Question

In \( A_3B \) crystal structure, A atoms occupied all octahedral as well as all tetrahedral voids and B atoms are at FCC centres. What is the formula of compound \( A_3B \)?

• A. \( A_3B \)
• B. \( A_{10}B_3 \)
• C. \( A_{15}B_{36} \)
• D. \( A_3B \)

Hint:

In FCC crystal structures, the number of atoms at the corners and faces of the unit cell is essential for determining the overall stoichiometry. For compounds involving voids, make sure to account for the number of atoms filling the octahedral and tetrahedral voids.

Answer 1

The Correct Option is A

In the \( A_3B \) crystal structure, A atoms occupy both the octahedral and tetrahedral voids, while B atoms are located at the face-centred cubic (FCC) lattice positions. In an FCC structure, each unit cell contains 4 atoms of type B. To satisfy the stoichiometry of the given crystal, the number of A atoms needs to be thrice the number of B atoms, leading to the formula \( A_3B \). Thus, the correct formula for the compound is \( A_3B \), which is option (1).

Answer 2

Step 1: Understand the crystal structure
In the given \( A_3B \) crystal structure, B atoms occupy the face-centered cubic (FCC) lattice points.
A atoms occupy all octahedral and tetrahedral voids present in the structure.

Step 2: Number of atoms at FCC positions
In an FCC lattice, there are 4 atoms of B per unit cell.

Step 3: Number of octahedral and tetrahedral voids
- Number of octahedral voids in FCC lattice = number of atoms in FCC = 4.
- Number of tetrahedral voids in FCC lattice = 2 × number of atoms in FCC = 8.

Step 4: Number of A atoms
A atoms occupy all octahedral and tetrahedral voids.
Total A atoms = 4 (octahedral) + 8 (tetrahedral) = 12.

Step 5: Calculate the formula ratio
B atoms = 4
A atoms = 12
Simplifying ratio A:B = 12:4 = 3:1

Step 6: Conclusion
Thus, the formula of the compound is \( A_3B \), consistent with the given formula.