In a solid AB_2,the coordination number of A | vedclass

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(D) In a $ccp$ lattice,the number of atoms per unit cell is $4$. The number of tetrahedral voids is $2 	imes (	ext{number of atoms}) = 2 	imes 4 =

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Equal to $A$ atoms

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(D) In a $ccp$ lattice,the number of atoms per unit cell is $4$. The number of tetrahedral voids is $2 	imes (	ext{number of atoms}) = 2 	imes 4 = 8$. Initially,there are $8$ $B$ atoms occupying these $8$ tetrahedral voids. Since half of the $B$ atoms are ejected,the number of $B$ atoms remaining is $8 / 2 = 4$. Therefore,the number of tetrahedral voids that remain filled is $4$,which is equal to the number of $A$ atoms $(4)$. Thus,the number of filled tetrahedral voids is equal to $A$ atoms.

In a solid $AB_2$,the coordination number of $A$ is $8$. It has a cubic close-packed $(ccp)$ lattice. Half of the $B$ atoms are,however,ejected from the solid. Now,the number of tetrahedral holes (voids) that remain filled is:

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