Determining the ormal Modes of Vibration Introduction at the end of last lecture you determined the symmetry and activity of the vibrational modes of ammonia Γ vib 3 ) = A 1 IR, pol) + EIR,depol) the vibrational modes or normal modes of each symmetry are shown below, Figure 1 ν 1 A 1 ) symmetric stretch ν A 1 ) symmetric bend ν 3a E) degenerate stretch ν 4a E) degenerate bend Figure 1 Vibrational modes traditionally a full normal mode analysis was very difficult, and simplified bench-top techniques would used to gain an understanding of the vibrations. For quick access to a subset of the vibrations the bench-top technique is still very useful computational chemistry packages are now widely available and it is routine to carry out an optimisation and frequency analysis to determine the normal modes and their intensities today we will cover both bench-top and computational techniques for determining the normal modes I will not be covering the mathematics of determining normal vibrations in detail, however I think) this is a fascinating area! The bench-top process of determining the normal vibrations The process is as follows: 1. define ALL symmetry elements, identify all symmetry operators. treat stretches, in-plane bends and out-of-plane bends separately 3. define vectors for each motion 4. determine the reducible representation for each set 5. reduce to the irreducible representation USE short-cuts!) 6. eliminate redundant components 7. use the projection operator to determine the form of the modes 8. add additional movements to ensure the CoM remains stationary Define ALL symmetry elements, identify all symmetry operations identify and label all the symmetry elements z z) y σ v 3) σ v ) σ v 1)=σ v z) E 3σ v A 1 1 1 1 A 1 1 1 E 1 Figure Defining symmetry elements and - stretching vectors unt / Lecture 3 1
IMPORTAT Determine the Form of the Vibrational Modes: Stretches First take a vector for each type of bond o use single arrowheads for stretches o define them relative to the symmetry elements and individually label them o here s i stands for bond stretch vector, Figure 3 determine the reducible representation, Figure 3 o how many vectors remain unmoved for each operation? o because we are using vectors we don t use the per atom table o go straight to determining the reducible representation reduce to the irreducible representation components o stretches will always have a totally symmetric mode o use short-cuts! stretching modes do not usually have redundant components In Class Activity complete the reduction table for the reducible representation of the stretches and then use the reduction formula to reduce the representation to its IR components E 3σ v Γs ) A 1 Γs ) A 1 Γs ) = Figure 3 Setup for determining the symmetry of the stretching modes The Projection Operator use the projection operator to determine the form of the vibrations. o you must be able to write this equation and define each of the symbols. h Q f The Projection Operator P Γ [ f ] = 1 h Q χ IR Q) Q[ f ] = number of operations in the group = a particular symmetry operation [ ] = operate on a function χ IR Q) = the character of the Irreducible Representation under Q Equation 1: The reduction formula o the reduction formula produced a number n IR ), while the projection operator produces a function the vibrational modes). An operator always acts on something, normally an atomic orbital function, or a vibrational basis function, hence the symbol is in square brackets [ f ] which means "operate on". unt / Lecture 3
this is most easily carried out using a projection table o each symmetry operation must be individually identified o until now we have not differentiated between operations that have been grouped together. For eample, we have just used " " however now we will need to compute the effect of each operation 1 and eplicitly. o this is why there is no "k" in the projection operator equation! o now it becomes obvious why each individual symmetry element had to be identified when setting the problem up E 3σ v σ v 3) σ v ) σ v 1) E 1 σ v 1) σ v ) σ v 3) Figure 4 Defining stretches and symmetry elements fill in the projection table o pick one of the basis vibrational vectors ) and work out how this function transforms under the symmetry operations o I've shown the first few for the E and the two operations in Figure 5 and filled in the start of the projection table in Figure 6 E 1 s Figure 5 Using the projection operator In Class Activity complete the projection table for the - stretches of ammonia E 1 σ v 1) σ v ) σ v 3) Q[ ] Figure 6 Filling in the projection table unt / Lecture 3 3
Form of the Vibrational Wavefunction or ormal Modes the full formula is P Γ [ f ] = 1 χ IR Q) Q[ f ] so far we have computed h Q f Q [ ] now we need to include the irreducible representation component χ IR Q). From our vibrational analysis we know the IR components are A 1 and E o we treat A 1 first, and the table is completed as shown in Figure 7 E 1 σ v 1) σ v ) σ v 3) Q[ ] A 1 1 1 1 1 1 1 χ A 1 Q)iQ[ ] Figure 7 Projection table for the A 1 vibration o finally we take care of the summation over Q the symmetry operations) in the equation to determine the final vector) form of the vibration, Figure 8 ν A 1 ) = 1 [ 6 + + + + + ] = 1 [ 6 + + ] == 1 [ 3 + + ] A 1 Figure 8 Finding the A 1 vibrational vector function draw out the vibrational mode o add vectors to eliminate the CoM motion, in this case an arrow is added to the atom to stop the whole molecule moving down. o the A 1 mode is a totally symmetric stretch we also know that there is a degenerate E vibration o degenerate means there are two vibrations with eactly the same energy, they give only one peak in the spectrum but with double the intensity!) o finding the first E mode is easy, finding the second requires generating an orthogonal function which is beyond the scope of this course In Class Activity find the equation and draw the vibration for one degenerate E mode E 1 σ v 1) σ v ) σ v 3) Q[ ] E χ E Q)iQ[ ] ν E) = Figure 9 Final stretching modes o the E mode is an asymmetric stretch unt / Lecture 3 4 E
IMPORTAT Consider the Breathing Motions 3 is not planar so there are no "in-plane" modes, however we can consider the modes that would be in-plane if 3 were planar, these are sometimes called the breathing motions the vector used for breathing motions has double arrowheads define the breathing mode vectors on a diagram, Figure 1 o these are represented as an epansion and contraction of the angles involved determine the reducible representation, and reduce it a' 1 a' 3 a' E 3σ v Γa ) 3 1 ν E) = 1 [ 6 a a a 1 3] Figure 1 Setup for the breathing modes o it is clear the breathing modes have the same reducible representation as the stretches, and hence will span the same IR Γa ) = A 1 + E eliminate redundant modes o the A 1 mode is redundant because if the molecule was planar all of the - - angles cannot close at once, this leaves only the E mode o this is a common feature of breathing modes; the totally symmetric mode is not allowed for physical reasons use the projection operator o the breathing vibrations have the same reducible representation and distribution of vectors as the stretching motions and therefore they will have similar functions, thus it is not necessary to repeat the projection process and we can write the equation for the E mode directly, Figure 1 sketch the wavefunction o the final sketch should contain only single headed arrows, Figure 11 o the form of these modes can be difficult to "sketch" from the projector method, thus highlighting a short coming of this method o from the equation we know a 1 is epanding doubling) and a and a 3 are contracting o * feels same forces on both sides and remains stationary, the other 's are moving breathing) in and out. o then to stop the whole molecule translating the and * must move. a 3 contracts * a a 1 contracts epands a 3 * a a 1 Figure 11 Final form of the breathing modes E * unt / Lecture 3 5
IMPORTAT Consider the Umbrella Motions the vector used for out-of-plane motions has single arrowheads define the breathing mode vectors on a diagram, Figure 1 o again we determine the reducible representation, and reduce it b 3 b 1 b - out-of-plane bends E 3σ v Γb X ) 3 1 Figure 1 Setup for the umbrella modes o it is clear that the umbrella vectors have the same reducible representation as the stretches, and hence will span the same IR Γb X ) = A 1 + E eliminate redundant modes o determine the allowed IR of the umbrella mode by considering the modes already identified o only two E modes are allowed, and we have already found both of them one is a stretch and the other an "in-plane" or breathing bend). o thus the active mode must be the A 1 symmetric bend. use the projection operator o because of the similarities with the breathing modes and stretches the form of this mode can be determined directly, Figure 13 Figure 13 Final form of the umbrella modes Complete the Assignment we have found the symmetry and vectors for all vibrational modes of 3 A 1 ν OOP bend A 1 ) = 1 [ 3 b 1 + b + b 3 ] ν 1 A 1 ) symmetric stretch ν A 1 ) symmetric bend ν 3a E) degenerate stretch Figure 14 Vibrational Modes ν 4a E) degenerate bend unt / Lecture 3 6
IMPORTAT we can add further detail to our interpretation of an eperimental spectrum or prediction of a spectrum yet to be obtained generally when assigning a spectrum: o stretches occur at higher wavenumbers than bends why!) o asymmetric modes generally occur at higher energies than symmetric modes but not always!) o group modes, ie those that belong to a functional group remain similar to those of the isolated molecule due to the local symmetry o if the symmetry is strongly broken, or the symmetry is lower more modes appear in the spectrum o rule of mutual eclusion molecules which have a centre of inversion have no coincidences between the infrared and Raman modes all of the information obtained can be summarised in a table and diagram Infrared Activity Symmetry Mode matri, cm -1 ) 3378 IR, depol 33 IR, pol 1646 IR, depol 16 IR, pol Table 1 if you are asked to describe the spectrum of 3 the following is an eample of the kind of information that is epected: o there are 6 possible vibrations for 3, all vibrations are present in both the Raman and Infrared spectra. The IR spectrum will contain 4 peak of which will be due to degenerate modes, and the Raman spectrum will also contain 4 peak of which will be polarised and the other two due to degenerate modes: Γ vib 3 ) = A 1 IR, pol) + EIR,depol) IMPORTAT IMPORTAT o the two highest wavenumber modes belong to the - stretches and will have A l and E symmetry, the A 1 mode will be Raman polarised o the two lowest wavenumber modes will belong to bends. The umbrella motion is totally symmetric and thus will be polarised in the Raman spectrum. The breathing motion is a degenerate asymmetric bend of E symmetry. a common error is to confuse the degenerate modes, these contribute 1 peak in the eperimental spectrum because the vibrations occur at the same energy, however there are two separate normal modes, and both count towards the 3-6 modes! when you are asked to "Assign the spectrum of X" you are epected to o determine the symmetry and activity of all vibrational modes Γ vib X) = n 1 Γ 1 IR?, pol?) + n Γ IR?, pol?) +! o and describe the epected spectrum as shown in the eample above) unt / Lecture 3 7
Practice! o you will have a problems class deriving the vibrational assignment and vectors of XeO)F 4 for further practice o the questions at the end of Chapter 5 in Group Theory for Chemists provide some nice eamples to practice on o one of your labs requires you to undertake a vibrational analysis o I give an additional problem for you to solve in the self-study questions at the end of these notes. The Schrödinger Equation a common area of confusion for students is in understanding when we are using the Schrödinger equation to solve for electronic wavefunctions MOs) and when we are solving the Schrödinger equation for nuclear solutions vibrations). the total Schrödinger equation includes translation, rotation, vibration and electronic components it actually includes more but this would be a very advanced QM course!) we assume each of these components can be solved separately total Ψ total = E total Ψ total Ψ total =ψ el ψ vib ψ rot ψ trans total = el + vib + rot + trans el ψ el = ε el ψ el vib ψ vib = ε vib ψ vib rot ψ rot = ε rot ψ rot trans ψ trans = ε trans ψ trans Equation components of the Schrödinger equaton translation is the movement of the CoM of the molecule in space, this can be free translation, Equation 3. trans =! d m d χ n = Ae ik + Be ik k = me! Equation 3 CoM components χ n = C cos k + Dsin k o A, B, C and D depend on the eact conditions the particle eperiences. In chemistry we generally assume the CoM is of the molecule is stationary and we ignore this part of the equation. The individual atoms can still translate but this now becomes part of the vibrational amiltonian o this is not a QM course so I m not going to discuss the CoM solutions any further the rotational part of the equation needs to include rotation of the nuclei and rotation of the electrons. Rotation of the electrons is included in the electronic amiltonian covered with Kim last term!). o Rotation of the nuclei has the following amiltonian Equation 4) where I q is the moment of inertia and J q is the angular momentum around each ais J q =I qq ω q. The solutions will involve spherical harmonic functions rot ψ rot θ,φ) = ε rot ψ rot θ,φ) Equation 4 rotational components rot = J + J y + J z I I yy I zz unt / Lecture 3 8
o there is a whole area of QM and spectroscopy devoted to rotational motions, particularly for diatomic molecules. Our time is limited and I have chosen not to cover this topic, if you are interested look up the rigid rotor and then eplore further! o ultimately rotational motion has an impact on Raman spectroscopy, as molecules that are not spherical ehibit anisotropic electronic polarizability and can ehibit rotational fine structure in their Raman spectra see the Raman spectrum of 3 from last lecture) net is the electronic part of the Schrödinger equation, Equation 5. This has a radial component dependent on L an associated Laguerre polynomial and an eponential. And an angular component, Φ lml φ) are the particle on a ring solutions covered by Kim last term and Θ lml θ) are associated Legendre functions or spherical harmonics also introduced by Kim last term. el = T e + V ne + V ee ψ el = R n r)y l,ml θ,φ)s s,ms σ ) Z R n r) = na Y l,ml θ,φ) = Θ lml θ)φ lml φ) 3 n l 1)! n[n + l)!] 3 ρ l l L _1 n+1 ρ)e ρ ρ = Z na)r a = 4πε! µe Φ lml φ) = Ae im lφ + Be im lφ Equation 5 electronic components m l = last but not least we come to the nuclear vibrational motions which involve movement of the nuclei, Equation 6 o bonds and angles have a harmonic potential, and this is called the harmonic oscillator O) form of the Schrödinger equation. The nuclear wavefunctions are rather comple equations, dependent on ermite polynomials and an eponential. Solving these equations is interesting, but part of an advanced QM course! o notice that this amiltonian has a kinetic energy term the first term) and a potential energy term the second term) O =! d m d + 1 k le! χ n = v v y )ep y y = α α =! mk 1 4 IMPORTAT Equation 6 vibrational components the important point to remember here is that we have nuclear χ) and electronic ψ) wavefunctions! Don t confuse the two. Computing the ormal Modes we are interested in the harmonic oscillator equation for the nuclear or vibrational motions, where did this potential 1/k ) in O come from? o a Taylor epansion is commonly used in physics, quantum mechanics and physical chemistry, it epands a function f) in terms of the value of f at =, the slope at = and the curvature at = and so on Figure 15 f ) = f ) + f!) + 1 f!!) +!! Equation 7 Taylor epansion unt / Lecture 3 9
f) f) f) f") f) f') Figure 15 Taylor epansion o Consider a Taylor epansion about the equilibrium position =) V) = V)! + dv d + 1 d V = d " $ # $ = Equation 8 O Taylor epansion + 1 d 3 V 3! d 3 3 +& " $$ # $$ o we set V)= vertical alignment) o at equilibrium the first derivative is zero slope is zero) o third order and higher terms are assumed to be very small and are ignored o this leaves only the second term and we associate the second derivative or curvature of the potential energy with the force constant k: V) = 1! d V $ " # d & V) = 1! k k = d V $ " # d & f)=f) f)=f)+f') f)=f)+f')+f")/ = Equation 9 so what does this mean? for eample a large force constant indicates a strong bond, and to stretch or contract a strong bond takes a large amount of energy, Figure 16 ER) large k R =steep curve R Figure 16 Force constants and curvature o however the above is a one-dimensional system, in a polyatomic molecule we have more than one bond distance and we have coordinates in 3 dimensions, y and z), where is the number of atoms " V V = V) + # $ i & ' i + 1 " V i $ # i ' i j +! i, j j & small k R =flat curve Equation 1 multidimensional O Taylor epansion R unt / Lecture 3 1
o we still have that V), the first derivative and the 3 rd and higher order terms are zero o the second term now is the generalised force constant k ij, and when all these terms are put together in a matri formulation, the matri is called the essian. V = 1 # k ij i j k ij = V & i, j $ i j ' Equation 11 o what this means is that vibrational motions within the molecule are coupled, when an atom is displaced via one vibration, this may influence the forces eperienced by another atom undergoing a different vibration. o ideally we would like to simplify the problem and decouple these "raw" motions, into normal coordinates or normal modes o first we transform to mass-weighted-coordinates q i = m i i V = 1 # V & K ij q i q j K ij = i, j $ q i q j ' = T + V T = 1 m i! i = 1 i i!q i Equation 1 o then we find those coordinates that diagonalise the generalised force constant matri, these are the normal coordinates or the vibrational modes as you know them, the vector diagrams we have derived are just pictorial representations of the vibrational modes, Q i. o thus in computing the normal coordinates or vibrational modes) we, or rather the computer program has diagonalised a large essian matri. # $ k 11 k 11! k 13 6) & k 1 k " ' k 3 6)1 " k 11 = V # $ 1 1 & ' # V & κ 11 = $ Q 1 Q 1 ' # κ 11! & κ " # $ κ 3 6)3 6) ' " V k 1 = # $ 1 & ' # V & κ 1 = $ Q 1 Q ' etc = etc Equation 13 multidimensional O Taylor epansion thus a spectrometer is not necessarily required, spectral information can also be obtained from a quantum chemical calculation, see the insert below which is the output from a calculation, Figure 17 o Frequencies tells us the wavenumber of the vibration o vibrations are listed in order of wavenumber and not the assignment number! o IR Inten indicates the intensity of the vibration: highest energy vibrations no. 4-6) are weakest, lowest energy vibrations no. 1-3) are strongest unt / Lecture 3 11
o the,y and z coordinates represent the displacement of that atom from its equilibrium position during the vibration o for eample the first vibration of A 1 symmetry ehibits large z-coordinate out-of-plane motion, this is the umbrella mode A 1 o one of the best advantages of carrying out calculations is that you can animate the vibrations home work!) Figure 17 Calculation output Key Points be able to determine vibrational modes using the bench-top projection method be able to use this information to assign and describe a spectrum be able to eplain the separation of the Schrödinger equation into different components for the total wavefunction and amiltonian. be able to write down the vibrational amiltonian and to be able describe, employing appropriate equations, how normal modes are computed be able to understand and interpret the output of a vibrational calculation Problems use the department computers and carry out an optimisation and then frequency analysis of the 3 molecule. Animate the vibrations. I suggest you use a 6-31G basis set, and submit the job directly to any computer ie do not use the hpc or scan portal). The job should take <3s to run. determine the symmetry and activity of the vibrational modes of cis F and then derive the stretching vibrations using the projector method. Confirm your answer by carrying out a frequency calculation, analyse and interpret your computational output. The eample from lecture one stated that we could determine which isomer cis or trans) of Pd 3 ) Cl was present from the IR spectrum. Prove this by determining the contribution to the spectrum for the Pd-Cl stretching modes for both the cis and trans complees. unt / Lecture 3 1