13-18 Hybrid Orbitals and Localized Orbitals
Chapter 13 Group Theory for solving the Schr¨odinger equation, we expect these symmetry patterns to appear in MOs for, say, ammonia, at extended H¨uckel, CNDO, INDO, MINDO, or ab initio
levels of computation.
13-18 Hybrid Orbitals and Localized Orbitals
The concept of a hybridized orbital is often encountered in the literature, especially in introductory discussions of bonding theory. While this concept is not essential to MO theory, it is used enough to justify a brief discussion.
In the previous section, we showed how one could transform from a basis set of STOs to a basis set of symmetry orbitals. Since these two sets are related through a unitary transformation, they are equivalent and must lead to the same MOs when we do a linear variation calculation. However, there are an infinite number of unitary transformations available, and so the set of symmetry orbitals is only one of an infinite number of possible equivalent bases. Of course, this set has the unique advantage of being a set of bases for representations of the symmetry group, which makes it easy to work with. Another set of equivalent basis functions are the hybrid orbitals. These have the distinction of being the functions that are concentrated along the directions of bonds in the system. Consider, for example, methane, which was discussed in detail in Chapter 10. The minimal basis set of valence STOs on carbon can be transformed to form four tetrahedrally directed hybrids: φ1 = 1 (s + px − py + pz), φ2 = 1 (s − px + py + pz)
2
2
φ3 = 1 (s − px − py − pz), φ4 = 1 (s + px + py − pz)
2
2
One of these hybrids, φ4, is shown in Fig. 13-17 and can be seen to point toward one of the hydrogen atoms. Because the square of each hybrid consists of one part s AO to three parts p AO, these are called sp3 hybrids (pronounced s-p-three). The reason for focusing on sp3 hybrids in this case is that they have physical appeal since they point along the C–H bonds and therefore seem to have a more natural relation to the electron-pair bond approach of G. N. Lewis. This is deceptive, however, because the set of four carbon sp3 orbitals is completely equivalent to the set of four carbon STOs.
The sum of squares of the hybrids is spherically symmetric just as is the sum of squares of STOs. Thus, even though each hybrid is directed toward a hydrogen, the electron density due to all four occupied hybrids is spherically symmetric. Furthermore, after the linear variation is performed, the MOs that are produced contain mixtures of hybrids to give us the exact same delocalized MOs produced from STOs. No single MO consists of just one hybrid and one hydrogen 1s AO, and therefore no single MO can be identified with one C–H bond. We conclude then that hybrid orbitals are one of an infinite number of choices of basis, that they have an appealing appearance because of their concentration in bond regions, but that no concentration of charge in the molecule results as a consequence of using hybrids rather than STOs. (Some concentration of charge in the bonds does result from overlap between basis functions on carbon and those on hydrogens, but this occurs to exactly the same degree for the various equivalent basis sets.)
Section 13-18 Hybrid Orbitals and Localized Orbitals
Figure 13-17 A sketch of the hybrid φ4 = 1 (s + p
2
x + py − pz ). The direction of the hybrid is coincident with that of a vector having the same x, y, z dependence as φ4. The other three hybrids are identical except that they point toward the other three H atoms. (The coordinate system here is rotated with respect to that used in Chapter 10.)
Another example is planar CH+(D
3
3h). As before, we can mix our minimal valence STOs on carbon to produce hybrids pointing toward the hydrogens. If one hydrogen is on the +y axis, the hybrids are
√2
ψ1 = 1 √ s + √ py, 3
3
ψ2 = 1 √ s + 1 √ px − 1 √ py,
3
2
6
ψ3 = 1 √ s − 1 √ px − 1 √ py
3
2
6
Each of these hybrids, when squared, is one part s to two parts p and is called an sp2 hybrid. The coefficients for the p AOs are determined from simple vector considerations: One merely calculates how x and y vectors must be combined to produce resultant vectors pointing toward the corners of an equilateral triangle. The resulting hybridized basis set for CH+ is ψ
3
1, ψ2, ψ3, plus the 2pz STO on carbon and a 1s STO on each hydrogen. As before, the sum of the squares of ψ1, ψ2, ψ3, and 2pz is spherically symmetric.
Chapter 13 Group Theory
We turn now to localized orbitals. We have been emphasizing that one can subject basis sets to unitary transformations without making any physical difference. A similar rule applies for filled molecular orbitals in a determinantal wavefunction. These too can be subjected to unitary transformations without affecting the total energy or total
electronic distribution for the system. (We have encountered this fact before. See, for instance, Appendix 7.) Thus, we have the capability of altering the appearance of the individual orbitals in the wavefunction without affecting the wavefunction itself.
Chemists tend to think of the electrons in molecules as being paired in bond regions, lone pairs, and inner shells, but MOs are delocalized and do not reflect this viewpoint.
But by carrying out unitary transformations, we can attempt to produce orbitals that are more localized without sacrificing any of the properties of the overall wavefunction. For methane, we could mix our four delocalized occupied MOs together to try to produce four new orbitals, each one concentrated in a different C–H bond region. One can do this, but it is important to realize that these localized orbitals are not eigenfunctions of an energy operator, for they have been produced by mixing eigenfunctions having different energies. Furthermore, the localization is never complete in any system of physical interest. Each localized orbital always contributes at least slightly to charge buildup in regions outside that of its primary localization. For instance, a localized C–H1 orbital in methane will have small “residual” components at H2, H3, and H4.
What we have been discussing in this section are various kinds of equivalent orbitals.
At the level of basis sets, we have indicated that a minimal valence basis set of STOs is equivalent to a set of symmetry orbitals and also to a set of hybrid orbitals (as well as an infinite number of other possibilities). At the level of molecular orbitals we have indicated that the set of occupied delocalized MOs is equivalent to an infinity of transformed sets, some of which will tend to be localized in regions chemists associate with bonds, lone pairs, or inner shells. One’s choice among the possibilities for basis
is a matter of taste. However, at the MO level, the delocalized MOs have two features that are sometimes advantageous. The first is that their energies are eigenvalues for some energy operator (the Fock operator in SCF theory). These are related in a simple way to ionization energies and electron affinities via Koopmans’ theorem. Hence, delocalized MOs are more appropriate when considering photoelectron spectra, etc. The second advantage of delocalized MOs is that they display in a clear way the symmetry requirements on the system because they are bases for representations. Hence, these MOs are the most appropriate to use when one is using MO phase relations to infer the nature of certain intra- or intermolecular interactions. (See Chapter 14 for examples.) When delocalized MOs are mixed to form localized orbitals, these energy and symmetry features become partially disguised.
13-19 Symmetry and Integration Throughout this book, the usefulness of symmetry to determine whether an integral vanishes has been emphasized. It should come as no surprise, therefore, that the formal mathematics of symmetry—group theory—is also useful for this purpose.
The basic idea we have used all along is that, if an integrand is antisymmetric for any symmetry operation, it must have equal positive and negative regions, which cancel on integration. If there is no symmetry operation for which the integrand is antisymmetric, then the integral need not vanish. (It still might vanish, but not because of symmetry.)
Section 13-19 Symmetry and Integration
The group theoretical equivalent of this is as follows. Suppose that we have an integral over the integrand f : f dv =?
The function f is identified as being related to some symmetry point group. (Examples are given shortly.) We want to know what representation f is a basis for. If f produces a representation containing A1, then f has some totally symmetric character and the integral need not vanish. But if f is devoid of A1 character, the integral vanishes by symmetry since all other representations are antisymmetric for at least one operation.
Our problem, therefore, is to decide which irreducible representations are present in the representation that is produced by the integrand f .
In quantum chemistry, the integrand of interest is often a product of wavefunctions (or orbitals) and operators. For example, the hamiltonian matrix ˆ H contains integrals of the form
H
ˆ
ij =
ψ∗
i H ψj dτ
We know that ˆ H is invariant for any symmetry operation of the group, and so ˆ H has A1 symmetry. ψi and ψj are assigned symmetries by comparing their behaviors under various operations with the group character table, as illustrated earlier. Thus, it is fairly easy to ascertain the symmetries of the various parts of the integrand. The problem is to determine the symmetry of the product ψ∗ ˆ
H ψ
i
j .
To develop a rule for products, let us consider the simplest case—the one-dimensional representations. Suppose that ψ1 and ψ2 are bases for one-dimensional representations 1 and 2. Then, for some symmetry operation R Rψ1 = χ1(R)ψ1, Rψ2 = χ2(R)ψ2 where χ is a character (1, −1, , or ∗). If we operate on the product ψ1ψ2 with R, we obtain7 Rψ1ψ2 = (Rψ1)(Rψ2) = χ1(R)ψ1χ2(R)ψ2 = χ1(R)χ2(R)ψ1ψ2 That is, the characters for the product ψ1ψ2 are equal to the products of the characters
for ψ1 and ψ2. We have demonstrated the rule for one-dimensional representations, but it can be proved for higher-dimensional cases as well. In group theory, the product of two functions, like ψ1ψ2, is referred to as a direct product to distinguish it from a product of symmetry operations, like σ3C+. The symbol for a direct product is ⊗.
3
For the C3v group, the characters for some direct products of bases for irreducible representations are shown in Table 13-25. The direct product x2 has as characters the product of characters of E times itself. These characters (4, 0, 1) do not agree with any of the irreducible representation character sets, and so E ⊗ E is reducible.
We can tell, in fact, that E ⊗ E is four-dimensional from the leading character. To resolve E ⊗ E, we employ the formula (13-7), which gives E ⊗ E = A1 ⊗ A2 ⊗ E, and fits the observation that E ⊗ E is four-dimensional. The other direct products listed 7That Rψ1ψ2 = (Rψ1)(Rψ2) is not always obvious to the student, but it should be evident that operating on (say reflecting) the function ψ1ψ2 gives the same result as reflecting ψ1 and ψ2 separately and then taking the product.
Chapter 13 Group TheoryTABLE 13-25 Characters for Direct Products of Bases for Irreducible Representations of C3v
C3v
E
3σ
2C3
A1
1
1
1
z
x2 + y2, z2
A2
1
−1
1
Rz
E
2
0
−1
(x, y), (Rx, Ry) (x2 − y2, xy)(xz, yz)
E ⊗ E
4
0
1
x2
y2
x2z2
xy
A2 ⊗ E
2
0
−1
Rzx
Rzy A2 ⊗ A2
1
1
1
R2z A1 ⊗ A2 ⊗ E
2
0
−1
zRzx
in Table 13-26 ( Rzx, R2z, zRzx, etc.) all give character sets indicative of irreducible representations. We see that A2 ⊗ E = E, A2 ⊗ A2 = A1, A2 ⊗ A1 ⊗ E = E.
There is a general rule that is illustrated by these examples: A direct product ofbases for two irreducible representations contains A1 character if and only if the two
irreducible representations are the same. That is, if fi is a basis for i and fj is a basis for j and fifj is a basis for i,j , where
≡
i, ˙
j
i ⊗ j = c1A1 ⊕ c22 ⊕ · · · ⊕ ci i ⊕ cj j ⊕ · · · then c1 = 0 if and only if i = j .
Another rule is that, if i is A1, then i,j = j ; that is, multiplying a function f2, by a totally symmetric function f1 gives a product with the symmetry of f2.
Now we are in a position to decide whether the integral of a product of functions and operators will vanish. For our examples, we will continue to use orbitals, operators, and coordinates from the ammonia molecule. Some of these quantities, segregated according to symmetry, are given in Table 13-26.
TABLE 13-26 Operators and Orbitals for Ammonia Classified by Symmetry
a1
e
N2s
N2px
N 2pz
N2py
√
√
(1/ 3)(1s1 + 1s2 + 1s3) (1/ 6)(2 · 1s1 − 1s2 − 1s3)
φ
1
√
φ4 MOs(see Table 13-23) (1/ 2)(1s2
− 1s3)
φ φ
7
2
ˆ
φ3
H
MOs
φ5
z
φ6
x
y
Section 13-19 Symmetry and IntegrationEXAMPLE 13-9 Indicate whether each of the following integrals must vanish due to symmetry.
1.
N
ˆ
2p H N
dv
z
2px
2.
N
ˆ
2p H 1 √ (1s
x
2 − 1s3)d v
2
SOLUTION
1.
N
ˆ
2p H N dv: The symmetries of the three functions in the integrand are respectively
z
2px
A1, A1, E. The direct product has symmetry E. There is no A1. The integral vanishes.
2.
N
ˆ
2p H 1 √ (1s
x 2 − 1s3) dv: The symmetries are E, A1, E. The direct product is therefore
2
E ⊗ E ⊗ A1 = A1 ⊕ A2 ⊕ E. Since A1 is present, the integral need not vanish.
These two examples are related to the block diagonalization of the matrix H, discussed in a previous section. The zero blocks in that matrix correspond to integrals between functions of different symmetry. Since ˆ H is of A1 symmetry, it has no influ ence on the symmetry of the integrand. If ψi and ψj have different symmetries, their direct product cannot contain A
ˆ 1 symmetry and the integral over ψi H ψj must vanish.
The reader can now understand how the computational procedure guarantees MOs of “pure” symmetry (i.e., bases of irreducible representations). If two basis functions ψi and ψj differ in symmetry, there will be a zero value for Hij . A zero Hij means that mixing ψi and ψj will produce no energy lowering. Hence, the variation procedure will not mix these functions together in the same MO, so the MO will not be of mixed symmetry.
EXAMPLE 13-10 Indicate whether each of the following integrals must vanish due to symmetry.
1.
φ1xφ4 dv
2.
φ3yφ5 dv
SOLUTION
1.
φ1xφ4 dv: Integrals of this sort are involved in calculating spectral intensities. The symme tries are A1 ⊗ E ⊗ A1 = E and the integral vanishes. This means a transition between states corresponding to an electron going from φ1 to φ4 (or φ4 to φ1) is forbidden for x-polarized light and an oriented molecule (see Section 12-9).
2.
φ3yφ5 dv: The symmetry here is E ⊗ E ⊗ E, which gives characters 8, 0, −1. This resolves into A1 ⊕ A2 ⊕ 3E and the integral need not vanish. Corresponding transitions are “y allowed.”
EXAMPLE 13-11 In Chapter 12 , we indicated that the π ∗ ← π transition in ethene is dipole-allowed and polarized parallel to the molecular axis. Verify this from the character table for ethene, after determining the group symbol.
Chapter 13 Group TheorySOLUTION Referring to Fig. 13-7, ethene yields the answers “no, no, no, yes, {no,” so choose any one. We arbitrarily take C2 along the C–C axis} “no, yes,” (go to D branch) “yes”, so D2h. Our choice of principal axis puts the C–C bond vertical. Using that we can identify the π MO as B2u and the π ∗ MO as B3g. The product of characters for B2uB3g = 1 1 − 1 − 1 − 1 − 1 1 1, which can be seen to have B1u symmetry. So π ∗(x, y, or z)π dv = 0 only if x, y, or z also has B1u symmetry. z does, so the transition is allowed and is z-polarized (i.e., it is a parallel transition.)
In this chapter we have seen how formal group theory can be used to characterize MO symmetries, construct symmetry orbitals, and indicate whether integrals vanish by symmetry. It is true that one can perform MO calculations and get correct results without explicitly considering symmetry or group theory, since the computational procedures satisfy symmetry considerations automatically. But group theory allows a much deeper understanding of the constraints that symmetry places on a problem and often leads to significant shortcuts in computation.
A notable feature of group theory is its hierarchy of concepts. At the lowest level are the symmetry operations and the basis functions they operate on. At the intermediate level are the representations for the group, produced from the basis functions. At the highest level are the characters, produced from the representations. The characters provide the “handles” that we actually work with, but our interest is often focused on the basis functions to which they are related. This tends to lend an air of unreality to the use of group theory. An aim of this chapter has been to avoid this feeling of unreality by dispensing with formal proofs, and instead illustrating relationships through investigation of examples. Further insight should come from solving the problems at the end of this chapter.
13-19.A Problems13-1. Do the following operations constitute a group? “come 90◦ to port” (P ) “come 90◦ to starboard” (S) “steady as she goes” (E).
13-2. The text indicates that every element in the group has an inverse if E appears in each column of the multiplication table. But E also appears once in each row.
What does this mean?
13-3. Consider the group of four operations of the drill soldier (Section 13-2).
a) To which symmetry point group is this set of four operations isomorphic (i.e., which group has the same product relationship)?
b) Based on the mathematical definition of class and Table 13-1, how many classes are there in this group?
c) Based on your physical intuition about kinds of operation, how many classes would you have anticipated for this group? If there is a discrepancy between (b) and (c), try to explain it.
13-4. The C3v (ammonia) group is of order six. This is the same as the number of ways one can place three hydrogens at the three corners of an equilateral triangle (3 · 2 · 1). When we consider the C4v group, we have 24 ways we can place four hydrogens at the corners of a square (4 · 3 · 2 · 1). But the C4v group only has order eight. Explain.
Section 13-19 Symmetry and Integration13-5. For each of the molecules (IV)-(VII), a) list the symmetry elements, b) calculate the number of symmetry operations for each element, c) obtain the order of the group, d) determine the group symmetry symbol, e) check your results for (a-c) against the appropriate character table in Appendix 11.
13-6. a) Demonstrate that U is a unitary matrix.
b) Demonstrate that U†AU is diagonal.
1
√
− 1 √
0
1
U =
2
2
, A =
1
√
1
√
1
0
2
2 13-7. Assign the following molecules to point groups, look up their character tables, and indicate in each case whether one could expect doubly degenerate MOs.
a) C6H6 b) CH2Cl2 c) B2H6 (see Problem 13-5) d) Staggered C2H6 e) Staggered CH3CCl3 13-8. Consider the planar molecule CO2− (VIII). The oxygen atoms are at the corners
3
of an equilateral triangle.
a) What is the point group of this molecule?
b) Using the appropriate character table, assign a symmetry symbol to each of MOs (IX)–(XII).
Chapter 13 Group Theory13-9. Table P13-9 gives the eigenvalues and eigenvectors resulting from an extended H¨uckel calculation of the allene molecule (XIII). The molecule is aligned as shown with respect to Cartesian axes. Ascertain the point group for this molecule. Using the character table for this group, assign a symmetry symbol to each MO. Is a transition from the highest occupied MO level to the lowest unoccupied level allowed by symmetry for any direction of polarization?
13-10. Consider the water molecule, oriented as shown in Fig. P13-10 with the y axis perpendicular to the molecular plane and the z axis bisecting the H–O–H angle.
a) Figure out as many nonredundant symmetry operations for this molecule as you can, and set up their multiplication table. Ascertain that you have a group of operations by checking closure, etc.
b) Use the functions z, Rz, x, and y as bases to set up a character table for this group.
TABLE P13-9 Extended H¨uckel Molecular Orbitals for Allene C1
C2
MO
Energy
no.
(a.u.)
2s
2pz
2px
2py
2s
2pz
2px
1
1.7868
0
0
1.58
0
−1.23
0
0.58
2
1.4916
1.52
0
0
0
−1.01
0
0.57
3
0.4927
−0.54
0
0
0
−0.51
0
−0.75
4
0.3681
0
0
0.67
0
0.36
0
0.83
5
0.3231
0
0
0
−0.34
0
0
0
6
0.3231
0
0.34
0
0
0
−0.11
0
7
−0.2619
0
0
0
0.85
0
0
0
8
−0.2619
0
0.85
0
0
0
−0.78
0
9
−0.4326
0
0.56
0
0
0
0.67
0
10
−0.4326
0
0
0
−0.56
0
0
0
11
−0.4881
0
0
−0.49
0
−0.08
0
0.41
12
−0.5558
0
−0.16
0
0
0
−0.04
0
13
−0.5558
0
0
0
0.16
0
0
0
14
−0.6221
0.43
0
0
0
−0.14
0
−0.32
15
−0.8102
0
0
−0.13
0
−0.44
0
−0.02
16
−0.9363
0.47
0
0
0
0.36
0
−0.02
Section 13-19 Symmetry and IntegrationFigure P13-10
Relation of the water molecule to cartesian axes.
c) Use the resulting table to find out the symmetries of all MOs that can contain 1s AOs on the hydrogens.
d) Produce the symmetry combinations of these AOs that can appear in the MOs of water.
13-11. Consider the trans-chlorobromotetramine cobalt (III) ion (XIV).
a) Find the appropriate symmetry elements for this molecule and set up their group multiplication table. (Ignore the hydrogens on the ammonias.) Check the multiplication table to be sure all requirements for a mathematical group are satisfied.
Atomic Orbital Coefficients
C5
2py
h3
h4
2s
2pz
2px
2py
h6
h7
0
0.30
0.30
1.23
0
0.58
0
−0.30 −0.30
0
0.24
0.24
−1.01
0
−0.57
0
0.24
0.24
0
0.61
0.61
−0.51
0
0.75
0
0.61
0.61
0
−0.58
−0.58
−0.36
0
0.83
0
0.58
0.58
1.27
−0.88
0.88
0
0
0
0.11
0
0
0
0
0
0
−1.27
0
0
0.88
−0.88
0.01
−0.21
0.21
0
0
0
−0.78
0
0
0
0
0
0
0.01
0
0
−0.21
0.21
0
0
0
0
−0.10
0
0
−0.16
0.16
0.10
0.16
−0.16
0
0
0
−0.68
0
0
0
0.17
0.17
0.08
0
0.41
0
−0.17 −0.17
0
0
0
0
−0.48
0
0
−0.44
0.44
−0.48 −0.44
0.44
0
0
0
−0.04
0
0
0
−0.24
−0.24
−0.14
0
0.32
0
−0.24 −0.24
0
−0.18
−0.18
0.44
0
−0.02
0
0.18
0.18
0
0.07
0.07
0.36
0
0.02
0
0.07
0.07
Chapter 13 Group Theory b) Use the following as bases for representations: z, Rz, x, y, x2 − y2, xy.
Make sure you get all the inequivalent irreducible representations allowed in the group by checking il2 = h.
i
c) Set up the character table. Now ascertain which symmetry orbitals contain 2s orbitals of nitrogen. Give the symmetry combinations of those AOs that appear in these symmetry orbitals.
13-12. Use the relationships that must exist among characters to complete the following tables. Include proper symbols for the representations, but do not include bases.
E
C2
σv
σ
v
E
2C3
3C2
σh
2S3
3σv (a)
(b)
13-13. The D5 group has four classes of operation and has order 10. How many inequivalent irreducible representations are there and what are their dimensions?
13-14. Consider the structure shown in Fig. P13-14.
Figure P13-14 Square pyramid.
a) Figure out the symmetry elements and operations for this molecule.
b) What is the group order and number of classes?
c) How many inequivalent irreducible representations are there and what are their dimensions?
d) Ascertain the group symbol and compare your answers with the character table in Appendix 11.
13-15. A group has the following representations: A1, A2, B1, B2, E1, E2. What is the group order and how many classes are there?
13-16. Find the matrices that transform s and p STOs into sp3 and sp2 hybrids (Sec tion 13-18). Demonstrate that these are unitary matrices.
13-17. It has been argued (Section 13-18) that sp2 hybrid orbitals are appropriate basis functions for CH+(D
3
3d). Could one use a basis set of sp3 hybrid orbitals for this system?