Sp3d2 hybridization

Thus the bonding in sulfur hexafluoride SF6 has for a long time been considered to involve two of the 3d orbitals of sulfur, with the sulfur in a sp3d2 hybridized state and... 

We need six orbitals to accommodate six electron pairs around an atom in an octahedral arrangement, as in SF6 and XeF4, and so we need to use two d-orbitals in addition to the valence s- and p-orbitals to form six sp3d2 hybrid orbitals (Fig. 3.18). These identical orbitals point toward the six corners of a regular octahedron. 

A transargononic structure for sulfur, with six bonds formed by sp3d2 hybrid orbitals, was suggested for sulfur in the octahedral molecule SF6 long ago, and also for one of the sulfur atoms, with ligancy 6, in binnite (Pauling and Neuman, 1934). Some transargononic structures of metal sulfides have been proposed recently by Franzen (1966). 

The Ni octahedra derive their stability from the interactions of s, p, and d electron orbitals to form octahedral sp3d2 hybrids. When these are sheared by dislocation motion this strong bonding is destroyed, and the octahedral symmetry is lost. Therefore, the overall (0°K) energy barrier to dislocation motion is about COCi/47r where = octahedral shear stiffness = [3C44 (Cu - Ci2)]/ [4C44 + (Cu - C12)] = 50.8 GPa (Prikhodko et al., 1998), and the barrier = 4.04 GPa. The octahedral shear stiffness is small compared with the primary stiffnesses C44 = 118 GPa, and (Cn - C12)/2 = 79 GPa. Thus elastic as well as plastic shear is easier on this plane than on either the (100), or the (110) planes. 

An extreme example of hybidization is the structure proposed for sulphur hexafluoride, SF6. The six S-F bonds are directed to the apices of a regular octahedron. An arrangement which would satisfy this number of covalent bonds is sp3d2 hybridization. The ground state of the sulphur atom is s2p4 and... 

Problem Evaluate the natural bonding angle(s) a for equivalent sp3d2 hybrids. Solution On putting X = 3 and p = 2 into Eq. (4.27), we obtain... 

A molecule exhibits sp3d2 hybridization in its bonding structure. The most probable geometric shape of this molecule is... 

The difference between d2sp3 and sp3d2 hybrids lies in the principal quantum number of the d orbitals. In d2sp3 hybrids, the principal quantum number of the d orbitals is one less than the principal quantum number of the s and p orbitals. In sp3d2 hybrids, the s, p, and d orbitals have the same principal quantum number. To determine which set of hybrids is used in any given complex, we must know the magnetic properties of the complex. 

Because [V(NH3)g]3+ is octahedral, the V3+ ion must use either d2sp3 or sp3d2 hybrid orbitals in accepting a share in six pairs of electrons from the six NH3 ligands. The preferred hybrids are d2sp3 because several 3d orbitals are vacant and d2sp3 hybrids have... 

The distortion of closed shell electrons can be very significant in the case of charged species. The Stemheimer antishielding factor has been determined for sp3d2 hybridized sulphur by evaluation of the ion-water contribution to the 33S relaxation rate in aqueous SO4, 83 The value found (y, — —43.9) is in good agreement with the value reported for S2 (yrf, — —52.2) in the crystalline state.84... 

In structures where there is sp, sp2, sp3, or sp3d2 hybridization, all of the positions are equivalent. For a molecule such as PF5, it is found that there are 10 electrons around the phosphoms atom, five from the P atom and one from each of five F atoms. The five pairs have minimum repulsion when they are directed toward the comers of a trigonal bipyramid, so the structure can be shown as... 

The hybrid orbital type d2sp3 refers to a case in which the d orbitals have a smaller principal quantum number than that of the s and p orbitals (e.g., 3d combined with 4s and 4p orbitals). The sp3d2 hybrid orbital type indicates a case where the s, p, and d orbitals all have the same principal quantum number (e.g., 4s, 4p, and 4d orbitals) in accord with the natural order of filling atomic orbitals having a given principle quantum number. Some of the possible hybrid orbital combinations will now be illustrated for complexes of first-row transition metals. 

For a complex containing a d4 ion, there are two distinct possibilities. The four electrons may either reside in three available 3d orbitals and a set of d2sp3 orbitals will be formed or else the empty 4d orbitals will be used to form a set of sp3d2 hybrids. The first case would have two unpaired electrons per complex ion, whereas the second would have four unpaired electrons per complex ion. Consequently, the magnetic moment can be used to distinguish between these two cases. An example of a d4 ion is Mn3+, for which two types of complexes are known. The first type of complex is illustrated by [Mn(CN)6]. The orbital population that results after the addition of six cyanide ions to form the complex [Mn(CN)6]3- may be shown as follows ... 

The molecule sulfur hexafluoride SF6 exemplifies one of the most common types of d orbital hybridization. The six bonds in this octahedrally-coordinated molecule are derived from mixing six atomic orbitals into a hybrid set. The easiest way to understand how these come about is to imagine that the molecule is made by combining an imaginary S6+ ion (which we refer to as the S(VI) valence state) with six F ions to form the neutral molecule. These now-empty 3s and 3p orbitals then mix with two 3d orbitals to form the sp3d2 hybrids. 

S(VI) represents the six-coordinated valence state of sulfur, which is considered to have lost its 3s and 3p electrons, making six equivalent sp3d2-hybridized orbitals available for accepting lone pairs contributed by the fluorine atoms. 

Combining these with suitable valence orbitals of a transiton metal, we form a o-bond between the sp2 orbital on carbon and one sp3d2 hybrid of o-symmetry (or d 2) on the metal centre (the classical 2VE dative c-bond). We are now left with the vacant p orbital, which has correct symmetry to interact with one metal t2 orbital (in contrast to CO, which... 

Aluminum has 3d orbitals relatively accessible, and not only may the valency of aluminum rise above four, but some d character may be present in the bonds of the tetravalent and also in the bonds of the trivalent aluminum compounds. At present only few organic aluminum compounds with five- and six-coordinated aluminum are known (sp3d and sp3d2 hybrids see Sections III,D and IV,C). The differences between the behavior of aluminum and boron compounds can partially be explained by the possibility of formation of these structures (trigonal bipyramid, octahedron). 

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