Model answers to the revision questions associated with

Model answers to the revision questions associated with
Prof Ashfold’s Chemistry 1A Spectroscopy course.
1.
Use eq. (1.1) from lecture notes:
2000 cm-1
0.15 nm
500 nm
9 GHz
2.
≡
≡
≡
≡
5 µm
2 × 1018 Hz
20000 cm-1
0.3 cm-1.
Infrared; vibrational transition
X-ray; change of nuclear configuration
visible; electronic transition
microwave; rotational transition.
Using eqs (1.1) and (1.2) from lecture notes:
10 µm
≡
3 × 1013 Hz
≡
1000 cm-1.
E = hν = 1.99 × 10-20 J molecule-1
= 1.99 × 10-20 NA/1000 = 11.99 kJ mol-1.
If the transition energy is doubled, so is the transition frequency, thus the wavelength
of the transition is halved.
3.
From eq. (1.5)
0.01 = exp (−κ 10-3 10-2)
∴ ln(0.01) = −10-5 κ = −2.303 × 10-5 ε
Thus ε = 2 × 105 dm3 mol-1 cm-1.
4.
Recall that a molecule must have a permanent dipole moment in order to show a pure
rotational spectrum. Thus the following molecules will be microwave active:
HCl, IBr, CH3Cl, CH2Cl2, H2O.
5.
From eq. (2.6) the energy separation between successive transitions in a pure rotation
spectrum is 2hB; thus the frequency separation is 2B (in Hz).
∴B = 2.1415 × 1010 / 2 = 1.07075 × 1010 Hz (0.35857 cm-1).
Then, given B = h / 8π2I,
I = h / 8π2B = 7.8375 × 10-46 kg m2.
From eq. (2.10),
I = µR2
and thus
R2 = I / µ
and
R = 1.7556 × 10-10 m.
= 7.8375 × 10-46 / 2.5428 × 10-27
From eq. (2.6), the frequency of a J → J + 1 transition is given by
ν = 2B(J + 1).
Thus for the J = 9 → J = 10 transition,
ν = 2 × (1.07075 × 1010 Hz) × 10
= 2.1415 × 1011 Hz
This frequency corresponds to a wavenumber of 7.1433 cm-1.
Recalling eq. (2.14)
Jmp =
k BT
1
= 16.58,
−
2 hB 2
i.e. Jmp = 17
Finally, to calculate number of rotations per second, use eq. (2.1) recalling that ω is in
radians per second and that there are 2π radians in a complete revolution. Thus, from
eq. (2.5), for J = 0:
Erot = BhJ(J + 1) = 0 and so ω = 0.
For J = 1,
Erot = 7.095 × 10-24 J,
ω = (2Erot / I)1/2 = 1.903 × 1011 rad s-1
ν = 3.03 × 1010 rotations s-1
For J = 10,
Erot = 3.9022 × 10-22 J,
ω = 1.411 × 1012 rad s-1, and ν = 2.25 × 1011 rotations s-1
6.
From eq. (2.10),
I (12CO) = (1.1386 × 10-26 kg) (1.2724 × 10-20 m2)
= 1.4487 × 10-46 kg m2.
From eq. (2.5), B = h / 8π2I = 5.7928 × 1010 Hz
For future reference, the same calculation for 13CO gives:
µ (13CO) = 1.1909 × 10-26 kg, I (13CO) = 1.5153 × 10-46 kg m2,
and
B = 5.5382 × 1010 Hz.
Rotational transition frequencies are given by eq. (2.6).
12
Transition
13
CO
CO
J=0→J=1
1.15856 × 1011 Hz
1.10764 × 1011 Hz
J=1→J=2
2.31712 × 1011 Hz
2.21528 × 1011 Hz
J=2→J=3
3.47568 × 1011 Hz
3.32292 × 1011 Hz
To resolve the respective J = 0 → J = 1 transitions, the spectral resolution would have
to better than the frequency separation between the respective lines (i.e. better than 5
GHz), but the isotopic ratio could also be deduced from the relative intensities of the
more widely separated J = 2 → J = 3 transitions (∆ν ~ 15 GHz).
7.
Using eq. (3.5),
Ev=0 = 2.6637 × 10-21 J; Ev=1 = 7.9670 × 10-21 J; Ev=2 = 1.3238 × 10-20 J.
The relevant transition frequencies are thus
for v = 0 → v = 1:
ν = 8.004 × 1012 Hz
( ~ν = 266.97 cm-1)
for v = 1 → v = 2:
ν = 7.955 × 1012 Hz
( ~ν = 265.35 cm-1)
The relative transition intensities will reflect the relative populations of the two levels
responsible for the respective absorptions, i.e. v = 0 and v = 1.
This population ratio is given by eq. (3.3), thus
N v =1
 − ( E v =1 − E v =0 ) 
= exp 
 = 0.278
N v =0
kT


8.
9.
Linear molecules have 3 translational degrees of freedom, 2 rotational degrees of
freedom and 3N-5 vibrational degrees of freedom.
Bent molecules have 3 translational degrees of freedom, 3 rotational degrees of
freedom and thus 3N-6 vibrational degrees of freedom.
Molecule
HBr
O2
OCS
SO2
BCl3
HCCH
CH4
CH3Br
C6H6
No. vibl.
modes
1
1
4
3
6
7
9
9
30
The fundamental frequency can be obtained using eq. (3.4):
µ(1H35Cl) = 1.626 × 10-27 kg
Thus
ν = 8.968 × 1013 Hz,
and the zero-point energy = ½hν = 2.971 × 10-20 J.
We need to recall eq. (2.5) to calculate the rotational constant, B:
B = h / 8π2I, with I = µRe2
Thus
I = 2.6433 × 10-47 kg m2
and
B = 3.1749 × 1011 Hz.
The next part of the question asks for wavenumbers of particular transitions, e.g. P(1)
is the transition v = 0, J = 1 → v = 1, J = 0. We need to use a version of eq. (3.7)
appropriate for cm-1 units, i.e.
~
~
~ν = ~ν + B
' J ' (J '+1) − B" J" (J"+1)
10
where ~ν10 is the separation (in cm-1) between the v = 0, J = 0 and v = 1, J = 0 levels,
~
~
B' and B" are the respective rotational constants (in cm-1) for molecules in the v = 0
and v = 1 levels, and J’ and J” are the quantum numbers of, respectively, the upper
and lower rotational states connected in the transition.
Given ν = 8.968 × 1013 Hz, ~ν10 = 2991.4 cm-1.
~
~
Given B = 3.1749 × 1011 Hz, B' = B" = 10.590 cm-1.
Thus:
Transition
~
B' J ' (J '+1) / cm-1
~
B" J" (J"+1) / cm-1
Transition
wavenumber / cm-1
P(1)
0
21.18
2970.22
P(2)
21.18
63.54
2949.04
P(3)
63.54
127.08
2927.86
R(0)
21.18
0
3012.58
R(1)
63.54
21.18
3033.76
Factors neglected in the model calculation are:
~
i)
Anharmonicity. Re’ will be slightly longer than Re”, and B' thus slightly
~
smaller than B" .
~
~
ii)
Centrifugal distortion. Both B' J ' (J '+1) and B" J" (J"+1) will increase with J
slightly less than shown in the calculation.
Both will lead to slight shifts from the calculated line positions.
10.
This question requires use of eq. (4.6). The necessary values for γ, the magnetogyric
ratios for the 1H and 13C nuclei, are obtained from
γ=
ge
2mP
Thus for 1H, γ = 2.6749 × 108 T s-1, and for 13C, γ = 6.724 × 107 T s-1. Both are spin ½
nuclei, so the necessary magnetic fields are simply given by
B = 2πν / γ
11.
12.
For 1H:
B = 0.7107 T
For 13C:
B = 2.8266 T
From Q. 10, we already know that γ = 2.6749 × 108 T s-1 for a 1H nucleus. Then use
eq. (4.6) to calculate the resonance frequencies:
B = 2.5 T:
ν = 106.4 MHz
B = 5.25 T:
ν = 223.5 MHz.
Possible structures are CH3CH2CH2Cl and CH3CHClCH3. The second of these
contains two equivalent CH3 groups, each of which is adjacent to one proton - the spin
of which can be either up or down. Thus we would expect this isomer to show a split
doublet, at small δ (since the CH3 protons are two C atoms away from the electron
withdrawing Cl atom), and a much weaker resonance at larger δ due to the proton on
the central C atom. If this interpretation is correct, the latter resonance should be split
into a septet of peaks, with relative intensities 1:6:15:20:15:6:1, as a result of
interaction with the six equivalent CH3 protons. This is very much in line with
spectrum (b).
CH3CH2CH2Cl would be expected to give three sets of resonances, with relative
integrated intensities of 3:2:2. The CH3 proton resonance should have a small
chemical shift (since it is far from the electron withdrawing Cl atom), and be split into
a 1:2:1 triplet as a result of spin-spin coupling with the protons of the neighbouring
CH2 group. The CH2 protons adjacent to the Cl should give a resonance at large δ
and, again, be split 1:2:1 as a result of spin-spin coupling with the protons the central
CH2 group. The protons in the central CH2 group would be expected to give a
resonance at intermediate δ and to have a complex splitting pattern. The resonance
should be split into a 1:3:3:1 quartet from spin-spin coupling with the CH3 protons and
each of the resulting peaks then further split in the ratio 1:2:1 as a result of coupling
with the terminal CH2Cl group protons. Such expectations are consistent with
spectrum (a).