Calculation of the Valence Angle W. H. DORE University of California, Berkeley, California
I
N THE recent paper by Gombert' on "The Valence half as long as the unit cube and having a volume oneAngle of the Carbon Atom," a rigid geometric proof eighth as large. The circles representing the carbon is given for the correctness of the commonly ac- atoms in this smaller cube have been blackened. Four cepted value for the tetrahedral valence angle. Be- of these atoms are on four nonadjacent comers of the lieving that the pedagogical significance of this dem- cube and there is one in its exact center. When the onstration should be emphasized, I am presenting herewith, in amplification, an alternative method of calculation together with some comments on the geometry involved. The method of approach diiers from that used by Gombert in that the geometry of the tetrahedron is here considered in relation to that of a surrounding cube. This way of looking a t the tetrahedron in its relation to an enveloping cube facilitates the calcnlations; it also aids greatly in visualizing certain crystal structures which involve tetrahedral groups of atoms. Many people find difficulty in grasping the spacial significance of the diamond structure as simultaneously a tetrahedral and a cubic arrangement of the carbon atoms. The unit crystal cell of the diamond is a cube having carbon atoms a t its eight comers and in the center of each of its six faces. There are four other carbon atoms within the body of the unit cube. Thus eighteen atoms must he shown in illustration of the unit cell, although statistically there are only eight atoms within the cell boundaries, the comer atoms and the face- comer atoms are joined by diagonals across the faces centered atoms being shared with neighboring cells. of the cube i t will he seen that the figure so produced is a true tetrahedron bounded by four equilateral triangles (see Figure 2). The central carbon atom is a t the center of the tetrahedron as well as of the smaller cuhe. Thus the lines joining the central carbon atom to the four surrounding carbon atoms (in Figure 1) show the true positions of the valence bonds. The value for the valence angle AOB (Figure 2) may be readily calculated from the relationship between the tetrahedron and its enveloping cube. If the length of the edge of the cube be designated as E, the face diagonal Then in triangle ABM,
M E tangent of angle A B M = T == = -AB
1 = 0.7071
-\/2B24 2
and angle A B M = 35'15'49" = angle BAO angle AOB = 180" - 2 X (35"15'4gV) = 109'28'22' ,
Figure 1represents the cubic unit cell of the diamond crystal with all of the atoms in their correct positions as established by X-ray analysis. In order to emphasize the simultaneous tetrahedral and cubic arrangement of these atoms, a smaller cube is lined in having an edge 1
A three-dimensional representation of the relation between a tetrahedron and its surrounding cuhe may be easily constructed from cardboard. An equilateral triande is cut out and subdivided into four smaller eauilateral triangles by cutting with a knife part way through the board. The three corner triangles are
GOMBERT, J. CAHM.EDUC.. 18, 336 (1941).
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then folded backward until their edges meet. The edges are then joined together by means of Scotch tape, forming a tetrahedron. If the tetrahedral edge has a length S, the related cube will have a face diagonal equal to S and the length of the cube edge E will be such that
s=&B
To construct the surrounding cube a strip of cardboard with length four times E and width E is subdivided into four squares by cutting part way through the board, then folded into an open cube and the edges joined. The tetrahedron may then be slipped into the space between the four walls of this open cube. Practical dimensions which give a neat fit are 4.8 cm. for S a n d 3.5 em. for E.
