Sp1 hybrid orbitals

In the boron trimethyl molecule the boron atom is surrounded by three pairs of valence electrons, which are involved in the formation of single covalent bonds to the three carbon atoms of the methyl groups. An electron-diffraction study10 has shown the molecule to be planar (except for the hydrogen atoms), as would be expected for sp1 hybrid orbitals. The —C distance is 1.56 i 0.02 A, which agrees reasonably well with the value 1.54 A calculated, with the electronegativity correction, by the use of 0.81 A for the boron single-bond radius.11... 

The electronic structure of dibromocarbene is shown next. The carbon atom has only six electrons in its valence shell. It is sp2 hybridized, with trigonal geometry. An unshared pair of electrons occupies one of the sp1 hybrid orbitals, and there is an empty p orbital extending above and below the plane of the atoms. A carbene has both a lone pair of electrons and an empty p orbital, so it can react as a nucleophile or as an electrophile. 

Noting that the bond angle of an sp hybridized atom is 109.5° and that of an sp1 hybridized atom is 120°, do you expect the bond angle between two hybrid orbitals to increase or decrease as the s-character of the hybrids is increased ... 

Similar, but different, redeployment is envisaged when a carbon atom combines with three other atoms, e.g. in ethene (ethylene) (p. 8) three sp2 hybrid atomic orbitals disposed at 120° to each other in the same plane (plane trigonal hybridisation) are then employed. Finally, when carbon combines with two other atoms, e.g. in ethyne (acetylene) (p. 9) two sp1 hybrid atomic orbitals disposed at 180° to each other (idigonal hybridisation) are employed. In each case the s orbital is always involved as it is the one of lowest energy level. 

These are all valid ways of deploying one 2s and three 2p atomic orbitals—in the case of sp2 hybridisation there will be one unhybridised p orbital also available (p. 8), and in the case of sp1 hybridisation there will be two (p. 10). Other, equally valid, modes of hybridisation are also possible in which the hybrid orbitals are not necessarily identical with each other, e.g. those used in CH2C12 compared with the ones used in CC14 and CH4. Hybridisation takes place so that the atom concerned can form as strong bonds as possible, and so that the other atoms thus bonded (and the electron pairs constituting the bonds) are as far apart from each other as possible, i.e. so that the total intrinsic energy of the resultant compound is at a minimum. 

Arynes present structural features of some interest. They clearly cannot be acetylenic in the usual sense as this would require enormous deformation of the benzene ring in order to accommodate the 180° bond angle required by the sp1 hybridised carbons in an alkyne (p. 9). It seems more likely that the delocalised 7i orbitals of the aromatic system are left largely untouched (aromatic stability thereby being conserved), and that the two available electrons are accommodated in the original sp2 hybrid orbitals (101) ... 

Both carbon atoms in ethene undergo sp2 hybridization. The C-H bonds involve overlap of sp1 carbon orbitals with Is orbitals of the H atoms. The carbon-carbon double bond involves the overlap of sp2 orbitals from each carbon to give the o bond and the side-on overlap of a p orbital from each carbon atom to give the n bond. 

Hybrid orbitals formed two sp three sp1 four spi one d five sp1d two d six sp dl... 

The halide (rimers consist or planar six-membered rings (Fig. I6.23).77 The bond angles are consistent with sp1 hybridization of the nitrogen and approximately sp3 hybridization of the phosphorus. Two of the sp3 orbitals of nitrogen, containing one electron each, are used for o bonding and the third contains a lone pair of electrons This leaves one electron left for the unhybridized p. orbital. 

The first X-ray crystal structure determination, carried out by Mills in cooperation with us 31) on pentacarbonyl[methoxy (phenyl) carbene ]-chromium(0), confirmed our originally postulated bonding concept. According to this concept, the carbene carbon atom is sp1 hybridized. It should therefore possess an empty p-orbital and be electron-deficient. 

Although it will not be important for our purposes, the dsp3 hybrid orbital set is different from the hybrids we have considered so far in that the hybrid orbitals pointing to the vertices of the triangle (often called the three equatorial hybrid orbitals) are slightly different in shape than the other two (the axial orbitals). This situation stands in contrast to the sp, sp1, and sp3 hybrid sets in which each orbital in a particular set is identical in shape to the others. 

The sp1 hybrid atomic orbitals have the following general form. 

See Fig. 10-7. The C s use sp1 hybrid atomic orbitals to form a bonds with each other and with the H s. The remaining p orbitals at right angles to the plane of the C s overlap laterally to form a n electron cloud. 

See Fig. 20-1. The four C s and the heteroatom Z use sp1 -hybridized atomic orbitals to form the o bonds. When Z is O or S, one of the unshared pairs of e s is in an sp2 HO. Each C has a p orbital with one electron and the heteroatom Z has a p orbital with two electrons. These five p orbitals are parallel to each other and overlap side-by-side to give a cyclic n system with six p electrons. These compounds are aromatic because six electrons fit Hiickel s 4n + 2 rule, which is extended to include heteroatoms. 

The first three geometries involve the tetrahedral, trigonal, and digonal hybrids discussed above and the fourth involves the use of pure s and p orbitals as discussed on page 149. The last structure contains three equivalent bonds at mutual angles of 60° and a fourth bond at an angle or approximately 145° to the others. It is impossible to construct s-p hybrid orbitals with angles less than 90°, and so structure V is ruled out. In this sense it may be said that hybridization does not allow structure V, but it may not be said that it chooses" one of the others. Carbon hybridizes sp, sp2, and sp1 in various compounds, and the choice of sp3 in methane is a result of the fact that the tetrahedral structure is the most stable possible. 

STRATEGY Use the VSEPR model to identify the shape of the molecule and then assign the hybridization consistent with that shape. All single bonds are cr-bonds and multiple i bonds are composed of a cr-bond and one or more TT-bonds. Because the C atom is attached to three atoms, we anticipate that its hybridization scheme is sp1 and that one unhybridized p-orbital remains. Finally, we form cr- and Tr-bonds by allowing the 1 orbitals to overlap. 

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