Voids

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:

The number of octahedral voids per lattice site in a lattice is ________. (Rounded off to the nearest integer)

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 ${\mathrm{CaF}}_2$ crystal, ${\mathrm{Ca}}^{2+}$ions are present in fcc arrangement, calculate the number of ${\mathrm{F}}^{-}$ ions in the unit cell.

A solid is formed and it has three types of atoms $\mathrm{X},\mathrm{Y}$ and $\mathrm{Z}, \mathrm{X}$ forms a $\mathrm{fcc}$ lattice, $\mathrm{Y}$ atoms occupying all tetrahedral voids and $\mathrm{Z}$ atoms occupying half of octahedral voids. The formula of solid is

In a compound, atoms of element $\text{'}\mathrm{B}\text{'}$ form cubic close packing lattice and those of element $\text{'}\mathrm{A}\text{'}$ occupy $\frac{1}{4}\mathrm{th}$ of tetrahedral voids. 

The formula of the compound would be :

A solid is formed by three elements $\mathrm{X},\mathrm{Y}$ and $\mathrm{Z}$. $\mathrm{X}$ atoms form a fcc lattice, $\mathrm{Y}$ atoms occupying all tetrahedral voids. $\mathrm{Z}$ atoms occupying half the octahedral voids. The formula of the solid is:

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

 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
 

A crystalline solid has three types of atoms $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}. \mathrm{X}$ forms fcc lattice, $\mathrm{Y}$ at each octahedral voids and $\mathrm{Z}$ at each tetrahedral voids. If all the atoms along one body diagonal and from one face has been removed. What will be the formulae of compound?

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

In a compound, atoms of element $\mathrm{Y}$ form $\mathrm{CCP}$ lattice and those of element $\mathrm{X}$ occupy $2/3^{\mathrm{rd}}$ of tetrahedral voids. The formula of the compound will be :-

The radius ratio of cations to anions in a closed pack tetrahedral arrangement is :-

A ion forms $\mathrm{HCP}$ and $\mathrm{B}$ ion is present in $\frac{1}{3}\mathrm{rd}$ octahedral voids. The simplest formula of the compound is :-

The distance between an octahedral and tetrahedral void in fcc lattice would be :-