Question

A compound contains elements X and Y. Y atoms form CCP lattice and atoms of X occupy 1/3rd of tetrahedral voids. What is the molecular formula of the compound?
A.$XY$
B.$\mathop X\nolimits_2 \mathop Y\nolimits_3 $
C.$\mathop X\nolimits_3 \mathop Y\nolimits_2 $
D.\[\mathop {XY}\nolimits_2 \]

Answer 1

Hint: As we have studied in three dimensional arrangement i.e CCP or HCP, The number of octahedral voids present in the lattice is equal to the number of lattice points and the number of tetrahedral voids is twice the number of lattice points.

Complete step by step answer:
Position of Tetrahedral Voids in CCP or FCC Lattice: Let a unit cell be divided into eight small cubes in which each small cube has atoms at alternate corners. It means there are 4 atoms in each small cube, which form a regular tetrahedron when joined together. In this tetrahedron there is a tetrahedral void. In whole there are 8 tetrahedral voids in all 8 tetrahedrons. Hence, we conclude that the number of tetrahedral voids is double the number of lattice points.

So now,
No. of tetrahedral voids in CCP = \[2 \times {\text{ }}no.{\text{ }}of{\text{ }}atoms{\text{ }}of{\text{ }}Y\]
Atoms of Y in CCP unit cell is always = 4
So tetrahedral void = \[2 \times {\text{ }}4{\text{ }} = {\text{ }}8\]

Number of tetrahedral voids occupied by element $x = \dfrac{1}{3} \times 8$
Now ratio of element X with that of $y = \dfrac{8}{3} + \dfrac{1}{4}$
 The ratio comes to be \[2:3\]
Hence the formula will be $\mathop X\nolimits_2 \mathop Y\nolimits_3 $.
Hence option B is correct.


Note:
In ionic solids generally bigger ions i.e., anions form CCP and smaller ions i.e., cations occupy the voids. In a given compound some fraction of tetrahedral or octahedral voids are occupied by cations.