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NEET AIPMT SOLVED PAPER MAINS 2012

  • question_answer
    Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is                                             

    A)  \[AB{{O}_{2}}\]               

    B)  \[{{A}_{2}}B{{O}_{2}}\]

    C)  \[{{A}_{2}}{{B}_{3}}{{O}_{4}}\] 

    D)  \[A{{B}_{2}}{{O}_{2}}\]

    Correct Answer: D

    Solution :

    According to ccp, Number of \[{{\text{O}}^{\text{2-}}}\] ions = 4 So, tetrahedral void = 8 and    octahedral void = 4 Since, A ions occupied \[\frac{\text{1}}{\text{4}}\text{th}\] of tetrahedral void.                                           \[\therefore \]Number of A ions\[=\frac{1}{4}\times 8=2\] Again, B ions occupied all octahedral void.        \[\therefore \]Number of B ions = 4 A : B : 0 = 2 : 4 : 4               = 1 : 2 : 2                \[\therefore \]Structure of oxide \[=A{{B}_{2}}{{O}_{2}}\]


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