Atom (A) is present in ccp form,atom (B) is p | vedclass

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(D) In a $ccp$ $(fcc)$ unit cell,there are $4$ atoms of $A$ (corners and face centers),$4$ octahedral voids $(B)$,and $8$ tetrahedral voids $(C)$.Alon

Vedclass · Accepted Answer

$A_5B_4C_8$

Vedclass · Answer

(D) In a $ccp$ $(fcc)$ unit cell,there are $4$ atoms of $A$ (corners and face centers),$4$ octahedral voids $(B)$,and $8$ tetrahedral voids $(C)$.Along one body diagonal,there are $2$ corners,$2$ tetrahedral voids,and $1$ octahedral void (at the body center).Original atoms: $A = 4$,$B = 4$,$C = 8$.After removing atoms along one body diagonal:$A$ atoms: $2$ corners are removed. Remaining $A = 4 - (2 	imes 1/8) = 4 - 1/4 = 15/4$.$B$ atoms: $1$ octahedral void (body center) is removed. Remaining $B = 4 - 1 = 3$.$C$ atoms: $2$ tetrahedral voids are removed. Remaining $C = 8 - 2 = 6$.Formula: $A_{15/4}B_3C_6$.Multiplying by $4$ to get the simplest integer ratio: $A_{15}B_{12}C_{24}$.Dividing by $3$: $A_5B_4C_8$.

Atom $(A)$ is present in $ccp$ form,atom $(B)$ is present in all the octahedral voids,and atom $(C)$ is present in all the tetrahedral voids. If all particles touching one of the body diagonal axes are removed,find the empirical formula of the solid.

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