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Chapter 9: Problem 89

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.

Short Answer

Expert verified
The angle made by connecting two alternating vertices of a cube to the center point can be found using analytical geometry. By assigning coordinates to the vertices of the cube and choosing the alternating vertices, we can find the center point of the cube and the position vectors from the center to the chosen vertices. After calculating the dot product and the magnitudes of the position vectors, we use the dot product formula to find the angle between them, which is approximately \(109.5^{\circ}\). This is the characteristic angle for tetrahedral molecules.

Step by step solution

01

Assign coordinates to the vertices of the cube

Let's assign coordinates to the vertices of the cube, starting at one corner and taking the cube's side as length 1. The vertices of the cube can be represented as: A (0, 0, 0), B (1, 0, 0), C (1, 1, 0), D (0, 1, 0), E (0, 0, 1), F (1, 0, 1), G (1, 1, 1), and H (0, 1, 1)
02

Choose the alternating vertices

Now we need to choose four alternating vertices to create a tetragon. Let's select vertices A, C, E, and G. This way we have: A (0, 0, 0), C (1, 1, 0), E (0, 0, 1), and G (1, 1, 1)
03

Find the center of the cube

The center of the cube can be found by taking the average of the coordinates of opposing vertices. The coordinates of the center point O are: O \(\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\)
04

Find the position vectors

We need the position vectors from the center of the cube (O) to the vertices A and G. Using the coordinates of O, A, and G, we find: \(\vec{OA} = \begin{bmatrix}-\frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}\) and \(\vec{OG} = \begin{bmatrix}\frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \end{bmatrix}\)
05

Calculate the dot product

Now, we can find the dot product of the two position vectors: \(\vec{OA} \cdot \vec{OG} = -\frac{1}{2}\times\frac{1}{2} + (-\frac{1}{2})\times\frac{1}{2} + (-\frac{1}{2})\times\frac{1}{2} = -\frac{1}{4}\)
06

Calculate the magnitudes of the position vectors

Next, we need to find the magnitudes of the position vectors \(\vec{OA}\) and \(\vec{OG}\): \(|\vec{OA}| = |\vec{OG}| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}\)
07

Use the dot product formula to find the angle

Now, we can use the dot product formula to find the angle between the two position vectors. The dot product formula is: \(\cos \theta = \frac{\vec{OA} \cdot \vec{OG}}{|\vec{OA}||\vec{OG}|}\) Substitute the known values into the formula: \(\cos \theta = \frac{-\frac{1}{4}}{\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}} = -\frac{1}{3}\) Now find the angle \(\theta\) by taking the inverse cosine: \(\theta = \cos^{-1} (-\frac{1}{3})\) \(\theta \approx 109.5^{\circ}\) The angle made by connecting two of the vertices to the center of the cube is approximately \(109.5^{\circ}\), which is the characteristic angle for tetrahedral molecules.

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Most popular questions from this chapter

Name the proper three-dimensional molecular shapes for each of the following molecules or ions, showing lone pairs as needed: $(\mathbf{a}) \mathrm{ClO}_{2}^{-}(\mathbf{b}) \mathrm{SO}_{4}^{2-}(\mathbf{c}) \mathrm{NF}_{3}(\mathbf{d}) \mathrm{CCl}_{2} \mathrm{Br}_{2}(\mathbf{e}) \mathrm{SF}_{4}^{2+}$

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Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.
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