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