VALENCE Hilary Term 2019

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1 VALENCE Hilary Term Lectures Prof M. Brouard Valence is the theory of the chemical bond Outline plan 1. The Born-Oppenheimer approximation 2. Bonding in H + 2 the LCAO approximation 3. Many electron molecules the orbital approximation, its strengths and weaknesses binding in H 2 splitting of degenerate configurations dissociation of H 2 4. Application of the variation principle to find LCAO s the secular equations

2 Cont. 5. Binding of the first row diatomics splitting into terms, and levels 6. Walsh diagrams bond angles in AH 2 systems 7. Hückel theory use of symmetry aromaticity bond order, electron densities and organic reactivity 8. Correlation Diagrams predicting chemical reactivity Links with other lecture courses Quantum Mechanics Spectroscopy Statistical Mechanics Reaction Rates Photochemistry (Year 3)

3 1: The Born-Oppenheimer approximation General Schrödinger equation for molecules For a molecular system the Hamiltonian is Ĥ = ˆT e + ˆT n + ˆV en + ˆV nn + ˆV ee ˆT Kinetic energy terms for the electrons (e) and nuclei (n); ˆV Potential energy terms; the various Coulomb interactions between the electrons and the nuclei. The Schrödinger equation for a molecule can be written ĤΨ(r,R) = EΨ(r,R) where r and R represent collectively the position vectors (with respect to the centre-ofmass) of the electrons and nuclei, respectively. This equation is simple to write down, but too complex to solve. We need to make a sequence of approximations.

4 The Schrödinger equation for H 2 Use coordinates expressed in the centre-of-mass (CM) frame (i.e. factor out kinetic energy of the CM) Use atomic units (e = 1, = 1, m e = 1) Ĥ = ˆT e + ˆT n + ˆV en + ˆV nn + ˆV ee A + r 2A 2 - r 1A R r 2 r 2B CM r 1-1 r 1B + B ˆT n = 1 2µ 2 R with µ = M AM B M A +M B ˆT e = r r 2 ˆV en = 1 r 1A 1 r 1B 1 r 2A 1 r 2B ˆV nn = 1 R ˆV ee = 1 r 12 1 r 1 r 2

5 The approximation: the electronic problem Electrons move much faster than nuclei, for the same force. Therefore, they adapt quickly to changes in nuclear motion. The Born-Oppenheimer approximation assumes that nuclei are stationary on the timescale of electron motion. We can therefore write the total wavefunction in a separable product form Ψ(r,R) = χ n (R)Ψ e (r R). χ n (R) wavefunction describing the motion of the nuclei; Ψ e (r R) the electronic wavefunction (depends parametrically on the coordinates of the nuclei). This separation is known as an adiabatic separation, and leads to adiabatic electronic states (see below). Define an electronic Hamiltonian Ĥ e = Ĥ ˆT n, which is a Hamiltonian for stationary nuclei.

6 Solution of the simplified Schrödinger equation for the electronic motion at fixed R, Cont. Ĥ e Ψ e (r R) = E e (R)Ψ e (r R) yields a set of electronic states, with energies E e (R) that depend on the positions of the nuclei, R. This is the equation on which we will focus for most of the remaining lectures. H 2 The electronic states with energies E e (R) represent the potential energy functions (curves), V(R), that nuclei experience at a given separation, R. V( R) 3 + u Note that E e (R) V(R) includes the Coulomb repulsion between the (fixed) nuclei ˆV nn. 1 + X( g ) H(1s)+H(1s) R

7 The approximation: the nuclear problem For each electronic state, E e (R), we can now attempt to solve the Schrödinger equation for the nuclear motion: Ĥ n χ n (R) = [ˆT n +E e (R) ] χ n (R), The electronic energies, E e (R), play the role of the potential energy, V(R), experienced by the nuclei at a given separation: = [ˆT n +V(R) ] χ n (R) = E n χ n (R). E n, are a set of energy levels associated with the vibrational and rotational motion of the nuclei in a specific electronic state. 3 g N 2 V( R) Each electronic state is described by a different potential energy curve, with its own set of rotationvibration states, obtained from the solution of the above Schrödinger for the nuclear motion. 1 + X( g ) R

8 Why is it an approximation? Write the full Schrödinger equation as ĤΨ(r,R) = EΨ(r,R) where E E n ĤΨ(r,R) = (Ĥ e + ˆT n )χ n (R)Ψ e (r R) = Ĥ e χ n (R)Ψ e (r R)+ ˆT n χ n (R)Ψ e (r R) Term 1 Term 2 Look at Term 1 Ĥ e χ n (R)Ψ e (r R) = χ n (R)Ĥ e Ψ e (r R) = χ n (R)E e (R)Ψ e (r R) = Ψ e (r R)V(R)χ n (R). Note that the electronic Hamiltonian does not operate on χ n (R). Look at Term 2 ˆT n χ n (R)Ψ e (r R) = χ n (R)ˆT n Ψ e (r R)+Ψ e (r R)ˆT n χ n (R) This term is more complicated.

9 Term 2 in more detail - an aside ˆT n χ n (R)Ψ e (r R) = χ n (R)ˆT n Ψ e (r R)+Ψ e (r R)ˆT n χ n (R) Note: A cross term in the derivatives has been neglected. It turns out not to contribute to the energy of the system E = Ψ Ĥ Ψ. ˆT n [χ n (R)Ψ e (r R)] d2 [χ n Ψ e ] dr 2 = d[χ nψ e +χ n Ψ e] dr = χ n Ψ e +2χ n Ψ e +χ n Ψ e When considering the energy of the system the middle term (in green) is zero: E = Ψ(r,R) Ĥ Ψ(r,R) = Ψ(r,R) Ĥ e + ˆT n Ψ(r,R) = + + χ n Ψ e ˆT n χ n Ψ e = χ n Ψ e χ n Ψ e However, d Ψ e Ψ e dr = d1 dr = 0 = Ψ e Ψ e + Ψ e Ψ e = 2 Ψ e Ψ e So the term in green goes to zero when considering the energy of the system.

10 Neglect of χ n (R)ˆT n Ψ e (r R) Term Neglect of the term in red is the essence of the Born-Oppenheimer approximation. Term 2 then becomes ˆT n χ n (R)Ψ e (r R) Ψ e (r R)ˆT n χ n (R) So combining remaining terms in 1 and 2 for the total energy gives ĤΨ(r,R) = Ψ e (r R)V(R)χ n (R)+Ψ e (r R)ˆT n χ n (R) = Ψ e (r R) [ˆT n +V(R) ] χ n (R) = Eχ n (R)Ψ e (r R) This all works provided χ n (R)ˆT n Ψ e (r R) is a relatively small term. Neglect of the term χ n (R)ˆT n Ψ e (r R) is reasonable provided Ψ e (r R) is a slowly varying function of the nuclear positions, R. Note further that ˆT e Ψ e (r R) µ/m e ˆT n Ψ e (r R). Because nuclei are much heavier than electrons, ˆT e Ψ e (r R) is usually much greater than ˆT n Ψ e (r R).

11 Breakdown of the Born-Oppenheimer approximation. The Born-Oppenheimer approximation may break down (non-adiabatic behaviour) when the nuclear motion is very fast (e.g., close to dissociation), or two electronic states have the same energy (e.g., at a curve crossing). V R + - Li + H ( ) ionic In the region of the crossing, the adiabatic electronic wavefunctions Ψ e (r R) (represented by the black lines) change rapidly in character. covalent Li + H R [See Problem sheet 1, question 5.]

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13 2: Solving the electronic problem A r A e - r B Low lying states of H + 2 Within the Born-Oppenheimer approximation, the Schrödinger equation for the electronic motion in the H + 2 molecular ion can be written with Ĥ e Ψ e (r R) = E e (R)Ψ e (r R) R CM B and Ĥ e = ˆT e + ˆV en + ˆV nn ˆV en = 1 r A 1 r B This equation can be solved exactly for H + 2. However, we need to develop a treatment suitable for larger systems.

14 LCAO approximation * A * B Suppose that the wavefunction (call it a molecular orbital) can be written as a linear combination of atomic orbitals (LCAOs) localized on atom A and B: Ψ e = c A φ A +c B φ B. In this example take the atomic orbitals φ i to be the 1s orbitals on atom i, although other orbitals could (should) be included in the summation. c i are the weighting coefficients. For H + 2, and other homonuclear diatomics, the electron density ρ(r) = Ψ e 2 = c 2 Aφ 2 A +c 2 Bφ 2 B +2c A c B φ A φ B must be equal at all points related by symmetry ( ). This requires that c 2 A = c2 B c A = ±c B = c

15 Normalization The wavefunction for the electron in H + 2 must be normalized, i.e. [ ] ρ(r)dr = c 2 φ 2 Adr+ φ 2 Bdr±2 φ A φ B dr = 1. However, the 1s atomic orbitals are already normalized individually φ 2 Adr = φ 2 Bdr = 1, such that the normalization constraint can be written [ ] 1/2 1 c =. 2(1±S AB ) S AB is known as the overlap integral, defined as S AB = φ A φ B dr The overlap integral is positive. A B

16 Wavefunctions Given the constant, c, the approximate wavefunctions for H + 2 can be written Ψ + e = φ A +φ B [2(1+S AB )] 1/2 Ψ e = φ A φ B [2(1 S AB )] 1/2. Two molecular wavefunctions are generated from the in-phase and out-of-phase superposition of the two atomic orbitals. The electron densities for the two states can be written (φ 2 A +φ2 B ) 2 φ A φ B ρ ± (r) = Ψ ± e 2 = φ2 A +φ2 B ±2φ Aφ B 2(1±S AB ) the sum of atomic electron densities. large where the orbitals overlap. A B A B In Ψ + e the electron density is focussed between the nuclei (and reduced elsewhere by (1+S AB )). In Ψ e the electron density is reduced between the nuclei (and increased elsewhere by (1 S AB )).

17 Orbital energies The molecular orbital energies can be obtained by evaluating the expression Substituting for the wavefunctions yields E ± e E ± e (R) = Ψ ± e Ĥ e Ψ ± e. E ± e (R) = H AA ±H AB ±H BA +H BB 2(1±S AB ), where we have defined the integrals H ij = φ i Ĥφ j dr. Because φ A and φ B are identical 1s orbitals, Therefore, the energy reduces to H AA = H BB and H AB = H BA. E ± e (R) = H AA ±H AB (1±S AB ).

18 Cont. Let s look at the integrals more closely: H AA = + + φ A [ Coulomb integral ˆT e 1 ] φ A dr r A ǫ 1s 1s orbital energy of H atom A ] φ A [ 1 r B φ A [ 1 R ] φ A dr J(R) attraction of 1s electron density on A to the B nucleus φ A dr 1 R Coulomb repulsion between nuclei J( R) A B Note that J(R) is negative. So H AA = H BB = ǫ 1s + 1 R +J(R) H AA can be thought of as the energy of the electron localized on A in the presence of B.

19 Cont. Similarly H AB = + φ A [ ˆT e 1 ] φ B dr r B ǫ 1s S AB Resonance integral φ A [ 1 r A ]φ B dr K(R) attraction of overlap density to nucleus A + φ A [ 1 R ] φ B dr 1 R S AB K(R) is also negative. K( R) A B So ( H AB = H BA = S AB ǫ 1s + 1 ) +K(R) R There is no classical equivalent to K(R).

20 Cont. Adding up the terms yields E ± e (R) = ǫ 1s + 1 R + J ±K 1±S AB. The integrals are performed explicitly in Green QM2 ( ) S AB (R) = e R 1+R+ R2 3 J(R) = [ ( 1 R e 2R 1+ 1 )] R K(R) = e R (1+R) In terms of bonding K(R) must dominate.

21 Cont. Contributions to bonding... E ± e (R) = ǫ 1s + 1 R + J ±K 1±S AB. In terms of bonding K(R) must dominate. nuclear repulsion nuclear repulsion total electronic energy R total electronic energy electronic binding energy R electronic binding energy The 1 ± S AB term in E ± e ensures that antibonding orbitals are more antibonding than bonding orbitals are bonding.

22 How bad is the LCAO approximation? Compare with the exact solution (within the Born-Oppenheimer approximation) for H % error in R e 46% error in D e (for the ground electronic state) V( R) 2 + X( g ) H A( u ) exact LCAO H(1s) + H + Provides a qualitative guide to the nature of the binding and order of the electronic states. Improve by increasing the number of basis functions in Ψ e, e.g., write Ψ e = c 1 φ 1sA +c 2 φ 1sB +c 3 φ 2sA +c 4 φ 2sB +c 5 φ 2pzA + Optimize the coefficients, c i, using the variation principle to minimize the energy. This would give the exact result in the limit of an infinite (or complete) basis set of orbitals. In addition, one can also improve the basis functions themselves, φ 1s = e ξr. Again, ξ can be optimized to minimize the energy. R

23 3: Many electrons Orbital symmetries Atomic orbitals are labelled according to the principal quantum number, n, and the orbital angular momentum quantum number, l. Electrons in a diatomic molecule experience a cylindrically symmetric electric field. This leads to a Stark splitting of the atomic l states into components with m l = 0 and m l = ±1, etc., where m l is the component of l along the symmetry axis of the field (the bond axis). 0 l = m l

24 Cont. The symmetries of the orbitals in diatomic molecules are labelled by the magnitude of the component of the angular momentum along the internuclear axis, λ. Looking down the symmetry axis λ = 0 (σ) λ = 1 (π) λ = 2 (δ) (cf. particle on a ring - see Problem sheet 2, question 3)

25 Cont. For H + 2 Ψ + e is the 1σ g orbital + g Ψ e is the 1σ u orbital + - u The 1 indicates that Ψ + e is the lowest orbital of that symmetry. The g and u symmetry labels refer to the inversion symmetry, Î, with respect to the molecular frame ÎΨ e (x,y,z) = ±Ψ e ( x, y, z) u - + g

26 The orbital approximation Treat many electron molecules in same way as many electron atoms. Define a molecular orbital (MO) as a one-electron wavefunction in a molecule. In some cases these MO s may be constructed from eigenfunctions of some effective one-electron Hamiltonian. Write the total wavefunction in the many electron case as a product of one electron MOs Ψ tot (r 1,r 2, r N ) = (φ a (r 1 )φ b (r 2 ) φ n (r N )) space [Φ(1)Φ(2) Φ(N)] spin. Here, the φ i are the molecular orbitals (e.g., the 1σ g orbital of H + 2 just discussed). The total wavefunction is associated with a particular electron configuration. Ψ tot is a product of a spatial wavefunction (which describes where the electrons are) and a spin part (which tells us whether the spin of each electron is up or down ).

27 Pauli exclusion principle The total wavefunction must be constructed so that it satisfies the Pauli exclusion principle. It must be antisymmetric with respect to exchange of any two identical electrons. For example, the ground state of the H 2 molecule has an electron configuration 1σ 2 g. The Pauli principle can only be satisfied if the electron spins are paired. The total wavefunction can be written, approximately Ψ tot (r 1,r 2 ) = (φ 1σg (r 1 )φ 1σg (r 2 )) 1 2 [α(1)β(2) α(2)β(1)]. In this case the spatial wavefunction is symmetric, and the spin function is antisymmetric: Ψ tot (r 2,r 1 ) = (φ 1σg (r 2 )φ 1σg (r 1 )) 1 2 [α(2)β(1) α(1)β(2)] = Ψ space Φ spin = Ψ tot (r 1,r 2 ) In general, a total (anti-symmetrized) wavefunction satisfying the Pauli Principle can be constructed using a Slater determinant (see Atkins MQM), which generates an antisymmetric sum of product terms in Ψ tot, rather than a single product.

28 Labelling of molecular states Recall that molecular orbitals are labelled according to λ. In the many electron case, different electronic states are generated according to how the electrons fill up the available orbitals. Molecular states are labelled with a term symbol: 2S+1 Λ. Λ is the magnitude of the component of the total electronic orbital angular momentum along the internuclear axis. The total component of the orbital angular momentum is obtained from the sum Λ = λ 1 +λ 2 +λ 3 + S is the total spin angular momentum quantum number, obtained from the vector sum S = s 1 +s 2 +s 3 + Note that the spin angular momentum is not coupled to the internuclear axis (i.e. is not affected so strongly by the electric field gradient) in the molecule.

29 Predictions of the orbital approximation Ground electronic states Molecule Config n Term No. of R e /Å D e /kjmol 1 Symbol bonding e s H + 2 (1σ g ) 1 2 Σ + g H 2 (1σ g ) 2 1 Σ + g He + 2 (1σ g ) 2 (1σ u ) 1 2 Σ + u He 2 (1σ g ) 2 (1σ u ) 2 1 Σ + g 0 see below MO picture provides qualitative rationale for experimental observations. Note: The electronic states are also labelled according to their inversion symmetry (i.e. g/u), and their symmetry (+/ ) with respect to reflection in the molecular plane see Problem sheet 2, question 3). σ orbitals, and therefore the states they lead to when occupied, are symmetric with respect to this operation. Doubly occupied orbitals are also totally symmetric (i.e. (g g) = g, (u u) = g, (+ +) = +, and ( ) = +).

30 Nature of the orbital approximation Does the orbital approximation for Ψ tot satisfy the Schrödinger equation? Look at H 2 ground state again ĤΨ tot = ( ĥ 1 +ĥ ) Ψ space Φ spin. r 12 ĥ i is the Hamiltonian for electron i in H + 2, and 1/r 12 is the electron-electron Coulomb repulsion. The Hamiltonian does not operate directly on the spin function, so we can write ĤΨ space = ( ĥ 1 +ĥ ) φ 1σg (r 1 )φ 1σg (r 2 ). r 12 This yields ĤΨ space = (ǫ 1σg +ǫ 1σg )Ψ space + 1 r 12 Ψ space. The last term is not a constant Ψ space.

31 Cont. To allow for this electron-electron repulsion term Ψ space must depend on r 12, the electronelectron separation. The electron positions described by Ψ space must be correlated - the orbital approximation does not allow for this electron correlation Note: Some improvements can be made by optimizing the orbitals so as to minimize the average inter-electron repulsion using the variation principle. This leads to the self-consistent field or Hartree-Fock orbitals (see below). However, these procedures do not eliminate problems associated with the neglect of electron correlation.

32 Consequences of the orbital approximation A. Dispersion forces He 2 is predicted to be unbound because anti-bonding orbitals are more antibonding than bonding orbitals are bonding. u He(1s) He(1s) g He 2 In reality the electron positions are correlated, leading to the van der Waals (dispersion) interaction, V(R) 1/r 6. Leads to a binding energy of ǫ/k B 11K. Within the orbital approximation, for example the ground state of H 2 is represented P(r 1,r 2 ) = Ψ e 2 = P 1σg (r 1 )P 1σg (r 2 ) i.e. a product of probabilities. The electron positions are intrinsically independent.

33 B. Dissociation Examine the large R behaviour of the electron density in the molecular orbital wavefunction for H 2 φ 1σg (r 1 )φ 1σg (r 2 ) 2 R = 1s A (r 1 ) 2 1s B (r 2 ) 2 + 1s B (r 1 ) 2 1s A (r 2 ) 2 + 1s A (r 1 ) 2 1s A (r 2 ) 2 + 1s B (r 1 ) 2 1s B (r 2 ) 2 (note that the cross terms vanish as R ) i.e. dissociation products are H + H + H + + H, and the estimate of the dissociation energy is very poor.

34 Valence bond description An alternative is to write a valence bond wavefunctions for the ground electronic state of H 2 Ψ VB (r 1,r 2 ) = N [1s A (r 1 )1s B (r 2 )+1s A (r 2 )1s B (r 1 )] This is not an orbital approximation. This completely removes the ionic terms in the wavefunction, leading to a more accurate description at the dissociation limit. However, the description of the molecule is poor for the same reason: the valence bond wavefunction is over-correlated.

35 Cont. A better approach is to mix the MO and VB wavefunctions Ψ e = N(λ)[Ψ MO +λψ VB ], where λ is optimized, using the variation principle, at each value of R. V( R) H 2 exact LCAO VB CI/MO+VB 1 + X( g ) H(1s) + H(1s) In this way the optimum mix of covalent and ionic character can be obtained at each R. R

36 Configuration interaction The same wavefunction is obtained using the MO wavefunction with configuration interaction. The configurations (1σ g ) 2 and (1σ u ) 2 both lead to electronic states of 1 Σ + g symmetry. Ψ 1 = φ 1σg (r 1 )φ 1σg (r 2 ) Ψ 2 = φ 1σu (r 1 )φ 1σu (r 2 ) Both of these electronic states correlate to the same H + H + H + + H products. 1 + g H 2 LCAO CI V( R) H H 1 + X( g ) H(1s) + H(1s) Better trial wavefunctions for these two 1 Σ + g states are obtained by mixing the two configurations Ψ ± = Ψ 1 ±λ Ψ 2. The optimized ground state wavefunction is the same as that obtained using the mixed valence bond/mo wavefunction. More details can be found in Green QM2. R

37 C. Configurations and terms The orbital approximation predicts that all terms arising from a given configuration will be degenerate. In reality, they can be split into different terms. The energy of a configuration is not obtained simply as the sum of the occupied orbital energies! Consider a (π) 2 configuration (see later for more discussion of B 2 and O 2 ). The π orbitals are two-fold degenerate, corresponding to an electron rotating about the bond axis in opposite directions, leading to m l = ±1. In reality, the above arrangements of the two electrons have different energies, because of differences in the inter-electron repulsion.

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39 4: Applications of the variation principle Secular equations Provides a general route to approximate molecular wavefunctions and orbital energies. Let Ψ be some trial (approximate) wavefunction. Then, according to the variation principle where E 0 is the true ground state. E E = Ψ Ĥ Ψ Ψ Ψ E 0, Apply this to LCAO molecular orbitals Ψ = N m c n φ n, n=1 where N is a normalization constant, and c n is the (real) coefficient for the nth basis function, φ n. Assume also that the basis functions, φ n, are normalized. For example, for the first row diatomics the φ n might be the basis set comprised of the 1s, 2s, 2p x, etc. orbitals on each atom.

40 Introduce simplifying notation ( )( ) Ψ Ψ = c n φ n c m φ m dτ = n m n,m where the overlap integral is defined S nm = φ n φ m dτ. c n c m S nm, Cont. Note that S nm = S mn. The numerator can be written ( ( ) Ψ Ĥ Ψ = c n φ n )Ĥ c m φ m dτ = m n,m where the matrix elements are defined H nm = φ n Ĥφ m dτ. n c n c m H nm, The Hermitian properties of Ĥ ensure that H nm = H mn.

41 Cont. To find the best estimate of E 0, minimize E with respect to the coefficients, c n. E c k = 1 Ψ Ĥ Ψ Ψ Ψ c k Ψ Ĥ Ψ Ψ Ψ 2 Ψ Ψ c k = 0 or Ψ Ĥ Ψ c k Ψ Ĥ Ψ Ψ Ψ Ψ Ψ c k = 0. The partial derivatives can be written explicitly Ψ Ĥ Ψ c k = 2 m c m H km and Ψ Ψ c k = 2 m c m S km, where the factor of 2 arises from the double summations in Ψ Ψ and Ψ Ĥ Ψ.

42 Cont. Therefore (H km ES km )c m = 0 for all k = 1 m. m In matrix notation this can be written (H ES)c = 0, which are known as the secular equations. The non-trivial solutions are given by det H ES = 0. Solution yields a polynomial with m roots, or eigenvalues, i.e. m MO energies, E, the lowest of which is the best estimate of the ground state energy. Once the eigenvalues, E, are known, the coefficients, c m, can be calculated by backsubstitution into the secular equations. For the lowest root one obtains the best estimate of Ψ 0.

43 Illustration for two orbitals Consider a heteronuclear diatomic molecule, e.g., a simple model of the HOMO of LiH Ψ c 1 φ Li +c 2 φ H, where φ Li could be the 2s orbital on Li, and φ H could be the 1s orbital on the H atom. The secular equation reduces to H 11 E H 12 ES 12 H 21 ES 21 H 22 E c 1 c 2 = 0 0 H 11 = α A and H 22 = α B are approximately the energies of the 2s orbital on Li ( 5eV = I.P. of Li) and the 1s orbital on H ( 13eV = I.P. of H). H 12 = H 21 = β 2eV at R e, known as the resonance integral, is the energy associated with the overlap density. S 12 = S 21 = S 0.4 at R e. Note that S 11 = S 22 = 1

44 Solving to get the energy levels. Multiplying out the secular determinant gives (α A E)(α B E) (β ES) 2 = 0, the solution of which is E = α A +α B 2βS ± (α A +α B 2βS) 2 4(1 S 2 )(α A α B β 2 ) 2(1 S 2 ) (1) To progress further, either 1. Stick the numbers into a computer 2. Set S = 0 (provides a useful guide to the general behaviour see below) 3. Set α A = α B = α and check that we get the homonuclear H + 2 case.

45 Check for the H + 2 case... E ± = α±β 1±S. This equation agrees with that for H + 2, when explicit results for α (= H AA) and β (= H AB ) are substituted.... and solving to get the wavefunctions. To get the orbitals substitute back into the secular equations. For example, for the ground state (Ψ + ) ( α α+β ) ( c A + β (α+β)s 1+S 1+S ) c B = 0, c A = c B. Normalization, as before (see p11), leads to c A = c B = 1 [2(1+S)] 1/2, as produced before using symmetry arguments. The same procedure can be used for the other root (i.e. Ψ ).

46 The two orbital case without overlap Using equation 1, setting S = 0 ( E = 1 2 α A +α B ± ) (α A +α B ) 2 4(α A α B β 2 ). When α A = α B = α E - = E ± = α±β E + = When α A α B E ± = α A +α B 2 ± α A α B 2 ( 1+ 4β 2 (α A α B ) 2 ) 1/2 B E - If α A α B = 2 β E +

47 Energies For small β/ α A α B the energies E ± can be approximated E ± = α A + β2 α A α B + α B β2 α A α B + Plot the MO energies versus (α A α B ) at fixed β. E - E/ A B extent of stabilization of lower orbital E + =0 This reveals that atomic orbitals of very different energies do not mix.

48 Wavefunctions For α A α B (so that orbital of A is lower than that of B), with and Ψ + = c + A φ A +c + B φ B Ψ = c A φ A +c B φ B, then c + A c + B = 2 +4β 2 + 2β with = α A α B. For LiH /β c /c + + A B c /c - - A B 3 A B)/

49 5: Bonding in diatomic molecules General observations Orbitals of different symmetry do not mix since H ij = S ij = φ i Ĥφ j dτ = β = 0 φ i φ j dτ = 0 Orbitals of very different energies do not mix since then /β is large (see above). The non-crossing rule. The energies of two states of the same symmetry can never be equal (since β 0 - see later).

50 First row homonuclear diatomics Expect the MO s to be made up of combinations of 1s, 2s, 2p x, 2p y, 2p z... atomic orbitals (with z lying along the bond). However, Rule 1: Two atomic orbitals will only both make a large contribution to an MO if their atomic energies are similar. Ignore mixing of 1s with 2s and 2p, and mixing of 2s with 2p z (depending on the 2s-2p atomic separation, see below), Rule 2: Orbitals on different atoms with zero overlap do not contribute to the same MO Hence, ignore mixing of 2s A, 2p z,a with 2p x,b and 2p y,b.

51 Leads to the following orbitals: 3σ u = 2p z,a +2p z,b 1π y,g = 2p y,a 2p y,b 1π x,g = 2p x,a 2p x,b 1π y,u = 2p y,a +2p y,b 1π x,u = 2p x,a +2p x,b 3σ g = 2p z,a 2p z,b 2σ u = 2s A 2s B 2σ g = 2s A +2s B 1s overlap is negligible at the optimum bond length for overlap of 2s and 2p. Expect overlap of 2s and 2p z orbitals to decrease less rapidly with R than for the 2p x,y orbitals.

52 MO diagram for O 2 and upwards If 2s and 2p z energy separation is large (i.e. E 2p E 2s bonding interaction) - limit of large penetration - obtain following MO diagram. 3 u 1 g 2p 2p 1 u 3 g 2 u 2s 2s 2 g 1s 1 u 1 g 1s

53 Mixing or hybridization of MOs Mixing of 2σ g and 3σ g, such that 3 g 2 g + 3 g Acquires non-bonding character. 3 g destabilized. 2 g 2 g - 3 g Acquires bonding character. 2 g stabilized. Ψ g = 2σ g λ3σ g develops bonding character Ψ + g = 3σ g +λ2σ g develops non-bonding character and Ψ u = 2σ u λ3σ u develops non-bonding character Ψ + u = 3σ u +λ2σ u develops anti-bonding character

54 Light atom picture If 2s and 2p z energy separation is small (i.e. E 2p E 2s bonding interaction) - limit of small penetration - this interaction leads to the light atom MO picture: 3 u 1 g 2p 3 g 2p 1 u 2s 2 u 2s 2 g 1s 1 u 1 g 1s An alternative, but equivalent description is to mix, or hybridize, the atomic orbitals before forming bond.

55 Properties of homonuclear diatomics Electron Excess Number Bond Diss n Term config. bonding unpaired length/ energy/ Symbol e s e s Å kjmol 1 Li 2 (2σ g ) Σ + g Be 2 (2σ g ) 2 (2σ u ) Σ + g B 2 [Be](1π u ) Σ g C 2 [Be](1π u ) Σ + g N + 2 [C](3σ g ) Σ + g N 2 [C](3σ g ) Σ + g O + 2 [N](1π g ) Π g O 2 [N](1π g ) Σ g F 2 [N](1π g ) Σ + g Bonding determined by mix of occupied orbitals. Note: [Be], etc. stands for the ground state electronic configuration of Be 2, etc.. See Problem sheet 2.

56 Splitting into terms - revisited r/pm ω e /cm 1 D e /ev E /ev I/eV N N NO NO O O [ ω e is the harmonic vibrational wavenumber, D e the bond dissociation energy, E the energy of the first excited electronic state, and I is the first ionization energy. ] Problem Sheet 2, Question 1

57 Splitting into terms - revisited Oxygen The ground electronic configuration of O 2 is [Be](3σ g ) 2 (1π u ) 4 (1π g ) 2. It is split in energy into three electronic states due to the effects of inter-electron repulsion. The states arising from the (π g ) 2 configuration can be worked out as follows: S = s 1 +s 2,s 1 +s 2 1, s 1 s 2 = 0 or 1 Λ = λ 1 +λ 2, λ 1 λ 2 = 0 (Σ) or 2 ( ) In principle there will be singlet and triplet Σ and states, but not all of these satisfy the Pauli exclusion principle.

58 Look at wavefunctions in more detail. For the state the electrons orbit about the bond in the same direction or Ψ = π +1 (1)π +1 (2)Ψ spin Ψ = π 1 (1)π 1 (2)Ψ spin. Exchange of the labels for electrons (1) and (2) tells us that the spatial part of the above wavefunctions are symmetric with respect to exchange of the electrons. To satisfy the Pauli exclusion principle, the spin functions must be anti-symmetric: Ψ spin = 1 2 [α(1)β(2) β(1)α(2)] The only possibility is that the state is a singlet state.

59 In the case of Σ, both singlet and triplet states are allowed Ψ3 Σ = 1 [π +1 (1)π 1 (2) π 1 (1)π +1 (2)] 2 α(1)α(2) 1 2 [α(1)β(2)+β(1)α(2)] β(1)β(2) Ψ1 Σ = 1 [π +1 (1)π 1 (2)+π 1 (1)π +1 (2)] { } 1 2 [α(1)β(2) β(1)α(2)] 2 The full term symbols are 3 Σ g, 1 g, and 1 Σ + g. Symmetry (direct product) tables can also be used to work most of this out π π = Σ + +[Σ ]+ The square brackets indicate that the spatial wavefunction for the Σ state is antisymmetric with respect to electron exchange. The +/ reflection symmetry label is determined using the procedures illustrated in problem sheet 2, question 3.

60 Energy ordering of the states The relative energies of the states can be worked out by applying Hund s rules, although the latter were really devised for atoms. The triplet state is lowest because of the formation of a Fermi hole in the triplet state. Notice that the spatial part of the wavefunction is zero when the coordinates of the two unpaired electrons are the same, Ψ3 Σspace = 1 2 [π +1 (1)π 1 (2) π 1 (1)π +1 (2)]. This spatial wavefunction therefore gives rise to reduced repulsion between the unpaired electrons. This effect is known as spin-correlation. The 1 state lies lower than 1 Σ because the electrons are able to avoid one another more in a state of high Λ. O 2 V( R) 1 + g 3 3 O( P) + O( P) 1 g 3 - X( g ) R

61 Splitting into terms - continued The low lying electronic states of C 2 Term Energy/ Vibrational Bondlength/ symbol cm 1 frequency/cm 1 Å 1 Σ + g Π u Π u Σ + u Problem Sheet 2, Question 2

62 Splitting into terms - continued Low lying states of C 2 3 u 1 g 3 g 1 u Things to note: u g < u < u < u The 3 Π u and 1 Π u states have similar properties (same electron configuration); The energy separation between the 3 Π u and 1 Π u states reflect the spin correlation energy (Fermi hole and Fermi heap); The 1 Σ + u state lies very high in energy; The ground 1 Σ + g and excited 1 Σ + u states have similar properties (suggesting that the σ g orbitals are largely non-bonding in character); The ground 1 Σ + g and excited 3 Π u states are very close in energy (due to the small gap between the 1π u and 3σ g and the reduced electron repulsion in the triplet state).

63 Splitting into terms - continued NO and O + 2 The ground state electronic configuration of NO (and O + 2 labels) [Be] (3σ g ) 2 (1π u ) 4 (1π g ) 1, ) is (nb., for NO drop the g/u leading to a 2 Π term (i.e. S = 1/2, Λ = 1). This state is split into two levels by the spin-orbit interaction: The coupling arises from the coupling between intrinsic magnetic moment of the electron and that generated by the orbital motion of the electron. Label states according to the full term symbol: 2S+1 Λ Ω, where Ω = Λ+Σ. Ω is the component of the total electronic angular momentum along the internuclear axis. Each spin-orbit state is therefore two-fold degenerate. For NO we have 2 Π 3/2 or 2 Π 1/2 spin-orbit states. The Ω = 1/2 state is the lowest in energy because the molecular orbital is less than half full (Hund s 3rd rule).

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65 6: The shapes of polyatomic molecules Symmetry adapted linear combinations (SALCs) AH 2 at linearity. H A H For 1st row element A, the valence MO s will involve 2s A, 2p x,a, 2p y,a 2p za and two 1s orbitals on the H atoms. Only orbitals of the same symmetry mix. In D h it s easy to write down the symmetry of the 2s and 2p orbitals of A (see below). 1s orbitals of H have no symmetry in the D h point group. Take linear combination of H 1s functions to symmetrize the H atom 1s basis functions: φ(σ g ) = 1s 1 +1s 2 φ(σ u ) = 1s 1 1s 2 See butadiene example later.

66 MO diagram for AH 2 molecules At linearity (D h ) the atomic orbitals have the symmetries 2s A 2p x,a,2p y,a σ g π u 2p za σ u (1s 1 1s 2 ) σ u (1s 1 +1s 2 ) σ g Expect: u g 2p A u 1s 1 +/- 1s2 2s A u g

67 Walsh diagram for AH 2 molecules What happens to the MO diagram on bending? Symmetry is lowered from D h to C 2v. H z O H x Descent in symmetry tables can be used to work out how the orbital symmetries in C 2v. 2 u 4a 1 3 g 2b 1 1b 2 1 u 1b 1 1 u 3a 1 2 g a o Bond angle, 90 o 2a 1 Stabilized because of weak bonding H H interaction, and mixing with 2p z. 1b 1 Destabilized because of reduction in overlap, and antibonding H H interaction. 3a 1 (p z ) Strongly stabilized by mixing with 4a 1 (3σ g ) - mixing strong because close in energy. 1b 2 (p y ) Remains non-bonding

68 Predicted bond angles for AH 2 molecules Molecule Config. θ BeH 2 (2a 1 ) 2 (1b 1 ) BH 2 (2a 1 ) 2 (1b 1 ) 2 (3a 1 ) CH 2 (2a 1 ) 2 (1b 1 ) 2 (3a 1 ) 1 (1b 2 ) Triplet NH 2 (2a 1 ) 2 (1b 1 ) 2 (3a 1 ) 2 (1b 2 ) H 2 O (2a 1 ) 2 (1b 1 ) 2 (3a 1 ) 2 (1b 2 ) See Photochemistry lecture course next year for applications to excited electronic states of AH 2 molecules.

69 7: Hückel theory for polyatomic molecules Introduction Approximate treatment of π electron systems in organic molecules: 1 2 Approximations π and σ frameworks completely separated 2. Trial wavefunctions can be used of the form: Ψ = i c i p π i 3. H ii = p i Ĥ p i = α same for all atoms 4. H ij = p i Ĥ p j = β i bonded to j Note that β is negative. = 0 otherwise. 5. S ij = p i p j = δ ij, where δ ij = 1 when i = j and δ ij = 0 when i j.

70 Ethene Optimize orbital coefficients as before using the variation principle. Know already that this leads to secular equations ( )( ) α E β c1 = 0. β α E c 2 So the energies are the solutions of the secular determinant α E β β α E = 0 E - = E + = Wavefunctions obtained by substituting the energies E ± back into secular equation: Ψ ± = 1 2 (p 1 ±p 2 ) Thus, with respect to two non-bonded p π electrons, ethene has a π stabilization energy of 2β (i.e., β for each electron). See Sheet 2, question 1 for a problem involving the allyl radical.

71 Butadiene There is nothing in Hückel theory that distinguishes cis from trans, or indeed linear, butadiene. As before, the trial wavefunction, Ψ = c 1 p 1 +c 2 p 2 +c 3 p 3 +c 4 p 4, is optimized using the variation principle. The secular equation can be written down directly α E β 0 0 β α E β 0 0 β α E β 0 0 β α E c 1 c 2 c 3 c 4 = 0. Multiplying out this leads to a quartic equation. Although this can be solved, here we will use of symmetry to simplify the algebra.

72 cont. Butadiene has a two-fold symmetry axis. Form symmetry adapted linear combinations (SALCs) of atomic orbitals χ 1 = 1 2 (p 1 +p 4 ) χ 2 = 1 2 (p 2 +p 3 ) χ 3 = 1 2 (p 2 p 3 ) χ 4 = 1 2 (p 1 p 4 ) χ 1 and χ 2 are symmetric (A), and χ 3 and χ 4 are antisymmetric (B) with respect to the C 2 operation. Let Ψ = c 1 χ 1 +c 2 χ 2 +c 3 χ 3 +c 4 χ 4 H ij and S ij vanish if i and j are of different symmetry: α E β 0 0 β α+β E α β E β 0 0 β α E c 1 c 2 c 3 c 4 = 0. Matrix becomes block diagonal.

73 Look at some of the matrix elements H 11 = χ 1 Ĥ χ 1 = 1 2 (p 1 +p 4 ) Ĥ 1 2 (p 1 +p 4 ) = 1 2 = α { p 1 Ĥ p 1 }{{} α + p 4 Ĥ p 4 }{{} α + p 1 Ĥ p 4 }{{} 0 H 14 = χ 1 Ĥ χ 4 = 1 2 (p 1 +p 4 ) Ĥ 1 2 (p 1 p 4 ) = 1 2 = 0 { p 1 Ĥ p 1 }{{} α p 4 Ĥ p 4 }{{} α p 1 Ĥ p 4 }{{} 0 H 33 = χ 3 Ĥ χ 3 = 1 2 (p 2 p 3 ) Ĥ 1 2 (p 2 p 3 ) = 1 2 { = α β p 2 Ĥ p 2 }{{} α + p 3 Ĥ p 3 }{{} α p 2 Ĥ p 3 }{{} β Note that, as before, the overlap integrals in the SALC basis are S ij = δ ij + p 4 Ĥ p 1 }{{} 0 + p 4 Ĥ p 1 }{{} 0 p 3 Ĥ p 2 }{{} β } } } cont.

74 cont. Now solve the upper 2 2 determinant. (α E)(α+β E) β 2 = 0 E ± = α+ 1± 5 2 β E 1 = α+1.62β E 3 = α 0.62β To get molecular orbital corresponding to E 1 substitute E 1 into secular equations 1.62βc 1 +βc 2 = 0 c 1 = c

75 cont. But wavefunction needs to be normalized Ψ 1 = N [ ] 1 2 (p 1 +p 4 ) (p 2 +p 3 ) Ψ 1 Ψ 1 = N 2 [ ] = 1 So Ψ 1 = 0.37p p p p 4 Similarly, find MO Ψ 3 associated with energy level E 3 Ψ 3 = 0.60p p p p 4

76 cont. From the lower 2 2 determinant one obtains E 2 = α+0.62β E 4 = α 1.62β with the normalized wavefunctions and Ψ 2 = 0.60p p p p 4 Ψ 4 = 0.37p p p p 4 E 4 4 E 3 3 E 2 2 E 1 1

77 Charge density on atoms Once the wavefunctions of the MOs are known the charge densities on each atom, q i, can be obtained from q i = k n k (c k i ) 2, where the sum over k is over all occupied molecular orbitals, and n k is the occupation number of MO k (either 0, 1 or 2 for non-degenerate orbitals). (c k i )2 is the square of the coefficient in the kth orbital on the ith atom. For butadiene, this reads (for atoms 1 and 2) q 1 = = 1 q 2 = = 1 and similarly for atoms 3 and 4. That the charge densities on all the atoms are equal is a general property of the ground states of alternant hydrocarbons. These also have a pairing of the MO energies E = α±xβ. See MKT for more details.

78 Resonance stabilization energy In Hückel theory the resonance stabilization energy provides an indication of the increased molecular stability that arises from π bonding. For the ground electronic state of butadiene the total electronic energy is E tot = 2(α+1.62β) + 2(α+0.62β) = 4α+4.48β Thus the resonance stabilization energy is E = E tot 4α = 4.48β This stabilization can be compared with that of 4β that would be generated from two isolated π bonds (i.e. in two isolated ethene molecules). E deloc = E 4β = 0.48β. This energy provides a measure of the delocalization energy, the extra stabilization that arises from delocalization of the electrons over the π system.

79 Bond order Provides an indication of the strength of the π bond between adjacent atoms i and j ρ ij = k n k c k ic k j. The sum is over MO s k, with n k the occupation number of those orbitals. For butadiene ρ 12 = 2( ) + 2( ) = 0.89 ρ 23 = 2( ) + 2( ) = 0.45 Suggests a partial π bond between the central atoms, so the terminal bonds might be shorter than the central bond. Bond lengths in excited states Similar treatment for excited electronic state. ρ 12 = 2( ) + ( ) + ( ) = 0.45 ρ 23 = 2( ) + ( ) + ( ) = 0.72 E 4 E 3 E 2 E 1 Outer bonds expand, inner bond contracts on excitation to the excited electronic state. (See Spectroscopy lectures, and the Franck-Condon principle - vibrational excitation accompanies electronic transition.)

80 ESR spectrum of the radical cation H H a 2 a 1 H H E 4 E 3 E 2 H H E 1 The strength of the hyperfine coupling to the protons is proportional to the unpaired electron density on the adjacent carbon atom. For the butadiene cation, [C 4 H 6 ] +, a 1 = a(c 2 1 )2 = a = 0.36a a 2 = a(c 2 2) 2 = a = 0.14a where a is a universal constant. The following spectrum is therefore predicted: 0.36a 0.14a electron coupling to 4 almost equivalent H s electron coupling to 2 equivalent H s

81 Susceptibility to electrophilic attack Rate of reaction is proportional to the barrier height. Estimate relative barriers for different pathways. E = E σ + E π # + H or E E + {HH Suppose that E σ is insensitive to position of attack. E π = E π E butadiene π E butadiene π Attack at central carbon Attack at terminal carbon = change in resonance stabilization = 4α+4.48β Eπ (1) 2α+2β E π (1) = 2α 2.48β a bit like ethene reactants + E + E products E π(2) 2α+2 2β a bit like allyl cation E π (2) = 2α 1.65β Attack at terminal atom preferred. Many approximate methods for calculating the localization energy have been devised (e.g., frontier orbitals) - see recommended texts.

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85 8: Applications of Hückel Theory Aromaticity π MOs of a cyclic polyene (N carbon atoms) N= α E β β β α E β β α E β β α E β β α E β β β α E c 1 c 2 c 3 c 4 c 5 c 6 x x x x x x c 1 c 2 c 3 c 4 c 5 c 6 = 0, with x = (α E)/β. A general row of the secular equations gives c n 1 + xc n + c n+1 = 0 For even membered rings (N = 2l where l is an integer) c 2l+n = c n

86 cont. Hence try: c k n = e ik2πn/2l coefficient on n th atom for k th MO where k = 0,±1,±2,,+l Substitute c k n back into equation for coefficients or or e ik2π(n 1)/2l +xe ik2πn/2l +e ik2π(n+1)/2l = 0 ) (e ikπ/l +x+e ikπ/l e ikπn/l = 0 x = 2cos kπ l. Therefore E k = α+2βcos kπ l k = 0,±1,±2,+l

87 cont. Hence, the resulting energy levels can be displayed diagrammatically E k = α+2βcos kπ l α+2βcosθ k = 0,±1,±2,+l, where θ = kπ/l correspond to the angles subtended by the vertices of a regular polyhedron. E = 4α+4β two ethenes. No extra resonance stabilization due to delocalization. Note unpaired electrons and degenerate ground state. E = 6α+8β, i.e. 2β lower than three ethenes - extra resonance stabilization due to delocalization. E = 8α+4(1+ 2)β, i.e. 4( 2 1)β lower than four ethenes - extra resonance stabilization but with unpaired electrons and degenerate ground state.

88 For odd membered rings (N = 2l +1) cont. Hence Therefore c k n = e ik2πn/(2l+1) c 2l+1+n = c n k = 0,±1,±2,,±l E k = α+2βcos 2πk 2l+1 k = 0,±1,±2,±l Origin of the 4n + 2 π-electron rule for stability of aromatic molecules.

89 Perturbation theory and steric interactions Find approximately the properties of a system from those of a system with a slightly different (simpler) Hamiltonian. General result - just quote First order energy E k = Ψ k Ĥ Ψ k E k change in energy of the k th orbital on perturbation Ψ k orbital k of the original (unperturbed) problem Ĥ difference between the original Hamiltonian and the full one (i.e. the perturbation)

90 Example 1 Find the π MO energies of having solved the butadiene problem. H H 1 2 H H 3 4 O The only major difference is in the energy of the oxygen atom p π orbital with respect to that of carbon, i.e. p 4 Ĥ p 4 = α O < α C However, we can estimate the energy lowering ( 0.6β) and can write α O = α C +0.6β Now use butadiene MOs in E k = Ψ k Ĥ Ψ k = i c k i p i Ĥ j c k j p j = i,j c k i ck j p i Ĥ p j (2)

91 cont. But the only matrix element which is affected is p 4 Ĥ p 4 such that All other terms in Eqn.(2) are zero. E k = (c k 4 )2 p 4 Ĥ p 4 = (c k 4 )2 0.6β. 2 (-0.37) x x (-0.60) x x 0.6 H H H H H H H H H H O

92 Example 2 In butadiene the π bond order between the central (2-3) atoms is less than the terminal (1-2, 3-4) bonds. Matrix element H 23 is less negative than H 12 and H 34. Let H 12 = H 34 = β H 23 = β +δ, where β is negative (as usual), but δ is positive. r = 1.3A H H H H r = 1.4A H H Find shift in the MO energies due to this effect. Perturbation to Hamiltonian only involves p 2 Ĥ p 3, therefore E k = (c k 2c k 3 +c k 3c k 2) p 2 Ĥ p 3 = 2c k 2c k 3δ. -2x x x x Tending to the energy level pattern for two ethenes.

93 Non-crossing rule Consider mixing of two orbitals φ 1 and φ 2 Ψ = c 1 φ 1 +c 2 φ 2. The orbitals φ i could be the MOs that we have just been considering for butadiene. We know that the energies of the mixed orbitals will be the solutions of the secular equations (see Problem sheet 1) E ± = 1 { } (H AA +H BB )± (H AA H BB ) 2 +4HAB 2. 2 So E + = E only if H AB = 0. In general, H AB = 0 only if the states have different symmetry (otherwise in region of crossing there will always be some small term in the Hamiltonian which mixes the two orbitals). H BB E - H AB H AA E + Bond length or reaction coordinate States of the same symmetry cannot cross.

94 Correlation diagrams Non-crossing rule is helpful in determining reaction pathways based on orbital symmetry arguments. Consider the ring opening of cyclobutene.? A B C A D B C D? A B D C Two categories of ring-opening. B A D C conrotatory A B C D A or B C 2 v A B C D disrotatory A B D C + or -

95 cont. The ground electronic state of cyclobutene (σ 2 π 2 ) smoothly correlates with the ground state (Ψ 2 1 Ψ2 2 ) of butadiene in the conrotatory mode, giving for thermal reaction. B C A 4 butadiene conrotatory A B cyclobutene disrotatory - - butadiene 4 D 3 B A A B B A From the first excited electronic state (σ 2 ππ ) of cyclobutene there is a smooth correlation with the excited state (Ψ 2 1 Ψ 2Ψ 3 ) of butadiene in the disrotatory mode, giving A for photon-initiated reaction. This is the basis of Woodward-Hoffmann rules. B D C