Calculating the Tetrahedral Bond Angle Using Spherical Polars and the Dot Product P. Glaister Department of Mathematics, University of Reading, Whiteknights, Reading, U.K.
A number of derivations of the tetrahedral bond angle have appeared recently. Kawa ( I ) inscribes a tetrahedron in a cube and uses the law of cosines, while Duffey (2)uses vector algebra as well as inscribing a tetrahedron in a cube. Apak and Tor (3)use analytical geometry. Dnffey (4)
has also recently pointed out that the vector derivation of Sutcliffe and Smith (5) was also given by Snatzke (6).A mechanical derivation by McCullough (7)and a proof using the centroid property of a tetrahedron by Cockburn (8) have also appeared. Following on from these a proof of the methane bond angle can be obtained using spherical polar coordinates to calculate the Cartesian coordinates of the hydrogen atoms, then using the dot product of vectors.
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Figure 1 . Methane system with rectangularcoordinates imposed.
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Journal of Chemical Education
Figure 2. Determining the spherical polar coordinates of each hydmgen atom.
The Methane Molecule
We begin by imposing a rectangular coordinate system on the methane molecule with the carbon atom at (0,0,0) and one hydrogen atom at (0,0,1).We assume that the hydrogen atoms are arranged symmetrically (see Fig. 1). Thus, each hydrogen atom is equidistant from the carbon atom (one unit, say), and any pair form an equal angle with the vertex at the origin. The spherical polar coordinates (r,B,$)of each hydrogen atom relative to the oriein can he determined very easily (see Fig. 2,. As can be seen in Firmre 2. the relationshio between Cartesian coordinates (&&) and spherical iolars (r,B,$) is given by
-
1.
rp,
6 .rp, cos rp -Tsm
Now, the dot product of two vectors a and b is given by a.b=abmsa
where a is the angle between them. However, because the hydrogen atoms are all a distance of one unit from the methane atom at (O,O, 01, the product of any two of the position vectors will express the cosine of the angle between them ($1. In other words, a = b = 1and a = $. Moreover, for position vectors, we get
and z=rco8+
where r is the radial coordinate; and 13 is the azimuthal angle Because each hydrogen atom is one unit from (0, 0, 01, the r value is 1.Because the hydrogen atom a t (O,0,1)has the same angular displacement from the other three hydrogen atoms, these three share the same 9,where $ will be the required bond angle. Finally, by the symmetry ofthe molecule, the 0 values of the other three hydrogen atoms must be evenly distributed between 0 and 2n, and these can be taken as 0,243, and 4d3. Therefore, the spherical polar coordinates of the three hydrogen atoms are
(1,4d3,0) They involve only the unknown bond angle @.The polar coordinates ofthe hydmgen atom at (0,0,1)are (1,0,0). From the relations above, the Cartesian coordinates of these three hydrogen atoms are (sin 0, 0 , cos 0)
where al, az, and as are the I, y, and z coordinates of A. Similarly, for B we have a. b = albl +azb2 + a3b3 Thus, taking the first two position vectors above, we get
using cos2++sin2+=1
-
which is a quadratic equation in cos 9 with the relevant root of cos $ = -113. This gives the bond angle as 0 = casl (-113) 109' 28' Literature Cited 1. Kaua,C. J. J Cham Edue. 1988, 65, 884. 2. Duffex G. H.J. Cham E d w . 1990,67,3&36. 3.Apak.R.;Tor,I . J. Chrm. Educ. 1991.68,970. 4.Duffey. G. H.J. Cham. E d w . 1892.69.171, 5 . Sukliffe, R. T.; Smith, S. J. J. Chem. Educ. 190!2,69,171. 6. Snatzke, G. J Cham. Educ. 1883,40,94. 7. McCdlough, T J Chrm. E d w . 1962.39.476, 8. Coekbum,B. L. J. Cham.Educ. 1363.40.94.
Volume 70 Number 7 July 1993
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