JEE Main & Advanced Sample Paper JEE Main - Mock Test - 33

  • question_answer
    A solid is formed and it has three types of atoms X, Y and Z. X forms a fee lattice with Y atoms occupying all tetrahedral voids and Z atoms occupying half of octahedral voids. The formula of solid is-

    A) \[{{X}_{4}}Y{{Z}_{2}}\]           

    B)        \[{{X}_{4}}{{Y}_{2}}Z\]

    C) \[X{{Y}_{2}}{{Z}_{4}}\]         

    D)        \[{{X}_{2}}{{Y}_{4}}Z\]

    Correct Answer: D

    Solution :

    [d] X is in f.c.c. lattice \[X=8\times \frac{1}{8}+6\times \frac{1}{2}=4\] T.H.O \[=4\times 2=8\] \[y=8\] O.H.V. = 4 \[Z=\frac{1}{2}\times 4=2\] \[{{X}_{4}}{{Y}_{8}}{{Z}_{2}}={{X}_{2}}{{Y}_{4}}Z\]


You need to login to perform this action.
You will be redirected in 3 sec spinner