Solid State

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}} $

Nearest neighbours are the atoms that are very close to a given atom. In body centered cubic unit cell, each atom is surrounded by eight atoms/ions present at the corner of the cell. So, the number of nearest neighbours in BCC is eight (8).

The hexagonal crystal system has lattice points at the twelve corners of the hexagonal prism and at the centres of the two hexagonal faces of the unit cell. The axial distances are $a=b \neq c$ and axial angles are $\alpha=\beta=90^{\circ}$ and $\gamma=120^{\circ}$.

The density is given by

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

Since their is $0.1 \%$ Schottky defect, so the $Z$ value will change.

$ Z=4-\frac{4 \times 0.1}{100}=3.996 $

Putting this value of $Z$ in above equation

$ \begin{aligned} \rho & =\frac{3.996 \times 55.8 \mathrm{~g}}{6.023 \times 10^{23} \times\left(400 \times 10^{-10} \mathrm{~cm}\right)^3} \\ & =5.8 \mathrm{~g} \mathrm{~cm}^{-3} . \end{aligned} $

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.

To answer this question, let's analyze each statement separately.

Statement I: Frenkel defects are vacancy as well as interstitial defects.

This statement is true. A Frenkel defect, also known as a dislocation defect, occurs in a crystalline solid when an atom or ion leaves its normal site and moves to an interstitial site, creating a vacancy defect at its original position and an interstitial defect at the new position. However, it's crucial to note that the vacancy and interstitial are created by the same atom or ion moving from one site to another within the crystal. So, the Frenkel defect involves both a vacancy and an interstitial defect, but they are correlated because they involve the displacement of the same atom or ion.

Statement II: Frenkel defect leads to color in ionic solids due to presence of F-centres.

This statement is false. The color in ionic solids is usually caused by F-centers (color centers), which are defects formed when an anion vacancy is occupied by one or more electrons. These electrons can absorb visible light, giving the crystal a characteristic color. However, the Frenkel defect, involving an atom or ion moving from its lattice site to an interstitial site, does not directly lead to the creation of F-centers. Instead, the coloration associated with F-centers is most commonly linked with Schottky defects (which involve paired vacancies of cations and anions but do not include displaced atoms to interstitial sites) or directly with anion vacancies that are not necessarily related to Frenkel defects.

Conclusion: Given the explanations above, the correct answer is Option C: Statement I is true but Statement II is false.

On heating the glass, it melts and takes up rounded shape at the edges, which has minimum surface area. This is due to the property of surface tension of liquids and not due to decrease in viscosity.

Viscosity generally decreases as the temperature increases.

Hence, both A and R are true but R is not the correct explanation of A.

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.