Variational principle

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Statement 1:

The complete set of exact eigenfunctions of H define an orthogonal complete basis set for the total space of wave functions.H is Hermitic. Let be Ya and Yb two normalized wavefunctions (associated with two different values Ea and Eb ). We have therefore (due to hermiticity)

< Ya |H| Yb > = Ea< Ya | Yb >

< Ya |H| Yb > = Eb < Ya | Yb >

Wherefrom (Ea - Eb) < Ya | Yb > = 0

and since EaEb

< Ya | Yb > = 0

If so, it is possible to express any function as a linear combination of the exact eigenfunctions, Yi.

demonstration:

Consequence:

Statement 2:

### Variational principle

The energy associated with Y a function is always above that of the eigenfunction of lowest energy: E0.Y is not a eigenfunction; it is associated with an energy that is an average energy (mean value)

A mean value is always intermediate relative to extreme:

Greater than the smallest!

> E0

Mean value
• If Y1 and Y2 are associated with the same eigenvalue o: O(aY1 +bY2)=o(aY1 +bY2)
• If not O(aY1 +bY2)=o1(aY1 )+o2(bY2)

we define = (a2o1+b2o2)/(a2+b2)

Dirac notations

Statement 2:

The energy associated with Y a function is always above that of the eigenfunction of lowest energy: E0.Ya and Yb are two eigenfunctions< Ya |H| = < Ya | Ea and |H| Yb > = Eb | Yb>

wherefrom < Ya |H| Yb > = Eadab

From statement 1, Y is a linear combination of Ya s

| Ya >= Saca| Ya >

let multiply the left by < Ya |; this leads to only one term ca= < Ya |Y >

and similarly c*a= < Y |Ya >

then | Y >= Sa< Ya |Y > | Ya >

### Variational principle

Statement 2, normalization:

Ya and Yb are two eigenfunctionsassociated to Ea and Eb

From statement 1, Y is a linear combination of Ya s

| Y >= Saca| Ya >

then

< Y | Y > =Sa,bca< Ya |Yb> cb

< Y | Y > = Sa,ca2< Ya |Ya> = Sa,ca2= 1

### Variational principle

Statement 2, Demonstration:

### Variational principle

< Y |H | Y > = Sa,bca < Ya |H | Yb > cb

< Y |H | Y > = Sa,bEa < Ya | Yb > cb

< Y |H | Y > = Sa,bEa cadab cb

< Y |H | Y > = Sa,Ea ca2 >Sa,E0 ca2 = E0

E > E0

An non-exact solution has always a higher energy than the lowest exact solution

Using Dirac notation

Ya and Yb are two eigenfunctions< Ya |H| = < Ya | Ea and |H| Yb > = Eb | Yb >

wherefrom < Ya |H| Yb > = Eada

From statement 1, Y is a linear combination of Ya s | Ya >= Saca| Ya >

Let multiplie the left by < Ya |; this leads to only one term ca= < Ya |Y >

and similarly c*a= < Y |Ya > then < Y | Y > = Sa,bca< Ya |Yb> cb

< Y | Y > = Sa,b< Ya |Y > < Ya |Yb> < Yb|Y > = Sa,b< Ya |Y > dab < Yb|Y >

< Y | Y > = Sa,|< Ya |Y >|2 =1

Demonstration < Y |H | Y > = Sa,b< Y | Ya > < Ya |H | Yb > < Y b|Y >

< Y |H | Y > = Sa,bEa < Y | Ya > < Ya | Yb > < Y b|Y >

< Y |H | Y > = Sa,bEa < Y | Ya > dab < Y b|Y >

< Y |H | Y > = Sa,Ea |< Ya |Y >|2 >Sa,E0|< Ya |Y >|2 = E0

< Y |H | Y > >E0

Variation principle

Given an approximate expression that depends on parameters, we must determine the parameters to minimize the energy.

Within LCAO, an MO is a linear combination of AOs: We have then to chose the coefficients so to minimize the energy.

Note that the variational principle is not restricted to cases of linear combinations.

Slater exponent for He

F = where Z* is a parameter

= - Z*2/2 u.a. = + Z*2/2 u.a.

=

using = Z* u.a.

= -Z* = -Z*2u.a.

Next slide

= Z* u.a.

demonstration

Slater exponent for He

F = where Z* is a parameter

Slater exponent for He

F = where Z* is a parameter

Search for the energy minimum

LCAO

Y is a linear combination of N fa s

| Y >= Saca| fa >

Atomic orbitals

Worse description than Y

an approximate function

Using parameters: ca

There are N coeffcients to determine

Let minimize E with respect to each of them

Therefore, there is a set of N equations, dE/dc = 0

This seems fine; however there is a problem; Which is the problem?

Lagrangian

There is a constraint due to normalization: the N coefficients are not independent.

We search the minimum of a function L of N+1 variables

Which “makes no difference with H”

L = - E [ - 1 ]

There is a set of linear expressions to derive

dL/dc = 0

Should be 0

Joseph Louis de Lagrange

French

1736 - 1831

Secular determinant

Solving | Hij-ESij| =0, we find the Eis.

This is a N degree equation of E to be solved.

From N AOs, we find N energies and N MOs.

We find simultaneously E and Y (eigenfunction and eigenvalues).

LCAO

Assuming no changein the AOs *, the MOs correspond to a unitary change of the basis set of AOs.

MOs are orthogonal and normalized.

Neglecting overlaps, Src2ir=1 for a MO, Yi

and Sic2ir=1 for a AO, fi ; the AOs are distributed among the MOs

* This is not the case for self consistent methods.

Sum over r atoms

for a given iorbital

Sum over iorbitals for a given r atom

Schrödinger equation for LCAO

HY = EY

H SiciFi = ESiciFi

Let multiply on the left by Fj, integrate and develop

Sici<|H |Fi > = ESici

Sici (Hij-E Sij ) = 0

This is the set of linear equation solved for

| Hij-ES ij |= 0 secular determinant

Erwin Rudolf Josef Alexander Schrödinger

Austrian

1887 –1961

The complete set of MOs are solution (not only the lowest).

Our first approach from the variational principle emphasized the solution of lowest energy

Every extreme (derivative=0) is a physical solution!

Hückel Theory

In the 1930's a theory was devised by Hückel to treat the p electrons of conjugated systems such as aromatic hydrocarbon systems, benzene and naphthalene.

Only p electron MO's are included because these determine the general properties of these molecules and the s electrons are ignored. This is referred to as sigma-pi separability.

The extended Hückel Theory introduced by Lipscomb and Hoffmann (1962) applies to all the electrons.

Erich Armand Arthur Joseph Hückel

German 1896-1980

Never awarded the Nobel prize!

Also known for Debye-Hückel theory

of electrolytic solutions

Hückel Theory

We consider only 2pZ orbitals

Two parameters:

The atomic 2p energy level:

E(2pZ) = a (~ -11.4 eV)

The resonance integral for adjacent 2p orbitals

Hij(C=C) = b (~ -3 eV)

E = a + x b  -x= (a – E)/ b

We chose defining b as negative

Sij = dij (the overlap is neglected).

Erich Armand Arthur Joseph Hückel

German 1896-1980

Never awarded the Nobel prize!

Also known for Debye-Hückel theory

of electrolytic solutions

SLATER KOSTER

Slater Koster for metal atoms : 9 orbitals s, p et d

Several overlaps

ss, sps, sds,

pps, ppp, pds, pdp,

dds, ddp, ddd.

Phys. Rev B 94 (1954) 1498

EHT Theory
• Only valence orbitals are described
• AOs are Slater orbitals
• Parameters are the Hii (atomic energy levels)
• The Slater exponents
• The Overlap, Sij, are rigorously calculated from the geometry and the AOs
• The resonance integrals, Hij, are derived from the Sij. Wolfsberg-Helmoltz formula

Hij = 1.75 (Hii+Hjj)/2 Sij

For linear and cyclic systems (with n atoms), general solutions exist.
• Linear:
• Cyclic:
• For linear polyenes the energy gap is given as:

Polyenes

Charles Alfred Coulson

(1910-1974), English

| Hij-ESij|= 0

1 means that the atoms 2 and 4 are connected, The determinant is symmetric relative to the diagonal

| Hij-ESij|= 0

This is an N degree equation of x

| Ei-x|= 0solving is a diagonalization problem

A set of first degree equations: E=Ei associated with Yi

The diagonalization is a unitary transformation

OMs are orthogonal thus Srcricrj= dij

If i=j, the sum of the coefficient’s squares for a given atom is 1 (normalization)

If i≠j, the sum of the products of coefficients is 0 (orthogonality)

The matrix of coefficients is unitary

wherefrom = Sicricsj= dij

If i=j, the sum of the coefficient’s squares for a given orbital is 1

AOs are distributed among all the MOs

If i≠j, the sum of the products of coefficients is 0

If all the orbitals were filled, there would be no interaction:

all the r-s bond indices should be zero.

C2H42 e in 2 orbitals as in H2We have solved the problem using symmetryand without solving the secular equation.

Inputs are what is in the secular determinant : a b and connectivity

Outputs are orbitals, energies, total energies, charges and bond indices

C2H4without overlap

N-1 linear equations; the last one is redundant.

C2H4including overlap 4e repulsion

E2 = (b -ES)2 E = ±(b -ES)

Mulliken

Overlap populations

2 OA interaction modifying a

-b2/ D

• (D/2 -E) (-D/2 -E) = b2 E2-(D2/4) = b2
• E2 = b2 +(D2/4)
• E = ±√[b2 +(D2/4)]
• If D = 0, E+ = b
• If b <
• E+ = (D/2) (1+4b2/ D2) 0.5
• E+ = (D/2) (1+2b2/ D2) = D/2 (1 +2b2/ D)
• E+ - E = b2/ D «2nd order Perturbation term

-D/2

D/2

b2/ D

The geometric mean of b and D/2

The topology is C1 bond C2, C2 bond to C3 and C3 bond to C4

A linear model contains the information with more symmetry (the topology does not distinguish between cis and trans)

Conservation of Orbital Symmetry

H C Longuet-Higgins E W Abrahamson

The Molecular orbitals are solution of the symmetry operators of the molecule.

MOs from different symmetry groups do not mix.

Hugh ChristopherLonguet-Higgins

1923-2004

Mind that symmetry 1 is for all the atoms

Not the reduced part

= 0

x2 - x - 1 = 0 x = (1 ± √5)/2 Golden numbers 1.618 and -0.618

Coefficients - (1 ± √5)/2 c1 + c2 = 0 and normalization c12 + c22 = ½

c = 0.3717 and 0.6015

-0.618

1.618

Golden ratio  Golden ratio conjugateF
Golden ratio

Golden ratio conjugate

a= √(b2+(2b)2) =b(1+√5)

a= b(1+√5)

b

2b

x2 = 1+x

The medial right triangle of this "golden" pyramid (see diagram), with sides              is interesting in its own right, demonstrating via the Pythagorean theorem the relationship or

Fibonacci recursion,irrational numbers (incommensurable)

Rabbit population, assuming that:

In the "zeroth" month, there is one pair of rabbit

In the first month, the first pair begets another pair

In the second month, both pairs of rabbits have another pair, and the first pair dies.

In the third month, the second pair and the new two pairs have a total of three new pairs, and the older second pair.

Spirale constructions,in nature,in music..Related ?The Doctrine of the Mean (中庸, py Zhōngyōng) is one of the Four Books, part of the Confucian canonical scriptures.
Pentagon, icosahedra

 =1.618

F=1/ =0.618

2

Recipe to build the same determinant

Express along the AOs and gather interactions with equivalent atoms

Mind that symmetry reduction makes the normalization incomplete.

Mind that symmetry 1 is for all the atoms

Not the reduced part

x2 + x - 1 = 0 x = - (1 ± √5)/2 Golden numbers 0.618 and -1.618

Coefficients (1 ± √5)/2 c1 + c2 = 0 and normalization c12 + c22 = ½

c = 0.3717 and 0.6015

0.618

-1.618

Recipe to build the same determinant

Express along the AOs and gather interactions with equivalent atoms

Mind that symmetry reduction makes the normalization incomplete.

Small 0.3717

Large 0.6015

Large 0.6015

Small 0.3717

Large 0.6015

Small 0.3717

The number of nodes increases, the amplitude is large at the middle for the extreme; it is large on the edges for the Frontier orbitals.

-.3717

-.6015

-1.618

0.3717

0.6015

0.6015

-.3717

-.618

0.6015

-.3717

-.6015

0.3717

0.618

-.3717

0.6015

0.6015

1.618

0.6015

0.3717

Bonding orders Ground state

0.894

0.894

0.447

No initial statement that distinguishes 1-2 from 2-3

C=C – C=C

Bonding orders Ground state

1.35Ả

1.35Ả

1.46Ả

No initial statement that distinguishes 1-2 from 2-3

Matches the Lewis formula and indicates the stabilization due to delocalization

C – C = C – C

Bonding orders Excited state

0.447

0.447

0.724

It is no use to iterate

The Hückel method is powerful since predictive.

It is possible to make the parameterization dependent on the differences in bond lengths choosing b12 =b34 larger than b23 or modifying a according to the charge.

This does not make the calculation predictive.

It is possible to make iterations (w technique) to make the results “self consistent”; this is not recommended. This does not allow controlling the parameters.

Annulenes

Secular equation:

b cr-1 -E cr + b cr+1 = 0

Annulenessolutions for CN

Rq (fr) = fr+s angle q = 2ps/N

The new coefficient for r is the old one multiplied by a constant

An annulene is a periodic system;

Solutions for annulenes are those of any periodic system: crystals (Bloch sums)

annulenesThe system allows solutions Rjq with angle (qj=2jp/N).

There is N rotations possible including identity (j=0 ou N) : the « simple » rotation q and the « multiple» rotations jq

Functions satisfying to rotations jq=2jp/N:

We build linear combinations of AOs,(Scrfr) satisfying rotation qj

(Scrfr) = e-ijqj (Scrfr) with cr (j) = N-1/2 eijrqj= e –i2pjr/N .

There are N solutions, with a quantum number: j.

.

i is the square root of -1; i2=-1

Rotation by 120° = 2p 2/6

(j=2, orbital y2)

C5xexp(-ijqj)=

N exp(i2p 10/6) x exp(-i2p 2/6) =exp(i2p 2/6)

N exp(i2p 4/6)

N exp(i2p 4/6)

N

N

N exp(i2p 2/6)

N

N exp(i2p 8/6)

N exp(i2p 8/6)

N

N exp(i2p 10/6)

N exp(i2p 10/6)

The 5th AO takes the place of the first

C5x exp(i2p 2/6)

annulenes

They are the LCAO functions satisfying the rotations qj

Yj= Scrfr

There are j solutions; the eigenvalues are e-ijsq

Benzene

annulenes

Two real solutions and 4 complexes

Degenerate by pairs

+

+

The complexe solutions are associated with rotations only; they are not eigenfunctions of the other symmetry operators and must be the degenerate

annulenes

Real solutions

Satisfying two mirror symmetry

annulenes

Orbitals from symmetry only

All the coefficients are equal c=1/2

E = 4*(< 1/2f2IHI1/2f3>+…)

= b

All the coefficients are equal c=1/√6

E=< 1/√6f1+.. .IHI1/√6f1+…>

= 12*(< 1/√6f1fIHI1/√6f2>+…) = 2b

annulenes

Orbitals from symmetry only

The total density in E orbital must respect symmetry

Degenerate orbitals: energy must be b

1/22+c2=2/6 → c= 1/√12

02+c2=2/6 → c= 1/√3

Normalization is satisfied 4*1/12+2*1/3=1

annulenes

Chosing another set of planes; rotation by 60°

Combining orbitals

YS’A’ = ½ [YSA + √3 YAS ]

YA’S’ = ½ [YSA - √3 YAS ]

New symmetry appears

While old ones desappears

Notations: x = ↔ E = a + x b units b (b <0) , origin a

f1f2f3f4

f1 -x 1 0 1

f2 1 -x 1 0 = 0

f3 0 1 -x 1

f4 1 0 1 -x

Two possibilities of using mirror symmetries

First set

1/2

1/2

1/2

1/2

Second set

1/2

1/√2

1/√2

1/2

The Jahn-Teller Theorem 1937

"any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy thereby removing the degeneracy"

There is a symmetry reduction (a geometry distortion) when the a set of HOMOs is not completely filled.

Hermann Arthur Jahn 1907-1979 English

Edward Teller 1908-2003

Hungarian-American

In an octahedral crystal field, the t2g orbitals occur at lower energy than the eg orbitals. The t2g are directed between bond axes while the eg point along bond axes.

4e-8e 0-6e 2e-8e

3e-5e-9e 3e-9e

The full octahedral symmetry takes place only when the t2g set is fully occupied.

C4H4 is unstable, it exists stabilized by asymmetric ligand (one eg orbital is stabilized).

It dimerizes easily.

Chain of Vn-benzenen+1: The HOMO gap vanishes for n=4; then there is a rotation of the successive benzene loosing the full D5h symmetry.

If the rings are eclipsed, there is no gap; the rotation of ring opens a gap.

Jahn-Teller effects are grouped into two categories.

The first arises from incomplete shells of degenerate orbitals. It includes the first-order Jahn-Teller effect and the pseudo Jahn-Teller effect.

The second arises from filled and empty molecular orbitals that are close in energy and is the second-order Jahn-Teller effect.

The two categories have quite different physical bases. As a result, geometric distortions produced by the first are quite small and normally lead to dynamic effects only.

In favorable cases, the second-order Jahn-Teller effect produces very large distortions, including complete dissociation of a molecule. This can occur even when the relevant molecular orbitals are separated in energy by as much as 4 eV.

The Jahn-Teller Theorem 1937

CuCl2 Cu2+ 9e

NiCl2 Ni2+ 8e

The shielding effect this has on the electrons is used to explain why the Jahn-Teller effect is generally only important for odd number occupancy of the eg level. The effect of Jahn-Teller distortions is best documented for Cu(II) complexes (with 3 electrons in the eg level) where the result is that most complexes are found to have elongation along the z-axis.

Apparent exceptions to the theorem are probably examples of what has been called the "dynamic Jahn-Teller effect". In these cases either the time frame of the measurement does not allow the distortion to be seen because of the molecule randomly undergoing movement or else the distortion is so small as to be negligible. For one of the copper complexes, the bond lengths are apparently identical. If the X-ray structure of the sample is redone at varying temperatures it is sometimes possible to "freeze" a molecule into a static position showing the distortions.
Alternant conjugated hydrocarbons

Definition: Atoms of conjugated molecules could be divided into two sets of atoms (starred and unstarred atoms) so that a member of one set is formally bonded only to members of the other set. Many compounds are alternant; they are not when their structure contains an odd-member ring.

Alternant conjugated hydrocarbons
• -E cr + Ss b cs = 0
• if Yj = Sr*crfr + Ss°csfs is an eigenfunction
• Y‘j = Sr*crfr - Ss°csfs is another one with Ej'=-Ej
• Orbital energy levels are symmetric relative to a level.
• If there is an odd number of orbital, one of them is nonbonding
Alternant conjugated hydrocarbons

Non bonding orbital, SOMO

The nonbonding orbital has non-zero coefficient only at the starred atoms, and the sum of the coefficient of the starred atoms attached to a given unstarred atom must equal zero.

The secular equation-E cr + becomesSs cs = 0 with E=0.

Applied on a starred atoms Ss°cs = 0 it gives again that cs = 0

Applied on a unstarred atoms Sr*cr = 0 it tells that the unstarred atoms belong to nodal planes.

It is easy to find the coefficients from there.

The SOMO is localized on the terminal atoms

Allyle anion

The charge is -1/2 on each terminal atom

Allyle cation

The charge is +1/2 on each terminal atom

species):

a = 1/√

a = -1/√

Dewar method for estimating aromaticity

Two bonds close the cycle

One bond makes the polyene

If the SOMO is symmetric, it is better to make 2 bonds:

If the SOMO is antisymmetric, it is better to make only one bond:

S or A

S 2n+1

A 2n-1

Polyene

S

Single

carbon

A

S

S

a

S

-a

4n+2 electrons aromatic

4n electrons antiaromatic

Dewar method for estimating aromaticity

Two bonds close the cycle

One bond makes the polyene

There is an energy gain

by forming a new cycle

The net charge of alternant hydrocarbons is zero.
• The electron density on r is 1
• | coeff2 of occupied MOs| = | coeff2 of unoccupied MOs |
• wherefrom Sini cri2=1
• demonstration :
• Socc cr,occ2 +Snb cr,nb2 +Svac cr,vacc2 =1
• 2 Socc cr,occ2 +Snb cr,nb2 = 1
• Sini cr,i2 = 1
The bond indices between atoms of the same set is zero.

(first order perturbation between atoms of the same set is zero).

OMs are orthogonal thus Srcricrj= dij

If i=j, the sum of the coefficient’s squares for a given atom is 1 (normalization)

If i≠j, the sum of the products of coefficients is 0 (orthogonality)

The matrix of coefficients is unitary

wherefrom = Sicricsj= dij

If i=j, the sum of the coefficient’s squares for a given orbital is 1

If i≠j, the sum of the products of coefficients is 0

• Socc cr,occcs,occ+Snb cr,nbcs,nb+vac cr,vacccs,vacc=drs
• 2 Socc cr,occcs,occ+Snb cr,nbcs,nb= 0
• Socc cr,occcs,occ+Snb cr,nbcs,nb= 0

Srnicricsj= 0

-.3717

-.6015

-1.618

0.3717

0.6015

0.6015

-.3717

-.618

0.6015

-.3717

-.6015

0.3717

0.618

-.3717

0.6015

0.6015

1.618

0.6015

0.3717

annulenes

Benzene is alternant

E = -2 b

E = -1

b

E = 1

b

b

E = 2

Importance of the non bonding orbital
• for radical species, it represents the unpaired electron
• For anions or cations it monitors the charge
• d= Sini cr,i2
• For anions d= 2 Socc cr,occ2 +2cr,nb2
• d= (2 Socc cr,occ2 +1cr,nb2 ) + cr,nb2
• d= 1 + cr,nb2
• For cations d= 2 Socc cr,occ2
• d= (2 Socc cr,occ2 +1cr,nb2 ) - cr,nb2
• d= 1 - cr,nb2
Heteroatoms:CO

x2-x-1=0

x= (1±√5)/2

O more electronegative

Antibonding, large amplitude on C

1 p electron for CO

Bonding, large amplitude on O

allyle

x3-2x=0

Solutions:

x=- √2

x=0

x= √2

Allyle coefficients

x=0

-xc1+c2=0 c2=0

c1-xc2+c3=0 c3=-c1 this orbital is found antisymm

c2-xc3=0 c2=0 no new information

x=√2

-√2 c1+c2=0 c2= √2 c1

c1-√2 c2+c3=0 c3 = c1 this orbital is found symm

c2-xc3=0 1/√2 c3= 1/√2 c1 = c2

These expressions have be normalized

1/2 -1/√2 1/2

Charges are controlled by the filling of the nonbonding MO.They appear on the terminal atoms.

The sum of charges is the global charge

-√2

0

√2

-1/√2 1/√2

1/2 1/√2 1/2

1/2 -1/√2 1/2

bond indices are independent from the filling of the nonbonding MO.

-√2

0

√2

-1/√2 1/√2

1/2 1/√2 1/2

Exercises

Find, using symmetry, the orbitals for

• the trimethylenemethane H2C=C(CH2)2

*Explain the origin of the solutions x=±1 for the pentadiene

*Explain the origin of the golden number solutions for the cyclopentadiene

Trimethylene-methane

Symmetry orbitals : MOs

F01/√2 [F0 +1/√3 (f1+ f2+f3)]

1/√3 (f1+ f2+f3)1/√2 [F0-1/√3 (f1+ f2+f3)]

1/√2 (f2-f3) 1/√2 (f2-f3)

√(2/3) f1+ 1√6 ( f2+f3) √(2/3) f1+ 1√6 ( f2+f3)

Energies 0 energies ± √3

}

Trimethylene-methane

1/√2 [F0 -1/√3 (f1+ f2+f3)]

√(2/3) f1+ 1√6 ( f2+f3) 1/√2 (f2-f3)

1/√2 [F0 +1/√3 (f1+ f2+f3)]

3

4

2

5

1

-x2(1-x)+2(1-x)+x=0 x3-x2-x+2=0

x=2 solution (x-2) (x2-x-1) x= (1±√5)/2

The antisymmetric solutions are those from butadiene

Pentagon, icosahedra

- =-1.618b

F=1/ =0.618b

2b

Pentagon, icosahedra

-1.618b

0.618b

2b

This level is bonding, preferably filled.

The anion is stable (aromatic)The cation is not (antiaromatic- Jahn-Teller situation)

Thorium+4

Exercise: naphtalene

1) Write the Huckel determinant for each symmetry group of the naphatlene.

2) Solve the SS secular determinant

3) Deduce from the SS energy levels those for AS

4) What must be the levels for SA and AA solutions ?

6) Explain from there what are the levels SS and AS at x=±1

naphtalene

SS -x(1-x) 2-3(1-x)=0 (1-x)(-x2+x+3)=0

x=1 solution (-x2+x+3)=0

x= (1±√13)/2x = 2.30277 and -1.30277

AS solutions have opposite energies (alternant)

-1, x=-2.30277 and +1.30277

The SA and AA solutions are the 4 MOs from butadiene

± 0.618 and ± 1.618

The levels at ±1 are duplication of the non bonding MO from the pentadienyle in-phase and out of-phase.

Exercises
• Calculate the energy for C3 and C5
• Linear or cyclic
• Positively charge and negatively charged.
• Calculate the MO energies for the biphenyle.
• Repeat the calculation assuming that the central bond is only half of a C=C bond.