Tetrahedral sp3 hybridization

Let us now look at several tetravalent molecules, and see what kind of hybridization might be involved when four outer atoms are bonded to a central atom. Perhaps the commonest and most important example of this bond type is methane, CH4. 

In the ground state of the free carbon atom, there are two unpaired electrons in separate 2p orbitals. In order to form four bonds (tetravalence), need four unpaired electrons in four separate but equivalent orbitals. We assume that the single 2s, and the three 2p orbitals of carbon mix into four sp3 hybrid orbitals which are chemically and geometrically identical the latter condition implies that the four hybrid orbitals extend toward the corners of a tetrahedron centered on the carbon atom. 

If lone pair electrons are present on the central atom, these can occupy one or more of the sp orbitals. This causes the molecular geometry to be different from the coordination geometry, which remains tetrahedral. 

In the ammonia molecule, for example, the nitrogen atom normally has three unpaired p electrons, but by mixing the 2s and 3p orbitals, we can create four sp3-hybrid orbitals just as in carbon. Three of these can form shared-electron bonds with 

Hybridization can also help explain the existence and structure of many inorganic molecular ions. Consider, for example, the zinc compounds shown here. At the top is shown the electron configuration of atomic zinc, and just below it, of the divalent zinc ion. Notice that this ion has no electrons at all in its 4-shell. In zinc chloride, shown in the third row, there are two equivalent chlorine atoms bonded to the zinc. The bonding orbitals are of sp character that is, they are hybrids of the 4s and one 4p orbital of the zinc atom. Since these orbitals are empty in the isolated zinc ion, the bonding electrons themselves are all contributed by the chlorine atoms, or rather, the chlor ide ions, for it is these that are the bonded species here. Each chloride ion possesses a complete octet of electrons, and two of these electrons occupy each sp bond orbital in the zinc chloride complex ion. This is an example of a coordinate covalent bond, in which the bonded atom contributes both of the electrons that make up the shared pair. 

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In addition to compounds with planar, sp2-hybridized carbons, compounds with tetrahedral, sp3-hybridized atoms can also be prochiral. An vp3-hybridizec) atom is said to be a prochirality center if, by changing one of its attached groups, it becomes a chirality center. The —GH2OH carbon atom of ethanol, for instance, is a prochirality center because changing one of its attached -H atoms converts it into a chirality center. 

Figure 3.17 Geometry of hybrid orbitals, (a) digonal sp hybrids oppositely directed along the same axis (b) trigonal sp2 hybrids pointing along three axes in a plane inclined at 120° (c) tetrahedral sp3 hybrids pointing towards the comers of a regular tetrahedron. (Reproduced with permission from R. McWeeny, Coulson s Valence, 1979, Oxford University Press, Oxford.)...

The secondary electronic effects in the ester function are essentially similar to the anomeric effect discussed previously for the acetal function, involving an n-o interaction. The only difference is that the central carbon is trigonal (sp2 hybridized) in esters and tetrahedral (sp3 hybridized) in acetals. 

Hydrocarbon skeletons are built up from tetrahedral (sp3), trigonal planar (sp2),-or linear (sp) hybridized carbon atoms. It is not necessary for you to go through the hybridization process each time you want to work out the shape of a skeleton. In real life molecules are not made from their constituent atoms but from other molecules and it doesn t matter how complicated a molecule might be or where it comes from it will have an easily predictable shape. All you have to do is count up the single bonds at each carbon atom. If there are two, that carbon atom is linear (sp hybridized), if there are three, that carbon atom is trigonal (sp2 hybridized), and, if there are four, that carbon atom is tetrahedral (sp3 hybridized). 

If you had drawn the molecule more professionally as shown in the margin, you would have to check that you counted up to four bonds at each carbon. Of course, if you just look at the double and triple bonds, you will get the right answer without counting single bonds at all. Carbon atoms with no 7E bonds are tetrahedral (sp3 hybridized), those with one ft bond are trigonal (sp2 hybridized), and those with two Jt bonds are linear (sp hybridized). This is essentially the VSEPRT approach with a bit more logic behind it. 

Notice how tire trigonal, planar sp2 hybridized carbon atom of the carbonyl group changes to a tetrahedral, sp3 hybridized state in the product. For each class of nucleophile you meet in this chapter, we will show you the HOMO-LUMO interaction involved in the addition reaction. 

Now the end carbon has a single unpaired electron. What do we do with it Before the bond broke, the end carbon was tetrahedral (sp3 hybridized). We might think that the single electron would still be in an sp3 orbital. However, since an sp3 orbital cannot overlap efficiently with a Jt bond, the single electron would then have to be localized on the end carbon atom. If the end carbon atom becomes trigonal (sp2 hybridized), the single electron could be in a p orbital and this could overlap and combine with the 7t bond. This would mean that the radical could be spread over the molecule in the same orbital that contained tire cation. 

We now apply our model to investigate the formation of four C-H bonds in the methane molecule CH4 of symmetry Tc (Magnasco, 2004a). For this molecule, tetrahedral sp3 hybridization is completely determined by molecular symmetry. We use the usual notation for the eight valence AOs, callings, x, y, zthe2s and 2p orbitals on C, and h, h2,h3, h4 the Is orbitals on the H atoms at the vertices of the tetrahedron (Figure 2.14). Molecular... 

Such a transformation can be used for relocalizing a given set of delocalized molecular orbitals in conformity with the chemical formula. For instance, the occupied orbitals of methane can be transformed into orbitals very close to simple two-center MO s constructed from tetrahedral sp3 hybrid orbitals and Is hydrogen orbitals 24,25,26) a. unitary transformation can hardly modify the wave function, except for an immaterial phase factor therefore, it leads to a description which is as valid as that in terms of the canonical delocalized Hartree-Fock orbitals. Of course, the localization obtained in this way is not perfect, but it is usually much better than is often believed. In the case of methane, the best localized orbitals are uniquely determined by symmetry 27> for less symmetric molecules one needs a criterion for best localization 28 29>, a problem on which we shall not insist here. A careful inspection reveals that there are three classes of compounds ... 

A carbon with four bonded groups is tetrahedral, sp3-hybridized, and has 109.5° bond angles. A carbon with three is trigonal, sp2-hybridized, and has 120° bond angles. A carbon with two bonded groups is linear, sp-hybridized, and has 180° bond angles. 

In a carbanion, there are four space occupying groups - three bonded groups and the non-bonding electron pair. As a result, it is tetrahedral, sp3-hybridized, and has 10 P bond angles. The non-bonding pair is in an sp3 hybrid orbital. 

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