Question

A compound made of elements A and B form fcc structure. Atoms of A are at the corners and atoms of B are present at the centres of faces of cube. What is the formula of the compound?

• A. AB
• B. $AB_{2}$
• C. $AB_{3}$
• D. $A_{2}B$

Hint:

Logic Tip: Standard Face-Centered Cubic (fcc) lattice arrangements where one element occupies corners and the other occupies face centers invariably result in an $XY_3$ stoichiometry.

Answer 1

The Correct Option is C

Concept:
In a cubic unit cell, the contribution of an atom depends on its location:

• An atom at a corner is shared by 8 adjacent unit cells, so its contribution to one unit cell is $\frac{1}{8}$.
• An atom at a face center is shared by 2 adjacent unit cells, so its contribution to one unit cell is $\frac{1}{2}$.

Step 1: Calculate the effective number of A atoms per unit cell.
Atoms of element A are present at the corners of the cube. A cube has 8 corners. Contribution of each corner atom = $\frac{1}{8}$ Total number of A atoms = $8 \times \frac{1}{8} = 1$
Step 2: Calculate the effective number of B atoms per unit cell.
Atoms of element B are present at the centres of the faces of the cube. A cube has 6 faces. Contribution of each face-centered atom = $\frac{1}{2}$ Total number of B atoms = $6 \times \frac{1}{2} = 3$
Step 3: Determine the empirical formula of the compound.
The ratio of atoms A to atoms B in the unit cell is: $$A : B = 1 : 3$$ Therefore, the formula of the compound is $AB_3$.