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: List the site contributions.
In a cubic cell, a corner atom is shared by 8 cells (contribution $1/8$) and a face-centre atom by 2 cells (contribution $1/2$).
Step 2: Count atoms of A at corners.
A sits at all 8 corners, so \[ n_A = 8 \times \frac{1}{8} = 1. \]
Step 3: Count atoms of B on faces.
B sits at all 6 face centres, so \[ n_B = 6 \times \frac{1}{2} = 3. \]
Step 4: Form the ratio.
The ratio $A : B = 1 : 3$.
Step 5: Write the empirical formula.
A ratio of $1 : 3$ gives the formula $\mathrm{AB_3}$.
Step 6: Conclude.
The compound is $\mathrm{AB_3}$, option (C).
\[ \boxed{\mathrm{AB_3}} \]