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.

Semester Dec – Apr 2010

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

MOLECULAR SPECIES THAT

ABSORB UV/VISIBLE RADIATION

Organic compounds

Inorganic species

Charge transfer

Definitions:Organic compound

Chemical 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 Transitions

3 types of electronic transitionsσ, п and n electronsd and f electronsCharge transfer electrons

What is σ, What is σ, and n and n electrons?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 () 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.

Examples of organic molecules containing п bonds.

CH2CH3

C

CC

C

CC

H

H

H

H

H

H

C C HH3C

C C

C CH H

H

H

H

H

ethylbenzene benzen

e

propyne

1,3-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.

NH2

C

R

O C C

H

H

H3C

Br

aminobenzene

Carbonyl compound

If R = H aldehyde

If R = CnHn 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 150 -250 nm region and are not very intense.

ε range: 100 – 3000 Lcm-1mol-1

n * transitions

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 184 2520

CH3NH2 215 600

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

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-1mol-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.

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.

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 Shift Descriptive Term

To Longer Wavelength Bathochromic

To Shorter Wavelength Hypsochromic

To Greater Absorbance Hyperchromic

To Lower Absorbance Hypochromic

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.

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

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

Important components in a UV-Vis spectrophotometer

Source lamp

Sample holder λ selector Detecto

r

Signal processor & readout

1 23 4

5

UV region:

- Deuterium lamp;

H2 discharge tube

Visible region:

- Tungsten lamp

Quartz/fused silica

Glass/quartz

Prism/monochromator

Prism/monochromator

Phototube,

PM tube, diode array

Phototube,

PM tube, diode array

UV-VIS INSTRUMENT

Single beamDouble beam

Single beam instrument

One radiation sourceFilter/monochromator (λ 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 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 Beer’s 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

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

Concentrationxi

Absorbanceyi

x2i y2i xiyi

5 0.201

10 0.420

15 0.654

20 0.862

25 1.084

xi yi xi2 yi

2 xiyi

N – is the number of points used

N = 5

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

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

Concentrationx

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.

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.

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, 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

Single-point standard addition method

Multiple additions method

Standard addition

- if Beer’s law is obeyed,

A = εbVstdCstd + εbVxCx

Vt Vt

= kVstdCstd + kVxCx

k is a constant equal to εb

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 addition

A1 = εbVxCx Vt A2 = εbVxCx

+ Vt

εbVsCs Vt

Eq. 1

Eq. 2

Absorbance of diluted sample

Absorbance 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

Standard, mL Absorbance

0.0 0.201

10.0 0.292

20.0 0.378

30.0 0.467

40.0 0.554

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