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 D216O and H218O isotopomers of water have Y = Z = 0, as for H216O, but X changes to D216O: X =

2mD(0.958 Å)cos(104.5°/2) = 0.118 Å 2mD + m1 6O

H218O: X =

2mH(0.958 Å)cos(104.5°/2) = 0.059 1 Å . 2mH + m1 8O

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 16O to 18O decreases X. Once we know the center of mass coordinates, 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 trigonometry used in Example 19.2 to locate the atoms, that the principal inertial coordinates of the atoms are 105.4° 105.4° (0.958 Å), cos (0.958 Å) – 0.118 Å, 0 2 2 D216O: = (±0.757 Å, 0.469 Å, 0) 16O : (a, b, c) = (0, –0.118 Å, 0) 105.4° 105.4° H: (a, b, c) = ±sin (0.958 Å), cos (0.958 Å) – 0.059 1 Å, 0 2 2 H218O: = (±0.757 Å, 0.527 Å, 0) 18O: (a, b, c) = (0, –0.059 1 Å, 0) . D: (a, b, c) = ±sin

1

With these (a, b, c) coordinates, we can use Eq. (19.3) to calculate the moments of inertia. Note that while the equilibrium geometry is identical for all isotopomers and the center of mass and (a, b, c) coordinates are not very different, the moments of inertia are quite different. 2 2 Ia = 2mDbD + m1 6ObO = 1.11 amu Å2 = 1.84 × 10–47 kg m2 2 D216O: Ib = 2mDaD = 2.31 amu Å2 = 3.84 × 10–47 kg m2 2 2 Ic = 2mD aD + bD2 + m1 6ObO = 3.42 amu Å2 = 5.68 × 10–47 kg m2 2 2 Ia = 2mHbH + m1 8ObO = 0.624 amu Å2 = 1.04 × 10–47 kg m2 2 H218O: Ib = 2mHaH = 1.16 amu Å2 = 1.92 × 10–47 kg m2 2 2 Ic = 2mH aH + bH2 + m1 8ObO = 1.78 amu Å2 = 2.96 × 10–47 kg m2 .

The key concept here is that the equilibrium structure of a molecule is independent of its isotopic composition. (We see later in this chapter that the average structure does depend on isotopic composition due to differences in zero-point energy motions. The equilibrium structure is the unique point of minimum B–O potential energy, while zero-point energies and the motions associated with them 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 N

Ia + Ib =

∑

i=1

mibi2

N

+

∑

miai2

i=1

N

=

∑

m i ai2 + bi2 = Ic .

i=1

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, we see 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 = 15 vibrations . Carbon suboxide is a linear five-atom molecule with 3N – 5 = 3·5 – 5 = 10 vibrations. The weakly-bound HFClF molecule is a nonlinear fouratom molecule with 3N – 6 = 3·4 – 6 = 6 vibrations, and ketene is a nonlinear five-atom molecule with 3N – 6 = 3·5 – 6 = 9 vibrations.

2

19.4

Equation (19.5) reads ∆Rvib ~ (me/M)1/4R, and with M = 1.15 × 10–26 kg for Li and 2.21 × 10–25 kg for Cs (both are atomic masses) and R = Re = 2.672 9 Å for Li and 4.47 Å for Cs, we predict ∆Rvib = 0.252 Å for Li2 and 0.201 Å for 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 harmonic force constant, 25.25 N m–1 for Li and 6.91 N m–1 for Cs. At the inner turning point, R–, (which is < Re) and at the outer turning point, R+, (> Re), the potential energy equals the total energy, which, in turn, is the zero-point energy V 0, 3.478 × 10–21 J for Li2 and 4.170 × 10–21 J for Cs2. Thus, we solve 1 V 0 = k R + – Re 2 2

1 V 0 = k R – – Re 2 2

and

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

2V0 k

2V0

, R– = R e –

k

, and ∆Rvib = 2

2V0 k

.

Numerically, this expression gives ∆Rvib = 0.332 Å for Li2 and 0.220 Å for Cs2, 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 2Σ+g 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 a σ MO with Λ = 0; g, because the MO is a σg MO that is symmetric with respect to inversion; and + because – states require participation from more than one non-σ MO. The first excited state finds this electron in the next highest energy + MO, the 1σ*u MO. This is a 2Σ+u state. For He2 , the three electrons have the 2 + MO configuration 1σ 2g1σ *1 u leading to a Σu state. The first excited state pro2 + motes one 1σg electron, leading to the configuration σ 1g1σ *2 u and a Σg state (which you should not be surprised to learn is not bound—work out the bond + order). The three-electron molecule HeH parallels He2 in its ground state (except we lose the g, u symmetry labels because HeH is heteronuclear). Its ground state is 1σ 22σ 1 with a 2Σ+ term symbol. The first excited state takes some thought to deduce. The 1σ MO is the He core MO, since the 1s He AO is much lower in energy than the H 1s, and the H 1s AO dominates the description of 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

3

1σ–2σ promotion energy is roughly the He ionization potential (24.6 eV) less the H electron affinity (0.75 eV), or about 23.8 eV. In contrast, the 2σ–3σ energy 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, much less than the 1σ–2σ excitation energy. Thus, the first excited state of HeH is 1σ23σ1 (rather than 1σ12σ2), which leads to a 2Σ+ term symbol. For HeH+, which is “protonated He,” the ground state is 1σ2 with a 1Σ+ term symbol. The first 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 (a singlet state) or 1 (a triplet state). Thus, 1σ12σ1 leads to two term symbols + (and two states, just as H + H leads to two states): 1Σ+ and 3Σ+ . Finally, Li2 has the ground state configuration 1σ 2g1σ *u22σ 1g and a 2Σ+g term symbol, just like H2+. The first excited state also parallels H2+. It has the configuration 1σ 2g1σ *u22σ *u1 and a 2Σ+u term symbol. The following table summarizes these arguments: Molecule

Ground state

H2+

1σg Σ g

He2+

1σg1σu

HeH

1σ 2σ

HeH+

1σ

Li+2

19.6

19.7

1 2 + 2

2

*1 2 + Σu 1 2 +

Σ

Σ

*2

*1 2 + Σu 1 *2 2 + 1σg1σu Σ g 2 1 2 +

1σu

2 1 +

2

First excited state

1σ 3σ 1

1 1 + 3 +

2

*2

1σ 2σ 1 2 +

Σ

Σ , Σ

*1 2 +

1σg1σu 2σg Σ g 1σg1σu 2σu Σ u A 4Π state has Λ = 1 and S = 3/2 so that Σ = –3/2, –1/2, 1/2, and 3/2. The possible 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 for Ω: 1/2, 3/2, and 5/2. On the other hand, the spin–orbit energy possibilities are AΣΛ = –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, and 5/2) in order to distinguish among the four spin–orbit energies. 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 = 4 264.02 cm–1. We also found that the H2 v = 1, J = 0 state is 4 158.54 cm–1 above the lowest energy v = 0, J = 0 state. This means we have an excess energy of 4 264.02 cm–1 – 4 158.54 cm–1 = 105.48 cm–1 that could perhaps excite H2 in 4

v = 1 to rotational states higher than J = 0. We use constants in Table 19.2 to find the energies of those states. First, we find the H2 rotational constant for the v = 1 state from Eq. (19.28): B 1 = Be – α e v +

3 1 = 56.260 cm–1 . = 60.853 0 cm–1 – 3.062 2 cm–1 2 2

Next, we write the rotational energy level expression in terms of the J quantum number, the rotational constant B1, and the centrifugal distortion constant De also 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 find F1(0) = 0 (of course), but F1(1) = 112.33 cm–1, which is already more than our 105.48 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 a diatomic is proportional to the reciprocal of the square of the internuclear separation: Be ~ Re–2. Thus, if the rotational constant for the B state of H2 is onethird as large as that for the ground state (the X 1Σ+g state), Be(B) = Be(X)/3, and we can write (since the reduced mass is the same for both states—both refer to H2) Be(X) R (B) 2 =3= e Be(B) Re(X)

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 the actual 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 energy of 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 of the two states given in the problem, 16 956.183 cm–1, and we can sketch a diagram like the one on the next page (which is actually quite accurately drawn using other information in the table, but a qualitative sketch would suffice here).

5

10 0

16 956.183 cm–1

De(A)

20 14 680.58 cm–1

V/103 cm–1

30

0

De(X) = 5890 cm–1

2.0

4.0

6.0

8.0 R/Å

10.

12.

14.

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 to the 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 in the 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 molecule rotates (7 × 10–9 s) × (6.67 × 1010 radians s–1) = 447 radians or, dividing by 2π radians per revolution, excited Na2 in this state rotates about 74 times in its 7 ns characteristic radiative lifetime. The harmonic vibrational constant for the B state is given as 124.090 cm–1 in Table 19.3, and since ωe = ¨(k/µ)1/2/[(100 cm 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 vibration per second, ν, through (see page 402 in the text) ν=

1 1 ω= 2π 2π

k . µ

We find ν = 3.72 × 1012 cycles s–1, or, in 7 ns, the v = 1 level of the Na2 B state 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 long time in the life of a molecule that has two heartbeats: a vibrational cycle tick and a 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-point energy (so that state (0, 0, 0, 0, 0, 0) has energy zero) is 6

E = ω 1v1 + ω 2v2 + ω 3v3 + ω 4v4 + ω 5v5 + ω 6 v 6 . Using the constants in the problem and systematically changing quantum numbers, 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, tetrahedral, octahedral, and icosahedral objects are spherical tops. Prolate symmetric tops include objects such as an American football, a cigar, and a blimp. Oblate symmetric tops include a Frisbee™, an innertube or automobile tire, and a dinner plate. Examples of molecules not mentioned in the text include the spherical tops cubane (C8H8), CCl4, W(CO)6, Ni(CO)4, etc., the prolate symmetric 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 coordinates equal zero. According to Eq. (19.3), the a moment of inertia is zero (because the b and c coordinates of each nucleus are zero), and Ib = Ic, as mentioned on page 717 in the text. The bond lengths mentioned in the problem allow us to construct the following diagram: 1.061 Å

1.203 Å

1.662 5 Å

1.061 Å

a

0.6015 Å

0 The center of mass of acetylene is the C≡C bond midpoint (for the H12C12CH isotopomer, which is what we assume we have here since it is by far the most abundant species in ordinary acetylene), and thus the a coordinate of each C atom is (1.203 Å)/2 = 0.6015 Å and that of each H atom is [1.061 Å + 0.6015 Å] = 1.662 5 Å. The moment of inertia expression depends on the atomic masses: mH = 1.673 7 × 10–27 kg and the atomic mass of 12C is, by definition of the atomic mass unit, exactly 12 amu, or 1.992 648 24 × 10–26 kg. The moment of inertia is (the factors of 2 account for the fact that we have two H atoms and two C atoms, of course)

7

2

Ib = 2mH 1.662 5 × 10–10 m + 2m1 2C 0.6015 × 10–10 m

2

= 2.366 × 10–46 kg m2 . The rotational constant follows from Eq. (19.33a): 2

¨ = 1.183 cm–1 . Be = –1 2Ibhc(100 cm m ) These numbers correspond to the equilibrium structure of acetylene, and they are thus quite close to the values for the ground vibrational state. The excited state spectrum, we are told, reveals that the energy difference between the J = 0 and J = 1 rotational levels of some excited vibrational state (call it v) is 2.327 5 cm–1. Since the energy expression for rotation of a linear polyatomic is the same as for a diatomic, Fv(J) = Bv J(J + 1), (Eq. (19.27) ignoring the centrifugal distortion term, about which we have no information), we see that this measured 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 a moment of inertia for this excited vibrational state equal to 2

¨ = 2.405 39 × 10–46 kg m2 . Iv = –1 2Bvhc(100 cm m ) If we assume the C≡C bond length is the same in this excited state as in the ground vibrational state, 1.203 Å, then the change in the moment of inertia must be due to a change (an increase) in the C–H bond length. If we call the C–H bond length R, we have Iv = 2.405 39 × 10–46 kg m2 2

2

= 2mH R + 0.6015 × 10–10 m + 2m1 2C 0.6015 × 10–10 m . 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 the effects of vibrational excitation on average bond length.)

8

SECTION 19.3 19.14 The plan here is to calculate first the spectroscopic constants for the 13CO isotopomer, 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 the transition energy. Table 19.2 lists the rotational constant Be and centrifugal distortion constant De for the common isotope 12CO, and the problem reminds us how these are related to reduced mass. Thus, we can write Be(13CO) Be(12CO)

=

µ( 12CO) µ( 13CO)

and

De(13CO) De(12CO)

=

µ( 12CO) µ( 13CO)

2

.

The nuclear masses given in the problem allow us to calculate the reduced masses: m 1 2C m 1 6O = 6.856 208 amu m1 2C + m1 6O m 1 3C m 1 6O µ( 13CO) = = 7.172 410 amu m1 3C + m1 6O µ( 12CO) =

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 mass always lowers rotational constants.) Next, we calculate the rotational energies for 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 correction constants αe and βe. Since we are interested in the rotational energy level spacing in a single vibrational state, these correction constants will cancel from our final answer, and we do not need to find their values for the 13CO isotopomer. Thus, we write simply F0(J) = BeJ(J + 1) – De[J(J + 1)]2 and use our 13CO constants to calculate F0(6) = 77.529 cm–1 and F0(5) = 55.380 cm–1 so that 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; we multiply an energy in cm–1 units by hc(100 cm m–1) to convert to joule energy units, then use ∆E = hν to find the frequency, or more directly ν/Hz = (∆E/cm–1) × (100 cm m–1) × (c/m s–1)

9

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 are quite close to the 12CO values shown in part (a) of that figure for the 5→6 absorption transition, but slightly smaller, as part (b) indicates for the 3→4 transition. 19.15 Our task here is to assign initial and final state quantum numbers to the features seen in a spectrum. This is often easy to do without knowing spectroscopic constants 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 what molecule is producing the spectrum, and even if we know the molecule, we may have to make guesses about its bond lengths, bond angles, etc.) Here, we know a great deal from Table 19.2 about the 7LiF isotopomer. In particular, we know 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 of closely spaced transitions separated by rather large gaps and since the spectrum is a microwave spectrum, we start our assignment with a rough calculation of the 6LiF rotational transition frequencies. We take the nuclear masses to be 6 amu and 19 amu, and calculate an approximate Be, using Eq. (19.33a) and the equilibrium bond length, ~1.56 Å = 1.56 × 10–10 m, listed in Table 19.2: 2

2

¨ ¨ = Be = 2Ie 100 cm m–1 hc 2µRe2 100 cm m–1 hc ¨

=

2

6 × 19 2 1.66 × 10–27 kg 1.56 × 10–10 m 100 cm m–1 hc 6 + 19 = 1.50 cm–1 . 2

Next, we calculate transition energies for various initial rotational quantum numbers J′′ using Eq. (19.41), ∆E ≅ 2Be(J′′ + 1), and construct this table: J′′ 0 1 2 3 4

∆E/cm–1 3.00 6.00 9.00 12.0 15.0

10

∆E/GHz 89.9 180. 270. 360. 450.

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 absorptions that change the rotational quantum number J from 2 to 3, from 3 to 4, and from 4 to 5, respectively. Now we must explain why there are pairs of lines instead of a single line for each rotational transition. The key clue is the uneven intensity of each member of each pair: the higher frequency line of each pair has four times the intensity of the lower frequency line. Since we are told that the spectrum is taken with 6LiF at high temperature, the answer must be that we are seeing the microwave spectra of two different vibrational states. At ordinary temperatures, ordinary molecules are almost entirely all in their ground vibrational state, but as the temperature is raised, more and more are vibrationally excited through the energetic collisions that characterize high temperatures. (The details of this excitation and its consequences are discussed in Chapters 22 and 23.) The more intense line of each pair corresponds to rotational transitions of molecules in the ground vibrational (v = 0) level, and the less intense line corresponds to transitions of molecules in the first excited (v = 1) vibrational level. Vibration-rotation coupling explains why the excited level transitions are at a smaller transition frequency: vibrational excitation almost always extends the average bond length in a diatomic, increasing the moment of inertia, lowering the rotational constant for that vibrational level, and thus lowering transition frequencies (or energies). 19.16 Figure 19.11 shows us that the R(1) and P(1) transitions both have J′′ = 1 initial 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 difference between the J = 2 and J = 0 levels of the v = 7 vibrational state, which we can express 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 this vibrational level is thus B7 = (30.042 cm–1)/6 = 5.173 7 cm–1. The reduced mass for H127I is µ= =

m H mI m H mI 1 amu × 127 amu 1.660 54 × 10–27 kg amu–1 = 1.647 6 × 10–27 kg 1 amu + 127 amu

so that the expression for B7 is

11

2

2

¨ ¨ = B7 = = 5.173 7 cm–1 –1 2 2Ie 100 cm m hc 2µR7 100 cm m–1 hc which we can solve for R7, the average bond length in the v = 7 vibrational level. We find R7 = 1.812 2 × 10–10 m = 1.812 2 Å, which is considerably longer 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 in highly excited vibrational levels. The molecule spends most of its time (speaking classically) with its bond extended far beyond its equilibrium position. The molecule is displaying the effects of the real internuclear potential energy function, which is not at all well approximated by a harmonic oscillator at high excitation energies. 19.17 The vibrational energy level expression for a diatomic, Eq. (19.26), depends on the harmonic vibrational constant ωe and anharmonic correction constants ωexe, etc. An accurate prediction of the overtone spectrum would require us to know as many anharmonic constants as possible, but since we are looking for a 400 cm–1 wide region, we can approximate the transition energy with just the constants listed in Table 19.2 for H35Cl: ωe = 2 990.946 cm–1 and ωexe = 52.819 cm–1. Thus, for this isotopomer, the hypothetical v = 0→2, J = 0→0 transition would occur at 1 12 1 12 ∆E = G(2) – G(0) = ω e 2 + – ω exe 2 + – ω e 0 + – ω exe 0 + 2 2 2 2 = 2ω e – 6ω exe = 5 664.978 cm–1 . Thus, we should center our scan at ~5 665 cm–1. Note that the harmonic approximation ∆E = 2ωe = 5 981.892 cm–1 is 317 cm–1 higher and would be a poor place to center our scan; anharmonic corrections are increasingly more important to consider as the vibrational quantum number v increases. (Table 19.2 also tells us the HCl, like all diatomic hydrides, has a large rotational constant, Be ≅ 10.6 cm–1. Thus, the vibration-rotation transitions—the P and R branches of the spectrum in the language of Figure 19.11—will extend over a significant region. A scan that extends 200 cm–1 away from the hypothetical Q branch will cover transitions from R(0) to about R(7), as you can verify if you include the rotational energy expression, Eq. (19.27), in the transition energy expression. Interestingly, the entire P branch falls within 200 cm–1 of our center frequency. The P(15) line is about 160 cm–1 from our scan center, but

12

higher P-branch transitions are closer to this center. This is called band head formation—the P branch lines march out from the band center, then pile up at a limit, the band head, then start marching back towards the center. This is due to the particular combination of centrifugal distortion and vibration-rotation interaction constant values H35Cl happens to have, but band head formation is quite common.) Now we consider DCl. Table 19.2 does not list vibrational constants for DCl, but ωe has a simple dependence on the reduced mass of the diatomic, as does any harmonic vibrational frequency constant: ωe ~ µ–1/2. An accurate calculation of the D35Cl ωe from the accurate H35Cl constant would entail an accurate nuclear mass for the 35Cl nucleus; here, we can take it to be simply 35 amu = 5.812 × 10–26 kg. The H nuclear mass is mp, the proton rest mass, 1.673 × 10–27 kg. Thus, the H35Cl reduced mass is µH 3 5Cl =

m p m 3 5C l = 1.626 × 10–27 kg , mp + m3 5Cl

which is quite close to the H atom mass. (This is a general result: µ ≅ the smaller mass if the two masses are very different. Recall our first discussion of the H atom in Chapter 12: the H atom reduced mass was quite close to the electron mass. See page 424 in the text.) For D35Cl, we can approximate the D nuclear mass quite closely by 2 amu = 3.321 × 10–27 kg. We find µD 3 5Cl =

m D m 3 5C l = 3.141 × 10–27 kg , mD + m3 5Cl

again quite close to the D mass and almost exactly twice the H35Cl reduced mass. Thus, to a good first approximation that is worth memorizing, substituting 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: µH 3 5Cl = 2 151.920 cm–1 . µD 3 5Cl

ω e(D35Cl) = ω e(H35Cl)

(Note that ωe(H35Cl)/ 2 = 2115 cm–1.) We found that we needed the ωexe anharmonic constant to locate the H35Cl overtone accurately, but how to we scale the H35Cl ωexe constant to yield ωexe for D35Cl? Table 19.2 lists this constant for various isotopes of molecular hydrogen, and if we recognize that µH2 = mH/2 and µD2 = mD/2 = 2µH2, we see that ωexe scales with µ–1: the H2 13

ωexe value is twice the D2 value (and three times the T2 value, etc.). (In general, each successively higher anharmonic constant depends on µ by one more factor of µ–1/2.) Thus, we can find µ 35 ω exe(D35Cl) = ω exe(H35Cl) H Cl = 27.342 cm–1 . µD 3 5Cl Our scan center for D35Cl is 2ωe – 6ωexe = 4 139.789 cm–1. Note that the simplest approximation, a frequency or wavenumber 2 lower than for H35Cl, gives 4006 cm–1, a quite acceptable first approximation that would allow us to find the real spectrum within our 400 cm–1 window. Finally, we can state that the H37Cl spectrum will occur at a lower frequency than the H35Cl spectrum due 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 versus 1.626 × 10–27 kg for H35Cl, but this is enough to shift the H37Cl spectrum about 4 cm–1 from the H35Cl v = 0 to 2 overtone spectrum. Most modern infrared spectrometers (that can operate in this region—it is a bit above the usual upper 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. The permanent dipole moment of a diatomic is (see Chapter 15 for details) the average of the dipole moment function, p(R), over the vibrational wavefunction. The only state of importance for an equilibrium gas sample is the ground vibrational state for the vast majority of ordinary gases like HF and CO, and we know that the v = 0 vibrational wavefunction is quite closely centered around Re with a very small spread. Thus, to a good first approximation, the observed dipole moment is simply the value of the dipole moment function evaluated at Re: 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 value 6.069 × 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 an interesting accident that the CO dipole moment function happens to be passing through zero in the vicinity of its equilibrium bond length. (Table 15.1 lists a permanent dipole moment magnitude of 0.374 × 10–30 C m in the –CO+ sense.) On the other hand, the discussion of the vibrational transition dipole moment on page 724 in the text shows that it is the slope of the dipole moment function, not its magnitude, that governs the transition probability. The graph of the dipole moment functions in the problem show that both HF and CO have very similar slopes in the vicinity of their equilibrium bond lengths.

14

19.19 The energy level expression for the rigid-rotor, harmonic oscillator diatomic is quite simple: E(v, J) = ω e v +

1 + Be J(J + 1) . 2

For a (hypothetical) Q branch transition, the v quantum number changes (usually by ±1, the strongly allowed selection rule), but the J quantum number does 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 Q branch would appear as a single feature, but it would be quite intense, since it would represent transitions for all the J states with significant population, in contrast to the P and R branches where each J state leads to a unique transition frequency. The R(0) line (for which J increases from 0 to 1) occurs at ∆E(R(0)) = E(v′, 1) – E(v″, 0) = ω e ∆v + 2B e while the P(1) line (J decreases from 1 to 0) occurs at ∆E(P(1)) = E(v′, 0) – E(v″, 1) = ω e ∆v – 2B e 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 to + 2 *2 4 2 be 1σ 2g 1σ *2 u 2σ g 2σ u 1πu 3σ g and that the lowest three electronic states of N2 have the electron configurations 2

*2

2

*2

1

2

*2

2

*2

2

2

*2

2

*1

2

1σg 1σu 2σg 2σu 1πu4 3σg 1σg 1σu 2σg 2σu 1πu3 3σg 1σg 1σu 2σg 2σu 1πu4 3σg

(the ground state) (the first excited state) (the second excited state).

In order of increasing energy, these are called the X, A, and B states, and molecular term symbols for them are not difficult to deduce. No fully occupied MO contributes to the term symbol (or, stated better, each represents a 1Σ+g contribution—S = 0 and Λ = 0—to the ion’s term symbols). The ground state term symbol is dominated by the single 3σg electron, which makes S = 1/2 and thus 2S + 1 = 2 (the state is a doublet state), but this electron is in a σ MO with Λ = 0, and since it is a g MO, so is the entire electronic state. Thus, the ground 15

state is a 2Σ+g 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 the πu MOs have one unpaired spin, and it must be a Π state as well because this unpaired electron carries Λ = 1, as do all electrons in π MOs. The state has the inversion symmetry of this unfilled π MO, which is u, so that the term symbol is 2Πu; the +/– superscript notation is used only for Σ states. The next state is also a doublet Σ state, but it has u symmetry from the unpaired electron in the 2σ*u MO. It is a 2Σ+u state. The photoelectron spectrum showed the greatest vibrational excitation in the A state (see Figure 14.13), and the Franck–Condon principle tells us that extensive vibrational excitation of an excited state from an excitation of the ground vibrational level of a lower energy electronic state (the N2 ground state here) means the two electronic states have significantly different bond lengths. It is unusual for an excited electronic state to have a bond length significantly shorter than the ground state, or for a molecular ion ground state 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, and + 1.074 2 Å, respectively), but the N2 A state bond length must be significantly longer (and experiment finds 1.175 Å). The photoelectron energies quoted in + Chapter 14 locate these N2 states above the v = 0 level of the N2 ground electronic state. (The lowest of these energies, 15.57 eV, is simply the N2 ionization potential.) What is not so obvious is that all three of these molecular ion electronic states dissociate to the same dissociation limit: a ground state N atom and a ground state N+ atomic ion. The graph on the next page shows all four potential energy curves, the neutral N2 ground state and the three ion states, accurately drawn from data derived from a number of spectroscopic experiments. 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 the slight differences among the N2 X state Re and those of the ion’s X and B states.

16

20 B 2Σu+ N+2 A 2Πu+ N+2 15

V/104 cm–1

X 2Σg+ N2+

10

5 X 1Σg+ N2

0

0.5

1.0

1.5 R/Å

2.0

2.5

19.21 Table 19.3 tells us Te for the A state is 14 680.58 cm–1 (see also Figure 19.2 and 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 the X state constants ωe = 159.124 cm–1 and ωexe = 0.7254 cm–1, we will have the transition energies, in cm–1, for the nine transitions of interest. We convert these 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 1 12 1 12 ∆E = Te + ω e′ v′ + – ω exe′ v′ + – ω e″ v″ + – ω exe″ v″ + 2 2 2 2 2 = 14 680.58 + 117.323 (22.5) – 0.3576 (22.5) – 159.124/2 + 0.7254/4 = 17 059.93 cm–1

17

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 to the table below: v′

λ/nm

∆E/cm–1

v″

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 with the wavelengths of the two atomic Na transitions given in the problem. (The sticks are drawn below with different heights just to help you sort them out; the stick heights do not represent intensities as they would appear experimentally.)

λ/nm

18

22→2

590

23→2

585

22→1 24→2

23→1

22→0 24→1

580

23→0

24→0

575

Na

595

600

Note how the various Na2 transitions (and remember that they would not appear as single, sharp features experimentally due to the many rotational transitions that accompany each vibrational transition) intermingle and appear near the Na atomic lines. 19.22 Difluroethylene exists in three isomeric forms: F

H C

F

F

F C

C H

F

C

H

H C

H

H

C F

1,1-difluoroethylene cis-difluoroethylene trans-difluoroethylene The first two isomers have a permanent dipole moment, which is required in order to have a microwave (rotational) absorption spectrum. The trans isomer, however, lacks a permanent dipole moment by symmetry and thus does not have a microwave spectrum. Since no transitions were seen, the sample must have been the trans isomer. 19.23 The symmetry of the AX3 planar symmetric top molecule indicates how we should place inertial axes with respect to the A–X bonds (and with the origin at the A atom, which is clearly the center of mass). If we call each A–X bond length R, we can make a diagram as shown below. At first, we will call the axes simply x, y, and z, but once we know the magnitudes of Ix, Iy, and Iz, we can confidently label them a, b, and c according to the magnitudes of each I. y X R x

A 60°

X

X

If we call the mass of the X atom m and note that the A atom sits at the coordinate origin so that its mass will not enter the moment of inertia calculation, the moment of inertia expressions in Eq. (19.3) become 19

3mR 2 Ix = ∑ m i y i2 + zi2 = mR 2 + 2mR 2 cos2 60° = 2 i

3mR 2 Iy = ∑ m i x i2 + zi2 = 2mR 2 sin2 60° = 2 i

Iz = ∑ m i x i2 + yi2 = mR 2 + 2mR 2 sin2 60° + cos2 60° = 3mR 2 . i

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, and this, 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 in this problem, however.) In BF3, since the B atom is at the center of mass if the F atoms have the same mass, isotopic substitution of the B atom does not change the moments of inertia. Using Eq. (19.33) to relate the rotational constants to moments of inertia, we see that the smaller rotational constant is the C constant and C = 0.173 cm–1 means 2

¨ Ic = = 1.62 × 10–45 kg m2 . 2hc 100 cm m–1 0.173 cm–1 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 and moment of inertia expressions for a diatomic:

Be =

¨

2

2µRe2hc 100 cm m–1

2

or

Re =

¨ . 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.

20

19.24 Equation (19.37b) reads Erot(J,K) = BJ(J + 1) + (C – B)K2, or, with the B and C constants in Figure 19.8(b), (0.345 cm–1)J(J + 1) + (–0.157 cm–1)K2. We can 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 transitions are J = 0 → J = 1, K = 0 (at 0.690 cm–1), J = 1, K = 0 → J = 2, K = 0 (at 2.070 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 six states, only three transitions are allowed, and two of these have the same transition energy so that only two transitions would be observed if the molecule was truly rigid. We can see this analytically if we recognize that the selection rules mean that absorption transitions only of the type J,K → (J + 1),K are allowed so 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 Erot expression, writing Erot(J,K) = BJ(J + 1) + (C – B)K 2 – DJJ 2(J + 1)2 – DJKJ(J + 1)K 2 – DKK 4 , the selection rules stay the same, of course, but the transition energy expression changes to ∆E = Erot(J + 1,K) – Erot(J,K) = 2B(J + 1) – 4DJ(J + 1)3 – 2DJK(J + 1)K 2 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 transition have difference transition energies. Note, however, that neither transition energy expression depends on C, and the more complete expression also does not contain the DK centrifugal distortion constant.

21

19.25 Acetylene has no permanent dipole moment, of course, but a vibrational motion that produces a dipole moment as the atoms move in a particular normal mode will be infrared active. The cis bend does just that. At the extremes of its motion, acetylene is bent in a way that leads to a net dipole moment; the C–H bond moments point in directions that lead to a net moment in a direction perpendicular to the C≡C bond and in the plane of the (bent) molecule. As long as the dipole moment function has a non-zero first derivative with respect to the atomic 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 how great the bending amplitude. This mode is not infrared active, but it is Raman active. It is always true in molecules with a center of symmetry that modes which are infrared active are not Raman active and vice versa. SECTION 19.4 1 +

19.26 Table 19.3 tells us that ωe for the Na2 ground state, X Σ g , is 159.124 cm–1, D e = 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), lets us calculate the Morse approximation to De from ωe and ωexe (which Table 19.3 tells us is 0.7254 cm–1): 2

De =

ωe

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 Morse centrifugal 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 =

4Be3 2 ωe

= 5.85 × 10–7 cm–1 ,

which we should compare to the observed De listed as 5.81 × 10–7 cm–1 in Table 19.3. The Morse vibration-rotation constant comes from Eq. (19.48e): 22

αe =

6

Be3 ωexe

ωe

1/2

– Be3 = 1.05 × 10–3 cm–1 ,

which we should compare to the tabulated value, 0.8736 × 10–3 cm–1. In general, the Morse expressions among spectroscopic constants such as Eq.’s (19.48d) and (19.48e) are more accurate than the Morse dissociation energy expression. The next problem has more to say about this. 19.27 If we solve Eq. (19.48e) for ωexe, we find

ωexe =

Be2 + ωe αe/6 Be3

2

.

Substituting this into Eq. (19.49) gives

De =

=

ω e2 Be3 4 Be2 + ω e αe/6 9ω e2 Be3 6Be2 + ω e αe

2

2

hc(100 cm m–1)

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, contrasting 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 predicting dissociation energies.

23

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 1 2 3 4 5

0 25.740 46.149 61.755 72.661 79.441

0.5 1.5 2.5 3.5 4.5 —

25.740 20.409 15.606 10.906 6.780 —

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

∆G(n)/cm–1

30 20 10 0 0

1

2

3 n

4

5

6

The line through the points is a least-squares fit straight line to them, and the area 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. The least-squares fit line is ∆G(n) = 27.744 cm–1 – (4.742 3 cm–1)n, and its y intercept 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 the intercepts from the graph (which is acceptable if the straight line through the points is drawn by hand rather than deduced from a least-squares analysis). As indicated in the problem, more advanced methods of extrapolating the data in a Birge-Sponer plot indicate that D0 is somewhat larger that the estimate derived from a linear fit: 84.75 cm–1 rather than ~81.2 cm–1. This tells us that the higher vibrational levels lead to ∆G(n) points that lie above the straight line 24

approximation. (Contrast Figure 19.16 for H2 where the high n points fall below the straight line extrapolation.) The experiment observed vibrational levels from v = 0 through 5, and our graph indicates that an unobserved v = 6 level is surely possible. Higher levels are difficult to predict without the more advanced methods that lead to the improved dissociation energy value. Finally, the zero-point energy is approximately one-half the v = 0 to 1 energy difference, 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 underestimates the zero-point energy. To include anharmonicity, we should fit the observed 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, one finds the zero-point energy is 14.95 cm–1 so that a more accurate estimate for the 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 symmetry of any potential energy function we might choose to represent it. Since an O–C–O bond angle of θ places the molecule in a configuration that is indistinguishable from an angle of 360° – θ (except for a rigid rotation of the molecule—see the diagram below), any potential energy function we write must have the property that V(θ) = V(360° – θ). 360° – θ O

θ C

O

=

C O

O

=

O

O C 360° – θ

If we write ∆θ = 180° – θ and imagine adding a cubic term to the potential energy function so that it becomes V(θ) = kb(∆θ)2/2 + b(∆θ)3 where b is a constant, we can write 1 1 V(θ) = kb (∆θ)2 + b(∆θ)3 = kb (180° – θ)2 + b(180° – θ)3 . 2 2 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

25

1 V(360° – θ) = kb [180° – (360° – θ)]2 + b[180° – (360° – θ)]3 2 1 = kb (θ – 180°)2 + b(θ – 180°)3 2 1 = kb [(–1)(180° – θ)]2 + b[(–1)(180° – θ)]3 2 1 = kb (180° – θ)2 – b(180° – θ)3 ≠ V(θ) . 2 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 odd power will not have the correct symmetry while any term containing an even power of ∆θ will have the correct symmetry. SECTION 19.5 19.30 The Doppler shift discussion in the text was centered around a moving absorber and a stationary light source. Here, we have a different physical situation: the moving molecule is emitting radiation toward a stationary receiver. We are told that the molecule is moving away from us, and in this case, the receiver sees a signal that is Doppler-shifted to greater wavelengths and thus to smaller frequency. (This shift may be familiar to you from optical astronomy. Atomic emissions 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 away from 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– = ν0

1–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 long as 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 calculate the Doppler frequency width ∆νD with T = 100 K and M = 29 g mol–1: ∆ν D = 7.16 × 10–7 ν 0

T/K M/g mol–1

1/2

= 0.883 MHz .

This width is substantially less than the Doppler shift, and thus the speed at which the cloud is moving away from us can be measured accurately. 26

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 expression. 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 = 1.6 × 10–2 cm–1 , 26

which is greater than the 0.004 cm–1 resolution of the instrument used to record Figure 19.20. A higher resolution spectrum would show no further detail; Figure 19.20 is “Doppler limited.” 19.32 For 14N2 with I = 1, there are (1 + 1)(2·1 + 1) = 6 symmetric nuclear-spin wavefunctions and 1(2·1 + 1) = 3 antisymmetric wavefunctions for a 6:3 = 2:1 intensity alternation. For 15N2 with I = 1/2, we find three symmetric and one antisymmetric wavefunction (just as in H2 or, for that matter, H12C12CH in Figure 19.20) for a 3:1 intensity alternation. For 14N15N, there is no intensity alternation for line to line because this molecule has distinguishable nuclei. As an aside, and as Figure 19.20 shows, these ratios are not exactly found in the observed spectra because another factor is guaranteed to give each transition a slightly 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 absorb light 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 of the Franck–Condon integral is 2 R – R e 2 + R – Re – δ . ψ′vibψ″vib ∝ exp – k 2¨ω

Expanding the argument to the exponential gives, with a little algebra, 2

2

R – Re 2 + R – Re – δ = 2 R – Re 2 + δ – 2δ R – Re , and the definition of q1,

27

q12 = k R – Re 2 ¨ω

q 2¨ω , R – Re 2 = 1 k

or

lets us write 2

2 R – Re 2 + δ – 2δ R – Re =

2q12¨ω 2 + δ – 2δq1 k

¨ω k

.

Thus, the Franck–Condon integral is ∞ 2 g0,0 = 1 exp – k R – R e 2 + R – Re – δ dq1 π1/2 –∞ 2¨ω ∞

2q12¨ω 2 + δ – 2δq1 = 1 exp – k k π1/2 –∞ 2¨ω =e

–kδ 2 /2¨ ω

π1/2

∞ –∞

exp –q12 +

¨ω k

dq1

2

k/¨ω δq1 dq1 = exp – k δ . ¨ω 2

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

0.01

0.05

0.1

0.5

g0, 0

0.990

0.779

0.368

1.38 × 10 –11

0.05

0.1

0.5

For k/¨ω = 800 Å–2, we find: δ/Å

0.01

g0, 0 0.961 0.368 0.0183 3.72 × 10 –44 We see that relative Re shifts of several hundredths Ångström (which are very common) have notable effects on the transition intensity while shifts of more than 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 construct the table at the top of the next page. 28

v1 v2

ª

v3

G(v 1 , v 2ª , v3)/cm–1 G(v 1 , v 2ª , v3) – G(0, 00, 0)/cm–1 2 532.25 0

0

0

0

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 states from the ground (0, 00, 0) state, and, as mentioned in the problem, two of these are seriously in error: (1, 00, 0) and (0, 20, 0). The theory behind the Fermi resonance 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 resonance interaction constant is We = –52.84 cm–1. Note that the Fermi resonance expression in the problem can be written Ea + Eb ± 4We2 + Ea – Eb 2 2

1/2

E + Eb = a ± 2

4We2 + ∆2 2

where ∆ = Ea – Eb. Writing the expression this way shows that the interaction leads to two new states that are equally above and below the average energy of the unperturbed states. The perturbed states’ energies are 3 816.09 cm–1 and 3 921.92 cm–1, and subtracting the zero-point energy, G(0, 00, 0) = 2 532.25 cm–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 contract together by the same amounts at all times. On the potential energy contour diagram 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, 0.1 Å) upper right-hand corner of the figure). In contrast, the antisymmetric stretch normal mode has one C–O bond contract while the other expands the same amount. 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 the top of the next page as two heavy diagonal lines.

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(RO–C – Re)/Å

0.05

0 1000

–0.05

2000 3000 4000

–0.05

0

0.05

(RC–O – Re)/Å If we sketch in a contour at the zero-point energy of 2532 cm–1 (the dashed contour in the figure above), we can read off the classical turning points of each normal mode from the intersection points of this contour with the appropriate straight line. For the symmetric normal mode, we see that one classical turning point has both bonds extended about 0.06 Å beyond equilibrium (i.e., from 1.160 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 one bond compressed about 0.06 Å while the other is extended about 0.06 Å. At the other turning point, the two bonds switch roles—the extended bond becomes the compressed bond and vice versa.

30