/*! 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 9 The corundum crystal structure, ... [FREE SOLUTION] | 91影视

91影视

The corundum crystal structure, found for \(\mathrm{Al}_{2} \mathrm{O}_{3},\) consists of an \(\mathrm{HCP}\) arrangement of \(\mathrm{O}^{2-}\) ions; the \(\mathrm{Al}^{3+}\) ions occupy octahedral positions. (a) What fraction of the available octahedral positions are filled with \(\mathrm{Al}^{3+}\) ions? (b) Sketch two close-packed \(\mathrm{O}^{2-}\) planes stacked in an \(A B\) sequence, and note octahedral positions that will be filled with the \(\mathrm{Al}^{3+}\) ions.

Short Answer

Expert verified
Answer: In the Al鈧侽鈧 crystal structure, \(\frac{2}{3}\) of the available octahedral positions are filled with Al鲁鈦 ions. To illustrate this, you would sketch two close-packed O虏鈦 planes stacked in an AB sequence and show Al鲁鈦 ions occupying 2 out of 3 octahedral positions between the layers, ensuring that from each pair of Al鲁鈦 ions, one is in the voids above the A layer and one is in the voids below the B layer. Leave the third octahedral position vacant.

Step by step solution

01

(a1) Relating ions in the Al鈧侽鈧 structure

For the Al鈧侽鈧 structure, we are given that O虏鈦 ions are arranged in an HCP unit cell, and Al鲁鈦 ions fill octahedral interstitial sites. The stoichiometry of Al鈧侽鈧 tells us that there are 2 Al鲁鈦 ions for every 3 O虏鈦 ions in the structure.
02

(a2) Calculate the number of octahedral positions

In an HCP arrangement, the number of octahedral positions is equal to the number of close-packed ions i.e. O虏鈦 ions. As the number of O虏鈦 ions is 3 (from the Al鈧侽鈧 stoichiometry), there are 3 octahedral positions.
03

(a3) Calculate the fraction of filled octahedral positions

We know there are 2 Al鲁鈦 ions and 3 octahedral positions available. Thus, the fraction of octahedral positions filled with Al鲁鈦 ions is \(latex \frac{2 \: Al^{3+}\: ions}{3 \: octahedral\: positions}=\frac{2}{3}\) So, \(\frac{2}{3}\) of the available octahedral positions are filled with Al鲁鈦 ions.
04

(b1) Sketching close-packed O虏鈦 planes

First, sketch two close-packed O虏鈦 planes stacked in an AB sequence: 1. Draw two parallel horizontal layers of O虏鈭 ions. 2. In each layer, arrange the ions in a hexagonal pattern, where each ion is surrounded by six neighbors in the same plane. A close-packed plane will resemble a hexagonal grid. 3. To stack the planes in an AB sequence, place the ions in the second layer (B layer) into the triangular gaps formed by the first layer (A layer) of O虏鈦 ions.
05

(b2) Marking octahedral positions for Al鲁鈦 ions

Now that we have the stacked close-packed O虏鈦 planes (AB), we need to mark the octahedral positions for the Al鲁鈦 ions. Since \(\frac{2}{3}\) of the octahedral positions are filled with Al鲁鈦 ions, in each sequence of 3 octahedral positions, 2 will be filled, and 1 will be vacant. 1. In the sketch, place Al鲁鈦 ions in 2 out of 3 octahedral positions (the halfway point between two O虏鈦 ions layers). 2. Make sure that from each pair of Al鲁鈦 ions, one is in the voids above the A layer and one is in the voids below the B layer. 3. Leave the third (remaining) octahedral position between the layers vacant. The sketch will now show two close-packed O虏鈦 planes stacked in an AB sequence with Al鲁鈦 ions occupying \(\frac{2}{3}\) of the octahedral positions between the layers.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

HCP Arrangement
When we delve into the intriguing world of crystal structures, it's essential to understand the arrangement of atoms or ions within the lattice. One common arrangement is the Hexagonal Close-Packed (HCP) structure. In this layout, ions are packed in a way that maximizes their number within a given volume, which results in an efficient use of space.

In the HCP arrangement, each ion lies at the corners of a hexagon and also at the center of the hexagon. Imagine a layer of marbles lying in a tray where each marble touches three others to its left and three to its right, forming a triangle both above and below it. Now, stack another layer exactly on top, and you will have portions of two opposite triangles of the first layer in between four marbles of the second layer. This is a small glimpse into how HCP is structured.

Furthermore, for materials like corundum (Al2O3), the oxygen ions (O2-) form the HCP structure. This specific arrangement influences various physical properties of the material like density, hardness, and slippage planes, making it fundamental in understanding how such materials will behave in different applications.
Octahedral Interstitial Sites

Where Atoms Find a Home in Crystals

Within the HCP structure, there exist little 'pockets' or interstices called octahedral interstitial sites. Imagine a void surrounded by six ions forming an octahedron 鈥 that's an octahedral interstitial site. In corundum's case, these sites serve as a home for aluminum ions (Al3+).

Now, imagine you're stacking oranges in a box: In between the layers, you can nestle smaller cherries in the gaps. These cherries are like interstitial atoms fitting into the gaps between the larger close-packed ions. Not every gap is filled; in Al2O3, only two-thirds of these octahedral sites contain an aluminum ion. This partial filling is crucial鈥攊t's what maintains the crystal's overall charge balance and stability, and consequently, determines its material properties.
Stoichiometry of Al2O3

Getting the Proportions Right

Stoichiometry, in the context of crystal chemistry, refers to the definitive ratio of atoms in a compound. For aluminum oxide or corundum (Al2O3), the stoichiometry is reflected in its formula: two aluminum ions (Al3+) for every three oxygen ions (O2-).

This isn't just a matter of counting atoms; it's about the entire structure's balance. Since oxygen ions form the close-packed layers and make up the majority of the crystal, you can think of them as the 'backbone' or framework. Aluminum ions, the 'guests,' are nestled within this framework in the octahedral sites. Thanks to the 2-to-3 ratio, the structure achieves electrical neutrality and the geometry that gives corundum its characteristic properties, such as hardness and resistance to corrosion. Understanding the stoichiometry of Al2O3 is not just about numbers; it's about the harmony of construction that allows the material to function in various applications, from abrasives to laser components.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The zinc blende crystal structure is one that may be generated from close- packed planes of anions. (a) Will the stacking sequence for this structure be FCC or HCP? Why? (b) Will cations fill tetrahedral or octahedral positions? Why? (c) What fraction of the positions will be occupied?

Iron oxide (FeO) has the rock salt crystal structure and a density of \(5.70 \mathrm{g} / \mathrm{cm}^{3}\) (a) Determine the unit cell edge length. (b) How does this result compare with the edge length as determined from the radii in Table \(12.3,\) assuming that the \(\mathrm{Fe}^{2+}\) and \(\mathrm{O}^{2-}\) ions just touch each other along the edges?

A circular specimen of \(\mathrm{MgO}\) is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is \(5560 \mathrm{N}\left(1250 \mathrm{lb}_{\mathrm{f}}\right),\) the flexural strength is \(105 \mathrm{MPa}(15,000 \mathrm{psi}),\) and the separation between load points is \(45 \mathrm{mm}(1.75 \text { in. })\).

Cite one reason why ceramic materials are, in general, harder yet more brittle than metals.

The unit cell for \(\mathrm{Fe}_{3} \mathrm{O}_{4}\left(\mathrm{FeO}-\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) has cubic symmetry with a unit cell edge length of \(0.839 \mathrm{nm} .\) If the density of this material is \(5.24 \mathrm{g} / \mathrm{cm}^{3},\) compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 12.3.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.