Solid State

Face diagonal $d$ of a cube is related to edge length ' $a$ ' by : $d=a \sqrt{2}$

In fec, $d=4 r$

$ \begin{aligned} & \begin{aligned} 4 r & =a \sqrt{2} \text { or } r=\frac{a \sqrt{2}}{4} \\ \text { put } a & =4.242 \mathop {\rm{A}}\limits^{\rm{o}} \\ r & =\frac{4.242 \times \sqrt{2}}{4}\mathop {\rm{A}}\limits^{\rm{o}} \approx 1.5 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} \end{aligned} $

Sodium crystallises in a body centred cubic (bcc) lattice with an atomic radius of $1.86\mathop {\rm{A}}\limits^{\rm{o}}$. Let, the edge length of the unit cell be $x \mathop {\rm{A}}\limits^{\rm{o}}$.

In a bcc lattice, the body diagonal is equal to $4 r$, where $r$ is the atomic radius. The body diagonal is also equal to $\sqrt{3} x$.

$ \begin{aligned}So,\,\, \sqrt{3} x & =4 \times 1.86 \\ \sqrt{3} x & =7.44 \\ x & =\frac{7.44}{\sqrt{3}}=\frac{7.44}{1.732} \approx 4.29 \mathop {\rm{A}}\limits^{\rm{o}} . \end{aligned} $

Therefore, the edge length $x$ is $4.29\mathop {\rm{A}}\limits^{\rm{o}}$.

The incorrect statement about crystals with a Schottky defect is : Density of the crystal increases. A Schottky defect involves an equal number of cations and anions missing from their lattice sites to maintain electrical neutrality. This defect is common in ionic compounds with similarly sized ions. Because ions are missing, the mass of the crystal decreases while the volume stays roughly the same, leading to a decrease in the crystal's density.

Density $(d)=\frac{Z \times M}{N_A \times a^3}$

Ratio of density of fcc and bcc,

$ \begin{aligned} \frac{d_{\mathrm{fcc}}}{d_{\mathrm{bcc}}} & =\frac{Z_{f c c} \times a_{b c c}^3}{Z_{b c c} \times a_{f c c}^3} \\ \Rightarrow \quad \frac{d_{f c c}}{d_{b c c}} & =\frac{4}{2} \times\left(\frac{3}{3.5}\right)^3 \\ & \approx 1.26 \end{aligned} $

Let the number of $B$ atom in $c c p=N$

Number of octahedral voids $=N$

Number of tetrahedral voids $=2 N$

Number of $A$ atoms from octahedrals voids $=\frac{N}{2}$

Number of $A$ atoms from tetrahedral voids $=\frac{1}{2} \times 2 N=N$

Total number of $A$ atoms $=\frac{N}{2}+N=\frac{3 N}{2}$

Ratio of $A: B=\frac{3 N}{2}: N$ or $3: 2$

Thus molecular formula $A_3 B_2$.

Using the formula to calculate molar mass,

$ M=\frac{d \times N_A \times a^3}{Z} $

For fcc crystal, $x=4$,

$ \begin{aligned} M & =\frac{2 \times 10^{23} \times 2.16 \times 10^{-22}}{4} \\ & =64.8 \mathrm{~g} / \mathrm{mol} \end{aligned} $

Step 1. Geometry of the simple cubic lattice

In a simple cubic unit cell:

• Atoms are at the corners only.

• Each corner atom is shared by 8 unit cells.

• The atoms just touch along the edge of the cube.

Hence, for a simple cubic structure:

$ \text{Edge length } a = 2r $

where $ r $ is the radius of the atom.

Step 2. Volume of the unit cell

$ V = a^3 = (2r)^3 = 8r^3 $

Step 3. Substitute the given symbol

Given that the radius of the atom $ r = x \, \text{pm} $,

$ V = 8x^3 \, \text{pm}^3 $

✅ Final Answer:

Option C: $ \boxed{8x^3} \, \text{pm}^3 $

Let’s analyze each option about interstitial compounds — these are formed when small atoms (like H, B, C, or N) occupy the interstitial spaces in metal lattices.

Option A

They have high melting points.

✅ True.

Interstitial compounds generally have very high melting points due to strong metallic and covalent character.

Option B

They lose electrical conductivity during the formation from metal.

❌ False.

The metal lattice is largely retained in interstitial compounds, and they generally remain good conductors of electricity (sometimes conductivity decreases slightly but not lost entirely).

Option C

They are chemically inert.

✅ True.

These compounds are usually chemically inert and stable.

Option D

They are very hard.

✅ True.

Interstitial compounds are very hard, often harder than the parent metal (e.g., steel from Fe and C).

✅ Correct answer: Option B

They lose electrical conductivity during the formation from metal.

Crystal system with,

$ \begin{aligned} & a \neq b \neq c \text { and } \alpha=\beta=\gamma=90 \\ & x=\text { orthorhombic } \end{aligned} $

Orthorhombic crystal has 4 bravais lattice.

So, $\quad y=4$

The volume of simple cube, $V=\alpha^3$

$ \begin{aligned} & V=6.4 \times 10^7 \mathrm{pm}^3 \\ & a=(V)^{1 / 3}=400 \mathrm{pm} \end{aligned} $

For simple cube $r=\frac{a}{2}=\frac{400}{2}=200 \mathrm{p} \mathrm{pl}$

Given Values:

The edge length $ a = 288\, \text{pm} $. The density ($ \rho $) is $ 7.2\, \text{g/cm}^3 $. The mass given is $ 208\, \text{g} $.

Step 1: Find volume of one unit cell

The edge length must be changed to centimeters: $ 1\, \text{pm} = 10^{-10}\, \text{cm} $. So: $ a = 288 \times 10^{-10}\, \text{cm} $.

The volume of one unit cell is: $ V = a^3 = (288 \times 10^{-10})^3 \text{ cm}^3 = 2.389 \times 10^{-23}\, \text{cm}^3 $

Step 2: Find total volume that 208 g will occupy

$ V_{\text{total}} = \frac{m}{\rho} = \frac{208}{7.2} = 28.89\, \text{cm}^3 $

Step 3: Find the number of unit cells in 208 g

$ \text{Number of unit cells} = \frac{V_{\text{total}}}{V_{\text{unit cell}}} = \frac{28.89}{2.389 \times 10^{-23}} = 1.209 \times 10^{24} $

Step 4: Find the total number of atoms

A body centered cubic (bcc) unit cell has 2 atoms per unit cell.

So, $ \text{Total atoms} = \text{Number of unit cells} \times 2 = 1.209 \times 10^{24} \times 2 = 2.42 \times 10^{24} $ which is the same as $ 24.2 \times 10^{23} $.

Mole fraction of $\mathrm{CdCl}_2=\frac{1 \times 10^{-4}}{100}$

Mole fraction of $\mathrm{CdCl}_2=1 \times 10^{-6}$

Number of cation vacancies

$ \begin{aligned} & =\text { Mole fraction of } \mathrm{CdCl}_2 \times N_A \\ & =1 \times 10^{-6} \times 6.023 \times 10^{23} \mathrm{~mol}^{-1} \\ & =6.023 \times 10^{17} \mathrm{~mol}^{-1} \end{aligned} $

$ \begin{aligned} &\text { Mass of one atom }\\ &\begin{aligned} &=\frac{\text { Atomic weight }}{N_A}=\frac{250}{6.02 \times 10^{23}} \mathrm{~g} \\ & \text { Volume }=\frac{\text { Mass }}{\text { Density }}=\frac{250 / 6.02 \times 10^{23}}{7.2} \\ & \Rightarrow \quad 5.76 \times 10^{-23} \mathrm{~cm}^2 \\ & a=\sqrt[3]{\text { volume }}=3.86 \times 10^{-8} \mathrm{~cm} \\ & a=2 r, r=\frac{a}{2}=1.93 \times 10^{-8} \mathrm{~cm} \text { or } 1.93 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} \end{aligned} $

$B$ (anion) forms a ccp

ccp has $=4$ atom/cell

Voids in ccp

Number of tetrahedral voids $=2 \times 4=8$

All tetrahedral voids are accupied by $A$ (cation)

$ \begin{gathered} A=8, B=4 \\ A: B=2: 1 \text { or } A_2 B \end{gathered} $

The relation between radius and edge length in BCC structure

$ \begin{aligned} \sqrt{3 a} & =4 r \\ a & =\frac{4 r}{\sqrt{3}}=\frac{4 \times 2.598}{\sqrt{3}}=6 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

$ \begin{aligned} \text { Volume } & =a^3=(6)^3 \times\left(10^{-10}\right)^3 \mathrm{~cm}^3 \\ & =216 \times 10^{-24} \mathrm{~cm}^3 \\ & =2.16 \times 10^{-22} \mathrm{~cm}^3 \end{aligned} $

The closest distance $r$ is given by

$ r=\frac{a}{\sqrt{2}} \Rightarrow r=\frac{4}{\sqrt{2}}=2 \sqrt{2} \mathop {\rm{A}}\limits^{\rm{o}}=x \mathop {\rm{A}}\limits^{\rm{o}}$

The density ( $d$ ) is given by, $d=\frac{Z M}{a^3 N_A}$

$ \begin{aligned} Z & =4, M=197 \mathrm{~g} \mathrm{~mol}^{-1}, \\ a & =4 \times 10^{-8} \mathrm{~cm}, N_A=6 \times 10^{23} \mathrm{~mol}^{-1} \\ d & =\frac{4 \times 197 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(4 \times 10^{-8} \mathrm{~cm}\right)^3 \times 6 \times 10^{23} \mathrm{~mol}^{-1}} \\ & \approx 20.52 \mathrm{~g} \mathrm{~cm}^{-3}=y \mathrm{gcm}^{-3} \\ \therefore \quad & x=2 \sqrt{2} \text { and } y=20.52 \end{aligned} $

To find the value of $ X $ in the metal deficient oxide sample $ M_x Y_2 O_4 $, where $ M $ is present in both +2 and +3 oxidation states and $ Y $ is in the +3 oxidation state, and given that the fraction of $ M^{2+} $ ions present in $ M $ is $\frac{1}{3}$, follow these steps:

1. Charge Balance Equation:

The total charge contributed by the cations ($ M $ and $ Y $) must balance the total negative charge contributed by the oxide ions $ O $:

$ x(M^{2+} + M^{3+}) + 2(Y^{3+}) = 4(-2) $

1. Oxidation States and Charges:

• Let the total number of $ M $ ions be $ x $.


• Given $\frac{1}{3}$ of $ M $ ions are $ M^{2+} $, the remaining $\frac{2}{3}$ are $ M^{3+} $.

1. Number of Ions:

• Number of $ M^{2+} $ ions = $\frac{x}{3}$.


• Number of $ M^{3+} $ ions = $\frac{2x}{3}$.

1. Total Positive Charge:

• Charge from $ M^{2+} $ ions = $ \left(\frac{x}{3}\right) \times 2 = \frac{2x}{3} $.


• Charge from $ M^{3+} $ ions = $ \left(\frac{2x}{3}\right) \times 3 = 2x $.


• Charge from $ Y^{3+} $ ions = $ 2 \times 3 = 6 $.

1. Total Positive Charge Calculation:

$ \frac{2x}{3} + 2x + 6 $

1. Total Negative Charge:

• Charge from 4 oxygen ions $ (O^{2-}) $:

$ 4 \times (-2) = -8 $

1. Setting Up the Charge Balance:

$ \frac{2x}{3} + 2x + 6 = 8 $

1. Solving for $ x $:

$ \frac{2x}{3} + 2x + 6 = 8 $

$ \frac{2x}{3} + 2x = 2 $

$ \frac{2x + 6x}{3} = 2 $

$ \frac{8x}{3} = 2 $

$ 8x = 6 $

$ x = \frac{6}{8} $

$ x = 0.75 $

Thus, the value of $ X $ is $ \boxed{0.75} $.

Option D: 0.75

In a body-centered cubic (bcc) lattice consisting of two types of atoms, X and Y, X atoms are located at the corners of the unit cell, and Y atoms are at the center. To determine the formula of this compound, we need to consider the contributions of atoms in the unit cell.

Corner Atom Contribution:

In a bcc lattice, each corner atom is shared by 8 adjacent cells. Therefore, the contribution of a corner atom to one unit cell is:

$ \text{Contribution of each corner atom} = \frac{1}{8} $

Effective Number of X Atoms:

Let's assume three corner X atoms are missing. Originally, all eight corners would contribute:

$ \text{Total contribution of corners (without missing atoms)} = 8 \times \frac{1}{8} = 1 \text{ atom} $

Since three corner atoms are missing, the effective number of atoms at the corners is:

$ X = 1 - \frac{3}{8} = \frac{5}{8} \text{ atoms} $

Center Atom Contribution:

In a bcc structure, the Y atom located at the center is not shared and contributes entirely to the unit cell:

$ Y = 1 \text{ atom} $

Ratio of Atoms:

To find the ratio of X to Y, we equate their contributions:

$ X:Y = \frac{5}{8}:1 $

By scaling this ratio to whole numbers, we multiply both by 8 to eliminate the fraction:

$ X:Y = 5:8 $

Therefore, the formula for the compound is $X_5Y_8$.

The molecular formula is $ AB_2O_4 $. Oxygen atoms form a ccp (cubic close-packed) arrangement.

Effective number of O atoms in ccp: 4

Number of tetrahedral voids: 8

Number of octahedral voids: 4

Atoms of $ A $ occupy tetrahedral voids, and atoms of $ B $ occupy octahedral voids.

Calculations:

Fraction of octahedral voids occupied by $ B $:

$ \frac{2}{4} \times 100 = 50\% $

Thus, $ y = 50\% $ (since 2 atoms of $ B $ occupy octahedral voids).

Fraction of tetrahedral voids occupied by $ A $:

$ \frac{1}{8} \times 100 = 12.5\% $

Therefore, $ x = 12.5\% $ (since 1 atom of $ A $ occupies tetrahedral voids).

Given,

Molar mass $ (M)=2.7 \times 10^{-2} \mathrm{~kg} / \mathrm{mol} $

Edge length $ (a)=405 \mathrm{pm} $

Density $ =2.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} $

Density is given for unit cell by formula, $ d=\frac{z \times M}{a^{3} \times N_{A}} $

(where, $ z $ is atom per unit cell)

$ \Rightarrow z=\frac{d \times a^{3} \times N_{A}}{M} $

Substitute all values (known) in Eq. (i), $ z=\frac{2.7 \times 10^{3} \times\left(405 \times 10^{-12}\right)^{3} \times 6.022 \times 10^{23}}{2.7 \times 10^{-2}} $ $ =4 $

Among the given statements, option (d) is correct, while all other are incorrect. Their correct forms are :

In Frenkel defect, the difference between cation and anion size is much larger,

Frenkel defect is a dislocation defect.

Schottky defect alter's the physical properties of solids.

In $A B_2 \mathrm{O}_4$, the number of atom/ions ratio, $A^{2+}: B^{3+}: O^{2-}=1: 2: 4 \quad\text {(spinel structure)}\quad \ldots \text { (i) }$

$\Rightarrow$ Considering their occupied positions,

$ \begin{array}{cccc} A^{2+} & : B^{3+} & : 0^{-} \\ \left(\frac{1}{8} \times \mathrm{Tv}\right) & \left(\frac{1}{x} \times \mathrm{OV}\right) & \left( \mathrm{CCP}\right) \end{array} $

$ \begin{aligned} & {\left[\begin{array}{l} \because \text { TV }=\text { Tetrahedral void } \\ \text { OV }=\text { octahedral void } \end{array}\right]} \\ & =\frac{1}{8} \times 8: \frac{1}{x} \times 4: 4 \\ & =1 \quad: \frac{4}{x} \quad: 4\quad \ldots \text { (ii) } \end{aligned} $

$\Rightarrow$ Comparing Eqs. (i) and (ii),

$ \begin{aligned} & \frac{4}{x}=2 \\ & \frac{1}{x}=\frac{2}{4}=\frac{1}{2} \end{aligned} $

In a cubic close-packed (ccp) lattice, there are 4 atoms per unit cell. Therefore, the number of octahedral voids is equal to the number of atoms, which is 4. The number of tetrahedral voids is twice the number of atoms, calculated as:

$ 2n = 2 \times 4 = 8 $

For element $ Y $:

$ n = 2 $

Thus, the number of octahedral voids is:

$ 2 \text{ (since } n = 2 \text{)} $

The number of tetrahedral voids is:

$ 2 \times 2 = 4 $

Atoms of $ X $ occupy 50% of the octahedral voids:

$ 0.5 \times 2 = 1 $

Given that $ X $ must account for one additional atom (since $ X_3 $ implies three atoms per formula unit), these additional $ X $ atoms occupy tetrahedral voids:

$ 3 - 1 = 2 \text{ atoms of } X $

The percentage of the tetrahedral voids occupied by $ X $ is:

$ \frac{2}{4} \times 100 = 50\% $

Thus, the percentage of unoccupied tetrahedral voids is:

$ 100\% - 50\% = 50\% $

Given the problem:

A compound is formed by atoms of $A$, $B$, and $C$. Atoms of $C$ form an hcp (hexagonal close-packed) lattice. Atoms of $A$ occupy $50\%$ of the octahedral voids, and atoms of $B$ occupy $\frac{2}{3}$ of the tetrahedral voids. We need to determine the molecular formula of the solid.

To solve this:

Atoms of $C$ form an hcp lattice. The effective number of atoms per unit cell for an hcp lattice is calculated by:

$ = \frac{1}{2} \times 12 + \frac{1}{2} \times 2 + 1 \times 3 = 6 $

Here:

12 atoms from corners contribute $\frac{1}{2}$ each.

2 atoms from faces contribute $\frac{1}{2}$ each.

3 atoms from the middle contribute 1 each.

Thus, the total is 6.

The number of octahedral voids in a lattice is equal to the number of atoms, so there are 6 octahedral voids. Atoms $A$ occupy $50\%$ of these octahedral voids:

$ \Rightarrow \frac{50}{100} \times 6 = 3 \text{ atoms of } A $

The number of tetrahedral voids is twice the number of atoms in the hcp lattice, which equals $2 \times 6 = 12$. Atoms $B$ occupy $\frac{2}{3}$ of these tetrahedral voids:

$ \Rightarrow \frac{2}{3} \times 12 = 8 \text{ atoms of } B $

Thus, the molecular formula of the solid is $A_3B_8C_6$.

Explanation of the Process

When zinc oxide is heated, it loses oxygen and forms excess zinc ions ($\mathrm{Zn}^{2+}$) and electrons ($\mathrm{e}^{-}$) which get trapped in the interstitial sites. This process can be represented as:

$\mathrm{ZnO \xrightarrow{heating} Zn^{2+} + \frac{1}{2}O_2 + 2e^-}$

This leads to a non-stoichiometric compound where there's a slight excess of zinc. The released electrons are responsible for the change in color.

Analyzing the Statements

Statement I: Zinc oxide colour changes to pale yellow

This is correct. When ZnO is heated, it typically turns pale yellow due to the creation of F-centers (electrons trapped in oxygen vacancies).

Statement II: The type of defect formed is 'metal deficiency'

This is incorrect. The defect formed is a 'metal excess' defect because there are more zinc ions than the stoichiometric ratio.

Statement III: Some $\mathrm{Zn}^{2+}$ and $\mathrm{e}^{-}$ are present in interstitial place

This is correct. As explained above, the excess zinc ions and electrons occupy interstitial sites in the crystal lattice.

Conclusion

Based on the analysis, statements I and III are correct, while statement II is incorrect.

Therefore, the correct answer is:

Option B: I, III, only

Option A: $\mathrm{SiO}_2$ is a covalent (network) solid with a very high melting point and acts as an insulator. This description is correct.

Option B: While $\mathrm{MgO}$ does have a high melting point and is an insulator, it is an ionic solid—not a covalent solid. The bonding in $\mathrm{MgO}$ is due to the electrostatic attraction between $\mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$ ions. Therefore, this set is incorrect.

Option C: $\mathrm{H}_2\mathrm{O}$ (ice) is a molecular solid. It is an insulator and has a relatively low melting point compared to network or ionic solids. This description is correct.

Option D: $\mathrm{Ag(s)}$ is a metallic solid, which is a conductor and has a high melting point. This description is correct.

Thus, the incorrect set is Option B.

Powder diffraction is a scientific techique using X-ray, neutron or electron diffraction on powder for structural characterisation of material.

The graph is drawn between intensity (on $Y$-axis) and degree of diffraction (2 $\theta$ ) on $X$-axis.

The density of $\beta$-Fe is given as $7.6 \, \text{g/cm}^3$. It crystallizes in a cubic lattice with a lattice parameter $a = 290 \, \text{pm}$. The molar mass of Fe is $56 \, \text{g/mol}$, and Avogadro's number is $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$.

To find the value of $Z$, the formula for the density of a crystalline solid is used:

$ d = \frac{m \times Z}{a^3 \times N_A} $

Rearranging for $Z$, we have:

$ Z = \frac{d \times a^3 \times N_A}{m} $

Substituting the given values:

Convert $a$ from picometers to centimeters: $a = 290 \, \text{pm} = 290 \times 10^{-10} \, \text{cm}$.

$ Z = \frac{7.6 \times (290 \times 10^{-10})^3 \times 6.022 \times 10^{23}}{56} $

This calculation yields:

$ Z = 2 $

Thus, the rank of the unit cell, $Z$, is 2.

In this problem, we are asked to determine the molecular formula of a compound that consists of atoms A (cations), B (cations), and O (anions) where the oxygen atoms form an hcp (hexagonal close-packed) lattice.

Oxygen Lattice: Given that the lattice is an hcp structure, each unit cell contains 6 oxygen atoms (anions).

Tetrahedral Holes: In an hcp lattice, there are 12 tetrahedral holes per unit cell. Atoms of A occupy 25% of these tetrahedral holes, which can be calculated as follows:

$ \frac{25}{100} \times 12 = 3 \text{ atoms of A per unit cell.} $

Octahedral Holes: There are 6 octahedral holes per unit cell. Atoms of B occupy 50% of the octahedral holes:

$ \frac{50}{100} \times 6 = 3 \text{ atoms of B per unit cell.} $

Molecular Formula Calculation: With the aforementioned counts of atoms per unit cell (3 A atoms, 3 B atoms, and 6 O atoms), the ratio of A : B : O becomes:

$ 3 : 3 : 6 = 1 : 1 : 2 $

Therefore, the molecular formula of the compound is $\text{ABO}_2$.

The correct answer is Option B: Quartz.

Here's why:

Quartz is famous for its piezoelectric properties. This means when mechanical stress is applied to quartz, it produces an electric charge.

These properties make quartz widely used in electronic devices such as oscillators, sensors, and frequency control applications.

The other materials listed (Tridymite, Zelolite, and Mica) are not primarily known for piezoelectric applications.

Thus, quartz is the most common and useful piezoelectric material among the options given.

Molecular solids are compounds of discreate molecules held together by intermolecular forces.

A network solid is a compound in which the atoms are bounded by covalent bonds in continuous network extending throughout the material. Molecular solids : $\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{SO}_2, \mathrm{HCl}$. Network solids : $\mathrm{SiO}_2$, AlN Thus, the number of molecualr and network solids are 4 and 2 respectively.

To find the distance between the layers in a crystalline solid that produces a diffraction peak at $2\theta = 60^\circ$, we first need to determine $\theta$. Given that $2\theta = 60^\circ$, it follows that $\theta = 30^\circ$.

Then, we use Bragg's law for diffraction, which is:

$ n\lambda = 2d \sin \theta $

Given:

Wavelength of X-rays, $\lambda = 1.54 \, \mathring{A} = 1.54 \times 10^{-10} \, \text{m}$

Order of reflection, $n = 1$

$\sin 30^\circ = 0.5$

Rearranging Bragg's law to solve for $d$, we have:

$ d = \frac{n\lambda}{2 \sin \theta} $

Substituting the given values into the equation:

$ d = \frac{1 \times 1.54 \times 10^{-10}}{2 \times 0.5} = \frac{1.54 \times 10^{-10}}{1} = 1.54 \times 10^{-10} \, \text{m} = 1.54 \times 10^{-8} \, \text{cm} $

Thus, the distance between the layers is $1.54 \times 10^{-8} \, \text{cm}$.

In a face-centered cubic (FCC) unit cell, atoms are present at the corners as well as at the centers of the faces. Hence, the diagonal of the face of the unit cell is equal to 4 times the radius of an atom.

This gives us the equation:

$\sqrt{2} a = 4r$

Which simplifies to:

$a = 2\sqrt{2}r$

In a body-centered cubic (BCC) unit cell, atoms are present at the corners and at the center of the unit cell. The body diagonal of the unit cell is equal to 4 times the radius of an atom.

This gives us the equation:

$\sqrt{3} a = 4r$

Which simplifies to:

$a = \frac{4}{\sqrt{3}}r$

Therefore, Option D is correct: $2\sqrt{2}r = a$ and $4r = \sqrt{3}a$

A compound is formed by two elements $\mathrm{X}$ and $\mathrm{Y}$. The element $\mathrm{Y}$ forms a cubic close-packed (CCP) arrangement, and element $\mathrm{X}$ occupies one third of the tetrahedral voids.

In a CCP structure, there are 4 atoms of $\mathrm{Y}$ in a unit cell. This means there are 8 tetrahedral voids in the CCP structure.

Since $\mathrm{X}$ occupies one third of the tetrahedral voids, there are $\frac{1}{3} \times 8 = \frac{8}{3}$ atoms of $\mathrm{X}$ in the formula unit.

The ratio of $\mathrm{X}$ to $\mathrm{Y}$ in the formula unit is $\frac{8}{3} : 4 = \frac{2}{3} : 1$. Multiplying both parts of the ratio by 3, we get $2 : 3$.

So, the formula of the compound is $\mathrm{X_2Y_{3}}$

As W-atoms are located at corners and each corner contributes $1 / 8$, and there are 8 corners in cube. So, total number of W -atoms $=8 \times \frac{1}{8}=1$

Na is located at centre and centre atom contributes fully to unit cell. So, total number of Na atoms $=1$

O -atom is located at edge centre and each such atom contributes $1 / 4$ and there are 12 edge centre in cube. So, total number of O -atoms $=\frac{1}{4} \times 12=3$ So, formula of compound is $\mathrm{NaWO}_3$.

Let ' $x$ ' ions of $M^{2+}$ be replaced by $M^{3+}$ ions in the given sample.

Number of $M^{2+}=(0.96-x)$

Since, oxide is neutral.

∴ Total charge on $M$-atoms $=$ Charge on O-atom

$ 2(0.96-x)+3 x=2 $

or, $1.92-2 x+3 x=2$

or   $ x=0.08 $

Fraction of $M^{3+}=\frac{0.08}{0.96}=0.083$

Fraction of $M^{2+}=\frac{0.88}{0.96}=0.916$

Atoms of oxygen forms ccp lattice,

$ Z_{\mathrm{O}}=4 $

Number of tetrahedral voids $=2 \times$ number of octahedral voids (or $2 \times$ number of atoms in ccp unit cell)

Number of atoms $A($ cation $)=\frac{1}{8}$

$ \text { tetrahedral voids }=\frac{1}{8} \times 2 \times 4=1 $

Number of atoms $B$ (cation)

$ \begin{aligned} & =\frac{1}{2} \text { of octahedral voids } \\ & =\frac{1}{2} \times 4=2 \end{aligned} $

The molecular formula of the compound is $A B_2 \mathrm{O}_4$.

As $X$ atoms are located at centre and central atom contributes fully to unit cell.

So, total number of $X$ atom $=1$

$Y$ atoms are located at corner and each corner contributes $\frac{1}{8}$, there are 8

corners but

1 corner is unoccupied, so total number of corners $=7$

∴ Total number of $Y$ atoms $=7 \times \frac{1}{8}=\frac{7}{8}$

So, formula of compound is $X Y_{7 / 8}$ or $X_8 Y_7$.

In tetragonal crystal system, two edge lengths are same while one is different and all axial angles are same i.e. $a=b \neq c, \alpha=\beta=\gamma=90^{\circ}$.

Let's break down the statements:

Statement (A) says that graphite is used as a dry lubricant in machines running at high temperatures. This is true because graphite maintains its lubricating properties even at high temperatures, making it suitable for such applications.

Statement (R) explains that the layers of graphite slip over each other when pressure is applied. This is also true. Graphite has a layered structure where the layers are weakly held together by van der Waals forces, allowing the layers to easily slide past one another, which is the key reason for its lubricating properties.

Since both statements are true and (R) correctly explains why graphite works as a dry lubricant, the correct answer is:

Option C: $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

To calculate the density of copper, given its face-centered cubic (fcc) unit structure, we will use the following information:

Edge length $ = x \, \text{Å} $

Atomic mass of copper (Cu) $ = 63.5 \, \text{u} $

Avogadro's number $ N_A = 6.0 \times 10^{23} \, \text{mol}^{-1} $

Number of atoms per fcc unit cell $ Z = 4 $

The formula to calculate the density $ d $ of a crystalline structure is given by:

$ d = \frac{Z \times M}{a^3 \times N_A} $

Where:

$ Z $ is the number of atoms per unit cell

$ M $ is the atomic mass

$ a^3 $ is the volume of the unit cell in cubic centimeters (since $ a = x \times 10^{-8} \, \text{cm} $ for edge length $ x \, \text{Å} $)

$ N_A $ is Avogadro's number

Substituting the given values into the equation:

$ d = \frac{4 \times 63.5}{(x \times 10^{-8})^3 \times 6.0 \times 10^{23}} $

Simplifying further:

$ d = \frac{254}{6 \times x^3 \times 10^{-24}} $

$ d = \frac{254 \times 10^{24}}{6 \times x^3} $

$ d = \frac{2540}{6 \times x^3} $

$ d = \frac{423}{x^3} \, \text{g/cm}^3 $

Therefore, the density of copper at the given temperature, expressed in grams per cubic centimeter, is approximately $\frac{423}{x^3} \, \text{g/cm}^3$.

Number of oxygen atoms $=\frac{1}{4} \times 12$

There are 12 edges in a cube

$ =\frac{1}{4} \times 12=3 $

∴ The formula becomes $\mathrm{NaWO}_3$.

(Number of W atoms $\frac{1}{8} \times 8=1$ )

Na present at body centre.

Given, density $(\rho)=3.70 \mathrm{~g} / \mathrm{cm}^3$

Molecular weight of $\mathrm{KBr}=120 \mathrm{~g} / \mathrm{mol}$

$ \begin{aligned} \rho & =\frac{Z \times \text { molecular weight }}{N_A \times V} \\ 3.70 & =\frac{4 \times 120}{6.022 \times 10^{23} \times V} \\ \Rightarrow \quad V & =\frac{480}{22.28 \times 10^{23}} \\ V & =215.4 \times 10^{-24} \mathrm{~cm}^3 \\ V & =a^3 \\ \therefore \quad a & =\frac{1}{3} \sqrt{V} \\ & =6 \times 10^{-8} \mathrm{~cm} . \end{aligned} $

The length of body diagonal in FCC lattice

$ \begin{array}{rlrl} & & =\sqrt{3} a \\ \therefore & & x & =\sqrt{3} a \\ & & \frac{x}{\sqrt{3}} & =a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

The distance between two octahedral void is the distance between an edge-center and body center which is

$ \sqrt{\left(\frac{a}{2}\right)^2+\left(\frac{a}{2}\right)^2}=\frac{a}{\sqrt{2}} $

From Eq. (i) $a=\frac{x}{\sqrt{3}}$

∴ Distance between two octahedral voids

$ =\frac{x}{\sqrt{3} \times \sqrt{2}}=\frac{x}{\sqrt{6}} $