Question

The FCC unit cell of a compound contains ions of $\mathrm{A}$ at the corner and ions of $\mathrm{B}$ at the centre of each face, what is the formula of the compound?

• A. $\mathrm{AB_2}$
• B. $\mathrm{A_2B}$
• C. $\mathrm{AB_3}$
• D. $\mathrm{AB}$

Hint:

In any cubic lattice system, remember the basic net totals: All 8 corners together contribute exactly 1 atom, while all 6 faces together contribute exactly 3 atoms. This directly gives a $1:3$ ratio for corner-to-face arrangements!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a crystalline solid with a face-centered cubic ($\mathrm{FCC}$) lattice geometry where atoms of element $\mathrm{A}$ occupy all the corner sites and atoms of element $\mathrm{B}$ occupy the centers of all the faces. We need to determine the empirical formula of this compound.

Step 2: Key Formula or Approach:
To find the empirical chemical formula, we calculate the effective number of atoms of $\mathrm{A}$ and $\mathrm{B}$ contained inside a single unit cell based on lattice site contributions:

• Contribution of an atom at a corner site = $\frac{1}{8}$

• Contribution of an atom at a face-centered site = $\frac{1}{2}$

Step 3: Detailed Explanation:
Let's find the effective number of atoms for each element:

• Number of $\mathrm{A}$ atoms: There are 8 corners in a cube, each containing an atom of $\mathrm{A}$. $$\text{Total } \mathrm{A} = 8 \text{ corners} \times \frac{1}{8} = 1$$

• Number of $\mathrm{B}$ atoms: There are 6 faces in a cube, each containing an atom of $\mathrm{B}$ at its center. $$\text{Total } \mathrm{B} = 6 \text{ faces} \times \frac{1}{2} = 3$$
The ratio of atoms of $\mathrm{A}$ to atoms of $\mathrm{B}$ in the unit cell is $1 : 3$. Therefore, the simplest empirical formula of the crystal compound is $\mathrm{AB_3}$.

Step 4: Final Answer:
The formula of the compound is $\mathrm{AB_3}$, which corresponds to option (C).