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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