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

Hint: 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."