Presentation on theme: "ULTRAVIOLET-VISIBLE SPECTROSCOPY"— Presentation transcript:
1 ULTRAVIOLET-VISIBLE SPECTROSCOPY Semester Dec – Apr 2010ULTRAVIOLET-VISIBLE SPECTROSCOPY
2 In this lecture, you will learn: Molecular species that absorb UV/VIS radiationAbsorption process in UV/VIS region in terms of its electronic transitionsImportant terminologies in UV/VIS spectroscopy
3 MOLECULAR SPECIES THAT ABSORB UV/VISIBLE RADIATION Inorganic speciesOrganic compoundsMOLECULAR SPECIES THAT ABSORB UV/VISIBLE RADIATIONCharge transfer
4 Definitions: Organic compound Inorganic species Charge transfer Chemical compound whose molecule contain carbon.E.g. C6H6, C3H4Inorganic speciesChemical compound that does not contain carbon.E.g. transition metal, lanthanide and actinide elementsCr, Co, Ni, etc..Charge transferA 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 ABSORPTIONIn 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 electronsd and f electronsCharge transfer electrons
10 Sigma ()electronElectrons 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 () electronElectrons 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. propyneethylbenzenebenzene1,3-butadiene
13 n electronElectrons 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 compoundIf R = H aldehydeIf R = CnHn ketoneaminobenzene2-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 levelsOccupied 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, ….bondsσ σ*vacuum UV region
18 n * transitionsSaturated 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 Compoundλmax (nm)εmaxH2O1671480CH3OH184150CH3Cl173200CH3I258365(CH3)2O2520CH3NH2215600
20 n * transitionsUnsaturated compounds containing atoms with unshared electron pairsThese 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 CHROMOPHOREUnsaturated organic functional groups that absorb in the UV/VIS region are known as chromophores.
25 AUXOCHROMEGroups 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 λbathochromic shift (red shift)- Absorption maximum shifted to longer λ
29 Terminology for Absorption Shifts Nature of ShiftDescriptive TermTo Longer WavelengthBathochromicTo Shorter WavelengthHypsochromicTo Greater AbsorbanceHyperchromicTo Lower AbsorbanceHypochromic
30 Absorption by Inorganic Species Involving d and f electrons absorption3d & 4d electrons- 1st and 2nd transition metal seriese.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
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
39 Single beam instrument One radiation sourceFilter/monochromator (λ selector)CellsDetectorReadout 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 amplifier2. Compensate for wide variations in source intensity with λ3. 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 absorbanceYou 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.
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 AbsorbanceConcentration, ppmHow 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 ConcentrationxiAbsorbanceyix2iy2ixiyi50.201100.420150.654200.862251.084 xi yi xi2 yi2 xiyiN = 5N – 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)NNN
52 Thus, the equation for the least-squares line is y = mx + b The slope of the line, m:m = SxySxxThe intercept, b:b = y - mxThus, the equation for the least-squares line isy = mx + b
53 Concentrationxy = mx + b510152025From 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 effectinvolves 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 AbsorbanceConcentration, ppm
58 How to produce standard addition curve? Add same quantity of unknown sample to a series of flasksAdd 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 methodMultiple additions method
60 - if Beer’s law is obeyed, A = εbVstdCstd + εbVxCx Vt Vt Standard addition- if Beer’s law is obeyed,A = εbVstdCstd εbVxCxVt Vt= kVstdCstd kVxCxk is a constant equal to εbVt
61 Standard Addition- Plot a graph: A vs VstdA = mVstd + bwhere 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 = kVxCxm kCstdCx = bCstdmVx
63 A1 = εbVxCx Vt A2 = εbVxCx + Vt εbVsCs Vt Standard Addition For single-point standard additionAbsorbance of diluted sampleA1 = εbVxCxVtEq. 1Absorbance of diluted sample + stdA2 = εbVxCx +VtεbVsCsVtEq. 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.
67 ExerciseThe 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, mLAbsorbance0.00.20110.00.29220.00.37830.00.46740.00.554Determine the concentration of chromium (in ppm) in the unknown.
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