vibrational theory bonds ~ springs e = ½ kx m1 mchem213/213-3-18.pdf · vibrational theory p.55 c...
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VIBRATIONAL THEORY p.55
C H
m1 m2
bonds ~ springs
E = ½ kx2
m1 m2
m1 m2
m2m1 stretch
equilibrium
compress
0 ror
xx
ro - x ro + x
x is + or -x2 is +
distance
m2m1
m1 m2
m1 m2
ro
ro
Graph ½ kx2 gives a parabola
SIMPLE HARMONIC OSCILLATOR
p. 55
vk
m mm m
o
12
1 2
1 2
for a SHOk = force constant = reduced mass
C------H O------H
when m1 is very large, does not change muchhowever (OH) > (CH) so k(OH) > k(CH)
3000cm-1 3400 cm-1
p. 55
CH) = 12/13 = 0.923
(OH) = 16/17 = 0.941
However, if the light element is different: p. 56
C----H C-----D1 2
CH CD
12 112 1
12 212 2
vk
m mm m
o
12
1 2
1 2
vv
v v
v v v
v cm cm
C D
C H
C H
C D
CD CH
CH
CD
CD CH CH
CD
1213
1424
12
300012
21001 1
k’s aresame
= 0.92 = 1.7
or CH (0.92/1.7)1/2
= 2200 cm-1 Huge difference
A spring (SHO) has all energies possible,MOLECULES DO NOT
only one fundamental frequency (spacings equal, h)
Ev = (v + ½)h
p. 57
normally, only a one level jump allowed p. 57
in IR, only lowest level is populated
hot lines, overtones weak
p. 58
v=0
v=1
v=2
v=3
Internuclear separationro average bond length
ground vibrational state
fundamental frequency of vibration, E = ho
1st overtonehot line,
Dissociation energy
REAL MOLECULES: Not a simple parabola!
o o E0→E2
Eo = 1/2 ho
E1 = 3/2 ho
E2 = 5/2 ho - a ‘bit’
E3 = 7/2 ho - a ‘bit’ more
Ev ≈ (v + ½)h
so in CO 0 = 2143 cm-1, 1 = 4250 cm-1 not 4286 cm-1
Reminder:Intensity of band depends on change in dipole during stretch or bend
CH3C CH CH3C CCH3weak 2150 cm-1 no IR band
CH3C Nstrong 2250 cm-1
p. 59
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 positionBUT not all movements of atoms in space correspond toa vibration:
p. 59
3 possible translations along x, y or z AND3 possible rotations around x, y and z do NOTchange the relative positions of the atoms
In general, if molecule has N atoms
there are only 3N-6 possible fundamental vibrations
[3N-5 if molecule is linear]
HO
H
HO
H
bent, triatomic3N-6 = 3(3)-6=33 fundamental frequencies
symmetric stretch3650 cm-1
HO
H
asymmetric stretch3750 cm-1
HO
H
bend1600 cm-1
Energy look at this one
p. 59/60
+-
During the vibration, the bond dipole changes
but it is the VECTOR SUM that is important,If this changes during vibration = IR active
p. 60
HO
H
HO
H
bent, triatomic3N-6 = 3(3)-6=33 fundamental frequencies
symmetric stretch3650 cm-1
HO
H
asymmetric stretch3750 cm-1
HO
H
bend1600 cm-1
Energy
p.60
3750 asym3650 sym
1600 bend
p. 60
(overlapping) Structure is ‘hairy’ due torotational fine structure:more on this shortly
CO2
= 0 = 0
= 0p. 61
C
linear, triatomic3N-5 = 3(3)-5=44 fundamental frequencies
symmetric stretch1330 cm-1
IR inactive (Raman active)
asymmetric stretch2550 cm-1
IR active
bends (degenerate)670 cm-1
IR active
Energy
O O
CO O CO O
CO O
CO O
CO2p. 61
asymmetricstretch
symmetricstretch bends
Energy
Z X Z Y where X is lighter than Y
Z X Z X Z X
wavenumber (cm-1)
>
>
>>
>
SUMMARYp. 62
Reduced mass effect:
Force constant effect:
(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-1 725 cm-1 650 cm-1 550 cm-1
(3) M-F M-Cl M-Br M-Iwhere Metal is more massive, e.g. Sn, Pb
600 cm-1 350 cm-1 225 cm-1 150 cm-1
From these, you can predict many others!
p. 62
POCl3 shows IR bands at 1290, 582, 486, & 267 cm-1
Problem:
O=PClCl
Cl
Cl > O, so P-O > P-Cl, so P-O is likely the highest = 1290 (plus P-O has some double bond character)
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
p. 62
Methane, CH4 likewise only shows 4 bands but has 9 {3N-6} fundamentals (others degenerate)
H
H H
HH
H H
HH
H H
HH
HH
H
asym stralwaysIR active
sym strif atoms sameNOT IR activedipoles cancel
sym bendNOT IRactive
bendIR ACTIVE
3020 cm-1 2914 (Raman) 1520 (Raman) 1305
p. 63
3020 cm-1 2914 (Raman) 1520 (Raman) 1305
p. 63
Vibrational ModeAssignments
Tabulated forMany CommonGeometries
Manualpages 65-68
ROTATIONAL LINES: the hairy bits
3020 cm-1 is 0but many lines
caused by many rotational energy levels,much closer spaced than vibrational levels
p. 63
m1 m2
ro
(b) rotation around thecenter of gravity,perpendicular to the bond
(a) rotationalong the axis ofthe bond
p. 69
Compared to stretching the bond, rotationaround the bond axes takes relatively littleenergy
m1 m2
ro
(b) rotation around thecenter of gravity,perpendicular to the bond
(a) rotationalong the axis ofthe bond
Eh
IJ J B J J
J rotational quantumnumber
B in Joulesh
I
B in cmh
I c
rotation
2
2
2
2
12
81 1
0 1 2 3
8
8
( ) ( )
, , , ...
( )
( )
I = r2
= reduced mass= m1m2 /(m1+m2)
note: E = hc/
I = Moment of inertia
B = constant for a particular bond
From quantum mechanics
p. 69
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
p. 70
Rotational energy level spacing increases with J
02B
6B
12B
20B
J=0J=1
J=2
J=3
J=4
E=2B
E=4B
E=6B
E=8B
E 2B 4B 6B 8B
J=0 J=1 J=1 J=2 J=2 J=3 J=3 J=4
E
Lines are equally spaced 2B apart
Rotational levels
Absorption spectrum
E = 8B
E = 6B
E = 4B
E = 2B
J0→J1 J1→J2 J2→J3 J3→J4
In spectrum of CO, lines are equally spaced p. 71
Notes:
Missing ‘middle’
Right side slightly larger
‘grass’ due to 13C
spacing hI
Ihspacing
rm m
m mr
rh
spacing
2 8
4
21
2
2
2
22 1 2
1 2
2
( )
( )
C-------Or
Spacing can give r
p. 71
Why this shape?
p. 71
p. 72
P Branch R Branch – high energy sidep. 71
Why the difference in intensities?
Energy levels are Boltzmanndistributed
more molecules in lowerenergy levels
But also, level J has a degeneracy of 2J+1 (from quantum)
J1
0
2J+1 sum up =
p. 71
Real molecules:
as bond stretches, r increases so I = r2, increases and B decreases: spacing (2B) decreases as go to higher J
as I gets large (in large molecules), B and hence spacinggets smaller and smaller, so only see rotational lines forsmall molecules
In larger molecules, there is an I value for each axis (x, y, z)giving rise to 3 sets of overlapping rotational lines
p. 71
H C N principal axis of rotation
vibration parallel to principal axis of rotationhas selection rule of J = +/- 1 (ie: J not 0)so no Q branch seen in the stretching regions (CH ~3300 cm-1)
H C N
H C Nbend perpendicular to principal axis of rotationhas selection rule of J = 0, +/- 1 so Q branch is seen in the bending region (~720 cm-1)
PQ
R
P R
3300 cm-1
720 cm-1
When we see the Q (middle) branch: p. 73
ASSIGNMENT 3
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
INORGANIC APPLICATIONS
Group frequencies used less, e.g. P=O ~ 1140-1300cm-1
GEOMETRY information from IR, consider the following:
Cl----Hg----ClHg
Cl Cl
Inactive in IR Active in IR
In fact we find:
413 (IR, asym stretch), 360 (Raman, sym stretch), 70 (bend)
so this suggests the molecule is linear
p. 75
Symmetric stretch is only IR active if bent
HN
HH F
BF
F
trigonal pyramid trigonal planar
manual, p 75
Raman onlyp. 75
Co
NH3
O
NH3NH3
NH3
H3N
C
O
O
Co
NH3
NH3NH3O
O NH3
CO
+
Br-
+
Cl-COO
O
2-
free carbonateIR shows asym at ~1500cm-1
but no sym (~1000cm-1) due to the symmetry of the CO3
2-
[Co(NH3)5CO3]Br IR shows a band in the sym region (~1100cm-1) which shows that the
CO32- fragment is no longer symmetric,
so it must be bonded to the metal center
[Co(NH3)4CO3]Cl IR shows a band in the sym region
(~1050cm-1) which shows that the CO32-
fragment is bonded to the metal, but in a different way than in [Co(NH3)5CO3]Br
as the sym has decreased
Structure of complexes
p. 76
METAL CARBONYLS
0 (CO) = 2143 cm-1, but Cr(CO)6 has 2100, 2000, 1985 cm-1
OCM
M OC
M OC
empty full
-bonding orbital
Donates electron density to metal … BUT
:
:
p. 77
empty antibonding (p*)filled d
puts electron density back on carbon
because this electron density is in an antibondingorbital, bond weakens, frequency decreases
p. 77
M OC ~ CM O
2100 – 2000 typically
C
O
M M
sometimes, CO can bridgetwo metals
then ~ 1850 cm-1
p. 78
terminal CO
bridging CO
CoC
CCo
O
O
CC
CCC
CO
O
OO
O
O
p. 78