Question

In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is :

• A. M₂A₃
• B. M₄A₃
• C. MA₃
• D. M₄A

Hint:

For any lattice with $N$ atoms, the number of octahedral voids is $N$ and tetrahedral voids is $2N$.

Answer 1

The Correct Option is B

To determine the formula of the binary compound, let's analyze the given information. In this binary compound:

• Atoms of element A form a hexagonal close-packed (hcp) structure.
• Atoms of element M occupy 2/3 of the tetrahedral voids in this hcp structure.

First, we need to understand the relationship between the atoms forming the hcp structure and its voids. In an hcp lattice:

• There are 2 atoms per unit cell.
• Each atom in the hcp structure creates two tetrahedral voids.
• Therefore, for 2 atoms in an hcp unit cell, there are 4 tetrahedral voids.

Now, let's apply the given condition that atoms of element M occupy 2/3 of the tetrahedral voids:

• The total number of tetrahedral voids in the unit cell of hcp = 2 × 2 = 4.
• Number of tetrahedral voids occupied by M = (2/3) × 4 = 8/3.

Since we need to find a whole number ratio for determining the chemical formula, let's equate the amounts of atoms A and M:

• For every cell with 2 atoms of A, there are (8/3) atoms of M.
• To convert this to a simple ratio, multiply through by 3:
• Ratio of M to A = 8 (from M) : 6 (from A), which simplifies to 4:3.

Therefore, the empirical formula for the compound is M₄A₃.

Thus, the correct answer is:

M₄A₃