Three-electron stabilization

Decomposition of Diazenes and Diazene iV-Oxides. Consideration of Three-Electron Stabilization. 

Stabilizing resonances also occur in other systems. Some well-known ones are the allyl radical and square cyclobutadiene. It has been shown that in these cases, the ground-state wave function is constructed from the out-of-phase combination of the two components [24,30]. In Section HI, it is shown that this is also a necessary result of Pauli s principle and the permutational symmetry of the polyelectronic wave function When the number of electron pairs exchanged in a two-state system is even, the ground state is the out-of-phase combination [28]. Three electrons may be considered as two electron pairs, one of which is half-populated. When both electron pahs are fully populated, an antiaromatic system arises ("Section HI). 

The negative charge is shared equally by all three oxygens Stabilization of nitrate ion by electron delocalization increases the equilibrium constant for its formation... 

The S-S bond dissociation energies of H2S2, H2S3 and H2S4 have been studied by Steudel et al. at the GCSD(T)//6-311 G(2df,p) level [42]. The calculated enthalpies AH° for the dissociation at the central bonds at 298 K are 247, 201 and 159 kJ mol respectively. The lower stability of the tri- and tetrasulfanes towards homolytic S-S cleavage is attributed to the stability of the generated HSS radical which is stabilized by the formation of a three-electron n bond. 

The particpation of one through three electrons in the orbital interaction gives rise to stabilization. The destabilization occurs when four electrons participate. 

In SnI reactions, the stability of the carbocation is the paramount issue. Recall that alkyl groups are electron donating. Therefore, 3° is best because the three alkyl groups stabilize the carbocation. 1° is the worst because there is only one alkyl group to stabilize the carbocation. This has nothing to do with sterics this is an argument of electronics (stability of charge). So we have two opposite trends, for completely different reasons ... 

For the F(n)-E(ir ) interaction only one electron is involved. For the E(tr)-F(n) interaction there are three electrons involved Two are lowered in energy and one is increased by the same amount. The others are lowered, so that there is a net one electron stabilized ... 

Resembles product of step stabilized by three electron-Because AG° is positive releasing groups... 

Both O and N have, as shown in Table 2.2, two electrons in the 1st shell that is thereby stabilized corresponding to the structure of He. A stabilization in the four orbits of the 2nd shell requires eight electrons. A stabilization of O and N corresponds to an electron configuration where each of these atoms obtains a structure comparable with Ne. The O and N atoms require an additional supply of two and three electrons, respectively, to establish the configuration of Ne, i.e., OX0 = -2 and OXN = -3. Sulfur is an atom with a configuration in the 3rd shell equal to the 2nd shell for oxygen, i.e., with six electrons. By accepting two more electrons, S will approach Ar in terms of stability. The oxidation level of S is, therefore, the same as for O, i.e., OXs = -2. 

Depending upon the numbers of electrons that are shared, double and triple bonds may be formed. For example to be stable the oxygen atom needs two more electrons and so it forms a double bond with another oxygen atom (O = O). The nitrogen atom needs to gain three electrons to reach stability and so it forms a triple bond with another nitrogen atom, (N = N). 

There are two other mechanistic possibilities, halogen atom abstraction (HAA) and halonium ion abstraction (EL), represented in Schemes 4.4 and 4.5, respectively, so as to display the stereochemistry of the reaction. Both reactions are expected to be faster than outer-sphere electron transfer, owing to stabilizing interactions in the transition state. They are also anticipated to both exhibit antiperiplanar preference, owing to partial delocalization over the C—C—Br framework of the unpaired electron in the HAA case or the electron pair in the EL case. Both mechanisms are compatible with the fact that the activation entropies are about the same as with outer-sphere electron donors (here, aromatic anion radicals). The bromine atom indeed bears three electron pairs located in two orthogonal 4p orbitals, perpendicular to the C—Br bond and in one s orbital. Bonded interactions in the transition... 

Pettifor s structure maps additional remarks. We have seen that in a phenomenological approach to the systematics of the crystal structures (and of other phase properties) several types of coordinates, derived from physical atomic properties, have been used for the preparation of (two-, three-dimensional) stability maps. Differences, sums, ratios of properties such as electronegativities, atomic radii and valence-electron numbers have been used. These variables, however, as stressed, for instance, by Villars et al. (1989) do not always clearly differentiate between chemically different atoms. 

Such strong inward pyramidalization of the nitrogen atoms is stabilized by the overlapping of the nitrogen electrons forming a three-electron two-center bond in the cation-radical discussed in... 

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