Problem 77 - Chemistry: The Central Science

Question

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.

Step-by-Step Solution

Answer

The angle between the two vectors connecting the vertices of the tetrahedron to the center of the cube can be found using analytical geometry. By assigning coordinates to the vertices of a cube with side length 1, we can determine the vertices of the tetrahedron and the center of the cube. We then find the vectors connecting the center of the cube to two adjacent vertices of the tetrahedron, and calculate the angle between these vectors using the dot product formula. The angle is found to be approximately \(109.5^{\circ}\), which is the characteristic angle for tetrahedral molecules.

Step 1: Find the coordinates of the tetrahedron vertices

First, let's assign coordinates to the vertices of the cube. Suppose the cube has side length 1. One possible set of vertices for the tetrahedron is: A (0, 0, 0), B (1, 1, 0), C (1, 0, 1), and D (0, 1, 1). These are four alternating corners of the cube.

Step 2: Find the center of the cube

To find the center of the cube, we can take the average of the coordinates of any two opposite vertices of the cube. Let's take the vertices with coordinates (0, 0, 0) and (1, 1, 1). The center of the cube, denoted as O, has coordinates \((\frac{1}{2}, \frac{1}{2}, \frac{1}{2})\).

Step 3: Find vectors connecting center to vertices

To find the vectors connecting the center of the cube to two adjacent vertices of the tetrahedron, we can subtract the coordinates of the center from the coordinates of the vertices. Let's find the vectors \(\vec{OA}\) and \(\vec{OB}\): \[ \vec{OA} = A - O = (0 - \frac{1}{2}, 0 - \frac{1}{2}, 0 - \frac{1}{2}) = (-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}) \] \[ \vec{OB} = B - O = (1 - \frac{1}{2}, 1 - \frac{1}{2}, 0 - \frac{1}{2}) = (\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}) \]

Step 4: Calculate the angle between vectors using the dot product

Recall that the dot product formula for two vectors \(\vec{u}\) and \(\vec{v}\) is given by: \[ \vec{u} \cdot \vec{v} = ||\vec{u}|| ||\vec{v}|| \cos{\theta} \] Where \(\theta\) is the angle between the two vectors. Therefore, the angle \(\theta\) can be calculated as follows: \[ \theta = \arccos{\frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| ||\vec{v}||}} \] Let \(\vec{u} = \vec{OA}\) and \(\vec{v} = \vec{OB}\). Calculate the dot product, magnitudes, and angle: \[ \vec{OA} \cdot \vec{OB} = (-\frac{1}{2})\cdot(\frac{1}{2}) + (-\frac{1}{2})\cdot(\frac{1}{2}) + (-\frac{1}{2})\cdot(-\frac{1}{2}) = -\frac{1}{4} \] \[ ||\vec{OA}|| = \sqrt{(-\frac{1}{2})^2 + (-\frac{1}{2})^2 + (-\frac{1}{2})^2} = \frac{\sqrt{3}}{2} \] \[ ||\vec{OB}|| = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2 + (-\frac{1}{2})^2} = \frac{\sqrt{3}}{2} \] \[ \theta = \arccos{\frac{-\frac{1}{4}}{(\frac{\sqrt{3}}{2})^2}} = \arccos{\frac{-1}{3}} \approx 109.5^{\circ} \] The angle between the two vectors connecting the vertices of the tetrahedron to the center of the cube is approximately \(109.5^{\circ}\), which is the characteristic angle for tetrahedral molecules.