Voids in Lattice

The distance between an octahedral and a tetrahedral void in $\mathrm{pm}$ in FCC lattice would be: 

(Consider edge length is $7 \mathrm{pm}$)

A molecule have radius ratio of $\left(\frac{{\mathrm{r}}^{-}}{{\mathrm{r}}^{+}}\right)$ is more than 0.732. Which of the following void is presented in the molecule.

A molecule have radius ratio of $\left(\frac{{\mathrm{r}}^{-}}{{\mathrm{r}}^{+}}\right)$ is 0.200 . Which of the following void is presented in the molecule.

The radius ratio of triangular planar radius is $\left(\frac{{\mathrm{r}}^{+}}{{\mathrm{r}}^{-}}=\right)$

Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are:

In a compound, atoms of element $\mathrm{Y}$ form cubic close packing lattice and those of element $\mathrm{X}$ occupy$\frac{2}{3} \mathrm{times}$  of tetrahedral voids. The formula of the compound will be

Octahedral voids are formed when the triangular voids in second layer exactly overlap with similar voids in the first layer.

The total number of tetrahedral voids in the face centred unit cell is

In a carbon allotrope having $\mathrm{FCC}$ lattice, two atoms are at $\left(0, 0, 0\right)$ and $\left(\frac{\mathrm{a}}{8}, \frac{\mathrm{a}}{8}, \frac{\mathrm{a}}{8}\right)$ coordinates. What is the ratio of $\mathrm{C}-\mathrm{C}$ bond length distance to the edge length of the unit cell?

 In a compound, oxide ions are arranged in cubic close packing arrangement. Cations $\mathrm{A}$ occupy $1/6$ of the tetrahedral voids and cations $\mathrm{B}$ occupy $1/3$ of the octahedral voids. The formula of the compound is
 

For the structure given below the site marked as S is a :-

An ionic compound is expected to have a tetrahedral structure if ${\mathrm{r}}_{+}/{\mathrm{r}}_{-}$ lies in the range of

The total number of tetrahedral voids in the face-centred unit cell is _____.

The number of tetrahedral sites per sphere in ccp structure is,

The number of octahedral sites per sphere in fcc structure is

The ratio of close packed atoms to octahedral holes in hexagonal close packing is

The ratio of close packed atoms to tetrahedral holes in cubic packing is

In a solid $\mathrm{XY}$, having the rock salt structure, $\mathrm{Y}$ atoms form the $\mathrm{FCC}$ arrangement and $\mathrm{X}$ atoms occupy all the octahedral voids. If the $\mathrm{X}$ atom at body centred octahedral void is missing, then the formula of the compound will be:

In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids?