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Presentation on theme: "ULTRAVIOLET-VISIBLE SPECTROSCOPY"— Presentation transcript:


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

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


36 Important components in a UV-Vis spectrophotometer
1 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.


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