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In spine! structure, oxides ions are cubical-closest packed, whereas I/8th of tetrahedral holes are occupied by cations $A^{2 +}$ and 1/2 of octahedral holes are occupied by cations $B^{3 +}$ ions. The general formula of the compound having spine! structure is

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$A_{4} B_{2} O$

${AB}_{2} O_{4}$

$A_{2} B_{4} O$

$A_{2} {BO}_{4}$

answer is B.

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Detailed Solution

The actual number $O^{2 -}$ in a unit cell

$= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$

Actual number of $A^{2 +}$ cells in a unit cell $= 8 \times \frac{1}{8} = 1$

Actual number of $B^{3 +}$ in a unit cell $= 4 \times \frac{1}{2} = 2$

has the general formula ${AB}_{2} O_{4}$ .

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