Bond Formation Using Atomic Orbitals

In writing the conventional Lewis structures for molecules, we assume that a covalent chemical bond between two atoms involves sharing of a pair of electrons from each atom. The following representation shows how atomic orbitals can be considered to be used in bond formation. 

we postulate that gle bonds a formed ing together of atomic nuclei through attractive forces exerted by electrons having paired spins (t4) in overlapping orbitals. This formulation is no particular improvement over what is implied by Lewis structures, except in so far as it provides further appreciation that the electrons involved must have paired spins. Because only two paired electrons can occupy a given orbital, a clear reason i s provided as to why two electrons are involved in single-bond formation rather than 3, 5 or 10. This type of bond is called, in molecular-orbital parlance, a a bon  

An important idea which is not clearly (if at all) implied in Lewis structures is The direction of a bond will be such 

F or an atom which forms two r bonds with orbitals to hydrogen we would expect the to be 90 . 

In the drawings here and later the shapes of the p orbitals will be represented as grossly elongated, tangent ellipsoids instead of tangent spheres. This representation is desirable in order to make the drawings clear and should not be taken for the correct orbital shape. 

Exercise 6-1 Write the ground-state configuration for a helium atom with two unpaired electrons (He ) that is in accord with the Pauli principle. Give your reasoning. 

Because two atomic orbitals can hold a maximum of four electrons, it is reasonable to ask why it is that two rather than one, three, or four electrons normally are involved in a bond. The answer is that two overlapping atomic 

6 Bonding in Organic Molecules. Atomic-Orbital Models 

3The designation sigma (cr) denotes that orbital overlap and electron density are greatest along the internuclear axis. 

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Although we have introduced references to bond formation with hybrid orbitals, we have not yet really tackled how to describe molecules using orbitals. The starting point for molecules, as for atoms, is the Schrodinger equation, and we can solve this to obtain electron wavefunctions or molecular orbitals. However, for molecules the electrons are attracted to all the nuclei in the molecule, not just one, and we have to include the repulsion between the nuclei in the energy. There are several methods and many programs available to calculate molecular orbitals. Nearly all employ two approximations. 

In the perfect lattice the dominant feature of the electron distribution is the formation of the covalent, directional bond between Ti atoms produced by the electrons associated with d-orbitals. The concentration of charge between adjacent A1 atoms corresponds to p and py electrons, but these electrons are spatially more dispersed than the d-electrons between titanium atoms. Significantly, there is no indication of a localized charge build-up between adjacent Ti and A1 atoms (Fu and Yoo 1990 Woodward, et al. 1991 Song, et al. 1994). The charge densities in (110) planes are shown in Fig. 7a and b for the structures relaxed using the Finnis-Sinclair type potentials and the full-potential LMTO method, respectively. 

So far, we have not considered whether terminal atoms, such as the Cl atoms in PC15, are hybridized. Because they are bonded to only one other atom, we cannot use bond angles to predict a hybridization scheme. However, spectroscopic data and calculation suggest that both s- and p-orbitals of terminal atoms take part in bond formation, and so it is reasonable to suppose that their orbitals are hybridized. The simplest model is to suppose that the three lone pairs and the bonding pair are arranged tetrahedrally and therefore that the chlorine atoms bond to the phosphorus atom by using sp hybrid orbitals. 

The carhon-carbon double bond in alkenes is more reactive than carbon-carbon single bonds and gives alkenes their characteristic properties. As we saw in Section 3.4, a double bond consists of a a-bond and a 7r-bond. Each carbon atom in a double bond is sp2 hybridized and uses the three hybrid orbitals to form three cr-bonds. The unhvbridized p-orbitals on each carbon atom overlap each other and form a Tr-bond. As we saw in Section 3.7, the carbon-carbon 7r-bond is relatively weak because the overlap responsible for the formation of the 7r-bond is less extensive than that responsible for the formation of the a-bond and the enhanced electron density does not lie directly between the two nuclei. A consequence of this weakness is the reaction most characteristic of alkenes, the replacement of the 77-bond by two new a-bonds, which is discussed in Section 18.6. 

A covalent bond between two atoms requires two electrons and two orbitals, one for each atom.f The factors determining the properties of the covalent bonds formed by an atom are primarily the number and nature of the orbitals (hybridised bond orbitals) available to the atom, and the number of electrons that it can use in bond formation without losing its electrical neutrality. The opportunities for stabilisation through resonance of covalent bonds among alternative positions are also important. 

The stability of sexivalent chromium, in the chromate ion and related ions, can also be understood. The chromic complexes, involving tervalent chromium, make use of d2sp3 bond orbitals, the three remaining outer electrons of the chromium atom being in three of the 3d orbitals, with parallel spins. The resonance energy of these three atomic electrons in a quartet state helps to stabilise the chromic compounds. However, if all of the nine outer orbitals of the chromium atom were available for bond formation, stable compounds might also be expected... 

The three structures of type 4 are unstable for two reasons the presence of electric charges of the same sign on adjacent atoms, and the use by the nitrogen atom of only three orbitals. The contribution of these structures to the normal state of the molecule is accordingly small, and we may take it to be zero. Also, structures of type 3, with a double bond and the transfer of the positive charge to a fluorine atom, are stabilized by the formation of an additional covalent bond with use of the fourth orbital and may accordingly make a greater contribution to the normal state moreover, there is an extra factor 2 for the six structures of type 3 over the three of type 1. 

There are 2.56 d orbitals available for bond formation. To form 5.78 bonds these would hybridize with the s orbital and 2.22 of the less stable p orbitals. In copper, with one electron more than nickel, there is available an additional 0.39 electron after the hole in the atomic d orbitals is filled. This might take part in bond formation, with use of additional Ap orbital. However, the increase in interatomic distance from nickel to copper suggests that it forms part of an unshared pair with part of the bonding electrons, thus decreasing the effective number of bonds. 

In this discussion of the transition elements we have considered only the orbitals (n— )d ns np. It seems probable that in some metals use is made also of the nd orbitals in bond formation. In gray tin, with the diamond structure, the four orbitals 5s5p3 are used with four outer electrons in the formation of tetrahedral bonds, the 4d shell being filled with ten electrons. The structure of white tin, in which each atom has six nearest neighbors (four at 3.016A and two at 3.17.5A), becomes reasonable if it is assumed that one of the 4d electrons is promoted to the 5d shell, and that six bonds are formed with use of the orbitals 4dSs5p35d. 

Remember from Chapter 6 that energy is released when a bond forms. Consequently, atoms that form covalent bonds tend to use all their valence s and p orbitals to make as many bonds as possible. We might expect the S p -hybridized aluminum atom to form a fourth bond with its unused 3 p orbital. A fourth bond does not form in A1 (C2 115)3 because the carbon atoms bonded to aluminum have neither orbitals nor electrons available for additional bond formation. The potential to form a fourth bond makes triethylaluminum a very reactive molecule. 

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