In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions `(A^(2+))` while one half of the octahedral voids occupied trivalent ions `(B^(3+))`. What is the formula of the oxide ?

Options

A. `XY_2O_4`
B. `X_4Y_5O_10`
C. `X_5Y_4O_10`
D. `X_2YO_4`

Answer

B

Explanation

No. of oxide ions per unit cell ` = 1/8 xx 8 + 1/2 = 4`
Number of tetrahedral voids per unit cell `= 2 xx 4 = 8`
Number or `X^(2+)` ions per unit cell = `(20)/(100) xx 8 = 8/5`
Number of octahedral voids per unit cell `= 1 xx 4 =4` .
Number of` Y^(3+)` ions per unit cell ` = (50 xx 4)/(100) = 2`
Hence, formula `X_(8//5) Y_2O_4` or `X_4Y_5O_10`