Question

A compound is formed by two elements A and B. The atoms of element B forms ccp structure. The atoms of A occupy 1/3rd of tetrahedral voids. What is the formula of the compound ?​

Answer 1

The atoms of element B form ccp structure. The number of tetrahedral voids generated is twice the number of B atoms.

Thus, number of tetrahedral voids = 2B

The atoms A occupy (1/3) of these tetrahedral voids.

Hence, number of A atoms = 2B×1/3

Ratio of A and B atoms = 2/3 B: 1B

= 2/3:1 = 2:3

Formula of compound = A_2B_3

Answer 2

Answer:

The formula of the compound is A₂B₃.

Explanation:

We know that,

In a face-centred lattice (FCC),
8 corners are present and each atom placed in the corner contributes 1/8 to the lattice.
Similarly, 6 faces are present and each face contributes 1/2 to the lattice.

∴ Total number of atoms present in an FCC lattice = $[(\frac{1}{2} \times 6) + (\frac{1}{8} \times 8)] = 3 + 1 = 4$

What are voids/interstices in a lattice?

Even after close packing, there remain some empty spaces left within a lattice. Such empty spaces are known as voids or interstices.

In 3D packing, two types of voids are present:

1. Tetrahedral Voids: The vacant space between 4 touching spheres (3 in one layer and 1 sphere of the upper or lower layer) forms a tetrahedral void.
   Since a sphere touches 3 spheres of its upper layer and 3 spheres of its lower layer, the total number of tetrahedral voids associated with 1 particular sphere is two.
   Thus, we can formulate a general formula stating that the number of tetrahedral voids in a close-packed 3D lattice is 2n, where n is the number of atoms present in the lattice.
2. Octahedral Voids: The vacant space between 6 touching spheres is known as an octahedral void.
   The number of octahedral voids present in a close-packed 3D lattice is n, where n is the number of atoms present in the lattice.

Calculation:

Thus, as per the given information,
Element B forms the ccp structure. Therefore, the number of atoms present in the lattice is 4.
Element A covers 1/3 parts of the tetrahedral voids present.
∴ Number of atoms of A present = $(\frac{1}{3} \times 8) = \frac{8}{3}$ [∵ total number of tetrahedral voids present in a FCC lattice = 2n = 2 × 4 = 8]

Conclusion:

∴ The formula of the compound

$= A_{\frac{8}{3} } B_{4}$
$= A_{8} B_{12}$
$= A_{2} B_{3}$

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