Chapter 8 The Simple Hiickel Method and Applications Q 8-1 The Importance of Symmetry Our discussions of the particle in a box, the harmonic oscillator, the hydrogen atom, and homonuclear diatomic molecules have all included emphasis on the role that symmetry plays in determining the qualitative nature of the eigenfunctions. When we encounter larger systems, detailed and accurate solutions become much more difficult to perform and interpret, but symmetry continues to exert strong control over the solutions. In this chapter, we will describe a rather simple quantum chemical method that was formulated in the early 1930s by E. Hiickel. One of the strengths of this method is that, by virtue of its crudeness and simplicity, the effects of symmetry and topology on molecular characteristics are easily seen. Also, the simplicity of the model makes it an excellent pedagogical tool for illustrating many quantum chemical concepts, such as bond order, electron densities, and orbital energies. Finally, the method and some of its variants continue to be useful for certain research applications. Indeed, it is difficult to argue against the proposition that every graduate student of organic and inorganic chemistry should be acquainted with the Hiickel molecular orbital (HMO) method. Q| 8-2 The Assumption of rr—n Separability The simple Hiickel method was devised to treat electrons in unsaturated molecules like ethylene and benzene. By 1930 it was recognized that unsaturated hydrocarbons are chemically more reactive than are alkanes, and that their spectroscopic and them namic properties are different too. The available evidence suggested the existence nl loosely held electrons in unsaturated molecules. We have already seen that, when atoms combine to form a linear molecule, w< can distinguish between MOs of type a, n, S,... depending on whether the MOs associated with an m quantum number of 0, ± I, ±2, ... Thus, in acetylene (C> minimal basis set of AOs on carbon and hydrogen lead to a and n MOs. Let us imagui that our acetylene molecule is aligned along the z Cartesian axis. Then the Px71 AOs on the carbons are antisymmetric for reflection through a plane contain11 molecular axis and the y axis. Similarly, the />,.jr-type AOs arc antisymme reflection through a plane containing the molccuiar axis and the v axis. The which are cr-type functions, are symmetric for reflection through any plane 0011 244 Section 8-2 The Assumption of a-n Separability 245 the molecular axis. It has become standard practice to carry over the a—n terminology to planar (but nonlinear) molecules, where m is no longer a "good" quantum number. In this expanded usage, a n orbital is one that is antisymmetric for reflection through the plane of the molecule, a rr orbital being symmetric for that reflection. Hiickel found that, by treating only the n electrons explicitly, it is possible to reproduce theoretically many of the observed properties of unsaturated molecules such as the uniform C-C bond lengths of benzene, the high-energy barrier to internal rotation about double bonds, and the unusual chemical stability of benzene. Subsequent work by a large number of investigators has revealed many other useful correlations between experiment and this simple HMO method for7r electrons. Treating only the 7r electrons explicitly and ignoring the a electrons is clearly an approximation, yet it appears to work surprisingly well. Physically, Hiickel's approximation may be viewed as one that has the n electrons moving in a potential field due to the nuclei and a "a core," which is assumed to be frozen as the n electrons move about. Mathematically, the a—ix separability approximation is (8-1) whet, £ , taken to be the electromc energy Eei plus the intemuclear repulsion Let us consider the implications of Eq. (8-1). We have already seen (Chapter 5) molecule should be ofThe^ V' (1.....n) = ^r(l.....k)f„(k+ 1.....n) and our hamiltonian should be separable into n and a parts: ill,2.....n)=Jex{\,2.....k) + Jt„{k + 1,... Equations (8-2) and (8-3) lead immediately to Eq. (8-1): f^*f*^+^fnxfBdx(\.....n) n) (8-2) (8-3) ,1) lU£**1>*dr{\.....k) fraJr*+adr(k + I.....„) ^'ng^.Buttheeo a ' " ^ Md ™« by Eq. (8-3) We n 1 T ™ ^ beCaUSe "is imPossible«° rigorously satisfy °" ^ZoZZT: *' *• S° ,hat individua"y depend comple ,y ^ a stl"sum 10 the correct total hamiltonian- ^ |-iEf+E''-*,(i)4(iw>,(i¥T(i) = E, />*,(lW>/(l)rfr(l) It follows that the total it energy En is a sum of one-electron energies: B„ = Et+Ej-+-<- + E, (8-10) (8-11) This means that the it electrons are being treated as though they are independent of each other, since E, depends only on 0, and is not influenced by the presence or absence of an electron in 4>j- However, this cannot be correct because it electrons in fact interact strongly with each other. Once again, such interactions will be roughly accounted for in an implicit way by the HMO method. The implicit inclusion of interelectronic interactions is possible because we never actually write down a detailed expression for the it one-electron hamiltonian operator lin (i). (We cannot write it down because it results from a it-o separability assumption and an independent it -electron assumption, and both assumptions are incorrect.) Hn (i) is considered to be an "effective" one-electron operator—an operator that somehow includes the important physical interactions of the problem so that it can lead to a reasonably correct energy value E,. A key point is that the HMO method ultimately evaluates E, via parameters that are evaluated by appeal to experiment. Hence, it is a semiem-pirical method. Since the experimental numbers must include effects resulting from all the interelectronic interactions, it follows that these effects are implicitly included to some extent in the HMO method through its parameters. It was pointed out in Chapter 5 that, when the independent electron approximation [Eqs. (8-8)—(8-11)] is taken, all states belonging to the same configuration become degenerate. In other words, considerations of space-spin symmetry do not affect the energy in that approximation. Therefore, the HMO method can make no explicit use of spin orbitals or Slater determinants, and so is normally taken to be a single product function as in Eq. (8-8). The Pauli principle is provided for by assigning no more than two electrons to a single MO. EXAMPLE 8-1 If O2 were treated by the HMO method, what would be the form of the wavefunction and energy for the ground state? SOLUTION ► The ground state configuration for 02 is lajla^2crJ2cr^3a^lit^xllt%» x ljTgiXljTgfy, where we have shown the degenerate members of tt levels explicitly and in their real forms. The HMO wavefunction is simply a product of the pi MOs, one for each of the six pi electrons: l7rU]j(l)l^U|X(2)ljr„iJ,(3)ljr„,y(4)ljrg|X(5)ljrgi^(6). The HMO energy is 2Ejz,u,x + 2E„uy + ExgX + En,g,y, which reduces to 4 £V,u +2£jr,g. Note that, because O2 is linear, there is no unique molecular plane containing the internuclear axis. Therefore this molecule has two sets of it MOs, one pair pointing in the x direction, the other pair pointing along y. For a planar molecule, only one of these pairs would qualify as jr MOs, as will be seen in the next section. Q| 8-4 Setting up the Hiickel Determinant 8-4.A Identifying the Basis Atomic Orbitals and Constructing a Determinant The allyL radical, C3H5, is a planar molecule1 with three unsaturated carbon centers (see Fig. 8-1). The minimal basis set of AOs for this molecule consists of a Is AO on each hydrogen and Is, 2s, 2px, 2py, and 2pz AOs on each carbon. Of all these AOs only the 2pz AOs at the three carbons are antisymmetric for reflection through the molecular plane. Figure 8-1 ► Sketch of the nuclear framework for the allyl radical. All the nuclei arc coplanar. The z axis is taken to be perpendicular to the plane containing the nuclei. The minimum energy conformation of the allyl system is planar. We will ignore the deviations from planarity resulting from vibrational bending of the system. 248 Chapter 8 The Simple Huckel Method and Applications Section 8-4 Setting up the Hückel Determinant 249 Figure 8-2 ► The three «type AOs in the minimal al basis set of the allyl radical. Following Huckel, we ignore all the rr-type AOs and take the three 2pz AOs as our set of basis functions. Notice that this restricts us to the carbon atoms: the hydrogens are not treated explicitly in the simple HMO method. We label our three basis functions x\' x2> x3 as indicated in Fig. 8-2. We will assume these AOs to be normalized. Suppose that we now perform a linear variation calculation using this basis set. We know this will lead to a 3 x 3 determinant having roots that are MO energies which can be used to obtain MO coefficient. The detcrminanlal equation is -ESu Hn - ES[2 Hz\ - ES2\ H22 - ESn «31 - ESm #32 - ESv. II23 - ES23 = 0 where Sij = J XiXjdv (8-12) (8-13) (8-14) Since Hij and SU are integrals over electron mdex.s supPr-ed in Eqs. (8-13) and (8-14). the space coordinates of a single electron 8-4.B The Quantity a We have already indicated that there is no way to write an explicit expression for Hi that is both consistent with our separability assumplions and physically correct. Bui. without an expression for H„, how can we evaluate the integrals Hip. The HM' method sidesteps this problem by carrying certain of the H,, integrals along as symbo s until they can be evaluated empirically by matching theory with experiment. Let us first consider the integrals H\\, H22 and Hr<. The interpretation consisten with these integrals is that H\ \, for instance, is the average energy of an electron in X\ experiencing a potential held due lo the entire molecule. Symmetry require H) i = #33. II22 should be different since an electron in AO X2 experiences a di « environment than it does when in xi or Xi- 1' seems likely, however, that H22 is not very different from H\\. In each case, we expect the dominant part of the potential to arise from interactions with the local carbon atom, with more distant atoms playing a secondary role. Hence, one of the approximations made in the HMO method is that all Ha are identical if Xi is on a carbon atom. The symbol a is used for such integrals. Thus, for the example at hand, H\\ = H22 = W33 =a. The quantity a is often called the coulomb integral2 8-4.C The Quantity 0 Next, wc consider the resonance integrals or bond integrals II\2, H23, and Hy?,. (The requirement that H„ be hermitian plus the fact that the x's and ll„ are real suffices to make these equal to H2\, Hyi, and H$i, respectively.) The interpretation consistent with these integrals is that H\2, for instance, is the energy of the overlap charge between Xi and X2- Symmetry requires that H\2 = tin in the allyl system. However, even when symmetry docs not require it, the assumption is made that all //,, are equal to the same quantity (called fi) when i and j refer to "neighbors" (i.e., atoms connected by a a bond). It is further assumed that Htj = 0 when i and j are not neighbors. Therefore, in the allyl case, Hn-H23 = P, #13=0. 8-4.D Overlap Integrals Since the x's are normalized, Sa = 1. The overlaps between neighbors arc typically around 0.3. Nevertheless, in the HMO method, all S,; (i ^ j) are taken to be zero. Although this seems a fairly drastic approximation, it has been shown to have little effect on the qualitative nature of the solutions. 8-4.E Further Manipulation of the Determinant Oui determinantal equation for the allyl system is now much simplified. It is -E B 0 0 a — E ß ■-E = 0 (8-15) Dividing each row of ihe determinant by 6 corresponds to dividing the whole determinant by B3. This will not affect the equality. Letting (a - E)/fi=x, we obtain the result X 1 0 1 x 1 0 1 x = 0 (8-16) j ^nn coulomb integral" for a is unfortunate since the same name is used for repulsion integrals of the 'ii'e '''1 0*2' c'2- ancl C3> 1101 tne'r aDS°lute values. So we anticipate using 0 °^ 'hese equations and obtaining absolute values by satisfying the normality 252 Chapter 8 The Simple Hiickel Method and Applications condition. Because we are neglecting overlap between AOs, the latter step corresponds to requiring ci + ci+ci = \ (8-22) The roots x are, in order of increasing energy, first. Then — */2.C\ + C2 -V2,0, +V2. Let us take x = —Jl = 0 CJ = 0 V2c3=0 (8-23a) (8-23b) (8-23c) Comparing Eqs. (8-23a) and (8-23c) gives ci = C3. Equation (8-23a) gives ci — -J2c\. Inserting these relations into the normality equation (8-22) gives (v^e,)+c? = l (8-24) cf + l 4cf = 1, (8-25) It makes no difference which sign we choose for c\ since any wavefunction is equivalent to its negative. (Both give the same ^r2.) Choosing c\ =+| gives 1 1 1 2' C2 = ^' C3 = 2 These coefficients define our lowest-energy MO, A similar approach may be taken for x = 0 and x U=0) 1 1 2»+ ^» + 5» : +V2. The results are I :X1 (8-26) (8-27) (8-28) (8-29) The ally] system MOs are sketched in Fig. 8-5. The lowest-energy MO, c/>i, has no nodes (other than the molecular-plane node common to all it MOs) and is said to be bonding in the C\ - C2 and C2 - C3 regions. It is reasonable that such a bonding MO should have an energy wherein the bond-related term /S acts to lower the energy, as is true here. 'Hie second-lowest energy MO, 2fa dv = 4 J (Xi - XiKXi ~ Xi) dv = i{J^>-jx^°-J^° + J^°} = i (8-37) Since S ^ 0, fa and fa are nonorthogonal. We can project out that part of fa that is orthogonal to fa by using the Schmidt orthogonalization procedure described in Section 6-10. We seek a new function 3 = 03 - Sfa where Therefore, fafadv. ^=fa - X-fa = ~ ix\+X2-2x-i) This function is orthogonal to fa but is not normalized. Renormalizing gives (8-38) (8-39) (8-40) (I) Ei = « + 2ß (jc = +1) U = -2) For this MO, the coefficients are all of the same sign, and so the AOs show phase agreement across all bonds and all interactions are bonding. To find fa and fa is trickier. We begin by inserting x = ±1 into our simultaneous equations. This gives c\ + C2 + c3 = 0 (three times) $+4+4=1 (8-34) (8-35) With three unknowns and two equations, an infinite number of solutions is possible. Let us pick a convenient one: c, = -c2, c3 = 0. The normalization requirement then gives c\ = 1 /*h, c2 = -1 / V2, c3 = 0. Let us call this solution fa ■ (8-36) We still need to find fa. There remain an infinite number of possibilities, so let us pick one- ci = 1/y/2, c2 = 0, c3 = -1 /y/l. We have used our experience with fa to choose c's that guarantee a normalized fa. Also, it is clear that fa is linearly independent of 03 = (XI + XI ■ ■2X3) (8-41) In summary, to produce HMO coefficients for degenerate MOs, pick any two independent solutions from the infinite choice available, and orthogonalize one of them to the other using the Schmidt (or any other) orthogonalization procedure. The MOs for the cyclopropenyl system as seen from above the molecular plane are sketched in Fig. 8-7. The MO fa can be seen to have both antibonding (C1-C2) and nonbonding (C1-C3, C2-C3) interactions, fa^ has antibonding (C1-C3, C2-C3) and bonding (C1-C2) interactions. The interactions are of such size and number as to give an equal net energy value (a — 0) in each case. Since nodal planes produce antibonding or nonbonding situations, it is not surprising that higher and higher-energy HMOs in a O0 GO 00 (a) n'T/L8'7 * The™Os for the cyclopropenyl system: (»)*, =(l/v1)0fl +X2 + X1); (b)fc = lWZ)Ufi - X2) (c) i = (1/V6)0ti +X2- 2X3). The nodal planes intersect the molecular plane at the dashed lines. T 256 Chapter 8 The Simple Hiickel Method and Applications system display more and more nodal planes. Notice that the MOs ij>i and 4>'{ have the 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 Müs in cyclic molecules. It is important to notice the symmetry characteristics of these MOs. \ 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 ij>2 and are neither symmetric nor antisymmetric for certain operations. (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 realMO) 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. [j 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 n-electron charge is distributed in the system by squaring the total wavefunction In the case of the neutral ally I radical, we have (taking \jr„ to be a simple product of MOs) 0*=0i(l)0i(2)02(3) (8-42) Hence, the probability for simultaneously finding electron 1 in dv(\), electron 2 in dv(2) and electron 3 in dv(i) is 02(1,2, 3)dv([)dv(2)dv{3) = ^(1)^(2)^(3)^(1)^(2)^(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 p for the allyl radical is (8-44) where we have suppressed the index for the electron. If we integrate p over all space, we obtain a value of three. This means we are certain of finding a total 7t charge corresponding to three jt electrons in the system. To find out how the n charge is distributed in the molecule, let us express p in terms of AOs. First, we write \ we would obtain 257 X\Xldv + 1 :4 + 2 XtXzdv 1 8-46) carbon I Uirak 1 ,i r' F "'"-S'aiion, as ocing "distributed" 4 at these figures for all the electm,^'u/ , 3' resPectIvely. If we accumulate Generalizing this approach gtves for the total ^electron density qt on atom i ali MOs «*4 (8-47) K Here k is the MO index, cn is the coefficient for an AO on atomin MO k, and «the "occupation number," is the number of electrons (0, 1, or 2) in MO k. (In those rare cases where c,* is complex, cfk in Eq. (8-47) musl be replaced by c*kcn.) If we apply Eq. (8-47) to the cyclopropenyl radical, we encounter an ambiguity. If the unpaired electron is assumed to be in MO 4>2 of Fig. 8-7, we obtain 91 = qi = g, 93 = g. On the other hand, if the unpai red electron is taken to be in 0j', q \ = qi = |, 43 = g. The 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 q\ =qi — I 93 = 1- The general rule is that, for purposes of calculating electron distributions, ihe electron occupation is averaged in any set of partially occupied, degenerate MOs. TABLE 8-1 ► HMO n Electron Densities in the Allyl Radical Chapter 8 The Simple Hiickel Method and Applications • /_ - A ' © G o oo 4f © © o 0O Figure 8-8 ► When the equilateral structure is distorted by decreasing K\2 and increasing Ä13, Ä23. the energies associated with '^ shift as shown. In actuality, the equilateral triangular structure for the cyclopropcnyl radical is unstable, 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 E\, E2, and £3 differently. In particular, a distortion of the sort depicted in Fig. 8-8 would have little effect on E\ but would raise E2 (increased antibonding) and lower E5 (decreased antibonding and increased bonding). Thus, there 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? The preference of the cyclopropenyl radical for a shape less symmetrical than what we might have anticipated is frequently called Jahn-Teller distortions Many times we are interested in comparing the ;r-electron distribution in the bonds instead of on the atoms. In the integrated expression (8-46) arc 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/v^ as indicating how much overlap charge is being placed in the C1-C2 ar>d C2-C3 bonds by an electron in 2-) Generalizing the argument gives, for py, the n-bond order between nearest-neighbor atoms i and j: all MOs P'j= nkclkcjk (8-48) where the symbols have the same meanings as in Eq. (8-46). Incases in which partially filled degenerate MOs are encountered, the averaging procedure described in connection with electron densities must be employed for bond orders as well. EXAMPLE 8-4 Calculate pa for (he cyclopropenyl radical, using data in Fig. 8-7. SOLUTION ► There are 2 electrons in 0, and the coefficients on atoms 1 and 3 are X so V3' this MO contributes 2 X (tf = 2/3. We allocate J electron ,0 fc. Since c3 = 0 in this MO the contribution to ^3 is zero. The remaining \ electron goes ,0 fc. yielding a contribution of V6 X v/6 = So Pl3 = |-5 = i. Q 8-9 Some Simplifying Generalizations Thus far we have presented the bare bones of the HMO method using fairly small systems as examples. If we try to apply this method directly to larger molecules, it is very cumbersome. A ten-carbon-atom system leads to a 10 x 10 HMO determinant. Expanding and solving this for roots and coefficients is tedious. However, there are some short cuts available for certain cases. In the event that the system is too complicated to yield to these, one can use computer programs which are readily available. For straight chain and monocyclic planar, conjugated hydrocarbon systems, simple formulas exist for HMO energy roots and coefficients. These are derivable from the very simple forms of the HMO determinants for such systems.5 We state the results without proof. For a straight chain of n unsaturated carbons numbered sequentially, x = -2cos[far/(» + l)], £ = 1,2,.. cik = [2/(n + \)]l!2sm[klTt/(n + 1)] where / is the atom index and k the MO index. 5SeeCoulson [2|. (8-49) (8-50) Chapter 8 The Simple Hiickel Method and Applications For a cyclic polyene of n carbons, T x = -2cos(2/t/«), k = 0, I, -1/2 exp[2jrifc(/-!)/«], / = V-1 (8-51) (8-52) The coefficients derived from Eq. (8-52) for monocyclic polyenes will be complex when the MO is one of a degenerate pair. In such cases one may take linear combinations of these degenerate MOs to produce MOs with real coefficients, if one desires. There is also a diagrammatic way to find the energy levels for linear and monocyclic systems.6 Let us consider monocycles first. One begins by drawing a circle of radius 2 Into this circle inscribe the cycle, point down, as shown in Fig. 8-9 for benzene. Project sideways the points where the polygon intersects the circle. The positions of these projections correspond to the HMO energy levels if the circle center is assumed to be at E = a (see Fig. 8-9). The number of intersections at a given energy is identical to the degeneracy. The numerical values for E are often obtainable from such a sketch by inspection or simple trigonometry. For straight chains, a modified version of the above method may be used: For an n-carbon chain, inscribe a cycle with 2n + 2 carbons into the circle as before. Projecting out all intersections except the highest and lowest, and ignoring degeneracies gives the proper roots. This is exemplified for the ally! system in Fig. 8-10. Examination of the energy levels in Figs. 8-9 and 8-10 reveals that the orbital energies are symmetrically disposed about E = a. Why is this so? Consider the allyl system. The lowest-energy MO has two bonding interactions. The highest-energy MO differs only in that these interactions are now antibonding. [See Fig. 8-5 and note that the coefficients in (j>\ and tfe are identical except for sign in Eqs. (8-27) and (8-29).] The role of the p terms is thus reversed and so they act to raise the orbital energy for 03 just as much as they lower it for 0i. A similar situation holds for benzene. As we will see shortly, the lowest energy corresponds to an MO without nodes between atoms, so this is a totally bonding MO. The highest-energy MO has nodal planes between all neighbor carbons, and so every interaction is antibonding. An analogous argument holds for the degenerate pairs of benzene MOs. These observations suggest that the energy of an E„-a-2» 1,- - E = a Figure 8-9 ► HMO energy levels for benzene produced by projecting intersections of a hexago with a circle of radius 2\ft\. See Frost and Musulin [3j. Section 8-9 Some Simplifying Generalizations 261 fj "or- v/20 Figure 8-10 ► HMO energy levels for the allyl system (n — 3) produced by projecting the intersections of an octagon (n = 2 x i + 2) with a circle of radius 2 MO should be expressible as a function of the net bond order associated with it, and this is indeed the case. The energy of the ;th MO is given by the expression -EE men I xkHnXidv k I -53) (8-54) neighburs E<=Ecl<*+ E ckic,,p k k.l (8-55) However, c\{ is qtj, the electron density at atom k due to one electron in MO (/>,-, and cificn is pu.i, the bond order between atoms k and / due to an electron in 4>i-Therefore, neighbors £.=522 E pus? (8_56) We have seen that the sum of electron densities must equal the total number of electrons present. For one electron in 0,-, this gives additional simrjlificntinn Theto;a/^_electron simplification. bonds Ei=a + 2fi'22'puti (8-57) k