Question

An alloy of copper, silver, and gold was found to have copper constituting CCP lattice. If silver atoms occupy the edge centers and gold is present at the body center, the alloy has a formula:

• A. $\mathrm{CuAgAu}$
• B. ${\mathrm{Cu}}_4{\mathrm{Ag}}_2\mathrm{Au}$
• C. ${\mathrm{Cu}}_4{\mathrm{Ag}}_3\mathrm{Au}$
• D. ${\mathrm{Cu}}_4{\mathrm{Ag}}_4\mathrm{Au}$

Answer 1

The correct option is C

${\mathrm{Cu}}_4{\mathrm{Ag}}_3\mathrm{Au}$

Explanation for the correct option :

(C) ${\mathbf{Cu}}_{\mathbf{4}}{\mathbf{Ag}}_{\mathbf{3}}\mathbf{Au}$

• Copper ($\mathrm{Cu}$) consist of CCP lattice. CCP(cubic closed packed) lattice is an alternative name for FCC(face-centered cubic) lattice. They occupy the corners and the face of a unit cell. Hence, the number of Copper atoms in a unit cell will be: $\frac{1}{8}\times 8+\frac{1}{2}\times 6=1+3=4$
• Silver($\mathrm{Ag}$) atoms in a unit cell occupy edges. A cube consists of 12 edges and each atom shares four edges. Therefore the number of silver atoms will be:$\frac{1}{4}\times 12=3$
• Gold ($\mathrm{Au}$) atoms are present at the center of the body and hence the number of atoms per unit cell will be:$1$

All together the formula will become: ${\mathrm{Cu}}_4{\mathrm{Ag}}_3\mathrm{Au}$

Explanation for the incorrect options :

Since, the required formula is ${\mathrm{Cu}}_4{\mathrm{Ag}}_3\mathrm{Au}$. Therefore, options (A), (B) and (D) stand incorrect.

Hence, the correct option is (C)${\mathbf{Cu}}_{\mathbf{4}}{\mathbf{Ag}}_{\mathbf{3}}\mathbf{Au}$