Square-planar point group

The cross terms (df)0(d,), with i j in Eq. (10.3) do not appear in the case of the isolated atom for which the electron density is the sum of the square of the atomic orbitals. In the molecular case, the cross terms will only be nonzero for orbitals belonging to the same representation of the point group of the molecule, like the eg orbitals in the case of trigonal site symmetry discussed above. In the square-planar point group D4h(4/m mm), the orbitals have alg, blg, b2g, and eg symmetry, and no such mixing occurs. 

The molecular structure of yellow crystalline Xe(OTep5)4 has been determined by X-ray analysis (see Pig. 18.3b) / the Xe atom is surrounded by a square-planar array of four O atoms, with the adjacent TeFj groups pointing, curiously, pair-wise up and down from this plane (Xe-O 203.9(5) and 202.6(5) pm, Te-O 188.5 pm). 

In addition, G and F matrix elements have been tabulated (see Appendix VII in Nakamoto 1997) for many simple molecular structure types (including bent triatomic, pyramidal and planar tetratomic, tetrahedral and square-planar 5-atom, and octahedral 7-atom molecules) in block-diagonalized form. MUBFF G and F matrices for tetrahedral XY4 and octahedral XY molecules are reproduced in Table 1. Tabulated matrices greatly facilitate calculations, and can easily be applied to vibrational modeling of isotopically substituted molecules. Matrix elements change, however, if the symmetry of the substituted molecule is lowered by isotopic substitution, and the tabulated matrices will not work in these circumstances. For instance, C Cl4, and all share full XY4 tetrahedral symmetry (point group Tj), but... 

Next, take the square planar molecule AB4 (say, PtClJ ), it is not linear, it does not have two or more Cn with 3 (though it does have one), its principal axis is CA and there are four axes perpendicular to this axis, the plane of the molecule is a ah plane and therefore it belongs to the point group. Notice that this molecule also possesses aA planes, hut the ah plane is enough to associate it with the point group. 

Determine what type of tt--bonding hybrid orbitals can be formed for the square planar AB4 molecule which belongs to the point group. 

Since a molecule with a center of symmetry, such as one belonging to point groups D h (n even). Cntl (n even). Dtlll (n odd), O, and //, cannot have a dipole moment, no matter how polar the individual bonds (Chapter 3), dipole moments have proved to be useful in distinguishing between two structures Much of the classic chemistry of square planar coordination compounds of the type MA2B2 was elucidated on the basis of cis isomers having dipole moments and traps isomers having none (see Chapter 12). ... 

If the molecule were to adopt a hypothetical square-planar structure in point group I>4h, the normal modes would be... 

Fig. 1 Schematic energy levels of the d orbitals in square-planar coordination of 3d metal phthalocyanine complexes. The ordering of the energy levels can be changed except for the highest orbital. The labels on the right are the irreducible representation in the D41, point group...

When a molecule is relatively small and/or belongs to a point group of relatively high symmetry, it is possible to elucidate the molecular structure by using the symmetry selection rules discussed in Section 1.14. Molecules of XY2 (linear or bent C2v), XY3 (planar D3h or pyramidal C3v), XY4 (square-planar D4h or tetrahedral Td) and XY5 (trigonal-bipyramidal D3h or tetragonal-pyramidal C4v) types may take one of the structures indicated in parentheses. Since the number of IR/Raman-active vibrations is different for each structure, the most probable structure can be chosen by comparing the number of observed IR/Raman bands with that predicted for each structure by symmetry selection rules. 

The ligands to be eliminated must be cis to one another for reductive elimination to occur. This is because the process is concerted. Two examples from palladium chemistry make this point clear. Warming in DMSO causes ethane production from the first palladium complex because the two methyl groups are cis in the square planar complex. The more elaborate second bisphosphine forces the two methyl groups to be trans and reductive elimination does not occur under the same conditions. Reductive elimination is one of the most important methods for the removal of a transition... 

Another sort of unconventional type of H-bond that has been proposed in the literature involves electron-rich transition metals as proton acceptor " ". These ideas are derived from spectroscopic data but the most explicit information arises from diffraction studies of crystals, where the evidence is largely geometrical. For example, an NH group approaches unexpectedly close to a Ft atom in a square-planar geometry, with R(Pt—H) = 2.262 A . The arrangement is nearly linear as the 9 (N—H -Pt) angle is 167°. The authors point to prior crystal structures in the literature where N—H and even C—H groups are positioned... 

---