Question

$\mathrm{A}$ forms $\mathrm{CCP}$ lattice, $\mathrm{B}$ occupies half of the octahedral voids and ‘$\mathrm{O}$’ occupies all the tetrahedral voids. Calculate the formula.

• A. ${\mathrm{A}}_2{\mathrm{BO}}_4$
• B. ${\mathrm{ABO}}_4$
• C. ${\mathrm{A}}_2{\mathrm{B}}_2\mathrm{O}$
• D. ${\mathrm{A}}_2{\mathrm{BO}}_2$

Answer 1

The correct option is A

${\mathrm{A}}_2{\mathrm{BO}}_4$

Cubic closed packed structure:

• $\mathrm{CCP}$ is a cubic closed packed structure in which the atoms are arranged at four corners and in the centre of each face of cube.
• So, the number of atoms per unit cell, $A=4$.
• Voids are unoccupied empty spaces present in between atoms arranged in the lattice structure.
• Tetrahedral voids are present in between four atoms having a tetrahedral arrangement.
• The number of tetrahedral voids present in $\mathrm{CCP}$, $O=2\times 4=8$.
• Octahedral voids are formed in between equilateral triangles with apices opposite to each other.
• In $\mathrm{CCP}$ lattice, octahedral voids, $B=4$, as only half of the voids are occupied thus $B=4\times \frac{1}{2}=2$.
• Hence, the formula for CCP lattice is $A_4{B}_2{O}_8$.
• By simplifying the formula we get ${\mathrm{A}}_2{\mathrm{BO}}_4$.
• Hence, option A is correct.