THEVALENCE ANGLEOFTHECARBONATOM PHILIP F. WEATHERILL University of Miehigon, Ann Arbor, Miehigon
RECENTLY Gombert' showed a method of calculating this angle, using a.solid model of the regular tetrahedron. Considered as a problem in spherical trigonometry, this calculation is extremely simple. Imagine the carbon atom a t the center of a sphere. The valence bonds will intersect the surface of the sphere a t four symmetrically placed points, forming four equal spherical triangles, the angles of which are all 120'. We wish to calculate the sides of these triangles, one of which, ABC,is shown in Figure 1. To find side AB,draw A D bisecting angle A. We now have the right-angled triangle ABD in which we know angle D B A (120") and angle B A D (60'). By Napier's Rule ("the sine of the middle part equals the product of the tangents of the adjacent parts") : cos side AB = cot 120" X cot 60" log cos AB = 9.7614394 9.7614394 = 9.5228788 sideAB = 109'28'16"
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