In the Classroom
An Evergreen: The Tetrahedral Bond Angle Marten J. ten Hoor J.W. Frisolaan 40, 9602 GJ Hoogezand, The Netherlands
A problem that keeps cropping up in the literature, and that is often solved in essentially the same way, may be called an Evergreen. The determination of the tetrahedral bond angle is such an Evergreen. In this Journal the subject was treated during 1941–1998 (1–27), in roughly periodic bursts. The determination of the tetrahedral bond angle seems to be a problem of all times, for in textbooks on general or organic chemistry its value is usually merely stated. It is then used in explanations of the decrease in bond angle in the series CH4, NH 3, H2O, and the deviation from its value can be used as a parameter to gauge the strain of small-ring molecules. Undoubtedly, instructors will be asked, time and again, how its value can be found. As three authors (5, 13, 26 ) offered two solutions each, the tetrahedral bond angle has been determined no fewer than 30 times, but not in 30 different ways. Apart from one very special solution based on spherical trigonometry (3), these solutions can be divided into two groups. In group I geometrical properties of a solid body are used, and group II comprises solutions obtained by application of vector algebra. Both groups may be divided into subgroups. In subgroup Ia a tetrahedron as such is considered (1, 6, 7, 10, 15, 24), but in subgroup Ib the properties of a cube (in which a tetrahedron may be inscribed) are used (2, 4, 11, 12, 13, 20, 23, 26 ). All methods of group II are based on the fact that the sum of four suitably chosen vectors is equal to the zero vector. This property may be taken for a consequence of the symmetry of the particular arrangement of these vectors (5, 14, 16–19), or may be explained by referring to the fact that a CX4 molecule has no permanent dipole moment (8, 9, 21, 22, 25, 27 ). In one article (26 ) the explanation is based on the fact that the center of mass of such a molecule coincides with the center of the carbon atom. Group II may also be divided into subgroups, according to whether the scalar (or dot) product of vectors is used (5, 9, 14, 16–19, 26 ) or not (5, 8, 21, 22, 25, 27 ). From the number of references given for each subgroup, it may be inferred that quite a few of the published solutions are essentially the same, or differ in a minor detail only. Bearing in mind Woolf ’s advice that “the shorter and more concrete a derivation the better” (23), I have selected two such derivations, one from each of the two groups mentioned above.
My favorite derivation from group I (4, 11, 13, 20, 23, 26) goes like this. Consider a cube with edges of length l – and face diagonals of length l √2. Six of these diagonals are the edges of an inscribed regular tetrahedron, ABCD. In Figure 1 only one of these diagonals is indicated. A line starting from the center of the tetrahedron (which coincides with the center of the cube) and ending at the midpoint of one of the face diagonals is perpendicular to that diagonal, bisects the tetrahedral angle θ = ∠AOB, and has the length 1⁄2 l. Now
1l 2 1 tan θ = 2 = 2 1l 2 2 and hence –
θ = 2 tan᎑1(√2 ) = 109.47122° = 109° 28′ 16′′
I have also given the outmoded notation (28) of an angle in degrees, minutes, and seconds because in some earlier work (1, 2, 6 ) a slightly larger value is given. My favorite derivation from group II (5, 8, 21, 22, 25, 27 ) is as follows. Imagine a ball-and-stick model of a CX4 molecule, but without X atoms, as in Figure 1 of ref 1. Imagine a vector of length l along each of the four sticks, pointing away from a common origin O that coincides with the center of the carbon atom, and let these vectors end in the points A, B, C, and D. The symmetry of this arrangement demands that the sum of the four vectors equal the zero vector. Imagine now a three-dimensional rectangular coordinate system with origin O, chosen in such a way that the vector OA coincides with the positive Z axis, whereas the vector OB lies in the YZ plane, as in Figure 2. The angle between vector OB and the negative Z axis is β, and its complement is α. Then θ = 90° + α = 180° – β
(1)
Z
A
A D O M C
β
l
B
B
Figure 1. Illustration of group I derivation.
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Y
α
O
Figure 2. Illustration of group II derivation.
Journal of Chemical Education • Vol. 79 No. 8 August 2002 • JChemEd.chem.wisc.edu
In the Classroom
and the vertical component of vector OB (as well as of the vectors OC and OD) has the length l sin α = l cos β. The sum of the lengths of these three vertical components equals the length of vector OA, hence 3l sin α = 3l cos β = l, from which α = sin᎑1(1⁄3) = 19.47122°, and β = cos᎑1(1⁄3) = 70.52878°. The tetrahedral bond angle can now be found from eq 1. To see why the sum of the four vectors would equal the zero vector, reference to the dipole moment of the CX4 molecule seems to appeal to many chemists (8, 21, 22, 25, 27 ). I refrained from using this argument because the underlying assumption that the vector sum of all bond dipole moments would yield the molecular dipole moment (though valid in this particular case) is not correct in general. On the basis of this assumption, a comparison of the polarity of the bonds of NH3 and NF3 leads to absurd results (29), and the method has been applied incorrectly in showing that the molecular dipole moment of trans-dichloroethene is zero (30). If one still wishes to use this argument, it is more appropriate to take a molecule like CCl4 (21, 22, 25) as an example. Then the vectors representing the bond dipole moments all point away from the carbon atom, and the angle between any two of them is indeed the tetrahedral bond angle. But in a molecule like CH4 (8, 27 ) these vectors all point toward the carbon atom, and then the angle between any two of them is not the tetrahedral bond angle, but another one of equal magnitude.
25. 26. 27. 28.
Literature Cited
29.
1. 2. 3. 4.
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30.
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JChemEd.chem.wisc.edu • Vol. 79 No. 8 August 2002 • Journal of Chemical Education
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