Chemistry
Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 10, Problem 57E

Titanium metal has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm3. Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint: In a body-centered arrangement of spheres, the spheres touch across the body diagonal.)

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

        The edge length of the unit cell of Titanium has to be determined.

Concept introduction:

        In packing of atoms in a crystal structure, the atoms are imagined as spheres. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing.

       In body-centered cubic unit cell, each of the six corners is occupied by every single atom. Center of the cube is occupied by one atom.

       Each atom in the corner is shared by eight unit cells and a single atom in the center of the cube remains unshared. Thus the number of atoms per unit cell in BCC unit cell is,

                  8×18atomsincorners+1atomatthecenter=1+1=2atoms       The edge length of one unit cell is given bya=4R3where  a=edgelength of unit cellR=radiusofatom.

Answer to Problem 57E

Answer

        The edge length of the unit cell of Titanium is 328 pm.

Explanation of Solution

Explanation

Calculate the mass of unit cell of Ti.

Average mass of one Ti atom=atomicmassofTiAvogadronumber=47.9g6.022×1023=7.95×10-23gEachunit cell contains2Ti atoms. therefore,Massofaunitcell=2×7.95×10-23g  =15.9×10-23g

            Each unit cell contains 2 Ti atoms. Therefore two times the average mass of one Ti atom gives mass of a unit cell.

Calculate the volume of unit cell of Ti.

knowndata:mass=15.9×10-23gdensity=4.50g/cm3density=massvolumevolume=massdensity=15.9×10-23g4.50g/cm3=35.3×10-24cm3

         Density of the Titanium is given and mass of the Titanium is calculated in the previous step. Substituting these two values in the equation density=massvolume and rearranging it, volume of the unit cell is obtained.

Determine the edge length of unit cell of Ti.

a3=35.3×10-24cm3a=33.53×10-24=3.28×10-8cm=328pm

          Volume of the bcc unit cell is calculated in the previous step. Cube root of the volume of a bcc unit cell, gives the edge length of bcc unit cell.

Conclusion

Conclusion

         The edge length of the unit cell of Titanium is determined.

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Chapter 10 Solutions

Chemistry

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  • The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69). Given that the density of cesium chloride is 3.97 g/cm3, and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent Cs+ and Cl ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of Cs+ is 169 pm, and the ionic radius of Cl is 181 pm.
    Calculate the percent of volume that is actually occupied by spheres in a body-centered cubic lattice of identical spheres You can do this by first relating the radius of a sphere, r, to the length of an edge of a unit cell, l. (Note that the spheres do not touch along an edge but do touch along a diagonal passing through the body-centered sphere.) Then calculate the volume of a unit cell in terms of r. The volume occupied by spheres equals the number of spheres per unit cell times the volume of a sphere (4r3/3).
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