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On semiempirical treatment of hydrocarbons
Debangshu Mukherjee
B.Sc Physics, IInd Year
Chennai Mathematical Institute
27.11.2009
Abstract
As, we do in case of small systems like H2+ or He atom, we write
down the explicit Hamiltonian equation and solve it to find out the wave
functions of the electrons, but however, we cannot use such a method in
case of big molecules, especially organic molecules. Our aim is to get an
insight into solving such systems upto a certain order of approximation.
The molecular orbitals of a molecule can be separated into σ and π MOs. This
comes from the valence electron approximation, which says that in a big atom,
we can consider the core electrons to be concentrated at the nucleus, but such
a treatment would reduce the valence electron to the 1s gerard state (even
state with lowest energy). So, we consider the inner orbitals to be orthogonal. However, a better approximation is to regard them as a continuous charge
distribution. This then reduces the effective Hamiltonian, which we can now
use.
1
Prescription for σ-bond electrons
First, we look into the simplest case of H2+ . We assume that a MO can be
written as linear combination of AOs. We can write Schrodinger’s Equation for
H2+ and explicitly solve it. We assume that the nucleus are stationary and they
are much much heavier than electrons. So, their momentum can be ignored
and the final wave function of the MO will not depend on the momentum of
the nuclei. i.e ψtot (qi , qα ) = ψelec (qi , qα )ψN (qi ). For a diatomic homonuclear
molecule, the wave function is:
ψ = a1 φ1 + a2 φ2
Now, we will use variational method to minimize H̄, where,
R ∗
ψ Hψdτ
H̄ = R ∗
ψ ψdτ
Substituting ψ in equation (1), we get,
R
(a1 φ∗1 + a2 φ∗2 )H(a1 φ1 + a2 φ2 )dτ
E= R
(a1 φ∗1 + a2 φ∗2 )(a1 φ1 + a2 φ2 )dτ
1
(1)
(2)
Simplifying this expression, taking into accout the following,
Z
Z
φ∗1 Hφ2 dτ = φ∗2 Hφ1 dτ
and
Z
φ∗1 φ2 dτ =
Z
φ∗2 φ1 dτ
we get,
E=
a21 H11 + 2a1 a2 H12 + a22 H22
a21 + 2a1 a2 S12 + a22
where,
Z
φ∗i Hφj dτ
Hij =
Z
φ∗i φj dτ
Sij =
Now, we know,
∂E
∂a1
= 0 and
∂E
∂a2
= 0 Finally, we get a determinant of the form
H11 − E
H21 − S21 E
H12 − S12 E =0
H22 − E whose solutions are:
+H12
−H12
E+ = H111+S
and E− = H111−S
by making the aasumption that H11 = H22
and H12 = H21 . It can be shown that E+ < E− by putting H = − 21 ∇2 −1/r2 (in
φ1 −φ2
electronic units). The wavefunction is: ψ+ = √φ1 +φ2 and ψ− = √
2−2S
(2+2S)
These are called the bonding(gerade) and anti-bonding(ungerade) MOs. i.e
symmetric and anti-symmetric. They look as follows:
In first case, the electron probability is mostly between the nucleii while in
second case, it is outside. So, (+) is more stable.
We can extend our calculations to p-orbitals and we find similar wave functions.
But, other orbitals have directional character. When, pz has formed σ bond,
there is lateral overlap which forms π bond. For heteronuclear atoms, we change
the wavefunction to: ψ+ = φ+ +λφ− and ψ− = φ+ −λφ− where λ is a weighteing
constant depending on electronegativity, since the electron is shifted to the
electronegative one and not at center. We can extend this same thing for bigger
molecules but, only for σ-bond electrons.
2
2
Semiempirical method: Free-electron MO method
We look into the ethylene molecule:
Figure 1: Ethylene molecule
Now, ψCH(σ)=a1 φ(1s) + a2 φ(sp2 ) and ψCC(σ) = a1 φ(sp2 ) + a2 φ(sp2 ).
This is because, the C gets sp2 hybridized to sp2 to form σ bonds with C and
H. We now look into π electrons. The effective Hamiltonian for π electrons is:
Ĥπ =
nπ
X
Ĥπcore (i) +
i=1
nπ X
X
1
r
i=1 j>i ij
where
1
Ĥπcore (i) = − ∇2i + V (i)
2
where V (i) is the P.E of the ith π electron in the field produced by nuclei and
σ electrons. We can now minimize the variational integral to find Eπ .
In FEMO(free-electron MO) treatment, effect of σ electrons is like particle in
a box. V = 0 at some place and ∞ otherwise. The interelectronic replusion
1/rij is ignored.So, we write the equations:
Ĥπ ψπ = Eπ ψπ
ψπ =
nπ
Y
φi
(3)
(4)
i=1
Ĥπcore (i)φi = ei φi
Eπ =
nπ
X
ei
(5)
(6)
i=1
Equation (4) is the strangest one. Actually, this comes from the Hartee-Fock
method, which assumes that the wavefunction of a manyparticle(fermoinic)
system is the product of the properly chosen wave functions of the individual particles. i.e ψ(x1 , x2 ) = χ(x1 )χ(x2 ). This is an ansatz and cannot be
proved. We also assume that for conjugated chain molecules, electrons move as
1-dimensional. So,
n2 h̄2
ei = i 2
8me l
3
By Pauli’s Principle, there can be at most 2 electrons in one M.O. The
π- electrons fill up the lowest 12 nπ free-electron π MOs. So, for a conjugated
molecule with nπ π electrons, highest occupied M.O is n = nπ /2. Lowest
frequency transition happens when an electron goes from highest occupied MO
to lowest empty MO. So, for lowest frequency emmission,
1
1
h̄
(nπ + 1)
=
(e n2π +1 − e n2π ) =
λ
h̄c
8me cl2
(7)
We apply this to conjugated polynes like CH2 = [CH − CH =]k CH2 . We take
l as the length of the carbon chain and l1 and l2 as the C-C single and double
bond lengths. Now, there can be rotations about the single bond producing the
states trans and cis as follows:
Figure 2: Configurations of ployenes
The FEMO treatment allows the molecule to move beyond the ends of the
carbon chain by a length of 43 l1 + 14 l2 at each end. So, we have,
3
1
3
1
l = kl1 + (k + 1)l2 + l1 + l2 = (k + )(l1 + l2 ) = (nC + 1)(l1 + l2 )
2
2
2
2
4
where nC is the number of carbon atoms. Now, nπ = nC . So, using equation
(7), we have
λ = 2me ch̄−1 (l1 + l2 )2 (nC + 1) = (64.6nm)(nC + 1)
(8)
Observed values of λ increases from 162.5 n.m to 447 n.m as k goes from 0 to 9.
It agrees qualitativelt, i.e predicts the increase of wavelength, but quantitative
result is poor as, they have 110% average absolute error. It fails for conjugated
ployenes, but, works well for polymethine ions
(CH3 )2 N+ =CH(-CH=CH)k -N̈(CH3 )2
3
Huckel’s MO Method
In this method, the π electron Hamiltonian is approximated by a simpler form
Ĥπ =
nπ
X
Ĥ (ef f ) (i)
i=1
Huckel assumed separability of σ and π electron functions based on the assumption that we can distinguish between σ, π, δ MO just by their m value.
Although this seems at weird approximation in first place, but, this eventually
explains uniform C-C bond length of benzene, high energy barrier to internal
rotation about double bonds and the unusual chemical stability of benzene.
Mathematically, this approximation means:
Etot = Eσ + Eπ
where Etot is the electronic energy Eel and internuclear interaction Vne . Also,
the wave function will be of product form. i.e:
ψ(1..n) = ψπ (1..k)ψσ (k + 1, .., n)
Our Hamiltonian is:
Ĥ(1, ..n) = Ĥπ (1, .., k) + Ĥσ (k + 1, .., n)
So, by variation, we get,
R ∗ ∗
ψπ ψσ (Ĥπ + Ĥσ )ψπ ψσ dτ (1, ..n)
R
Ē =
= Eπ + E σ
ψπ∗ ψσ∗ ψπ ψσ dτ (1, ..n
From these equations, it is clear that we just cannot minimize Eπ by varying
ψπ . It is impossible to define Ĥπ and Ĥσ so that they depend on completely
different group of electrons. Writing the operators explicitly,
Ĥπ (1, .., k) = −
k
k
k
k
1X X
1X 2 X
∇i +
Vne (i) +
2 i=1
2 i=1
i=1
j=1,j6=i
n
n
n
1 X 2 X
1 X
Ĥσ (k + 1, .., n) = −
∇i +
Vne (i) +
2
2
i=k+1
k+1
5
1
rij
n
X
i=k+1 j=k+1,j6=i
1
+ Vnn
rij
Observe,
Ĥ − Ĥπ − Ĥσ =
k
n
X
X
i=1 j=k+1
1
rij
Hence, we assume that there is some form of Ĥ ef f which somehow incorporates
the average effect of π electrons. So, our equation is:
Ĥ ef f (i)φi = ei φi
(9)
Now, we assume that MOs are LCAOs i.e (Linear Combination of Atomic
Orbitals)
nC
X
cri fr
(10)
φi =
r=1
where fr is a 2pπ AO on the rth carbon atom and nC is the number of carbon
atoms. We substitute (10) in the variation integral
R ∗
φ Hφdτ
H̄ = R ∗
φ φdτ
We define overlap integral as
Z
Sjk = fj∗ fk dτ
Z
and Hjk =
fj∗ Ĥfk dτ
Also, making the aasumptions that cj ’s are independent i.e
Sji =
∗
Sij
= Sij
and Hji =
∗
Hij
∂cj
∂ci
= δij and also,
= Hij
Moreover, the wave functions φ will be normalised. So, Sii = 1. Using these
information, finally we get n equations with n unknowns which finally gives the
determinant:
det(Hij − Sij W ) = 0
(11)
H11 -S11 H̄
H21 -S21 H̄
.
.
.
Hn1 -Sn1 H̄
H12 -S12 H̄
H22 -S22 H̄
.
.
.
Hn2 -Sn2 H̄
...
...
...
...
...
...
H1n -S1n H̄
H2n -S2n H̄
.
.
.
Hnn -Snn H̄
=
0
(12)
ef f
Now, we define terms like Hiief f =α (Coulomb Integral) and terms like Hij
=
β (Bond Integral) where Ci and Cj are bonded. We assume that
H
=
0
when
PnC ij 2
Ci and Cj are not bonded. Also, The MO are normalized i.e r=1
|cri | = 1.
So, in butadiene,
Figure 3: Butadiene molecule
6
ef f
ef f
ef f
= β and
= H34
= H23
By Huckel’s assumption, here Hiief f = α and H12
ef f
ef f
ef f
H13 = H14 = H24 = 0.
The secular equation looks like:
α − ek
β
0
0
β
α − ek
β
0
α−ek
β ,
Making the transformation x =
x
1
0
0
1
x
1
0
0
1
x
1
0
β
α − ek
β
0
0
1
x
=
0
0
β
α − ek
=
0
we get,
0
(13)
In general, a determinant of the form:
a
c
0
.
0
0
b
a
c
.
.
.
0
b
a
.
.
.
0
0
b
.
.
.
.
.
.
.
.
.
.
.
.
.
0
0
.
.
.
.
c
0
.
.
.
.
a
c
0
0
0
.
b
a
=
Πnj=1 [a − 2(bc)1/2 cos(
jπ
)]
n+1
For Butadiene, we get,x= -1.618,-0.618,0.618,1.618. Finally, putting this x, we
solve for ci ’s. The normalized HMO energies are:
φ1 = 0.372f1 + 0.602f2 + 0.602f3 + 0.372f4
φ2 = 0.602f1 + 0.372f2 − 0.372f3 − 0.602f4
φ3 = 0.602f1 − 0.372f2 − 0.372f3 + 0.602f4
φ4 = 0.372f1 − 0.602f2 + 0.602f3 − 0.372f4
For butadiene, the MO look like
Figure 4: MO for Butadiene: Nodal planes are indicated by dashed line
7
On calculating the MO energies, we can find that the energy of φ1 is minimum. This must be the lowest-energy π MO. Its energy is α+1.618β, so β must
be negative. In fact, in Butadiene, bond energy is negative which implies very
stable bond, we can observe, it has resonating structure due to alternate double
bonds
and form a continuous π cloud. Moreover, we approximate 2pπ AO by
R ∗ ef
fi Ĥ f fi dτ = α. If, energy is less, then it is bonding or else antibonding.
4
Analysis of conjugated polyenes
For conjugated polyene, with nC carbon atoms, the secular equation looks like
equation (13) but of order nC . So,
x = 2cos[
jπ
]
nC + 1
and
ej = α + 2βcos[
jπ
]
nC + 1
Highest occupied and lowest vacant π orbitals have j = 21 nC and j = 12 nC + 1.
So, longest wavelength is
1
4β
π
= − sin
λ
h̄c
2nC + 2
Here, observed value of Butadiene λ is 217n.m and |β| = 4.62eV. But, calculated
value of λ differs by 44%.
4.1
Benzene
For benzene, C6 H6 i.e
the HMO equation is:
x
1
0
0
0
1
1
x
1
0
0
0
0
1
x
1
0
0
0
0
1
x
1
0
0
0
0
1
x
1
1
0
0
0
1
x
=
0
The values of x = -1,+1,+2,+1,-1,-2.
So, the energies ei = α + 2β, α + β, α + β, α − β, α − β, α − 2β. There are two
doubly degenerate levels as can be seen.
8
The energy diagram for the benzene MO is
Figure 5: Energy diagram of benzene
HMO coefficients can be determined purely from symmetry operations. Now,
symmetry operator commutes with H, so, their eigenfunctions must be same.
Now, (ÔC6 )6 = I. Eigenvalues are e2πik/6 , where k= 0,1,..,5. So, we write,
taking φi as LCAO,
e2πik/6 φj = ÔC6 φj
6
X
crj e2πik/6 fr =
r=1
6
X
crj ÔC6 fr =
r=1
6
X
crj fr−1 =
r=1
6
X
cr+1,j fr
r=1
Equating coefficients of corressponding AOs , we have
cr+1,j = e2πik/6 crj
Using normalization,
P6
i=1
|crj |2 = 1, we can say,
1
|crj | = √
6
The Huckel MOs are
√
φ1 = 1/ 6(f1 + f2 + f3 + f4 + f5 + f6 )
√
φ2 = 1/ 6(f1 + eiπ/3 f2 + e2iπ/3 f3 − f4 + e4iπ/3 f5 + e5iπ/3 f6 )
√
φ3 = 1/ 6(f1 + e−iπ/3 f2 + e−2iπ/3 f3 − f4 + e−4iπ/3 f5 + e−5i pi/3 f6 )
√
φ4 = 1/ 6(f1 + e2iπ/3 f2 + e4iπ/3 f3 + f4 + e2iπ/3 f5 + e4iπ/3 f6 )
√
φ5 = 1/ 6(f1 + e−2iπ/3 f2 + e−4iπ/3 f3 + f4 + e−2iπ/3 f5 + e−4iπ3 f6 )
√
φ6 = 1/ 6(f1 − f2 + f3 − f4 + f5 − f6 )
9
The MO orbitals of benzene looks like follows.
Figure 6: MO for benzene
4.2
Monocyclic conjugated ployenes
We can now generalize to monocyclic conjugated polyenes and the HMO energies
and coefficients are:
2πk
ek = α + 2βcos
nC
1
2πi(r − 1)k
crk = √ exp[
]
nC
nC
nC
2πi(r − 1)k
1 X
exp[
φk = √
]fr
nC r=1
nC
Let us consider the Huckel energy level for monocyclic polyenes Cn Hn . Lowest
shell consists of nondegenerate level and holds 2 electrons all others are doubly
degenerate and holds 4 electrons. If, nC is even, the highest π MO is nondegenerate. Or else, there is unpaired electron which makes the system unstable.
So, all electrons must be paired. So, nπ = 4m + 2 where m is integer. This is
Huckel’s famous 4m + 2 rule.
4.3
Napthalene
We can extend this to more complex systems like napthalene where 2 benzene
rings are fused. For napthalene, secular equation is difficult to solve, as we get
a 10 × 10 matrix. We identify te point group of napthalene.(D2h )
10
Figure 7: Napthalene
So, we write the MO waves in terms of the AOs. Exploiting the symmetry,
we get a 3×3 matrix whixh is much more easier to solve.So, in conjugated
hydrocarbons,
1
|β|
=
∆x
λ
h̄c
where ∆x is difference in x values. Now, observed λ1 is 34700 cm−1 . The error
reduces to 14.4%. In fact by the Pariser-Parr-Pople Method, which takes intoi
accout the electronic energy too, can account very nicely for the λ value and
thus we can get more closer to the actual system by approximation.
11
5
Citations
1. Ira.N.Levine, Quantum Chemistry (Prentice Hall Press)
2. James.E.House, Fundamentals of Quantum Chemistry (Elsevier Academic
press)
3. Lowe & Peterson Quantum Chemistry(Elsevier Academic press)
4. Atkins and Friedman, Molecular Quantum Mechanics (Oxford University
Press)
5. Pauling and Wilson, Introduction to Quantum Mechanics
6. www.wikipedia.org
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