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Symmetry descent paths

Scheme 1 JT symmetry descent paths of Oh parent group and its subgroups (upper lines in rectangles) for IRs (bottom lines in rectangles) Eg (a), Tig and Tgg (b). Analogous schemes may be obtained for ungerade IRs ( , Tiu, T2u) replacing subscripts g by the u ones where appropriate. For continuation see Schemes 2 Td and T groups), 4 Sn group) and 6 (Du and Dg groups)... Scheme 1 JT symmetry descent paths of Oh parent group and its subgroups (upper lines in rectangles) for IRs (bottom lines in rectangles) Eg (a), Tig and Tgg (b). Analogous schemes may be obtained for ungerade IRs ( , Tiu, T2u) replacing subscripts g by the u ones where appropriate. For continuation see Schemes 2 Td and T groups), 4 Sn group) and 6 (Du and Dg groups)...
Scheme 2 JT symmetry descent paths of Td parent group and its subgroups (upper lines in rectangles) for IRs (bottom lines in rectangles) E (a), Tj and T2 (b)... Scheme 2 JT symmetry descent paths of Td parent group and its subgroups (upper lines in rectangles) for IRs (bottom lines in rectangles) E (a), Tj and T2 (b)...
We performed MP2/cc-pVTZ geometry optimization of cyclo-CsHs radical using Gaussian03 software [24], We have found Cs stable structure ( A electronic state) and PES saddle points of C2v ( Bi and A2 electronic states) and Cs i A" electronic state) symmetries (see Table 4 and Fig. 1) in agreement with step-by-step descent method (because the original Oh plane of the parent Dsh group is not conserved in the non-planar cyclopropenyl radical). Two symmetry descent paths of Scheme 6b may be employed ... [Pg.63]

It is evident that electronic state of D2h symmetry group cannot arise due to JT effect (by splitting Eg electronic state of D4h group) because it is not allowed by the symmetry descent path (see Scheme 5c)... [Pg.74]

Planar C2v structure with Ai electronic state (C model) cannot be explained by JT symmetry descent from parent group and must be explained by JT symmetry descent of parent D h symmetry group (see Fig. 4) with double electron degeneracy by the symmetry descent path (Scheme 6b)... [Pg.74]

The method of step-by-step symmetry descent does not explain the mechanisms that are responsible for JT distortions. Some opponents argue that its predictions are far too wide on account of selectivity ( all is possible ). On the other hand, this treatment is based exclusively on group theory and does not account for any approximations used in the recent solutions of Schrddinger equation. Chemical thermodynamics does not solve the problems of chemical kinetics but nobody demands to do it as well. Thus we cannot demand this theory to solve also the mechanistic problems despite the epikernel principle solves it. The problem of too wide predictions can be reduced by minimizing the numbers and lengths of symmetry descent paths (see the applications in this study). [Pg.75]

The method of step-by-step descent in symmetry indicates two descent paths to JT stable immediate subgroups of D2d (see Scheme 5d) [22]... [Pg.71]

This theorem has been proven using an earlier result of Pechukas along a steepest descent path the point symmetry group (as well as the framework group) of nuclear configurations may change only at a critical point, where it must have all those point symmetry elements (framework group elements, resp.) that are present at non-critical points of the path [5]. [Pg.99]

The proof of this statement follows from the fact that if the point symmetry g(K) of a nuclear configuration K e C(X,i)), different from the critical point, K K(X,i), is the same as the point symmetry g(K(X,i)) at the critical point K(X,i) of the same catchment region C(X,i), then this symmetry g(K(X,i)) must be present along the entire steepest descent path from the point K to the critical point g(K(X,i)). Since the steepest descent path must enter the open neighborhood N(K(A,i)) in order to reach the critical point K(X,i), the open neighborhood N(K(X,i)) must also contain a continuum of points with the same symmetry g(K(X,i)). Consequently, if N(K(X,i)) does not have points K K(X,i) with symmetry g(K(X,i)), then the catchment region C(X,i) caruiot have such points either, and the point symmetry g(K( i,i)) is unique within the entire catchment region C(X4). [Pg.36]

This rule has been pointed out in ref. [3] (for an illustration see Figure 4 in ref. [3]). A steepest descent (relaxation) path may lead to and terminate at a boundary point of Gy, however, actual crossing of the boundary cannot happen. The above symmetry boundary non-crossing rule is equally valid for the inverted potential energy hypersurfaces -E(K). [Pg.100]

A correct description of the reaction coordinate (IRC) has been obtained by applying the method of the steepest descent from the saddle point to the neighboring minimum of the PES [19,27,32] (see Sect. 1.3.4.2) for the reaction H +CH4 (X = Y = H). The calculations confirm that along all the path of Eq. (5.2) the 3 symmetry of the reacting system is retained. Therefore the PES in the reaction zone is defined by only four independent parameters (Fig. 5.1). In earlier calculations of this reaction, for example in Ref. [22], where an extended basis of orbitals was employed and polarization functions were included, the reaction path and the transition state structure characteristics were computed in the reaction coordinate regime. The distance r2 (Fig. 5.1) was chosen as such a coordinate. The calculation led to the erroneous conclusion... [Pg.119]


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See also in sourсe #XX -- [ Pg.60 , Pg.61 , Pg.62 , Pg.63 , Pg.64 , Pg.65 , Pg.66 , Pg.74 ]




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Symmetry descent

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