A) \[A{{B}_{2}}\]
B) \[{{A}_{2}}B\]
C) \[{{A}_{4}}{{B}_{3}}\]
D) \[{{A}_{3}}{{B}_{4}}\]
Correct Answer: D
Solution :
There were 6 A atoms on the face-centres removing face-centred atoms along one of the axes means removal of 2 A atoms. Now, number of A atoms per unit cell \[=\underset{(\text{corners})}{\mathop{8\times \frac{1}{8}}}\,+\underset{(\text{face}-\text{centred})}{\mathop{4\times \frac{1}{2}=3}}\,\] Number of B atoms per unit cell \[=\underset{(\text{edge}\,\text{centred})}{\mathop{12\times \frac{1}{4}}}\,\]+ \[\underset{(\text{body}\,\text{centred})}{\mathop{\underset{{}}{\mathop{1=4}}\,}}\,\] Hence the resultant stoichiometry is \[{{A}_{3}}{{B}_{4}}\]You need to login to perform this action.
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