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FOR 2020 - PG-3- PHYSICAL CHEMISTRY

PHYSICAL CHEMISTRY -IIIFOR PG

( MADRAS UNIVERSITY )

Unit I: Thermodynamics - I:-

Unit II: Spectroscopy - II:-

Rotational spectroscopy.

Vibrational spectra

Raman spectra:.

Unit III: Spectroscopy - III:-

Resonance spectroscopy-

ESR

Mass spectra

Mossbauer spectroscopy

Quadrupole interactions

Unit IV: Electrochemistry of solution

Unit V: Quantum Chemistry – IV

BY

Dr. C.SEBASTIAN ANTONY SELVAN

ASST. PROFESSOR OF CHEMISTRY

R. V. GOVT.ARTS COLLEGE

CHENGALPATTU

9444040115

SEP- 2019

C.SEBASTIAN, AP / CHEM. R.V.GOVT.ARTS COLLEGE,CHENGAIPATTU MOB: 9444040115


Long live scientist Doberiner And his Law of Triads too

Long live Scientist Newland And his Law of Octaves too -2

1. Lithium Sodium PotassiumRubidium Cesium Francium -2Which are first A group metals-2They are known as Alkali metals-2( long)

2. Beryllium Magnesium CalciumStrontium Barium Radium -2Which are second A group metals -2They are known as Alkaline earths -2( long)

3. Boron belongs to third A groupAluminium Gallium Indium Thalium -2Are the members of Boron family -2They form first group of p- block elements-2(long)

4.Carbon Silicon GermaniumStannum Plumbum form fourth A group-2Fourth B consists of three metals-2 Titanium Zirconium Hafnium – 2( long)

5. Phosphorous Arsenic Antimony Bismuth Belong to fifth A Nitrogen family – 2Sulphur Selenium Tellurium Polonium – 2 Belong to sixth A Oxygen family-2( long)

6. Fluorine Chlorine Bromine IodineAstatine are the five halogens-2They do belong to seventh A group-2They form salt with strong bases.-2( long)

7. Helium Neon Argon Krypton Xenon Radon are the rare gases-2They are known as noble gases -2Which are basically inert nature-2( long)

8.Elements forming colour compounds All are found in d- block series-2Scandium is the first member-2Yttrium forms the first of second row -2( long)

9 Vanadium Niobium Tantalum

Chromium Molybdenum hard Tungsten-2Manganese Technecium Rhenium-2Are the essential d- block elements-2(long)

10.Eight group consists of nine metals Which are Ferric Iron Cobalt Nickel – 2Ruthenium Rhodium Palladium-2Osmium Iridium Platinum-2( long)

11.Copper Silver Gold are theSo called essential coinage metals-2Zinc Cadmium Mercury are their neighbours-2They do belong to transition elements-2 ( long)

12.Elements following Lanthanum All but fourteen are Lanthanides-2Cerium Prasodium Neodymium-2Are the first three lanthanides - 2( long)

13.Promithyum Samarium EuropiumGadalonium Terbium Dysprocium-2Holmium Erbium Thulium Yutterbium – 2Luetecium are the rest of lanthanides-2. ( long)

14.Actinides are also fourteen numbersWhich are following element Actinium – 2Thorium Protactinium Uranium -2Are the first three actinides- 2 ( long)

15.Neptunium Plutonium AmericiumCurium Berkelium Californium -2Einsteinium Fermium Mendelevium-2Nobelium Lawrencium are the actinides-2 ( long)

16. Rutherfordium Dubnium SeaborgiumBohrium Hassium Meitnerium - 2Darmastidium Roentgenium Copernicium - 2Are the newly found d - block elements

17. Hydrogen resembles alkali metals



Halogens also same as hydrogenSo we are in need of your help in Predicting the position of Hydrogen

Composed by: Dr. C.SEBASTIAN A NTONY SELVAN. ASST. PROF in CHEMISTRY, R.V GOVT.ARTS COLLEGE, CHENGALPATTU , MOB: 9444040115 . OCT - 2017

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UNIVERSITY OF MADRASM.Sc., DEGREE COURSE IN CHEMISTRY

REGULATIONS (w.e.f. 2015-2016)

THIRD SEMESTER

Course Components/Title of the paper

Inst.

Hour

s/

week

Credi

ts

Exam

Hour

s

MARKS

CI

A

EX

T

TOTA

L

Core Paper – IX: Organic Chemistry-III 6 4 3 25 75 100

Core Paper – X :Inorganic Chemistry-III 6 4 3 25 75 100

Core Paper – XI: Physical Chemistry-III 6 4 3 25 75 100

Extra Disciplinary Paper – II: Materials

Science/ Bioorganic Chemistry/ Research

Methodology/Bioinorganic Chemistry

4 3 3 25 75 100

Elective Paper –III : Organic Chemistry

Practical-II*4 3 6 40 60 100

Elective Paper –IV: Inorganic Chemistry

Practical-II*4 3 6 40 60 100

Soft Skill Paper – III -- 2 3 40 60 100

30 23



Core Paper - XI - Physical Chemistry – III (90 Hours)Unit I: Thermodynamics - I:-

Partial molar properties - Partial molar free energy (Chemical potential) - Partial molar volume and partial molar heat content - their significance and determination of these quantities. Variation of chemical potential with temperature and pressure. Thermodynamics of real gases - gas mixture - fugacity definition - determination of fugacity variation of fugacity with temperature and pressure -thermodynamics of ideal and non ideal binary solutions-dilute solutions-excess functions for non-'ideal solutions and their determination-the concepts of activity and activity coefficients-determination of standard free energies. Choice of standard states - determination of activity and activity coefficients for non-electrolytes.Unit II: Spectroscopy - II:-

Rotational spectroscopy of a rigid rotar – non-rigid rotor-diatomic and polyatomic molecules. Vibrational spectroscopy-harmonic oscillator-anhormonicity –Vibration – rotation spectra of diatomic vibrating molecules selection rules-P,Q and R branches.

Vibrational spectra of polyatomic molecules- fundamental vibrations – normal modes of vibration- overtones, combination and difference bands- Fermi resonance.

Raman spectra: Classical theory of Raman effect and molecular polarisability – pure rotational Raman spectra – Vibrational Raman spectra – Rotational fine structure – Rule of mutual exclusion – Polarization of light and Raman effect.Unit III: Spectroscopy - III:-

Resonance spectroscopy-Zeeman effect-equation of motion of spin in magnetic fields-chemical shift-spin-spin coupling-NMR of simple AX and AMX type molecules- H1 , 13C, 19F, 31P NMR spectra - a brief qualitative discussion of Fourier transform spectroscopy.

ESR: principle, spin-orbit coupling. Hyperfine interaction. McConnell reactions. Mass spectra: Theory and instrumentation, McLaffetry rearrangement fragmentation pattern for simple aliphatic and aromatic alkanes, alcohols, aldehydes and ketones.- Mossbauer spectroscopy- Doppler effects, isomer shift, electron-neutron hyperfine interactions. Quadrupole interactions and Magnetic interactions.Unit IV: Electrochemistry of solution:-

Mean ionic activity and activity coefficient: concept of ionic strength, Debye-Huckel theory of strong electrolytes-activity coefficient of strong electrolytes-determination -Debye Huckel limiting law at appreciable concentration of electrolytes - Debye Huckel Bronsted equation-qualitative and quantitative verification.

Redox reaction: cell potential, Galvanic cell, Electrolytic cell, Nernst equation for cell potential of electrolyte. Electrode equilibrium-thermodynamic electrodes and electrode potential, electrochemical cells and electromotive force.Unit V: Quantum Chemistry – IV:-

Approximation methods –perturbation and variation method –application to hydrogen ,helium atoms –R.S. coupling and term symbols for atoms in the ground state – Slater orbital and HF –SCF methods Born – Heimer approximation –valence bond theory for hydrogen molecule –LACO –MO theory for di and polyatomic molecules –concept of hybridization –



Huckel theory for conjugated molecules (ethylene , butadiene and benzene)- semi empirical methods .Text Books:-

Elective Paper II - Physical Chemistry Parcticals I

UNIT I THERMODYNAMICS - I

1. .Partial molar properties

2. Partial molar free energy (Chemical potential)

3. Partial molar volume

4. partial molar heat content

5. Significance

6. Determination of these quantities.

7. Variation of chemical potential with temperature and pressure.

8. Thermodynamics of real gases

9. Gas mixture

10. Fugacity definition

11. Determination of fugacity

12. Variation of fugacity with temperature and pressure

13. Thermodynamics of ideal binary solutions

14. Thermodynamics of non ideal binary solutions

15. Dilute solutions

16. Excess functions for non-'ideal solutions and their determination

17. Concepts of activity and activity coefficients

18. Determination of standard free energies.

19. Choice of standard states



20. Determination of activity and activity coefficients for non-electrolytes.

120 PARTIAL MOLAR PROPERTIES ( THERMODYNAMICS OF OPEN SYSTEM )

The partial molar property is defined as the “ variation in the property of a open

system , with a change in number of moles , at constant temperature and pressure”

If X denotes the property of a open system ,at constant temperature and pressure ,with

number of moles of constituents n1, n2, n3… then

partial molar property X = (∂ X∂ n 1

) T, P, n2, n3 where X denotes U,H,S,A, and G

partial molar internal energy ( U ¿=( ∂U∂n 1

) T, P, n2,n3.

partial molar entropy (S¿=( ∂ S∂ n 1

) T, P, n2,n3...

partial molar enthalpy (H ) = (∂ H∂ n1

) T, P, n2,n3...

partial molar Helmholtz work function ( A ) = (∂ A∂ n 1

) T, P, n2,n3...

partial molar Free energy (G ) = (∂ G∂ n 1

) T, P, n2,n3...

partial molar volume V = (∂V∂ n 1

) T, P, n2,n3...

220 PARTIAL MOLAR FREE ENERGY (CHEMICAL POTENTIAL)

The chemical potential of a given substance is the change in the free energy of the

open system with a change in number of moles , at constant temperature and pressure” .

It is expressed as μi = (∂ G∂ n 1

) T, P, n2,n3...

320 PARTIAL MOLAR VOLUME



The partial molar volume of a given substance is the change in the volume of the open

system with a change in number of moles , at constant temperature and pressure” .

It is expressed as V = (∂V∂ n 1

) T, P, n2,n3...

Significance:

1. Molar volume is not additive but Partial Molar volume is additive .

2. The additive relation is known as Euler relation

Problem: The partial molar volume of alcohol in a mixture of alcohol and water is 45.5 ml,

nA = 2.0 mol, nB = 2.0 mol . Find the partial molar volume of water if the total volume of the

mixture is 101.5

Solution:

V = 101.5 ml

V A = 45.5 ml,

nA = 2.0 mol,

nB = 2.0 mol

V B = ?

V B = V−nA V A

nB

= 101.5−91.0

2

= 5.25 ml

Problem 2: The partial molar volume of alcohol and water in a mixture of alcohol and

water is 45.5 ml and 5.25 ml respectively. If, nA = 2.0 mol, nB = 2.0 mol find the total

volume of the mixture

Solution

V A = 45.5 ml,

V B = 5.25 ml

nA = 2.0 mol,

nB = 2.0 mol

V = ?



= 2 ( 45.5 ) + 2 ( 5.25)

= 101 .5 ml

Problem 3 :Partial molar Volume of ethanol and water is 57.4 ml and 16. 9 ml respectively

Find the total volume of the mixture when one mole of each is mixed. .Find also the

volume of mixing

Density of ethanol = 0.789 gm/ cm3

Solution:

Mole fraction = 0.5

Partial molar Volume of ethanol V A= 57.4 ml

Partial molar Volume of water V B= 16 .9 ml

V = nA V A + nBV B

Volume of mixture = 1( 57.4) + 1 ( 18)

= 74.3 ml

To find volume of mixing:

volume of mixing = volume of the mixture - volume of separated components

Volume of one mole of pure ethanol V = mass

density

= 46. 070. 789

= 58.3 ml

Volume of one mole of pure water

= 181

= 18 ml

Total volume of Pure components before mixing

= 18 + 58.3

= 76.3 ml

volume of the mixture = 74.3ml

The difference between the volume of the mixture and the separated components

= 74.3- 76.3

= - 2.0 ml



FOR ONLINE TEST -09-08-2020 SUNDAY

. The partial molar volume of alcohol in a mixture containing one mole of alcohol and one

mole of water is 50 ml, Find the partial molar volume of water if the total volume of the

mixture is 101

Ans

V = v1V1+ n2V2

101= 1 ( 50) +1( V2)

V2 = 51

The partial molar volume of alcohol and water in a mixture of alcohol and water is 40 ml

and 20 ml respectively. If, nA = 2.0 mol, nB = 2.0 mol find the total volume of the mixture

Ans : 2(40) + 2(20)

= 120 ml

Partial molar Volume of ethanol and water is 53 .1 ml and 16. 9 ml respectively Find the

volume of mixing when one mole of each is mixed. Total volume of Pure components

before mixing is 72 ml

V = 1(53.1) + 1( 16.9)

= 70 ml

Volume of mixing = 70 -72 = -2 ml

420 PARTIAL MOLAR HEAT CONTENT

The partial molar heat content of a given substance is the change in the heat content

of the open system with a change in number of moles , at constant temperature and

pressure” .

It is expressed as H= (∂ H∂ n 1

) T, P, n2,n3...

520 SIGNIFICANCE

6

20 DETERMINATION OF THESE QUANTITIES.



DETERMINATION OF PARTIAL MOLAR VOLUME

1. DIRECT METHOD

Consider a two component system

1. Determine the volume V, for various mixtures of two components by keeping the number

of moles n1 constant and varying the other ( n2) .

2. The volume of the solution is plotted against molality (m2)

3. The slope at a given concentration represents the partial molar volume of the solute at

the concentration.

4. Partial molar volume of the solvent can be obtained from Gibbs Duhem equation.

Gibbs –Duhem equation is ∑ ¿dμi= 0

Where ni – number of moles. μi - chemical potential

2. METHOD OF INTERCEPT:

By this method, it is possible to determine the partial molar properties of both

constituents of a binary mixture of any composition.

Let’ Va’ be the average volume , n1 and n2 are the number of moles of solvent and solute

of a binary mixture then

Va = V

(n1+n2)

Rearranging,

V = Va (n1+ n2)

Differentiating with respect to n2

(∂ V∂ n2

) = Va ( 0+1) + (n1+ n2)(∂ V a

∂ n2)


n1 0.1 0.1 0.1 0.1 0.1

n2 0.1 0.2 0.3 0.4 0.5

V = n1+ n2 0.2 0.3 0.4 0.5 0.6


= Va + (n1+ n2)(∂ V a

∂ n 2)

At constant composition ‘∂’ can be replaced by ‘d’

= Va + (n1+ n2)(dV a

dn 2)

= Va + (n1+n2)

d n2 dV a ---------------------

1

The mole fraction of the solvent x1 = n1

(n1+n2)

differentiating with respect to n2, [ d( 1x ) = -

1x2 ]

d x1

d n2 = -

n1

(n1+n2)2

= n1

(n1+n2)× −1

(n1+n2)

= x1 × −1

(n1+n2) [ x1 =

n1

(n1+n2) ]

= −x1

(n1+n2)

Rearranging, (n1+n2)

d n2 = -

x1

d x1

substituting in 1 (∂ V∂ n2

) = Va - x1

d x1 dV a

V 2 = V a - x1 dV a

d x1

-------------------2

The values of Va for mixtures of various compositions are plotted against x1 . Let O be the

point corresponding to the mole fraction of the solvent in solution, at which the partial

molar volume is required. Through O, a tangent LM to the curve is drawn. The slope d V a

d x1

is found out from the graph

V 2 = V a - x1 ( slope)

Using this relation the partial molar volume can be found out.



The slope of LM = d V a

d x1 -------------------3

Slope = Tan < NOL

= ln

ON ------------------------4

∴ dV a

d x1 =

lnON [ from 3 and 4]

LN = ONd V a

d x1 [ Rearranging]

= x1 d V a

d x1

V 2 = V a - x1 d V a

d x1- [ from 1]

= AL – AN

= LN

Same way BM = V 1

Procedure:

1. Prepare 10 set of binary Mixture with varying weight fraction

2. Find out the density of each mixture

3. Convert the density in to specific volume

4. Plot specific volume against weight fraction of A.

5. Draw a tangent through the point at which pmv is required.

6. The intercept of the tangent along Wa = 1 is noted.

7. Multiply this with the molecular weight of A

8. This will give the pmv of A

9. The intercept of the tangent along Wb = 1 is noted.

10. Multiply this with the molecular weight of B



11. This will give the pmv of B

3. BY DENSITY MEASUREMENT

𝛒 = n1 M 1+n2 M 2

1000V

= n1 M 1+M 2

1000 V [ consider 1 mole of solute]

1000V 𝛒 = n1 M 1+M 2 [ rearranging]

n1=1000 V ρ−M 2

M 1

dn1=1

M 1 [ 1000( Vdρ + ρdV)- 0 ] [Differentiating ]

d n1

dV =

1M1

[ 1000( V(d ρdV ) + ρ] [dividing by dV]

dVd n1

= M1

¿¿ [ taking reciprocal]

V 1 = M1

¿¿ --------1 [

dVd n1

= V 1 ]

V = 1c

dV = −1c2 dc [ d(

1c ) =

−1c2 dc ]

dVd ρ =

−1c2

d cd ρ [ dividing by

dρ]

d ρdV = - c2

d ρdc [ taking reciprocal]

V d ρdV = - V× c2

d ρdc [multiplying by V]

V d ρdV = - c

d ρdc [ V=

1c ]

substituting in 1 we get

V 1 = M1

¿¿



determine density of the solution at various concentrations and plot of ρ versus ‘c’ gives the

slope d ρdc from which V 1 can be calculated

graph:

DETERMINATION USING APPARENT MOLAR PROPERTY:

Apparent molar property of a solute(φ ¿ is given as

φ = G−n1 G1

n2

G 1−the property per moleof pure constituent1

G−the property per mole of solution

n1 – number of moles of solvent

n2 - number of moles of solute.

φ n2 = G−n1G1 [ rearranging]

G=n1G1+φ n2

differentiating with respect to n2,

∂ G∂ n2

= 0 + n2 ∂ φ∂ n2

+ φ

G = n2 ∂ φ∂ n2

+ φ [∂ G∂ n2

= G]

for dilute solution n2 = m. therefore the above equation becomes

G = m ∂ φ∂ m + φ

= m ∂ φ∂ m + φ

= ∂ φ

∂( lnm) + φ m

dm = d(ln m )



plot of φ against ln m gives the partial molar property

Problem

Determine the pmv of water and ethanol at mole fraction of alcohol 0.6 from the following

data. The solution gives intercepts 1.1 and 0.9 on the Walc = 1 axis and Wwater = 1 axis

respectively.

Solution:

partial molar volume of ethanol = 1.1 × molecular weight of ethanol

= 1.1 × 46

= 50.6 cm3 / mole

partial molar volume of water = 0.9 × molecular weight of water

= 0.9 × 18

= 16.2 cm3 / mole

Problem : Two moles of solute is dissolved in 10 moles of solvent whose entropy is 10

units. if the entropy of solvent is 20 apparent molar is

Solution:

φ = S−n1 S1

n2

= 20−(10)(0.2)

10

720 VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE AND

PRESSURE

VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE

(Relationship between chemical potential and partial molar entropy)



The chemical potential is given by μ = (∂ G∂ n

) T, P, n2,n3...

Differentiating with respect to T at constant pressure and number of moles

( ∂ μ∂ T )P , N = ( ∂2G

∂ n ∂T) -----------------1

For an open system, G = H-TS

= (E + PV) – TS [ H = E +PV]

Differentiating

dG = dE + PdV + VdP – TdS – SdT

= dq + VdP – TdS – SdT [ from I law dE + PdV = dq]

= TdS + VdP – TdS – SdT [ from II law dq = TdS]

dG = VdP – SdT ---------------------2

but we know that G = f ( T, P, n1,n2,n3....)

The total differential is given by

dG = (∂ G∂T

) P, n2,n3... dT +( ∂ G∂ P

) T, n2,n3. dP +( ∂ G∂ n 1

) T, P,n2 dn1 +( ∂ G∂ n 2

) T, P,n1 dn2 ---------------3

equating the coefficients of dT in 2 and 3 we get

( ∂ G∂T

) P, n2,n3... = - S

Differentiating both sides with respect to n

( ∂2G∂ T ∂ n

) = - ( ∂ S∂ n

) T, P

Substituting in 1 [ ( ∂2G∂ T ∂ n

) = ¿)]

( ∂ μ∂ T ) P,N = - ( ∂ S

∂ n) T, P

= - S

= - partial molar entropy of component i.

At constant pressure [ ‘∂’ becomes ‘d’ ]

dμdT = - S

dμ = - S dT

This shows that when temperature increases, chemical potential decreases.

liquid



gas

μ solid

Temperature Tm Tb

The graph of chemical potential versus temperature for a substance in solid, liquid

and gaseous state shows that at the melting point the chemical potential of solid and

liquid are same. Similarly at the boiling point, the chemical potential of liquid and

vapour are same.

VARIATION OF CHEMICAL POTENTIAL WITH PRESSURE

(Relationship between chemical potential and partial molar volume)

The chemical potential is given by μi = (∂ G∂ n 1

) T, P, n2,n3...

Differentiating μ with respect to P at constant temperature and number of moles

( ∂ μ∂ P ) T,P = ( ∂2G

∂ n 1∂ P) -----------------1

For an open system, G = H-TS

= E + PV – TS [ H = E +PV]

dG = dE + PdV + VdP – TdS – SdT

= dq + VdP – TdS – SdT [ from I law dE + PdV = dq]

= TdS + VdP – TdS – SdT [ from II law dq = TdS]

dG = VdP – SdT ---------------------2

but we know that G = f ( T, P, n1,n2,n3....)

The total differential is given by

dG = (∂ G∂T

) P, n2,n3... dT +( ∂ G∂ P

) T, n2,n3. dP +( ∂ G∂ n 1

) T, P,dn1+( ∂ G∂ n 2

) T, P, dn2 ------------3

equating the coefficients of dP in 2 and 3 we get

(∂ G∂ P

) T, n2,n3... = V

Differentiating with respect to n, ( ∂2G∂ P ∂n

) = (∂ V∂ n

) T, P



Substituting in 1 [ ( ∂2G∂ T ∂ n

) = ¿

)]

( ∂ μ∂ P ) T,n1,n2.. = ( ∂ V

∂ n) T, P

= V

= partial molar volume of component i.

820 THERMODYNAMICS OF REAL GASES-

920GAS MIXTURE

1020 FUGACITY

It is the pressure term for real gas. It is represented by the letter ‘f ’

Relation between fugacity and free energy:

For a closed system, G = H-TS

= (E + PV) – TS [ H = E +PV]

Differentiating dG = dE + PdV + VdP – TdS – SdT

= dq + VdP – TdS – SdT [ from I law]

= TdS + VdP – TdS – SdT [ from II

law]

dG = VdP – SdT

At constant temperature, ( dG) T = VdP

= nRT dPP [ PV = nRT]



= nRT d(lnP) [ f=P]

= nRT d(lnP)

∫ dG = ∫f 1

f 2

nRT d (lnf ) [Integration between fugacities f1 and

f2 ]

G = nRT ¿¿

= nRT( ln f2 – ln f1)

= nRT ln f 2

f 1

This is the relation between fugacity and free energy:

Problem 1. Calculate the free energy change accompanying the compression of one mole

of a gas at 27 o C, from 20 to 200 atm. The fugacitities at 20 and 200 atm pressure are 50 and

100 atm respectively.

Solution:

Temperature T = 27 o C

= 27 +273

= 300K

Initial fugacity f1 = 50 atm

Final fugacity f2 = 100 atm

free energy change ∆G = ?

The relation between fugacity and free energy∆ G = nRT ln f 2

f 1

∆G = nRT ln f 2

f 1

= (1) (8.314) ( 300) ln 10050

= 2494.2 ln 2

1120DETERMINATION OF FUGACITY

There are two methods

1. GRAPHICAL METHOD

Fugacity is determined using the value of departure from ideal behavior (𝛂)

α = V id - V ------------------1



= RTP - V [ PV = RT]

V = RTP - α

Multiplying by dP

VdP = RTP dP - α dP

dG = VdP at constant temperature. therefore

dG = RTP dP - α dP -----------------------2

the relation between fugacity and dG is given by dG = RT d ln f ---------------------------3

Comparing 2 and 3

RT dlnf = RT dPP – α dP

dividing by RT

dlnf = dPP –

αRT dP

Upon integration

∫ dlnf = ∫ dPP - ∫

0

P αRT

dP

lnf = ln P – 1

RT ∫0

P

α dP

Rearranging

lnf - ln P = – 1

RT ∫0

P

α dP

ln fp = –

1RT ∫

0

P

α dP

The integral in this equation can be evaluated by measuring the area under the

curve obtained when α is plotted against P. In this method, for different pressure value ,

departure from ideal behavior (𝛂) is determined using the formula

α = RTP – V

and a graph is plotted with pressure along x- axis and 𝛂 along y – axis .The area under the

curve gives the value of ∫0

P

αdP using this value, fugacity can be calculated by the formula



ln fp = -

1RT ∫

0

P

αdP

= - 1

RT × area under the curve

2 FROM DENSITY MEASUREMENTS

if 𝛂 is independent of pressure the this method can be used.

ln fp = - ∫

0

P αRT

dP [ from method 1]

= - α

RT ∫0

P

dP [if 𝛂 is independent of pressure ]

= - αPRT

fp = e

−αPRT

= 1 - αPRT [ e− x=1−x ¿

= RT −αP

RT

= PVRT [ α =

RTP – V]

= P

RTV

= P

P id

f = P2

P id

Pid = ρ2

M RT

ρ2

P2 =

ρ1

P1



Problem: Find the fugacity of N2 at 100 K and 100 atm pressure. The area under the curve

obtained when α is plotted against P is 83.14 units

Solution:

area under the curve = 83.14

temperature = 100 K

ln fp = –

1RT ∫

0

P

α dP

= - 1

RT × area under the curve

= - 1

8.314×100 × 83.14

ln fp = - 0.1

Taking exponential on both sides

fp = e−0.1

f = P× e−0.1

f = 100 e−0.1 atm

1220VARIATION OF FUGACITY WITH TEMPERATURE AND PRESSURE

VARIATION OF FUGACITY WITH TEMPERATURE: The relation between free energy change and fugacity is given by

∆G = RT ln f

rearranging

∆ GT = R ln f

∆ G = G2 - G1

G2

T -

G1

T = R lnf

Differentiating with respect to temperature we get



∂(G2

T)

∂ T - ∂(

G1

T)

∂ T =

∂(R ln f )∂ T

From Gibbs – Helmholtz equation we know that ∂( G

T)

∂T = -

HT2

Therefore the above equation becomes,

- H2

T 2 - ( - H 1

T 2 ) = R ∂( ln f )∂ T

- H 2

T 2 + H1

T 2 = R ∂( ln f )

∂ T

H ¿

T 2 ¿ = R ∂(lnf )

∂T

Rearranging

H ¿

R T 2 ¿ = ∂(lnf )

∂T

∂(lnf )∂T

= H ¿

R T 2 ¿

At constant pressure ‘∂’ becomes ‘d’

d (lnf )

dT =

H ¿

R T 2 ¿

d ( lnf) = ( H2-H1 ) dT

R T 2

If the fugacity changes from f1 to f2 during temperature changes from T1 to T2 the

above integration can be done within the limits.

∫f

f 2

d (lnf ) = H1−H 2

R ∫

T 2

T 1

¿¿ ) dT

ln f = H 1−H 2

R [

−1T ] ∫ dT

T2 = −1T

Applying limits

ln f 2 – ln f 1 = H 1−H 2

R [

−1T 1

- (−1T 2

) ]

ln f 2f 1 =

H 1−H 2

R [ -

1T1

+1T2

]



= - H 1−H 2

R [

1T2

- 1T1

]

ln f 2f 1 = -

H 1−H 2

R [

T1−T 2

T 1 T2 ] .

using this expression, variation of fugacity with temperature can be calculated if the fugacity

is known at one temperature

VARIATION OF FUGACITY WITH PRESSURE:

The influence of pressure on the fugacity of gas at constant temperature can be derived as

follows

μ = RT ln f

Differentiating with respect to pressure

(∂ μ∂ P

¿ T = RT ∂lnf∂ p

(∂ μ∂ P

¿ T = V

Substituting in the above equation,

V = RT ∂lnf∂ p

Rearranging ∂lnf∂ p =

VRT

At constant temperature ‘∂’ becomes ‘d’

d lnfdp =

VRT

dlnf = V

RT dP

If the fugacity changes from f1 to f2 during pressure changes from P1 to P2 the

above integration can be done within the limits.

∫f 1

f 2

dlnf = V

RT ∫P 2

P 1

dP

ln f = V

RT [ P ] ∫ dP = P

applying limits

ln f 2 – ln f 1 = V

RT [ P2-P1 ]



ln f 2f 1 =

VRT [ P2-P1 ]

using this expression, variation of fugacity with pressure can be calculated if the fugacity is

known at one pressure

Problem The fugacity of oxygen at 100 K was found to be 95 atm. Calculate its value at 200

K . The increase in molar heat content is 19.8 cal/deg/mol [ given ln 95 = 4.55, e4.55 = 90.0

R = 1.98 cal/mol]

Solution:

The fugacity at 100 K = 95 atm

The fugacity at 200 K = ?

∆H = 19.8 cal/deg/mol

T 1 = 100 K

T 2 = 200 K

ln f 2f 1 = -

H 1−H 2

R [

T1−T 2

T 1 T2 ]

ln f 295 = -

19.81.98 [

200−10020000 ] .

= - 10

200

ln f2 - ln 95 = - 10

200 [ ln (mn ) = ln m – ln n ]

ln f2 = ln 95 - 10

200

= ln 95 -0.05

=4.55 – 0.05

= 4.5

f 2 = e4.5

f2 = 90.0 atm

Problem 5: Calculate the difference in heat content of one mole of oxygen gas at 1 atm.

Pressure and the same gas when behaving ideally at 25 oC.[ given for oxygen Tc = 154.3 K,

Pc = 49.7 atm,

Solution:

P1 = 1atm

T = 25oC = 298 K

R = 1.987 cal/deg/mol



H-H * = 9 ×1.987 ×154.3

128 × 49.7 ( 1 – 18 (154.3)2

(298)2 ) ( -1)

= 1.66 cal /mol

Problem6: The fugacity of oxygen at 100 atm was found to be 95 atm. Calculate its value at 200 atm . The molar volume at 100 K was found to be 0.198 litSolution:

ln f 2f 1 =

VRT [ P2-P1 ]

ln f 295 =

0.1981.98× 100 [ 200 - 100 ]

= 0.1

ln f2 - ln 95 = 0.1 [ ln (mn ) = ln m – ln n ]

ln f2 = ln 95 +0.1 = 4.55 +0.1 = 4.65 f2 =104.58 atm

1320 THERMODYNAMICS OF IDEAL SOLUTIONS

IDEAL SOLUTION: A solution is said to be ideal if it obeys Roault’s law through out the composition range of the system.

Raoult’s law states “ Partial vapour pressure of any constituent of solution is equal to product of mole fraction and its vapour pressure in the pure state.” Partial pressure = mole fraction × vapour pressure p1 = x1 × p0

THERMODYNAMICSa. Free energy change of mixing:

∆ Gmix = RT ∑ x i ln x i

Proof:

Suppose a solution is formed by mixing n1 moles of A and n2 moles of B. The free

energy of the solution is given by

G = n1 GA+¿ n2 GB

Where GA and GB are the partial molal free energies ( chemical potential)of the

constituents.

∆ Gmix = free energy of solution – ( sum of free energies of the constituents)



= G – (n1 GA+¿ n2 GB)

Substituting the value og ‘G’

∆ Gmix = (n1 GA+¿ n2 GB ) – (n1 GA+¿ n2 GB)

= n1 ¿ – GA ¿+¿ n2 ¿) ------------1

chemical potential μ = μ0+ RT ln a

Where ‘a’ is activity

or

GA = GA + RT ln a A

Rearranging

GA - GA = RT ln a A Similarly

GB - GB = RT ln aB

Substituting in the above equation(1)

∆ Gmix = n1RT ln a A+¿ n2RT ln aB

= RT (n1ln a A+¿ n2 ln aB)

If the solution is ideal , activity is equal to mole fraction

∆ Gmix = RT (n1ln x A+¿ n2 ln xB)

Dividing by ¿¿ n2)

∆ Gmix

(n1+n2) = RT (

n1

(n1+n2) ln x A+¿

n2

(n1+n2) ln xB)

Since x A = n1

(n1+n2) , xB =

n2

(n1+n2)

∆ Gmix

(n1+n2) = RT ( x Aln x A+¿ xB ln xB)

If the total amount of the two components is one (n¿¿1+n2)¿ = 1


b. Volume change of mixing:

∆ V mix = 0Proof: The change in chemical potential with pressure is equal to molar volumed μd P = V

∆ Gmix = ∆ V mix

Differentiating the ∆ Gmix with respect to pressure we get



∂(∆ Gmix)∂ P

= 0

Therefore ∆ V mix = 0c. Enthalpy change of mixing:

∆ Hmix = 0

Proof:


Dividing by ‘T’

∆ Gmix

T = R ∑ x i ln x i -------------2

Differentiating the ( ∆ Gmix

T¿with respect to Temperature we get

ddT (∆ Gmix

T )= 0 -------------3

Differentiating the ∆ Gmix

Tby U / V model we get

d ( UV ) =

V dUdT

−U dVdT

V 2

ddT (∆ Gmix

T ) = T d

dT (∆ Gmix )−∆ Gmix (1 )

T 2 ----------4

Comparing 3 and 4

T ddT ( ∆Gmix )−∆ Gmix = 0

T ddT ( ∆Gmix )=∆ Gmix ----------------- 5

Applying Gibbs – Helmholtz equation ∆ G - ∆ H = T ∂(∆ G)∂ T

Introducing the subscript ‘mix’

∆ Gmix - ∆ Hmix = Td

dT ( ∆Gmix )

Substituting from 5 ∆ Gmix - ∆ Hmix = ∆ Gmix

Therefore ∆ Hmix = 0



1420 NON IDEAL BINARY SOLUTIONS

Ideal solution : benzene and toluene, n-heptane and n- hexane , ethyl bromide and ethyl

iodide, chloro benzene and bromo benzene

For ideal solution ∆ V mix = 0 For non - ideal solution ∆ V mix ≠ 0

A solution is said to be non- ideal if it does not obey Roault’s law

Raoult’s law states “ Partial vapour pressure of any constituent of solution is equal to product of mole fraction and its vapour pressure in the pure state.” Partial pressure = mole fraction × vapour pressure p1 = x1 × p0

Example : CS2 and acetone , acetone and chloroform

1520 DILUTE SOLUTIONS

A dilute solution may be defined as one in which solvent obeys Raoult’s law and

the solute satisfies Henry’s law.

According to Henry’s law f2 = K x2 where K is Henry’s constant

The properties of dilute solution are called colligative properties. These are

1. Lowering of vapour pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

1620 EXCESS FUNCTIONS FOR NON - IDEAL SOLUTIONS AND THEIR

DETERMINATION

EXCESS FUNCTION: It is defined as the difference between the partial molar property of a

component in a real mixture and that of the component in an ideal mixture.

If X is any component in the mixture, then

Excess Function = (X) real mixture - (X) ideal mixture

EXCESS QUANTITIES:

1. Excess molar volume

2. . Excess molar enthalpy

3. . Excess heat capacity



4. . Excess chemical potential

1720 CONCEPTS OF ACTIVITY AND ACTIVITY COEFFICIENTS

ACTIVITY

Activity of a substance in any given state is defined as the ratio of the fugacity of the substance in the mixture to the fugacity of the same substance in the pure state.

Activity = fugacity∈mixture

fugacity∈ pure state

a = f 1

f 0

Problem 7: The fugacity of a pure gas is 20 atm and the same gas in a mixture is 40 atm . Calculate the activity of the gas.

Solution:



a = 4020

= 2

Activity coefficient:

For real gas the activity is proportional to pressure a ∝ P

a = 𝛄 P

𝛄 is called activity coefficient.

Activity coefficient can be defined as the ratio between activity and pressure of the real gas.

activity coefficient 𝛄 = activitypressure

Problem 8: The activity of a gas at 20 atm pressure is 5 . What is its activity coefficient?

Solution:

Activity coefficient 𝛄 = activitypressure

= 5

20

= 0.25



Problem 9. The fugacity of a pure gas is 5 atm and the same gas in a mixture is 20 atm

Calculate the activity coefficient of the gas at 40 atm pressure

Solution:

Activity coefficient = activitypressure



a = 205

= 4

γ = 4

40

= 0.1

1820 DETERMINATION OF STANDARD FREE ENERGIES.

There are three methods

1. From equilibrium constant

∆ G0 = RT ln K eqm

2. From EMF measurements

∆ G0 = - nE0 F

3. From entropy measurements

∆ G0 = ∆ H 0 - T ∆ S0

∆ S0 can be calculated from III law of thermodynamics using heat capacity

measurements



1920 CHOICE OF STANDARD STATES

1. FOR GASES :

1. Since the fugacities of real gases can be measured, the standard sate can be fixed

in terms of fugacity.

2. The fugacity corresponding to 1 atm, pressure is taken as standard state .

2. COMPONENTS IN SOLUTION:

The standard state for solvent is based on Raoult’s law and that for solute is based

on Henry’s law

A. SOLVENT:

1. The standard state for solvent is chosen as a total pressure of 1 atm.

Let f 0 be the fugacity at 1atm. pressure. f 0 = p

On addition of solute the fugacity changes in to f 1

activity is given by a = f 1

f 0

According to Raoult’s law f1 = fo x1

substituting in the above equation we get

a = f o x1

f 0

= x1

3. The activity of solvent is equal to its mole fraction.

4. For a non ideal solution, f 0 ≠ p



B. SOLUTE:

1. If the solute and solvent are completely miscible in all proportions, the standard

state of solute is taken in the same manner as for the solvent.

2. When the solute has a limited solubility , two systems are used.

1. RATIONAL SYSTEM:

1. If the composition is expressed in mole fraction , the system is known as rational

system.

2. Here the standard state is the fugacity , at which the mole fraction of the solute is unity.

3. At this stage it behaves ideally and obeys Henry’s law.

4. According to Henry’s law f2 = K x2 where K is Henry’s constant

5.At the standard state, (x2 = 1) let f 2 = f 0 then ,the above equation becomes f 0 = K

substituting in the above equation we get

a = Kx 2K

= x2

2 PRACTICAL SYSTEM:

1. If the composition is expressed in molarity or molality , the system is known as

practical system.

2. Under these conditions, Henrys law becomes f2 = K m2

3. Choice standard fugacity is such that m2 → 0 , a2

m2 → 1



Lim [a2

m2 ] = 1

4. If the activity is plotted against mole fraction of the solute , standard state is the state at

which m2 = 1

2020 DETERMINATION OF ACTIVITY AND ACTIVITY COEFFICIENTS FOR

NON-ELECTROLYTES

By Nernst distribution law

Nernst distribution law states that when a solute is distributed two immiscible solvents, the ratio of concentrations of the solute in solvent – 1 and solvent -2 is a constant.

Let a solute be distributed between a pair of two immiscible solvents A and B . The molalities of solutes in the two solvents A and B are mA and mB respectively. For a system in equilibrium fA = fB . From the definition of activity



Let aA and aB represents the activity of the solute in solvent A and B respectively. Then

aA = f A

f A0 , aB =

f B

f B0

Dividing

aA

aB =

f B0

f A0

Therefore aA = aB × f B

0

f A0 ------------------------------------------1

f B0 = mB

f A0 = mA

Substituting in equation 1 we get



aA = aB × mB

mA ----------------2

According to Henry’s law

f A = K A mA

f B = K BmB

At equilibrium f A = f B , Therefore

K A mA = K BmB

Rearranging

mB

mA =

K A

K B

Substituting in 2 , it becomes

aA = aB ×K A

K B where

K A

K B ratio of Henry’s law

The values of Henry’s law constant is given by K A

K B = lim

mB → 0

mA

mB

Substituting in the above equation we get aA = aB × limmB → 0

m A

mB

The limiting value of the equation is evaluated by plotting mA

mB against mB and extrapolating

the graph to zero concentration.

limmB → 0

mA

mB

From the graph limmB → 0

mA

mB can be found out . From which we can determine the

activity of the solute by knowing the activity of the same solute in solvent B( mB)

activity coefficient 𝛄m = a2

x2



= p2

x2 ×

p2¿

x2¿

To find p2

¿

x2¿ , a graph is plotted by

p2

x2 versus x2 and the straight line obtained is

extrapolated to

x2 = 0. This gives the value of ‘k’.

p2 can be obtained from the graph for the particular mole fraction. substituting in 2 we can

calculate the activity of solute.

Problem 10 : Mercurous chloride is distributed in benzene and water. The concentration in

benzene and water are given below.

Aqueous layer 0.2 0.1 0.05 0.03 0.01

Benzene layer 0.02 0.01 0.004 0.00

2

?

The plot of concentration ratio against concentration in benzene gives an intercept of

12.2. Plot the data and Calculate the activity of mercurous chloride in water when the

concentration in benzene is 0.01.

Solution:

limmB → 0

mA

mB = 12.2

aB = 0.01

aA = aB × limmB → 0

mA

mB

= 0.01 × 12.2 = 0.122

Problem: 0.5 molal aqueous solution of non – volatile solute has a vapour pressure of 15.5

mm of mercury. At the same temperature, the vapour pressure of pure water is 31.0 mm of

mercury. Find the activity of and activity coefficient of water

Solution:



a = 15.531

= 0.5

activity coefficient 𝛄m = am

= 0.5

100018

= 0.5

55.5

= 0.009

activity coefficient 𝛄x = ax

= 0.5

100018

+0.5

= 0.556

= 0.05

Problem: In an experiment ,the partial pressure of methanol in aqueous solution at various

mole fraction was found out. The plot of p2

x2 versus x2 on extrapolation to x2 = 0, gives a

value of 180 . At a mole fraction of 0.6 , the p2

x2 value from the graph was found to be 15.

Calculate the activity and activity coefficient of methanol

Solution:

p2¿

x2¿ = 180

p2 corresponds to 0.6 = p2

x2 × x2

= 15 × 0. 6

= 9

a2 = p2

p2¿

x2¿

= 9

180



= 0. 05

activity coefficient 𝛄m = a2

x2

= 0.050.6

= 0.080

Problem One molal solution of sucrose has a vapour pressure of 20.5 mm of mercury at

100K. At the same temperature the vapour pressure of pure solvent is 20.8 mm. calculate

the activity and activity coefficient of solvent in the given solution.

a = p0

p1

=20.520.8

= 0.98



UNIT II SPECTROSCOPY – II

A. Rotational spectroscopy

1. of a rigid rotar

2. non-rigid rotor

3. diatomic

4. polyatomic molecules.

B. Vibrational spectroscopy-

1. harmonic oscillator-

2. anhormonicity

3. –Vibration – rotation spectra of diatomic vibrating molecules

4. selection rules-

5. P,Q and R branches.

6. Vibrational spectra of polyatomic molecules-

7. fundamental vibrations

8. normal modes of vibration

9. overtones,

10. combination and difference bands-

11. Fermi resonance.

C. Raman spectra:

1. Classical theory of Raman effect

2. molecular polarisability

3. pure rotational Raman spectra

4. Vibrational Raman spectra

5. Rotational fine structure



6. Rule of mutual exclusion

7. Polarization of light and Raman effect.

A. ROTATIONAL SPECTROSCOPY ( MICROWAVE SPECTRA)

These spectra result from transition between the rotational energy levels of a gaseous

molecules, which posses a permanent dipole moment on the absorption of radiation falling

in the microwave region.

RADIO WAVE REGION

MICROW WAVE REGION

IR REGION

VISIBLEREGION

UVREGION

X-RAYREGION

wavelength 1-1000 m 1-100 cm 1-𝛍m – 1 m 380- 780 nm 1-380 nm 0.1-10 nm

Thus these spectra are shown by molecules possessing a permanent dipole moment

example: like HCl, CO, NO and H2O vapour. Homonuclear diatomic molecules such as

H2,Cl2 O2, N2 etc and linear poly atomic molecules such as CO2 do not posses dipole

moment and hence they do not show microwave spectra.

14 ROTATIONAL SPECTROSCOPY OF A RIGID ROTAR

The system consisting of two spherical particles attached together, separated by finite

fixed distance and capable of rotating about an axis passing through the centre of mass

constitutes the rigid rotator.

A diatomic molecule consists of two masses bound together. The distance between

the masses, or the bond length, (l) can be considered fixed because the level of vibration in

the bond is small compared to the bond length.



Figure 1.11.1: Rigid Rotor Model of a Diatomic Molecule

Reduced Mass

The system can be simplified using the concept of reduced mass which allows it to be treated

as one rotating body. The system can be entirely described by the fixed distance between the

two masses instead of their individual radii of rotation.

RELATION BETWEEN MOMENT OF INERTIA AND INTER NUCLEAR

DISTANCE

Consider a rotating molecule as rigid rotator with free axis. Let m1 and m2 be

the masses of the atoms connected by a rigid bar of length r

r = r1 + r2

r1 r2

r

This molecule rotates end over end about a point C . This is defined by balancing equation

m1r1 = m2r2 ------------------------------------------------------1

The moment of inertia about C is given by

I = m1r12 + m2r2 2

= ( m1r1 ) r1 + (m2r2 ) r2 [ splitting the square terms]

= (m2r2 ) r1 + (m1r1) r2 [ using 1]

= r1 r2 (m1 + m2 )

from 1b m1r1 = m2r2

m1r1 = m2 ( r - r1 ) [ from 1 r2 = r - r1 ]

= m2 r - m2 r1


r1


m1r1 + m2 r1 = m2 r [ rearranging]

r1 ( m1 + m2 ) = m2 r

∴ r1 = m2 r

m1+m 2

Similarly, m2r2 = m1r1 [ from 2]

m2r2 = m1 ( r – r2 ) [ from 1 r1 = r - r2 ]

= m1 r - m1 r2

m2r2 + m1 r2 = m1 r [ rearranging]

r2 ( m1 + m2 ) = m1 r

∴ r2 = m1 r

m1+m 2

Substituting the values of r1 and r2 in equation 1

I = (m2 r

m1+m 2) (

m1 rm1+m 2

¿ (m1 + m2 )

= (m1 m2

m1+m 2) r2

I = μ r2 where μ is reduced mass μ = m1m2

m1+m 2

This shows, the relation between atomic masses and the bond length.

Q: Derive the relation between atomic mass and bond length of a diatomic molecule.

EXPRESSION FOR ROTATIONAL ENERGY LEVEL:

The expression for rotational energy is obtained by solving the Schrodinger equation

Schrodinger equation in Cartesian co-ordinates is ∂ 2Ψ∂ x2 +

∂ 2Ψ∂ y 2 +

∂ 2Ψ∂ z 2 +

8π 2mh2 ( E – V)

Ψ = 0

This can be converted into spherical polar co-ordinates using x = r sin θ cosφ

y = r sin θ sinφ z = r cos θ and the transformed equation is

1r2

∂∂ r ( r2

∂Ψ∂ r ) +

1r2 sinθ

∂

∂θ ( sin θ ∂Ψ∂ θ ) +

1r2sin2θ

∂ 2Ψ∂ φ 2 +8π 2m E

h2 Ψ = 0

In this equation the wave function Ψ is a function of r, θ,φ

For a rigid rotator r = 1 and potential energy is zero

1

r2 sinθ

∂∂θ ( sin θ

∂Ψ∂ θ ) +

1r2sin 2θ

∂ 2Ψ∂ φ 2 +8 π 2m E

h2 Ψ = 0

1

sinθ ∂

∂θ ( sin θ ∂Ψ∂ θ ) +

1sin2θ

∂ 2Ψ∂ φ 2 +8π 2m E

h2 Ψ = 0



1

sinθ ∂

∂θ ( sin θ ∂Ψ∂ θ ) +

1sin2θ

∂ 2Ψ∂ φ 2 +βΨ = 0 where β = 8π 2m E

h2

1

sinθ [ sin θ ∂2Ψ∂ θ2 +

∂Ψ∂ θ cos θ ]+

1sin2θ

∂ 2Ψ∂ φ 2 +βΨ = 0

∂2Ψ

∂ θ2 + cosθsin θ

∂Ψ∂ θ +

1sin2θ

∂ 2Ψ∂ φ 2 +βΨ = 0

This equation can be separated into two equations , each involving a single independent

variable. and solving these we get

sin2 θ d2 Xdu2 -2 cos θ

dXdu + ( β - m2

sin2θ¿ X = 0

( 1- u2) d2 X

d u2 -2u dXdu + ( β - m2

(1−u2)¿ X = 0 ----------------1 where u = cosθ

This resembles associated Legendre differential equation which is

( 1-x2)∂2 y∂ x2 - 2x

dy∂ x + ( J (J+1) - m2

(1−x 2)¿ X = 0 ---------------------2

comparing 1 and 2 we get

J(J + 1 ) = β

8 π 2m Eh2 = J(J + 1 ) [ β = 8 π 2m E

h2 ]

∴ E = h2

8π 2m J (J+1)

This is the expression for rotational energy level

To find the allowed Energy Levels:


The rotational energy is E = h2

8 π 2 IJ(J+1) joules ,

In terms of wave number ∈ = Ehc cm -1 [ E = hϑ,

= h2

8 π2 IJ (J+1)

hc cm -1

= h

8 π 2 I cJ(J+1) cm -1

∈ = B J(J+1) cm -1 where B is rotational constant B =

h8 π 2 I c



when J = 0 , ∈ = 0 (0+1)

= 0

when J = 1 , ∈ = B ( 1+1)

= 2B cm -1

when J = 2 ∈ = 2B ( 2+1)

= 6B cm -1 ,

When J = 3 , ∈ = 12B, When J = 4 , ∈ = 20B

Thus the allowed rotational energy levels are 0. 2B, 6B,12B,20B

To find rotational transitions:(Selection rule):

All transitions in which J changes by one unit are allowed transitions( intense) and

other transitions are forbidden( weak)

∆ J = ± 1 Where J represents Rotational quantum

number. ∆ J = +¿ 1 corresponds to absorption spectrum while ∆ J = −¿1 corresponds

to emission spectrum

If the molecule is raised from the J= 0 to J = 1 by absorbing energy

ϑ J=0 → J = 1 = 2B ( 0+1) cm-1

= 2B cm -1

an absorption line will appear at 2B cm -1

If now the molecule is raised from the J= 1 to J = 2 by absorbing more energy

ϑ J=1 → J = 2 = 2B ( 1+1) cm-

= 4B cm -1 ,

an absorption line will appear at 4B cm -1

To raise the molecule from the state J to J+1, we would have

ϑ J → J+1 = ϑ J+1 - ϑ J

= B (J+1)(J+2) – B J (J+1)

= B (J+1) ( J+2- J) [ taking common factor

outside]

= 2B ( J+1) cm-1



This shows that the distance between the successive spectral line is same which is equal to

2B

Problem-.Calculate the bond length in CO if the atomic mass of C is 12.000, and that of O is

15.999. The moment of inertia of CO is 14.5695 kg m2

Solution:

Moment of inertia I = μ r 02 where μ is reduced mass

∴ r0 = √Iμ

I = 14.5695 kg m2

Reduced mass μ = m1 m2

m1+m 2

= (12×15.999)

12+15.999

= 191.38827.999

= 6. 857

r0 = 14.5695

6.857 = 2. 1247

Problem : The pure rotational spectrum of CN consists of a series of equally spaced

lines separated by 3. 8 cm -1. Calculate the inter nuclear distance of the molecule. reduced

mass = 1.07 × 10 -26 Kg



Data:

μ = 1.07 × 10 -26 Kg

spacing between the lines 2B = 3.8 cm -1.

∴ B = 1.9 = h

8 π 2 I c

I = h

8 π 2 B c

= (6.62 ×10−34)

8 × (3.14 )2 (1.9 )(3 ×1010)

= 1.47 × 10 -46 Kg m2

I = μ r 2

∴ r = √ Iμ

= 117 pm

Problem- The bond length of NO molecule is 1.151 × 10 -10 m. Calculate the frequency in cm

-1 for the pure rotational levels in the spectrum of NO corresponding to the following

changes in the rotational quantum number 0 →1, 1 →2 , and 2→ 3

Reduced mass μ = m1 m2

m1+m 2

= 14+1614 ×16 ×

16.023× 1023

= 1.24 × 10 -26 Kg

Moment of inertia I = μ r 2

= (1.24 × 10 -26 ) ( 1.151 × 10 -10 ) 2

= 1.426 × 10 -46 Kg m2

Rotational constant B = h

8 π 2 I c

= 6.62× 10−34

8 ×3.141× 3.1411.426 × 10−46× 3 ×108 × 10 -2 cm -1

= 1.956 cm -1

To obtain the frequency

ϑ J → J+1 = 2B ( J+1) cm-1

ϑ 0 → 1 = ( 2 ) ( 1.956) ( 0 + 1)

= 3.912 cm-1



ϑ 1→ 2 = ( 2 ) ( 1.956) ( 1 + 1)

= 7.824 cm-1

ϑ 2 → 3 = ( 2 ) ( 1.956) ( 2 + 1)

= 11.736 cm-1

Frequency separation:

The rotational spectrum of a rigid diatomic molecule consists of a series of lines at 2B,

4B, 6B, 8B etc . these lines are equally spaced by an amount of 2B , called frequency

separation.

Problem The reduced mass of CO, whose bond length 1.13 A , is 1.14 × 10 -26 Kg.

Calculate the energy of the molecule in the first excited state.

Data: r = 1.13 A

μ = 1.14 × 10 -26 Kg

J= 1

E = h2

8π 2 IJ(J+1) joules ,

I = μ r 2

= (1.14 × 10 -26 Kg ) ( 1.13 × 10 -10 m )2

= 1.46 × 10 -46 Kg –m2

E = (6.62 ×10−34)2

8 × (3.14 )2¿¿ [ 1( 1+1)]

= 7.61 × 10 -23 joule

24 ROTATIONAL SPECTROSCOPY OF NON RIGID ROTOR

The non – rigid rotator is one in which

1.The bond between the atom is elastic ( non- rigid)

2.The distance between the atoms is not constant.

Consider a diatomic molecule as non - rigid rotator. A rotating molecule having a

permanent dipole moment generates an electric field which interacts with the electric

component of the Microwave.

TO FIND THE ALLOWED ENERGY LEVELS:




The rotational energy is given by E = BJ (J+1) - DJ2 (J+1)2

D is called centrifugal distortion constant D = h3

32 π4 I 2r2 kc cm -1`

when J = 0 , ∈ = 0

when J = 1 ∈ = B(1) ( 1+1) - D (1)2 (1+1)2

= 2B - 4D

when J = 2 ∈ = B(2) ( 2+1) - D (2)2 (2+1)2

= 6B - 36D

when J = 3 ∈ = B(3) ( 3+1) - D (3)2 (3+1)2

=126B - 108D

Thus the allowed rotational energy levels are 2B-4D 6B-36D,12B -108D

To raise the molecule from the state J to J+1, we would have

E J → J+1 = E J+1 - E J

= [ B(J+1) (J+1+1 ) – D(J+1) 2 (J+1+1)2 ] - [ B J (J+1) - DJ2 (J+1)2 ]

= [ B(J+1) (J+2 ) – D(J+1) 2 (J+2)2 ] - [ B J (J+1) - DJ2 (J+1)2 ]

= [ B(J+1) (J+2 ) – B J (J+1) ] – [ D(J+1) 2 (J+2)2 - DJ2 (J+1)2 ]

[ grouping]

= B(J+1) [ (J+2) – J ] – [ D(J+1) 2 [ (J+2)2 -J2 ] [ taking common factor

outside]

= B(J+1) [2 ] – [ D(J+1) 2 [ J2 + 4J +4 -J2]

= 2B(J+1) – [ D(J+1) 2 [4J +4 ]

= 2B(J+1) – [ D(J+1) 2 4(J +1) ]

= 2B(J+1) – 4D(J+1)3



This shows that at higher J value the rotational energy levels have lesser energy than that of

rigid

rotator.

Selection rule :

All transitions in which J changes by one unit are allowed transitions( intense) and

other transitions are forbidden( weak) ∆ J = ± 1 Where J represents Rotational

quantum number. The spectrum of non- rigid rotator is similar to that of rigid molecule ,

except that each spectral line will undergo displacement towards low frequency value.

34 ROTATIONAL SPECTROSCOPY OF DIATOMIC MOLECULES

44 ROTATIONAL SPECTROSCOPY OF POLY ATOMIC MOLECULES

1. LINEAR POLY ATOMIC MOLECULES

Example Carbon oxy sulphide (COS ) , HCN,

1. It can be treated as diatomic molecules

2. The transition from one rotational state to other is governed by



a. ∆E = hϑ ( Bohr quantum condition)

b. ∆J =± 1

c. molecule must have permanent diplole moment

To find the allowed Energy Levels:


The rotational energy is E = h2


In terms of wave number ∈ = Ehc cm -1 [ E = hϑ,

= h2

8 π2 IJ (J+1)

hc cm -1

= h

8 π 2 I cJ(J+1) cm -1

∈ = B J(J+1) cm -1 where B is rotational constant B =

h8 π 2 I c

3. The rotational spectra of linear poly atomic molecule consists of set of equally spaced

lines.

The spacing between the spectrum gives 2B value from which B can be found out.

Rotational constant B = h

8 π 2 I c

4.For a molecule with ‘n’ atoms, there are (n-1) unknown internuclear distances

For a triatomic molecule there are 2 bond distances to be determined. but from the spectrum,

only one moment of inertia can be obtained. Therefore microwave spectra of (n-1) different

isotopic species are recorded.

TYPES OF ROTATIONS

There are three types of rotation

1. Rotation about bond axis( moment of inertia Ia )

2. End-over- end on the plane (moment of inertia Ib)

3. End-over- end at right angle to the plane (moment of inertia Ic)

For Linear molecules

1.Mmoment of inertia about the molecular axis is zero Ia = 0

2. Ib = Ic

3.Example : CO,HCl, HCN, N2O,C2H2, OCS



2. NON- LINEAR POLY ATOMIC MOLECULES

Classification of molecules based on inertia moments

Depending on the relative size of the inertia moments, rotors can de divided into four classes.

1.Linear rotors

For a linear molecule IA << IB = IC.

For most purposes IA can taken to be zero. For a linear molecule the separation of lines in the rotational spectrum can be related directly to the moment of inertia of the molecule. Since the moment of inertia is quadratic in the bond lengths, the microwave spectrum yields the bond lengths directly, provided the atomic masses are known.

Examples of linear molecules are obviously the diatomics such as oxygen (O=O), carbon monoxide (C≡O), and nitrogen (N≡N). But also many triatoms are linear: carbon dioxide (O=C=O), hydrogen cyanide (HC≡N), and carbonyl sulfide (O=C=S). Examples of larger linear molecules arechloroethyne (HC≡CCl), and acetylene (HC≡CH).

2.]Symmetric tops

A symmetric top is a rotor in which two moments of inertia are the same.

There are two classes of symmetric tops, oblate symmetric tops (frisbee or disc shaped) with IA = IB < IC and prolate symmetric tops (cigar shaped) with IA < IB = IC.

Symmetric tops have a three-fold or higher rotational symmetry axis.

Examples of oblate symmetric tops are: benzene (C6H6), cyclobutadiene (C4H4), and ammonia (NH3). Prolate tops are: chloroform(CHCl3) and methylacetylene (CH3C≡CH).

3.Spherical tops

A spherical top molecule is a special case of a symmetric tops with equal moment of inertia about all three axes IA = IB = IC.

The spherical top molecules have cubic symmetry.

Examples of spherical tops are: methane (CH4), phosphorus tetramer (P4), carbon tetrachloride (CCl4), ammonium ion (NH4

+), and uranium hexafluoride (UF6).


http://knowino.org/w/index.php?title=Uranium_hexafluoride&action=edit&redlink=1

http://knowino.org/w/index.php?title=Ammonium_ion&action=edit&redlink=1

http://knowino.org/w/index.php?title=Carbon_tetrachloride&action=edit&redlink=1

http://knowino.org/w/index.php?title=Carbon_tetrachloride&action=edit&redlink=1

http://knowino.org/w/index.php?title=Phosphorus_tetramer&action=edit&redlink=1

http://knowino.org/w/index.php?title=Methane&action=edit&redlink=1

http://knowino.org/w/index.php?title=Cubic_symmetry&action=edit&redlink=1

http://knowino.org/w/index.php?title=Methylacetylene&action=edit&redlink=1

http://knowino.org/w/index.php?title=Chloroform&action=edit&redlink=1

http://knowino.org/w/index.php?title=Ammonia&action=edit&redlink=1

http://knowino.org/w/index.php?title=Cyclobutadiene&action=edit&redlink=1

http://knowino.org/w/index.php?title=Benzene&action=edit&redlink=1

http://knowino.org/w/index.php?title=Acetylene&action=edit&redlink=1

http://knowino.org/w/index.php?title=Chloroethyne&action=edit&redlink=1

http://knowino.org/w/index.php?title=Carbonyl_sulfide&action=edit&redlink=1

http://knowino.org/w/index.php?title=Hydrogen_cyanide&action=edit&redlink=1

http://knowino.org/w/index.php?title=Carbon_dioxide&action=edit&redlink=1

http://knowino.org/w/index.php?title=Carbon_dioxide&action=edit&redlink=1

http://knowino.org/wiki/Nitrogen

http://knowino.org/w/index.php?title=Carbon_monoxide&action=edit&redlink=1

http://knowino.org/w/index.php?title=Carbon_monoxide&action=edit&redlink=1

https://m.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Oxygen.html

http://knowino.org/w/index.php?title=Bond_length&action=edit&redlink=1


4.Asymmetric tops

A rotor is an asymmetric top if all three moments of inertia are different.

Most of the larger molecules are asymmetric tops. For such molecules a simple interpretation of the microwave spectrum usually is not possible.

.

Examples of asymmetric tops: anthracene (C14H10), water (H2O), and nitrogen dioxide (NO2).

On the basis of moment of inertia of molecules, non – linear molecules are classified as

1. Spherical top molecules: ( Ia = Ib = Ic)

1. In these molecules all the three moment of inertia are equal.

2. As these molecules are not polar, no rotational spectrum will be obtained.

3. Example CH4, CCl4 , SF6

2.. Symmetric top molecules:

1. These molecules have equal moments of inertia about two axis of rotation while the

moment of inertia about the remaining axis is not equal to either of the other two.

2. If Ia = Ib < Ic the molecules are said to be prolate

Example: CHCl3, CHCI3, CH3F

3 . If Ia = Ib > Ic the molecules are said to be oblate

Example NH3,BF3,

3. Asymmetric top molecules

All the three moments of inertia are different ( Ia ≠ Ib ≠ Ic )

Example H2O,SO2,

Differences between Micro wave spectra and IR spectra

Microwave spectra IR

1 characteristics of the molecule as a

whole

IR is of functional group of absorbing molecule

2 The substance must be gas for IR it may be solid ,liquid or gas

3 It is absorption spectra IR is absorption or emission

Limitations of Micro wave spectra

1. Molecule must have permanent dipole moment.


http://knowino.org/w/index.php?title=Nitrogen_dioxide&action=edit&redlink=1

https://m.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Water.html

http://knowino.org/w/index.php?title=Anthracene&action=edit&redlink=1


2. Sample should be gas

B. VIBRATIONAL SPECTROSCOPY-

111 HARMONIC OSCILLATOR

The vibratory motion of a diatomic molecule is similar to that of vibration of

harmonic oscillator. In such an oscillator restoring force (the force tending to restore the

atom to its original state ), is directly proportional to displacement. of the atom. If F is the

restoring force and x’ is the displacement of the atom then

F ∝ x

F = - k x here ‘k’ is known as force constant. This is known as Hooks law

EXPRESSION FOR ENERGY LEVELS:

The potential energy of such system is k x2

2

.Energy of such system is obtained by solving Schrodinger equation

E = hγ ( v + 12)

Where v – vibrational quantum number and γ – vibrational frequency.

To convert the energy from joules to cm – 1

G(v) = Ehc

= (v+½)hγ

hc

= ( v+ ½ ) we

Where we is called equilibrium vibrational frequency.

1. The lowest energy of the oscillator , is called ground state energy or zero point energy and

is obtained by putting n = 0, in the above expression.

E0 = 12 h γ E1 =

32 h γ,

E2 = 52 h γ, E3 =

72 h γ

3. The difference in energies of successive energy levels are E1 – E0 = h γ ,



E2 – E1 = h γ ,

This shows the successive energy levels are equally spaced. The separation between

two adjacent energy levels being hγ.

Diagram:

The vibrational frequency is given by γ = 1

2 πc (kμ) ½ cm -1

where μ is reduced masses and k is force constant

selection rule

The selection rule for vibrational transition in the SHO is dv = ± 1

i.e the vibrational quantum number changes by unity.

vibrational transition

Using the selection rule , the frequency of vibrational transition is given by

∆γ = G ( v → v+ 1 )

= ( v + 1 + ½ ) we - ( v+ ½ ) we

= we

This transition from v= 0 to v= 1 is called fundamental vibrational frequency.



211 ANHORMONICITY

PURE VIBRATIONAL SPECTROSCOPY -ANHORMONIC OSCILLATOR

Real molecules do not obey Hooks law because the bonds in the real molecule is

considered to be perfectly elastic. So they can be considered as anharmonic oscillator. In

such a case the potential energy is given by

V = De ( 1- eβx ) 2

Where De is dissociation energy and β is rotational constant.

This is known as Morse equation. Energy of such system is obtained by solving

Schrodinger equation.

E = hγ ( v + 12) - hγ ( v +

12) 2 xe

Where v – vibrational quantum number and γ – vibrational frequency. xe is anharmonicity

constant.

selection rule

The selection rule for the transition in the anharmonic oscillator is

dv = ± 1, ± 2, ± 3,….

The transition frequency is given by

∆γ = v [ 1- (v+1) xe ] cm – 1

where ‘v’ represents the vibrational quantum number of the final state.



The transition from v= 0 to v= 1 is called fundamental vibrational frequency.

∆γ = ± 1

∆γ = (1) ( 1- (1+1) xe ) cm – 1

= ( 1- 2 xe ) cm – 1

311 VIBRATION – ROTATION SPECTRA OF DIATOMIC VIBRATING

MOLECULES

Rotational–vibrational spectroscopy is concerned with infrared and Raman

spectra of molecules in the gas phase.

Transitions involve changes in both vibrational and rotational states

When such transitions emit or absorb photons , the frequency is proportional to the difference

in energy levels

Since changes in rotational energy levels are typically much smaller than changes in

vibrational energy levels, changes in rotational state are said to give fine structure to the

vibrational spectrum.

For a given vibrational transition, the same theoretical treatment as for pure rotational

spectroscopy gives the rotational quantum numbers, energy levels, and selection rules.

In linear and spherical top molecules, rotational lines are found as simple progressions at both

higher and lower frequencies relative to the pure vibration frequency.

In symmetric top molecules the transitions are classified as parallel when the dipole

moment change is parallel to the principal axis of rotation, and perpendicular when the

change is perpendicular to that axis.

411SELECTION RULES-

The selection rule has two consequences.

1. Both the vibrational and rotational quantum numbers must change. The transition :∆ v =± 1, ∆J = 0 (Q-branch) is forbidden

2. The energy change of rotation can be either subtracted from or added to the energy change of vibration, giving the P- and R- branches of the spectrum, respectively.

selection rule is ∆ v =± 1, ∆J = ±1


https://en.wikipedia.org/wiki/Molecular_dipole_moment

https://en.wikipedia.org/wiki/Molecular_dipole_moment

https://en.wikipedia.org/wiki/Selection_rule

https://en.wikipedia.org/wiki/Quantum_number

https://en.wikipedia.org/wiki/Rotational_spectroscopy

https://en.wikipedia.org/wiki/Rotational_spectroscopy

https://en.wikipedia.org/wiki/Energy_level

https://en.wikipedia.org/wiki/Frequency

https://en.wikipedia.org/wiki/Photon

https://en.wikipedia.org/wiki/Quantum_state

https://en.wikipedia.org/wiki/Rotational_transition

https://en.wikipedia.org/wiki/Molecular_vibration

https://en.wikipedia.org/wiki/Gas_phase

https://en.wikipedia.org/wiki/Molecule

https://en.wikipedia.org/wiki/Raman_spectroscopy

https://en.wikipedia.org/wiki/Raman_spectroscopy

https://en.wikipedia.org/wiki/Infrared_spectroscopy


511P,Q AND R BRANCHES IN VIBRATIONAL SPECTRA

VIBRATION – ROTATION SPECTRA OF DIATOMIC MOLECULES

When a molecule absorbs radiation, which is enough to cause change in vibrational

energy , its rotational energy may also change..Hence, the absorption band contains , number

of closely spaced lines . Such type of spectrum is referred to as vibration- rotation

spectrum.

The rotational energy is given by E = BJ (J+1) - DJ2 (J+1)2

where B is rotational constant B = h

8 π 2 I c

Vibrational energy is given by E = hγ ( v + 12) - hγ ( v +

12) 2 xe

Where v – vibrational quantum number

γ – vibrational frequency.

xe - anharmonicity constant

The total energy of the vibrating rotator is given by the sum of the vibrational and

rotational energies. Etotal = Evib+ Erot

= hγ ( v + 12) xe + BJ (J+1) - hγ ( v +

12) 2 - DJ2 (J+1)2

The first term corresponds to a anharmonic oscillator

The second term corresponds to rigid rotor,

The third term corresponds to harmonicity

The fourth term corresponds to centrifugal distortion.

Thus, the vibrational spectrum does not occur in a single line

but a number of lines appear on either side of the expected position.

These are named as P,Q and R branches.

P- BRANCH:

When transition occurs such a way that v → v+1 and J → J+1, The energy change is

∆E = [ hγ ( v +1 + 12) + B (J+1)( J+1+1) ] - [ hγ ( v +

12) + B J(J+1) ]

Rearranging

= [ hγ ( v +1 + 12) - hγ ( v +

12) ] + [ B (J+1)( J+2 ) - B J(J+1) ]



= [ hγ ( v +1 + 12 - v -

12) ] + B (J+1) [ J+2 - J]

= hγ(1) + B (J+1) (2)

= hγ +2 B (J+1)

These lines are named as P- branch

selection rule is ∆ v = 1, ∆J = +1

R- BRANCH:

When transition occurs such a way that v → v+1 and J+1 → J

The energy change is

∆E = hγ ( v +1 + 12) - hγ ( v +

12) + B J(J+1) - B (J+1)( J+2)

Rearranging

= hγ ( v +1 + 12 - v-

12) + B (J+1) [ J –J-2]

= hγ - 2B (J+1)

These lines are named as R- branch

selection rule is ∆ v = 1, ∆J = -1

Q- BRANCH:

In linear polyatomic molecules, selection rule is ∆ v = 1, ∆J = 0

. The frequency for such transition is γ = γ0

Vibration type Selection rule Frequency of transition Band contour



1 Parallel ∆ v = 1, ∆J = ±1 γ = γ0 ±2mBJ P,R

2 Perpendicular ∆ v = 1, ∆J = 0 γ = γ0 P,R,Q

611VIBRATIONAL SPECTRA OF POLYATOMIC MOLECULES-

The molecule having ‘n’ atoms have ‘3n ‘ degrees of freedom.

Translational ( T) + rotational(R ) + vibrational (V) = 3n

Linear:

A linear poly atomic molecule having ‘n’ atoms can have 3 translational degrees of

freedom. 2 rotational degrees of freedom.

Translational ( T) + rotational(R ) + vibrational (V) = 3n

3+2 + Vibrational degree of freedom = 3n

= 3n -5

For example carbon dioxide is a linear molecule.

number of vibrational modes = 3n-5

= 3(3) – 5

= 4

Non – linear:

A non linear molecule poly atomic molecule having ‘n’ atoms can have 3

translational degrees of freedom and 3 rotational degrees of freedom

3+3 + Vibrational degree of freedom = 3n

Vibrational degree of freedom = 3n -6

For example water is a non –linear molecule

number of vibrational modes = 3n-6 [ ‘n’ is the number of atoms]

= 3(3) – 6

= 3

Problem: Find the number of vibrational modes of oxygen and benzene

Solution:

Since oxygen is a linear molecule number of vibrational modes of oxygen and benzene

= 3n-5

= 3(2) – 5

= 1

vibrational modes of benzene = 3n-6

= 3(12)- 6



= 30

711FUNDAMENTAL VIBRATIONS



The excitation from the ground state(v= 0) to the lowest excited state (v= 1) is known

as fundamental vibration and the frequency at which it takes place is called fundamental

vibrational frequency

The transition from v= 0 to v= 1 is called fundamental vibrational frequency.

E = hγ ( v + 12) - hγ ( v +

12) 2 xe

Where ‘xe’ is anharmonicity constant

∆E = hγ { [ ( 1 + 12) - ( 1 +

12) 2 xe ] - [ ( 0 +

12) - ( 0 +

12) 2 xe ] }

= hγ { [ 32) - (

32 ) 2 xe ] - [ (

12) - (

12) 2 xe ] }

= hγ ( 32 -

94

xe - 12 +

14

xe )

= hγ { ( 32 -

12) - (

94 -

14 ) xe }

= hγ ( 1 - 2 xe ) cm – 1

fundamental vibrational frequency = ( 1 - 2 xe ) cm – 1

811NORMAL MODES OF VIBRATION

1. Stretching vibrations:

These are the movements of atoms along the bond axis such that the inter atomic

distance is increases or decreases

a. Symmetrical stretching vibrations:

When the stretching and compression occur in a symmetrical fashion we call it

symmetrical stretching. Here there is no change in bond angle.

For example in CH2 , the H- atoms move away from the central carbon atom without

change in bond angle.

b. Asymmetrical stretching



When one bond is compressing while the other is stretching , it is called un

symmetrical stretching

2. Bending Vibration:

If a vibration causes change in bond angles between bonds with a common atom , it

is called bending vibration. It is of two types.

A. In plane bending.

1. Scissoring:

Here the two atoms joined to a central atom move towards each other or away from

each other, then it is called Scissoring.(Like scissor)

Towards each other away from each other

2. Rocking:

Here the both atoms joined to a central atom move same side in the same plane

B. Out of plane bending

1.. Wagging:

Here the both atoms joined to a central atom move same side in out of plane



2. Twisting :

Here the structural unit rotates about the bond in out of plane

911OVERTONES,

If transition takes place from ground level( v= 0) to any state other than v= 1 , then

the bands obtained are called overtones. In overtones vibrational quantum number changes

by ±2, ±3 etc

The transition frequency is given by

∆γ = v [ 1- (v+1) xe ] cm – 1

The transition from v= 0 to v= 2 is called first overtone.( second anharmonic)

∆γ = ± 2

E = hγ ( v + 12) - hγ ( v +

12) 2 xe

Where ‘xe’ is anharmonicity constant

∆E = hγ { [ ( 2 + 12) - ( 2 +

12) 2 xe ] - [ ( 0 +

12) - ( 0 +

12) 2 xe ] }

= hγ { [ 52) - (

52) 2 xe ] - [ (

12) - (

12) 2 xe ] }

= hγ ( 52 -

254

xe - 12 +

14

xe )

= hγ { ( 52 -

12) - (

254 -

14 ) xe }

= hγ { 2 - 6 xe }

Frequency of first overtone = 2 ( 1 - 3 xe ) cm – 1

The transition from v= 0 to v= 3 is called second overtone.( third anharmonic)

∆γ = ± 3

∆E = hγ { [ ( 3 + 12) - ( 3 +

12) 2 xe ] - [ ( 0 +

12) - ( 0 +

12) 2 xe ] }

= hγ { [ 72) - (

72 ) 2 xe ] - [ (

12) - (

12) 2 xe ] }

= hγ ( 72 -

494

xe - 12 +

14

xe )

= hγ { ( 72 -

12) - (

494 -

14 ) xe }



= hγ { 3 - 12 xe }

Frequency of first overtone = 3 ( 1 - 4 xe ) cm – 1

The transition from v= 0 to v= 4 is called third overtone.

∆γ = ± 4

= 3(1- 5xe ) cm

As the frequencies of the first and second overtone bands are 2 to 3 times the frequency of

the fundamental , they appear in the region of shorter wavelength as compared to the

fundamental band.

1011COMBINATION AND DIFFERENCE BANDS-

When two vibrational frequencies of a molecule couple to give a new frequency and

such frequency is IR active then the new frequency is called combinational frequency.

If γ1 and γ2 are the two frequencies combined then the frequency of combination

band is

γcomb = γ1 + γ2



If the observed frequency of the combination is equal to the difference between them,

then the band is called difference band

If γ1 and γ2 are the two frequencies combined then the frequency of difference band

is

γdiff = γ1 - γ2

1111FERMI RESONANCE.

Resonance means for small vibration large amplitude

The shifting of energies and intensities of absorption bands in an IR spectra or

Raman spectra is called Fermi resonance.

Causes: it takes place due to mixing of fundamental vibration mode and overtone

Condition: the modes should have equal energy

Fermi resonance in CO2

CO2 has four vibrational modes ( 3n-5)

1. Asymmetric stretching : it shows a band at 1337 cm -1 .

2. Symmetric stretching : It does not make chang in dipole moment. Therefore no band

appears for this mode

3. The two bending vibrations in CO2 are equivalent and absorb at same frequency 667

cm -1.

4. The overtone = 2x

= 2 ( 667)

= 1334 cm -1



since the overtone has same energy (1334 cm -1 ) as the fundamental mode( 1337 cm -1 ) ,

mixing of modes take place , causing the shift towards higher frequency.

Due to the effect of Fermi resonance the first band shifts towards higher frequency and give

rise to two bands at 1285 and 1388 cm -1

CO2

In CO2, the bending vibration ν2 (667 cm−1) has symmetry Πu. The first excited state of ν2 is

denoted 0110 (no excitation in the ν1 mode, one quantum of excitation in the ν2 bending mode



with angular momentum about the molecular axis equal to ±1, no excitation in the ν3 mode)

and clearly transforms according to the irreducible representation Πu. Putting two quanta into

the ν2 mode leads to a state with components of symmetry (Πu × Πu)+ = Σ+g+ Δ g. These are

called 0200 and 0220, respectively. 0200 has the same symmetry (Σ+g) and a very similar

energy to the first excited state of v1 denoted 100 (one quantum of excitation in the

ν1 symmetric stretch mode, no excitation in the ν2 mode, no excitation in the ν3 mode). The

calculated unperturbed frequency of 100 is 1337 cm−1, and, ignoring anharmonicity, the

frequency of 0200 is 1334, twice the 667 cm−1 of 0110. The states 0200 and 100 can therefore

mix, producing a splitting and also a significant increase in the intensity of the 0200

transition, so that both the 0200 and 100 transitions have similar intensities.

C. RAMAN SPECTRA:

17 CLASSICAL THEORY OF RAMAN EFFECT



When mono chromatic light falls on a polarisable substance, the light gets scattered.

When external electric field is applied, it induces a dipole moment( μ ) in the molecule

which is given by

μ = α E where α – Polarisability E – Electric field --------------1

The electric field is given by

E = E0 Sin γt

Polarisability α = α0 + ∂ α∂ q Sin γmt


𝛍 = ( α0 + ∂ α∂ q Sin γmt ) E0 Sin γt [substituting in 2 ]

= α0 E0 Sin γt + α0 ∂ α∂ q E0 Sin γmt Sin γt

= α0 E0 Sin γt + ½ ∂ α∂ q [cos(γm –γ) + cos(γm + γ) [ Sin A SinB = ½ [ cos (A-B)

- cos (A+B)]

. 1. If the vibration does not change the polarisability of the molecule then ∂ α∂ q = 0, Then the

above equation becomes 𝛍 = α0 E0 Sin γt . The dipole oscillates only at the frequency of

incident radiation.

Raman shift is given by (γm –γ) and ( γm +γ)

Thus the oscillating dipole has three distinct frequency components

Hence Raman spectra of a vibrating molecule consists of an intense band at

incident frequency and two weak bands one above and one below that of the intense band.

stokes lines Rayleigh scattering antistoke lines

γex > γ γex = γ γex < γ

Raman scattering



27MOLECULAR POLARISABILITY

applications of electric field on polarisable molecule causes dipole moment in the

system. The magnitude of this induced moment will be proportional to the electric field

applied.

Dipole moment μ = α E

Where α is proportionality constant called polarisability and is defined as the electric dipole

induced in the molecular system by a unit electric field.

37PURE ROTATIONAL RAMAN SPECTRA

The selection rule for rotational Raman transitions is ∆J =0, ±2 where ∆J =0

corresponds to Rayleigh scattering. The rotational energy E = h2


------------------------1h

In terms of wave number ∈ = Ehc cm -1 [ E = hϑ],

= h2

8 π2 IJ (J+1)

hc cm -1



= h

8 π 2 I cJ(J+1) cm -1

= B J(J+1) cm -1 where B is rotational constant B = h

8π 2 I c

Imagine the molecule to be in the J= 0 state ( ground state). Let the incident radiation be

absorbed to raise it to the J= 1 state. The energy absorbed will be ∈ J=2 - ∈ J=0 = B 2(2+1) - B (0)(0+1)

= 6B – 0

= 6B cm -1 and therefore ,

ϑ J=0 → J = 2 = 6B cm -1 , in other words an absorption line will appear at 6B cm -1 If now the

molecule is raised from the J= 1 to J = 3 by absorbing more energy∈ J=2 - ∈ J=0 = B 3(3+1) - B (1)(1+1)

= 12 B – 2B

= 10 B

and therefore , ϑ J=0 → J = 2 = 10B cm -1 , so, there will be a line at 10B cm-1

The energy separation between successive Raman lines = 10 B – 6B

= 4B

In general , to raise the molecule from the state J to J+2, we would have

ϑ J → J+2 = B (J+2)(J+2 +1 ) – B J (J+1)

= B (J+2)(J+3 ) – B J (J+1)

= B [ J2 + 5J +6 – J2 - J ]

= B [4J +6 ]

Lines arising from these energy changes are referred to as S branch lines( stokes lines)

If the frequency of the incident light is γ ex

For stokes lines γ s J →J + 2 = γ ex - B [4J +6 ]

For antistoke lines γ A J + 2 → J = γ ex + B [4J +6 ]



O – BRANCH ( ANTISTOKES ) Q-BRANCH S- BRANCH ( STOKES LINES)

Rotational Raman lines of linear molecules have nearly double the spacing of the

microwave rotational spectral lines for the same molecule.



47 VIBRATIONAL RAMAN SPECTRA



VIBRATIONAL – ROTATIONAL RAMAN SPECTRA: [ O, Q and S BRANCHES

IN RAMAN SPECTRA]

When a molecule absorbs radiation, which is enough to cause change in vibrational

energy , its rotational energy may also change..Hence the absorption band due to vibrational

energy change will contain number of closely spaced lines due to change in rotational

energy. Such type of Raman spectrum is referred to as vibration- rotation Raman spectrum.

The total energy of the vibrating rotator is given by the sum of the vibrational and

rotational energies. Etotal = Evib+ Erot

= hγ ( v + 12) + B J(J+1)

Thus the vibrational spectrum does not occur in a single line but a number of lines

appear on either side of the expected position. These are named as O,Q and S branches.

selection rule for Raman spectra is ∆ v = 1, ∆J = ± 2

O- BRANCH:

When transition occurs such a way that v → v+1 and J → J+2, The energy change is

∆E = hγ ( v +1 + 12) + B (J+2)( J+2+1) - hγ ( v +

12) + B J(J+1)

= hγ ( v +1 + 12) - hγ ( v +

12) + B (J+2)( J+3) - B J(J+1)

= hγ ( v +1 + 12 - v-

12) + B[ J2 + 5J + 6 - J2 - J]

= hγ + B (4J+ 6)

These lines are named as O- branch

selection rule is ∆ v = 1, ∆J = +2

Q- BRANCH:

In linear polyatomic molecules, apart from oscillations of dipole parallel to the bond

axis , they produce oscillatory dipoles perpendicular to the bond axis.

In such a case selection rule is ∆ v = 1, ∆J = 0

. The frequency for such transition is γ = γ0

∆E = hγ ( v +1 + 12) + B (J)( J+1) - hγ ( v +

12) - B J(J+1)

= hγ



S- BRANCH:

When transition occurs such a way that v+1 → v and J+2 → J The energy change is

∆E = hγ ( v + 12) - hγ ( v +1 +

12) + B J(J+1) - B (J+2)( J+3)

= - hγ ( v +1 + 12 - v-

12) + B[ J2 + J - J2 - 5J - 6 ]

= - hγ - B (4J+6)

These lines are named as S- branch

selection rule is ∆ v = 1, ∆J = -2

. Spectral lines are displaced by 6B, 10B, to higher frequency than fundamental frequency.

∆ v = +1 ∆ v = +1 ∆ v = +1

∆J = +2 ∆J = 0 ∆J = -2

57ROTATIONAL FINE STRUCTURE

Fine structure, is the splitting of the main spectral lines of an atom into two or more

components, each representing a slightly different wavelength.

Fine structure is produced when an atom emits light in making the transition from

one energy state to another.

67RULE OF MUTUAL EXCLUSION


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If a molecule has centre of symmetry, the Raman active vibrations are IR inactive and

Vice versa. If there is no centre of symmetry then some vibrations (not all) may be both

Raman and IR active.

77POLARIZATION OF LIGHT AND RAMAN EFFECTRAMAN SPECTRA

Light waves which have vibrations in all directions are called unpolarised light.

When passed through Nicol prism, this light can be changed in to one which have vibrations

in only one direction. These are called polarised light.The phenomenon of conversion of

unpolarised light in to polarized light is known as polarization.



IR RAMAN

1. Spectra is due to absorption of light Due to scattering of light

2. Permanent dipole moment is necessary condition Polarisability is necessary condition

3. Water can not be used as solvent because it is opaque Can be used

4. Homo nuclear diatomic molecules like H2,O2,N2,F2,I2

are IR in active because they do not have Permanent

dipole moment

Homo nuclear diatomic molecules like

H2,O2,N2,F2,I2 .. are Raman active

because they are polarisable.



RAMAN EFFECT

when a light beam is deflected by molecules there occurs a change in the wavelength of light. This is known as Raman effect

. When a beam of light passes through, transparent sample of a chemical compound, a small fraction of the light emerges in directions other than that of the incident (incoming) beam. A small part, has wavelengths different from that of the incident light; its presence is a result of the Raman effect


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UNIT IIISPECTROSCOPY – III

1. Resonance spectroscopy

2. Zeeman effect

3. Equation of motion of spin in magnetic fields

4. Chemical shift

5. Spin-spin coupling

6. NMR of simple AX and AMX type molecules-

7. H1 NMR

8. 13C - NMR

9. 19F- NMR

10. 31P NMR spectra

11. A brief qualitative discussion of Fourier Transform spectroscopy

12. ESR: principle

13. Spin-orbit coupling.

14. Hyperfine interaction.

15. McConnell relation.

16. Mass spectra: Theory and instrumentation,

17. McLaffetry rearrangement

18. Fragmentation pattern for simple aliphatic and aromatic alkanes

19. Fragmentation pattern for alcohols

20. Fragmentation pattern for aldehydes and ketones.-

21. Mossbauer spectroscopy-

22. Doppler effect

23. Isomer shift

24. Electron-neutron hyperfine interactions.

25. Quadrupole interactions and Magnetic interactions.

125 .Resonance spectroscopy



The reinforcement by the synchronous vibration of a neighboring object is

called resonance

If the spectra is obtained by the resonance technique it is called resonance

spectra

1. Nuclear Magnetic Resonance spectra[ NMR]

2. Electron Spin Resonance spectra[ESR]

3. Nuclear Quadrapole Resonance spectra[NQR]

225 . Zeeman effect:

1. The splitting of spectral lines, in the presence of applied magnetic field is called

Zeeman effect.

2. This effect arises / as a result of interaction between / applied magnetic field / and the

magnetic moment /associated with electron spin.

3. The field splits the line with a given J value in to ( 2J+1) levels.

For example if J = 1

The number of lines observed = 2 (1) + 1

= 3

If J = 2 , The number of lines observed = 2 (2) + 1

= 5



4 If the term is singlet , it is called normal Zeeman effect.

5. If it is multiplet ( doublet, triplet) it is called anomalous Zeeman effect. The anomalous

effect appears on transitions where the net spin of the electrons is an odd half-integer, so that

the number of Zeeman sub-levels is even.

6. When the spectral lines are absorption lines, the effect is called inverse zeeman effect.

7. Transition is governed by selection rule ∆m = 0 ,± 1

Selection rules.

Transition of electrons are not observed between all pairs of energy levels. Some

transitions are "forbidden" ( i.e., highly improbable) while others are "allowed" by a set of

rules called selection rules

Any electron transition which involves the emission of a photon must involve a

change of 1 in the angular momentum. total angular momentum may change by either zero or

one


https://en.wikipedia.org/wiki/Electron

https://en.wikipedia.org/wiki/Spin_(physics)


8. At even-higher field strength, when the strength of the external field is comparable to the

strength of the atom's internal field, electron coupling is disturbed and the spectral lines

rearrange. This is called the Paschen-Back effect

Find the number of transitions occur when ‘l’ changes from ‘1’ to ‘2’ under magnetic field

Solution: ‘l = 1 ’ level splits in to -1,0,1‘l = 2 ’ level splits in to -2, -1,0,1,2

Possible transitions 1. +2 to +1 ∆m = 2-1

= 12. +1 to +1 ∆m = 1-1

= 03. +1 to 0 ∆m = 1-0

= 14. 0 to +1 ∆m = 0-1

= -15. 0 to 0 ∆m = 0-0

= 06. 0 to -1 ∆m = 0-(-1)

= +17. -1 to -1 ∆m = -1-(-1)

= 08. -1 to 0 ∆m = -1- 0

= -19. -2 to -1 ∆m = -2- (-1)

= -1

There are nine transitions consistent with the selection rule.


https://en.wikipedia.org/wiki/Zeeman_effect#Strong_field_(Paschen%E2%80%93Back_effect)


3

25EQUATION OF MOTION OF SPIN/ IN MAGNETIC FIELDS

1. When an object rotates , this action generates , an intrinsic angular momentum.

2. If there were no friction in air , the object would spin forever.

3. This intrinsic angular momentum is, a vector and thus spin is also a vector.

4. If the object is spinning from right to left, then the spin vector is pointing up;

Imagine a current loop lying in a plane

If the loop has current I and an enclosed area A, then the magnetic moment is the

product of the current and area , with the direction n̂ parallel to the normal direction of the

plane.

μ→

= I × A× n̂

To derive an equation of motion

we can relate the time derivative of the angular momentum to the torque



In the presence of a magnetic field, the torque is the cross-product of the magnetic moment µ

→ and the magnetic field B →

we can write down a first order differential equation of the magnetic moment which describes

the motion of a spin:.

This represents part of the Bloch equation and leads to the description of the precession of a

spin in the presence of a magnetic field. This equation tells us that the rate of change of µ →

depends on B → and that its motion is perpendicular to both µ → and B → .

-------------------------- NOT FOR EXAMINATION--------------------

The torque due to a constant magnetic field is to be calculated for a circular loop with radius R and current I centered in the x-y plane . Let the field lie in the y-z plane and have magnitude B. The differential torque on d → can be written as:

Using double cross-product formula The “BAC-CAB” rule is A → × (B → × C → ) = B → (A → ⋅ C → ) − C → (A → ⋅ B → )

The magnetic field and the cylindrical unit vectors shown in Fig



------------------1

An integration of the above equation over the polar angle φ, with φ^ from equation 1, gives the total torque.

----------------2

The average value of sin2 φ or cos2 φ over any multiple of π/2 is 1/2.

Equation 2 is exactly µ → × B → in view of the fact that the magnetic dipole moment for the circular loop of Figure is

-------------------------- NOT FOR EXAMINATION--------------------

425 CHEMICAL SHIFT

The shift in the positions of NMR signals which arise due to shielding or

deshielding of protons is called chemical shift.

Chemical shift (δ ) = change∈the frequency

operating frequency∈mega cycles

= (γ sample – γ reference)

operating frequency∈mega cycles.

TetraMethylSilane (TMS) is used as reference, whose formula is ( CH3)4Si

It is designated as “τ” or in ppm

Factors affecting the chemical shift:

A. Intramolecular factors:

1. Electronegativity( shielding and deshielding)



If electro negative atom is present in a compound, it pulls the electron cloud

towards itself and hence the proton gets deshielded ( less shielded) therefore the peak will

be shifted towards high value

For example the methyl proton in CH3I gives peak with a chemical shift value of 2.1ppm

whereas CH3F will give at 4.2 ppm. This is due to higher electro negativity of F

NMR of CH3I NMR of CH3F

2. Substitutional effect:

If the electro negative atom is more , then the proton will be deshielded more and

chemical shift will be more

For example the CH3 protons in methyl chloride gives peak with a chemical shift value

of 3.3ppm

whereas CHCl3 will give at 7.3 ppm. This is due to higher the presence of 3

chlorine atolms

3. Anisotropic effects of chemical bonds.

Different properties in different directions is called anisotropy.

For example polarisation, magnetisation, hardness, conductivity



Anisotropic effects are the result of a local induced magnetic field experienced by a nucleus

resulting from circulating electrons.

Anisotropy is found in Acetylene, ethylene and benzene

NMR of ETHYLENE Chemical shift = 5.25 ppm NMR of ETHANE Chemical shift =

0.86ppm



NMR of acetylene and Methane NMR of BENZENE Chemical shift =

2.05ppm

Chemical shift = 1.8 ppm( acetylene)

0.23ppm ( for methane)



4. Vanderwals deshielding:

The electron cloud of the bulky group will tend to repel the electron cloud and the

proton is deshielded.

B. Intermolecular factors

1. Hydrogen bonding A hydrogen atom involved in hydrogen bonding / shares its electrons with two

electronegative elements. As a result, it is itself deshielded and comes in to resonance at low

field.

For example, the OH group in chloro phenol shows peak at 6.3 ppm whereas that

of o- hydroxyl acetophenone shows at 12.3 ppm. This is due to Hydrogen bonding



2. Solvents The solvents used in NMR should be chemically inert, magnetically isotropic,

devoid of hydrogen atom and should dissolve the sample to a reasonable extent.[C6H6 , C6D6]

Problem -4. An NMR signal for a compound is found to be 180 Hz downward from TMS

peak using a spectrometer operating at 60 MHz. Calculate its chemical shift.

Chemical shift = shift ¿TMS ¿operating frequency × 10 6

= 180

60× 106 × 10 6

= 3ppm

Problem -5 A compound shows a proton NMR peak at 300 Hz downfield from TMS peak

in a spectrometer with chemical shift of 3pp. . Find the operating frequency

Solution:

Chemical shift = 300

x×106 ×106

3 = 300

x×106 ×106

X = 100 MHz



525 SPIN – SPIN COUPLING ( SPLITTING) INVOLVING H1

The interaction between the spins of neighbouring nuclei in a molecule is

known as spin- spin coupling. This leads to splitting of signals in to multiplet. This splitting

of signal is known as spin- spin splitting.

The multiplicity depends upon the number of nearby protons. If there are ‘n’ adjacent protons , then the number of signals will be ( n+ 1). This is known as (n+1) rule.

The (n+1) Rule, in 1H states that “ if a given nucleus is coupled

to ‘n’ number of nuclei that are equivalent , the multiplicity of the of the peak is (n+1)”

The relative intensities of the individual lines of a multiplet correspond to the numerical coefficient of the binomial expression (1 + x) n

If n = 2 , then (1 + x) 2 = 1 + 2x + x2

intensity ratio 1: 2:1

If n =3 , then (1 + x) 3 = 1 + 3x +3 x2 + x3

intensity ratio 1: 3:3:1

It is given by Pascal’s Triangle

SIGNAL PEAK AREA RATIO

singlet 1

doublet 1 1

triplet 1 2 1

quartet 1 3 3 1

quintet 1 4 6 4 1

Example :

Consider the NMR of 1,1,2- trichloro ethane.

It has two environmentally different hydrogen atoms

Therefore two signals are expected.

But the spectra shows a triplet and a doublet.

The Hb proton interacts with two Ha proton and hence its signal is split in to3 peaks in accordance with ( n+1) rule with relative intensity 1:2:1



Similarly the Ha proton interacts with Hb proton and hence its signal is split in to2 peaks in accordance with ( n+1) rule with relative intensity 1:1

If there are three protons the intensity will be

Intensity is 1:3:3:1



NUMBER OF SIGNALS

1.

2.

3.

4.

5.



6.Which has multiplet in nmr?

I. benzene II. iso butylene III. 2-chloro propylene

7. The spectrum of isopropyl chloride shows

8. The spectrum of CH3-CH2-O- COOH shows

9.. A triplet and a quartet appears in

I. benzene II ethyl chloride III 2-chloro propylene



Coupling constant:

The spacing of adjacent lines in the multiplets due to spin- spin coupling of

protons, is known as coupling constant J.

What is coupling constant? How it is calculated? Explain with suitable example.

Coupling constant:

1. The distance between the peaks in a given multiplet is a measure of the magnitude of

splitting effect.

2. It is referred to as coupling constant and is denoted by the symbol J.

3. Numerical value of J is expressed in Hz or cps. (cycles per second).

4. Unlike the chemical shifts, the value of J is independent of the applied field strength and

depends only upon the molecular structure.



5. For a pair of mutually coupled protons, the coupling constant due to splitting by one

proton has the same value as the coupling constant due to the splitting by the second

proton In other words, mutually coupled protons show the same magnitude of the

splitting of each other signals.

The coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and

Hb hydrogen sets, and this interaction is of the same magnitude in either direction.

Determination:

The size of coupling constant is determined by the number and kind of intervening

chemical bonds and the spatial relations between the protons

(a) For protons attached to the same carbons atom (i.e., germinal protons), the values of

J varies from 0-20 Hz depending upon the bond angle and overall structure of the

molecule.

(b) For protons attached to adjacent carbon atoms (i.e., vicinal protons), J varies from 2-

18 Hz depending upon the spatial positions and the structure of the molecules as a whole

Calculation of Coupling constant:

For the simple case of a doublet, the coupling constant is the difference between two peaks.

J is measured in Hz, not ppm.

First convert the peaks from ppm into hertz.

Suppose we have one peak at 4.260 ppm and another at 4.247 ppm.

To get Hz, just multiply these values by the field strength in mHz.

If we used a 500 mHz NMR machine, our peaks are at 2130 Hz and 2123.5 respectively.

The J value is just the difference.

In this case it is 2130 - 2123.5 = 6.5 Hz.

Problem

An NMR spectrum obtained from a spectrometer operated at 500 MHz shows peaks at

0.864 and 0.849 ppm find the coupling constant

Solution:

J =( 0.864 – 0.849 )× 500

= 0.15 × 500

= 7.5 Hz



TRIPLETTriplet has three peaks. We can take difference of any of the two peaks because the proton

couples equally with both the neighboring protons

Problem

An NMR spectrum obtained from a spectrometer operated at 500 MHz shows peaks at

3.585, 3.568 and 3.55ppm Find the coupling constant

Solution:

J =( 3.585 – 3.568 )× 500

= 0.017 × 500

= 8.5 Hz

Or

J =( 3.568 - 3.551 )× 500

= 0.017 × 500

= 8.5 Hz

QUARTET

The coupling constant for quartet is calculated just like triplet. Taking difference of any two

consecutive peaks will give the J value for quartet.

DOUBLE DOUBLETA double doublet has two coupling constants because it has two doublets. The coupling tree of the signal is quite helpful in identifying the peaks which will be used for calculation of the coupling constants in this case.


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We have four lines in this tree. The two coupling constants are denoted as J1 and J2.

Its quite simple to calculate J2 by taking difference of the line 1 and 2 or line 3 and 4.

J1 is the coupling constant for the two blue line in the tree which are in fact not visible in the

spectrum.

But if you look at this tree closely, you will find that the difference between line 1 and 3 or 2

and 4 is actually equal to the difference between the two blue lines, which in fact is J1.

Problem

An NMR spectrum obtained from a spectrometer operated at 500 MHz shows peaks at 2.

837, 2.808, 2.825 and 2,796 ppm Find the coupling constant

Solution:

J1 =( 2.837 – 2.808 )× 500

= 0.029 × 500

= 14.5 Hz

Or

J1 =( 2.825 - 2.796 )× 500

= 0.029 × 500

= 14.5 Hz

J2 =( 2.837 – 2.825 )× 500

= 0.012 × 500

= 6 Hz

Or

J1 =( 2.808 - 2.796 )× 500

= 0.012 × 500

= 6 Hz

Long range coupling.

In NMR spectroscopy, coupling between nuclei that are separated by more than three

bonds. Coupling of Ha with Hd (if present) is long range coupling, because these protons are

separated by four sigma bonds.



LABELLING THE PROTONS ( SPIN SYSTEM NOTATION)-POPOLE METHOD

To designate the protons the following conventions are used.

1. The protons separated by small chemical shift is designated by the letters A,B,C

2. Those far away from chemical shift by X,Y,Z

3. Those intermediate as M,N,O

625NMR OF SIMPLE AX –SYSTEM ( n+1 rule obeys)

If the chemical shift between the protons is large compared to the coupling between

them (νAX >> JAX), we label them as HA and HX.

Some molecules that give AB or AX patterns are shown below (spectra are all at 300 MHz):

Disubstituted alkenes



The splitting follows (n+1 ) rule and it is called first order spectrum

peak numbers and intensities follow the polynomial coefficient scheme

∆𝛄 >> J

Example:

It contains two types of protons designated as Ha and Hb.

In its spectrum a simple pattern of two doublets appears.

Only two signals corresponding to Ha and Hb are expected.



But due to spin – spin interaction of Ha on Hb, the Hb signal is split up into a doublet

with intensities 1:1 ratio.

Similarly due to spin – spin interaction of Hb on Ha, the Ha signal is split up into a

doublet with intensities 1:1 ratio.

Coupling constant J is the same in both cases No distortion occurs

-----------NOT FOR EXAMINATION-----------

Energy Levels of AX Spectra The four energy levels for an AX system are given by the equation below, by substituting

the four possible spin combinations of mA and mX (++, +-, -+, --):

E = -(mAνA + mXνX) + mAmXJAX

There are four states: αα, αβ, βα, ββ.

We will use the convention: αα is the lowest energy state (α is aligned with the field, m = +½)

and ββ is the highest energy state (β is aligned against the field, m = -½).

The first term in the equation is the chemical shift part, the second term the coupling part.

If the coupling is a small perturbation, then the energy is simply the sum of the two parts.

In energy level terms, this means that the energy separation of the αβ and βα states is large

compared to J.



725 .NMR OF SIMPLE AMX – SYSTEM( n+1 rule fails)

If three protons are involved and they all have different chemical shifts , the

system is called AMX system.

Here the (n+1) rule fails. since the protons attached to a single carbon are chemically

non –equivalent .



Each proton has different chemical shift values, coupling constant different from

those of each of

other protons

Because HA and HB are non equivalent and HC is coupled differently to HA than to HB

the (n+1)

rule fails

Proton A couples with X which splits the signal into a doublet. But A also couples

with M so that each line of A is further split in two giving four lines in all. Thus it shows a

double doublet ( JAM and JAX)

Similarly the M signal is split in to two by coupling with X and each line is further

split in to two by coupling with A . Two coupling constants are seen JMX and JAX and M

appears as a double doublet. Lastly the X signal is split in to a double doublet by two

successive couplings JAM and JAX. Totally 12 lines are observable.

J BC = 14 Hz , J BC = 6.3 Hz , J AB = 1.5 Hz

Other Example styrene oxide, vinyl acetate, furfural



Styrene oxide

.Vinyl acetate:



825 13 C NMR [ CMR]

1.Why 13C shows spectra but 12C does not ?

All nuclei possessing spin value ½ , resonate with radio frequency and give rise to

NMR spectra. 12 C has spin value zero and hence it will not show NMR spectra whereas 13

C has spin value ½ therefore it shows NMR spectra.

2. Number of signals

For every non – equivalent carbon atom there will be a signal in the NMR. For example



1H NMR of CH3CH2OH 13 C- NMR of CH3CH2OH

Ethanol shows two signals, since it contains two non – equivalent carbon atoms.

3. No carbon- carbon coupling :

The abundance of 13C is very low, the chance for 13C nuclei adjacent to each other

in the same molecule is very small. Therefore carbon- carbon coupling is not possible

4. Multiplicity and intensity

Spin spin coupling between C13 – C13 nuclei will not be observed due to low natural

abundance of C13. But there is coupling between carbons and their attached hydrogen.

Due to coupling between carbon and hydrogen the signal is split up in to multiplet.

The number of signals obey (n+1) rule. Thus a methyl group appears as quartet, a

methylene as triplet, a methane as doublet, and a quarternary carbon as singlet

.The intensity obeys Pascal’s triangle.

.5. Chemical shift:

The chemical shift values are measured in ppm downfield from TMS. The range of

chemical shift is 200 ppm.



6. Operating frequency:

CMR requires only one fourth of transmitter frequency. For PMR if 60 Hz is

required for CMR only (604 = 15) Hz is sufficient.

7. Double resonance ( decoupling)

To make the spectra simpler, the coupling due to protons should be removed. This

process is called decoupling. This is achieved by resonating the substance twice with

different frequencies. And this process is known as double resonance.

In completely proton decoupled spectra, all carbon resonance appear as singlets.

8.Regions of interest in 13C

Comparison of H 1 and C13



H 1 C13

1 100 % natural abundance 1.1 %

2 Chemical shift ranges from 0 – 10 ppm Up to 400 ppm

3 Coupling between H-H is more abundant Ignored because molecule having more than one C13 nuclei is less

4 High intense lines with sharp peak Low intensity with broad peak

5 Number of protons is measured by peak area

Due to the presence of Nuclear Overhaser Effect (NOE) and relaxation time , it is not so

6 It is used to assign the number of protons present on each site

To find the nature of binding of carbon with other functional groups

925 . F19 NMR

FLUORINE-19 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

Fluorine -19 nuclear magnetic resonance spectroscopy (fluorine NMR or 19F NMR) is an

analytical technique used to detect and identify fluorine-containing compounds

Operational details

19F has

1. a nuclear spin (I) of ½

2. a high magnetogyric ratio ( 40.05 MHz/ tesla) . [ the ratio between magnetic momemt

and its angular momentum ]

3. 19F comprises 100% of naturally occurring fluorine.

Consequently, this isotope is highly responsive to NMR measurements.



Chemical Shift

19F NMR chemical shifts vary from 300 to 700 ppm

The very wide spectral range can cause problems in recording spectra, such as poor data

resolution

The reference compound for 19F NMR spectroscopy, is CFCl3 (0 ppm),

Typical chemical shift ranges for 19F

Spin-spin coupling

1. Fluorine atoms can also couple with each other.



19F-19F coupling constants are generally larger than 1H-1H coupling constants.

Hydrogen couples with fluorine, which is seen in 19F spectrum

It is also possible to record decoupled 19F{1H}

Long range 19F-19F coupling, (2J, 3J or 4J of even 5J) are also observed

Other nuclei can also couple with fluorine

Applications

19F NMR spectroscopy is used to analyze

1. the structure of organofluorine compounds.

2.The pharmaceuticals that contain C-F bonds.

3. to analyse fluoride salts.

1025P31 NMR

PHOSPHORUS-31 NUCLEAR MAGNETIC RESONANCE



Phosphorus-31 NMR spectroscopy is an analytical chemistry technique that

uses nuclear magnetic resonance (NMR) to study chemical compounds that contain

phosphorous

.

1. 31P has an isotopic abundance of 100%

2. a relatively high gyromagnetic ratio. [ 17.24 MHz/ tesla) [the ratio between

magnetic momemt and its angular momentum ]

3. The 31P nucleus also has a spin of ½,

Operational aspects

With a gyromagnetic ratio 40.5% of that for 1H, 31P NMR signals are observed near

202 MHz on an 11.7 Tesla magnet

Chemical shifts

Chemical shifts are referenced to 85% phosphoric acid, which is assigned the chemical

shift of 0,

The chemical shifts ranges from 250 to – 250 ppm, which is much wider than for 1H NMR.

Typical chemical shift ranges



Coupling constants

1.One-bond coupling is illustrated by PH3 where J(P,H) is 189 Hz.

2.Two-bond couplings, are an order of magnitude smaller.

3. Spectra are recorded with protons decoupled.

Applications in chemistry

1. 31P-NMR spectroscopy is useful to assign structures of phosphorus-containing

compounds

2. used for studies of phospholipid bilayers and biological membranes

1125Fourier Transform NMR (FTNMR)



In conventional NMR ( continuous wave CW NMR) the spectrum is presented in

time domain and the intensity of radiation is a function of time

I = f(t).

To measure NMR in time domain , one has to scan a very broad band of

frequency which is time consuming / but in frequency domain, one can apply all the

frequencies in a very short time.

If the NMR spectrum is presented in frequency domain, it is called FT-NMR.

Here the intensity of radiation is a function of frequency

I= f(ϑ )

f(γ) can be converted into f( t) by the following formula

f(t) = ∫−∞

+∞

e2 πiγt f(γ) dγ This is known as Fourier transform. Conversely

f(γ) = ∫−∞

+∞

e−2 πiγt f(t) dt This is known as inverse Fourier transform.

In this technique, the sample is subjected to a high power pulse of radio frequency

radiation. It causes all the spin active nuclei to resonate all at once. As soon as the radio

frequency pulse terminates, the magnetisation vector starts decaying. This is known as Free

Induction Decay (FID) This is modulated .The signal detected as the nuclei return to the

equilibrium and recorded as an array of numbers in a ccomputer.



.



The band of frequency required is obtained by applying a square wave pulse of RF

power in x-y plane. The square wave is synthesised by superposition of several sinusoidal

waves. The pulse is supplied by the radiofrequency field which has Larmour frequency of

100 MHz. If the length of pulse is tp the sample receives the range of frequiencies of abour

1/tp + γ. The Fourier components add to or subtract from this carrier frequency.

As soon as the radio frequency pulse terminates, the magnetisation vector in the x-y

plane starts decaying. This is called free induction decay[FID]. In order to improve signal

to noise [S/N ]ratio several consecutive pulses are applied thereby enhancing the intensity of

the signal. The FIDs are collected in the computer and averaged before being Fourier

transformed to give time spectrum.

[NOT FOR EXAMINATION]

F[f(x)] = 1√2 π ∫

−∞

+∞

f ( x)eisx dx

Problem :Find the Fourier Transform of f(x) = cos x if 0<x<1

Solution:

F[f(x)] = 1√2 π ∫

−∞

+∞

f ( x)eisx dx

= 1√2 π∫0

1

cos x eisx dx

= 1√2 π∫0

1

cos x ¿

= 1√2 π∫0

1

cos x cos sx+i cos x sin sx dx

= 1√2 π

× 12∫0

1

cos (s+1 ) x+¿cos ( s−1 ) x+isin ( s+1 ) x+sin ¿¿¿

=1

2√2π [

sin (s+1 ) x( s+1 )

+ sin (s−1 ) x

( s−1 )- i

cos (s+1 ) x(s+1)

- i cos (s−1 ) x

(s−1) ]

=1

2√2π [

sin (s+1 )( s+1 )

+ sin (s−1 )

( s−1 )- i

cos (s+1 )(s+1)

- i cos (s−1 )(s−1)

] – [ 0+0 - i1

(s+1) - i 1

(s−1) ]



?



1225 ELECTRON SPIN RESONANCE (ESR) SPECTROSCOPY

(OR) ELECTRON PARAMAGNEIC RESONANCE (EPR) SPECTROSCOPY

It is a branch of absorption spectroscopy in which radiation of micro wave frequency ( 10 9 – 10 13 Hz) is absorbed by paramagnetic substance.

It’s use is therefore confined to 1. atoms with unpaired electrons, 2. free radicals, 3. ions, 4. triple state molecules, 5. molecular fragments, 6. complexes with unpaired d and f electrons.

s.no ESR NMR

1. It arises due to spinning of electron about nucleus. It arises due to interaction between

proton and magnetic field

2. Absorption region is microwave Absorption region is Radio wave

3. Spectra is shown only by compounds having unpaired electron ( paramagnetic substances, free radicals, ions, triple state molecules, molecular fragments, complexes with unpaired d and f electrons.

Spectra is shown by all compounds

4. No Chemical shift( if there is neighbouring free electron then the molecule may not be paramagnetic)

Due to the presence of neighbouring proton there occurs chemical shift

5. Required frequency is 28000 Hz Required frequency is 40 Hz

7. Standard used is DPPH TMS

DPPH - 2,2-diphenyl-1-picrylhydrazyl radical



DPPH contains 1.53 × 1021 unpaired electrons per gram

It has splitting factor g = 2.0036

In benzene solution it gives 5 lines with intensity ratio 1:2:3:2:1

1325 PRINCIPLE OF ESR

ESR is based on the fact that paramagnetic substances exhibit characteristic magnetic properties. Such magnetic characteristics is due to spinning action of unpaired electrons about the nucleus.

The spinning motion of electron so generates a magnetic field. The magnetic moment due to spinning is given by

μs = - g eh

4 πmc √s (s+1) -----------------1

‘g’ is called Lande’s splitting factor. eh

4 πmc is called Bohr magneton (β)

√s (s+1) is called spin angular momentum Ms

Therefore the above equation becomes μs = - gβMs -----------------2 when an electron is placed in an external magnetic field of strength B the magnetic energy of interaction will be E = - μs B . Substituting the value of μs from 2 we get E = + g β Ms B

For spin Ms = + ½ , E ½ = + ½ g β B

For spin Ms = - ½ , E- ½ = - ½ g β B

∆E = E ½ - E- ½

= + ½ g β B – (- ½ g β B)



= g β B

. If the electron in magnetic field (B) is irradiated with microwave radiation of

frequency ϑ, the energy hϑ will be absorbed by the electron. When this energy , coincides

the energy level separation between the two states, the resonance absorption takes place

causing the excitation of electrons.

∴ hϑ = g β B This is the condition for resonance.

The selection rule in ESR is ∆ MI = 0 ,∆ Ms = ± 1

ms = - ½ ms = ±1/2

ms = + ½

No field external magnetic field

Since the ESR spectrum is not sharp (broad), it is difficult to locate the peak in the absorption mode. Therefore spectrum in derivative mode is recorded.

ESR spectrum in absorption mode ESR spectrum in derivative mode



1425HYPERFINE SPLITTING( ELECTRON – NUCLEUS COUPLING)

The magnetic interaction between the electron spin and the nuclear spin ( I ≠ 0) in the

same molecule is called hyperfine coupling and the splitting of resonance lines into

component lines is called hyperfine splitting.

Hyperfine splitting of Hydrogen atom:

1. Hydrogen atom has one proton and one electron with s = ½ .

2. There are two possible orientations for the magnetic moments of the nucleus Ms = ±

½

3. In the absence of magnetic field, the single electron of spin Ms = ½ gives rise to

doubly degenerate energy state.

4. When a magnetic field is applied, the degeneracy is removed and the two energy

levels , one corresponding to Ms = - ½ and the other corresponding to Ms = + ½

will be obtained.

5. When the proton interacts, each energy state is further split up in to two energy levels

corresponding to MI = + ½ and MI = - ½ where MI is the nuclear spin angular

momentum quantum number.

6. Thus corresponding to two energy levels, four different energy levels are obtained.

7. The selection rule in ESR is ∆ MI = 0 ,∆ Ms = ± 1

For Ms =+ ½ , MI = - ½ + ½ = 0

For Ms = - ½ , MI = - ½ + ½ = 0

Applying selection rule it is found that ESR spectrum of Hydrogen consists

of two peaks. The expression for energy is given by



E = ms gβH + A msmI Where A – hyperfine coupling constant.(HCC)

E ½ . ½ = ½ gβH + ¼ A,

E ½ .- ½ = ½ gβH - ¼ A

E- ½ .- ½ = - ½ gβH + ¼ A,

E- ½ . ½ = - ½ gβH - ¼ A

ESR – spectra of H - ATOM

2 lines are observed



In deutrium atom 3 lines are observed

For Methyl radical

4 lines are observed



Hyperfine splitting constant (hcc)

Rules for predicting number of hyperfine lines:

1. When an electron interacts with ‘n’- equivalent nuclei of spin I Number of ESR lines = ( 2nI + 1 )

2. The relative intensity of lines is proportional to the coefficient of binomial expansion of ( 1+x)n [ pascal’s triangle] . but it is restricted to equivalent protons.

For example The number of hyperfine lines in H atom = 2(1) ( ½ ) + 1 = 2 with intensity ratio 1: 1The number of hyperfine lines in D atom = 2(2) ( ½ ) + 1 = 3 with intensity ratio 1:2:1The number of hyperfine lines in CH2 radical = 2(2) ( ½ ) + 1 = 3 with intensity ratio 1:2:1The number of hyperfine lines in CH3 radical = 2(3) ( ½ ) + 1 = 4 with intensity ratio 1:3:3:1

The number of hyperfine lines in C6H5 radical = 2(6) ( ½ ) + 1 = 7



The number of hyperfine lines in benzo quinone radical = 2(4) ( ½ ) + 1 = 5

When an electron interacts with ‘n’- equivalent nuclei of spin Ii and a set of equivalent nuclei ‘m’ of spin Ij , then Number of lines = ( 2nIi + 1 ) ( 2mIj +1 )This is applicable for nuclei with I -= ½ For example

Naphthalene ion Naphthalene ion has two sets of 4 equivalent protons each.

Number of lines = ( 2) (4) ( ½ )+ 1 )× ( 2) (4) ( ½ )+ 1 )

= 25 ?



Anthracene ion

It has 3 sets of equivalent protons . Set α and β contains 4 equivalent protons each and set γ contains 2 equivalent protons

Number of lines = [ ( 2) (4) ( ½ )+ 1 )] ×[( 2) (4) ( ½ )+ 1)] × [ ( 2) (2) ( ½ )+ 1]

= 75 lines ?

1525 MC-CONNELL RELATION

It gives the relation between hyperfine coupling constant (hcc)and the spin density.



For a free atom hcc is proportional to the spin density

A∝ ρ

A = Q ρ where A – the dependence of hcc, ρ - spin density, Q- proton

hyperfine constant. This is known as McConnell relation

Use:

It is used to calculate the electron density at each carbon.

1. For hydrogen

Since the electron spends its time in 1s orbital only ρ = 1

Hence For hydrogen atom A = Q , standard value of Q= 500 G ( gauss)

2. For methyl radical (CH3)

The separation between the lines (A) is found to be 23 gauss

Applying McConnell relation

ρ = AQ

= 23

500

= 0.046

= 4.6 %

Electron density on carbon = 100- 3(4.6 )

= 100 - 13. 8

= 86.4 %

This implies that the unpaired electron spends 4.6 % of its time in is orbital of each

hydrogen atom.

And 86.4 % in the vicinity of carbon atom.

3. The esr spectrum of benzene radical C6H6.

In benzene the spin polarisation leads to small spin density in both C and H atomic

orbital.

standard value of Q= 22.5 G ( gauss)

Shows a septet with coupling constant 3.75 G

ρ = AQ

= 3.7522.5

= 0.0076

There are 6 hydrogens with this electron density making a total of



6×0.0076 = 0.046

= 5%

The remaining 95 % of electron’s time is spent equally on the 6 carbon atoms

On each carbon = 956

= 15.8

This shows that magnitude of spin density on the ring proton is 5 % as large as the pz

spin density on the adjacent ring carbon.

1625THEORY ( PRINCIPLE) OF MASS SPECTRA

In mass spectra , the sample is vaporised and sent to ionisation chamber where the

molecule forms molecular ion( M+). This is further subjected to fragmentation by applying

high voltage The fragments having charge are accelerated by electric plates . the

accelerated charged particle passes through radial magnetic field.

Depending on the mass of the fragment it follows the radial path. After this the

rays are sent through slits so the only particular ion comes out of the slit. It is detected and

recorded.

Expression for radius:



The potential energy due to electric field E = Ee .

This energy is converted into kinetic energy ½ m v2

Therefore ½ m v2 = Ee

v2 = 2Eem ------------------1

The accelerated charged fragments are subjected to radial magnetic field(B)

The force experienced by the fragment = Bev

This is equal to centrifugal force m v2

r where r is the radius of the circular motion

Therefore m v2

r = Bev

v = Berm

Squaring on both sides

v2 = B2e2r2

m2 ----------------------2

from 1 and 2 B2e2r2

m2 = 2 Ee

m

r2 = 2 EmB2 e

= 2EB2 ¿ )

r2 ∝ me

This indicates that the radius of the curvature of the different ions depends on me

which shows that the fragment having low me value has lesser radius than the one having

higher me .

me = B

2r2

2 E

Thus the ions are separated on the basis of m/e values. The mass spectrum can be obtained

either by changing B at constant E or changing E at constant B.

INSTRUMENTATION

Mass spectrometer consists of the following components.



1. Vapourisation chamber2. Ionisation chamber3. Accelerator4. Mass analyser

1.Vapourisation chamber

The sample is introduced into an inlet chamber and volatilized.

2.Ionisation chamber

There are two methods of ionization

A. Electron ionization.( EI-MS)

In this technique , electrons are accelerated from a hot tungsten filament to an anode.

These electrons strike the stream of molecules creating cations

The product is a cation radical .

Advantage: it gives rise to fragment ions

Disadvantage: absence of molecular ion

B. Chemical ionization( CI-MS)

In this method, sample molecules are combined with a stream of ionised reagent gas

like methane, ammonia and isobutene. When the sample molecules collide with the

ionised reagent the molecule gets ionised.



Thus in positive ion CI spectra molecular weight information is obtained from

protonation of sample molecule. The observed m/e value is one unit more than the true

molecular weight.

3.Accelerator

The resulting positively charged particles are accelerated by the accelerating plates

into a bent chamber surrounded by a magnetic field

4.Mass analyser:

Dempster’s mass spectrometer is used . the positive ions accelerated by electric field

travel in a circular path under a magnetic field B on a collector

Each of the ions of different masses follows a circular path whose radius is given by

r2 = 2 EmB2 e

. Ions with larger me follow a path with smaller radius.

4. RecorderA graphic presentation is used which represents the intensities of the signals at various m/e values.



1725 Mc - LEFERTY REARRANGEMENT

Compounds having 𝛄- H atom and unsaturation centre undergo H- transfer

reaction in such a way that the α−β bond is broken and the β−γ bond is formed. The γ

– hydrogen atom is transferred to unsaturation centre. This type of rearrangement is called

McLeferty rearrangement

Example 1: butanal

m/e = 72 [4(12) + 1(16) +8(1) ] m/e = 28 + m/e = 44

peak at = 72 – 28 = 44

Example 2: pentanoic acid



m/e = 102 [5(12) + 2(16) +10(1) ] m/e = 42 + m/e =

60

peak at = 102 – 42 = 60

If even mass number molecular ion yield fragments of even mass number and odd

mass number molecular ion yield fragments of odd mass number, it indicates

rearrangement

Amide

Primary amides show a base peak due to the McLafferty rearrangement.

Example 1 : 3-Methylbutyramide (C5H11NO) with MW = 101.15

Problem : Pentanoic acid undergoes McLafferty rearrangement. One of the peak could

have been appeared at m/e value[ 4 marks]

Solution :

Ans = 60



Problem : 3-Methylbutyramide undergoes McLafferty rearrangement. One of the peak

could have been appeared at m/e value[ 4 marks]

Peak at = 101- 42= 59

Problem : The mass spectra of a compound [ MWt = 72] shows peak at 44

[ CH2CHOH] with the elimination of ethylene molecule .The compound could be

Solution :

Mol.wt = 48 + 8+16 = 72

The compound is Butanal

1. It involves the cleavage of a β - bond followed by a γ - hydrogen transfer

2. The rearrangement leads to elimination of neutral molecules

3. It takes place through a six membered transition state.

4. The ion formed is called MRion

Example 4. Problem :The MR ion formed in 2- hexanone is



Example 5: Problem :The MR ion formed in 1- pentene

Example 6: Problem :The MR ion formed in n-propyl benzene

Example 7. Problem :The MR ion formed in methyl-(1-phenyl propyl) ketone

Example 8 Double McLafferty rearrangement

Double MR is reported in certain ketone. For example in 4- heptanone



FRAGMENTATION PATTERN

In mass spectrometry, fragmentation is the dissociation of energetically unstable

molecular ions formed from passing the molecules in the ionization chamber of a mass

spectrometer.

The fragments of a molecule cause a pattern in the mass spectrum used to determine

structural information of the molecule.

Parent peak ( molecular ion peak)1. It is the peak with its mass equal to that of molecular formula.2. This is formed by removal of electron ( not atom) from parent molecule.

M + e → M+ + 2eCH3OH → CH3OH + + 2e

3. It is the peak of highest mass number except isotopic peak.

2. Base peak : 1. It is the tallest peak in the spectra.2. Some times parent peak may be the base peak .

3. Meta stable peak:1. These are low intensity , broad peaks.2. These occur at non – integer m/e value.3. Some ions decompose before completing the mass analysis which gives this peak. 3. Mass of meta stable ion = ¿¿


https://en.wikipedia.org/wiki/Mass_spectrum

https://en.wikipedia.org/wiki/Dissociation_(chemistry)


4.Isotopic peak ( M+1) peak:

1825 FRAGMENTATION PATTERN OF ALKANES

1. The molecular ion in n-alkanes is always present.

2. The molecular ion peak has low intensity for long chain compounds

3. These fragment ions are surrounded by clusters of smaller peaks due to the presences of 13C , (M+1) + and loss of hydrogen (M-1)+ ,and (M-2)+ etc.

4. The fragmentation pattern contains clusters of peaks 14 mass units apart (which represent loss of (CH2) 5. EXAMPLE : mass spectrum of Pentane (C5H12) with MW = 72, shows peaks at m/z values corresponding to 57,43 and 29



FRAGMENTATION PATTERN OF BRANCHED CHAIN ALKANES

1. Bond cleavage occurs at the site of branching 2. Greater number of fragments result from a branched chain compound compared to

straight chain compound. This is due to more sites available for cleavage.3. Largest substituent at a branch is eliminated readily as a radical4. Parent ion peak is not observed.



FRAGMENTATION PATTERN FOR SIMPLE AROMATIC HYDROCARBONS

1. The aromatic molecules like benzene and toluene are accompanied by M+1 and M+2

peaks

due to 13 C and or D

Benzene displays characteristics peaks at m/e = 78, 77, 53,51 and 39



Toluene displays peaks corresponds to the rupture of benzylic bond ( β- cleavage in the side chain)It has strong molecular ion peak The base peak at m/e = 91 corresponds to the ion C7H7 + which is stabilized by the formation of tropylium ion It also shows a peak due to loss of CH3 group.



mass spectrum of Hexane (C6H14) with MW = 86, shows peaks at m/z values corresponding to 57,43 and 29

1925FRAGMENTATION PATTERN FOR SIMPLE ALCOHOLS

1. Primary and secondary alcohols give weak molecular ion peaks.

2. Characteristic ions result via fission of the C-C bond next to the oxygen atom

3. H radical also be ejected

4. For example Methanol shows three peaks at me = 29,31 and 32

5.The peak at me = 32 is due to molecular ion peak

6. The peak at me = 31 arise from loss of H atom and forming hydromethyl ( CH2=OH)



5. The peak atme = 29 arises from loss of two more H- atoms.

7. The minor peak atme = 15 arises from the loss of OH from the molecular ion peak.

AlcoholAn alcohol's molecular ion is small or non-existent. Cleavage of the C-C bond next to the oxygen usually occurs. A loss of H2O may occur as in the spectra below.

3-Pentanol (C5H12O) with MW = 88.15



2025FRAGMENTATION PATTERN OF ALDEHYDES AND

KETONES

1. ALIPHATIC ALDEHYDES:

1. Aldehydes give molecular ions of low intensity

2. The C-C and C-H bond adjacent to oxygen atom gets broken and give corresponding

peaks

3. Elimination of H atom results ( acylium ion) M-1 peak

4. M- R peak is also found out.

5. When 𝛄 – hydrogen is present McLafferty rearrangement takes place . This leads to the

fission of 𝛂- 𝛃 C- C

bond to give peaks corresponding to the products formed.

For example fragmentation of 1- butanal,

m/e = 72 [4(12) + 1(16) +8(1) ] m/e = 28 + m/e = 44



For example fragmentation of 2-methyl butanal

For example fragmentation of 3-methyl butanal

AldehydeCleavage of bonds next to the carboxyl group results in the loss of hydrogen (molecular ion less 1) or the loss of CHO (molecular ion less 29).

3-Phenyl-2-propenal (C9H8O) with MW = 132.16

2. AROMATIC ALDEHYDES

1. These compounds give strong peaks for the molecular ion.

2. They undergo cleavage at the bond 𝛃 – to the ring . This accounts the base peak.

3. Aromatic aldehydes are characterized by M-1 peak due to loss of hydrogen.

4. This eliminates CO to give phenyl ion

5. This further eliminates C2H2 to give C4H3.



For example fragmentation of Benzaldehyde

3.ALIPHATIC KETONES:

1. The major fragmentation involves the fission of C-C bonds adjacent to the oxygen atom.

2. The loss of larger alkyl group results the base peak.

3. For methyl ketone acylium ion ( CH3CO) ion is the base peak.

4. Alkyl ions are formed from acylium ion by the loss of CO.

5. When 𝛄 – hydrogen is present McLafferty rearrangement takes place . This leads to the

fission of 𝛂- 𝛃 C- C bond to give peaks corresponding to the products formed



For example fragmentation of Acetone

For example fragmentation of 2- pentanone

4. AROMATIC KETONES

1. These compounds give strong peaks for the molecular ion.

2. They undergo cleavage at the bond 𝛃 – to the ring . this accounts the base peak.

3. Aromatic aldehydes are characterized by M-1 peak due to loss of hydrogen.

4. This eliminates CO to give phenyl ion

5. This further eliminates C2H2 to give C5H5.

For example fragmentation of Benzophenone

2125MOSSBAUER SPECTRA

It is the study of γ rays emission and subsequent re- absorption in solids

Principle

If a nucleus at rest emits a gamma ray, the energy of the gamma ray is

slightly less than the natural energy of the transition, but in order for a nucleus at rest

to absorb a gamma ray, the gamma ray's energy must be slightly greater than the natural

energy, because in both cases energy is lost to recoil.



This means that nuclear resonance (emission and absorption of the same gamma ray

by identical nuclei) is unobservable with free nuclei, because the shift in energy is too great

and the emission and absorption spectra have no significant overlap.

Nuclei in a solid crystal, however, are not free to recoil because they are bound in

place in the crystal lattice.

When a nucleus in a solid emits or absorbs a gamma ray, some energy can still be lost

as recoil energy, but in this case it always occurs in discrete packets called phonons

. Any whole number of phonons can be emitted, including zero, which is known as a

"recoil-free" event. Mössbauer found that a significant fraction of emission and absorption

events will be recoil-free, which is quantified using the Lamb–Mössbauer factor.

it means that gamma rays emitted by one nucleus can be resonantly absorbed by a sample

containing nuclei of the same isotope, and this absorption can be measured.

Certain atomic nuclei bound in a crystal emit gamma rays which can be used

as a probe of energy levels in other nuclei.

Sodium nitroprusside is a common reference material

In the resulting spectra, gamma ray intensity is plotted as a function of the source velocity

. At velocities corresponding to the resonant energy levels of the sample, a fraction of the

gamma rays are absorbed, resulting in a drop in the measured intensity and a corresponding

dip in the spectrum.

The number, positions, and intensities of the dips (also called peaks; dips in transmitted

intensity are peaks in absorbance) provide information about the chemical environment of the

absorbing nuclei and can be used to characterize the sample.


https://en.wikipedia.org/wiki/Lamb%E2%80%93M%C3%B6ssbauer_factor

https://en.wikipedia.org/wiki/Phonon

https://en.wikipedia.org/wiki/Crystal


2225DOPPLER EFFECTS

The Doppler effect is the change in frequency or wavelength of a wave in relation

to observer who is moving relative to the wave source.


https://en.wikipedia.org/wiki/Observer_(physics)

https://en.wikipedia.org/wiki/Wave

https://en.wikipedia.org/wiki/Wavelength

https://en.wikipedia.org/wiki/Frequency


2325 ISOMER SHIFT

Isomer shift (δ) is the shift in the resonance energy of a nucleus due to the transition of

electrons within its s orbital.

The whole spectrum is shifted in either a positive or negative direction depending upon the s

electron charge density.

This change arises due to alterations in the electrostatic response between the non-zero

probability s orbital electrons and the non-zero volume nucleus of the orbit.

Only electrons in s orbitals demonstrate non-zero probability because their 3D spherical

shape incorporates the volume taken up by the nucleus.

However, the p, d, and other electrons may influence the s electron density through

a screening effect.

Isomer shift can be expressed using the formula

CS = K [Re2 - Rg

2 ] [Ψs2(0)]a - [Ψs

2(0)]b

Where CS – Chemical Shift, K is a nuclear constant,

the difference between Re2 and Rg

2 is the effective nuclear charge radius difference between

excited state and the ground state,

the difference between [Ψs2(0)]aand [Ψs

2(0)]b is the electron density difference on the nucleus

(a = source, b = sample).

.

2425 QUADRUPOLE INTERACTIONS ( Electron-Neutron Hyperfine Interactions.)


https://en.wikipedia.org/wiki/Screening_effect

https://en.wikipedia.org/wiki/File:Mossbauer_Isomer_Shift_and_Quadrupole_Splitting_for_57Fe.svg


For nucleus having I > ½ ( I = 1, 3/2, 2, 5/2 ….) , the distribution of electric charge ,

is unsymmetrical which leads to quadrapole moment ( four charges).These nuclei are

referred to quadrapole nuclei

Symmetric distribution of charge unsymmetric distribution of charge

( dipole I = 0, ½ ) (quadrapole, I > ½ )

Quadrupole splitting reflects the interaction between the nuclear energy levels and

surrounding electric field gradient (EFG).

Nuclei in states with non-spherical charge distributions, i.e. all those with angular quantum

number (I) greater than 1/2, produce an asymmetrical electric field which splits the nuclear

energy levels.

In the case of an isotope with a I=3/2 excited state, such as 57Fe or 119Sn, the 3/2 to 1/2

transition is split into two substates mI=±1/2 and mI=±3/2.

These appear as two specific peaks in a spectrum, sometimes referred to as a 'doublet'.

Quadrupole splitting is measured as the separation between these two peaks and reflects the

character of the electric field at the nucleus.

The Quadrupole splitting can be used for determining oxidation state, spin state, site

symmetry and the arrangement of ligands.

The ground to excited state transitions appear as two specific peaks in a spectrum,

[ "doublet".]

Quadrupole splitting is measured as the separation between these two peaks and reflects

the character of the electric field at the nucleus.



2425 MAGNETIC INTERACTIONS

Magnetic splitting (hyperfine splitting) is a result of the interaction between the nucleus and

any surrounding magnetic field.

A nucleus with spin, I, splits into 2I + 1 sub-energy levels in the presence of magnetic field.

For example, a nucleus with spin state I= 3/2 will split into 4 non-degenerate sub-states with

mI values of +3/2, +1/2, -1/2 and −3/2.

The selection rule of magnetic dipoles means that transitions between the excited state and

ground state can only occur where mI changes by 0 or 1.

Magnetic splitting of the nuclear energy levels and the corresponding Mössbauer spectrum

This gives six possible transitions for a 3/2 to 1/2 transition.

This gives 6 possible transitions for a 3/2 to 1/2 transition. Therefore only 6 peaks

can be monitored in a spectrum produced by a hyperfine splitting.

The extent of splitting is proportional to the magnetic field strength at the nucleus.


https://en.wikipedia.org/wiki/File:Mossbauer_-Magnetting_Splitting.jpg


Therefore, the magnetic field can be readily determined from the spacing between the

outer peaks.

Combination of all

The three Mössbauer parameters: isomer shift, quadrupole splitting, and hyperfine

splitting can often be used to identify a particular compound by comparison to spectra for

standards.

In some cases, a compound may have more than one possible position for the Mössbauer

active atom. For example, the crystal structure of magnetite (Fe3O4) supports two

different sites for the iron atoms. Its spectrum has 12 peaks, a sextet for each potential

atomic site, corresponding to two sets of Mössbauer parameters.

Many times all effects are observed: isomer shift, quadrupole splitting, and magnetic

Zeeman effect.

In such cases the isomer shift is given by the average of all lines.

The quadrupole splitting when all the four excited substates are equally shifted (two

substates are lifted and other two are lowered) is given by the shift of the outer two lines

relative to the inner four lines (all inner four lines shift in opposite to the outer most two

lines).

In addition, the relative intensities of the various peaks reflect the relative concentrations

of compounds in a sample and can be used for semi-quantitative analysis.

Also, since ferromagnetic phenomena are size-dependent, in some cases spectra can

provide insight into the crystallite size and grain structure of a material.

.


https://en.wikipedia.org/wiki/Magnetite

https://en.wikipedia.org/wiki/File:Mossbauer_Isomer_Shift_and_Quadrupole_Splitting_for_57Fe.svg


UNIT IVELECTROCHEMISTRY OF SOLUTION

1. Mean ionic activity and activity coefficient:

2. Concept of ionic strength,

3. Debye-Huckel theory of strong electrolytes

4. Activity coefficient of strong electrolytes-

5. Determination

6. Debye Huckel limiting law at appreciable concentration of electrolytes –

7. Debye Huckel Bronsted equation-qualitative and quantitative verification.

8. Redox reaction:

9. Cell potential,

10. Galvanic cell,

11. Electrolytic cell,

12. Nernst equation for cell potential of electrolyte.

13. Electrode equilibrium

14. Thermodynamic electrodes and electrode potential,

15. Electrochemical cells and electromotive force.



115MEAN IONIC ACTIVITY AND ACTIVITY COEFFICIENT:

215 CONCEPT OF IONIC STRENGTH,

Ionic strength:

It is the measure of electrical intensity, due to the presence of ions in the solution. It is given by half of the sum of all the terms obtained by multiplying the molality of each ion by the square of its valency.

μ = ½ ( m1z1 2 + m2 z22 + m3z3 2+.................)

1.Calculate the ionic strength of 0.15 molal KCl solution

m+ = 0.15 m- = 0.15 , z+ = 1, z- = 1

μ = ½ ( m1z1 2 + m2 z22 + m3z3 2+.................)

μ = ½ ( (0.15)(1) 2 + (0.15)(1) 2 )

= 0.152. .Calculate the ionic strength of 0.25 molal K 2SO4 solution

m+ = (2 ×0.25 ) m- = 0.15 , z+ = 1, z- = 2

μ = ½ ( m1z1 2 + m2 z22 + m3z3 2+.................)

μ = ½ ( (2 ×0.25 )(1) 2 + (0.25)(2) 2 ) = 0.752. .Calculate the ionic strength of solution which is 0.1 M KCl and 0.2M molal K 2SO4

m1 = 0. 1 + (2 ×0.2 ) = 0.5

m-2 = 0.1 z1 = 1, z2 = 1 , z3 = 2

μ = ½ ( m1z1 2 + m2 z22 + m3z3 2+.................)

μ = ½ ( ( 0.5¿ (1) 2 + ( 0.1¿ (1) 2 + (0.2)(2) 2 ) = 0.7

315 DEBYE-HUCKEL THEORY OF STRONG ELECTROLYTES

DEBYE- HUCKEL THEORY OF STRONG ELECTROLYTE:

According to this theory,



1.The strong electrolytes are completely ionised in solution at all concentrations.

2.If the solution is very dilute, the distance between the ions is large and hence the

electrostatic force is very less

3.If the solution is of high concentration, the inter ionic force will be more and the

electrolyte will remain in pairs such as A+B - ,known as ionic doublet and hence

conductance decreases

4. Each ion is surrounded by ions of opposite charge giving rise to an ionic atmosphere

Thus a cation is surrounded by a group of anions as represented below.

6. The movement of ion is slow down by three factors namely viscous force, Asymmetric

effect and Electrophoretic effect.

1. Viscous effect( frictional force):

It is due to viscous drag of the solvent on the movement of ions.

The frictional force F1 = Ki u where K - coefficient of friction of solvent.

2.Asymmetric effect ( relaxation effect ):

On passing current, the movement of ion becomes slow. This is known as

asymmetric effect or relaxation effect.

Relaxation force F 2 = ∈2 z kwV

6 D KT

∈ - constant, z - charge per unit volume, k - Reciprocal of thickness,

w - 2- √2, V - Applied potential, D- Dielectric constant, K - Frictional

coefficient, T - Temperature

3.Electrophoretic effect:

The ionic atmosphere associated with water of hydration t slows down the movement

of ions. This effect is known as electrophoretic effect.



Electrophoretic force F3 = ∈ z k6 πη Ki V

where Ki - coefficient of frictional resistance

The driving force due to the applied electric field is balanced by the three forces

∴ ∈Z V = F1 + F2 +F3

using this we can derive

ƛ = ƛ 0 - { 82.4

η (DT )1/2 + 8.2× 105( DT )3 /2

ƛ0} √ C

. = ƛ 0 - ( A+ B ƛ0 ) √ C where A = 82.4

η (DT )1/2 ,and B = 8.2× 105( DT )3 /2

This is known as Debye- Huckel – Onsager( DHO) conductance equation

ƛ = - ( A+ B ƛ0 )√ C + ƛ 0

The plot of ƛ versus √ C is a straight line.

from the graph equivalent conductance at infinite dilution is = 0.25 units

Testing:

1. Plot of ƛ versus √c should be a straight line

2. The slope should be equal to A+ B ƛ0

Problem: When the observed equivalent conductance were plotted against the square root

of the corresponding concentration the the straight line obtained had a slope of 0.05 and if ƛ0



is equal to 0.01 and B is equal to 0.2, the value of A is

Solution:

A+ B ƛ0 = 0.05

A+ (0.2)( 0.01) = 0.05

A+ 0.002 = 0. 0.05

A = 0.05 – 0.002

= .048

Problem: What will be the slope expected if the Onsager constants were found to beA=

60.0 and B = 1.5 The equivalent conductance at infinite dilution was 1.5 × 10−2

Solution:

A+ B ƛ0 = 60 .0 + 1.5 × 1.5 × 10−2

= 60 + 2.25 × 10−2

= 60 +0.225

= 60.225

For bi-bi valant solute(CuSO4) the plot was not straight line but is concave to the axis of

the latter parameter.

In non- aqueos solutions, if the dielectric constant is very low, the deviation is more. For

example the onsager slope of KI in water ( D= 78.5) was found to be 73 ( calculated = 80)

but that in acetone ( D= 20 ) the slope was found to be 1000 ( calculated 638)

This is due to the fact that in non – aqueous solvents of low dielectric constant , the solute

behaves incompletely dissociated

415ACTIVITY COEFFICIENT OF STRONG ELECTROLYTES-



515DETERMINATION

615DEBYE HUCKEL LIMITING LAW AT APPRECIABLE

CONCENTRATION OF ELECTROLYTES –

Debye - Huckle limitng law ( DHLL) :

The relation between activity coefficient and ionic strength is called DHLL. This is given by log f = - Az2√ μ

where f = activity coefficient, A – constant, z- valency of the ionThis shows that the activity coefficient of an ion

1. Decreases with increasing ionic strength of the solution.2. Decreases when the valency is higher3. Decreases when the dielectric constant is lower4. Decreases when the temperature is lower

similarly mean activity coefficient (f ± ) is given by log f ± = - A z+ z - √ μ

This is called limiting law because the assumptions made during derivation is applicable at infinite dilution only.

Derivation:

The chemical potential and activity coefficient are related as

RT lnf = 𝛍

= δGδN where G – free energy and N – total number of

ions

The change in free energy is equal to the work done to bring an ion of charge ‘ze’ from the

ionic atmosphere to centre of the ion. Therefore

RT lnf = δWδN ------------------------------1

The work done is given by W = - N z2 e2

3 D× k where k = ( 4 π∈2

D KT ∑ ni z i

2 ¿ ½

But n = cN

1000



W = - N z2 e2

3 D× ( 4 π∈2

D KT ×

N1000 ∑ ci z i

2¿ ½

Let A = - z2 e2

3 D× ( 4 π∈2

D KT ×

11000 ∑ ci z i

2¿ ½

W = N ×A ×√N

δWδN = A ×

δδN ( N

32 )

= A × 32 × N

12

= - z2 e2

3 D× ( 4 π∈2

D KT ×

11000 ∑ ci zi

2¿ ½ × 32 × N

12

= - z2 e2

2 D× ( 4 π∈2

D KT ×

11000 ∑ ci zi

2¿ ½ × N12

= - z2 e2

2 D× ( 4 π∈2

D KT ×

N1000 ∑ ci zi

2¿ ½

𝜇 = ½ × N

1000 ∑ ci z i2

Therefore the above equation becomes

δWδN = - z2 e2

2 D× ( 4 π∈2

D KT × 2μ ) ½


RT lnf = - z2 e2

2 D× ( 8 π∈2

D KT ) ½ ×√μ

log f = 1

2.303 RT - z2 e2

2D× ( 8 π∈2

D KT ) ½ ×√μ

let A = 1

2.303 RT - e2

2 D× ( 8 π∈2

D KT ) ½

log f = - A z2 √μ

This is known as Debye - Huckle limitng law .

For appreciable concentration

The ionic size factor ( a) should be introduced and hence the above equation becomes

log f = - A z+¿ z−¿

1+aB √ μ¿¿ μ ½

This is the expression for mean ionic coefficient.When polarisation of solvent molecules is involved the equation becomes

log f = - A z+¿ z−¿

1+ aB √ μ¿¿ μ ½ + C √μ



This is known as Huckel equation and C is called salting – out constant. This can be reduced as log f± = -A z+¿ z−¿¿ ¿ μ ½ + C μ

This is known as Debye – Huckel – Bronsted equation

715 DEBYE HUCKEL BRONSTED EQUATION-QUALITATIVE AND

QUANTITATIVE VERIFICATION

Qualitative verification of DH equations:According to limiting law equation the plot of log f against √μ should be a straight

line with slope - Az+ z_ ( I). If the ionic size factor is introduced the plot will be as II. The addition of salting out factor results further increase of activity coefficient by an amount proportional to ionic strength the result is the log f vs √μ curve is of the form III.

log f

√u √μ

Quantitative tests of the Debye –Huckel limiting equation:

1. Influence of valences on activity coefficient:

Debye - Huckle limitng law is

log f = - Az2√ μ where f = activity coefficient

To study the influence of valances on activity coefficient, for number of electrolytes of different valence types activity coefficients in aqueous solution are experimentally determined. The values of ‘ log f’ are plotted against square root of ionic strength.

Since the Debye Huckel constant A for water at 25 oC, is 0.509 the limiting slopes of the plots should be – 0.509.


III

II

I

KCl (uni-uni valent)

CaCl2 ( bi-univalent)

ZnSO4 ( bi-bi valent)


log f

√μIt is evident from the graph that the experimental results are in harmony as

expected by Debye – Huckel limiting law.

2.Influence of dielectric constant:According to Debye’s limiting law

log f = - 1.823 × 10 6 × z2

( DT )32× √𝛍 where D is dielectric constant of the medium

For electrolytes of same valence types, the limiting slope of the plot of ‘log f ‘

against √𝛍 at constant temperature should be inversely proportional to D32 .. The activity

coefficient of a given electrolyte in a number of different media of varying dielectric

constant is plotted against square root of ionic strength.

- log f

√μThe experimental results obtained are in agreement with the theoretical

requirements.

3.Influence of temperature:


KCl (uni-uni valent)

CaCl2 ( bi-univalent)

ZnSO4 ( bi-bi valent)

D = 9

D = 20

D = 80


It is not easy to vary the temperature, without changing the dielectric constant and so

these factors may be considered together. The slope of plot of log f against √μ should vary

as 1

( DT )32 where T is the absolute temperature at which the activity coefficients are

measured. The experimental results obtained for liquid NH3 at – 75 0 C are in agreement

with the theoretical requirements.

815REDOX REACTION:

A redox (oxidation-reduction) reaction is any chemical reaction in which the

oxidation

number of a molecule, atom, or ion changes by gaining or losing an electron.

An example of a redox reaction is the thermite reaction, in which iron atoms

in ferric oxide lose (or give up) O atoms to Al atoms, producing Al2O3.

Fe2O3(s)+2Al(s)→Al2O3(s)+2Fe(l)

915CELL POTENTIAL,

Electrode potential is defined as the tendency of the metal to undergo

oxidation(loss of electron) or reduction(gain of electron), when it is in contact

with its own salt solution. If the metal undergoes oxidation, the electrode

potential is called oxidation potential. If the metal undergoes reduction then, the

potential developed is called reduction potential.

Standard electrode potential:

It is defined as the tendency of the metal to undergo oxidation or reduction when

it is in contact with its salt solution with 1M concentration and 250C and at 1atm

pressure. It is denoted as E0.



1015 GALVANIC CELL,

These are cells in which electricity is generated due to chemical reaction . It

differs from chemical cells in the following way,

Half-cell. Each half of an electrochemical cell, where oxidation occurs and the half

where reduction occurs, is called the half cell.

Cell diagram or Representation of a Cell

A cell diagram is an abbreviated symbolic depiction of an electrochemical cell. For this

purpose, we will consider that a cell consists of two half-cells. Each half-cell is again made of

a metal electrode in contact with metal ion in solution.

IUPAC Conventions.

1. Anode half-cell is written on the left and cathode half-cell on the right.

2. The two half-cells are separated by a double vertical line (salt bridge) in between

3 The double vertical line represents the salt bridge, permitting ion flow while preventing the

electrolyte from mixing



4.The electrode in anode half-cell is on the left, while in cathode it is on the right of the

metal ion.

5. a single vertical line (|) represents a phase boundary between metal electrode and ion

solution (electrolyte).

Thus the two half-cells in a voltaic cell are indicated as

Zn | Zn2+ | | Cu2+ | Cu

Anode Half-Cell Cathode Half-Cell

(5) The symbol for an inert electrode, like the platinum electrode is often enclosed in a

bracket. For example,

Mg | Mg2+ | | H+ | H2(Pt)

(6) The value of emf of a cell is written on the right of the cell diagram.

Thus a zinc-copper cell has emf 1.1 V and is represented as

Zn | ZnSO4 | | CuSO4 | Cu E = + 1.1 V

S.NO CHEMICAL CELL ELECTROCHEMICAL CELL

1 Those cells in which chemical reaction

takes place, because of electric current

are called chemical cells.

Those cells in which electricity is generated

due to chemical reaction ,is called as

electrochemical cells.

2 Electrical energy is converted to

chemical energy.

Chemical energy is converted to

electrochemical energy.

3 Positive electrode – Anode

Negative electrode – Cathode

Positive electrode – Cathode

Negative electrode – Anode

4. Example: electrolytic cell Example: Denial cell, Lechanche cell

1115 ELECTROLYTIC CELL,

An electrolytic cell is an electrochemical cell that drives a non-

spontaneous redox reaction through the application of electrical energy.

They are often used to decompose chemical compounds, in a process

called electrolysis.

. Electroplating (e.g. of copper, silver, nickel or chromium) is done using an electrolytic cell.


https://en.wikipedia.org/wiki/Electroplating

https://en.wikipedia.org/wiki/Electrolysis

https://en.wikipedia.org/wiki/Redox

https://en.wikipedia.org/wiki/Electrochemical_cell


1215 NERNST EQUATION FOR CELL POTENTIAL OF ELECTROLYTE.

(Effect of concentration on Electrode potential )

For any redox reaction the change in free energy is given by Vant Hoff

equation as

∆G = ∆G0 + RT ln[ P][ R]

w here ∆G is change in free energy,

∆G0 is change in free energy at standard condition

R is gas constant,

T is temperature.

If ‘ n’ electrons are transferred and if F denotes the faraday of charge. ,E denotes the

potential then the free energy during the reaction is given by ∆G = - nEF. where the negative

sign indicates that the free energy decreases.

Similarly ∆G0 = -nE0F where E0 is the standard electrode potential.

Substituting in equation 1 we get,

-nEF = -nE0F + RT ln [ P][ R]

Dividing equation 2 by –nF we get

E = E0 - RTnF ln

[ P][ R]

This is the general Nernst equation for any redox reaction.

Case -1

Consider a redox reaction: Mn+ + ne- M



Applying Nernst equation we get,

E = E0 - RTnF ln [ M ]

¿ ¿¿

= E0 – 2.303 RTnF log [ M ]

¿¿¿

Since [M] >> [Mn+], assuming [M] = 1. We get

E = E0 – 2.303 RTnF log

1M n+¿¿

Therefore E = E0 + 2.303 RTnF log [Mn+]

At 250C, T = 298 ; R = 8.314 ; F = 96450. Substituting all these values in above

equation we get

E = E0 + 0.0591

n log [Mn+]

This is the Nernst equation for a reduction reaction at 250C.

Case -2

For oxidation reaction M Mn+ + ne- , the Nernst equation becomes

E = E0 - RTnF ln ¿¿¿

Since [M] >> [Mn+] and assuming [M] = 1. At 250C

E = E0 – 0.0591

n log [Mn+]

APPLICATIONS:

1.To determine the electrode potential ,at any given concentration.

Problem : What is the potential of a half-cell consisting of zinc electrode in 0.01 M

ZnSO4 solution 25oC. Eo = 0.763 V. [0.8221 V]

2. To determine the emf of a cell.

Ecell = Eright – E left

Problem : Calculate the emf of the cell. Zn | Zn2+ (0.001 M) || Ag+ (0.1 M) | Ag.The

standard potential of Ag/Ag+ half-cell is + 0.80 V and Zn/Zn2+ is–0.76 V. [1.56 V]

Problem : Calculate the E.M.F. of the zinc - silver cell at 25oC when [Zn2+] = 0.10 M

and [Ag+] = 10 M. (Eo cell at 25oC = 1.56 volt]



Problem : Calculate the standard e.m.f. of the reaction Fe3+ + 3e– Fe(s). Given the

e.m.f. values of Fe3+ + e Fe2+ and Fe2+ + 2e Fe(s) as + 0.771V and –0.44 V

respectively.

3.Calculation of Equilibrium constant for the cell reaction

The Nernst equation for a cell is

Ecell = Eocell – 0.059

n log K

log K = n × E0

0.059

Problem : The standard electrode potentials of the half cells Ag+ / Ag and Fe3+,

Fe2+ / Pt are

0.7991 V and 0.771 V respectively. Calculate the equilibrium constant of the

reaction :

Ag(s) + Fe3+ Ag+ + Fe2+

1315 ELECTRODE EQUILIBRIUM

1415THERMODYNAMIC OF ELECTRODES AND ELECTRODE

POTENTIAL,

An electrochemical cell contains three open thermodynamic systems that, in

dynamic equilibrium, equalize their electrochemical potentials with that of their

surrounding by forming an electric-double-layer-capacitor at the interface of the

electrolyte with each of the two electrodes.

Since the electrode/electrolyte interfaces are heterojunctions, the

electrochemical potentials or Fermi levels of the two materials that contact the

electrolyte at the two electrodes determine the voltage of a cell.

The voltage is the sum of the voltages of the two interfacial electric-double-

layer capacitors at the two electrode/electrolyte interfaces.



1515 ELECTROCHEMICAL CELLS.[ CONCENTRATION CELLS]

Those galvanic cells in which electrical energy is produced only due to

difference in concentration, are called concentration cells

Let the concentration of the electrolyte in right side electrode is C2 and that of

left side electrode is C1 then

Eright = E0 + 0.0591

n log [C2]

Eleft = E0 + 0.0591

n log [C1]

Ecell = Eright – E left

= 0.0591

n log [C2]- + 0.0591

n log [C1]

emf = 0.059

n log C2

C1

1616 ELECTROMOTIVE FORCE

Electromotive force is the electric potential generated by either a electrochemical

cell or a changing magnetic field. It is also known as voltage.


https://www.thoughtco.com/voltage-2699022

https://www.thoughtco.com/electrical-energy-definition-and-examples-4119325


UNIT VQUANTUM CHEMISTRY – IV

1. Approximation Methods

2. Perturbation Method

3. Application Of Perturbation Method To Hydrogen

4. Application Of Perturbation Method To Helium Atom

5. Variation Method



6. Application Of Variation Method To Hydrogen

7. Application Of Variation Method To Helium Atom

8. R.S. Coupling

9. Term Symbols For Atoms In The Ground State

10. Slater Orbital

11. Hf –Scf Methods

12. Born – Heimer Approximation

13. Valence Bond Theory For Hydrogen Molecule

14. Laco –Mo Theory For Di Atomic Molecules

15. Polyatomic Molecules

16. Concept Of Hybridization

17. Huckel Theory For Conjugated Molecules

1. Ethylene

2. Butadiene

3. Benzene

18. Semi Emprical Methods

118 APPROXIMATION METHODS

Need for approximation methods:

Schrodinger wave equation can be solved for systems having only one electron. But

if an atom or molecule has many interacting electrons , we can not solve Schrodinger

equation. Because there will be more than one potential energy terms. In such a case we

must go for approximation methods.

218 PERTURBATION METHOD



This method involves the determination of eigen functions( Ψ ‘ ) , and eigen values

( E’ ) of the of the perturbed system , in terms of those of the unperturbed (Ψ0 and E 0 )

This method is applied if

1.The system differs only slightly from the unperturbed ( known)system.

2.Energy , wave function and Hamiltonian for the unperturbed system ( E0 , Ψ 0 , and H0 ) are

known.

Example :

1.Normal Hydrogen atom is unperturbed system

When placed in an electric ( Stark effect), it causes perturbation.

When placed in magnetic field (Zeemann effect) it is perturbed system

2.A simple harmonic oscillator is an unperturbed system On the other hand an unharmonic

oscillator is a perturbed system.

There are two types

1. Time independent perturbation method

2. Time dependent perturbation method

Time independent perturbation method

Suppose we have a system with Time independent Hamiltonian H and we are unable

to solve the Schrodinger equation HΨ = E Ψ for the eigen value and eigen function.

There are two cases

1. Non – degenerate case

2. Degenerate case

Time independent non- degenerate perturbation method

Consider a system whose .Energy , wave function and Hamiltonian are known. Let

them be E0 , Ψ 0 , and H0 . If a small perturbation is applied, then

The Hamiltonian is decomposed in to two parts as

H = H0 + λ H’

where H0 is the perturbed part and λ H’ is the perturbation.

Let the eigen values and eigen functions of the unperturbed problem be E10 , E2

0 , E30 , and

Ψ 10 , Ψ 2

0 ,Ψ 30



It is assumed that the eigen functions and the energies can be expressed in the form of

power series as

Ψ = Ψ n0 + λ Ψ n

1 + λ 2 Ψ n2 .........Ψ n

k

E = En0 + λ En

1 + λ2 En2 ..........En

k

Where Ψ nk and En

k are the ‘ k th ‘ order correction in wave function and energy

The Schrodinger equation is given by

HΨ = E Ψ

Substituting the values, and omitting the higher terms

{ H0 + λ H’ }{ Ψ n0 + λ Ψ n

1 + λ 2 Ψ n2 }= { En

0 + λ En1 + λ2 En

2 }{Ψ n0 + λ Ψ n

1 + λ 2 Ψ n2 . }

Expanding

H0Ψ n0

+ H0 λ Ψ n1 + H0 λ 2 Ψ n

2 + λ H’ Ψ n0 + λ 2H’ Ψ n

1 + λ 3 H’ Ψ n2

= En0 Ψ n

0 + En0λ Ψ n

1 +En0 λ2Ψ n

2 + λ En1 Ψ n

0 + λ2 En1 Ψ n

1 + λ3 En1 Ψ n

2 + λ2 En1 Ψ n

0 + λ3 En1 Ψ n

1 + λ4 En2

Ψ n2

Neglecting the higher terms (λ3 and λ4 terms )

H0Ψ n0

+ H0 λ Ψ n1 + H0 λ 2 Ψ n

2 + λ H’ Ψ n0 + λ 2H’ Ψ n

1

= En0 Ψ n

0 + En0λ Ψ n

1 +En0 λ2 Ψ n

2 .+ λ En1 Ψ n

0 + λ2 En1 Ψ n

1 + λ2 En1 Ψ n

0

rearranging

H0Ψ n0

+ λ [ H0Ψ n1 + H’Ψ n

0 ] + λ 2 [ H0 Ψ n2 + H’ Ψ n

1 ]

= En0 Ψ n

0 + λ [ En0 Ψ n

1 + En1 Ψ n

0 ] + λ2 [ En0 Ψ n

2 .+ En1 Ψ n

1 + En1 Ψ n

0

]

Comparing constant term , the coefficient of λ and coefficient of λ 2 we get the Schrodinger

equation

H0Ψ n0

= En0 Ψ n

0 : unperturbed system.

H0Ψ n1 + H’Ψ n

0 = En0 Ψ n

1 + En1 Ψ n

0 : first order perturbation



H0 Ψ n2 + H’ Ψ n

1 = En0 Ψ n

2 .+ En1 Ψ n

1 +En1 Ψ n

0 : Second order perturbation

Evaluation of First order energy:

The Schrodinger equation for first order perturbation is

H0Ψ n1 + H’Ψ n

0 = En0 Ψ n

1 + En1 Ψ n

0

The unknown function Ψ n1

can be expanded in terms o known functions

Ψ 10 , Ψ 2

0 ,Ψ 30 ,…….Ψ m

0

Ψ n 1 = ∑0

m

Cm Ψ m0

Substituting in the above equation,

H0 ∑0

m

Cm Ψ m0 + H1 Ψ n

0 = En0 ∑

0

m

Cm Ψ m0 + En

1 Ψ n0

H0 Ψ m0 ∑

0

m

Cm+ H1 Ψ n0 = En

0 Ψ m ∑0

m

Cm0 + En

1 Ψ n0

[Rearranging]

since H0 Ψ m0 = Em

0 Ψ m 0 the above equation becomes

Em0 Ψ m 0

∑0

m

Cm + H1 Ψ n0 = En

0 Ψ m ∑0

m

Cm + En1 Ψ n

0

Em0 Ψ m

0 ∑0

m

Cm - En0 Ψ m ∑

0

m

Cm + H1 Ψ n0 = En

1 Ψ n0

[Rearranging]

Ψ m0 ∑

0

m

Cm { Em0 −En

0 } + H1 Ψ n0 = En

1 Ψ n0

------------------5

Multiplying the above equation by Ψ n0 from the left and integrate

∫Ψ n0Ψ m

0 ∑0

m

Cm {Em0 −En

0 } dτ+ ∫Ψ n0 H 1 Ψ n

0 dτ = ∫Ψ n0 En

1 Ψ n0

dτ

[ ∫Ψ n0 En

1 Ψ n0

=En1 ∫Ψ n

0Ψ n0

but ∫Ψ n0 H1 Ψ n

0dτ ≠ H 1∫Ψ n0Ψ n

0dτ ]




∫Ψ n0Ψ m

0 ∑0

m

Cm {Em0 −En

0 }) dτ + ∫Ψ n0 H1 Ψ n

0 dτ = En1∫Ψ n

0 Ψ n0 dτ

but ∫Ψ n0Ψ m

0 = 0,

∫Ψ n0Ψ n

0 = 1


0 + ∫Ψ n0 H1 Ψ n

0 = En1

∴ En1

= ∫Ψ n0 H 1 Ψ n

0

This is the expression for first order perturbed energy .This shows that the first order

perturbation energy for a non-degenerate system is just the perturbation function

averaged over the corresponding unperturbed state of the system

Thus the first order energy is

En = En0 + λ ∫Ψ n

0 H1 Ψ n0 dT

Evaluation of First order wave function:

From equation 5

Ψ m 0 ∑0

m

Cm { Em0 −En

0 }+ Ψ n0 H1 = Ψ n

0 En1

Multiplying the above equation by Ψ m0 from the right and integrate

∫Ψ m0 ∑

0

m

Cm {Em0 −En

0 }Ψ m0 dτ + ∫Ψ n

0 H1 Ψ m0 dτ = ∫ Ψ n

0 En1 Ψ m

0 dτ

[ ∫Ψ n0 En

1 Ψ n0

=En1 ∫Ψ n

0Ψ n0 but ∫Ψ n

0 H1 Ψ n0dτ

≠ H1∫Ψ n0Ψ n

0dτ ]


∑0

m

Cm {Em0 −En

0 } ∫Ψ m0 Ψ m

0 dτ + ∫Ψ n0 H1 Ψ m

0 dτ = En1 ∫ Ψ n

0 Ψ m0 dτ

But ∫Ψ m0 Ψ m

0 = 1,



∫ Ψ n0 Ψ m

0 = 0


∑0

m

Cm {Em0 −En

0 } + ∫Ψ m0 H 1 Ψ n

0 = 0

∴ Cm = - ∫Ψ m

0 H 1Ψ n0 dT

{Em0 −En

0 }

Substituting in 4, we get Ψ n = - ∑m=0

m ∫Ψ m0 H 1Ψ n

0d T

{Em0 −En

0 } Ψ m0

This shows that the first order correction in wave function

1. is always negative

2. can be determined from the eigen function and eigen values of unperturbed system.

Thus the first order wave function is Ψ n = Ψ n0 - λ ∑

m=0

m ∫Ψ m0 H 1Ψ n

0 d T

{Em0 −En

0 } Ψ m

0

318 APPLICATION OF PERTURBATION METHOD TO HYDROGEN

Perturbation theorem is applied to hydrogen atom to find the corrected ( perturbed )

energy and the corresponding perturbed wave function It involves the following steps.

1. Making perturbation:

Normal hydrogen atom is unperturbed but when it is kept in electric field it is

subjected to perturbation.

2.Writing the expression for wave function

The value of Ψ for hydrogen atom is identified Ψ =( e−r

√ π )2. Identifying the perturbed Hamiltonian:

The application of electric field changes the Hamiltonian of hydrogen atom

H= H0 + H ’ . In this H’ is the perturbed Hamiltonian.

3. Calculation of first order perturbed energy.

The perturbed energy is calculated by the following formula

The first order perturbed energy is given by E1 = ∫Ψ 0 H ’Ψ 0 dτ

In spherical co-ordinates E = ∫Ψ 0 H ’Ψ 0 r2 dr∫0

π

sin θ d θ∫0

2π

dφ



4. Adding with the ground state energy:

The perturbed energy is added with the ground state energy

For example When a uniform electric field along Z- axis is applied, the perturbation term

is given as

H ‘ = Fz.

The first order perturbed energy is given by E1 = ∫Ψ 0 H ’Ψ 0 dτ

In spherical co-ordinates E = ∫Ψ 0 H ’Ψ 0 r2 dr∫0

π

sin θ d θ∫0

2π

dφ and z = r cos θ

= ∫0

∞

( e−r

√ π )(F rcosθ)( e−r

√ π )r2dr∫0

π

sin θ d θ∫0

2 π

dφ

= Fπ ∫

0

∞

e−2 r (rcos θ)r2 dr∫0

π

sin θ dθ∫0

2 π

dφ

= eFπ ∫

0

∞

e−2 r r3 dr∫0

π

cos θ sin θ d θ∫0

2 π

dφ

= eFπ ∫

0

∞

e−2 r r3 dr 12∫0

π

sin 2θ d θ∫0

2 π

dφ [ sin 2x = 2 sin x cos x ]

= eFπ [

3!24 ] [ ½ (

−cos2 θ2 ) ¿0

π ( 2 π )

∫0

∞

xn e−ax dx= n!(a)n+1

= eFπ [

3!24 ] [ ½ ( 0)

= 0

This shows that there is no first order effect.

418 APPLICATION OF PERTURBATION METHOD TO HELIUM ATOM

Helium atom contains two electrons and one Helium nucleus. In this atom, besides an

attractive interaction between electron and nucleus there is a repulsive interaction between

the two electrons.



Let the distance between nucleus and electron-1 and electron-2 be r1 and r2

respectively. Let r12 represents the inter electronic distance. The potential energy for a system

of two electrons, and a nucleus of charge +e is

V = - 2r1

- 2r2

+ 1r12

The Hamiltonian(H) for He atom (in a.u) H = - 12 ∇1

2 -

12 ∇1

2 - 2r1

- 2r2

+ 1r12

The first two terms are operator for kinetic energy , the third and fourth terms are potential

energy of attraction between electron and nucleus and the last term is the potential energy of

inter electron repulsion. Because of this term, Schrodinger equation for the atom can not be

solved and we use approximation method

Perturbation method separates the Hamiltonian in to two parts namely H0 and H1

H = Ho +¿ H1

Ho = -{ 12 [ ∇1

2 + ∇1

2 ] -

2r1

- 2r2

is the Hamiltonian for

unperturbed system

H1 = 1

r12

The wave function of unperturbed( zeroth order) system is given by

Ψ 0 = √ Z3

π e−Z r 1× √ Z3

π e−Z r 2

= Z3

π×e−Z r 1e−Zr 2

The corresponding zeroth order energy is E0 = - Z2 (a.u)

The first order perturbation energy is the average value of the perturbation function

H’ = 1r12

, over the unperturbed state of the system

E1 = ∫Ψ 0 H ’Ψ 0 dτ



= ∫ Z3

π× e−Z r1 e−Z r2 ×( 1

r12)× Z3

π× e−Z r1 e−Z r 2 dτ

= ( Z3

π )2

∫¿¿ dτ

The volume element in spherical co ordinates is dτ = r12dr1sinθ1dφ1 r2

2dr2sinθ2dφ2

upon integration we get E 1 = 58 Z ,

The unperturbed energy is given by E 0 = - Z2

Therefore the total energy E = E 0 + E 1

= - Z2 + 58 Z (a.u)

= - 4 + 58 ×2 (a.u)

= - 114 × (27.2 ) eV

= -74.8 eV

This is the ground state energy of Helium atom. Experimental value is -78.98 Ev

518 VARIATION METHOD

Advantage over perturbation method:

It does not require that there should be a similar problem that has been solved

previously.

Method:

This method says that with any trial function Ψ, the expectation value of energy E

will be equal to or greater than the true value E0, which is the lowest energy eigen value

of the Hamiltonian of the system E ≥ E0

i.e if we choose number of wave functions, Ψ 1, Ψ2, Ψ3 etc and calculate the values

E1,E2,E3 , corresponding to them, then each of the values E , will be greater than the energy

E0.

Steps:

1. Choose series of trial functions that depend some arbitrary parameters α, β etc

called variational parameter.

2. Calculate E in each case, by postulate of quantum mechanics.



E = ∫Ψ H Ψ ¿dτ

∫Ψ Ψ ¿ dτ

3. Minimize E with respect to the parameter used.

4. Pick up the lowest value that would be closest to the true value .

5. The trial function corresponding to that value will be the best function.

Proof:

Consider the wave function Ψ, which is a linear combination of normalised and

orthogonal eigen functions φ1 and φ2 with normalization constants a1 ,a2 respectively.

Ψ = a1 φ1 +a2 φ2

From postulates of Quantum mechanics,


∫Ψ Ψ ¿ dτsubstituting the values of Ψ , we get

E = ∫(a 1φ 1+a 2φ 2) H ¿¿¿

Since Ψ is real function, Ψ * can be replaced by Ψ itself∴ E = ∫(a 1 φ 1+a 2 φ 2) H ¿¿¿

= a1

2∫φ 1 H φ 1 dτ+a22∫ φ 2 H φ2 dτ+a1 a 2∫φ 1 H φ 2dτ +a 1 a2∫φ 2 H φ 1 dτ

a12∫φ1

2d τ+a22∫φ2

2d τ+2a1 a2∫ (φ1φ 2)d τ

∫φ1 H φ 1dτ = E1

∫φ 2 H φ 2dτ = E2

∫φ1 H φ 2dτ = 0

∫φ 2 H φ 1dτ = 0

∫φ12 dτ = 1

∫φ22 d τ = 1 ,

∫φ1φ 2 dτ = 0

Substituting in the above equation E = a1

2 E1+a22 E 2

a12+a2

2

Subtracting E0 on both sides E - E0 = a1

2 E1+a22 E 2

a12+a2

2 - E0

= a1

2 E1+a22 E 2−E0(a1

2+a22)

a12+a2

2 [ taking LCM]

Taking only Denominator, E - E0 = a12 E 1+a2

2 E 2−E0(a12+a2

2)



= a12 E 1+a2

2 E 2−E0 a12−E0 a2

2 ¿

= a12 ( E1 - E0 ) + a2

2 ( E2 - E0 )

Since E0 is the lowest energy of the system, ( E1 - E0 ) and ( E2 - E0 ) are positive .

Therefore

E - E0 > 0

E > E0

This shows that the variation method provides an upper bound to the ground

state energy of the system.

SECULAR EQUATION:

Consider the wave function Ψ, which is a linear combination of normalised and

orthogonal eigen functions φ1 and φ2 with normalization constants a1 ,a2 respectively.

Ψ = a1 φ1 +a2 φ2

From postulates of Quantum mechanics,


∫Ψ Ψ ¿ dτsubstituting the values of Ψ , we get

E = ∫(a 1 φ 1+a 2 φ 2) H ¿¿¿ ---------------1

Since Ψ is real function, Ψ * can be replaced by Ψ itself∴ E = ∫(a 1φ 1+a2 φ 2) H ¿¿¿

= a1

2 H11+a22 H22+2 a1 a2 H 12

a12 S11+a2

2 S22+2 a1 a2 S12

E ( a12 S11+a2

2 S22+2a 1 a 2 S12¿ = a12 H11+a2

2 H22 +2 a1 a 2H12

Differentiating with respect to a1

∂ E∂ a1

( a12 S11+a2

2 S22+2 a1 a2 S12¿ + E ( 2a1 S11+0+2a2 S12 ¿

= 2a1 H11+0 +2 a 2 H12

∂ E∂ a1

= 0


E ( 2a1 S11+2a2 S12¿ = 2a1 H 11+2 a 2H 12

E (a1 S11+a2 S12 ¿ = a1 H 11+a2 H12 [ dividing by 2 ]

Ea1 S11+E a2 S12 = a1 H 11+a 2 H 12

a1¿ ) + a2¿¿ - E S12¿ = 0



Similarly, a1¿ ) + a2¿¿ - ES22¿ = 0

These are called secular equations.

This shows that a trial function depends linearly on the variation parameter leads to

secular equation and secular determinant.

618 APPLICATION OF VARIATION METHOD TO HYDROGEN

Variation theorem is applied to hydrogen atom to find the minimum ground state

energy and the corresponding acceptable wave function It involves the following steps.

1. Selection of wave function

Many acceptable wave functions , in terms of variation parameter are chosen to calculate the

energy

2. Writing Hamiltonian:

Hamiltonian operator of hydrogen atom in spherical co-ordinates is

H = −12r 2

ddr

¿ ) - 1r ( in atomic units).

3. Calculation of energy in terms of variation parameter.

The expectation value is calculated by quantum mechanics


∫Ψ Ψ ¿ dτ

4. Estimation of variation parameter:

After getting energy in terms of variation parameter, the expression should be to

minimised. To do this the function should be differentiated with respect to the parameter

and equated to zero. From this the value of the parameter can be obtained.

5. Calculation of energy:

The value of parameter should be substituted in the energy expression to obtain the

exact value of energy.

For example the acceptable wave functions for hydrogen atom with ‘a’ as variation parameter

are

Ψ = e−ar, Ψ =e−a r2

, Ψ = r e−ar

Hamiltonian operator of hydrogen atom in spherical co-ordinates is



H = −12r 2

ddr

¿ ) - 1r ( in atomic units).



∫Ψ Ψ ¿ dτ

= ∫

0

∞

Ψ H Ψ r2 dr∫0

π

sinθ dθ∫0

2π

dφ

∫Ψ Ψ ¿r2 dr∫0

π

sin θ dθ∫0

2 π

dφ d𝛕 = [r2 dr∫

0

π

sin θ d θ∫0

2 π

dφ¿

= ∫

0

∞

(e−ar )(−12 r2

ddr (r2 d

dr )−1r )( e−ar ) r2 dr∫

0

π

sin θ d θ∫0

2 π

dφ

∫ ( e−ar ) (e−ar ) r2 dr∫0

π

sinθ d θ∫0

2 π

dφ

Similarly for other functions also energy is calculated.

For the wave functions Ψ =e−ar , the expectation value of energy is E=¿is ( a2

2 - a )

For Ψ = e−a r 2

is E=¿ 3∝2 -√ 8 α

π where α is the parameter

For Ψ = r e−ar the expectation value of energy is E=α2

6 -

α2 .

To find the parameter ’a’ , E should be minimized .

Its first order derivative with respect to the parameter ’a’ is

dEda = a – 1

This should be equal to zero.

dEda = 0

0 = a – 1

∴ a = 1

Substituting the value of a , we get

E = ½ - 1

= - 0.5

Similar treatment should be done for other energies and the function which gives lowest

energy is the best wave function.



EXAMPLE -2

The ground state energy of H- atom can be calculated using variation method.

Consider the wave function is Ψ = e−ar, where ‘a’ is the variation parameter. The

Hamiltonian for H- atom in spherical co ordinate is H = (−12 r2

ddr (r2 d

dr )−1r )



∫Ψ Ψ ¿ dτ

= ∫

0

∞

Ψ H Ψ r2 dr∫0

π

sinθ dθ∫0

2π

dφ

∫Ψ Ψ ¿r2 dr∫0

π

sin θ dθ∫0

2 π

dφ d𝛕 = [r2 dr∫

0

π

sin θ d θ∫0

2 π

dφ¿

= ∫

0

∞

(e−ar )(−12 r2

ddr (r2 d

dr )−1r )( e−ar ) r2 dr∫

0

π

sin θ d θ∫0

2 π

dφ

∫ ( e−ar ) (e−ar ) r2 dr∫0

π

sinθ dθ∫0

2 π

dφ

Since the 𝛉 term and 𝛗 term are constants, they can be cancelled. Therefore the above

equation becomes, E = ∫

0

∞

(e−ar )(−12 r2

ddr (r2 d

dr )− 1r )( e−ar ) r2 dr

∫ ( e−ar ) (e−ar ) r2 dr

= a2

2 - a [Upon integration]



dEda = a – 1


dEda = 0

0 = a – 1

∴ a = 1




E = ½ - 1

= - 0.5 this is the true value.

Problem : Use the functions Ψ = e−ar to calculate the ground state energy of H-atom by

variation

method. Given ∫ e−ar H e−ar dτ = a−28 a2 and ∫ e−ar e−ardτ =

14 a3

Solution:



∫Ψ Ψ ¿ dτ

= a−28 a2 × 4 a3

1

= (a−2 )a

2

= a2

2 - a



dEda = a – 1


dEda = 0

0 = a – 1

∴ a = 1


E = ½ - 1

¿−0.5

718 APPLICATION OF VARIATION METHOD TO HELIUM ATOM



Helium atom contains two electrons and one Helium nucleus. In this atom, besides an

attractive interaction between electron and nucleus there is a repulsive interaction between

the two electrons.

Let the distance between nucleus and electron-1 and electron-2 be r1 and r2

respectively. Let r12 represents the inter electronic distance. The potential energy for a system

of two electrons, and a nucleus of charge +e is

V = - 2r1

- 2r2

+ 1r12

The Hamiltonian(H) for He atom (in a.u) H = -{ 12 [ ∇1

2 + ∇1

2 ] -

2r1

- 2r2

+ 1r12

The first two terms are operator for kinetic energy , the third and fourth terms are

potential energy of attraction between electron and nucleus and the last term is the potential

energy of inter electron repulsion

The variational method approximation requires that a trial wavefunction with one or

more adjustable parameters be chosen.

1. Choosing Trial Wave function

The calculations begin with a wave function in which both electrons are placed in a

hydrogenic orbital with scale factor α. The wave function is given by

Ψ(1,2)=Φ(1)Φ(2)

= a3

π e−α (r 1+r2)

2. Calculation of energy in terms of variation parameter

E = ∫Ψ H Ψ dτ

= ∫Ψ [−{12[∇1

2+∇22]− 2

r1− 2

r2+ 1

r12]Ψ dτ

= −12 ∫Ψ ∇1

2Ψ dτ −12 ∫Ψ ∇2

2Ψ dτ - ∫Ψ [ 2r1 ¿

¿ ]Ψ dτ - ∫Ψ [ 2r2

]Ψ dτ + ∫Ψ [ 1r12

]Ψ dτ

Substituting the values and on integration we get



E = a2 - 278 a

Where ‘a ‘ is variation parameter

3.Determination of variation parameter

E = a2 - 278 a

Differentiating with respect to ‘a’

dEda = 2a –

278

To minimize this put dEda =0

2a = 278 -

∴ a = 2716

Substituting we get

E = (2716

¿¿2 – (278 ) ×(

2716

¿

= 2716 (

2716 -

278 )

= 2716 (

−2716 )

= −729256

= - 2.8476 a.u and the experimentally determined ground state energy is Eexp=−2.903

818 R.S. COUPLING [RUSSELL – SAUNDERS COUPLING ( LS –

COUPLING)

Combination of orbital angular momenta(L) and spin (S) of individual electrons is

called

RS – coupling. Total electronic angular momentum (J) is given by the.

J = L + S

Where L = total orbital angular momentum vector = ∑ Li

S = total spin angular momentum vector = ∑ Si

The value of L is given by L = ( l1 + l2), ( l1 + l2 -1)…. (l1-l2) where l1 and l2 are

azimuthal quantum numbers This series is known as Clebsch – Gordon series.



918 TERM SYMBOLS FOR ATOMS IN THE GROUND STATE

It is represented by writing the spin multiplicity as superscripts to the left of the

spectral state (term letters) with angular momentum quantum number (J) value as subscript

to the right.

TERM SYMBOL = spin multiplicity spectral state J

To find the Spectral state

Value of ‘ L’ 0 1 2 3 4 5 6

TERM LETTER S P D F G H I

For example if the value of L is 0, then the term letter is S.

if the value of L is 1, then the term letter is P and so on.

1. Find the ground state term symbol for d1 ion

Solution:

+2 +1 0 -1 -2

↑

Spin multiplicity = 2s + 1

= 2 ( ½ ) + 1

= 2 ( doublet)

L = 2

J = 2 + ½

= 5/2

Therefore the ground state symbol 2 D 5/2

2.. Find the ground state term symbol for d2 ion

Solution:

+2 +1 0 -1 -2

↑ ↑

s = ½ +½



= 1

Spin multiplicity = 2s + 1

= 2 (1 ) + 1

= 3 ( Triplet)

L = 2(1) + 1(1)

= 2 + 1

= 3

Spectral state is F

J = L+S, L+S -1, L+S-2…

= 4,3,2,1

Therefore the ground state symbol 3 F

1018 SLATER ORBITAL [ SLATER TYPE ORBITALS]

Hydrogen like orbitals are given by

Ψ nlm = N r F(r) e−Zr

n Y lm ( θ,φ)

Here N is the normalisation constant, Rnl is the radial function and Y lm

( θ,φ) is the angular part ( spherical harmonics)Slater suggested an orbital by taking into consideration ,the shielding effect of the

nucleus by the electrons.it is given by

Ψ’ nlm = N r n-1 e−Z ' r

n ' Y lm ( θ,φ)

Where n’ is the effective principal quantum number and Z’ is the effective nuclear charge.Functions of these oritals are called ‘Slater Type Orbitals’ (STO)

1. STO differ from H- like orbitals only in radial part.2. Unlike H-like orbitals, STOs have no radial nodes3. STO’s are not mutually orthogonal

SLATER RULES

1. The effective principal quantum number (n’) is related to the quantum number (n) as

follows


n 1 2 3 4 5 6

n' 1 2 3 3.7 4 4.2


2. The effective nuclear charge (Z’) is related with actual atomic number as

Z’ = Z-S, where ‘S’ is screening constant

3. Screening constant is calculated as follows.

1. The orbitals are grouped as

(1s), (2s,2p), (3s,3p), (3d),(4s,4p),(4d)

2. No contribution from electrons in groups higher than , the group in question

For example For 3s or 3p electrons , no contribution from 3d,4s,4p,4d

3. Each electron in the same group contributes 0.35

4. Electron in 1s group contributes 0.30 only

5. In case of ‘s’ or ‘p’ group, each electron in the next lower group contributes 0.85

6. For electron in ‘d’ or ‘f’ the contribution from each inner electron is 1.0

7. Each electron in groups lower by two or more numbers from the one in question

contributes 1.0

1. Determine the effective nuclear charge for the 1s electron in He.

Electronic configuration of He : 1s2. There are two electrons. The effective nuclear charge

on one electron = 1×(0.30) [ rule 4]

S = 0.3

Z’ = Z-S [ Z is the atomic number of He =2]

= 2-0.3

= 1.7

The effective nuclear charge for the 1s electron in He = 1.7 ( no unit)

2. Determine the effective nuclear charge of 2s and 2p electrons in C.

Electronic configuration of C : 1s2. 2s2 2p 2

These are grouped as (1s) (2s,2p)

There are 4 electrons in set 2

. The effective nuclear charge on one electron = 3×(0.35) [ rule 3 for remaining 3 electrons

+ 2(0.85)[ rule 5 for 2 electrons]

= 2.75

Z’ = Z-S [ Z is the atomic number of C =6]

= 6- 2.75



= 3.25

The effective nuclear charge for the 2s electron in C = 3.25 ( no unit)

Problem1 :Find the expression for STO for 2s, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.Solution:

Slater Type Orbitals is given by Ψ’ nlm = N’ r n-1 e−Z ' r

n ' Y lm ( θ,φ)

For 2s orbital, n= 2 , l=0. ∴

Ψ’ 200 = N r 2-1 e−Z ' r

n ' Y lm ( θ,φ)

= N r 2-1 e−Z ' r

n ' ×1

√4 π [ Y 0

0 ( θ,φ) = 1

√4 π]

To find Z’ ( effective nuclear charge of 2s electrons in N).

Electronic configuration of N : 1s2. (2s 2p ) 5 , These are grouped as (1s) (2s,2p)

There are 5 electrons in set 2

. The effective nuclear charge on one electron

= 4×(0.35) [ rule 3 for remaining 4 electrons + 2(0.85) [ rule 5 for

2 electrons]

= 3.10

Z’ = Z-S [ Z is the atomic number of N =7]

= 7- 3.1

= 3.9

n'= n=2

Substituting, we get

Ψ’ 200 = N r 2-1 e−3.9 r

2 1

√4 π

= N r e−3.9r

2 1

√4 π This is the expression for STO of N atom∴ Ψ’ 200 = N × (2 - zr ) × e –( zr

2 ) 1√4 π

Slater Type Orbital ΨS 200 = N× r ×e−3.9r

2 × 1

√4 π

Hydrogen like orbital Ψ 200 = N× (2 - 3.9 r ) × e –( 3.9 r2 ) 1

√4 π

Comparison:

There is no node in STO, but in HLO node occurs at 2

3.9



Problem 2 Find the expression for STO for 2px, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.Solution:The angular wave function is

Θ l,m = √ 2 l+12

× (l−m ) !(l+m ) !

× Plm ( cos θ ) ×

1√2 π

e imφ

Plm ( cos θ ) =

12l× l!

( 1- cos 2 θ ) m/2 ( d

d (cosθ) ) l+m [cos 2 θ -1 ) l

For 2pz orbital.,l l=1, m=0

P10 =

1211 !

( 1- cos 2 θ ) 0 ( d

d (cosθ) ) [cos 2 θ -1 ) 1× e0

= 12 (

dd (cosθ) ) [cos 2 θ -1 )

= 12 [ 2 cos θ ]

= cos θ

∴ Θ l,m = √ 2(1)+12

× cos θ × 1

√2 π

= √ 32

×1

√2 π cos θ

= √ 3

√ 4 π cos θ

Slater Type Orbital ΨS 200 = N× r ×e−3.9r

2 × √ 3

√ 4 π cos θ

Hydrogen like orbital Ψ 200 = N× (2 - 3.9 r ) × e –( 3.9 r2 ) √ 3

√ 4 π cos θ

Comparison:

There is no node in STO, but in HLO node occurs at 2

3.9

Problem 3 Calculate the average value of ‘r’ for 1s electron of Lithium atom using Slater type orbitals.Solution:

Average of r = ∫Ψ rΨ dT

∫ΨΨ dT

= ∫(N ’rn−1 e−Z 'r

n ' )r ¿¿¿ STO Ψ’ nlm = N’ r n-1 e−Z ' r

n ' Y lm ( θ,φ)



= ∫r2 n−1 e

−2 Z 'rn ' r2 dr

∫ ’ r2 n−2 e−2Z ' r

n ' r2 dr [Y l

m ( θ,φ) will be cancelled]

= ∫ r2n+1 e

−2 Z 'rn ' dr

∫ ’r2 n e−2 Z' r

n ' dr

= (2 n+1)!¿¿

× ¿¿ Formula : ∫0

∞

xn e−ax dx = n!

an+1

= (2n+1 )× 2n !¿¿

× 1

2n! (n+1) ! = (n+1) n!

= (2n+1 )× n

2 z

For lithium atom n = 3, for is orbital Z = z’ – 1(0.3) = 3-0.3 = 2.7

Average of r = 215.4

=3.6

Problem 4 Calculate the first ionisation energy for Li atom on the basis of Slater rules.

Solution:

I.P of Li = E of Li + - E of Li

E= ∑ 12

z2

n2 in a.u = ∑ 13.6 z2

n2 e.v

1118 HF –SCF METHODS HARTREE – FOCK SELF CONSISTENT FIELD

METHOD.

Hamiltonian for’ n’ electron atom is H = - ½ ∑ ∇2 - ∑ zr + ½ ∑ ∑

1rij

The

inter electronic repulsion term (½ ∑ ∑ 1rij

) does not permit separation of Schrodinger

equation in to one electron equations. Hence the product wave function will not yield correct

energy. The best possible wave function is obtained by using a set of orbitals , which are the

solutions of n simultaneous Schrodinger equations.

Hφ = Eφ, H = - 12 ∇2 -

zr + Vi



Where Vi represents the potential energy of electron, due to repulsion by all other electrons..

To calculate Vi, Hartree assumed that each electron in the atom moves , in a potential field

of all other electrons. He solved the above equation by iterative technique as follows

Consider the wave function as product of one electron orbitals.

Ψ = (φ1) ( φ2) ( φ3)….

Let electron-1 is moving through the cloud of other electrons. The potential energy

of interaction of electron-1 with all other electrons is given by

V1 = ∑j=2

n ∫φi2d τ

r 1 j where φ is one electron orbital.

On solving the above expression we obtain an improved orbital φ1' .

Now the electron-2 may be considered to move in the field . The new set ( φ1) is

now used to recalculate the potential energy . The potential V2 is calculated as above The

potential energy of interaction of electron-2 with all other electrons is given by

V2 = ∑j=1,3. .

n ∫φ 1' dτr 1 j

where φ is one electron orbital.

On solving the above expression we obtain an improved orbital φ2'

.. Finally we have the wave function as Ψ = (φ1’) ( φ2’) ( φ3’)…. The orbitals are denoted

by ( φ) and are now consistent with the potential field . These are called Self Consistent

Field orbitals ( SCF)Using this value the hamiltonian and hence the energy are calculated.

The cycle of iteration is continued till we get the same value as in the earlier step .

Ψ = φ2 × φ2

Use φ2

solve


[ - ½ ∇12 - ½ ∇2

2 - Zr1

- Zr1

+ ∫ (φ2 )2 d τr12

] φ1 = E φ1


Use φ1;

NO

yes

YES

NOT FOR EXAMINATION

Apply iterative method F( x) = x2 – 6x +5x2 – 6x +5 = 0 6x = x2 +5

x = x2+5

6

x1= f(0) = (0 )2+56

= 0. 83

x2 = f( 0.833) = (0.83 )2+56

= 0.68+5

6 =

0.94

x2 = f( 0.94) = (0.94 )2+56

= 0.88+5

6 =

0.98

x2 = f( 0.98) = (0.98 )2+56

= 0.96+5

6 =

0.99

x2 = f( 0.99) = (0.99 )2+56

= 0.92+5

6 =

0.99till we get value equal to previous value


STOP

φ2; ( improved orbital)

Are the

values

same ?

[ - ½ ∇12 - ½ ∇2

2 - Zr1

- Zr1

+ ∫ (φ1; )2d τ

r12

] φ2 = E φ2

φ1; ( improved orbital)


Hartree Fock ‘s theory: Hartree’s method suffers from a drawback that the orbital product

wave function is not antisymmetric. Fock introduced anti symmetrised wave function and

followed the same iterarative procedure. The orbitals obtained are called HFSCF AO ‘s

The antisymmetrised product wave function is expressed as determinant.`

Ψ = 1

√n! φ1(1) φ 1(2) φ2 (3) φ2 (4)... φ ( n)



Where φ1(1) is a spin orbital with e1 and α – spinφ 1(2) is the spin orbital with e2 and β –

spin in the same orbital φ1. He introduced Fock’s operator ( F) in place of H to include

electron repulsion and electron exchange.

The Fock operator is defined as F = - ½ ∑ ∇2 - ∑ zr + ∑ [ 2 Ji – Ki]

Ji is called coloumb operator and Ki is called exchange operator

1218 BORN – HEIMER APPROXIMATION

The Schrodinger equation includes both nuclear and electronic motions. To separate

the part of Schrodinger equation due to electronic motion, Max Born and Robert

Oppenheimer assumed that nucleus is assumed to be stationary ,when the electrons move.

This is because nuclei are 1000 times heavier than electrons.The electronic wave function Ψe

(r,R) depends on ‘r’(distance between electrons) as well as ‘R’( distance between nuclei)

and the nuclear function depends ‘R’ only.

Ψ = Ψe (r,R) Ψ(R)

Substituting in Schrodinger equation HΨ = E Ψ, we get

H Ψe (r,R) Ψ(R) = E Ψe (r,R) Ψ(R)

For a molecule consisting N nuclei ( u, v, w....) and n ( i ,j, k.... ) electrons , the

complete Hamiltonian operator is given by

H = nuclear kinetic energy + electronic kinetic energy + nuclear potential energy + nuclear –

electron attraction potential energy + electronic repulsion potential energy.

H = - h2

8π 2 ∑u

N 1mu

∇u2 - h2

8 π 2 ∑i

n 1me

∇e2 +

∑u , v

zu z v e2

( 4 π∈0 ) ruv

+¿∑i ,u

zu e2

( 4 π∈0 ) r iu

+¿∑i , j

e2

( 4π∈0 ) rij

¿¿

Applying BO approximation we can neglect the first term and hence Hamiltonian becomes

H = - h2

2 ∑

i

n 1me

∇e2 + ∑

u , v

zu z v e2

( 4 π∈0 ) ruv

+¿∑i ,u

zu e2

( 4π∈0 ) r iu

+¿∑i , j

e2

( 4 π∈0 ) rij

¿¿

Using this we can solve Schrodinger equation.

1318 VALENCE BOND THEORY FOR HYDROGEN MOLECULE

(HEITLER – LONDON THEORY)



Consider the hydrogen molecule is made up of HA and HB. . Let A and B represent the

two H-atoms , 1 and 2 represent the two electrons and S represents the ‘s’ orbital.

Let us assume that electron 1 is associated with atom A and electron 2 is associated with atom B then the wave function of this state is given by Φ1 = SA(1)SB(2) -----------------1

If the electron 1 is associated with atom B and electron 2 is associated with atom A then the wave function of this state is given by Φ2 = SA(2)SB(1) --------------------2

The wave functions of Hydrogen molecule is given by the Linear Combination of Atomic Orbitals( LCAO). which is Ψ = a1 Φ1 + a2 Φ2 where a1 and a2 are normalization constants

Substituting , Ψ = a1 [SA(1)SB(2) ] + a2 [SA(2)SB(1) ] --------------3

To find the normalization constants’a1 ‘ and ‘a2’:

The condition for normalised wave function is

∫−∞

∞

ΨΨ * dτ = 1

∫−∞

∞

{a1[S A (1) S B (2)]+a2[S A (2) SB (1)] } {a1 [ S A (1 ) S B (2 ) ]+a2 [S A (2 ) S B (1 ) ] }dτ = 1

∫−∞

∞

a1 [S A(1)S B(2)]+a2[S A(2)S B(1)]2 dτ = 1

a12 ∫

−∞

∞

[S A(1)S B(2)]¿¿2 dτ + a22 ∫

−∞

∞

[S A(2)S B(1)]¿¿2 dτ + 2 a1 a2

∫−∞

∞

[S A(1)S B(2)]S A (2)S B(1)¿ dτ1 dτ2 = 1

∫−∞

∞

[S A(1)S B(2)] dτ1 = ∫−∞

∞

[S A(2)S B(1)] dτ2 = 1,

∫−∞

∞

[S A(1)S B(2)]S A (2)S B(1)¿ dτ1 dτ2 = S2

Therefore the above equation becomes a12 (1)+ a2

2 (1) + 2 a1 a2 S2 = 1

Since a1 = a2 , a12 + a1

2 + 2 a12 S2 = 1



2a12 + 2 a1

2 S2 = 1

2a12 ( 1 + S2 ) = 1

a12 =

12 (1+S2)

a1 = 1

√2 (1+S2 )

Since a1 = a2 a2 = 1

√2 (1+S2 )

therefore the function becomes

, Ψ = 1

√2 (1+S2 ) [SA(1)SB(2) ] + 1

√2 (1+S2 ) [SA(2)SB(1) ]

, = 1

√2 (1+S2 ) { [SA(1)SB(2) ] + [SA(2)SB(1) ] }

This is for symmetric wave function. For antisymmetric wave function we have

, Ψ = 1

√2 (1+S2 ) { [SA(1)SB(2) ] - [SA(2)SB(1) ] }

There is only one symmetric wave function

A(1)B(2) + B(1) A(2) [ α (1)β (2) - α (2)β (1)]

This is called singlet state because S= 0, 2S+ 1 = 2(0) + 1 = 1

There are three antisymmetric wave functions

A(1)B(2) - B(1) A(2) [ α(1) α (2)]

A(1)B(2) - B(1) A(2) [ β (1) β (2)]

A(1)B(2) - B(1) A(2) [ α(1)β (2) + α (2)β (1)]

This is called triplett state because S= 1, 2S+ 1 = 2(1) + 1 = 3



To find energy:

Variation treatment leads to two secular equations

a1 ( H11 – E) + a2 ( H12 – ES 2 ) = 0

a1 (H21 – ES2 ) + a2 ( H22 – E ) = 0

on solving the 2× 2 determinant, we get

E + = H 11+H 12

2 (1+S2 ) , E - = H 11−H12

2 (1−S2 )

To find H11 and H12:

The Hamiltonian is given by

H = - ½ ∇12 -

1r A 1

- ½ ∇22 -

1rB2

- 1

r A 2 -

1rB 1

+ 1r12

+ 1R

Where 1

r A 1,

1r A 2

, 1

rB1 ,

1rB2

, 1

r12 and

1R represents the distances.

Let HA = - ½ ∇12 -

1r A1

, HB = - ½ ∇22 -

1rB 2

, H’ = - 1

r A2 -

1rB 1

+ 1

r12 +

1R

Solving we get

E + = 2 EH + J+ K

(1+S2 ) + 1R

E - = 2 EH + J−K

(1−S2 ) + 1R



First two terms in the right side constitute electrostatic energy and the third term is nuclear

repulsion energy.

∆ E+ = J+ K

(1+S2 )

It is found out that K is more negative than J. Therefore ∆ E+ is negative for al values of

R. it represents the stable state. ∆ E- is positive representing repulsive state

The ∆ E+ plot shows that two H- atoms attract each other as they approach , passes through

a state of minimum energy and begin to repel when they are very close.The plot ∆ E- never

shows minimum

MO THEORY VB - THEORY



1418 LCAO –MO THEORY FOR DI ATOMIC MOLECULES

LINEAR COMBINATION OF ATOMIC ORBITAL- MOLECULAR ORBITAL

(LCAO-MO)

-TREATMENT OF DIATOMIC MOLECULES

According to LCAO theory, the molecular orbitals are formed by linear

combination of atomic orbitals (AOs) .

HYDROGEN MOLECULE ( H2)

Consider the hydrogen molecule is made up of HA and HB. The wave functions of

Hydrogen molecule is given by the Linear Combination of Atomic Orbitals( LCAO). which

is

Ψ = a1 SA + a2 SB ----------------------------------1

Where SA represents the s- orbital of hydrogen A and SB represents the s- orbital of hydrogen

B.and a1 and a2 are normalization constants.

The secular equations are

a1 ( HAA – ESAA) + a2 ( HAB – ESAB) = 0 e1

e2

a1 (HBA – ESBA ) + a2 ( HBB – ESBB) = 0

The equations have non-trivial solution only if the secular determinant is equal to zero.

HAA – ESAA HAB – ESAB

HBA – ESBA HBB – ESBB 0

Since both electrons are indistinguishable, HAA = HBB

Since H is a Hermitian operator HAB = HBA

SAA = SBB = 1 , SBA = SAB = 0 ( overlap integral)

Substituting in the above determinant,

HAA – E HAB – ES

HAB – ES HAA – E 0

( HAA – E ) 2 = ( HAB – ES ) 2

∴ HAA – E = ± HAB – ES


H A HB


Let E = E1 , when positive value is taken, therefore the above equation becomes

HAA – E1 = + HAB – E1S

Rearranging, HAA – HAB = E1 - E1S

= E1 ( 1- S )

∴ E1 = H AA – H AB

(1−S)

Let E = E2 , when negative value is taken, therefore the above equation becomes

HAA – E2 = - ( HAB – E2S)

Rearranging, HAA + HAB = E2 + E2S

= E2 ( 1 + S )

∴ E2 = H AA+– H AB

(1+S )

Thus the energy levels are E1 = H AA – H AB

(1−S),

E2 = H AA+– H AB

(1+S )

To find the normalization constants, let us apply the condition for normalization

∫ΨΨ dt = 1

∫(a1 S A+a2 SB)( a1 SA+a2 SB ) dt=¿¿ 1

∫ ( a1 S A+a2 SB )2dt=¿¿ 1

∫ ( a1 S A )2 dt + ∫ ( a2 SB )2 dt + 2 ∫ a1 SA a2 SB dt=¿¿ 1

a12∫ ( SA )2 dt + a2

2∫ ( SB )2 dt + 2 a1a2∫S A SB dt=¿¿ 1

∫ ( S A )2 dt = 1, ∫ ( SB )2 dt = 1, ∫ SA SB dt=¿¿ S


a12+ a2

2+ 2 a1a2 S = 1

Since both are electrons, a1=a2 = A

A 2 + A2 + 2A2S = 1

2A2 +2A2S = 1

2A2 ( 1 +S) = 1

A2 = 1

2(1+S )



A = 1

√2(1+S¿)¿ substituting in 1, we get

Let Ψ + be the for the symmetric wave function

Ψ+ = 1

√2(1+S¿)¿ SA +1

√2(1+S¿)¿ SB

= 1

√2(1+S¿)¿ ( SA + SB )

Similarly the anti symmetric wave function is given by

Ψ- = 1

√2(1−S¿)¿ ( SA - SB )

1518 POLYATOMIC MOLECULES



1618 CONCEPT OF HYBRIDIZATION

Definition:

Bond formation involves the overlapping of 2s and 2p orbitals .This linear

combination of orbitals of the same atom is called hybridization. Thus hybridization is the

mixing of orbitals to produce equivalent number of orbitals called hybrid orbitals.

A combination of one ‘s’ orbital and one ‘p’ orbital is called ‘sp’ hybridization.

Similarly a combination of one ‘s’ orbital and two ‘p’ orbitals is called ‘sp 2’ hybridization

1. s-p hybridization:

The combination of one s- orbital and one p- orbitals , giving two hybrid orbitals Ψ1 and

Ψ2 may be expressed as

Ψ1 = a1 s + b1 p ----------------1

Ψ2 = a2s + b2p ------------------2

The values of a1,b1, a2 and b2 can be determined by the following considerations.

Ψ1 and Ψ2 are normalized , orthogonal and equivalent.

Since the two hybridized orbitals are equivalent, the share of s functions is equal∴ a12 = a2

2 = ½ -------------------3

a12 = ½

∴ a1 = 1√2

a22 = ½

∴ a2 = 1√2

Ψ1 is normalized

a12 + b1

2 = 1

(12 ) + b1

2 = 1

∴ b12 =

12



b1 = 1√2

Ssince Ψ1 and Ψ2 are orthogonal

a1b1 + a2b2 = 0

(1√2

) ( 1√2

) + ( 1√2

) (b2) = 0

12 + (

1√2

) (b2) = 0

( 1√2

) (b2) = - ½

∴ b2 = - 12 ×

√ 21

= - 1

√2√ 2 ×

√ 21

= 1√2

The wave functions are Ψ1 = 1√2

s + 1√2

p = 1√2

( s+p)

Ψ2 = 1√2

s + −1√2

p = 1√2

( s- p)

Directional characteristics:

Ψ1 = 1√2

( s+p), Ψ2 = 1√2

( s- p)

The angular function of 2s orbital = 1

√ 4 π *

The angular function of 2pz orbital.,l l=1, m=0 = √ 34 π

cos θ **

Substituting Ψ1 = 1√2

(1

√ 4 π+ √ 3

√ 4 π cos θ) = 1

√ 4 π (1+√ 3 cosθ ¿ ¿√2 )

Ψ2 = 1√2

(1

√ 4 π− √ 3

√ 4 π cos θ) = 1

√ 4 π (1−√ 3cosθ ¿ ¿√2 )

The value of maximum in either case is 1+√ 3

√ 2 = 1.932 which is greater than that for pure 2s

orbital(f=1) and a 2p orbital ( √ 3 = 1.732)

The functions are maximum θ = 0, θ = π Therefore the angle between the two functions is 180 o



sp 2 hybridization:

The combination of one s- orbital and two p- orbitals , giving three hybrid orbitals Ψ1 , Ψ2 and Ψ 3 may be expressed as

Ψ1 = a1 s + b1 px + c1 py ----------------1Ψ2 = a2 s + b2 px + c2 py ---------------2 Ψ3 = a3 s + b3 px + c3 py -----------------3

The values of coefficients can be determined by the following considerations.Ψ1 , Ψ2 and Ψ3 are normalized , orthogonal and equivalent. Since the three hybridized orbitals are equivalent, the share of s functions is equal

∴ a12 = a2

2 = a32 =

13 ----------4

a1 = a2 = a3 = 1

√ 3

Let c1 may be assigned to fixed direction. Therefore c1 = 0

To find Ψ1

Since Ψ1 is normalized,

a12 + b1

2 + c12 = 1

(1

√ 3 )2 + b12 = 1

(13 ) + b1

2 = 1

∴ b12 = 1 -

13

= 23

b1 = √23

Thus the wave function is

To find Ψ2

Since Ψ1 and Ψ2 are orthogonal a1a2 + b1 b2 + c1c2 = 0

(1√3

) ( 1√3

) + ( √ 2√3

) b2 = 0

13 + (

√ 2√3

) b2 = 0

( √ 2√3

) b2 = - 13

b2 = −13 ×

√ 3√ 2

= −1

√3 √ 3

∴ b2 = -−1√ 6

Since Ψ2 is normalized a2

2 + b22 + c2

2 = 1

(1

√ 3 )2 +(−1√ 6 ) 2 + c2

2 = 1

(13 ) + (

16 ) + c2

2 = 1

(2+1

6 ) + c22 = 1

(12 ) + c2

2 = 1

∴ c22 = 1 –

c2 = 1√2



Ψ1 = 1

√ 3 s + √23 px

Ψ2 = 1

√ 3 s - ( 1

√ 6 ) px + 1√2

py


To find Ψ3


(1√3

) ( 1√3

) + ( √ 2√3

) b3 = 0

13 + (

√ 2√3

) b3 = 0

b3 = −13 ×

√ 3√ 2

= −1

√ 3 √ 3 ×

√ 3√ 2

∴ b3 = −1√ 6


(1√3

) ( 1√3

) +(−1√ 6 )(

−1√ 6 ) +

1√2

c3 = 0

13 +

16 +

1√2

c3 = 0

12 +

1√2

c3 = 0

c3 = −12 ×

√ 21

= −1

√ 2 √ 2 × √ 21

∴ c3 = −1√ 2



Ψ1 = 1

√ 3 s + √23 px

Ψ2 = 1

√ 3 s - ( 1

√ 6 ) px + 1√2

py

Ψ3 = 1

√ 3 s - ( 1

√ 6 ) px - 1√2

py


2 √ π

,

2px orbital. = (√ 3

2√ π sin θ cos φ) and that

of 2py orbital. = ( √ 3

2√ π sin θ sin φ)

Substituting

Ψ1 = 1

√ 3 ( 1

2 √ π¿ + √

23 (

√ 32 √ π sin θ cos

φ)

Ψ2 = 1

√ 3 ( 1

2√ π ) - ( 1

√ 6 ) (√ 3

2√ π sin θ

cos φ) + 1√2

( √ 3

2 √ π sin θ sin φ)

Ψ2 = 1

√ 3 ( 1

2√ π ) - ( 1

√ 6 ) (√ 3

2√ π sin θ

cos φ) - 1√2

( √ 3


Let us assume that Ψ1 points towards x-axis. The direction of Ψ2 can be determined as

follows

Ψ2 = 1

√ 3 ( 1

2 √ π ) - ( 1

√ 6 ) (√ 3

2 √ π sin θ cos φ) + 1√2

( √ 3


= ( 1

2√ π ) [ 1

√ 3 - ( 1

√ 6 ) (√3 sin θ cos φ) + 1√2

(√3 sin θ sin φ)

If the two functions are in the xy plane , then 𝛉 = 90. Sin 90 = 1., Substituting


Ψ3 = 1

√ 3 s - ( 1

√ 6 ) px - 1√2

py


Ψ2 = ( 1

2 √ π ) [ 1

√ 3 - ( 1

√ 6 ) (√3 cos φ) + 1√2

(√3 sin φ)

= ( 1

2 √ π ) [ 1

√ 3 - ( 1

√ 2√ 3 ) (√3 cos φ) + 1√2

(√3 sin φ)

= ( 1

2 √ π ) [ 1

√ 3 - 1

√ 2 cos φ + √ 3√2

sin φ)

Let f = ( 1

√ 3 - 1

√ 2 cos φ + √ 3√2

sin φ)

= ( 1

√ 3 - 1

√ 2 cos φ + √ 3√2

( 1- cos 2φ) ½ )

Let x = cos φ

Then f = ( 1

√ 3 - 1

√ 2 x + √ 3√2

( 1- x2 ) ½

Differentiating with respect to ‘x’

dfdx = -

1√ 2 +

√ 3√2

× ½ ( 1- x2 ) -½ ( -2x)

Put dfdx = 0

0 = - 1

√ 2 + √ 3√2

× ½ ( 1- x2 ) -½ ( -2x)

1

√ 2 = √ 3√2

× ( 1- x2 ) -½ ( - x)

1 = - √3 ( 1 – cos 2 φ) – ½ cos φ

= - √3 ( sin 2 φ) – ½ cos φ

= - √3 1

sin φ cos φ

- 1

√ 3 = cosφsin φ

Taking reciprocal

- √3 = sin φcosφ

Tan φ = - √3

∴ φ = 120

Thus the function f2 is found to be at an angle of 120 o with respect to f1

Similarly the function f3 is found to be at an angle of 240 o with respect to f1



Θ l,m = √ 2 l+12

× (l−m ) !( l+m ) !

× Plm(cos θ) ×

1√2 π

e imφ but Plm(cos θ)=

12l× l!

( 1- cos 2 θ ) m/2 (

dd (cosθ)

) l+m [cos 2 θ -1 ) l

*angular function of s orbital (l=0,m=0)

P00 ( cos θ ) =

120× l!

( 1- cos 2 θ ) 0 ( d

d (cosθ) ) 0 [cos 2 θ -1 ) 0 = 1

∴ Θ 0,0 =√ 2(0)+12

×(1)× 1

√2 π = √ 1

2 ×

1√2 π

= 1

√ 4 π

** angular function of p orbital( l=1, m=0)

P10 = ½ (

ddθ ) [cos 2 θ -1 ) l× e0 = ½ [ 2 cosθ ] = cosθ

∴ Θ 1,0 = √ 2(1)+12

× (1) × 1

√2 π = √ 3

2 ×

1√2 π

= 1

√ 4 π cosθ

3.sp3 hybridization:

The combination of one s- orbital and three p- orbitals , giving three hybrid orbitals Ψ1 , Ψ2 Ψ 3 and Ψ 4 may be expressed as

Ψ1 = a1 s + b1 px + c1 py + d1pz ----------------1Ψ2 = a2 s + b2 px + c2 py + d2pz ---------------2 Ψ3 = a3 s + b3 px + c3 py + d3pz -----------------3Ψ4 = a4 s + b4 px + c4 py + d4pz

The values of coefficients can be determined by the following considerations.Ψ1 , Ψ2 , Ψ3 and Ψ 4 are normalized , orthogonal and equivalent. Since the four hybridized orbitals are equivalent, the share of s functions is equal

∴ a12 = a2

2 = a32 = a4

2 = 14 ----------4

a1 = a2 = a3 = a4 = 12

Let c1 may be assigned to fixed direction. Therefore c1 = 0

To find Ψ1

Since Ψ1 is normalized, a1

2 + b12 + c1

2 + d12 = 1

(12 )2 + b1

2 + c12 + d1

2 = 1

14 + b1

2 + c12 + d1

2 = 1

b12 + c1

2 + d12 = 1 -

14



b12 + c1

2 + d12 =

34

----------------------------------5 The first of the four hybrids can be chosen arbitrarily to lie in any direction. let it be along x- direction. along this direction c1 = 0, d1 = 0.

∴ b12 + 0 +0 =

34

b1 = √34


To find Ψ2

Since Ψ1 and Ψ2 are orthogonal a1a2 + b1 b2 + c1c2 + d1d2 = 0

(12 ) (

12) + (

√ 3√4

) b2 + 0 +0 = 0

14 + (

√ 32 ) b2 = 0

( √ 32 ) b2 = -

14

b2 = −14 ×

2√ 3

= −12 √ 3

Since Ψ1 is normalized, a2

2 + b22 + c2

2 + d22 = 1

Since along the xy- plane , the orbital py has no contribution, c2 = 0,

(12 )2 + (

−12√ 3

¿2 + 0 + d22 = 1

14 +

112 + d2

2 = 1

412 + d2

2 = 1

13 + d2

2 = 1

d22 = 1 -

13

= 23


Ψ1 = 12 s + √3

4 px


d2 = √ 23


To find Ψ3


(12 ) (

12) + (

√ 3√4

) b3 + 0 +0 = 0

14 + (

√ 32 ) b3 = 0

( √ 32 ) b3 = -

14

b3 = - 14 ×

2√ 3

b3 = −12 √ 3


(12 ) (

12) + (

−12 √ 3 ) (

−12√ 3 ) + 0 + (√

23 )d3 = 0

14 +

112 + (√

23 )d3 = 0

4

12 + (√ 23 )d3 = 0

13 + (√

23 )d3 = 0

d3 = −13 ×

√ 3√ 2

= −1

√ 3 √ 3×

√ 3√ 2

= −1√ 6

Since Ψ3 is normalized, a32 + b3

2 + c32 + d3

2 = 1

(12 )2 + (

−12 √ 3

¿2 + c32 +(

−1√ 6 ) 2 = 1

14 +

112 + c3

2 + 16 = 1

612 + c3

2 = 1

12 + c3

2 = 1, ∴ c3 = √ 12



Ψ2 = 12 s - (

12 √ 3 ) px +

√ 2√3

pz

Ψ3 = 12 s - ( 1

2 √ 3 ) px + 1√2

py - 1

√ 6 pz


To find Ψ4


(12 ) (

12) + (

√ 3√4

) b4 + 0 +0 = 0

14 + (

√ 32 ) b4 = 0

( √ 32 ) b4 = -

14

b4 = - 14 ×

2√ 3

b4 = −12 √ 3


(12 ) (

12) + (

−12 √ 3 ) (

−12√ 3 ) + 0 + (√

23 )d4 = 0

14 +

112 + (√

23 )d4 = 0

4

12 + (√ 23 )d4 = 0

13 + (√

23 )d4 = 0

d4 = −13 ×

√ 3√ 2

= −1

√ 3 √ 3×

√ 3√ 2

= −1√ 6


(12 ) (

12) + (

−12 √ 3 ) (

−12√ 3 ) + (√

12 ) c4 + (

−1√ 6 )(

−1√ 6 ) = 0

14 +

112 + (√

12 ) c4 + (

16 ) = 0

612 + (√

12 ) c4 = 0

12 + (√

12 ) c4 = 0

(√ 12 ) c4 = -

12



c4 = - 12 ×

√ 21

= - 1

√2 √ 2 ×

√ 21

∴ c4 = - 1

√ 2


Thus the wave function of hybrid orbitals are

Ψ1 = 12 s + √

34 px

Ψ2 = 12 s - (

12 √ 3 ) px +

√ 2√3

pz

Ψ3 = 12 s - (

12 √ 3 ) px +

1√2

py - 1

√ 6 pz

Ψ4 = 12 s - (

12 √ 3 ) px -

1√2

py - 1

√ 6 pz



2√ π ,

2px orbital. = (√ 3

2 √ π sin θ cos φ) , 2py orbital. = ( √ 3

2√ π sin θ sin φ) , 2pz orbital = ( √ 3

2 √ π

cos θ), Substituting the values

Ψ1 = 12 (

12√ π ) + √

34 (

√ 32 √ π sin θ cos φ)

Ψ2 = 12 (

12√ π ) - (

12√ 3 ) (

√ 32√ π sin θ cos φ) +

√ 2√3

( √ 3

2√ π cos θ)

Ψ3 = 12 (

12 √ π ) - (

12 √ 3 ) (

√ 32 √ π sin θ cos φ) +

1√2

( √ 3

2 √ π sin θ sin φ) - 1

√ 6 ( √ 3

2√ π

cos θ)

Ψ4 = 12 (

12√ π ) - (

12 √ 3 ) (

√ 32 √ π sin θ cos φ) -

1√2

( √ 3

2 √ π sin θ sin φ) - 1

√ 6 ( √ 3

2 √ π

cos θ)

Let Ψ1 be along the x- axis , the direction of Ψ2 which lies in the xz – plane can be

determined as follows


Ψ4 = 12 s - ( 1

2 √ 3 ) px - 1√2

py - 1

√ 6 pz


Ψ2 = 12 (

12√ π ) - (

12√ 3 ) (

√ 32√ π sin θ cos φ) +

√ 2√3

( √ 3

2√ π cos θ)

= (1

2√ π ) [12 - (

12 √ 3 ) (√ 3 sin θ cos φ) +

√ 2√3

√3 cos θ)]

= (1

2√ π ) [ 12 - (

12 sin θ cos φ) + √2cos θ)]

Let f = 12 -

12 sin θ cos φ + √2cos θ]

In the xz- plane φ = 0 for positive lobe and φ = 180 for negative lobe and cos 180 = -1.

Therefore for negative lobe, the above equation becomes f = 12 +

12 sin θ + √2cos θ

Differentiating with respect to ‘θ’

dfdθ =

12 cos θ + √2 ( - sin θ)

Put dfdx = 0

12 cos θ = √2 sin θ

sinθcosθ =

12√ 2

Tan θ = 19 o 28 ‘

Therefore the function f2 makes an angle of 19 o 28 ‘ or 90 + 19 o 28 ‘ = 109 o 28 ‘

1718 HUCKEL THEORY FOR CONJUGATED MOLECULES

:

This is an empirical method for organic compounds containing conjugated carbon

chain linear as well as cyclic. The theory is based on the following assumptions.

1. The wave function of Huckel Molecular Orbital(HMO) is taken as a linear

combination of atomic orbitals(AOs)

Ψi = ai1 p1 + a i2 p2 + ai3p3 +…………….

For example, Ψ1 = a11 p1 + a12 p2 + a13p3 +…………….



2. .There will be ‘n’ energy levels corresponding to ‘n’ HMOs , each being expressed in

terms of α and β

3. Energy is calculated by E = ∫Ψ HΨ dT

∫ΨΨ dT

4. Variation treatment leads to ‘n’ secular equations, where ‘n’ is number of C atoms.

For example for ethylene two secular equations and for butadiene there are four

secular equations.

a1 ( H11 – ES11) + a2 ( H12 – ES12) + …………………….. + an ( H1n – ES1n ) = 0

a1 ( H21 – ES21) + a2 ( H22 – ES22) + ……………………+ an ( H2n– ES2n ) = 0

……………………………………………………………………………………

……………………………………………………………………………………..

a1 ( Hn1 – ESn1) + a2 ( Hn2 – ESn2) +…………………… + an ( Hnn – ESnn) = 0

5. He introduce some approximations

a. All integrals of type H ii = α

For example H11 = H22 = H33 = α

b. All integrals of type in which i and j are directly bonded, H ij = β

For example H12 = H21 = β, if C1 and C2 are directly bonded.

c. All integrals of type in which i and j are not directly bonded, H ij = 0

For example H14 = H41 = 0, if C1 and C4 are not directly bonded.

d. Integrals of type S ii = 1

For example S11 = S22 = S33 = 1.

e. Integrals of type S ij = 0

For example S12 = S23 = S34 = 0.

Thus for a linear conjugated chain the secular determinant takes the form

α – E(1) β 0 0

β α – E(1) β 0

0 β α – E(1) β = 0

0 0 β α – E(1)

Dividing by β

α – Eβ 1 0 0



1 α – E

β 1 0

0 1 α – E

β 1 = 0

0 0 1 α – E

β

Put x = α – E

β the above determinant becomes

x 1 0 0

1 x 1 0 = 0

0 1 x 1

0 0 1 x

5. On expanding the n× n determinant, a polynomial of n th degree in x which has ‘n’ real

roots x1,x2,x3….will be obtained.

6. The negative root corresponds to ‘ bonding level’. The positive root corresponds to ‘ anti

bonding level’. When ‘n’ is odd, Ei = α corresponds to ‘non bonding level

7. By inserting the values of E, in the secular equations, the values of HMO coefficients are

obtained.

1. APPLICATION OF HMO TO ETHYLENE:



The skeleton framework of ethylene is C1= C2.

1. Atomic orbitals

Since there are two carbon atoms in ethylene there are two atomic orbitals ( φ1 and

φ2 )

2. Molecular orbitals

The MO may be written as the linear combination of atomic orbitals

Ψ = a1 φ1+ a2 φ2

where a1 and a2 are Huckel’s coefficients

3. The secular equations

Since there are two carbon atoms there will be two secular equations.. They are

a1 ( H11 – ES11) + a2 ( H12 – ES12) = 0

a1 (H21 – ES21 ) + a2 ( H22 – ES22) = 0

4. The secular determinant

secular determinant is formed by leaving the HMO coefficient’s

¿ = 0

5. Apply Huckel’s approximation to find energy equation

H11 = H22 = α, ( Coloumb integral)

H12 = H21 = β ( exchange integral)

S11 = S22 = 1 ,

S21 = S12 = 0 ( overlap integral)

Substituting in the above determinant,

|α−E ββ α−E|=0

dividing by β

α−Eβ 1

1 α−E

β = 0

Put α−E

β = x

This is the energy equation.

6. Solve the secular determinant find the polynomial

x 1



1 x = 0

polynomial is x 2 - 1 = 0

7. To solve the polynomial:

x 2 - 1 = 0

x 2 = 1

x = ± 1

8. Substitute the value in energy

equation.

when x = +1, let E = E1

α−Eβ = x

α−E1

β=+1

α−E 1=β

∴ E1 = α- β

when x = -1, let E = E2

α−E

β = x

α−E2

β = -1

α−E 2=−β

∴ E2 = α + β

Thus the two HMO energies of ethylene are α + β , α – β

The energy level E2 is called bonding HMO( Huckel MolecularOrbital ) because it has lower

energy ( since β is negative ) and E1 is called anti bonding HMO

π – BOND ENERGY:



π – bond energy = total energy – energy in the absence of bonding

= 2(α + β) - 2 α

= 2 β

To determine the HMO co efficients:

a1 ( H11 – ES11) + a2 ( H12 – ES12) = 0

a1 (α- E) + a2 β = 0

since α−E2

β=−1 [Take any one value]

(α- E) = - β [ Take E2 as E]

substituting in equation 1∴ a1 (- β) + a2 β = 0

a1 β = a2 β

a1 = a2

By normalization condition

a12 + a2

2 = 1

a12 + a 1

2 = 1 [a1 = a2 ]

2 a12 = 1

∴ a1 = 1

√ 2 and a2 = 1

√ 2

Thus the wave functions are

Ψ1 = 1

√ 2 φ1+ 1

√ 2 φ2 or = 0.707 φ1+ 0.707 φ2

Ψ2 = 1

√ 2 φ1 - 1

√ 2 φ2 or = 0.707 φ1 - 0.707 φ2



α + β represents bonding energy level because β is negative

The antibonding wave function of ethylene is α – β

The total energy of ethylene in terms of α and β is 2α + 2β

Number of nodal plane in Ψ1 of ethylene is zer0

Number of nodal plane in Ψ2 of ethylene is one

Charge density:

The Charge density is given ρ = 1- qr = 1 – 1 = 0

Π – bond order

This is given by pij = ∑ n( Aij× Bij) Where n is number electrons, Aij –

P12 = 2 × 1

√ 2× 1

√ 2 = 1

Free valency:

For C1: F1 = √3 - P12 = √3 - 1 = 1. 732 -1 = 0. 732

For C1: F2 = √3 - P12 = √3 - 1 = 1. 732 -1 = 0. 732

2 APPLICATION TO BUTADIENE:



The carbon skeleton of butadiene is

C1-C2-C3-C4

There are four AO’s to be combined to form MO’s

Ψ = a1 φ1+ a2 φ2 + a3 φ1+ a4 φ2 -------------------------------1

Where φ1 , φ2, φ3 and φ4 are the atomic orbitals and the secular equations are

a1 ( H11 – ES11) + a2 ( H12 – ES12) + a3 ( H13 – ES14) + a4 ( H14 – ES14 ) = 0

a1 ( H21 – ES21) + a2 ( H22 – ES22) + a3 ( H23– ES23) + a4 ( H24– ES24 ) = 0

a1 ( H31 – ES31) + a2 ( H32 – ES32) + a3 ( H33 – ES33) + a4 ( H34 – ES34) = 0


1. Hii = Hjj = α, ( Coloumb integral) for example H11 = H22 = α

2. Hij = Hji = β ( exchange integral)

3. Hij = Hji = 0 if there is no direct link between i and j

4. Sii = Sjj = 1 ,

5. Sij = Sji = 0 ( overlap integral)

Using this value, the secular determinant is

α – E(1) β 0 0

β α – E(1) β 0

0 β α – E(1) β = 0

0 0 β α – E(1)

Dividing by β

α – Eβ 1 0 0

1 α – E

β 1 0

0 1 α – E

β 1 = 0

0 0 1 α – E

β

Put x = α – E

β the above determinant becomes



x 1 0 0

1 x 1 0 = 0

0 1 x 1

0 0 1 x

x 1 0 1 1 0

x 1 x 1 - 1 0 x 1 = 0

0 1 x 0 1 x

x { [ x ( x2 -1 ) - 1( x-0) ] } - [ 1 ( x2 – 1) - 1 ( 0)] = 0

x4 – x2 - x 2– x2 +1 = 0

x4 –3 x2 +1 = 0 --------------1

put y = x2 then the above equation becomes,

y2 – 3y + 1 = 0

y = +3 ±√9−42

= +3 ±√52

The roots are y 1 = +3+√52

y 2 = +3−√52

To find x: y1 = x2

∴x2 = +3+√52

= 2¿¿ [ multiplying and dividing by 2]

= 1+5+2√54

= ¿¿

x1 = + 1+√52

= 1. 618

x2 = - 1+√52

= - 1.618

similarly

y 2 = +3−√52

x3 = + 0.618 x 4 = - 0.618



∴ The four roots of equation 1 are x1 = + 1.618, x2 = - 1.618 , x3 = + 0.618 x 4 =

- 0.618

when x = + 1.618 , let energy be E1,

x = α – E

β

1.618 = α – E1

β

E1 = α - 1.618 βwhen x = - 1.618 let energy be E2

then -1.618 = α – E2

β

∴ E2 = α + 1.618 β

when x = + 0.618 let energy be E3,

then, 0.618 = α – E3

β

E3 = α - 0.618 βwhen x = - 0.618 let energy be

E4 , then, 0.618 = α – E3

β

E4 = α + 0.618 β

To determine the HMO co efficients:


a1 ( H21 – ES21) + a2 ( H22 – ES22) + a3 ( H23– ES23) + a4 ( H24– ES24 ) = 0

a1 ( H31 – ES31) + a2 ( H32 – ES32) + a3 ( H33 – ES33) + a4 ( H34 – ES34) = 0


the above equations can be rewritten as

a1 (α- E) + a2 β + a3 (0) + a4 (0) = 0 since there is no direct link between 1-3 and 1-4

a1 β + a2 (α- E) + a3 β + a4 (0) = 0

a1 (0) + a2 β + a3 (α- E) + a4 β = 0

a1 (0) + a2 (0) + a3 β + a4 (α- E) = 0

dividing by β, and put x = α – E

β a1x + a2 = 0 a1 + a2x + a3 = 0

a2 + a3x + a4 = 0

a3 + a4 x = 0 By normalization condition

a12 + a2

2 + a32 + a4

2 = 1

put x = -1.618, we will get. a1 = a4 = 0.372 and a2 = a3 = 0.602

Thus the wave function is Ψ1 = 0.372 φ1 + 0.602 φ2 + 0.602 φ2 + 0.372 φ2

put x = -0. 618, we will get, a1 = 0.602, a2 = 0.372, a3 = - 0.372 and a4 =- 0.602



Thus the wave function is ,Ψ2 = 0.602 φ1 + 0.372 φ2 - 0.372 φ2 - 0.602 φ2

put x = +0.618, we will get ,a1 = 0.602, a2 = - 0.372, a3 = - 0.372 and a4 =

+0.602

Thus the wave function is ,Ψ2 = 0.602 φ1 - 0.372 φ2 - 0.372 φ2 + 0.602 φ2

put x = +1.618, we will get , a1 = 0.372 , a2 = - 0.602, a3 = + 0.602 and a4 = -

0.372

Thus the wave function is ,Ψ2 = 0.372 φ1 - 0.602 φ2 + 0.602 φ2 - 0.372 φ2

WAVE FUNCTIONS:

Number of nodal plane in Ψ1 of butadiene is zero



3 APPLICATION OF HMO TO BENZENE

The HMO wave function for benzene is Ψ1 = a1 p1 + a2 p2 + a3p3 + a4p4 + a5p5 + a6p6

the secular equations in terms of ‘x’ are

a1x + a2 +a6 = 0

a1 + a2x + a3 = 0

a2 + a3x + a4 = 0

a3+a4x+ a5 = 0

a4+a5x+ a6 = 0

a1 +a5 + a6x =0

The corresponding determinant is

x 1 0 0 0 1

1 x 1 0 0 0

0 1 x 1 0 0

0 0 1 x 1 0

0 0 0 1 x 1

1 0 0 0 1 x

Expanding of this polynomial will lead to x6 – 6 x4 +9x2 -4 = 0

put y = x2, the above equation becomes y3 -6y2+9y-4 = 0

put y = 1, 1-6+9-4 =0 ∴ (y -1)is a factor

y=1 ia a root. To find the other roots 1 1 -6 9 -4 0 1 -5 4 -------------------------------- 1 -5 4 0

The equation becomes y2 – 5y +4 = 0

( y-4)( y-1) = 0

y= 4 and y = 1

y = 1, y=1 and y = 4∴ x2 = 1, x2 = 1 and x2 = 4

x= ± 1, x= ±1 and x = ±2

Therefore the six roots are +1, +1, -1,-1,+2,-2

When x1 = -2, E1 = α +2β bonding



When x2 = -1, E2 = α +β degenerate bonding

When x3 = -1, E3 = α +β

When x4 = +1, E4 = α –β degenerate non- bonding

When x5 = +1, E5 = α -β

When x6 = +2, E6 = α -2β non-bonding

ALITER:

Expression for energy

E = α + 2 β Cos ( 2 πrN ) Where N – number of π electrons, r = 0,1,2…N

For benzene N = 6

When r = 0

E1 = α + 2 β Cos (0)

= α + 2 β (1) [cos (0) = 1 ]

= α + 2 β

When r = 1

E2 = α + 2 β Cos (2 π6 )

= α + 2 β Cos (60) [ π= 180 ]

= α + 2 β (12 ) [cos (60) = ½

= α + β

When r = 2

E2 = α + 2 β Cos (4 π6 )

= α + 2 β Cos (120) [ π= 180 ]

= α + 2 β ( - 12 ) [cos (120) = -

½

= α - β

When r = 3

E3 = α + 2 β Cos (6 π6 )

= α + 2 β Cos (π) [ π= 180 ]



= α + 2 β (-1 ) [cos (π ) = -1

= α - 2β

When r = 4

E3 = α + 2 β Cos (8 π6 )

= α + 2 β Cos (240) [ π= 180 ]

= α + 2 β (-½ ) [cos ( 240 ) =-

½

= α - β

When r = 5

E3 = α + 2 β Cos (10 π

6 )

= α + 2 β Cos (300) [ π= 180 ]

= α + 2 β (½ ) [cos (300 ) =½

= α + β

Thus the energy levels are

α +2β , α + β , α + β , α - β, α - β , α -2β

E1, E2 and E3 levels corresponds to bonding MOs. Two of these energy levels are

degenerate.

The 6 electrons of benzene occupy these bonding MOs of low energy. E4, E5 and E6 levels

corresponds to anti bonding MOs .Two of these energy levels are also degenerate.

Total energy = 2 (α +2β) + 2 (α +β) + 2(α +β)

= 6α +8β

Since benzene has 3 double bonds the total energy of benzene should be equal to 3 times that



of compound ‘having one’ double bond∴ Expected energy of benzene = 3 (energy of ethylene)

= 3(2α + 2β) [ energy of ethylene = 2α

+ 2β]

= 6α + 6β)

Difference in energy = 6α +8β - 3 [2α +2β]

= 2β

This is due to delocalization of π electrons and this energy is known as delocalization

energy.

The value of β is –75 KJ/ mol. Therefore delocalization energy of benzene = –150 KJ/mol.

(36 Kcal/mol) This leads to stability of benzene.

.

The wave functions are

1818 SEMI EMPRICAL METHODS



In HFSCF method, large number of integrals need to be evaluated. For example, even in a small molecule like CH4, we have to evaluate 1035 two electron integrals. Hence semi empirical methods with drastic approximations have been developed.

1. Neglect of Valence Electron: In this method all inner shell electrons are ignored. For example in C,N, O only 2s 2p atomic orbitals are considered. This is effected by replacing the nuclear charge Zp by effective nuclear charge Z’p in the Hamiltonian operator. Hamiltonian H =

The Fock operator =

2. Zero Differential Overlap ( ZDO) : Any product of atomic orbitals within an integral is zero if they are different. Φ μa ×φ μb = 0 if μa ≠ μb Here μa and μb refer to atomic orbitals of atomic centre Aand B.This condition is known as Zero Differential Overlap ( ZDO) condition.

3. Neglect of Diatomic Differential Overlap. (NDDO) In this scheme All overlap integrals, in which μa ≠ μb and all 2-electron integrals are taken as zero.

∫ μa (1 ) μb (1 ) 1r12

μa (2 ) μb (2 )dt 1dt 2 = 0

This is known as Neglect of Diatomic Differential Overlap. (NDDO)

4. Intermediate Neglect od Differential Overlap ( INDO) In this scheme S μa μb = 0 if μa ≠ μb S μa μb = 0

∫ μa μbdτ = 0

This is known as Intermediate Neglect od Differential Overlap ( INDO)5. Complete Neglect of Differential Overlap ( CNDO)

In this scheme Both Overlap integral and non-coloumbic integral are zero.

S μa μb = 0 if μa ≠ μb and This is known as Complete Neglect of Differential Overlap ( CNDO)Hartree – Fock –Roothan (HFR) method

.

ASSIGNMENT- 1 Class : II. M.Sc DATE :

SUB: PHYSICAL CHEMISTRY –III UNIT I: THERMODYNAMICS –I



PART – A ( 2 marks)

1. What are partial molar quantities?

2. Define chemical potential.

3. Define partial molar enthalpy.

4. List three partial molar quantities with their mathematical representation

5. What is fugacity?

6. Differentiate activity and activity coefficient.

7. What are non-ideal solutions?

8. What are dilute solutions?

9. Define excess functions.

10. List out any four excess functions

PART – C( 5marks)

1. Show that the rate of change of chemical potential of a constituent ‘i’ with pressure , at

constant temperature and composition is equal to the partial molar volume of ‘i’ .

2.Variation of chemical potential of a component ‘i; with temperature, at constant pressure

and composition is equal to partial molar entropy of ‘i’. – Prove.

3. The partial molar volume of alcohol in a mixture of alcohol and water is 45.5 ml, nA =

2.0 mol, nB = 2.0 mol . Find the partial molar volume of water if the total volume of the

mixture is 101.5

PART- C (10 marks)

1. Explain how fugacity of a real gas experimentally determined?

2 Arrive the expression to determine the variation of activity with temperature.

3. What do you mean by partial molar volume. How is it determined ?

4 Discuss the variation of fugacity with temperature and pressure.

5. a. How is activity and activity coefficients of a non-electrolyte determined?

b. Write note on Choice of Standard states.

*******

ASSIGNMENT- 2Class : II. M.Sc DATE :SUB: PHYSICAL CHEMISTRY –III UNIT II: SPECROSCOPY – 1

.1. What is rotational spectra?2. Which shows micro wave spectra : CO, CO2, O2, HCN, OCS, H2, CCl4, 3. Give the expression for the energy levels of non- rigid rotator.



4. Write down the selection rule for micro wave spectra.5. In what way rigid rotator differs from non- rigid rotator?6. What are symmetric top molecules?7. List some examples for assymetric top molecules.8. Point out the limitations of micro wave spectra9. What is degeneracy of rotational level ?10. How do hot bands differ from overtones?11. .What are combination bands?12. How does Raman scattering differ from Rayleigh scattering?13. .Give the Condition for Raman activity14. .Define polarisability.15. Explain the micro wave spectra of non – rigid rotator16. Discuss the micro wave spectra of poly atomic molecules17. Show that the frequency separation of micro wave spectrum is twice that of rotational

constant.18. .Deduce an expression for the potential energy of a vibrating molecule.19. Sketch the potential energy curve of a vibrating molecule under

1. Simple harmonic .2. Anharmonic.20. .Arrive the expression for fundamental and overtones21. .Explain the origin of P and R branches using diatomic vibrator.22..Discuss Fermi resonance phenomena occurring CO2 molecule.23. .Explain the classical theory of Raman effect.24. .Show that the spectral lines on Rotational Raman spectra are separated by 4B .25.Explain how inter nuclear distance distances in poly atomic molecules determined?26.Derive the expression for energy of rotational level and show that the rotational lines are equally spaced.17. The pure rotational spectrum of CO consists of a series of equally spaced lines separated by 3. 8 cm -1. Calculate the inter nuclear distance of the molecule. reduced mass = 1.07 × 10 -26 Kg

ASSIGNMENT- 3 Class : II. M.Sc DATE :SUB: PHYSICAL CHEMISTRY –III UNIT III: SPECROSCOPY – III

PART-B (2 marks)1. What is zeemann effect?2. What is hyper fine inter action?3. What is Doppler effect?4. Define isomer shift

PART – C ( 5 marks)



1. Derive the expression for the equation of motion of spin in magnetic fields2. Define chemical shift. Discuss the factors affecting chemical shift.3. Explain spin – spin coupling in NMR4. Discuss the NMR of AX and AMX type molecules.5. Write note on 13C NMR.6. Explain the fragmentation pattern for simple aliphatic and aromatic alkanes7. Write down Mc – Connel relation and discuss its application.8. Explain the principle behind the Mass spectroscopic technique.9. . With neat diagram explain the components of mass spectrometer10. Write note on McLaffetry rearrangement11. . Discuss the fragmentation pattern of alkanes.12. Explain the fragmentation pattern of . aromatic hydrocarbon 13. Write note on a. parent peak b. base peak c. isotopic peak 14. Account for the following peaks at the m/ei. . – in toluene ii . -------in methanol iii. –in 2-butanone

iv. – in toluene v. benzaldehyde.15. Mass spectra of 1-butanal gives peaks at the following m/e values.

Show the fragmentation pattern that explains these observations.16. The molecular ion peak at an m/e value of -. Assign a structure that would be expected to

give rise to the above peak.17. How will you distinguish 2- pentanone and 3- pentanone using Mass Spectral

studies. Draw the mass spectrum of the fragments formed.

. PART – D ( 10 marks)1. Discuss briefly about 31 P NMR and 19 F NMR2. Write a brief account of FT-NMR3. Discuss the principle behind ESR4. Discuss the fragmentation pattern for alcohols, aldehydes and ketones.5. Write note on electron-neutron hyperfine interactions. 6. Discuss briefly about Quadrupole interactions and Magnetic interactions7. Discuss the theory and instrumentation of Mass spectra

ASSIGNMENT- 5

Class : II. M.Sc DATE :

SUB: PHYSICAL CHEMISTRY –III UNIT V : QUANTUM CHEMISTRY –

IV

:

PART – A

1. Identify the perturbation term in the Hamiltonian of Helium atom 2. Mention the conditions to apply perturbation theory 3. Identify H0 and H1 of an oscillator governed by potential energy V = ½ kx2 + ax3 + bx



4. A hydrogen atom subjected to electric field has Hamiltonian H = - 12 ∇2 -

2r1

+rcosθ

.Separate the perturbed part from unperturbed part.5. Write down the second order perturbed Schrodinger wave equation 6. What is the need for approximation methods?7. List out the steps followed in variation method.8. Give two examples for perturbation9. In what way variation method differs from perturbation method.10. Write down the expression for first order perturbation energy and wave function11. Find the term symbol for N.12. .Verify whether an energy state with term symbol 2P 5/2 can exist13. Deduce the atomic term symbol for boron.14. Determine the electronic configuration of the element whose ground state term symbol is4S

3/2.

15. The term symbol of a particular atomic state is 6S5/2. Suggest a possible electronic configuration

16. . What is screening constant?.17. . Calculate the screening constant of 2s electron of Carbon

20. Differentiate Slater type orbitals and Hydrogen like orbitals. 21. What are Slater type orbitals?22.Determine the effective nuclear charge of 1s electron in F.23. What is Born – Heimer approximation ?25. Write down the Hamiltonian for Hydrogen molecule

26.Write the Hamiltonian for helium atom and mention the terms involved.

27. What are resonance and coulomb integrals?

28.What are exchange integrals?

29. What are the three important approximations of Huckel LCAO-MO theory?30. What do you mean by hybridization?.31..Mention the degenerate energy levels of benzene in terms of α and β32. Give the expression for wave function and energy of ethylene in terms of α∧β.

33. Write down the secular determinant for H2 molecule. 34. Write down the Secular determinant for the He atom35. Write down the secular determinant of benzene in terms of α∧β

36. Give the expression for wave functions of H2 molecule by VB theory.37. Draw the wave functions of butadiene

PART- B (5 marks )

1. Discuss the application of perturbation theorem to Hydrogen atom.2. Derive the expression for first order perturbation energy and wave function.



3. Arrive the secular equation using variation theorem.4. Discuss the application of variation theorem to hydrogen atom.5. Show that the variation method provides an upper bound to the ground state energy of the

system.

6. The average energy of H –atom in terms of variation parameter ’a’ is ( a2

6 -

a2 ) . Find the

variation parameter and the true energy

7..For the selected wave function Ψ = r e−ar the expectation value of energy is E=α2

6 -

α2 .

Find the value of 𝛂 and hence calculate the energy8. Find the expression for STO for 2s, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.9.. What do you mean by Effective nuclear charge. Calculate the Effective nuclear charge of

2s electron of Nitrogen and Calcium11. Write note on HFSCF method12. Explain Born-Oppenheimer approximation13. Outline the salient features of VB(Heitler-London) theory as applied to Hydrogen molecule14 Apply HMO theory for ethylene molecule and determine its energy and wave function . 15.Give the assumptions of Huckel molecular orbital theory

16. Use 2s and 2pz atomic orbitals to construct two equivalent sp hybrid orbitals and determine the angle between the hybrid orbitals.

PART- C ( 10 marks )

1.Show that the first order perturbation energy for a non-degenerate system is just the perturbation function averaged over the corresponding unperturbed state of the system. Derive the expression for the eigen function of the perturbed system2. Discuss the first order perturbation theory for a non –degenerate level .Using this theory, solve the Schrodinger wave equation for ground state of Helium and obtain the expression for eigen energy3.A Hydrogen atom is exposed to an electric field of strength F so that its perturbed

Hamiltonian is F r cosθ. Show that there is no first order effect. Given Ψ 0¿ = e−r

√ π and

∫0

∞

e−2 r r3 dr = 384. Apply variation method to Helium atom and calculate its ground state energy. Compare your result with that obtained from perturbation method.9. Consider Ψ as a linear combination of two eigen functions φ1 and φ 2 with normalization

constant a1 and a2 respectively. Find the expectation value of energy and show that it is lesser than the lowest energy E0

12.The expectation energy of H-atom were calculated as a2

6 -

a2 and

3a2 - 2√ 2 a

π when the

wave functions are Ψ = re−ar and Ψ = e−ar 2

respectively.



a. Minimize this average energy and find the variation parameter ‘a’ b. Find the actual energyc. Which is better wave function ?. Comment your result

13 Explain VB theory for H2 molecule considering Ψ as a linear combination of two eigen functions φ1 and φ 2 with normalization constant a1 and a2 respectively

14. Write down the secular equations and set up the secular determinant for butadiene . Determine the energies of the HMOs in terms of 𝛂 and 𝛃 and calculate the free valencies of the terminal carbons.

15.Give the secular equations for benzene . Calculate the energies and HMO coefficients Determine its π- bond order

16. Write note on Semi Empirical methods.

R.V.GOVT.ARTS COLLEGE ,CHENGALPATTU -603001DEPARTMENT OF CHEMISTRY

MODEL EXAM OCT 2014Class : II. M.Sc DATE:SUB: PHYSICAL CHEMISTRY – III MAX. MARK : 75

PART – A answer any ten questions ( 1 × 10 = 10 )

1. The fugacity of a pure gas is 20 atm and the same gas in a mixture is 40 atm . Calculate

the activity of the gas.

2. Define partial molar properties.

3. What is Fermi resonance?

4. Give the selection rule for electronic spectra.

5. Define spin – spin coupling.



6. An NMR signal for a compound is found to be 180 Hz downward from TMS peak using a spectrometer operating at 60 MHz. Calculate its chemical shift.

7. Calculate the ionic strength of 0.1M of BaCl2.

8. What is mean ionic activity ?

9. List out the steps followed in variation method.

10. Write down the expression for first order perturbation energy and wave function.

11. Write down the secular determinant for hexa triene.

12. . What are Slater type orbitals?

PART – B answer any five questions ( 5 × 5 = 25 )

13. What is chemical potential? Discuss the variation of chemical potential with temperature and pressure.

14. Derive Gibbs – Duhem- Margus equation

15. Discuss the various types of transitions in organic compounds

16. What is Zeemann effect? Explain.

17. Explain Debye-Huckel limiting law

18. Apply first order perturbation result to calculate the ground state energy of Helium

19. Determine energy and wave function of ethylene .Calculate its free valency, charge density and π- bond order.

PART – C answer any four questions ( 10× 4 = 40 )

20. a. Derive the expression to show the variation of fugacity with temperature and pressure.b. Calculate the free energy change accompanying the compression of one mole of a gas at 27 o C, from 20 to 200 atm. The fugacitities at 20 and 200 atm pressure are 50 and 100 atm respectively

c. Write a note on choice of standard states

20. a. Explain the principle behindRaman spectra

b. The bond length of NO molecule is 1.151 × 10 -10 m. Calculate the frequency in cm -1

for the pure rotational levels in the spectrum of NO corresponding to the following

changes in the rotational quantum number 0 →1, 1 →2 , and 2→ 3

22. a. Explain Fourier transform spectroscopy



b. An AX type spectrum has 4 lines δ = 4.8, 4.7 , 1.1 and 1.0 ppm measured from TMS with an NMR spectrometer operating at 100 MHz. Estimate the spin – spin coupling constant and chemical shift.

23. a. Discuss about Debye-Huckel theory of strong electrolytes. b. How is Debye Huckel Bronsted equation verified?

24. Use the trial wave function Ψ = e−arto find the energy eigen values for the ground state of H- atom using variation theorem. Hamiltonian in spherical co-ordinates is

H =−12r 2

ddr

¿) - 1r [given ∫

0

∞

e−2 ar r2dr = 1

4 a3 ∫0

∞

e−2 arr dr = 1

4 a2]

1. Write down the secular equations and set up the secular determinant for butadiene

and Determine the energies of the HMOs in terms of 𝛂 and 𝛃 and calculate the free

valencies of the terminal carbons.

**************

R.V.GOVT.ARTS COLLEGE ,CHENGALPATTU -603001DEPARTMENT OF CHEMISTRY MODEL EXAM OCT 2016

CLASS : II M.SC MER3C - PHYSICAL CHEMISTRY - III MARK : 75DATE : 19.10.2016 DURATION : 3 HRS

PART – A ( 10 × 1 = 10) 1. Define chemical potential 2.. What is fugacity 3. How does Raman scattering differ from Rayleigh scattering.? 4. How do hot bands differ from overtones? 5. Define isomer shift 6. An NMR signal for a compound is found to be 180 Hz downward from TMS peak using a spectrometer operating at 60 MHz. Calculate its chemical shift. 7. Write down Nernst equation 8. Define Galvanic cell. 9. What is screening constant. Calculate the screening constant of 2s electron of Carbon 10. Differentiate Slater type orbitals and Hydrogen like orbitals.11. What is hybridization?12. Differentiate base peak and molecular ion peak

PART – B (5 × 5 = 25) [ Answer any five]13. Show that the variation of chemical potential with pressure is equal to partial molar Volume.14. Explain the classical theory of Raman effect.



15. Write note on Mc – Lafeerty rearrangement 16 Calculate the magnetic field strength required to observe the NMR lines for fluorine at 60 MHz. given gN for Flourine is 5.3 and nuclear magneton (uN) = 5 × 10 27 J/T17. How is activity coefficient determined by emf method ?18. Discuss the application of perturbation theorem to Hydrogen atom.19. Apply HMO theory to butadiene and arrive the expression for its wave functions

PART – C ( 4 × 10 = 40 ) [ Answer any four]20.a. Discuss the variation of fugacity with temperature and pressure. b. Derive Gibbs Dugam Margus equation21a. Explain the origin of P,Q and R branches using diatomic vibrator. b. The pure rotational spectrum of CN consists of a series of equally spaced lines separated by 3.8 cm -1. Calculate the inter nuclear distance of the molecule. reduced mass = 1.07 × 10 -26 Kg22. Explain hyperfine splitting 23 Derive Debye Huckel limiting law. 24.a. Discuss the application of variation theorem to Hydrogen atom. b. Derive the expression for first order perturbation energy and wave function25. Write note on a Choice of standard state b. Fourier transform spectroscopy

*************R.V.GOVT.ARTS COLLEGE ,CHENGALPATTU -603001

DEPARTMENT OF CHEMISTRY MODEL EXAM OCT 2019

CLASS : II M.SC MER3C - PHYSICAL CHEMISTRY - III MARK : 75DATE : --.10.2019 DURATION : 3 HRS

PART – A ( 10 × 1 = 10) 1. Define chemical potential 2.. What is fugacity 3. How does Raman scattering differ from Rayleigh scattering.? 4. How do hot bands differ from overtones? 5. What is Mc-Lefferty rearrangement? 6.. Define base peak in Mass spectra. 7. Write down Nernst equation 8. Define Galvanic cell. 9. What is screening constant. Calculate the screening constant of 2s electron of Carbon 10. Differentiate Slater type orbitals and Hydrogen like orbitals.11. What is hybridization?12. Differentiate base peak and molecular ion peak

PART – B (5 × 5 = 25) [ Answer any five]13. Show that the variation of chemical potential with pressure is equal to partial molar Volume.14. Explain the classical theory of Raman effect.15. Write note on Zemann effect. 16 Explain the hyperfine splitting pattern of methyl radical.17. How is activity coefficient determined by emf method ?



18. Discuss the application of perturbation theorem to Hydrogen atom.19. Apply HMO theory to butadiene and arrive the expression for its wave functions

PART – C ( 4 × 10 = 40 ) [ Answer any four]20.a. Discuss the variation of fugacity with temperature and pressure. b. How is fugacity of a gas determined21a. Explain the origin of P,Q and R branches using diatomic vibrator. b. The pure rotational spectrum of CN consists of a series of equally spaced lines separated by 3.8 cm -1. Calculate the inter nuclear distance of the molecule. reduced mass = 1.07 × 10 -26 Kg22. Explain the following

a. 13C NMR b. Doppler effect c. Fragmentation of alcohols 23 Derive Debye Huckel limiting law. 24.a. Discuss the application of variation theorem to Helium atom. b. Derive the expression for first order perturbation energy and wave function25. Write note on a Choice of standard state b. Fourier transform spectroscopy

*************

Calculate the magnetic field strength required to observe the NMR lines for fluorine at 60 MHz. given nuclear g- factor (gN ) for flourine is 5.3 and nuclear magneton (uN) = 5 × 10 - 27 J/T

Solution:

From Bohr frequency condition, hϑ = μN gN Bz

Bz = , hϑ

μ N g N

= , ( 6 .626× 10−34 ) ×60× 106

5.3× 5× 10−27

. = 39.756 × 10−27 ,26.5 ×10−27

= 1.5 T

Calculate the NMR frequency of a proton in the magnetic field of 1.5 T given that nuclear g- factor (gN ) is 5.3 and nuclear magneton (uN) = 5 × 10 - 27 J/T

Solution:

From Bohr frequency condition, hϑ = μN gN Bz

ϑ = μ N g N B z ,

h



= .5.3×5× 10−27 ,× 1.56 .626 × 10−34

. = 39.75× 10−27 ,6.626 × 10−34

= 6 × 10 7 Hz

= 60 MHz

Find the value of nuclear magneton (uN )

Solution:

nuclear magneton (uN ) = eh

4 π m p

= 1.602×10−19× 6.626 × 10−34

4× 3.141× 1.67 ×10−27

= 10.61× 10−53

20.98× 10−27

= 5.05 ×10−27 J/T

UNIT – I – THERMODYNAMICS CODE: 43211234

MARKS : 13,14,15,16,20,22,23 25,26,29 (10 × 3 = 30) Others 1 mark each (20 × 1 =

20 )

DATE : 10 -08-2020 TIME : 1.30 pm -2.00 pm

1. The term partial molar property is applicable to

I. open system II . closed system III isolated system

a. I and III only b. II and III only

c. III only d. I only

2 . Variation in property with change in number of moles at constant T and P is called

a. ideal property b. real property

c. partial molar property d. entropy

3. Chemical potential is

a. partial molar internal energy b. Partial Molar Free Energy

c. partial molar enthalpy d. partial molar volume

4. Which represents chemical potential ?

a. (∂ G∂ n 1

) T, P, n2,n3... b. (∂ H∂ n 1

) T, P, n2,n3



c. ∂ S

∂ n1¿ T, P, n2,n3 d. (

∂ U∂ n 1

) T, P, n2,n3

5. Variation of chemical potential with temperature is equal to

a. partial molar entropy b. partial molar internal energy

c. Partial Molar Free Energy d. partial molar volume

6. When temperature increases, chemical potential

a. increases b. decreases.

c. remains same d. not predicable

7. Chemical potential of a solid at its melting point is X. The chemical potential of its

liquid is

a. greater than X b. lesser than X

c. equal to X d. twice of X

8.Chemical potential of a liquid at its boiling point is Y. Then the chemical potential of its

vapour is

a. greater than Y b. lesser than Y

c. twice of Y d. equal to Y

9. Variation of chemical potential with pressure is equal to

a. partial molar internal energy b. partial molar entropy

c. Partial Molar Free Energy d. partial molar volume

10. Gibbs –Duhem equation is

a. ∑ ¿dGi= 0 b ∑ μi dGi= 0

c. ∑ ¿dμi= 0 d. ∑ μi dni= 0

11. Which is true ?

a. molar volume is additive but partial molar volume is not additive

b. molar volume is not additive but partial molar volume is additive

c. both molar volume and partial molar volume is not additive

d. both molar volume and partial molar volume is additive

12. The partial molar volume of NaCl solution and water are 49.5 and 48.5 respectively.

The total volume of the mixture when one mole of each is mixed is



a. equal to 98 b. lesser than 98

c. greater than 98 d. equal to 1.0

13. The partial molar volume of alcohol in a mixture containing one mole of alcohol and

one mole of water is 50 ml, Find the partial molar volume of water if the total volume of

the mixture is 101

a. 51 b. 60

c. 20 d.100

14.The partial molar volume of alcohol and water in a mixture of alcohol and water is 40

ml and 20 ml respectively. If, the mixture contains 2 moles of each find the total volume of

the mixture

a. 51 b. 120

c. 20 d.100

15. Partial molar volume of ethanol and water is 53 .1 ml and 16. 9 ml respectively Find the

volume of mixing when one mole of each is mixed. Total volume of pure components before

mixing is 72 ml

a. +2 b. -12

c. -2 d. -7

16. The variation of chemical potential of a solid with temperature is shown below.

Find the chemical potential of the compound in the liquid state at its melting point

a. 0.2 b. 0.5

c. -2 d. 0.3

17. Which is true ?

I. ( ∂ μ∂ P ) T,n1,n2.. = (

∂ V∂ n

) T, P II. ( ∂ μ∂ P ) T,n1,n2. = V III . ( ∂ μ

∂ T ) P,N = - (∂ S∂ n

) T, P

a. I only b. II only

c. III only d. all the three



18. ( ∂ μ∂ T )P , N is equal to

a. V b. S

c. - S d. G

19. The Relation between fugacity and free energy change is given by

a. ∆ G = nT ln(d ln f ) b. ∆ G = nRT (d ln f )

c. ∆ f = nR ln (d ln G ) d. ∆ f = nRT ln (d lnG )

20 In the determination of fugacity by graphical method the area under the curve of α

versus P is 83.14 units: Find the fugacity of N2 at 100 K and 10 atm pressure e−0.1 = 0.9048

a. 9.048 b. 90.48

c. 904.8 d. 0.301

21. For an ideal binary mixture which is true ?

I.∆ V mix = 0 II. ∆ Hmix = 0 III. ∆ Gmix = RT ∑ x i ln x i IV. ∆ Gmix = 0

a. I ,II and III only b. I and III only

c. II and III only d. I ,II and IV only

22. The fugacity of a pure hydrogen gas was found to be is 25 atm and the when it is mixed

with argon it increases to 50 atm . Calculate the activity of the gas.

a. 70 b. 2

c. 90 d. 25

23.The fugacity of a pure gas is 10 atm and the same gas in a mixture is 20 atm Calculate the

activity coefficient of the gas at 20 atm pressure

a. 70 b. 0. 2

c. 0.1 d. 20

24. In rational system the composition is expressed in

a. molarity b. molality

c. normality d. mole fraction

25. In the determination of activity for non-electrolytes by Nernst distribution law CS2

(A)and H2O (B) were used. Plot of mA

mB against mB gave a straight line which on

extrapolating to zero concentration. gave 0.2 and if the activity of B is 20 what is the

activity of the component A.

a. 20 b. 0. 2



c. 0.1 d. 4

26. The following graph shows the determination of activity for non-electrolytes by Nernst

distribution law

If the activity of B is 50 what is the activity of the component A.

a. 20 b. 0. 2

c. 2.5 d. 4

27. Departure from ideal behavior (𝛂) is determined using the formula

a. α = RTP |+ V b. α =

RTP – V

c. α = RTV – P d. α =

RP – V

28. ASSERTION(A) The standard sate of real gas can be fixed in terms of fugacity.

REASON ( R) Fugacities of real gases can be measured

a. Both A and R are correct and R is the correct explanation

b. A is correct and R is wrong

c. A is correct and R is not the correct explanation

d. Both A and R are wrong

29.. Match the following

1 (X) real mixture - (X) ideal mixture A Variation of chemical potential with pressure

2 p1 = x1 × p0 B Roault’s law

3 RT ∑ x i ln x i C Excess function

4 ( ∂ V∂ n

) T, P D Free energy change of mixing

5 ( ∂ S∂ n

¿ T, P E Decrease of chemical potential

with temperature

a. 1-C,2-B,3-D,4-A, 5-E

b. 1-C,2-D,3-B,4-E 5-A

c. 1-E,2-D,3-C,4-B, 5-A



d. 1-A ,2-B,3-C,4-D, 5-E

30 .Which is not true ?

a. Fugacity in pure state multiplied with activity gives fugacity of mixture

b. For gases fugacity corresponding to one Kelvin is taken as standard state .

c. In practical system of standard state composition is expressed in molarity

d. The ratio between activity and pressure of the real gas is called activity coefficient.

*****************

KEY

13. The partial molar volume of alcohol in a mixture containing one mole of alcohol and

one mole of water is 50 ml, Find the partial molar volume of water if the total volume of

the mixture is 101

V B = V−nA V A

nB

= 101 – (1 )(50)

1

= 51

14.The partial molar volume of alcohol and water in a mixture of alcohol and water is 40

ml and 20 ml respectively. If, the mixture contains 2 moles of each find the total volume of

the mixture

V = nA V A + nB V B

= 2 (20) + 2 (40 )

= 120

15. Partial molar volume of ethanol and water is 53 .1 ml and 16. 9 ml respectively Find the

volume of mixing when one mole of each is mixed. Total volume of pure components before

mixing is 72 ml

volume of mixing = ( 53 .1 + 16. 9) – 72

= -12

16. The variation of chemical potential of a solid with temperature is shown below.



Find the chemical potential of the compound in the liquid state at its melting point

chemical potential of the compound in the liquid state at its melting point = 0.3

20 In the determination of fugacity by graphical method the area under the curve of α

versus P is 83.14 units: Find the fugacity of N2 at 100 K and 10 atm pressure e−0.1 = 0.9048

ln fp = -

1RT × area under the curve

= - 1

8.314×100 × 83.14

= - 0.1

taking exponential f = p × e−0.1

= 100 × 0.9048

= 9.048

22. The fugacity of a pure hydrogen gas was found to be is 25 atm and the when it is mixed

with argon it increases to 50 atm . Calculate the activity of the gas.

Activity = 5025

= 2

23.The fugacity of a pure gas is 10 atm and the same gas in a mixture is 20 atm Calculate the

activity coefficient of the gas at 20 atm pressure

Activity = 2010

= 2

activity coefficient = ap



= 2

20

= 0.1

25. In the determination of activity for non-electrolytes by Nernst distribution law CS2

(A)and H2O (B) were used. Plot of mA

mB against mB gave a straight line which on

extrapolating to zero concentration. gave 0.2 and if the activity of B is 20 what is the

activity of the component A.


mA

mB

= 20 × 0.2 = 0.4 26. The following graph shows the determination of activity for non-electrolytes by Nernst

distribution law

If the activity of B is 50 what is the activity of the component A


mA

mB

= 50 × 0.5 = 2.5 .

UNIT – 3 – SPECTROSCOPY - 3 CODE: 43211234

MARKS : 100 [ Each caries 1 mark except mentioned ] DURATION : 60 mts

DATE : 24 -08-2020 TIME : 9.30 pm -10.30

pm

1. Which is not resonance spectra?

a. NMR b. ESR



c. NQR d. Mossbauer

2. Zeemann effect is splitting a spectral line into several components by

a. Electric field b. increasing intensity

c. Magnetic field d. Internal pressure

3. Which is not true?

a. Normal Zeeman effect is due to interaction between external magnetic field and

orbital magnetic moment

b. Anomalous Zeeman effect is due to interaction between external electric field and

Combined orbital and intrinsic magnetic moment

c. Normal Zeeman effect is observed when the spin is zero

d. Only nine transitions consistent with the selection rule for l = 1 to l= 2

4. Number of allowed transitions from the energy level l = 1 to l= 2 is

a. 9 b. 5

c.4 d.12

5.The energy levels corresponding to the j value 1 and 2 are split into – number of levels

respectively

a. 3, 5 b. 5,4

c.4,8 d.12,8

6. Zeeman effect splits the energy state P1/2 and P3/2 in to --- number of states

a. 3, 5 b. 2,4

c.4,8 d.12,8

7. Number of transition allowed from the energy level P3/2 to S1/2 is

a. 9 b. 6

c.4 d.12

8. Number of transition allowed from the energy level P1/2 to S1/2 is

a. 9 b. 6

c.8 d.4

9.Transition of electrons from the energy level P3/2 to S1/2, when magnetic field is applied

is

a. Normal Zeeman effect b. Stark effect

c. Inverse Zeeman effect d. Anomalous Zeeman effect

10.The compound which has high value of chemical shift ( in )

a. CH3Cl b. CH3I

c. CH3F d. CH3Br



11.Deshielding of proton in CH3F is more than that of CH3I. this is because

a. Fluorine is small atom b. Fluorine has high electronegativity

c. Fluorine is a gas d. Fluorine does not form oxy acid

12.Which is most deshielded?

a. CHCl3 b.CH3Cl

c. CH2Cl2 d. CH4

13. Anisotropic induced magnetic field effects are found in

I. alkane II. alkene III alkyne

a. II and III only b. II only c III only d. I and III only

14. Chemical shift value of ethane was 0.86 but that of ethylene is 5.25.This is due to

a. Hydrogen bonding b. Anisotropic

c. Inductive effect d. Electromeric effect 15.Which is not true in 13C –NMR

I . Chemical shift ranges from 0 – 10 ppm II. 100 % natural abundance

III. Coupling is significance


c. III only d. II and III only

16.In FTNMR, spectrum is presented in

a. time domain b. velocity domain

c. position domain d. frequency domain

17. In FTNMR

a. frequency domain converted into time domain

b. time domain is converted into amplitude domain

c. amplitude domain is converted into time domain

d. time domain is converted into frequency domain

18.The reference compound for 19F NMR spectroscopy is

a. CCl3F b. CH2F2

c.. CHF3 d. CF4

19. The number of NMR signals in Methyl cyclopropane

a. 1 b.3

c. 2 d.4

20 Which is not true ?



a. In PMR chemical shift ranges from 0 – 100 ppm while in CMR it ranges from 0- 10

ppm

b. In PMR coupling between H-H is more abundant while in CMR it is ignored

c. PMR gives high intense lines while CMR gives low intense peak

d. In PMR 100 % natural abundance while in CMR it is only 1.1 %

21. The PMR signal in 1,1,2- trichloroethane contains [ 4 marks]



a. doublet( 2H) , triplet(1H) b.doublet(1H) , triplet(2H)

c. doublet(2H) , triplet(3H) d. quartett(2H) , triplet(1H)

22.A compound has MF C2H6O. Its PMR is given below.

The compound could be [ 4 marks]

a. Ethanol b. acetone

c propanone d. ethyl methyl ketone

23.Which one of these spectra corresponds to that of CHCl2CH2Cl [ 4 marks]

24.The compound with MF C5H12 which shows one PMR signal is [ 4 marks]

a. ethyl iso propyl ketone b. 2, 2- dimethyl butane

c 2- methyl butane d. 2, 2- dimethyl propane

25. Match the following?[ [ 4 marks]

1 CH3CH3 A Doublet, multiplet

2 CH3CH2Cl B Singlet, quartet triplet

3 Cl-CH2-O -CH2-CH3 C Triplet, quartet

4 CH3-CH- CH3

CH3

D singlet

a. 1-A ,2-B,3-C,4-D b. 1-D,2-B,3-C,4-A



c. 1-C,2-D,3-B,4-A d.1-D,2-C,3-B,4-A

26. Match the following?[ 4 marks]

compound Number of NMR signals

1 1-Chloro propane A 1

2 Ethoxy acetic acid, B 2

3 1,1 dibromo ethane C 3

4 1,2 dibromoethane D 4

a. 1-A ,2-B,3-C,4-D b. 1-C,2-B,3-D,4-A

c. 1-C,2-D,3-B,4-A d.1-C,2- D,3-A,4-B

27. A compound shows a proton NMR peak at 300 Hz downfield from TMS peak in a

spectrometer operating at 100 MHz. the chemical shift of the compound is ?[ 4 marks]

a. 9 ppm b. 3ppm

c.8 ppm d.4 ppm

28.Which shows one NMR signal? [ 4 marks]

a. propanone b. Ethyl chloride

c. Methanol d. 1-Chloro propane

29. The compound does not show two NMR signals is[ 4 marks]

a. ethyl benzene, b. iso butylene

b. 1,1 dibromo ethane d. 2- chloro propane


spectrometer with chemical shift of 3pp. .The operating frequency of the spectrometer is [ 4

marks]

a. 50 MHz b.100 MHz

c. 2 MHz d.4 MHz

31. An NMR spectrum obtained from a spectrometer operated at 500 MHz shows peaks at 0.864

and 0.849 ppm find the coupling constant[ 4 marks]

a. 50 Hz b.100 Hz

c. 7.5Hz d.4 Hz

32. An NMR spectrum obtained from a spectrometer operated at 500 MHz shows peaks at 3.585,

3.568 and 3.55ppm Find the coupling constant[ 4 marks]

a. 50 Hz b.100 Hz



c. 7.5Hz d. 8.5 Hz

33. An NMR spectrum obtained from a spectrometer operated at 500 MHz shows peaks at

2. 837, 2.808, 2.825 and 2,796 ppm Find the coupling constants[ 4 marks]

a. 50,7.5 Hz b.100, 2.5 Hz

c. 7.5, 69 Hz d.14.5,6 Hz

34. The compound which shows three NMR signals is [ 4 marks]

I. 1-Chloro propane II. iso propanol III diethyl succinate

a.I only b. II only c I,II,III d. III only

35.. Which has multiplet in nmr? [ 4 marks]

I. benzene II. iso butylene III. 2-chloro propane

a. I only b. III only

c II only d. I and II only

36. The spectrum of isopropyl chloride shows[ 4 marks]

a. doublet, multiplet b. doublet, doublet

c. triplet, quartet d. doublet

37. The spectrum of CH3-CH2-O- COOH shows[ 4 marks]

a. triplet, quartet, singlet b. triplet, doublet,multiplet

c. triplet, quartet, doublet d. quartet, triplet,multiplet

38.. A triplet and a quartet appears in [ 4 marks]



c III only d. I,II

39. The compound showing one PMR signal with MF C8H18 [ 4 marks]

a. ethyl benzene, b. 2- iso propyl pentane

c. 2,2,3,3 – tetramethyl butane d. 2- ethyl hexane

40.. The organic compound with MF C5 H12 shows the following NMR signals. Triplet δ =

0.98 ( 6H) , sextet t δ = 1.28 (4H), pentet t δ = 1.28(2H) [ 4 marks]

a. 2-ethyl propane b. 3- methyl butane

c. 2,2,-dimethyl propane d. n – pentane



KEY

19.

22 . Establish the structure of the following compounds showing one PMR signal C2H6O

Solution:

Only one signal shows all 6 hydrogens should have same environment

24. Establish the structure of the following compounds showing one PMR signal C5H12

Solution:


2,2 – Dimethyl propane

26. number of signals


spectrometer with chemical shift of 3pp. . Find the operating frequency



Solution:

Chemical shift = 300

x×106 ×106

3 = 300

x×106 ×106

X = 100 MHz


0.864 and 0.849 ppm find the coupling constant

Solution:

J =( 0.864 – 0.849 )× 500

= 0.15 × 500

= 7.5 Hz


3.585, 3.568 and 3.55ppm Find the coupling constant

Solution:

J =( 3.585 – 3.568 )× 500

= 0.017 × 500

= 8.5 Hz


2. 837, 2.808, 2.825 and 2,796 ppm Find the coupling constant

Solution:

J1 =( 2.837 – 2.808 )× 500

= 0.029 × 500

= 14.5 Hz

J2 =( 2.837 – 2.825 )× 500

= 0.012 × 500

= 6 Hz

34. The compound which shows three NMR signals is



35.Which has multiplet in nmr?

I. benzene II. iso butylene III. 2-chloro propylene

36. The spectrum of isopropyl chloride shows

37. The spectrum of CH3-CH2-O- COOH shows

38.. A triplet and a quartet appears in




39. Establish the structure of the following compounds showing one PMR signal

C8H18,

Solution:

DBE = C+1- H2

= 9 – 9

= 0


2,2,3,3 – tetramethyl butane

40.. The organic compound with MF C5 H12 shows the following NMR signals. Triplet δ =

0.98 ( 6H) , sextet t δ = 1.28 (4H), pentet t δ = 1.28(2H)

Solution:

1. Triplet 6H shows there should be a carbon containing 3 H-atom having a

neighbour CH2 or CH – CH - CH

2. CH3 – CH2 – CH3

3. Pentet (2 H) shows there should be carbon having a neighbour with 4. CH2 – C- CH2

5. Sextet t δ = 1.28 (4H), shows CH2 – C- CH3

Therefore the structure could be CH3 – CH2 – CH2 – CH2 –CH3

n – pentane

UNIT – 3 – ESR SPECTRA CODE: 43211234



MARKS : (12+ 8 = 20 ) [ Each caries 1 mark except mentioned ] DURATION : 20

mts

DATE : 31 -08 -2020 TIME : 9.30 pm -10.00

pm

This paper contains 14 questions

1.Which is not true in ESR?

I. It is an absorption spectra. II. Absorption is in Ultra Violet region

II. Spectra is shown only by paramagnetic substances

a. I only b. I and II only c. I and III only d. II only

2.The magnetic energy of interaction is given by

a. E = gBMs b. E = βBMs

c.. E = gβBMs d. E = gβB

3.The number of hyperfine lines in Deuterium atom is

a. 1 b. 3

c.. 2 d. 4



4. Absorption region in ESR spectroscopy is

a. microwave region b. radio wave region

c. γ-ray region d. X-ray region

5. ESR is observed in ,

I. free radicals, II. triple state molecules III. Diamagnetic molecules

a. I and II only b. II and III only

c. III only d. I and III only

6.. The selection rule in ESR is

a. ∆MI = ±1 ,∆Ms = ± 1 b. ∆MI = 0 ,∆Ms = ± 1

c. ∆ MI = 0 ,∆ Ms = 0 d. ∆MI = ±1 ,∆Ms = ± 2

7 . McConnell relation is

a. AH = Q CH b. A H =eh

4 π mc

c. AH = ρ Q CH d. g = hϑβ B

8. Chemical structure of the standard used in ESR is

9..Match the following ions with number of ESR lines [ 2 marks]

1 1.4-Benzoquinone radical anion A 7

2 Naphthalene ion B 75

3 Antracene ion C 25

4 Radical anion of benzene D 5

5 Deuterium E 3



a. 1-A ,2-B,3-C,4-D, 5-E b. 1-D,2-B,3-C,4-A,5-E

c. 1-C,2-D,3-B,4-A,5-E d.1-D,2-C,3-B,4-D,5-E

10. The separation between the lines in the ESR spectra of methyl radical was found to be

23 gauss and if Q= 500 G the electron density on carbon is [ 2 marks]

a. 50% b. 75%

c.. 86.4% d. 95.5 %

11.The spin density of benzene is 22.5 G .It shows a septet with coupling constant 3.75 G

The time at which the electron spend on each carbon atom is [ 2 marks]

a. 50% b. 15.8 %

c.. 26% d. 95. %

12.Match the following atom/ ions with number of ESR lines [ 2 marks]

1 H-atom A 7

2 CH2 radical B 4

3 CH3 radical C 3

4 Radical anion of benzene D 2

a. 1-D,2-C,3-B,4-A b. 1-D,2-B,3-C,4-A,

c. 1-C,2-D,3-B,4-A, d. 1-A ,2-B,3-C,4-D,


a. In ESR absorption region is radio wave while in NMR absorption region is

microwave

b. ESR arises due to spinning of electron about nucleus. NMR arises due to interaction

between proton and magnetic field

c. ESR Spectra is shown only by compounds having unpaired electron while NMR

Spectra is shown by all compounds

d. No Chemical shift in ESR while in NMR there occurs chemical shift

14.The standard used in ESR is

a. 1,2-diphenyl-1-picrylhydrazyl radical b. 2,2-diphenyl-1-picrylhydrazyl radical

c. 1,1-diphenyl-2-picrylhydrazyl radical d. 2,2-diphenyl-3-picrylhydrazyl radical

15. Number of unpaired electrons per gram in DPPH is

a.0 .53 × 102 0 b. 2 .53 × 101 2



c. 1.53 × 1021 d. 1 .53 × 1013

16. Number of line in benzene solution of DPPH is

a. 2 b. 3

c. 4 d.5

key

Number of ESR lines

benzo quinone radical = ( 2nI + 1 )

2(4) ( ½ ) + 1 = 5

Naphthalene ion

Number of lines = ( 2nIi + 1 ) ( 2mIj +1 ).

Naphthalene ion has two sets of 4 equivalent protons each.

Number of lines = ( 2) (4) ( ½ )+ 1 )× ( 2) (4) ( ½ )+ 1 )

= 25 ?

Anthracene ion

It has 3 sets of equivalent protons . Set α and β contains 4 equivalent protons each and set γ

contains 2 equivalent protons

Number of lines = [ ( 2) (4) ( ½ )+ 1 )] ×[( 2) (4) ( ½ )+ 1)] × [ ( 2) (2) ( ½ )+ 1]

= 75 lines ?

Radical anion of benzene

The number of hyperfine lines in C6H5 radical = 2(6) ( ½ ) + 1

= 7

The number of hyperfine lines in D atom = 2(2) ( ½ ) + 1

= 3

2. The separation between the lines in the ESR spectra of methyl radical was found to be 23

gauss and if Q= 500 G the electron density on carbon is

The separation between the lines (A) is found to be 23 gauss

Applying McConnell relation



ρ = AQ

= 23

500

= 0.046

= 4.6 %

Electron density on carbon = 100- 3(4.6 )

= 100 - 13. 8

= 86.4 %

3.The spin density of benzene is 22.5 G .It shows a septet with coupling constant 3.75 G

The time at which the electron spend on each carbon atom is

standard value of Q= 22.5 G ( gauss)

Shows a septet with coupling constant 3.75 G

ρ = AQ

= 3.7522.5

= 0.0076

There are 6 hydrogens with this electron density making a total of

6×0.0076 = 0.046

= 5%

The remaining 95 % of electron’s time is spent equally on the 6 carbon atoms

On each carbon = 956

= 15.8

4. The number of hyperfine lines in H atom = 2(1) ( ½ ) + 1

= 2 The number of hyperfine lines in CH2 radical = 2(2) ( ½ ) + 1 = 3 The number of hyperfine lines in CH3 radical = 2(3) ( ½ ) + 1

= 4

Radical anion of benzene

The number of hyperfine lines in C6H5 radical = 2(6) ( ½ ) + 1



= 7

UNIT – 3 – MASS SPECTRA CODE: 43211234

MARKS : [9+ 44 ] 53 [ Each caries 1 mark except mentioned ] DURATION : 60

mts

DATE : 07-09-2020 TIME : 9.30 pm -10.30

pm

This paper contains 20 questions

1. McLafferty rearrangement takes place in

I. 3-methylbutramide II. pentanoic acid III. butanal

a. II and III only b. II only

c III only d. I , II and III

2. Necessary requirement for the occurrence of McLafferty rearrangement is that the

compound must have

I. 𝛄- H atom II. unsaturation III at least one co ordinate bond

a. II and III only b. II only

c I and II only d. I and III only

3. CH3CH2CH2CHO undergoes McLafferty rearrangement. One of the product formed

could be

a. propene b. ethene

c. ethyne d. propanone

4. Pentanoic acid undergoes McLafferty rearrangement. One of the peak could have been

appeared at m/e value[ 4 marks]

a. 60 b. 28

c. 26 d. 15

5. 3-Methylbutyramide undergoes McLafferty rearrangement. One of the peak could have been


a. 59 b. 28

c. 26 d. 15

6. The mass spectra of a compound [ MWt = 72] shows peak at 44 [ CH2CHOH] with the

elimination of ethylene molecule .The compound could be [ 4 marks]



a. 3-Methylbutyramide b. Butanal

c. pentanoic acid d. butanol

7. Which is not true in McLafferty rearrangement?

a. It involves the cleavage of a β - bond followed by a γ - hydrogen transfer

b. The rearrangement leads to elimination of neutral molecules

c. It takes place only in aldehydes

d. It takes place through a six membered transition state.

8.The MR ion formed in 2- hexanone corresponds to the m/e is [ 4 marks]

a.59 b. 28

c. 26 d. 58

9.The MR ion formed in 1- pentene [ 4 marks]

a. CH3CH b. CH2CHCH2

c. CHCH d. CH3CHCH2

10 The MR ion formed in n-propyl benzene corresponds to the m/e is [ 4 marks]

a. 59 b. 28

c. 92 d. 58

11.The MR ion formed in methyl-n- propyl ketone [ 4 marks]

a. CH3COH b. CH3COHCH2

c. CHCOH d. CH3CH2COHCH2

12.Double MR is found in

a. CH3 CH2CH2 CO CH2CH2CH3 b. CH3 CH2 CO CH2CH2CH3

c. CH3 CO CH2CH2CH2CH3 d.CH3 CO CH2CH2CH2CH3

13. Double MR in 4- heptanone corresponds to the m/e is [ 4 marks]

a.58 b. 28

c. 26 d. 68

14.Match the following?[ [ 4 marks]

compound m/e of MR ion

1 . Butanal A 42



2 Pentanoic acid B 59

3 3-Methylbutyramide C 60

4 2- hexanone D 44

5 1- pentene E 58

a.1-A ,2-B,3-C,4-D, 5-E b. 1-D,2-C,3-B,4-E , 5-A

c. 1-C,2-D,3-B,4-A , 5-E d.1-D,2-C,3-B,4-A, 5-E


compound MR ion

1 4- heptanone A CH3CHCH2

2 methyl-(1-phenyl propyl) ketone B C6H5CH2

3 n-propyl benzene C CH3COHCH3

4 1- pentene D CH3COHCH2

a.1-A ,2-B,3-C,4-D, b. 1-D,2-C,3-B,4-E ,

c. 1-C,2-D,3-B,4-A , d.1-D,2-C,3-B,4-A,

16. The ionised reagent not used in Mass Spectra is

a. methane b. ammonia

c. isobutene. d. chloroform

17.The radius of the curvature of the different ions in mass spectrometer is related with (me

¿

as

a. r2 = 2 mB2e

b. r2= EmB2e

c. r2 = 2EB2e

d. r2 = 2 EmB2 e

18. ASSERTION(A): Chemical ionization technique is preferred to Electron impact

technique

RESAON ( R) : Electron impact technique will not produce molecular ion peak



a. A is correct but R is wrong

b. Both A and R are wrong

c. A is correct and R is the correct explanation for A

d. A is correct and R is not the correct explanation for A


a. In chemical ionization the observed m/e value is one unit more than the true molecular

weight.

b. In positive ion CI spectra molecule is decomposed

c. Methane is an ionised reagent gas

d. Electron ionization gives rise to fragment ions

20 Which is not true?

a. In Dempster’s mass spectrometer Ions with larger me follow a path with greater

radius.

b. positive ion CI spectra molecule gets protonated

c. Electron impact technique will not produce molecular ion peak

d. Mass spectra is recorded with intensities of the signals against mass values.

Key

1. McLafferty rearrangement takes place in 3-methylbutramide pentanoic acid .

butanal[ all the 3 compounds ]

2. Necessary requirement for the occurrence of McLafferty rearrangement is that the

compound must have 𝛄- H atom and unsaturation

3. . CH3CH2CH2CHO undergoes McLafferty rearrangement. One of the product formed

could be



Ans : ethene

4 Pentanoic acid undergoes McLafferty rearrangement. One of the peak could have been


Solution :

Ans = 60

5. 3-Methylbutyramide undergoes McLafferty rearrangement. One of the peak could have

been appeared at m/e value[ 4 marks]

Peak at = 101- 42= 59

6: The mass spectra of a compound [ MWt = 72] shows peak at 44 [ CH2CHOH] with the

elimination of ethylene molecule .The compound could be

Solution :



Mol.wt = 48 + 8+16 = 72

The compound is Butanal

7. McLafferty rearrangement

a. It involves the cleavage of a β - bond followed by a γ - hydrogen transfer

b. The rearrangement leads to elimination of neutral molecules

c. It takes place through a six membered transition state.

8.The MR ion formed in 2- hexanone is

9.The MR ion formed in 1- pentene

10.The MR ion formed in n-propyl benzene

11. The MR ion formed in methyl-(1-phenyl propyl) ketone



12. Double MR is found in 4- heptanone[ CH3 CH2CH2 CO CH2CH2CH3]

13. . Double MR in 4- heptanone

Match the following?[ [ 4 marks]

compound m/e of MR ion

1 . Butanal A 42

2 Pentanoic acid B 59

3 3-Methylbutyramide C 60

4 2- hexanone D 44

5 1- pentene E 58

b. 1-D,2-C,3-B,4-E , 5-A


compound MR ion

1 4- heptanone A CH3CHCH2

2 methyl-(1-phenyl propyl) ketone B C6H5CH2



3 n-propyl benzene C CH3COHCH3

4 1- pentene D CH3COHCH2

c. 1-C,2-D,3-B,4-A ,


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