A solid A B has the $B^{-} i o n s$ arranged as below. If the $A^{+} i o n s$ occupy half of the octahedral sites in the structure. The formula of solid is:

A B

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A B_{2}

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A_{2} B

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A_{3} B_{4}

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Solution

The correct option is B $A B_{2}$
$A^{+}$ occupy half of the octahedral sites $= \frac{1}{2} \times 4 = 2$
$B^{-}$ is in FCC lattice $= 4$
Formula of solid $\Rightarrow A_{2} B_{4} = A B_{2}$

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Similar questions

A solid A⁺B^- has the B^- ions arranged as below. If the A⁺ ions accupy half of the tetrahedral sites in the structure. The formula of solid is:

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A solid A⁺B^- has the B^- ions arranged as below. If the A⁺ ions accupy half of the tetrahedral sites in the structure. The formula of solid is:

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