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Given an alloy of Cu , Ag , and Au in which Cu atoms constitute the CCP arrangement. If the hypothetical formula of the alloy is Cu4Ag3Au . What are the probable locations of Ag and Au atoms?
A) Ag - all tetrahedral voids; Au - all octahedral voids
B) Ag - 38 the tetrahedral voids; Au - 14 the octahedral voids
C) Ag - 12 octahedral voids; Au - 12 tetrahedral voids
D) Ag - all octahedral voids ; Au - all tetrahedral voids

Answer
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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
 18(Number.of.corners)+12×618(8)+12×61+34copper.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 38 tetrahedral voids and Au has 14 octahedral voids.
Hence, the correct option is B) Ag - 38 the tetrahedral voids; Au - 14 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."