Orbital Hybridization is the concept of mixing of 2 single atomicto form new hybrid orbitals to describe atomic bonding properties in a molecule. Hybridised orbitals are very useful in the explanation of the shape of molecular orbitals of molecules.
In predicting bond angles in small molecules, we find we can do a great deal with the simple idea that unlike charges produce attractive forces while like charges produce repulsive forces. We will have electron-nuclear attractions, electron-electron repulsions, and nucleus-nucleus repulsions. Let us first consider the case of a molecule with just two electron-pair bonds, as might be expected to be formed by combination of beryllium and hydrogen to give beryllium hydride, H:Be:H. The problem will be how to formulate the bonds and how to predict what the H—Be—H angle, θ, will be:
If we proceed as we did with the H-H bond, we might try to formulate bond formation in BeH2 by bringing two hydrogen atoms in the (Is)1 state up to beryllium in the (ls)2(2s)2 ground state (Table 6-1). But there is a problem in the ground-state configuration of beryllium, the 2s orbital is full and cannot accommodate any more electrons. The way around this is to “promote” one of the 2s2 electrons of beryllium to a 2p orbital. The resulting beryllium atom, (\s)’i(2s)t (2p)1, called the valence state, then could form a cr bond with a (Is)1 hydrogen by overlap of the Is and 2s orbitals as shown in 1 (also see Figure 6-5):
We might formulate a second a bond involving the 2p orbital, but a new problem arises as to where the hydrogen should be located relative to the beryllium orbital. Is it as in 2, 3, or some other way? The Be and H nuclei will be farther apart in 2 than they will be in 3 or any other similar Arrangement, so there will be less internuclear repulsion with 2. We therefore expect the hydrogen to locate along a line going through the greatest extension of the 2p orbital. According to this simple picture, beryllium hydride should have two different types of H-Be bonds — one as in 1 and the other as in 2. This is intuitively unreasonable for such a simple compound. Furthermore, the H-Be-H bond angle is unspecified by this picture because the 2s Be orbital is spherically symmetrical and could form bonds equally well in any direction. However, if we forget about the orbitals and only consider the possible repulsions between the electron pairs, and between the hydrogen nuclei, we can see that these repulsions will be minimized when the H-Be-H bond angle is 180°. Thus arrangement 5 should be more favorable than 4, with a H—Be—H angle less than 180°: Unfortunately, we cannot check this particular bond angle by experiment because BeH2 is unstable and reacts with itself to give a high-molecular-weight solid. However, a number of other compounds, such as (CH3)2Be, BeCl2, (CH3)2Hg, HgF2, and (CH3)2Zn, are known to have cr bonds involving (s)H/7)1 valence states. Measurements of the bond angles at the metal of these substances in the vapor state has shown them to be uniformly 180°. How are the s and p orbitals deployed in this kind of bonding?-It turns out that stronger bonds are formed when the degree of overlap of the orbitals - is high. The degree of overlap will depend on the sizes of the orbital and, particularly, on how far out they extend from the nucleus. Figure 6-7 shows how far 2s and 2p orbitals extend relative to one another. Bonding with these orbitals as in 1 and 2 does not utilize the overlapping power of the orbitals to the fullest extent. With 1 we have overlap that uses only part of the 2s orbital, and with 2, only a part of the 2p orbital. Molecules such as BeH2 can be formulated with better overlap and equivalent bonds with the aid of the concept of orbital hybridization. This concept, published independently by L. Pauling and J. C. Slater in 1931, involves determining which (if any) combinations of s and p orbitals may overlap better and make more effective bonds than do the individual ,v and p orbitals. The mathematical procedure for orbital hybridization predicts that an s and a p orbital of one atom can form two stronger covalent bonds if they combine to form two new orbitals called sp-hybrid orbitals (Figure 6-8). Each sp-hybrid orbital has an overlapping power of 1.93, compared to the pure s orbital taken as unity and a pure p orbital as 1.73. Bond angles of 180° are expected for bonds to an atom using sp-hybrid orbitals and, of course, this also is the angle we expect on the basis of our consideration of minimum electron-pair and internuclear repulsions. Henceforth, we will proceed on the basis that molecules of the type X:M:X may form sp-hybrid bonds. On the basis of repulsion between electron pairs and between nuclei, molecules such as BH3, B(CH3)3, BF3, and A1C13, in which the central atom forms threecovalent bonds using the valence-state electronic configuration
(S)1 (px)1 (py)1, are expected to be planar with bond angles of 120°. For example, Any departure from the planar arrangement will be less stable because it will increase internuclear and interelectronic repulsion by bringing nuclei closer together and the electron pairs closer together. The (S)1, (px)1, and (py)1 orbitals used in bonding in these compounds can be hybridized to give three equivalent sp2 orbitals (Figure 6-9). These sp2 orbitals have their axes in a common plane and are at 120° to one another. The predicted overlapping power is 1.99. With atoms such as carbon and silicon, the valence-state electronic configuration to form four covalent bonds has to be (s)1 (px)1 (py)1 (pz)1. Repulsion between the electron pairs and between the attached nuclei will be minimized by formation of a tetrahedral arrangement of the bonds. The same geometry is predicted from hybridization of one 5 and three p orbitals, which gives four sp3-hybrid orbitals directed at angles of 109.5° to each other. The predicted relative overlapping power of sp3-hybrid orbitals is 2.00 (Figure 6-10).
Molecular Shape and Orbital Hybridization video