1. m1 VIBRATIONAL THEORY p.55 bonds ~ springs E = ½ kx2 m2 x is + or -C H
E = ½ kx2 m1 m2
x is + or -
x2 is +

2. Graph ½ kx2 gives a parabola
SIMPLE HARMONIC OSCILLATOR

3. for a SHO k = force constant m = reduced mass m(CH) = 12/13 = 0.923
p. 55
for a SHO
k = force constant
m = reduced mass
m(CH) = 12/13 = 0.923
m(OH) = 16/17 = 0.941
O------H
C------H
3000cm-1
3400 cm-1
when m1 is very large, m does not change much
however n(OH) > n(CH) so k(OH) > k(CH)

4. However, if the light element is different: p. 56
C----H
C-----D 1 2
= 0.92
= 1.7
k’s are
same
or nCH (0.92/1.7)1/2
= 2200 cm-1
Huge difference

5. only one fundamental frequency
A spring (SHO) has all energies possible,
MOLECULES DO NOT
p. 57
Ev = (v + ½)hn
only one fundamental frequency
(spacings equal, hn)

6. normally, only a one level jump allowed p. 57
hot lines,
overtones weak
in IR, only lowest level is populated

7. REAL MOLECULES: Not a simple parabola!
Dissociation energy
Ev ≈ (v + ½)hno
v=3
E3 = 7/2 hno - a ‘bit’ more
v=2
E2 = 5/2 hno - a ‘bit’
hot line,
n<no
n<2no E0→E2
1st overtone
v=1
E1 = 3/2 hno
fundamental frequency of vibration, DE = hno
v=0
Eo = 1/2 hno
ground vibrational state
Internuclear separation r
average bond length o

8. so in CO n0 = 2143 cm-1, n1 = 4250 cm-1 not 4286 cm-1
p. 59
so in CO n0 = 2143 cm-1, n1 = 4250 cm-1 not 4286 cm-1
Reminder:
Intensity of band depends on
change in dipole during stretch or bend

9. Polyatomics X-----Y has one stretching vibration ONLY
What about X---Y---Z ?
Separate atoms EACH need an x, y, z co-ordinate
So N atoms require 3N coordinates to specify position
BUT not all movements of atoms in space correspond to
a vibration:
3 possible translations along x, y or z AND
3 possible rotations around x, y and z do NOT
change the relative positions of the atoms

10. In general, if molecule has N atoms
there are only 3N-6 possible fundamental vibrations
[3N-5 if molecule is linear]
p. 59/60
look at this one

17. During the vibration, the bond dipole changes- +
but it is the VECTOR SUM that is important,
If this changes during vibration = IR active

18. p.60

19. Structure is ‘hairy’ due to rotational fine structure:
p. 60
3750 asym
3650 sym
1600 bend
(overlapping)
Structure is ‘hairy’ due to
rotational fine structure:
more on this shortly

20. CO2
p. 61
Dm = 0
Dm = 0
Dm = 0

21. p. 61
CO2

22. p. 62
SUMMARY
Reduced mass effect:
Force constant effect:

23. M-F M-Cl M-Br M-I where Metal is more massive, e.g. Sn, Pb
(1) C≡X C=X C-X X=C,O,N
~2200 cm-1 ~1650 cm-1 ~1100 cm-1
(2) C-F C-Cl C-Br C-I
1050 cm cm cm cm-1
M-F M-Cl M-Br M-I where Metal is more massive, e.g. Sn, Pb
600 cm cm cm cm-1
From these, you can predict many others!

24. Problem: POCl3 shows IR bands at 1290, 582, 486, & 267 cm-1 Cl
Cl > O, so P-O > P-Cl, so P-O is likely the
highest = 1290 (plus P-O has some double
bond character)
O=P Cl Cl
stretches > bends
asym str > sym str
582 = asym str; 486 sym str; 267 is a bend
Note: 3N-6 = 9, so there are 5 other fundamentals
*** not always obvious what these are, i.e. Raman
(IR inactive) or degenerate or combination bands

25. Methane, CH4 likewise only shows 4 bands
p. 63
Methane, CH4 likewise only shows 4 bands
but has 9 {3N-6} fundamentals (others degenerate)
sym str
if atoms same
NOT IR active
dipoles cancel
sym bend
NOT IR
active
bend
IR ACTIVE
asym str
always
IR active
3020 cm-1
2914 (Raman)
1520 (Raman)
1305

26. p. 63
3020 cm-1
2914 (Raman)
1520 (Raman)
1305

27. Vibrational Mode Assignments Manual pages 65-68 Tabulated for
Many Common
Geometries
Manual
pages
65-68

28. ROTATIONAL LINES: the hairy bits
p. 63
ROTATIONAL LINES: the hairy bits
3020 cm-1 is n0
but many lines
caused by many rotational energy levels,
much closer spaced than vibrational levels

29. p. 69
Compared to stretching the bond, rotation around the bond axes takes relatively little energy

30. B = constant for a particular bond
From quantum mechanics
I = Moment of inertia
I = mr2
= reduced mass
= m1m2 /(m1+m2)
B = constant for a particular bond
note: E = hc/l

31. Rotational energy level spacing increases with J
Erot = B J (J+1)
J = 0, 1, 2, 3, 4,....
J=0, E0=0 J=1, E1=2B J=2, E2=6B J=3, E3=12B
DE = 8B
DE = 6B
DE = 4B
DE = 2B
J0→J1
J1→J2
J2→J3
J3→J4
Rotational energy level spacing increases with J

32. In spectrum of CO, lines are equally spaced
Notes:
Missing ‘middle’
Right side slightly larger
‘grass’ due to 13C

33. p. 71
C O r
Spacing can give r

34. p. 71
Why this shape?

35. p. 72

36. R Branch – high energy side
P Branch
R Branch – high energy side
p. 71

37. Why the difference in intensities? Energy levels are Boltzmann
p. 71
Why the difference
in intensities?
Energy levels are Boltzmann
distributed
more molecules in lower
energy levels
But also, level J has a degeneracy of 2J+1 (from quantum)
2J+1
sum up = 1 J

38. p. 71
Real molecules:
as bond stretches, r increases so I = mr2, increases and
B decreases: spacing (2B) decreases as go to higher J
as I gets large (in large molecules), B and hence spacing
gets smaller and smaller, so only see rotational lines for
small molecules
In larger molecules, there is an I value for each axis (x, y, z)
giving rise to 3 sets of overlapping rotational lines

39. When we see the Q (middle) branch:
p. 73
ASSIGNMENT 3

40. Summary: Rotational fine structure
During vibrational transitions, changes in rotational state can also occur
Rotational state changes have a selection rule ΔJ = ± 1 or ΔJ = 0, ±1 depending on molecular symmetry and type of vibration
Rotational levels are spaced increasingly far apart according to E = B(J)(J+1)
Many J levels are occupied at room temperature and the number of equal energy levels (the degeneracy) of a given J is actually 2J+1
Thus transition V=0 to V=1 is accompanied by a change in J but since there are many starting J states occupied, you get an equally spaced progression of absorptions to the higher energy side (R branch) if ΔJ = +1, to the lower energy side (P branch) if ΔJ = -1 and right at the fundamental frequency ν0 (Q branch) if ΔJ = 0
The Q branch will NOT be observed if the vibration is along the principle axis of rotation (bond axis) of a linear molecule

41. INORGANIC APPLICATIONS
Group frequencies used less, e.g. P=O ~ cm-1
GEOMETRY information from IR, consider the following: Hg
Symmetric stretch is only IR active if bent
Cl----Hg----Cl Cl Cl
Active in IR
Inactive in IR
In fact we find:
413 (IR, asym stretch), 360 (Raman, sym stretch), 70 (bend)
so this suggests the molecule is linear

42. p. 75
Raman only
manual, p 75

43. Structure of complexes

44. p. 77
METAL CARBONYLS
n0 (CO) = 2143 cm-1, but Cr(CO)6 has 2100, 2000, 1985 cm-1 :
empty
full :
s-bonding orbital
Donates electron density to metal … BUT

45. empty antibonding (pp*) filled dp
puts electron density back on carbon
because this electron density is in an antibonding
orbital, bond weakens, frequency decreases

46. sometimes, CO can bridge two metals then n ~ 1850 cm-1
p. 78
2100 – 2000 typically
sometimes, CO can bridge
two metals
then n ~ 1850 cm-1

47. p. 78
bridging CO
terminal CO