In a cubic close packed structure of mixed oxide ,one eighth of tetr

1.	In a cubic close packed structure of mixed oxide ,one eighth of tetrahedral voids are occupied by divalent X2+ ions, while one half of the octahedral voids are occupied by trivalent ions (Y3+)What is the formula of the compound?
Answer» BThe formula of compound will be AB₂O₄ Explanation: We all know that oxygen like bigger molecules occupy the lattice points. The effective number of atoms in a CCP unit cell is 8 x 1/8 + 6 x 1/2 = 4 Also the total number of tetrahedral voids (X is presnt) in CCP is 8 so no. of divalent ions = 8/8 = 1 Number of Octahedral voids (Y is present) are 4 so 1/2 of no. of divalent ion will be 4/2 = 2 Number of oxide = present at corners + at face centre = 8 x 1/8 + 6 x 1/2 = 4 So formula of compound will be AB₂O₄

In a cubic close packed structure of mixed oxide ,one eighth of tetrahedral voids are occupied by divalent X2+ ions, while one half of the octahedral voids are occupied by trivalent ions (Y3+)What is the formula of the compound?

Answer»

BThe formula of compound will be AB₂O₄

Explanation:

We all know that oxygen like bigger molecules occupy the lattice points. The effective number of atoms in a CCP unit cell is 8 x 1/8 + 6 x 1/2 = 4

Also the total number of tetrahedral voids (X is presnt) in CCP is 8 so no. of divalent ions = 8/8 = 1

Number of Octahedral voids (Y is present) are 4 so 1/2 of no. of divalent ion will be 4/2 = 2

Number of oxide = present at corners + at face centre = 8 x 1/8 + 6 x 1/2 = 4

So formula of compound will be AB₂O₄

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