II-1

Empirical connection for energy – Account for anharmonicity / rotational-vibration

1)eXe – positive generally reduce spacing in upper  levels (~2)

Dissociation:neutral – finite number of levels U ~ 1 / R6

ionic – infinite number levels U ~ 1 / R, large R

2)e – vibration-rotation coupling – positive generally – bond together with

B = Be – e ( + ½) decrease rotation spacing at high  increase I → decrease B

3)De – centrifugal distortion → J increase → R increase → B decrease

always positive:De = 4Be2 / 2→ small value

 ~ 103 – 102 Be

4)Y00 – shift of all energy levels – constant

only impact if zero-point energy issue (e.g. kinetic isotope effect)

Units – typically all in cm-1

If Dissociation energy/potential represented as Morse

UKK = De [1 – e-a(R - Re)]2

De = UKK (Re)

Now can takeR = 0 + (B′ – B″)J2 + (3B′ – B″)J + 2B

 = 0 + (B′ – B″)J2 - (B′ – B″)J

Let B = B – e ( + ½)ignore De term

R = 0 + (e (′ – ″)J2 ) + 2Be (J+1) – e (3′ – ″ + 1)J - 2e (′ + ½)

R = 0 + 2[Be – ] (J + 1) – (J+1)2

 = 0 + 2(Be – 2e) [Be – ]J – eJ2J = 1,2,3, …

can analyze spectra by introducing

M = J+1 - R branch

-J - P branch

 = 0 + (2Be – 2e)m – em2quadratic fit of n vs m

so if measure all frequencies and make a table can also measure

differences between them(’s)

consider (m) = (m+1) – (m)

 = (2Be – 2e) – 2em – e = -2em + (2Be – 3e)

this last is a linear equation in m hence, plot (m) vs m

slope:-2e

intercept: 2Be - 3e

See attached data:

for B → 1 spectra:

HCI35HCI37

(as determined by aBe10.492 cm -110.456 cm -1

physical chemistrye0.2847 cm -10.2875 cm -1

lab student)re1.280 Å1.279 Å

Geometry!

[INSERT HANDOUT (full page) – wavenumbers spectra]

[INSERT HANDOUT (full page) – PQR contour spectra]

Rotation - Vibration picture

Put vibrations and rotation together

connect for arrhenius and centrifugal

R branch: J = J" , J + 1 = J'J" → JJ = +1

" = 0' = 1 = +1

R = (0 – 2eXe) + [2Be - 2e] (J+1) – e (J+1)2 – 4D (J+1)3

 = (0 – 2eXe) + [2Be - 2e]J – eJ2 + 4D (J)3

since e, De almost always positive:

Intensity profile → same as seen for rotational states i.e. {error in last one!}

again –see rise in intensity from 2J+1 and fall off from exp

–can use as a thermometer

Differences

(R) = [2Be – 2e] – e[2J+1]  (2B – 3) – 2J

(R) = 2e – 4D [6J + 12]plot (R) vs. J

slope:-2e det m!

intercept: 2Be - 3e

Note – Units – all previous stuff0 = 0 / c

cm-1Be = B / LeE/ h =  etc

sample of increased resolution (ability to sep  + ')

CO Anharmonicity example – big loss intensity,20 – 1 0 – 2

Discuss HCl spectra:O → 1 and O → 2 spectra

a)note:doubling of peaks due to Cl35 and Cl37

HCl35 – higher energy and bigger

isotope mass(Cl mw ~ 35.5) 75% 35, 25% 37

different – same k

if had DCl where peak? – ~ 2100cm-1

b)note:splitting of peaks (J levels) bigger in P branch than in R branch

Vibration-rotation coupling due to diff. in B and ex → most obvious In O → 2 spectra

c)note:splitting of 35-37 changes slightly due to diff of B35 and B37

d)note: 0 (O  2)  20 (O  1)

arrheniusO → 1 035 = 2885.9 cm-1037 = 2883.6 cm-1

O → 2035 ~ 5668037 ~ 5664

 (2 – 2 1)-104 cm-1-103 cm-1

e)could use formula from class and determine B0 + B1 on B0 + B2from difference in peaks – to describe would need series of Bi’s

more useful:E = h0 ( + ½) + BeJ (J+1) – e( + ½) J (J+1)

evaluateR = 0 + (2Be – 3e) + (2Be – 4e) J – eJ2J = 0,1,2, …

[INSERT HANDOUT (full page) – HCl rot-vib spectra]

Note all the corrections:Xe, e, De

Do not change selection rules: = 1

J = 1J = 0, 1

but do change energy/spacingRaman linear

Examples:

CO – very regular spacing – less rotational/vibratonal coupling

→deep potential well

→do see spread in P branch

→weak bonds – isotopes(13C ~ 1%)

HCN – linear molecules act sameway butI now depend on 2 bond dist.

rotational spectra same J = 1, spacing 2B, … , B = 2 / 2I

vibration spectra – if vibration maintains linear geometry – same

P + R branchJ = 1, i = 1, i = 0

(ex – HCN str  H–C N~3500 cm-1)

–if distant then not linear/ get P Q R branch

Q branch:J = 0,  = 1

Since no change or J then

Q= 0 + (B′ – B″)J + (J+1)

= 0 + (-e) J (J+1)

so lines very close, and shift with J depending on e

H – C  C – HAcetylene – linear – center sym / H’s equivalent
H – C  C – D – no center – H  D – separate stretch

ex:3C-D (mostly) stretch – P and R branch okay – extra peak – overlap ??

5 + 4 – combination – anharmonic 5 = 1 ,4 = 1

Note → bend → x + y → x iy → angular momentum → (call  sign)

HCCH – 5 + 4 – similar freq (up due D  H) but J’s ??

intensity modulated by nuclear statistics (H I = ½)

[INSERT HANDOUT (full page) – spectra]

[INSERT HANDOUT (full page) – HCN spectra]

[INSERT HANDOUT (full page) – spectra]

Polyatomic Rotations – again assume rigid

→orientation now 3 angles ()

so there will be 3 degrees ??? and 3 quantum numbers

See McHaleEuler angles do rotation

Ch 8.2.2 –rotate about Z – like before

 –rotate about Y – concept like before, but order important

 –rotate around molecular axis

Harmonic oscillation in -like ??

The x – with ?? axis → ?? energy ?? ??

Can have angular momentum about each axis but not orthogonal rotated system

but total angular momentum conserved

L2 = Lx2 + Ly2 + Lz2 = Lx2 + Ly2 + Lz2

molecule form lab form

formation use:

Ĵ2, ĴZ2, Ĵz2 form set converting operators so must be function DJMK (, , )

that is eigen value

DJNM = e+iMJNM () e+cK

see parallel in ,  behavior

Now have eigen value relations for angular momentum but to get energy will need mass effect

moment of matter → convert to primary axis – diagonalize

Convention

To get energy exact parallel linear case

Let A = (cm-1)B =C =

Cases:a)Spherical Top → Ia = Ib = Ic = I,ex: SF6, CH4, C60

H = J2 / 2I

EJ = J (J+1) = J (J+1) Abut:M = J, J-1, … , -J

K= J, J-1, … , -J

(2 J + 1)2 degenerate

No 0 → no -wave on far IR

0 – spherically symmetric → no Rotational Raman

b)Symmetrical Top →Ia < Ib = IcProlate

Ia = Ib < IcOblate

a – principle axis (equivalent z)

Energy dependent on J, K

= J (J+1) B + K2 (A-B)degeneracy: 2 (2I + 1)ex K = 0

EobJK = J (J+1) B + K2 (C-B)(same they swap ??)

Difference comes from K2 term -- levels included with K for prolate (A > B) but decrease w/K to oblate (C < B)

Now modify selection rules (no energy)

-waveJ = ±1, M = 0, ±1, K = 0

RamanJ = 0,±1, ±2

c)Asymmetric tops → most cases Ia≠ Ib≠ Ic

need final approx solution to full

K not good QN but still characterize state

extrapolate between prolate (-1)/oblate (+1) F =

[INSERT HANDOUT (full page) – Methene graph]

[INSERT HANDOUT (full page) – More spectra]

Polyatomics – review

The major issue is handling a 3-D structure → motions in 3D –3 axis rotation

–3 vibration degrees

shape – becomes an issue

Rotations: McHale, Ch 8

Struve, Ch 5

Trot=ij → rotation

=½ • I• Iij – i central tensor

Change coordinate with Euler (handout), etc

No potential →

a,b,c principle axis

Spherical Top:Ia = Ib = IcJ = B J (J+1)

Symmetrical Top:prolateIa < Ib = IcEJK = B J (J + 1) + (A – B) K2K = 0

oblateIa = Ib < IcEJK = B J (J + 1) + (C – B) K2

Asymmetrical Top:Ia Ib Ic

Note none of this depends on explicit form of , just angular momentum properties

-wave (for IR) J = ±1 (+1 –abs) / K = 0 / 0 ≠ 0

Raman (Stokes & acti)J = 0,±1, ±2 / K = 0 /  – asymmetry

except when K = 0 or J = 0

A B C

Examples:CH45.2412

C6H66.18960.0948

NH39.44430.196

C2H62.6810.6621

CH3Cl5.090.443901

H2CO9.4057.2951.134

H2O27.87714.5129.285

High resolution  low pressure 10-3 - 10-4 torr / long path – low ?? / Stark modulate

in/out resource w/Klystron

[INSERT HANDOUT (full page) – Raman spectra]

[INSERT HANDOUT (full page) – K dispersion]

Intensity ~ degeneracy • population (e -EJ/KT) but since separate

small compared to kT  n = nJ" – nJ'

is operative, also ~ 

so actual map will differ from (2J+1) eJ(J+1)/KT

see this pattern in diff J" → J'J = 1

J = 2 in Raman

Selective rules work Spherical/ Symmetry with quantum number state

for Asymmetry  K not good quantum number

but J = ±1 goodM = 0, ±1

Intensity Variation patterns – can be interference or nuclear spin statistics

Fermior spin half integers: , …

with f must be anti-symmetry with r/ to exchange (parity)

2 nucleii:(2I + 1)2 degeneracyMI = I, I -1, I -2, … , -I

states: (2I + 1) (I + 1) symmetry + (2I + 1) I asymmetry

egH2:½, ½, , ,  + – symmetrical

 - – asymmetrical

Rotational states:YJM = (-1)J YJM

Vibrational state:X = X– even

soJ = 0  asymmetrical spin (1)– weakerJ = 0  J = 1

J = 1  symmetrical spin (3)– strongerJ = 1  J = 2

Boson  integer spin states must be symmetric

exI = 0  all symmetrical J = odd not exist

Separating oblate/prolate/symmetry/asymmetry  need establish symmetry

Clear first step find axes of rotation molecule equivalent

This basis for symmetry ?? but more general