(A) In an $fcc$ unit cell,the number of atoms $A$ at the corners is $8 \times \frac{1}{8} = 1$.Initially,the number of atoms $B$ at the face centers i

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(A) In an $fcc$ unit cell,the number of atoms $A$ at the corners is $8 \times \frac{1}{8} = 1$.Initially,the number of atoms $B$ at the face centers is $6 \times \frac{1}{2} = 3$.If one atom $B$ is removed from a face center,the remaining number of atoms $B$ is $3 - \frac{1}{2} = 2.5 = \frac{5}{2}$.The ratio of $A:B$ is $1 : \frac{5}{2}$,which simplifies to $2 : 5$.Therefore,the formula of the compound is $A_2B_5$.

Class 12 Chemistry Solid State Crystal packing

Medium

In an $fcc$ crystal lattice,atoms $A$ are at the corners of the cube and atoms $B$ are at the face centers. If one atom is removed from one of the face centers,find the formula of the compound.

A
$A_2B_5$
B
$A_4B_5$
C
$A_8B_5$
D
$A_2B_3$

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Solid State

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Crystal packing

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EasyJEE MAIN 2022

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In an $fcc$ crystal lattice,atoms $A$ are at the corners of the cube and atoms $B$ are at the face centers. If one atom is removed from one of the face centers,find the formula of the compound.

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$Ag$ crystallises in $fcc$ lattice. What is the total number of tetrahedral voids present in $540 \ g$ of $Ag$ metal (in $N_A$)? $(N_A = \text{Avogadro number}; Ag \text{ atomic weight} = 108 \ g \ mol^{-1})$

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The molecular formula of a crystal is $AB_2O_4$. Oxygen atoms form $ccp$ lattice. Atoms of $A$ occupy $x \%$ of tetrahedral voids and atoms of $B$ occupy $y \%$ of octahedral voids. $x$ and $y$ are respectively:

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