Question

Given an alloy of $Cu$ , $Ag$ , and $Au$ in which $Cu$ atoms constitute the CCP arrangement. If the hypothetical formula of the alloy is $C{u_4}A{g_3}Au$ . What are the probable locations of $Ag$ and $Au$ atoms?
A) $Ag$ - all tetrahedral voids; $Au$ - all octahedral voids
B) $Ag$ - $\dfrac{3}{8}$ the tetrahedral voids; $Au$ - $\dfrac{1}{4}$ the octahedral voids
C) $Ag$ - $\dfrac{1}{2}$ octahedral voids; $Au$ - $\dfrac{1}{2}$ tetrahedral voids
D) $Ag$ - all octahedral voids ; $Au$ - all tetrahedral voids

Answer 1

Three layers (ABCABC...) of hexagonally organised atoms make up a CCP structure. Since they contact six atoms in their layer, plus three atoms in the layer above and three atoms in the layer below, atoms in a CCP configuration have a coordination number of $12$ .

Complete answer:
We know that Copper atoms ( $Cu$ ) form a CCP (Cubic Close Packed) structure, which is equivalent to an FCC (Face Centered Cubic) arrangement, according to the question.
Amount of copper atoms in a unit cell
\Rightarrow \dfrac{1}{8}\left( {Number.of.corners} \right) + \dfrac{1}{2} \times 6 \\\ \Rightarrow \dfrac{1}{8}\left( 8 \right) + \dfrac{1}{2} \times 6 \\\ \Rightarrow 1 + 3 \\\ \Rightarrow 4copper.atoms \\\
The number of octahedral voids (spaces) equals the number of atoms in the FCC arrangement, which is $4$ .
The Ag atoms appear to be at the centres of the cube's sides since the structure is oriented.
Since Ag has three atoms and a cube has eight corners, the fraction would be $Ag$ has $\dfrac{3}{8}$ tetrahedral voids and $Au$ has $\dfrac{1}{4}$ octahedral voids.
Hence, the correct option is B) $Ag$ - $\dfrac{3}{8}$ the tetrahedral voids; $Au$ - $\dfrac{1}{4}$ the octahedral voids.

Note:
Cubic Close Packed (CCP) or Face Centered Cubic (fcc) The same lattice is known by two different names. This cell is created by injecting another atom into each face of a simple cubic lattice, thus the term "face centred cubic."