D2sp3 hybrid orbitals

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... 

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. 

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 ... 

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. 

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. 

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... 

Valence bond theory describes the bonding in complexes in terms of two-electron, coordinate covalent bonds resulting from the overlap of filled ligand orbitals with vacant metal hybrid orbitals that point in the direction of the ligands sp (linear), sp3 (tetrahedral), dsp2 (square planar), and d2sp3 or sp3d2 (octahedral). 

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 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. 

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 ... 

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. 

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... 

But as we assume the pz orbital to be involved mainly in a-bonding by means of d2jp3-hybrid orbitals to the axial carbonyl group, we consider only the two dv orbitals. The a bonds to the metal are formed by overlap of nitrogen lone-pair orbitals and two metal d2sp3 AO s. These may be pointing in the direction of the a bonds as in 44 or in the direction shown in 45 which corresponds to the linear combination of both d orbitals (6). 

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 ... 

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