/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 The vertices of a tetrahedron co... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 91

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
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

01

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#
02

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 路 B = |A| |B| cos(胃) where |A| and |B| are the magnitudes of the vectors, and 胃 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 胃. OA 路 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(胃) Rearranging and solving for 胃: cos(胃) = -0.25 / ((sqrt(3)/2)(sqrt(3)/2)) cos(胃) = -1/3 胃 = arccos(-1/3) 鈮� 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.

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Most popular questions from this chapter

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \AA\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D} .\) (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}\) is a planar molecule that has the following carbocarbon bond lengths: (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(C-\) bond length in butadiene \((1.48\) A) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length \((1.54 \hat{\mathrm{A}}) .\) Does this imply that the middle \(\mathrm{C}-\mathrm{Cbond}\) in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\)? (\mathbf{d} ) Based on your answer for part ( c ),discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\) C bond.

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}_{2}\) or \(\mathrm{Be}_{2}{ }^{+}\) be expected to exist? Explain.

The nitrogen atoms in \(\mathrm{N}_{2}\) participate in multiple bonding, whereas those in hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the nitrogen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{N}-\mathrm{N}\) bond?
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