Question

A crystal is made up of particles A, B, and C, A forms fcc packing, B occupies all octahedral voids and C occupies all tetrahedral voids. If all the particles along one body diagonal are removed, then the formula of the crystal would be:

• A. $AB{C}_2$
• B. $A_{2}B{C}_2$
• C. $A_8{B}_4{C}_5$
• D. $A_5{B}_4{C}_8$

Answer 1

The correct option is D $A_5{B}_4{C}_8$

In fcc, unit cell number of atoms present $=4$

Now, $2\times$ Number of octahedral voids $=$ Number of tetrahedral voids

And Number of octahedral voids $=$ Number of atoms present in cubic unit cell.

So, $A=4,B=4,C=8$

On the body diagonal, there is one octahedral void (at body center) and two tetrahedral voids (at ${\left(\frac{1}{4}\right)}^{th}$ of the distance from each corners). So, after removal of atoms on body diagonal, we have:

$A=6\times \frac{1}{2}+\frac{(8-2)}{2}=3+\frac{3}{4}=\frac{15}{4}$

(face) (corners)

$B=4-1=3$; $C=8-2=6$

So, $ABC\Rightarrow ABC$

$\frac{15}{4}\times 43\times 46\times 4$ $\frac{15}{5}\frac{12}{4}\frac{24}{8}$

So, formula of crystal is $A_5{B}_4{C}_8$