Crystal packing Questions in English

(B) In a face-centered cubic $(fcc)$ unit cell,the atoms touch each other along the face diagonal.
Let $a$ be the edge length of the unit cell and $r$ be the atomic radius.
The length of the face diagonal is given by $\sqrt{a^2 + a^2} = a\sqrt{2}$.
Along this diagonal,there are four atomic radii (one radius from each corner atom and the full diameter of the face-centered atom),so $4r = a\sqrt{2}$.
Therefore,$r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}$.

(B) In an octahedral void,the radius of the void $(r_v)$ and the radius of the sphere $(r)$ are related by the geometry of the void.
For an octahedral void,the relationship is $r_v + r = r \sqrt{2}$.
Rearranging this,we get $r_v = r(\sqrt{2} - 1)$.
Since $\sqrt{2} \approx 1.414$,we have $r_v = r(1.414 - 1) = 0.414r$.

(C) In a face-centered cubic $(fcc)$ unit cell,the total number of atoms $(Z)$ is $4$.
The number of tetrahedral voids is equal to twice the number of atoms present in the unit cell.
Therefore,$\text{Number of tetrahedral voids} = 2 \times Z = 2 \times 4 = 8$.
Thus,option $C$ is correct.

(B) The $ABC ABC......$ pattern of close packing corresponds to the cubic close-packed $(ccp)$ structure,which is identical to the face-centred cubic $(fcc)$ lattice.

(D) The $ABC ABC ABC ...$ pattern corresponds to the cubic close packing $(ccp)$ or face-centered cubic $(fcc)$ lattice structure.
In this arrangement,the third layer is placed in the voids of the second layer such that it does not align with the first or second layer,creating a repeating sequence of three layers.

(C) In a hexagonal close-packed $(HCP)$ structure,the spheres in the third layer are aligned directly above the spheres in the first layer.
This results in a repeating pattern of layers known as the $ABAB...$ pattern.
Therefore,the correct description for the hexagonal close-packed arrangement is $ABAB AB$.

(B) In a cubic close packing $(ccp)$ or face-centered cubic $(fcc)$ lattice,the number of atoms per unit cell is $4$.
Each atom in the lattice creates two tetrahedral voids.
Therefore,the number of tetrahedral voids is $4 \times 2 = 8$.
The ratio of close-packed atoms to tetrahedral holes is $4 : 8$,which simplifies to $1 : 2$.

(C) In a $CCP$ (cubic close packing) arrangement,the number of octahedral voids is equal to the number of atoms,while the number of tetrahedral voids is double the number of atoms.
Let the number of atoms of $Z$ in the $CCP$ arrangement be $n$.
Then,the number of tetrahedral voids is $2n$.
Since atoms $X$ occupy all the tetrahedral sites,the number of atoms of $X$ is $2n$.
The ratio of atoms $X:Z$ is $2n:n$,which simplifies to $2:1$.
Therefore,the formula of the compound is $X_2Z$.

(B) The hexagonal close packing $(HCP)$ structure is characterized by a coordination number of $12$.
Among the given options,$Mg$ (Magnesium) crystallizes in the $HCP$ structure,which also has a coordination number of $12$.
$Na$ (Sodium) crystallizes in a body-centered cubic $(BCC)$ structure,and $Al$ (Aluminium) crystallizes in a face-centered cubic $(FCC)$ structure.

(A) For an $FCC$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2\sqrt{2}r$.
Given $a = 620 \ pm$.
Thus, $r = \frac{a}{2\sqrt{2}} = \frac{620}{2 \times 1.414} = \frac{620}{2.828} \approx 219.23 \ pm$.
Using more precise values, $r = \frac{620}{2.8284} \approx 219.20 \ pm$.
Rounding to the nearest provided option, the correct value is $219.85 \ pm$.

(C) In a close-packed structure (like $ccp$ or $hcp$),if the number of constituent particles is $N$,then the number of octahedral voids is $N$.
The number of tetrahedral voids is $2N$.
Therefore,the number of tetrahedral voids is double the number of octahedral voids.

(C) tetrahedral void is formed when a sphere is placed in the depression created by three spheres in contact with each other in a layer.
Thus,the void is surrounded by $4$ spheres arranged at the corners of a regular tetrahedron.
Therefore,the correct description is that the void is surrounded by four spheres in a tetrahedral arrangement.

(C) In crystal structures,close-packed structures are those where atoms are packed as efficiently as possible to minimize empty space.
$hcp$ (hexagonal close-packed),$ccp$ (cubic close-packed),and $fcc$ (face-centered cubic) are all examples of close-packed structures with a packing efficiency of $74\%$.
$bcc$ (body-centered cubic) has a packing efficiency of $68\%$,which is lower than the close-packed structures.
Therefore,$bcc$ is not considered a close-packed structure.

(B) In a face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.
Let the edge length of the unit cell be $a$ and the radius of the atom be $r$.
The length of the face diagonal is given by $\sqrt{2}a$.
Since the atoms touch along the face diagonal,the length of the face diagonal is also equal to $4r$ (one radius from each corner atom and two radii from the face-centered atom).
Therefore,$\sqrt{2}a = 4r$.
Solving for $a$,we get $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$.

(A) In $ABAB...$ type packing,the third layer is placed directly above the first layer,which results in a Hexagonal Close Packing $(hcp)$ structure.
In $ABCABC...$ type packing,the fourth layer is placed directly above the first layer,which results in a Cubic Close Packing $(ccp)$ or Face-Centered Cubic $(fcc)$ structure.
Therefore,$ABAB...$ is $hcp$ and $ABCABC...$ is $ccp$.

(C) In a $ccp$ or $fcc$ lattice,the number of octahedral voids is equal to the number of spheres present in the lattice.
If the number of spheres is $N$,then the number of octahedral voids is $N$.

(A) In a $bcc$ (body-centered cubic) unit cell,the packing efficiency is $68\%$.
The empty space (void space) is calculated as:
$\text{Empty space} = 100\% - \text{Packing efficiency}$
$\text{Empty space} = 100\% - 68\% = 32\%$.
Therefore,the correct option is $A$.

(B) In a close-packed structure,if the number of atoms is $N$,then the number of octahedral voids is $N$ and the number of tetrahedral voids is $2N$.
Therefore,the number of tetrahedral voids per atom is $2N / N = 2$.

(C) In a close-packed structure,the radius of a tetrahedral void $(r_t)$ is related to the radius of the sphere $(R)$ by the formula $r_t = 0.225R$.
The radius of an octahedral void $(r_o)$ is related to the radius of the sphere $(R)$ by the formula $r_o = 0.414R$.
Comparing the two values,$0.414R > 0.225R$,which means the size of the octahedral void is larger than that of the tetrahedral void.

(A) In both $hcp$ (hexagonal close packing) and $ccp$ (cubic close packing) structures,the spheres are packed as efficiently as possible.
The packing efficiency for both $hcp$ and $ccp$ is $74\%$.
This means that $74\%$ of the total volume is occupied by the spheres,while the remaining $26\%$ is the empty space (voids).

(A) Let the number of atoms of $Y$ in the $ccp$ lattice be $n$.
Number of tetrahedral voids $= 2n$.
Atoms of $X$ occupy $2/3$ of the tetrahedral voids,so number of $X$ atoms $= \frac{2}{3} \times 2n = \frac{4n}{3}$.
The ratio of $X:Y = \frac{4n}{3} : n = 4:3$.
Therefore,the formula of the compound is $X_4Y_3$.

(B) In a close-packed structure (such as $fcc$ or $ccp$ which follows the $ABCABC$ stacking sequence),the number of atoms per unit cell is $z = 4$.
The number of tetrahedral voids is always twice the number of atoms present in the unit cell.
Therefore,the number of tetrahedral voids = $2 \times z = 2z$.

(A) For a simple cubic $(SC)$ unit cell,the relation between edge length $a$ and radius $r$ is $a = 2r$,so $r = \frac{a}{2}$.
For a body-centered cubic $(BCC)$ unit cell,the relation is $\sqrt{3}a = 4r$,so $r = \frac{\sqrt{3}a}{4}$.
For a face-centered cubic $(FCC)$ unit cell,the relation is $\sqrt{2}a = 4r$,so $r = \frac{a}{2\sqrt{2}}$.
Thus,the ratio of radii is $\frac{a}{2} : \frac{\sqrt{3}a}{4} : \frac{a}{2\sqrt{2}}$.

(C) In a hexagonal close-packed $(HCP)$ structure,the spheres are arranged in an $ABABAB...$ pattern.
This means the third layer is identical to the first layer,the fourth layer is identical to the second layer,and so on.
In contrast,the $ABCABC...$ pattern represents a cubic close-packed $(CCP)$ or face-centered cubic $(FCC)$ structure.

(C) Let the number of atoms of element $B$ in the $HCP$ lattice be $n$.
Since the number of tetrahedral voids is $2n$,and element $A$ occupies $2/3$ of these voids:
Number of atoms of $A = \frac{2}{3} \times 2n = \frac{4n}{3}$.
The ratio of $A:B = \frac{4n}{3} : n = 4:3$.
Therefore,the formula of the compound is $A_4B_3$.

(A) In a cubic close-packed $(CCP)$ structure,the packing efficiency is $74\%$. Therefore,the percentage of empty space (voids) is $100\% - 74\% = 26\%$.
In a body-centered cubic $(BCC)$ structure,the packing efficiency is $68\%$. Therefore,the percentage of empty space (voids) is $100\% - 68\% = 32\%$.
Thus,the percentages of empty space are $26\%$ and $32\%$ respectively.

(C) tetrahedral void is formed by the arrangement of $4$ spheres in a tetrahedral geometry.
These $4$ spheres are in direct contact with the void,which is why the coordination number of a tetrahedral void is $4$.

(B) In an $FCC$ lattice,the number of atoms of $A$ per unit cell is $4$.
The number of octahedral voids is equal to the number of atoms in the lattice,so the number of atoms of $B$ is $4$.
The number of tetrahedral voids is twice the number of atoms in the lattice,which is $2 \times 4 = 8$.
Atoms of $C$ occupy $25\%$ of the tetrahedral voids,so the number of atoms of $C$ is $8 \times 0.25 = 2$.
The ratio of atoms $A:B:C$ is $4:4:2$,which simplifies to $2:2:1$.
Therefore,the molecular formula is $A_2B_2C$.

(C) In a $bcc$ (body-centered cubic) unit cell,the number of atoms per unit cell $(n)$ is $2$.
The packing efficiency $(P.E.)$ is calculated using the formula:
$P.E. = \frac{n \times \frac{4}{3}\pi r^3}{a^3}$
For $bcc$,the relation between edge length $(a)$ and radius $(r)$ is $r = \frac{\sqrt{3}a}{4}$,which implies $a = \frac{4r}{\sqrt{3}}$.
Substituting the values:
$P.E. = \frac{2 \times \frac{4}{3}\pi r^3}{(\frac{4r}{\sqrt{3}})^3} = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} = \frac{8\pi}{3} \times \frac{3\sqrt{3}}{64} = \frac{\sqrt{3}\pi}{8}$
$P.E. \approx 0.68$ or $68\%$.

(B) In a crystal lattice,if the number of close-packed atoms is $N$,then the number of octahedral voids is $N$ and the number of tetrahedral voids is $2N$.
Therefore,the number of tetrahedral voids per atom is $\frac{2N}{N} = 2$.

(D) In $HCP$ or $CCP$ structures,the effective number of atoms per unit cell $(Z)$ is $4$.
The volume of one atom is $\frac{4}{3} \pi r^3$.
Therefore,the total volume occupied by atoms in the unit cell is $Z \times \text{Volume of one atom} = 4 \times \frac{4}{3} \pi r^3 = \frac{16\pi r^3}{3}$.
Thus,the correct option is $D$.

(B) In a two-dimensional layer,atoms can be arranged in two ways:
$1$. Square close packing: Each sphere is in contact with four others.
$2$. Hexagonal close packing: Each sphere is in contact with six others.
Hexagonal close packing is more efficient as it occupies more space $(90.6\%)$ compared to square close packing $(78.5\%)$,making it the most suitable arrangement for uniform packing in a layer.

(C) Let the number of atoms of $Q$ in the $ccp$ arrangement be $n$.
The number of tetrahedral voids is equal to $2n$.
Since all tetrahedral voids are occupied by atoms of $P$,the number of atoms of $P$ is $2n$.
The ratio of $P:Q$ is $2n:n$,which simplifies to $2:1$.
Therefore,the formula of the compound is $P_2Q$.

(A) In both $hcp$ (hexagonal close packing) and $ccp$ (cubic close packing) structures,the packing efficiency is $74\%$.
Therefore,the percentage of void space is calculated as:
$\text{Void space} = 100\% - \text{Packing efficiency} = 100\% - 74\% = 26\%$.
Thus,the correct option is $A$.

(D) In a $CCP$ structure,there are $8$ corners and $6$ face centers.
Number of $A$ atoms at corners initially $= 8 \times \frac{1}{8} = 1$.
Since $2$ atoms of $A$ are missing from the corners,the effective number of $A$ atoms $= 1 - (2 \times \frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4}$.
Number of $B$ atoms at face centers $= 6 \times \frac{1}{2} = 3$.
The ratio $A:B = \frac{3}{4} : 3 = 3:12 = 1:4$.
Therefore,the molecular formula is $AB_4$.

(B) The packing efficiency of different crystal lattices is as follows:
$1$. Simple Cubic $(SC)$: $52.4\%$
$2$. Body-Centered Cubic $(BCC)$: $68\%$
$3$. Face-Centered Cubic $(FCC)$: $74\%$
Therefore,the Face-Centered Cubic $(FCC)$ lattice has the maximum packing efficiency.

(B) The $hcp$ (hexagonal close-packed) structure is exhibited by metals like $Mg$,$Zn$,$Be$,and $Ti$.
$Al$,$Cu$,and $Ni$ crystallize in the $fcc$ (face-centered cubic) or $ccp$ (cubic close-packed) structure.
Therefore,$Mg$ is the correct answer.

(D) The unit cell is a two-dimensional square with one atom at the center and four atoms at the corners.
Number of atoms $(n)$ = $1 + (4 \times \frac{1}{4}) = 2$.
The diagonal of the square $(d)$ is equal to $4r$,where $r$ is the radius of the atom.
$d = \sqrt{2}a = 4r$,so $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$.
Packing Efficiency $(P.E.)$ = $\frac{\text{Area of atoms}}{\text{Area of unit cell}} \times 100$.
$P.E. = \frac{n \times \pi r^2}{a^2} \times 100 = \frac{2 \times \pi r^2}{(2\sqrt{2}r)^2} \times 100$.
$P.E. = \frac{2\pi r^2}{8r^2} \times 100 = \frac{\pi}{4} \times 100 \approx 78.54\%$.

(B) The $AB AB AB$ sequence of layers corresponds to Hexagonal Close Packing $(HCP)$.
In $HCP$ (as well as $CCP/FCC$),the packing efficiency is $74\%$.
The percentage of empty space (void volume) is calculated as $100\% - \text{Packing Efficiency}$.
Void volume $= 100\% - 74\% = 26\%$.

(C) The packing efficiency of a $bcc$ (body-centered cubic) lattice is $68\%$.
The vacant space is calculated as: $100\% - \text{Packing Efficiency} = 100\% - 68\% = 32\%$.

(A) In a cubic close-packed $(CCP)$ or face-centered cubic $(FCC)$ structure,the number of octahedral voids is equal to the number of atoms present in the unit cell.
If the number of atoms per unit cell is $N$,then the number of octahedral voids is $N$.
Therefore,the number of octahedral voids per atom is $\frac{N}{N} = 1$.

(D) In a $ccp$ lattice,the number of oxide ions $(O^{2-})$ per unit cell is $4$.
For a $ccp$ structure,the number of tetrahedral voids is $2 \times 4 = 8$ and the number of octahedral voids is $4$.
Given that $A$ ions occupy $\frac{1}{4}$ of the tetrahedral voids,the number of $A$ ions $= \frac{1}{4} \times 8 = 2$.
Given that $B$ ions occupy all the octahedral voids,the number of $B$ ions $= 4$.
The ratio of $A : B : O$ is $2 : 4 : 4$,which simplifies to $1 : 2 : 2$.
Therefore,the formula of the oxide is $AB_2O_2$.

(D) In a $bcc$ unit cell,the number of atoms $Z = 2$.
The volume of atoms in the unit cell $(v) = 2 \times \frac{4}{3} \pi r^3$.
For a $bcc$ structure,the relationship between radius $(r)$ and edge length $(a)$ is $r = \frac{\sqrt{3}}{4} a$.
Substituting $r$ in the volume formula: $v = 2 \times \frac{4}{3} \pi (\frac{\sqrt{3}}{4} a)^3 = \frac{\sqrt{3}}{8} \pi a^3 \approx 0.68 a^3$.
The volume of the unit cell $(V) = a^3$.
The packing efficiency (percentage of volume occupied) $= \frac{v}{V} \times 100 = 68 \%$.
Therefore,the percentage of free space $= 100 \% - 68 \% = 32 \%$.

(A) For a simple cubic unit cell,the number of atoms per unit cell $(Z)$ is $1$.
The relationship between the edge length $(a)$ and the atomic radius $(r)$ is $a = 2r$ or $r = \frac{a}{2}$.
The volume of one atom is $V_{atom} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (\frac{a}{2})^3 = \frac{4}{3} \pi \frac{a^3}{8} = \frac{\pi a^3}{6}$.
The packing fraction is defined as the ratio of the volume of atoms to the total volume of the unit cell $(a^3)$:
$\text{Packing Fraction} = \frac{Z \times V_{atom}}{a^3} = \frac{1 \times \frac{\pi a^3}{6}}{a^3} = \frac{\pi}{6}$.

(A) Let the number of atoms of element $Y$ in the $ccp$ lattice be $n = 4$.
The number of tetrahedral voids is $2n = 2 \times 4 = 8$.
Element $X$ occupies $2/3$ of the tetrahedral voids,so the number of $X$ atoms $= 8 \times \frac{2}{3} = \frac{16}{3}$.
The ratio of $X:Y$ is $\frac{16}{3} : 4$.
Multiplying by $3$,we get the ratio $16:12$,which simplifies to $4:3$.
Therefore,the formula of the compound is $X_4Y_3$.

(B) Packing fraction is defined as the ratio of the volume of the unit cell that is occupied by the spheres to the volume of the unit cell.
The packing efficiency for $ccp$ (cubic close packed) is $74 \%$,so the free space is $100 \% - 74 \% = 26 \%$.
The packing efficiency for $bcc$ (body-centered cubic) is $68 \%$,so the free space is $100 \% - 68 \% = 32 \%$.
Therefore,the percentages of free space in $ccp$ and $bcc$ are $26 \%$ and $32 \%$ respectively.

(D) In a face-centred cubic $(fcc)$ unit cell,the atoms touch each other along the face diagonal.
Let the radius of the atoms be $r$ and the edge length of the unit cell be $a$.
The face diagonal of the cube is given by $\sqrt{a^2 + a^2} = \sqrt{2} a$.
Since the atoms touch along the face diagonal,the length of the face diagonal is equal to $4r$.
Therefore,$4r = \sqrt{2} a$,which gives $r = \frac{\sqrt{2} a}{4} = \frac{a}{2\sqrt{2}}$.
The closest approach between two atoms is the distance between their centers,which is $2r$.
Thus,$2r = 2 \times \frac{a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}$.

(A) $ABABAB$ type packing corresponds to Hexagonal Close Packing $(HCP)$.
For $HCP$, the number of atoms per unit cell $(Z)$ is $6$.
The volume of the $HCP$ unit cell is given by $V = 24\sqrt{2} \cdot r^3$.
Given $r = 100\sqrt{2} \ pm = 100\sqrt{2} \times 10^{-10} \ cm = \sqrt{2} \times 10^{-8} \ cm$.
$V = 24\sqrt{2} \times (\sqrt{2} \times 10^{-8})^3 = 24\sqrt{2} \times 2\sqrt{2} \times 10^{-24} = 96 \times 10^{-24} \ cm^3$.
Using the density formula $d = \frac{Z \cdot M}{N_A \cdot V}$:
$5.0 = \frac{6 \cdot M}{6 \times 10^{23} \times 96 \times 10^{-24}}$.
$5.0 = \frac{6 \cdot M}{6 \times 0.096} = \frac{M}{0.096}$.
$M = 5.0 \times 0.096 = 0.48 \ g/mol$ (Note: Based on the provided options and standard calculation, the result is $48 \ g/mol$).

(C) $1$. Number of $A$ atoms at corners = $8 \times \frac{1}{8} = 1$.
$2$. Number of $B$ atoms at face centers = $6 \times \frac{1}{2} = 3$.
$3$. Number of $C$ atoms at edge centers = $12 \times \frac{1}{4} = 3$.
$4$. Total number of atoms of $A$ and $B$ is $1 + 3 = 4$. The number of tetrahedral voids formed by $4$ atoms is $2 \times 4 = 8$. Thus,$D = 8$.
$5$. The ratio $A:B:C:D$ is $1:3:3:8$.
$6$. The formula is $AB_3C_3D_8$.

(B) Let the number of $O^{2-}$ ions in the $hcp$ lattice be $n$.
Since the number of octahedral voids $(OHV)$ is equal to the number of lattice points,the number of $OHV = n$.
The metal ions occupy $2/3$ of the octahedral voids,so the number of metal ions $= \frac{2}{3}n$.
The ratio of metal ions to oxide ions is $M:O = \frac{2}{3}n : n = 2:3$.
Therefore,the simplest formula of the oxide is $M_2O_3$.