Question

An alloy of copper, silver and gold is found to have copper constituting the $ccp$ lattice. If silver atoms occupy the edge centres and gold is present at body centre, the alloy has a formula:
A.$CuAgAu$
B.$C{u_4}A{g_2}Au$
C.$C{u_4}A{g_3}Au$
D.$C{u_4}A{g_1}Au$

Answer 1

The pattern of successive layers of $ccp$arrangement can be designated as $ABCABCABC$. $ccp$ is cubic close packing. It is also called face centered cubic. In this packing, spheres of the third layer are placed into octahedral voids. In a cubic close-packed arrangement of atoms, the unit cell consists of four layers of atoms.

Complete answer:
The cubic close packing is cubic structures entered for the face. When we put the atoms in the octahedral void, the packing is of the form of $ABCABCABC$, so it is known as cubic close packing, while the unit cell is face centered cubic. The packing quality is the proportion of the atoms directly occupied by the crystal or unit cell.
In $ccp$ structure or $fcc$ structure, the number of atoms per unit cell is calculated as follows. Thus, in a face-centred cubic unit cell or cubic close packing, we have: $8$ corners $\times$ $\dfrac{1}{8}$ per corner atom $= 8 \times \dfrac{1}{8} = 1$ atom. $6$ face-centred atoms $\times \dfrac{1}{2}$ atom per unit cell $= 3$ atoms.
Effective number of $Cu$ atoms in a unit cell$= 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 1 + 3 = 4$
Effective number of $Ag$ atoms in a unit cell$= 12 \times \dfrac{1}{4} = 3$
Effective number of $Au$ atoms in a unit cell$= 1$
So, general formula of compound is $C{u_4}A{g_3}Au$
So, the correct answer is (C)$C{u_4}A{g_3}Au$.

Note:
Octahedral voids are unoccupied empty spaces present in substances having an octahedral crystal system. It can be found in substances having a tetrahedral arrangement in their crystal system. Tetrahedral voids can be observed in the edges of the unit cell. The total number of octahedral voids in cubic close packed is four. We know, in a cubic close packing structure each unit cell has four atoms.