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"Valence and Molecular Structure," Lectures 1 and 2.

"Valence and Molecular Structure," Lectures 1 and 2. 1957.
Produced for the Institutes Program of the National Science Foundation. Robert and Jane Chapin, producers.

Lecture 2, Part 6. (6:38)


Linus Pauling: While I’m on the subject, I might mention diamond. Here is a model not built on such a large scale, so, small enough so that I’m able to lift it. A model of diamond. Each carbon atom is attached by carbon-carbon single bonds. Two, four others that surround this tetrahedrally. And, the distance found experimentally between the carbon atoms is 1.54 angstrom, just the distance that is found in ethane, and in many other substances in which there are carbon-carbon single bonds.

We can understand too why diamond is so hard. Here are these chemical bonds, strong bonds, which connect all of the atoms in the crystal together into a single molecule. In order to break a crystal of diamond, as for example if you tried to scratch it with some other substance, it would be necessary to carry out a chemical reaction, the reaction of breaking carbon-carbon bonds. This is so hard to do, the bonds are so strong, that the substance is very hard, the hardest substance known. I may mention, also, that the double bond, which is the stronger bond, has length 1.33 angstrom, the triple bond, still stronger, has length 1.20 angstrom. As the bond between the carbon atoms gets stronger, the atoms are pulled more and more closely together.

There is one aspect of the carbon atom, the bonds formed by the carbon atom that I should mention. The question is this: You remember that the orbitals for the carbon atom are the 1s orbital, the 2s orbital, and the 3 2p orbitals. Now, these are the orbitals in the valence shell, and one might well ask ‘is not one bond formed by the carbon atom not different from the other three?’ The answer to this is a simple one. No, the four bonds are equivalent. And, we can describe these bonds in terms of four equivalent oribtals. Instead of the s orbitals and the p orbitals, we can formulate, as hybrid orbitals, four that are directed toward the four corners of a regular tetrahedron, and these orbitals are perfectly satisfactory in providing an explanation of the four equivalent tetrahedral bonds.

I think that it is interesting that if it had happened, as it might well have happened, that chemists discovered quantum theory, wave mechanics, rather than the physicists, then we would be saying that the s orbital and the p orbital that the chemists are interested in, are hybrid orbitals, formed of the four equivalent tetrahedral orbitals of the carbon atom.

Let us consider now some substances, some molecules, for which the classical valence theory was not satisfactory. An example is benzene, the benzene molecule C6H6, may be represented, according to classical theory, by this structure, a six-membered carbon ring. In order that the carbon atom be quadrivalent, we must have, we must have in this six-membered carbon ring, not only the bonds from carbon to hydrogen and carbon to carbon, but also double bonds.

But, chemists found a hundred years ago that, if two of these hydrogen atoms were replaced by, say, chlorine atoms, one did not obtain two substances, a substance in which the chlorine atoms were around carbon atoms held together by a single bond, and another substance in which chlorine atoms surround carbon atoms held together by a double bond, instead, only one substance could be obtained of this sort.

The explanation of this is given by the theory of resonance. This theory states that sometimes, instead of writing a single valence bond structure to represent a molecule, one must write two or more valence bond structures and lump them together. Benzene is described now as being a resonance hybrid of these two, those are identical. I’ll have to draw the double bonds differently. There’s the second one, with the chlorine atoms here for dichlorobenzene. We write these two Kekule structures, a single structure of this sort was first written by Kekule nearly a hundred years ago. We write these two Kekule structures and say that the two structures together provide a satisfactory description of the benzene molecule.

A similar sort of structure can be written for graphite. This represents a portion of the graphite crystal. A very soft substance, the molecule is to be thought of as being infinite in size, a very large layer consisting of hexagonal rings. If I start to represent this structure, I can show carbon atoms attached together in rings. Now, I need to have a double bond on this carbon atom in order that there will be a quadrivalent carbon atom. But, the double bond does not to be here. It may be there or here or here. There are a great many structures that I can draw to represent the molecule of graphite.

The physical properties of graphite I nicely explained by this structure. On this scale, a scale of this model, the layers, these giant two dimensional molecules, are about this far away form one another and very loosely attached to one another so that they can slide back and forth relative to one another.


Creator: Linus Pauling
Associated: Robert Chapin, Jane Chapin, National Science Foundation
Clip ID: 1957v.1-15

Full Work

Creator: National Science Foundation
Associated: Linus Pauling, Robert Chapin, Jane Chapin

Date: 1957
Genre: video
ID: 1957v.1
Copyright: More Information

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