332

CHAPTER 14

The Origins of Chemical Bonding

SECTION 14.1

14.1 A proton (of mass mp) in a 1-D box of length L = 10a0, the system in part (a),has a Hamiltonian we write in terms of the ordinary length x as

H = – ¨ 2

2mp d2

dx2 .

To write this in atomic units, we substitute ¨ = 1 and express lengths in multi-ples of the Bohr radius a0 and masses in multiples of the electron rest mass, me.We define m = mp/me and ξ = x/a0 so that, formally, we first write

H = – ¨ 2

2mema02 d2

dξ2

and then set ¨ = 1, me = 1, and a0 = 1, yielding the atomic units version of H:

H = – 1

2m d2

dξ2

where we must remember that m is the proton mass measured in multiples of meand ξ is the distance in multiples of a0. Part (b) considers an electron in a har-monic oscillator with the ordinary Hamiltonian

H = – ¨ 2

2me d2

dx2 + 1

2 kx2 .

333

The kinetic energy term is treated just as for the 1-D box in part (a), but whatshould we take for the atomic unit of force constant? Two possibilities come tomind. The first equates the atomic unit of energy from Table 14.1, e2/(4πε0a0),to the energy quantum of the harmonic oscillator, ¨ω with ω equal to k/me ifthe mass is the electron mass. Solving this equality for k gives

k = e8me

3

(4πε0)4¨ 6 .

The second possibility equates the harmonic oscillator potential energy, kx2/2,to an appropriate measure of energy. The harmonic oscillator Virial Theoremsays that the average harmonic oscillator potential energy, <V>, equals one-halfthe total energy. You can show that equating one-half the atomic unit of energyto ka0

2/2, the potential energy expressed with the atomic unit of length, gives thesame expression for k as above. Thus, we define κ to be the force constantmeasured in multiples this atomic unit of force constant and ξ to be the distancein units of a0, and we write

H = – 12

d2

dξ2 +

12

κξ2

where we have no variable for the mass because our mass here is the electronmass and atomic units set me = 1. For part (c), an H atom orbiting in a plane ata radius of 20a0, note that we have, formally, the same scenario discussed inExample 12.6 on page 419 in the text, which describes an electron orbiting in aplane at a constant radius. The ordinary Hamiltonian is purely kinetic energy:

H = L2

2mR2 =

Lz2

2mR2 = –

¨ 2

2mH(20a0)2 d2

dφ2

where R = 20a0 is the orbiting radius, m = mH, the hydrogen atom mass, andthe Lz

2 is given in Eq. (12.32c). If we define µ = mH/me, the H atom mass in

multiples of the electron mass, and then set me = 1, ¨ = 1, and a0 = 1, we havethe atomic units version:

H = – 1

2µ(20)2

d2

dφ 2 = –

1800µ

d2

dφ 2 .

334

14.2 The HeH+ B–O Hamiltonian has terms for the kinetic energy of each electron (1and 2) and for the potential energy between He and H nuclei a fixed distance Rapart, between each electron and each nucleus, and between the two electrons:

HB-O = – 12

∇12 –

12

∇22 –

2r1He

– 2

r2He –

1r1H

– 1

r2H +

1r12

+ 2R

.

The first two terms are electron kinetic energy terms, the next two are electron–He nucleus (Z = 2) attractive potential energy terms, the next two are electron–H nucleus (Z = 1) attractive potential energy terms, and the final two are elec-tron–electron and nucleus–nucleus repulsive potential energy terms, respec-tively.

14.3 The solution to this problem requires a mixture of straightforward math andsome physical insight. First, the math. The elliptical coordinate representationof the Laplacian operator is given in the problem, but the H2

+ Hamiltonian, Eq.(14.4), contains the variables r1A and r1B (as well as the constant internucleardistance R), and our first step is to express these two distance variables in ellip-tical coordinates. The elliptical distances ξ and η are defined at the bottom ofpage 503 in the text:

ξ = r1A + r1B

R and η =

r1A – r1B

R .

Solving these for r1A and r1B gives

r1A = R ξ + η

2 and r1B = R

ξ – η2

.

We write the Schrödinger equation as Hψ – Eψ = 0, write the wavefunction asψ(ξ, η, φ) = Ξ(ξ)Η(η)Φ(φ), substitute the expressions above for r1A and r1Binto the Hamiltonian, Eq. (14.4), and find

Hψ – Eψ = HΞ(ξ)Η(η)Φ(φ) – EΞ(ξ)Η(η)Φ(φ) = 0

= – 12

∇12Ξ(ξ)Η(η)Φ(φ) +

1R

– 1

r1A –

1r1B

– E Ξ(ξ)Η(η)Φ(φ)

335

= – 12

∇12Ξ(ξ)Η(η)Φ(φ) +

1R

– 2

R(ξ + η) –

2

R(ξ – η) – E Ξ(ξ)Η(η)Φ(φ)

= – 12

∇12Ξ(ξ)Η(η)Φ(φ) +

1R

– 4ξ

R ξ 2 – η 2

– E Ξ(ξ)Η(η)Φ(φ) .

Before introducing the elliptical Laplacian, we can take two simplifying steps atthis point. First, since R is a constant, we can define E′ = E – 1/R. Next, wenote that the factor multiplying the elliptical Laplacian, 4/R2(ξ2 – η2), also mul-tiplies (within a factor of R) the middle term in parentheses in the last expres-sion above. When we introduce the elliptical Laplacian, divide the equationthrough by –4/R2(ξ2 – η2), and introduce E′, we have

12

∂∂ξ

ξ 2 – 1

∂∂ξ

+ ∂

∂η 1 – η 2

∂

∂η +

ξ 2 – η 2

ξ 2 – 1 1 – η 2

∂2

∂φ2 Ξ(ξ)Η(η)Φ(φ)

+ Rξ + R2 ξ 2

– η 2

4E′ Ξ(ξ)Η(η)Φ(φ) = 0 .

Next, we note that the factor multiplying ∂2/∂φ2 can be factored:

ξ 2 – η 2

ξ 2 – 1 1 – η 2

= 1

ξ 2 – 1

+ 1

1 – η 2 .

This is important, because in order to separate variables, we must write theSchrödinger equation as a sum of terms each of which contains only a singleindependent coordinate variable. While this factorization helps, note that eachfactor still multiplies ∂2/∂φ2, and to eliminate φ, we need some physical insight.Due to the cylindrical symmetry of the system, the Hamiltonian has no potentialenergy term that depends on φ; this coordinate appears only in the (kineticenergy operator) Laplacian. Thus, physically, motion in the φ direction is thatof a free particle, and we have solved this problem already—it is the particle-on-a-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 be1/ 2π , but we can ignore it here) and m is any integer: m = 0, ±1, ±2, … .

336

This lets us evaluate the second derivative of Φ: ∂2Φ/∂φ2 = –m2Φ, and whenwe substitute this into the full Schrödinger equation, we find that, since m is aconstant, we have separated the equation:

12

∂∂ξ

ξ 2 – 1

∂∂ξ

+ ∂

∂η 1 – η 2

∂

∂η –

m2

ξ 2 – 1

– m2

1 – η 2 Ξ(ξ)Η(η)Φ(φ)

+ Rξ + R2ξ 2

E′4

– R2η 2

E′4

Ξ(ξ)Η(η)Φ(φ) = 0 .

Complete separation follows (and note that we have only the functions Ξ and Ηto worry about) if we divide by ΞΗΦ (after expanding the first term above sothat the correct functions are operated on by the appropriate derivatives):

1

2Ξ d

dξ ξ 2

– 1 dΞ

dξ –

m2

ξ 2 – 1

+ Rξ + R2ξ 2

E′4

+ 1

2Η

d

dη 1 – η 2

dΗ

dη –

m2

1 – η 2 –

R2η 2E′

4 = 0 .

The equation above has grouped together all those terms that depend on ξ andall those that depend on η. For any value of the constants m, R, and E′, eachgroup is a unique ordinary differential equation for Ξ(ξ) and Η(η). Their solu-tions are known, albeit involved.

14.4 The integral in question here is

S = 1sA 1sB dτ = 1π

e–r1A e–r1B dτ

where we wish to use elliptical coordinates. We can express the two distancesin these coordinates using the expressions derived in the previous problem:

r1A = R ξ + η

2 and r1B = R

ξ – η2

so that, using the expression for the volume element dτ given in the problem,we have

337

S = 1π

e–r1A e–r1B dτ

= R 3

8π

ξ = 1

∞

η = –1

1

ξ2 – η 2 e–R(ξ + η)/2 e–R(ξ – η)/2 dξdηdφ

φ = 0

2π

= R 3

4

ξ = 1

∞

ξ2 – η 2 e–Rξ dξdη

η = –1

1

where the simple integration over φ has been performed to go from the secondto the final expression above. Expanding the double integral gives us twodouble integrals, and integration over η is straightforward:

S = R 3

4

ξ = 1

∞

ξ2 – η 2 e–Rξ dξdη

η = –1

1

= R 3

4

ξ = 1

∞

ξ2 e–Rξ dξdη

η = –1

1

–

ξ = 1

∞

η 2 e–Rξ dξdη

η = –1

1

= R 3

4 2 ξ2 e–Rξ dξ

ξ = 1

∞

– 23

e–Rξ dξ

ξ = 1

∞

.

The integrals given with the problem show us how to integrate over ξ:

S = R 3

4 2 ξ 2

e–Rξ dξ

ξ = 1

∞

– 23

e–Rξ dξ

ξ = 1

∞

= R 3

4 2 e–Rξ –

ξ 2

R –

2ξ

R 2 –

2

R 31

∞

– 23

e–Rξ

–R 1

∞

= e–R R 3

2

1R

+ 2

R 2 +

2

R 3 –

R 2

6 = e–R 1 + R +

R 2

3 .

338

14.5 Note that diagram (a) looks like an atomic d orbital, (b) looks like two atomic porbitals 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 thesediagrams again as models of various bonding situations. We can construct thetable below for the eigenvalues of the various symmetry operators:

σh σv i

(a) 1 1 1

(b) –1 1 1

(c) 1 –1 1

(d) 1 1 1

14.6 We take the z axis to lie along the internuclear axis with z = 0 centered betweenthe nuclei that are spaced at R = 2 au:

z = 0

R = 2 au AB

z = R/2z = –R/2

The g and u wavefunctions are given in Eq. (14.7) on page 508 with the over-lap integral S given earlier on that page and derived in Problem 14.4. We cantake the 1s atomic orbitals as e–r where r is the distance from the nucleus, ne-glecting overall normalization, since we are considering only relative compar-isons at various points along z. For atom A, r is |z – R/2| (since z = R/2 shouldcorrespond to r = 0 for this atom and r is always positive), and similarly, r is |z+ R/2| for atom B. Thus, the wavefunctions are

Ψg = e– z – R/2 + e– z + R/2

2(1 + S) and Ψg =

e– z – R/2 – e– z + R/2

2(1 – S)

where

S = e–R 1 + R + R 2

3 = 0.5865 (for R = 2).

Graphs of Ψg and Ψu are shown at the top of the next page (with zero for the gwavefunction’s axis at the bottom of the graph and that for u in the middle):

339

Ψg

0z

R/2 R 2R–R/2–R–2R

Ψu

If R = 2, then S = 0.5865, and at the bond midpoint, z = 0 and Ψg = 0.4131; atthe A nucleus, z = R/2 and Ψg = 0.6374. These numbers give the ratio

Ψg(z = 0)

Ψg(z = R/2) =

0.41310.6374

= 0.6481 .

At a point R/2 beyond the A nucleus, z = R and Ψg = 0.2345, and we have

Ψg(z = R)

Ψg(z = R/2) =

0.23450.6374

= 0.3679

so that the electron density (which is proportional to the square of the wave-function) is about a factor of 3 larger (since (0.6481/0.3679)2 ≅ 3) at the bondmidpoint than at a distance R/2 on the “outside” of the bond.

14.7 In atomic units, an electron half-way between two protons spaced 1.2 au apartis a system with a total potential energy

V = 1

1.2 –

10.6

– 1

0.6 = –

31.2

= –2.5 au

where the first term is the proton-proton repulsion and the second and third arethe electron-proton attractions. This is significantly lower than –1.34 au, aspredicted in the problem. Next, we consider the geometry at the top of the nextpage and ask for that value of r that reproduces the –1.34 au potential energyvalue.

340

1.2 au

rr

+ +

–

We solve

V = 1

1.2 –

2r = –1.34 au so that r = 0.92 au .

The diagram above is drawn to scale for this value of r. The potential energy istelling us that the electron is not so tightly localized between the nuclei.

14.8 The molecular Virial Theorem (Equations (14.13b) and (14.13c)) tell us

<T> = –<E> – R d<E>

dR and <V> = 2<E> + R

d<E>dR

so that for <E> = –0.5 + 2.887 exp(–1.050 R), we have, by direct substitution,

<T> = 0.5 – 2.887e–1.050R + 3.031Re–1.050R

<V> = –1.0 + 5.774e–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 theexact curves shown with the problem.

2 4 6R/au

–0.6

–0.4

–0.2

Ene

rgy/

au

–0.8

0

0 8

<T>

<E>u

<V>

341

SECTION 14.2

14.9 Example 13.5 discusses the ground and first excited states of the halogenatoms, and the key to explaining the photoelectron spectrum of the rare gases isto recognize that the rare gas cations from Ne through Rn are isoelectronic to thehalogens F through At and thus have the same qualitative energy level structure.For example, 21.22 eV photoionization of Ar has two possible outcomes:

Ar(3s23p6 S01 ) + hν →

Ar+(3s23p5 P3/22 ) + e–

Ar+(3s23p5 P1/22 ) + e–

In accord with Hund’s Rules and in analogy with the halogens, the ground stateof Ar+ is 2P3/2 and the first excited state is 2P1/2. If the photon has 21.22 eVenergy and the photoelectron has a 5.46 eV kinetic energy, then 21.22 eV –5.46 eV = 15.76 eV represents the ionization energy of Ar to one of thesestates. If the electron kinetic energy is 5.28 eV, the same calculation shows that15.94 eV is needed to reach the other ion state. Thus, the photoelectron withthe greater kinetic energy corresponds to photoionization to the ground 2P3/2state, while the smaller kinetic energy corresponds to the 2P1/2 excited state ionproduct. Note from the data given in the problem that the energy separationbetween these two states follows the same general periodic trend as do the halo-gen atom state separations: 0.18 eV for Ar+, 0.66 eV for Kr+, and 1.31 eV forXe+. (Ne+ has a similar energy level structure—and the separation of the cor-responding two states is smaller—but 21.22 eV light is not energetic enough toionize Ne, which has a 21.564 eV ionization potential.)

14.10 We have ten electrons in HF, and the ground-state electron configuration is1σ22σ23σ21π4. (If we followed the analogy with CO on the MO ordering, wemight write 1σ22σ23σ24σ21π2, but this would place two electrons in the 1πMO, and we would expect them to enter as a triplet, one in each of the twodegenerate 1π MOs. HF is a closed-shell, singlet molecule. Hence the 4σ MOin 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 stateconfiguration is 1σ22σ23σ21π3. The 1π electrons are not particularly bondingin character, and thus the ion state has nearly the same bond length as the neu-tral. (Experiment shows it to be 1.001 Å compared to 0.9168 Å for neutralHF.) The features spaced by about 0.35 eV in the photoelectron spectrum aredue to simultaneous ionization to this electronic state and vibrational excitationof the molecular ion. The broad 2.13 eV feature is the result of removing a 3σ

342

electron, leaving the excited configuration 1σ22σ23σ11π4. The 3σ MO isstrongly bonding, and removing an electron from it produces an ion state that isless strongly bound and has a much longer bond length (1.224 Å). The breadthof this feature is due to extensive internal excitation of this excited ion state.

14.11 We can use isoelectronic analogies to guide our ground-state electron configu-ration assignments for the ions: NO+ is isoelectronic to N2 and NO– is isoelec-tronic to O2. We consult Table 14.3 (neglecting the u and g notation since ourmolecules are heteronuclear) and follow the discussion in Example 14.2, NO+

has the 14-electron configuration 1σ22σ23σ24σ21π45σ2, NO has the 15-electron configuration 1σ22σ23σ24σ21π45σ22π1, and NO– has the 16-electronconfiguration 1σ22σ23σ24σ21π45σ22π2. We assign NO+ a bond order of 3,paralleling N2, assign NO– a bond order of 2, paralleling O2, and NO has afractional 2.5 bond order, as did O2

+ in Example 14.3. We can use the energydata in the problem to construct the diagram below, which confirms that thebond energies of NO+, NO, and NO– follow their bond orders. The bondenergy of NO+ is De(NO+) = 13.617 eV + 6.496 eV – 9.25 eV = 10.86 eV,and that of NO– is De(NO–) = 6.496 eV + 0.024 eV – 1.461 eV = 5.059 eV, inaccord with the bond orders. (And finally, N does not have a positive electronaffinity because of the stability afforded by its half-filled 2p subshell in theground-state configuration of the neutral atom: 1s22s22p3.)

N+ + O+ + 3e–

N+ + O + 2e–

N + O+ + 2e–

N + O + e–

N + O–

NO–NO + e–

NO+ + 2e–

0

–5

–10

–15

–20

–25

–30

–35

Ene

rgy/

eV

14.5

34 e

V

13.6

17 e

V

13.6

17 e

V

1.461 eV

0.024 eV9.25

eV

6.49

6 eV

De(NO+)

De(NO–)De(NO)

10.8

6 eV

5.059 eV

343

14.12 Since both He and Be have closed shell (for 1s2 He) or closed subshell (for1s22s2 Be) ground-state configurations, we should expect HeBe to have noformal bond, in analogy to He2 or Be2 (see Table 14.3). We can write aground-state MO configuration easily enough: 1σ22σ23σ2. Now we confrontthe less-obvious question of bond order. The lowest energy AO of He and Bemust 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 en-ergy than the Be 2s AO (by about 15.3 eV, the difference in ionization poten-tials), and excitation energies for He (discussed in Chapter 13 on page 475 inthe text) tell us that the He 2s AO is about 20 eV above the 1s, or 4.7 eV abovethe Be 2s. These large separations in AO energies leads us to predict that, eventhough these orbitals have the correct symmetry to mix, they will not do so toany appreciable extent. Thus, we conclude that the HeBe 1σ is essentially thepure Be 1s AO, the 2σ is essentially pure He 1s, and the 3σ MO is the Be 2sAO. The lack of any appreciable mixing supports our suspicion that the bondenergy should be quite small, and all three MOs are essentially nonbonding.Formally, this six-electron molecule is isoelectronic to Li2, but the electron dis-tribution and its resultant bonding is obviously quite different! The first excitedstate comes from the MO excitation 1σ22σ23σ2 → 1σ22σ23σ14σ1, and the onlyquestion is the predominant atomic orbital contribution to the 4σ MO: is it He 2sor Be 2p? We know the He 2s orbital is about 4.7 eV above the Be 2s, and weshould expect the Be 2p orbital to be a comparable energy above Be 2s. (Thisis not an obvious energy to guess! Atomic spectroscopy shows that the 2s2p3P1 state of Be is about 2.73 eV above the 2s2 1S0 ground state, and the 2s2p1P1 state is about 5.28 eV above 1S0.) Consequently, the 4σ MO should be asignificant mixture of He 2s and Be 2p (in a 2pσ orientation) and lead toconsiderably stronger bonding in this excited state with 4σ not as stronglylocalized on either atom as the lower energy MOs clearly are.

14.13 As is often the case, isoelectronic analogies are useful here. The anion CN– isisoelectronic to CO (or N2) and has the closed-MO ground-state configuration(see Example 14.2) 1σ22σ23σ24σ21π25σ2. The radical CF is a 15-electronspecies like NO (see Problem 14.11) with the ground-state configuration1σ22σ23σ24σ21π45σ22π1 and one unpaired electron. The CH radical is aseven electron species with the 1σ22σ23σ24σ1 configuration and one unpairedelectron, while CH+ loses the 4σ electron (predominantly from the C atom) togive a 1σ22σ23σ2 ground-state, closed-MO configuration.

14.14 The interesting dication He22+ is isoelectronic to H2 and dissociates to two He+

ions (rather than to He + He2+, which is higher in energy by 29.8 eV, as you

344

can verify from the first and second ionization potentials of He discussed inChapter 13). We assign a simple 1σg

2 configuration to ground state He2

2+ and abond order of 1, suggesting that it is bound, but we must remember that, unlikeH2, He2

2+ has a residual nuclear charge on each atom that is not offset by elec-tron charge. If we bring two naked protons together, they repel, of course, andthe uncompensated nuclear charges on He2

2+ are roughly just that: two +echarges repelling each other. We can easily plot the pure Coulombic repulsionas a function of internuclear distance; it is just e2/(4πε0)R, as shown by theheavier line in the graph below (with energy converted to eV units). We seefrom the graph that at distances characteristic of an expected He2

2+ bond length,around 1 Å or less, the nuclear repulsion energy has risen nearly 15 eV abovethe asymptotic energy zero at large R. It is unreasonable to expect the He2

2+

bond energy to be greater than (or even anywhere close to) 15 eV. Conse-quently, the binding well in He2

2+ is a depression in the nuclear repulsion curve(the lighter curve in the graph, drawn to follow an ab initio calculation for theHe2

2+ ground state). This means He22+ is only metastably bound. The two He+

ions that comprise it constantly repel each other, and even if trapped in thebinding well, they can eventually tunnel through the potential energy hump thatmomentarily binds them, escaping to R = ∞.

1 2 3R/Å

5

10

15

400

20

Ene

rgy/

eV

SECTION 14.3

14.15 We place the zero of energy at the energy of H(1s) + H(1s) infinitely apart sothat the ion pair H+ + H– infinitely apart has the energy IP(H) – EA(H) =13.598 – 0.7542 eV = 12.844 eV, the difference between the H ionizationpotential 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 –e2/4πε0R = –14.4 eV/(R/Å) (as derived in

345

footnote 17 on page 549 in the text) shifted up by 12.844 eV. With this energyscale, the energy of the neutral pair H(1s) + H(ns) is 13.598(1 – 1/n2). If n =1, this energy is zero, and the ion pair energy falls to zero at R such that

– 14.4 eV

R/Å + 12.844 eV = 0 or R = 1.12 Å .

This is the first ion pair–neutral line crossing point. Repeating this calculation,substituting 13.598(1 – 1/n2) for zero in the expression above with n = 2, 3,and 4, gives crossing points at R = 5.44 Å, 19.0 Å, and 151 Å, respectively. Ifn = 5, the neutral energy is above the ion pair asymptote at 12.844 eV. Thesevarious energies, crossing points, and ion-pair attraction curve are shown in thegraph 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 inenergy than the H(2s) + H(2s) asymptotic limit, and we can calculate that thislimit lies at 2[13.598(1 – 1/22)] eV = 20.397 eV. Thus, H+ + H– → H(2s) +H(2s) is endothermic by 20.397 eV – 12.844 eV = 7.553 eV.

10 20R/Å

0

5

10

Ene

rgy/

eV

300

14

H(1s) + H(1s)

H(1s) + H(2s)

H(1s) + H(3s) H(1s) + H(4s) H+ + H–H(1s) + H(5s)

H+H– ionattraction

14.16 We can adapt the general methodology used for H3+ on page 528 of the text to

H2 if we let φi in Eq. (14.14) represent the H 1s AO on nucleus i. The generalvariational equations for this LCAO approximation are

c1(H11 – ES11) + c2(H12 – ES12) = 0c1(H21 – ES21) + c2(H22 – ES22) = 0 .

If we apply the Hückel approximations to these equations, we have

c1(α – E) + c2β = 0 and c1β + c2(α – E) = 0 .

346

The secular determinant equation is

α – E β

β α – E = 0 ,

and if we define x = (α – E)/β, this equation becomes

x 11 x

= 0 or x2 – 1 = 0 or x = ±1 .

If x = ±1, then E = α ± β. For x = –1, or E = α + β, (the ground state), wesolve the variational equations for c1 and c2, including the requirement that c1

2 +c2

2 = 1:

–c1 + c2 = 0 , c1 – c2 = 0 , c12 + c2

2 = 1 so that c1 = c2 = 1

2 .

The ground-state wavefunction is thus

ψ = 1

2 1sA + 1sB ,

which looks like Ψg for H2+ (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 equationsare solved by c1 = –c2 = 1/ 2 for an excited state wavefunction

ψ = 1

2 1sA – 1sB ,

which looks like Ψu for H2+ (Eq. (14.7b) with S = 1 again).

14.17 Whether we are considering the neutral radical, the anion, or the cation, theconnectivity of the carbon atoms stays the same, and so does the Hückel seculardeterminant. This determinant is exactly the same as in Eq. (14.19) for linearH3:

x 1 01 x 10 1 x

= 0 .

347

Expanding it gives x(x2 – 2) = 0, or x = 0, ± 2. Since the energy, E, equals α– xβ, the three Hückel π MO energies are, in increasing energy,

E ′ = α + 2β , E ′′ = α, and E ′′′ = α – 2β .

The allyl cation has two π electrons, the neutral has three, and the anion hasfour. Thus, their total π electron MO energies in this approximation are

cation: E = 2E′ = 2α + 2 2βneutral: E = 2E′ + E ′′ = 3α + 2 2βanion: E = 2E′ + 2E ′′ = 4α + 2 2β .

14.18 As suggested in the problem, we label the three C atoms in the linear allyl radi-cal a, b, and c. With this notation, the three variational equations are (contrastEq. (14.15) for H3

+ and the discussion of the secular determinant in the previousproblem)

cax + cb = 0 , ca + cbx + cc = 0 , and cb + ccx = 0 ,

and the normalization equation ca2 + cb

2 + cc2 = 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 = – 2, wefind that these equations are satisfied by ca = cc = 1/2 and cb = 1/ 2. For x = 0,ca = –cc = 1/ 2 and cb = 0, and for x = + 2, we find ca = cc = 1/2 and cb =–1/ 2. Thus, the three π MOs are identical in form to those shown on page532 in the text and graphed in Figure 14.17 (and reproduced on the cover of thetext):

ψ′ = 12

pa + 2pb + pc

ψ′′ = 1

2 pa – pc

ψ′′′ = 12

pa – 2pb + pc .

These π MOs look very much like Figure 14.17 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 thetop of the next page. Solid and dashed contours denote regions of oppositealgebraic sign, as in Figure 14.17.

348

Ψ′ Ψ′′ Ψ′′′

14.19 The trimethylene methane radical has a Lewis electron-dot structure that we candraw (in one of three equivalent ways, related by the three equivalent positionsof the C=C double bond) as a diradical, in accord with the Hückel prediction:

C

CCC

HH

H H

HH

The central C atom is bonded to the three others, but they in turn are bondedonly to that central atom, not to each other. Thus, its Hückel secular determi-nant is the one that reflects this connectivity. If we let the upper left-hand entryin this determinant represent the central atom, the correct choice is

x 1 1 11 x 0 01 0 x 01 0 0 x

.

14.20 With the simplifications stated in the problem, f = 1 and βCX = βCC = β, the Xatom’s α value is simply αX = αC + β = α + β. The secular determinant equa-tion for the π MO energies is thus

α – E β

β α + β – E = 0 or

x 11 x + 1

= 0

349

if we define x = (α – E)/β as usual. Expanding the determinant gives x2 + x –1 = 0, which has the solutions

x = ± 52

– 12

or E = α + 1 ± 5

2β =

α – 0.618βα + 1.618β

.

The ethylene π MO Hückel energies are E = α ± β, in contrast to α + 1.618βand α – 0.618β here. Since β < 0, the lower energy MO here is that with E =α + 1.618β and is stabilized by 0.618β in comparison to the lower energyethylene MO. The next problem shows in more detail why this is true.

14.21 If we substitute x = (α – E)/β (in the form E = α – βx) into the variationalequations for c1 and c2, we find

c1x + c2 = 0c1 + c2(1 + x) = 0 ,

and we also require c12 + c2

2 = 1. The two values for x from the previous prob-lem, x = ± 5/2 – 1/2, will give us two sets of c1 and c2 values, one for thelower energy bonding π MO and one for the excited antibonding MO. Becausethe two equations above that are linear in c1 and c2 both equal zero (they aresaid to be homogeneous—see Example 13.3 and footnote 5 on page 461 in thetext), we can use either one of them and the nonlinear normalization equation,c1

2 + c22 = 1, to find our sets of c1 and c2 values. For example, we substitute

one x value, x = 5/2 – 1/2, in the first linear equation and write

c15 – 12

+ c2 = 0 or c2 = – 5 – 12

c1 .

We substitute this expression for c2 into the normalization equation and find

c12 + c2

2 = c12 + c1

2 5 – 12

2 = c1

2 5 – 5

2 = 1 or c1 =

2

5 – 5 = 0.8507 .

Substituting this value for c1 into the expression for c2 gives c2 = –0.5257. Ifwe repeat these steps using the other value for x, we find c1 = 0.5257 and c2 =0.8507. Now we must interpret these values. The table on the next pagecorrelates each MO energy with its corresponding coefficients. The moreelectronegative atom is atom 2, and we see that the bonding MO (which has nonode—c1 and c2 have the same algebraic sign—and which also has the lower

350

energy) is more localized on atom 2 (c2 > c1) while the antibonding MO (onenode—c1 and c2 have opposite signs) is more localized on the lesselectronegative atom 1.

MO type x E c1 c2

bonding – 5/2 – 1/2 α + 1.618β 0.5257 0.8507antibonding 5/2 – 1/2 α – 0.618β 0.8507 –0.5257

14.22 The construction described in the problem is shown below. (Some simpletrigonometry will verify that the inscribed hexagon vertices fall exactly at α +2β, α + β, α – β, and α – 2β, as shown in the diagram.)

Ene

rgy

→

α + 2β

α + β

α

α – β

α – 2β

↑↓

↑↓↑↓

With six π electrons, benzene fills the lowest three MOs (with two paired elec-trons in each of these three MO, as shown) for a total π energy 6α + 8β, asstated in the text, and thus with a stabilization, or resonance, energy 2| β | incomparison with a true cyclohexatriene. If the lowest energy MO has no nodesand the highest has three, then the MO next above the lowest must have onenode and the third MO must have two. The energy diagram shows these MOsto be doubly degenerate, and the pictures we can draw with these nodal patternsreflect this degeneracy—there are two pictures for each. Shading is used toindicate different algebraic signs, and dashed lines locate nodes.

SECTION 14.4

14.23 The BH2 radical’s ground state MO configuration comes from the atomic con-figurations B 1s22s22p1 and H 1s1 for a total of seven electrons. If the molecu-

351

lar ground state had been linear, it would have had the MO configuration1σg

22σg

21σu

*21πu

1, but the Walsh correlation diagram in Figure 14.20 shows usthat the bent molecular orbital configuration 1a1

22a121b2

23a11 will be lower in en-

ergy due to the strong tendency for the 3a1 MO to bend the molecule. This MOcontrols the geometry of the ground state. The first excited state configurationcomes from excitation of the highest energy electron to the next available MO.In the bent ground state, the 3a1 electron is promoted to the 1b1 MO, but thisMO does not have a strong preference for bending, and Figure 14.2 now showsus that the 1b2 MO governs the geometry, leading to a linear excited state.

14.24 We consult Figure 14.19(b) and note that the 1b1 MO is indifferent to the bondangle—it is essentially the A atom’s 2p orbital that points perpendicular to themolecular plane. As the H–A–H bond angle is reduced to 0, this MO becomesa 2pA π MO in the resultant diatomic. Similarly, the 1b2 MO becomes the other2pA π MO as the 1sH contributions to 1b2 cancel at zero bond angle.

14.25 From Table 14.5, we see that the bond angle of the dihydrides from water toH2Te systematically decrease toward 90°. This is suggestive of bonding to thecentral 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 atomsdecreases (these atoms are farther apart), and the valence electrons on the centralatom have systematically larger principal quantum numbers, a fact which causesa greater energy mismatch between the A valence atomic orbital energies andthose of H. This mismatch gives the highest energy MOs in the molecularground state a more purely atomic A orbital character. (See also Problem 14.34for another way to understand this change in geometry.)

14.26 There are, of course, many possible answers here, but the following list con-tains fairly well known species. Good examples of linear 16 electron moleculesinclude OCS (a direct analog to CO2) and NO2

+ (an isoelectronic analog). For17 electron species, which are bent, both BF2 and O3

+ are reasonable examples.The nitrosyl halides such as NOF and NOCl are good examples of bent 18electron species, as is the nitrite ion NO2

–. The 19 and 20 electron species arerather scarce. The 19 electron BrO2 is a direct analog to ClO2, and the ozoneanion O3

– also falls in this class. For 20 electrons, the nitrogen dihalide anionssuch as NF2

–, NCl2–, NClF–, etc., are known. Other 22 electron, linear trihalide

anions are known besides I3–, but not all, and the rare gas dihalides FXeCl and

XeCl2 are known.

352

14.27 For the dsp2 square planar hybrids, we should mix in the dx2 – y2 orbital (if wetake the plane of the hybrids to be the x–y plane, since this d orbital pointsalong the x and y directions. (See Figure 12.15(d) on page 436 in the text.)The dsp3 trigonal bipyramidal hybrids involve the dz2 orbital. We can think ofthese hybrids as ordinary sp2 trigonal planar hybrids added to a pair of hybridspointing along the z direction and composed of the dz2 and pz orbitals. Finally,the d2sp3 octahedral hybrids involve both dz2 and dx2 – y2.

14.28 In dimethyl zinc (or cadmium or mercury—all three are known), the central Znatom has the ground state configuration 3d104s2. Promotion to the 3d104s14p1

excited configuration followed by sp hybridization of the 4s and 4p orbitalsleads to the observed linear HC3–Zn–CH3 bond angle.

14.29 We have 16 valence electrons in carbon dioxide and thus the ground-state MOconfiguration for the linear geometry is 1σg

21σu

*22σg

23σg

22σu

*24σg

23σu

*21πu

41πg4,

as discussed in the text on page 540 and shown in Figure 14.21(a). Excitationto the first excited state involves promotion of one electron from the 1πg non-bonding MO to the next higher energy MO. If the molecule stays linear, this isthe antibonding 2πu MO, but Figure 14.22 shows that one component of thisMO correlates on bending to the 6a1 MO, which strongly prefers to be bent. Itis this MO that governs not only the first excited state geometry, but also theanion geometry, since it is the LUMO of ground state CO2 and thus is the or-bital that accepts an extra electron when CO2

– is formed. Since CO2– wants to be

bent, but CO2 is linear, when an extra electron is initially attached to CO2, theincipient anion finds itself in an energetically unfavorable geometry—linear in-stead of bent—and the extra energy associated with this unfavorable geometryis greater than that gained through electron attachment. Thus, the electron is notattached and the electron affinity is negative. You should recognize that electronattachment into the first excited state of CO2 (into the “hole” left in the 1πg non-bonding MO on excitation) could lead to a stable anion, since the electron wouldbe attaching itself into an already bent molecule.

SECTION 14.5

14.30 We can draw the energy level diagram for the HeH+ system that is shown onthe next page using the data in the problem and ionization energies for He and Hfrom previous chapters. This diagram shows that HeH+ in its ground statedissociates into the He (1s2 1S0) + H+ pair (because the first ionization potentialof He is greater than the ionization potential of H) while the first excited state ofHeH+ dissociates into He+ (1s1 2S1/2) + H (1s1 2S1/2), the charge-exchanged

353

analog of the ground state dissociation pair. The excited states of He combinewith a proton to form even higher energy HeH+ excited states. We can also seethis if we look at the MOs of these states. We have only two electrons, andthus the ground state MO is simply 1σ2. The energy difference between the Heand H 1s AOs (He 1s is much lower) tells us that the 1σ MO is almost a pureHe 1s orbital; ground state HeH+ is not much more than a proton stuck to a Heatom. The first excited state, 1σ12σ1, places an electron in the 2σ MO, whichis predominantly the H atom 1s orbital. This state is a hydrogen atom bound tothe He+ ion. (The ground state binding energy of HeH+ is 2.0 eV, by theway.)

He2+ + H+ + 2e–

He (1s2 1S0) + H+

He2+ + H– (1s2 1S0)

He+ (1s1 2S1/2) + H (1s1 2S1/2)

He* (1s2s 1S0) + H+ He* (1s2s 3S1) + H+

0

–10

–20

–30

–40

–50

–60

–70

–80

Ene

rgy/

eV

14.31 The water ground state MO configuration ends with the 1b12 MO occupancy,

and excitation to the 1b114a1

1 configuration tells us we should consult Figure14.20 to see how the 4a1 MO energy changes with bond angle. We see that thisMO wants to be linear, and thus we can conclude that states derived from thisexcited configuration should be less bent than the ground state.

14.32 If we consult the dihydride Walsh correlation diagram in Figure 14.20, we cancorrelate the occupied MOs of bent ground-state H2O, 1a1

22a121b2

23a121b1

2, to thelinear configuration 1σg

22σg

21σu

*21πu

4. Note that this is very similar to the HFMO configuration discussed in Problem 12.10.

354

GENERAL PROBLEMS

14.33 The π electrons in bombykol are localized through the two C–C double bonds,exactly as in butadiene. All the other carbon atoms are sp3 hybridized andbonded through the molecule’s σ MO framework. We would expect the π MOsto include the molecules’ HOMO, just as in butadiene, since the framework σMOs are lower in energy.

14.34 Normalization of either the α or β hybrids ensure that a12 + a2

2 = 1 if the s, px,py, and pz orbitals are individually orthonormal. If α and β are to be orthogo-nal, we must also have

αβ dτ = a1s + a2 px cosθ2

+ py sinθ2

a1s + a2 px cosθ2

– py sinθ2

dτ = 0 .

Expanding the integrand and invoking orthonormality of s, px, and py gives

αβ dτ = a12 + a2

2 px cosθ2

+ py sinθ2

px cosθ2

– py sinθ2

dτ

= a12 + a2

2 cos2 θ2

– sin2 θ2

= 0 .

The trigonometric identities

cos2 θ2

= 1 + cos θ

2 and sin2

θ2

= 1 – cos θ

2

let us to write

αβ dτ = a12 + a2

2 cos2 θ2

– sin2 θ2

= a12 + a2

2 cos θ = 0 ,

and solving

a12 + a2

2 cos θ = 0 and a12 + a2

2 = 1

simultaneously for a12 and a2

2 gives the two expressions stated in the problem:

a12 =

cos θcos θ – 1

and a12 =

11 – cos θ

.

355

In a pure sp2 hybrid, θ = 120°, and a12, the s character, is 1/3, as we should

expect. For θ = 105.2°, we find a12 = 0.2077. It is also interesting to note that

θ = 90° gives a12 = 0 (no s character) and a2

2 = 1 (pure p character), as weshould expect for orbitals directed at 90°. (See also Problem 14.25.)

14.35 Continuing from the previous problem, if a32 + a4

2 = 1 and 2a22 + a3

2 = 1, thevalue for a1 from the previous problem lets us find

a32 =

1 + cos θ1 – cos θ

and a42 =

2 cos θcos θ – 1

.

Again, if θ = 120° (pure sp2), a32 = 1/3, but for θ = 105.2°, a3

2 = 0.5845.

14.36 If we have N double bonds, we have 2N π electrons to pair into energy levelsin a 1-D box of length L = 2NR where R = 1.4 Å, the C–C bond length. Thismeans we put two electrons in the state n = 1, and so on, until we reach thelevel n = N where n is the usual particle-in-a-box quantum number. The energyfor this level is the HOMO energy, which, using the energy expression fromChapter 12 for this system, is

EHOMO = ¨ 2

π2

2mL2 n2 =

¨ 2π2

2me N 2

4N 2R 2 =

¨ 2π2

8meR2

where me 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:

ELUMO = ¨ 2

π2

2me N + 1 2

4N 2R 2 =

¨ 2π2

8meR2 1 + 2

N + 1

N 2 = EHOMO 1 + 2

N + 1

N 2 ,

and the HOMO–LUMO excitation energy is thus

ELUMO – EHOMO = EHOMO 2N

+ 1

N 2 .

The numerical value for this excitation energy (for N values of chemicalinterest, i.e., in the range 1 to maybe a dozen or so, has a magnitude of a feweV, in rough qualitative agreement with experiment. In particular, many dyemolecules have conjugated systems in them that can be adjusted in length inorder to control the dye’s color.

356

14.37 If we number the carbon atoms in cyclobutadiene as shown below

β1

β2

1

2 3

4

and construct the Hückel secular determinant with rows and columns numberedthe same way, we have

α – E β2 0 β1

β2 α – E β1 0

0 β1 α – E β2

β1 0 β2 α – E

= 0 ,

and with the definitions for x and b given in the problem, x = (α – E)/β2 and b= β1/β2, this determinant becomes

x 1 0 b 1 x b 00 b x 1b 0 1 x

= 0 .

Expanding the determinant (see Example 14.4 for the expansion of a 4 × 4 de-terminant) gives the polynomial specified in the problem: x4 – 2(b2 + 1)x2 + (b2

– 1)2 = 0. If b = 1, this polynomial reduces to that of square cyclobutadiene inExample 14.4: x4 – 4x2 = 0. If b = 0, then β1 = 0 and there is no bond be-tween atoms 2 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 simplyx4 – 2x2 + 1 = (x2 – 1)2 = 0, and a single ethylene secular polynomial is x2 – 1= 0. Since β represents the Hamiltonian matrix element Hij, i ≠ j, we might ex-pect β2 to be more negative than β1, since β2 represents the coupling betweenthe more closely spaced and more strongly interacting double-bonded carbons.We should expect nonsquare cyclobutadiene to be a singlet, since we know theb = 0 limit gives us two ethylenes, which are singlets. As b → 1, the MO en-ergy pattern approaches that of square cyclobutadiene where the middle two en-ergy levels become degenerate, as in Example 14.4. Only then does the pos-sibility of a triplet state appear. For example, if b has the intermediate value

357

0.5, the secular polynomial factors into (x + 3/2)(x – 3/2)(x + 1/2)(x – 1/2) = 0,or into four equally spaced MO energies.

14.38 The novel PsH molecule has only one heavy particle—the proton—and thus theBorn–Oppenheimer approximation is of no use. This “molecule” is moreatomic-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 positronbound to it in no obvious, fixed locations. The energy-level diagram suggestedin the problem is shown below. (The Ps ionization energy (Ps → e– + e+) ishalf that of H. See Problem 12.46.)

0

–5

–10

–15

–20

–25

Ene

rgy/

eV

H+ + 2e– + e+

Ps + H+ + e–

H + e– + e+

Ps + H

H– + e+

PsH

This diagram shows us that PsH is stable toward the dissociation PsH → H– +e+.