Question

In an alloy of copper, silver and gold, the copper atoms constitute the ccp lattice and silver atoms occupies half of the tetrahedral voids and gold is present at the octahedral voids of the unit cell.
What will be the formula of the compound if all the atoms from one of the faces of the cube are removed?

• A. $C{u}_{8}A{g}_{4}A{u}_3$
• B. $C{u}_{3}A{g}_{3}A{u}_4$
• C. $C{u}_{8}A{g}_{3}A{u}_3$
• D. $C{u}_{3}A{g}_{4}A{u}_3$

Answer 1

The correct option is D $C{u}_{3}A{g}_{4}A{u}_3$

In fcc lattice, the octahedral voids are present at 12 edge centre and at the body centre of the unit cell.
There are two tetrahedral voids on the every body diagonal of fcc. Thus, totally 8 tetrahedral voids present inside the unit cell.
Removing the face of a cube will result in removal of 4 corner atoms and 1 face centre and 4 edge centre atoms.
Thus,
$\text{Number of Cu atoms}=4\times \frac{1}{8}+\frac{5}{2}=3$
$\text{Number of Ag atoms}=4$
$\text{Number of Au atoms}=\frac{8}{4}+1=3$
Thus, the formula of the unit cell is
$C{u}_{3}A{g}_{4}A{u}_3$
Thus, correct option is (d).