Department of Chemistry

University of Cambridge

Part II Chemistry 2004-2005

High-Resolution Molecular SpectroscopyElectronic Spectroscopy

Lecturer: Professor Jacek Klinowski

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Electronic spectra of diatomic molecules

The Born-Oppenheimer approximationA diatomic molecule may undergo electronic, vibrational and rotational transitions.The Born-Oppenheimer approximation (“Since the energies of the various motions arevery different, motions of a diatomic molecule may be considered as independent”)allows us to express the total molecular energy as:

Etotal ≈ Eelec + Evib + Erot

or equivalently using term values:T ≈ Te + G(v) + F(J)

where Te is the electronic energy which depends only on the electronic configurationand nuclear repulsion. A change in the total energy of a molecule (in wavenumbers) is:

Δεtotal = Δεelec + Δεvib + Δεrot cm-1

where the approximate orders of magnitude of the three quantities are:Δεelec ≈ Δεvib x 103 ≈ Δεrot x 106

Vibrational changes thus produce a “coarse structure”, and rotational changes a “finestructure” on the electronic transitions. Unlike in vibrational and rotationalspectroscopies, all molecules give electronic spectra.

Vibrational coarse structureIgnoring rotational changes, we have

Δεtotal = Δεelec + Δεvib cm-1

orεtotal = εelec + (v + 1/2)ϖe – xe(v + 1/2)2 ϖe cm-1 (v = 0, 1, 2,…)

where ϖe is the equilibrium vibrational frequency expressed in wavenumbers and xe isthe anharmonicity constant. The lower energy states are marked with a double prime,and the higher states by a single prime. The figure below shows the vibrational coarsestructure in an electronic transition from the ground state (v’ = 0), to a higher state. Byconvention, X is used to denote the electronic ground state, and A, B, C etc. the excitedstates in order of increasing energy. The electronic energy is measured from the bottomof the respective potential energy curves, while the lowest molecular energy level isoffset upwards by the zero point energy of that vibration. In general v” ≠ v’.

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Vibrational transitions occur providing the electronic transition dipole moment and theFranck-Condon factors are non-zero. In the infrared, the selection rules for v arise fromthe symmetry of the different vibrational wavefunctions, and are strictly Δv = ±1 for thesimple harmonic oscillator. For an anharmonic oscillator, this strict rule begins to breakdown and overtones are allowed, albeit with rapidly declining intensity. For vibrationaltransitions between different electronic levels, this rule breaks down completely andthere is now no longer any restriction on Δv, so that every v’ ← v” transition has someprobability, giving rise to many spectral lines. However, absorption spectra from theground state are much simpler: virtually all the molecules are in their lowest vibrationalstate (v” = 0), so that the only transitions we observe are (0, 1), (0, 2), (0, 3) etc.

Note that the upper state is given first within the brackets. Each set of transitions in aband is called a v’ progression, since the value of v’ increases by unity for each line inthe set. The lines in a band are closer together at high frequencies because of the

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anharmonicity of the upper state vibrations, which causes vibration energy levels toconverge. Note that, in general, the spacings of vibrational energy levels in the twoelectronic states are different.To quantify:

Δεtotal = Δεelec + Δεvib. = (ε’ – ε”) ++ [(v’ + 1/2)ϖe’ – xe’(v’ + 1/2)2 ϖe’ – (v” + 1/2)ϖe” – xe”(v” + 1/2)2 ϖe”]cm-1

If a sufficient number (at least five) lines can be resolved in the band, the values of xe’,ϖe’, xe”, ϖe” and (ε’ – ε”) can be calculated. Information on the excited electronic states isparticularly valuable, since such states are extremely unstable, and therefore short-lived.

Vibrational band analysis: the Deslandres tableThe Deslandres table is a convenient means of verifying vibrational assignments. Thedifferences between rows and columns should be constant, corresponding to the energydifference between a particular pair of vibrational levels in either the upper state (rows)or the lower state (columns):

A portion of the Deslandres table for the 1Π ← 1Σ electronic transition of CO is givenbelow. Vibrational wavenumbers are shown in bold, with the differences in plaincharacters.

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v” = 0 v” = 1 – 0 1 v” = 2 – 1 2 v” = 3 – 2 3v’ = 0 64748 2143 62605 2117 60488 2090 58398

v’ = 1 – 0 1480 1480 1480 14801 66228 2143 64085 2117 61968 2090 59878

v’ = 2 – 1 1440 1440 1440 14402 67668 2143 65525 2116 63409 2090 61319

v’ = 3 – 2 1402 1402 1402 14023 69070 2143 66927 2117 64810 2090 62720

Intensity of vibrational-electronic spectra: the Franck-Condon principleThe vibrational lines in a progression are not of the same intensity. Intensities of thevibrational bands are determined by three factors:

(1) The intrinsic strength of the transition (the

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˜ B coefficient).(2) The populations of the levels involved.(3) The overlap of the vibrational wavefunctions (the Franck-Condon factor).The Franck-Condon principle states: “Since electronic transitions occur very rapidly(≈ 10-15 s), vibration and rotation of the molecule do not change the internuclear distanceappreciably during the transition”. The intensity of a transition is greatest for thelargest of the vibrational wavefunctions. The figure shows the probability distributionfor a diatomic molecule. The nuclei are most likely to be found at distances apart givenby the maxima of the curve for each vibrational level.

In view of the Franck-Condon principle, electronic transitions occur “vertically” on apotential energy diagram. Consider three distinct situations.

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(1) If the internuclear distances in the upper and lower states are equal (re” ≈ re’), themost probable transition is (0, 0), as indicated by the vertical line in the figure below.However, there is a non-zero probability of (1, 0), (2, 0), (3, 0) etc. transitions. Thesuccessive lines will therefore have rapidly diminishing intensities.

(2) If the excited electronic state has a slightly larger nuclear separation than the groundstate (re” > re’), the most probable (and thus most intense) transition is (2, 0). Theintensities of the neighbouring transitions are lower.

(3) When the excited electronic state has a considerably larger nuclear separation thanthe ground state (re” >> re’), the vibrational state to which the transition takes placehas a high v’ value. Further, transitions can occur to levels where the molecule hasenergy in excess of its dissociation energy. From such states the molecule willdissociate without any vibrations and, since the fragments which are formed maytake up any value of kinetic energy, the transitions are not quantized and acontinuum results.

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The Franck-Condon factor: quantum mechanical treatmentThe Franck-Condon factor can be derived quantum mechanically from the fact that theintensity of any transition is proportional to the square of the transition dipole moment:

Mfi = ∫ Ψ*f µ Ψi dτN

where µ is the dipole moment operator and both Ψf and Ψi are functions of the electronicand vibrational states of the upper and lower levels, Ψe’v’ and Ψe”v” respectively. Therotational component of the total wavefunction is ignored at this stage.Separating the dipole moment operator into electronic and a nuclear components:

µ = µe + µN

and writing Ψe’v’ as Ψe’Ψv’ (the Born Oppenheimer approximation), the transition dipolemoment becomes:

Mfi = ∫ ∫ Ψ*e’Ψ

*v’(µe + µN)Ψe”Ψv” dτedτN

= ∫ Ψ*e’ µeΨe”dτe ∫ Ψ*

v’Ψv” dτN

+ ∫ Ψ*e’Ψe”dτe ∫ Ψ*

v’ µN Ψv” dτN

Note that, neglecting anharmonicity, vibrational wavefunctions within one electronicstate are orthogonal, but between different electronic states they are not. Since Ψ*

e’ andΨe” are distinct electronic states (which are orthogonal), the second term is zero, and thetransition dipole moment can be written as:

Mfi = Relec

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v' v"Where the electronic dipole moment Relec = ∫ Ψ*

e’ µeΨe”dτe

and

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v' v" = ∫ Ψ*v’ Ψv” dτN

The intensity of the transition is proportional to Mfi2, and thus to

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v' v" 2, which is

the Franck-Condon factor, consistent with the classical approach.

Dissociation energyThe figure below shows two ways in which electronic excitation can lead to dissociation.The figure on the left describes the case for re” >> re’. The dashed lines of the Morsecurve represent the dissociation of the normal and excited molecule, the dissociationenergies being Do’ and Do”. The total energy of the dissociation products from the upperstate is greater by an amount Eex than that of the products of dissociation in the lowerstate. This is the excitation energy of the products of dissociation. The lowerwavenumber limit of the spectral continuum must represent just sufficient energy to

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cause no more than dissociation. The dissociation products separate with zero kineticenergy, and we have:

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˜ v continuum limit = Do” + Eex cm-1

and we can therefore determine Do” if we know Eex. Thermochemical measurementsoften lead to an approximate value of Do”, and hence, since Do” + E ex is accuratelymeasured spectroscopically, an approximate value of Eex is obtained. Secondly, if morethan one spectroscopic dissociation limit is found (corresponding to dissociation intotwo or more different states of products), the separation between the excitation energiesoften corresponds with the separations between only one set of excited states of theatoms observed spectroscopically. This gives the nature and the energies of the excitedproducts.

When the internuclear distances in both states are such that transitions near to thedissociation limit are of negligible probability, no continua appear in the spectra.However, it is still possible to determine the dissociation energy by noting how thevibrational lines converge. The separation between the neighbouring vibrational levels:

Δε = εv+1 – εv = ϖe[1 – 2xe(v + 1)] cm-1

This separation decreases with increasing v, and the dissociation limit is reached whenΔε tends to zero. The maximum value of v is vmax, where

ϖe[1 – 2xe(vmax + 1)] = 0so that

vmax = (1/2xe) – 1Since xe is of the order of 10-2, vmax is about 50.

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The Birge-Sponer extrapolationThe expression for the separation of vibrational levels given above does not applyexactly at high values of v because higher order anharmonicity terms are thensignificant. However, it is still possible to obtain the dissociation energy by using theBirge-Sponer extrapolation, involving plotting Δε versus the vibrational quantumnumber v. The plot is not a straight line, but by extrapolating it to Δε = 0 we obtain vmax,while the area under the plot gives the dissociation energy directly.

To summarize, provided that transitions are assigned correctly, the two methodsavailable for the determination of the dissociation energies Do” and Do’ are:(1) The Birge-Sponer extrapolation(2) If the Morse potential is assumed to be correct, the equilibrium vibration

frequencies νe’ and ν e” and the anharmonicity constants

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˜ v exe’ and

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˜ v exe” can bedetermined for the two levels using:

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˜ v (v” = 1 ← v” = 0) =

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˜ v e” – 2

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˜ v exe”

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˜ v (v” = 2 ← v” = 1) =

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˜ v e” – 4

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˜ v exe”

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De =˜ v e

2

4 ˜ v exe

A minimum of five entries in the Deslandres table (the bordered region) is required forthe determination of the dissociation energies of both electronic states.

Rotational fine structure of electronic-vibrational transitionsIf we ignore centrifugal distortion and use the Born-Oppenheimer approximation, thetotal energy of a diatomic molecule is:

εtotal = εelec + εvib +

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˜ B J(J + 1) cm-1

while changes in the total energy are

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Δεtotal = Δ(εelec + εvib) + Δ[

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˜ B J(J + 1)] cm-1

and the wavenumber of a spectroscopic line corresponding to such a change is:

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˜ v spect. =

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˜ v v’,v” + Δ[

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˜ B J(J + 1)] cm-1

where

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˜ v v’,v” represents the wavenumber of electronic-vibrational transitions.The selection rule for J depends on the type of electronic transition which the moleculeundergoes. If both the lower and upper electronic states are 1Σ states (with noelectronic angular momentum about the internuclear axis, S = 0), the selection rule is:

ΔJ = ±1 only for 1Σ ↔ 1Σ transitionswhereas for all other transitions (provided that at least one of the states has S ≠ 0) therule is:

ΔJ = 0 or ±1In the latter case there is an added restriction that a state with J = 0 cannot undergo atransition to another J = 0 state: J = 0 ↔ J = 0 is not allowed. Once more conservation ofangular momentum can be used to explain these selection rules. However, forelectronic transitions bond orders are expected to change and thus bond lengths arelikely to be markedly different between different electronic states (e.g. CO). Thespreading out or bunching up of transitions which occurs in vibrational-rotationalspectra may thus be more marked for electronic transitions, and “band heads” nowbecome evident at low J. Frequently these are more visible in spectra than the bandorigin, resembling Q branches in IR spectra.Thus for transitions between 1Σ states, only P and R branches will appear, while forother transitions Q branches will appear in addition. We have for the difference of thetwo electronic states:

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˜ ν spect =

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˜ v v ',v" +

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˜ B ’J’(J’ + 1) –

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˜ B ”J”(J” + 1) cm-1

giving the P, R and Q branches as follows:(1) P branch: ΔJ = –1, J” = J’ + 1

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˜ ν P=

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˜ v v ',v" – (

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˜ B ’ +

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˜ B ”)(J’ + 1) + (

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˜ B ’ –

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˜ B ”) (J’ + 1)2 cm-1

where J’ = 0, 1, 2,…(2) R branch: ΔJ = +1, J’ = J” + 1

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˜ ν R=

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˜ v v ',v" + (

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˜ B ’ +

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˜ B ”)(J” + 1) + (

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˜ B ’ –

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˜ B ”) (J’ + 1)2 cm-1

where J” = 0, 1, 2,…These two equations can be combined into:

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˜ ν P,R =

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˜ v v ',v" + (

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˜ B ’ +

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˜ B ”) m + (

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˜ B ’ –

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˜ B ”) m2 cm-1

where m = ±1, ±2,… with positive m values for the R branch (ΔJ = +1) and negativevalues for the P branch (ΔJ = –1). Note that m cannot be zero (as this would correspond

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to J’ = –1 for the P branch), so that no line from the P and R branches appears at the

band origin,

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˜ v v ',v" . The figure shows the resulting spectra for

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˜ B ’ <

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˜ B ” and a 10%

difference in the magnitude of

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˜ B ’ and

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˜ B ”. P branch lines appear at the lowwavenumber side of the band origin, and the spacing between the lines increases withm. The R branch appears on the high wavenumber side, and the line spacing decreasesrapidly with m. The point at which the R branch separation decreases to zero is knownas the band head.

(3) Q branch: ΔJ = 0, J’ = J” ≠ 0 (since J = 0 ↔ J = 0 is not allowed)

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˜ ν Q=

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˜ v v ',v" + (

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˜ B ’ –

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˜ B ”)J” + (

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˜ B ’ –

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˜ B ”) J”2 cm-1

The Q branch lines lie on the low wavenumber side of the origin.

The Fortrat diagramRewrite the expressions for the P, R and Q lines with continuously variable parameters pand q:

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˜ ν P,R =

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˜ v v ',v" + (

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˜ B ’ +

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˜ B ”) p + (

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˜ B ’ –

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˜ B ”) p2

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˜ ν Q=

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˜ v v ',v" + (

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˜ B ’ –

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˜ B ”) q + (

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˜ B ’ –

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˜ B ”) q2

Each equation represents a Fortrat parabola, where p takes both positive and negativevalues, while q is always positive. The band head is clearly at the vertex of the P, Rparabola. We calculate the position of the vertex by differentiation:

d

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˜ ν P,R /dp =

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˜ B ’ +

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˜ B ” + (

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˜ B ’ –

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˜ B ”) p = 0

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or

p = – (

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˜ B ’ +

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˜ B ”)/2(

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˜ B ’ –

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˜ B ”) for band head

If

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˜ B ’ <

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˜ B ”, the band head occurs at positive p values (i.e. in the R branch). Conversely,

if

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˜ B ’ >

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˜ B ”, the band head is found at negative p values (i.e. in the P branch). A 10%

difference between

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˜ B ’ and

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˜ B ” gives p = 11.

PredissociationA phenomenon called predissociation arises when the Morse curves of a molecule intwo different excited states intersect, with one of the states being stable (with aminimum on the energy curve), and the other continuous. Suppose a transition takesplace from some lower state into the vibrational levels shown bracketed on the left.

Now if a transition takes place into the levels labelled a, b or c, a normal vibrational-electronic spectrum occurs, complete with rotational fine structure. Two such bands

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appear on the left of the spectrum below. However, for a transition into the levelslabelled d, e or f, the molecule may “cross over” onto the continuous curve anddissociate. Such transition (known as “radiationless transfer”) is faster than the timetaken by the molecule to rotate (ca. 10-10 s), but usually slower than the vibrational time(ca. 10-13 s). Thus predissociation will occur before the molecule rotates (destroying therotational fine structure), while the vibrational structure is preserved. If the cross-overis faster than the vibrational time, a complete continuum will be observed.

Electronic structure of diatomic molecules

Molecular orbital theoryAn atomic orbital is the space within which an electron belonging to the orbital spends95% of its time. The orbital ψ is described by three quantum numbers: n, l and m (or lz).The energy of the orbital also depends on electron spin. In molecular orbital theoryorbitals embrace two or more nuclei. The same rules: (1) lowest energy first; (2)maximum of two paired electrons per orbital; (3) parallel spins in degenerate orbitals,apply to the filling of molecular orbitals. In the LCAO theory, the approximate shape ofmolecular orbitals is made up by taking sums and differences of the atomic orbitals. Forthe hydrogen molecule we have the sum:

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ψH2 =ψ1s +ψ1s

This bonding orbital (called 1sσ, as it is produced from two s atomic orbitals) is a simplesymmetrical ellipsoid. It does not change sign upon inversion about the centre ofsymmetry, which is marked by the subscript g (German gerade = even). The orbital isthus known as 1sσg.The difference:

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ψH2 =ψ1s −ψ1s

In this antibonding orbital (called 1sσ∗) the charge is concentrated outside the nuclei,which repel one another. This orbital does change sign upon inversion, which ismarked by the subscript u (German ungerade = odd). It is thus known as 1sσu*.

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The energies of the two s orbitals depend on the internuclear distance as shown below.

Using the same procedure we obtain the two 2pσ orbitals by combining atomic 2pzorbitals:

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which results in bonding (2pπu) and antibonding (2pπg*) orbitals.When two 2py orbitals combine, the result is as follows:

The figure below shows the various orbitals in order of increasing energy. For lightdiatomic molecules, such as Li2, there is considerable overlap and interaction between2pπu, 2pσg, and 2sσu* orbitals. For H2 this overlap does not occur, so that the two casesare considered separately.

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Consider some specific homonuclear diatomic molecules:Hydrogen, H2. In the ground state there are two 1s electrons occupying the molecular1sσg orbital with opposing spins. Their energy is lower than in two separate atoms, sothat the molecule is stable.Helium, He2. There is a total of four electrons. Only two can go to 1sσg, the otherswould have to go to 1sσu*. However, since more energy would be absorbed by theformer than evolved by the latter, the molecule is unstable.Nitrogen, N2. Given a total of 14 electrons, omitting the asterisks, the configuration is:1sσg

21sσu22sσg

22sσu22pσg

22pπu4

1sσg2 and 1sσu

2 as well as 2sσg2 and 2sσu

2 contributions cancel each other, and we are leftwith six electrons, in bonding 2pσg and 2pπu orbitals. This results in a triple bond.Oxygen, O2. There are two more electrons than in the nitrogen molecule. The lowestavailable orbital is the antibonding 2pπg*. Two electrons in this orbital cancel thecontribution from two electrons in 2pπu, and we are left with four electrons, i.e. a doublebond. Further, since the antibonding 2pπg* orbitals are degenerate, the electrons occupyone each to satisfy electron repulsion. However, according to Hund’s rule, their spinsmust be parallel. The unpaired electrons make the oxygen molecule paramagnetic.

Electronic angular momentum in diatomic moleculesThe total energy of an electron depends mainly on its distance from the nucleus(represented by the quantum number n) and the orbital and spin angular momenta(quantum numbers l and s), and on the way these are coupled together (quantumnumber j).The same applies to electrons in molecules. However, in order to discuss thecomponents of the vector l we need to specify some reference direction called the zdirection. The angular momentum of a diatomic molecule is referenced to theinternuclear axis. Thus, l can be thought of as precessing about the internuclear axis,but with a well defined component (in units of h) along it. The direction of the angular

momentum along the internuclear axis does not affect the overall energy, so that forml > 0 there is a degeneracy of two.When referring to molecules, the equivalent Greek symbols are used, lower case forindividual electrons, upper case for the sum over the molecule. Thus, the angularmomentum of electron i along the internuclear axis is λi (i.e. σ electrons have λ = 0, πelectrons λ = 1 etc.), and the total molecular orbital angular momentum is:

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Λ = λii∑ (equivalent to the single atom case where

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L = lii∑ )

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The same argument applies to spin angular momentum, with the projection of spinangular momentum along the inter-nuclear axis denoted Σ (equivalent to S in an atom).This Σ is not to be confused with a Σ state, for which Λ = 0. Thus for a diatomicmolecule, the total angular momentum J (neglecting nuclear spin) is now the vector sumof the orbital (Λ), spin (Σ) and rotational (R) angular momentum:

Note that the rotational angular momentum is perpendicular to the inter-nuclear axis(since Iz = 0). The total angular momentum thus has a projection Ω (in units of h) along

the inter-nuclear axis. There are always 2S+1 spin components, where 2S+1 is the spinmultiplicity.Each electronic state can thus be described by a term symbol of the form:

(spin multiplicity)[Component of L](net angular momentum component)

or 2S+1ΛΩNO, which in its ground state has an electronic configuration …(2pσ)2(2pπ)4(2pπ∗), has asingle unpaired electron. Its spin multiplicity is thus 2 (a doublet state), and it has totalorbital angular momentum of 1 (an unpaired π electron). Its term symbol is thus 2Π, ormore completely 2Π1/2 or 2Π3/2, depending on whether the spin and orbital angularmomentum components are parallel or anti-parallel. Clearly NO may also possess

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rotational angular momentum, but this does not affect the term symbol which refers tothe electrons alone.Two additional subscripts and superscripts are used in term symbols. The parityindicates whether a wavefunction has the same sign (g) or the opposite sign (u) underthe process of inversion. Heteronuclear molecules do not have centres of inversion andthus g and u are not applicable. The second parameter indicates the symmetry of theelectronic wavefunction with respect to reflection in a plane containing the nuclei, andis applied only to Σ wavefunctions. If an electronic orbital wavefunction is symmetricwith respect to reflection (Σ+) it must, by the Pauli principle, be associated with an anti-symmetric electron spin wavefunction, and vice-versa.For example, the ground state electronic configuration of O2 is …(2pσg)2(2pπu)4(2pπ∗g)2.Hund’s rules indicate that the lowest energy state is the triplet (which is symmetric in

spin), and the term symbol is thus 3

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Σg−.

Labelling of electronic states in atoms and moleculesorbital momentum spin

momentumtotal

momentumatoms

single electron l (symbol s, p, d for l = 0, 1, 2,…) s jsingle electron(z-component) lz sz jz

several electrons L (symbol S, P, D for L = 0, 1, 2,…) S Jseveral electrons(z-component) Lz Sz Jz

moleculessingle electron l s ja

single electron(axial component)

λ (symbol σ, π, δ for λ = 0, 1, 2,…) σ ω

several electrons L S Ja

several electrons(axial component)

Λ (symbol Σ, Π, Δ for Λ = 0, 1, 2,…) Σ Ω

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The spectrum of molecular hydrogenThere are two electrons. We thus expect to find singlet and triplet states, depending onwhether the electron spins are paired or parallel. In the ground state both electrons arein the same 1sσg orbital, and must form a singlet state. We have λ1 = λ2 = 0 and Λ = 0:the state is thus 1Σ. We also specify the orbital geometry: since both electrons are in thesame orbital, we write 1Σg. Finally, since the wavefunction of the electron is unchangedupon reflection in the plane of symmetry, the full term symbol for the ground state of H2

is (1sσg)2 1Σg+.

We consider the three possible excited states: (1sσg2sσg), (1sσg2pσg) and (1sσg2pπu).

(1sσg2sσg). Since both electrons are σ electrons, Λ = λ1 + λ2 = 0. Since we are consideringsinglet states, S = 0. Both constituting orbitals are even and symmetrical, so we have(1sσg2sσg) 1Σg

+.

(1sσg2pσg). We have again a 1Σ state, since both electrons are σ, but the overall state isnow odd (a combination of an electron arising from an even 1s state with an electronfrom an odd p state gives and overall odd state). Thus (1sσg2pσg) 1Σu

+.

(1sσg2pπu). Now Λ = λ1 + λ2 = 1, since one electron is in a π state and one originates froma 2p orbital, the overall state is (1sσg2pπu) 1Πu.

The energies of the three states are 1Σu+ < 1Πu < 1Σg

+.Similar states are obtained by excitation to 3s and 3p states, 4s and 4p states, etc. Inaddition, for n = 3, 4,… electrons can also be excited to the nd orbital. In this case wehave for the energies: (1sσndδ) 1Σg

+ < (1sσndπ) 1Πg < (1sσndδ) 1Δg.

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Transitions between the various energy levels follow the selection rules:

1. ΔΛ = 0, ±1This is rationalised in terms of conservation of angular momentum of molecule plusphoton which has a spin angular momentum of h. For ΔΛ = 0, rotational angular

momentum changes by one (ΔJ = ± 1). Thus transitions Σ ↔ Σ, Σ ↔ Π and Π ↔ Π,etc. are allowed, but Σ ↔ Δ, for example, is not allowed.2. ΔS = 0.Transitions which change the multiplicity are weak for molecules formed of light atoms,where spin-orbit coupling is small. For heavier atoms, spin-orbit coupling effects arelarger, and transitions for which ΔS ≠ 0 can be seen.3. ΔΩ = 0, ±1.Σ+ states can go only to into other Σ+ states (or to Π states), while Σ– can go only to Σ– (orΠ). Thus Σ+ ↔ Σ+ and Σ– ↔ Σ–

g ↔u (where applicable).

Consider now some triplet states of molecular hydrogen (S = 1). Both electrons occupythe same orbital, so that the state of the lowest energy will be either (1sσg2sσg),(1sσg2pσg) or (1sσg2pπu). The first two are 3Σ states, and the third is 3Π. Following theabove selection rules, the order of the energies will be:

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(1sσg2pσg) 3Σu+ < (1sσg2pπu) 3Πu < (1sσg2sσg) 3Σg

+.Note that, because of the ΔS = 0 selection rule, transitions between singlet and tripletstates are not allowed.For CO, the allowed electronic electric dipole transitions are:

The O2 molecule

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The only allowed electronic electric dipole transition from the ground state is B ← X.As O2 is homonuclear, no vibration-rotation spectrum would be expected within the Xstate. Thus O2 would not be expected to absorb at wavelengths longer than ~ 200 nm.Given the shapes of the two potential energy curves, this absorption would be expectedto be very temperature dependent (stronger at higher temperatures).Hückel molecular orbital theory has been used to obtain a semi-quantitative picture ofthe π-electronic structure of a number of organic molecules. Radiative transitions mayoccur between these different electronic states.

The electronic structure of CH2O

The non-bonding state, n, corresponds to the lone pairs of electrons on the oxygen atom.In its electronic ground state the HOMO (the X state) is the n-state. Promotion ofelectrons leads to A and a states etc. The lowest energy electron promotion is to the π*state, leading to a π* ← n transition, which occurs at ~290 nm. The other transitionsshown occur at shorter wavelengths (higher energies), e.g. the σ* ← n occurs at 190 nm.

Change of molecular shape on absorptionConsider a hydride H2A, where A is a polyvalent atom and the H–A bonds are at 90o toone another. Two of the p orbitals of A are involved, leaving the third (“non-bonding”)orbital unaffected. We label the bonding orbitals as a1 and a2. In the linear molecule weencounter “orbital hybridization”, whereby atom A mixes its s orbital with one of its porbitals, forming two strong-bonding sp hybrid orbitals. In this configuration there are

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now two non-bonding p orbitals, N1 and N 2, and two bonding orbitals formed byoverlap between sp hybrids and hydrogen 1s, called b1 and b2. Given that the energy ofa non-bonding orbital is higher than that of a bonding orbital and that the sp orbital islower in energy than a p bonding orbital, we plot the energy changes for a smoothtransition from 90o to 180o bonding.

We see that (1) the energy of N1 remains constant; (2) the energy of the bonding orbitala1 decreases as it passes over into the stronger b1 orbital; (3) bonding orbital a2 becomesnon-bonding N2, thus increasing in energy; (4) bonding orbitals b1 and b2 are formed byabsorption of the non-bonding s into a1; (5) 90o to 180o represent maxima and minima onthe energy curves.Consider BeH2. Be has the electronic ground state configuration 1s22s2, with two outerbonding electrons. Each hydrogen atom contributes a further electron, so that BeH2 hasfour electrons to place into molecular orbitals. The most stable state will be for twoelectrons to go into b1 and two into b2, thus producing a linear molecule. However,when the molecule is electronically excited, the next available orbitals for the excitedelectron are N2 or N1. With a configuration b12b2

1N21 the most stable state will be at a

bond angle α. Thus the excited state is bent.Linear molecules such as CO2 and HC≡CH become bent on excitation.

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Electronic spectra of polyatomic moleculesElectronic spectra of polyatomic molecules are very complex and difficult to interpret.However, the vibrational frequencies of a particular atomic grouping within a molecule,such as CH3, C=O or C=C, are fairly insensitive to the nature of the rest of the molecule,as are the bond length and dissociation energy. We may therefore, as an approximation,discuss the spectrum of each bond in isolation. Such bonds are said to be “localized”. Ifwe recognize the lines from such a bond in the spectrum, we may assume that itsvibrational frequency gives a good estimate of the dissociation energy of the molecule.We note that molecules can be investigated spectroscopically in their very short-livedexcited states.Conjugated molecules are those in which there is an alternation of single and doublebonds along a chain of carbon atoms. The π molecular orbital energy level diagrams ofconjugated molecules can be constructed using a set of approximations suggested byHückel in 1931. In this approach, the π orbitals are treated separately from the σorbitals, and the latter form a rigid framework which determines the general shape ofthe molecule. All the carbon atoms are treated identically.The effect of bond conjugation is to shift the π* ← n absorption to lower energies. Whentwo isolated double bonds are brought into conjugation, both levels are shifted to givebonding and anti-bonding orbitals:

The effect is that the LUMO π* orbital is shifted downwards, and the π* ← n absorptionshifts to lower energy. Repeated conjugation increases this effect. When sufficientconjugation is present, the energy of the LUMO π* orbital is lowered to the point whereν2 corresponds to the visible region of the spectrum (~500 nm rather than 290 nm), afeature which is exploited in tuneable dye lasers.

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Absorption by a C=C double bond excites a π electron into an antibonding π* orbital.The chromophore activity is due to an π*← π transition, corresponding to absorption at180 nm. When the double bond is part of a conjugated chain, the energies of themolecular orbitals lie closer together, and the π*← π transition moves to longerwavelengths. It may even lie in the visible region if the conjugated system is longenough.

An important example is the photochemical mechanism of vision. The retina of the eyecontains “visual purple”, a combination of a protein with 11-cis-retinal, which acts as achromophore and is a receptor of photons entering the eye. 11-cis-retinal itself absorbsat 380 nm, but in combination with the protein the absorption maximum shifts to ca. 500nm and tails into the blue. The conjugated double bonds are responsible for the abilityof the molecule to absorb over the entire visible region. They play another importantrole: in its electronically excited state the conjugated chain can isomerize by twistingabout an excited C=C bond, forming 11-trans-retinal. The primary step in vision isphoton absorption followed by isomerization, which triggers a nerve impulse to thebrain.

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Re-emission of energy by an excited moleculeFollowing the absorption of a photon, the processes which may occur are:

Dissociation. The excited molecule breaks into two fragments.Re-emission. This is simply a reverse of absorption.Fluorescence. The spontaneously emitted fluorescent radiation ceases immediatelyafter the exciting radiation is switched off. The initial radiation takes the molecule intoan excited electronic state. The excited molecule gives up its energy non-radiatively viacollisions with the surrounding molecules, thus going down the vibrational levels to thelowest vibrational level of the electronically excited state. At that point, thesurrounding molecules may be unable to accept the larger energy difference needed tolower the molecule to the ground electronic state. They may survive long enough to

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undergo spontaneous emission at a larger wavelength (because of the energy lostduring the collisions). The fluorescence spectrum has a vibrational structurecharacteristic of the lower electronic state.

Phosphorescence. The initial radiation takes the molecule into an excited electronicstate. Because the energy curves of the singlet and triplet state cross one another, themolecule may undergo intersystem crossing and become a triplet state. This happenswhen the molecule contains a heavier atom, such as S, because the spin-orbit coupling isthen large. The molecule deposits its energy via collisions with the surroundingmolecules, but becomes trapped at the lowest vibrational energy level of the triplet. Itcannot radiate its energy, because return to the ground state is spin-forbidden.However, the radiative transition is not totally spin-forbidden because of the spin-orbitcoupling, so that the molecule is able to radiate weakly.

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Phosphorescence proceeds on a much longer time scale than fluorescence.