Classify – Group Theory
Classify – Group Theory Γxyz ( Γatoms = Γ3N ( count atoms NOT momentum
vibrations Γ3N-6 = Γ3N - Γtrans - Γrot
as character: χ3N = χxyz ( χatom
then reduce to get linear combination inved [pic]
Can categorize – subspaces
stretches χstr = ( bonds that do not ???
bends, etc etc.
again reduce idea these pretty different energies
but these may or may not span the space
must pick carefully – include all motion
Group Theory provides test – do you get all representation
Alternate may use Projection Operator
pΓ(r) = (ci ri will give linear combination of equivalent “r”, could be bends, ???
Example 2:
go on to CH4–Td
[pic]
|Td |E |8C3 |3C2 |6S4 |6σd | |
|A1 |1 |1 |1 |1 |1 | |
|A2 |1 |1 |1 |-1 |-1 | |
|E |2 |-1 |2 |0 |0 | |
|T1 |3 |0 |-1 |1 |-1 |RxRyRz |
|T2 |3 |0 |-1 |-1 |1 |x y z |
Again χ3N = χxyz ( χatoms
|χxyz |3 |0 |-1 |-1 |1 |
|χatom |5 |2 |1 |1 |3 |
|χ3N |15 |0 |-1 |-1 |3 |
|-(χtrans + χrot) |6 |0 |-2 |0 |0 |
|χ3N-6 |9 |0 |1 |-1 |3 |
Reduce χ3N-6
9A1 = [pic] [ 1 • 9 • 1 + 8 • 0 • 1 + 3 • 1 • 1 + 6 • 1 • (1) + 6 • 3 • 1 ] = [pic] = 1
9A2 = [pic] [ 1 • 9 • 1 + 3 • 1 • 1 + 6 • -1 • -1 + 6 • 3 • -1 ] = 0
9E = [pic] [ 2 • 9 • 1 + 3 • 2 • 1 ] = [pic] = 1
9T1 = [pic] [ 3 • 9 + 3 • -1 • 1 + 6 • 1 • -1 + 6 • -1 • 3 ] = 0
9T2 = [pic] [ 3 • 9 + 3 • -1 • 1 + 6 • -1 • -1 + 6 • 1 • 3 ] = [pic] = 2
thus Γ3N-6 = A1 + E + 2T2 see how 1 + 2 + 2 • 3 = coordinates
less obvious:
NH3: χN-H = 3 0 1
reduce: A1 + E
χHNH = 3 0 1 (Note reflection in place bisect
reduce: A1 + E angle gives +1)
recall χ3N-6 = 2A1 + 2E 2 - 1D + 2 - 2D
( 6 dimensional
3N - 6 = 12 - 6 = 6
so these 3N–H + 3HNH span the space
bit harder
CH4 – see attached Handout
χ3N-6 = A1 + E + 2T2 1D + 2D + 2 • 3D
( 9 dimensional
3N - 6 = 15 - 6 = 9
Now could choose C-H str – 4
H-C-H bend – 6
10 – problem
since more internal coordinate then 3N-6 these cannot be all independent
χC-H = 4 1 0 0 2 implies 2A1 + E + 2T2
reduces to A1 + T2 get one too many A1 coordinates
one is not independent or in this
χHCH = 6 0 2 0 2 χHCH = α12 + α13 + α21 + α23 + α24 + α34 = 0
reduces to A1 + E + T2 can’t all open at once
Now that we have a way of getting at a system of coordinates we must look at how to use them
Vibrations of polyatomics – solve 3N-dimensional (R)
[ TN + Ukk (R) ] χυ (R) = Eυ χυ (R)
now only interested in relative motion
can remove C of M + rotation degree freedom
get 3N-6 independent coordinate but express as function of R’s still
Normally express as Cartesian displacement
coordinate ( derivation from equilibrium in rotating frame
δ1 = Δx1, δ2 = Δy1 … δ3N = Δ = zN
for vibration problem mass weighted Cartesian displacement coordinate easier
q1 = m1½ Δx1, q2 = m1½ Δy2 … q3n = mN½ ΔzN
Classically: [pic]
Potential normally done in Harmonic Approximation (same as diatomic, more coordinates)
[pic]
Now same as for diatomic: Ue – constant / just shift potential E
[pic] for minimum
1st non-zero / non-constant term is quadratic
[pic] (qi • qj)
but of course there are more – anharmonic terms
If keep just this and TN: [pic]
This is coupled – multidimensional – can’t separate as written
TN is diagonal: q = [pic] TN = [pic]
TN = [pic] (direct product ???)
In this form:
VN = [pic] VN = ½ q1 q2 … qN [pic][pic]
goal – change coordinates T = [pic]
V = [pic]
both terms diagonal, to span space Qi = [pic]
need transformation L [pic] ( diagonal matrix, λi on diagonal
writing trans form:
[pic] = matrix of eigenvectors of [pic] , L-1 = LT
λi = eigenvalues of [pic]
secular determinant:
solve det ([pic] - δjk λm) = 0 U(( ( 3N x 3N, 3N λm values but 6 ( zero
plug λm into secular equations:
[pic](Ujk - δjk λm) ℓkm = 0
[pic] = { ℓkm }
Qi = [pic]ℓkm qk and inverse qj = ( ℓjk Qk
or q = [pic] Q Q = [pic]T q
Put it all together
2Vvib = [pic] = (LQ)T U(( (LQ) = QT LT U(( L Q
= QT ( Q or 2V = ( λm Qk2 ( diagonal
same idea:
2T = [pic] ( diagonal (LT L = 1)
can separate – solve one coordinate at a time (H = [pic])
Classical: F = ma = [pic] = [pic] = -λ Q
wave equation: [pic] + [pic] = 0 ( Qk = Bk sin (λ½ t + bk)
Quantum Mechanics
H = TN + VN
= ½ ( Qk2 + ½ ( λk Qk2
= [pic]
each one is a 1-D harmonic oscillator problem
Know solution: Hvib = [pic]hk hk χk = Ek χk
Evib = [pic](υk + ½) h νk Ek = (υk + ½) h νk
Yvib = [pic] χυk (Qk) χk = Nke [pic] Hυk (αk½ Qk)
αk = [pic] [Note: can’t simply write k,m now]
recall: summed H ( product w/f
( Total energy sum independent vibrational energies
{Note zero potential E, ½ [pic]h υk ( non ???}
( Product function makes determinant [pic] easier
use Group Theory Γψvib = [pic]Γχυk
so need know representation of each vibration (keep doing that) and take product ( representation of full w/f
look at what changes ( unchanged – no contributions
Selection rules IR [pic] how determine?
expand: μ = μe + [pic] Qk + [pic] + …
constant but vector leads to ΔJ = (1,0 rotation
eg: [pic][pic]
This term only non-zero pure rotation, orientation is independent
2nd term ( vibrational excite
[pic]
still orientation effect ( ΔJ = (1,0
vibration (harmonic oscillator) ( Δυk = (1
but only υk change Δυj = 0 j ( k
and [pic] dipole moment must change along coordinate Qk
to do this Qk and μ must have same symmetry
Group Theory language: [pic] [pic] Γμ ( ΓQk ( Γx
so look in table
representations for x,y,z and vibrational IR allowed (assume χ(( = υ = ??)
Raman Spectra selection α = αe + [pic] Qk + [pic] Qj
same idea αe ( pure rotation, transform as x2, y2, z2 xy, yz, zx, ΔJ = 0, (1, (2
[pic] ( polarizability must change to see vibrational transition
Δυk = (1, Δυj = 0 ( exact same [pic]
( Γvib ( Γα ( Γx2,y2,z2,xy, yz,xz ( see Character Table
Harmonic Approximation
Rotation effects – see Handout – Banwell depends on symmetry
Linear Molecules – (review) – McHale, Ch 9 & 10.6
11 vibrations ( stretch along axis
C(( ( A1 ((+) ( ( M = 0, in terms of angular momentum
D(( ( A1U ((u+), A1g
Δυ = (1, ΔJ = (1 IR Just like diatomic DJ = 0 possible
Δυ = (1, ΔJ = 0 Raman due to K = 0 if electron angular momentum
( vibrations ( distort molecule from linear (bend)
C(( - E1 (x,y) ; D(h - E1u ((u) ( IR allowed
C(( - E1, E2 ; D(h - E1g,2g ((g, Δg) ( Raman allowed
Δυ = (1, ΔJ = 0, (1 – IR P,Q,R branches
Δυ = (1, ΔJ = 0, (1, (2 – Raman O,P,Q,R,S branches
Isotopes – spin of nuclei – total w/f fermion – asymmetry (-1)
get intensity alteration: J – {even, odd exchange symmetry: bosom – symmetry (+1)
Note – pure rotation, this would only be Raman
– vibration / rotation – see change symmetry but population effect remains
Spherical top molecules
A1 ( not allowed ( IR
Totally symmetric modes: Δυ = (1, ΔJ = 0, (1, (2
Asymmetric modes (T2) Δυ = (1, ΔJ = 0, (1; ΔJ = 0, (1, (2
Sort of like diatomic but degeneracy in K = (2J + 1)
[ADD Infra-red spectroscopy]
[ADD Banwell-Fund. Molecule]
[ADD Infra-red spectroscopy]
Summary
IR selection rules [pic] ( 0 ΔυK = (1, ΔυJ = 0 j ( K
[pic] ( 0 ΔJ = 0, (1 ΔM = 0, (1 ΔK = 0
[pic] ( 0 ΓQK ( Γμ = Γxyz
Raman – same except: ΔJ = 0, (1, (2 since operator Y2(1(2
ΓQK ( Γα = Γx2,y2,z2,xz,yz,xz
IR – dipole moment change / Raman – polarizability change
center of symmetry – IR/Raman u + g – exclusive
Linear – A1 modes (E) – IR P,R branch, ΔJ ( 0
Note: D(h – no IR for symmetry stretches / need symmetry A((1u
Raman – can have ΔJ = 0, (2 θ,Q,S branch
E modes (() – IR P,Q,R ΔJ = (1, 0
Assume start ground state υ = 0 ( χ0 = total symmetry / if higher temperature
can start υ = 1 on higher ( hot bend still Δυ ( 0
Isotopes ( if center of symmetry, i, then spin ½ – asymmetry
get alternating intensity J {odd, even
due to population ½ + ½ = 0
Raman Polarization 2 photon ( can measure scalar ( or ll to excitation
Spherical top ( = [pic] ( polarized ( total symmetry
Total symmetry A1 mode Raman Δυ = (1 ΔJ = 0, (1, (2
IR – not allowed (xyz ~ T)
Asymmetry Molecules T2 Δυ = (1 ΔJ = 0, (1 – IR
ΔJ = 0, (1, (2 – Raman
like linear but ΔK = 0, K-degeneracy (2J + 1) – affect intensity
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Symmetrical Tops
parallel vibration ( Γvib = Γz [pic] Δυ = (1, ΔJ = 0,(1, ΔK = 0
note this is same situation as t-linear
perpendicular vibration ( Γvib = Γxy (E) Δυ = (1, ΔJ = 0,(1, ΔK = (1
( this will give multiple sets of P,Q,R lines, differences depend on (A–B)
ex: CH3I A >> B ( Q branch separation much greater than R,P
Handout CH3I
Raman ( Δυ = (1, ΔJ = 0,(1, (2, ΔK = 0 symmetry ΔK = (1 asymmetry
Asymmetry tops ( like prolate, like oblate
band contours distinguish types of modes
A B C – A – strong center, weak wing
B – minimum center, strong wing
C – strong center and wing
Anharmonicity ( Δυ = 0,(1,(2 … Δυi = (1, Δυj = (1
Intensity – Group Theory just tells us on/off, but allowed transition may be weak or strong
IR – [pic] ( bigger dipole change
for motion of change C = 0 strong, C = C weak
Raman – [pic] ( large polarizability – delocalized parts – aromatics, heavy atoms
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