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

Organic compounds

Inorganic species

Charge transfer

Definitions:Organic compoundChemical compound whose molecule contain carbon.E.g. C6H6, C3H4

Inorganic 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

NOTE: Transition metals - groups IIIB through IB

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.

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

Type of Transitions3 types of electronic transitions, and n electronsd and f electronsCharge transfer electrons

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

Examples of organic molecules containing bonds.ethylbenzenebenzenepropyne1,3-butadiene

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.

Examples of organic molecules with non-bonding electrons.aminobenzeneCarbonyl compoundIf R = H aldehydeIf R = CnHn ketone

2-bromopropene

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 diagramOccupied levelsUnoccupied levels

Never observed in normal UV/Vis work.The absorption maxima are < 150 nm.The energy required to induce a * 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 *

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. range: 100 3000 Lcm-1mol-1

Some examples of absorption due to n * transitions

Compoundmax (nm)maxH2O1671480CH3OH184150CH3Cl173200CH3I258365(CH3)2O1842520CH3NH2215600

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

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

Some examples of absorption due to n * and * transitions

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

AUXOCHROMEGroups 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.

Effect of Multichromophores on AbsorptionMore 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 :i) Solvent effects (shift to shorter : blue shift)ii) Structural details of the molecules

Other Factor that Influenced Absorption

Important terminologieshypsochromic shift (blue shift)- Absorption maximum shifted to shorter

bathochromic shift (red shift)- Absorption maximum shifted to longer

Terminology for Absorption Shifts

Nature of ShiftDescriptive TermTo Longer WavelengthBathochromicTo Shorter WavelengthHypsochromicTo Greater AbsorbanceHyperchromicTo Lower AbsorbanceHypochromic

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.

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

Absorption involved transfer of electron from the donor to an orbital that is largely associated with the acceptor.an electron occupying in a 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

Important components in a UV-Vis spectrophotometerSource lampSample holder selectorDetectorSignal processor & readout12345UV region: - Deuterium lamp; H2 discharge tube

Visible region: - Tungsten lampQuartz/fused silicaGlass/quartzPrism/monochromatorPrism/monochromatorPhototube,PM tube, diode arrayPhototube,PM tube, diode array

UV-VIS INSTRUMENTSingle beamDouble beam

Single beam instrument

One radiation sourceFilter/monochromator ( selector)CellsDetectorReadout device

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 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 amplifier2.Compensate for wide variations in source intensity with 3.Continuous recording of transmittance or absorbance spectra

Double beam instrument

Quantitative AnalysisThe fundamental law on which absorption methods are based on Beers law (Beer-Lambert law).

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.

Quantitative AnalysisCalibration curve methodStandard 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 CurveHow 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

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 LineN is the number of points usedN = 5

ConcentrationxiAbsorbanceyix2iy2ixiyi50.201100.420150.654200.862251.084 xi yi xi2 yi2 xiyi

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

The slope of the line, m:m = Sxy Sxx

The intercept, b:

b = y - mx

Thus, the equation for the least-squares line isy = mx + b

From the least-squares line equation, you can calculate the new y values by substituting the x value.Then plot the graph.

Concentrationxy = mx + b510152025

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 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

Concentration, ppmAbsorbance

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

Single-point standard addition methodMultiple additions method

Standard addition- if Beers law is obeyed,

A = bVstdCstd + bVxCx Vt Vt

= kVstdCstd + kVxCxk is a constant equal to b Vt

Standard Addition - Plot a graph: A vs Vstd

A = mVstd + b

where the slope m and intercept b are:

m = kCstd;b = kVxCx

Cx can be obtained from the ratio of these two quantities: m and b

b = kVxCx m kCstd

Cx = bCstd mVx

Standard Addition

For single-point standard additionA1 = bVxCx Vt A2 = bVxCx + Vt bVsCs Vt Eq. 1Eq. 2Absorbance of diluted sampleAbsorbance of diluted sample + std

Standard AdditionFor single-point standard addition

Dividing the 2nd equation by the first & then rearrange it will give:

Cx = A1 Cs Vs (A2 A1 ) Vx

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

Example 14-2 (page 376)

Solution:

Cx = (0.428) (0.03 mg PO43-/mL) (5.00mL) (0.538 0.428)(2.00mL)

= 0.292 mg PO43- / mL sample

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

Standard, mLAbsorbance0.00.20110.00.29220.00.37830.00.46740.00.554

***********