D2sp3 hybridization

For metal ions having configurations d°, d1, d1, or d. there will always be two of the d orbitals empty to form a set of d2sp3 hybrids. Therefore, we expect complexes of these metal ions to be octahedral in which the hybrid orbital type is d2sp3. If we consider Cr3+ as an example, the formation of a complex can be shown as follows ... 

However, we should emphasize that the NBO/NRT concepts of hybridization, Lewis structure, and resonance differ in important respects from previous empirical usage of these terms. In earlier phases of valence theory it was seldom possible to determine, e.g., the atomic hybridization by independent theoretical or experimental procedures, and instead this term became a loosely coded synonym for the molecular topology. For example, a trigonally coordinated atom might be categorized as sp2-hybridized or an octahedrally coordinated atom as d2sp3-hybridized, with no supporting evidence for the accuracy of these labels as descriptors of actual... 

A persistent feature of qualitative models of transition-metal bonding is the supposed importance of p orbitals in the skeletal hybridization.76 Pauling originally envisioned dsp2 hybrids for square-planar or d2sp3 hybrids for octahedral bonding, both of 50% p character. Moreover, the 18-electron rule for transition-metal complexes seems to require participation of nine metal orbitals, presumably the five d, one s, and three p orbitals of the outermost [( — l)d]5[ s]1[ p]3 quantum shell. 

A large number of halogen carbonyls has been prepared some of them have not the 18-electron configuration in PtCl2(CO)2, for example, there are only sixteen electrons. The carbonyls always have eighteen electrons, but the number of ligands is variable. It is therefore doubtful that they have d2sp3 hybridization. 

AB6, Octahedral. This is perhaps the best known case, where the AIlf + Eg + TXu set of orbitals is made up of, (dzi, and (pr, pv, pz) and d2sp3 hybrids are often cited in the chemical literature. 

Construct wave functions for six equivalent octahedral (d2sp3) hybrid orbitals, using dX2 y2, d, s, px, py, and pz valence orbitals. 

In octahedral complexes, the metal ion uses either sp3d2 or d2sp3 hybrid orbitals. To see the difference between these two kinds of hybrids, let s consider... 

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... 

In Section 7.2.1, we have seen that many hybridization schemes involve d orbitals. In fact, we do not anticipate any technical difficulty in the construction of hybrids that have d orbital participation. Let us take octahedral d2sp3 hybrids, directed along Cartesian axes (Fig. 7.1.10), as an example. From Table 7.1.5,... 

The postulated existence of the oxysulfide MoOS can be ruled unlikely on the basis of the structures of Mo02 und MoS2. Molybdenum dioxide has a three-dimensional structure with distorted octahedral coordination (d2sp3 hybridization) about the molybdenum atoms12while MoS2 has a layer structure with a trigonal prismatic coordination about the molybdenum atoms (d4sp hydridization). Consequently, any existence of a MoOS species may be expected to be only transient. 

The basic idea in this approach is that six atomic orbitals are used to form six hybrid orbitals that are directed toward the corners of an octahedron, the known structure of complexes containing Ti3+. Hybrid orbital types that meet these requirements are d2sp3 and sp3d2. In the case of the Ti3+ ion, two of the 3d orbitals are empty so the dx yi and dzi, the 4s orbital, and the three 4p orbitals can form a set of d2sp3 hybrids. Moreover, the orbitals will be empty as is required for the metal to accept the six pairs of electrons from the six ligands. The electrons in the complex ion [Ti(H20)6]3+ can now be represented as follows ... 

Other d1, d2, or d3 metal ions would behave similarly because they would still have two empty 3d orbitals that could form a set of d2sp3 hybrids. Thus, complexes of Ti2+, V3+, and Cr3+ would be similar except for the number of electrons in the unhybridized 3d orbitals. Addition of six ligands such as NH3 or H20 will result in an octahedral complex in which the six pairs of donated electrons are arranged as shown for Cr3+, a d3 ion ... 

If only octahedral complexes are considered, it can be seen that two of the 3d orbitals are required for the formation of d2sp3 hybrid orbitals. The remaining three orbitals can hold a maximum of six electrons, so it is impossible to have two of the 3d orbitals vacant if there are seven or more electrons in the metal ion. As a result, d7, ds, d9, or octahedral complexes are always of the outer orbital (high-spin) type. 

Consider the Ni2+ ion that has the configuration 3ds. It is readily apparent that if an octahedral complex is formed, it must be of the outer orbital type because two of the 3d orbitals cannot be vacated to form d2sp3 hybrids. However, a set of sp3 hybrids could be formed without involving the 3d orbitals. It is not surprising to find that tetrahedral complexes of Ni2+ (such as [Ni(NH3)4]2+) are known that have magnetic moments of about 3 BM corresponding to two unpaired electrons. This type of complex can be represented as... 

This requires an octahedral arrangement of pairs and, in turn, an octahedral set of six hybrid orbitals. This leads to d2sp3 hybridization, in which two d orbitals, one s orbital, and three p orbitals are combined (see Fig. 14.23). Note that six electron pairs around an atom are always arranged octabedrally, requiring d2sp3 hybridization of the atom. Each d2sp3 orbital on the sulfur atom is used to bond to a fluorine atom. Since there are four pairs on each fluorine atom, the fluorine atoms are assumed to be sp3 hybridized. 

The Lewis structure for XeF4 has six pairs of electrons around xenon that are arranged octahedrally to minimize repulsions. An octahedral set of six atomic orbitals is required to hold these electrons, and the xenon atom is assumed to be d2sp3 hybridized. 

Compounds of coordination number jive are comparatively rare. The complexes of coordination number six are mostly octahedral, and contain d2sp3 hybrid bonds. 

For an octahedral complex the jr-bonding leads in the (d2sp3) hybridization model to an increase in the number of electrons in the group of d orbitals which may be treated by Slater s rules, as well as to an increase of electrons in the next outer shell. As it is probable that the jr-coordinated electrons may not be completely assigned to the transition metal ion, and as the more accurate tables of Clementi 108> contain only values for neutral atoms, the resulting effect can... 

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