A) \[A & {{ & }_{2}}{{B}_{2}}\]
B) \[A{{B}_{3}}\]
C) AB
D) \[{{A}_{3}}B\]
Correct Answer: B
Solution :
A atoms are at corners of a cube So, the number of A atoms per unit cell \[=8\times \frac{1}{8}=1\]. Similarly, B atoms are present at face centres of a cube. So, the number of S atoms per unit cell \[=6\times \frac{1}{2}=\] Hence, the compound is \[A{{B}_{3}}\]
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