$A$ compound is formed by two elements $A$ and $B$. Atoms of the element $B$ (as anion) make $ccp$ lattice and those of element $A$ (as cation) occupy all tetrahedral voids. The formula of the compound is
- A$A_4 B_3$
- B$AB$
- C$AB_2$
- D$A_2 B$
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The anions $(A)$ of a $C_{p}A_{q}$ molecule form an $fcc$ lattice. Cations $(C)$ are positioned at the body center and half of the edge centers. The formula of the molecule is:
In a close-packed lattice,how does the size of an octahedral void compare to that of a tetrahedral void?
Easy
View SolutionThe ratio of densities if the same element undergoes $FCC$ and $HCP$ close packing is:
Difficult
View SolutionCalculate the total volume occupied by all particles in an $fcc$ unit cell if the volume of the unit cell is $6.4 \times 10^{-23} \ cm^3$.
Calculate the packing efficiency in a metal crystal for a face-centred cubic $(FCC)$ unit cell (assuming that atoms are touching each other).
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