Crystal packing Questions in English

(C) In an $hcp$ lattice,the number of atoms of $O$ is $N = 6$.
Number of octahedral voids $= N = 6$.
Number of tetrahedral voids $= 2N = 12$.
Atoms of $A$ occupy $25\%$ of tetrahedral voids: $A = 0.25 \times 12 = 3$.
Atoms of $B$ occupy $50\%$ of octahedral voids: $B = 0.50 \times 6 = 3$.
Atoms of $O$ in $hcp$ lattice $= 6$.
The ratio of atoms $A:B:O$ is $3:3:6$,which simplifies to $1:1:2$.
Therefore,the molecular formula is $ABO_2$.

(A) Let the number of atoms of $C$ in the $hcp$ lattice be $n = 4$.
Number of octahedral voids $= n = 4$.
Number of tetrahedral voids $= 2n = 8$.
Atoms of $A$ occupy $50\%$ of octahedral voids,so $A = 4 \times 0.5 = 2$.
Atoms of $B$ occupy $\frac{2}{3}$ of tetrahedral voids,so $B = 8 \times \frac{2}{3} = \frac{16}{3}$.
The ratio of $A:B:C$ is $2 : \frac{16}{3} : 4$.
Multiplying by $3$ to get whole numbers,we get $6 : 16 : 12$.
Dividing by $2$,the simplest ratio is $3 : 8 : 6$.
Thus,the molecular formula is $A_3 B_8 C_6$.

(C) In a $ccp$ lattice,let the number of $Y$ atoms be $N = 4$.
Number of octahedral voids $= N = 4$.
Number of tetrahedral voids $= 2N = 8$.
Atoms of $X$ occupy $50\%$ of octahedral voids $= 0.5 \times 4 = 2$.
Atoms of $X$ occupy $x\%$ of tetrahedral voids $= \frac{x}{100} \times 8 = 0.08x$.
Total atoms of $X = 2 + 0.08x$.
Given formula is $X_3Y_2$,so the ratio $X:Y = 3:2$.
Since $Y = 4$,$X = \frac{3}{2} \times 4 = 6$.
$2 + 0.08x = 6$ $\Rightarrow 0.08x = 4$ $\Rightarrow x = 50$.
Percentage of occupied tetrahedral voids $= 50\%$.
Percentage of unoccupied tetrahedral voids $= 100 - 50 = 50\%$.

(D) In an $FCC$ unit cell,the number of atoms $(n)$ is $4$.
Number of octahedral voids $= n = 4$.
Number of tetrahedral voids $= 2n = 8$.
Given that $A$ atoms occupy all octahedral and all tetrahedral voids:
Total number of $A$ atoms $= 4 + 8 = 12$.
Given that $B$ atoms are at the $FCC$ lattice points:
Total number of $B$ atoms $= 4$.
The ratio of $A:B$ is $12:4$,which simplifies to $3:1$.
Therefore,the formula of the compound is $A_3 B$.

(A) In a $BCC$ unit cell,the number of $B^{-x}$ ions per unit cell is $2$.
The number of tetrahedral voids in a crystal is twice the number of atoms per unit cell.
Since $B^{-x}$ ions form the $BCC$ lattice,the number of tetrahedral voids $= 2 \times 2 = 4$.
$A^{+y}$ ions occupy all the tetrahedral voids,so the number of $A^{+y}$ ions $= 4$.
The ratio of $A:B = 4:2$,which simplifies to $2:1$.
Therefore,the formula of the compound is $A_2 B$.

(C) In an $fcc$ unit cell,the number of atoms $(Z)$ is $4$.
The number of octahedral voids is equal to $Z = 4$,and the number of tetrahedral voids is equal to $2Z = 8$. Thus,statement $(a)$ is incorrect.
Tetrahedral voids are located on the body diagonals,not at the edge centers. Thus,statement $(b)$ is incorrect.
In an $fcc$ lattice,octahedral voids are located at the body center $(1)$ and at the centers of each of the $12$ edges $(12 \times 1/4 = 3)$,totaling $4$. Thus,statement $(c)$ is correct.
The packing efficiency of both $fcc$ and $hcp$ lattices is $74 \%$. Thus,statement $(d)$ is incorrect.

(D) In an $fcc$ unit cell,anions $(A)$ are present at $8$ corners and $6$ face centers.
Effective number of $A = (\frac{1}{8} \times 8) + (\frac{1}{2} \times 6) = 1 + 3 = 4$.
Cations $(C)$ are present at the body center $(1)$ and half of the edge centers ($12 \times \frac{1}{2} = 6$ edges).
Effective number of $C = 1 + (6 \times \frac{1}{4}) = 1 + 1.5 = 2.5 = \frac{5}{2}$.
The ratio of $C:A = \frac{5}{2} : 4 = 5 : 8$.
Thus,the formula of the molecule is $C_{5}A_{8}$.

(B) The packing efficiency of $hcp$,$ccp$,and $fcc$ lattices are all equal to $74\%$.
Since $fcc$ and $ccp$ are the same arrangement (cubic close packing),their packing efficiencies are identical.
Therefore,the statement that the packing efficiency of $fcc$ lattice $>$ packing efficiency of $ccp$ lattice is incorrect.

(A) $Ag$ crystallises in $fcc$ lattice. Each unit cell contains $4$ atoms and $8$ tetrahedral voids.
Number of moles of $Ag = \frac{540 \ g}{108 \ g \ mol^{-1}} = 5 \ mol$.
Number of $Ag$ atoms = $5 N_A$.
Since $Ag$ crystallises in $fcc$,the number of unit cells = $\frac{\text{Total atoms}}{4} = \frac{5 N_A}{4} = 1.25 N_A$.
Number of tetrahedral voids = $8 \times \text{Number of unit cells} = 8 \times 1.25 N_A = 10 N_A$.

(B) In an $hcp$ lattice,the number of atoms per unit cell is $6$.
Since anions $A$ form the $hcp$ array,the number of anions $A = 6$.
The number of tetrahedral voids is equal to $2 \times$ (number of atoms in the lattice) $= 2 \times 6 = 12$.
Given that $2/3$ of the tetrahedral voids are occupied by cations $C$,the number of cations $C = \frac{2}{3} \times 12 = 8$.
The ratio of $C : A = 8 : 6$,which simplifies to $4 : 3$.
Therefore,the formula of the crystal is $C_4 A_3$.

(C) Cubic crystal systems are classified into three types based on the arrangement of lattice points:
$(a)$ Simple Cubic $(SC)$: Lattice points are arranged only at the corners of the cube.
$(b)$ Body-Centered Cubic $(BCC)$: Lattice points are arranged at the corners and at the center of the cube.
$(c)$ Face-Centered Cubic $(FCC)$: Lattice points are arranged at the corners and at the center of each face.
Therefore,in a face-centered unit cell,the lattice points are present at the corners and the face centers of the unit cell.

(A) In a close-packed structure of oxide ions $(O^{2-})$,let the number of $O^{2-}$ ions be $N = 4$.
Number of octahedral voids $= N = 4$.
Number of tetrahedral voids $= 2N = 8$.
$A$ occupies all octahedral voids,so number of $A$ atoms $= 4$.
$B$ occupies $(2/3)$rd of tetrahedral voids,so number of $B$ atoms $= 8 \times (2/3) = 16/3$.
The ratio of $A : B : O$ is $4 : (16/3) : 4$.
Multiplying by $3$ to get the simplest integer ratio: $12 : 16 : 12$.
Dividing by $4$: $3 : 4 : 3$.
Thus,the molecular formula is $A_3 B_4 O_3$.

(A) Let the number of atoms of $Y$ in the $ccp$ lattice be $n = 4$.
Number of octahedral voids = $n = 4$.
Number of tetrahedral voids = $2n = 8$.
Atoms of $X$ occupy half of the octahedral voids and half of the tetrahedral voids.
Number of $X$ atoms = $\frac{1}{2} \times 4 + \frac{1}{2} \times 8 = 2 + 4 = 6$.
The ratio of $X : Y$ is $6 : 4$,which simplifies to $3 : 2$.
Therefore,the formula of the compound is $X_3 Y_2$.

(A) 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 $2N$.
Given,number of moles of metal atoms $= 0.5 \ mol$.
Number of octahedral voids $= 0.5 \ mol$.
Number of tetrahedral voids $= 2 \times 0.5 \ mol = 1.0 \ mol$.
Total number of voids $= \text{octahedral voids} + \text{tetrahedral voids} = 0.5 \ mol + 1.0 \ mol = 1.5 \ mol$.
Thus,the total number of voids is $1.5 \ mol$ and the number of tetrahedral voids is $1.0 \ mol$.
Therefore,option $(A)$ is the correct answer.

(B) In an $hcp$ lattice,the number of octahedral voids is equal to the number of atoms forming the lattice.
Let the number of $Y$ atoms be $N$.
Then,the number of octahedral voids is also $N$.
The formula of the compound is $XY_3$,which means for every $1$ atom of $Y$,there are $3$ atoms of $X$ occupying the octahedral voids.
However,in an $hcp$ lattice,there is only $1$ octahedral void per atom of $Y$.
This implies that the stoichiometry $XY_3$ is impossible for a standard $hcp$ lattice where $X$ occupies octahedral voids,as there are only $N$ octahedral voids available for $N$ atoms of $Y$.
Assuming the question implies that $X$ occupies a fraction of the available octahedral voids,and given the formula $XY_3$ is provided as a premise,the fraction of occupied voids is $3$ per $1$ atom of $Y$,which is physically impossible in a simple $hcp$ lattice.
If we interpret the question as: $N$ atoms of $Y$ provide $N$ octahedral voids,and $X$ occupies $1/3$ of the available voids to result in $XY_{1/3}$ or similar,the provided formula $XY_3$ suggests an error in the problem statement.
Assuming the intended question was $X_{1/3}Y$,the fraction occupied is $1/3$,and the fraction unoccupied is $1 - 1/3 = 2/3$.

(D) In a hexagonal close-packed $(HCP)$ structure,the number of oxide ions per unit cell is $6$.
The number of octahedral voids in an $HCP$ structure is equal to the number of atoms,which is $6$.
Given that two out of every three octahedral holes are occupied by metal ions,the number of metal ions $= \frac{2}{3} \times 6 = 4$.
The ratio of metal ions to oxide ions is $4 : 6$,which simplifies to $2 : 3$.
Therefore,the formula of the metal oxide is $M_2O_3$.

(B) In a cubic close packing $(CCP)$ or face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.
The length of the face diagonal is given by $\sqrt{2} a$,where $a$ is the edge length of the unit cell.
Since the face diagonal consists of the radius of the corner atom,the diameter of the face-centered atom,and the radius of the other corner atom,we have: $4r = \sqrt{2} a$.
Therefore,the radius $r$ is given by $r = \frac{\sqrt{2} a}{4} = \frac{a}{2 \sqrt{2}}$.

(B) Let the number of atoms of $B$ in the $ccp$ lattice be $n$.
Since $B$ forms a $ccp$ lattice,the number of octahedral voids is $n$ and the number of tetrahedral voids is $2n$.
Cations of $A$ occupy $50 \%$ of octahedral voids,so number of $A$ atoms in octahedral voids $= 0.5 \times n = 0.5n$.
Cations of $A$ occupy $50 \%$ of tetrahedral voids,so number of $A$ atoms in tetrahedral voids $= 0.5 \times 2n = n$.
Total number of $A$ atoms $= 0.5n + n = 1.5n$.
The ratio of $A:B = 1.5n : n = 1.5 : 1 = 3 : 2$.
Therefore,the molecular formula of the solid is $A_3B_2$.

(D) In a $ccp$ lattice,if the number of atoms is $N$,then the number of octahedral voids is $N$ and the number of tetrahedral voids is $2N$.
Given the formula $AB_2O_4$,let the number of $O$ atoms be $4$.
Since $O$ forms a $ccp$ lattice,$N = 4$.
Number of tetrahedral voids $= 2 \times 4 = 8$.
Number of octahedral voids $= 4$.
Atoms of $A$ occupy $\frac{1}{8}$ of tetrahedral voids,so number of $A$ atoms $= \frac{1}{8} \times 8 = 1$.
Atoms of $B$ occupy octahedral voids. From the formula $AB_2O_4$,there are $2$ atoms of $B$.
Fraction of occupied octahedral voids $= \frac{\text{Number of } B \text{ atoms}}{\text{Total octahedral voids}} = \frac{2}{4} = \frac{1}{2}$.
Fraction of vacant octahedral voids $= 1 - \frac{1}{2} = \frac{1}{2}$.

(A) In a $ccp$ lattice,if the number of oxygen atoms is $4$,then:
Number of octahedral voids $= 4$
Number of tetrahedral voids $= 8$
Given the formula $AB_2O_4$,there are $1$ atom of $A$,$2$ atoms of $B$,and $4$ atoms of $O$.
For atom $A$ in tetrahedral voids: $x \% \text{ of } 8 = 1 \implies x = (1/8) \times 100 = 12.5 \%$.
For atom $B$ in octahedral voids: $y \% \text{ of } 4 = 2 \implies y = (2/4) \times 100 = 50 \%$.
Thus,$x = 12.5 \%$ and $y = 50 \%$.

(B) Number of $X$ atoms per unit cell $= 1 \times 1 = 1$.
Number of $Y$ atoms per unit cell $= (8 - 1) \times \frac{1}{8} = \frac{7}{8}$.
Ratio of $X : Y = 1 : \frac{7}{8} = 8 : 7$.
Therefore,the empirical formula is $X_8 Y_7$.

(C) Let the number of oxygen atoms in the $ccp$ lattice be $1$.
Since the number of octahedral voids is equal to the number of atoms,the number of octahedral voids is $1$.
The number of tetrahedral voids is twice the number of atoms,so the number of tetrahedral voids is $2$.
Atoms of $A$ occupy $\frac{1}{8}$ of the tetrahedral voids,so the number of $A$ atoms $= \frac{1}{8} \times 2 = \frac{1}{4}$.
Atoms of $B$ occupy half of the octahedral voids,so the number of $B$ atoms $= \frac{1}{2} \times 1 = \frac{1}{2}$.
The ratio of $A:B:O$ is $\frac{1}{4} : \frac{1}{2} : 1$.
Multiplying the ratio by $4$,we get $1 : 2 : 4$.
Thus,the molecular formula of the compound is $AB_2O_4$.

(C) In an $hcp$ lattice,the number of atoms per unit cell is $6$.
The number of tetrahedral voids is twice the number of atoms,which is $2 \times 6 = 12$.
Given that $A$ occupies $2/3$ of the tetrahedral voids,the number of atoms of $A = \frac{2}{3} \times 12 = 8$.
The ratio of atoms $A : B = 8 : 6 = 4 : 3$.
Therefore,the formula of the compound is $A_4 B_3$.

(C) In a simple cubic structure,atoms are present only at the corners. The effective number of atoms per unit cell is $Z = 1$.
Volume of one atom $= \frac{4}{3} \pi r^3$,where $r$ is the radius of the atom.
Volume of the unit cell $= a^3$,where $a$ is the edge length.
In a simple cubic unit cell,the atoms touch along the edge,so $a = 2r$.
Packing fraction $= \frac{Z \times \text{Volume of one atom}}{\text{Volume of unit cell}} = \frac{1 \times \frac{4}{3} \pi r^3}{(2r)^3} = \frac{4 \pi r^3}{3 \times 8 r^3} = \frac{\pi}{6}$.

(A) In a hexagonal close-packed $(HCP)$ structure,the number of atoms per unit cell $(N)$ is $6$.
For $1$ mole of the compound,the number of atoms is $N_A = 6.022 \times 10^{23}$.
In any close-packed structure,the number of octahedral voids is equal to the number of atoms $(N)$,and the number of tetrahedral voids is equal to $2N$.
Total number of voids = (Number of octahedral voids) + (Number of tetrahedral voids) = $N + 2N = 3N$.
For $1$ mole,total voids = $3 \times N_A = 3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{24}$.
Rounding to two significant figures,we get $1.8 \times 10^{24}$.

(A) In an $FCC$ lattice,the number of atoms per unit cell is $4$.
Since $Z$ forms the $FCC$ lattice,the number of $Z$ atoms $= 4$.
The number of octahedral voids is equal to the number of atoms in the lattice,so the number of $X$ atoms $= 4$.
The number of tetrahedral voids is twice the number of atoms in the lattice,so the number of tetrahedral voids $= 2 \times 4 = 8$.
Given that $Y$ occupies $\frac{1}{3}$ of the tetrahedral voids,the number of $Y$ atoms $= \frac{1}{3} \times 8 = \frac{8}{3}$.
The ratio of atoms $X:Y:Z = 4 : \frac{8}{3} : 4$.
Multiplying by $3$ to get the simplest whole number ratio: $12 : 8 : 12$,which simplifies to $3 : 2 : 3$.
Therefore,the formula of the compound is $X_3Y_2Z_3$.

(B) In an $hcp$ lattice,the number of atoms per unit cell is $6$.
Let the number of atoms of $Z$ be $N = 6$.
The number of octahedral voids is equal to the number of atoms,so the number of $X$ atoms $= N = 6$.
The number of tetrahedral voids is $2N = 12$.
Atoms of $Y$ occupy half of the tetrahedral voids,so the number of $Y$ atoms $= \frac{1}{2} \times 12 = 6$.
The ratio of atoms $X : Y : Z$ is $6 : 6 : 6$,which simplifies to $1 : 1 : 1$.
Therefore,the molecular formula of the solid is $XYZ$.

(B) In a close-packed structure ($CCP$ or $HCP$),if there are $n$ spheres (atoms or ions) in the packing,then:
Number of octahedral voids $= n$
Number of tetrahedral voids $= 2n$
In a $CCP$ unit cell,the number of atoms per unit cell is $4$.
Therefore,the number of octahedral voids $= 4$.
The number of tetrahedral voids $= 2 \times 4 = 8$.
Thus,the number of tetrahedral and octahedral voids are $8$ and $4$ respectively.

(A) The packing efficiency is the percentage of total space occupied by the constituent particles in a crystal lattice.
For simple cubic $(sc)$ lattice,the packing efficiency is $52.4 \%$.
For body-centred cubic $(bcc)$ lattice,the packing efficiency is $68 \%$.
For cubic close packing $(ccp)$ or face-centred cubic $(fcc)$ lattice,the packing efficiency is $74 \%$.
Therefore,the order of packing efficiency is $sc < bcc < ccp$.

(A) In a close packed structure ($hcp$ or $ccp$):
$(I)$ The number of octahedral voids is equal to the number of particles $(N)$ present in the close packing.
$(II)$ The number of tetrahedral voids is equal to $2N$,where $N$ is the number of particles.
Given that there are $X$ atoms,the number of octahedral voids is $X$ and the number of tetrahedral voids is $2X$.