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} $

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} $

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}} $

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} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Density }\left(\rho_\alpha\right) \text { for } \alpha \text {-form }=\frac{M Z_1}{N_0\left(a_1\right)^3} \\ & \text { and density }\left(\rho_\beta\right) \text { for } \beta \text {-from }=\frac{M Z_2}{N_0\left(a_2\right)^3} \end{aligned} \end{aligned} $

Given, side length $\left(a_1\right)$ for $(\alpha)=2 \mathrm{pm}$ side length $\left(a_2\right)$ for $(\beta)=4 \mathrm{pm}$ where, $M=$ molar-mass

$N_0=$ Avagadro number

$Z_1$ and $Z_2=$ number of particles in $\alpha$ and $\beta$ form.

$a_1$ and $a_2$ are side lengths.

Here, $Z_1=4, Z_2=2, a_1=2, a_2=4$

$ \begin{aligned} \therefore \quad \frac{\rho_\alpha}{\rho_\beta} & =\frac{M Z_1 / N_0\left(a_1\right)^3}{M Z_2 / N_0\left(a_2\right)^3}=\frac{Z_1 \times\left(a_2\right)^3}{Z_2 \times\left(a_1\right)^3} \\ \frac{\rho_\alpha}{\rho_\beta} & =\frac{4 \times(4)^3}{2 \times(2)^3}=\frac{256}{16}=16 \end{aligned} $

Hence, option (b) is the correct answer.

At atmospheric pressure ice crystallises in the hexagonal form, but at very low temperature water condense to cubic form.

Miller indices (hkl) of crystal are the reciprocal of intercepts along crystallographic axes ( $a, b, c$ ). Here, $h=\frac{1}{1}, k=\frac{1}{\frac{1}{2}}=2, l=\frac{1}{3}$

$ \text { } \quad \begin{aligned} So,\,\,\,\,\,\,\,\,\, h k l & =1: 2: \frac{1}{3} \\ & =3: 6: 1 \text { or }(361) \\ \Rightarrow \quad(h k l) & =(361) \end{aligned} $

∵ For bcc lattice, radius ratio for $r^{+} / r^{-}$is 0.7 , i.e. radius ratio of cation by anion ranges from 0.732 to 1.0 . Also, for bcc $r^{-}$ions are at corners,

$ \text { Thus, } \quad 4 \times r^{-}=\sqrt{3} a \quad(\because \sqrt{3} a=1.7 \times 200) $

$ \begin{aligned}and\,\,\, 4 \times r^{-} & =1.7 \times 200 \\ r^{-} & =\frac{1.7 \times 200}{4}=85 \mathrm{pm} \end{aligned} $

$\therefore \quad \frac{r^{+}}{r^{-}}=\frac{70}{85}=0.82$, which is close to 0.7. Hence, option (a) is the correct answer.

Given, copper crystallises in ccp manner

$ \text { radius }=1.41 \mathop {\rm{A}}\limits^{\rm{o}}=141 \mathrm{pm} $

Atomic weight of copper $=64$

Avogadro's number $=6 \times 10^{23}$, Density $=$ ?

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

Here, edge length of unit cell can be calculate as

$ \begin{aligned} & r=a / 2 \sqrt{2} \Rightarrow a=r \times 2 \sqrt{2} \\ & a=141 \times 2 \times 1.414 \\ & a=398.74 \mathrm{pm} \text { or } 398.74 \times 10^{-10} \mathrm{~cm} \end{aligned} $

Volume of unit cell

$ =\left(398.74 \times 10^{-10} \mathrm{~cm}\right)^3 $

For $\mathrm{ccp}=2=4, N_a=6.0 \times 10^{23}$

$ \begin{aligned} & f=\frac{4 \times 64}{\left(398.74 \times 10^{-10} \mathrm{~cm}\right)^3 \times 6.0 \times 10^{23}} \\ & f=6.73 \approx 6.67 \end{aligned} $

Hence, option (a) is correct.