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CHEMISTRY CHE521, PHYSICAL CHEMISTRY AND QUANTUM MECHANICS

PG SEMESTER II, PAPER :- V

M.Sc. Chemistry: Second Semester

Physical Chemistry and quantum Mechanics

Paper: V Paper Code: CHE 521

UNIT-I

Introduction Quantum mechanics

Postulates of quantum mechanics …………………………………………………………………………………… Eigen Function and Eigen value………………………………………………………………………………………… Operators……………………………………………………………………………………………………………………….. Linear and Hermitian Operator……………………………………………………………………………………… Theorem Of Operator……………….…………………………………………………………………………………… Angular Momentum Operators………………………………………………………………………………………

UNIT-II

Solvable Systems

Concept of associated Legendre polynomials………………………………………………………………. Quantum mechanical Treatment of rigid diatomic molecule……………………………………….. Eigen function and Eigen values……………………………………………………………………………………. Application of schrodinger wave equation to one electron Sepration of R, θ and ϕ equation Solution of R, θ and ϕ equation of system (H-like atom) Radial wave functions and angular wave functions Most probable distance of electron in H-like atom Energy of electron in ground state Concept of s,p and d orbitals Concept of Hermite polynomials Quantum mechanical treatment of simple harmonic oscillator………………………………..…… Energy of electron in ground state Eigen functions and Eigen values……………………………………………………………………………………

UNIT-III

Approximate method

Perturbation method ……

First order perturbation

Application to He-atom: Variation theorem

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Secular equations

Slater determinants

Singlet and Triplets states in He-atom

Coulomb and exchange integrals Symmetric and antisymmetric wave function

Anti-symmetric and Pauli’s exdusion principle

UNIT-IV

Chemical Bonding

LCAO-MO theory…………………………………………………………………………………………………………. Application of LCAO-MO theory to H2

+ ion and H2 molecule………………………………………… Heitler Lordon theory(VBT) and its application to H2 molecule…………………………………….

UNIT-V

Huckel Molecular orbital theory

HMOT applied to ethylene…………………………………………………………………………………………….. Allyl system and Butadiene: Calculation of bond order………………………………………………….. DE Free valence and charge densities for system………………………………………………………

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Prof. Ratan Kumar Choudhary

Professor of Chemistry

Specialization in Physical Chemistry

L. N. Mithila University, Darbhanga.

Subject CHEMISTRY

Paper V

Paper Code CHE521

Topic Physical Chemistry and quantum

Mechanics

5



UNIT-I

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Postulates of quantum mechanics – Postulate: 1 The state of a submicroscopic system is described by a function that

depends on the coordinates of the particles (s) and on the time. The function which is called state function or wave function contains all

the information that can be obtained about the system. The has the important property that (r, t) *(r, t) is the probability

that the particle lies in the volume element d located at r at time t, where d = dx, dx.dy or dx.dy.dz. The wave function must be singled value, continuous and finite.

Postulate: 2 To every physically ‘observable’ measurable property (i.e. position, momentum, energy, etc.), there corresponds a linear Hermitian operator. The operators corresponding to the above quantities are as follow:- Observable Operator Symbol Operator

Position (x) x x

Position (r) r r

Momentum (P) p

Kinetic Energy (T) T

Potential energy V(r) )(ˆ rV V(r)

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Total energy (E) H

Or

)(2

22

rVVm

h

Angular momentum lx xl

ly yl

lz zl

Postulate: 3

The possible values of any physical quantity of a system (i.e. energy, angular momentum, etc) that can result from measurements of the physically observable property are given by the eigen values ‘a’ in the operator equation.

A = a

Where A is the operator corresponding to that physical quantity and ( ) a well behave function is the eigen function. According to this

postulates, the value of dynamical variables can be quantized. If the system is in an eigen state of A with eigen value ‘a’ then any measurement of the quantity A will yield ‘a’.

Although measurements must always yield an eigen value, the state does not have to be an eigen state of A initially. An arbitrary state can be explained in the complete set of eigen vectors of A ( A i = ai i)

as =

n

i iiC where n nay go to infinity.

Ci = Coefficient Postulate: 4

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If a system in a state described by a normalized wave function , then

the average value of the observable corresponding to A is given by,

dAA ˆ*

Postulate: 5 The time dependent Schrödinger equation exist for every system which is given by

t

hiH tr

),(ˆ

Where H is the Hamiltonian (i.e. energy) operator of the system. If the Hamiltonian is independent of time, we have the possibility of states of definite energy E. For such states the state function must satisfy. EH ˆ

Postulate: 6 The total wave function must be antisymmetric with respect to the interchange of all coordinates of one fermion with those of another. Electronic spin must be included in this set of coordinates. The pauli exclusion principle is a direct result of this antisymmetric principal. Fermions – A subatomic particles, such as a nucleon, which has half integral spin and follows the statistical description given by Fermi and Dirac. Eigen function (A well beheave function) and Eigen value For a stationary or standing wave in a stretched string, the amplitude function f(x) can have significance only for certain definite value of . These functions must satisfy the following condition.

a) f (x) must be equal to zero at each end of the string as amplitude of vibration is zero at the points where string is fixed.

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b) f (x) must be single valued and also finite i.e. at every point on the vibrating string, the amplitude has a definite value at any given instant.

There will be many expression for f (x) or satisfying the Schrödinger

equation all of which, how ever are not acceptable. Only those value of wave function are acceptable and are of significance which

correspond to some definite values of the total energy E. Such values of total energy E is called Eigen values which can also be called proper or characteristic values. The corresponding values of wave function f(x) or are called Eigen functions or also the wave

functions. In this case of stretched string, the eigen functions of Schrödinger equation will be those which must satisfy the following conditions. i) must be singled valued. That means for each value of the variable x,

y and z there is only one definite value of the function .

ii) must be finite.

iii) must be continuous. That means there must not exist any sudden

changes in as its variable is changed.

iv) must become zero at infinity.

Eigen value comes from the German, “Eigen wert” which means “proper of characteristics value”. Similarly, eigen function comes from “Eigen Function” the meaning of which is “proper or characteristic function” Corresponding to each eigen value there is an eigen function. The solution of the Schrödinger equation for a given energy Ei involves also finding the specific function i which describe the energy state. The

solution of time independent Schrödinger equation takes the form, iii EH ˆ

The eigen value concept is not limited to energy. When applied to a general operator Q, it can take the form,

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iii qQ ˆ

Where, Q = Operator

i = Eigen function

iq = Eigen value

if the function i is an eigen function for that operator. The eigen value

iq may be describe and in such cases we can say that the physical

variable is ‘quantized’ and that the index is play the role of a “quantum number” which characterizes the state. Operators; Angular momentum operators; Linear operator; Hermitian operator; Theorem of operators Operators:- The mathematical formulation of quantum mechanics is built upon the concept of an operators. It is a mathematical instruction or procedure to be carried out on a function so as to get another function. It is written in the form, (Operator).(Function) = (Another function) The function on which the operation is carried out is called “operand”. The operator when written alone is meaning less or has no significance. The dot (.) used in the left hand side does not mean that the function is multiplied with the operator. e.g. Operator . Operand/function = New function

i) xxdx

d22 ; here

dx

d = Operator

x2 = operand 2x = the result of the operation.

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ii) dx

d (x4) = 4x3; dx

d = Operator, x4= Operand & 4x3= the result of

operation

iii) Cxdxx 443 Here ∫ ( )dx which stands for integration with respect

to x is termed us operator. Cx 4

4

the result of the operation when the

operator applies on operand x3. Similarly, taking the square root, cube root or multiplication by a constant K etc. are different operators which can be carried on any function. In case the symbol used for the operator is not self explanatory, a suitable letter or some symbol for the operation may be used with putting the symbol caret over it.

Algebra of operators:- The operators follow certain rules similar to those of algebra which includes addition, substraction, multiplication, square and commutation. Addition or substraction (Sum and difference ) of two operators If A & B are two different operators and f(x) is operand then, ( A + B )fx = A (fx) + B (fx) Similarly, ( A - B ) fx = A (fx) - B (fx)

e.g. if D = dx

d

then, 53)5(ˆ5.3ˆ 333 xxDxD

Or, 53)5( 33 xxdx

d = 3x2 + (3x3 – 15 )= 3x3 + 3x2 -15

Multiplication (Product) of two operators Multiplication of two operators means operations by the two operators one after the other. But the operation will be carried out from right to left. e.g. If A & B are two different operators and f(x) is the operand, then the expression ( A B f(x)) implies that at first f(x) will be operated by the

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operator B to get the result say f ′(x) and then f ′(x)will be operated by the operator A to get the final result say f ′′(x) as, A B f(x)implies that, B f(x) = f ′(x) Then, A f ′(x)= f ′′(x) So we can say that A B f(x)= f ′′(x). If the same operation is to be done a number of times in continuations, it to be shown by a power of operator. It can also be called square of an operator, i.e. the product of the operator with itself. e.g. A f(x)= A 2f(x) It may be noted that, usually A B f(x) B A f(x)

Let us suppose A = x & B = dx

d & f(x)= x2

Then, A B f(x)= x. dx

d (x2) = x.2x=2x2

But, B A f(x)=dx

d .x.(x2) = dx

d x3=3x2

Hence A B f(x) B A f(x) Because in both the case L.H.S & R.H.S the result as the order of using operation is from right to left. Commutation or Commutator:- There is a major difference between the algebra (mathematical) and operator algebra. In ordinary algebra the number obey the commutative law of multiplication, but the operators do not necessarily do so. e.g. ab = ba if a & b are numbers , but BA ˆˆ AB ˆˆ as explained in case of multiplication of operators. If two operators BA ˆ&ˆ are said to be commute then they must have to satisfy the following relation with reference to a function.

BA ˆˆ f(x) = AB ˆˆ f(x)

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We can define the commutator [ BA ˆ,ˆ ] of the operators A and B as the

operator BA ˆˆ - AB ˆˆ . If BA ˆˆ = AB ˆˆ then [ BA ˆ,ˆ ] = 0 which is the condition for commutation and in

such case [ BA ˆ,ˆ ] denotes the commutator of BA ˆ&ˆ or can say that BA ˆ&ˆ

commute and if BA ˆˆ AB ˆˆ then BA ˆ&ˆ do not commute.

e.g. The operator 3 and dx

d make no difference, the order in

which we apply.

033,3

dx

d

dx

d

dx

d

Let A be dx

d and B be 2

2

dx

d and f(x) =sin x

Then, BA ˆˆ f(x) = xxdx

d

dx

dcossin

2

2

And AB ˆˆ f(x) = xxdx

d

dx

dcossin

2

2

So, BA ˆˆ f(x) = AB ˆˆ f(x). Hence BA ˆ&ˆ commute. The significance of commuting operators lies in the fact that the observables corresponding to their eigen values can be determined precisely simultaneously. Linear operators:- An operator A is said to be a linear operator if for the two function f and g when it satisfy the following relation. )()(ˆ xgxfA = )(ˆ)(ˆ xgAxfA

Or, )()(ˆ21 xgCxfCA = )(ˆ)(ˆ

21 gfACxfAC

Where C1 & C2 are real or complex constants, and f(x) and g(x) are function of x.

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In other words an operator A is said to be linear if the application on the sum of two functions gives the result which is equal to the sum of the operations on the two functions separately. The multiplicative, differential and integral operators are linear, while operators corresponding to taking square, square root, logarithm etc are non-linear because we may not get the same result with the operator on the sum of two functions and the sum of operations on two functions separately. e.g. Differential:

dx

d)2( 2 xx =

dx

d 2x +2dx

d x = x2 + 2 (Linear)

Integral:

Kxx

xdxdxxdxxx 23

22

22)2( (Linear)

Square root: 5169 but 743169 (Non-linear) Logarithm: log (x + y) log x + log y (Non-linear) Hermitian operator:- The quantum-mechanical operators that represent physical quantities are linear. These operators must meet an additional requirement. A quantum mechanical operator satisfying the following condition is known as Hermitian condition. Let if an operator A has two eigen functions and , and if

dAdA ˆˆ ; When and are real.

Or, dAdA*

* ˆˆ ; When and are complex, * is the

complex conjugate of .It is the volume element of space in which the

function is defined , then the operator A is called Hermitian operator. Let A be the linear operator representing the physical property A. the average value of A is

(A) = dA ˆ*

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Where is the state function of the system, since the average

value of a physical quantity must be a real number, we demand that, (A) = (A)*

Or,

∫ Ψ ∗ A Ψ d τ = ∫ Ψ( A Ψ )*d τ This equation holds for any function that can represent a possible

state of the system. That means it must hold for all well-behaved function . Thus a linear operator that satisfies for all well-behaved

function is called ‘Hermitian operator”. e.g. let =eix and = sin x are the two acceptable eigen functions, then,

xdxedxxdx

dedA ixix sin)(sinˆ

2

2*

dxeixdxe

dx

dxdA ixix )(sin)(sinˆ 2*

2

2*

= dxxe ixsin

The two integrals are the same, therefore, 2

2

dx

d is a Hermitian operator.

Anti Hermitian:- An operator A is said to be Anti Hermitian, if A *= - A . When a Hermitian is multiplied by ik, where K is a real constant it becomes anti- Hermitian and vice-versa. Properties or Theorem of Hermitian Operator:- There are some important theorem about the eigen function and eigin value of Hermitian operators. 1. The eigen value of Hermitian operators are real. Let us consider the eigen value equation aA ˆ ………(i); Where A is Hermitian operator its eigen

function and ‘a’ is the eigen value. Then we have to prove that ‘a’ is real

The complex conjugate of above equation (i) is ***

ˆ aA ……….(ii)

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Now let us multiply the equation (i) by * and equation (ii) by , and

then on integrating we get, dadadA *** ˆ

Or, dadadA *****

ˆ

Since, A is Hermitian

dAdA*

* ˆˆ

Therefore, da* = da **

Or, a = a* hence ‘a’ is real. 2. The eigen function corresponding to the non-degenerate (different) eigen values of a Hermitian operator are orthogonal. Let 1 & 2 are the two eigen function of a Hermitian operators A

corresponding to two eigen values a1 & a2 respectively. The condition of orthogonality is, 0,,0,0 2

*

1

*

2121 dordord

The eigen value equations are, A 1 = a1 1 and A

2 = a2 2

Let us multiply the first equation A1 = a1 1 by *

2 and then on

integration, dadadA 1

*

211

*

21

*

2ˆ ………(i)

We know that A is Hermitian, then,

dadAdA*

221

*

211

*

2ˆˆ

= da*

21

*

2

= da*

212 ………(ii)

Thus from equation (i) & (ii) dada

*

2121

*

21

Or, 0)(*

2121 daa ……..(iii)

Since, 21 aa

Hence 0*

21 d and 1 & 2 are orthogonal.

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If, however a1 = a2 i.e. Ψ1 & Ψ2 are degenerate, then the integral d*

21

must not be zero and in such a case 1 & 2 may not be orthogonal.

3. If a Hermitian operator A commutes with an arbitrary operator B and 1 & 2 are the two eigen functions of A with non degenerated

eigen function then, 0|ˆ| 21 B

Let us suppose, 2211ˆˆ bAandaA

Since, 0ˆˆˆˆ]ˆ,ˆ[ ABBABA

We have, 212121 |ˆˆ||ˆˆ|||ˆ,ˆ|| ABBABA

= 2121ˆ|ˆ||ˆ|ˆ ABBA

= 0|ˆ|)( 21 Bba

Since )( ba 0, as in the case of non-degenerated eigen functions,

Ψ 1 & Ψ2 a b Thus, 0|ˆ| 21 B

4. If two Hermitian operators BA ˆ&ˆ possess a common eigen function, then they commute. According to theorem we have, ;ˆˆ bBandaA both BA ˆ&ˆ

possessing common eigen function .

Therefore, aBbAABBABA ˆˆˆˆˆˆ]ˆ,ˆ[

= 0 abba

Thus the Hermitian operators BA ˆ&ˆ commute. 5. If two Hermitian operators BA ˆ&ˆ commute then they must have common eigen function. This is reverse of theorem – 4 described above. According to theorem, we have 0]ˆ,ˆ[ BA , now let us assume that aA ˆ

. Now we have to prove that is an eigen function of B also.

We have, 0)ˆ()ˆ(ˆˆˆˆˆ]ˆ,ˆ[ BaBAABBABA

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or, )ˆ()ˆ(ˆ BaBA

This implies that B is an eigen function of A with the eigen value ‘a’. It

can happen under following conditions only. i) B is a multiple of i.e. bB ˆ , thus is an eigen function of B with

the eigen value b. ii) B and are degenerate eigen function of A .

Although we have taken BA ˆ&ˆ to be Hermitian operators but the theorem 4 & 5 are valid for only linear operator .Theorem 5 assumes special significance when an arbitrary operator A commutes with the Hermitian operators H of a system. Then in such case , which is an

eigen function of H will be also an eigen function of A.

Angular Momentum Operator:- Theory of angular momentum:-

To understand angular momentum operator, we must have knowledge about the theory of the angular momentum which is an important physical quantity associated with rotational motion as well as in the quantum mechanical treatment of atomic structure. In a circular

motion the momentum p is replaced by L (angular momentum) and F

is replaced by N (𝑡𝑜𝑟𝑞𝑢𝑒). For a particle of mass ‘m’ rotating around a point o, the angular momentum ( L

) is defined as the vector product,

( L

) = pr

………..(i)

Where r , a vector denotes the distance of the

particles from O and p another vector denotes the linear

momentum of the particle. The vector L

is directed outwards from o at right angles to the

plane of the rotation i.e. the plane of the vectors pandr .

The angular momentum vector

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In Cartesian coordinate, a vector is expressed in terms or unit vectors kandji

,, and thus the Cartesian components are,

r = xⅈ + yj + zk

p = pxi + pyj + pzk - --------------(ii)

l = lxi +lyj + lzk

And the vector product, prL

can be written as,

prL

=

zyx ppp

zyx

kji

…………(iii)

or, kypxpjxpzpizpypL xyzxyz

)()()( ………..(iv)

Now by comparing equation (ii) & (iv) we have,

xyz

zxy

yzx

ypxpL

xpzpL

zpypL

……….(v)

Here Lx, Ly and Lz are the components of L along x, y, and z axes. Classical mechanics puts no restriction on the magnitude of the angular momentum and its components Lx, Ly and Lz can be determined simultaneously precisely. But in quantum mechanics, however, simultaneous determination of two physically observable quantities depends on, whether the corresponding operators commute. Angular Momentum operators:-

Now we know, from equation (v)

xyz

zxy

yzx

ypxpL

xpzpL

zpypL

By replacing px, py and pz in above equation bytheir operators, the operators for Lx, Ly and Lz may be written as,

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xL =

yz

zyhi

yL =

zx

xzhi …….(vi) Where

2

hh

zL =

xy

yxhi

The total angular momentum L = iLx +j Ly+kLz is a vector operator.So,

2222 ˆˆˆˆˆˆ zLyLxLLLL ……..(vii)

The most identifying features of these operators which distinguish them from the corresponding classical observable are the following commutation relation.

yLhixLzL

xLhizLyL

zLhiyLxL

ˆˆ,ˆ

ˆˆ,ˆ

ˆˆ,ˆ

………(viii)

The operators xL , yL and zL so not commute among themselves but

each of them commutes with 2L . The commutation relations shown by (viii) indicates that no two components of angular momentum can be measured simultaneously with an arbitrary precision. However, square of the total angular momentum as well as Lx, Ly and Lz can be simultaneously specified. i.e. 0]ˆˆ[ 2 qLL Where q = x, y, z.

In this way, it can be shown that 2L commutes with xL , yL and zL

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UNIT-II

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Solvable Systems:- Legendre Polynomials (Mathematical concept):- The legendre polynomials, sometimes called legendre functions of the first kind, legendre coefficient or zonal harmonics are the solutions to the legendre differential equation. The associated legendre polynomials m

l

m

l PandxP )( are solutions to

the associated legendre differential equation, where l is a positive integer and m = 0, ……. l. The first few lengendre polynomials are P0(x) = 1

P1(x) = x

P2(x) = )13(2

1 2 x

P3(x) = )35(2

1 2 xx

P4(x) = )33035(8

1 24 xx

P5(x) = )157063(8

1 35 xxx

P6(x) = )5105315231(16

1 246 xxx

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When ordered from smallest to largest powers and with the denominators factored out, the triangle of nonzero coefficient is 1, 1, –1, 3, 5, 3, –30……. The leading denominators are 1, 1, 2, 2, 8, 8, 16, 16, 128, 128, 256, 256 … The first few powers in terms of Legendre polynomials are, x = P1(x)

x2 = )](2)([3

120 xPxP

x3 = )](2)(3[5

131 xPxP

x4 = )](8)(20)(7[35

1420 xPxPxP

x5 = )](8)(28)(27[63

1531 xPxPxP

x6 = )](16)(72)(110)(33[231

16420 xPxPxPxP

A closed form for these is given by,

xp

nn nlln

nlx l

l n

n

,... 2/)1(2, !!1!2

12

!)12(

The legendre polynomials can also be generated using Gram-Schmidt ortho-normalisation in the open interval (-1, 1) with the weighting function 1

.

3

11.)(

1.

1

2

1

1

1

1

2

1

1

2

1

1

3

2

1

1

1

11

0

xdx

dxx

dxx

dxxxxxp

xdx

xdxxxp

xp

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xdxx

dxx

x

dxx

dxxx

xxp

1

1

2

1

1

2

2

2

1

1

2

2

1

1

2

2

3

3

1

3

1

3

1

3

1

xx5

33

Normalisation so that 1xp gives the expected Legendre polynomials.

The “shifted” Legendre polynomials are a set of functions analogous to the Legendre polynomials, but defined as the interval (0, 1). They obey the orthogonality relationship.

1

012

1)( mn

n

nm dxxpxp

The first law are

1123020

166

12

1

23

3

2

2

1

0

xxxxp

xxxp

xxp

xp

The Legendre polynomials are orthogonal over (-1, 1) with weighting factor 1 and satisfy,

1

1 12

2mn

n

mn dxxpxp

(where mn is called Kronecker delta)

Now let us consider the equation,

021 2 nxydx

dyx ....... (i)

If we write this in the form

)1(

22x

nxdx

y

dy

……(ii)

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It on integration gives,

y= c (1–x2) ……(iii)

Where c is a constant, If we differentiate equation (i) (n+1) times, the result will be,

(1–x2) 0)1(2)1(

)1(

2

)2(

n

n

n

n

n

n

dx

ydnn

dx

ydx

dx

yd ………(iv)

The equation (iv) may be written as,

(1–x2) 0)1(22

2

znndx

dzx

dx

zd …….(v)

Where n

n

n

xdx

dc

dx

ydz )1( 2

2

2

……..(vi)

This equation (v) is known as Legendre’s equation and the particular

solution,

n

nnn xdx

dn

nxpz )1(

!2

1)( 2 …..(vii)

Is called Legendre polynomials of degree n. (p0(x) is defined

to be unity)

The Associated Legendre polynomials:- If the equation,

(1 – x2) 02 nxydx

dy is differentiated (m+n+1) times, then the the

equation becomes,

(1 – x2) 0))(1()1(21

1

2

2

nm

nm

nm

nm

nm

nm

dx

ydmnnm

dx

ydxm

dx

yd .....(viii)

which may be written as,

(1–x2) 0))(1()1(22

2

zmnnmdx

dzxm

dx

zd

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Where, nxdx

dcxp

dx

d

dx

ydz

nm

ym

nm

m

nm

nm

)1()( 2

…….(ix)

Now if z is substituted by

22)1(m

xuz

…….(x)

Then 22

21

1

2 m

xx

mxu

dx

du

dx

dz

or,

u

x

xmm

x

m

dx

du

x

mx

dx

ud

dx

zd22

2

222

2

2

2

)1(

)2(

11

222)1(

m

x

So that the differential equation for u is

(1–x2) 01

)1(22

2

2

2

u

x

mnn

dx

dux

dx

ud ……..(xi)

The equation (xi) is known as the associated Legendre equation, and the function u, denoted by u = )(xPm

n is called the associated Legendre

polynomial of change n and order m. From equation (ix) and (x) we see that,

)(xPm

n = )(/

)1()1( 22/2 xPdx

mxzx nm

dm

m

now applying equation (vii)

)(xPm

n = n

mn

mn

nx

dx

d

x

mx1

!2

2/1 22

……..(xii)

It is apparent that )()(0 xPxP nn and also since )(xPn is a polynomial of

degree n, )(xPm

n is zero if m > x.

The function )(xPm

n and )(xPm

l are orthogonal in the internal –1 1 x

if l n . Quantum Mechanics of some simple system

27



There are certain simple systems for which the Schrödinger equation can be solved exactly. These systems are an idealization of naturally occurring systems, but the consideration of them furnish an insight into the method of quantum mechanics and give results which are useful in the discussion of many problems of physical and chemical interest. e.g. Free particle, the particle in a box, the rigid rotater, the harmonic oscillator etc. Upto Graduation level we have already learned systems like free particle and the particle in one dimensional and three dimensional boxes. The free particle is a simplest imaginable system would be a particle of mass in moving in the x-direction under the influence of no forces. All possible values of Energy Ex from 0 to are possible and we have thus a continuous spectrum of energy values for this system. The Particle in a box:- In this system we considered a particle of mass m constrained to move in a fixed region of space, which for simplicity we have taken a rectangular box with edges of box a, b and c and volume v = a b c. The P.E (V) have been taken equal to zero inside the box and at the boundaries of the box and in the remaining of space we considered v . The value of eigen function xyz and Eigen value Exyz

were evaluated by solving Schrödinger equation and the final results were therefore,

.sin.sin.sin8

zc

ny

b

nx

a

n

v

zyxxyz

& Exyz =

2

2

2

2

2

22

8 c

n

b

n

a

n

m

h zyx

Where zyx nandnn , = 1, 2, 3, 4…....

1. Rigid Rotor (or Rotator) and Rotational Spectroscopy of diatomic molecules: -

It is simple problem after the solution of Schrödinger equation for the H-atom. It is a case of two-particle system with the particles held at a

28



fixed distance from each other by a rigid massless rod of length r. It is simply called two-particle rigid rotor or rotator. It is a suitable model for diatomic molecules which we may consider a dumb bell which consists of two balls of masses m1 & m2 representing the atoms connected by a rigid rod of length ‘r’ that represents the chemical bond between the atoms. This is known as the rigid rotator model. We consider the rotation of this rigid rotator about an axis perpendicular to its own axis and passing through the centre of gravity as shown below:-

Now let us ignore the masses of electrons and the vibration of the nuclei and thus a diatomic molecule may be compared to a rigid rotator. The rotational motion of such a diatomic rotator, corresponds to the motion of two masses in such a way that the centre of masses and the inter nuclear distance remain unchanged. Let the centre of mass be considered to be at the origin of the coordinate system. If during rotation the two masses m1 & m2 moves with linear velocities v1 & v2 respectively and the distance r1 & r2 remains unchanged, then the K.E.(T) due to rotation is given by

2

2

2

2

2

1

2

1

2

22

2

112

1

2

1

2

1

2

1rmrmvmvmT

= Irmrm 22

22

2

11

2

2

1

2

1 Where = angular velocity

I = Moment of Inertia

29



Rotation of rigid diatomic molecule can occur in a plane (with fixed axis of rotation) or in a space (i.e. with free axis) Rotation in Plane (Fixed axis) If a diatomic molecule rotates in the xy plane rotational axis concides with z-axis. The angular momentum will have only the z-components Lz (Lx = Ly = 0 because the molecule cannot move out of plane). The rotational energy of the molecule will be equal to the Kinetic Energy (i.e. v = 0). The Hamiltonian for such a system will be,

2

2

2

22

82

ˆˆ

d

d

I

h

I

LH z

and the corresponding Schrödinger equation may be written as,

2

2

2

2

2 8M

h

E

d

d ………(i)

The angle is the angle of rotation in the plane and is the reduced

mass given

by = )( 21

21

mm

mm

The eigen function corresponding to equation (i) are known to be,

ime2

1

and the eigen values as,

I

hME

2

22

8 , where M = 0, ......3,2,1

& I = r2

In other words, a diatomic molecule rotating in a plane may be treated as a single particle. Rotation in Space (Free Axis)

30



Rotation in space means that the axis of rotation is in free to take any

direction in space. In this case, there are two angular variables and

is the angle which the axis of rotation makes with z-axis, and is the angle in the xy plane between the x axis and the projection of the rotation axis in the xy plane. Since the inter nuclear distance does not change, the potential energy v is zero. Then, the rotation energy, which is the same as the kinetic energy , is given by,

I

LE

2

2

……(ii)

Where L is the angular momentum. The Hamiltonian is given by the equation,

2

2

22

22

sin

1.sin.

sin

1

82

ˆˆ

I

h

I

LH

The Schrödinger equation may be written as,

),(8

,.sin

1sin

sin

12

2

22

h

IE

……….(iii)

This is the same as the equation, obtained by particle on a sphere as,

E

mr

h

2

2

222

2

sin

1.sin.

sin

1

8

I

hll

I

hE

2

2

2

2

8)1(

8

……….(iv) Where l = 0, 1, 2, 3, ….

The eigen functions ( , ) are the spherical harmonics

im

lMl emNP )(cos||),(,

where N = Normalization factor) Thus, a diatomic molecule rotating in space with free axis may, therefore, be treated as a particle on a sphere. The wave functions in such type of particle on a sphere may, therefore called, “spherical harmonics”.

31



Simple Harmonic Oscillator Classical concept:- When a particle oscillates about its mean position along a straight line under the action of a force which (a) is directed towards the mean position and (b) is proportional to the displacement at any instant from this position, the motion of the particle is said to be simple harmonic and oscillating particles is simple harmonic oscillator or a linear harmonic oscillator. The H.O. is an important example of periodic motion because it seems as an exact or an approximate model for many problems in classical or quantum mechanics. At temp. above the 00K the atom in a crystal are temporarily displaced from their normal positions in the structure due to absorption of thermal energy. These atoms in a molecule execute vibratory motion along a bond with bond length and bond angles changing periodically. Consequently, inter atomic forces obeying Hook’s law act on displaced atom. The vibration execute by such atom are similar to those of a simple harmonic oscillator. If the equilibrium position of a linear harmonic oscillator of mass m be the origin ‘o’, and the straight line of its motion be the x-axis, and If 𝑥 be the displacement from ‘o’, then the restoring force,

32



F = –Kx …….(i) where k = a positive constant According to Newton’s second law of motion,

F = ma

2

2

&dt

xd

onacceleratia

or, F = 2

2

dt

xdm ………(ii)

Combining Hook’s law and Newton’s equation of motion i.e. equation (i) & (ii),

2

2

dt

xdm = –kx or, x

m

k

dt

xd

2

2

Putting = m

k , xdt

xd 2

2

2

……..(iii)

We may obtain the general solution for equation (iii) tBtAx cossin ………(iv)

or, 𝑥 = 𝐴 sⅈn (𝜔𝑡 +𝜋

2) = 𝐴 sⅈn(𝜔𝑡 − 𝜃) = 𝐴 cos𝜔𝑡………(v)

The equation (v) shows that the particle performs simple harmonic oscillations with frequency

= m

k

2

1

2 …….(vi)

m

k

mk 22 The potential energy (V) of oscillation at any point x is given by

V = 2

00 2

1kxkxdxFdx

xx

kxF

kxF

or, 222

2

1

2

1xmkxV ………(vii)

mkand

m

kas

2

The kinetic energy (T) at displacement x is given by

2

2

1

dt

dxmT

dt

dxvas

=1

2𝑚𝐴2 𝜔2 𝑐𝑜𝑠2(𝜔𝑡 + 𝜃)

33



222

2

1xAm …….(vii)

Thus the total energy E according to classical mechanics is E = T + V

=

22222

2

1

2

1xmxAm

= 222222

2

1

2

1

2

1xmxmAm

2

2

1kA ……..(ix) (k = m2)

In equation (ix) A represents the amplitude of vibration. Equation (ix) shows that the oscillator can have any positive value of E. The path of motion will be a parabola which also represents the variation of potential energy. At extreme points ( ax ) of oscillation, all the energy is potential

energy ,2

1 2

ka while at equilibrium point (x = 0), all energy is kinetic

energy. This figure shows that at any point within the extremes, the vertical line PQ represents K.E and QR the P.E and the line PR represents the total energy. Classical treatment of a harmonic oscillator gives different results from those obtained from quantum mechanical treatment. However, classical treatment is restricted to large particles (bodies) and the quantum mechanical to electronic, atomic and molecular particles. Quantum mechanical treatment of Simple Harmonic Oscillator:- Since the potential energy of oscillation at any points x is

34



22

2

1xmV = 2𝜋2𝜈2𝑚𝑥2 from equation (vi & vii)

The Schrödinger Equation for such a motion may be written as,

028 222

2

2

2

2

mxh

m

dx

d ………(x)

Before we attempt to find the expression for and E, let us simplify the equation (x) by making the following substitution,

h

m

h

mE

2

2

2 4,

8 ………..(xi)

Now substituting & in equation (x) it becomes

022

2

2

xdx

d ……….(xii)

Then by making the following changes of variables

Let y =√𝛽 𝑥; 2

2

2

2 1

dx

d

dy

d

so that

x =2

2

2

2

2 dy

d

dx

dand

y

……..(xiii)

Substituting from equation (xiii) for value of x & 2

2

dx

d in equation (xii) we

have,

02

2

2

ydy

d

Or, 02

2

2

y

dy

d ……….(xiv)

The equation (xiv) is a simple equation in the sense that it is free of

quantities like , m, h, etc and since the variable y is dimensionless, the equation is purely algebraic. Solution of the equation:- It is not easy to solve equation (xiv). However, an asymptotic solution

can be obtained for very large values of y. when

y , the equation

(xiv) reduces to,

35



02

2

2

ydy

d …….(xv)

The solution of equation (xv) can be shown to be

2

2y

e

Hence 𝜓 = 𝑒− 𝑦2

2 …..(xvi) (as for large value of y, the approximate solution of equation (xv) is 𝜓 =

𝑒− 𝑦2

2 since y hence we reject the form 2

2y

e

) This suggest that the general solution for equation (xiv) contains the

factor 2

2y

e

and the form of the solution may be written as,

2

2

)(

y

eyF

………(xvii)

Where F(y) is the same function of y whose form has to be determined by finding the differential equation for it. Writing F for F(y), and differentiating equation (xvii) with respect to y, we have,

𝑑2𝜓

𝑑𝑦2 = {𝑑2𝐹

𝑑𝑦2 − 2𝑦 𝑑𝐹

𝑑𝑦+ (𝑦 − 1)𝐹} 𝑒

−𝑦2

2 ………(xviii)

Hence by substituting for and 2

2

dy

d

from equation (xvii) and (xviii) respectively in the equation (xiv)

0)1(2 2222

2

222

FeyeFydy

dFy

dy

Fdyy

or, 012 22

22

y

eFdy

dFy

dy

Fd

………(xix)

Since 2

2y

e is not zero, except for y the expression within big brackets

be equal to zero

or, 0122

2

F

dy

dFy

dy

Fd

……..(xx)

Thus is the differential equation for function F(y). Equation (xx) is known in mathematics as Hermit’s differential equation. The solution for

36



equation of this type is obtained by assuming that the function F can be expanded as power of series in y i.e. ........4

4

3

3

2

210 yayayayaaF ……..(xxi)

This leads to dy

dF......432 3

4

2

321 yayayaa ………(xxii)

or 2

2

dy

Fd ......433221 2

432 yayaa ……(xxiii)

Substituting these equations into equation (xx), we get, .......)32()22()12(......433221 3

3

2

21

2

322 yayayayayaa

.....111 2

210 yayaa

= 0

……(xxiv) The equation (xxiv) can be satisfied only if the coefficient of each power of y is zero. Thus,

For y0 : 1×2a2 + 01 a

= 0

For y1 : 2×3a3 + 0121 1

a

For y2 : 3×4a4 + 0221 2

a

and so on.

For yk : (k+1)(k+2)ak+2+ 021

kak

Which leads to the formula for determination of the coefficients,

ak+2 = 0)2)(1(

21

kakk

k

………..(xxv)

where k is an integer. This is known as recursion formula. It permits calculation of the coefficient ak+2 of the term yk+2 in term of the coefficient ak of yk.

37



We can obtain a2, a4, a6...... etc in terms of a0 (even set) and a3, a5, a7… etc in term of (odd set). The power series (xxi) may therefore be written as,

oddk

k

k

evenk

k

k

k

k

k yayayayF0

)( ………(xxvi)

It can be shown that if the power series F(y) is infinitely long, the higher order terms will dominated and the function F(y) will behaved like the Taylor series expansion of,

......

2

.....2

14

22)

n

yyye

ny

If an and an+2 be the co-efficient of yn and yn+2, then

2

2

22

1

22

22

nnn

n

a

a

n

n

When n is very large,

na

a

n

n 22

In other words, for large values of the exponent n, the recursion

formula (xxv) for the power series F(y) reduces to na

a

n

n 22

Thus means that for large values of the exponent n, F(y) will behave like

𝑒(𝑦2) This makes the solution (xiii) for as,

= 𝑒(𝑦2).

2

2y

e =

2

2y

e …………. (xxvii) The function (xxii) is unacceptable as it rapidly tends to infinity y increases. Thus difficulty can be overcome if a condition is found when the saries (xxi) breaks off after a finite number of terms. This is possible if for certain value of K (say n)

38



``

12 n

……………. (xxviii)

So that the numerator in equation (xxv) vanishes, n may be 0, 1, 2, 3, etc Hermit polynomials :- The resulting expression with finite number of terms is a polynomial H(y), rather than the power series of the type (xxi).The complete solution of equation (xiv) is then,

y)= NH(y)

2

2y

e …………. (xxix) H(y) is usually called Hermit polynomial; N is normalization factor. The

function y) in equation (xxix) is finite and single valued for any values

of y, being controlled by the factor .

2

2y

e , and decreases continuously as y increases: hence it is an acceptable function. The hermit polynomials are defined, in general as follows:-

Hn(y) = (-1)n 𝑒(𝑦2)2

)( 2

dy

ed yn

……….. (xxx)

Where n is the degree of polynomial, that is the highest power of y in it.

The Complete Eigen Function From equation (xxix) and (xxx), the complete wave functions of the harmonic oscillator can be derived corresponding to definite

permissible values of

or n:

e.g. the polynomial of 2nd degree is divided as follow:-

H2(y) = (-1)2 𝑒(𝑦2)2

)( 2

dy

ed yn

= 2442 2)(2)()( 222

yeyee yyy

39



Normalization factor N in equation (xxix) can be determined from the condition

1)(22

2

dyeyHN

dxx

y

nn

………..(xxxi)

(because dxdyxy , )

The integral on evaluation leads to,

ndyeyH nyn 2)( )(2 2

…………(xxxii)

Therefore from equation (xxxi),

122

nN n

or, 2

1

2

nN

n ………(xxxiii)

For 0, N0 = 4

1

1, N1 = 4

1

2

1

e.g. Let us derive H4(y) and H5(y) and functions (x) and 5(x) using

equations (xxix) and (xxx)

(y) = NH(y) 2

2y

e

…………(xxix)

Hn(y) = (-1)n 2

)( )(2

2

dy

ede

yny

............(xxx)

H4(y) = 12 – 48y2 + 16y4

H5(y) = 120y + 160y2 + 32y5

40



2, N2 = 4

1

8

1

3, N3 = 4

1

48

1

and so on.

The Energy (Eigen Values) We have seen that satisfactory wave function for harmonic oscillator are obtained only if for a certain value n of the exponent of y in equation, ........4

4

3

3

2

210 yayayayaaF

12 n

;

If we use equation

= 2

28

h

mE ; = h

m 24

For and

122

nh

E

or, E = hn

2

1

Where n = 0, 1, 2, 3, ……. This shows that, i) the energy of the harmonic oscillator is quantized, contrary to the classical view, and this result emerges from the solution of Schrödinger equation. ii) The energies are determined by the parameter ‘n’ which is called vibrational quantum number of the oscillator.

iii) The lowest value of the energy (when n = 0) is h2

1 ; ‘this is the zero

point energy’. In classical view it is zero. iv) As ‘n’ increases by unity, the energy increases by h so that the energy levels are equally spaced as shown in the figure given below:- n = 4 E4 = 9/2 h

n = 3 E3 = 7/2 h

41



n = 2 E2 = 5/2 h n = 1 E1 = 3/2 h

n = 0 E0 = 1/2 h

The wave functions are symmetric for (even) i.e. x0, x2, x4 …....and anti-symmetric for (odd) i.e. x, x3, x5………….

42



UNIT-IV

43



Dr. Ashok kumar Gupta

Professor & Head

University Department of Chemistry

L.N.M.U Darbhanga,

Specialization in physical chemistry

44



Introduction – Atom consists of a nucleus and electrons and a molecule is an aggregate of same or different atoms containing two or more nuclei and electrons. The quantum mechanical treatment of molecule needs non relativistic Hamiltonian of molecule containing n electron and N nuclei. Nucleus is regarded as positively charged point mass. As we know,

VTH ˆˆˆ VandTH ˆˆ,ˆ are Hamiltonian, kinetic energy and potential energy

operators respectively. T and V are Kinetic energy and potential energy of system consisting of electrons and nuclei. T = K.E. of nuclei + K.E. of electrons = Tn + Te (Say) V = P.E. of system due to interaction of electrons and electrons, proton and proton (i.e. nuclei) and electrons and nuclei. i.e. V= Vee + Vnn +Vee Now writing for their operators we have,

en TTT ˆˆˆ

and nenneenennee VVVVVVVV ˆˆˆˆ

For a single particle in three dimensional space K.E. operator T is expressed as follows:

T =−h2

8π2m ∇2 =

ℏ2

2m∇2 ℏ=

h

2π

Here, the molecule consist of N number of nuclei. Let us their masses are M1, M2…MN with position co-ordinates (i = 1, 2…..N) Then,

45



22

2

2

2

1

1

2 1........

11

2ˆ

N

N

nMMM

hT

N

i

i

i

nM

hT

1

22 1

2ˆ ………(1)

N

i

i

i

nM

T1

21

2

1ˆ …………..(1)* (in au) [ h = 1au]

Now,

22

2

2

1

2 1........

11

2ˆ

nemmm

hT

Where, m is the mass of electrons with position co-ordinates (i = 1, 2,…..n)

i.e. 22

2

2

1

2

.......2

ˆne

m

hT

n

j

jem

hT

1

22

2ˆ ……….(2)

n

j

jem

T1

2

2

1ˆ (in au, since m = 1au) ……….(2)*

Now, ......................ˆ

10

2

240

2

230

2

130

2

120

2

n

eeeerk

e

rk

e

rk

e

rk

e

rk

eVV

............,,....., 124231312 nrrrrr are inter electronic distances.

00 4k is permittivity of free space.

Incompact form:

n

lk kl

eeeerk

eVV

0

2

2

1ˆ ………..(3)

n

lk kl

eeeerk

VV1

2

1ˆ

0

(in au, since, e=1au) ………(3)*

...............4

.......ˆ

10

2

1

240

2

2

230

2

32

130

2

31

120

2

21 n

Nnnnn

Rk

ezz

Rk

ezz

Rk

ezz

Rk

ezz

Rk

ezzVV

............,,....., 124231312 NRRRRR are inter electronic distances.

N

qp pq

qp

nnnnR

ezz

kVV

2

02

1ˆ …………(4)

N

qp pq

qp

nnnnR

zz

kVV

02

1ˆ (in au) ……………(4)*

46



Vee and Vnn represent inter electronic and inter nuclear repulsive potential energies. Vne is electron-nuclear attractive potential energy.

...............ˆ

220

2

2

210

2

2

130

2

1

120

2

1

110

2

1

rk

ez

rk

ez

rk

ez

rk

ez

rk

ezVV nene

i.e. i j ij

inene

r

ez

kVV

2

0

1ˆ …………….(5)

i.e i j ij

inene

r

z

kVV

0

1ˆ (in au) …………….(5)

ijr

1 is the distance of its nucleus and Jth electron

It may be ..........1

.........1

,1

,1

.........1

,1

,1

..........1

,1

,1

.......11

,1

321333231232221131211 NnNNN rrrrrrrrrrrrr

We can write the Hamiltonian operator of a molecule in following way: (using 1, 2, 3, 4, & 5)

nenneeennenneeen VVVTTVVVTTH ˆˆˆˆˆˆˆˆ

n

j

n

lk i j ij

i

oqp pq

qp

oklo

j

N

i

i

i r

z

kR

zz

krkmMH

1

2

1

2 1

2

11

2

1

2

11

2

1ˆ (in au)

However, ko = 1 (in au) …………… (6) Once, Hamiltonian is obtained, we need suitable were function of molecular system to deal the system quantum mechanically through time independent Schrödinger were equation: EH ˆ ………….(7)

Let the wave function denoted by ),( Rr representing r and R for

collective 3n electronic and 3N nuclear co-ordinates respectively. Now, RrERrVVVTTRrH nenneeen 11

ˆˆ),(ˆ …….(8)

The equation (8) cannot be solved analytically due to presence of term Vee. More over, the mobility of nuclei creates another problem. Born and Oppenheimer overcame the complications using approximate method. It is called Born Oppenheimer approximation. The Born-Oppenheimer Approximation

47



Since, we know that electrons are much lighter than nuclei. Even the

lightest nuclei i.e. proton is 1836

1 times heavier than an electron, the

latter moves appreciably faster than the former and instantaneously adjust to new nuclear position. Born and Oppenheimer proposed, on the assumption that electrons move independently of nuclei move, the solution of Schrödinger wave equation for a system of mobile electrons and nuclei by merely solving the equation of electronic motion for different fixed inter nuclear position. This is Born-Oppenheimer approximation. Let Rr, the composite function of electronic and nuclear motion be

separable into R(r) and (R).

R(r) represents the electronic function of r for given R (inter nuclear

distance) and (R) the nuclear function of R.

i.e. (r,R) = R(r). (R) ……….(9)

Let us designate R(r) as e and (R) as n

i.e. (r,R) = e . n ……….(9)*

e satisfies the electronic Schrödinger equation i.e. eeeeneeeee EVVTH ˆˆ ………(10)

Putting the explict values of nT and eT in equation (8) we have

n

j

neneeneeej

N

i

i

i

EVVTmM 1

2

1

2 ˆ2

11

2

1 ……….(11)

We know that, d2(xy) = xd2y + yd2x + zdxdy einieinnienei 2

222 ………..(12)

ejnnej 22

………….(13)

From 11, 12 and 13 we have

48



nenenenneee

n

j

jn

einieinnie

N

i i

EVVVmM

1

222

1 22

1

2

1

………….(14) Square bracketed terms (1st one of L.H.S. of equation 14) describe the coupling of electronic and nuclear motions. It includes nuclear mass Mi in the denominators of coupling terms. These terms are regulates on the basis of the Born-Oppenheimer Approximation that 0 ei . Since

e is generally a slowly varying function of the nuclear co-ordinates. This assumption works quite satisfactory for the ground state of molecules.

Now dividing remaining term by n we have,

n

j

eeennneeej

N

i

ni

in

e EVVVM 1

2

1

2

2

11

2

1

02

11

2

1

1

2

1

2

ennee

n

j

eneeej

N

i

ni

in

e VEVVM

……..(15)

eee

n

j

eneeej EVV

1

2

2

1 ………(16) (from equation 10)

Equation (15) reduce to

01

2

1

1

2

enneee

N

i

ni

in

e VEEM

…………(17)

Dividing equation (17) by n

e

on both sides,

01

2

1

1

2

nnnnne

N

i

ni

i

VEEM

i.e. nnnne

N

i

i

i

EVEM

1

21

2

1 …………..(18)

i.e. nn

N

i

i

i

EREM

)(1

2

1

1

2 ………..(19)

49



Equation (18) and (19) describe the nuclear motion (vibrational and rotational) in potential field of E(R) = Ee + Vnn.

Born-Oppenheimer Approximation is utilized to solve the equation (8). It involves two steps. In the first step equation (10) is solved for various values of R i.e. various nucleus conformations. It gives the value of Ee. Now adding Vnn to it we get E(R) which is the function of R. solution of equation (19) gives the total energy of the molecule. A plot of E(R) versus R is customarily called potential energy curve of diatomic molecules and multidimensional potential energy surface for polyatomic molecules. For a diatomic molecule R is internuclear distance and for polyatomic molecules R includes bond angle, bond lengths and dihedral angles, all possible internal co-ordinates. Solution of Schrödinger Equation (electronic) of molecules: As the equation (10), the the electronic Schrödinger equation follows, eeeeneeeee EVVTH ˆˆ

The solution of which is one of most fundamental problems in quantum chemistry. Vnn is inter-nuclear repulsion energy. It can be added to Ee at the end of calculation to get E(R). The inclusion of Vee terms makes solution analytically difficult.

Explicit form: eee

n

j

n

lk kloi j ij

i

o

jee Erkr

z

kH

1

2 1

2

11ˆ

K0 = 1(in au)

eee

n

lk kl

n

j i j ij

ijee E

rr

zH

1

2

1

2

1ˆ

1

2 ....…(20)

50



The first bracketed part is identified as electronic Hamiltonians (for separate entities formed from system at convenience) and second part as inter-nuclear repulsion energy.

i.e. eee

n

lk kl

e

n

j i j ij

ij E

rr

z

.2

1

2

1

1

2 …………(21)

Solutions of systems like H2+, H2 etc for evaluation of their electronic

wave functions e and energies Ee or E led to achievement of solution of bigger molecules. Two approximate methods have been in use for solution of electronic Schrödinger equation since early days. These are known as Molecular Orbital Theory (MOT) and Valence Bond Theory (VBT). MOT was proposed independently by Hund and Milliken in 1928. Huckel, Lennard-Jones, Longuet-Higgins and Coulson also contributed significantly to development of MOT. Pioneering work on the H2 molecule by Heitler and London in 1927 laid the foundation of VBT, further developed by Pauling and Slater. In both MOT and VBT the approximate wave function of electrons in molecular system are expressed in term of atomic wave function which are atom-centered or some other functions having same symmetry as that of atomic wave function/ orbital. The atom-centered functions from together the basis set for the molecules. The Molecular Orbital Theory (MOT)

On the basis of the Born-Oppenheimer Approximation a molecule is treated as an entity where the electron are not centered to particular nucleus / nuclei rather these are delocalized over all the nuclei i.e. each electron remains under influence of potential field due to all the nuclei. As we have seen,

51



eee

n

j

n

lk kli j ij

ij E

rr

z

1

2 1

2

1

2

1 ……(from 21)

Writing the equation (21) for jth electron (j = 1, 2, 3, ….., n) we have,

)()()(1

2

1 2jjEj

rr

zeee

n

kj jk

n

i ij

ij

……….(22)

In brief, )()( jjH e = )()( jjE ee …………(23)

Where, H(j) is the Hamiltonian of single electron j. Ee(j) and e(j) are the

energy and eigen function of the same e(j) is decentralised over all nuclei, hence, it is called as MOECULAR ORBITAL. The complete

electronic wave function e will be the product of e(j)’s. More correctly anti-symmetric product of spin-orbital wave functions i.e.

2)........4()3()2()1(N eeeee Ne

onsoandeeee )......1()1()1(),1()1()1( .

LCAO–MO

We need MO function e in order to solve the equation (21) and (22). LCAO-MO approximation method provides an approximate trial function in order to solve aforesaid equations. The atomic orbital wave functions are chosen as basic functions to construct trial function by linear combination of atomic orbital wave functions centered on each nucleus. The linear combination a.o’s with coefficients give one electron molecular orbital wave function often called LCAO-MO. Larger will be the number of basic functions, more accurate will be the MO wave function. Such as,

n

j

jjnne aaaai1

2211 ....... ……….(24)

52



For example, one electron MO wave function for H2

+ ion as depicted below,

A and B are nuclei. Let A and B are atomic

wave function assuming the electron residing at A and B separately.

e = a1 A + a2 B

For H2 molecule, there will be two MO wave function e(1) an e(2) which can be given as follows.

For electron (1) the atomic wave functions are A(1) and B(1) & for

electron (2) these are denoted by A(2)

and B(2).

Thus,

e(1) = 11 21 BA aa

e(2) = 22 21 BA aa

The complete wave function (without considering anti-symmetrisation) 22212.1 2121 BABAEE aaaa

Application of LCAO–MO approach LCAO – MO approach may be applied to deal with H2

+ ion and H2 molecule quantum mechanically. The hydrogen molecule ion (H2

+) – H2

+ ion is only molecular system to be solved through Schrödinger equation as H-atom is solvable. H-atom needs spherical co-ordinates while H2

+ ion needs elliptical co-ordinates for an exact solution.

53



It will be more convenient to use LCAO-MO approach using approximate trial function in the form of LCAO-MO functions. Let us depict H2

+ ion through following figure.

A and b are two nuclei of H-atom i.e. protons associated with electron e at arbitrary distances rA and rB from A and B. R is inter-nuclear distance.

The LCAO-MO function is given as

= BA aa 21 …………(25)

In ground state 1s(A) and 1s(B). a1 and a2 are corresponding coefficients. Hamiltonian for H2

+ is given by

R

e

r

e

r

e

mMM

hH

BAe

B

B

A

A

222222

2 111

2ˆ

……….(26)

MA = MB = M are masses of nuclei A and B. me is the mass of electron.

22 11B

B

A

A MMmay be neglected with respect to 21

em

because,

MA = MB = M >> m

lightertimesabout

1836

1

Equation (26) reduces as follows considering above fact and the atomic units (Where, 1,1,1 emh e )

Rrr

HBA

111

2

1ˆ 2 …………(27)

Now let us apply the variation method using LCAO-MO functions as trial function (eq. 25) and Hamiltonian (eq. 27), we have to set

.sec,021

equationsulargetwea

E

a

E

54



a1(HAA – ESAA) + a2(HAB – ESAB) = 0 …………(28) a1(HBA – ESBA) + a2(HBB – ESBB) = 0 …………(29) where,

BBBBABBA

BAABAAAA

BBBBABBA

BAABAAAA

SS

SS

HHHH

HHHH

|,|

|,|

ˆ,ˆ

ˆ,ˆ

H is Hermitian operators, hence, HAB = HBA

A and B are normal functions, hence, SAA = SBB = 1 Let AB | = ABA = SAB = BA | = S (Say)

Equation (28) and (29) now take the form a1(HAA – E) + a2(HAB – ES) = 0 …………(30) a1(HAB – ES) + a2(HBB – E) = 0 …………(31) These are homogenous linear equations and their solutions will be non trivial only when the following determinant will be equal to zero.

0HH

HH

BBAB

ABAA

EES

ESE …………(32)

Since, two H-nuclei are same hence, HAA=HBB

Therefore,

0HH

HH

AAAB

ABAA

EES

ESE …………(33)

i.e. ( EAAH )2 = ( ESABH )2

i.e. ( EAAH ) = ( ESABH ) …………(34)

We will have to values E, E1 and E2

S

HHE ABAA

11 …………..(35)

S

HHE ABAA

12 ………….(36)

Calculation of HAA and HAB

55



HAA =< 𝜙𝐴 | �� | 𝜙𝐴 >

i.e. A

BA

AAARrr

H

111

2

1 2

AAA

B

AA

A

ARrr

111

2

1 2

= AAA

B

AAHARr

E |11

|

= AAA

B

AAAHRr

E |11

|

Since, A is normal, AA | = 1

Say A

B

Ar

1 = j

Thus, HAA = EH – j + R

1 ………..(37)

Now, HAB = proceduresimilarabovefollowingH BA ,|ˆ|

We have, HAB = EHs – k +R

s ………….(38)

Where, k = BAR

1 and sBA |

Putting the value of HAA and HAB from equation (37) and (38) into equation (35) and (36) we have,

Rs

kjEE H

1

11

…………..(39)

Rs

kjEE H

1

12

………….(40)

j and k along with S are function of R. calculations od these three integrals are avoided here. Only results are given here.

56



RRR eR

RSeRkeRR

j

31&1;11

1 22

HAA and HAB are (–) ive in value hence, form (35) & (36) it is obvious that E1 < E2. The MO wave function corresponding to eigen value E1 is called bonding MO and that corresponding to E2 is anti bonding MO.

In G.S., A = 1sA

B = 1sB

Molecular orbital wave functions

Since, LCAO-MO function: = BA aa 21

In ground state, = BA sasa 11 21

Calculation of a1 and a2 for bonding & anti bonding MO functions Putting the value of E = E1 into equation (28) we have,

0.11

21

S

S

HHHa

S

HHHa ABAA

ABABAA

AA

i.e. 0)()1()()1( 2211 SHHaSHaHHASHa ABAAABABAAAA

i.e. 0)()( 21 AAABABAA SHHaHSHa

i.e. aaa 21 (say) …………(41)

Similarly by putting the value of E = E2 in equation (31) we have, )(21 Sayaaa ………….(42)

Now, bonding MO function 1ans anti-bonding Mo function 2 may be expressed as,

57



)11(1 BA ssa …………(43)

)11(2 BA ssa …………(44)

Since, 1 and 2 MO functions are normalized i.e. 2211 |1|

i.e. 11 | = 111|112 BABA ssssa

Or, = 11|121|11|1 222 BABBAA ssassassa

i.e. 2a2 + 2a2 s = 1

i.e. s

a22

1

………..(45)

Similarly 1| 22 , we may have, s

a22

1

………..(46)

Thus, BA sss

1122

11

………..(47)

BA sss

111

2 ………..(48)

These are expressions of BMO and anti BMO functions. We can summarise the results we get till now,

Bonding MO: BA sss

1122

11

Rs

kjE

s

HHE H

ABAA 1

111

Anti bonding MO: BA sss

1122

12

Rs

kjE

s

HHE H

ABAA 1

111

Probability density 2

1 Represent the probability density of finding the electron in the

molecule in bonding state.

58



(+)

o Re

BABA SSSSS

1.211122

1 222

1

………..(49)

21 AS and 21 BS represent electron densities around the nuclei A and B

respectively. The electron clouds are spherically symmetrical. 21 AS

BA rr

B eS

21 represent elliptical electron cloud.

If a graph is plotted between 2

1 and inter nuclear distance (R), following

curve is obtained in accordance with equation (49).

This show that electron densities is non bonding state which is equal (21 AS 21 BS ) increases to BABA SSSS 1.2111 22 . This shows that electron density

increases by BA SS 1.21 in between two nuclei A and B in bonding state.

Considering BA SS 11 , increases in electron density in between nuclei A

and B is

2222 12

11

2

12111

)1(2

1AAAAA sssss

s

i.e.

s

ss

sss

s

sxAAA

1

111

1

211

12

14 2222

where,

22 1

2

11

2

1AA ss is the electron density in non bonding state. If we

` put the value of S = 0.46, we have,

22 137.01

1AA s

s

ss

(0.37) 21 AS is the increases in

electron density between two nuclei in bonding state.

R

59



(-)

f a graph is plotted for E1 vs R, we get a curve with maximum at R=Re = 1.32 A0. the plot is shown below.

This is equilibrium bond length (inter-nuclear distance) of H2+ ion.

However, the experimental value of Re is 1.06A0.

In anti bonding state, the plot of E2 vs R fives following curve with no minimum. It is repulsive state.

Improvements were done by several authors. C.A. Coulson modified is

basic functions as rzrz

13

where z is variable parameters which can be

obtained by minimizing energy i.e. 0zd

dE .

60



Plot of 2 and 22 gives following

curves.

In anti bonding state, as we see, the electron density decreases in between nuclei A and B. As we see,

BABA SSSS

S1.2111

22

1 222

2

0~ BA SS 11

Now, let us designate 1 and 2 by and *, bonding and anti bonding MO’s.

Three dimensional plots of and *gives the idea that and *MO’s are symmetrical with respect to rotation around the inter nuclear axis A-B as well as reflection across the vertical plase containing same axis.

Bonding MO is symmetric with respect to inversion, hence, called g (sigma gerade) and anti bonding MO is anti symmetric with respect to

same, hence called u (sigma un-gerade).

61



Hydrogen Molecule

H2 molecule consists of two hydrogen nuclei (protons) and two electron. Let us designate them A and B and 1 and 2 respectively. It is schematically shown as below.

Hamiltonian of molecule H follows as

R

e

r

e

r

e

r

e

r

e

r

e

m

h

m

h

M

h

M

hH

ABBA

B

B

A

A

2

12

2

2

2

1

2

2

2

1

22

2

22

1

22

22

2

2222ˆ

……….(*)

MA = MB = M = (mass of H-nucleus), m = mass of electron

M >>> m

electronen heavier th th time

1836

1 isproton

Hence, 2

22

1

22

22

2

2222

m

h

m

h

M

h

M

hB

B

A

A

The first two terms may be neglected to rest items in Hamiltonian

R

e

r

e

r

e

r

e

r

e

r

e

m

h

m

hH

ABBA

2

12

2

2

2

1

2

2

2

1

22

2

22

1

2

22ˆ ………..(49)

In atomic unit, the electronic Hamiltonian H may be expressed as

Rrrrrr

HBABA

111111

2

1

2

1ˆ

121221

2

2

2

1 …………(50)

Equation (3) may be rearranged in following way:

62



Rrrrrr

HBABA

1111

2

111

2

1ˆ

1222

2

2

11

2

1

1st and 2nd brackets terms represent electronic Hamiltonians of H2+

molecule ions composed of nuclei A and B with electron (1) and (2) respectively. Let them designate as H (1) and H (2)

Thus, Rr

HHH11

)2(ˆ)1(ˆˆ

12

………..(51)

R

1 is inter-nuclear repulsion which does not change during rapidly

moving electron. Hence, this may be left for some time.

Thus, 12

1)2(ˆ)1(ˆˆ

rHHH …………(52)

Once calculation is done on the basis of H (in equation 52) we may add

R

1 (the inter nuclear repulsion energy0 to it.

LCAO-MO approach –

In ground state atomic wave function may be designate as 1SA(1) and 1SB(1) for one of two molecular orbital functions.

1 = a11sA(1)+ a21sB(1) ………..(*)

1SA(2)and a11SB(2)are linearly combined in the same way to give

another molecular orbital (2), where,

)2(1)2(12 21 BA sasa ………(**) 21 aa

The complete MO wave function is given by

63



MO = (1) . (2)

As we have seen in case of H2+ ion containing two nuclei (A&B) and only

one electron (1) or (2) may be designated as g(1) and g(2) considering the ground state of H2 molecule.

Thus, MO = g(1)g(2)

As we know, 111122

11 BAg ss

s

………….(53)

212122

12 BAg ss

s

………….(54)

As we are dealing with two electrons in H2 molecule, we must follow Pauli’s rule of anti-symmetry of wave functions (spin-orbital). Hence,

MO must be anti-symmetrised by multiplication of orbital wave function with symmetry with anti-symmetry spin wave function.

12212

1 symmetricisgg )2()1(

1221 is anti-symmetric spin wave function composed from

and spins.

Thus, 12212

1.21 ggMO as …………(55)

221122112

1 ggggMO as

21212

1ggggMO as ………….(56)

{As per rotation, gggg and . }

MO(as) is determinantal from follows as

64



)2()2(

)1()1(

2

1

gg

gg

MO as

……………(57)

Or simply as, 212

1ggMO as …………(58)

g and g contain electrons of different spin i.e. and

Eigen values (energy of system) Enrgy of electronic system is given by

E=

)(|)(

)(||)(

asas

asHas

MOMO

MOMO

………….. (59)

)(|)( asas MOMO =1 thus |)(asMO is normalized

Thus E = )(||)( asHas MOMO

)1()2(21)2()1(2

1ˆ)1()2(21)2()1(2

1 gggg H

i.e.

E= )1()2(21|)1()2(21)2()1(ˆ)2()1(2

1 gggg H ..

1st part refers to space and 2nd one the spatial part. Let us solve one by one.

Spin part:

1|12|2

2|21|122|12|22|21|1

= 1 × 1 + 1 × 1 – 0 × 0 – 0×0 = 2 …………(61)

Note:

|01|

|01|

dsds

and

dsds

Thus, equation (60) reduces to

)2()1(ˆ)2()1( gggg HE …………..(62)

(60)

65



i.e. )2()1(1

2ˆ1ˆ)2()1(12

ggggr

HHE

)2()1(1ˆ)2()1(

)2()1(2ˆ)2()1()2()1(1ˆ)2()1(

12

gggg

gggggggg

rH

HH

)2()1(1

)2()1(

)1(|)1()2(2ˆ)2()2(|)2()1(1ˆ)1(

12

gggg

gggggggg

r

HH

2

12

22

1111

22ggHH r

EEE ………(63)

Since, 2|211|1 gggg

2)2(2ˆ)2(&11)1(1ˆ)1(22

HggHgg EHEH

Since, 222

21HHH

EEE

And say, Jr

gg 2

12

22

11

2221 gg and represent electron densities due to electron (1) and (2) in

MOg . Thus, integral J represents total coulombic repulsion energy

between two electronic charges in the MO. Thus, JEE

H

2

2 …………(64)

Aw we have seen,

s

kjEE HH

12

2

Hence, J

s

kjEE H

1

22 ……….(65)

Now we add R

1 the inter-nuclear repulsion to E to get total ground

energy of H2 molecule.

i.e. R

Js

kjEE H

1

1

22

…………(66)

66



Excited state (singlet and triplet) of H2 molecule – Schematic representation of ground state and excited states may be seen in the following way. Ground state snit symmetric MO spins wave function may be

)2()1())2()1() gggg iii

The anti symmetric combination with normalization constant

2

1

may be )2(1)2(12

1gggg …….*

(Excited state) MO – spin function (anti symmetric) for ‘a’ configuration:

)1(2)2(12

1ugug …………..(67)

For ‘b’ configuration, )1(2)2(12

1ugug …………(68)

(c) and (d) are equivalent with and spin and their combinations gives two different MO spins functions.

)1(2)2(12

1

2

1)1(2)2(1

2

1

2

1ugugugug

2

1 as an addition factor comes from , spins combinations

i.e. )1(2)2(12

1)1(2)2(1

2

1ugugugug ………..(69)

and, )1(2)2(12

1)1(2)2(1

2

1ugugugug ………..(70)

Equations (67), (68), and (69) represent triplet state, say 3 with Ms =

1,–1 and 0. However, equation (70) represent singlet state, say |with Ms = 0

67



All these above facts may also be explained by using symmetric and anti symmetric MO’s and spin functions. Symmetric MO is

)1(2)2(12

1ugug

Anti-symmetric MO )1(2)2(12

1ugug

Symmetric spin wave functions are 21,21 & 12212

1 .

Only anti symmetric MO spin function may be obtained by product of symmetric MO and anti symmetric spin wave function, or the product of anti symmetric MO and symmetric spin wave functions. Thus, we have four combinations –

A) 21)1(2)2(12

1 ugug

B) 21)1(2)2(12

1 ugug

C) 12212

1)1(2)2(1

2

1 ugug

D) 12212

1)1(2)2(1

2

1 ugug

Equation (67), (68) and (69) are same as (A), (B) and (C) and (69) is same as (70) Calculation of energy of excited singlet and triplet electronic states –

Energy of excited electronic states 1 (singlet state) – 1E Now,

12212

1|)1(221

2

1

)1()2()2()1(2

1ˆ)1()2()2()1(2

11

ugugugug HE

…………(71)

68



1221|)1(2212

1

)1()2()2()1(ˆ)1()2()2()1(2

11

ugugugug HE

………….(72) Let us solve the spin part of (72) i.e. 1221|)1(221

21|1212|2112|1221|21

1122

22111|12|22|21|1

……….(73)

= 1+1 – 0 – 0 = 2, since 1|| iiii ……….(74)

From (72) and (74), we have,

)1()2()2()1(ˆ)1()2()2()1(2

11

ugugugug HE

i.e. )2()1(ˆ)1()2()1()2(ˆ)2()1(

)1()2(ˆ)1()2()2()1(ˆ)2()1(2

11

ugugugug

ugugugug

HH

HHE

There are four integral parts, let them be designate as I, II, III & IV

Integral part I: )2()1(1

2ˆ1ˆ)2()1(2

1

12

ugugr

HH

21

12

11|1

2|2ˆ|22|211ˆ|12

1

2

12

2

uggg

uuuugg

r

HH

2121

2

1

2

1

2

1ugug JEE ............(75)

Integral part II: )1()2(1

2ˆ1ˆ)2()2(2

1

12

ugugr

HH

21

22

1

2|21|1ˆ|12

11|122ˆ|2

2

1

2

12

2

ug

gguuuugg

r

HH

122

11

2

12

2

1ugug JEE ............(76)

69



Integral part III: )1()2(1

2ˆ1ˆ)2()1(2

1

12

ugugr

HH

)2(21

)1(12|2ˆ|2

1|12

11|1ˆ|12|2

2

1

12

uguggu

ugugug

rH

H

)2(21

)1(12

12|2|20

2

11|1|10

2

1

12

ugugguuugr

EE =

ugk

2

1 …………(77)

Similarly, integral part IV = ug

k 2

1 ………..(78)

Summing all the integral parts I, II, III and IV we have,

ugugugkJEEE 1 ………..(79)

Here, ggg

EEE )2()1( (Electron 1 and 2 are

indistinguishable)

uuuEEE )2()1(

ugJ represents coulombic repulsion energy between electron in g

and u MO’s. ug

k represents the exchange energy of electron between

g and u MO’s. Energy of electronic triplet states (3E)

12212

1|)1(221

2

1

)1()2()2()1(2

1ˆ)1()2()2()1(2

13

ugugugug HE

.............(80) Following same procedure as above we will have,

70



o

o

ugugug

kJEEE 3 ……….(81)

Limitation of MO theory Let us calculate inter-nuclear distance and bond dissociation energy using LCAO-MO procedure discussed above. Equation (66) can be used to calculate ground state energy. Let us plot calculated values of E vs R, keeping EH = 0, we may have following curves.

Calculated value of equilibrium inter-nuclear distance (Re = 0.84 A) Expt. Value = 0.74 A

called value of DE = 2.68eV expt. value of DE = 4.75eV

71



Covalent and ionic contribution to MO wave function

In ground state, MO function is g(1)g(2)

2121111122

1BABA ssss

s

(ignoring spin part)

211121112111211122

1ABBABBAA ssssssss

s

…….(82)

I II First two term in I represent both electron (1) and (2) are associated with either nucleus A or nucleus B representing ionic character.

)(

)2,1(

)()()(

)2,1(,

BAorBA

2nd two term in II represent electrons (1) and (2) are separately associated with nuclei A and B, respectively covalent character.

)1()2()2()1(, BAorBA

Equation (82) may be put into compact form:

s22

1

[ionic +covalent]…………………(83)

(83) shows ground state MO is combination of ionic & covalent in equal magnitude. It is unlikely because there should be greater contribution

of covalent character to MO. This is the short coming of LCAO-MO theory on account of which low value of bond dissociation energy has been obtained theoretically.

72



The Valence Bond Treatment (H2 molecule) The valence bond method is the quantum mechanical treatment of Lewis theory of electron pair bond formation. Heitler and London did not propose any new type of molecular orbital in which electrons are associated with all the nuclei present in a molecule. Rather, the atomic orbitals were retained i.e. electrons were assumed to be centred on their respective nuclei and on this basis, molecular wave function was framed. Let us designate two nuclei by A and B and two electrons by (1) and (2) in H2 molecule. When nuclei A and B are at infinite distance there may be two following possibilities.

)1(1)2(1)2(1)1(1

)1(...).........2()2(...).........1(

BABA ssssorbitalsAt

III

BABA

Molecular wave function (as trial function on the basis of VBT) for the inter-nuclear distance AB appreciably smaller, may be constructed in

following manner: 1 = )2(1)1(1 BA ss and 2 = )1(1)2(1 BA ss , where 1 and

2 are the product of BA sands 11 for two possibilities I and II. Now,

molecular function may be obtained by linear combination of 1 and 2 with different weighting factors C1 and C2 respectively. i.e. )1(1)2(1)2(1)1(1 211221 BABA ssCssCCC ……….(84)

Using variation method, we get secular equations.

22112211

22112211

|

)(ˆ

CCCC

CCHCCE

………….(85)

And solving for

21

0C

E

C

E ………….(86)

Following secular equations are obtained. C1(H11 – ES11) + C2(H12 – ES12) = 0 ………….(87) C1(H21 – ES21) + C2(H22 – ES22) = 0 ………….(88)

73



These are homogenous linear equations. Non-trivial solution is possible only when determinant called as secular determinant is zero.

i.e.

0

HH

HH

22222121

12121111

ESES

ESES …………...(89)

Where, 22221221

21121111

ˆ&ˆ

ˆ,ˆ

HHHH

HHHH

Again, )ˆ(21122211 operator Hermitian is HHHandHH

22221221

21121111

|&|

|,|

SS

SS

2

21122211 1 SSSandSS

2

12

.)2(1|)2(1)1(1|)1(1

)1(1)2(1|)2(1)1(1

SSSssss

ssssS

BABA

BABA

Determinant (89) thus reduces to

0HH

HH

11

2

12

2

1211

EES

ESE ………….(90)

Solving (90) we have,

22

12

2

11 ESHEH

i.e. 2

1211 ESHEH …………...(91)

i.e. 2

1211

1 S

HHE

…………..(92)

E have two value say E+ and E – given by expressions

2

1211

1 S

HHE

…………(93)

2

1211

1 S

HHE

…………(94)

Evaluation of H11 and H12

H11 = )2(1)1(1|ˆ|)2(1)1(1 BABA ssHss

74



)2(1)1(1111111

2

1

2

1)2(1)1(1

121221

2

2

2

1 BA

BABA

BA ssRrrrrr

ss

)2(1)1(111111

2

11

2

1)2(1)1(1

12122

2

2

1

2

1 BA

BABA

BA ssRrrrrr

ss

)2(1)1(11

2

1)2(1)1(1

1

2

1 BA

A

BA ssr

ss

)2(1)1(11

)2(1)1(1)2(1)1(11

)2(1)1(112

BA

B

BABA

A

BA ssr

ssssr

ss

)2(1)1(11

2

1)2(1)1(1

2

2

2 BA

B

BA ssr

ss

)2(1)1(11

)2(1)1(121

BABA ssr

ss

)2(1)1(11

)2(1)1(1 BABA ssR

ss ……………(95)

There are six integrals in equation (95)

1st: )2(1|)2(1)1(11

2

1)1(1

1

2

1 BBA

A

A sssr

s

= )1,(11, AEAE HH ………….(a)

2nd: )1(1|)2(1)2(11

2

1)2(1

2

2

2 AAB

B

B sssr

s

)2,(11, BEBE HH …………(b)

3rd: )(211

)1(1 1

22

2

sayjsr

s B

A

A ………….(c)

4th: )(211

)1(1 2

22

1

sayjsr

s B

B

A ………….(d)

5th: )(211

)1(1 2

12

2 sayjsr

s BA ………….(e)

6th: R

ssR

BA

1)2(1|)2(1s)1(1|)1(1s

1BA …………(f)

75



EH(A,1) is the energy of 1s electron (1) centred on nucleus A. EH(B,1) is the energy of 1s electron (2) centerd on nucleus B. both represent H-atom and both are same. Let EH(A,1) = EH(B,2) = EH Both – 1j and – 2j are same say – j , respectively attractive force between

nuclei A and B and 1s and 1s electrons, while j represents repulsive force between 1sA and 1sB electrons.

Thus, R

jER

jjEH HH

12

12211 …………(96)

Where, Jjj 2

Now let us evaluate H12

12H )1(1)2(1|ˆ|)2(1)1(1 BABA ssHss

)1(1)2(111111

2

11

2

1)2(1)1(1

12122

2

2

1

2

1 BA

BAAB

BA ssRrrrrr

ss

)2(1)1(11

2

1)2(1)1(1

1

2

1 AB

B

BA ssr

ss

)1(1)2(11

2

1)2(1)1(1

2

2

2 BA

A

BA ssr

ss

)1(1)2(11

)2(1)1(11

BA

A

BA ssr

ss

)1(1)2(11

)2(1)1(1)1(1)2(11

)2(1)1(1122

BABABA

B

BA ssr

ssssr

ss

)2(1|)2(1s)1(1|)1(1s1

AA BB ssR

……………(97)

There are six integrals. 1st: )2(1|)1(11,)2(1)1(1 ABHBA ssBEss

= 21|2111|111, BABAH ssssBE

= 221, SESBE HH ………….( a )

76



2nd: )2(1)2(12,)2(1)1(1 BAHBA ssAEss

21|2111|112, BABAH ssssAE

222, SESAE HH …………(b)

3rd: )2(1|)2(1111

)1(1

1

BAB

A

A sssr

s

sk' ………….( c )

4th: )1(1|)1(1211

)2(12

BAB

B

A sssr

s

sk' ………….( d )

5th: )(21211

)1(1)1(112

saykssr

ss BABA ………….( e )

6th: R

S 2

…………( f )

krepresents the energy of exchange of positions of electrons (1) and (2) between nuclei A and B. k represents energy due to overlap cloud 1SA1SB due to electrons (1) and (2).

Thus, R

SkSkSEH H

22

12 22

R

SkSEH H

22

12 2 ………….(98)

Putting the values of H11 and H12 from (97) and (98) into (93) and (94) equations, we have,

RS

KJEE H

1

12

2

…………(99)

RS

KJEE H

1

12

2

…………(100)

J and K are found to be negative. Hence, E – > E+

Molecular functions of H2 molecule – Since, 02

122111 ESHCEHC secular equation

Putting , ,1 2

1211

S

HHEE

we find , C1 = C2 = C

77



,1 2

1211

S

HHEE

we find, C1 = – C2 = C

Thus, we have,

)101...(..........

11212111

11212111

BABA

BABA

ssssC

ssssC

+ and – must be normal i.e. 1||

We will have for +, 22 22

1&

22

1

SCfor

S

Thus,

)103.....(..........1121211122

1

)102....(..........1121211122

1

2

2

BABA

BABA

ssssS

ssssS

Limitation of VBT:

Let us plot E (E+ and E –) vs R putting EH

= 0

We have following energy curveMinimum in E+ curve corresponds to stable H2. from curve, we have, Re=0.80 Å (expt. = 0.74 Å)

De = dissociation energy = 3.14eV Expt. De = 3.74 eV If we compare these results with MOI ones, we find VBT gives better treatment than MOT.

78



Now let us discuss short comings of VBT and modifications. The symmetric molecular function of H2 is

1121211122

1

2BABA ssss

S

It represent electrons 1 and 2 are equally centred on nuclei A and B of

H2 molecule i.e. III

BABA (1).H .(2)H and (2).H .(1)Hboth structure I and II

represent exchange of electron between A and B. these are covalent structures.

However, it lacks ionic structures like IVIII

HHandHH BABA ::

Structure III and IV show both electrons are simultaneously associated with either A or B i.e.

BABA HHandHH . Let us constitute functions

(2)(1)1s1s (2)(1)1s1s BBAA for these ionic structures.

Now it is matter of consideration that how much this ionic part contributes to the complete wave function. Since, contribution of covalent structures to the stability of H2 Molecule is much greater than

any one. Let us consider fraction of ionic structure, contribute to complete function i.e.

+ = cov+ ion ………….(104)

Where,

)2(1)1(1)2(1)1(1

)1(1)2(1)2(1)1(1cov

BBAAion

BABA

ssss

ssss

Here, is a variable parameter. By using variation method is to be

found equal to 6

1 i.e. ~16%. It shows that only 6

1 th contribution is due

to ionic structure to complete function + i.e. sym. By using modified equation (104) we find much more improved dissociation energy in comparision to that obtained from covalent wave function.

79



Concept of resonance The exchange of electrons between nuclei of H-atoms in H2 molecule may lead to following structures

IIIIII

HHHHHH)()()()(

::

Complete wave function may be considering these structures

= C11 + C22 + C33

It may be generalized for n structures as

= C11 + C22 + C33……….+ Cnn …………(105)

Now energy (E) calculated from will be lower than energy E1, E2…….En

obtained form 1, 2……….n respectively. let E1 corresponding to 1 is the lowest of E1, E2, E3……En. Then, (E1 – E) will represent resonance energy which is utilised in the stabilization of molecule. The above structures are called resonating structures. ELECTRON DENSITIES –

Ψ±2 presents probability densities due to .

2121111211121211122

1 22222

BABABABA ssssssssS

……..(106) The density of one electron is obtained by integrating above equation (106) over another electron’s co-ordinates.

1)1(2)1(11122

11 22

2

2

2

BABA sssssS

d

80



Similarly,

2)2(2)2(12122

12 22

2 BABA sssssS

Now, 21

BABA sssssS

4121222

1 22

2

BABA sssssS

2111

1 22

2

………….(107)

Based on MOT,

BABA BsssS

2111

1 22

…………(108)

This shows, BVMO

This shows that MO function piles up more (–) ive charge between nuclei. Spin consideration to molecular function According to Pauli’s principle of anti-symmetry each orbital/ space wave functions must get anti symmetrised by multiplying the formers with appropriate spin wave functions.

Since, 1121211122

1)(

2BABA ssss

Ssym

Ground state

1121211122

1)(

2BABA ssss

Sasy

Excited state

Spin wave functions are

12212

1)( i Anti symmetric

12212

1)(

21)(

21)(

iv

iii

ii

Symmetric

81



Ground state + may be anti-symmetrised by multiplication with anti-symmetric spin function.

Excited state – may be anti-symmetrised spin functions. Thus,

0,012212

1

1,12)1(

1,121

0,012212

1

3

3

3

1

MsS

MsS

MsS

MsS

g

e

e

g

1g represent singlet ground state and three degenerate 3e triplet excited states.

82



UNIT-V

83



C C C C C

2PZ 2PZ 2PZ 2PZ

2PZ

The Huckel Molecular Orbital Theory: A number of semi empirical methods have been developed for quantum mechanical treatment of polyatomic molecules which take cognizance of the fact that inner shells (orbitals) of constituent atoms in a molecule remain practically unchanged during formation of a molecule. Most of the integrals in semi empirical methods are evaluated approximately and sometimes by fitting experimental data HMO theory is the oldest and the most popular semi empirical method to deal with planar conjugate hydrocarbons quantum mechanically such as ethylene, allyl systems,1:3 butadiene, cyclo propenylsystems. Cyclobutadiene, benzene napthalene etc., they contain usually alternate single and double bond in their lewis structures. Single c-c bond containing 𝜎-bond and c=c double bod containing 𝜎 and 𝜋 bonds. In Huekel theory only the 𝜋 –electrons are considered. Essential features of the Huckel theory- Let us considers conjugated planar organic molecule in which all atoms

lie in xy plane .

XY Plane

(i) There is no interaction between 𝜎 and 𝜋 electons based on the fact that 2pz orbitals are perpendicular xy plane containg 𝜎 bonds.

H

H

H

H

84



N

p= 1

(ii) Each c atom contributes a 2pz orbitals to the basis set denoted by a set {∅p} we have.

Ψi = 𝛴 Cpi Φp (1)

Where Ψi is a Huckel MO and N is the number of c-atoms. In neutral molecule it is equal to number of 𝜋 electons. The H-atoms are excluded as they do not contain a 2pz orbital. The atomic orbitals are assumed to be orthonormal i.e <∅p|∅q>= 𝛿pq.

(iii) The HMo’s are assumed to satisfy the one-electrons system equations,

��𝛹I = ∈I 𝛹I -----------------------------(2)

Where �� is a some sort of effective one electron Hamiltonian operator which is not explicitly defined and ∈I is the orbital energy. Multiplying both by ψi at the both sides of equation (2) and integrating we have,

∈I = < Ψi|��| Ψi >----------------------------------(3)

Ψi is normalised. For which, we get ∑ 𝑐𝑛=1𝑝=1

2pi=1-------------(4)

The Huckel MOs span the appropriate irreducible representations of molecular point group.

85



Orbital energies ∈I are determined variationally by solving eqn(2) /(3)

Det H-∈ I =0-------------------------(5)

Where matrix H is

H11 H12---------------H1n

H21 H22-------------- H2n -----------------(6)

----- ---- --- ------

------- ----- ---------------

Hn1 Hn2-------------Hnn

I is unit matrix 1 0 ------------- 0 0 1 ------------- 0

- - - - - - - - - - - -- 0 ----------------------------(7) - - - - - - - - - ------- 0 0 0 -------------- 1

Where the matrix element Hpq = <∅p|��| ∅𝑞 >all diagonal elements Hpp are set to be ∝ called the coulomb integral. The off diagonal elements Hpq are set equal to zero for p-q =+2,+3 ---- ie for non adjacent c-atoms and these are set equal to 𝛽 Where p-q= ±1 ie for adjacent c-atoms in chain as well as structure molecule. 𝛽 is called lresonance integral. Both ∝ 𝑎𝑛𝑑 𝛽 are negative. Since,

86



�� is not defined ∝ 𝑎𝑛𝑑 𝛽 can not be evaluated theoretically, they are treated as empirical parameters. Thus for a linear conjugated chain organic molecule, secular determinant follows as-

(∝ −∈) 𝛽 0 --------------------0

𝛽 (∝ −∈) 𝛽 ----------------------0

0 𝛽 (∝ −∈) 𝛽 -----------------------0 =0-----------(8)

------ -------- ---------------- -------------

------- ---------- ------------------- --------

0 0 ---------------- 𝛽 ∝ −∈) I

Where, H11=H22=H33=-----------------------Hnn=∝ H12=H21=H23=H32=-----------------= 𝛽 H13= H31=H24=H42=H14=----------=0 Dividing both sides by 𝛽 on both of equation (8) we have following determined.

X 1 0 --------------------------0 1 x 1 -------------------------- 0 0 1 x 1 ------- - ------ 0 ----- ------ -------- ---------- =0-------------(9) ------- ---------- ----------- 0 0 0 0 ------1 x

Where, 𝛼−𝐸

𝛽 = x

87



C - C

However, for a cyclic conjugated organic molecule when ist and nth carbon atoms are bonded, the determinant charges

x 1 0 ------------------------ -01

1 x 1 ------------------------- 00

0 1 x 1 ----------------- 00 =0-------------(11)

--- --- ----- ----- - - - - -

----- ----- ------- -- - - -

1 0 0 0 -------- 0 1 x

The Huckel theory is an independent particle model. Thus, the Hamiltonian operator and the wave function of a system of N 𝜋 −𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 are given by

�� (1,2------N)= �� (1) + �� (2) +------------��(n) -------------(12) Ψ (1,2, --------N)= Ψ (1). Ψ (2) --------------- ΨN (N)-------------(13) In the framework of the Huckel theory the schrodinger equation for system is given by

��(1,2-----------N) Ψ (1,2-------------N) = E𝜋 Ψ (1,2------) -----------(14) Where, Eπ = ∑ nii=1 i ------------------------ 15 Where, ni is the number of electrons in Ψi.

Illustrative examples: HMO’s and orbital energies of conjugates system: -

ETHYLENE

2PZ 2PZ

88



E

2

E

1

HMO may be expressed as Ψ= c1∅1+ c2∅2--------------------(16)

∅1 and ∅2 are 2pz orbitals of carbon atoms c1 and c2. There are two atoms (two 𝜋 electrons)here, there will be two HMO’s Ψ1 = C11∅1+ C21∅2------------------------ (17) Ψ2= C22 ∅1 + C22∅2------------------------(18) (see eq.(11) ) Huckel secular determinant will be

X 1

1 X

Where , α−E

β= x

Solving eq (19) we have, x2-1=0 i.e x = ±1------------------------ (20) Corresponding HMO energies for x=1 and x=-1 are, thus E1 = ∝ + 𝛽 ----------------------(21) E2 = ∝ − 𝛽----------------------(22) Obviously, E2>E1 since both ∝ 𝑎𝑛𝑑 𝛽 are negative. Thus, E1 and E2 correspond to bonding and antibonding state.

Non-bonding

state

Fig.

=0-------------------(19)

89



𝜋 − electron energy= Eπ = ∑ ni2i=1 i

=2(∝ + 𝛽) +0x(∝ − 𝛽) =2∝ +2 𝛽 ---------------------------------(23)

𝜋 bond energy = E 𝜋- energy of non-bonded electrons =2(∝ + 𝛽)-2∝ =2 𝛽 ------------------------------(24) HMO coefficient (C11,C21C12 and C22) Secular equations are C11x + C21 =0 C11x + C21x =0 for bonding M.O. x=for bonding M.O. x=−1 Ie C11 - C21 =0 ie. C11 = +C21

Since, C112 + C21 2 =1, we have, C11 = C21 = 1/√2

Thus, bonding MO (ψ1)= 1/√2 (∅ 1 + ∅ 2) -----------------------(25)

Similarly for antibonding M.O. where x=1, we have, -C12=C22= 1/√2

Thus antibonding M.O. ( ψ2) = 1/√2 (∅ 1 - ∅ 2) --------------(26) The pertinent expressions are reproduced without derivation for linear polyenes we have,

Xk= -2cos (𝑘𝜋

𝑁+1); K=1,2,3 -------------------*

Cpk= Dk sin (𝑝𝑘𝜋

𝑁+1 ); p=1,2,3--------------------**

Dk is a constant obtained from normalising ψk .

Here , for ethylene , Putting xk = x1 for N=2, x1 =-2cos 60 =-1

90



Xk=x2 for N=2, x2=-2cos 120=+1

Putting Cpk=C11=D1 sin 60 = √3/2 D1 = √3/2 D1 where D1 = const.

C21=D1 sin 120 = √3/2 D1 = √3/2 D1 where D2 = const.(another)

Cpk=C12=D1 sin 120 = √3/2 D2

C22=D2 sin 240 = - √3/2 D2

Thus,

Ψ1 = √3/2 D1 ∅ 1 + √3/2 D1∅ 2=√3/2 . D1 (∅ 1 + ∅ 2)

< ψ1/ψ1 > =1 ¾ D12 (1+1+0) i.e D1 = √2/3, than C11=√3/2 x √2/3 =

1/√2.

Similarly , C21 = 1/√2, C12 = 1/√2. and C22 = 1/√2. We have the same M.O.S and 𝜋- energy for the same as above. HMO’s of ethylene may be graphically represented both two dimensionally and three dimensionally.

Ψ1 Ψ2

Ψ1= 1

√2 (∅ 1 + ∅ 2) Ψ2=

1

√2 (∅ 1 - ∅ 2)

Here is no node. Here is a node in between C1 and C2.

C1

C2 C1 C2

91



al

l

It has no centre of inversion it has a centre of inversion.

Thus Ψ1 is Ψu (ungerade) and Ψ2 is Ψg (gerade)

Electron Density

Electron density at pth C-atom, qp is the sum of electron densities contributed by different electrons is each HMO.

q p = ∑njCjp2 --------------------------------------(27)

in ethylene Ψ2 is vacant, hence , q1=2xC112 = 2x(1/√2)=1

, q2=2xC212 = 2x(1/√2)=1

CHARGE DENSITY

In conjugate 𝜋 −system, a seutral corbon is associated with an electron density1. Hence, the net charge density is defined as

𝝃r=1-qp------------------------(28) For ethylene, 𝜉1=1-1=0=𝝃2 ----------------------(29) 𝜋-bond order - 𝜋 bonding order represents the degree of 𝜋-bonding between adjacent p and q atom. It is defined as ppq = ∑njcjp. cjq-------------------------------------(30).

92



pj

pq= 𝜋 bond order due to ith HMO electrons.

C1--------C2

Here, P12 = P112= 2x1/√2 x1/√2+0x1/√2x 1/√2=1---------------(31)

ALLYL Systems

Allyl systems contain three c-atoms may be allyl radical, allyl carbocation or allyl corbanion i.e. C3 H5., C3H5

(+) and C3H5(-) , represented

by skeleton ,

2pz 2pz 2pz

HMO function may be Ψ1 = C11 ∅1 + C21 ∅2 + C31 ∅3 --------------------(32) Ψ2 = C12 ∅1 + C22 ∅2 + C32 ∅3 ----------------------------(33) Ψ2 = C13 ∅1 + C23 ∅2 + C33 ∅3 ----------------------------(34) For matrix form: - C11 C21 C31 ∅1 Ψ1

C12 C22 C32 ∅2 = Ψ2 -----------(35)

C13 C23 C33 ∅3 Ψ3

Scoular determinant is

x 1 0 1 x 1 =0 -------------------(36) 0 1 x

C C C

93



i.e x - = x(x2-1)-x= x3-2x =0------------(37)

We have now, x(x2-2) ie x=0, √2 and -√2

Now, E1=∝ +√2𝛽 for x= -√2 -----------------(38) E2=∝ for x=0 --------------------(39)

E3=∝-√2 𝛽 for x=√2 -----------------(40) Thus, E1<E2<E3 They correspond to bonding, nonbonding and antibonding states respectively.

E3-------------∝ −√2𝛽 E2 ----------------- −∝

E1----------------∝ +√2𝛽 𝜋- electron energy, 𝜋- bond energy and delocalisation energy (DE) of allyl systems: - C3H5

+, C3H5- systems contain 2,3 and 4 𝜋- electrons respectively.

E3---------------------∝ −√2𝛽 --------- ----------- --------------------- E2---------------------∝ --------↑------------ ------------↑↓ ---------

E1-----------↑↓ -------∝ +√2𝛽 -------- ↑↓--------- -----------↑↓ ---------- C3H5 C3H5 C3H5 Fig. HMO levels of allyl systems: 𝜋 –electron Energy E 𝜋(C3H5

+) = 2(∝ +√2𝛽)------------------------------------(41)

E 𝜋 (C3H5.) = 2(∝+√2𝛽)+ ∝ = 3∝+2√2𝛽-----------------(42)

E 𝜋 (C3H5-) = 2(∝+√2𝛽) + 2 ∝ = 4∝ +2√2𝛽------------(43)

X

1

1

x

X

0

0

x

94



𝜋 bond energy E 𝜋 BE (C3H5+) = 2(∝+√2𝛽) − 2 ∝= 2√2𝛽 -----------(44)

E 𝜋, BE (C3H5.) = 3∝+ 2√2𝛽-3∝ = 2√2𝛽 ----------(45)

E 𝜋, BE (C3H5-)= 4∝+ 2√2𝛽-4∝ = 2√2𝛽 ----------(46)

DE=E 𝜋 –energy of 𝜋 electrons is localised state. -------------------------------------- C__________C___________C C= C C E 𝜋 E 𝜋(ethylene) + E 𝜋 (c) =2(∝+𝛽)+ ∝n= 2 ∝+ 2𝛽+n∝ (n=0,1,2)

DE (C3H5+) = 2(∝+√2𝛽) − (2∝+ 2𝛽 + 0 x∝)

=2∝+2√2𝛽 −2∝- 2𝛽= 2√2𝛽-2𝛽= 2(√2 − 1) 𝛽= 0:828 𝛽

DE (C3H5.) = 3 ∝+2√2𝛽-(2∝+2𝛽 + 1 x ∝) =2(√2 − 1) 𝛽=0:828 𝛽 --(47)

DE (C3H5-)= 4 ∝-2√2𝛽-(2∝+2𝛽 + 2 x ∝) = 2(√2 − 1) 𝛽=0:828 𝛽

We see that DE is all the allyl system is same and 0.828 𝛽 Expressions for HMOs Secular equations are C11x + C21 = 0 ------------------(48) C12 x + C21x + C32=0---------------(49) Where, x=∝ −E/ 𝛽 C21+C31x=0----------------(50)

For bonding Mo. where; x=-√2.

Putting x= √2 for bonding HMO ψ1

C11x + C21=0= -√2 C11+C21 =0

i,e √2 C11 = + C21 ------------------------(51)

C11+ C21 (-√2)+C31=0------------------(52)

C21+ C31 x = C21-√2 C31=0

i,e √2 C31 = C21 -------------------------(53)

From (51) and (53), C11=C31

95



Since, C112+ C21

2 + C312 =1

C112 (1+√2) 2 +1) =1 ie C11 = ½ = C31 -------------------(54)

Hence, C21= √2/2 ------------------------------------(55)

From eqn (32), (54) and (55) we have, ψ1=1/2∅1+√2/2∅2+1/2∅3-----------(55) Writing secular equation for non-bonding and anti-bonding MOs by

putting x=0 2√2 and solving using normalising condition of coefficients we will have,

Ψ2=1

√2 (∅1 - ∅3 ) ---------------------(56) nonbonding Mo

Ψ3 = 1

√2 ∅1 -

√2

2∅2+

1

2 ∅3 -------------(57) anti bonding Mo.

Two dimension and three-dimension pictorial representation of HMOs-

96



Electron Density

qp=∑𝑛𝑗 𝑐𝑗2𝑝 Allyl carbocation: 𝜓1=1

2∅1+

√2

2∅2 +

1

2 ∅3

𝜓2 = 1

√2 ∅1 __

1

√2 ∅3

𝜓3 =1

2 ∅1 -

√2

2∅2+

1

2 ∅3

Ed at c1 = q1 = 2x(1

2) 2 +0 x(

1

√2)2 +0x(

1

2) 2 =

1

2

Ed at c2=q2=2x(√2

2 )2 +0x02+0x(-

√2

2)2=1 ----------------------(58)

Ed at c3=q3 =2x(1

2) 2+0x(

1

√2)2 +0x (

1

2) 2 =

1

2

Allyl radical: 𝜓1 =1

2∅1+

√2

2 ∅2 +

1

2∅3 & so-on as above.

Ed at c1 = q1=2x(1

2) 2 +1 x(

√2

2 )2 +0x(

1

2) 2 = 1

Ed at c2=q2=2x(√2

2 )2 +1x02+0x(-

1

√2 )2=1 -----------------(59)

Ed at c3=q3 =2x(1

2) 2+1x(

1

√2)2 +0x (

1

2) 2 =1

Allyl carbanion:

Ed at c1 = q1=2x(1

2) 2 +2x(

1

√2 )2 +0x(

1

2) 2 = 3/2

Ed at c2=q2=2x(√2

2 )2 +2x02+0x(-

√2

2)2=1 ---------------(60)

Ed at c3=q3 =2x(1

2) 2+2x(−

1

√2)2 +0x (

1

2) 2 =3/2

CHARGE DENSITY Allyl carbocation: 𝝃1 = ½-1=-1/2= 𝝃3, 𝝃2=1-1=0 Allyl radical: 𝝃1 =1-1=0= 𝝃2= 𝝃3 ------------------(61)

Allyl carbanion: 𝝃1=3/2-1=1/2= 𝝃3, 𝝃2= 1-1=0

97



C C C C

𝜋 − 𝐵𝑜𝑛𝑑 ORDER(P): PPq == ∑𝑛𝑗 𝑐𝑗𝑝 .Cjq

Carbocation P12=2x1/2x√2

2 =

√2

2 =0.707

P23=2x√2

2 x

1

2 =0.707 bonding HMO certain two

electrons only.

Allyl radical p12=2x1

2 x

√2

2+1x

1

√2x0=0.707=p23 bonding &nonbonding

Mo, certain 221 electrons.

Allyl ciborium ion P12=2x1

2x√2

2+2x

1

√2x0=0.707 = P23

1, 3- BUTADIENE

CH2= CH-CH= CH2 may be depicted through following

Skeleton

2pz 2pz 2pz 2pz

HMO functions for conjugate systems of four carbon atoms and four 𝜋-electons may be given by 𝜓1 = C11∅1 + C21∅2+ C31∅3+ C41∅4 ------------------------(62) 𝜓2 = C12∅1 + C22∅2+ C32∅3+ C42∅4 ------------------------(63) 𝜓3 = C13∅1 + C23∅2+ C33∅3+ C43∅4 ------------------------(64) 𝜓4 = C14∅1 + C24∅2+ C34∅3+ C44∅4 ------------------------(65) Eq. (62),(63),(64) and (65) may be put in following matrix form:

98



- -

C11 C21 C31 C41 ∅1 𝜓1

C12 C22 C32 C42 ∅2 = 𝜓2

C13 C23 C33 C43 ∅3 𝜓3 ----------------------------------------(66)

C14 C24 C34 C44 ∅4 𝜓4

Huckel secular determinant of the system is given by

X 4 0 0

1 X 1 0

0 1 X 1 =0------------------------------(67)

0 0 1 X

Let us solve the determinant,

i.e. x { x(x2-1) - (x) } - { (x2-1) – 0} = 0 i.e. x4 – 3x2 + 1 = 0 -----------------------------------(68)

i.e. (x2)2 - 3x2 +1 = 0 i.e. x2 = 3 ± √9−4

2 =

3+√5

2and

3−√5

2 ------69

X 1

O

1 X

1

O 1

X

1 1

O

O X

1

O 1

X

X - = O

X 1

1 X

1 1

0 X

X 1

1 X

0 1

0 X

- X

=O -

99



E4 --------------- 𝛼 − 1.618𝛽

E3 ----------------𝛼 − 0.618𝛽

E2 ---------------- 𝛼 + 0.618𝛽

E1 ----------------- 𝛼 + 1.618𝛽

c-c-c-c c=c-c=c

i.e. x2 = 3±√5

2 =

(√5)2+12±2⋅1⋅√5

4 =

(√5±1)2

4

i.e. x= √5+1

2, -

√5+1

2, √5−1

2 and

√5−1

2

= 1.618, -1.618 and – 0.618 X = 1.6180, -0.618, 0.618, +1.618 -------------------70

Putting the value from (70) into x= 𝛼−𝐸

𝛽

We have four values of E i,e

Order : E1 < E2 < E3 < E4

HMO level with π- electron π electron energy

Eπ= 2E1 + 2E2

Eπ= 2(α+1.618β) + 2(α+0.618β) Eπ= 4α + 4.48β -------------------------------------(75)

π bond energy = (4α + 4.48β )- 4α= 4.48β ------------------ (76) delocalisation energy (DE) it is difference of π electron energy of delocalised structure (i) and localised structure (ii)

------π----- π π

Eπ= 4α + 4.48β Eπ= 2(α+1.618β) + 2(α+0.618β)

E1 = 𝛼 + 1.618𝛽 ----------- (71)

E1 = 𝛼 + 1.618𝛽 ----------- (72)

E1 = 𝛼 − 1.618𝛽 ----------- (73)

E1 = 𝛼 − 1.618𝛽 ----------- (74)

100



√7.24

DE= (4α + 4.48β )- (4α-4β) DE= 0.48β --------------------------------------(77) HOM function Secular equation are C11x + C21 = 0 C11x + C21 + C31 = 0 C21x + C31 + C41 = 0 ------(78)for HOM Ψ1 where x=-1.618

C11x + C21 = 0 Putting the value x = -1.618 is (78) we have, C11(÷1.618) + C21 = 0 i.e. C21 = 1.618C11 (79) C31 + (-1.618)C41 = 0 i.e. C31 = 1.618 C41 (80) C31 = -C11 - C21x = - C11 – 1.618 X (1.618 )C11 = C11 { (1.618)2

- 1}=1.618 C11

i.e C31 = 1.618 C11 (81); from (79) & (81) we have, C21 = C31 (82) Now,

i.e 1.618 C41 = + C31x = C21 = (from 80) i.e C11 = C41 (from 79 & 83)

(83)

Now from normalisation condition, C11

2 + C221 + C2

41 + C241 = 1

i.e C211 + (1.618)2 C2

11+ (1.618)2 C211 + C2

11 = 0 [from 79,80,81,82,83,& 84] i.e C11 = 1 = 0.372 = C41 (85)

101



Again, C21 = 1.618 x 0.372 = 0.602 = C31 (86) Thus, we have,Ψ1=0.372Φ1 + 0.602Φ2 +0.602Φ3 + 0.372Φ4 (87) Putting x=-0.618,0.618 and +1.618 we will have Ψ2 , Ψ3 , & Ψ4

HMO functions Ψ2 = 0.602Φ1 + 0.372Φ2 -0.372Φ3 - 0.602Φ4 (88) Ψ3 = 0.602Φ1 + 0.372Φ2 -0.372Φ3 + 0.602Φ4 (89) Ψ4= 0.372Φ1 - 0.602Φ2 +0.602Φ3 -0.372Φ4 (90) Graphical representation of HMO’s

C

1

C2 C3 C4 Ψ1 C

1

C2 C3 C4 Ψ2

C

1

C2 C3 C4 Ψ3 C

1

C2 C3 C4 Ψ4

102



3D-representation of HMO’s

ELECTRON DENSITY q1 = ed at C1 = 2 X (0.372)2 + 2 X (0.602)2 =1.0 q2 = ed at C2 = 2 X (0.602)2 + 2 X (0.372)2 =1.0 (91) q3 = ed at C3 = 2 X (0.602)2 + 2 X (-0.372)2 =1.0 q1 = q2 = q3 = q4 q4= ed at C4 = 2 X (0.372)2 + 2 X (-0.672)2 =1.0 qi = 4.0 qi = 4.0 confirms the presence 4π bonds in a molecule. CHARGE DENSITY ε1 = 1.0 -1.0 -0 = ε2= ε3= ε4 (92)

π-bond order p12 = 2 x 0.372 x 0.602 + 2 x 0.602 x 0.372 =0.896 p23 = 2 x 0.602 x 0.372 + 2 x 0.372 x(-0.372)=0.448

103



p34 = 2 x 0.602 x 0.372+2x(-0.372)x(-0.602)=0.896 Thus, p12 = p34 > p23

This shows C1 – C2 and C3 – C4 the terminal bonds have more π- double –bond character than the central C2 – C3.

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