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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
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

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{\star}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \(^{4} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of M, draw the Lewis structure of a CO molecule, with the carbon nearest the M. The CO bond axis should be on the \(x\) -axis. (c) Draw the CO \(\pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M},\) with phases. Can you see how they will overlap with the \(\pi_{2 p}^{\star}\) orbital of CO? (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?(\mathrm{f})\) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMO-LUMO gap; see Exercise 9.104 ). Sometimes you can visually tell which HOMO-LUMO gap is larger for one molecule than another. Suppose you have samples of two crystalline powders-one is white, and one is green. Which one has the larger HOMO-LUMO gap?

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN},(\mathbf{b}) \mathrm{SO}_{3}^{2-},(\mathbf{c}) \mathrm{SF}_{4}\) \((\mathbf{d}) \mathrm{PF}_{6},(\mathbf{e}) \mathrm{NH}_{3} \mathrm{Cl}^{+},(\mathbf{f}) \mathrm{N}_{3}^{-}\)

(a) The PH \(_{3}\) molecule is polar. Does this offer experimental proof that the molecule cannot be planar? Explain. (b) It turns out that ozone, \(\mathrm{O}_{3},\) has a small dipole moment. How is this possible, given that all the atoms are the same?
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