In a solid AB,A atoms are in ccp arrangement | vedclass

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(A) In a $ccp$ arrangement,the number of $A$ atoms per unit cell is $4$ (corners + face centers).The number of octahedral sites in a $ccp$ unit cell i

Vedclass · Accepted Answer

$3$

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(A) In a $ccp$ arrangement,the number of $A$ atoms per unit cell is $4$ (corners + face centers).The number of octahedral sites in a $ccp$ unit cell is equal to the number of atoms,which is $4$.When two atoms from opposite faces are removed,we remove two $A$ atoms from the face centers.New number of $A$ atoms = $4 - (2 	imes \frac{1}{2}) = 4 - 1 = 3$.The $B$ atoms occupy all octahedral sites. There are $4$ octahedral sites (one at the body center and $12$ at the edges,each shared by $4$ unit cells: $1 + 12 	imes \frac{1}{4} = 4$).Since the removal of face-centered $A$ atoms does not affect the octahedral sites,the number of $B$ atoms remains $4$.Thus,the stoichiometry is $A_{3} B_{4}$.The value of $x$ is $3$.

In a solid $AB$,$A$ atoms are in $ccp$ arrangement and $B$ atoms occupy all the octahedral sites. If two atoms from the opposite faces are removed,then the resultant stoichiometry of the compound is $A_{x} B_{y}$. The value of $x$ is ..... [nearest integer]

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