JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
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) M₄A
- (d) MA₃
Answer and solution
Answer (b) M₄A₃
Solution
Chemistry ma’am · 3 steps
Count the tetrahedral voids first: there are two for every atom in the hcp lattice.
- Voids per atom. Let the hcp lattice have N atoms of A. Any close-packed lattice of N atoms has 2N tetrahedral voids and N octahedral voids.
- Fill the voids. M occupies 2/3 of the 2N tetrahedral voids. So the number of M atoms is 2/3 × 2N = 4N/3.
M : A = 4N/3 : N = 4 : 3
- So. The formula is M₄A₃, which is option (b).
The idea to keep. In hcp or ccp, N atoms give 2N tetrahedral and N octahedral voids. Multiply the fraction occupied by the right void count, then take the whole-number ratio.
Answer from the NTA final answer key. Solution by PrepWiser.