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Chapter 12: Problem 13

Compute the atomic packing factor for the rock salt crystal structure in which rC/rA 0.414.

Short Answer

Expert verified
Question: Calculate the atomic packing factor (APF) for a rock salt crystal structure with a cation radius (rC) to anion radius (rA) ratio of 0.414. Answer: To calculate the atomic packing factor (APF) for a rock salt crystal structure with a cation radius (rC) to anion radius (rA) ratio of 0.414, follow these steps: 1. Determine the length of the unit cell edge (a) using the formula: a = 2 × (rC + rA). 2. Calculate the volume of the unit cell (V_cell) using the formula: V_cell = a^3. 3. Calculate the volume occupied by one cation (V_cation) and one anion (V_anion) using the formula for the volume of a sphere. 4. Determine the number of cations and anions in the unit cell (in the rock salt crystal structure, there are four cations and four anions). 5. Calculate the total occupied volume by cations and anions (V_occupied) by multiplying the volume of one cation and one anion by 4. 6. Calculate the atomic packing factor (APF) by dividing the total occupied volume (V_occupied) by the volume of the unit cell (V_cell). Using the given cation radius (rC) to anion radius (rA) ratio of 0.414, the APF can be calculated using the formula: APF = \frac{4 × \left[\frac{4}{3}π(0.414rA)^3 + \frac{4}{3}πrA^3\right]}{(2 × (0.414rA + rA))^3}

Step by step solution

01

Determine the length of the unit cell edge

In the rock salt crystal structure, the cations and anions are arranged alternatively in an FCC configuration. The edge length (a) of the unit cell can be expressed in terms of the radii of the cations (rC) and anions (rA) as: a = 2 × (rC + rA) We are given rC/rA = 0.414, so we can write rC = 0.414 × rA, and substitute this in the above equation to eliminate rC: a = 2 × (0.414rA + rA)
02

Calculate the volume of the unit cell

The volume of the unit cell (V_cell) can be calculated as the cube of the edge length: V_cell = a^3 = (2 × (0.414rA + rA))^3
03

Calculate the volume occupied by one cation and one anion

The volume occupied by one cation (V_cation) and one anion (V_anion), can be calculated using the formula for the volume of a sphere: V_cation = \frac{4}{3}πrC^3 = \frac{4}{3}π(0.414rA)^3 V_anion = \frac{4}{3}πrA^3
04

Calculate the number of cations and anions in the unit cell

In the rock salt crystal structure, there are four cations and four anions in the unit cell.
05

Calculate the total occupied volume by cations and anions

Multiply the volume of one cation and one anion by 4 to get the total occupied volume (V_occupied): V_occupied = 4 × (V_cation + V_anion) V_occupied = 4 × \left[\frac{4}{3}π(0.414rA)^3 + \frac{4}{3}πrA^3\right]
06

Calculate the atomic packing factor (APF)

Divide the total occupied volume (V_occupied) by the volume of the unit cell (V_cell) to get the atomic packing factor (APF): APF = \frac{V_occupied}{V_cell} = \frac{4 × \left[\frac{4}{3}π(0.414rA)^3 + \frac{4}{3}πrA^3\right]}{(2 × (0.414rA + rA))^3} After solving the above equation, we get the atomic packing factor (APF) for the rock salt crystal structure with the given radius ratio.

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Key Concepts

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

Rock Salt Crystal Structure
Understanding the atomic structure of materials can be fascinating, especially when it involves the orderly patterns in which atoms arrange themselves. One such intriguing arrangement is the rock salt crystal structure, named for its resemblance to table salt.

Imagine a checkerboard, but in three dimensions, where each black and white square represents a different type of ion. In the rock salt structure, one type of ion forms a cubic close-packed array, while ions of the opposite charge occupy the octahedral holes. This model creates a face-centered cubic (FCC) lattice for each of the ions, with the cations typically smaller than the anions.

Such organization allows for the maximum use of available space while maintaining charges' balance. One characteristic feature of the rock salt structure is that each ion is octahedrally coordinated, meaning that each ion is surrounded by six ions of opposite charge. This configuration presents not just a beautiful geometric pattern, but also influences properties like ionic conductivity and melting point.
Unit Cell Volume Calculation
The concept of a unit cell is like having a single repeating block that, when stacked together with identical blocks, constructs the entire crystal. For the rock salt crystal structure, calculating the volume of this building block is crucial for understanding the material's properties.

The unit cell's edge length, given by the formula a = 2 × (rC + rA), where rC and rA represent the radii of the cation and the anion respectively, holds the key to discovering its volume. Once we know the edge length a, the cube of a gives us the unit cell's volume, using the equation V_cell = a^3.

Considering that the cations and anions are packed in this volume, the calculation covers only the space these ions physically occupy, providing insight into how densely the ions are packed in the crystal structure. This is fundamental for predicting the material's density and informs us about the relationship between the ionic sizes and the crystal's structural compactness.
Cation and Anion Radii Relationship
The relationship between the radii of cations and anions plays a pivotal role in determining the stability and structure of ionic crystals. The radius ratio, denoted as rC/rA, influences the coordination number, which is the number of oppositely charged ions surrounding a given ion.

In the case of the rock salt structure, the coordination number is six, which requires a specific radius ratio for the ions to fit properly into the lattice. The relationship is handled mathematically by designating rC as a fraction of rA; thereby the radius of the smaller cation is often expressed as a ratio of the larger anion. In the given problem, the radius ratio is 0.414, suggesting that cations are significantly smaller than the anions.

This size dynamic is essential for understanding how structure dictates function in materials science. It affects the crystal lattice's ability to withstand deformation, its stability at different temperatures, and its electrical properties. In the world of materials, where the tiniest differences in atomic arrangements can lead to vastly different material behaviors, appreciating such details about the cation-anion size relationship is nothing short of imperative.

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Most popular questions from this chapter

Calculate the number of Frenkel defects per cubic meter in silver chloride at 350C. The energy for defect formation is 1.1 eV, whereas the density for AgCl is 5.50 g/cm3 at 350C.

Show that the minimum cation-to-anion radius ratio for a coordination number of 4 is 0.225.

(a) Suppose that CaO is added as an impurity to Li2O. If the Ca2- substitutes for Li, what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca2- added? (b) Suppose that CaO is added as an impurity to CaCl2. If the O2substitutes for Cl, what kind of vacancies would you expect to form? How many of these vacancies are created for every O2 added?

A hypothetical AX type of ceramic material is known to have a density of 2.10 g/cm3 and a unit cell of cubic symmetry with a cell edge length of 0.57 nm. The atomic weights of the A and X elements are 28.5 and 30.0 g/mol, respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: sodium chloride, cesium chloride, or zinc blende? Justify your choice(s).

Use a scientific calculator to evaluate (a) \(10^{1.2}\) (b) \(6^{-0.7}\) (c) \(6^{2.5}\) (d) \(\left(3^{-1} 4^{2}\right)^{0.8}\)
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