GENERAL CHEMISTRY CHAPTER TEN THEORIES OF CHEMICAL BONDING 1 INTRODUCTION. In the preceding chapter, we introduced a quantum chemical description of the atom in which the electrons are assigned to atomic orbitals. We now carry the concept of electron orbitals into our descriptions of molecules. There have been two major theories of chemical bonding, They both offer ways of describing the manner in which atomic orbitals combine when chemical bonds take place. 1It is helpful to look at a diagram that shows how the energy of a pair of atoms, that have unpaired electrons so that they can form a bond, varies with the distance separating them. At large distances, there is no interaction between the two atoms and we think of this as a suitable energy zero. If we try to superimpose one atom on the other, there will be a very substantial repulsion, as we see at very small separations. Between the two extremes is an energy minimum that corresponds to the formation of a stable molecule. Bond formation is associated with the overlapping of atomic orbitals each of which contains only one electron. In this geometry, the added proton - proton and electron - electron repulsions are more than offset by the extra attractions between protons and electrons. 22 Valence Bond Theory. The figure shows the overlap region occurring directly between the two protons. In all theories of covalent bonding, the overlapping of two atomic orbitals creates a region of enhanced electron probability (electron density). If we think of the electrons as a form of molecule - making glue, putting most it between the two atoms results in the best adhesion. When the overlap region is on the line joining the two nuclei, we refer to the bond as being a (σ) sigma bond. Sigma bonds can also be formed by the head-on overlap of an s with a p orbital or by two p orbitals. 32.1 Hybrid Orbitals. We next need to look back at the description that we furnished for the methane (CH4) molecule. The Lewis structure shows four bond pairs, forming a stable octet around the central carbon atom. Valence Shell Electron Pair Repulsion theory leads us to expect the four electron pairs to be located at the four corners of a tetrahedron and the four carbon hydrogen bonds to be in a tetrahedral arrangement. 4More recently, we found that the ground state electron configuration of carbon is [He] 2s22p2, with the 2p electrons occupying different orbitals This only provides two unpaired electrons and we would expect there to be four in order to account for the four C-H bonds in methane. We can get the four unpaired electrons by promoting one of the occupants of the 2s orbital to the vacant spot in the third 2p orbital. That gives an electron configuration of [He] 2s2p3. The excitation energy involved in reaching this state is more than compensated for by the possibility of forming two more C-H bonds. There remains one more problem with two distinct types of atomic orbitals, s and p, we would expect there to be two types of C-H bonds. One bond would involve the carbon 2s orbital and the other three the 2p orbitals. We know, however, that experiment tells us that all four bonds are the same. We find that we can mix the wave functions for the 2s and the three 2p wave functions to create four sp3 hybrids. 5The hybrid orbitals have two important properties, both of which are evident from the following figure. For s and p orbitals, the electron probability is symmetrically distributed about the nucleus. The sp3 hybrid orbitals are said to be directed orbitals. That is consistent with the electron probability being mostly on one-side of the nucleus. That means that there is a concentration of electron density in the direction of the bonding partner (in this case hydrogen). The other factor is that the four sp3 hybrid orbitals point towards the corners of a tetrahedron. We encountered other molecules with stable octets. Ammonia is an example of a central atom with three bond pairs and one lone pair. The ground state configuration of nitrogen is [He]2s22p3. That provides the three unpaired electrons necessary for three bonds. It does not, however, conform with the pyramidal shape. A better description is obtained from invoking hybrid orbitals. 6It will be recalled that the HNH angle is less than the perfect tetrahedral value of 109.5o. The recipes for the two kinds of hybrid orbitals are not quite the same. The following figure shows again the trigonal pyramidal structure of ammonia. Similar descriptions can be furnished for the hybrid orbitals of the oxygen in water. 7We looked, earlier, at the pair of molecules BF3 and BeF2 in which we described the central atom as being electron deficient because it has less than a stable octet of electrons. Boron has three bond pairs and beryllium two bond pairs. Neither atom has lone pairs in those particular molecules. We can extend our use of hybrid orbitals to elaborate our discussions of those two cases. Boron has a ground state electron configuration [He] 2s22p, with only one unpaired electron. By promoting one of the 2s electrons to a vacant 2p orbital we have a situation in which boron can form three covalent bonds. Those bonds can only be equivalent if the three valence shell orbitals are hybridized. The following figure shows the nature of sp2 hybrid orbitals. 8The figure shows that the three hybrid orbitals are directed towards the corners of an equilateral triangle with the boron atom sitting at its center. The remaining p orbital is not occupied. For beryllium, we have: 9The sp hybrids on the beryllium atom are seen to point in opposite directions which is consistent with the known linearity of the molecular structure. A very different situation arises when we have five bond pairs as in the case of PF5. The ground state electron configuration for an isolated phosphorus atom is [Ne] 3s2 3p3, which would provide three unpaired electrons which is acceptable for PF3 but not PF5. To get five unpaired electrons, we would need to excite one of the 3s electrons into a 3d orbital. That gives a configuration of [Ne] 3s 3p3 3d. 10There is no three dimensional shape with five equivalent bonds. The most satisfactory structure is that of a trigonal bipyramid, as shown in the diagram. We see that the five bonds are split into two types. Two of the bonds are axial and the other three form a trigonal