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

Step 1: In a hexagonal close-packed (hcp) structure, let the number of atoms of element A be $n$.
Step 2: The number of tetrahedral voids (TVs) is $2n$.
Step 3: Element M occupies $2/3$ of the TVs. Number of M atoms $= \frac{2}{3} \times 2n = \frac{4}{3}n$.
Step 4: The ratio of M to A is $\frac{4}{3}n : n$, which simplifies to $4 : 3$. Therefore, the formula is $M_4A_3$.