WEBVTT

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hello and welcome to the day four of the sixth
week of the lecture series so we have learnt

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about the s l cs how to found the s l cs using
position of it so now the question is like

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what what we will do with this s l cs ok what
information that this symmetric constrain

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toward the you know the linear termination
of atomic orbitals will you know give us

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so now in in case of this chemical bonding
or bond formation so like the vales band

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theory that was put forward by and many
others they consider this you know the bond

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to be its like a two center bond you know
so the the you know interaction between

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the atomic orbitals of two atoms a molecule
will make the you know where functioning localized

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between the you know within the bond
but you know when the molecular orbital

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theory came in to the existence then that
give you know much more freedom because in

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management theory it taken consider any
kind of interaction between two forms that

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are found ok

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so in molecular orbital theory we have
this you know idea that the wave functions

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for in molecular orbital they you know extent
up to the whole you know all about molecule

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ok so essentially you can fill the presence
of the orbitals all about the molecule so

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that in one sense it helpful for us because
you know in that case we can actually you

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know use the use that are you know symmetry
adopted and what that we help us that will

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help us in you know solving for various
problem including the energy of the orbitals

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of the occupancy of the orbitals of the bond
orders all this things without going

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in to detail you know our heart core
computation ok

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so in one way we can save a lot of time
and without doing too much of work we can

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say about the energies of molecule how
will be energies of the orbitals and all ok

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and certain rules associated with this
will help us in assigning different

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you know particles many different you know
say diff different wave functions to

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different molecular orbitals now one of
the you know the commonest approach

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to you know reduce the motion of an molecular
orbital in an express it and practical

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form is the approach of linear combination
of economic orbital in short l c a u we

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so what you approach we met that its you
know atomic orbitals you know they can

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combine in a linear fashion and they ultimately
you know give us the molecular orbital so

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all the you know in combined orbitals they
are spread over all the molecule it seems

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they are quite delocalized so this is an approximation
and here we write molecular orbital as

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the linear combination of four atomic orbitals
which is clear for his name and how would

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we do that huh so if we have the you know
the function for the atomic orbital ith orbital

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in a molecule as phi i and you know psi k
is the kth molecular orbital then i can express

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psi k in terms of the all phi i s right with
a particular coefficient which is denoted

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as c i k so this particular you know combination
is nothing but an linear combination right

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now this phi i s are the that you know that
they represent the atomic wave function or

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the atomic orbitals here so this act as a
basis function so all the atomic orbitals

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that is the you know complete set of the phi
i s act as the basis set ok and we we allows

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pref you know would like to work with an ortho
normal basis functions so sake of ortho normal

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functions as my basis so therefore this phi
i s are you know waiting in such a way

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that they are normalized all right and any
q atomic orbitals so atomic wave functions

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you know they found the basis of universal
presentations and automatically they will

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become ortho gonal so if you make them ortho
normalized also then they will become ortho

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normal

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so we will get as set of ortho normal basis
function so by using this molecular orbitals

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approach a pareticular form of the wave equation
ok using this wave functions in form this

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you know wave function atomic wave function
we get the wave function to be molecular orbital

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and then we can get and by of equation which
has a particular form and which has

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call secular equation we will see in a while
what is that and that can be developed

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so how it can be developed so the wave equation
can be written in an usual way so by using

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equation so h psi equals to e psi

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so i can write h psi minus e psi which is
nothing but i can write in this form where

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both hamiltonian and energy is use doesnt
operator and is equals to zero ok this is

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something which we are all familiar with now
l c o expression for the psi that is the

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molecular orbital is now introduce by giving
this expression so now if i considered only

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two term l c a o m o thus above equation that
is here what i get is this because i am

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considering this i to be one and two so
i have already have two terms i could consider

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term so for that i will have like you know
c one h minus e phi one plus c two then

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h minus e phi two then c three and so on what
i i take the simplest example were already

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two terms will exist

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so as if i have two basis functions phi one
and phi two and for that i can re write this

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equation as this fine so i can this particular
equation to be as a now if i multiply this

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particular equation by phi one and integrate
over all the spatial coordinates meaning by

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over d tau then what do you get so what i
just say is written here so i multiplied both

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the sides by five one and then i integrated
over d tau so that will remain equals to

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zero now in this you know expression what
you can see is that there are several integrals

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because you can separate this different quantities
like you know h minus e when you expand then

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you can get several integrals

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so we will use certain notations for simplicity
the notation are h one one which stands for

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psi i h phi i integrated d tau where h i j
is the integral which is given as this

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the two wave functions it define phi i and
phi j and operated may still the same hamiltonian

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and then there is a another integral which
is the integration over d tau for the product

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of the two functions phi i and phi j and that
is given by this intri term s i j and this

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s i j is know as for overlap integral so here
we explicitly say what are you know

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which particular integral is you know
gives what

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so this h i i so it can be h one one or h
two two then in that phi one h phi one integration

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over d tau or phi two h phi two integration
of d tau that gives the energy of the atomic

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orbital phi i because here there is no interaction
between two different orbital the same atomic

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orbital for which i have the wave function
as phi one so that i am considering so this

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give me the energy of the atomic orbital whereas
h i j that gives me the energy of interaction

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between pairs of atomic orbitals which is
not too difficult to understand those i

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have two different wave functions corresponding
to two different atomic orbitals and they

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are interacting energetically and this integral
h i j will give me the energy of this particular

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interaction and as i said s i j is called
overlap integral ok so how much two orbitals

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overlap will be given by this s i j integral

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so now i have the you know energy equation
that i can write in this fraction right using

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those notations so i got this particular energy
equation considering only by multiplying

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that you know that secular equation by phi
i and integrating so i could also do the

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same thing with phi two that means i could
multiply this legulation by phi two and integrate

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over all the space and in that case what i
could get i would get something like this

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right so you can see the similarity between
this two equation that we got

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now this two equation if i put them together
so i have got this two set of equations now

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these two equations form a system of homogeneous
linear equation in c one and c two which are

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the coefficient for linear combination so
for this homogeneous linear equations

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you can always have the trivial solutions
as c one and c two both are zero now if

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in order to get nontrivial solution what
we need to have for you know by crammers rule

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you should have the determinant of the
matrix form by the coefficients of this c

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ones and c twos to be equals to zero right
so then only i will have nontrivial solution

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and here we have this secular equation
in form of this determinant

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so this is the secular determinant it in
terms of this integrals h one one or h one

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two you know and we have a term particularly
this you know s one two so this e s one two

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that e s two one that we can get because
you know you if you go back and see

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this integrals here so you can change
you know inter change your position of phi

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one and e ok so this is has nothing to do
because its like a just in a [scar/scalar]

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scalar quantity so therefore you have like
e an integration of phi one phi two

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so therefore you have e s one two right in
that way you get this term ok and the other

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terms already we have explained so one can
get the values of this h one one also h

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i i s or h i j s or s i j s one can also
try to estimate or compute ok with the

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help of some sort of approximation and the
reason that approximation one can solve the

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secular equation and get the value of e which
is of at most importance now all the integrals

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that we discussed right so like this you know
h one one h one two s one two s two one all

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the individual you know integrals if
you have to solve them be it computationally

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or some otherwise this is fairly time consuming
so we need to actually do some approximation

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so that our job is easier we save lot of time
right

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so one of such approximation is huckel
approximation so this huckel approximation

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i will explain that shortly this is
a very interesting you know approximation

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but this serves our propers a pretty good
way ok so for us huckel approximation will

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be a pretty valid approximation and that will
help us in various different ways we will

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see that soon so what is this huckel approximation
so huckel approximation says that all the

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you know h i j s are equal to alpha i j for
vibrate this you know i equals to j ok so

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what does that mean it means that you know
this interaction energy is zero until and

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unless this you know atomic orbital wave function
that we are considering the phi i that is

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phi one phi two phi three

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so this interaction energy will be zero except
for the case why this know i th and j th orbitals

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are adjacent to each other so i have like
you know one orbital in the next orbital the

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next orbital and so on so this you know if
i can see that the first one and the third

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one then the interaction energy is assume
to be zero under huckel approximation ok otherwise

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you know if they are adjacent they will
have a value of alpha i j all right so

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if i consider special case i equals to j then
i may i have like h i i so this h i i is alpha

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and this alpha is taken to be the zero
of the energy ok

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so we will talk about that in detail later
when we will take particular example and you

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know explain that and the overlap integral
is taken to be zero under huckel approximation

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so with these approximations we will try to
see now that how it simplifies the treatment

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of a little bit complex problem so for
example you know pi system of the molecule

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naphthalene so what we will do you will use
all the p pi orbitals as main phases and we

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number this phi pi orbitals as which shown
here so here you have a p pi orbital here

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you have phi pi orbital so we have numbered
we have numbered them has one two three four

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five six seven eight and nine ten and you
see the number are down in slightly different

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way so that is becomes here we are considering
the equi balance of the atoms here

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so if i take the you know this ten bases
function you know involving ten phi pi

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orbitals on the carbon atoms of than a prime
molecule and then we do the treatment as i

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you know said in last few minutes then what
we can do we can form ultimately as secular

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determinants so in the previous examples we
said that ok we choose only two function phi

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one and phi two so we had a two by two secular
determinant right so here we have ten basis

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function here and then ultimately i should
get ten by ten secular determinant and then

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correspondingly i get the secular equation
as this right now there are you know there

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is a three step procedure for setting up
you know symmetry factored secular equation

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so this is like you know we have huge ten
by ten determinants which is really kind of

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scale so can we bring symmetry restrictions
here and simplify my life

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so first what we do we use a state of atomic
orbitals as sate as a basis for a representation

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of the group and then we reduce this representation
ton its reduce able components in the next

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step we combine the basis you know functions
here orbitals in two linear combinations

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that means this we form the s a l cs by utilizing
the position operated and then we list all

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the s a l cs so that all those belonging to
a given representation occur together in the

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list so we we tabulate them and we will
use this list to level the rows and columns

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of the secular determinant only the elements
of the secular determinant that lie at the

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intersection of a row and column belonging
to the same irreducible representation can

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be non zero and these non zero element will
be in blocks along the principal diagonal

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so similar way as we did when we learnt
about how to diagonalize the you know matrix

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by symmetry transformation i am saying this
analogically its similar so the secular determinant

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will therefore be symmetry factor all right

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so lets take a take the example of this particular
molecule naphthalene so we said that

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you know we have ten p pi orbitals as my
basis so the point group of the naphthalene

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is d two each and here is the for this particular
point d two h now using the concept of unset

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of atom you should be able to form the
representation ok so directly writing the

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characters we get the representation as gamma
and this shown here so we have the you know

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non zero characters for identity for c
two i for sigma x y ok and rest of the characters

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are zero fine so what we are suppose to do
we are suppose to suppose to reduce in

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the next day so now at the beginning we
say that this p pi orbital corresponding to

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octane carbon atoms with naphthalene they
are not all the equivalent but they had you

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know they form some subsets which you know
contains the equivalent you know the

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basis function c f

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so in we have essentially three different
sets in the first set we have the p pi

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orbital comprising you know function phi
one phi four phi five phi eight and we have

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another set having phi two phi three phi
six phi twelve and the third set con contains

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the phi nine and phi ten now what you can
do you can you know either form the basis

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you know representation as we just said in
terms of the gamma right here or what you

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can do you can consider the individual sets
just like the example that we get in the previous

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class for p f three

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so if you do that if you take this basis functions
as a basis set and then another set another

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set then you form a representation and then
reduce to get this irreducible representation

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now whatever way you do the results should
be the same so what you can do you can form

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this this way and all together you can
find out how many irreducible represent which

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type of reducible representations occurring
in how many time you should get the same result

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if you take the reducible representation
comprising the from by using the basis

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having a set having ten p pi orbitals you
will get the same result ok

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so what you have here you have total
two a u orbitals two a that two orbitals

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with a symmetry three orbitals with one u
symmetry and two orbitals with b two g and

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three orbitals with the three g symmetry
so next is we have to find out the s a l cs

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right corresponding to each one of this irreducible
representation so we have u u or u one u and

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b to g and b three g so we have learnt how
to get those and all these irreducible representation

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that we found by reducing this irreducible
representation found out of this basis sets

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three basis sets then we found the projection
operators for those irreducible representation

25:18.919 --> 25:27.610
and find the you know s a n cs that
we can form and normalize them

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so you get the you know s a l cs using the
same concepts that we have used earlier so

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all together out of ten basis set basis
function you should get ten s a l cs and we

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see that we get ten s a l cs so s a l cs
that we got we are writing in terms

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of this psi so psi one to psi ten are our
s a l cs now we can you know get the

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secular determinant in terms of blocks and
we get two three by three blocks and two

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two by two blocks so essentially we have two
three by three determinant and two two by

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two determinant and lets start one by one

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so b three g had three by three determinant
and we take that and solve the equation right

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so when we write the secular determinant
and then the secular equation for sebi three

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three and you try to solve that you get you
know it is all in terms of the the notation

26:44.759 --> 26:51.470
that are used under huckel approximation that
is either alphas or betas now i mention that

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alpha is taking to be the energy zero so what
is the energy zero so what is the energy zero

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here so in in case of molecular orbital when
we we draw the molecular orbitals so as

27:04.899 --> 27:10.720
a function energy so energy follows in a vertical
relation and then we have different molecular

27:10.720 --> 27:18.080
orbitals form so now the energy of energy
equals to zero will lay somewhere where essentially

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you have the atomic orbitals having their
energy so main that there is no interaction

27:26.510 --> 27:34.889
between this atomic orbitals at that particular
energy at that particular energy so two

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atoms are not at all forming a bond

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so that energy level is taken as a zero so
an under huckel approximation this alpha is

27:43.910 --> 27:52.879
taken to be that zero ok so and beta is the
unit of energy so beta will give me one

27:52.879 --> 28:03.280
o taking those terminologies and s one
two ys two one these are you know zero

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under approximation we get the results as
eight eight eight equals to alpha h nine one

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zero so nine ten is equals to beta and
h eight ten has root to beta so we you

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know can keep going and calculating all the
other secular determinants for other irreducible

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representation and then put the conditions
that is impose by the huckel approximation

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and you ultimately get the total
you know energies ok

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so energies corresponding to which irreducible
representation essentially meaning that for

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say a u you you have two s ls meaning that
two orbitals will be formed ok so when you

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solve for the determinants ok if it is two
dimensional you get two roots if it three

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dimensional you get three so you get two or
three different energies write so ones you

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solve for all the determinants that you get
you get these energies and then you have to

29:29.210 --> 29:40.940
arrange these energies with a energy scale
fine so here we have done that is after

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finding this you know energies corresponding
to a particular you know orbital with a

29:47.400 --> 29:48.769
particular symmetry

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so then we just add over this energies
from lower to higher so energy zero is here

29:56.620 --> 30:05.110
ok in the unit of beta of course so so
as you go higher you know you have this

30:05.110 --> 30:10.059
negative because beta itself is a negative
where about there is an interaction you get

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you know the energy has negative
so when you fill up the molecular orbital

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that you form you follow you know the hound
rule and the exclusion principal so when

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you follow the hound rule and exclusion principal
you should be able to fill up the all the

30:30.190 --> 30:36.950
electrons into the orbitals ok so with
this we will stop today and we will come back

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tomorrow with some other applications of
this symmetry and the group theory that we

30:41.999 --> 30:44.000
have learnt so for till then
thank you so much