Sp2 hybrid orbital

Figure 1.13 An sp hybridized carbon. The three equivalent sp2 hybrid orbitals (green) lie in a plane at angles of 120° to one another, and a single unhybridized p orbital (red/blue) is perpendicular to the sp2 plane.

When we discussed sp3 hybrid orbitals in Section 1.6, we said that the four valence-shell atomic orbitals of carbon combine to form four equivalent sp3 hybrids. Imagine instead that the 2s orbital combines with only two of the three available 2p orbitals. Three sp2 hybrid orbitals result, and one 2p orbital remains unchanged- The three sp2 orbitals lie in a plane at angles of 120° to one another, with the remaining p orbital perpendicular to the sp2 plane, as shown in Figure 1.13. 

Humulene. structure of, 202 Hund s rule, 6 sp Hybrid orbitals. 17-18 sp2 Hybrid orbitals, 15. sp3 Hybrid orbitals, 12-14 Hydrate, 701... 

We use different hybridization schemes to describe other arrangements of electron pairs (Fig. 3.16). For example, to explain a trigonal planar electron arrangement, like that in BF, and each carbon atom in ethene, we mix one s-orbital with two /7-orbitals and so produce three sp2 hybrid orbitals ... 

Multiple bonds are formed when an atom forms a tr-bond by using an sp or sp2 hybrid orbital and one or more ir-bonds by using unhybridized p-orbitals. The side-by-side overlap that forms a ir-bond makes a molecule resistant to twisting, results in bonds weaker than tr-bonds, and prevents atoms with large radii from forming multiple bonds. 

First, the VB part of the description of benzene. Each C atom is sp2 hybridized, with one electron in each hybrid orbital. Each C atom has a p.-orbital perpendicular to the plane defined by the hybrid orbitals, and it contains one electron. Two sp2 hybrid orbitals on each C atom overlap and form cr-bonds with similar orbitals on the two neighboring C atoms, forming the 120° internal angle of the benzene hexagon. The third, outward-pointing sp2 hybrid orbital on each C atom forms a hydrogen atom. The resulting cr-framework is the same as that illustrated in Fig. 3.20. 

Both non-bonding and bonding electron pairs occupy sp2-hybrid orbitals while again a p-orbital is unoccupied. In this case, the bonding angle should be 120° (geometry B). 

An oxygen atom can also form a double bond to carbon thus in propanone (acetone), Me2C=Q , the oxygen atom could use three sp2 hybrid orbitals one to form a a bond by overlap with an sp2 orbital of the carbon atom, and the other two to accommodate the two lone pairs of electrons. This leaves an unhybridised p orbital on both oxygen and carbon, and these can overlap with each other laterally (cf. C=C, p. 9) to form a n bond ... 

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) ... 

In some molecules that involve sp2 hybrid orbitals on the central atom, the bond angles deviate considerably from 120°. For example in F2CO, the bond angle is 108° ... 

The planar framework has a bonds as just shown, which involve sp2 hybrid orbitals on the boron atoms. This leaves one unhybridized p orbital that is perpendicular to the plane. The B2H6 molecule can be considered being made by adding two H+ ions to a hypothetical B21142 ion that is isoelectronic with C2H4 because each carbon atom has one more electron than does a boron atom. In the B2I l42 ion, the two additional electrons reside in a tt bond that lies above and below the plane of the structure just shown. When two H+ ions are added, they become attached to the lobes of the n bond to produce a structure, the details of which can be shown as... 

The strength of a Lewis acid is a measure of its ability to attract a pair of electrons on a molecule that is behaving as a Lewis base. Fluorine is more electronegative than chlorine, so it appears that three fluorine atoms should withdraw electron density from the boron atom, leaving it more positive. This would also happen to some extent when the peripheral atoms are chlorine, but chlorine is less electronegative than fluorine. On this basis, we would expect BF3 to be a stronger Lewis acid. However, in the BF3 molecule, the boron atom uses sp2 hybrid orbitals, which leaves one empty 2p orbital that is perpendicular to the plane of the molecule. The fluorine atoms have filled 2p orbitals that can overlap with the empty 2p orbital on the boron atom to give some double bond character to the B-F bonds. 

From the sketch, we can see that two p orbitals and one s orbital are being hybridized, forming three new sp2 hybrid orbitals. See the following figure in Section 8-6. The three sp2 hybrid orbitals are arranged as so ... 

Let s begin assigning valence electrons by half-filling three sp2 hybrid orbitals on the sulfur atom (atom A) and one sp2 hybrid orbital on each of the oxygen atoms (atoms B, C and D)... 

Next, we half-fill the lone unhybridized 3p orbital on sulfur and the lone 2p orbital on the oxygen atom with a formal charge of zero (atom B). Following this, the 2p orbital of the other two oxygen atoms (atoms C and D), are filled and then lone pairs are placed in the sp2 hybrid orbitals that are still empty. At this stage, then, all 24 valence electrons have been put into atomic and hybrid orbitals on the four atoms. Now we overlap the six half-filled sp2 hybrid orbitals to generate the cr-bond framework and combine the three 2p orbitals (2 filled, one half-filled) and the 3p orbital (half-filled) to form the four 7t-molecular orbitals, as shown below ... 

The half-filled sp2 hybrid orbitals overlap to form the ar bonding structure, and a hexagonal array of atoms. The 2p2 orbitals then overlap to form the n bonding orbitals. There are as many n electrons in a sample of BN as there are in a sample of graphite, assuming both samples have the same number of atoms. 

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