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Basics of VB theory and XMVB program

Computer exercises

Exercise 1 : Starting up with the H<math>{}_2</math> molecule Two Gamess and XMVB input files for the H<math>{}_2</math> molecule are provided in the Exercise folder on the tutorial machines : the file h2-atom.xmi input uses the fragment specification in terms of atoms (frgtyp=atom) ; the file h2-sao.xmi input uses the fragment specification in terms of symmetry-adapted orbitals (frgtyp=sao). There are VBSCF calculations with the 6-31G(d,p) basis set. Just inspect these inputs, run the gamess-xmvb program (using : vbrun h2-atom and : vbrun h2-sao, and analyze the outputs. Then these input files could serve you as templates for the next exercises. Exercise 2 : HF molecule : weights and bond energy Compute a VBSCF three structure wave function for the HF molecule, using the frgtyp=sao specification, automatic guess (guess=auto), and boys keyword. Which structure(s) should be kept in further BOVB calculations ? Using VBSCF orbitals as guess orbitals : Compute a L-BOVB wave function on a selected subset of structures ; Compute a VBCISD wave function, freezing the 1s core orbital of fluorine in the VBCI calculation (NCOR=1 option), and printing only structures which have a coefficient superior to 0.01 (ctol=0.01 option) ; Compare structure weights at the VBSCF, L-BOVB and VBCI levels Compute bond energies at the L-BOVB and VBCISD levels. Remark We do not recommend to use the frgtyp=atom specification together with the automatic guess (guess=auto). With frgtyp=atom you should specify orbital guess from HF MOs through an extra $gus section (see manual, and next exercises). Hints To compute the bond energy at the BOVB level, you can simply use the ROHF energies computed with Gamess for the separate H and F atoms, as the L-BOVB wave function dissociate to uncorrelated H+F fragments. To compute the bond energy at the VBCISD level, you should however compute the separate fragments at this level of theory. >> general guidelines for BOVB calculations Exercise 3 : F<math>{}_2</math> molecule and charge-shift resonance energy Compute a VBSCF wave function for the F<math>{}_2</math> molecule, using the cc-pvtz basis set, and with inactive orbitals localized on only one of the fluorine atoms ; first using the frgtyp=sao specification and automatic guess (guess=auto) ; second using the frgtyp=atom specification and providing HF MOs as guess orbital through an extra $Gus section in the xmvb input D-BOVB level : Compute a L-BOVB wave function using VBSCF orbitals as guess orbitals ; Starting from the previous solution, compute a D-BOVB solution, by allowing only the inactive orbitals to delocalize onto the two atoms, while the active orbitals are kept frozen. Compare total energy with the previous level. π-D-BOVB level : Recompute a π-D-BOVB solution for the F<math>{}_2</math> molecule (see : >> see "high symmetry case" in the "general guidelines for BOVB calculations". Compare the total energy and weights with the previous level. We want to calculate the charge-shift resonance energy (RE_CS) for the F<math>{}_2</math> molecule. For that, we have to compute a VB wave-function corresponding to a single covalent structure, and take the energy difference with the full (covalent+ionic) wave-function. Compute a purely covalent wave function for F<math>{}_2</math> at the VBSCF level. What would be the L-BOVB solution ? Compute a purely covalent wave function for F<math>{}_2</math> at the D-BOVB level. Deduce the RE_CS at the VBSCF, L-BOVB and D-BOVB. Compare it with the estimated exact bond energy : 39.0 kcal/mol). ***** If you have not finished these first three exercises by the end of tutorial session 1, we recommend that you move directly to tutorial 2 ***** Exercice 4 (optional) : Solvent effect on C(Me)<math>{}_3</math>-Cl weights C(Me)<math>{}_3</math>-Cl at equilibrium geometry : Compute a VBSCF wave function using frgtyp=atom and a $Gus section to specify guess orbitals. The active electron pair will be the C-Cl bond, and all inactive orbitals should be localized either on Cl or on the C(Me3) fragment. Which structures should be kept in further BOVB calculations ? Redo the VBSCF calculation reading the orbital file obtained at the previous step as guess file, and now requesting a boys localization (keyword : boys). Compare the VBSCF orbitals obtained with and without the boys keyword (you can use the moldendat utility and them molden to display them). Compute a L-BOVB wave function. Compute a D-BOVB wave function, by freezing the active orbitals, and delocalizing all inactive orbitals onto the whole molecule (see also : >> general guidelines for BOVB calculations). Starting from guess orbitals obtained at equilibrium geometry, redo the D-BOVB calculation for the large inter fragment distance. How does the weights of the different structures evolve when the molecule is stretched ? Redo the D-BOVB calculations at equilibrium geometry and large distance using VBPCM for water. How does the weights change with solvation effects ? Hints and remarks It is strongly advisable to use the boys keyword at the VBSCF step, as it will provide more physically meaningful orbitals for the next BOVB calculations (the same remark would hold with VBCI) To reperform D-BOVB at large interatomic distances (question n°2), you have to proceed in two steps : starting from the orbitals converged at equilibrium distances (question n°1.3) as a guess : reperform a L-BOVB first at large interatomic distances ; then, starting from the orbitals just obtained as a guess do a D-BOVB calculation. Same remark for solvent calculations : you can start from converged gas phase L-BOVB orbitals as guess, but you'll have first to reperform L-BOVB and then D-BOVB.

Paper exercise (optional - homework)

Exercise 5 : The lone pairs of H2O (for further reading, see S. Shaik and P.C. Hiberty, "The Chemist's Guide to VB theory", Wiley, Hoboken, New Jersey, 2008, pp. 107-109) This exercise aims at comparing two descriptions of the lone pairs of H<math>{}_2</math>O : (i) the MO description in terms of non-equivalent canonical MOs and (ii) the « rabbit-ear » VB description in terms of two equivalent hybrid orbitals. <math>\Psi_{\textrm{MO}}</math>   <math>\Psi_{\textrm{VB}}</math> Focusing on the lone pairs only, write the four-electron single-determinants <math>\Psi_{\textrm{MO}} </math> and <math>\Psi_{\textrm{VB}} </math> . Expand <math>\Psi_{\textrm{VB}} </math> into elementary determinants containing only <math>n</math> and <math>p</math> orbitals, eliminate determinants having two identical spinorbitals, and show the equivalence between <math>\Psi_{\textrm{VB}}</math> and <math>\Psi_{\textrm{MO}}</math>. We now remove one electron from H<math>{}_2</math>O. Write the two possible VB structures <math>\Phi_1</math> and <math>\Phi_2</math> in the VB framework. The two ionized states are the symmetry-adapted combinations and . From the sign of the hamiltonian matrix element <math>\langle \Phi_1 \vert \hat{H} \vert \Phi_2 \rangle</math>, give the energy ordering of the two ionized states. By expanding the two ionized states into elementary determinants (dropping the normalization constants), show that they are equivalent, respectively, to the MO configurations <math>\vert nn\bar{p}\vert</math> and <math>\vert pp\bar{n}\vert</math>. Appendix Hamiltonian matrix element between determinants differing by one spin-orbital : <math>\langle \cdots i \cdots \vert \hat{H} \vert \cdots j \cdots\rangle = \beta_{ij}</math> Don’t forget to put the orbitals of the two determinants in maximal correspondence before applying the rule. Answer 1. <math> \Psi_{\textrm{MO}}=\vert n\bar{n}p\bar{p}\vert </math>   ;   <math> \Psi_{\textrm{VB}}=\vert \left( n-\lambda p\right)\left( \bar{n}-\lambda\bar{p}\right)\left( n+\lambda p\right)\left(\bar{n}+\lambda\bar{p}\right)\vert </math> 2. <math> \begin{matrix} \Psi_{\textrm{VB}} &=&\vert \left( n\bar{n} -\lambda p\bar{n} -\lambda n\bar{p} +\lambda^2 p\bar{p} \right) \left( n\bar{n} +\lambda p\bar{n} +\lambda n\bar{p} +\lambda^2 p\bar{p} \right) \vert \end{matrix} </math> <math> \begin{matrix} \Psi_{\textrm{VB}} &=&\vert n\bar{n}n\bar{n} \vert + \lambda\vert n\bar{n}p\bar{n} \vert + \lambda\vert n\bar{n}n\bar{p} \vert + \lambda^2 \vert n\bar{n}p\bar{p} \vert - \lambda \vert p\bar{n}n\bar{n} \vert -\lambda^2 \vert p\bar{n}p\bar{n} \vert -\lambda^2 \vert p\bar{n}n\bar{p} \vert -\lambda^3 \vert p\bar{n}p\bar{p} \vert \end{matrix} </math> <math> \begin{matrix} - \lambda \vert n\bar{p}n\bar{n} \vert -\lambda^2 \vert n\bar{p}p\bar{n} \vert -\lambda^2 \vert n\bar{p}n\bar{p} \vert -\lambda^3 \vert n\bar{p}p\bar{p} \vert + \lambda^2 \vert p\bar{p}n\bar{n} \vert +\lambda^3 \vert p\bar{p}p\bar{n} \vert +\lambda^3 \vert p\bar{p}n\bar{p} \vert +\lambda^4 \vert p\bar{p}p\bar{p} \vert \end{matrix} </math> After eliminating all determinants having two orbitals with the same spin, there remains : <math> \begin{matrix} \Psi_{\textrm{VB}} &=& \lambda^2 \vert n\bar{n}p\bar{p} \vert -\lambda^2 \vert p\bar{n}n\bar{p} \vert -\lambda^2 \vert n\bar{p}p\bar{n} \vert + \lambda^2 \vert p\bar{p}n\bar{n} \vert \end{matrix} </math> After permuting the columns and changing signs accordingly, there remains : <math> \Psi_{\textrm{VB}}=4\lambda^2\vert n\bar{n}p\bar{p} \vert=\Psi_{\textrm{MO}} </math> (if one includes normalization factors). 3. <math>\Phi_1=\vert \left( n-\lambda p \right)\left( \bar{n}-\lambda\bar{p} \right)\left( n+\lambda p \right) \vert </math>, <math>\Phi_2=\vert \left( n+\lambda p \right)\left( \bar{n}+\lambda\bar{p} \right)\left( n-\lambda p \right) \vert </math> Permuting the first and third orbitals in <math>\Phi_2</math> and changing the sign accordingly, we get <math>-\Phi_2</math> that has maximum orbital correspondence with <math>\Phi_1</math> : <math>-\Phi_2</math> =<math>\vert \left( n-\lambda p \right) \left( \bar{n}+\lambda\bar{p} \right) \left( n+\lambda p \right) \vert </math>. 4. <math>\Phi_1</math> and <math>-\Phi_2</math> differ by only one orbital, <math>\left( \bar{n}-\lambda \bar{p} \right)</math> in <math>\Phi_1</math> which becomes <math>\left( \bar{n}+\lambda \bar{p} \right)</math> in <math>-\Phi_2</math>. Therefore the matrix element <math> \langle \Phi_1 \vert \hat{H} \vert -\Phi_2 \rangle </math> is a simple <math>\beta</math> integral, necessarily negative. Hence, the lowest ionized state is <math> \frac{1}{\sqrt{2}}\left(\Phi_1-\Phi_2\right)</math> while the higher ionized state is <math> \frac{1}{\sqrt{2}}\left(\Phi_1+\Phi_2\right)</math>. 5. <math>\Phi_1=\vert n\bar{n}n\vert -\lambda\vert p\bar{n}n\vert -\lambda\vert n\bar{p}n\vert +\lambda^2\vert p\bar{p}n\vert+\lambda\vert n\bar{n}p\vert-\lambda^2\vert p\bar{n}p\vert-\lambda^2\vert n\bar{p}p\vert+\lambda^3\vert \bar{p}p\vert</math> <math>\Phi_1=+2\lambda^2\vert p\bar{p}n \vert +2\lambda\vert n\bar{n}p \vert</math> In the same way, one shows that <math>\Phi_2=-2\lambda^2\vert p\bar{p}n \vert +2\lambda\vert n\bar{n}p \vert</math>. It follows that : <math>\left(\Phi_1-\Phi_2\right)\propto \vert n\bar{n}p \vert </math> (lowest ionized state in MO theory) <math>\left(\Phi_1+\Phi_2\right)\propto \vert p\bar{p}n \vert </math> (higher ionized state in MO theory). It is concluded that 1) VB theory yields two ionization potentials for H<math>{}_2</math>O, in agreement with experiment, and 2) that these ionization potentials are exactly the same as the ones found in elementary MO theory.