Hybrid orbitals overlap

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. 

Figure 10-4 shows the hybridization that occurs in ethylene, H2C=CH2. Each carbon has sp2 hybridization. On each carbon, two of the hybrid orbitals overlap with an s-orbital on a hydrogen atom to form a carbon-to-hydrogen covalent bond. The third sp2 hybrid orbital overlaps with the sp2 hybrid on the other carbon to form a carbon-to-carbon covalent bond. Note that each carbon has a remaining p-orbital that has not undergone hybridization. These are also overlapping above and below a line joining the carbons. 

The empty sp hybrid orbital overlaps with a filled orbital of F which holds two electrons. 

The bond angles of the carbon atoms in benzene are 120°. All carbon atoms are s p -hybridized, and each carbon atom has a single unhybridized p orbital perpendicular to the plane of the ring. The carbon p -hybridized orbitals overlap to form the ring of the benzene molecule. Because the C—C bond lengths are 1.39 A, the p orbitals are close enough to overlap efficiently and equally all round the ring. 

Another factor which affects the most stable arrangement of the atom in a molecule is the variation of bond energy with hybridization. The directed lobes of s-p hybrid orbitals overlap more effectively than the undirected s orbitals, the two-lobed p orbitals, or the diffuse if orbitals. The increased overlap results in stronger bonds... 

According to eq. (3.15), the magnitude of one-bond car bon-13 —nitrogen-15 coupling constants correlates with the product of s characters of the hybrid orbitals overlapping in the C-N bond [135] ... 

When one of these hybrid orbitals overlaps an orbital of another atom, a a-bond is formed in the area where the orbitals coincide, ir-bonds are formed by the overlap of pure p orbitals. 

Similarly, in hydrogen cyanide, HCN, we assume that the carbon is sp-hybridized, since it is combined with only two other atoms, and is hence in a divalent state. One of the sp-hybrid orbitals overlaps with the hydrogen 1 s orbital, while the other overlaps end-to-end with one of Vr VP VP the three unhybridized p orbitals of the nitrogen atom. 

C-2 is j/7 -hybridized. The three jp -hybridized orbitals overlap with orbitals on 0-1, C-3, and C-7 to form three cr bonds that lie in the same plane approximately 120° from each other. The p orbital, perpendicular to this plane, is parallel to the p orbital on 0-1 so these p orbitals can overlap to produce the C—O tt bond. 

The empty sp hybrid orbital overlaps with a filled orbital of F which holds two electrons, F -)- BF3 — BF4 (coordinate covalent bonding)... 

FIGURE 10.10 The linear geometry of BeCl2 can be explained by assuming that Be is sp-hybridized. The two sp hybrid orbitals overlap with the two chlorine 3p orbitals to form two covalent bonds. 

The sp hybrid orbitals overlap with the 5p orbitals of I to form three covalent A1—I bonds. We predict that the AII3 molecule is trigonal planar and all the I All angles are 120°. 

These hybrid orbitals overlap the 4p orbitals of Br to form five covalent P—Br bonds. Since there are no lone pairs on the P atom, the geometry of PBr5 is trigonal bipyramidal. 

In Figure 4.6b we illustrate how the tetrahedral structure of CH4 relates to a cubic framework. This relationship is important because it allows us to describe a tetrahedron in terms of a Cartesian axis set. Within valence bond theory, the bonding in CH4 can conveniently be described in terms of an sp valence state for C, i.e. four degenerate orbitals, each containing one electron. Each hybrid orbital overlaps with the li atomic orbital of one H atom to generate one of four equivalent, localized 2c-2e C—H (T-interactions. 

We can give the same VB description for the hybridization of the central atoms in polyatomic ions. In NH4+ and 804 , the N and S atoms, respectively, form four sp hybrid orbitals directed toward the corners of a regular tetrahedron. Each of these sp hybrid orbitals overlaps with an orbital on a neighboring atom (H in NH4+, O in 804 ) to form a bond. 

The hybridized orbitals overlap with the normal or hybridized orbitals in the other atom or atoms. 

Hybrid orbitals overlap better than pure atomic orbitals. 

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