Symmetry Group Theoryp 4up

Symmetry and Group Theoryfor Computational Chemistry

ApplicationsChemistry 5325/5326

Angelo R. RossiDepartment of Chemistry

The University of Connecticut

[email protected] - February, 2012

Symmetry and Group Theory

Symmetry Selection Rulesfor Infrared and Raman Spectra

January - February, 2012 2

Infrared SpectraA Fundamental Transition consists of a transition from a molecule in a vibrational ground state (initialvibrational state wave function, ψi) to a vibrationally excited state (final vibrational state wave function, ψf )where the molecule absorbs one quantum of energy in one vibrational mode.

A vibrational transition in the infrared occurs when the molecular dipole moment (µ) interacts with incidentradiation which occurs with a probability which is proportional to the transition moment:

∫ψiµψfdτ

• A transition is said to be forbidden in the infrared if the value of this integral is zero because theprobability of that transition is zero and no absorption will be observed.

• The integral will be zero unless the direct product of ψiµψf contains the totally symmetricrepresentation which has the character +1 for all symmetry operations for the molecule underconsideration.

The vector µ can be split into three components, µx, µy , and µz along the Cartesian coordinate axes, andonly one of the three integrals needs to be non-zero:

∫ψi

µxµyµz

ψfdτ




Infrared Spectra

The vibrational ground state wave function, ψi belongs to the totally symmetricrepresentation.

The symmetry properties of the components of the dipole moment (µx, µy, µz) are thesame as those of the translation vectors along the same axes: Tx, Ty, TzTx, Ty, TzTx, Ty, Tz.

The symmetry of the vibrationally excited state wave function, ψf are the same as thesymmetry which describes the vibrational mode.

Therefore, it is necessary to form direct products of the totally symmetricrepresentation of ground state vibrational wave function (ψi), the irreduciblerepresentations of each of the translation vectors (Tx, Ty, TzTx, Ty, TzTx, Ty, Tz), and the irreduciblerepresentation of the excited vibration under consideration.




Infrared SpectraConsider the vibrations of the tetrahedral molecule, ruthenium tetroxide (Td symmetry),

where there are vibrations of A1 , E, and T2 , and deduce the infrared activity of each of them.

1. ψi has A1 symmetry.

2. The character table for Td shows that Tx, Ty, TzTx, Ty , TzTx, Ty, Tz together have T2 symmetry.

The direct products are then

A1 vibration: A1 ⊗ T2 ⊗ A1 = T2E vibration: A1 ⊗ T2 ⊗ E = T2 ⊗ E = T1 + T2T2 vibration: A1 ⊗ T2 ⊗ T2 = T2 ⊗ T2 = A1 + E + T1 + T2

Thus, the T2 vibrations are infrared active because the direct products produce an A1 representation, but the E and T2 vibrations will notappear in the infrared spectrum.

An important result of the above analysis is that if an excited vibrational mode has the same symmetry as the translation vectors, Tx, Ty, TzTx, Ty, TzTx, Ty , Tz ,for that point group, then the totally symmetric irreducible representation is present and a transition from the vibrational ground state to thatexcited vibrational mode will be infrared active.




Raman SpectraThe probability of a vibrational transition occurring in Raman scattering is proportional to:

∫ψiαψfdτ

where α is the polarizability of the molecule.The Raman effect depends on a molecular dipole induced by the electromagnetic field of the incidentradiation and is proportional to the polarizability of the molecule which is a measure of the ease with whichthe molecular electron distribution can be distorted.α is a tensor, i.e. a 3 x 3 array of components

αx2 αxy αxzαyx αy2 αyzαzx αzy αz2

so there will be six distinct components

∫ψi

αx2αy2αz2αxyαyzαzx

ψfdτ

where one non-zero integral is needed to have an allowed Raman transition.




Raman SpectraFor the ruthenium tetroxide molecule with Td symmetry, the components of polarizability have the following symmetries:

A1 x2 + y2 + z2

E 2z2 − x2 − y2, x2 − y2

T2 xy, yz, zx

The vibrations are A1 , E, and T2 , and it is possible to deduce the infrared activity of each of them. The direct products are then

A1 ⊗

A1E

T2

⊗ A1 = A1, E, T2; A1 vibration is possible.

A1 ⊗

A1E

T2

⊗ E = E, (A1 + A2 + E), (T1 + T2); E vibrations are possible.

A1 ⊗

A1E

T2

⊗ T2 = T2, (T1 + T2), (A1 + E + T1 + T2); T2 vibrations are possible.

Thus, all of the vibrations in the RuO4 molecule are Raman active.

A summary of the infrared and Raman activity is given as

A1 : x2 + y2 + z2 Raman OnlyE: 2z2 − x2 − y2, x2 − y2 Raman OnlyT2 : (Tx, Ty , TzTx, Ty , TzTx, Ty, Tz), (xy, yz, zx) Infrared and Raman Active


The Infrared Spectrum of Pd(NH 3)2Cl2


One can determine which isomer (cis or trans) is present in a sample from the IR spectrum by determiningthe contribution to the spectrum of the Pd-Cl stretching modes for both the cis and trans complexes.

The trans isomer exhibits a single Pd-Cl stretching vibration (νPd-Cl) around 350 cm−1, while the cis isomerexhibits two stretching modes.

M-X VibrationsIR Raman

trans isomer, D2h B2u Au

cis isomer, C2v A1, B2 A1, B2

Active M-X Stretching Modes for ML 2X2 Complexes

IR Spectrum of cis and trans Pd(NH3)2Cl2 from Kazauo Nakamoto, Infrared and Raman Spectra of Inorganic and Coordination Compounds,5th Edition (1997), Part B, p. 10, Fig. III-5, John Wiley & Sons, New York


The Infrared Spectrum of trans Pd(NH3)2Cl2


Determine the contribution of the Pd-Cl stretching modes in the trans complex.

D2h E C2(z) C2(y) C2(x) i σ(xy) σ(xz) σ(yz)

ΓvibPd−Cl 2 0 0 2 0 2 2 0

ΓvibPd−Cl = Ag +B3u

• IR active modes have the same symmetry as translational vectors (Γ(IR) = B1u, B2u.B3u) whileRaman modes have the same symmetry as binary functions (Γ(Raman) = Ag , B1g , B2g , B3g).

• Trans Pd(NH3)2Cl2 has a center of symmetry, and the rule of mutual exclusion indicates that novibrational modes will be present in both the Raman and IR spectra.

• Thus, there should be one polarised active Raman mode and one active infrared mode ΓvibPd−Cl =Ag(polarised) +B3u(IR)


The Infrared Spectrum of cis Pd(NH3)2Cl2


Determine the contributions to Pd-Cl stretching modes for the cis Pd(NH3)2Cl2 complex.

C2v E C2 σ(xz) σ′(yz)ΓvibPd−Cl 2 0 0 2

ΓvibPd−Cl = A1 +B2

• IR active modes have the same symmetry as translational vectors (Γ(IR) = A1, B1.B2) whileRaman modes have the same symmetry as binary functions (Γ(Raman) = A1, A2, B1, B2).

• Two infrared active modes and two Raman active modes, one of which will be polarised areexpected. Both modes will be present in both the IR and Raman spectra:ΓvibPd−Cl = A1(IR, polarised) +B2(IR, depolarized)

Summary1. For the higher symmetry trans Pd(NH3)2Cl2 complex, a single mode Pd-Cl stretching vibration is

expected in the IR spectrum, while two Pd-Cl stretching modes are expected in the IR spectrum forthe lower symmetry cis Pd(NH3)2Cl2 complex

2. The next step will be to construct and optimize the cis and trans Pd(NH3)2Cl2 complexes, obtain thevibrational modes, and compare the results with the experimental spectrum.


The Methane Molecule


The methane molecule belongs to the Td Point Group.

Symmetry Operations for the CH4 Tetrahedral Molecule


Reducible Representations of 4 x H1s Orbitals on Methane


As an example, consider the irreducible representations spanned by the set of 4 H 1s basis functions in themethane molecule.

Character Table for Td Symmetry

Reducible Representations of the 4 x H1s on CH4


The Reduction Formula and Basis Functions


The reduction formula

n(i) =1

h

∑

R

χr(R)χi(R)

provides a procedure for reducing the representation spanned by a set of basis functions in a molecule.

• n(i) - number of times the ith irreducible representation occurs in the representation r that is being reduced.

• h - the order of the group, i.e. the number of operations in the group.

• ∑R

- summation of all R operations in the group. If there is more than one operation in a given class, each operation must be included

in the summation.

• χr(R) - character of the reducible representation r under the symmetry operator R.

• χi(R) - character of the irreducible representation i under the symmetry operator R

n(A1) = 124

[(4 × 1 × 1) + (1 × 1 × 8) + (0 × 1 × 3) + (0 × 1 × 6) + (2 × 1 × 6)] = 1

n(A2) = 124

[(4 × 1 × 1) + (1 × 1 × 8) + (0 × −1 × 3) + (2 × −1 × 6) + (2 × 1 × 6)] = 0

n(E) = 124

[(4 × 2 × 1) + (1 × −1 × 8) + (0 × 2 × 3) + (0 × 0 × 6) + (2 × 0 × 6)] = 0

n(T1) = 124

[(4 × 3 × 1) + (1 × 0 × 8) + (0 × −1 × 3) + (0 × 1 × 6) + (2 × −1 × 6)] = 0

n(T2) = 124

[(4 × 3 × 1) + (1 × 0 × 8) + (0 × −1 × 3) + (0 × −1 × 6) + (2 × 1 × 6)] = 1

Γ(H1s) = A1 + T2


Symmetry Adapted Linear Combinations of Orbitals (SALCs)


SLACs and MO Diagram for the Methane MoleculeInspection of the character table for the Td point group shows that the p orbitals on the central C atom of CH4 transform as T2 . It’s easy to showthat the the T2 SLAC of the 4 H 1s functions match to the three orthogonal T2 C 2p orbitals.

In addition, the A1 SLAC of the H 1s orbitals match with the C2s orbital.

This yields the interaction diagram below for CH4 with occupied MOs of two different energies as shown below:

MO Diagram for CH 4 Showing the SLACs

The above description of the MO diagram is confirmed below in the photoelectron spectrum for CH4 .

The Photoelectron Spectrum of CH 4


The Cyclopropenium Ion


The cyclopropenium ion (C3H3) provides an excellent example of how todetermine the symmetry of the MOs of a molecular species from GroupTheory.

❚❚

✔✔

✐+1

23

H

H✧✧ H

❜❜

The C3H3 ion belongs to the D3h point group.


The Cyclopropenium Ion: Secular Determinant


For C3H3, the secular determinant in the Huckel approximation becomes

∣∣∣∣∣∣

α− ǫ β ββ α− ǫ ββ β α− ǫ

∣∣∣∣∣∣= 0

This is only a 3× 3 determinant which is not difficult to evaluate explicitly.

However, group theory can be used to simplify this, and the same ideascan be applied to larger molecules leading to substantial reduction ineffort.


The Cyclopropenium Ion: D3h Character Table


The D3h Character Table

D3h EEE 2C3C3C3 3C2C2C2 σhσhσh 2 S3S3S3 3σvσvσvA′

1 1 1 1 1 1 1 x2 + y2, z2

A′2 1 1 -1 1 1 -1 Rz

E′ 2 -1 0 2 -1 0 (x, y) x2 − y2, xyA′′

1 1 1 1 -1 -1 -1A′′

2 1 1 -1 -1 -1 1E′′ 2 -1 0 -2 1 0 (Rx, Ry) xz, yz

φ1 φ1 φ2, φ3 −φ1 −φ1 φ2, φ3 φ1φ2 φ2 φ3, φ1 −φ2 −φ2 φ3, φ1 φ2φ3 φ3 φ1, φ2 −φ3 −φ3 φ1, φ2 φ3Γreducible 3 0 −1 −3 0 1


The Cyclopropenium Ion: Reducible Representations


To use group theory to simplify MOs, we must first determine the irreduciblerepresentations spanned by the AOs.

For example, a CmHm polyene would require matrices of size m×m to beconstructed, but a simple set of rules for determining the characters that may bederived from these matrices are:

1. If an orbital is unchanged after applying a symmetry operation, a value of +1 perorbital is assigned to the character.

2. If the orbital changes sign, a value of -1 per orbital is assigned to the character.

3. If the orbital changes location, a value of 0 per orbital is assigned to thecharacter.


The Cyclopropenium Ion: Reducible Representations


Applying these rules to C3H3 for the reducible representation yields

D3h EEE 2C3C3C3 3C2C2C2 σhσhσh 2 S3S3S3 3σvσvσvΓreducible 3 0 -1 -3 0 1

aA′1

= 112

[(1)(3) + (2)(0) + (3)(−1) + (1)(−3) + (2)(0) + (3)(1)] = 0

aA′2

= 112

[(1)(3) + (2)(0) + (−3)(−1) + (1)(−3) + (2)(0) + (−3)(1)] = 0

aE′ = 112

[(2)(3) + (−2)(0) + (0)(−1) + (2)(−3) + (−2)(0) + (0)(1)] = 0

MOs of C3H3 cannot have A′ symmetry.

aA′′1

= 112

[(1)(3) + (2)(0) + (3)(−1) + (−1)(−3) + (−2)(0) + (−3)(1)] = 0

aA′′2

= 112

[(1)(3) + (2)(0) + (−3)(−1) + (−1)(−3) + (−2)(0) + (3)(1)] = 1

aE′′ = 112

[(2)(3) + (−2)(0) + (0)(−1) + (−2)(−3) + (2)(0) + (0)(1)] = 1

Thus, Γreducible = A′′2 + E′′.


The Cyclopropenium Ion: Secular Determinant in Terms of SLACs


To apply Huckel theory, we can automatically separate out the A′′2 symmetry sub-block of the secular

determinant from the E′′ symmetry sub-block.

Secular Determinant in Terms of SLACs

ψ1 = ψA′′2

= 1√3(φ1 + φ2 + φ3)

ψ2 = ψ(1)E′′ = 1√

6(2φ1 − φ2 − φ3)

ψ3 = ψ(2)E′′ = 1√

2(φ2 − φ3)

∣∣∣∣∣∣

< ψ1|H|ψ1 > −ǫ 0 00 < ψ2|H|ψ2 > −ǫ < ψ2|H|ψ3 >0 < ψ2|H|ψ3 > < ψ3|H|ψ3 > −ǫ

∣∣∣∣∣∣= 0


Energy-Level Diagram for C 3H3


❚❚

✔✔

✐+1

23

H

H✧✧ H

❜❜

α + 2βa′′2

α− βe′′


Using the Projection Operator Method to GenerateMOs as Bases for Irreducible Representations


It is not always possible to deduce the form of an SALC by matching the cental atom orbital of appropriatesymmetry.

However, the projection operator method can be used to a generate symmetry adapted molecular orbital(MO) from atomic orbitals (AOs) that form a basis for a given representation.

A projection operator projects a certain part a symmetry orbital from another orbital.

ψSALC = Piφa =∑

R

Rφaχi(R)

where

• ψSALC is the wave function for the SALC belonging to the ith irreducible representation.• Pi is the projection operator for the ith irreducible representation.• φa is one of the basis functions.• ∑

Rrepresents a summation over all operations in the group.

• Rφa indicates the atomic basis function generated by the operation R on φa.• χi(R) is the character of the ith irreducible representation under R.


Using the Projection Operator Method to Generate MOs That areBases for Irreducible Representations


A simple set of rules for determining the MO which forms a basis for agiven symmetry representation can be used:

1. Choose an AO on an atom and determine into which AO it istransformed by each symmetry operator of that group.

2. Multiply the AO of the transformed species by the character of theoperator in the representation of interest for each symmetryoperator.

3. The resulting linear combination of AOs forms an MO that is abasis for that representation.

The method will be applied to the cyclopropenium ion which belongs tothe D3h point group.




Obtain the projection operator for the A′′2 symmetry representation:

PA′′2 =

1

12[E+C3+C

23 −C(1)

2 −C(2)2 −C(3)

2 −σh−S3−S23 +σ

(1)v +σ(2)

v +σ(3)v ]

Project out MOs for the A′′2 symmetry representation

PA′′2 φ1 = 1

12(φ1 + φ2 + φ3 + φ1 + φ3 + φ2 + φ1 + φ2 + φ3 + φ1 + φ3 + φ2) =

13(φ1 + φ2 + φ3)

PA′′2 φ2 = 1

12(φ2 + φ3 + φ1 + φ3 + φ2 + φ1 + φ2 + φ3 + φ1 + φ3 + φ2 + φ1) =

13(φ1 + φ2 + φ3)

PA′′2 φ3 = 1

12(φ3 + φ1 + φ2 + φ2 + φ1 + φ3 + φ3 + φ1 + φ2 + φ2 + φ1 + φ3) =

13(φ1 + φ2 + φ3)

The normalized symmetry adapted linear for A′′2 is given as

ψA′′2=

1√3(φ1 + φ2 + φ3)




Obtain the projection operator for the E ′′ symmetry representation:

PE′′=

2

12[2E − C3 − C2

3 − 2σh + S3 + S23 ]

Project out MOs for the E ′′ symmetry representation

PE′′φ1 = 2

12(2φ1 − φ2 − φ3 + 2φ1 − φ2 − φ3) =

13(2φ1 − φ2 − φ3)

PE′′φ2 = 2

12(2φ2 − φ1 − φ3 + 2φ2 − φ1 − φ3) =

13(2φ2 − φ1 − φ3)

PE′′φ3 = 2

12(2φ3 − φ1 − φ2 + 2φ3 − φ1 − φ2) =

13(2φ3 − φ1 − φ2)

From the above, it is possible to form two linear independent functions:

ψ(1)E′′ = PE′′

φ1 = 1√6(2φ1 − φ2 − φ3)

ψ(2)E′′ = PE′′

φ2 − PE′′φ3 = 1√

2(φ2 − φ3)


Franck-Condon Factors


Within the Born-Oppenheimer approximations, the electric-dipole transition momentwhich provides the probility of a transition from the ground-electronic state, ψi, to anexcited electronic state, ψf , is given by

µf←i = −e∑

i

〈ψf |rrri|ψi〉︸︷︷︸

Electric DipoleTransition Moment

〈νf |νi〉︸︷︷︸Franck-CondonOverlap IntegralS(νf , νi)

The intensity of a transition is proportional to the square of the transition moment,|S(νf , νi)|2The upper and lower vibrational wavefunctions, νf and νi, are approximatelyharmonic-oscillator-like functions, so that the largest overlap of probabilities for theground-state and excited-state vibrations will occur for a vibrational quantum numbergreater than 0.

This leads to the fact that the absorption probability is largest into that vibrational stateof the excited electronic states whose probability is largest directly above theequilibrium internuclear distance.


Electronic Absorption Spectra of Molecules


Potential energy curves and absorption spec-tra for three electronic transitions (verticallines).

(a) The minimum in the potential in the ex-cited state coincides with the groundstate.

(b) The minimum in the potential in the ex-cited state is at a larger internuclear dis-tance than in the ground state.

(c) The absorption may raise the excitedstate to a higher energy than its disso-ciation energy so that the absorption iscontinuous.


The Franck-Condon Principle


The Franck-Condon principle states that inan electronic transition, the overlap of theground vibrational wavefunction in the lowerelectronic state.

The various vibrational wavefunctions in theupper electronic state is greatest for the vi-brational level whose classical turning pointis at the equilibrium separation in the lowerstate.


Fluorescence and Phosphorescence


Excited-state molecules in a crystal, in solution, or in a gas undergo frequent collisionswith other molecules in which they lose energy and return to the lowest vibrationalstate of S1.

Internal conversion is a nonradiative transition induced by collisions and involves adecay from a higher vibrational state to the ground vibrational state of the samemultiplicity.

Once in the lowest vibrational state of S1, the molecule can undergo a radiativetransition to a vibrational state in S0 (fluorescence), or it can make a nonradiativetransition to an excited vibrational state of T1 in a process called intersystemcrossing which can lead to phosphorescence .

Because internal conversion to the ground vibrational state of S1 is generally fast inconparison with fluorescence to S0, the vibrationally excited-state molecule will relaxto the ground vibrationally state of S1 before undergoing fluorescence.

As a result of the relaxation, the fluorescence spectrum is shifted to lowerenergies relative to the absorption .


Absorption and Fluorescence

January - February, 2012 29 Symmetry and Group Theory

Absorption and Phosphorescence


Absorption from S0 leads to a population of excitedvibrational states in S1.

The molecule has a finite probability of making atransition to an excited vibrational state of T1, if ithas the same geometry in both states and if thereare vibrational levels of the same energy in bothstates.

The dashed arrow indicates the coincidence of vi-brational levels in T1 and S1.

The initial excitation to S1 occurs to a vibration of

maximum overlap with the ground state of S0.


Absorption, Fluorescence, and Phosphorescence

January - February, 2012 31 Symmetry and Group Theory

Lifetimes for Fluorescence and Phosphorescence


In fluorescence , the radiation is emitted during a transition betweenelectronic states of the same spin or multiplicity .

In phosphorescence the radiation is emitted in a transition betweenstates of different multiplicities .

• The lifetime of excited singlet states is short, i.e. usually between10−6 s and 10−9 s.

• The lowest triplet state of a molecule with a singlet ground state islong (10−4s to 100 s) because the transitions between states ofdifferent multiplicity are forbidden, and the rate is slow leading to alonger lifetime.


Effect of Solute-Solvent Equilibrium on Absorption andFluorescence Transiton


The small difference on the 0-0 bands for absorption and fluorecence isdue to the difference in the solvation of the initial and final states.


Calculations of Excited States


Acetaldehyde


Jahn-Teller Effects


Start by assuming a particlar configuration of the molecule in a certain point group.

The Schrodinger equation is assumed to be solved giving rise to eigenvalues (E0, E1, . . . , En), and to thecorresponding wave functions (ψ0, ψ1, . . . , ψn).

After distortion by displacement of the normal coordinate, Q, from the original position, the change inenergy is given by:

E = E0 +

⟨ψ0

∣∣∣∣∂U

∂Q

∣∣∣∣ψ0

⟩Q

︸︷︷︸FOJT

+

1

2

⟨ψ0

∣∣∣∣∂2U

∂Q2

∣∣∣∣ψ0

⟩−

∑

k

⟨ψ0

∣∣∣ ∂U∂Q∣∣∣ψk

⟩2

Ek − E0

Q2

︸︷︷︸SOJT

where U is the nuclear-nuclear and nuclear-electronic potential energy

A first-order Jahn-Teller distortion (FOJT) states that a nonlinear molecule with an incompletely filled,degenerate HOMO level is susceptible to a structural distortion which removes the degeneracy. Thisstructural distortion is related to a degenerate electronic state with partially filled highest occupied molecularorbitals which are degenerate.

A structural distortion arising from a second-order energy change in the HOMO is termed a second-orderJahn-Teller (SOJT) distortion.


First-Order Jahn-Teller Effects


Consider a d8 transition metal complex which has the electronic configuration t62ge∗ 2

A distortion coordinate is needed such that the matrix element < ψ0|∂U∂Q |ψ0 > 6= 0.Then, Γe ⊗ Γe =

1 A1g +3 A2g +

1 Eg

If ΓEgis the symmetry species of the degenerate state, and ΓQ is the distortion

coordinate, then the symmetry representation of the distortion coordinate mustcontain the totally symmetric representation, A1g:

ΓEg⊗ ΓQ ⊗ ΓEg

⊂ ΓA1g

which implies that the distortion is ΓQ = ΓEg as shown below:


First-Order Jahn-Teller Effects: Ni(H 2O)62+


Ni(H2O)62+ complex is a d8 complex wth the electronic configuration t62ge∗ 2 which splits into

1A1g +3 A2g +1 Eg statesThe triplet state will probably lowest in energy, and the doubly degenerate singlet state is next in energy.The 1Eg state is Jahn-Teller unstable and splits apart in energy during an e vibration.Stretching of two trans ligands will lead to a drop in the energy of the dz2 orbital while the shrinking of fourequatorial bonds leads to destabilization of the dx2-y2 orbital, as shown in the graph on the left.The energetic behavior of the two states closest in energy as a result of this distortion is shown in the graphon right.If the splitting between the singlet and triplet states is small, the singlet state crosses the triplet statesomewhere along the reaction coordinate. If the splitting is large, it is expected to find an octahedral tripletas in Ni(H2O)62+.


Second Order Jahn-Teller Effects


The symmetry species of the distortion which allows the mixing of the HOMO andLUMO is derived from group theory.

A distortion coordinate is needed such that the matrix element< ψHOMO|∂U∂Q |ψLUMO > 6= 0.

If ΓHOMO and ΓLUMO are the symmetry species of the HOMO and LUMO, respectivelyand ΓQ is the distortion coordinate, then the symmetry representation of the distortioncoordinate must contain the totally symmetric representation, a1:

ΓHOMO ⊗ ΓQ ⊗ ΓLUMO ⊂ Γa1

This then implies that the distortion coordinate Q has the symmetry representation:

ΓQ = ΓHOMO ⊗ ΓLUMO




Why is NH 3 Pyramidal?

ΓHOMO ⊗ ΓLUMO = Γa′′2⊗ Γa′1

⊂ Γa′′2

so the distortion coordinate is a′′2 , in the z-direction.

Calculations show that the inversion barriers are 5.8kcal/mol, 30 kcal/mol, and 46 kcal/mol while theHAH angles are 106.7◦, 93.3◦, and 92.1◦ for NH3,PH3, and AsH3, respectively.

The 2a′1 orbital is stabilized by mixing with the a′′2 or-bital. The magnitude of the stabilization is inverselyproportional to the energy gap between the 2a′1 anda′′2 orbitals: Einv ∝ e(a′′2 )− e(2a′1) ∝ 1

∆e.




An MO diagram of the σ bonding in the SF4 constructed in a hypothetical tetrahedral geometry is shown below.

1. The bonding a1 and t2 levels are occupied as in the CF4 molecule, but there is also an occupied antibonding orbital, a∗1 .2. In addition, there is an unoccupied antibonding t∗2 level at higher energy.3. Consider a distortion to a lower symmetry than Td . During descent of symmetry, the a1 orbital remains totally symmetric. But, the

threefold degeneracy of the t∗2 level will be lifted.4. Provided that one of the lowest unoccupied molecular orbital (LUMO, t∗2 ) orbitals is also totally symmetric, the highest occupied

antibonding molecular orbital (the HOMO, a∗1 ) can mix together to stabilize the lower level.


Distortion of SF 4 from T d to Lower Symmetry


Distortions to C2v or C3v symmetry would lower the electronic energy, but distortion toeither D2d or D4h symmetry would not.

The C2v and C3v structures are similar to those predicted by VSEPR arguments.


Can Ni(CO) 4 Distort from T d to D 4h Symmetry?


−→


Orbital Correlation Diagram for the Distortion of Ni(CO) 4 from T d toD4h Symmetry


Tetrahedral(Td) Square Planar(D4h)

b1g (x2-y2)

(x2-y2,xz,yz) t2

a1g (z2)

(z2, xy) e eg (xz,yz)

b2g (xy)

The above graph is called an orbital correlation diagram. It is related to a Walshdiagram which is a plot of orbital energies of a molecular transformation as a functionof some variable.


Distortion of Ni(CO) 4 from T d to D 4h Symmetry


Td D4h

Unfilled Orbitals

t2 (4px, 4py, 4pz) a2u + eu

a2 (4s) a1g

Filled Orbitals

t2 (dx2-y2, dxz, dyz) b1g + eg

e (dz2, dxy) a1g + b2g

During the descent to lower symmetry there is no lower unoccupiedorbital that can mix and stabilize the occupied orbitals.

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