8-8 Charge Distributions from HMOs
Chapter 8 The Simple H ¨uckel Method and Applications system display more and more nodal planes. Notice that the MOs φ2 and φ have the
3
same number of nodal planes (one, not counting the one in the molecular plane) but that these planes are perpendicular to each other. This is a common feature of some degenerate, orthogonal MOs in cyclic molecules.
It is important to notice the symmetry characteristics of these MOs. φ1 is either symmetric or antisymmetric for every symmetry operation of the molecule. (It is antisymmetric for reflection through the molecular plane, symmetric for rotation about the threefold axis, etc.) This must be so for any nondegenerate MO. But the degenerate MOs φ2 and φ are neither symmetric nor antisymmetric for certain operations. (φ
3
2 is antisymmetric for reflection through the plane indicated by the dashed line in Fig. 8-7, but is neither symmetric nor antisymmetric for rotation about the threefold axis by 120◦.) In fact, one can easily show that, given a cycle with an odd number of centers, each with one AO of a common type, there is but one way to combine the AOs (to form a real MO) so that the result is symmetric or antisymmetric for all rotations and reflections of the cycle. Hence, an HMO calculation for a three-, five, seven-, . . . membered ring can give only one nondegenerate MO. However, for a cycle containing an even number of centers, the analogous argument shows that two nondegenerate MOs exist.
8-8 Charge Distributions from HMOs
Now that we have a method that provides us with orbitals and orbital energies, it should be possible to get information about the way the π -electron charge is distributed in the system by squaring the total wavefunction ψπ . In the case of the neutral allyl radical, we have (taking ψπ to be a simple product of MOs) ψπ = φ1(1)φ1(2)φ2(3)
(8-42)
Hence, the probability for simultaneously finding electron 1 in dv(1), electron 2 in dv(2) and electron 3 in dv(3) is
ψ2 π (1, 2, 3)d v(1)d v(2)d v(3) = φ2 1 (1)φ2 1 (2)φ2 2 (3)d v(1)d v(2)d v(3)
(8-43)
For most physical properties of interest, we need to know the probability for finding an
electron in a three-dimensional volume element dv. Since the probability for finding an electron in dv is the sum of the probabilities for finding each electron there, the one-electron density function ρ for the allyl radical is ρ = 2φ2 +
1
φ22
(8-44)
where we have suppressed the index for the electron. If we integrate ρ over all space, we obtain a value of three. This means we are certain of finding a total π charge corresponding to three π electrons in the system.
To find out how the π charge is distributed in the molecule, let us express ρ in terms of AOs. First, we write φ2 and φ2 separately:
1
2
φ2 = 1 + 1 + 1 + 1 √
1
χ 2
χ 2
χ 2 χ1χ2 + 1 √ χ2χ3 + 1χ1χ3 4 1
2 2
4 3
2
2
2
φ2 = 1 + 1 −
2
χ 2
χ 2
χ1χ3
(8-45)
2 1
2 3
Section 8-8 Charge Distributions from HMOs
If we were to integrate φ2 we would obtain
1
−→1
−→1
−→1
−→0
φ21 dv = 1 χ 2
χ 2
χ 2 χ1χ2 dv
4
1 dv + 1
2
2 dv + 1
4
3 dv + 1
√2
−→0
−→0
+ 1 √
χ2χ3 dv + 1 χ1χ3 dv
2
2 = 1 + 1 + 1 = 1
(8-46)
4
2
4
Thus, one electron in φ1 shows up, upon integration, as being “distributed” 1 at
4 carbon 1, 1 at carbon 2, and 1 at carbon 3. We say that the atomic π -electron densities
2
4
due to an electron in φ1 are 1 , 1 , 1 at C
4
2
4
1, C2, and C3, respectively, If we accumulate these figures for all the electrons, we arrive at a total π -electron density for each carbon.
For the allyl radical, Table 8-1 shows that each atom has a π -electron density of unity.
Generalizing this approach gives for the total π -electron density qi on atom i all MOs
qi =
nkc2ik
(8-47)
k
Here k is the MO index, cik is the coefficient for an AO on atom i in MO k, and nk, the “occupation number,” is the number of electrons (0, 1, or 2) in MO k. (In those rare cases where cik is complex, c2 in Eq. (8-47) must be replaced by c∗ c
ik ik ik .)
If we apply Eq. (8-47) to the cyclopropenyl radical, we encounter an ambiguity. If the unpaired electron is assumed to be in MO φ2 of Fig. 8-7, we obtain q1 = q2 = 7 , q .
6
3 = 4 6
On the other hand, if the unpaired electron is taken to be in φ, q , q . The
3
1 = q2 = 5
6
3 = 8 6
HMO method resolves this ambiguity by assuming that each of the degenerate MOs is occupied by half an electron. This has the effect of forcing the charge distribution to show the overall symmetry of the molecule. In this example, it follows that q1 = q2 = q3 = 1. The general rule is that, for purposes of calculating electron distributions, the electron occupation is averaged in any set of partially occupied, degenerate MOs.
TABLE 8-1 HMO π Electron Densities in the Allyl Radical Carbon atom
Electron
1
2
3
1 in φ
1
1
1
1
4
2
4
2 in φ
1
1
1
1
4
2
4
3 in φ
1
2
0
1
2
2
–
–
–
Sum
1
1
1
Chapter 8 The Simple H ¨uckel Method and ApplicationsFigure 8-8 When the equilateral structure is distorted by decreasing R12 and increasing R13, R23, the energies associated with φ1, φ2, φ shift as shown.
3
In actuality, the equilateral triangular structure for the cyclopropenyl radical is unsta ble, and therefore the above-described averaging process is only a theoretical idealization. It is fairly easy to see that a distortion from equilateral to isosceles form will affect the MO energies E1, E2, and E differently. In particular, a distortion of the
3
sort depicted in Fig. 8-8 would have little effect on E1 but would raise E2 (increased antibonding) and lower E (decreased antibonding and increased bonding). Thus, there
3
is good reason for the cyclopropenyl radical to be more stable in an isosceles rather than equilateral triangular form. This is an example of the Jahn–Teller theorem, which states, in effect, that a system having an odd number of electrons in degenerate MOs
will change its nuclear configuration in a way to remove the degeneracy.3 The preference of the cyclopropenyl radical for a shape less symmetrical than what we might have anticipated is frequently called Jahn–Teller distortion.4 Many times we are interested in comparing the π -electron distribution in the bonds instead of on the atoms. In the integrated expression (8-46) are cross terms that vanish under the HMO assumption of zero overlap. But the overlaps are not actually zero,
√
especially between AOs on nearest neighbors. Hence, we might view the factors 1/ 2 as indicating how much overlap charge is being placed in the C1–C2 and C2–C3 bonds by an electron in φ1. The C1–C3 bond is usually ignored because these atoms are not nearest neighbors and therefore have much smaller AO overlap. Since S12 = S23 = Sij for neighbors i and j in any π system (assuming equal bond distances), we need not include Sij , explicitly in our bond index. If we proceed in this manner, two electrons in 3Linear systems are exceptions to this rule. Problems are also encountered if there is an odd number of electrons and spin-orbit coupling is substantial. The reader should realize that the above statement of the theorem is a little misleading inasmuch as it makes it sound like the molecule finds itself in a symmetric geometry that produces denerate MOs and then “distorts” to a lower-energy geometry. It is actually we who have guessed a geometry that is too symmetric. When our calculations reveal that this results in degenerate orbital energies containing an odd number of electrons, we are alerted that we have erred in our assumption, and that the molecule is really in a less symmetric, lower energy geometry.
4See Salem [1, Chapter 8].