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In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids are occupied by divalentions, while one-half of octahedral voids are occupied by trivalent, the what will be formula of oxide? 1.) XY 2 O 4 2.) X 2 YO 4 3.) X 4 Y 5 O 10 4.) X 5 Y 4 O 10 i know the answer but i need full explaination why it can’t be fourth option?

In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids are occupied by divalentions, while one-half of octahedral voids are occupied by trivalent, the what will be formula of oxide?
1.) XY2O4              2.) X2YO4
3.) X4Y5O10        4.) X5Y4O10
i know the answer but i need full explaination why it can’t be fourth option?

Grade:12

2 Answers

Vasantha Kumari
askIITians Faculty 38 Points
9 years ago

XY2O4 is the correct option.

Here, we are given the number of oxide ions(O) per unit cell =1 and that of tetrahedral voide per ion in lattice=2

Now, the number of divalent cation (X) = 1/8 times 2 which is equal to ¼ and number of octahedral voids per ion in lattice =1.

Thus, the trivalent cation number (Y) = 1 times ½ = ½.

Our formula now is X14Y12O = XY2O4

Thanks & Regards,

Vasantha Sivaraj,

askIITians faculty

Dheeraj Kakade
26 Points
7 years ago
In ccp anions occupy primitives of the cube while cations occupied voids. In ccp there are two tetrahedral voids and one octahedral holes. 15 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+) ∴ number of divalent cations in tetrahedral voids = 5 1 2 × . Since half of the octahedral voids are occupied by trivalent cations (Y3+) ∴ number of trivalent cations = 2 1 1× . So the formula is the compounds is (X) (Y) (O) 1 2 1 5 1 2× or 1 2 1 5 X2Y O , or X4Y5O10 ∴ (C)

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