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Version du 12 juillet 2012 à 12:05

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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


2/ Computer exercise

VBSCD for H—H + H. -> H. + H—H at VBSCF then VBCISD level.

In this exercise the VBSCD for H—H + H. -> H. + H—H at VBSCF then VBCISD level will be computed with 6-31G**. Computations for reactant and transition state are requested and other points are optional for advanced users.

2.1 Computations

Compute the Energies and Wavefunctions at Reactant and Transition State with Different Sets of VB Structures

  1. Write the following VB structure sets, compare and see the difference:
    1. all structures;
    2. minimal structures for reactant;
    3. minimal structures for product.
  2. Perform VBSCF and VBCISD calculations for reactant:
    1. Perform a VBSCF calculation with "orbtyp=hao" and "boys";
    2. Perform a VBCISD calculation with VBSCF orbital as initial guess.
    3. Perform VBSCF and VBCISD calculations with minimal structures for reactant and product.
  3. Perform VBSCF and VBCISD calculations for transition state with the same procedure as in step 2.


2.2 Analysis: Wavefunctions and Energies

  1. Compute the Barrier height of the <math>\textrm{S}_{\textrm{N}}2</math> reaction at VBSCF and VBCISD levels. See the difference of the barrier heights.
  2. Compare the energies for reactant and product structures at reactant and transition state, by both VBSCF and VBCISD. What's the difference of the energies at different points? Why?
  3. Compute the resonance energies at both reactant and transition state points, see the difference of the resonance energies.


2.3 Optional : Compute all points and draw the VBSCDs at VBSCF and VBCISD levels.

| class="collapsible collapsed wikitable" |- !BLW within GAMESS (Version: MAR-25-2010 R2) |- | a- VB Structures used in the computations

Structures.png


Total 8 Structures of The System


Reactant Structures.png


Structures of The Reactant State


Product Structures.png


Structures of The Product State


b- Computational results

Weights of VB Structures of H-H-H Abstract Reaction at Reactant Geometry
S1 S2 S3 S4 S5 S6 S7 S8
VBSCF 0.803 0.003 0.096 0.001 0.095 0.000 0.000 0.000
VBCISD 0.770 0.005 0.110 0.003 0.111 0.000 0.000 0.001


Weights of VB Structures of H-H-H Abstract Reaction at Transition State Geometry
S1 S2 S3 S4 S5 S6 S7 S8
VBSCF 0.344 0.344 0.096 0.035 0.025 0.025 0.096 0.035
VBCISD 0.358 0.358 0.059 0.036 0.046 0.046 0.059 0.036


Energies (a.u.) and Barriers (kcal/mol) of H-H-H Abstract Reaction
VBSCF VBCISD
Reactant -1.64637 -1.66241
Transition State -1.60706 -1.63827
Barrier 24.7 15.1


Energies(a.u.) and Resonance Energies (B, in kcal/mol) of H-H-H Abstract Reaction at Reactant Geometry
VBSCF VBCISD
All Structures -1.64637 -1.66241
Reactant -1.64617 -1.66208
Product -1.40873 -1.41851
B 0.1 0.2


Energies (a.u.) and Resonance Energies (B, in kcal/mol) of H-H-H Abstract Reaction at Transition State Geometry
VBSCF VBCISD
All Structures -1.60706 -1.63827
Reactant -1.54798 -1.56655
Product -1.54798 -1.56655
B 37.1 45.0



Optional : VBSCDs for H-H-H abstract reaction by VBSCF and VBCISD

H3-VBSCD-VBSCF.png H3-VBSCD-VBCISD.png
  VBSCF   VBCISD |}

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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