Given information

• The anions $X$ form a cubic close packed (ccp) arrangement.
• In a ccp (fcc) unit cell, the number of $X$ atoms per unit cell is: $8\times \frac{1}{8} + 6\times \frac{1}{2} = 1+3=4$
• So, total number of $X$ atoms in the unit cell is 4.

Count the metal atoms $M$ from the shown unit cell In a ccp lattice, there are:

• 8 tetrahedral voids per unit cell
• 4 octahedral voids per unit cell

From the given figure (standard representation for this question), the metal atoms $M$ occupy half of the tetrahedral voids.

Therefore, number of $M$ atoms in the unit cell: $\frac{1}{2}\times 8 = 4$

Find the ratio $M:X$ $M:X = 4:4 = 1:1$

Hence, the empirical formula should be: $\boxed{MX}$

Check options

• A: $MX$ ✅
• B: $MX_2$ ❌
• C: $M_2X$ ❌
• D: $M_5X_{14}$ ❌

Comparison with stored correct answer

• Derived answer: $MX$
• Stored correct answer: $MX_2$
• These do not match.

The stored answer would be correct only if the number of $M$ atoms were 2 while $X=4$, i.e. if only one-fourth of tetrahedral voids or half of octahedral voids were occupied. But for the shown/standard ccp tetrahedral occupation pattern, the formula comes out as $MX$.