Rotational Symmetry of a Methane Molecule and the Bond Angle The rotational symmetry of a methane molecule can be used to great advantage to calculate the bond angle. Suppose that the carbon atom is at the origin 0 of Cartesian coordinates and the hydrogen atoms are a t points A, B, C, andD. If the vector from 0 t a d is (k,k,k) (of length and from 0 to B is (a,b,c), then by the 120' rotational symmetry of the molecule, the vectors fmm 0 to C and 0 to D are (b,c,a) and (c,a,b), respectively. Now if the angle between two vedors v and u is 8 then u u = uu em 9 using the dot product of vectors. Thus
a)
.
(k,k,k). fa,b,c) = 3k2 cos 9
and
(a,b,c). (b,c,a) = 3k2 ms 8
(1)
where 9 is the common bond angle between OA, OB, OC, and OD and (a2+ b2 + c2)ln = a k is the common bond length. Working out the dot pmducts in eq 1we have
and, hence, squaring
using eq 2 again. Dividing eq 3 by 3k4 then gives the quadratic equation 3 cos
= 1+ 2 cos e
whose negative root gives the required bond angle 0 = em-'(-113) = 109' 28'.
P. Glalster Department of Mathematics University of Reading Whiteknights, Reading. U.K.
Volume 70 Number 5 May 1993
351
