Crystal packing Questions in English

(C) In a $ccp$ lattice,the number of $A$ atoms per unit cell is $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$.
$B$ atoms occupy the edge centers. There are $12$ edges in a unit cell,and each edge center is shared by $4$ unit cells. Thus,the number of $B$ atoms is $12 \times \frac{1}{4} = 3$.
$C$ atoms are present at the body center. There is only $1$ body center per unit cell,which is not shared by any other unit cell. Thus,the number of $C$ atoms is $1 \times 1 = 1$.
Therefore,the ratio of $A:B:C$ is $4:3:1$,and the formula of the alloy is $A_4B_3C$.

(C) In a hexagonal closest packing $(hcp)$,the effective number of atoms per unit cell is $6$.
The number of octahedral voids is equal to the number of atoms in the packing,which is $6$.
The cation $(C)$ occupies $2/3$ of the octahedral voids,so the number of $C$ atoms $= 6 \times (2/3) = 4$.
The number of anion $(A)$ atoms is $6$.
Thus,the ratio of $C:A$ is $4:6$,which simplifies to $2:3$.
Therefore,the formula of the compound is $C_{2}A_{3}$.

(D) The cubic close packing (ccp) is equivalent to the face-centered cubic (fcc) structure.
In an fcc unit cell,the number of atoms per unit cell is $Z = 4$.
The number of tetrahedral voids is given by the formula $2 \times Z$.
Therefore,the number of tetrahedral voids $= 2 \times 4 = 8$.
These $8$ tetrahedral voids are located entirely within the body of the unit cell.

(D) In an $FCC$ unit cell,the number of $X$ atoms at corners and face centers is $4$.
There are $8$ tetrahedral voids in an $FCC$ unit cell.
Since $Y$ atoms occupy all tetrahedral sites,the number of $Y$ atoms is $8$.
The ratio of $X : Y$ is $4 : 8$,which simplifies to $1 : 2$.
Therefore,the formula of the compound is $XY_2$.

(A) In an $FCC$ unit cell,there are $8$ tetrahedral voids located at the corners of the smaller cube formed by joining the centers of these voids.
Since each of these $8$ tetrahedral voids is shared by $8$ such cubes,the effective number of tetrahedral voids within the new cube is $8 \times \frac{1}{8} = 1$.
Additionally,there is an octahedral void located at the body center of the original $FCC$ unit cell,which also corresponds to the body center of the new smaller cube.
Therefore,the new cube contains $1$ effective tetrahedral void and $1$ octahedral void.

(C) In an $FCC$ lattice,the distance between nearest neighbours is $d = \frac{a_{edge}}{\sqrt{2}} = a$. Thus,the edge length of the unit cell is $a_{edge} = a\sqrt{2}$.
The triangular voids in $ABCABC...$ packing are the tetrahedral voids $(THV)$.
The distance between the centres of the nearest triangular voids of two consecutive layers corresponds to the distance between two adjacent tetrahedral voids.
In an $FCC$ unit cell,the distance between two adjacent tetrahedral voids is $\frac{a_{edge}}{2}$.
Substituting $a_{edge} = a\sqrt{2}$,we get: $\text{Distance} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.

(C) In a $CCP$ lattice,the number of oxide ions $(O^{2-})$ per unit cell is $4$.
The number of octahedral voids $(O.V.)$ is $4$ and the number of tetrahedral voids $(T.V.)$ is $8$.
Given the formula $AB_2O_4$,there are $1$ bivalent ion $(A^{2+})$ and $2$ trivalent ions $(B^{3+})$ per $4$ oxide ions.
$A^{2+}$ ions occupy octahedral voids: $1$ $O.V.$ is occupied by $A^{2+}$.
$B^{3+}$ ions occupy both $T.V.$ and $O.V.$ in equal numbers. Let $x$ be the number of $B^{3+}$ ions in $T.V.$ and $x$ be the number of $B^{3+}$ ions in $O.V.$.
Total $B^{3+} = x + x = 2x = 2$,so $x = 1$.
Thus,$B^{3+}$ ions occupy $1$ $T.V.$ and $1$ $O.V$.
Total $O.V.$ occupied = (occupied by $A^{2+}$) + (occupied by $B^{3+}$) = $1 + 1 = 2$.
Total $O.V.$ available = $4$.
Unoccupied $O.V.$ = $4 - 2 = 2$.
Fraction of unoccupied $O.V.$ = $2/4 = 1/2$.

(A) In an $FCC$ unit cell:
Number of $X$ atoms (at corners and face-centres) $= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$.
Number of $Y$ atoms (in octahedral voids) $= 4$.
Number of $Z$ atoms (in tetrahedral voids) $= 8$.
Along one body diagonal,there are two corner $X$ atoms,one octahedral void $(Y)$ at the body centre,and two tetrahedral voids $(Z)$.
Removing these atoms:
Number of $X$ atoms $= 4 - (2 \times \frac{1}{8}) = 4 - 0.25 = 3.75 = \frac{15}{4}$.
Number of $Y$ atoms $= 4 - 1 = 3$.
Number of $Z$ atoms $= 8 - 2 = 6$.
The ratio $X : Y : Z = \frac{15}{4} : 3 : 6$.
Multiplying by $4$,we get $15 : 12 : 24$.
Dividing by $3$,we get $5 : 4 : 8$.
Therefore,the simplest formula is $X_5Y_4Z_8$.

(C) The $ABCABC$ packing corresponds to a face-centered cubic $(FCC)$ structure.
For an $FCC$ unit cell,the relation between edge length $(a)$ and atomic radius $(r)$ is $a = 2\sqrt{2}r$,so $r = \frac{a}{2\sqrt{2}}$.
Given $a = 10 \ \mathring{A} = 10 \times 10^{-8} \ cm$.
Radius $r = \frac{10 \times 10^{-8}}{2\sqrt{2}} \ cm$.
Volume of a single metal atom $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (\frac{10 \times 10^{-8}}{2\sqrt{2}})^3 \ cm^3$.
Mass of a single metal atom $m = \frac{\text{Atomic mass}}{N_A} = \frac{60.23}{6.023 \times 10^{23}} = 10^{-22} \ g$.
Density of the atom $\rho = \frac{m}{V} = \frac{10^{-22}}{\frac{4}{3} \pi (\frac{10 \times 10^{-8}}{2\sqrt{2}})^3} \approx 0.54 \ g/cm^3$.

(D) In $2D$ square packing,the coordination number is $4$ and the packing fraction is $\frac{\pi}{4} \approx 0.785$.
In $2D$ hexagonal packing,the coordination number is $6$ and the packing fraction is $\frac{\pi}{2\sqrt{3}} \approx 0.906$.
Since the packing fraction of hexagonal packing $(0.906)$ is higher than square packing $(0.785)$,the void fraction $(1 - \text{packing fraction})$ of hexagonal packing is lower than that of square packing.
In $2D$ hexagonal close packing,each atom is in contact with $6$ other atoms,creating $6$ trigonal planar voids around it.

(A) In a $bcc$ (body-centered cubic) unit cell,the packing efficiency is $68\%$.
Therefore,the percentage of vacant space is calculated as:
$\text{Vacant space} = 100\% - \text{Packing efficiency}$
$\text{Vacant space} = 100\% - 68\% = 32\%$

(B) The limiting radius ratio $(r/R)$ determines the coordination number and the type of void formed in a crystal lattice.For a trigonal void,the limiting radius ratio is $0.155$.For a tetrahedral void,the limiting radius ratio is $0.225$.For an octahedral void,the limiting radius ratio is $0.414$.For a cubic void,the limiting radius ratio is $0.732$.Therefore,the range for a tetrahedral void is $0.225$ to $0.414$.

(A) In a $CCP$ structure,the number of oxide ions $(O^{2-})$ per unit cell is $4$.
Number of tetrahedral voids $(TV)$ = $2 \times 4 = 8$.
Number of octahedral voids $(OV)$ = $4$.
Number of $A^{2+}$ ions = $\frac{1}{5} \times 8 = \frac{8}{5}$.
Number of $B^{3+}$ ions = $\frac{1}{2} \times 4 = 2$.
The ratio of $A : B : O$ is $\frac{8}{5} : 2 : 4$.
Multiplying the ratio by $5$,we get $8 : 10 : 20$.
Simplifying the ratio by dividing by $2$,we get $4 : 5 : 10$.
Thus,the formula of the oxide is $A_4B_5O_{10}$.

(D) In an antifluorite structure,the lattice sites of the $ccp$ (cubic close packing) arrangement are occupied by the anions.
In the case of $K_2PtCl_6$,the complex ion $(PtCl_6)^{2-}$ acts as the anion.
Therefore,the corners and face centers of the cubic unit cell are occupied by $(PtCl_6)^{2-}$ ions.

(C) In a hexagonal closest packing $(HCP)$ lattice,the number of octahedral voids is equal to the number of atoms present in the lattice.
Let the number of anions $(A)$ be $N$.
Then,the number of octahedral voids is also $N$.
The cations $(C)$ occupy $2/3$ of the octahedral voids,so the number of cations is $(2/3)N$.
The ratio of $C:A$ is $(2/3)N : N$,which simplifies to $2:3$.
Therefore,the formula of the compound is $C_2A_3$.

(B) In a cubic-closest packed $(CCP)$ structure,the number of oxide ions $(O^{2-})$ per unit cell is $4$.
The number of tetrahedral voids is $2 \times 4 = 8$.
The number of octahedral voids is $4$.
Given that $1/8$ of tetrahedral voids are occupied by $A^{2+}$ cations:
Number of $A^{2+} = 8 \times (1/8) = 1$.
Given that $1/2$ of octahedral voids are occupied by $B^{3+}$ cations:
Number of $B^{3+} = 4 \times (1/2) = 2$.
Therefore,the ratio of $A : B : O$ is $1 : 2 : 4$.
The general formula is $AB_{2}O_{4}$.

(A) An octahedral void is formed by the arrangement of $6$ spheres in an octahedral geometry.
Therefore,it is surrounded by $6$ spheres.

(C) Let the number of oxide ions $(O^{2-})$ in the $HCP$ lattice be $N = 6$.
The number of $OHV$ (Octahedral Voids) is equal to the number of ions,so $OHV = 6$.
The number of $THV$ (Tetrahedral Voids) is twice the number of ions,so $THV = 2 \times 6 = 12$.
Cations $A$ occupy $\frac{1}{6}$ of $THV = \frac{1}{6} \times 12 = 2$.
Cations $B$ occupy $\frac{1}{3}$ of $OHV = \frac{1}{3} \times 6 = 2$.
The ratio of atoms $A : B : O$ is $2 : 2 : 6$,which simplifies to $1 : 1 : 3$.
Therefore,the empirical formula is $ABO_3$.

(D) In an $FCC$ unit cell,there are $8$ corners and $6$ face centres.
Initially,the number of atoms of $A$ at corners $= 8 \times \frac{1}{8} = 1$.
The number of atoms of $B$ at face centres $= 6 \times \frac{1}{2} = 3$.
One body diagonal passes through $2$ corners and the body centre. However,in an $FCC$ lattice,there are no atoms at the body centre.
When atoms along one body diagonal are removed,$2$ corner atoms are removed.
Remaining atoms of $A = 1 - (2 \times \frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4}$.
Since no face-centred atoms lie on the body diagonal,the number of $B$ atoms remains $3$.
The ratio of $A:B = \frac{3}{4} : 3 = 3 : 12 = 1 : 4$.
Therefore,the empirical formula is $AB_4$.

(B) In a close-packed structure,the number of octahedral voids is equal to the number of atoms present in the lattice.
Given,number of moles of the compound $= 0.3 \, mol$.
Number of atoms $= 0.3 \times 6.022 \times 10^{23} \, \text{atoms/mol} = 1.8066 \times 10^{23} \, \text{atoms}$.
Since the number of octahedral voids is equal to the number of atoms,the number of octahedral voids $= 1.8066 \times 10^{23}$.

(A) Let the number of oxide ions $(O^{2-})$ in the $CCP$ lattice be $N = 4$.
The number of tetrahedral voids is $2N = 2 \times 4 = 8$.
The number of octahedral voids is $N = 4$.
Cations $A$ occupy $1/6^{th}$ of the tetrahedral voids: $A = \frac{1}{6} \times 8 = \frac{4}{3}$.
Cations $B$ occupy $1/3^{rd}$ of the octahedral voids: $B = \frac{1}{3} \times 4 = \frac{4}{3}$.
The ratio of $A : B : O$ is $\frac{4}{3} : \frac{4}{3} : 4$.
Multiplying by $3/4$,we get the ratio $1 : 1 : 3$.
Therefore,the formula of the compound is $ABO_3$.

(D) In a face-centred cubic $(FCC)$ unit cell:
Number of $A$ atoms at corners = $8 \times \frac{1}{8} = 1$.
Total number of face centres = $6$.
Number of $B$ atoms present = $6 - 3 = 3$.
Contribution of $B$ atoms at face centres = $3 \times \frac{1}{2} = \frac{3}{2}$.
Ratio of $A:B = 1 : \frac{3}{2} = 2 : 3$.
Therefore,the simplest formula of the compound is $A_2B_3$.

(B) In the diamond crystal structure,the carbon atoms form a face-centered cubic $(FCC)$ lattice.
There are $8$ tetrahedral voids in an $FCC$ unit cell.
In diamond,only $4$ of these $8$ tetrahedral voids are occupied by carbon atoms.
Therefore,the percentage of occupied tetrahedral voids is $(4/8) \times 100 = 50\%$.

(D) Let the number of $O^{2-}$ ions in the crystal lattice be $N$.
Then,the number of octahedral voids $= N$.
The number of tetrahedral voids $= 2N$.
Number of $A^{2+}$ ions occupying $1/8$ of tetrahedral voids $= \frac{1}{8} \times 2N = \frac{1}{4}N$.
Number of $B^{3+}$ ions occupying $1/2$ of octahedral voids $= \frac{1}{2} \times N = \frac{1}{2}N$.
The ratio of $A^{2+} : B^{3+} : O^{2-}$ is $\frac{1}{4}N : \frac{1}{2}N : N$.
Multiplying by $4$,we get the ratio $1 : 2 : 4$.
Therefore,the formula of the mixed oxide is $AB_2O_4$.

(D) In an $HCP$ lattice,the number of $B^{2-}$ ions per unit cell is $N = 6$.
The number of tetrahedral voids is $2N = 2 \times 6 = 12$.
The $A^{x+}$ ions occupy one-fourth of the tetrahedral voids,so the number of $A^{x+}$ ions is $\frac{1}{4} \times 12 = 3$.
The formula of the compound is $A_3B_6$,which simplifies to $AB_2$.
For the compound to be electrically neutral,the total positive charge must equal the total negative charge: $3 \times (+x) + 6 \times (-2) = 0$.
$3x - 12 = 0$,which gives $3x = 12$,so $x = 4$.

(D) hexagonal close-packed $(HCP)$ unit cell contains $6$ atoms per unit cell.
The number of tetrahedral voids in a crystal lattice is equal to twice the number of atoms present in the unit cell.
Therefore,the number of tetrahedral voids $= 2 \times 6 = 12$.

(C) In a cubic close packed $(ccp)$ or face-centered cubic $(fcc)$ structure,the number of atoms per unit cell is $Z = 4$.
There is one octahedral void at the body center of the cube and $12$ octahedral voids at the edges,each shared by $4$ unit cells.
Number of octahedral voids at the body center $= 1$.
Number of octahedral voids at the edges $= 12 \times \frac{1}{4} = 3$.
Total number of octahedral voids per unit cell $= 1 + 3 = 4$.
Since there are $4$ atoms per unit cell,the number of octahedral voids per atom $= \frac{4}{4} = 1$.

(A) In an $hcp$ lattice,the number of tetrahedral voids is twice the number of atoms forming the lattice.
Let the number of $B$ atoms forming the $hcp$ lattice be $N$.
Then,the number of tetrahedral voids $= 2N$.
Given the formula is $A_2B_3$,the ratio of atoms is $A:B = 2:3$.
If $B$ forms the $hcp$ lattice ($N$ atoms),then $A$ atoms occupy the tetrahedral voids.
Fraction of tetrahedral voids occupied by $A = \frac{\text{Number of } A \text{ atoms}}{\text{Number of tetrahedral voids}} = \frac{2N/3}{2N} = \frac{1}{3}$.
Thus,$B$ forms the $hcp$ lattice and $A$ occupies $\frac{1}{3}$ of the tetrahedral voids.

(B) In a $ccp$ (cubic close-packed) structure,the number of atoms per unit cell is $Z = 4$.
$1$. Number of $B$ atoms $= 4$.
$2$. Number of octahedral voids $= 4$. Since '$A$' occupies half of the octahedral voids,the number of $A$ atoms $= 4 \times \frac{1}{2} = 2$.
$3$. Number of tetrahedral voids $= 2 \times Z = 2 \times 4 = 8$. Since oxygen atoms occupy all the tetrahedral voids,the number of oxygen atoms $= 8$.
The ratio of atoms $A:B:O$ is $2:4:8$,which simplifies to $1:2:4$.
Therefore,the formula of the bimetallic oxide is $AB_2O_4$.

(A) In a $bcc$ unit cell,the body diagonal is $\sqrt{3}a$,where $a$ is the edge length.
Along the body diagonal,the atoms are in contact: $2r_A + 2r_B = \sqrt{3}a$.
Given $r_B = 2r_A$,so $2r_A + 2(2r_A) = \sqrt{3}a$,which gives $6r_A = \sqrt{3}a$,or $a = 2\sqrt{3}r_A$.
The packing efficiency is given by the ratio of the volume of atoms to the volume of the unit cell.
Volume of atoms = $8 \times (\frac{1}{8}) \times \frac{4}{3}\pi r_A^3 + 1 \times \frac{4}{3}\pi r_B^3 = \frac{4}{3}\pi r_A^3 + \frac{4}{3}\pi (2r_A)^3 = \frac{4}{3}\pi r_A^3 (1 + 8) = 12\pi r_A^3$.
Volume of unit cell = $a^3 = (2\sqrt{3}r_A)^3 = 8 \times 3\sqrt{3} r_A^3 = 24\sqrt{3} r_A^3$.
Packing efficiency = $\frac{12\pi r_A^3}{24\sqrt{3} r_A^3} \times 100 = \frac{\pi}{2\sqrt{3}} \times 100 \approx \frac{3.14159}{3.464} \times 100 \approx 90.6\%$.
Since the packing efficiency depends only on the ratio of radii and the structure type,it is independent of the absolute edge length. Thus,the packing efficiency in solid $2$ is approximately $90\%$.

(C) In a zinc blende structure,the anions ($B$ ions) form a $ccp$ (cubic close-packed) lattice.
Number of $B$ ions per unit cell $= 4$.
The number of tetrahedral holes in a $ccp$ lattice is twice the number of atoms,which is $2 \times 4 = 8$.
Given that $A$ ions occupy $25\%$ of the tetrahedral holes.
Number of $A$ ions $= 25\% \times 8 = 0.25 \times 8 = 2$.
The ratio of $A:B = 2:4 = 1:2$.
Therefore,the formula of the solid is $AB_2$.

(D) In a $ccp$ $(fcc)$ unit cell,there are $4$ atoms of $A$ (corners and face centers),$4$ octahedral voids $(B)$,and $8$ tetrahedral voids $(C)$.
Along one body diagonal,there are $2$ corners,$2$ tetrahedral voids,and $1$ octahedral void (at the body center).
Original atoms: $A = 4$,$B = 4$,$C = 8$.
After removing atoms along one body diagonal:
$A$ atoms: $2$ corners are removed. Remaining $A = 4 - (2 \times 1/8) = 4 - 1/4 = 15/4$.
$B$ atoms: $1$ octahedral void (body center) is removed. Remaining $B = 4 - 1 = 3$.
$C$ atoms: $2$ tetrahedral voids are removed. Remaining $C = 8 - 2 = 6$.
Formula: $A_{15/4}B_3C_6$.
Multiplying by $4$ to get the simplest integer ratio: $A_{15}B_{12}C_{24}$.
Dividing by $3$: $A_5B_4C_8$.

(A) The density of a unit cell is given by the formula: $\rho = \frac{Z \times M}{N_A \times a^3}$.
For both $FCC$ (Face-Centered Cubic) and $HCP$ (Hexagonal Close Packing),the coordination number is $12$ and the packing efficiency is the same,which is $74\%$.
Since the element is the same,the molar mass $(M)$ and Avogadro's number $(N_A)$ are constant.
For $FCC$,the number of atoms per unit cell $(Z)$ is $4$ and the edge length $(a)$ is $2\sqrt{2}r$.
For $HCP$,the number of atoms per unit cell $(Z)$ is $6$,the height $(h)$ is $4\sqrt{2/3}r$,and the base area is $6\sqrt{3}r^2$.
In both structures,the atoms occupy the same volume fraction of the space. Because the packing efficiency is identical for both $FCC$ and $HCP$ $(74\%)$,the density of the crystal lattice formed by the same element in both arrangements will be identical.
Therefore,the ratio of densities is $1:1$ or $1$.

(C) In a $2D$ hexagonal close packing,each sphere is in contact with $6$ other spheres.
Consider a unit cell area formed by the centers of the spheres. The area of the hexagon formed by the centers of $6$ spheres around a central sphere is $6 \times (\frac{\sqrt{3}}{4} \times (2r)^2) = 6\sqrt{3}r^2$.
However,the effective area occupied by the spheres within this unit is $3 \times (\pi r^2)$.
The packing fraction is calculated as: $\text{Packing Fraction} = \frac{\text{Area occupied by spheres}}{\text{Total area of the unit cell}}$.
For $2D$ hexagonal packing,the packing fraction is $\frac{\pi}{2\sqrt{3}} \approx 0.906$.

(B) In an $FCC$ unit cell,the atoms touch along the face diagonal.
The relationship between edge length $a$ and atomic radius $r$ is $4r = a\sqrt{2}$,which implies $r = \frac{a}{2\sqrt{2}}$.
Along the edge of the unit cell,two atoms are present at the corners,each contributing a radius $r$ to the edge length.
The total length covered by atoms along the edge is $2r$.
The uncovered length along the edge is $a - 2r$.
The fraction of the edge not covered is $\frac{a - 2r}{a} = 1 - \frac{2r}{a}$.
Substituting $r = \frac{a}{2\sqrt{2}}$ into the expression:
Fraction $= 1 - \frac{2(a / 2\sqrt{2})}{a} = 1 - \frac{1}{\sqrt{2}}$.
Since $\frac{1}{\sqrt{2}} \approx 0.707$,the fraction $= 1 - 0.707 = 0.293$.

(B) In a face-centred cubic $(FCC)$ unit cell,there are $8$ corners and $6$ face centres.
Each corner atom contributes $1/8$ to the unit cell.
The total number of atoms per unit cell is $Z = (8 \times 1/8) + (6 \times 1/2) = 4$.
The number of tetrahedral voids is equal to $2Z$,which equals $2 \times 4 = 8$.
These $8$ tetrahedral voids are located on the body diagonals of the cube,at a distance of $1/4$ of the body diagonal length from each corner.

(D) The packing efficiency of a crystal lattice is the fraction of the total volume occupied by the atoms.
Void fraction is calculated as: $\text{Void fraction} = 1 - \text{Packing efficiency}$.
For a simple cubic lattice,the packing efficiency is $\frac{\pi}{6} \approx 0.524$.
Therefore,the void fraction is $1 - 0.524 = 0.476$.
Thus,the simple cubic lattice has a void fraction of $0.476$.

(A) In a face-centred cubic $(FCC)$ lattice:
Number of atoms of $A$ at corners = $8 \times \frac{1}{8} = 1$.
Total number of face-centred positions in an $FCC$ unit cell is $6$.
Since one atom of $B$ is missing from one of the face-centred positions,the number of atoms of $B$ present = $6 - 1 = 5$.
Contribution of each face-centred atom to the unit cell = $\frac{1}{2}$.
Number of atoms of $B = 5 \times \frac{1}{2} = \frac{5}{2}$.
The ratio of atoms $A : B = 1 : \frac{5}{2} = 2 : 5$.
Therefore,the formula of the compound is $A_2B_5$.

(A) The $HCP$ unit cell consists of $6$ atoms per unit cell.
The base of the $HCP$ unit cell is a regular hexagon with side length $a = 2r$.
The area of the base is $6 \times \frac{\sqrt{3}}{4} a^2 = 6 \times \frac{\sqrt{3}}{4} (2r)^2 = 6\sqrt{3} r^2$.
The height of the $HCP$ unit cell is $h = 4r \sqrt{\frac{2}{3}}$.
Volume $=$ Area of base $\times$ Height $= (6\sqrt{3} r^2) \times (4r \sqrt{\frac{2}{3}}) = 24\sqrt{3} \times \sqrt{\frac{2}{3}} r^3 = 24\sqrt{2} r^3$.

(C) In the antifluorite structure,such as $Na_{2}O$,the anions $(O^{2-})$ form a $CCP$ (cubic close-packed) arrangement.
The cations $(Na^+)$ occupy all the tetrahedral voids present in the lattice.

(B) In a $ccp$ lattice,the number of effective atoms per unit cell is $4$.
Therefore,the number of $B$ atoms = $4$.
The number of octahedral voids = $4$.
The number of tetrahedral voids = $2 \times 4 = 8$.
Cation $A$ occupies $50\%$ of octahedral voids and $50\%$ of tetrahedral voids.
Number of $A$ atoms = $(0.50 \times 4) + (0.50 \times 8) = 2 + 4 = 6$.
The ratio of $A:B = 6:4 = 3:2$.
Thus,the formula of the crystalline solid is $A_3B_2$.

(D) In a $fcc$ unit cell:
Number of $A$ atoms (corners) = $8 \times \frac{1}{8} = 1$.
Number of $B$ atoms (face centers) = $6 \times \frac{1}{2} = 3$.
Number of $C$ atoms (tetrahedral voids) = $8$.
Number of $D$ atoms (octahedral voids) = $4$.
Along one body diagonal,there are $2$ corners and $2$ tetrahedral voids.
Removing these:
$A = 1 - (2 \times \frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4}$.
$B = 3$ (no change).
$C = 8 - 2 = 6$.
$D = 4$ (no change).
Formula: $A_{3/4}B_3C_6D_4$.
Multiplying by $4$: $A_3B_{12}C_{24}D_{16}$.

(B) For a tetrahedral void,the radius ratio is $\frac{r}{R} = 0.225$.
For an octahedral void,the radius ratio is $\frac{r}{R} = 0.414$.
Since $0.225 < 0.414$,the size of the tetrahedral void is smaller than that of the octahedral void.

(D) In a $ccp$ lattice,the number of atoms per unit cell is $4$.
The number of tetrahedral voids is $2 \times (\text{number of atoms}) = 2 \times 4 = 8$.
Initially,there are $8$ $B$ atoms occupying these $8$ tetrahedral voids.
Since half of the $B$ atoms are ejected,the number of $B$ atoms remaining is $8 / 2 = 4$.
Therefore,the number of tetrahedral voids that remain filled is $4$,which is equal to the number of $A$ atoms $(4)$.
Thus,the number of filled tetrahedral voids is equal to $A$ atoms.

(B) In a cubic close-packed $(CCP)$ or face-centered cubic $(FCC)$ system,the arrangement of layers follows the $ABCABC...$ pattern.
In this system,the number of atoms per unit cell $(Z)$ is $4$.
The number of tetrahedral voids in a crystal lattice is always twice the number of atoms present in the unit cell.
Therefore,if the number of atoms per unit cell is $Z$,the number of tetrahedral voids is $2Z$.

(C) In an $FCC$ unit cell,the number of effective atoms is $n = 4$.
Number of octahedral voids $= n = 4$.
Number of tetrahedral voids $= 2n = 8$.
Atom $A$ is in the $FCC$ lattice,so the number of $A$ atoms $= 4$.
Atom $B$ occupies all octahedral voids $(4)$ and half of the tetrahedral voids $(8 / 2 = 4)$.
Total number of $B$ atoms $= 4 + 4 = 8$.
The ratio of $A:B = 4:8 = 1:2$.
Therefore,the formula of the compound is $AB_2$.

(C) In structure $I$,the central void is surrounded by $6$ atoms (at the center of the unit cell),which represents an octahedral void.
In structure $II$,the void is at the edge center,surrounded by $6$ atoms,which also represents an octahedral void.
In structure $III$,the void is surrounded by $4$ atoms in a tetrahedral arrangement,which represents a tetrahedral void.
Therefore,$I$ and $II$ represent octahedral holes.

(C) In a zinc blende $(ZnS)$ structure,the anions $(B^{-})$ form a face-centered cubic $(fcc)$ lattice.
The number of $B^{-}$ ions per unit cell is $4$.
The number of tetrahedral voids in an $fcc$ lattice is $2 \times (\text{number of atoms}) = 2 \times 4 = 8$.
Since the formula is $AB$,the number of $A^{+}$ ions per unit cell must be $4$.
Therefore,the fraction of tetrahedral voids occupied by $A^{+}$ ions is $\frac{4}{8} = 0.5$,which is $50 \%$.

(D) In a closed packed array of $N$ spheres,the number of tetrahedral voids (holes) is $2N$ and the number of octahedral voids is $N$.
For example,an $fcc$ unit cell has $4$ atoms per unit cell.
It contains $4$ octahedral voids and $8$ tetrahedral voids,which follows the $N$ and $2N$ relationship respectively.

(A) In a $CCP$ (cubic close packing) arrangement:
$1$. Number of atoms at corners $(B)$ = $8 \times (1/8) = 1$.
$2$. Number of atoms at face centres $(A)$ = $6 \times (1/2) = 3$.
$3$. Total number of tetrahedral voids = $2 \times (\text{number of atoms in } CCP) = 2 \times 4 = 8$.
$4$. Number of $C$ atoms occupying $50\%$ of tetrahedral voids = $0.50 \times 8 = 4$.
Therefore,the ratio of atoms is $A:B:C = 3:1:4$.
The molecular formula is $A_3 B C_4$.