Number of Atoms in A Unit Cell

In face centred cubic (FCC) crystal lattice, edge length is 400 pm. The diameter of greatest sphere which can be fit into the interstitial void without distortion of lattice is

A substance ${\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}$ crystallizes in a face centred cubic (FCC) lattice in which atoms $\text{A}$ occupy each corner of the cube and atoms $\text{B}$ occupy face centres of the cube composition of the substance ${\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}$ is

In a compound, atoms of an element $\mathrm{Y}$ form $\mathrm{CCP}$ lattice and those of element $\mathrm{X}$ occupies ${\frac{2}{3}}^{\mathrm{rd}}$ of the tetrahedral voids. The formula of the compound can be

In a mixed oxide of $\text{'}\mathrm{A}\text{'}$ and $\text{'}\mathrm{B}\text{'} \text{'}\mathrm{A}\text{'}$ occupies all the octahedral voids while $\text{'}\mathrm{B}\text{'}$ occupies $(2/3{)}^{\mathrm{rd}}$ the tetrahedral voids. The molecular formula of this oxide is ______

In a cubic-close packed structure containing $\mathrm{X},\mathrm{Y}$ and $\mathrm{Z}$ atoms, if $\mathrm{Z}$ occupies all the face centres, $\mathrm{X}$ occupies all the corners and $\mathrm{Y}$ occupies the body centre of the cube, what is the formula of this compound?

In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atom B is missing from one of the face centered points, the formula of the ionic compound is :

In chromium chloride $\left(\mathrm{C}\mathrm{r}\mathrm{C}{\mathrm{l}}_3\right),\mathrm{ }\mathrm{C}{\mathrm{l}}^{-}$ ions have cubic close packed arrangement and $\mathrm{C}{\mathrm{r}}^{3+}$ ions are present in the octahedral holes. The fraction of the total number of holes occupied is

A crystal made up of particles X, Y, and Z. X forms FCC packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. If all particles along one body diagonal are removed, then the formula of the crystal is

A crystal is made of particles $\mathrm{X}, \mathrm{Y} \mathrm{and} \mathrm{Z}$. $\mathrm{X}$ forms FCC packing $\mathrm{Y}$ occupies all the octahedral voids of $\mathrm{X}$ and $\mathrm{Z}$ occupies all the tetrahedral voids of $\mathrm{X}$. If all the particles along one body diagonal are removed then the formula of the crystal would be-

The no. of particles present per BCC unit cell is :

Unit cell of a metal has edge length of $288 \mathrm{pm}$ and density of $7.86 \mathrm{g} {\mathrm{cm}}^{–3}$. Determine the type of crystal lattice. [Atomic mass of metal $= 56 \mathrm{g} {\mathrm{mol}}^{–1}$] 

The number of atoms per unit cell of body centred cube is: 

The total number of body centred lattices possible among the $14$ Bravais lattices is

A cubic structure is formed where atoms of element $\mathrm{X}$ are occupied at corner of cube and also at face centers. Atoms of element $\mathrm{Y}$ are present at body center and at the edge centers. If all the atoms are removed along a plane passing through the middle of the cube (bisecting the four edges), the formula will become

Which of the following type of cubic lattice has maximum number of atoms per unit cell

$\mathrm{Na}$ and $\mathrm{Mg}$ crystallize in $\mathrm{BCC}$ and $\mathrm{FCC}$ type crystals respectively, then the number of atoms of $\mathrm{Na}$ and $\mathrm{Mg}$ Present in the unit cell of their respective crystal is:

The cubic unit cell of$\mathrm{Al} (\mathrm{molar} \mathrm{mass}=27 \mathrm{g} {\mathrm{mol}}^{-1})$has an edge length of $405 \mathrm{pm}$. Its density is $2.7 \mathrm{g} {\mathrm{cm}}^{-3}$. The cubic unit cell is

The edge length of face centred cubic unit cell is $508 \mathrm{pm}$. If the radius of the cation is $110 \mathrm{pm}$, the radius of the anion is