Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 12 The Solid State - Study Questions - Page 467c: 42



Work Step by Step

Note: $M$ is the atomic weight, $N_A$ is the Avogadro's number, $\rho$ is the density (BCC): $8 \times 1/8$ atoms in the corners and one in the center, so 2 atoms per unit cell; Cube's diagonal $l.\sqrt3=4r\rightarrow l=4r/\sqrt3; V=l^3$ $\rho=(2\times M/N_A)\div(64.r^3/3\sqrt3)=(3\sqrt3/32)\times(\dfrac{M}{N_A.r^3})$ (FCC) $8 \times 1/8$ atoms in the corners and $6\times1/2$ in the faces, 4 atoms per unit cell Face's diagonal: $l.\sqrt2=4.r\rightarrow l=4r/\sqrt2; V=l^3$ $\rho=(4\times M/N_A)\div(64.r^3/2\sqrt2)=(\sqrt2/8)\times(\dfrac{M}{N_A.r^3})$ (Primitive) $8\times 1/8$ one atom per cell, $l=2.r$ $\rho=(M/N_A)\div(8.r^3)=(1/8)\times(\dfrac{M}{N_A.r^3})$ The factor $\dfrac{M}{N_A.r^3}$ is $50.9415/6.022\times10^{23}.(132\times10^{-10})^3=36.78\ g/cm^3$ The density divided by this factor is $6.11/36.78=0.166$ which is closer to $3\sqrt3/32$, so the unit cell is in a body-centered cubic arrangement.
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