6.1: Spatial Trigonometry-The Bond Angle in Methane

When the going gets tough, the tough lower their standards. This idea, the theme of the whole book, underlies the final street-fighting tool of reasoning by analogy. Its advice is simple: Faced with a difficult problem, construct and solve a similar but simpler problem—an analogous problem. Practice develops fluency. The tool is introduced in spatial trigonometry (Section 6.1); sharpened on solid geometry and topology (Section 6.2); then applied to discrete mathematics (Section 6.3) and, in the farewell example, to an infinite transcendental sum (Section 6.4).

Spatial trigonometry: The bond angle in methane

The first analogy comes from spatial trigonometry. In methane (chemical formula \(CH^{4}\)), a carbon atom sits at the center of a regular tetrahedron, and one hydrogen atom sits at each vertex. What is the angle θ between two carbon–hydrogen bonds?

Angles in three dimensions are hard to visualize. Try, for example, to imagine and calculate the angle between two faces of a regular tetrahedron. Because two-dimensional angles are easy to visualize, let’s construct and analyze an analogous planar molecule. Knowing its bond angle might help us guess methane’s bond angle.

Question

Should the analogous planar molecule have four or three hydrogens?

Four hydrogens produce four bonds which, when spaced regularly in a plane, produce two different bond angles. In contrast, methane contains only one bond angle. Therefore, using four hydrogens alters a crucial feature of the original problem. The likely solution is to construct the analogous planar molecule using only three hydrogens.

Three hydrogens arranged regularly in a plane create only one bond angle: \(θ = 120^{◦}\). Perhaps this angle is the bond angle in methane! One data point, however, is a thin reed on which to hang a prediction for higher dimensions. The single data point for two dimensions (\(d = 2\)) is consistent with numerous conjectures for example, that in d dimensions the bond angle is \(120^{◦}\) or \((60d)^{◦}\) or much else.

Selecting a reasonable conjecture requires gathering further data. Easily available data comes from an even simpler yet analogous problem: the one-dimensional, linear molecule \(CH_{2}\). Its two hydrogens sit opposite one another, so the two C–H bonds form an angle of \(θ = 180^{◦}\).

Question

Based on the accumulated data, what are reasonable conjectures for the three-dimensional angle \(θ_{3}\)?

The one-dimensional molecule eliminates the conjecture that \(θ_{d} = (60d)^{◦}\). It also suggests new conjectures for example, that \(θ_{d} = (240 − 60d)^{◦}\) or \(θ_{d} = 360^{◦}/(d + 1)\). Testing these conjectures is an ideal task for the method of easy cases. The easy-cases test of higher dimensions (high d) refutes the conjecture that \(θ_{d} = (240 − 60d)^{◦}\). For high d, it predicts implausible bond angles—namely, \(θ = 0\) for \(d = 4\) and \(θ < 0\) for \(d > 4\).

Fortunately, the second suggestion, \(θ_{d} = 360^{◦}/(d + 1)\), passes the same easy-cases test. Let’s continue to test it by evaluating its prediction for methane namely, \(θ_{3} = 90^{◦}\). Imagine then a big brother of methane: a \(CH_{6}\) molecule with carbon at the center of a cube and six hydrogens at the face centers. Its small bond angle is \(90^{◦}\). (The other bond angle is \(180^{◦}\).) Now remove two hydrogens to turn \(CH_{6}\) into \(CH_{4}\), evenly spreading out the remaining four hydrogens. Reducing the crowding raises the small bond angle above \(90^{◦}\) and refutes the prediction that \(θ_{3} = 90^{◦}\).

Problem 6.1 How many hydrogens?

How many hydrogens are needed in the analogous four- and five-dimensional bond-angle problems? Use this information to show that \(θ_{4} > 90^{◦}\). Is \(θ_{d} > 90^{◦}\). for all \(d\)?

The data so far have refuted the simplest rational-function conjectures \((240 − 60d)^{◦}\) and \(360^{◦}/(d+1)\). Although other rational-function conjectures might survive, with only two data points the possibilities are too vast. Worse, θd might not even be a rational function of d.

Progress requires a new idea: The bond angle might not be the simplest variable to study. An analogous difficulty arises when conjecturing the next term in the series 3, 5, 11, 29, ...

Question

What is the next term in the series?

At first glance, the numbers seems almost random. Yet subtracting 2 from each term produces 1, 3, 9, 27, ... Thus, in the original series the next term is likely to be 83. Similarly, a simple transformation of the \(θ_{d}\) data might help us conjecture a pattern for \(θ_{d}\).

Question

What transformation of the \(θ_{d}\) data produces simple patterns?

The desired transformation should produce simple patterns and have aesthetic or logical justification. One justification is the structure of an honest calculation of the bond angle, which can be computed as a dot product of two C–H vectors (Problem 6.3). Because dot products involve cosines, a worthwhile transformation of \(θ_{d}\) is \(\cos θ_{d}\).

This transformation simplifies the data: The \(\cos θ_{d}\) series begins simply −1, −1/2, . . . Two plausible continuations are −1/4 or −1/3; they correspond, respectively, to the general term \(−1/2^{d−1}\) or \(−1/d\).

Question

Which continuation and conjecture is the more plausible?

Both conjectures predict \(\cos θ < 0\) and therefore \(θ_{d} > 90^{◦}\) (for all \(d\)). This shared prediction is encouraging (Problem 6.1); however, being shared means that it does not distinguish between the conjectures.

Does either conjecture math the molecular geometry? An important geometric feature, apart from the bond angle, is the position of the carbon. In one dimension, it lies halfway between the two hydrogens, so it splits the H–H line segment into two pieces having a 1:1 length ratio.

In two dimensions, the carbon lies on the altitude that connects one hydrogen to the midpoint of the other two hydrogens. The carbon splits the altitude into two pieces having a 1:2 length ratio.

Question

How does the carbon split the analogous altitude of methane?

In methane, the analogous altitude runs from the top vertex to the center of the base. The carbon lies at the mean position and therefore at the mean height of the four hydrogens. Because the three base hydrogens have zero height, the mean height of the four hydrogens is h/4, where h is the height of the top hydrogen. Thus, in three dimensions, the carbon splits the altitude into two parts having a length ratio of \(h/4 : 3h/4\) or 1 : 3. In d dimensions, therefore, the carbon probably splits the altitude into two parts having a length ratio of \(1:d\) (Problem 6.2).

Because \(1 : d\) arises naturally in the geometry, \(\cos θ_{d}\) is more likely to contain \(1/d\) rather than \(1/2^{d - 1}\). Thus, the more like of the two \(\cos θ_{d}\) conjectures is that

\[cosθ_{d} = -\frac{1}{d}. \label{6.1} \]

For methane, where \(d = 3\), the predicted bond angle is \(\arccos(−1/3)\) or approximately \(109.47^{◦}\). This prediction using reasoning by analogy agrees with experiment and with an honest calculation using analytic geometry (Problem 6.3).

Multiple problems

Problem 6.2 Carbon’s position in higher dimensions

Justify conjecture that the carbon splits the altitude into two pieces having a length ratio \(1 : d\).

Problem 6.3 Analytic-geometry solution

In order to check the solution using analogy, use analytic geometry as follows to find the bond angle. First, assign coordinates (\(x_{n}\), \(y_{n}\), \(z_{n}\)) to the \(n\) hydrogens, where \(n = 1 . . . 4\), and solve for those coordinates. (Use symmetry to make the coordinates as simple as you can.) Then choose two C–H vectors and compute the angle that they subtend.

Problem 6.4 Extreme case of high dimensionality

Draw a picture to explain the small-angle approximation \(\arccos x ≈ \pi/2 − x\). What is the approximate bond angle in high dimensions (large \(d\))? Can you find an intuitive explanation for the approximate bond angle?