In a compound, oxide ions have a cubic close packing arrangement. Cations A are present in half of the tetrahedral voids and cation B occupies the octahedral voids. The simplest formula of the compound is:

A B_{2} O_{4}

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A_{2} B O_{4}

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A B O_{2}

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A B O

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Solution

The correct option is D $A B O$
There are 4 oxide ions per unit cell here.
Tetrahedral voids = $2 \times$ number of oxide ions = 8
Number of 'A' ions = half the tetrahedral voids, 4 ions
Octahedral voids = number of oxide ions = 4
Number of 'B' ions = all the octahedral voids, 4 ions
$A_{4} B_{4} O_{4} = A B O$

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In a compound, oxide ions have a cubic close packing arrangement. Cations A are present in half of the tetrahedral voids and cation B occupies the octahedral voids. The simplest formula of the compound is:

The correct option is D ABOThere are 4 oxide ions per unit cell here. Tetrahedral voids = 2× number of oxide ions = 8 Number of 'A' ions = half the tetrahedral voids, 4 ions Octahedral voids = number of oxide ions = 4 Number of 'B' ions = all the octahedral voids, 4 ions A4B4O4=ABO