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.
29
(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