Question

An ionic compound $XY$ forms a crystal structure in which $Y$ atoms form ccp lattice and $X$ atoms are so positioned that each $X$ atom touches $6$ $Y$ atoms. If the nearest distance between two $X$ atoms is $2\sqrt{2}\mathring{A}$, then the density (g/cm${}^3$) of $XY$ crystal is:

(Given : Atomic mass of $X$ $=14$ amu, $Y$ $=50$ amu, Avogadro's number $=$ $6\times {10}^{23}$)

• A. $5\sqrt{6}$
• B. $6.66$
• C. $0.88$
• D. $5.67$

Answer 1

The correct option is B $6.66$

Density of the unit cell,
$d=\frac{ZM}{V{N}_A}$
Here, $Z=4,M=50+14=64,$ and distance between two nearest octahedral voids in fcc lattice $=2\sqrt{2}=a/\sqrt{2}$

$a=4\mathring{A}=4\times {10}^{-8}cm$ , $N_A=6.023\times {10}^{23}$
Note: The distance between two nearest octahedral voids in fcc lattice is nothing but the distance b/w the two adjacent edge centre of a cube.

By putting values,

$d=\frac{4\times 64}{a^3{N}_A}$

$d=6.66g/c{m}^3$