In a solid,B^{-} ions occupy corners of a cub | vedclass

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(D) In a $ccp$ structure,the number of effective atoms per unit cell is $4$. Since $B^{-}$ ions form the $ccp$ structure,the number of $B^{-}$ ions $=

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$AB$

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(D) In a $ccp$ structure,the number of effective atoms per unit cell is $4$. Since $B^{-}$ ions form the $ccp$ structure,the number of $B^{-}$ ions $= 4$.In a $ccp$ lattice,the number of tetrahedral voids is twice the number of atoms,so the number of tetrahedral voids $= 2 	imes 4 = 8$.$A^{+}$ ions occupy half of the tetrahedral voids,so the number of $A^{+}$ ions $= 8 	imes (1/2) = 4$.The ratio of $A^{+} : B^{-}$ is $4 : 4$,which simplifies to $1 : 1$.Therefore,the formula of the solid is $AB$.

In a solid,$B^{-}$ ions occupy corners of a cube forming $ccp$ structure. If $A^{+}$ ions occupy half the tetrahedral voids,the formula of the solid is:

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