d 2 dx 2. H = 2 2m p 0 dξ 2 H = 2m e ma2 H = 1 2m dξ 2

Transcription

1 CHAPTER 14 The Origins of Chemical Bonding SECTION A proton (of mass m p ) in a 1-D box of length L = 10a 0, the system in part (a), has a Hamiltonian we write in terms of the ordinary length x as H = m p d dx. To write this in atomic units, we substitute = 1 and express lengths in multiples of the Bohr radius a 0 and masses in multiples of the electron rest mass, m e. We define m = m p /m e and ξ = x/a 0 so that, formally, we first write H = d m e ma 0 dξ and then set = 1, m e = 1, and a 0 = 1, yielding the atomic units version of H: H = 1 m d dξ where we must remember that m is the proton mass measured in multiples of m e and ξ is the distance in multiples of a 0. Part (b) considers an electron in a harmonic oscillator with the ordinary Hamiltonian H = m e d dx + 1 kx. 33

2 The kinetic energy term is treated just as for the 1-D box in part (a), but what should we take for the atomic unit of force constant? Two possibilities come to mind. The first equates the atomic unit of energy from Table 14.1, e /(4πε 0 a 0 ), to the energy quantum of the harmonic oscillator, ω with ω equal to k/m e if the mass is the electron mass. Solving this equality for k gives k = e 8 m e 3 (4πε 0 ) 4 6. The second possibility equates the harmonic oscillator potential energy, kx /, to an appropriate measure of energy. The harmonic oscillator Virial Theorem says that the average harmonic oscillator potential energy, <V>, equals one-half the total energy. You can show that equating one-half the atomic unit of energy to ka 0 /, the potential energy expressed with the atomic unit of length, gives the same expression for k as above. Thus, we define κ to be the force constant measured in multiples this atomic unit of force constant and ξ to be the distance in units of a 0, and we write H = 1 d dξ + 1 κξ where we have no variable for the mass because our mass here is the electron mass and atomic units set m e = 1. For part (c), an H atom orbiting in a plane at a radius of 0a 0, note that we have, formally, the same scenario discussed in Example 1.6 on page 419 in the text, which describes an electron orbiting in a plane at a constant radius. The ordinary Hamiltonian is purely kinetic energy: H = L mr = L z mr = d m H (0a 0 ) dφ where R = 0a 0 is the orbiting radius, m = m H, the hydrogen atom mass, and the L z is given in Eq. (1.3c). If we define µ = mh /m e, the H atom mass in multiples of the electron mass, and then set m e = 1, = 1, and a 0 = 1, we have the atomic units version: H = 1 d µ(0) dφ = 1 d 800µ dφ. 333

3 14. The HeH + B O Hamiltonian has terms for the kinetic energy of each electron (1 and ) and for the potential energy between He and H nuclei a fixed distance R apart, between each electron and each nucleus, and between the two electrons: H B-O = r 1He r He r 1H r H r 1 R. The first two terms are electron kinetic energy terms, the next two are electron He nucleus (Z = ) attractive potential energy terms, the next two are electron H nucleus (Z = 1) attractive potential energy terms, and the final two are electron electron and nucleus nucleus repulsive potential energy terms, respectively The solution to this problem requires a mixture of straightforward math and some physical insight. First, the math. The elliptical coordinate representation of the Laplacian operator is given in the problem, but the H + Hamiltonian, Eq. (14.4), contains the variables r 1A and r 1B (as well as the constant internuclear distance R), and our first step is to express these two distance variables in elliptical coordinates. The elliptical distances ξ and η are defined at the bottom of page 503 in the text: ξ = r 1A + r 1B R and η = r 1A r 1B R. Solving these for r 1A and r 1B gives r 1A = R ξ + η and r 1B = R ξ η. We write the Schrödinger equation as Hψ Eψ = 0, write the wavefunction as ψ(ξ, η, φ) = Ξ(ξ)Η(η)Φ(φ), substitute the expressions above for r 1A and r 1B into the Hamiltonian, Eq. (14.4), and find Hψ Eψ = HΞ(ξ)Η(η)Φ(φ) EΞ(ξ)Η(η)Φ(φ) = 0 = 1 1 Ξ(ξ)Η(η)Φ(φ) + 1 R 1 1 E Ξ(ξ)Η(η)Φ(φ) r 1A r 1B 334

4 = 1 1 Ξ(ξ)Η(η)Φ(φ) + 1 R R(ξ + η) R(ξ η) E Ξ(ξ)Η(η)Φ(φ) = 1 1 Ξ(ξ)Η(η)Φ(φ) + 1 R 4ξ R ξ η E Ξ(ξ)Η(η)Φ(φ). Before introducing the elliptical Laplacian, we can take two simplifying steps at this point. First, since R is a constant, we can define E = E 1/R. Next, we note that the factor multiplying the elliptical Laplacian, 4/R (ξ η ), also multiplies (within a factor of R) the middle term in parentheses in the last expression above. When we introduce the elliptical Laplacian, divide the equation through by 4/R (ξ η ), and introduce E, we have 1 ξ ξ 1 ξ + η 1 η η + + Rξ + R ξ η E Ξ(ξ)Η(η)Φ(φ) = 0. 4 ξ η ξ 1 1 η φ Ξ(ξ)Η(η)Φ(φ) Next, we note that the factor multiplying / φ can be factored: ξ η ξ 1 1 η = 1 ξ η. This is important, because in order to separate variables, we must write the Schrödinger equation as a sum of terms each of which contains only a single independent coordinate variable. While this factorization helps, note that each factor still multiplies / φ, and to eliminate φ, we need some physical insight. Due to the cylindrical symmetry of the system, the Hamiltonian has no potential energy term that depends on φ; this coordinate appears only in the (kinetic energy operator) Laplacian. Thus, physically, motion in the φ direction is that of a free particle, and we have solved this problem already it is the particle-ona-ring problem (see page 506). This means Φ(φ) must be Φ(φ) = N e imφ where N is some piece of the overall normalization constant (it happens to be 1/ π, but we can ignore it here) and m is any integer: m = 0, ±1, ±,. 335

5 This lets us evaluate the second derivative of Φ: Φ/ φ = m Φ, and when we substitute this into the full Schrödinger equation, we find that, since m is a constant, we have separated the equation: 1 ξ ξ 1 ξ + η 1 η η m + Rξ + R ξ E 4 R η E 4 ξ 1 Ξ(ξ)Η(η)Φ(φ) = 0. m 1 η Ξ(ξ)Η(η)Φ(φ) Complete separation follows (and note that we have only the functions Ξ and Η to worry about) if we divide by ΞΗΦ (after expanding the first term above so that the correct functions are operated on by the appropriate derivatives): 1 d Ξ dξ + ξ 1 dξ m dξ 1 d Η dη ξ 1 + Rξ + R ξ E 4 1 η dη dη m 1 η R η E 4 = 0. The equation above has grouped together all those terms that depend on ξ and all those that depend on η. For any value of the constants m, R, and E, each group is a unique ordinary differential equation for Ξ(ξ) and Η(η). Their solutions are known, albeit involved The integral in question here is S = 1s A 1s B dτ = 1 π e r 1A e r 1B dτ where we wish to use elliptical coordinates. We can express the two distances in these coordinates using the expressions derived in the previous problem: r 1A = R ξ + η and r 1B = R ξ η so that, using the expression for the volume element dτ given in the problem, we have 336

6 S = 1 π e r 1A e r 1B dτ = R 3 8π = R 3 4 ξ = 1 ξ = 1 1 π ξ η e R(ξ + η)/ e R(ξ η)/ dξdηdφ η = 1 φ = 0 1 η = 1 ξ η e Rξ dξdη where the simple integration over φ has been performed to go from the second to the final expression above. Expanding the double integral gives us two double integrals, and integration over η is straightforward: 1 S = R 3 4 ξ η e Rξ dξdη ξ = 1 η = 1 1 = R 3 4 ξ e Rξ dξdη ξ = 1 η = 1 ξ = 1 = R 3 4 ξ e Rξ dξ 3 ξ = 1 ξ = 1 1 e Rξ dξ η = 1. η e Rξ dξdη The integrals given with the problem show us how to integrate over ξ: S = R 3 4 = R 3 4 ξ e Rξ dξ 3 ξ = 1 = e R R 3 e Rξ ξ ξ = 1 R ξ R R R + R + R 3 R 6 e Rξ dξ 3 e Rξ R 1 = e R 1 + R + R

7 14.5 Note that diagram (a) looks like an atomic d orbital, (b) looks like two atomic p orbitals next to each other, (c) looks like a d orbital pointing to four s orbitals, and (d) resembles the top diagram on the cover of the text. We will meet these diagrams again as models of various bonding situations. We can construct the table below for the eigenvalues of the various symmetry operators: σ h σ v i (a) (b) (c) (d) We take the z axis to lie along the internuclear axis with z = 0 centered between the nuclei that are spaced at R = au: z = R/ z = 0 z = R/ B R = au A The g and u wavefunctions are given in Eq. (14.7) on page 508 with the overlap integral S given earlier on that page and derived in Problem We can take the 1s atomic orbitals as e r where r is the distance from the nucleus, neglecting overall normalization, since we are considering only relative comparisons at various points along z. For atom A, r is z R/ (since z = R/ should correspond to r = 0 for this atom and r is always positive), and similarly, r is z + R/ for atom B. Thus, the wavefunctions are Ψ g = e z R/ + e z + R/ (1 + S) and Ψ g = e z R/ e z + R/ (1 S) where S = e R 1 + R + R 3 = (for R = ). Graphs of Ψ g and Ψ u are shown at the top of the next page (with zero for the g wavefunction s axis at the bottom of the graph and that for u in the middle): 338

8 Ψ g Ψ u R R R/ 0 z R/ R R If R =, then S = , and at the bond midpoint, z = 0 and Ψ g = ; at the A nucleus, z = R/ and Ψ g = These numbers give the ratio Ψ g (z = 0) Ψ g (z = R/) = = At a point R/ beyond the A nucleus, z = R and Ψ g = 0.345, and we have Ψ g (z = R) Ψ g (z = R/) = = so that the electron density (which is proportional to the square of the wavefunction) is about a factor of 3 larger (since (0.6481/0.3679) 3) at the bond midpoint than at a distance R/ on the outside of the bond In atomic units, an electron half-way between two protons spaced 1. au apart is a system with a total potential energy V = = 3 =.5 au 1. where the first term is the proton-proton repulsion and the second and third are the electron-proton attractions. This is significantly lower than 1.34 au, as predicted in the problem. Next, we consider the geometry at the top of the next page and ask for that value of r that reproduces the 1.34 au potential energy value. 339

9 r r 1. au + + We solve V = 1 1. = 1.34 au so that r = 0.9 au. r The diagram above is drawn to scale for this value of r. The potential energy is telling us that the electron is not so tightly localized between the nuclei The molecular Virial Theorem (Equations (14.13b) and (14.13c)) tell us <T> = <E> R d<e> dr and <V> = <E> + R d<e> dr so that for <E> = exp( R), we have, by direct substitution, <T> = e 1.050R Re 1.050R <V> = e 1.050R 3.031Re 1.050R. A graph of these functions, shown below with the same shifts for the <T> and <V> curves as were used in the text, compares favorably to the graph of the exact curves shown with the problem. Energy/au 0 <T> 0. <E> u <V> R/au 340

10 SECTION Example 13.5 discusses the ground and first excited states of the halogen atoms, and the key to explaining the photoelectron spectrum of the rare gases is to recognize that the rare gas cations from Ne through Rn are isoelectronic to the halogens F through At and thus have the same qualitative energy level structure. For example, 1. ev photoionization of Ar has two possible outcomes: Ar(3s 3p 6 1 S 0 ) + hν Ar + (3s 3p 5 P 3/ ) + e Ar + (3s 3p 5 P 1/ ) + e In accord with Hund s Rules and in analogy with the halogens, the ground state of Ar + is P 3/ and the first excited state is P 1/. If the photon has 1. ev energy and the photoelectron has a 5.46 ev kinetic energy, then 1. ev 5.46 ev = ev represents the ionization energy of Ar to one of these states. If the electron kinetic energy is 5.8 ev, the same calculation shows that ev is needed to reach the other ion state. Thus, the photoelectron with the greater kinetic energy corresponds to photoionization to the ground P 3/ state, while the smaller kinetic energy corresponds to the P 1/ excited state ion product. Note from the data given in the problem that the energy separation between these two states follows the same general periodic trend as do the halogen atom state separations: 0.18 ev for Ar +, 0.66 ev for Kr +, and 1.31 ev for Xe +. (Ne + has a similar energy level structure and the separation of the corresponding two states is smaller but 1. ev light is not energetic enough to ionize Ne, which has a ev ionization potential.) We have ten electrons in HF, and the ground-state electron configuration is 1σ σ 3σ 1π 4. (If we followed the analogy with CO on the MO ordering, we might write 1σ σ 3σ 4σ 1π, but this would place two electrons in the 1π MO, and we would expect them to enter as a triplet, one in each of the two degenerate 1π MOs. HF is a closed-shell, singlet molecule. Hence the 4σ MO in HF and in all the hydrides from LiH through HF lies above the 1π MO.) Photoelectrons around 5.18 ev must come from the HOMO so that the ion state configuration is 1σ σ 3σ 1π 3. The 1π electrons are not particularly bonding in character, and thus the ion state has nearly the same bond length as the neutral. (Experiment shows it to be Å compared to Å for neutral HF.) The features spaced by about 0.35 ev in the photoelectron spectrum are due to simultaneous ionization to this electronic state and vibrational excitation of the molecular ion. The broad.13 ev feature is the result of removing a 3σ 341

11 electron, leaving the excited configuration 1σ σ 3σ 1 1π 4. The 3σ MO is strongly bonding, and removing an electron from it produces an ion state that is less strongly bound and has a much longer bond length (1.4 Å). The breadth of this feature is due to extensive internal excitation of this excited ion state We can use isoelectronic analogies to guide our ground-state electron configuration assignments for the ions: NO + is isoelectronic to N and NO is isoelectronic to O. We consult Table 14.3 (neglecting the u and g notation since our molecules are heteronuclear) and follow the discussion in Example 14., NO + has the 14-electron configuration 1σ σ 3σ 4σ 1π 4 5σ, NO has the 15- electron configuration 1σ σ 3σ 4σ 1π 4 5σ π 1, and NO has the 16-electron configuration 1σ σ 3σ 4σ 1π 4 5σ π. We assign NO + a bond order of 3, paralleling N, assign NO a bond order of, paralleling O, and NO has a fractional.5 bond order, as did O + in Example We can use the energy data in the problem to construct the diagram below, which confirms that the bond energies of NO +, NO, and NO follow their bond orders. The bond energy of NO + is D e (NO + ) = ev ev 9.5 ev = ev, and that of NO is D e (NO ) = ev ev ev = ev, in accord with the bond orders. (And finally, N does not have a positive electron affinity because of the stability afforded by its half-filled p subshell in the ground-state configuration of the neutral atom: 1s s p 3.) 0 N + + O + + 3e Energy/eV ev ev 9.5 ev ev N + + O + e N + O + + e D e (NO + ) NO + + e N + O + e ev D e (NO) 0.04 ev N + O NO + e ev NO ev D e (NO ) ev 34

12 14.1 Since both He and Be have closed shell (for 1s He) or closed subshell (for 1s s Be) ground-state configurations, we should expect HeBe to have no formal bond, in analogy to He or Be (see Table 14.3). We can write a ground-state MO configuration easily enough: 1σ σ 3σ. Now we confront the less-obvious question of bond order. The lowest energy AO of He and Be must be the 1s AO of Be, because the nuclear charge of Be is twice that of He. Ionization potentials (see Table 7.3) show us that the He 1s AO is lower in energy than the Be s AO (by about 15.3 ev, the difference in ionization potentials), and excitation energies for He (discussed in Chapter 13 on page 475 in the text) tell us that the He s AO is about 0 ev above the 1s, or 4.7 ev above the Be s. These large separations in AO energies leads us to predict that, even though these orbitals have the correct symmetry to mix, they will not do so to any appreciable extent. Thus, we conclude that the HeBe 1σ is essentially the pure Be 1s AO, the σ is essentially pure He 1s, and the 3σ MO is the Be s AO. The lack of any appreciable mixing supports our suspicion that the bond energy should be quite small, and all three MOs are essentially nonbonding. Formally, this six-electron molecule is isoelectronic to Li, but the electron distribution and its resultant bonding is obviously quite different! The first excited state comes from the MO excitation 1σ σ 3σ 1σ σ 3σ 1 4σ 1, and the only question is the predominant atomic orbital contribution to the 4σ MO: is it He s or Be p? We know the He s orbital is about 4.7 ev above the Be s, and we should expect the Be p orbital to be a comparable energy above Be s. (This is not an obvious energy to guess! Atomic spectroscopy shows that the sp 3 P 1 state of Be is about.73 ev above the s 1 S 0 ground state, and the sp 1 P 1 state is about 5.8 ev above 1 S 0.) Consequently, the 4σ MO should be a significant mixture of He s and Be p (in a pσ orientation) and lead to considerably stronger bonding in this excited state with 4σ not as strongly localized on either atom as the lower energy MOs clearly are As is often the case, isoelectronic analogies are useful here. The anion CN is isoelectronic to CO (or N ) and has the closed-mo ground-state configuration (see Example 14.) 1σ σ 3σ 4σ 1π 5σ. The radical CF is a 15-electron species like NO (see Problem 14.11) with the ground-state configuration 1σ σ 3σ 4σ 1π 4 5σ π 1 and one unpaired electron. The CH radical is a seven electron species with the 1σ σ 3σ 4σ 1 configuration and one unpaired electron, while CH + loses the 4σ electron (predominantly from the C atom) to give a 1σ σ 3σ ground-state, closed-mo configuration The interesting dication He + is isoelectronic to H and dissociates to two He + ions (rather than to He + He +, which is higher in energy by 9.8 ev, as you 343

13 can verify from the first and second ionization potentials of He discussed in Chapter 13). We assign a simple 1σ g configuration to ground state He + and a bond order of 1, suggesting that it is bound, but we must remember that, unlike H, He + has a residual nuclear charge on each atom that is not offset by electron charge. If we bring two naked protons together, they repel, of course, and the uncompensated nuclear charges on He + are roughly just that: two +e charges repelling each other. We can easily plot the pure Coulombic repulsion as a function of internuclear distance; it is just e /(4πε 0 )R, as shown by the heavier line in the graph below (with energy converted to ev units). We see from the graph that at distances characteristic of an expected He + bond length, around 1 Å or less, the nuclear repulsion energy has risen nearly 15 ev above the asymptotic energy zero at large R. It is unreasonable to expect the He + bond energy to be greater than (or even anywhere close to) 15 ev. Consequently, the binding well in He + is a depression in the nuclear repulsion curve (the lighter curve in the graph, drawn to follow an ab initio calculation for the He + ground state). This means He + is only metastably bound. The two He + ions that comprise it constantly repel each other, and even if trapped in the binding well, they can eventually tunnel through the potential energy hump that momentarily binds them, escaping to R =. Energy/eV SECTION R/Å We place the zero of energy at the energy of H(1s) + H(1s) infinitely apart so that the ion pair H + + H infinitely apart has the energy IP(H) EA(H) = ev = ev, the difference between the H ionization potential and its electron affinity (using data from Chapter 7). As H + + H approach to some finite distance R, their energy falls due to their attraction, following the Coulombic function e /4πε 0 R = 14.4 ev/(r/å) (as derived in 344

14 footnote 17 on page 549 in the text) shifted up by ev. With this energy scale, the energy of the neutral pair H(1s) + H(ns) is (1 1/n ). If n = 1, this energy is zero, and the ion pair energy falls to zero at R such that 14.4 ev R/Å ev = 0 or R = 1.1 Å. This is the first ion pair neutral line crossing point. Repeating this calculation, substituting (1 1/n ) for zero in the expression above with n =, 3, and 4, gives crossing points at R = 5.44 Å, 19.0 Å, and 151 Å, respectively. If n = 5, the neutral energy is above the ion pair asymptote at ev. These various energies, crossing points, and ion-pair attraction curve are shown in the graph below (although the n = 4 crossing point is off the scale of this graph). Finally, we can tell from the graph that the ion pair state must be lower in energy than the H(s) + H(s) asymptotic limit, and we can calculate that this limit lies at [13.598(1 1/ )] ev = ev. Thus, H + + H H(s) + H(s) is endothermic by ev ev = ev. 14 H(1s) + H(3s) H(1s) + H(4s) H(1s) + H(5s) H + + H Energy/eV 10 5 H + H ion attraction H(1s) + H(s) R/Å H(1s) + H(1s) We can adapt the general methodology used for H 3 + on page 58 of the text to H if we let φ i in Eq. (14.14) represent the H 1s AO on nucleus i. The general variational equations for this LCAO approximation are c 1 (H 11 ES 11 ) + c (H 1 ES 1 ) = 0 c 1 (H 1 ES 1 ) + c (H ES ) = 0. If we apply the Hückel approximations to these equations, we have c 1 (α E) + c β = 0 and c 1 β + c (α E) =

15 The secular determinant equation is α E β β α E = 0, and if we define x = (α E)/β, this equation becomes x 1 1 x = 0 or x 1 = 0 or x = ±1. If x = ±1, then E = α ± β. For x = 1, or E = α + β, (the ground state), we solve the variational equations for c 1 and c, including the requirement that c 1 + c = 1: c 1 + c = 0, c 1 c = 0, c 1 + c = 1 so that c1 = c = 1. The ground-state wavefunction is thus ψ = 1 1s A + 1s B, which looks like Ψ g for H + (Eq. (14.7a) with S, the overlap integral, equal to 1). For the excited state, x = +1, or E = α β so that the variational equations are solved by c 1 = c = 1/ for an excited state wavefunction ψ = 1 1s A 1s B, which looks like Ψ u for H + (Eq. (14.7b) with S = 1 again) Whether we are considering the neutral radical, the anion, or the cation, the connectivity of the carbon atoms stays the same, and so does the Hückel secular determinant. This determinant is exactly the same as in Eq. (14.19) for linear H 3 : x 10 1 x 1 01x =

16 Expanding it gives x(x ) = 0, or x = 0, ±. Since the energy, E, equals α xβ, the three Hückel π MO energies are, in increasing energy, E = α + β, E = α, and E = α β. The allyl cation has two π electrons, the neutral has three, and the anion has four. Thus, their total π electron MO energies in this approximation are cation: E = E = α + β neutral: E = E + E = 3α + β anion: E = E + E = 4α + β As suggested in the problem, we label the three C atoms in the linear allyl radical a, b, and c. With this notation, the three variational equations are (contrast Eq. (14.15) for H 3 + and the discussion of the secular determinant in the previous problem) c a x + c b = 0, c a + c b x + c c = 0, and c b + c c x = 0, and the normalization equation c a + cb + cc = 1 also holds, as always. We must solve this series of equations three times, once for each MO energy (i.e., once for each value of x as found in the previous problem). For x =, we find that these equations are satisfied by c a = c c = 1/ and c b = 1/. For x = 0, c a = c c = 1/ and c b = 0, and for x = +, we find c a = c c = 1/ and c b = 1/. Thus, the three π MOs are identical in form to those shown on page 53 in the text and graphed in Figure (and reproduced on the cover of the text): ψ = 1 p a + p b + p c ψ = 1 p a p c ψ = 1 p a p b + p c. These π MOs look very much like Figure when viewed from the top, (i.e., along the symmetry axis of the p orbital), and from the side, (i.e., looking at the symmetry plane of these orbitals), they appear as shown at the top of the next page. Solid and dashed contours denote regions of opposite algebraic sign, as in Figure

17 Ψ Ψ Ψ The trimethylene methane radical has a Lewis electron-dot structure that we can draw (in one of three equivalent ways, related by the three equivalent positions of the C=C double bond) as a diradical, in accord with the Hückel prediction: H C H H C C C H H H The central C atom is bonded to the three others, but they in turn are bonded only to that central atom, not to each other. Thus, its Hückel secular determinant is the one that reflects this connectivity. If we let the upper left-hand entry in this determinant represent the central atom, the correct choice is x x 00 10x 0 100x With the simplifications stated in the problem, f = 1 and β CX = β CC = β, the X atom s α value is simply α X = α C + β = α + β. The secular determinant equation for the π MO energies is thus α E β β α + β E = 0 or x 1 1 x + 1 = 0 348

18 if we define x = (α E)/β as usual. Expanding the determinant gives x + x 1 = 0, which has the solutions x = ± 5 1 or E = α + 1 ± 5 β = α 0.618β α β. The ethylene π MO Hückel energies are E = α ± β, in contrast to α β and α 0.618β here. Since β < 0, the lower energy MO here is that with E = α β and is stabilized by 0.618β in comparison to the lower energy ethylene MO. The next problem shows in more detail why this is true If we substitute x = (α E)/β (in the form E = α βx) into the variational equations for c 1 and c, we find c 1 x + c = 0 c 1 + c (1 + x) = 0, and we also require c 1 + c = 1. The two values for x from the previous problem, x = ± 5/ 1/, will give us two sets of c 1 and c values, one for the lower energy bonding π MO and one for the excited antibonding MO. Because the two equations above that are linear in c 1 and c both equal zero (they are said to be homogeneous see Example 13.3 and footnote 5 on page 461 in the text), we can use either one of them and the nonlinear normalization equation, c 1 + c = 1, to find our sets of c1 and c values. For example, we substitute one x value, x = 5/ 1/, in the first linear equation and write c c = 0 or c = 5 1 c 1. We substitute this expression for c into the normalization equation and find c 1 + c = c1 + c1 5 1 = c1 5 5 = 1 or c 1 = 5 5 = Substituting this value for c 1 into the expression for c gives c = If we repeat these steps using the other value for x, we find c 1 = and c = Now we must interpret these values. The table on the next page correlates each MO energy with its corresponding coefficients. The more electronegative atom is atom, and we see that the bonding MO (which has no node c 1 and c have the same algebraic sign and which also has the lower 349

19 energy) is more localized on atom (c > c 1 ) while the antibonding MO (one node c 1 and c have opposite signs) is more localized on the less electronegative atom 1. MO type x E c 1 c bonding 5/ 1/ α β antibonding 5/ 1/ α 0.618β The construction described in the problem is shown below. (Some simple trigonometry will verify that the inscribed hexagon vertices fall exactly at α + β, α + β, α β, and α β, as shown in the diagram.) α β Energy α β α α + β α + β With six π electrons, benzene fills the lowest three MOs (with two paired electrons in each of these three MO, as shown) for a total π energy 6α + 8β, as stated in the text, and thus with a stabilization, or resonance, energy β in comparison with a true cyclohexatriene. If the lowest energy MO has no nodes and the highest has three, then the MO next above the lowest must have one node and the third MO must have two. The energy diagram shows these MOs to be doubly degenerate, and the pictures we can draw with these nodal patterns reflect this degeneracy there are two pictures for each. Shading is used to indicate different algebraic signs, and dashed lines locate nodes. SECTION The BH radical s ground state MO configuration comes from the atomic configurations B 1s s p 1 and H 1s 1 for a total of seven electrons. If the molecu- 350

20 lar ground state had been linear, it would have had the MO configuration 1σ g σg 1σu * 1πu 1, but the Walsh correlation diagram in Figure 14.0 shows us that the bent molecular orbital configuration 1a 1 a1 1b 3a1 1 will be lower in energy due to the strong tendency for the 3a 1 MO to bend the molecule. This MO controls the geometry of the ground state. The first excited state configuration comes from excitation of the highest energy electron to the next available MO. In the bent ground state, the 3a 1 electron is promoted to the 1b 1 MO, but this MO does not have a strong preference for bending, and Figure 14. now shows us that the 1b MO governs the geometry, leading to a linear excited state We consult Figure 14.19(b) and note that the 1b 1 MO is indifferent to the bond angle it is essentially the A atom s p orbital that points perpendicular to the molecular plane. As the H A H bond angle is reduced to 0, this MO becomes a p A π MO in the resultant diatomic. Similarly, the 1b MO becomes the other p A π MO as the 1s H contributions to 1b cancel at zero bond angle From Table 14.5, we see that the bond angle of the dihydrides from water to H Te systematically decrease toward 90. This is suggestive of bonding to the central atom s p orbitals, which naturally point at a 90 angle from each other. As the central atom becomes larger, the repulsion between the two H atoms decreases (these atoms are farther apart), and the valence electrons on the central atom have systematically larger principal quantum numbers, a fact which causes a greater energy mismatch between the A valence atomic orbital energies and those of H. This mismatch gives the highest energy MOs in the molecular ground state a more purely atomic A orbital character. (See also Problem for another way to understand this change in geometry.) 14.6 There are, of course, many possible answers here, but the following list contains fairly well known species. Good examples of linear 16 electron molecules include OCS (a direct analog to CO ) and NO + (an isoelectronic analog). For 17 electron species, which are bent, both BF and O 3 + are reasonable examples. The nitrosyl halides such as NOF and NOCl are good examples of bent 18 electron species, as is the nitrite ion NO. The 19 and 0 electron species are rather scarce. The 19 electron BrO is a direct analog to ClO, and the ozone anion O 3 also falls in this class. For 0 electrons, the nitrogen dihalide anions such as NF, NCl, NClF, etc., are known. Other electron, linear trihalide anions are known besides I 3, but not all, and the rare gas dihalides FXeCl and XeCl are known. 351

21 14.7 For the dsp square planar hybrids, we should mix in the d x y orbital (if we take the plane of the hybrids to be the x y plane, since this d orbital points along the x and y directions. (See Figure 1.15(d) on page 436 in the text.) The dsp 3 trigonal bipyramidal hybrids involve the d z orbital. We can think of these hybrids as ordinary sp trigonal planar hybrids added to a pair of hybrids pointing along the z direction and composed of the d z and p z orbitals. Finally, the d sp 3 octahedral hybrids involve both d z and d x y In dimethyl zinc (or cadmium or mercury all three are known), the central Zn atom has the ground state configuration 3d 10 4s. Promotion to the 3d 10 4s 1 4p 1 excited configuration followed by sp hybridization of the 4s and 4p orbitals leads to the observed linear HC 3 Zn CH 3 bond angle We have 16 valence electrons in carbon dioxide and thus the ground-state MO configuration for the linear geometry is 1σ g 1σu * σg 3σg σu * 4σg 3σu * 1πu 4 1πg 4, as discussed in the text on page 540 and shown in Figure 14.1(a). Excitation to the first excited state involves promotion of one electron from the 1π g nonbonding MO to the next higher energy MO. If the molecule stays linear, this is the antibonding π u MO, but Figure 14. shows that one component of this MO correlates on bending to the 6a 1 MO, which strongly prefers to be bent. It is this MO that governs not only the first excited state geometry, but also the anion geometry, since it is the LUMO of ground state CO and thus is the orbital that accepts an extra electron when CO is formed. Since CO wants to be bent, but CO is linear, when an extra electron is initially attached to CO, the incipient anion finds itself in an energetically unfavorable geometry linear instead of bent and the extra energy associated with this unfavorable geometry is greater than that gained through electron attachment. Thus, the electron is not attached and the electron affinity is negative. You should recognize that electron attachment into the first excited state of CO (into the hole left in the 1π g nonbonding MO on excitation) could lead to a stable anion, since the electron would be attaching itself into an already bent molecule. SECTION We can draw the energy level diagram for the HeH + system that is shown on the next page using the data in the problem and ionization energies for He and H from previous chapters. This diagram shows that HeH + in its ground state dissociates into the He (1s 1 S 0 ) + H + pair (because the first ionization potential of He is greater than the ionization potential of H) while the first excited state of HeH + dissociates into He + (1s 1 S 1/ ) + H (1s 1 S 1/ ), the charge-exchanged 35

22 analog of the ground state dissociation pair. The excited states of He combine with a proton to form even higher energy HeH + excited states. We can also see this if we look at the MOs of these states. We have only two electrons, and thus the ground state MO is simply 1σ. The energy difference between the He and H 1s AOs (He 1s is much lower) tells us that the 1σ MO is almost a pure He 1s orbital; ground state HeH + is not much more than a proton stuck to a He atom. The first excited state, 1σ 1 σ 1, places an electron in the σ MO, which is predominantly the H atom 1s orbital. This state is a hydrogen atom bound to the He + ion. (The ground state binding energy of HeH + is.0 ev, by the way.) 0 10 He + + H + + e He + + H (1s 1 S 0 ) 0 Energy/eV He * (1ss 1 S 0 ) + H + He * (1ss 3 S 1 ) + H + He + (1s 1 S 1/ ) + H (1s 1 S 1/ ) He (1s 1 S 0 ) + H The water ground state MO configuration ends with the 1b 1 MO occupancy, and excitation to the 1b 1 1 4a1 1 configuration tells us we should consult Figure 14.0 to see how the 4a 1 MO energy changes with bond angle. We see that this MO wants to be linear, and thus we can conclude that states derived from this excited configuration should be less bent than the ground state If we consult the dihydride Walsh correlation diagram in Figure 14.0, we can correlate the occupied MOs of bent ground-state H O, 1a 1 a1 1b 3a1 1b1, to the linear configuration 1σ g σg 1σu * 1πu 4. Note that this is very similar to the HF MO configuration discussed in Problem

23 GENERAL PROBLEMS The π electrons in bombykol are localized through the two C C double bonds, exactly as in butadiene. All the other carbon atoms are sp 3 hybridized and bonded through the molecule s σ MO framework. We would expect the π MOs to include the molecules HOMO, just as in butadiene, since the framework σ MOs are lower in energy Normalization of either the α or β hybrids ensure that a 1 + a = 1 if the s, px, p y, and p z orbitals are individually orthonormal. If α and β are to be orthogonal, we must also have αβ dτ = a 1 s + a p x cos θ + p y sinθ a 1 s + a p x cos θ p y sinθ dτ = 0. Expanding the integrand and invoking orthonormality of s, p x, and p y gives αβ dτ = a 1 + a p x cos θ + p y sinθ p x cosθ p y sinθ dτ = a 1 + a cos θ sin θ = 0. The trigonometric identities let us to write cos θ = 1 + cos θ and sin θ = 1 cos θ αβ dτ = a 1 + a cos θ sin θ = a 1 + a cos θ = 0, and solving a 1 + a cos θ = 0 and a1 + a = 1 simultaneously for a 1 and a gives the two expressions stated in the problem: a 1 = cos θ cos θ 1 and a 1 = 1 1 cos θ. 354

24 In a pure sp hybrid, θ = 10, and a 1, the s character, is 1/3, as we should expect. For θ = 105., we find a 1 = It is also interesting to note that θ = 90 gives a 1 = 0 (no s character) and a = 1 (pure p character), as we should expect for orbitals directed at 90. (See also Problem 14.5.) Continuing from the previous problem, if a 3 + a4 = 1 and a + a3 = 1, the value for a 1 from the previous problem lets us find a 3 = 1 + cos θ 1 cos θ and a 4 = cos θ cos θ 1. Again, if θ = 10 (pure sp ), a 3 = 1/3, but for θ = 105., a3 = If we have N double bonds, we have N π electrons to pair into energy levels in a 1-D box of length L = NR where R = 1.4 Å, the C C bond length. This means we put two electrons in the state n = 1, and so on, until we reach the level n = N where n is the usual particle-in-a-box quantum number. The energy for this level is the HOMO energy, which, using the energy expression from Chapter 1 for this system, is E HOMO = π ml n = π 4N R = π 8m e R m e N where m e is the electron mass. Note that this expression is independent of N. The LUMO energy is the energy of the level with n = N + 1: E LUMO = π N + 1 m e 4N R = π 8m e R 1 + N + 1 N = E HOMO 1 + N + 1 N, and the HOMO LUMO excitation energy is thus E LUMO E HOMO = E HOMO N + 1 N. The numerical value for this excitation energy (for N values of chemical interest, i.e., in the range 1 to maybe a dozen or so, has a magnitude of a few ev, in rough qualitative agreement with experiment. In particular, many dye molecules have conjugated systems in them that can be adjusted in length in order to control the dye s color. 355

25 14.37 If we number the carbon atoms in cyclobutadiene as shown below β 1 3 β 1 4 and construct the Hückel secular determinant with rows and columns numbered the same way, we have α E β 0 β 1 β α E β β 1 α E β β 1 0 β α E = 0, and with the definitions for x and b given in the problem, x = (α E)/β and b = β 1 /β, this determinant becomes x 1 0 b 1 x b 0 0 b x 1 b 0 1 x = 0. Expanding the determinant (see Example 14.4 for the expansion of a 4 4 determinant) gives the polynomial specified in the problem: x 4 (b + 1)x + (b 1) = 0. If b = 1, this polynomial reduces to that of square cyclobutadiene in Example 14.4: x 4 4x = 0. If b = 0, then β 1 = 0 and there is no bond between atoms and 3 and atoms 1 and 4; we have only two ethylene molecules. The secular polynomial agrees with this interpretation. For b = 0, it is simply x 4 x + 1 = (x 1) = 0, and a single ethylene secular polynomial is x 1 = 0. Since β represents the Hamiltonian matrix element H ij, i j, we might expect β to be more negative than β 1, since β represents the coupling between the more closely spaced and more strongly interacting double-bonded carbons. We should expect nonsquare cyclobutadiene to be a singlet, since we know the b = 0 limit gives us two ethylenes, which are singlets. As b 1, the MO energy pattern approaches that of square cyclobutadiene where the middle two energy levels become degenerate, as in Example Only then does the possibility of a triplet state appear. For example, if b has the intermediate value 356

26 0.5, the secular polynomial factors into (x + 3/)(x 3/)(x + 1/)(x 1/) = 0, or into four equally spaced MO energies The novel PsH molecule has only one heavy particle the proton and thus the Born Oppenheimer approximation is of no use. This molecule is more atomic-like than molecule-like, rather like He. In the context of Figure 14.1, PsH looks like a fixed, heavy proton with two electrons and one positron bound to it in no obvious, fixed locations. The energy-level diagram suggested in the problem is shown below. (The Ps ionization energy (Ps e + e + ) is half that of H. See Problem 1.46.) 0 H + + e + e + 5 Ps + H + + e Energy/eV H + e + e + H + e Ps + H PsH This diagram shows us that PsH is stable toward the dissociation PsH H + e