# Spectrophotometry Spectroscopy is the study of

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Spectrophotometry Spectroscopy is the study of
interaction of spectrum of light with a substance to be analysed, for its identification (i.e qualitative analysis) as well as determination of its amount (i.e quantitative analysis).

Light It is one of the different forms of radiant energy, such as, sunlight, radio waves and X-rays . Thy are called electromagnetic radiations (EMR) due presence of both electric and magnetic components

Dual nature of light Light exhibits wave property during its propagation and energy particle during its interaction with matter. The double nature of light (waves and particles) is known as dualism. 1. Wave property EMR display the property of continuous waves and can be described by the characteristics of wave motion. Such wave motion in conveniently classified according to the wavelength.

Wavelength () It is the linear distance measured along the line of propagation between crest of one wave to the next. Unit of Length Micron () = 1 x10-6m = 1 x 10-4cm = 1x 10-3 mm. Millimicron (m) = or nanometer (nm) = 1 x 10-9m = 1 x 10-7 cm = 1 x 10-6mm. Angstrom (Ao) = 1 x 10-10m = 1 x 10-8cm = 1 x 10-7 mm.

Frequency [] Relation between  &   = C /  Wave number ( `)
is the number of waves/ second or number of cycles occurring/ second (CPS) or Hertz (Hz) or waves vibration/ second. Relation between  &  C =    = C /  C= 3.0 x 1010 cm/s where c is the velocity of light cm/s. Wave number ( `) is the number of waves/ cm `= 1/  Its unit is cm-1

2-Particle property (light as energy)
Light consist of energy packets, known as photons. The energy (E) of photons is proportional to the frequency i.e related to c and . It can be expressed by max plank relation: E = h  where h = max plank constant = 3.63 x erg., sec.) ( = C / ) i.e E   or E    1/  Therefore energy of a beam of EMR increases as wave length decreases.

The shorter the wave length, the greater the energy of the photons and the more powerful the radiation. U.V range (200nm-400nm) which contain shorter , carrying more energy photon than a beam of visible range . Visible range (400nm-800nm) having high energy more than I.R range (>800nm).

Example: The λ of the sodium D line is 589nm . What are the frequency( ) and wave umber ( `) Solution:  = C/λ C=3.0 x 108 m/s λnm----λm (nm x 10-9 )  = 3x108m/s / 589nm x 10-9 = 5.09x1014 s-1 `= 1 / λ (`= cm-1 ) λ nm----- λ cm ( nm x 10-7 cm ) `= 1 / 589 x10-7 = 1.7 x 104 cm-1

Second Lectures

For analytical purposes we use the region of
I.R, visible and U.V radiations . -UV radiation region is classified into : far UV from (10nm-200nm) and near UV from ( 200nm-380nm- -Visible radiation region ( nm) consist of colored radiations, which are, violet, indigo, blue, green, yellow, orange and red . -IR radiation region ( 0.75um-1000um ) is classified into : near IR , mid IR and far IR .

Interaction of a substance with EMR
When a molecule interact with radiant energy The molecule is said to be excited , because the outer valence electrons undergo transition from original energy level ground state (E g) to an excited state (Es). Excited state Es Ground state E g The transition energy is given by the following equation : E = Es E g = h 

When a molecule in the ground state absorbs EMR, 3 energy state transition will take place.
These types of transition are : 1) Electronic ) Vibrational ) Rotational . 1-When the molecule absorb Visible and U.V region . Raising electrons to a higher energy level (Electronic transition energy ) 2-When the molecule absorb I.R region. Raising the Vibration of molecule (Vibrational transition energy) 3-When the molecule absorb F.I.R and Microwave regions. Increasing rotation of the molecule (Rotational transition energy)

The total energy E total = E electronic + E vibrational + E rotational
The relative value of E electronic : E vibrational: E rotational is : : : 10 (1) Photons of Far IR and Microwave spectrum (E↓) cause only rotation of the molecules. (2) Photons of IR spectrum (E >E far IR and microwave ) cause vibration and rotation of the molecule (3) Photons of UV-Visible spectrum (E↑>EIR >E far IR) cause electronic transition which is accompanied by vibrational and rotational transion .

The relation between structure and energy absorbed
The energy absorbed by a molecule in the ground state is dependent on the nature of the bonds within a molecule -The outermost electrons in organic molecules may be: strong (σ) bonds, weaker () bonds or nonbonding (n) 1-Sigma () electrons: they are bonding electrons posses the lowest energy level ( the most stable single bonds). 2-Pi () electrons :the bonding electrons constituting the pi bonds (the weaker double or treble bonds) 3-Non-bonding (n) : electrons: don’t participate in bonding, they usually occupy the highest level of ground state (hetero atoms like N,O,S )

In the excited state ; - electrons occupy an anti bonding energy level denoted as * and the transition is termed -* transition. --electrons occupy the anti bonding * level, and the transition is termed - * . -while n electrons occupy either * or *,and the transition is termed n- * or n- *

absorb at <180nm in the far UV.
-Compounds containing only -electrons ( →σ* transition) are the saturated hydrocarbons which absorb at <180nm in the far UV. -They are transparent in the near UV ( nm) making them ideal solvents for other compounds to be studied in this region. -n-electrons absorption in saturated compounds (n→σ* transition ) containing hetero atoms or halogens. -The majority of these compounds show no bsorption in the near UV,eg. Methanol at 177nm, tri ethylamine at 199nm or chloroform at 173nm. -Alcohols and ethers absorb at wavelength shorter than 185nm and so they are useful as common solvents at > 200nm.

-- * transition in unsaturated compounds containing double or triple bonds are easily exited than in single covalent bonds. -Also , the energy required for - * in conjugated compounds is smaller than in non-conjugated ones. -So conjugated compounds absorb at longer wave length than un-conjugated one . -n→ * transition occur in compounds containing non-bonding electrons adjacent to unsaturated centers (double or triple bonds)

Suggest which of these compounds absorb in UV-Visible spectrum ?
Compound: Type of transition: Answer 1) CH3-CH3 : 2) CH2=CH2 : 3) CH3-O-H : 4) CH3-C=O : | CH3

Lecture III 1) Absorption Spectrum . 2) Chromophores and Auxochromes .
3) Bathochromic shift and Hypsochromic shift .

1) Absorption spectrum according to the electronic transition that occur in each organic molecule, absorption spectrum is obtained by plotting Absorbance (A) as a function of wavelength (). It has characteristic shape with the  of maximum absorbance (max). It is characteristic for each molecule according to its structure and the type of transitional energy Therefore it is used for identification of a chemical substance (qualitative analysis). Also max is used for quantitative measurement, in order to increase sensitivity and to minimize error of the analytical method.

2) Chromophores and Auxochromes
Chromophres : Are unsaturated groups responsible for - * and n→ * electronic transitions. e.g. C=C , C=O , N=N and N=O ( 200nm-800nm) 3)Bathochromic & Hypsochromic shift -Bathochromic shift (or red hift) It is the shift of max to a longer wavelength due to substitution with certain functional groups (e.g. –OH and –NH2), when two or more chromophores are present in conjugation, change in pH and effect of the medium (solvent). - Hypsochromic shift (or blue shift) It is the shift of max to a shorter wavelength due to removal of conjugation by changing pH or polarity of the solvent.

- Hypochromic effect -Auxochromes
- Hyperchromic effect an increase in the intensity of absorption usually due to introduction of an auxochrome - Hypochromic effect It involves a decrease in the intensity of absorption -Auxochromes Are saturated groups posses unshared electrons, and does not absorb in near UV or visible radiations e.g. OH,NH2. But when attached to chomophoric molecule, increase both its wave length and intensity of absorption maximum . Because auxochrome inters into resonance interaction with the chromophore , thus increase the extent of conjugation, shift the absorption maximum to longer wave length

Example 1): effect of conjugation on absorption spectrum
Changes in Absorption spectrum Example 1): effect of conjugation on absorption spectrum Increase in conjugation, increas absorbance of light to higher , bathochromic shift with hyperchromic effect.

Lecture IV Effect of pH on absorption spectrum.
2)Polychromatic and Monochromatic light. 3)Theory of light absorption

Example 2) Effect of pH The spectra of compounds containing acidic (phenolic-OH) or basic (-NH2) groups are dependent on the pH of the medium. The U.V spectrum of phenol in acid medium, benzenoid form while in alkaline medium is the phenate anion ,quinonoid form The free pair of of electrons of O2 increasing the elocalization of the -electrons, leading to the formation of conjugated system. So , electrons become more energetic and need less energy to be excited, therefore absorb longer  bathochromic shift ; red shift) with hyperchromic effect Phenol

Its spectrum exhibits bathochromic shift and hyperchromic effect in alkaline medium due to its conversion to the quinonoid species Aniline While in acid medium (anilinium ) lost the free pair electrons of N decrease the conjugation . Its spectrum in acid medium exhibit hypsochromic shift and hypochromic effect due to its conversion to the benzenoid species.

Polychromatic light Monochromatic light
A beam of light containing several wavelengths , e.g. white light Monochromatic light A beam of light containing radiation of only one discrete wavelength

Therefore, under experimental conditions.
Theory of light absorption : When a monochromatic light having intensity (Io) is allowed to pass through absorbing medium ; Some is absorbed (I a), reflected (I r), transmitted (It), refracted (If) and scattered (Is). Is = zero for clear solution, while If and Ir may be canceled by means of control cuvette containing the solvent in which the substance to be anaylsed is dissolved Therefore, under experimental conditions. Io = Ia It or Ia = Io - It

Lecture V 1-Beer-Lambert Law 2-Absorptivity , Molar absorptivity and A1%.1cm

% Transmittance %T = 100 x T Transmittance T = It / Io
The diagram below shows a beam of monochromatic radiation of radiant power Io, directed at a sample solution. Absorption takes place and the beam of radiation leaving the sample has radiant power I. % Transmittance %T = 100 x T Transmittance T = It / Io Absorbance A = log10 I0 / It A = log10 1 / T A = log10 100 / %T A = 2 - log10 %T

The relationship between absorbance and transmittance is illustrated in the following diagram:
So, if all the light passes through a solution without any absorption, then absorbance is zero, and percent transmittance is 100%. If all the light is absorbed, then percent transmittance is zero, and absorption is infinite. Path length / cm 0.2 0.4 0.6 0.8 1.0 %T 100 50 25 12.5 6.25 3.125 Absorbance 0.3 0.9 1.2 1.5

log Io/It  b (thickness )
lamberts’ law: When a monochromatic light enter absorbing medium, its intensity is decreased exponentially with the increase of thickness of the absorbing medium (i.e solution) (b) at constant concentration (C) log Io/It  b (thickness ) log Io / It = K b C b It C b Log Io/It A

Beers’ law log Io/It  C (concentration) log Io / It = K C
When a monochromatic light enter absorbing medium, its intensity is decreased exponentially with the increase of concentration of the absorbing medium (C) at constant pathlength (b) log Io/It  C (concentration) log Io / It = K C b C Log Io/It b C It A

a: is a constant, known as absorptivity
Beer-lambert’s law: Absorbance (A ) = log Io/It log Io/It  b  C log Io/It  b C log Io/It = a b C A = a b C a: is a constant, known as absorptivity which is the absorbance, when thickness of solution is unity (1cm) and concentration is unity (1gm/L)

 = a x mol. wt a =  / mol.wt Molar absorptivity or epsilon ()
If the unit of concentration is 1M i.e. conc.= mol/L (a) is known as molar absorptivity or epsilon () As, a=A/b c =A/1cm . gm/L = L.gm-1.cm-1 molar absorptivity ,  =A/b c= A/1cm.mol/L Its unit is L mol-1 cm-1 (mol=gm/ mol. weight) i.e. Its unit = L.MWt.cm-1gm-1.  = a x mol. wt a =  / mol.wt

 = A (1% - 1cm) x mol. Wt / 10 A (1% - 1cm):
If unit of concentration is 1% ( i.e. 1gm/100 ml ) A (1% - 1cm) = A/ 1cm.gm% = 100. cm-1.gm-1. a= L.gm-1. cm-1 A (1% - 1cm) = a x 10 a= A (1% - 1cm) / 10 (a =  / mol. Wt = A (1% - 1cm) / 10 )  = A (1% - 1cm) x mol. Wt / 10 A (1% - 1cm) =  x 10 / mol. Wt

ε = A 1% 1cm x M.wt / 10 = 1030 x 147 / 10 = 15141 litre.cm-1.mol-1
Both  and A 1%, 1cm are characteristic for each substance at the same max , pH and type of solvent and are used for quantitative purpose Problem 1: A sample solution give absorbance equals 0.6 , A1% 1cm = 1030 and its molecular weight = 147. Calculate the sample concentration in Mol/L Solution 1: ε = A 1% 1cm x M.wt / 10 = x 147 / 10 = litre.cm-1.mol-1 A =ε b c = x 1 x C =15141 x C C = 0.5 / = x 10-5 mol / L

Lecture VI 1-Isosbestic point 2-Colorimetry .

Isosbestic point -At different pH, the spectrum will be shifted to different max but all spectra intersect at certain  which is known as isosbestic point - At isosbestic point, the same absorbance is given for the same concentration at different pH,i.e. absorbance is not pH dependent but concentration dependent -Thus solution ; its max affected by pH , must be buffered at specific pH or measurements are carried out at the isosbestic point.

Colorimetry When white light passes through a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. e.g orthophenanthrolene which reacts with ferrous (Fe2+) in buffered medium (acidic pH) to produce intense red colour. Complementary colors are diametrically opposite each other. Thus, absorption of nm light renders a substance yellow, and absorption of nm light makes it red.

wavelength region, nm color complementary color Violet Yellow-green Blue Yellow Blue-green Orange Green-blue Red Green Purple Colored substances appear colored because they selectively absorbed some of wavelengths of visible light and transmitted other wavelengths or colors (apparent color), Red substances absorb the blue- green wavelengths from the visible region, so the transmitted light appears red Blue substances absorb the yellow wavelengths, so the transmitted light appears blue.

Requirements for substances to be measured colorimetricaly:
Substance must be coloured e.g CuSO4, organic dyes,…. 2-If the substance to be analysed is colourless, it must react with certain reagent (known as chromogen) to produce equivalent coloured product. 3-If there is no suitable chromogen, the substance must be converted to a certain derivative which has a suitable chromogen.

Quiz I)Discuss shortly -Interaction of a substance with EMR
-Factors affecting absorption spectrum II) Solve the following problem 1-A 5.00x10-4M sample solution is measured in a cell with 1 cm bath length ; its absorbance at 592nm equals a-What is the molar absorptivity at 582nm. If a solution of unknown concentration of the same sample has an absorbance at the same wave length. b-What is its concentration

III) Complete the answer in Exercise 1
2-Calculate the wave legnth in um , and in Angestron 3-Calculate the frequancy and energy of this wave length III) Complete the answer in Exercise 1

2-If the substance to be analysed is colourless, it must react with certain reagent (known as chromogen) to produce equivalent coloured product. Orthophenanthrolene reacts with ferrous (Fe2+)in buffered medium (acidic pH) to produce intense red color. 2+ 3

3-If the sample is colorless and there is no suitable chromogen, the substance must be converted to a certain derivative which can be react with suitable reagent producing color . -Esters are first converted to hydroxamic acid derivative through the reaction with hydroxylamine. Hydroxamic acid derivative gives purple color on addition of ferric (Fe3+) due to the formation of iron chelate . O O R –C– O Et + H2N – OH R –C– NH – OH + Et OH Hydroxamic acid derivative + Fe3+ then measuring the absorbance at 520nm.

Requirements for ideal chromogen
Chromogen is a compound containing chromophoric group Requirements for ideal chromogen 1-Should be colorless or easily separated from the colored product 2-It Should be selective. 3-Its reaction to produce colored product, should be of known mechanism and proceed stoichiometrically. 4-The full development of color must be rapid. 5-It must produce only one color of specified max.

Requirements for coloured product
1-Should be of intense color, to increase the sensitivity 2-Should be unaffected by pH or the pH must be specified and maintained by suitable buffer or the measurement is carried out at  of isosbestic 3-Should be stable with time 4.The reaction of its formation, must be rapid and quantitative. 5-The colored product, should obey Beer-lambert’s law, i.e on plotting A versus C at fixed b, we obtain straight line passing through the origin.

Instrumentation The instrument used, usually consists of 5 basic components 1-Radiant energy source. 2-Dispersing system (or monochromator). 3-Sample compartment (cuvette). 4-Detector 5-Recorder (meter).

1. Radiant energy source (source of light)
In visible range In U.V range Tungsten lamp Deuterium lamp (D2) (or hydrogen lamp.) 2. Dispersing system (monochromator) It convert polychromatic light to monochromatic light (definite range of .) a) Filters b) Prisms C) Grating act by selective absorption of unwanted  and transmit the complementary color, which is needed to be absorbed by the sample to be analysed. Filters may be : gelatin, liquid and tinted glass Disadvantages:Filter transmit a wide band of nm. which is not exactly monochromatic and used only in visible range .

The dispersion power of the prism is ∝ 1/ 
b) Prisms Act by refraction of light In visible range In U.V range glass prism quartz or fused silica prism. The dispersion power of the prism is ∝ 1/  . As they are transparent to UV light Cannot be used for UV as it absorb all UV light The dispersion power of the prism is ∝ 1/  i.e. angle of refraction ↑ as  ↓

The grating disperses the light beam into almost single .
C) Grating Grating consists of a large number of parallel lines ruled very close to each other on a highly polished surface e.g aluminum or aluminized glass (600 line/mm). Each ruled groove functions as a scattering center for light rays falling on its edge. The grating disperses the light beam into almost single .

It works in both UV and visible spectral region by dispersion.
The scattering power of a grating is ↑ as the number of groves ↑ . The resolution power of grating is > prism > filter. d) Associated optics are used to control light intensity Collimating lenses slit of variable width mirrors and diafragms helps to narrow band width alignment of the beam. act as condensers They should be suitable for the spectral range, i.e glass for visible range and quartz or fused silica for U.V range.

3.Sample compartment (Cuvette)
Transparent surface Opaque surface glass for visible range It is made of → quartz or fused silica for U.V range. 4. Detector a) photocell (Photovoltaic cell) e.g Barrier layer cell b) Phototube (photomultiplier or photoemissive tube)

a) photocell (Photovoltaic cell) e.g Barrier layer cell
Light falling on cell Transparent metal layer of Ago (Collecting electrode) Photosenitive semiconductor of selenium Metal base Plate of iron - +

b) Phototube (photomultiplier or photoemissive tube)

5. Recorder (meter) Electric signal produced in detector
is fed to a sensitive galvanometer, its scale is graduated in absorbance or/and transmittance units. Commercial instruments 1. Filter photo-electric colorimeter

2. Compensating two-photocell colorimeter
In this type, fluctuation in intensity of EMR source is automatically cancelled.

3. Prism spectrophotometer