Solid State

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} $

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 $1.54 \times 10^{-8} \, \text{cm}$.

Given, $d=8.92 \mathrm{~g} \mathrm{~cm}^{-3}$

(Edge length) $a=3.61 \times 10^{-8} \mathrm{~cm}$

$M=\text { ? }$

For fcc, $Z=4$

$\begin{aligned} d & =\frac{Z \times M}{a^3 \times N} \\ 8.92 & =\frac{4 \times M}{\left(3.61 \times 10^{-8}\right)^3 \times 6.022 \times 10^{23}} \\ \Rightarrow \quad M & =\frac{8.92 \times 47.045 \times 10^{-24} \times 6.022 \times 10^{-23}}{4} \\ & =63.178 \mathrm{u} \end{aligned}$

(a) There are equal number of octahedral and tetrahedral voids in a unit cell for fcc lattice is incorrect statement. The number of octahedral and tetrahedral void in unit cell for fcc lattice is 4 and 8 respectively.

(b) The tetrahedral voids are present at the edge centers is incorrect as they are present on each body diagonal, i.e. diagonal of cubic unit cell.

(c) Octahedral voids are present at the body center and edge centers. Statement ($c$) is correct.

(d) Its packing efficiency ( PE ) is higher than hcp lattice PE is incorrect.

The packing efficiency of hcp is $74 \%$ and that of fcc is also $74 \%$. It is higher in comparison to packing efficiency of simple cube and bcc which have values 52.4 and $68 \%$ respectively.

A. Frenkel defects is a defect in which an atom/ion occupies a normally vacant site other than its own. This defect is possible if the cations are smaller in size as compared to the anions. Size of $\mathrm{Ag}^{+}$ is smaller than $\mathrm{Cl}^{\ominus}$ so, it shows Frenkel defect.

B. Schottky defects is a type of defect in which both cations and anions remain missing from their lattice sites in equal numbers. This is possible if size of both ions is nearly same. Size of $\mathrm{Na}^{+}$ is equal to size of $\mathrm{Cl}^{\ominus}$ so, it shows Schottky defect.

C. Vacancy defects is a defect in which crystals have vacant sites.

D. Metal deficiency defects, in this solids have less number of metals, FeO shows metal deficiency defect due to variable oxidation state of Fe.

Hence, the correct match is

Molecular solids are the solids in which atoms or molecules are held together by London dispersion forces, dipole-dipole interactions or hydrogen bonds. They have low melting points and flexibility and are poor conductors. HCl has dipole-dipole forces.

$\mathrm{H}_2 \mathrm{O}$ has hydrogen bonding.

$\mathrm{CCl}_4$ has London dispersion forces.

Thus, these are molecular solids.

$\mathrm{SiO}_2$ is a covalent or network solid.

Network solid or covalent solid is a compound in which the atoms are bonded together by covalent bonds in a continuous network. Out of given solids, network solids are; AlN, diamond and $\mathrm{SiO}_2$.

Ionic solid are composed of ions (cations and anions) held together by electrostatic forces. Out of given solids, ionic solids are; $\mathrm{NaCl}, \mathrm{MgO}, \mathrm{CaF}_2$, ZnS.

If molten NaCl contains an impurity of $\mathrm{SrCl}_2$ then, on crystallisation, some of the sites of $\mathrm{Na}^{+}$ ions are occupied by $\mathrm{Sr}^{2+}$. Each $\mathrm{Sr}^{2+}$ will replace two $\mathrm{Na}^{+}$ ions. One site is occupied by one $\mathrm{Sr}^{2+}$ but other remains vacant. Thus, a cationic vacancy is produced.

Photographic plate of film consists of a glass plate or thin strip of celluloid which is coated with the thin layer of an emulsion of silver bromide $(\mathrm{AgBr})$ dispersed in gelatin.

Given:

A metal crystallizes with a face-centered cubic (fcc) lattice.

The edge length (unit cell edge) = $ x \, \mathrm{pm} $

We need to find: diameter of the metal atom in pm.

Step 1: Relationship between atomic radius and edge length in fcc

In an fcc lattice, atoms touch each other along the face diagonal.

• Along the face diagonal, there are 4 radii across it.

That means:

$ \text{Face diagonal} = 4r $

But from geometry:

$ \text{Face diagonal} = \sqrt{2} \times a $

where $ a $ = edge length of the cube (given as $ x $).

Step 2: Equate the two expressions

$ \sqrt{2} a = 4r $

$ r = \frac{a}{2\sqrt{2}} $

Step 3: Diameter = $ 2r $

$ \text{Diameter} = 2r = \frac{a}{\sqrt{2}} $

Step 4: Substitute $ a = x $

$ \boxed{\text{Diameter} = \frac{x}{\sqrt{2}} \, \mathrm{pm}} $

✅ Correct Option:

Option B — $ \frac{x}{\sqrt{2}} $

In the face centred unit cell, the lattice points are present at the corners and face centres of unit cell.

In face centered cubic unit cell (fcc), each particle is at the corner and at face centeres. Corner atoms contribute $1 / 8$ per unit cell. Face center atoms contribute $=1 / 2$

Now, number of atoms, $Z=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$

Hence, fcc crystal contains 4 atoms in each unit cell.