Presentation on theme: "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."— Presentation transcript:
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
MOLECULAR SPECIES THAT ABSORB UV/VISIBLE RADIATION Organic compounds Inorganic species Charge transfer
Definitions: Organic compound Chemical compound whose molecule contain carbon. E.g. C 6 H 6, C 3 H 4 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.
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. aminobenzene Carbonyl compound If R = H aldehyde If R = C n H n ketone 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 Occupied levels Unoccupied 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 C- C, C-H, C-O, C-X, ….bonds vacuum UV region σ σ*
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 -1 mol -1 n * transitions
Some examples of absorption due to n σ* transitions Compound λ max (nm)ε max H2OH2O CH 3 OH CH 3 Cl CH 3 I (CH 3 ) 2 O CH 3 NH
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 -1 mol -1 n * 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 -1 mol -1 * transitions
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.
Factors that influenced the λ: i) Solvent effects (shift to shorter λ: blue shift) ii) Structural details of the molecules Other Factor that Influenced Absorption
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 ShiftDescriptive Term To Longer WavelengthBathochromic To Shorter WavelengthHypsochromic To Greater AbsorbanceHyperchromic To Lower AbsorbanceHypochromic
Absorption by Inorganic Species Involving d and f electrons absorption 3d & 4d electrons - 1 st and 2 nd 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 -, H 2 O, NH 3 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
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. Single beam instrument
Double beam instrument Double-beam instrument with beams separated in space
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 Double beam instrument
Quantitative Analysis The fundamental law on which absorption methods are based on Beers 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
Standard Calibration Curve 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. Concentration, ppm Absorbance
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. A
Finding the Least-Squares Line Concentration x i Absorbance y i x2ix2iy2iy2ixiyixiyi x i y i x i 2 y i 2 x i y i N – is the number of points used N = 5
The calculation of the slope and intercept is simplified by defining three quantities S xx, S yy and S xy. S xx = (x i – x) 2 = x i 2 – ( x i ) 2 ……(1) S yy = (y i – y) 2 = y i 2 – ( y i ) 2 ……(2) S xy = (x i – x) (y i – y) 2 = x i y i – x i y i …(3) N N N
The slope of the line, m: m = S xy S xx The intercept, b: b = y - mx Thus, the equation for the least-squares line is y = mx + b
Concentration x y = mx + b 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 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.
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 Concentration, ppm Absorbance
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 Beers law is obeyed, A = εbV std C std + εbV x C x V t V t = kV std C std + kV x C x k is a constant equal to εb V t
Standard Addition - Plot a graph: A vs V std A = m V std + b where the slope m and intercept b are: m = kC std ;b = kV x C x
C x can be obtained from the ratio of these two quantities: m and b b = kV x C x m kC std C x = bC std mV x
Standard Addition For single-point standard addition A 1 = εbV x C x V t A 2 = εbV x C x + V t εbV s C s V t Eq. 1 Eq. 2 Absorbance of diluted sample Absorbance of diluted sample + std
Standard Addition For single-point standard addition Dividing the 2 nd equation by the first & then rearrange it will give: C x = A 1 C s V s (A 2 – A 1 ) V x
Example Standard Addition (single point addition) 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. Example 14-2 (page 376)
Solution: C x = (0.428) (0.03 mg PO 4 3- /mL) (5.00mL) (0.538 – 0.428)(2.00mL) = mg PO 4 3- / mL sample
Standard, mLAbsorbance 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. Determine the concentration of chromium (in ppm) in the unknown. Exercise