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<big> <math>B = 0.5BDE </math></big>
 
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Version du 12 juillet 2012 à 12:06

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Valence Bond State correlation diagrams

Exercise 1 : Computation of state correlation Diagrams for a 3 centers / 4 electrons system

In this exercise the <math>\textrm{S}_{\textrm{N}}2</math> reaction Cl<math>{}^{-}</math> + CH3Cl -> ClCH3 + Cl<math>{}^{-}</math> will be studied in both vacuum and solution. Valence Bond State Correlation Diagrams (VBSCD) will be constructed at <math>\pi</math>-D-BOVB level. There are two parts in this exercise: basic part and optional part. The basic part is performed with MCP-DZP basis set in which the inner orbitals in Cl and C are described with MCP pseudo potential. The optional part is performed with 6-31+G* basis set, using the general specification for the xmvb input (expert users). Only reactant and transition state will be computed in this exercise, which is sufficient to build the VBSCD diagrams.

>> general guidelines for BOVB calculations


Exercise 3 (paper exercise) : Conical intersection in H3 radical

(for further reading, see S. Shaik and P.C. Hiberty, "The Chemist's Guide to VB theory", Wiley, Hoboken, New Jersey, 2008, pp. 157-161, exercises 6.11-6.14 pp. 174-176, and answers to the exercises pp. 188-192.

Consider three hydrogen atoms Ha, Hb, Hc, with respective atomic orbitals a, b and c, and the two VB structures Image 1.png] and P.png] .

The Ha-Hb and Hb-Hc distances are equal. Image 3.png

  1. By using the thumb rules recalled below, where squared overlap terms are neglected, derive the expression of the energies of R and P, and of the reduced Hamiltonian matrix element between R and P for the 3-orbital/3-electrons reacting system [Ha--Hb--Hc]•.
  2. From the sign of this latter integral when θ > 60°, derive the expressions of the ground state Ψ and of the first excited state Ψ* of the H3• system. One may drop the normalization constants for simplicity. What bonding scheme does the excited state represent ?
  3. Show that the reduced Hamiltonian matrix element is largest in the collinear transition state geometry, and drops to zero in the equilateral triangular structure.
  4. Show that R and P VB structures are degenerate in the equilateral triangular structure, and that Ψ≠ and Ψ* are also degenerate in this geometry.
  5. We now extend the above conclusions to the allyl radical. What are the bonding schemes corresponding to the ground state and first excited state ? What geometrical distortion would make these two states degenerate ? What would be the end product of a photochemical excitation of allyl radical to its first excited state ?

Appendix : Thumb rules for the calculations of effective Hamiltonian matrix elements between determinants.

  • Energy of a determinant D : <math><D|H|D> = -2 \sum_{i<j}^{ } \beta_{ij} S_{ij}</math> (if orbitals i and j have parallel spins)
  • Matrix element between determinants differing by spin inversion of two spin-orbitals :

<math> <D|H|D'>=<|...i\overline{j}...||H||...\overline{i}j...|>= -2 \beta_{ij} S_{ij}</math> (for <math>D</math>, <math>D'</math> differing by spin inversion of two spin-orbitals)

>> Answer

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