Question

Element X crystallizes in 12 coordination fcc lattice. On applyng high temperature it changes to 8 coordination bcc lattice. The ratio of the density of the crystal lattice before and and after applying high temperature is :

• A. $1:2$
• B. $3:2$
• C. $\sqrt{2}:\sqrt{3}$
• D. $2(\sqrt{2}{)}^3:(\sqrt{3}{)}^3$

Answer 1

The correct option is D $2(\sqrt{2}{)}^3:(\sqrt{3}{)}^3$

For a 12 coordination FCC lattice :

Density of the unit cell :
${\rho }_1=\frac{Z_{fcc}\times M}{N_A\times a_1^3}$
Here M is the molecular mass
For FCC:
$a=2\sqrt{2}r{Z}_{fcc}=4$
For a 8 coordination BCC lattice :
Density of the unit cell :
${\rho }_2=\frac{Z_{bcc}\times M}{N_A\times a_2^3}$
For BCC:
$a=\frac{4r}{\sqrt{3}}{Z}_{bcc}=2$

$\frac{{\rho }_1}{{\rho }_2}=\frac{4}{2}\times \frac{{\left(\frac{4r}{\sqrt{3}}\right)}^3}{(2\sqrt{2}r{)}^3}=\frac{4\sqrt{2}}{3\sqrt{3}}\frac{{\rho }_1}{{\rho }_2}=\frac{2(\sqrt{2}{)}^3}{(\sqrt{3}{)}^3}$