1. ULTRAVIOLET-VISIBLE SPECTROSCOPY
Semester Dec – Apr 2010
ULTRAVIOLET-VISIBLE SPECTROSCOPY

2. 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

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

4. Definitions: Organic compound Inorganic species Charge transfer
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

5. NOTE: Transition metals - groups IIIB through IB

6. 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.

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

8. Type of Transitions 3 types of electronic transitions
σ, п and n electrons
d and f electrons
Charge transfer electrons

9. What is σ,  and n electrons?

10. 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.

11. 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.

12. Examples of organic molecules containing п bonds.
propyne
ethylbenzene
benzene
1,3-butadiene

13. 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.

14. Examples of organic molecules with non-bonding electrons.
Carbonyl compound
If R = H aldehyde
If R = CnHn ketone
aminobenzene
2-bromopropene

15. 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).

16. Electronic energy levels diagram
Unoccupied levels
Occupied levels

17.  * transitions σ σ* 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 C-C, C-H, C-O, C-X, ….bonds
σ σ*
vacuum UV region

18. 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 nm region and are not very intense.
ε range: 100 – 3000 Lcm-1mol-1

19. Some examples of absorption due to n σ* transitions
Compound
λmax (nm)
εmax
H2O
167
1480
CH3OH
184
150
CH3Cl
173
200
CH3I
258
365
(CH3)2O
2520
CH3NH2
215
600

20. 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

21.  * 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

22. Some examples of absorption due to n * and  * transitions

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

25. 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.

26. 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.

27. 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

28. Important terminologies
hypsochromic shift (blue shift)
- Absorption maximum shifted to shorter λ
bathochromic shift (red shift)
- Absorption maximum shifted to longer λ

29. 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

30. 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.

31. Absorption spectra of some transition-metal ions

32. 4f & 5f electrons
- Ions of lanthanide and actinide elements
- Their spectra consists of narrow, well-defined characteristic absorption peaks

33. Typical absorption spectra for lanthanide ions

34. 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

35. INSTRUMENTATION

36. Important components in a UV-Vis spectrophotometer1 2 5 3 4
Source lamp
Sample holder
Signal processor & readout
λ selector
Detector
UV region:
- Deuterium lamp;
H2 discharge tube
Quartz/fused silica
Phototube,
PM tube, diode array
Prism/monochromator
Visible region:
- Tungsten lamp
Glass/quartz
Prism/monochromator
Phototube,
PM tube, diode array

37. UV-VIS INSTRUMENT
Single beam
Double beam

38. Single beam instrument

39. Single beam instrument
One radiation source
Filter/monochromator (λ selector)
Cells
Detector
Readout device

40. 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.

41. Double beam instrument
Double-beam instrument with beams separated in space

42. 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

43. Quantitative Analysis
The fundamental law on which absorption methods are based on Beer’s law (Beer-Lambert law).

44. 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.

46. Quantitative Analysis
Calibration curve method
Standard addition method

47. 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

48. 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 .

49. 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 least-squares method.

50. 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

51. Syy =  (yi – y)2 =  yi2– ( yi)2 ……(2)
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

52. Thus, the equation for the least-squares line is y = mx + b
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

53. 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.

54. 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.

56. 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

57. Standard Addition Plot
Absorbance
Concentration, ppm

58. 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

59. Standard Addition Methods
Single-point standard addition method
Multiple additions method

60. - if Beer’s law is obeyed, A = εbVstdCstd + εbVxCx Vt Vt
Standard addition
- if Beer’s law is obeyed,
A = εbVstdCstd εbVxCx
Vt Vt
= kVstdCstd kVxCx
k is a constant equal to εb Vt

61. Standard Addition
- Plot a graph: A vs Vstd
A = mVstd + b
where the slope m and intercept b are:
m = kCstd ; b = kVxCx

62. Cx can be obtained from the ratio of these two quantities: m and b
b = kVxCx
m kCstd
Cx = bCstd
mVx

63. A1 = εbVxCx Vt A2 = εbVxCx + Vt εbVsCs Vt Standard Addition
For single-point standard addition
Absorbance of diluted sample
A1 = εbVxCx Vt
Eq. 1
Absorbance of diluted sample + std
A2 = εbVxCx + Vt
εbVsCs Vt
Eq. 2

64. 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

65. 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 Calculate the concentration of phosphate in milligrams per millimeter of the specimen.

66. Solution: Cx = (0. 428) (0. 03 mg PO43-/mL) (5. 00mL) (0. 538 – 0
Solution: Cx = (0.428) (0.03 mg PO43-/mL) (5.00mL) (0.538 – 0.428)(2.00mL) = mg PO43- / mL sample

67. 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.