Section 12-10
Section 12-10 Spectroscopic Selection Rules
norbornene, which has but one double bond. Experimental measurements support this contention. Also, since the highest occupied MO of norbornadiene is antibonding between carbons 3 and 5 and also 6 and 2, whereas the lowest empty MO is bonding between these carbons, excitation of an electron from the former MO to the latter should promote formation of quadricyclene (Fig. 12-7d). This compound is a common product in the photochemistry of norbornadienes.
12-10 Connection Between Time-IndependentPerturbation Theory and Spectroscopic
Selection Rules
Molecules may change to higher-or lower-energy states under the influence of incident light. Such processes are called, respectively, absorption and induced emission. Perturbation theory can be used to study such transitions induced by an external oscillating electromagnetic field. Here we briefly describe a rather simple connection between selection rules and the perturbation theory we have discussed in this chapter.
Let a molecule initially be in a state with wavefunction ψi. We are interested in the probability of a transition occurring to some final state with wavefunction ψf . A timedependent perturbation treatment (not given here, but see Section 6-16) indicates that such a transition is probable only when the external field frequency ν satisfies the conservation of energy relation ν = |Ef − Ei|/ h
(12-86)
Even when this condition is satisfied, however, we may find experimentally that the transition is so improbable as to be undetectable. Evidently some factor other than satisfaction of Eq. (12-86) is also involved.
Since the molecule is being subjected to light, it experiences fluctuating electric and magnetic fields. For ordinary (as opposed to magnetic) spectroscopy, the effects of the magnetic field are negligible compared to those of the electric field. Therefore, we ignore the former and imagine the molecule at a particular instantaneous value
of an external electric field. As we have seen from earlier sections, this field causes polarization through the admixture of unperturbed wavefunctions ψ1, ψ2, etc., with ψi to form a perturbed wavefunction φi. For a normalized φi, we have, therefore, φi = ciψi + c1ψ1 + c2ψ2 + · · · + cf ψf + · · ·
(12-87)
At a later time, when the perturbation is over, the system returns to an unperturbed state. The probability of returning to the initial state is given by c∗c i i . The probability of going instead to the final state described by ψf is given by c∗ c f f . If cf is zero, there is no tendency for a transition to that final state to occur and the transition is said to be forbidden. If cf is nonzero, the transition is allowed.
Since the coefficient cf is given to first order by [see Eq. (12-18)], ψf |H|ψi cf =
(12-88)
Ei − Ef where H is the perturbation operator for the electric field component of the light, we can say at once that a transition between two states is forbidden (to first order) if ψf |H|ψi vanishes.
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory
One example of a forbidden transition is that between a singlet state and a triplet state of a system. We can see this at once since H is an electric field and does not interact with spin magnetic moment. Therefore the orthogonality between ψi and ψf due to spin functions is uninfluenced by H and the integral ψf |H |ψ˙ must vanish.
1
In effect, the perturbation H causes the initial singlet state to become polarized by mixing in other singlet state functions, but gives no impetus for mixing in states of different multiplicity. Transitions between states of different multiplicity are said to be spin forbidden.
Another example of a forbidden transition is that between two different s-type states of a hydrogen atom. Such states have spherically symmetric wavefunctions, but H (the electric field) is antisymmetric for reflection through a plane (to within an additive constant), and so ψf |H |ψi must vanish for reasons of symmetry. It is easy to generalize this argument to other states of the hydrogen atom (p to p, d to d, etc.) and also to certain other atoms (e.g., the alkali metals) electronically similar to hydrogen.
Molecular electronic transitions can be understood from the same standpoint. The intense π ∗ ← π transition in ethylene is a simple example.9,10 Let us imagine that the molecule is oriented as shown in Fig. 12-8. Suppose we could somehow orient all our molecules this way (in a host matrix or in a crystal) and that we then subjected the sample to plane-polarized light. We will consider what should happen for light polarized in each of the directions x, y, z. First, let us consider the integral π|z|π∗. The ethylene molecule has three reflection planes of symmetry, so we can examine the symmetry of the integrand with respect to each of these three reflections. Remember that, if the integrand is antisymmetric for any one of these reflections, the integral vanishes. The function π can be seen, from inspection of Fig. 12-8, to be symmetric for reflection in the xz and yz planes, antisymmetric for reflection in the xy plane. These observations are shown in Table 12-2, along with similar conclusions regarding the symmetries of the functions π ∗, x, y, and z. The symbols “s” and “a” stand for “symmetric” and “antisymmetric.” Our integral π|z|π∗ can now be seen to have an integrand that is symmetric for reflection in the xz plane but antisymmetric for reflection in the xy
Figure 12-8 π and π ∗ MOs for ethylene. The C–C bond is coincident with the x axis and all nuclei lie in the xy plane.
9Here, π∗ refers to the antibonding πg MO, not to complex conjugation.
10In this discussion, we consider molecules of high symmetry, so that reflection planes of symmetry exist. For molecules of lower symmetry, one must resort to actual calculation of the cf in Eq. (12-88).
Section 12-10 Spectroscopic Selection RulesTABLE 12-2 Symmetries of Functions under Reflection through Cartesian Coordinate Planes
Symmetry operation
Function
xy reflection
xz reflection
yz reflection
π
a
s
s
π ∗
a
s
a
z
a
s
s
x
s
s
a
y
s
a
s
π z π ∗ a a a = a s s s = s s s a = a π x π ∗ a s a = s s s s = s s a a = s π y π ∗ a s a = s s a s = a s s a = a and yz planes. Hence, this integral vanishes, and this transition is forbidden insofar as light polarized perpendicular to the molecular plane is concerned. According to Table 12-2, symmetry also causes π|y|π∗ to vanish. However, symmetry does not force π|x|π∗ to vanish. Therefore, the π∗ ← π transition is allowed and is polarized along the C–C axis. Indeed, because of the significant spatial extension of the π and π ∗ MOs (compared to AOs), the integral π|x|π∗ is relatively large, which means that the transition is not only allowed, but is intense.
It is physically reasonable that the π ∗ ← π transition should be polarized along the C–C axis. If some π ∗ character is mixed with π , it is easy to see that this results in a shift of π charge from one carbon to the other—a shift along the x axis. Conversely, an electric field in the x direction will cause polarization along the x axis, hence mix π ∗ character into the π MO. When the perturbation is removed, a finite probability exists that the molecule will go to the state wherein π ∗ is occupied.
In the above example, we imagined all ethylene molecules to be identically oriented.
Under those conditions, we would observe a maximum in π ∗ ← π absorption when our incident light was polarized parallel to the molecular axis and zero absorption (ideally) when the polarization axis was perpendicular to the molecular axis. If the light is unpolarized, or if the ethylene is randomly oriented, as in liquid or gaseous states, the absorption is isotropic and allowed because there is always a certain degree of “overlap” between the molecular axes of most of the molecules and the direction of the electric field due to the light.
The example discussed above illustrates a simple and powerful rule for recognizing whether or not a given atomic or molecular orbital dipolar transition is allowed (assuming electron spin agreement): A dipole transition is allowed between orbitals that have
opposite symmetry for one and only one reflection plane.11 The polarization of the
transition will be perpendicular to that reflection plane.
11This means that, for an allowed transition, one orbital has one and only one nodal plane where the other does not.
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation TheoryEXAMPLE 12-8 Selection rules for allowed dipolar transitions in a hydrogen atom include l = ±1, ml = 0, ±1. Show that these rules are equivalent to the symmetry rule stated above.a Consider the following cases: 2s ← 1s, 2px ← 1s, 3dxy ← 2s, 3dxz ← 2pz, 3dxy ← 2pz.
aThere is yet another way to rationalize these selection rules. Photons are bosons with spin quantum number of 1, hence they carry spin angular momentum. The total angular momentum of the atom or molecule that absorbs a photon must change to obey the conservation requirement consistent with the vector addition of orbital and photon angular momenta. The l rule handles that for one-electron atoms. Furthermore, for a unit change of l, the z-component of the absorber’s orbital angular momentum can change by +1, 0, or −1 a.u. The ml rule takes care of that. This rationalizes the atomic selection rules, but doesn’t yield information on polarization or intensity as readily as the more physical approach described in the text, and it is less easily applied to molecules. Generally, however, the more ways one understands a thing, the better.
SOLUTION The 2s ← 1s transition is forbidden by the l rule (because l = 0) and also by the symmetry rule (because neither wavefunction has a nodal plane). This makes it easy to see that the l rule applies because it forces one of the orbitals to have one and only one more nodal plane than the other. The 2px ← 1s transition is allowed by the l rule. Since 2px is a linear combination of 2p+1 and 2p−1, the ml rule is also satisfied, so the transition is allowed. When we examine symmetry, we note that there is only one plane of disagreement–the yz plane, so this approach also indicates that the transition is allowed, and that it is polarized in the x direction (perpendicular to the yz plane). The 3dxy ← 2s transition is forbidden by the l rule since l = 2.
The symmetry approach indicates that a transition is forbidden because the 3dxy AO has two nodal planes, whereas 2s has none. The 3dxz ← 2pz transition has l = 1. 3dxz is a combination of 3d+1 and 3d−1, and 2pz is 2p0, so ml = ±1. Therefore, this is an allowed transition. The symmetry approach shows 3dxz with two nodal planes (xy and yz) and 2pz with one (xy). Therefore, there is only one plane of symmetry disagreement (yz), so the transition is allowed and is x polarized.
The 3dxy ← 2pz transition has l = +1. 3dxy is a linear combination of 3d+2 and 3d−2, and 2pz is 2p0, so ml = ±2. Therefore, the transition is forbidden. The symmetry analysis shows 3dxy having two nodal planes (xz and yz), and 2pz having one (xy). Therefore, they disagree in symmetry in three planes, and the transition is forbidden. This last case makes it clear why the ml rule is needed: When satisfied, it guarantees that all but one of the nodal planes in the two AOs coincide. (If ml = ±2, then, taking the z axis to be vertical, one AO has two more vertical nodal planes than the other, so there must be at least two reflection planes of symmetry disagreement.)
12-10.A Problems12-1. Prove that the energy to first order for the lowest-energy state of a perturbed system is an upper bound for the exact energy of the lowest-energy state of the perturbed system, that is, that E0 + W (1) ≥ W
0
0.
12-2. A one-dimensional box potential is perturbed as sketched in Fig. P12-2. From a consideration of the first three wavefunctions of the unperturbed system, ψ1, ψ2, Figure P12-2
Section 12-10 Spectroscopic Selection Rules ψ3, which will have its energy increased most, to first order, and which least?
No explicit calculation is necessary to answer this question.
12-3. Evaluate by inspection the first-order contributions to the energies for the states shown in Fig. P12-3. In every case the unperturbed state is a particle in the indicated state in the one-dimensional, infinitely deep, square well.
Figure P12-3 a) H = sine function as shown, ψi = ψ1. b) H as shown, ψi = ψ1. c) H as shown; ψi = ψ1. d) H as shown; ψi = ψ2. e) H = cosine as shown, ψ˙ι = ψ2. f) H as shown; ψi = ψ2.
12-4. a) Evaluate W (1) for the particle in a box with the perturbing potential shown in
2
Fig. P12-4.
b) For the perturbation above: 1) What sign would you expect for c(1)? Describe the effect of this on φ
21
1; on W (2).
1
2) What sign would you expect for c(1)? Describe the effect of this on φ
12
2; on W (2).
2
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation TheoryFigure P12-4 12-5. Calculate c(1) for the perturbed particle in a wire example discussed in Sec 41
tion 12-3. Show that c(1) is only about 2% as large as c(1). Use
41
21
π
y sin my sin ny dy = (−1)m+n − 1 2mn/(m2 − n2)2, m, n = 1, 2, . . . , m = n
0 12-6. An electron moves in a harmonic potential, V = 1 kx2. What is the effect, to first
2
order, on the energies of superimposing an electric field, V = Ex? Explain your reasoning.
12-7. Calculate the energy to first order of He+ in its lowest-energy state. Use the hydrogen atom in its ground state as your zeroth-order approximation. Use
atomic units. Predict the signs (plus, zero, minus) of c(1) and c(1) . Explain
2s,1s
2p0,1s
your reasoning.
12-8. The previous problem gives an energy to first order of −3/2 a.u. for He+, using H as a starting point. Now reverse the process and try to get the ground-state energy of H to first order, using He+ as starting point. Discuss the reasonableness of your answer in terms of a lower bound.
12-9. A hydrogen atom in its ground state is perturbed by imposition of a uniform z-directed electric field of strength F atomic units: Field = −F z = −F r cos θ.
a) What is the first-order change in energy experienced by the atom? Show your logic.
b) For the first-order correction to the ground-state wavefunction, we can con sider c(1) and c(1) to be the coefficients for mixing in 2s and 2p
2s,1s
2p
z char z ,1s acter. Predict the signs of these coefficients and sketch their effects on the ground-state wavefunction.
c) Predict the sign of W (2) and explain your reasoning.
1s
Section 12-10 Spectroscopic Selection Rules12-10. What is the H¨uckel MO π -electron energy to first order of the molecule (I) if C4 is perturbed by H = 0.1β? (See Appendix 6 for H¨uckel data.)
12-11. Consider the fulvene molecule (II) in the HMO approximation (seeAppendix 6).
At which carbon will a perturbation involving α affect the total π -electron energy the least? Calculate to first order the energy of fulvene with a6 = a + 0.5β. If a computer and H¨uckel program are available to you, calculate the energy for this perturbed molecule directly and compare with your first-order result.
12-12. Produce an expression for πkk, the atom–atom self-polarizability, from Eq. (12-70). Can this, like πlk, have either sign? What can you infer about the sign of δqk when δαk is positive? Discuss the physical sense of this.
12-13. Using data from Example 12-6, calculate the self-atom polarizability, π1,1 for carbon 1 in naphthalene.
12-14. Consider the hexatriene system. (See Appendix 6 for data.) One of the two central atoms is perturbed so that α becomes replaced by α + hβ on that one atom only.
a) What is the energy change, to first order, of the total π energy of hexatriene?
b) Which of the occupied MOs is perturbed the most (i.e., has the largest first order contribution to the energy) by this perturbation?
12-15. Answer the following questions from examination of HMO data for benzcy clopentadienyl radical, C9H7, given in Appendix 6.
a) What is the total π energy to first order if atom 5 is replaced by a new atom X which contributes the same number of π electrons a carbon did and has αx = α + 0.5β and also has βCX = 0.9β?
b) For the positive ion of this system, how do you think the self-atom polariz ability should compare qualitativeIy at atoms 1 and 2? Make your prediction by inspection, rather than computation, and explain your thinking.
12-16. Use simple H¨uckel MOs for butadiene to calculate to first order the change in π energy that would result from closing cis-butadiene to cyclobutadiene (III).
Repeat the approach for closing hexatriene to benzene. (IV). Which of these two systems benefits most from cyclic as opposed to linear topology? For each
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory system, compare your energy to first order with the actual H¨uckel energy (see Appendix 6 for data).
12-17. The unperturbed H¨uckel MO energy levels of the allyl system are sketched in Fig. P12-17. The system is perturbed so that H22 = α + cβ where c is positive.
a) Sketch the effects of this perturbation, to first order, on the energy levels.
b) Calculate to first order the perturbed MOs φ(1), φ(1), φ(1) in terms of the
1
2
3
unperturbed MOs ψ1, ψ2, ψ3. Sketch the results in a manner that makes clear the nature of the change in each MO.
Figure P12-17 12-18. Calculate the atom–atom polarizabilities π1,2 and π1,3 for the allyl cation. What do your results indicate will happen to the π -electron densities at atoms 2 and 3 if atom 1 becomes more attractive?
12-19. The cyclopropenyl system is perturbed so that the H¨uckel matrix element H22 = a + cβ, where c is positive.
a) Ascertain, by calculation or inspection, the appropriate zeroth-order degen erate wavefunctions for this situation.
b) Sketch the effects of the perturbation on the orbital energies to first order.
12-20. Given the HMO data in Appendix 6 for the cyclopentadienyl radical, calculate to first order all the orbital energies you would obtain if atom 2 were changed so that α2 became α + 0.5β.
12-21. Imagine two allyl radicals coming together to form benzene. Each allyl has three π MOs and benzene has six.
a) Sketch an energy level diagram showing how interaction between allyl orbitalscanberelatedtotheMOenergiesofbenzene,atthesimpleH¨uckellevel.
b) Show with sketches how the three lowest-energy zeroth-order allyl combi nation orbitals are related to the three lowest-energy benzene MOs.
c) Estimate to first order the energies of the benzene-like MOs coming from the nonbonding MOs of the allyl pair. Compare these to the relevant MO energies of benzene. (Remember that, when combining fragments, you must renormalize the functions) d) What is the energy to first order of the lowest MO of the allyl combination, and how does it compare to the lowest benzene orbital energy?
12-22. Discuss the possible transitions π3 ← π2 and π4 ← π2 in butadiene (Here π1 is the lowest-energy MO, etc.)
a) Are they dipole-allowed transitions?
b) If yes, what is the polarization?
Section 12-10 Spectroscopic Selection Rules
425
TABLE P12-24a
Level
Energy
c1
c2
c3
c4
c5
c6
c7
−2.01β
−0.238
0.500
−0.406
0.354
0.336
0.354
−0.406
−1.26β
−0.397
0.500
−0.116
−0.354
0.562
−0.354
−0.116
−1.00β
0
0
−0.500
0.500
0
−0.500
0.500
0.00
−0.756
0
0.378
0
−0.378
0
0.378
1.00β
1.26β
2.10β
a Coefficients for 2pz basis function forming the four highest MOs are listed.
Overlap has been assumed to be negligible.
12-23. Calculate the dipole moment in the z direction for the states φ(0) ± of Eq. (12-81).
Now construct a variational function of the form φ = cos(α)2s + sin(α)2pz and maximize the z component of the dipole moment as a function of α. Are the states φ(0) ± those of maximum dipole magnitude? Discuss why this is reasonable.
12-24. The benzyl radical, C7H7(V), has the H¨uckel energies and ground-state config uration given in Table P12-24.
The radical is trapped and oriented in an external reference system as shown.
Light polarized in the x direction is beamed on the system, and the frequency varied until an absorption is observed. Assuming this to result from excitation of the unpaired electron, to which level has the electron been promoted?
12-25. The simple H¨uckel energies, occupation numbers, and coefficients for MOs in naphthalene (VI) are listed in Table P12-25. Assume that a single crystal of naphthalene is oriented so that each molecule is aligned with respect to an external coordinate system (VI).
a) Light polarized in the x direction is beamed on the crystal. Assuming that the electron is excited from the highest occupied MO, to which empty MOs could it be promoted by the x-polarized light?
b) Which transitions from the highest occupied MO would be allowed for y-polarized light?
c) Which ones would be allowed for z-polarized light?
d) Are any transitions from the highest occupied MO forbidden for nonpolar ized light?
0
1
.41
.35
.46
0
0
0
0
0
0
−
−
.42
.42
.40
.26
.30
0
0
0
0
0
0
−
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
−
−
−
.42
.42
.40
.26
.30
0
0
0
0
0
0
−
−
−
ficients
Coef
.41
.35
.46
0
0
0
0
0
0
−
−
.42
.42
.40
.26
.30
0
0
0
0
0
0
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
−
−
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
123456789
.42
.42
.40
.26
.30
0
0
0
0
0
0
−
−
−
−
no.
2
2
2
2
2
2
0
0
0
0
Occupation
P12-25
gies
β
−
.303
.618
.303
.000
.618
.618
.000
.303
.618
.303
ABLE
2
1
1
1
0
0
1
1
1
2
T
Ener
−
−
−
−
−
+
+
+
+
+
426
norbornene, which has but one double bond. Experimental measurements support this contention. Also, since the highest occupied MO of norbornadiene is antibonding between carbons 3 and 5 and also 6 and 2, whereas the lowest empty MO is bonding between these carbons, excitation of an electron from the former MO to the latter should promote formation of quadricyclene (Fig. 12-7d). This compound is a common product in the photochemistry of norbornadienes.
12-10 Connection Between Time-IndependentPerturbation Theory and Spectroscopic
Selection Rules
Molecules may change to higher-or lower-energy states under the influence of incident light. Such processes are called, respectively, absorption and induced emission. Perturbation theory can be used to study such transitions induced by an external oscillating electromagnetic field. Here we briefly describe a rather simple connection between selection rules and the perturbation theory we have discussed in this chapter.
Let a molecule initially be in a state with wavefunction ψi. We are interested in the probability of a transition occurring to some final state with wavefunction ψf . A timedependent perturbation treatment (not given here, but see Section 6-16) indicates that such a transition is probable only when the external field frequency ν satisfies the conservation of energy relation ν = |Ef − Ei|/ h
(12-86)
Even when this condition is satisfied, however, we may find experimentally that the transition is so improbable as to be undetectable. Evidently some factor other than satisfaction of Eq. (12-86) is also involved.
Since the molecule is being subjected to light, it experiences fluctuating electric and magnetic fields. For ordinary (as opposed to magnetic) spectroscopy, the effects of the magnetic field are negligible compared to those of the electric field. Therefore, we ignore the former and imagine the molecule at a particular instantaneous value
of an external electric field. As we have seen from earlier sections, this field causes polarization through the admixture of unperturbed wavefunctions ψ1, ψ2, etc., with ψi to form a perturbed wavefunction φi. For a normalized φi, we have, therefore, φi = ciψi + c1ψ1 + c2ψ2 + · · · + cf ψf + · · ·
(12-87)
At a later time, when the perturbation is over, the system returns to an unperturbed state. The probability of returning to the initial state is given by c∗c i i . The probability of going instead to the final state described by ψf is given by c∗ c f f . If cf is zero, there is no tendency for a transition to that final state to occur and the transition is said to be forbidden. If cf is nonzero, the transition is allowed.
Since the coefficient cf is given to first order by [see Eq. (12-18)], ψf |H|ψi cf =
(12-88)
Ei − Ef where H is the perturbation operator for the electric field component of the light, we can say at once that a transition between two states is forbidden (to first order) if ψf |H|ψi vanishes.
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory
One example of a forbidden transition is that between a singlet state and a triplet state of a system. We can see this at once since H is an electric field and does not interact with spin magnetic moment. Therefore the orthogonality between ψi and ψf due to spin functions is uninfluenced by H and the integral ψf |H |ψ˙ must vanish.
1
In effect, the perturbation H causes the initial singlet state to become polarized by mixing in other singlet state functions, but gives no impetus for mixing in states of different multiplicity. Transitions between states of different multiplicity are said to be spin forbidden.
Another example of a forbidden transition is that between two different s-type states of a hydrogen atom. Such states have spherically symmetric wavefunctions, but H (the electric field) is antisymmetric for reflection through a plane (to within an additive constant), and so ψf |H |ψi must vanish for reasons of symmetry. It is easy to generalize this argument to other states of the hydrogen atom (p to p, d to d, etc.) and also to certain other atoms (e.g., the alkali metals) electronically similar to hydrogen.
Molecular electronic transitions can be understood from the same standpoint. The intense π ∗ ← π transition in ethylene is a simple example.9,10 Let us imagine that the molecule is oriented as shown in Fig. 12-8. Suppose we could somehow orient all our molecules this way (in a host matrix or in a crystal) and that we then subjected the sample to plane-polarized light. We will consider what should happen for light polarized in each of the directions x, y, z. First, let us consider the integral π|z|π∗. The ethylene molecule has three reflection planes of symmetry, so we can examine the symmetry of the integrand with respect to each of these three reflections. Remember that, if the integrand is antisymmetric for any one of these reflections, the integral vanishes. The function π can be seen, from inspection of Fig. 12-8, to be symmetric for reflection in the xz and yz planes, antisymmetric for reflection in the xy plane. These observations are shown in Table 12-2, along with similar conclusions regarding the symmetries of the functions π ∗, x, y, and z. The symbols “s” and “a” stand for “symmetric” and “antisymmetric.” Our integral π|z|π∗ can now be seen to have an integrand that is symmetric for reflection in the xz plane but antisymmetric for reflection in the xy
Figure 12-8 π and π ∗ MOs for ethylene. The C–C bond is coincident with the x axis and all nuclei lie in the xy plane.
9Here, π∗ refers to the antibonding πg MO, not to complex conjugation.
10In this discussion, we consider molecules of high symmetry, so that reflection planes of symmetry exist. For molecules of lower symmetry, one must resort to actual calculation of the cf in Eq. (12-88).
Section 12-10 Spectroscopic Selection RulesTABLE 12-2 Symmetries of Functions under Reflection through Cartesian Coordinate Planes
Symmetry operation
Function
xy reflection
xz reflection
yz reflection
π
a
s
s
π ∗
a
s
a
z
a
s
s
x
s
s
a
y
s
a
s
π z π ∗ a a a = a s s s = s s s a = a π x π ∗ a s a = s s s s = s s a a = s π y π ∗ a s a = s s a s = a s s a = a and yz planes. Hence, this integral vanishes, and this transition is forbidden insofar as light polarized perpendicular to the molecular plane is concerned. According to Table 12-2, symmetry also causes π|y|π∗ to vanish. However, symmetry does not force π|x|π∗ to vanish. Therefore, the π∗ ← π transition is allowed and is polarized along the C–C axis. Indeed, because of the significant spatial extension of the π and π ∗ MOs (compared to AOs), the integral π|x|π∗ is relatively large, which means that the transition is not only allowed, but is intense.
It is physically reasonable that the π ∗ ← π transition should be polarized along the C–C axis. If some π ∗ character is mixed with π , it is easy to see that this results in a shift of π charge from one carbon to the other—a shift along the x axis. Conversely, an electric field in the x direction will cause polarization along the x axis, hence mix π ∗ character into the π MO. When the perturbation is removed, a finite probability exists that the molecule will go to the state wherein π ∗ is occupied.
In the above example, we imagined all ethylene molecules to be identically oriented.
Under those conditions, we would observe a maximum in π ∗ ← π absorption when our incident light was polarized parallel to the molecular axis and zero absorption (ideally) when the polarization axis was perpendicular to the molecular axis. If the light is unpolarized, or if the ethylene is randomly oriented, as in liquid or gaseous states, the absorption is isotropic and allowed because there is always a certain degree of “overlap” between the molecular axes of most of the molecules and the direction of the electric field due to the light.
The example discussed above illustrates a simple and powerful rule for recognizing whether or not a given atomic or molecular orbital dipolar transition is allowed (assuming electron spin agreement): A dipole transition is allowed between orbitals that have
opposite symmetry for one and only one reflection plane.11 The polarization of the
transition will be perpendicular to that reflection plane.
11This means that, for an allowed transition, one orbital has one and only one nodal plane where the other does not.
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation TheoryEXAMPLE 12-8 Selection rules for allowed dipolar transitions in a hydrogen atom include l = ±1, ml = 0, ±1. Show that these rules are equivalent to the symmetry rule stated above.a Consider the following cases: 2s ← 1s, 2px ← 1s, 3dxy ← 2s, 3dxz ← 2pz, 3dxy ← 2pz.
aThere is yet another way to rationalize these selection rules. Photons are bosons with spin quantum number of 1, hence they carry spin angular momentum. The total angular momentum of the atom or molecule that absorbs a photon must change to obey the conservation requirement consistent with the vector addition of orbital and photon angular momenta. The l rule handles that for one-electron atoms. Furthermore, for a unit change of l, the z-component of the absorber’s orbital angular momentum can change by +1, 0, or −1 a.u. The ml rule takes care of that. This rationalizes the atomic selection rules, but doesn’t yield information on polarization or intensity as readily as the more physical approach described in the text, and it is less easily applied to molecules. Generally, however, the more ways one understands a thing, the better.
SOLUTION The 2s ← 1s transition is forbidden by the l rule (because l = 0) and also by the symmetry rule (because neither wavefunction has a nodal plane). This makes it easy to see that the l rule applies because it forces one of the orbitals to have one and only one more nodal plane than the other. The 2px ← 1s transition is allowed by the l rule. Since 2px is a linear combination of 2p+1 and 2p−1, the ml rule is also satisfied, so the transition is allowed. When we examine symmetry, we note that there is only one plane of disagreement–the yz plane, so this approach also indicates that the transition is allowed, and that it is polarized in the x direction (perpendicular to the yz plane). The 3dxy ← 2s transition is forbidden by the l rule since l = 2.
The symmetry approach indicates that a transition is forbidden because the 3dxy AO has two nodal planes, whereas 2s has none. The 3dxz ← 2pz transition has l = 1. 3dxz is a combination of 3d+1 and 3d−1, and 2pz is 2p0, so ml = ±1. Therefore, this is an allowed transition. The symmetry approach shows 3dxz with two nodal planes (xy and yz) and 2pz with one (xy). Therefore, there is only one plane of symmetry disagreement (yz), so the transition is allowed and is x polarized.
The 3dxy ← 2pz transition has l = +1. 3dxy is a linear combination of 3d+2 and 3d−2, and 2pz is 2p0, so ml = ±2. Therefore, the transition is forbidden. The symmetry analysis shows 3dxy having two nodal planes (xz and yz), and 2pz having one (xy). Therefore, they disagree in symmetry in three planes, and the transition is forbidden. This last case makes it clear why the ml rule is needed: When satisfied, it guarantees that all but one of the nodal planes in the two AOs coincide. (If ml = ±2, then, taking the z axis to be vertical, one AO has two more vertical nodal planes than the other, so there must be at least two reflection planes of symmetry disagreement.)
12-10.A Problems12-1. Prove that the energy to first order for the lowest-energy state of a perturbed system is an upper bound for the exact energy of the lowest-energy state of the perturbed system, that is, that E0 + W (1) ≥ W
0
0.
12-2. A one-dimensional box potential is perturbed as sketched in Fig. P12-2. From a consideration of the first three wavefunctions of the unperturbed system, ψ1, ψ2, Figure P12-2
Section 12-10 Spectroscopic Selection Rules ψ3, which will have its energy increased most, to first order, and which least?
No explicit calculation is necessary to answer this question.
12-3. Evaluate by inspection the first-order contributions to the energies for the states shown in Fig. P12-3. In every case the unperturbed state is a particle in the indicated state in the one-dimensional, infinitely deep, square well.
Figure P12-3 a) H = sine function as shown, ψi = ψ1. b) H as shown, ψi = ψ1. c) H as shown; ψi = ψ1. d) H as shown; ψi = ψ2. e) H = cosine as shown, ψ˙ι = ψ2. f) H as shown; ψi = ψ2.
12-4. a) Evaluate W (1) for the particle in a box with the perturbing potential shown in
2
Fig. P12-4.
b) For the perturbation above: 1) What sign would you expect for c(1)? Describe the effect of this on φ
21
1; on W (2).
1
2) What sign would you expect for c(1)? Describe the effect of this on φ
12
2; on W (2).
2
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation TheoryFigure P12-4 12-5. Calculate c(1) for the perturbed particle in a wire example discussed in Sec 41
tion 12-3. Show that c(1) is only about 2% as large as c(1). Use
41
21
π
y sin my sin ny dy = (−1)m+n − 1 2mn/(m2 − n2)2, m, n = 1, 2, . . . , m = n
0 12-6. An electron moves in a harmonic potential, V = 1 kx2. What is the effect, to first
2
order, on the energies of superimposing an electric field, V = Ex? Explain your reasoning.
12-7. Calculate the energy to first order of He+ in its lowest-energy state. Use the hydrogen atom in its ground state as your zeroth-order approximation. Use
atomic units. Predict the signs (plus, zero, minus) of c(1) and c(1) . Explain
2s,1s
2p0,1s
your reasoning.
12-8. The previous problem gives an energy to first order of −3/2 a.u. for He+, using H as a starting point. Now reverse the process and try to get the ground-state energy of H to first order, using He+ as starting point. Discuss the reasonableness of your answer in terms of a lower bound.
12-9. A hydrogen atom in its ground state is perturbed by imposition of a uniform z-directed electric field of strength F atomic units: Field = −F z = −F r cos θ.
a) What is the first-order change in energy experienced by the atom? Show your logic.
b) For the first-order correction to the ground-state wavefunction, we can con sider c(1) and c(1) to be the coefficients for mixing in 2s and 2p
2s,1s
2p
z char z ,1s acter. Predict the signs of these coefficients and sketch their effects on the ground-state wavefunction.
c) Predict the sign of W (2) and explain your reasoning.
1s
Section 12-10 Spectroscopic Selection Rules12-10. What is the H¨uckel MO π -electron energy to first order of the molecule (I) if C4 is perturbed by H = 0.1β? (See Appendix 6 for H¨uckel data.)
12-11. Consider the fulvene molecule (II) in the HMO approximation (seeAppendix 6).
At which carbon will a perturbation involving α affect the total π -electron energy the least? Calculate to first order the energy of fulvene with a6 = a + 0.5β. If a computer and H¨uckel program are available to you, calculate the energy for this perturbed molecule directly and compare with your first-order result.
12-12. Produce an expression for πkk, the atom–atom self-polarizability, from Eq. (12-70). Can this, like πlk, have either sign? What can you infer about the sign of δqk when δαk is positive? Discuss the physical sense of this.
12-13. Using data from Example 12-6, calculate the self-atom polarizability, π1,1 for carbon 1 in naphthalene.
12-14. Consider the hexatriene system. (See Appendix 6 for data.) One of the two central atoms is perturbed so that α becomes replaced by α + hβ on that one atom only.
a) What is the energy change, to first order, of the total π energy of hexatriene?
b) Which of the occupied MOs is perturbed the most (i.e., has the largest first order contribution to the energy) by this perturbation?
12-15. Answer the following questions from examination of HMO data for benzcy clopentadienyl radical, C9H7, given in Appendix 6.
a) What is the total π energy to first order if atom 5 is replaced by a new atom X which contributes the same number of π electrons a carbon did and has αx = α + 0.5β and also has βCX = 0.9β?
b) For the positive ion of this system, how do you think the self-atom polariz ability should compare qualitativeIy at atoms 1 and 2? Make your prediction by inspection, rather than computation, and explain your thinking.
12-16. Use simple H¨uckel MOs for butadiene to calculate to first order the change in π energy that would result from closing cis-butadiene to cyclobutadiene (III).
Repeat the approach for closing hexatriene to benzene. (IV). Which of these two systems benefits most from cyclic as opposed to linear topology? For each
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory system, compare your energy to first order with the actual H¨uckel energy (see Appendix 6 for data).
12-17. The unperturbed H¨uckel MO energy levels of the allyl system are sketched in Fig. P12-17. The system is perturbed so that H22 = α + cβ where c is positive.
a) Sketch the effects of this perturbation, to first order, on the energy levels.
b) Calculate to first order the perturbed MOs φ(1), φ(1), φ(1) in terms of the
1
2
3
unperturbed MOs ψ1, ψ2, ψ3. Sketch the results in a manner that makes clear the nature of the change in each MO.
Figure P12-17 12-18. Calculate the atom–atom polarizabilities π1,2 and π1,3 for the allyl cation. What do your results indicate will happen to the π -electron densities at atoms 2 and 3 if atom 1 becomes more attractive?
12-19. The cyclopropenyl system is perturbed so that the H¨uckel matrix element H22 = a + cβ, where c is positive.
a) Ascertain, by calculation or inspection, the appropriate zeroth-order degen erate wavefunctions for this situation.
b) Sketch the effects of the perturbation on the orbital energies to first order.
12-20. Given the HMO data in Appendix 6 for the cyclopentadienyl radical, calculate to first order all the orbital energies you would obtain if atom 2 were changed so that α2 became α + 0.5β.
12-21. Imagine two allyl radicals coming together to form benzene. Each allyl has three π MOs and benzene has six.
a) Sketch an energy level diagram showing how interaction between allyl orbitalscanberelatedtotheMOenergiesofbenzene,atthesimpleH¨uckellevel.
b) Show with sketches how the three lowest-energy zeroth-order allyl combi nation orbitals are related to the three lowest-energy benzene MOs.
c) Estimate to first order the energies of the benzene-like MOs coming from the nonbonding MOs of the allyl pair. Compare these to the relevant MO energies of benzene. (Remember that, when combining fragments, you must renormalize the functions) d) What is the energy to first order of the lowest MO of the allyl combination, and how does it compare to the lowest benzene orbital energy?
12-22. Discuss the possible transitions π3 ← π2 and π4 ← π2 in butadiene (Here π1 is the lowest-energy MO, etc.)
a) Are they dipole-allowed transitions?
b) If yes, what is the polarization?
Section 12-10 Spectroscopic Selection Rules
425
TABLE P12-24a
Level
Energy
c1
c2
c3
c4
c5
c6
c7
−2.01β
−0.238
0.500
−0.406
0.354
0.336
0.354
−0.406
−1.26β
−0.397
0.500
−0.116
−0.354
0.562
−0.354
−0.116
−1.00β
0
0
−0.500
0.500
0
−0.500
0.500
0.00
−0.756
0
0.378
0
−0.378
0
0.378
1.00β
1.26β
2.10β
a Coefficients for 2pz basis function forming the four highest MOs are listed.
Overlap has been assumed to be negligible.
12-23. Calculate the dipole moment in the z direction for the states φ(0) ± of Eq. (12-81).
Now construct a variational function of the form φ = cos(α)2s + sin(α)2pz and maximize the z component of the dipole moment as a function of α. Are the states φ(0) ± those of maximum dipole magnitude? Discuss why this is reasonable.
12-24. The benzyl radical, C7H7(V), has the H¨uckel energies and ground-state config uration given in Table P12-24.
The radical is trapped and oriented in an external reference system as shown.
Light polarized in the x direction is beamed on the system, and the frequency varied until an absorption is observed. Assuming this to result from excitation of the unpaired electron, to which level has the electron been promoted?
12-25. The simple H¨uckel energies, occupation numbers, and coefficients for MOs in naphthalene (VI) are listed in Table P12-25. Assume that a single crystal of naphthalene is oriented so that each molecule is aligned with respect to an external coordinate system (VI).
a) Light polarized in the x direction is beamed on the crystal. Assuming that the electron is excited from the highest occupied MO, to which empty MOs could it be promoted by the x-polarized light?
b) Which transitions from the highest occupied MO would be allowed for y-polarized light?
c) Which ones would be allowed for z-polarized light?
d) Are any transitions from the highest occupied MO forbidden for nonpolar ized light?
0
1
.41
.35
.46
0
0
0
0
0
0
−
−
.42
.42
.40
.26
.30
0
0
0
0
0
0
−
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
−
−
−
.42
.42
.40
.26
.30
0
0
0
0
0
0
−
−
−
ficients
Coef
.41
.35
.46
0
0
0
0
0
0
−
−
.42
.42
.40
.26
.30
0
0
0
0
0
0
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
−
−
.26
.26
.41
.17
.42
.23
0
0
0
0
0
0
−
−
123456789
.42
.42
.40
.26
.30
0
0
0
0
0
0
−
−
−
−
no.
2
2
2
2
2
2
0
0
0
0
Occupation
P12-25
gies
β
−
.303
.618
.303
.000
.618
.618
.000
.303
.618
.303
ABLE
2
1
1
1
0
0
1
1
1
2
T
Ener
−
−
−
−
−
+
+
+
+
+
426