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Tetrahedral Bond Angle from Elementary Trigonometry
The alternative approach of using the scalar
(or dot) product of vectors enables the determination
of the bond angle in a tetrahedral molecule in a simple
way. There is, of course, an even more straightforward
derivation suitable for students who are unfamiliar
with vectors, or products thereof, but who do know some
elementary trigonometry. The starting point is the
figure showing triangle OAB. The point O is the center of
a cube, and A and B are at opposite corners of a face
of that cube in which fits a regular tetrahedron. The
required bond angle alpha = AÔB; and using Pythagoras'
theorem, AB = 2(square root 2) is the diagonal of a face of the cube.
Hence from right-angled triangle OEB, tan(alpha/2) = (square root 2) and therefore alpha =
2tan-1(square root 2) is approx. 109°
28' (see Fig. 1).
Figure 1. Tetrahedral bond angle from elementary trigonometry.
Methane Bond Angle and Mass Centers
A simple proof of the methane bond angle worth using in the classroom because it appeals to students is one that uses the following mass center approach.
For a molecule consisting of n atoms of mass mi the center of mass G relative to an origin O has position
(1)
where ri is the position of each atom relative to
O. Since the methane molecule is symmetric, the mass center
is at the carbon atom. Thus, if the origin is taken to be
at the carbon atom, then from eq 1
(2)
rH1 + rH2 +
rH3 + rH4 = 0
Squaring eq 2 gives 4r2 + 12r2cosq = 0; that is, 1
+ 3cosq = 0, where r is the C-H bond length and q is the required H-C-H bond angle. Hence
q = cos-1(-1/3) as usual.
See Letter re: this article.
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