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JEE Main & Advanced Chemistry The Solid State / ठोस प्रावस्था Question Bank Crystal structure and Coordination number

  • question_answer
    In a solid 'AB' having the \[NaCl\] structure, 'A' atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is           [IIT Screening 2001]

    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 \[{{A}_{3}}{{B}_{4}}\]


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