Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 11

Iron titanate, \(\mathrm{Fe} \mathrm{Ti} \mathrm{O}_{3},\) forms in the ilmenite crystal structure that consists of an HCP arrangement of \(\mathrm{O}^{2-}\) ions. (a) Which type of interstitial site will the \(\mathrm{Fe}^{2+}\) ions occupy? Why? (b) Which type of interstitial site will the \(\mathrm{Ti}^{4+}\) ions occupy? Why? (c) What fraction of the total tetrahedral sites will be occupied? (d) What fraction of the total octahedral sites will be occupied?

Short Answer

Expert verified
Calculate the fraction of occupied tetrahedral and octahedral sites. Answer: In the ilmenite crystal structure, Fe²⁺ ions prefer occupying the octahedral interstitial sites, while Ti⁴⁺ ions prefer occupying the tetrahedral interstitial sites. The fraction of occupied tetrahedral sites is 25%, and the fraction of occupied octahedral sites is 50%.

Step by step solution

01

Understanding crystal structures and terminology

The HCP (Hexagonal Close-packed) structure consists of layers of spheres (in this case O²– ions) arranged in a hexagonal pattern, creating octahedral and tetrahedral interstitial sites for other ions to fill. The size and charges of the interstitial ions will determine their preferred site.
02

Identifying preferred interstitial site for Fe²⁺ ions#a)

For Fe²⁺ ions, we will determine which interstitial site is preferred by looking at the size and charge of the ion. Fe²⁺ ions have a smaller size and charge than Ti⁴⁺ ions; therefore, they prefer occupying the octahedral interstitial sites, as these sites provide a stable electric environment with an appropriate size to accommodate their smaller radius.
03

Identifying preferred interstitial site for Ti⁴⁺ ions#b)

For Ti⁴⁺ ions, we again look at their size and charge. Due to their larger size and charge, Ti⁴⁺ ions will prefer the larger tetrahedral interstitial sites, providing them with a more stable electric environment and enough space to accommodate their larger ionic radius.
04

Calculate the fraction of occupied tetrahedral sites#c)

To calculate the fraction of occupied tetrahedral sites, we need to know the total number of tetrahedral sites and the number of Ti⁴⁺ ions accommodated in these sites. In this crystal structure, the formula indicates there is 1 Ti⁴⁺ ion for every 2 O²– ions. In HCP, there are 2 tetrahedral sites per O²– ion. So, for 2 O²– ions, there are 4 tetrahedral sites. Since there is 1 Ti⁴⁺ ion occupying 1 of these sites, the fraction of occupied tetrahedral sites is \(\frac{1}{4}\) or 25%.
05

Calculate the fraction of occupied octahedral sites#d)

To calculate the fraction of occupied octahedral sites, we need to know the total number of octahedral sites and the number of Fe²⁺ ions accommodated in these sites. The crystal structure formula indicates there's 1 Fe²⁺ ion for every 2 O²– ions. In an HCP structure, there's 1 octahedral site per O²– ion. So, for 2 O²– ions, there are 2 octahedral sites. Since there is 1 Fe²⁺ ion occupying 1 of these sites, the fraction of occupied octahedral sites is \(\frac{1}{2}\) or 50%.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 modulus of elasticity for titanium carbide (TiC) having 5 vol\% porosity is \(310 \mathrm{GPa}\left(45 \times 10^{6} \mathrm{psi}\right)\) (a) Compute the modulus of elasticity for the nonporous material. (b) At what volume percent porosity will the modulus of elasticity be \(240 \quad\) GPa \(\left(35 \times 10^{6} \mathrm{psi}\right) ?\)

Calculate the number of Frenkel defects per cubic meter in silver chloride at \(350^{\circ} \mathrm{C}\). The energy for defect formation is \(1.1 \mathrm{eV}\), while the density for AgCl is \(5.50 \mathrm{g} / \mathrm{cm}^{3}\) at \(\left(350^{\circ} \mathrm{C}\right)\).

(a) A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of \(300 \mathrm{MPa}(43,500 \mathrm{psi})\). If the specimen radius is \(5.0 \mathrm{mm}(0.20 \mathrm{in.})\) and the support point separation distance is \(15.0 \mathrm{mm}\) \((0.61 \text { in. }),\) predict whether or not you would expect the specimen to fracture when a load of \(7500 \mathrm{N}\left(1690 \mathrm{lb}_{\mathrm{f}}\right)\) is applied? Justify your prediction. (b) Would you be \(100 \%\) certain of the prediction in part (a)? Why or why not?

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 \(45 \mathrm{mm}(1.75 \text { in. })\).

For a ceramic compound, what are the two characteristics of the component ions that determine the crystal structure?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Fluids

Read Explanation

Magnetism

Read Explanation

Force

Read Explanation

Oscillations

Read Explanation

Fundamentals of Physics

Read Explanation
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.

Sign-up for free