Packing in Voids of Ionic Solids

In which of the following crystals, the alternate tetrahedral voids are occupied?

Ionic radii of cation ${\mathrm{A}}^{+}$ and anion ${\mathrm{B}}^{-}$ are $102$ and $181 \mathrm{pm}$ respectively. These ions are allowed to crystallise into an ionic solid. This crystal has cubic close packing for ${\mathrm{B}}^{-}$ . ${\mathrm{A}}^{+}$ is present in all octahedral voids. The edge length of the unit cell of the crystal $\mathrm{AB}$ is _____ $\mathrm{pm}$. (Nearest Integer)

In $\mathrm{BeO}$ (Zinc Blende structure), ${\mathrm{Mg}}^{2+}$ is introduced in available tetrahedral voids. Then ions are removed from a single body diagonal of the unit cell. What will be the molecular formula of the unit cell?

The structure of $\mathrm{CsCl}$ is different from $\mathrm{NaCl}$ though both have similar formula. Give reason.

The edge length of ${\mathrm{M}}^{+}{\mathrm{X}}^{-}$ ($\mathrm{fcc}$ structure) is $7.2 \stackrel{\mathrm{o}}{\mathrm{A}}$. Assuming ${\mathrm{M}}^{+}-{\mathrm{X}}^{-}$ contact along the cell edge, calculate the radius of ${\mathrm{X}}^{-}$ ion $\left({\mathrm{r}}_{{\mathrm{M}}^{+}}=1.6 \stackrel{\mathrm{o}}{\mathrm{A}}\right)?$

The edge length of unit cell of sodium chloride is $564\mathrm{pm}$. If the size of ${\mathrm{Cl}}^{-}$ion $181\mathrm{pm}$, the size of ${\mathrm{Na}}^{+}$ion would be

Number of unit cells in $10 \mathrm{g}$ $\mathrm{NaCl}$

In a $\mathrm{NaCl}$ structure, if positions of $\mathrm{Na}$ atoms and $\mathrm{Cl}$ atoms are interchanged, then in the new unit cell

If radius of an octahedral void is $\mathrm{r}$ and atomic radius of atoms assuming cubical close packing is $\mathrm{R}$. Then the relation between $\mathrm{r}$ and $\mathrm{R}$ is

If radius of a metal atom $\left(\mathrm{A}\right)$ is $5 \mathrm{pm}$ and radius of an electronegative atom $\left(\mathrm{B}\right)$ is $20 \mathrm{pm}$, then in the unit cell

A student finds a crystal of $\mathrm{NaCl}$ having mass $58.5 \mathrm{mg}.$ He uses some of it to prepare $0.05\mathrm{M},10 \mathrm{mLNaCl}$ solution. How many unit cells are remained in the crystal ? $\left({\mathrm{N}}_{\mathrm{A}}=6\times {10}^{23}\right)$

A crystal is made of particles $\mathrm{A}$ and $\mathrm{B}$. A forms $\mathrm{F}\mathrm{C}\mathrm{C}$ packing and $\mathrm{B}$ occupies all the octahedral voids. If all the particles along the plane as shown in figure are removed, then, the formula of the crystal would be :

A solid compound $\mathrm{XY}$ has $\mathrm{NaCl}$ structure. If the radius of the cation is $100 \mathrm{pm}$, the radius of the anion $\left({\mathrm{Y}}^{-}\right)$ in pm will be

(Round of the answer up to four significant figures)

If the radius ratio is in the range of $0.414-0.732$, then the coordination number will be

If in $\mathrm{NaCl}$ crystal all ions present along one axis passing through centre of two opposite faces are removed then formula of the crystal will be:

What is the maximum possible radius of the anion, If the radius of cation is $200 \mathrm{pm}$ in a solid $\mathrm{AB}$ (Rock-salt type) ?

In $\mathrm{NaCl}$ solid, if all the ions are removed along one of the axes joining the centres of any two diagonally opposite edges in a unit cell, then the resulting stoichiometry of the solid is

The radius of the ${\mathrm{Na}}^{+}$ is $95 \mathrm{pm}$ and that of ${\mathrm{Cl}}^{-}$ ion is $181 \mathrm{pm}.$ Predict the co-ordination number of ${\mathrm{Na}}^{+} :$

In an ionic compound ${\mathrm{A}}^{+}{\mathrm{X}}^{-}$, the radii of ${\mathrm{A}}^{+}$ and ${\mathrm{X}}^{-}$ are $1 \mathrm{pm}$ and $2 \mathrm{pm}$ respectively. The volume of unit cell of crystal $\mathrm{AX}$ is $\dots \dots ..$ and structure is similar to$\dots \dots \dots$

In zinc blende $\left(\mathrm{ZnS}\right)$, distance between two closest ${\mathrm{Zn}}^{2+}$ ions is $5\stackrel{\circ }{\mathrm{A}}$ then distance between two next closest ${\mathrm{S}}^{2-}$ ions is -