   # The radius of tungsten is 137 pm and the density is 19.3 g/cm 3 . Does elemental tungsten have a face-centered cubic structure or a body-centered cubic structure? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 9, Problem 60E
Textbook Problem
151 views

## The radius of tungsten is 137 pm and the density is 19.3 g/cm3. Does elemental tungsten have a face-centered cubic structure or a body-centered cubic structure?

Interpretation Introduction

Interpretation:

The lattice structure of elemental tungsten has to be identified and justified.

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=edge length of unit cellR=radiusofatom

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

Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

8×18atomsincorners+6×12atomsinfaces=1+3=4atoms The edge length of one unit cell is given bya=2R2where  a=edge length of unit cellR=radiusofatom

### Explanation of Solution

Explanation

Calculate the density of tungsten by assuming its structure as FCC.

givendata:R=137pm=137×10-12ma=2R2=2×137×10-12m×1.414=387.4pmvolume of unit cell=a3=(387.4×10-12)3m3=5.8×10-23cm3Average mass of one 'W' atom=atomicmassof'W'Avogadronumber=183.84g6.022×1023=30.53×10-23gEachFCCunit cell contains4'W' atoms. therefore,Massofaunitcell=4×30.53×10-23g  =122.12×10-23g

density=massvolume=122.12×10-23g5.8×10-23cm3=21

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