Three electron bond problem

A molecule AB composed of the two monovalent positive atoms A and B (for example, an Hj molecule) approaches the surface of a semiconductor as depicted in Fig. 7a. We observe that in such a configuration the molecule AB, as can be shown, does not form a weak bond with the surface. The author investigated the behavior of a free electron in the semiconductor and the energy of the system as a function of the distance b, which figures in the formulas as a parameter. The problem was treated as a three-electron one (one electron in each of the atoms A and B -)- a free lattice electron). 

Alkenes undergo addition reactions at the double bond. The tt electrons of alkenes are a nucleophilic site and they react with electrophiles by three mechanisms (see Problem 3.40). 

The obvious deficiency of crystal-field theory is that it does not properly take into account the effect of the ligand electrons. To do this a molecular-orbital (MO) model is used in which the individual electron orbitals become a linear combination of the atomic orbitals (LCAO) belonging to the various atoms. Before going into the general problem, it is instructive to consider the simple three-electron example in which a metal atom with one ligand atom whose orbital contains two electrons. Two MO s are formed from the two atomic orbitals... 

Before turning to the problem of calculating the spin Hamiltonian, we should consider how the results can be interpreted with respect to the bonding. From the spin Hamiltonian parameters, we shall obtain the values for Cji of the antibonding orbitals, and we need to know what relation these have to the Cj, of the bonding states. One approach to this problem, which seems fruitful, is the gross-population analysis developed by Mulliken (26). Consider first our three-electron example. In this case we split up the term... 

Answer. For two octahedral clusters fused on an edge, the eve count is (2 x 26 — 14) = 38 whereas the observed count is (10 Ga + 6 R) = 30 -I- 6 = 36. Thus, we cannot assume non-cluster bonding lone pairs on the bare Ga atoms. With the mno rule, m = 2, n = 10 and o = 0 giving m + n = 12 sep. Each of the two Ga atoms shared between the clusters contributes all three valence electrons. Hence, we have 6 RGa + 2 Ga(shared) + 2 Ga(unshared) = (12 + 6 + 2x)/2 = 12 sep, where x is the contribution of the unshared cluster Ga atoms. Clearly x = 3 in this cluster, which suggests there are no formal lone pairs on these two Ga vertices. Indeed, the structure shows the Ga-Ga distances between the apical RGa and Ga centers (broken lines in the drawing) are about 0.2 A shorter than the other Ga-Ga distances. Electron counting identifies the cluster bonding problem but does not solve it. We will have more to say about this cluster type below. 

The basic problem of using NO gas with clusters is that this reagent often cleaves the metal-metal bond as well as substituting for a carbonyl ligand. Since NO donates three electrons, displacement of precisely one CO and one M—M bond is feasible. Equations (4) (75), (5) (76), and (6) (77) are excellent examples of this. 

Slater s method which has been described in Chapter i6 for the problem of three electrons may be applied to systems with any number of electrons. Each electron may exist in the field of any of the nuclei, and resonance among the different states, representing different electronic distributions, will occur. With four atoms having four valency electrons, 4 = 24 different arrangements of the electrons between the nuclei are possible, with 2 = 16 different spin states. In all there will therefore be 24 X 16 = 384 different complete wave functions. But as we have already seen, states with different values of the spin quantum number do not interact with one another. This permits a considerable reduction in the number of states to be considered. We are only concerned with stable configurations in which all the electrons in pairs neutralize their spin by the formation of a covalent bond, i.e. S—0 and S =0. 

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