# In a cubic arrangement of atoms of  and $C$, atoms of $A$ are present at the corners of the unit cell,$\mathrm{B}$ atoms are at face centers and $\mathrm{C}$ at tetrahedral voids. If one of the atom from one corner is missing in the unit cell, then the simplest formula of compound will be:

1. A

${A}_{7}{B}_{3}{C}_{8}$

2. B

${A}_{7}{B}_{24}{C}_{64}$

3. C

${A}_{718}{B}_{3}{C}_{4}$

4. D

${A}_{7}{B}_{48}{C}_{64}$

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### Solution:

Number of atoms of $A$ in unit cell$=7×\frac{1}{8}=\frac{7}{8}$

Number of atoms of $B$ in unit cell$=6×\frac{1}{2}=3$

Number of atoms of $C$ in unit cell$=8×1=8$

$\therefore$Simplest formula$={A}_{7/8}{B}_{3}{C}_{8}$

$={A}_{7}{B}_{24}{C}_{64}$

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