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In a face centered cubic arrangement of A and B atoms in which ‘A’ atoms are at the corners of the unit cell and ‘B’ are at the face centers. One of the ‘A’ atoms is missing from one corner in the unit cell. The simplest formula of compound is:
$A.$ ${A_7}{B_3}$
$B.$ ${A_7}{B_8}$
$C.$ $A{B_3}$
$D.$ ${A_7}{B_{24}}$

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Last updated date: 13th Jun 2024
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Answer
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HINT- We have to keep in mind that, atoms on a corner are shared by eight unit cells and hence contribute only $\dfrac{1}{8}$ atom per unit cell, while atoms on a face are shared by two unit cells, each counts as $\dfrac{1}{2}$ atom per unit cell. Thus, the total number of atoms in each unit cell is $8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 1 + 3 = 4$

Complete step by step solution:
Now moving onto the question, In a face centered cubic arrangement of A and B atoms in which ‘A’ atoms are at the corners of the unit cell.
There are a total 8 corners and each contributes one eight to the unit cell.
One of the ‘A’ atoms is missing from one corner in the unit cell. So, there are 7 ‘A’ atoms with total contribution $7 \times \dfrac{1}{8} = \dfrac{7}{8}$
‘B’ atoms are at the face centers. There are six face centers. Each contributes one half to the unit cell.
Total contribution= $6 \times \dfrac{1}{2} = 3$
A:B= $\dfrac{7}{8}:3 = 7:24$
The simplest formula of compound is ${A_7}{B_{24}}$

Hence, the correct option is $D.$

NOTE- FCC stands for Face Centered Cubic. It is a type of atomic arrangement and is relatively “tightly” packed. FCC is formally defined as a cubic lattice with the face positions fully equivalent to each of the eight corners. Face Centered Cubic structures are, as the name implies, based around “faces” of a cube with an atom at each of the corners and one at the center of each face. The face centered cubic has a coordination number of 12 and contains 4 atoms per unit cell. Since there are eight corners, there are eight tetrahedral voids in a fcc unit cell.