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What fraction of the total volume of a cubic closest packed structure is occupied by atoms? (Hint: \(\left.V_{\text {sphene }}=\frac{4}{3} \pi r^{3} .\right)\) What fraction of the total volume of a simple cubic structure is occupied by atoms? Compare the answers.

Short Answer

Expert verified
In the cubic closest packed (ccp) structure, the fraction of the total volume occupied by atoms is approximately 74.05%, while in the simple cubic (sc) structure, the fraction is approximately 52.36%. Therefore, the cubic closest packed structure has a higher packing efficiency.

Step by step solution

01

Calculate the volume occupied by atoms in a cubic closest packed structure

Cubic closest packed structures have a coordination number of 12, meaning that each sphere is surrounded by 12 others. In a cubic closest packing, the spheres touch along the face diagonals. To find the volume of the unit cell, we can break it down into smaller shapes. In a ccp structure, the two spheres along a face diagonal will have their centers at vertices of a right triangle. Two of its sides will have lengths equal to four times the radius (4r) of the spheres, and its face diagonal, which connects the centers of two spheres, will have length equal to two times the diameter (4r) of the spheres. Using the Pythagorean theorem, we have: \( (4r)^2 + (4r)^2 = (4r)^2 \) \( 2*(4r)^2 = (4r)^2 \) \( (4r)^2 = 8r^2 \) Now we can find the volume of a single unit cell: \( V_\text{ccp} = a^3 = (4r)^3 = 64r^3 \) The ccp unit cell contains 4 spheres, so the total volume occupied by atoms is: \( V_\text{spheres} = 4 * \frac{4}{3}\pi r^3 \) Therefore, the fraction of the cubic closest packed structure occupied by atoms is: \( \frac{V_\text{spheres}}{V_\text{ccp}} = \frac{4 * \frac{4}{3}\pi r^3}{64r^3} \)
02

Calculate the volume occupied by atoms in a simple cubic structure

In a simple cubic structure, there is one atom per unit cell. The atom touches along the edges, and thus, the length of the edge (a) is equal to twice the radius of the atom (2r). The volume of the unit cell in a simple cubic structure is: \( V_\text{sc} = a^3 = (2r)^3 = 8r^3 \) Since there is one atom per unit cell, the total volume occupied by atoms is: \( V_\text{sphere} = \frac{4}{3}\pi r^3 \) Therefore, the fraction of the simple cubic structure occupied by atoms is: \( \frac{V_\text{sphere}}{V_\text{sc}} = \frac{\frac{4}{3}\pi r^3}{8r^3} \)
03

Comparing the fractions

Now we can compare the fractions of the two structures: For ccp: \( \frac{4 * \frac{4}{3}\pi r^3}{64r^3} = \frac{\pi}{3} \approx 0.7405\) For sc: \( \frac{\frac{4}{3}\pi r^3}{8r^3} = \frac{\pi}{6} \approx 0.5236 \) Based on the calculation, the cubic closest packed structure has a higher packing efficiency (74.05%) compared to the simple cubic structure (52.36%).

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Key Concepts

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

Cubic Closest Packed
In understanding the cubic closest packed (ccp) structure, envision a type of arrangement that combines several atoms or spheres in a balanced way to utilize space efficiently. Multiple layers of spheres are stacked to ensure minimal voids between them. The ccp has each sphere touching 12 other spheres—six in its current layer, three in the layer above, and three in the layer below. This creates a repeating three-dimensional pattern and heavily contributes to the densest possible atomic packing in a crystalline structure. This arrangement leads to a particular packing efficiency and a unique coordination number.
Simple Cubic Structure
The simple cubic (sc) structure is one of the simplest forms of crystal lattice structures. In this arrangement, atoms are positioned at the vertices of a cube. Since each atom is shared by eight adjacent cubes, the simple cubic structure only contains a single atom per unit cell. The distinctive feature is that each atom directly touches its neighbors on each side of the cube, but beyond this, the sc structure has relatively large amounts of empty space when compared to more densely packed structures such as ccp. This leads to a lower packing efficiency and coordination number, reflective of the fewer atoms each central atom is immediately surrounded by.
Packing Efficiency
Packing efficiency denotes the fraction of space in a crystal lattice that is actually occupied by the constituent particles—atoms, ions, or molecules. For the ccp structure, the packing efficiency is roughly 74%, meaning the atoms occupy a significant portion of available space. However, in the sc structure, the packing efficiency drops to about 52%, indicating a notable amount of unfilled space within the crystal lattice. It's an essential metric for understanding how substances with each structure will interact with their environment, as the more efficiently a structure is packed, the less prone it is to penetration by other substances and the higher its material density will be.
Coordination Number
The coordination number of an atom in a crystal lattice refers to the number of nearest-neighbor atoms to a reference atom. In the ccp structure, this number is 12 due to how the spheres are packed, granting any chosen sphere a total of 12 immediate neighbors. In stark contrast, the sc structure's coordination number is noticeably lower at 6 because an atom in a simple cubic array is only immediately surrounded by six others—each face of the cube has one neighboring atom. The coordination number is a crucial factor in determining the mechanical, thermal, and electrical properties of the material.

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Most popular questions from this chapter

Rationalize the difference in boiling points for each of the following pairs of substances: a. \(n\) -pentane \(\quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(36.2^{\circ} \mathrm{C}\) neopentane CC(C)(C)C \(9.5^{\circ} \mathrm{C}\) b. HF \(20^{\circ} \mathrm{C}\) \(\mathrm{HCl} \quad-85^{\circ} \mathrm{C}\) c. HCl \(-85^{\circ} \mathrm{C}\) \(\mathrm{LiCl} \quad 1360^{\circ} \mathrm{C}\) d. \(n\) -pentane \(\quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(36.2^{\circ} \mathrm{C}\) \(\begin{array}{lll}n \text { -hexane } & \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} & 69^{\circ} \mathrm{C}\end{array}\)

Nickel has a face-centered cubic unit cell. The density of nickel is \(6.84 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate a value for the atomic radius of nickel.

You have three covalent compounds with three very different boiling points. All of the compounds have similar molar mass and relative shape. Explain how these three compounds could have very different boiling points.

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