Research: Science & Education
Two Comments on Bond Angles P. Glaister Department of Mathematics, University of Reading, Whiteknights, Reading, UK
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 α = AÔB; and using Pythagoras’ theorem, AB = 2√2 is the diagonal of a face of the cube. Hence from right-angled triangle OEB, tan(α /2) = √2 and therefore α = 2 tan{1(√2) . 109° 28′ (see Fig. 1).
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 m i the center of mass G relative to an origin O has position
OG =
1086
m 1r 1 + m 2r 2 + … + m nr n m1 + m2 + … mn
(1)
2(2) 1/2 E
A
B
1 α
O Figure 1. Tetrahedral bond angle from elementary trigonometry.
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 r H 1 + rH 2 + rH 3 + rH 4 = 0
(2)
Squaring eq 2 gives 4r2 + 12r2cosθ = 0; that is, 1 + 3cosθ = 0, where r is the C–H bond length and θ is the required H–C–H bond angle. Hence θ = cos{1({1/3) as usual.
Journal of Chemical Education • Vol. 74 No. 9 September 1997
