( R) MOLECULES: VALENCE BOND DESCRIPTION VALENCE BOND APPROXIMATION METHOD. Example: H 2. at some fixed R, repeat for many different values

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1 Lecture #9 page MOLECULES: VLENCE OND DESCRIPTION VLENCE OND PPROXIMTION METHOD Example: H H = + + r r r r r R Determine E ( R) at some fixed R, repeat for many different values Electronic Schrodinger equation for fixed parameter R: H ( r; R) ψ ( r; R) = E ( R) ψ ( r; R) Spatial wavefunction ψ constructed with H atom s O s on nuclei & ( ) ( ) ( ) ( ) ψ = N s s + s s where N is a normalization constant Symmetric w.r.t. e- interchange, spin part must be antisymmetric. Total wavefunction = N s ( ) s ( ) + s ( ) s ( ) α() β( ) α( ) β() Spin wavefunctions are not functions of position no, subscripts Normalization r r σ σ ( ) ( ) + ( ) ( ) α() β( ) α( ) β() = N d d d d s s s s Spin wavefunctions are orthonormal, so just as for He atom,

2 Lecture #9 page dσdσ ( ) ( ) ( ) ( ) α β α β = ( ) ( ) + ( ) ( ) ( ) ( ) ( ) ( ) = ( + 0) dσdσ α β β α α β α β = Spatial part ( ) ( ) ( ) ( ) r r + = r r () ( ) + () ( ) + i () () ( ) ( ) = ( + + ) N d d s s s s N d d s s s s s s s s N S where ( ) ( ) ( ) ( ) S = drs s = dr s s Different from He atom ss state since O s are centered at different atoms Normalization: ( ) N + S = N = ( + S ) Valence bond wavefunction for H, with normalized spatial and spin parts: = s ( ) s ( ) + s ( ) s ( ) α() β( ) α( ) β() ( + S ) Overlap integral S(R) plays important role Overlap of s orbitals on H nuclei at and.0 Overlap integral for H atom s Os vs. internuclear separation S 0.5 R R/a 0 t large R, S( R) 0 s R 0, S( R) No overlap between s orbitals on separated atoms Complete overlap between s orbitals on same atom SR ( e) for H molecule equilibrium bond length

3 Lecture #9 page 3 SR ( ) can be evaluated analytically (McQ problem 9-3) ( ) Now calculate E R for fixed R: * E ( ) ( ) R = drdr dσdσ r; R H ( r; R) ( r ; R) H only operates on spatial coordinates spin part of ( S ) 3 R R SR ( ) = e + R+ 3 * can be omitted E ( ) ( ) ( ) ( ) ( ) R = d d s s + s s H s( ) s( ) s( ) s( ) + r r + Two pairs of equal terms E ( ) () ( ) R = drdr s s H s() s( ) ( + S ) + dr () ( ) dr s s H s( ) s( ) ( + S ) Regroup terms in H : or H H = r r r r r = h + h + U = r r r r r = h + h + U R R Hamiltonian for e- # on nucleus Then

4 Lecture #9 page 4 ( + S ) ( + S ) ( ) () ( ) E ( ) () ( ) R = drdr s s h + h + U s s ( ) ( ) ( ) + dr () ( ) dr s s h + h + U s s ut h () s () s ( ) = h ( ) s ( ) s ( ) = E s ( ) s ( ) s h () s ( ) s () = h ( ) s ( ) s ( ) = E s ( ) s ( ) s So where E s = = H atom O energy ( ) () ( ) () ( ) drdr h s s + h s s = = ( ) ( ) ( ) () ( ) drdr h s s + h s s = S =S () ( ) ( ) ( ) ( ) drdr s s U s s J R Coulomb integral () ( ) ( ) ( ) ( ) drdr s s U s s K R Exchange integral Ground state ectronic energy for H ( ) ( ) ( ) ( ) ( ) J( R) + K( R) + S + E R = S + J R + K R = + This is energy rative to complety separated nuclei and ectrons It includes the value for the H atom O energies Ground state ectronic energy for H rative to separated H atoms

5 Lecture #9 page 5 ( R) E = ( ) + K( R) + J R Coulomb integral terms all describe Coulomb interactions, which would be present classically J R d d s s r r r R ( ) = r r ( ) + + ( ) () ( ) () ( ) s s s s = dr dr + drdr + r r r R nalytical solution (McQ. problem 9-4) R 5 3 R J( R) = e + R R Exchange integral has no similar classical analog or interpretation K( R) = d r d r s( ) s( ) + + s( ) s( ) r r r R Electron is on nucleus and, so is ectron Pury QM effect due to indistinguishability of the two ectrons How big are the two contributions? Coulomb ( J) and exchange ( K) integrals for s orbitals on H atoms separated by distance R J Most of the bond energy arises from exchange, a pury QM effect! K R/ a 0

6 Lecture #9 page 6 What about the antisymmetric spatial wavefunction? () ( ) ( ) ( ) ψ = N s s s s Spin part must be symmetric in this case Three possibilities α() α( ) = N s ( ) s ( ) + s ( ) s ( ) β() β( ) α( ) β( ) + α( ) β( ) is a triplet state - 3 spin states have M =, 0, S The other wavefunction is a singlet state with M = 0 S Treatment analogous to above shows N = ( S ) E ( R) J R = + Energy rative to separated H atoms E ( R) ( ) K( R) + J R = ( ) K( R) + J( R ) - negative, small in magnitude K( R ) - negative, larger in magnitude than J( R ) for all R triplet state is unstable for all R rative to separated atoms Valence bond treatment predicts stable singlet ground state

7 Lecture #9 page 7 6 H molecule potential energy curves Valence bond treatment 4 E - (triplet) E - E (ev) H E + (singlet) Exp't R/a 0 ( R ) E gives equilibrium bond length, dissociation energy e V Exp't V ( Z =.66 ) R.6 au = e ( R ) E e -0.6 au = ev 3.80 ev ev ν vib 4080 cm cm cm - Valence bond treatment is good qualitativy, not accurate quantitativy ccuracy improved somewhat by variational optimization of effective nuclear charge Z