Transcript UV-Visible Spectroscopy

Semester Dec – Apr 2010
In this lecture, you will learn:
 Molecular species that absorb UV/VIS radiation
 Absorption process in UV/VIS region in terms of its
electronic transitions
 Important terminologies in UV/VIS spectroscopy
Inorganic
species
Organic
compounds
MOLECULAR SPECIES THAT
ABSORB UV/VISIBLE RADIATION
Charge transfer
Definitions:
 Organic compound
 Chemical compound whose molecule contain carbon.
 E.g. C6H6, C3H4
 Inorganic species
 Chemical compound that does not contain carbon.
 E.g. transition metal, lanthanide and actinide elements
 Cr, Co, Ni, etc..
 Charge transfer
 A complex where one species is an electron donor and the other is an
electron acceptor.
 E.g. iron(III) thiocyanate complex
NOTE: Transition metals - groups IIIB through IB
UV-VIS ABSORPTION
 In UV/VIS spectroscopy, the transitions which result in
the absorption of EM radiation in this region are
transitions btw electronic energy levels.
Molecular absorption
- In molecules, not only have
electronic level but also
consist of vibrational and
rotational sub-levels.
- This result in band spectra.
Type of Transitions
 3 types of electronic transitions
 σ, п and n electrons
 d and f electrons
 Charge transfer electrons
What is σ,  and n electrons?
Sigma ()electron
 Electrons involved in single bonds such as those
between carbon and hydrogen in alkanes.
 These bonds are called sigma (σ) bonds.
 The amount of energy required to excite electrons in σ
bond is more than UV photons of wavelength. For this
reason, alkanes and other saturated compounds
(compounds with only single bonds) do not absorb UV
radiation and therefore frequently very useful as
transparent solvents for the study of other molecules.
For example, hexane, C6H14.
Pi () electron
 Electrons involved in double and triple bonds
(unsaturated).
 These bonds involve a pi (п) bond.
 For example: alkenes, alkynes, conjugated olefins and
aromatic compounds.
 Electrons in п bonds are excited relatively easily; these
compounds commonly absorb in the UV or visible
region.

Examples of organic molecules containing п bonds.
H
CH2CH3
H
H
ethylbenzene
H3C
C
C
C
C
C
C
H
C
C
H
propyne
H
H
C
benzene
H
H
C
H
H
C
H
C
H
1,3-butadiene
n electron
 Electrons that are not involved in bonding between
atoms are called n electrons.
 Organic compounds containing nitrogen, oxygen,
sulfur or halogens frequently contain electrons that are
nonbonding.
 Compounds that contain n electrons absorb UV/VIS
radiation.

Examples of organic molecules with non-bonding electrons.
NH2
C
H3C
O
R
aminobenzene
Carbonyl compound
If R = H
aldehyde
If R = CnHn
ketone
H
C
Br
C
H
2-bromopropene
Absorption by Organic Compounds
 UV/Vis absorption by organic compounds requires
that the energy absorbed corresponds to a jump from
occupied orbital unoccupied orbital.
 Generally, the most probable transition is from the
highest occupied molecular orbital (HOMO) to the
lowest unoccupied molecular orbital (LUMO).
Electronic energy levels diagram
Unoccupied levels
Occupied levels

* transitions
 Never observed in normal UV/Vis work.
 The absorption maxima are < 150 nm.
 The energy required to induce a σ
σ* transition is
too great (see the arrow in energy level diagram)
 This type of absorption corresponds to breaking of CC, C-H, C-O, C-X, ….bonds
σ
σ*
vacuum UV region
n
* transitions
 Saturated compounds must contain atoms with
unshared electron pairs.
 Compounds containing O, S, N and halogens can
absorb via this type of transition.
 Absorptions are typically in the 150 -250 nm region
and are not very intense.
 ε range: 100 – 3000 Lcm-1mol-1
Some examples of absorption due to n
Compound
σ* transitions
λmax (nm)
εmax
H2O
167
1480
CH3OH
184
150
CH3Cl
173
200
CH3I
258
365
(CH3)2O
184
2520
CH3NH2
215
600
n
* transitions
 Unsaturated compounds containing atoms with
unshared electron pairs
 These result in some of the most intense absorption in
200 – 700 nm region.
 ε range: 10 – 100 Lcm-1mol-1

* transitions
 Unsaturated compounds to provide the  orbitals.
 These result in some of the most intense absorption in
200 – 700 nm region.
 ε range: 10oo – 10,000 Lcm-1mol-1
Some examples of absorption due to n
* and 
* transitions
CHROMOPHORE
 Unsaturated organic functional groups that absorb in
the UV/VIS region are known as chromophores.
AUXOCHROME
 Groups such as –OH, -NH2 & halogens that attached to
the doubley bonded atoms cause the normal
chromophoric absorption to occur at longer λ (red shift).
These groups are called auxochrome.
Effect of Multichromophores on Absorption
 More chromophores in the same molecule cause
bathochromic effect ( shift to longer ) and
hyperchromic effect(increase in intensity)
 In the conjugated chromophores * electrons are
delocalized over larger number of atoms causing a
decrease in the energy of  * transitions and an
increase in  due to an increase in probability for
transition.
Other Factor that Influenced Absorption
 Factors that influenced the λ:
i) Solvent effects (shift to shorter λ: blue shift)
ii) Structural details of the molecules
Important terminologies

hypsochromic shift (blue shift)
- Absorption maximum shifted to shorter λ

bathochromic shift (red shift)
- Absorption maximum shifted to longer λ
Terminology for Absorption Shifts
Nature of Shift
Descriptive Term
To Longer Wavelength
Bathochromic
To Shorter Wavelength
Hypsochromic
To Greater Absorbance
Hyperchromic
To Lower Absorbance
Hypochromic
Absorption by Inorganic Species
Involving d and f electrons absorption

3d & 4d electrons
- 1st and 2nd transition metal series
e.g. Cr, Co, Ni & Cu
- Absorb broad bands of VIS radiation
- Absorption involved transitions btw filled and unfilled
d-orbitals with energies that depend on the ligands, such
as Cl-, H2O, NH3 or CN- which are bonded to the metal
ions.
Absorption spectra of some transition-metal ions

4f & 5f electrons
- Ions of lanthanide and actinide elements
- Their spectra consists of narrow, well-defined
characteristic absorption peaks
Typical absorption spectra for lanthanide ions
Charge Transfer Absorption
 Absorption involved transfer of electron from the
donor to an orbital that is largely associated with the
acceptor.
 an electron occupying in a σ or  orbital (electron
donor) in the ligand is transferred to an unfilled
orbital of the metal (electron acceptor) and vice-versa.
 e.g. red colour of the iron(III) thiocyanate complex
INSTRUMENTATION
Important components in a UV-Vis
spectrophotometer
1
Source
lamp
2
Sample
holder
3
4
λ selector
Detector
UV region:
- Deuterium lamp;
H2 discharge tube
Quartz/fused
silica
Prism/monochromator
Glass/quartz
Prism/monochromator
Phototube,
PM tube, diode
array
Visible region:
- Tungsten lamp
Phototube,
PM tube, diode
array
5
Signal
processor
& readout
UV-VIS INSTRUMENT
 Single beam
 Double beam
Single beam instrument
 Single beam instrument
 One radiation source
 Filter/monochromator (λ selector)
 Cells
 Detector
 Readout device
Single beam instrument

Disadvantages:


Two separate readings has to be made on the
light. This results in some error because the
fluctuations in the intensity of the light do occur
in the line voltage, the power source and in the
light bulb btw measurements.
Changing of wavelength is accompanied by a
change in light intensity. Thus spectral scanning
is not possible.
Double beam instrument
Double-beam instrument with beams separated in space
Double beam instrument

Advantages:
1. Compensate for all but most short-term fluctuations in the
radiant output of the source as well as for drift in the
transducer and amplifier
2. Compensate for wide variations in source intensity with λ
3. Continuous recording of transmittance or absorbance
spectra
Quantitative Analysis
 The fundamental law on which absorption methods
are based on Beer’s law (Beer-Lambert law).
Measuring absorbance
 You must always attempt to work at the wavelength of
maximum absorbance (max)
 This is the point of maximum response, so better
sensitivity and lower detection limits.
 You will also have reduced error in your measurement.
Quantitative Analysis


Calibration curve method
Standard addition method
 Calibration curve method
 A general method for determining the concentration
of a substance in an unknown sample by comparing
the unknown to a set of std sample of known
concentration
Absorbance
Standard Calibration Curve
Concentration, ppm
 How to measure the concentration of unknown?
 Practically, you have measure the absorbance of your unknown. Once you
know the absorbance value, you can just read the corresponding
concentration from the graph .
A
How to produce standard
calibration curve?
 Prepare a series of
standard solution with
known concentration.
 Measure the absorbance
of the standard
solutions.
 Plot the graph A vs
concentration of std.
 Find the ‘best’ straight
line by using leastsquares method.
Finding the Least-Squares Line
Concentration
xi
Absorbance
yi
5
0.201
10
0.420
15
0.654
20
0.862
25
1.084
 xi
 yi
x2i
 xi2
N=5
N – is the number of points used
y2i
 yi2
xiyi
 xiyi
 The calculation of the slope and intercept is simplified by
defining three quantities Sxx, Syy and Sxy.
 Sxx = 
(xi – x)2 =  xi2– ( xi)2
……(1)
N
 Syy =  (yi – y)2 =  yi2– ( yi)2
……(2)
N
 Sxy =  (xi – x) (yi – y)2 =  xiyi –  xi  yi …(3)
N
 The slope of the line, m:
m = Sxy
Sxx
 The intercept, b:
b = y - mx
 Thus, the equation for the least-squares line is
y = mx + b
Concentration
x
y = mx + b
5
10
15
20
25
 From the least-squares line equation, you can calculate the
new y values by substituting the x value.
 Then plot the graph.
 Most linear regression implementations have an option to
“force the line through the origin,” which means forcing the
intercept of the line through the point (0,0). This might
seem reasonable, since a sample with no detectable
concentration should produce no response in a detector,
but must be used with care.
 HOWEVER, forcing the plot through (0,0) is not always
recommended, since most curves are run well above the
instrumental limit of detection (LOD). Randomly, adding a
point (0,0) can skew the curve because the instrument’s
response near the LOD is not predictable and is rarely
linear. As illustrated next page, forcing a curve through the
origin can, under some circumstances, bias results.
 Standard addition method
 used to overcome matrix effect
 involves adding one or more increments of a standard
solution to sample aliquots of the same size.
 each solution is diluted to a fixed volume before
measuring its absorbance
Standard Addition Plot
Absorbance
Concentration, ppm
How to produce standard addition curve?
1. Add same quantity of unknown sample to a series of flasks
2. Add varying amounts of standard (made in solvent) to each flask, e.g.
0,5,10,15mL)
3. Fill each flask to line, mix and measure
Standard Addition Methods
Single-point standard
addition method
Multiple additions
method
Standard addition
- if Beer’s law is obeyed,

A = εbVstdCstd
Vt
= kVstdCstd
+
+
εbVxCx
Vt
kVxCx
k is a constant equal to
εb
Vt

Standard Addition
- Plot a graph: A vs Vstd
A = mVstd + b
where the slope m and intercept b are:
m = kCstd
;
b = kVxCx

Cx can be obtained from the ratio of these two
quantities: m and b
b
m
= kVxCx
kCstd
Cx
= bCstd
mVx
Standard Addition
 For single-point standard addition
Absorbance of
diluted sample
Absorbance of
diluted sample
+ std
A1 = εbVxCx
Vt
A2 = εbVxCx +
Vt
Eq. 1
εbVsCs
Vt
Eq. 2
 Standard Addition
 For single-point standard addition
Dividing the 2nd equation by the first & then rearrange it
will give:
Cx =
A1 Cs Vs
(A2 – A1 ) Vx
Example Standard Addition (single point addition)
Example 14-2 (page 376)
A 2.00-mL urine specimen was treated with reagent to
generate a color with phosphate, following which the
sample was diluted to 100 mL. To a second 2.00mL
sample was added exactly 5.00mL of a phosphate
solution containing 0.03 mg phosphate /mL, which was
treated in the same way as the original sample. The
absorbance of the first solution was 0.428, while the
second one was 0.538. Calculate the concentration of
phosphate in milligrams per millimeter of the
specimen.
Solution:
Cx = (0.428) (0.03 mg PO43-/mL) (5.00mL)
(0.538 – 0.428)(2.00mL)
= 0.292 mg PO43- / mL sample
Exercise
The concentration of an unknown chromium solution was determined by
pipetting 10.0mL of the unknown into each of five 50.0 mL volumetric
flasks. Various volumes of a standard containing 12.2 ppm chromium were
added to the flasks and then the solutions were diluted to the mark.
Standard, mL
0.0
Absorbance
0.201
10.0
0.292
20.0
0.378
30.0
0.467
40.0
0.554
Determine the concentration of chromium (in ppm) in the unknown.