CHAPTER 19 Molecular Spectroscopy of Small Free Moleculeschem81/pdfs/19.pdf · Molecular...

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1 CHAPTER 19 Molecular Spectroscopy of Small Free Molecules SECTION 19.1 19.1 We can follow Examples 19.1 and 19.2 here, making mass substitutions as needed. From Example 19.1, we see that the centers of mass for the D 2 16 O and H 2 18 O isotopomers of water have Y = Z = 0, as for H 2 16 O, but X changes to D 2 O: 16 X = 2m D (0.958 Å)cos(104.5°/2) 2m D + m O 16 = 0.118 Å H 2 O: 18 X = 2m H (0.958 Å)cos(104.5°/2) 2m H + m O 18 = 0.059 1 Å . Note that increasing the mass from H to D increases X (i.e., moves the center of mass farther from the origin at the center of the O atom) while increasing the O mass from 16 O to 18 O decreases X. Once we know the center of mass coor- dinates, we can find the coordinates of the O and H or D atoms in the principal inertial axis coordinate system, (a, b, c). We find, following the simple trigo- nometry used in Example 19.2 to locate the atoms, that the principal inertial coordinates of the atoms are D = ±sin 105.4° 2 (0.958 Å), cos 105.4° 2 (0.958 Å) – 0.118 Å, 0 =(±0.757 Å, 0.469 Å, 0) O 16 = (0, –0.118 Å, 0) H = ±sin 105.4° 2 (0.958 Å), cos 105.4° 2 (0.958 Å) – 0.059 1 Å, 0 =(±0.757 Å, 0.527 Å, 0) O 18 = (0, –0.059 1 Å, 0) . : (a, b, c) : (a, b, c) : (a, b, c) : (a, b, c) D 2 16 O: H 2 18 O:

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CHAPTER 19

Molecular Spectroscopy of Small Free Molecules

SECTION 19.1

19.1 We can follow Examples 19.1 and 19.2 here, making mass substitutions asneeded. From Example 19.1, we see that the centers of mass for the D216O andH218O isotopomers of water have Y = Z = 0, as for H216O, but X changes to

D2 O:16 X = 2mD(0.958 Å)cos(104.5°/2)

2mD + m O16 = 0.118 Å

H2 O:18 X = 2mH(0.958 Å)cos(104.5°/2)

2mH + m O18 = 0.059 1 Å .

Note that increasing the mass from H to D increases X (i.e., moves the centerof mass farther from the origin at the center of the O atom) while increasing theO mass from 16O to 18O decreases X. Once we know the center of mass coor-dinates, we can find the coordinates of the O and H or D atoms in the principalinertial axis coordinate system, (a, b, c). We find, following the simple trigo-nometry used in Example 19.2 to locate the atoms, that the principal inertialcoordinates of the atoms are

D = ±sin105.4°

2 (0.958 Å), cos

105.4°2

(0.958 Å) – 0.118 Å, 0

= (±0.757 Å, 0.469 Å, 0)

O16 = (0, –0.118 Å, 0)

H = ±sin105.4°

2 (0.958 Å), cos

105.4°2

(0.958 Å) – 0.059 1 Å, 0

= (±0.757 Å, 0.527 Å, 0)

O18 = (0, –0.059 1 Å, 0) .

: (a, b, c)

: (a, b, c)

: (a, b, c)

: (a, b, c)

D216O:

H218O:

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With these (a, b, c) coordinates, we can use Eq. (19.3) to calculate the momentsof inertia. Note that while the equilibrium geometry is identical for all isotopo-mers and the center of mass and (a, b, c) coordinates are not very different, themoments of inertia are quite different.

D2 O:16

Ia = 2mDbD2 + m O16 bO

2 = 1.11 amu Å2 = 1.84 × 10–47 kg m2

Ib = 2mDaD2 = 2.31 amu Å2 = 3.84 × 10–47 kg m2

Ic = 2mD aD2 + bD

2 + m O16 bO2 = 3.42 amu Å2 = 5.68 × 10–47 kg m2

H2 O:18

Ia = 2mHbH2 + m O18 bO

2 = 0.624 amu Å2 = 1.04 × 10–47 kg m2

Ib = 2mHaH2 = 1.16 amu Å2 = 1.92 × 10–47 kg m2

Ic = 2mH aH2 + bH

2 + m O18 bO2 = 1.78 amu Å2 = 2.96 × 10–47 kg m2 .

The key concept here is that the equilibrium structure of a molecule is indepen-dent of its isotopic composition. (We see later in this chapter that the averagestructure does depend on isotopic composition due to differences in zero-pointenergy motions. The equilibrium structure is the unique point of minimum B–Opotential energy, while zero-point energies and the motions associated withthem vary with mass.)

19.2 Let the plane of the molecule be the a–b plane so that ci = 0 for every atom.Then, from the definitions of Ia, Ib, and Ic given in Eq. (19.3), we have

Ia + Ib = mibi2∑

i = 1

N

+ miai2∑

i = 1

N

= mi ai2 + bi

2∑i = 1

N

= Ic .

By convention (see Example 19.2), we choose a, b, and c so that Ia ≤ Ib ≤ Ic.With this definition and the moment of inertia values listed in the problem, wesee that both N3H and HFCO are planar. For each, Ia + Ib = Ic.

19.3 Allene is a nonlinear seven-atom molecule. Thus, it has 3N – 6 = 3·7 – 6 = 15vibrations . Carbon suboxide is a linear five-atom molecule with 3N – 5 = 3·5– 5 = 10 vibrations. The weakly-bound HFClF molecule is a nonlinear four-atom molecule with 3N – 6 = 3·4 – 6 = 6 vibrations, and ketene is a nonlinearfive-atom molecule with 3N – 6 = 3·5 – 6 = 9 vibrations.

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19.4 Equation (19.5) reads ∆Rvib ~ (me/M)1/4R, and with M = 1.15 × 10–26 kg forLi and 2.21 × 10–25 kg for Cs (both are atomic masses) and R = Re = 2.672 9Å for Li and 4.47 Å for Cs, we predict ∆Rvib = 0.252 Å for Li2 and 0.201 Åfor Cs2. Now we compare this to a harmonic oscillator model’s prediction.The potential energy function is V(R) = k(R – Re)2/2 where k is the harmonicforce constant, 25.25 N m–1 for Li and 6.91 N m–1 for Cs. At the inner turn-ing point, R–, (which is < Re) and at the outer turning point, R+, (> Re), thepotential energy equals the total energy, which, in turn, is the zero-point energyV0, 3.478 × 10–21 J for Li2 and 4.170 × 10–21 J for Cs2. Thus, we solve

V0 = 12

k R+ – Re2 and V0 =

12

k R– – Re2

for R+ and R– and calculate ∆Rvib = R+ – R– for each molecule. We find

R+ = Re + 2V0

k , R– = Re –

2V0

k , and ∆Rvib = 2

2V0

k .

Numerically, this expression gives ∆Rvib = 0.332 Å for Li2 and 0.220 Å forCs2, which are in reasonable agreement with the very approximate Eq. (19.5).

SECTION 19.2

19.5 The H2+ molecule has a single electron, and in the ground state, it must be in the

1σg MO. This leads to a Σg+2 term symbol: a doublet, because there is only one

electron so that S = 1/2 and 2S + 1 = 2; Σ, because the unpaired electron is in aσ MO with Λ = 0; g, because the MO is a σg MO that is symmetric with respectto inversion; and + because – states require participation from more than onenon-σ MO. The first excited state finds this electron in the next highest energyMO, the 1σu

* MO. This is a Σu+2 state. For He2

+, the three electrons have theMO configuration 1σg

21σu*1 leading to a Σu

+2 state. The first excited state pro-motes one 1σg electron, leading to the configuration σg

11σu*2 and a Σg

+2 state(which you should not be surprised to learn is not bound—work out the bondorder). The three-electron molecule HeH parallels He2

+ in its ground state(except we lose the g, u symmetry labels because HeH is heteronuclear). Itsground state is 1σ22σ1 with a Σ+2 term symbol. The first excited state takessome thought to deduce. The 1σ MO is the He core MO, since the 1s He AO ismuch lower in energy than the H 1s, and the H 1s AO dominates the descriptionof the 2σ MO. Now we ask the following question: which takes less energy,promoting a 1σ electron to 2σ or promoting the 2σ electron to the 3σ MO? The

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1σ–2σ promotion energy is roughly the He ionization potential (24.6 eV) lessthe H electron affinity (0.75 eV), or about 23.8 eV. In contrast, the 2σ–3σenergy difference is roughly the 1s–2s excitation energy of atomic hydrogen,which is three-fourths the H atom ionization energy, or about 10.2 eV, muchless than the 1σ–2σ excitation energy. Thus, the first excited state of HeH is1σ23σ1 (rather than 1σ12σ2), which leads to a Σ+2 term symbol. For HeH+,which is “protonated He,” the ground state is 1σ2 with a Σ+1 term symbol. Thefirst excited MO is the 2σ, which we have seen is largely the H atom 1s AO,and the MO configuration is therefore 1σ12σ1 (so that this state resembles He+

stuck to H). Now we have two unpaired electrons so that S could be either 0 (asinglet state) or 1 (a triplet state). Thus, 1σ12σ1 leads to two term symbols(and two states, just as H + H leads to two states): Σ+1 and Σ+3 . Finally, Li2

+

has the ground state configuration 1σg21σu

*22σg1 and a Σg

+2 term symbol, justlike H2

+. The first excited state also parallels H2+. It has the configuration

1σg21σu

*22σu*1 and a Σu

+2 term symbol. The following table summarizes thesearguments:

Molecule Ground state First excited state

H2+ 1σg

1 Σg

+21σu

*1 Σu

+2

He2+ 1σg

21σu

*1 Σu

+21σg

11σu

*2 Σg

+2

HeH 1σ22σ1

Σ+21σ2

3σ1 Σ+2

HeH+ 1σ2 Σ+1

1σ12σ1

Σ+1, Σ+3

Li2+ 1σg

21σu

*22σg

1 Σg

+21σg

21σu

*22σu

*1 Σu

+2

19.6 A 4Π state has Λ = 1 and S = 3/2 so that Σ = –3/2, –1/2, 1/2, and 3/2. Thepossible values for Ω = |Σ + Λ| are |1 – 3/2| = 1/2, |1 – 1/2| = 1/2, |1 + 1/2| =3/2, and |1 + 3/2| = 5/2, and we see that there are only three distinct values forΩ: 1/2, 3/2, and 5/2. On the other hand, the spin–orbit energy possibilities areAΣΛ = –3A/2, –A/2, A/2, and 3A/2, a total of four different energies. Thus,we need Σ + Λ as a label (which ranges over the four values –1/2, 1/2, 3/2, and5/2) in order to distinguish among the four spin–orbit energies.

19.7 In Example 19.5, we found that the total internal energy in CO in the v = 2, J =1 state is G(2) – G(0) + F2(1) = 4 260.26 cm–1 + 3.775 1 cm–1 = 4 264.02cm–1. We also found that the H2 v = 1, J = 0 state is 4 158.54 cm–1 above thelowest energy v = 0, J = 0 state. This means we have an excess energy of4 264.02 cm–1 – 4 158.54 cm–1 = 105.48 cm–1 that could perhaps excite H2 in

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v = 1 to rotational states higher than J = 0. We use constants in Table 19.2 tofind the energies of those states. First, we find the H2 rotational constant forthe v = 1 state from Eq. (19.28):

B1 = Be – αe v + 12

= 60.853 0 cm–1 – 3.062 2 cm–1 32

= 56.260 cm–1 .

Next, we write the rotational energy level expression in terms of the J quantumnumber, the rotational constant B1, and the centrifugal distortion constant Dealso listed in Table 19.2:

F1(J) = B1J(J + 1) – De J(J + 1) 2

= 56.260 cm–1 J(J + 1) – 4.71 × 10–2 cm–1 J(J + 1) 2 .

We are looking for the largest J value such that F1 ≤ 105.48 cm–1. We findF1(0) = 0 (of course), but F1(1) = 112.33 cm–1, which is already more than our105.48 cm–1 excess. Thus, we can excite H2 to v = 1, J = 0, but not to v = 1,J = 1 or higher.

19.8 We see from Eq. (19.21) (or Eq. (19.33a)) that the rotational constant for adiatomic is proportional to the reciprocal of the square of the internuclear sepa-ration: Be ~ Re

–2. Thus, if the rotational constant for the B state of H2 is one-third as large as that for the ground state (the X Σg

+1 state), Be(B) = Be(X)/3,and we can write (since the reduced mass is the same for both states—both referto H2)

Be(X)

Be(B) = 3 =

Re(B)

Re(X)

2 or Re(B) = 3 Re(X) = 3 (0.741 Å) = 1.283 Å .

(The actual rotational constants are 60.853 cm–1 and 20.015 cm–1 so that theactual Re value for the B state is 1.293 Å.)

19.9 Table 19.3 tells us that the dissociation energy of the ground state (the X state)is 5890 cm–1 (see footnote a) and that the difference between the lowest energyof the X state and the lowest energy of the A state is Te(A) = 14 680.58 cm–1.Add to this information the energy separation between the dissociation limits ofthe two states given in the problem, 16 956.183 cm–1, and we can sketch a dia-gram like the one on the next page (which is actually quite accurately drawnusing other information in the table, but a qualitative sketch would suffice here).

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2.0 4.0 6.0 8.0 10. 12. 14.

10

20

0

30

0R/Å

16 9

56.1

83 c

m–1

14 6

80.5

8 cm

–1

De(A)

De(X) = 5890 cm–1

V/1

03 c

m–1

We see that Te(A) + De(A) = De(X) + 16 956.183 cm–1 or, solving for De(A),we find the A state’s dissociation energy is 8166 cm–1.

19.10 If Na2 has J = 1, it has a rotational angular momentum [J(J + 1)]1/2¨ =1.491 × 10–34 J s. Equation (12.30) relates the classical angular velocity ω tothe angular momentum L and the moment of inertia I: L = Iω. Since I = µR2

for a diatomic and since µ = 11.494 885 2 amu for Na2 and R = 3.422 8 Å inthe B state (both from Table 19.3), we can calculate I = 2.235 × 10–45 kg m2

so that ω = L/I = 6.67 × 1010 s–1. Thus, in 7 ns = 7 × 10–9 s, the moleculerotates (7 × 10–9 s) × (6.67 × 1010 radians s–1) = 447 radians or, dividing by2π radians per revolution, excited Na2 in this state rotates about 74 times in its 7ns characteristic radiative lifetime. The harmonic vibrational constant for the Bstate is given as 124.090 cm–1 in Table 19.3, and since ωe = ¨(k/µ)1/2/[(100cm m–1)hc] (see the top of page 706 in the text), we can calculate (k/µ)1/2,which is related to the classical harmonic oscillator frequency in cycles of vibra-tion per second, ν, through (see page 402 in the text)

ν = 1

2π ω =

12π

.

We find ν = 3.72 × 1012 cycles s–1, or, in 7 ns, the v = 1 level of the Na2 Bstate vibrates through (7 × 10–9 s) × (3.72 × 1012 cycles s–1) = 2.6 × 104

cycles. Note that vibration is much faster than rotation, and that 7 ns is a longtime in the life of a molecule that has two heartbeats: a vibrational cycle tick anda rotational cycle tick, both of which are quite fast.

19.11 The energy of state (v1, v2, v3, v4, v5, v6) referenced to the zero-pointenergy (so that state (0, 0, 0, 0, 0, 0) has energy zero) is

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E = ω1v1 + ω 2v2 + ω 3v3 + ω 4v4 + ω 5v5 + ω 6v6 .

Using the constants in the problem and systematically changing quantum num-bers, we find the following states have energy less than 3000 cm–1: (1, 0, 0, 0,0, 0), (0, 1, 0, 0, 0, 0), (0, 0, 1, 0, 0, 0), (0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 0, 1),(0, 1, 1, 0, 0, 0), (0, 1, 0, 1, 0, 0), (0, 1, 0, 0, 0, 1), (0, 0, 1, 1, 0, 0), (0, 0,1, 0, 0, 1), (0, 0, 0, 1, 0, 1), (0, 2, 0, 0, 0, 0), (0, 0, 2, 0, 0, 0), (0, 0, 0, 2,0, 0), (0, 0, 0, 0, 0, 2), and (0, 0, 0, 3, 0, 0).

19.12 There are, of course, many possible answers here. All spherical, cubical, tetra-hedral, octahedral, and icosahedral objects are spherical tops. Prolate symmet-ric tops include objects such as an American football, a cigar, and a blimp.Oblate symmetric tops include a Frisbee™, an innertube or automobile tire, anda dinner plate. Examples of molecules not mentioned in the text include thespherical tops cubane (C8H8), CCl4, W(CO)6, Ni(CO)4, etc., the prolate sym-metric tops CH3Cl, NH3, etc., and the oblate symmetric tops C6F6, CCl3F,etc.

19.13 We follow the standard notation and coordinate system for a linear molecule,aligning the nuclei along the a axis, so that each atom’s b and c coordinatesequal zero. According to Eq. (19.3), the a moment of inertia is zero (becausethe b and c coordinates of each nucleus are zero), and Ib = Ic, as mentioned onpage 717 in the text. The bond lengths mentioned in the problem allow us toconstruct the following diagram:

1.203 Å 1.061 Å1.061 Å

0.6015 Å1.662 5 Å

a

0

The center of mass of acetylene is the C≡C bond midpoint (for the H12C12CHisotopomer, which is what we assume we have here since it is by far the mostabundant species in ordinary acetylene), and thus the a coordinate of each Catom is (1.203 Å)/2 = 0.6015 Å and that of each H atom is [1.061 Å + 0.6015Å] = 1.662 5 Å. The moment of inertia expression depends on the atomicmasses: mH = 1.673 7 × 10–27 kg and the atomic mass of 12C is, by definitionof the atomic mass unit, exactly 12 amu, or 1.992 648 24 × 10–26 kg. Themoment of inertia is (the factors of 2 account for the fact that we have two Hatoms and two C atoms, of course)

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Ib = 2mH 1.662 5 × 10–10 m2 + 2m C12 0.6015 × 10–10 m

2

= 2.366 × 10–46 kg m2 .

The rotational constant follows from Eq. (19.33a):

Be = ¨ 2

2Ibhc(100 cm m–1) = 1.183 cm–1 .

These numbers correspond to the equilibrium structure of acetylene, and theyare thus quite close to the values for the ground vibrational state. The excitedstate spectrum, we are told, reveals that the energy difference between the J = 0and J = 1 rotational levels of some excited vibrational state (call it v) is 2.327 5cm–1. Since the energy expression for rotation of a linear polyatomic is thesame as for a diatomic, Fv(J) = Bv J(J + 1), (Eq. (19.27) ignoring the centrifu-gal distortion term, about which we have no information), we see that this mea-sured energy difference is

2.327 5 cm–1 = Fv(1) – Fv(0) = 2Bv

or Bv = (2.327 5 cm–1)/2 = 1.163 75 cm–1. This number corresponds to amoment of inertia for this excited vibrational state equal to

Iv = ¨ 2

2Bvhc(100 cm m–1) = 2.405 39 × 10–46 kg m2 .

If we assume the C≡C bond length is the same in this excited state as in theground vibrational state, 1.203 Å, then the change in the moment of inertia mustbe due to a change (an increase) in the C–H bond length. If we call the C–Hbond length R, we have

Iv = 2.405 39 × 10–46 kg m2

= 2mH R + 0.6015 × 10–10 m2 + 2m C12 0.6015 × 10–10 m

2 .

Solving this expression for R gives R = 1.096 Å, slightly, but significantly,longer than in the ground vibrational state. (See also Problem 19.16 for theeffects of vibrational excitation on average bond length.)

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SECTION 19.3

19.14 The plan here is to calculate first the spectroscopic constants for the 13CO iso-topomer, then to use those constants in Eq. (19.27) to find the energies of the J= 6 and J = 5 rotational levels, and finally to subtract those energies to find thetransition energy. Table 19.2 lists the rotational constant Be and centrifugaldistortion constant De for the common isotope 12CO, and the problem remindsus how these are related to reduced mass. Thus, we can write

Be( C13 O)

Be( C12 O) =

µ( C12 O)

µ( C13 O) and

De( C13 O)

De( C12 O) =

µ( C12 O)

µ( C13 O)

2

.

The nuclear masses given in the problem allow us to calculate the reducedmasses:

µ( C12 O) = m C12 m O16

m C12 + m O16 = 6.856 208 amu

µ( C13 O) = m C13 m O16

m C13 + m O16 = 7.172 410 amu

which give Be(13CO) = 1.846 2 cm–1 and De(13CO) = 5.593 7 × 10–6 cm–1.(Note that both constants are smaller than for 12CO. Increasing isotopic massalways lowers rotational constants.) Next, we calculate the rotational energiesfor J = 6 and 5 from Eq. (19.26). Since we are in the ground vibrational state,v = 0, and we write F0(J) = B0J(J + 1) – D0[J(J + 1)]2. Equations (19.28) and(19.29) relate the constants for vibrational state v to the equilibrium constants,and we see that B0 and D0 equal Be and De except for vibration-rotation correc-tion constants αe and βe. Since we are interested in the rotational energy levelspacing in a single vibrational state, these correction constants will cancel fromour final answer, and we do not need to find their values for the 13CO iso-topomer. Thus, we write simply F0(J) = BeJ(J + 1) – De[J(J + 1)]2 and use our13CO constants to calculate F0(6) = 77.529 cm–1 and F0(5) = 55.380 cm–1 sothat the transition energy is ∆E6→5 = 22.149 cm–1. A transition energy in cm–1

units is easily converted to a transition frequency in Hz (or GHz) units; wemultiply an energy in cm–1 units by hc(100 cm m–1) to convert to joule energyunits, then use ∆E = hν to find the frequency, or more directly

ν/Hz = (∆E/cm–1) × (100 cm m–1) × (c/m s–1)

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so that 22.149 cm–1 corresponds to ν = 6.640 1 × 1011 Hz or 664.01 GHz.A quick check with Figure 19.10 shows that these numbers are correct; they arequite close to the 12CO values shown in part (a) of that figure for the 5→6absorption transition, but slightly smaller, as part (b) indicates for the 3→4transition.

19.15 Our task here is to assign initial and final state quantum numbers to the featuresseen in a spectrum. This is often easy to do without knowing spectroscopicconstants with great accuracy. (After all, the first time a spectrum is recorded,these constants may be very poorly known. We may not even be sure whatmolecule is producing the spectrum, and even if we know the molecule, wemay have to make guesses about its bond lengths, bond angles, etc.) Here, weknow a great deal from Table 19.2 about the 7LiF isotopomer. In particular, weknow Re, which is isotopically invariant, and from Re and the nuclear masses,we can calculate Be for 6LiF. Since the features in the spectrum are pairs ofclosely spaced transitions separated by rather large gaps and since the spectrumis a microwave spectrum, we start our assignment with a rough calculation ofthe 6LiF rotational transition frequencies. We take the nuclear masses to be 6amu and 19 amu, and calculate an approximate Be, using Eq. (19.33a) and theequilibrium bond length, ~1.56 Å = 1.56 × 10–10 m, listed in Table 19.2:

Be = ¨ 2

2Ie 100 cm m–1 hc =

¨ 2

2µRe2 100 cm m–1 hc

= ¨ 2

2 6 × 196 + 19

1.66 × 10–27 kg 1.56 × 10–10 m2 100 cm m–1 hc

= 1.50 cm–1 .

Next, we calculate transition energies for various initial rotational quantumnumbers J′′ using Eq. (19.41), ∆E ≅ 2Be(J′′ + 1), and construct this table:

J′′ ∆E/cm–1 ∆E/GHz

0 3.00 89.91 6.00 180.2 9.00 270.3 12.0 360.4 15.0 450.

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11

The spectrum shows pairs of lines at ~270 GHz, ~360 GHz, and ~450 GHz,and the table above shows that these pairs are somehow associated with absorp-tions that change the rotational quantum number J from 2 to 3, from 3 to 4, andfrom 4 to 5, respectively. Now we must explain why there are pairs of linesinstead of a single line for each rotational transition. The key clue is the unevenintensity of each member of each pair: the higher frequency line of each pair hasfour times the intensity of the lower frequency line. Since we are told that thespectrum is taken with 6LiF at high temperature, the answer must be that we areseeing the microwave spectra of two different vibrational states. At ordinarytemperatures, ordinary molecules are almost entirely all in their ground vibra-tional state, but as the temperature is raised, more and more are vibrationallyexcited through the energetic collisions that characterize high temperatures.(The details of this excitation and its consequences are discussed in Chapters 22and 23.) The more intense line of each pair corresponds to rotational transitionsof molecules in the ground vibrational (v = 0) level, and the less intense linecorresponds to transitions of molecules in the first excited (v = 1) vibrationallevel. Vibration-rotation coupling explains why the excited level transitions areat a smaller transition frequency: vibrational excitation almost always extendsthe average bond length in a diatomic, increasing the moment of inertia, lower-ing the rotational constant for that vibrational level, and thus lowering transitionfrequencies (or energies).

19.16 Figure 19.11 shows us that the R(1) and P(1) transitions both have J′′ = 1 ini-tial rotational quantum numbers, but the R(1) transition ends in a level with J′ =2 while the P(1) transition ends with J′ = 0. Thus, the difference R(1) – P(1) =13 899.315 cm–1 – 13 868.273 cm–1 = 30.042 cm–1 is the energy differencebetween the J = 2 and J = 0 levels of the v = 7 vibrational state, which we canexpress as F7(2) – F7(0) = B7[2(2 + 1)] – B7[0(0 + 1)] = 6B7 (using Eq.(19.27) in the rigid rotor approximation). The rotational constant for thisvibrational level is thus B7 = (30.042 cm–1)/6 = 5.173 7 cm–1. The reducedmass for H127I is

µ = mHmI

mHmI

= 1 amu × 127 amu1 amu + 127 amu

1.660 54 × 10–27 kg amu–1 = 1.647 6 × 10–27 kg

so that the expression for B7 is

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B7 = ¨ 2

2Ie 100 cm m–1 hc =

¨ 2

2µR72 100 cm m–1 hc

= 5.173 7 cm–1

which we can solve for R7, the average bond length in the v = 7 vibrationallevel. We find R7 = 1.812 2 × 10–10 m = 1.812 2 Å, which is considerablylonger than the equilibrium bond length Re = 1.609 2 Å listed in Table 19.2.The reason for this increase can be traced to the nature of the nuclear motion inhighly excited vibrational levels. The molecule spends most of its time (speak-ing classically) with its bond extended far beyond its equilibrium position. Themolecule is displaying the effects of the real internuclear potential energy func-tion, which is not at all well approximated by a harmonic oscillator at high exci-tation energies.

19.17 The vibrational energy level expression for a diatomic, Eq. (19.26), depends onthe harmonic vibrational constant ωe and anharmonic correction constants ωexe,etc. An accurate prediction of the overtone spectrum would require us to knowas many anharmonic constants as possible, but since we are looking for a 400cm–1 wide region, we can approximate the transition energy with just the con-stants listed in Table 19.2 for H35Cl: ωe = 2 990.946 cm–1 and ωexe = 52.819cm–1. Thus, for this isotopomer, the hypothetical v = 0→2, J = 0→0 transi-tion would occur at

∆E = G(2) – G(0) = ωe 2 + 12

– ωexe 2 + 12

2 – ωe 0 +

12

– ωexe 0 + 12

2

= 2ωe – 6ωexe = 5 664.978 cm–1 .

Thus, we should center our scan at ~5 665 cm–1. Note that the harmonicapproximation ∆E = 2ωe = 5 981.892 cm–1 is 317 cm–1 higher and would be apoor place to center our scan; anharmonic corrections are increasingly moreimportant to consider as the vibrational quantum number v increases. (Table19.2 also tells us the HCl, like all diatomic hydrides, has a large rotational con-stant, Be ≅ 10.6 cm–1. Thus, the vibration-rotation transitions—the P and Rbranches of the spectrum in the language of Figure 19.11—will extend over asignificant region. A scan that extends 200 cm–1 away from the hypothetical Qbranch will cover transitions from R(0) to about R(7), as you can verify if youinclude the rotational energy expression, Eq. (19.27), in the transition energyexpression. Interestingly, the entire P branch falls within 200 cm–1 of our cen-ter frequency. The P(15) line is about 160 cm–1 from our scan center, but

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higher P-branch transitions are closer to this center. This is called band headformation—the P branch lines march out from the band center, then pile up at alimit, the band head, then start marching back towards the center. This is due tothe particular combination of centrifugal distortion and vibration-rotation inter-action constant values H35Cl happens to have, but band head formation is quitecommon.) Now we consider DCl. Table 19.2 does not list vibrational con-stants for DCl, but ωe has a simple dependence on the reduced mass of thediatomic, as does any harmonic vibrational frequency constant: ωe ~ µ–1/2. Anaccurate calculation of the D35Cl ωe from the accurate H35Cl constant wouldentail an accurate nuclear mass for the 35Cl nucleus; here, we can take it to besimply 35 amu = 5.812 × 10–26 kg. The H nuclear mass is mp, the protonrest mass, 1.673 × 10–27 kg. Thus, the H35Cl reduced mass is

µH C35 l = mpm C35 l

mp + m C35 l = 1.626 × 10–27 kg ,

which is quite close to the H atom mass. (This is a general result: µ ≅ thesmaller mass if the two masses are very different. Recall our first discussion ofthe H atom in Chapter 12: the H atom reduced mass was quite close to the elec-tron mass. See page 424 in the text.) For D35Cl, we can approximate the Dnuclear mass quite closely by 2 amu = 3.321 × 10–27 kg. We find

µD C35 l = mDm C35 l

mD + m C35 l = 3.141 × 10–27 kg ,

again quite close to the D mass and almost exactly twice the H35Cl reducedmass. Thus, to a good first approximation that is worth memorizing, substitut-ing D for H lowers any H–X vibrational frequency by a factor quite close to

2. (See page 726 in the text.) We can do better than this approximation here,however, and we now go on to calculate the D35Cl harmonic constant:

ωe(D C35 l) = ωe(H C35 l) µH C35 l

µD C35 l = 2 151.920 cm–1 .

(Note that ωe(H35Cl)/ 2 = 2115 cm–1.) We found that we needed the ωexeanharmonic constant to locate the H35Cl overtone accurately, but how to wescale the H35Cl ωexe constant to yield ωexe for D35Cl? Table 19.2 lists thisconstant for various isotopes of molecular hydrogen, and if we recognize thatµH2 = mH/2 and µD2 = mD/2 = 2µH2, we see that ωexe scales with µ–1: the H2

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14

ωexe value is twice the D2 value (and three times the T2 value, etc.). (In gen-eral, each successively higher anharmonic constant depends on µ by one morefactor of µ–1/2.) Thus, we can find

ωexe(D C35 l) = ωexe(H C35 l) µH C35 l

µD C35 l = 27.342 cm–1 .

Our scan center for D35Cl is 2ωe – 6ωexe = 4 139.789 cm–1. Note that thesimplest approximation, a frequency or wavenumber 2 lower than for H35Cl,gives 4006 cm–1, a quite acceptable first approximation that would allow us tofind the real spectrum within our 400 cm–1 window. Finally, we can state thatthe H37Cl spectrum will occur at a lower frequency than the H35Cl spectrumdue to the increased reduced mass of H37Cl over H35Cl. The shift is not great,since the reduced masses are quite close: 1.629 × 10–27 kg for H37Cl versus1.626 × 10–27 kg for H35Cl, but this is enough to shift the H37Cl spectrumabout 4 cm–1 from the H35Cl v = 0 to 2 overtone spectrum. Most moderninfrared spectrometers (that can operate in this region—it is a bit above the usualupper limit of most routine instruments) have resolutions better than 4 cm–1.

19.18 Table 19.2 tells us that Re for HF is 0.9168 Å and 1.128 3 Å for CO. Thepermanent dipole moment of a diatomic is (see Chapter 15 for details) the aver-age of the dipole moment function, p(R), over the vibrational wavefunction.The only state of importance for an equilibrium gas sample is the ground vibra-tional state for the vast majority of ordinary gases like HF and CO, and weknow that the v = 0 vibrational wavefunction is quite closely centered aroundRe with a very small spread. Thus, to a good first approximation, the observeddipole moment is simply the value of the dipole moment function evaluated atRe: p(Re). The graph in the problem shows us that p(Re) for HF is quite large,about 6 × 10–30 C m (Table 15.1 lists the accurate experimental value6.069 × 10–30 C m). In contrast, the CO dipole moment function is negative(i.e., in the sense –CO+) in the vicinity of Re and quite close to zero. It is aninteresting accident that the CO dipole moment function happens to be passingthrough zero in the vicinity of its equilibrium bond length. (Table 15.1 lists apermanent dipole moment magnitude of 0.374 × 10–30 C m in the –CO+sense.) On the other hand, the discussion of the vibrational transition dipolemoment on page 724 in the text shows that it is the slope of the dipole momentfunction, not its magnitude, that governs the transition probability. The graphof the dipole moment functions in the problem show that both HF and CO havevery similar slopes in the vicinity of their equilibrium bond lengths.

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19.19 The energy level expression for the rigid-rotor, harmonic oscillator diatomic isquite simple:

E(v, J) = ω e v + 12

+ BeJ(J + 1) .

For a (hypothetical) Q branch transition, the v quantum number changes(usually by ±1, the strongly allowed selection rule), but the J quantum numberdoes not. Thus, the transition energy for all J states of the molecule is the same:ωe∆v where ∆v is the vibrational quantum number change magnitude. The Qbranch would appear as a single feature, but it would be quite intense, since itwould represent transitions for all the J states with significant population, incontrast to the P and R branches where each J state leads to a unique transitionfrequency. The R(0) line (for which J increases from 0 to 1) occurs at

∆E(R(0)) = E(v′, 1) – E(v″, 0) = ωe∆v + 2Be

while the P(1) line (J decreases from 1 to 0) occurs at

∆E(P(1)) = E(v′, 0) – E(v″, 1) = ωe∆v – 2Be

which is exactly as far below the Q line as the R(0) line is above it.

19.20 Chapter 14 deduced that the N2 ground electronic state MO configuration has tobe 1σg

21σu*22σg

22σu*21πu

43σg2 and that the lowest three electronic states of N2

+

have the electron configurations

1σg2 1σu

*2 2σg

2 2σu

*2 1πu

4 3σg1 (the ground state)

1σg2 1σu

*2 2σg

2 2σu

*2 1πu

3 3σg2 (the first excited state)

1σg2 1σu

*2 2σg

2 2σu

*1 1πu

4 3σg2 (the second excited state).

In order of increasing energy, these are called the X, A, and B states, andmolecular term symbols for them are not difficult to deduce. No fully occupiedMO contributes to the term symbol (or, stated better, each represents a Σg

+1

contribution—S = 0 and Λ = 0—to the ion’s term symbols). The ground stateterm symbol is dominated by the single 3σg electron, which makes S = 1/2 andthus 2S + 1 = 2 (the state is a doublet state), but this electron is in a σ MO withΛ = 0, and since it is a g MO, so is the entire electronic state. Thus, the ground

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16

state is a Σg+2 state (and we are sure that the state is a + state because the π MOs

are still fully occupied). The first excited state must also be a doublet, since theπu MOs have one unpaired spin, and it must be a Π state as well because thisunpaired electron carries Λ = 1, as do all electrons in π MOs. The state has theinversion symmetry of this unfilled π MO, which is u, so that the term symbolis Πu

2 ; the +/– superscript notation is used only for Σ states. The next state isalso a doublet Σ state, but it has u symmetry from the unpaired electron in the2σu

* MO. It is a Σu+2 state. The photoelectron spectrum showed the greatest

vibrational excitation in the A state (see Figure 14.13), and the Franck–Condonprinciple tells us that extensive vibrational excitation of an excited state from anexcitation of the ground vibrational level of a lower energy electronic state (theN2 ground state here) means the two electronic states have significantly differ-ent bond lengths. It is unusual for an excited electronic state to have a bondlength significantly shorter than the ground state, or for a molecular ion groundstate to have a shorter bond than its neutral molecular parent. The bond lengths(Re values) of the N2 ground state, the N2

+ X state, and the N2+ B state must all

be comparable (other experiments show that they are 1.097 68 Å, 1.116 Å, and1.074 2 Å, respectively), but the N2

+ A state bond length must be significantlylonger (and experiment finds 1.175 Å). The photoelectron energies quoted inChapter 14 locate these N2

+ states above the v = 0 level of the N2 ground elec-tronic state. (The lowest of these energies, 15.57 eV, is simply the N2 ioniza-tion potential.) What is not so obvious is that all three of these molecular ionelectronic states dissociate to the same dissociation limit: a ground state N atomand a ground state N+ atomic ion. The graph on the next page shows all fourpotential energy curves, the neutral N2 ground state and the three ion states,accurately drawn from data derived from a number of spectroscopic experi-ments. The vertical dashed line is drawn at the N2 equilibrium bond length,1.097 68 Å, to point out the significantly longer A state bond length and theslight differences among the N2 X state Re and those of the ion’s X and Bstates.

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17

1.0 1.5 2.0

5

10

15

2.50.5R/Å

V/1

04 c

m–1

20

0

X 1Σg N2+

X 2Σg N2+ +

A 2Πu N2+ +

B 2Σu N2+ +

19.21 Table 19.3 tells us Te for the A state is 14 680.58 cm–1 (see also Figure 19.2and Problem 19.9), and if we add to this the vibrational energy for v′ = 22, 23,and 24 using the A state constants ωe = 117.323 cm–1 and ωexe = 0.3576 cm–1

in Eq. (19.26) and subtract the vibrational energy for v″ = 0, 1, and 2 using theX state constants ωe = 159.124 cm–1 and ωexe = 0.7254 cm–1, we will have thetransition energies, in cm–1, for the nine transitions of interest. We convertthese transition energies ∆E into vacuum wavelengths in nm through λ/nm =107/∆E/cm–1. For example, the v″ = 0 to v′ = 22 calculation is

∆E = Te + ωe′ v′ +

12

– ωexe′ v′ +

12

2 – ωe

″ v″ + 12

– ωexe″ v″ +

12

2

= 14 680.58 + 117.323 (22.5) – 0.3576 (22.5)2 – 159.124/2 + 0.7254/4

= 17 059.93 cm–1

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18

so that

λ = 107nm cm–1

17 059.93 cm–1 = 586.17 nm .

Repeating this calculation for the other vibrational quantum numbers of interest(a spreadsheet computer program speeds this calculation enormously!) leads tothe table below:

v ′ v″ ∆E/cm–1 λ/nm

22 0 17 059.93 586.17

22 1 16 902.26 591.64

22 2 16 746.04 597.16

23 0 17 160.81 582.72

23 1 17 003.13 588.13

23 2 16 846.91 593.58

24 0 17 260.96 579.34

24 1 17 103.29 584.68

24 2 16 947.10 590.07

These transition wavelengths are shown in a stick spectrum below along withthe wavelengths of the two atomic Na transitions given in the problem. (Thesticks are drawn below with different heights just to help you sort them out; thestick heights do not represent intensities as they would appear experimentally.)

580 585 590 595575 600λ/nm

24→0

23→0

24→1 22→

0

23→1

24→2 22→

1

23→2

22→2

Na

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19

Note how the various Na2 transitions (and remember that they would not appearas single, sharp features experimentally due to the many rotational transitionsthat accompany each vibrational transition) intermingle and appear near the Naatomic lines.

19.22 Difluroethylene exists in three isomeric forms:

C CH

HF

F

1,1-difluoroethylene

C CF

HH

F

cis-difluoroethylene

C CH

FH

F

trans-difluoroethylene

The first two isomers have a permanent dipole moment, which is required inorder to have a microwave (rotational) absorption spectrum. The trans isomer,however, lacks a permanent dipole moment by symmetry and thus does nothave a microwave spectrum. Since no transitions were seen, the sample musthave been the trans isomer.

19.23 The symmetry of the AX3 planar symmetric top molecule indicates how weshould place inertial axes with respect to the A–X bonds (and with the origin atthe A atom, which is clearly the center of mass). If we call each A–X bondlength R, we can make a diagram as shown below. At first, we will call theaxes simply x, y, and z, but once we know the magnitudes of Ix, Iy, and Iz, wecan confidently label them a, b, and c according to the magnitudes of each I.

X X

X

x

y

R

60°

A

If we call the mass of the X atom m and note that the A atom sits at the coordi-nate origin so that its mass will not enter the moment of inertia calculation, themoment of inertia expressions in Eq. (19.3) become

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20

Ix = mi yi2 + zi

2∑i

= mR2 + 2mR2cos260° = 3mR2

2

Iy = mi xi2 + zi

2∑i

= 2mR2sin260° = 3mR2

2

Iz = mi xi2 + yi

2∑i

= mR2 + 2mR2 sin260° + cos260° = 3mR2 .

Thus, AX3 is an oblate symmetric top with Ix = Iy = Ia = Ib and Iz = Ic with Ic =2Ia = 2Ib. (Problem 19.2 showed that Ia + Ib = Ic for a planar molecule, andthis, along with Ia = Ib for an oblate symmetric rotor, is enough to prove that Ic= 2Ia here. We need the explicit expressions for the moments of inertia later inthis problem, however.) In BF3, since the B atom is at the center of mass if theF atoms have the same mass, isotopic substitution of the B atom does notchange the moments of inertia. Using Eq. (19.33) to relate the rotational con-stants to moments of inertia, we see that the smaller rotational constant is the Cconstant and C = 0.173 cm–1 means

Ic = ¨ 2

2hc 100 cm m–1 0.173 cm–1 = 1.62 × 10–45 kg m2 .

The bond length, using our expressions above and the mass of the F atom, is

R = Ic

3m =

1.62 × 10–45 kg m2

3 3.155 × 10–26 kg = 1.31 × 10–10 m = 1.31 Å .

In contrast, the 11BF bond length can be found from the rotational constant andmoment of inertia expressions for a diatomic:

Be = ¨ 2

2µRe2hc 100 cm m–1

or Re = ¨ 2

2µBehc 100 cm m–1 .

With Be = 1.516 cm–1 and the 11BF reduced mass, we find Re = 1.26 Å,somewhat shorter than in BF3.

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19.24 Equation (19.37b) reads Erot(J,K) = BJ(J + 1) + (C – B)K2, or, with the B andC constants in Figure 19.8(b), (0.345 cm–1)J(J + 1) + (–0.157 cm–1)K2. Wecan use this expression to construct the table below:

J K Erot/cm–1

0 0 0

1 0 0.690

1 ±1 0.533

2 0 2.070

2 ±1 1.913

2 ±2 1.442

From the selection rules ∆J = ±1, ∆K = 0, we see that the allowed transitionsare J = 0 → J = 1, K = 0 (at 0.690 cm–1), J = 1, K = 0 → J = 2, K = 0 (at2.070 cm–1 – 0.690 cm–1 = 1.380 cm–1), and J = 1, K = 1 → J = 2, K = 1(also at 1.913 cm–1 – 0.533 cm–1 = 1.380 cm–1). Thus, among these sixstates, only three transitions are allowed, and two of these have the same transi-tion energy so that only two transitions would be observed if the molecule wastruly rigid. We can see this analytically if we recognize that the selection rulesmean that absorption transitions only of the type J,K → (J + 1),K are allowedso that the transition energies are ∆E = Erot(J + 1,K) – Erot(J,K) = 2B(J + 1),independent of K. If we include the effects of centrifugal distortion in the Erotexpression, writing

Erot(J,K) = BJ(J + 1) + (C – B)K2 – DJJ2(J + 1)2 – DJKJ(J + 1)K2 – DKK4 ,

the selection rules stay the same, of course, but the transition energy expressionchanges to

∆E = Erot(J + 1,K) – Erot(J,K)

= 2B(J + 1) – 4DJ(J + 1)3 – 2DJK(J + 1)K2

which does depend on K. Including centrifugal distortion shows that the J = 1,K = 0 → J = 2, K = 0 transition and the J = 1, K = 1 → J = 2, K = 1 transitionhave difference transition energies. Note, however, that neither transitionenergy expression depends on C, and the more complete expression also doesnot contain the DK centrifugal distortion constant.

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19.25 Acetylene has no permanent dipole moment, of course, but a vibrational motionthat produces a dipole moment as the atoms move in a particular normal modewill be infrared active. The cis bend does just that. At the extremes of itsmotion, acetylene is bent in a way that leads to a net dipole moment; the C–Hbond moments point in directions that lead to a net moment in a direction per-pendicular to the C≡C bond and in the plane of the (bent) molecule. As long asthe dipole moment function has a non-zero first derivative with respect to theatomic motion of the normal mode in question, the vibration is infrared active.In contrast, the trans bend always has a zero dipole moment no matter howgreat the bending amplitude. This mode is not infrared active, but it is Ramanactive. It is always true in molecules with a center of symmetry that modeswhich are infrared active are not Raman active and vice versa.

SECTION 19.4

19.26 Table 19.3 tells us that ωe for the Na2 ground state, X Σg+1, is 159.124 cm–1,

De = 5890 cm–1 = 1.17 × 10–19 J, and µ = 11.495 amu = 1.909 × 10–26 kg,and Eq. (19.48a) can be rearranged to

β = 2πcωe(100 cm m–1) µ

2De

1/2 = 8.561 × 109 m–1 .

On the other hand, the Morse dissociation energy expression, Eq. (19.48b), letsus calculate the Morse approximation to De from ωe and ωexe (which Table19.3 tells us is 0.7254 cm–1):

De = ωe

2

4ωexe

hc(100 cm m–1) = 1.733 × 10–19 J ,

(or 8726 cm–1, quite a bit larger than the true dissociation energy). The Morsecentrifugal distortion constant expression, Eq. (19.48d), depends on ωe and Be(which Table 19.3 tells us is 0.154 707 cm–1) so that

De = 4Be

3

ωe2

= 5.85 × 10–7 cm–1 ,

which we should compare to the observed De listed as 5.81 × 10–7 cm–1 inTable 19.3. The Morse vibration-rotation constant comes from Eq. (19.48e):

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23

αe = 6

ωe

Be3ωexe

1/2 – Be

3 = 1.05 × 10–3 cm–1 ,

which we should compare to the tabulated value, 0.8736 × 10–3 cm–1. Ingeneral, the Morse expressions among spectroscopic constants such as Eq.’s(19.48d) and (19.48e) are more accurate than the Morse dissociation energyexpression. The next problem has more to say about this.

19.27 If we solve Eq. (19.48e) for ωexe, we find

ωexe = Be

2 + ωeαe/62

Be3

.

Substituting this into Eq. (19.49) gives

De = ωe

2Be3

4 Be2 + ωeαe/6

2 hc(100 cm m–1)

= 9ωe

2Be3

6Be2 + ωeαe

2 hc(100 cm m–1) .

If we drop the factor hc(100 cm m–1) so that our dissociation energy is in cm–1

units, and using data in Table 19.2, we can construct the following table, con-trasting this new expression to Eq (19.49):

Molecule De(expt.)/cm–1 De(Eq. (19.49))/cm–1 De(new)/cm–1

F2 13 370 18 695 17 519

CO 90 544 88 578 83 804

HF 49 390 47 635 40 237

H2 38 298 39 911 30 832

The moral here is that neither Morse expression is particularly accurate at pre-dicting dissociation energies.

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19.28 Following the discussion in the text leading to Figure 19.16, we define n = v +1/2 and construct the table below from the data given in the problem:

v G(v)/cm–1 n ∆G(n)/cm–1

0 0 0.5 25.740

1 25.740 1.5 20.409

2 46.149 2.5 15.606

3 61.755 3.5 10.906

4 72.661 4.5 6.780

5 79.441 — —

Next, we plot ∆G(n) versus n, as in Figure 19.16:

2 4

10

20

3 510 6n

30

0

∆G(n

)/cm

–1

The line through the points is a least-squares fit straight line to them, and thearea under this line (the shaded area) is the Birge–Sponer approximation to D0.We can estimate this area from the x and y axes intercepts read from the graph:area = (x intercept) × (y intercept)/2 ≅ (5.8) × (28 cm–1)/2 = 81.2 cm–1. Theleast-squares fit line is ∆G(n) = 27.744 cm–1 – (4.742 3 cm–1)n, and its y inter-cept is 27.744 cm–1 while its x intercept is (27.744 cm–1)/(4.742 3 cm–1) =5.850. The area under the least-squares line is thus (5.850) × (27.744 cm–1)/2= 81.16 cm–1, in good agreement with the value deduced from reading theintercepts from the graph (which is acceptable if the straight line through thepoints is drawn by hand rather than deduced from a least-squares analysis). Asindicated in the problem, more advanced methods of extrapolating the data in aBirge-Sponer plot indicate that D0 is somewhat larger that the estimate derivedfrom a linear fit: 84.75 cm–1 rather than ~81.2 cm–1. This tells us that thehigher vibrational levels lead to ∆G(n) points that lie above the straight line

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25

approximation. (Contrast Figure 19.16 for H2 where the high n points fallbelow the straight line extrapolation.) The experiment observed vibrationallevels from v = 0 through 5, and our graph indicates that an unobserved v = 6level is surely possible. Higher levels are difficult to predict without the moreadvanced methods that lead to the improved dissociation energy value. Finally,the zero-point energy is approximately one-half the v = 0 to 1 energy differ-ence, 12.87 cm–1, for a De value of 84.75 cm–1 + 12.87 cm–1 = 97.62 cm–1.This approximation does not take anharmonicity into account and thus underes-timates the zero-point energy. To include anharmonicity, we should fit theobserved vibrational energies data to the variant of Eq. (19.26) that takes the v= 0 level as the energy origin: ωev – ωexev2 + … . When this is done, onefinds the zero-point energy is 14.95 cm–1 so that a more accurate estimate forthe dissociation energy is De = 84.75 cm–1 + 14.95 cm–1 = 99.70 cm–1.

19.29 The key to the question here lies in the symmetry of CO2 and thus in the sym-metry of any potential energy function we might choose to represent it. Sincean O–C–O bond angle of θ places the molecule in a configuration that is indis-tinguishable from an angle of 360° – θ (except for a rigid rotation of themolecule—see the diagram below), any potential energy function we write musthave the property that V(θ) = V(360° – θ).

O OC

θ

O OC

360° – θ

O OC

360° – θ

= =

If we write ∆θ = 180° – θ and imagine adding a cubic term to the potentialenergy function so that it becomes V(θ) = kb(∆θ)2/2 + b(∆θ)3 where b is aconstant, we can write

V(θ) = 12

kb (∆θ)2 + b(∆θ)3 = 12

kb (180° – θ)2 + b(180° – θ)3 .

On the other hand, we can also write this expression for a bond angle of 360° –θ, and by symmetry, this should equal the expression above forV(θ). We find

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26

V(360° – θ) = 12

kb [180° – (360° – θ)]2 + b[180° – (360° – θ)]3

= 12

kb (θ – 180°)2 + b(θ – 180°)3

= 12

kb [(–1)(180° – θ)]2 + b[(–1)(180° – θ)]3

= 12

kb (180° – θ)2 – b(180° – θ)3 ≠ V(θ) .

We see that the quadratic term is the same, but the cubic term has changed sign.This argument can be generalized to show that any term containing ∆θ to an oddpower will not have the correct symmetry while any term containing an evenpower of ∆θ will have the correct symmetry.

SECTION 19.5

19.30 The Doppler shift discussion in the text was centered around a moving absorberand a stationary light source. Here, we have a different physical situation: themoving molecule is emitting radiation toward a stationary receiver. We are toldthat the molecule is moving away from us, and in this case, the receiver sees asignal that is Doppler-shifted to greater wavelengths and thus to smaller fre-quency. (This shift may be familiar to you from optical astronomy. Atomicemissions from distant stars are frequently red-shifted to longer wavelength,and this shift is used to calculate the speed at which the star is moving awayfrom us.) Consequently, we use Eq. (19.50b) with β = v/c = (11 000 m s–1)/(299 792 458 m s–1) = 3.67 × 10–7 and ν0 = 664.01 GHz to find

ν– = ν01 – β

1 + β

1/2

= (664.01 GHz) × 1 – 3.67 × 10–7

1 + 3.67 × 10–7

1/2

= 663.99 GHz .

This is a small shift, about 20 MHz, but it can be measured accurately as longas the Doppler width of the emission is less than 20 MHz. We use Eq. (19.51)(dividing ∆E and E0 by h to convert the expression to frequency units) to calcu-late the Doppler frequency width ∆νD with T = 100 K and M = 29 g mol–1:

∆νD = 7.16 × 10–7 ν0 T/K

M/g mol–1

1/2 = 0.883 MHz .

This width is substantially less than the Doppler shift, and thus the speed atwhich the cloud is moving away from us can be measured accurately.

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19.31 Figure 19.20 shows a series of absorption features all around 6570 cm–1,which we can take to represent E0 in Eq. (19.51), the Doppler width expres-sion. With T = 300 K (it is a room temperature spectrum) and M = 26 g mol–1

(it is a spectrum of acetylene, HCCH), Eq. (19.51) gives

∆E ≅ 7.16 × 10–7 6570 cm–1 300

26 = 1.6 × 10–2 cm–1 ,

which is greater than the 0.004 cm–1 resolution of the instrument used to recordFigure 19.20. A higher resolution spectrum would show no further detail; Fig-ure 19.20 is “Doppler limited.”

19.32 For 14N2 with I = 1, there are (1 + 1)(2·1 + 1) = 6 symmetric nuclear-spinwavefunctions and 1(2·1 + 1) = 3 antisymmetric wavefunctions for a 6:3 = 2:1intensity alternation. For 15N2 with I = 1/2, we find three symmetric and oneantisymmetric wavefunction (just as in H2 or, for that matter, H12C12CH inFigure 19.20) for a 3:1 intensity alternation. For 14N15N, there is no intensityalternation for line to line because this molecule has distinguishable nuclei. Asan aside, and as Figure 19.20 shows, these ratios are not exactly found in theobserved spectra because another factor is guaranteed to give each transition aslightly different intensity whether or not nuclear spin statistics play a role.This factor is the variation with J of the number of molecules available to absorblight in a sample at equilibrium, as discussed in detail in Chapter 23.

GENERAL PROBLEMS

19.33 Using the two wavefunctions given in the problem, we see that the integrand ofthe Franck–Condon integral is

ψvib′ ψvib

″ ∝ exp – k

2¨ω R – Re

2 + R – Re – δ 2 .

Expanding the argument to the exponential gives, with a little algebra,

R – Re2 + R – Re – δ 2

= 2 R – Re2 + δ2

– 2δ R – Re ,

and the definition of q1,

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q12 = k

¨ω R – Re

2 or R – Re2 =

q12¨ωk

,

lets us write

2 R – Re2 + δ2

– 2δ R – Re = 2q1

2¨ωk

+ δ2 – 2δq1

¨ωk

.

Thus, the Franck–Condon integral is

g0,0 = 1π1/2

exp – k

2¨ω R – Re

2 + R – Re – δ 2 dq1

–∞

= 1π1/2

exp – k

2¨ω

2q12¨ωk

+ δ2 – 2δq1

¨ωk

dq1

–∞

= e–kδ2/2¨ω

π1/2 exp –q1

2 + k/¨ω δq1 dq1–∞

∞ = exp – k

¨ω δ

2

2 .

For k/¨ω = 200 Å–2, we can calculate g0, 0 for various δ values and constructthe table below:

δ/Å 0.01 0.05 0.1 0.5

g0, 0 0.990 0.779 0.368 1.38 × 10–11

For k/¨ω = 800 Å–2, we find:

δ/Å 0.01 0.05 0.1 0.5

g0, 0 0.961 0.368 0.0183 3.72 × 10–44

We see that relative Re shifts of several hundredths Ångström (which are verycommon) have notable effects on the transition intensity while shifts of morethan a few tenths Ångström effectively reduce the transition probability to zero.

19.34 If we follow Example 19.7, but use the full vibrational energy expression, Eq.(19.31), with the anharmonic constants given in the problem, we can constructthe table at the top of the next page.

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29

v1 v2 ª v3 G(v1, v2ª, v3)/cm–1 G(v1, v2

ª, v3) – G(0, 00, 0)/cm–1

0 0 0 0 2 532.25 0

1 0 0 0 3 866.18 1 333.93

0 1 1 0 3 199.72 667.47

0 2 0 0 3 871.83 1 339.58

0 2 2 0 3 867.95 1 335.70

The final column predicts the transition energies of these four excited statesfrom the ground (0, 00, 0) state, and, as mentioned in the problem, two of theseare seriously in error: (1, 00, 0) and (0, 20, 0). The theory behind the Fermiresonance interaction between these two states greatly improves the agreement.In the notation used in the problem, the unperturbed energies are Ea = E(1, 00,0) = 3 866.18 cm–1 and Eb = E(0, 20, 0) = 3 871.83 cm–1, and the Fermi reso-nance interaction constant is We = –52.84 cm–1. Note that the Fermi resonanceexpression in the problem can be written

Ea + Eb ± 4We2 + Ea – Eb

2 1/2

2 =

Ea + Eb

2 ±

4We2 + ∆2

2

where ∆ = Ea – Eb. Writing the expression this way shows that the interactionleads to two new states that are equally above and below the average energy ofthe unperturbed states. The perturbed states’ energies are 3 816.09 cm–1 and3 921.92 cm–1, and subtracting the zero-point energy, G(0, 00, 0) = 2 532.25cm–1, from these energies gives the predicted transition energies 1 283.84 cm–1

and 1 389.67 cm–1 which are in excellent agreement with the observed values.

19.35 In the symmetric stretch normal mode, both C–O bonds expand or contracttogether by the same amounts at all times. On the potential energy contour dia-gram in the problem, this motion lies along a line with unit positive slope (i.e.,extending from the (–0.1 Å, –0.1 Å) lower left-hand corner to the (0.1 Å, 0.1Å) upper right-hand corner of the figure). In contrast, the antisymmetric stretchnormal mode has one C–O bond contract while the other expands the sameamount. This motion follows a line with unit negative slope on the diagram(i.e., from the (–0.1 Å, 0.1 Å) upper left-hand corner to the (0.1 Å, –0.1 Å)lower right-hand corner in the figure. These lines are shown in the figure at thetop of the next page as two heavy diagonal lines.

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30

–0.05

0

0.05

–0.05 0 0.05

(RO

–C –

Re)

(RC–O – Re)/Å

1000

20003000 4000

If we sketch in a contour at the zero-point energy of 2532 cm–1 (the dashedcontour in the figure above), we can read off the classical turning points of eachnormal mode from the intersection points of this contour with the appropriatestraight line. For the symmetric normal mode, we see that one classical turningpoint has both bonds extended about 0.06 Å beyond equilibrium (i.e., from1.160 Å to about 1.22 Å), while at the other, both are compressed about 0.05 Å(to 1.11 Å). In the antisymmetric mode, one classical turning point has onebond compressed about 0.06 Å while the other is extended about 0.06 Å. Atthe other turning point, the two bonds switch roles—the extended bondbecomes the compressed bond and vice versa.