Добавил:
researchgate.net Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:

Yuriy Kruglyak. Quantum Chemistry_Kiev_1963-1991

.pdf
Скачиваний:
28
Добавлен:
28.02.2018
Размер:
16.43 Mб
Скачать

resonance integral of a bond (weakening of a stronger bond or strengthening of a weaker bond) leads to two local states appearing in the forbidden zone.

The recent theoretical results [111, 131, 134] provide an evidence in favor of the electron-correlation nature of the polyene-spectrum gap. But it appears likely that the question still remains doubtful (see, e.g., [135 – 137]). The above mentioned differences in the properties of local states can be used to study experimentally whether the energy gap is due to electron correlations or its appearance is a consequence of the bond alternation.

The results obtained so far seem to be useful in the study of the following question. In contrast to polyenes, the first optical transition frequency in the symmetric cyanide dyes tends to zero when the conjugated chain of the dye is lengthened [138]. Nevertheless, the long conjugated chains of cyanide dyes and polyenes differ by their end groups only. Then, it is natural to correlate the above difference in the optical spectra of these two classes of molecules with the effect of nitrogen atoms of the end groups of cyanide dyes. Indeed, the insertion of nitrogen atoms into the polyene chain can give rise to a local state near the bottom of an empty zone. As a consequence, the first optical transition corresponds to the transition of an electron from this local level to an empty zone. The energy of this transition is small for long chains. Then, the extrapolation of experimental data can give zero value (or nearly zero value) of the first transition frequency. Let us also note that the

conjugated chains of cyanide dyes consist of an odd number of atoms N. But,

the

number of π-electrons Ne is even: Ne = N ±1. If Ne = N +1 then the local

state

considered above is occupied in the ground state. If Ne = N 1 then there is a hole in a valence zone of cyanide dye and the explanation of optical experiments is trivial.

5.3. Appendix

We first deal with the derivation of main relations used in § 5.1, namely, will consider the sum in (172):

(i)

|Ck(σj)(v) |2

 

4

 

sin2kv[εq(i)

+aτσ (1)v ]

 

 

1

G σ (v,v;zqσ )

 

=

 

 

 

 

 

 

 

 

 

 

 

 

k, j εk( j) zq(σi)

N k

 

 

 

2π

 

(i) dεq(i)

N

0

 

2

2

 

(i)

 

 

 

 

 

 

εk

εq

 

Θqσ

εq

 

 

 

 

(A1)

 

 

 

 

 

N

dq

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[εq(i) +aτσ (1)v ]S(i)(q,σ),

where we have used (173). To calculate S(i) (q,σ) we shall use the method developed by Lifshits [20, 21]. Let us denote

S(i)(q,σ) = S(i)(q,σ) + S(i)(q,σ)

(A2)

1

2

 

and evaluate each sum separately, namely:

490

 

(i)

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

2π Θ(qiσ) (dεq(i)

/dq)εq(i) sin2kv

 

 

 

 

2

 

 

 

 

sin2qv

 

 

 

 

 

S1

(q,σ) =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2π (i)

dεq(i)

π

 

 

 

 

 

 

 

 

 

 

 

 

 

(i)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kq

 

 

 

 

2

2

2

 

 

2

 

 

 

 

 

 

(i)

 

 

(i) dεq

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(εk

εq )

 

εk εq

 

 

Θqσ

 

 

 

 

 

 

 

 

 

 

Θ

qσ

ε

q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dq

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N

dq

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2sin2qv

 

2

 

 

 

 

 

 

 

π Θ(qiσ)

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ε

(i)

(dε

(i)

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

π

 

 

(i)

 

 

 

 

(i)

 

 

 

 

 

 

 

 

 

q

 

q

 

/dq) N

 

 

k

q

(k

 

q)

 

 

 

 

 

 

 

 

 

π Θ

qσ

 

 

 

 

 

N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k q

N

Θqσ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2sin2qv

 

 

 

 

 

 

 

π Θ(qiσ)

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π Θ(qiσ)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= −2sin

 

qvctg

 

 

 

 

 

 

 

 

 

 

ε

(i)

(dε

(i)

 

 

 

 

 

 

 

nπ(nπ

 

 

 

 

(i)

)

π

 

(i)

 

ε

(i)

(dε

(i)

/dq)

 

 

q

 

q

/dq)

0

 

π Θ

qσ

 

Θ

qσ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

q

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(i)

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

sin2kv

 

 

 

1

 

 

 

 

1coskv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

S2

 

(q,σ) =

 

 

kq εk2 εq2

=

 

 

C

 

dk

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

N

2πβ2

cosk cos2q

 

N

 

 

 

 

,

(A3)

(A4)

where denotes the principal value of a corresponding contour integral taken from

C

0 to π . In order to evaluate (A4) we need to calculate

I

 

 

coskv

 

dk = I1

+ I2,

cosk cos2q

 

 

 

 

 

C

 

 

 

 

 

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

I1 =

1

dx

 

 

 

eivx

,

 

2

cos x cosq

 

 

 

1

C

eivx

 

 

 

I2 =

dx

 

 

.

 

2

 

 

cos x cosq

 

 

 

 

C

 

 

 

 

The integrals (A6) can be evaluated by the residue theory. The integral taken along the contour C1, and I2 – along contour C2 (fig. 4). Calculations give

I = iπ res

eivz eivz

|z=q =π sinvq .

cos z cosq

2

sin q

(A5)

(A6)

I1 is

(A7)

Fig. 4. The contours for the evaluation of integrals (A6).

491

The substitution of (A7) into (A5) and (A4) results in the relation

S2(i)(q,σ) = − 2β1 2 sin2sin2vqq .

Equation (174) can be obtained from (A3), (A8), (A1), and (172). The eigenfunctions of (170) are defined as [20, 21]

 

 

 

 

ϕq(σi) (µ) = −tτq(σi)

C( j)(µ)C( j)(ν)

.

 

 

 

 

 

 

 

 

 

 

kσ

( j)

k(σi)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k,

j

 

εk

 

zqσ

 

 

 

 

 

 

 

 

The sum in (A9) is calculated just like as S1(i) (q,σ) .

 

 

 

 

 

 

 

 

Let us evaluate a normalization constant τq(σi) , namely:

 

 

 

 

 

 

 

N

 

 

 

(i)

 

 

 

 

 

 

 

2

2

 

 

 

 

 

π

 

 

2

 

1

 

(i)

 

 

 

 

 

 

 

(i)

 

 

 

 

 

(i)

 

2

Cqσ (ν)sin qν

 

 

 

 

 

 

[ϕqσ

(µ)]

=

 

 

 

 

 

 

tτqσ

 

 

 

k q

 

 

Θqσ

 

 

 

 

dε

(i)

/dq

 

N

N

 

 

µ=

 

 

 

q

 

 

 

 

k

 

 

 

 

 

 

 

N

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(i)

(ν)sin qν

 

 

2

 

 

 

 

 

 

 

 

 

 

 

(i)

 

2

 

C

 

(i)

 

 

 

 

 

 

 

 

 

 

 

 

qσ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 2N(tτqσ )

 

 

 

tτqσ

 

=1.

 

 

 

 

 

 

 

 

 

(dε(i)/dq)sinπ Θ(i)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

 

 

 

qσ

 

 

 

 

 

 

 

Substituting τq(σi) from (A10) into (A9) one obtains (176 – 178). It follows from (A10) that

 

 

 

C( j)(ν)

2

d z(iσ)

1

d

 

[C( j)(ν)]2

 

1

d z(iσ)

1

(tτq(σi) )2 =

 

 

 

kσ

 

=

q

 

kσ

 

= −

 

q

.

k, j

 

 

dt

dt k, j

z(i)

t2

dt

 

 

ε( j) z(i)

 

 

ε( j)

 

 

 

 

 

 

k

qσ

 

 

 

 

 

k

qσ

 

 

 

 

 

(A8)

(A9)

(A10)

(A11)

Taking also into account that according with (A9) and (A10) ϕq(σi) (ν) =τq(σi) , one obtains

(180).

Now let us consider functions G0σ (v, µ; z) , where

 

 

 

| z |

 

(d,

 

 

),

 

 

 

 

 

1+d 2

 

 

 

 

2 | β |

 

 

 

 

 

 

 

 

 

i.e., for states splitting off zones. Using (164) and (165) one obtains

 

[Ck(σj)(ν)]2

 

z +

(1)ν dτσ π

 

G0σ (v,ν;z) = −k, j

 

 

=

 

 

 

 

0 dk (1cosνk) / (α +cosk)

(A12)

εk( j) z

 

 

π | β |

 

where

 

 

 

 

 

 

 

 

z = z / 2 | β |, d = a / 2 | β |,

α =1+2(d 2 z2 ).

 

492

The integral in (A12) is calculated as the integral (A5) except the poles of the

integrand

are in the

complex

plane k

on the

lines

Re k = 0

(z2 >1+ d 2 ) and

Re k = π

(| z | < d ) . Having carried out the calculations one obtains

 

 

π

 

cosνk

(1)ν

πeνq0 / sh q , (α > 0)

 

 

dk

 

 

 

 

 

0

 

(A13)

 

 

 

=

 

νQ0

 

 

 

 

α +cosk

πe

/ shQ ,

(α < 0)

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ch q =1+2(d 2 z2 ),

 

(z2 < d 2 )

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

chQ = 2(z2

d 2 ) 1. (z2 >1+d 2 )

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

Using (A13) one can calculate all functions G0σ (v, µ; z)

with z2

>1+ d 2 or z2 < d 2 .

In particular, one can obtain equations for local energies

 

 

 

 

 

1tG0σ (v,ν;zpz ) = 0

 

 

(A14)

and for corresponding functions

 

 

 

 

 

 

 

 

 

 

 

ϕpσ (µ) = tτpσG0σ (µ,ν;zpz ) .

 

(A15)

The relations (191), (194), (196) – (200) | λ | 1, then it follows from (A13) and (A14) that

results from (A14) and (A15). If q0Q0 1. Using (191) and (178) one

can see that if | λ | 1 then

τ2σ t2

(1e2q )sh2q, (q = q ,Q )

p

0

0

hence

|ϕpσ (µ) |2 λ2 .

(A16)

Finally we turn now to cumulenes which have two orthogonal π-systems, as compared with polyenes, and will end with the thorough discussion of the physical nature of the forbidden zone in quasi-one-dimensional electron systems.

6. Basics of π-Electron Model of Cumulenes

Cumulene molecules have the general formula H2C=(С=)N-2СH2 and contain a linear chain of N carbon atoms. The inner N – 2 atoms are characterized by diagonal hybridization sp and are in the valence state didiπxπy. Hybridization of the end-C- atoms should be close to trigonal sp2, and these atoms can be in valence state trtrtrπx or trtrtrπy. Properties of cumulenes are discussed in several reviews [139 – 142]. Even cumulenes (EC) with the ethylene as the first member of ECs are known to be planar with symmetry D2h. In odd cumulenes (OC) with the allene [143] as the first

493

member of OCs the two end-groups are perpendicular to one another with symmetry D2d. Both experimental facts are in accordance to valence bond theory.

The ease of cis-trans isomerization for the ECs or of stereoisomerization for the OCs is determined by the barrier height of internal rotation of the CH2 end-groups. Rotation of one of the CH2 groups by 180returns the cumulene molecule to its initial state. It is a natural suggestion that the barrier height is determined by the energy of such a molecular conformation in which one of the CH2 groups is turned by 90in comparison with the most stable conformation. In the following under barrier height V we shall imply the difference between energies of the lowest singlet states of the molecular conformations with symmetry D2h and D2d.

The barriers V in cumulenes were considered theoretically in [144, 145, 8, 9]. Popov [145] used a simple Huckel method which leads to the conclusion that with an increase of the number of C atoms the barrier tends to zero which is actually simply obvious from physical point of view. σ-Bonds of cumulene chains have cylindrical symmetry and their energy does not depend upon the angle of rotation of the endgroups. Therefore if direct interaction of the end-groups is neglected the barrier height is determined by the energy change of the π-electrons with the change of the molecular conformation.

Cumulenes CNH4 have 2N – 2 π-electrons. In accordance with the simple MO theory 2N – 2 levels can contain either N – 1 bonding levels and equally many antibonding levels in ECs or N – 2 bonding and equally many antibonding levels plus 2 nonbonding levels in OCs. In the former 2N – 2 π-electrons occupy all N – 1 bonding levels; in the later – all N – 2 bonding levels and the two remaining electrons occupy the two-fold degenerate nonbonding level. The first distribution is energetically more favorable than the second one. This is achieved for even N for planar conformations and for odd N for twisted conformations. This may be considered as a simple explanation of the known experimental fact [142] that the stable conformation of the ECs is planar, but that of the OCs is twisted with perpendicular arrangements of planes of the end-groups. This very interesting property of the cumulenes was in fact first explained by van’t Hoff [146] in 1877 using the tetrahedral model of the carbon atom.

Let us choose the coordinate system in a way so that in the conformation D2h π- AOs of the subsystem with N AOs are directed along x-axis and with N – 2 AOs – along y-axis. The z-axis passes through the C atoms. Conformation D2d is formed by a rotation of one of the end-AO’s by 90. In this case the number of AOs which are directed along the x- and y-axis equals N – 1 in both cases.

In the conformation D2h πx-states have symmetry b2g and b3u, and πy-states – b2u and b3g. In the conformation D2d all π-MOs transform according to the irreducible representation e. Therefore in this conformation the frontier MOs (pair of nonbonding

494

orbitals) is degenerated by symmetry. Accidental degeneration of the frontiers MOs in the conformation D2h remains in the Pariser – Parr – Pople (PPP) [147, 148] approximation also, for in this case zero differential overlap approximation is used. It is removed by alternation of the bond lengths.

The lowest electronic configuration of the cumulene molecule in its unstable conformation has a multiplet structure with states 3A2, 1B1, 1A1, and 1B2 for ECs and 3Au, 1Au, 1Ag, and 1A'g for OCs. We shall see later that when electronic interaction is accounted for the lowest states become 3A2, 1B1, resp. 3Au, 1Au. The states 1A1, 1B2, resp. 1Ag, 1A'g correspond to electron transfer between the perpendicular x- and y- subsystems of π-AOs. The molecule in its stable conformation, which is 1Ag for ECs and 1A1 for OCs has a closed shell. The degeneration of the frontier π-MOs is removed for inorganic cumulenes with alternating atoms of different electronegativity. To a smaller degree the same is true if the difference in the hybridization between the parameters of inner and outer C atoms is taken into account. But even in this case the lowest singlet state may be 1Au if the orbital energy splitting does not exceed the splitting of even and odd states.

In the following we shall neglect the difference in hybridization between outer and inner C atoms. This approximation is sufficiently good because the integrals for sp2 and sp states are almost equal [149].

Let us the x- and y-MOs in the conformation D2h write down as a linear combination of the π-AOs xν and yν with the chain of AOs yν denoted by primed symbols

ϕi = Cνi xν , ϕi′= Cνi yν

ν

ν

The summation is extended over all AOs of the chain. In the same manner it is possible to set up the components of the degenerate pairs of the MOs in the conformation D2d.

Let Aˆi+ be the creation operator for an electron i of orbital state ϕi and spin state α , and Aˆi+ be the same for spin state β . Degenerate orbital pairs of open shell will be denoted by the symbols k and k, and orbitals of closed shell by j and j. Then the wave functions of states with closed shell Ψc may be written as

Ψc (1A ,1A ) ≡ Ψc,

 

 

 

 

 

 

1

g

 

 

 

 

,

Ψ

c

 

ˆ+ ˆ+

ˆ+

ˆ+

 

0

 

 

 

= Aj Aj

Aj

Aj

 

 

 

 

 

j

j

 

 

 

 

where 0 is the vacuum state.

Wave functions of states with open shell Ψo will be written as follows:

495

o

 

3

 

 

3

 

 

 

 

1

 

 

 

ˆ+ ˆ+

ˆ+

ˆ+

c

 

Ψ

(

 

A2,

 

 

 

Au ) =

 

 

 

 

 

 

 

(Ak

Ak

+ Ak

Ak

)Ψ

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

o 1

 

1

 

 

 

 

 

 

1

 

 

 

ˆ+

ˆ+

ˆ+

ˆ+

c

 

 

Ψ

(

B1,

 

 

Au ) =

 

 

 

 

 

 

 

(AkAk

AkAk )Ψ

,

 

 

 

 

 

 

 

 

 

 

 

 

2

 

o 1

 

 

1

 

 

 

 

 

1

 

 

 

ˆ+

ˆ+

ˆ+ ˆ+

c

 

Ψ

(

B2,

 

 

Ag ) =

 

 

 

 

 

 

 

(Ak

Ak

Ak

Ak)Ψ

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

o

1

 

1

 

 

 

 

 

 

1

 

 

 

ˆ+

ˆ+

ˆ+

ˆ+

c

 

 

Ψ

(

 

A1,

 

 

Ag) =

 

 

 

 

 

 

 

(Ak Ak

+ AkAk)Ψ

.

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

For these states the z-component of the total spin MS = 0 . Two other components of the triplet state 3 A2 or 3 Au with MS = ±1 are described by the functions

Aˆk+Aˆk+Ψc and Aˆk+Aˆk+Ψc .

Let us introduce the standard notations:

Hk = ϕk H coreϕk dτ,

Jij = ϕi ϕj r1 ϕiϕjdτ1dτ2,

12

Kij = ϕi ϕj r1 ϕiϕjdτ1dτ2.

12

Then the energy of states with closed shell will be:

Ec (1 A1, 1 Ag ) = 2H j +2H j

+(2J j1 j2

K j1 j2 ) + (4J j1 j2

2K j1 j2) + (2J j1j2K j1j2) + Ecore ,

j

j

j1 j2

j1 j2

j1j2

where Ecore is the core total energy. If we denote

 

E1 = Ec + Hk + Hk

+ (2J jk K jk

+2J jkK jk) + (2J jk K jk +2J jkK jk) ,

 

j

 

 

j

where Ec means an expression which has the same structure as Ec (1 A1, 1 Ag ) above, the

sums being taken over the closed shell only, the energy of the states with open shell are:

Eo (3A , 3A ) = E + J

kk

 

K

kk

,

 

 

 

 

 

 

2

u

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eo (1B ,1A ) = E + J

kk

+ K

kk

,

 

 

 

 

 

 

1

u

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eo (1B

,1A ) = E

+

1

(J

kk

+ J

kk

) K

kk

,

2

g

1

 

2

 

 

 

 

 

 

 

 

 

 

Eo (1A ,1A) = E

+

1

(J

kk

+ J

kk

) + K

kk

.

1

g

1

 

2

 

 

 

 

 

 

 

 

 

Usually

496

Pµνo

Jij < 12 (Jii + J jj )

holds. This means that among the lower singlet states the lowest are 1B1 and 1 Au . Reducing the MOs to AOs the integrals over the AOs

κλ µν = xκ (1)xµ (2) r1 xλ (1)xν (2)dτ1dτ2

12

will have to be calculated. Zero differential overlap

κλ µν = κλ µν =δκλδµν κκ µµ =δκλδµνγκµ will be used in this context.

Core integrals Hµν with µ ν will be accounted for only in case of neighbouring atoms and renamed βµν (βµµ 0) . Integrals between AOs πx and πy Hµν

are zero for symmetry reasons. Integrals

Hµµ will be calculated in the Goeppert-

Mayer and Sklar approximation [150], neglecting penetration integrals

Hµµ = −Iµ γµν γµν+γµµ,

ν

ν

Hµµ= −Iµ γ

µν γµν+γµµ.

ν

ν

Here Iµ is ionization potential of π-electron in the corresponding valence state and in the outer field of neighbouring neutral atoms. It is obvious that Iµ= Iµ as well as γµµ= γµµ . The summation runs over all AOs πx resp. πy .

Let us introduce the following notations for density matrix elements in AO representation:

Pµνc = Cµ jCν j , Pµνo = CµkCνk , PµνT = 2Pµνc + Pµνo , j

and analogous expressions for the primed densities. For the states with closed shell is equal to zero.

Using these notations and under the assumption of the approximations mentioned above we obtain

J jk =Pµµc Pννo γµν , j µν

K jk =Pµνc Pµνo γµν ,

j

µν

Jkk

= PµoµPννo γµν .

 

µν

497

In the zero differential overlap approximation all exchange integrals of the type Kijare zero. When the necessary substitutions are done we get the following

expressions for the energy of states with closed shell:

Ec (1A1,1Ag ) = (γνν Iν )PννT +(γννIν)PνTν

 

 

ν

 

 

 

 

ν

 

 

 

 

 

 

1 PµµT

PννT

PννT

1 (PµνT

)2 γµν + PµνT

 

+

βµν

µν

 

2

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

µν+ PµTνβ

+ 1 PµTµPνTνPνTν1 (PµTν)2 γ

µν

 

 

2

 

 

4

 

 

 

′′

 

 

 

 

 

 

+(PµµT PνTνPνTνPµµT )γµνµν

. (201)

µν

Further simplifications will follow if we take into account that for alternant

hydrocarbons it holds that PT = PT′ ′ =1[151]. This is also true for the SCF method in

νν ν ν

the PPP approximation, which is assumed, if the ionization potentials and integrals are put equal for all C atoms [148, 152] including the end-atoms:

Iν = IνI, γνν =γννγ .

This assumption seems to be not far from the truth for organic cumulenes.

If the alternant properties of cumulenes are taken into account then the energy of the states with closed shell can be divided up as follows:

Ec (1A ,1A ) = Ec

+ Ec + E

 

+ Ecore ,

 

 

 

 

 

1

g

x

y

 

int

 

 

 

 

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

µν

1 PµνT

 

2

 

 

 

Exc = (γνν Iν ) +PµνT

βµν

γ

 

γµν

,

(202a)

ν

µν

 

 

2

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

γµν1 PµTν

2

 

 

 

 

Eyc = (γννIν) + PµTνβµν

 

γµν

 

,

(202b)

ν

µν

2

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eint = −γµν.

µν

The energy Exc represents the π-electron energy of a hypothetical compound with the same space structure as the corresponding cumulene with closed shell but having only one system of AOs of the type πx . The same is true for the energy Eyc . Eint represents the energy of the static electron interaction of the two chains and does not depend upon the MO coefficients.

498

Analogous transformations for the states with open shell 1B1 and 1 Au lead to the following result:

 

Eo (1B ,1A ) = Eo

+ Eo + E

+ Ecore ,

 

 

 

 

 

 

 

1

u

x

y

int

 

 

 

 

 

 

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

γµν 1 PµνT

 

2

1 Pµνo

 

2

 

 

 

Exo = (γνν Iν ) +PµνT

βµν

γµν

 

γµν

,

(203a)

ν

µν

 

2

 

2

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eoy = (γννIν) +

 

PµTνβµν

ν

µν

1γ

µν

1 PT 2

γ

µν

1 Po 2

γ

 

2

 

2

µν

 

 

2

µν

 

 

 

 

 

 

 

 

 

 

µν. (203b)

As we see, division into two chains is possible also in this case, but now each chain is in a doublet state and has an open shell structure as in organic free radicals.

However, for the open shell states 1 A1 , 1B2 , 1 Ag , and 1 Agdivision of the π-electron

system in two subsystems is not possible despite of the fact that rule PT = PT′ ′ is

νν ν ν

satisfied.

The energy Eint is not the same for different cumulene conformations. A simple consideration yields

Eint (D2d ) Eint (D2h ) = −γαω,

where α and ω are the indices of the end-atoms.

Let us note one incorrectness of the Goeppert-Mayer and Sklar approximation [150] when one calculates the interaction energy of positive core charges ED . In fact,

if we try to find ED in this approximation by the method of Dewar and Gleicher [153]

ED = γµν,

µ<ν

where the summation is taken over all AOs of the two chains, one gets different interaction energies for different conformations:

ED (D2d ) ED (D2h ) =γαωγαω .

However on physical grounds the interaction energies of positive charges in different core conformations of cumulenes can not be different. These differences are small, of course, and decrease rapidly with increasing chain length.

If one accepts the differences mentioned then the barrier height V may be found from the relation

V = Ex (D2d ) + Ey (D2d ) Ex (D2h ) Ey (D2h ) γαω .

(204)

The last term will then result from compensations of charges of Eint

and Ecore .

499

Соседние файлы в предмете Квантовая химия