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.

Short answer

In order to demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is ${109.5}^{\circ }$, we found the coordinates of vertices and the center of the cube. We then formed two vectors connecting the vertices to the center and used dot product formula to find the angle between these vectors. After calculations, we found that the angle is approximately ${109.5}^{\circ }$, which is the characteristic angle for tetrahedral molecules.

Step-by-step solution

Choose vertices (0,0,0), (1,1,0), (1,0,1), and (0,1,1) as the vertices of the tetrahedron. These are the alternating corners of the cube. #Step 2: Finding the Coordinates of the Cube Center#

The center of a cube with side length 1 has coordinates (0.5, 0.5, 0.5). #Step 3: Finding the Vectors Connecting the Vertices to the Center#

Let's denote the vertices of the tetrahedron as A(0,0,0), B(1,1,0), C(1,0,1), and D(0,1,1) and the center of the cube as O(0.5, 0.5, 0.5). We'll calculate the vectors from vertices A and B to the center, which will be OA and OB, respectively. Vector OA can be found as follows: OA = O - A = (0.5 - 0, 0.5 - 0, 0.5 - 0) = (0.5, 0.5, 0.5) Vector OB can be found as follows: OB = O - B = (0.5 - 1, 0.5 - 1, 0.5 - 0) = (-0.5, -0.5, 0.5) #Step 4: Finding the Angle Between the Vectors Using the Dot Product Formula#

The dot product of two vectors A and B is defined as: A · B = |A| |B| cos(θ) where |A| and |B| are the magnitudes of the vectors, and θ is the angle between them. We'll find the dot product of the OA and OB vectors and their magnitudes, and then solve for the angle θ. OA · OB = (0.5)(-0.5) + (0.5)(-0.5) + (0.5)(0.5) = -0.25 |OA| = sqrt((0.5)^2 + (0.5)^2 + (0.5)^2) = sqrt(3)/2 |OB| = sqrt((-0.5)^2 + (-0.5)^2 + (0.5)^2) = sqrt(3)/2 Now, we can substitute these values into the dot product formula: -0.25 = (sqrt(3)/2)(sqrt(3)/2)cos(θ) Rearranging and solving for θ: cos(θ) = -0.25 / ((sqrt(3)/2)(sqrt(3)/2)) cos(θ) = -1/3 θ = arccos(-1/3) ≈ 109.5^{\circ} So, the angle between the vectors connecting two vertices of the tetrahedron to the center of the cube is approximately ${109.5}^{\circ }$, which is the characteristic angle for tetrahedral molecules.