ULTRAVIOLET-VISIBLE SPECTROSCOPY

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ULTRAVIOLET-VISIBLE SPECTROSCOPY

• Semester Dec Apr 2010

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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

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MOLECULAR SPECIES THAT ABSORB UV/VISIBLE
RADIATION
Inorganic species
Organic compounds
Charge transfer
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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

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NOTE Transition metals - groups IIIB through IB
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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.

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Molecular absorption
- In molecules, not only have electronic level
but also consist of vibrational and rotational
sub-levels. - This result in band spectra.
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Type of Transitions

• 3 types of electronic transitions
• s, ? and n electrons
• d and f electrons
• Charge transfer electrons

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What is s, ? and n electrons?
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Sigma (?)electron

• Electrons involved in single bonds such as those
  between carbon and hydrogen in alkanes.
• These bonds are called sigma (s) bonds.
• The amount of energy required to excite electrons
  in s 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.

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Pi (?) electron

• Electrons involved in double and triple bonds
  (unsaturated).
• These bonds involve a pi (?) bond.
• For example alkenes, alkynes, conjugated olefins
  and aromatic compounds.
• Electrons in ? bonds are excited relatively
  easily these compounds commonly absorb in the UV
  or visible region.

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• Examples of organic molecules containing ? bonds.

propyne
ethylbenzene
benzene
1,3-butadiene
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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.

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• Examples of organic molecules with non-bonding
  electrons.

Carbonyl compound If R H aldehyde If R
CnHn ketone
aminobenzene
2-bromopropene
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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).

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Electronic energy levels diagram
Unoccupied levels
Occupied levels
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? ? transitions

• Never observed in normal UV/Vis work.
• The absorption maxima are lt 150 nm.
• The energy required to induce a s s
  transition is too great (see the arrow in energy
  level diagram)
• This type of absorption corresponds to breaking
  of C-C, C-H, C-O, C-X, .bonds

s s
vacuum UV region
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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.
• e range 100 3000 Lcm-1mol-1

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Some examples of absorption due to n s
transitions
Compound ?max (nm) emax
H2O 167 1480
CH3OH 184 150
CH3Cl 173 200
CH3I 258 365
(CH3)2O 184 2520
CH3NH2 215 600
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n ? transitions

• Unsaturated compounds containing atoms with
  unshared electron pairs
• These result in some of the most intense
  absorption in 200 700 nm region.
• e range 10 100 Lcm-1mol-1

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? ? transitions

• Unsaturated compounds to provide the ? orbitals.
• These result in some of the most intense
  absorption in 200 700 nm region.
• e range 10oo 10,000 Lcm-1mol-1

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Some examples of absorption due to n ? and
? ? transitions
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CHROMOPHORE

• Unsaturated organic functional groups that absorb
  in the UV/VIS region are known as chromophores.

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AUXOCHROME

• Groups such as OH, -NH2 halogens that attached
  to the doubley bonded atoms cause the normal
  chromophoric absorption to occur at longer ? (red
  shift). These groups are called auxochrome.

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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.

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Other Factor that Influenced Absorption

• Factors that influenced the ?
• i) Solvent effects (shift to shorter ? blue
  shift)
• ii) Structural details of the molecules

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Important terminologies

• hypsochromic shift (blue shift)
• - Absorption maximum shifted to shorter ?
• bathochromic shift (red shift)
• - Absorption maximum shifted to longer ?

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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
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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.

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Absorption spectra of some transition-metal ions
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• 4f 5f electrons
• - Ions of lanthanide and actinide elements
• - Their spectra consists of narrow, well-defined
  characteristic absorption peaks

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Typical absorption spectra for lanthanide ions
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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 s 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

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INSTRUMENTATION
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Important components in a UV-Vis spectrophotometer
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Source lamp
Sample holder
Signal processor readout
? selector
Detector
UV region - Deuterium lamp H2 discharge tube
Phototube, PM tube, diode array
Quartz/fused silica
Prism/monochromator
Visible region - Tungsten lamp
Prism/monochromator
Glass/quartz
Phototube, PM tube, diode array
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UV-VIS INSTRUMENT

• Single beam
• Double beam

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Single beam instrument
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• Single beam instrument
• One radiation source
• Filter/monochromator (? selector)
• Cells
• Detector
• Readout device

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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.

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Double beam instrument
Double-beam instrument with beams separated in
space
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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 ?
• 3. Continuous recording of transmittance or
  absorbance spectra

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Quantitative Analysis

• The fundamental law on which absorption methods
  are based on Beers law (Beer-Lambert law).

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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.

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Quantitative Analysis

• Calibration curve method
• Standard addition method

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• 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

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Standard Calibration Curve
Absorbance
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 .

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How to produce standard calibration curve?
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• 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
  least-squares method.

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Finding the Least-Squares Line
Concentration xi Absorbance yi x2i y2i xiyi
5 0.201
10 0.420
15 0.654
20 0.862
25 1.084
? xi ? yi ? xi2 ? yi2 ? xiyi
N 5
N is the number of points used
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• 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)
• Syy ? (yi y)2 ? yi2 (? yi)2 (2)
• Sxy ? (xi x) (yi y)2 ? xiyi ? xi ? yi
  (3)

N
N
N
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• 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

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Concentration x y mx b
5
10
15
20
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• From the least-squares line equation, you can
  calculate the new y values by substituting the x
  value.
• Then plot the graph.

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• 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 instruments 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.

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• 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

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Standard Addition Plot
Absorbance
Concentration, ppm
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How to produce standard addition curve?

• Add same quantity of unknown sample to a series
  of flasks
• Add varying amounts of standard (made in solvent)
  to each flask, e.g. 0,5,10,15mL)
• Fill each flask to line, mix and measure

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Standard Addition Methods
Single-point standard addition method
Multiple additions method
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• Standard addition
• - if Beers law is obeyed,
• A ebVstdCstd ebVxCx
• Vt Vt
• kVstdCstd kVxCx

k is a constant equal to eb
Vt
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• Standard Addition
• - Plot a graph A vs Vstd
• A mVstd b
• where the slope m and intercept b are
• m kCstd b kVxCx

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• Cx can be obtained from the ratio of these two
  quantities m and b
• b kVxCx
• m kCstd
• Cx bCstd
• mVx

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• Standard Addition
• For single-point standard addition

Absorbance of diluted sample
A1 ebVxCx Vt
Eq. 1
Absorbance of diluted sample std
A2 ebVxCx Vt

ebVsCs Vt
Eq. 2
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• 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

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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.

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• Solution
• Cx (0.428) (0.03 mg PO43-/mL) (5.00mL)
• (0.538 0.428)(2.00mL)
• 0.292 mg PO43- / mL sample

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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 Absorbance
0.0 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.