UNIT III

PHOTOCHEMISTRY AND SPECTROSCOPY

PART A

1. Define Photochemistry.Photochemistry is the study of chemical reactions that are caused by absorption of light radiations.

2. What are the laws of Photochemistry? 1. The Grotthus-Draper law. 2. The Stark- Einstein law of photochemical equivalence. 3. Lamberts law. 4. Beer-Lamberts law.

3. State Grotthus – Draper law.This law states that “when light falls on any substance, only the fraction of incident light

which is absorbed by the substance can bring about a chemical change”. It is also called the principle of photochemical activation.

4. State Stark – Einstein law.This law states that “In the primary photochemical process (first step) each reacting

molecule is activated by one quantum of effective light”. This law is also called the principle of quantum activation.

5. Define Beer - Lamberts law.This law states that “When a beam of monochromatic radiation is passed through a

homogenous absorbing solution, the decrease in the intensity of the radiation dI with the thickness of the absorbing solution dx is directly proportional to the intensity of the incident radiation I as well as the concentration of the solution C.

6.Define Quantum yield or Quantum Efficiency. Quantum yield (Ф) is defined as “the number of molecules of the substance undergoing

photochemical reaction per quantum of radiation absorbed‟‟. The quantum yield for photochemical reaction is defined as

Ф = No. of molecules reacting in a given time

No. of quanta (photons) of light absorbed in the same time

7.What are primary and secondary photochemical reactions? Primary process – the reacting molecules undergo activation by absorption of light.

A + h ν A*Secondary process – the activated molecules undergo a photochemical change.

A* B

8. Define Photo-Sensitization.In some photochemical reactions, the reactant molecules do not absorb radiation and no

chemical reaction occurs. Certain reactions are made sensitive by the presence of small quantity of foreign substance which can absorb light and stimulate the reaction without taking part in it.The substance which absorbs light and induces a photochemical reaction without undergoing chemical reaction is known as photosensitiser and the phenomenon is known as Photo-sensitization.Examples:-Atomic Photosensitisers – Mercury, cadmium, ZincMolecular Photosensitisers – Benzophenone, Sulphur dioxide, Uranyl sulphate.

9. What is quenching?During Photosensitization the foreign substance (sensitizer), absorbs light and gets

excited. When the excited foreign substance collides with another substance it gets converted in to some other product due to the transfer of its energy to the colliding substance. This process is known as quenching.

10. What is fluorescence?Certain substances (atoms or molecules) when exposed to radiation of short

wavelength (high frequency), emit light of different frequencies compared to those of incident radiations. This process is called fluorescence and stops as soon as the radiation is cut off. Actually decay period is very short i.e. 10-9 to 10-4 sec. The substance which exhibits fluorescence is called fluorescent substance.Examples: - Fluorite (naturally occurring CaF2), Petroleum, Organic dyes like Eosin, fluorescein, Chlorophyll, Quinine sulphate solution, Vapours of Sodium, Iodine, Mercury.

11. What is Phosphorescence?Many substance (or molecules) when exposed to radiations of short wavelength (high

frequency) continue to emit light for sometime (10-4 to few seconds) even after incident light is cut off. This phenomenon is called phosphorescence and is chiefly caused by Ultraviolet or Visible light. The substance which exhibits phosphorescence is called phosphorescent substance. Examples: - Zinc Sulphide, Alkaline earth sulphides (CaS, BaS, SrS).

12. Define internal conversion and intersystem crossing.Internal conversion: - It is a type of transition which involves the return of the activated

molecule from higher excited state to the lower excited state.S3 S2 T3 T2S2 S1 T2 T1

The energy of the activated molecule is dissipated in the form of heat through molecular collisions. This process occurs in less than about 10-11 seconds. This is also known as non-radiative transitions

Intersystem crossing: - The process in which the energy of the activated molecule is lost through transition between states of different spin multiplicity.

S2 T2S1 T1

Even though these transitions are forbidden they occur at relatively slow rates.

13.What is chemiluminescence? The phenomenon of emission of visible light as a result of chemical reaction at a temperature at which a black body normally does not emit light is known as chemiluminescence

SPECTROSCOPYPART A

1.What are the differences between atomic spectra and molecular spectra?

Atomic Spectra Molecular SpectraInteraction of electromagnetic radiation Interaction of electromagnetic radiation withwith atoms. molecules.Line spectrum is obtained Complicated spectrum is obtainedIt is due to electronic transition in an It is due to vibrational, rotational andelement. electronic transition in a molecule.

2. What is Finger Print region? Mention its uses.The vibration spectral region at 1400 – 700 cm -1 gives very rich and intense absorption

bands which is unique for each molecule. This region is termed as Finger Print Region.Uses: It can be used to detect the presence of functional group and also to identify and characterize the molecule just as a finger print can be used to identify a person.

3. What are the differences between Chromophore and Auxochrome?S.No Chromophore Auxochrome1. This group is responsible for the It does not impart colour, but when

colour of the compound. conjugated with a chromophore colouris produced.

2. It does not form salt. It forms salt.3. It contains at least one multiple It contains lone pair of electrons.

bond.eg: -NO2, -NO , -N=N- eg: -OH , -NH2 , -NR2

4. Mention three applications of UV-Visible Spectroscopy.i. Quantitative analysis: By using Beer-Lamberts law, the concentration of unknown solution can be determined.ii. Qualitative analysis: It is used for identification of aromatic compounds and conjugated olefins. iii. Detection of impurities: It is used for detecting impurities in organic compound

5. Name the transition responsible for molecular spectra.The molecular spectrum arises due to the following three

transitions. Rotational transitions,Vibrational transitions, Electronic transitions.

6. Define the term – Bathochromic shift, Hypsochromic shiftBathochromic shift: shifting of absorption band towards a longer wavelength. Hypsochromic shift: shifting of absorption band towards a shorter wavelength.

7.Sate two applications of IR Spectroscopy.1.To identify the compounds: The IR spectrum of unknown compound with that of a known compounds and from the resemblance of the two spectra , the nature of the compound can be established.2.Detection of impurities: The presence of impurities can be easily detected. They absorb strongly in that part of the spectrum whereas the main component does not absorb.

PHOTOCHEMISTRY AND SPECTROSCOPY PART - B

1.Illustrate the Stark-Einstein Law of photochemical equivalance.2.Compare fluorescence with Phosphorescence.3.Explain the principle and instrumentation of IR spectroscopy.4.Write down the various Electronic Transitions.5. Explain Chemiluminescence and Photosensitisation with suitable examples.6.Derive the expression for Beer- Lambert’s law and mention its limitations.7.How quantum efficiency is determined experimentally? Explain.8.Explain the photo physical processes with the help of Jablonski diagram..9.What are the causes for high and low quantum yields? Define the same.10. Explain the principle and instrumentation of UV – Visible spectroscopy.

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PART – B

1. Illustrate the Stark –Einstein law of Photochemical Equivalence.This law is also called as principle of quantum activation.This law states that “In the primary photochemical process each reacting molecule is

activated by one quantum of effective light”. This law implies a 1:1 relationship between the number of reacting molecules and the number of quanta of light absorbed and is applicable only for the primary process.

Primary step: A+ hν A*

Secondary Step: A* B

Overall reaction: A + hν B

The molecule A absorbs a photon of light and gets activated (primary step).The activated molecule (A*) then decomposes to yield B (secondary step).

Suppose ν is the frequency of radiation absorbed then the corresponding quantum of energy absorbed per molecule will be hν.The Energy absorbed per mole of the reacting substance is called one Einstein is given by

`E = NAhν =NAhc/λ

NA = Avagadro number = 6.023X 1023 /moleh = Planks constant = 6.625X10-34 J secC = velocity of light = 3.08X 108 m/secE = 6.023x1023 X 6.626x10-34 X 3x108 = 0.1196 J/ mol

λ λ (m)

also we know 1 calorie = 4.184 J

E = 0.1196 J/mol = 2.859x10-2 cal/mol

4.184 J/cal x λ λ

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2. Derive the expression for Beer – Lamberts law. State its disadvantages.Beer-Lambert‟s law states that when a beam of monochromatic radiation is passed

through a solution of an absorbing substance, the rate of decrease of intensity of radiation (dI) with thickness of the absorbing solution (dx) is proportional to the intensity of incident radiation, I, as well as concentration of the solution, C.It is mathematically represented as - dI/dx = kIC, where k = molar absorption co–efficient. Upon integration with in the limits, we get

I x

dI/I = - kCdx

I0 0

ln I/I0 = - kCx or 2.303log I/I0 = -

kCx log I0/I = k /2.303 * Cx

A = Cx (Beer- Lamberts law)

Where, ε = k/2.303 = molar absorptivity co–efficient. Hence, the absorbance is directly proportional to molar concentration, C, and thickness (or) path length, x.Application of Beer – Lambert LawDetermination of unknown concentration:Let the absorbance and the concentration of standard solution be As and Cs,

respectively, As = εCsx …… (1),and the absorbance and the concentration of unknown solution be Au and Cu,

respectively. Au = εCux ……. (2)Dividing equation (1) by equation (2), we get

As = Au

Cs Cu

Cu = Au * Cs

As

Thus the concentration of unknown solution is calculated.Limitations:

Applicable only for monochromatic light and dilute solutions. Deviation occurs if the solution undergoes association or dissociation. Valid only at constant temperature. It is not applied to suspensions. Deviations may occur if the solution contains impurities.

3. Outline the experimental determination of quantum yield. There are two types of determination to find out the quantum yields of photochemical reactions.

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Determination of number of moles of the light absorbing substance that react in a given time.

Determination of number of photons of light of required wavelength absorbed by the same substance in the same time.

Determination of number of moles reacting :The number of molecules reacted in a given time can be determined by the usual analytical techniques used in chemical kinetics.

The rate of the reaction is measured by pipetting small quantities of sample from the reaction mixture from time to time and the concentration of reactants are continuously measured by the usual volumetric methods (or) change in some physical property such as refractive index / absorption / optical rotation. From the data the amount and number of molecules reacted can be calculated.Experimental determination of amount of photons absorbed:

A photochemical reaction takes place by the absorption of photons of light by the reacting molecules. In order to study the rate of reaction it is essential to determine the intensity of light absorbed by the reacting molecule.

An experimental arrangement for the study of rate of a photochemical reaction is shown in the figure.

Light radiation from a suitable source (tungsten filament or mercury vapour lamp) is rendered parallel by a lens. The parallel beam is then passed through a monochromator, which yields a beam of the desired wavelength only. The monochromatic light then enters the reaction cell containing the reacting mixture (made of quartz) which is immersed in a thermostat. Finally, the light transmitted from the cell is made to fall on a detector which measures the intensity of the transmitted radiation.The intensity of radiation is generally measured with the help of actinometer. It contains 0.05 M oxalic acid and 0.01 M uranyl sulphate in water.On exposure to radiation the following reaction takes place.

UO22+ + hν (UO2

2+)*

UO22+ + COOH CO2 + CO + H2O + UO2

2+

COOH

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The standard solution of uranyl oxalate is exposed to light radiation of definite frequency for a certain period of time. The extent of reaction can be determined by titrating the remaining oxalic acid with standard KMnO4 solution. The amount of oxalic acid consumed in the photochemical reaction is a measure of intensity of radiation (i.e. number of photons absorbed).

3. Explain the mechanism of photochemical decomposition of HI. Comment on the reason for low quantum yield.

The photochemical decomposition of HI is found to take place in the wavelength range of 2070Å - 2820Å.The following mechanism has been suggested for the photochemical decomposition of HI.One molecule of HI absorbs one quantum of radiation (primary process)

HI + hν k1 H + I Rate = k1Iabs

H + HI k2 H2 + I Rate = k2 [H] [HI]I + I k3 I2 Rate = k3 [I]2

The overall rate of decomposition of HI is given by adding the rate equation (i) & (ii)

-d[HI] = k1Iabs + k2 [H] [HI] (1)

dt

Applying steady state approximation to [H] ,we have

d [H] = k1Iabs - k2 [H] [HI] =0

dt

k1Iabs = k2 [H] [HI] (2)

Substituting equation (2) in equation (1) , we get

-d[HI] = k1Iabs + k1Iabs = 2 k1Iabs

dt

The quantum yield of the reaction is given by

Ф = Rate of disappearance of HI = -d [HI]/ dt = 2 k1Iabs = 2.0Rate of absorption of light k1Iabs k1Iabs

Actually the quantum yield of the reaction falls below 2.0 after sometime. This is due to the fact that sufficient I2 is produced in step (iii), the following reaction, regenerating, HI is also set up. H + I2 HI + I

4. Explain the mechanism of photo physical process with the help of Jablonski diagram.Spin multiplicity:-Most molecules posses an even number of paired electrons in ground state. The spin multiplicity of the state is given as 2S + 1, where S is the total spins of the electrons When the spins are paired

Then S = s1+ s2 = +1/2 + (-1/2) = 0

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Hence 2S + 1 = 2x0 + 1 =1, the spin multiplicity is 1.

The molecule is in singlet ground state.

Spin Orientation due to absorption of light

On absorption of a photon of energy hν, one of the paired electrons goes to a higher energy level (i.e. excited state). The spin orientation of the two electrons may be either parallel or antiparallel. If the spins are parallel, thenS = s1+ s2 = +1/2 + (+1/2) = 1Hence 2S + 1= 2x1 +1 =3, the spin multiplicity is 3. The molecule is in triplet excited state.If the spins are antiparallel, thenS = s1+ s2 = +1/2 + (-1/2) = 0Hence 2S+ 1= 2x 0+ 1 = 1, the spin multiplicity is 1.The molecule is in singlet excited state. Depending upon the energy hυ of a photon, electron can jump to any higher electronic state and hence we get a series ofSinglet excited states S1, S2, S3, Triplet excited state T1, T2, T3…S1, S2, S3… etc is called first singlet excited state, second singlet excited state, and third singlet excited state respectively.T1, T2, T3… etc is called first triplet excited state, second triplet excited state, and third triplet excited state respectively.According to Quantum mechanics the singlet excited state has higher energy than correspondingtriplet excited state. Thus E (S1) > E (T1); E (S2) > E (T2) ; E(S3) > E(T3)Consequences of light absorption: When light photon (hν) is absorbed by a molecule, the electron of the absorbing molecule may jump from S0 (Singlet ground state) to singlet excited state S1 or S2or S3.Depending upon the energy of the light photon absorbed, for each singlet excited state (S1, S2, S3 etc) there is a corresponding triplet state respectively. (T1, T2, T3etc).The molecule is said to be activated, whether it is in singlet excited state or triplet excited state.M0 + hν M*M0 = Molecule in the ground state M*=Molecule in the activated state.The activated molecule returns to the ground state by dissipating its energy (hν) through any of the following type of process.1.Non-radiative transitions:-Such a transition is from the higher excited states (S1, S3 or T2, T3) to the first excited state (S1 or T1).Since such a transition does not involve the emission of radiation, it is referred to as radiation less or non radiative transitions.

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2. Internal Conversion: - These transitions involve the return of the activated molecule from the higher excited state to the first excited states. The energy of the activated molecule is dissipated in the form of heat through molecular collisions.This process occurs in less than about 10-11s

S3 S1 T3 T1

S2 S1 T2 T1Inter system crossing: - (ISC)The process in which the energy of the activated molecule is lost through transition between state of different spin multiplicity. Even though these transitions are forbidden, they occur at relatively slow ratesS2 T2S1 T1Radiative transition:-These transitions involve the return of the activated molecule from the singlet excited state S1 and the triplet excited state T1 to the singlet ground state S0. So, such a transition is accompanied by emission of radiation.i) When a molecule in the S1 state returns to the ground state S0, emission of radiation occurs in about 10-8 s and this process is known as fluorescence. ii) When a molecule in the T1 state returns to the ground state S0, emission of radiation occurs at a rather slow rate. This process is called phosphorescence. Since the T1 to S1, transition is spectroscopically forbidden the life time of phosphorescence is much longer (about 10-3 and higher).

5. Explain the mechanism of energy transfer in photochemical reactions. Photosensitization and Quenching:- In some photochemical reactions, the reactant molecules do not absorb radiation and no chemical reaction occurs. Certain reactions are made sensitive by the presence of small quantities of foreign substance which can absorb light and stimulate the reaction without taking part in it. The substance which absorbs light and induces a photochemical reaction without undergoing chemical reaction is known as photosensitiser and the phenomenon is known as Photo-sensitization.

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Examples:-Atomic Photosensitisers – Mercury, cadmium, ZincMolecular Photosensitisers – Benzophenone, Sulphur dioxide, Uranyl sulphate.During Photosensitization the foreign substance (sensitizer), absorbs light and gets excited. When the excited foreign substance collides with another substance it gets converted in to some other product due to the transfer of its energy to the colliding substance. This process is known as quenching.Mechanism of photosensitization and Quenching:-

Consider a general donor – acceptor system in which only the donor D i.e. the sensitizer, absorbs the incident photon and the triplet state of the donor is higher than the triplet state of the acceptor A .i.e. the reactant. Absorption of the photon produces the singlet excited state of the donor,1D which via intersystem crossing (ISC) gives the triplet state of the donor 3D. This triplet excited state then collides with the acceptor producing the triplet excited state of the acceptor 3A and the ground state of the donor. If 3A gives the desired product, the mechanism is called photosensitization.However if the desired products result from the excited state of the donor (3D), then A is called the quencher and the process is known as quenching.The reactions for photosensitization and quenching may be represented as below,

D + hν 1D1D 3D

3D + A D + 3A3A products (photosensitization)3D products (quenching)

The triplet excited state of the sensitizer or donor 1D must be higher in energy level than the triplet excited state of the reactant or acceptor (3A) so that the energy available is enough to raise the reactant molecule to its triplet state.

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Examples of photosensitized reactions:-Dissociation of H2 molecule:

Irradiation of a mixture of hydrogen gas and mercury vapour with light of wavelength 253.7 nm brings about dissociation of hydrogen molecule in to hydrogen atom.Hg + hν Hg*Hg* + hν H2* + Hg

H2* 2HHere mercury acts as photosensitiser.Photosynthesis in plants:-

The most outstanding example for photosensitization is photosynthesis of carbohydrates in plants from H2O and CO2in which the green colouring matter of plants acts as photosensitiser. CO2 and H2O do not absorb any radiation in the visible region, but chlorophyll absorbs the radiation. It absorbs visible light in the wavelength range of 400-700 nm. The energy of light absorbed by chlorophyll is transferred to CO2 and H2O which then react to form (CH2O) n (cellulose /sugar).

6. Explain the mechanism of chemiluminescence with examples.Chemiluminescence is defined as the production of light by a chemical reaction and is thus

the reverse of photochemical reaction. In order for a reaction to be chemiluminescent it must furnish sufficient energy to raise at least one of the reactants or intermediates to an electronically excited state so that as it returns to the ground state, it emits energy in the form of radiation.Since the emission occurs at ordinary temperature, the emitted radiation is also known as “cold light”.Examples of chemiluminescence reactions:1. The glow of phosphorus and its oxide, in which the oxide in its electronic excited state emits radiation. 2. Oxidation of 5-mainophthalic hydrazines or cyclic hydrazines by H2O2 emits green light. 3.‟Cold light‟ emission by glow worms. This is an example of bioluminescence due to aerial oxidation of luciferon (a protein) in the presence of enzyme (luciferase).

.

PART B

1. What is principle of UV Spectroscopy? Explain its components and working. Principle:

UV – Visible spectra arises from the transition of valence electrons from the ground state to excited state by absorbing light from UV (100 – 400 nm) or visible region (400 – 750nm). It is a type of absorption spectra.Components:Radiation source: Hydrogen or Deuterium lamps are radiation source for UV region and tungsten filament lamp for visible region.Monochromators: The monochromators is used to select a particular wavelength.Cells: The cells are made up of quartz glass are used to hold the sample which should fulfill the following conditions.

They must be uniform in construction. The material should be inert to solvents.They must be transparent to UV and Visible light.Detectors: The converts the radiation in to current which is directly proportional to the concentration of the solutionEg.. They are Photo multiplier tube or Photocells.Recorder: The recorder records the signal from the detector and shows a display.Working:The Radiation from the source is allowed to pass through the monochromator. It allow particular wavelength of light to pass through the exit slit. The beam of radiation coming out of the slit is split into two parallel beams.

Block diagram

One half of the beam passes through the sample cell and another half passes through reference cell (only solvent). The detectors compare the intensities of the transmitted two beams of light. If the compound absorb light at a particular wavelength, the intensity of the transmitted beam through sample (I) will be less than that of reference beam (Io). The recorder produces the graph,

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which is a plot of wavelength of light Vs absorbance of light. This graph is known as adsorption spectrum.8. Explain the principle, components and working of IR spectroscopy.Principle:

IR Spectra is produced by the absorption of energy by a molecule in the infrared region and the transitions occur between vibrational levels.Range of IR radiation:

Near Infrared = 12500 – 4000 cm-1

Infrared = 4000 – 667 cm-1

Far Infrared = 667 – 50 cm-1

Components:Radiation source: The main source of radiation is Nichrome wire or Nernst glower which is a filament containing oxides of Zr, Th, Ce held together with a binder.Monochromators: It allows the light of the required wavelength to pass through light of other Sample cell: The cells must be transparent to IR radiation are used to hold the sample. Detector: Detectors are used to convert thermal radiant energy into electrical energy. Recorder: The recorder record the signal coming out from the detectorBlock Diagram:

Working:The radiation emitted by the source is split into two parallel beams. One of the beams passes

through the sample and the other passes through the reference sample. When the two beams of light recombine they produce a signal which is measured by detector. The signal from the detector is recorded by the recorder.

9. Explain the molecular vibration in IR spectrum. Calculate the fundamental modes of vibration for non-linear and linear molecule.There are two kinds of fundamental vibrations in molecule.Stretching vibrations: During stretching, the distance between two atoms decreases or increases, but bond angle remains unaltered.Bending vibrations: During bending, bond angle increases or decreases but bond distance remains unaltered.Types of Stretching and bending vibrationsThe number of fundamental modes of vibration of a molecule can be calculated as follows. Non-linear molecule has 3N-6fundamental modes of „vibrations and a linear molecule has 3N-5 fundamental modes of vibrations where N is the number of atoms in a molecule..Stretching vibrations: It classified into two types

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(i) Symmetric stretching: The atoms of the molecule are moving in the same direction. (ii) Asymmetric stretching: The atoms of the molecule are moving in opposite direction.

Asymmetric stretching requires more energy than the Symmetric stretching. Bending vibrations: It classified into two types.In-plane bending Out-of plane bendingFundamental modes of vibration for non-linear and linear molecule1.Methane (CH4) (N=5)

It is a non-linear molecule. Hence, 3N-6 = 3x5-6 = 9 Fundamental vibrational modes.2.Benzene (C6H6) (N=12)

It is a non-linear molecule. Hence, 3N-6 = 3x12-6 = 30 Fundamental vibrational modes.2. Water (H2O) (N=3) It is a non-linear molecule. Hence, 3N-6 = 3x9-6 = 3 Fundamental

vibrational modes.

(i) Symmetric stretching (ii) Asymmetric stretching (ii) Bending vibration

υ1 = 3652 cm-1 υ2 = 3756 cm-1 υ1 = 1596 cm-1

4. Carbon dioxide (CO2) (N=3)

It is a linear molecule. Hence, 3N-5 = 3x3-5 = 4 Fundamental vibrational modes.

(i) (ii) (iii) (iv)

Symmetric stretching Asymmetric stretching In-plane Bending Out-plane bending

υ1 = 1340 cm-1 υ2 = 2350 cm-1 υ3 = 666 cm-1 υ4 = 666 cm-1

10. Discuss the applications of UV-Spectroscopy.Qualitative analysis: Many types of compounds can be identified by comparing the UV spectrum of the sample with that of similar compounds available in reference books. Testing the purity of the sample: UV Visible spectroscopy is the best method for detecting impurities in organic compounds because

The bands due to impurities are very intense. Saturated compounds have weak absorption band and unsaturated compounds

have strong absorption band. The kinetics of chemical reaction: The progress of a chemical reaction can be easily followed by examining the UV-spectra of test solution at different time intervals. The concentration of either the reactant or product can be followed.

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Determination of calcium in blood serum. Calcium in blood serum in indirectly determined by converting the Ca present in 1 ml of serum as calcium oxalate, and dissolving the calcium oxalate in dilute sulphuric acid and treated with ceric sulphate solution. The absorption of the excess Ce+4ion is measured at 315nm. The amount of Ca in blood serum is indirectly calculated from the amount of Ce+4ion consumed by the calcium oxalate.Quantitative analysis: UV absorption spectroscopy is used for the quantitative determination of compounds based on Beer-Lambert‟s law.

11. Discuss the applications of IR spectroscopy.Identification of organic compounds. : IR spectroscopy is very useful in identification of organic compounds.Identification of functional groups: The region from 4000 – 1400 cm-1 in IR spectrum is useful for functional group analysis.

Group Frequency rangeC=O 1750 – 1600 cm-1

Free OH 3650 – 3300 cm-1

H bonded OH 3300 – 2800 cm-1

-C = C- 2250 – 2100 cm-1

-C = N 2260 cm-1

-NH2 3350 cm-1

-N = O 1600 – 1500 cm-1

-COOH 3300 – 2700 cm-1

Testing the purity of a sample: Presence of an impurity can be detected by their characteristic peaks.Determination of symmetry of a molecule: The symmetry of a molecule whether linear or non-linear can be determined from the IR spectrum.Example: IR spectra of H2O gives 3 peaks at 667 cm-1, 1330cm-1, 2349 cm-1. Since non-linear molecule should exhibit (3N-6) = 3 peaks, the compound is non linear.Study of hydrogen bonding molecule: Intermolecular and Intramolecular hydrogen bonding can be differentiated by taking a series of spectra of a compound at different concentrations. Intermolecular hydrogen bonding decreases with dilution and the intensity of such peaks will also decrease. Intramolecular hydrogen bonding on the other hand will show no such change. This can be explained by taking the ortho and para nitrophenols.Study of progress of a chemical reactionThe rate the reaction can be determined by taking IR spectra at regular intervals of time Example: Oxidation of secondary alcohol to ketoneSecondary alcohol gives the absorption band at 3570cm-1 due to - OH stretching slowly disappears and a new band appears at 1725cm-1 due to C=O stretching.

12. What are the different types of electronic transactions? Draw the energy level diagram for various transitions.n→π* transitions:n→π* transitions are shown by unsaturated molecule containing hetero atoms like N, O & S.

It occurs due to the transition of non-bonding lone pair of electrons to the antibonding

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orbitals.This transition shows a weak band, and occurs in longer wavelength with low intensity. Example aldehyde & ketone with no double bond shows a band in the range of 270-300nm and if it is existing with the double bond show a band in the range of 300-350nm.

n→π* transitions:n→π* transitions are shown by unsaturated molecule containing hetero

atoms like N, O & S. It occurs due to the transition of non-bonding lone pair of electrons to the antibonding orbitals.This transition shows a weak band, and occurs in longer wavelength with low intensity. Example aldehyde and ketone with no double bond shows a band in the range of 270-300nm and if it is existing with the double bond show a band in the range of 300-350nm.σ→σ* transitions:

This type of transitions occurs in the compounds having only single bonds. Energy required for this transition is more and absorption band occurs in the far UV region (120-136 nm) Example: Methane, Ethanen→σ* transitions:

These transitions occur in saturated compounds having lone pair of electrons. This transition occurs along with n→σ* transition .The energy required for this transition is less than σ→σ* so absorption band occurs at longer wavelength in the near UV region (180-200nm). Example: (CH3)3N for n→σ* transition- at 227nm, for σ→σ* transition at 99nm.π→π* transitions:

These transitions occur in unsaturated compounds i.e. the compounds having double bond and triple bonds. Example: Ethylene molecule shows intense band at 174nm and weak band at 200nm. Both are due to π→π* transitions. According to selection rule, the intense band at 174nm is due to allowed transistion.If there is any alkyl substitution in olefins shifts the absorption band to longer wavelength. This effect is known is batho-chromic effect or red shift.

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