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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
By assigning coordinates to the cube's vertices and choosing a tetrahedron within the cube, we calculate the center of the cube as \(O=(0.5,0.5,0.5)\). We determine vectors connecting the center to the tetrahedron vertices: \(\vec{OA}=(-0.5,-0.5,-0.5)\), \(\vec{OD}=(-0.5,-0.5,0.5)\), and \(\vec{OE}=(0.5,0.5,-0.5)\). Using the formula with the dot product and vectors' length, we find the angle between the vectors \(\vec{OA}\) and \(\vec{OD}\) as \(\theta \approx 109.5^{\circ}\). This angle applies to any pair of vectors connecting the center to the tetrahedron vertices, demonstrating the characteristic angle for tetrahedral molecules.

Step by step solution

01

Assigning coordinates to the cube and tetrahedron vertices

Let's assign coordinates to the vertices of the cube, considering the cube's edge of length 1. We place the origin of the coordinate system at one vertex of the cube, with three edges coming out along the positive x, y, and z axes: \( A(0,0,0) \), \( B(1,0,0) \), \( C(0,1,0) \), \( D(0,0,1) \), \( E(1,1,0) \), \( F(1,0,1) \), \( G(0,1,1) \), and \( H(1,1,1) \). Now we choose four alternating corners of the cube to be the vertices of the tetrahedron, for instance, A, D, E, and G.
02

Calculating the center of the cube

To find the center of the cube, we will average the coordinates of two opposite vertices, for example, the vertices A and H: \[ O = (\frac{A_x + H_x}{2},\frac{A_y + H_y}{2},\frac{A_z + H_z}{2}) \] Which simplifies to \( O = (0.5,0.5,0.5) \).
03

Step 3:Finding the vectors connecting the center to the tetrahedron vertices

We will find the vectors by subtracting the center coordinates O from the coordinates of the tetrahedron vertices A, D, and E: \[ \vec{OA} = A - O = (0,0,0) - (0.5,0.5,0.5) = (-0.5,-0.5,-0.5) \] \[ \vec{OD} = D - O = (0,0,1) - (0.5,0.5,0.5) = (-0.5,-0.5,0.5) \] \[ \vec{OE} = E - O = (1,1,0) - (0.5,0.5,0.5) = (0.5,0.5,-0.5) \]
04

Calculating the angle between the vectors

We will use the formula with the dot product and vectors' length to find the angle \(\theta\) between the vectors \(\vec{OA}\) and \(\vec{OD}\) first: \[ cos(\theta) = \frac{\vec{OA}\cdot\vec{OD}}{\|\vec{OA}\|\|\vec{OD}\|} \] We first calculate the dot product and the length of vectors: \[ \vec{OA}\cdot\vec{OD} = (-0.5)(-0.5) + (-0.5)(-0.5) + (-0.5)(0.5) = 0.75, \] \[ \|\vec{OA}\| = \sqrt{(-0.5)^2 + (-0.5)^2 + (-0.5)^2} = \frac{\sqrt{3}}{2}, \] \[ \|\vec{OD}\| = \|\vec{OA}\| = \frac{\sqrt{3}}{2}\] Plug these into the formula: \[ cos(\theta) = \frac{0.75}{(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})} = \frac{0.75}{\frac{3}{4}} = 1 - \frac{1}{3} =\frac{2}{3} \] Now, calculate \(\theta\): \[ \theta = arccos(\frac{2}{3}) \approx 109.5^{\circ} \] Since all vertices of the tetrahedron are symmetric, we can conclude that the angle between any of the vectors connecting the center of the cube to its vertices will be approximately \(109.5^{\circ}\), which is the characteristic angle for tetrahedral molecules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Analytical Geometry
Analytical geometry, also known as coordinate geometry, provides a powerful framework for studying geometric figures through algebraic equations. By using a coordinate system to represent geometric objects, it bridges the gap between algebra and geometry.

This field allows us to analyze shapes, measure distances, and calculate angles using coordinates. In the context of a tetrahedron within a cube, analytical geometry helps in assigning coordinates to the vertices and determining geometric relationships, such as finding the angle formed by vectors extending from the cube's center to its corners.
Cube Coordinate System
In the cube coordinate system, the vertices of a cube are assigned coordinates based on its edges. This system simplifies calculations as we assume one corner of the cube is at the origin (0,0,0), and its adjacent edges align with the positive x, y, and z axes.

For a standard unit cube, with sides of length 1, the coordinates of its vertices become easy to handle. In this exercise, vertices of a cube are given as:
  • A(0,0,0), B(1,0,0), C(0,1,0), D(0,0,1),
  • E(1,1,0), F(1,0,1), G(0,1,1), and H(1,1,1).
Choosing specific vertices like A, D, E, and G gives us the vertices of a tetrahedron within the cube. These vertices are crucial for performing further geometric calculations.
Vector Dot Product
The vector dot product, also known as the scalar product, is a key concept in vector mathematics. It is used to determine the angle between two vectors and gives a measure of their directional alignment.

The formula for the dot product of vectors \(\vec{A}\) and \(\vec{B}\)is \(\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z\). In terms of angle calculation, if the vectors have a shared point, such as the center of a cube, the cosine of the angle \(\theta\) between them is determined by \[cos(\theta) = \frac{\vec{A}\cdot\vec{B}}{\|\vec{A}\|\|\vec{B}\|}\].

For finding the tetrahedral angle, the dot product helps in quantifying how aligned two vectors are, thus allowing us to find the cosine of the angle between the vectors originating from the cube's center to its alternating corners.
Tetrahedral Angle
The tetrahedral angle is a unique and essential concept in geometry, particularly in the study of molecular shapes. It refers to the angle formed between any two vectors that connect the center of a tetrahedron to its vertices. This angle is famously known as \(109.5^{\circ}\).

In our exercise, the vertices of a tetrahedron are selected from alternating corners of a cube. By calculating vectors from the center of the cube to these vertices, and using the dot product formula, we find the characteristic tetrahedral angle. Recognizing this angle is crucial in both geometry and chemistry, where it describes the shape of molecules such as methane.

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Most popular questions from this chapter

(a) An AB \(_{6}\) molecule has no lone pairs of electrons on the A atom. What is its molecular geometry? (b) An AB \(_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four B atoms). What is the electron-domain geometry around the A atom? (c) For the AB \(_{4}\) molecule in part (b), predict the molecular geometry.

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(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}^{-}\) localized or delocalized?

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