Question

A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy $\dfrac{1}{3}$ rd of tetrahedral voids. What is the formula of the compound?

Answer 1

Hint: The structure of metals can be explained when atoms are packed together. The atoms in metals are considered as spheres of identical size. Close packing is the most efficient arrangement of spheres of identical size and to fill available spaces in which each sphere touches six neighboring spheres.

Complete step by step answer:
Close packing is of two types-hexagonal packing and cubic close packing. Close packing can be explained by considering particles as spheres of identical size arranged in a row, i.e. bottom layer A. When second layer of particles is placed on layer A, each sphere of layer B will rest in the hollow void or hole in bottom layer. The spheres of the second layer will produce two types of voids-tetrahedral and octahedral voids. Tetrahedral voids falls above a sphere in the bottom layer. Octahedral void falls above a void in the bottom layer.
If tetrahedral voids are covered, the third layer is identical to the bottom layer. This is called hcp arrangement. If octahedral voids are covered, the third layer is not identical to the bottom layer. This is called ccp arrangement.
The number of atoms in face-centered cubic unit cell, ${\text{Z = 4}}$
So the number of tetrahedral voids $ = 2{\text{Z}} = 8$
According to the question, ${\text{N}} \to {\text{4}}$ and ${\text{M}} \to \dfrac{1}{3} \times 8 = \dfrac{8}{3}$
Thus the formula will be ${{\text{N}}_4}{{\text{M}}_{\dfrac{8}{3}}}$.
So we have to convert the subscripts to whole numbers. So multiply with three on both, we get ${{\text{N}}_{12}}{{\text{M}}_8}$
On simplifying, we get the formula of the compound as ${{\text{N}}_3}{{\text{M}}_2}$

Note:
Wherever a sphere of the second layer is above the void of the first layer, a tetrahedral void is formed. The number of tetrahedral voids generated is twice the number of octahedral voids generated. The octahedral void in fcc is formed by atoms at three edges and one center.