Hybridization sigma bond

MMVB is a hybrid force field, which uses MM to treat the unreactive molecular framework, combined with a valence bond (VB) approach to treat the reactive part. The MM part uses the MM2 force field [58], which is well adapted for organic molecules. The VB part uses a parametrized Heisenberg spin Hamiltonian, which can be illustrated by considering a two orbital, two electron description of a sigma bond described by the VB determinants... 

You may recall that we discussed the bonding in ethene in Chapter 7. The double bond in ethene and other alkenes consists of a sigma bond and a pi bond. The ethene molecule is planar. There is no rotation about the double bond, since that would require breaking the pi bond. The bond angle in ethene is 120°, corresponding to sp2 hybridization about each carbon atom. The geometries of ethene and the next member of the alkene series, QHg, are shown in Figure 22.6. 

The most important alkyne by far is the first member of the series, commonly called acetylene. Recall from Chapter 7 that the C2H2 molecule is linear, with 180° bond angles. The triple bond consists of a sigma bond and two pi bonds each carbon atom is sp-hybridized. The geometries of acetylene and the next member of the series, C3H4, are shown in Figure 22.7. 

Valence bond theory (Chapter 7) explains the fact that the three N—O bonds are identical by invoking the idea of resonance, with three contributing structures. MO theory, on the other hand, considers that the skeleton of the nitrate ion is established by the three sigma bonds while the electron pair in the pi orbital is delocalized, shared by all of the atoms in the molecule. According to MO theory, a similar interpretation applies with all of the resonance hybrids described in Chapter 7, including SO S03, and C032-. 

The Lewis dot formalism shows any halogen in a molecule surrounded by three electron lone pairs. An unfortunate consequence of this perspective is that it is natural to assume that these electrons are equivalent and symmetrically distributed (i.e., that the iodine is sp3 hybridized). Even simple quantum mechanical calculations, however, show that this is not the case [148]. Consider the diiodine molecule in the gas phase (Fig. 3). There is a region directly opposite the I-I sigma bond where the nucleus is poorly shielded by the atoms electron cloud. Allen described this as polar flattening , where the effective atomic radius is shorter at this point than it is perpendicular to the I-I bond [149]. Politzer and coworkers simply call it a sigma hole [150,151]. This area of positive electrostatic potential also coincides with the LUMO of the molecule (Fig. 4). 

Figure 1.17 The hypothetical formation of the bonding molecular orbitals of ethane from two sp -hybridized carbon atoms and six hydrogen atoms. All of the bonds are sigma bonds. (Antibonding sigma molecular orbitals — are called a orbitals — are formed in each instance as well, but for simplicity these are not shown.)... 

Carbon atoms (1) and (4) use sp3 hybrid orbitals to form four sigma bonds, three by overlap with the hydrogen Is orbitals and one by overlap with an sp2 orbital from the central carbon (2). The two carbon atoms involved in the double bond undergo sp2 hybridization. They form C-H bonds by overlapping with Is orbitals of the H atoms. The C=C double bond is formed similarly to that described in (a). 

The left-most C atom (in the structure drawn below) is sp3 hybridized, and the C-H bonds to that C atom are between the sp3 orbitals on C and the Is orbital on H. The other two C atoms are sp hybridized. The right-hand C-H bond is between the sp orbital on C and the Is orbital on H. The c a C triple bond is composed of one sigma bond formed by overlap of sp orbitals, one from each C atom, and two pi bonds, each formed by the overlap of two 2p orbitals, one from each C atom (that is a 2py—2py overlap and a 2pz—2pz overlap). 

Table 3.18. Parameters e and (cf. Eqs. (3.93a)-(3.93c)) for describing hybridization k of sigma bonds in homonuclear diatomics...

Next, formation of n sigma bonds to the ligands requires formation of n equivalent sd"-1 bonding hybrids hi, I12,..., h , leaving k + 1 — n unhybridized singly occupied d spin-orbitals and 9 — k + n unoccupied d spin-orbitals,... 

In each case the central metal atom of H MCH2 is expected to form a skeleton of n + 1 sigma bonds (n to H, one to C) using approximate sd" hybridization. 

Sigma The sigma bond ctww is isotropic (s-like shape) around the bond axis and has the profile (Fig. 4.24(a)) to be expected from the sd/J hybrids discussed in the preceding section (cf. Fig. 4.9). 

From the above considerations, we can recognize that an ideal hypersaturated coordination complex would arise from what may be denoted as a 3cu/3a/3n metal configuration, with three orthogonal cu bonds (3cu one each in the x,y, and z directions) built from three parent sigma bonds (3 a from sd2 hybrids at 90° angles), and with three lone pairs (3n pure d orbitals) in the duodectet of the parent... 

Solution The electron count is 16e (eight from Ru, two from each phosphine, one from each Cl, and two from methylene, assumed to make a Ru=C double bond). Two cu bonds result, leaving four electrons to make the ctruc and 7TruC bonds and two lone pairs. Skeletal sigma-bonding of the parent Lewis-like structure requires three sd2 hybrids (90° angles),... 

Similarly, C — H sigma bonds in the C2H6 molecule are formed by the end to end overlap of sp3 hybrid orbitals of the carbon atoms with the Is orbitals of the hydrogen atoms. The C—Co bond is formed by the end to end overlap of the sp3 hybrid orbitals of the C atoms. So in the C2H6 molecule there are six C — H o bonds and one C—Co bond making seven o bonds in total. 

When carbon atoms undergo sp3 hybridization, the hybrid orbitals form sigma bonds. 

---