Solid State

• In a ccp (fcc) unit cell, number of atoms per cell, n = 4

• Edge length

$a=400\text{ pm}=400\times10^{-12}\text{ m}=4.0\times10^{-8}\text{ cm}$

• Volume of cell

$V=a^3=(4.0\times10^{-8}\text{ cm})^3=6.4\times10^{-23}\text{ cm}^3$

• Mass of one atom

$m_{\rm atom}=\frac{M}{N_A}=\frac{105.6\text{ g/mol}}{6.0\times10^{23}\text{ mol}^{-1}} =1.76\times10^{-22}\text{ g}$

• Mass of unit cell

$m_{\rm cell}=n\;m_{\rm atom}=4\times1.76\times10^{-22}=7.04\times10^{-22}\text{ g}$

• Density

$\rho=\frac{m_{\rm cell}}{V} =\frac{7.04\times10^{-22}}{6.4\times10^{-23}} \approx11\;\text{g/cm}^3$

Answer: 11 g·cm⁻³.

In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by :

$\sqrt{3}a = 4r$

Given the unit cell edge length (a) of sodium metal as 4 Å :

$a = 4 ~\mathop A\limits^o$

We can now solve for the radius (r) of the sodium atom :

$4r = \sqrt{3}a$

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

$r = \frac{\sqrt{3} \times 4 ~\mathop A\limits^o}{4}$

$r = \sqrt{3} ~\mathop A\limits^o$

Now, we can approximate the numerical value :

$r \approx 1.732 ~\mathop A\limits^o$

To express the radius as a multiple of 10⁻¹ $\mathop A\limits^o$ :

$r = 1.732 \times 10^{1} \times 10^{-1} ~\mathop A\limits^o = 17.32 \times 10^{-1} ~\mathop A\limits^o$

So, the radius of the sodium atom is 17.32 × 10⁻¹ $\mathop A\limits^o$.

Given:

• Atomic substance A with molar mass $M = 12 \, \text{g mol}^{-1}$


• Cubic crystal structure with edge length $a = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m}$


• Density $\rho = 3.0 \, \text{g mL}^{-1}$


• Avogadro's number $N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}$

First, calculate the volume of the unit cell $V = a^3 = (300 \times 10^{-12} \, \text{m})^3 = 27 \times 10^{-30} \, \text{m}^3$.

Next, calculate the mass of the unit cell using the given density. Density is mass over volume, so mass

$m = \rho \cdot V = 3.0 \, \text{g mL}^{-1} \cdot 27 \times 10^{-30} \, \text{m}^3 \cdot \frac{1 \, \text{mL}}{10^{-6} \, \text{m}^3} = 81 \times 10^{-24} \, \text{g}$.

Then, calculate the number of moles in one unit cell. The molar mass is mass over number of moles, so

$n = \frac{m}{M} = \frac{81 \times 10^{-24} \, \text{g}}{12 \, \text{g mol}^{-1}} = 6.75 \times 10^{-26} \, \text{mol}$.

Finally, calculate the number of atoms in one unit cell. The Avogadro constant is the number of atoms per mole,

so number of atoms $ = n \cdot N_A = 6.75 \times 10^{-26} \, \text{mol} \cdot 6.02 \times 10^{23} \, \text{mol}^{-1} = 4.06$.

Rounding to the nearest integer, we get 4 atoms. So, there are 4 atoms present in one unit cell of substance A.

Body-centered unit cells can be found in the following crystal systems among those listed:

1. Cubic: Body-centered cubic (BCC) is one of the lattice structures that cubic systems can have.


2. Tetragonal: A body-centered tetragonal system is also a possibility.


3. Orthorhombic: The orthorhombic system can have a body-centered orthorhombic lattice.

Therefore, the number of crystal systems from the list where a body-centered unit cell can be found is 3.

$\begin{aligned} & d_1 \text {, Density of fcc lattice of metal } M=\frac{4 \times M}{N_0\left(a_{\mathrm{fcc}}\right)^3} \\\\ & d_2 \text {, Density of bcc lattice of metal } M=\frac{2 \times M}{N_0\left(a_{\mathrm{bcc}}\right)^3} \\\\ & \frac{d_1}{d_2}=\frac{4}{2}\left(\frac{a_{\mathrm{bcc}}}{a_{\mathrm{fcc}}}\right)^3=2\left(\frac{2.5}{2}\right)^3=3.90 \simeq 4\end{aligned}$

$\mathrm{d=\frac{z\times m}{a^3}}$

$\mathrm{4=\frac{z\times72}{6\times10^{23}(5\times10^{-8})^3}}$

$\mathrm{4=\frac{z\times72}{6\times125\times10^{-1}}}$

$=\mathrm{z \approx4}$

Number of voids

$=0.02\times3\times6.02\times10^{23}$

$\simeq36\times10^{21}$

In cubic close packing, octahedral voids form at edge centers and body center of the cube

a = 2(rA⁺ + rB^–)

a = 2 (102 + 181)

a = 566 pm

$\rho=\frac{Z M}{N_{A} a^{3}} \Rightarrow M=\frac{9.03 \times 6.02 \times 10^{23} \times\left(4 \times 10^{-8}\right)^{3}}{4}$

$ \begin{aligned} &=\frac{9.03 \times 6.02 \times 64 \times 10^{-1}}{4} \\\\ &=86.9 \mathrm{~g} \mathrm{~mol}^{-1} \\\\ &\approx 87 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned} $

$M$ is body certred cubic, $\therefore Z=2$

Let mass of 1 atom of $M$ is $A$

Edge length $=300 \,\mathrm{pm}$

Density $=6 \mathrm{~g} / \mathrm{cm}^3$

$ \therefore 6 \mathrm{~g} / \mathrm{cm}^3=\frac{\mathrm{Z} \times \mathrm{A}}{\left(300 \times 10^{-10}\right)^3}=\frac{2 \times \mathrm{A}}{27 \times 10^{-24}} $

$\mathrm{A}=81 \times 10^{-24} \mathrm{~g}$

$\therefore$ Atomic mass $=48.6 \mathrm{~g}$

$\therefore$ Mole in $180 \mathrm{~g}=\frac{180}{48.6}=3.7$ moles

Atoms of $\mathrm{M}=3.7 \times 6 \times 10^{23}$ $=22.22 \times 10^{23}$ atoms

$\mathrm{Fe}_{0.93} \mathrm{O}$

Let the number of $\mathrm{O}^{-2}$ ions be 100 and the number of $\mathrm{Fe}^{+2}$ ions be $\mathrm{X}$ The number of $\mathrm{Fe}^{+3}$ ions be $(93-\mathrm{X})$

$ \begin{aligned} &\therefore \,X(2)+(93-X) 3=200 \\\\ &279-X=200 \\\\ &X=79 \\\\ &\therefore \quad \% \text { of } F^{+2} \text { ions } =\frac{79}{93} \times 100 \\\\ &\simeq 85 \% \end{aligned} $

$A$ atoms are in CCP contribution of $A$ is

$A=4$

If atoms from opposite faces are removed then

$ \begin{aligned} & A=4-x \times \frac{1}{x} \\\\ & A=3 \end{aligned} $

Value of $x=3$

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

$ 43.1=\frac{4 \times 58.5}{a^{3} \times 6.02 \times 10^{23}} $

$ \begin{aligned} & a^{3}=0.9 \times 10^{-23} \\\\ & =9 \times 10^{-24} \end{aligned} $

$a=2.08 \times 10^{-8} \mathrm{~cm}$

$=2.08 \times 10^{-10} \mathrm{~m}$

for $\mathrm{NaCl}$, distance between $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}=\frac{a}{2}$

$=1.04 \times 10^{-10} \mathrm{~m}$

Since $X$ occupies hop lattice, Number of particles of type $X$ in a unit cell $=6$

Number of particles of type $Y=\frac{2}{3} \times 12=8$

$\therefore$ Percentage of element $X=\frac{6}{14} \times 100$

$ \begin{aligned} &=\frac{300}{7} \\\\ &=42.85 \\\\ &\simeq 43 \% \end{aligned} $

Anions forms CCP or FCC(A^$-$) = 4A^$-$ per unit cell Cations occupy all octahedral voids (B⁺) = 4B⁺ per unit cell

cell formula $\to$ A₄B₄

Empirical formula $\to$ AB $\to$ (x = 1)

$d\left( {{{gm} \over {cc}}} \right) = {{4 \times {M \over {{N_A}}}} \over {{{(a\,cm)}^3}}}$

$7.62 = {{4 \times M/6.022 \times {{10}^{23}}} \over {{{(0.4518 \times {{10}^{ - 7}}cm)}^3}}} $

$\Rightarrow $ M = 105.8 g/mol

Carbon atoms occupy FCC lattice points as well as half of the tetrahedral voids.

Therefore number of carbon atoms per unit cell = 8

For every Sr⁺² ion, 1 cationic vacancy is created. Hence, no. of Sr⁺² ion = Number of cationic vacancies

Since mole percentage of SrBr₂ dropped is 10^$-$5 to that of total moles of KBr.

Hence,

No. of cationic vacancy $ = {{{{10}^{-5}}} \over {100}} \times {1 \over {119}} \times {N_A}$

$ = {1 \over {119}} \times {10^{ - 7}} \times 6.022 \times {10^{23}}$

$ = 5 \times {10^{ - 2}} \times {10^{ - 7}} \times {10^{23}} = 5 \times {10^{14}}$

For HCP structure Z = 6

Tetrahedral Void = (Z $\times$ 2) = 12

Octahedral Void = [Z $\times$ 1] = 6

No. of void per unit cell = 18

No. of unit cell = $\left( {{{0.581 \times {N_A}} \over {70 \times 6}}} \right)$

No. of void = $\left( {{{0.581 \times 6.023 \times {{10}^{23}}} \over {70 \times 6}}} \right) \times 18$

$ = 14.99 \times {10^{21}}$

For BCC unit cell, $\sqrt 3 a = 4R$

$a = {{4R} \over {\sqrt 3 }} = 27$

$R = {{27\sqrt 3 } \over 4}$

For FCC unit cell

$\sqrt 2 a = 4R$

$ \Rightarrow $ $a = {4 \over {\sqrt 2 }}\left( {{{27\sqrt 3 } \over 4}} \right)$

$ \Rightarrow $ $a = 27\sqrt {{3 \over 2}} $

$ \Rightarrow $ $a = 33.12 \approx 33$

Let us assume, the crystal has fcc or ccp lattice which has octahedral voids.

Number of lattice sites occupied = 8 corner + 6 face centres = 14

Number of octahedral voids = 12 edge centres + 1 body centre = 13

Number of octahedral void(s) per lattice site

$ = {{13} \over {14}} = 0.928 \simeq 1$

Density of copper, $d = {{Z \times M} \over {{a^3} \times {N_A}}}$

Given, Z = 4, for fcc lattice,

M = 63.54 g mol^$-$1

= 63.54 $\times$ 10^$-$3 kg mol^$-$1,

a = 3.596 $\mathop A\limits^o $ = 3.596 $\times$ 10^$-$10 m,

N_A = 6.022 $\times$ 10²³ mol^$-$1

On putting given values, we get

$ \Rightarrow d = {{4 \times (63.54 \times {{10}^{ - 3}})} \over {{{(3.596 \times {{10}^{ - 10}})}^3} \times (6.022 \times {{10}^{23}})}}$ kg/m³

$ = 9076.26 \simeq 9077$ kg/m³

Coordination number is the number of nearest neighbours of a central atom in the structure.

bcc has a coordination number of 8 and contains 2 atoms per unit cell.

This is because each atom touches four atoms in the layer above it, four in the layer below it and none in its own layer.

Molar mass of an element (M) = 27 gm mol^–1

Edge length of a cubic unit cell (a) = 405 pm = 4.05 × 10^–8 cm

density of the element (d) = 2.7 gm/cc

d = ${{Z \times M} \over {{N_A} \times {{\left( a \right)}^3}}}$

$ \Rightarrow $ 2.7 = ${{Z \times 27} \over {6 \times {{10}^{23}} \times {{\left( {4.05 \times {{10}^{ - 8}}} \right)}^3}}}$

$ \Rightarrow $ Z = 4

The element has fcc unit cell

$ \therefore $ $\sqrt 2 $a = 4r

$ \Rightarrow $ r = ${{1.414 \times 405} \over 4}$

= 143 pm

= 143 × 10^–12 m

The unit cell of initial structure of ionic solid MX looks like

In NaCl type of solids cations (Na⁺) occupy the octahedral voids while anions (Cl^$-$) occupy the face centre positions.

However, as per the demand of problem the position of cations and anions are swapped.

We also know that (for 1 unit cell)

(A) Total number of atoms at FCC = 4

(B) Total number of octahedral voids = 4 (as no. of atoms at FCC = No. of octahedral voids)

Now taking the conditions one by one

(i) If we remove all the anions except the central one than number of left anions.

= 4 $-$ 3 = 1

(ii) If we replace all the face centred cations by anions than effective number of cations will be = 4 $-$ 3 = 1 Likewise effective number of anions will be = 1 + 3 = 4

(iii) If we remove all the corner cations then effective number of cations will be 1 $-$ 1 = 0

(iv) If we replace central anion with cation then effective number of cations will be 0 + 1 = 1 Likewise effective number of anions will be 4 $-$ 1 = 3

Thus, as the final outcome, total number of cations present in Z after fulfilling all the four sequential instructions = 1 Likewise, total number of anions = 3

Hence, the value of ${{Number\,of\,anions} \over {Number\,of\,cations}} = {3 \over 1} = 3$

Given:

(i) Edge length of face centred cubic (FCC) unit cell (a) $=400 \mathrm{pm}$

(ii) Density of the substance $(\rho)=8 \mathrm{~g} \mathrm{~cm}^{-3}$

(iii) Mass of the crystal $(m)=256 \mathrm{~g}$

To Find: The value of $\mathrm{N}$ in $\mathrm{N} \times 10^{24}$

Formula: Density of the cell $(\rho)$

$ \begin{aligned} & =\frac{\text { Mass of crystal }(\mathrm{M}) \times \begin{array}{l} \text { No. of atoms in unit cell } \\ \end{array}}{\text { Volume of unit cell } \times \text { No. of atoms of pure substance in mass m }} \\ \end{aligned} $

Calculations: A face-centred cubic (F.C.C.) lattice has atoms of pure substance at the corner of the unit cell and/or atom each at the centre of the face of the unit cell.

Since there are eight atoms at the corners and one atom at each of the six faces (i.e., six atoms in total):

Total number of atoms present per unit cell $(z)$

$ =\frac{1}{8} \times 8+\frac{1}{2} \times 6=4 $

Density of the cell $(\rho)$

$ \begin{aligned} & =\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right)^3 \times \mathrm{N}^1} \\\\ & 8 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right) \times \mathrm{N}^1} \\\\ & \mathrm{~N}^{\prime}=\frac{256 \mathrm{~g} \times 4}{8 \mathrm{~g} \mathrm{~cm}^{-3} \times(400)^3 \times 10^{-30}} \\\\ & \mathbf{N}^{\prime}=2 \times 10^{24} \text { atoms } \end{aligned} $

Hence, the crystal of pure substance of mass $256 \mathrm{~g}$ contains $2 \times 10^{24}$ atoms.

Now, $\mathrm{N}^{\prime}=\mathrm{N} \times 10^{24}$

Therefore, value of $\mathrm{N}=2$.

(i) The shape of a regular octahedron :

When the octahedron is truncated from all edges.

(ii) Each face of truncated octahedron is hexagonal.

Since, there are 8 faces, there are eight hexagons.

AlCl₃ exists as a close packed lattice of chloride ions Cl^– with Al³⁺ occupying octahedral holes. Hence, coordination number of Al³⁺ is = 6.

Given,

$\mathrm{M}_{w}=75 \mathrm{~g} \mathrm{~mol}^{-1}$

$a=5$ $\mathop A\limits^o $ = 5 $\times$ 10$^{-8}$ cm

$\rho=2$ g cm$^{-3}$

$\mathrm{N_A=6\times10^{23}}$

To Find: Radius of metal atom ($r$)

To find the value of $r$, we need to find out the type of unit cell. For that, the value of Z is required. On the basis of the value of Z the type of unit cell can be identified and using suitable formula, radius of metal atom will be determined.

Step 1 : Determination of $Z_{\text {eff }}$

The formula for calculating $\mathrm{Z}_{\text {eff }}$ will be obtained from the formula for density.

The formula for calculating density is,

$\rho=\frac{\mathrm{Z} \times \mathrm{M}_{w}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}$

Here, $\rho$ is the density of unit cell, Z is the number of atoms present in a unit cell, $\mathrm{M}_{w}$ is the molecular weight, $a$ is the edge length of a unit cell and $\mathrm{N}_{\mathrm{A}}$ is the Avogadro's number.

On rearranging this formula,

$\mathrm{Z}=\frac{\rho \times a^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{M}_{w}}$ ...... (i)

Substituting the respective values in equation (i),

$Z=\frac{2 \times\left(5 \times 10^{-8}\right)^{3} \times\left(6 \times 10^{23}\right)}{75}$

$\therefore \quad \mathrm{Z}=1.98 \approx 2$

The value of Z is 2 for BCC (body centered cubic) unit cell.

Step 1: Determination of radius of metal atom.

For $\mathrm{BCC}$, the radius of atom is given by,

$4 r=\sqrt{3} a$

Here, $r$ denotes the radius of the atom.

$\begin{array}{ll} \therefore & r=\frac{\sqrt{3}}{4} a \\ \therefore & r=\frac{\sqrt{3}}{4} \times 5=2.165 ~\mathop A\limits^o \\ \therefore & r=216.5 ~\mathrm{pm} \end{array}$

The radius of metal atom is $216.5 ~\mathrm{pm}$.

Types of Voids in FCC:

In a face-centered cubic (FCC) structure, there are two types of empty spaces, called voids: tetrahedral voids and octahedral voids.

Ratio for Void Sizes:

For a tetrahedral void: $\frac{r}{R}=0.225$

For an octahedral void: $\frac{r}{R}=0.414$

Here, $r$ is the radius of the smaller atom that goes into the empty site (the interstitial atom), and $R$ is the radius of the atom making up the structure.

Choosing the Correct Void:

The largest atom that can fit into the FCC structure without changing its shape fits in the octahedral void, because this void is bigger than the tetrahedral one. So, we use the octahedral void ratio: $\frac{r}{R}=0.414$

Finding the Radius ($R$) of the Main Atom:

In FCC, the relation between the radius $R$ and the edge length $a$ is: $R=\frac{a}{2\sqrt{2}}$

The question says $a = 400$ pm.

Calculate $R$:

$R=\frac{400}{2\sqrt{2}}$

Find $r$ (radius of the maximum atom that fits in an octahedral site):

Substitute the value of $R$ into the octahedral void ratio:

$r=0.414\times\frac{400}{2\sqrt{2}}=58.55~\text{pm}$

Calculate Diameter:

The diameter of the atom is twice the radius:

Diameter = $2r = 2 \times 58.55 = 117.1~\text{pm}$

Final Answer:

The largest atom that can fit into the empty site in an FCC lattice (without changing the structure) has a diameter of 117.1 pm.

Note:

Only the octahedral void is considered for calculating the maximum possible diameter, because it is the largest void in FCC structures.

AB has a rock salt (NaCl) structure. This type of crystal structure possesses fcc unit cell and contains four formula units per unit cell, i.e., Z = 4.

In case of a rock salt structure, the edge-length (a) of the unit cell = 2 $\times$ (radius of cation + radius of anion)

Therefore, the edge-length (a) of the unit cell of AB crystal = 2 $\times$ Y^1/3 nm = 2Y^1/3 $\times$ 10^$-$9 m.

We know, $\rho = {{Z \times M} \over {N \times {a^3}}}$

Given : M = 6.022 Y g mol^$-$1 = 6.022 $\times$ 10^$-$3 Y kg mol^$-$1

$\therefore$ $\rho = {{4 \times 6.022 \times {{10}^{ - 3}}Y} \over {6.022 \times {{10}^{23}} \times {{(2{Y^{1/3}} \times {{10}^{ - 9}})}^3}}} = 5.0$ kg m^$-$3

(1) Density of the crystal = 5.0 kg m^$-$3

(2) The observed density (= 20 kg m^$-$3) is higher than that of the calculated density. This indicates that the crystal structure of AB is likely to have non-stoichiometric defect in the form of metal excess or metal deficiency defect or to have impurity defect in the form of substitutional impurity defect or interstitial impurity defect.