A) \[{{\operatorname{AB}}_{3}}\]
B) \[{{\operatorname{AB}}_{4}}\]
C) \[{{\operatorname{A}}_{2}}{{B}_{5}}\]
D) \[{{\operatorname{AB}}_{2}}\]
Correct Answer: B
Solution :
A form corner points and two atoms of A are missing from corner \[\therefore \]Atoms at comer (A) = \[6\times \frac{1}{8}=\frac{3}{4}\] Atoms at face centre (B) = \[6\times \frac{1}{2}=3\] \[\therefore \]\[{{A}_{3/4}}{{B}_{3}}\]i.e., \[A{{B}_{4}}\]
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