In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids one occupied by X ions while one half of the octahedral voids are occupied by Y ions. Then formula of the oxide will be

Question

In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids one occupied by X ions while one half of the octahedral voids are occupied by Y ions. Then formula of the oxide will be (A) X5Y4O10 (B) X4Y5O10 (C) XY2O4 (D) X2YO4

Tardigrade Admin · Accepted Answer

Number of O2 - ions = n Number of tetrahedral voids = 2n Number of octahedral voids = n Number of X ions =(2 n/5) Number of Y ions =(n/2) beginmatrix X:Y:O (2 n/5):(n/2):n 4:5:10 endmatrix Simplest formula =X4Y5O10 .

Q. In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids one occupied by $X$ ions while one half of the octahedral voids are occupied by $Y$ ions. Then formula of the oxide will be

$X_{5} Y_{4} O_{10}$

$X_{4} Y_{5} O_{10}$

$X Y_{2} O_{4}$

$X_{2} Y O_{4}$

Number of $O^{2 -}$ ions $= n$
Number of tetrahedral voids $= 2 n$
Number of octahedral voids $= n$
Number of $X$ ions $= \frac{2 n}{5}$
Number of $Y$ ions $= \frac{n}{2}$
$X : Y : O \frac{2 n}{5} : \frac{n}{2} : n 4 : 5 : 10$
Simplest formula $= X_{4} Y_{5} O_{10}$ .

Q. In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids one occupied by X ions while one half of the octahedral voids are occupied by Y ions. Then formula of the oxide will be

X5​Y4​O10​

X4​Y5​O10​

XY2​O4​

X2​YO4​