Physical Chemistry (II)

Lecture 20

CHEM 3172-80

Lecturer: Hanning Chen, Ph.D.04/10/2017

Electronic Transition

Quiz 20

5 minutes

Please stop writing when the timer stops !

Electronic Ground and Excited States

HA HB

bonding orbitalσ 1s

anti-bonding orbital

σ 1s*

Both bonding and anti-bonding orbitals are the eigenfunctions of the time-independent Schrödinger equation

ground state

excited state

~ eV

transitions between electronic wavefunctions

Molecular Vibrationd

~ 10 TetraHertz (1013 / s)

ΔEvib = !ω :∼ 0.1 eVMolecular Rotation

~ 100 GigaHertz (1011 / s)

ΔErot = 2B(J +1) :∼ 0.001 eVtransitions between nuclear wavefunctions

Detection of Electronic Transitions

ΔEelec :∼ eVThe electronic transitions can be detected by a wide variety of spectroscopy techniques,

including X-ray, Ultraviolet , Visible and Infrared absorption

typical light absorption spectrum

“bio-labels”

multiple peaks widened by thermal fluctuation and vibronic coupling

System Setup for Light Absorption

absorbing sample with a molar concentration

c[ ]I0 I1

λ λ

no change on the transmitted wavelength, only the reduction of the transmitted intensity

L0(light path)

Light absorbed by the infinitesimal thickness, dL, of the sample:

dL

dI = −k[c]IdLk : molar absorption coefficient I : light intensity at the slab I ≠ I0 !!!

0 L0

some photons have already been absorbed before reaching the slab !!!

I < I0

I

the Beer-Lambert LawWhat is the light intensity profile along the light path ?

dII= −k[c]dL

Integration of L from 0 to L0,

dIII0

I1

∫ = −k[c]dL0

L0

∫L = 0→ I = I0L = L0 → I = I1

ln I1I0

= −k[c]L0

Beer-Lambert law: I1 = I0e−k[c]L0 The light intensity drops exponentially (NOT linearly)

along the light path

Transmittance: T = I1I0

= e−k c[ ]L0 optical density: OD = − logT = − log10 e−k c[ ]L0

= 0.43k[c]L0large optical density means strong light absorption

Electronic Spectra of Diatomic MoleculesTerm symbols were used before to specify the quantum states of a multi-electron atom

Let us start with the ground state H2 molecule:

1sA 1sB

Λ: total orbital angular momentum quantum number

σ 1s

σ 1s*

projection of orbital angular momentum onto the molecular axis

λ = 0 λ = +1 or -1 λ = +2 or -2δ − bond

number of nodal planes

Term Symbols

For ground state H2 molecule: λ1 = 0,λ2 = 0 λ! = λ1 + λ2 + ...+ λn( )!total projection: λi : contribution from the i th electron

λtotal = 0 + 0 = 0

symbol

0 1 2 3λtotalΣ Π Δ Φ

orbital angular momentum term for ground state H2 molecule is Σspin angular momentum term: 2S+1

two paired electrons: s1 =12, s2 = − 1

2stotal =

12− 12= 0 2S+1 = 1

so the term symbol should be 1Σ , but the full dress is actually 1Σgwhere does the “g” come from?

Parity of Molecular Orbitalssymmetric with respect to an inversion center

gerade

anti-symmetric with respect to an inversion center

ungerade

gerade

gerade

gerade

ungerade

ungerade

ungerade

ϕ x, y, z( ) = +ϕ −x,−y,−z( )

ϕ x, y, z( ) = −ϕ −x,−y,−z( )

Ψ =ϕ1ϕ2...ϕnThe wavefunction parity only depends on the number of ungerade electrons.

Nungerade = odd→ ungerade Nungerade = even→ gerade

Ground State O2

2pA 2pB

2sA 2sB

1sA 1sBσ g1s

σ u*1s

σ g2s

σ u*2s

π u2p

π g*2p

σ u*2p

σ g2p

unpaired electrons

λ = −1 λ = +1

λ = 0

λ = 0

Ground State O2

In total, we have 8 valence electrons

total orbital projection:

The closed subshells have no “net” contribution to the term symbols : 1Σg

λtotal = 2 × 0 + 2 × (−1)+ 2 × (+1)+1× (−1)+1× (+1) = 0

total spin projection: stotal =12+ 12= 1 (triplet state)

parity of the molecular orbital: gerade or ungerade ?

the total number of electrons at the ungerade orbitals: 4 gerade

the full dress of ground state O2 is3Σg

what happens if the two unpaired electrons become paired ?

π g*2p π g

*2p

Excited State O2

2pA 2pB

2sA 2sB

1sA 1sBσ g1s

σ u*1s

σ g2s

σ u*2s

π u2p

π g*2p

σ u*2p

σ g2p

paired electrons

λ = −1 λ = +1

λ = 0

λ = 0

Excited State O2

In total, we have 8 valence electrons

total orbital projection: λtotal = 2 × 0 + 2 × (−1)+ 2 × (+1)+ 2 × (−1) = −2

total spin projection: stotal =12− 12= 0 (singlet state)

the total number of electrons at the ungerade orbitals: 4 gerade

the full dress of excited state O2 is 1Δg

λtotal = 2→ Δ

what is the term symbol for O2+

total orbital projection: λtotal = 2 × 0 + 2 × (−1)+ 2 × (+1)+1× (−1) = −1

total spin projection: stotal =12 (doublet state) 2Πg

the total number of electrons at the ungerade orbitals: 4 gerade

Reflection Symmetry of Molecular Orbitals

Oxzσ = +σ

Oxzπ x = +π x

Oxzπ x* = +π x

*

no change

π x

− : reflection symmetry

π y

+ : reflection symmetry

Oxzπ y = −π y

Oxzπ y* = −π y

*

Ψ total −( ) = Ψ spin +( )Ψ spatial −( )Ψ total −( ) = Ψ spin −( )Ψ spatial +( )

Σ−

Σ+

only important

for Σ

terms

Total Angular Momentum

L

S

ΛA BΣΩ = Λ + ΣTotal angular momentum:

addition of two vectors

For the ground state O23Σg

spin component: Σ= +1,0,−1{ }orbital component: Λ= 0{ }

Ω = +1,0,−1{ }3Σ1,

3Σ0

Splitting of energy levels due to the spin-orbit coupling

Ω = +1 and Ω = −1 are energetically degenerate

Selection Rules for Electronic Transitions

ΔΛ = 0,±1 ΔS = 0 ΔΣ = 0 ΔΩ = 0,±1ΔS = 0 : no spin flipping ΔΣ = 0 : no spin reorientation

photon is a spin-1 boson particle

µif = si

*∫ s f dr ψ i*!r∫ ψ f drtransition dipole moment:

ΔΛ = 0,±1 ΔΩ = 0,±1 conservation of angular momentum

For terms:Σ only Σ+ ↔ Σ+ and Σ− ↔ Σ− are allowed

For molecules with a center of inversion,

only u↔ g is allowed

(Laporte selection rule)

(parity symmetry must change)

(no reflection symmetry change)

Frank-Condon Principleonly the vertical transitions are allowed !

because the lighter electrons move much much faster than

the heavier nuclei

light-driven events are ultrafast

µif = vi

*∫ vf dr ψ i*!r∫ ψ f dr

nuclear wavefunction overlap electronic transition dipole

v '' = 0→ v ' = 0 forbidden

vv ''=0*∫ vv '=0dr→ 0 due to the shift of

the energy minima

Rotational StructureSimilar to the vibrational-rotational spectrum,

a rotational transition can accompany an electronic transition

ΔJ = −1: P branch ΔJ = +1: R branchΔJ = 0 : Q branch

EP = Ee − (B '+ B '')J + (B ''− B ')J2

EQ = Ee + (B ''− B ')J(J +1)

ER = Ee + (B '+ B '') J +1( ) + (B ''− B ') J +1( )2

B '(B '') for the initial(final) electronic state B''−B' < 0

ER = Ee + (B '+ B '') J +1( ) + (B ''− B ') J +1( )2+ −

when (B '+ B '') J +1( ) + (B ''− B ') J +1( )2 = 0→ J = B '+ B ''B '− B ''

−1 λR : shortest wavelength(band head)

For abnormal contraction upon electronic excitation, B''−B' > 0

J = B '+ B ''B ''− B '

λP : longest wavelength

(red shift) (blue shift)

(band tail)

Review of Homework 6Review of Homework 1912.16 Consider the molecule CH3Cl. (a) To what point group does the molecule belong? (b) How many normal modes of vibration does the molecule have? (c) What are the symmetries of the normal modes of vibration for this molecule?(d) Which of the vibrational modes of this molecule are infrared active ? (e) which of the vibrational modes of this moleculeare Raman active?

C3v

3N − 6 = 3× 5 − 6 = 9 vibrational normal modes

E 1 1 11 1 -1

2 -1 0

A1A2E

E 2C3 3σ vC3v

Γ3N 15 0 3= (6 − 3)A1 :

161×1×15 + 2 ×1× 0 + 3×1× 3( ) = 4 A2 :

161×1×15 + 2 ×1× 0 + 3× (−1)× 3( ) = 1

A3 :161× 2 ×15 + 2 × (−1)× 0 + 3× 0 × 3( ) = 5

Γ3N = 4A1 +A2 + 5E

Review of Homework 6Review of Homework 19

Γ translation : 3 0 1Γ translation = A1 + E

Γ rotation : 3 0Γ rotation = A2 + E

Γvib = Γ3N − Γ rot − Γ trans = 3A1 + 3E

x,y( )→ E z→ A1 all modes are IR-active

x2 + y2,z2 → A1 xy,x2 − y2( ),(yz,zx)→ E

all modes are Raman-active too

−1

Homework 20

Reading assignment: Chapters 13.1 and 13.2

Homework assignment: Exercises 13.17 Problems 13.1

Homework assignments must be turned in by 5:00 PM, April 11th, Tuesday

to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall