Question

In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids are occupied by divalent (X++) ions while one- half of the octahedral voids are occupied by trivalent ions (Y+3), then the formula of the oxide is.

• A. XY2O4
• B. X2YO4
• C. X4Y5O10
• D. X5Y4O10

Answer 1

Answer: (C)

X4Y5O10

Answer 2

In ccp anions occuply primitives of the cube while cations occupied voids. In ccp there are two tetrahedral voids and one octahedral holes per anion.

For one oxygen atom there are two tetrahedral holes and one octahedral hole.

Since one fifth of the tetrahedral voids are occupied by divalent cations (X2+)

$\therefore$ Number of divalent cations in tetrahedral voids = $2 \times \frac{1}{5}$

Since half of the octaheadral voids are occupied by trivalent cations (Y3+)

$\therefore$ number of trivalent cations in octaheadral voids = $1 \times \frac{1}{2}$

So the formula is the compound is $(X)_{2 \times \frac{1}{5}}(Y)_{\frac{1}{2}}(O)_{1}$

or $(X)_{\frac{2}{5}}(Y)_{\frac{1}{2}}(O)_{1}$,

or X4 Y5 O10