mahalakshmi - vidyarthiplus (v+)

Mr.N.S & Dr.S.P./Chemistry MAHALAKSHMI ENGINEERING COLLEGE, TRICHY-621213

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI - 621 213

Sub. Code: CY 6151 Semester : I

Subject : ENGINEERING CHEMISTRY-I Unit –III: PHOTO CHEMISTRY & SPECTROSCOPY

2 Marks Questions and Answers

01. Define the term photochemistry.

It deals with the chemical reaction caused mainly visible and UV region of electromagnetic radiation

i.e., in the wavelength range of 2000-8000Ao. Ex: photo synthesis , photochemistry.

02. What is the difference between the thermal and photochemical reactions?

Photochemical reaction Thermal reaction

1. Absorption of light radiations Absorption or liberation of heat

2. Temperature does not affect the reaction Affects the reaction

3. Reaction takes place in presence of light. Both in dark and as well as in light.

4. ΔG- decreases or increases during the reaction. Decreases during the reaction

5. zero order 0,1,2,3,2/3,3/2,etc.

03. Mention the laws in photochemistry.

1. Grotthuss – Draper Law.

2. Stark- Einstein Law.

3. Beer-Lambert’s Law.

04. State Grotthuss – Draper Law.

i. This is known as I law of photochemistry.

ii. Only the light, that is absorbed by the substance can bring about a chemical change in the

substance. i.e light energy is needed for the chemical reaction.

05. State Stark – Einstein Law.

i. This is known as I law of photochemistry.

ii. Each molecule taking a part in a photochemical reaction absorbs one quantum of radiation. (one

photon)

06. State Beer Lambertz law.

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 the concentration of the solution’C’

A=εCl

07. What are the limitations of Beer-Lambert’s law?

It is applicable for only dilute solutions.

It is not obeyed, if the radiation used is not monochromatic.

Deviation may occur, if the solution contains impurities.

08. A solution of thickness 2 cm transmits 40% incident light. Calculate the concentration of the

solution given thatε = 5842 dm3 mol-1 cm-1

% T = 40 %, T= 0.4, l = 2 cm ; A= -log T = - log 0.40 = 0.398.

A = ∑Cl; C = A/∑l ; 0.398/5842 x 2 ; = 3.4x10-5mol.dm-3

Prepared by:

Mr.N.S.,Dr.S.P., AP/Chemistry.

www.Vidyarthiplus.com



09. Calculate the concentration in µg/ml of a solution of organic compound whose molecular mass

is 211.2, in 011 M HCL giving an absorbance at its λ max281 nm of 0.612 in a 4 cm ell. The major

absorptivity at 281 nm is 5372 mole-1cm-1.

A = ∑Cl ;C = A/∑l; 0.612/5372x4 = 2.848x10-5mol/lit = 6.015µg/ml

10. What do you mean by quantum efficiency or quantum yield?

It is defined as the number of molecules of reactants reacted or products formed per quantum of

light is absorbed.

11. How quantum yield are determined?

i. Determination of number of molecules reacted.

ii. Determination of number of photon absorbed.

12. What is chemical actinometer?

It is a device used to measure the amount of radiation absorbed by the reactants in a

photochemical reaction.

13. Define photo process.

It is the process to analyse non radiative transitions. It involves two transitions.

1. Internal conversion. 2. Inter System crossing

14. What is Fluorescence?

When a molecule or atom absorbs radiation of higher frequency (shorter wavelength) it gets excited

then the excited atom or molecule re-emits the radiation of the same frequency within short time 10-8 sec.

This process is called fluorescence. It will stop as soon as the incident radiation is cut-off. Ex: Fluorite,

uranium.

15. What is Phosphorescence?

When a molecule or atom absorbs radiation of higher frequency the emission of radiation is

continuous for some time even after the incident radiation is cut-off. This process is called

phosphorescence.

16. What are the difference between fluorescence and phosphorescence?

Fluorescence Phosphorescence

1. Its decay period is much long Its decay period is very short

2. Transition between states of different multiplicity. Transition between states of same multiplicity.

3. It is not observed in solution at room temperature. It can be observed in solution at room temperature.

4. It is shown by some elements in vapour state. It is rarely observed in vapopurs and gases.

17. What do you mean by chemiluminescence?

When light is emitted at ordinary temperature as a result of chemical reactions, the phenomenon is

Known as chemiluminescence. It is known as a reverse of photochemical reaction.

Ex: fire flies – luciferon – cold light.




18. Define photosensitization.

The foreign substance, which absorbs the radiation and transfers the absorbed energy to the

reactants is called a photosensitization.

19. Define electromagnetic spectrum.

The entire range over which electromagnetic radiation exists is known as electromagnetic

spectrum.

20. What are the factors affecting absorption of radiation?

(i). The nature of the absorbing molecules. (ii). The concentration of molecules.

(iii). The length of the path of the radiation through the matter.

21. Give four types of shifts.

Shifts: (i) Bathochromic-Red shift – Higher wavelength, lower frequency.

(ii) Hypsochromic – Blue shift – Lower wave length, Higher frequency.

(iii) Hyperchromic – An increase in intensity

(iv) Hypochromic – An decrease in intensity

PART – B(16 Marks Questions with Answers)

1. Explain photochemical laws in photochemistry.

Laws of photochemistry:

1. Grotthuss- Draper law

2. Stark- Einstein law.

3. Lambert-Beer law

1. Grotthuss- Draper law:

This is known as I law of photochemistry.

It is known as principle of photochemical activation.

“Only the light, that is absorbed by the substance can bring about a chemical change in the

substance”. i.e., light energy is needed for the chemical reaction.

2. Stark- Einstein law:

This is known as II law of photochemistry.

Photochemical equivalence / principle of quantum activation.

“Each molecule taking part in a photochemical reaction absorbs one quantum of radiation(one

photon)”. It is a 1:1 relationship between the number of reacting molecules & the quantum of light absorbed

in the primary process.

3. Lambertz-Beer law:

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 the concentration of the solution ‘C’.

-dI/dx = k I C

A= Є C x

Thus the absorbance(A) is directly proportional to molar concentration (C) & thickness (or) path

length(x).




2. What are the conditions & reasons of high & low quantum yield?

Quantum yield (Ф):

It is defined as the number of molecules of reactants reacted or products formed per quantum of

light absorbed.

High quantum yield:

When 2 or more molecules are decomposed – per photon, the quantum yield

Ф > 1 & the reaction has a high quantum yield.

Low quantum yield:

When the number of molecules decomposed is less than one per photon, the quantum yield Ф < 1

& the reaction has a low quantum yield.

Conditions for Quantum yield:

All the reactant molecules should be initially in the same energy state & hence equally reactive.

The reactivity of the molecules should be temperature independent.

The molecules in the activated state should be largely unstable & decomposed to form the

products.

Reasons for high Quantum yield:

i).Absorption of radiation in the primary step produces atoms or free radicals which initiate a series of chain

reaction.

ii).Formation of intermediate products will act as a catalyst.

iii).If the reactions are exothermic, the heat evolved may activate other molecules without absorbing the

additional quanta of radiation.

iv).The activated molecules may collide with other molecules & activate them which in turn activate other

reactants.

Example: formation of HCl.

Reaction of H2 & Cl2:

The product comes from two steps .

(i) Primary reaction.

(ii) Secondary reaction.

i).Primary reaction:

Reactant Absorbs one photon Dissociates into atoms Photochemical reaction.

ii).Secondary reaction:

Atoms come from primary reaction decompose Product

In above reaction, one chlorine molecule absorbs a photon & dissociated into two chlorine atoms.

Formation of HCl in secondary step as follows.

Primary reaction: Cl2 + hγ 2 Cl (1)

Secondary reaction: Cl + H2 HCl + H + Heat (2)

H + Cl2 HCl + Cl + Heat (3)

The chlorine atom used up in step 2, is formed again in step 3. This will propagate the chain reaction as

heat is evolved in step 2, 3. This will lead the formation of HCl & absorption of one photon energy.

Therefore quantum yield Φ = 106/1 = 106.

Reasons for low Quantum yield:

I. The activated molecules may get deactivated before forming the products.

II. The activated molecules may lose their energy by colliding with non-activated molecules.

III. Reactant molecules may not obtained the require energy of the reaction.




IV. The primary reaction may be reversible in nature.

V. Reunion of atoms will result low quantum yield.

Ex: dimerisation of anthracene to dianthracene.

2C14H10 + hγ C28H20.

The quantum yield = 2, but actually it is found to be 0.5.

3. How can you determine the quantum yield?

Determination of quantum yield: It is determine by two ways.

Determination of number of molecules reacted.

Determination of number of photons absorbed.

(i). Determination of number of molecules reacted:

1) It can be determine by the usual analytical technique.

2) Chemical kinetics used.

3) Determined by the number of molecules in a given time.

Experiment:

Known amount of reactant samples taken from reaction mixture time to time. It is measured by

usually volumetric analysis method or optical rotation. From the data, the amount & hence the number of

molecules reacted can be calculated.

(ii). Determination of number of photons absorbed:

1) Determination is done by suitable experimental set up like UV-absorption light.

2) The light radiation source gives required radiation. Ex: sunlight, Hg vapour lamp.

3) Radiation passed through lens & it produces beams in parallel.

4) Beams are then passed through monochromator to convert a single wavelength.

5) Then it is allowed to the cell.

6) Light is absorbed & falls on detector, it displays intensity of radiation.




4. Explain photo process internal conversion(IC) & intersystem crossing(ISC).

Photo process:

It is the process to analyse non-radiative transitions. It involves two transitions.

1) Internal conversion.

2) Inter system crossing.

1).Internal conversion (IC):

These transitions involve the return of the activated molecule from the higher excited states to the

first excited states, namely

S3 S1; S2 S1 / T3 T1; T2 T1

The energy of the activated molecule is given out in the form of heat through molecular collisions. This

process is called internal collisions(IC). It occurs in less than 0.1 second.

2). Inter system crossing (ISC):

The molecule may also lose energy by another process called inter system crossing (ISC0. These

transitions involve the return of the activated molecules from the states of different spins (i.e.,) different

multiplicity.

S2 T2; S1 T1

The transitions are forbidden, occurs relatively at slow rates.

5. Explain the terms fluorescence & phosphorescence.

Fluorescence:

When a molecule or atom absorbs radiation of higher frequency (shorter wave length) it gets

excited. Then the excited atom or molecule re-emits the radiation of the same frequency within

short time 10-8 sec. This process is called fluorescence. It will stop as soon as the incident radiation

is cut off.

Ex: fluorite (CaF2); petroleum; uranium; organic dyes.

Types:

Two types: 1. Resonance fluorescence

2. Sensitized fluorescence

1. Resonance fluorescence:

Mercury vapour shows this at λ=253.7nm.

2. Sensitized fluorescence:

Hg vapour is mixed with vapours of Thalium, silver, zinc or Pb.

Phosphorescence:

When a molecule or atoms absorbs radiation of higher frequency, the emission of radiation

is continuous for some time even after the incident radiation is cut off. This process is called

phosphorescence.

Ex: ZnS, CaS, BaSO4, etc.

Difference between fluorescence & phosphorescence:

Fluorescence Phosphorescence

1 .Its decay period is much long. Its decay period is very short.

2. Transition between states of different multiplicity.

Transition between states of same multiplicity.

3. It is not observed in solution at room temperature.

It can be observed in solution at room temperature.

4. It is shown by some elements in vapour state.

It is rarely observed in vapour & gases.




6. Give the mechanism for fluorescence & phosphorescence using Jablonski diagram.

Mechanism for fluorescence & phosphorescence using Jablonski diagram:

When an electron jumps from the ground state to any of the higher electronic states, depending

upon the energy of the photon absorbed, we can get a series of

i).Singlet excited states – S1, S2, S3, etc. These are known as first, second & third singlet excited

state etc.

ii).Triplet excited states –T1, T2, T3, etc. These are known as first, second & third triplet excited

state, etc.

Generally the energy of the singlet excited state is higher than the corresponding triplet excited state.

Hence the energy sequence can be written as, ES1 > ET1; ES2 >ET2; ES3 > ET3; etc.

When a molecule absorbs light radiation, the electron may jump from S0 to S1, S2, S3

singlet state depending upon the energy of light radiation as shown in the diagram. For each singlet excited

state there is a corresponding triplet excited state.

S1 T1; S2 T2; S3 T3

The molecule whether in the singlet or triplet excited state is said to be activated.

A0 + hγ A*.

A0 Ground state molecule; A* Excited state molecule.

The activated molecules return to be ground state by emitting the energy through the following

process.

1). Non- radioactive transitions.

a) Internal conversion(IC).

b) Inter system crossing(ISC).

2).Radioactive transitions.

Excited state Ground state.

a). Fluorescence: S1 S0.

An allowed transitions occurs only about 10-8 sec.

b). Phosphorescence: T1 S0.

Triplet excited state Ground state.

It is a forbidden transition, occurs slowly.

c). Quenching (stopping) :

The fluorescence may be stopped when the excited molecule collides with a normal molecule before

it fluoresces.




7. Explain Photosensization with neat diagram.

The foreign substance, which absorbs the radiation & transfers the absorbed energy to the

reactants is called a photosensitizer. This process is called photosensitization.

Ex: 1. Atomic photosensitizers – Hg, Cd, Zn.

2. Molecular photosensitizers – SO2.

Examples for reactions: Dissociation of H2 molecule and Photosynthesis in plants.

Mechanism:

Mechanism of Photosensitisation

It can be explained by , General donor = D and Acceptor = A

Donor – D- absorbs the incident photon.

S0 S1 or 1D

Donor Inter system crossing (ISC) gives triplet excited state. T1 or 3D.

Triplet state donor > Triplet state acceptor 3A.

When the triplet excited state of acceptor 3A gives the desired products. The mechanism is called

photosensitisation.

From excited state to directly donor 3D, called quenching process.

D + hγ 1D

1D ISC 3D.

3D + A D + 3A.

3A Products.

3D Products.

The energy available is enough to excite the reactant molecule to its excited state.

From the diagram, …. line indicates the transfer of energy from sensitizer to the reactant.




8. Explain in detail on Electronic, Vibrational and rotational transitions.

(A). Electronic Transition:

(i). Electronic energy arises due to the motion of e-s.

(ii). It results from the transition of e- from ground state energy level to an excited state energy level of a

molecule due to the absorption of radiations in the visible region 12500-25000cm-1.

(iii). Electronic transitions in a molecule are accompanied by vibrational and rotational transitions, the

electronic spectra of molecules are highly complex.

(B). Vibrational Transitions:

(i). Vibrational energy arises due to the to and fro motion of the molecule.

(ii). Stretching, Contracting and Bending of covalent bonds in a molecule.

(iii). It results between the vibrational energy levels of a molecule, due to the absorption of radiation in the

IR region 500-400 cm-1.

(iv). It is accompanied by rotational transition, so it is called as vibrational – rotational spectra.

(v). It is shown in CO2, H2O molecules.

(vi). Change in dipole moment of the molecule, vibrational motion takes place.

(C). Rotational Transition:

(i). It arises when the molecule rotates about an axis perpendicular to the internuclear axis.

(ii). It results from the transitions between the rotational energy levels of a molecule, due to the absorption

of radiation in the microwave region 1 – 100 cm-1.

(iii). Shown in the molecules like HCl, CO, H2O, NO which possess a permanent dipolemoment.




9. Explain the principle, instrumentation and block diagram of UV spectrum.

Principle: When energy is absorbed by a molecule in the UV region, it bring about some changes in the

electronic energy of the molecule resulting from transitions of the valance electrons.

UV Wavelength : 100-400 nm

Visible Wavelength: 400-750 nm

In UV-Visible, four types of electronic transitions occurs.

(i) σ → σ* (ii) n → σ* (iii) n → π* (iv) π → π*

Block Diagram:

Instrumentation:

It is a device which detects the percentage of transmittance of light radiation when light of certain

intensity and frequency range is passed through the sample.

Thus the instrumentation compares and records the intensity of transmitted light with that of incident

light.

The various components of UV spectrophotometer as follows:

(a) Radiation Source: Hydrogen and Deuterium lamps are commonly used radiation source in UV of

wavelength 200-750 nm. It must be stable and provide continuous and constant radiation.

(b) Filter: It allows only the light of required wave length to pass through and absorb the rest.

(c) Slits: Entrance: It provides a narrow source of light.

Exit : Narrow band for observation.

(d) Cell: One for the sample solution and other for the solvent.

(e) Detector: It consists of photoelectric cells which converts the light radiation into current signal.

(f) Recorder: Finally recorder records the % of absorbance and it shows in display.

10. What are the applications of UV Spectroscopy?

Applications of UV Spectrum

(i) Predicting relationship between different groups.

(ii) Qualitative analysis

(iii) Detection of impurities.

(iv) Qualitative analysis

(v) Determination of molecular weight

(vi) Dissociation constants of Acids and Bases.

(vii) Determination of Calcium in blood serum

(viii) Studying kinetics of chemical reactions.




11. Explain the principle, instrumentation and block diagram of IR spectrum.

Principle: IR spectra is produced by the absorption of energy by a molecule in the IR region and the

transitions occur between vibrational levels.

Range of IR region: (i) Near IR = 12,500 – 4,000 cm-1

(ii) IR = 4,000 – 667 cm-1

(iii) Far IR = 667 – 50 cm-1

Instrumentation:

It is a device which detects the % of transmittance of light radiation when lioght of certain intensity

and frequency range is passed through the samples.

Thus the instrument compares and records the intensity of transmitted light with that of incident

light.

The various components of IR spectrophotomenter as follows:

Block Diagram:

(a) Radiation Source: Nichrome wire and Nernst glower are the main sources of IR radiation.

They are heated by electrically at 1200 – 2000OC, they glow and

produce IR radiation.

(b) Monochromator: It allows only the light of required wavelength to pass through and absorb the rest.

(c) Slits: Entrance : It provide a narrow source of light

Exit: Narrow band for observation.

(d) Cell: One for the sample solution other for the solvent.

(e) Detector: It consists of photoelectric cells which converts the light radiation into current signal.

(f) Recorder: Finally recorder records the % of absorbance and it shows in display.

Two types of fundamental Vibrations in IR

(i) Stretching Vibrations: During stretching, the distance between two atoms decreases or increases, but

the bond angle remains unaltered.

(ii) Bending Vibrations: During Bending, the bond angle decreases or increases, but the bond distance

remains unaltered.




12. What are the applications of IR spectroscopy?

Applications of IR Spectrum

(i) Identification of the compound cab be established.

(ii) Detection of functional groups

(iii) Testing purity of a sample

(iv) Study of progress of a chemical reaction.

(v) Determination of shape of a molecule.

(vi) To study tautomerism.

(vii) Industrial applications

(viii) Isomers can be identified in the finger print region.

(ix) Determination of Hydrogen bonding in a molecule.

(x) Determination of Aromaticity.



mahalakshmi - vidyarthiplus (v+)

Documents