measurements in capillary liquid chromatography.
Monochromator Stray Light
All gratings and mirrors scatter a small fraction of the input light. The scattered radiation gets spread throughout the monochromator, with some reaching the exit slit. The scatter is composed of all frequencies entering the monochromator. The scatter is then transferred by lenses or mirrors though the sample and to the detector. The expression for transmission has to be modified to take this into account.
0 0
stray
stray
I IT
I I
l
l
Since the stray light occurs at all wavelengths it will be absorbed to a different extent than l. Usually a large fraction of stray light is not absorbed at all. This makes it difficult to measure absorbance values above ~2. High-performance spectrophotometers are ordinarily constructed with double monochromators to reduce the amount of stray light.
3. Stray Light The stray light striking detector is a potential source of error.
Apparent A is decreased as a result
•Thus a negative deviation is expected
•Deviations are expected near the limits of the instrument
Components
•Visible radiation is the most serious stray light problem
•For Uv-Vis spectrometers
Sources of Stray Light and Its Effect of on Absorbance
Source: M. R. Sharpe, Anal. Chem. 56, 339A-356A (1984).
Apparent deviation due to stray radiation
Sample Fluorescence
• When a sample fluoresces, some of the emitted photons reach the detector. This gives an abnormally high transmission.
fluorI IT
I
l
l
•The effect of fluorescence can be reduced by moving the sample away from the detector. •The effect of fluorescence can be eliminated by using two monochromators that scan synchronously.
source
wavelength
selector #1
sample cell
detector
Sample Scatter • Samples that scatter light lose radiation in addition to that absorbed. This causes an abnormally low transmission.
0
scatterI IT
I
l
l
• Since most scattered
light is in the forward direction (toward the detector), its effect can be reduced by moving the detector very close to the sample. • Scattered light is proportional to 1/l4. This makes it easy to identify.
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
200 250 300 350 400 450 500 550 600 650 700 750 800
wavelength (nm)
abso
rba
nce
Sample Cell Reflections
• For a quartz sample cell each air/quartz interface reflects ~3.5% of the light, while each quartz/water interface reflects ~0.2%. These reflective losses represent light that never reaches the detector. The measured transmission is lower than expected.
0
reflectI IT
I
l
l
• The effect of reflection can be reduced by using a reference cell that contains only solvent. The two cells must have identical optical properties (called matched cells) for the reflective losses to cancel. • This problem is sufficiently acute that expensive spectrophotometers have a mechanism for "flattening" and zeroing the baseline when running solvent versus solvent.
Source Width
_spectrum 35 _source 5
400 450 500 550 6000
0.5
1
wavelength (nm)
abso
rption
_spectrum 35 _source 12
400 450 500 550 6000
0.5
1
wavelength (nm)
abso
rption
_spectrum 35 _source 35
400 450 500 550 6000
0.5
1
wavelength (nm)
abso
rption
_spectrum 35 _source 70
400 450 500 550 6000
0.5
1
wavelength (nm)
abso
rption
• The spectral width of the source determines the shape of the measured absorption spectrum. The two curves are "convoluted" mathematically (source; original spectrum; measured spectrum)
Source Width (2)
0
0.2
0.4
0.6
0.8
1
1.2
0 0.00002 0.00004 0.00006 0.00008 0.0001
concentration (M)
abso
rptio
n
source << peak
source = peak
Atruep/ (p2 + s
2)1/2
Ameasured
Atrue
• Beer's Law only holds for monochromatic light. With non-monochromatic light, the slope of the calibration curve decreases. In addition the curve is non-linear for high absorption values.
Slit Width
• Spectral slit width : it is the spread of the image
along the frequency, wavenumber or wavelength scale
• It is proportional to the mechanical slit width
• If the absorption band is sharp or if the measurement
is made at the steep slope of the spectral band may
be different over the spectral band width and
deviation may be noticed
• Typical bandwidth of a spectrometer is of the order of 1 nm
Molecular absorption bands are broader than 1 nm thus the
Effect of spectral bandwidth is negligible when A is measured
at lmax
Effect of Bandwidth on Spectral Detail
• Absorbance increases (significantly)
as slit width decreases.
• However, decrease in slit width leads to a
(second-order) reduction in power of radiant
energy; so at very narrow slit widths, high
S/N can lead to loss of spectral detail.
Source: Skoog, Holler, and Nieman, Principles of Instrumental Analysis, 5th edition, Saunders College Publishing.
Deviation Due to Chemical Factors
•Sources are: Dissociation, association, complex formation,
polymerization or solvolysis.
•For example dissociation of benzoic acid and potassium
•dichromate
•Complexes of low stability can be studied by
spectroscopy.
•Consider the formation of a complex MnLp,
where M is a metal ion and L is a ligand:
nM + pL MnLp •The molar ratio of the two components of a complex
is important.
•In a quantitative determination, an excess of ligand
should be added to force the equilibrium toward
completion.
Determination of Stoichiometry
Mole-
Ratio
Method
Conc. Of one
component (M) is
kept constant and
that of the other is
varied to get a
series of [L]/[M]
ratios
Molar-ratio method, showing different curves. (a) Component L does not absorb at the wavelength of maximum absorption for the complex, for example, Fe(III)- Tiron.
(b) Component L absorbs slightly at the wavelength of maximum absorption for the complex, for example, Zn-Pan. (c) An excess of component L causes a decrease in absorbance of the complex, for example, Bi-xylenol orange .
Continuous Variation (Job’s( Method
• Conc. of both components is varied.
• But the total number of moles of both components is
kept constant
• This causes a change in mole ratios
mL of 2.5 m M P mL of 2.5 mM X X/P mol X/(mol X + mol P)
1 9 9/1 0.9
2 8 4/1 0.8
Job curves of thio-Michler's ketone comlexes of
mercury and of palladium
Studies of Chemical Equilibria
Can be determined spectrophotometrically
][A[HA]
If HA and A- are known. These can be determined
by converting both forms into any of them through
addition of acid or base
When [HA] = [A-]; pka = pH
The peak belongs to A- since intensity is reduced by decreasing pH
We may write the equation:
•This gives the segmoid curve shown above.
•This curve can be obtained without prior knowledge of Ka
or by measuring the absorbance of HA in various buffer
solutions
