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

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

In $\mathrm{NaCl}$ crystal, the number of nearest neighbours of each ${\mathrm{Na}}^{+}$ ions are:-

An ionic solid $\mathrm{PQ}$ crystallises in rock salt structure with density $4.0\mathrm{ }\mathrm{gm}/{\mathrm{cm}}^3.$. If the radius of cation and anion is $83$and $167 \mathrm{pm}$ respectively, then the molar mass of solid is $\left[{\mathrm{N}}_{\mathrm{A}}=6\times {10}^{23}\right]\mathrm{ }$

The radius of ${\mathrm{Ag}}^{+}$ ion is $126 \mathrm{pm}$ and ${\mathrm{I}}^{-}$ ion is $216 \mathrm{pm}$. The co-ordination number of ${\mathrm{Ag}}^{+}$ ion is:-

The tetrahedral voids formed by $\mathrm{ccp}$ arrangement of ions ${\mathrm{Cl}}^{-}$ in rock salt structure are