Answer the following in brief.

Question 13 Marks

A compound forms hexagonal close packed (hcp) structure. What is the number of (i) Octahedral voids (il) Tetrahedral voids (iii) Total voids formed in 0.4 mol of it?

Answer

In the close packing of spheres, certain hollows left vacant, these holes in the crystal s is called interstitial voids.
There are two types of voids : octahedral voids and tetrahedral voids. Therefore, the total number of voids is equal to sum of octahedral voids and tetrahedral voids.
Number of octahedral voids is equal to number of atoms (lattice points) present in the crystal.
Number of tetrahedral voids is equal to twice of number of number of atoms.

We have given 0.4 moles of compound.
We know that number of atoms in 1 mole = 6.023×10²³ atoms
Then number of atoms in 0.4 moles = 0.4×6.023×10²³
= 2.41 × 10²³ atoms
(a) Number of octahedral voids = Number of atoms in 0.4mol of compound
Number of octahedral voids = 2.41 × 10²³
(b) Number of tetrahedral voids = 2× Number of atoms in 0.4mol crystal
= 2× 2.41 × 10²³
= 4.82 × 10²³
(c) Total number of voids = octahedral voids + tetrahedral voids
Total voids = 2.41 × 10²³ + 4.82 × 10²³
Total voids = 7.23 × 10²³ Voids.

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Question 23 Marks

Silver crystallises in $\text{F.C.C.} ($face$-$centered cubic crystal$)$ structure. The edge length of the unit cell is found to be $408.7 \ pm.$ Calculate density of the unit cell. $[$Given: Molar mass of silver is $108 \ g \ mol^{-1}].$

Answer

Given: Edge length $(a) = 408.7 \ pm = 408.7 \times 10^{–12} m = 408.7 \times 10^{-10} \ cm,$
Molar mass/Atomic mass of silver $= 108 \ g \ mol^{-1}$
To find: Density $(d)$
Formulae:
$1.$ Mass of one atom $=\frac{\text { Atomic mass }}{\text { Avogardo Number }}$
$2.$ Volumeofunitcell $=a^3$
$3.$ Density $=\frac{\text { mass of unit cell }}{\text { Volume of unit cell }}$
Calculation: For $fcc$ lattice, number of atoms per unit cell is $4.$
Mass of one atom of silver $=\frac{\text { Atomic mass }}{\text { Avogadro number }}$
$=\frac{108}{6.023 \times 10^{23}}=17.9 \times 10^{-23} g$
Mass of unit cell $=4 \times 17.9 \times 10^{-23}=71.7 \times 10^{-23} g$
Volume of unit cell $=a^3=\left(408.7 \times 10^{-10} \ cm \right)^3=6.827 \times 10^{-23} \ cm ^3$
Density $=\frac{\text { mass of unit cell }}{\text { Volume of unit cell }}$
$=\frac{71.7 \times 10^{-23} g}{6.827 \times 10^{-23} \ cm ^3}=10.5 g$

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Question 33 Marks

A unit cell of iron crystal has edge length $288 \ pm$ and density $7.86 g \ cm^{-3} .$ Find the number of atoms per unit cell and type of the crystal lattice. $[$Given: Molar mass of iron $=56 g \ mol ^{-1}$; Avogadro's Number, $N_A = 6.022 \times 10^{23} .]$

Answer

Edge length of unit cell $(a) = 288 \ pm$
volume of unit cell $= (a)^3$
$=(288)^3 = 2.389 \times 10^{-23} \ cm^3$
density of iron $= \rho = 7.86g \ cm^3$
Mass of iron unit cell $(M) =$ Density $\times$ Volume
$= 7.86 \times 2.389 \times 10^{-23}$
$= 18.778 \times 10^{-23} g$
molar mass of iron $= 56 \ g \ mol^{-1}$
moles of iron in a unit cell $=$ mass of iron in unit cell / molar mass of iron
$= 18.778 \times 10^{-23} / 56$
$= 3.353 \times 10^{-24}$ molnumber of atoms per unit cell $=$ moles of iron in a unit cell $\times$ Avogadro's number
$= 3.353 \times 10^{-24} \times 6.022 \times 10^{23}$
$= 2.0191$
$≈ 2$ atoms per unit cell
In a body centered cubic structure $(bcc),$ the total number of atoms equals $2.$

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Question 43 Marks

Calculate the number of atoms present in $2$ gram of crystal which has face—centered cubic $(fcc)$ crystal lattice having edge length of $100 \ pm$ and density $10 \ g \ cm^{-3}$

Answer

Given: Density $(d) = 10 \ g \ cm^{−3}$
Edge length $(a) = 100 \ pm = 100 \times 10^{−10} \ cm.$
Mass of crystal $= 2 g$
To find: Number of atoms
Formula: Density $=$ Mass/Volume
Calculation: Density $=$ Mass/Volume
$\therefore$ Volume $=$ Mass/Density
$\therefore$ Volume $=\frac{2 g }{10 gcm ^{-3}}=0.2 \ cm ^3$
Volume of unti cell $= a^3 = (100 \times 10^{–10} \ cm)^3 = 1 \times 10^{−24} \ cm^3$
Number of unit cells in $2 g$ of crystal $=$ total volume/volume of unit cell
$\frac{(0.2 \ cm )^3}{1 \times 10^{-24} \ cm ^4}$
The given unit cell is of $fcc$ type, therefore, it contains $4$ atoms.
$0.2 \times 10^{24}$ unit cells will contain $4 \times 0.2 \times 10^{24} = 0.8 \times 10^{24}$ atoms
$= 8 \times 10^{23}$ atoms
Number of atoms present in $2g$ of crystal is $8 \times 10^{23}$ atoms

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Question 53 Marks

Unit cell of a metal has edge length of $288 \ pm$ and density of $7.86 g \ cm^{-3}.$ Determine the type of crystal lattice.$($ Atomic mass of metal $= 56 g \ mol^{-1})$

Answer

bcc

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Question 63 Marks

Define Anisotropy. Distinguish between crystalline solids and amorphous solids.

Answer

The ability of crystalline solids to change values of physical properties when measured in different directions is called anisotropy.

	Crystalline solid	Amorphous solid
(i)	They have a definite geometrical shape	They have an irregular shape
(ii)	They have a sharp melting point	They melt over a range of temperature
(iii)	They are anisotropic	They are isotropic
(iv)	They are pure solid	They are supercooled liquid.
(v)	They have long-range order of regular pattern of arrangement of constituent particles.	They have short-range order of regular pattern of arrangement of constituent particles.
(vi)	They have definite heat of fusion.	They do not have a definite heat of fusion.

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Question 73 Marks

The density of iron crystal is $8.54 \ g \ cm^{-3}.$ If the edge length of unit cell is $2.8 \mathring A$ and atomic mass is $56 \ g \ mol^{-1} ,$ find the number of atoms in the unit cell. $[$Given: Avogadro's number $= 6.022 \times 10^{23}, 1 \mathring A = 1 \times 10^{-8} \ cm]$

Answer

Given $a =2.8 \mathring A =2.8 \times 10^{-8} \ cm$
Molar Mass $= 56 \ gm$
Density $=8.54 \ g \ cm ^{-3}$
Volume $=(a)^3$
$=\left(2.8 \times 10^{-8}\right)^3$
$=21.95 \times 10^{-24} \ cm ^3$
Density of unit cell $=\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}$
$\therefore$ Mass of unit cell = density $\times$ volume
$=21.95 \times 10^{-24} \times 8.54$
$=187.47 \times 10^{-24}$
$56 \ gm$ of $Fe$ contains $6.022 x 10^{23}$ atoms
$187.47 \times 10^{-24} \ gm \ Fe$ contain $=\frac{6.022 \times 10^{23} \times 187.47 \times 10^{-24}}{56}$
$=20.15 \times 10^{-1}$
$= 2.01$
$= 2$

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Question 83 Marks

Silver crystallises in $\text{FCC}$ structure. If density of silver is $10.51 \ g \ cm^{-3},$ calculate the volume of unit cell.
$[$Atomic mass of silver $(Ag) = 108 \ g \ mol^{-1} ]$

Answer

Density of $Ag = 10.51 \ g/cm^3$
Vol. of unit cell $=$ ?
Mass of one atom of silver
$=\frac{\text { molar mass of silver }}{ N _{ A }}$
$=\frac{108}{6.022 \times 10^{23}}$
$=17.93 \times 10^{-23}$
Mass of unit cell of silver $=$ Atoms in unit cell $\times$ Mass of $1$ atom
$=4 \times 17.93 \times 10^{-23}$
$=71.72 \times 10^{-23}$
$\therefore$ Density of $Ag =\frac{\text { mass of unit cell }}{\text { Vol.of unit cell }}$
$\therefore$ Vol of unit cell $=\frac{\text { mass of unit cell }}{\text { density of } Ag }$
$=\frac{71.72 \times 10^{-23}}{10.51}$
col.of.unit cell $= 68.24 \times 10^{-23} \ cm^3$

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Question 93 Marks

Niobium crystallises as body centered cube $( BCC )$ and has density of $8.55 kg / dm ^{-3}$. Calculate the atomic radius of niobium. (Given: Atomic mass of niobium =93).

Answer

Crystal structure of Niobium = bcc
Density (d) = 8.55 kg dm-3
Atomic mass of Niobium = 93
$d=\frac{n \times M}{V \times N_A}$
No. of atoms per unit cell (n) in bcc = 2
Total volume of unit cell = a3
$d=\frac{2 \times 93}{a^3 \times 6.022 \times 10^{23}}$
$a^3=\frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}}=36.12 \times 10^{-24}$
$a=\sqrt[3]{36.12 \times 10^{-24}}$
a= 3.305 x 10-8
In BCC unit cell,
$r=\frac{\sqrt{3} \times a}{4}=\frac{\sqrt{3} \times 3.305 \times 10^{-8}}{4}=1.431 \times 10^{-8}$

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Question 103 Marks

Face centered cubic crystal lattice of copper has density of $8.966 \ g \ cm^{-3}$. Calculate the volume of the unit cell. Given molar mass of copper is $63.5 \ g \ mol^{-1}$ and Avogadro number $N_A$ is $6.022 \times 10^{23}.$

Answer

Data: $d =8.966 g \ cm ^{-1}$
$M.M.$ of copper $=63.5 \ g \ mol ^{-1}$
$N _{ A }=6.022 XX10^{23 }$
Solution : $d=\frac{Z \times M}{V \times N_A}$
$Z=4(\because F C C)$
$V=\frac{Z \times M}{d \times N_A}$
$=\frac{4 \times 63.5}{8.966 \times 6.023 \times 10^{23}}$
$V=4.704 \times 10^{-23} \ cm ^3$

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Question 113 Marks

Classify the following solids: $(1)$ Diamond $(ii) \ NaCl \ \ (iii) \ P_4$ molecule $(iv)$ Brass .
What is Schottky defect?

Answer

$(a) \ P4 -$ Molecular solid $(b)$ Brass $-$ Metallic solid $(c)$ Diamond $-$ Covalent solid $(d) \ NaCl -$ Ionic solid
Schottky defect is defined as the one in which equal number of cations and anions are missing from their lattice positions in an ionic compound.

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Question 123 Marks

The density of silver having atomic mass $107.8 \ g \ mol^{-1}$ is $10.8 g \ cm^{-3}.$ If the edge length of cubic unit cell is $4.05 \times 10^{-8} \ cm,$ find the number of silver atoms in the unit cell.
$(N_A =6.022 \times 10^{23}, 1 \mathring A =10^{-8} \ cm)$

Answer

Given:
Density $(d) = 10.8 g \ cm^{-3}$
Edge length $(a) = 4.05 \times 10^{- 8} \ cm$
Molar mass $= 107.8 \ g \ mol^{-1}$
Avogadro's number $(N_A) = 6.022 \times 10^{23}$
To find:
Number of atoms in the unit cell
Formula:
$a.$ Mass of one atom $=\frac{\text { Atomic mass }}{\text { Avogadro number }}$
$b.$ Volume of unit cell $=a^3$
$c.$ Density $=\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}$
Calculation:
$a)$ Mass of one $Ag$ atom $=\frac{\text { Atomic mass of } Ag }{\text { Avogadro number }}$
Avogadro number
$=\frac{107.8}{6.022 \times 10^{23}}$
$=1.79 \times 10^{-22} g$
$b)$ Volume of unit cell $=a^3$
$=\left(4.05 \times 10^{-8}\right)^3$
$=6.64 \times 10^{-23} \ cm ^3$
$c)$ Density $( d )=\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}$
$=\frac{\text { Number of atoms in unit cell } \times \text { Mass of one atom }}{\text { Volume of unit cell }}$
$10.8=\frac{\text { Number of atoms in unit cell } \times 1.79 \times 10^{-22}}{6.64 \times 10^{-23}}$
Number of atoms in unit cell $=\frac{10.8 \times 6.64 \times 10^{-23}}{1.79 \times 10^{-22}}$
$=40.06 \times 10^{-1}=4.0 \approx 4$
$\therefore$ The number of atoms in the unit cell of silver is $4.$

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Question 133 Marks

A metal crystallises into two cubic faces namely face centered (FCC) and body centered (BCC), whose unit cell edge lengths are 3.5 Å and 3.0 Å respectively. Find the ratio of the densities of FCC and BCC.

Answer

Edge length of FCC unit cell (a1) = 3.5 Å
Edge length of BCC unit cell (a2) = 3.0 Å
Density of unit cell $=d=\frac{Z \times M}{N_A \times a^3} gcm ^{-3}$
Density of FCC unit cell $=d_1=\frac{4 \times M}{N_A \times(3.5 A)^3}$
Density of BCC unit cell $=d_2=\frac{2 \times M}{N_A \times(3.0 A)^3}$
Ratio of densities of FCC and BCC unit cell is,
$\frac{d_1}{d_2}=\frac{4 \times M}{N_A \times(3.5 A)^3} \times \frac{N_A \times(3.0 A)^3}{2 \times M}$
$\frac{d_1}{d_2}=\frac{54}{42.875}=1.259 \approx 1.26$

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Question 143 Marks

Determine the density of Cesium chloride which crystallizes in $\text{BCC}$ type structure with the edge length $412. 1 \ pm.$ The atomic masses of $Cs$ and $C1$ are $133$ and $35.5$ respectively. Predict the co$-$ordination number of $Cs^+$ ion if $r_{cs+} = 1.69 A$ and $= r_{CI}-= 1.81 A.$

Answer

Given: Edge length $(a) = 412.1 \ pm = 4.12 x 10^{-8} \ cm$
Molar mass $= 133 + 35.5 = 168.5 \ g \ mol^{-1}$
To find: Density $(d)$
Formulae: Mass of one molec $=$ Molar mass/Avogadro number
Volume of unit cell $= a^3$
Density $=$ Massof unit cell/Volumeof unit cell
Calculation: In the bcc type unit cell of $CsCl,$ there is one $Cs^+$ ion at the body centre
position and $8 Cl^-$ ions are at the $8$ corners
$\therefore$ Number of $Cs^+$ in unit cell $= 1$
Number of $Cl^-$ in unit cell $= 1/8 \times 8 = 1$
Hence, the unit cell contains one $CsCl$ molecule
Mass of one $CsCl$ molecule $=$ Molar mass/Avogadro number $= 168.5 \ g \ mol^{-1}/6.023 \times 10^{23} mol^{-1}$
$= 2.798 \times 10^{-22} g$
$\therefore$ Mass of unit cell $= 1 \times 2.798 \times 10^{-22}g = 2.798 \times 10^{-22}g$
Volume of unit cell $= a^3 = (4.12 \times 10^{–8}cm)^3= 6.993 \times 10^{-23}cm^3$
$\therefore$ Density $=$ Massof unit cell/Volumeof unit cell $= 2.798 10^{-22}/6.993 \times 10^{-23} \ cm^3 = 4.0 g \ cm^{–3}$
The radius ratio $(r^+/r^–)$ defines the coordination number of the cation
$r_{C s^{+}}=1.69 \mathring A , r_{C l^{-}}=1.81 \mathring A $
$\therefore \frac{r_{C s^{+}}}{r_{C l^{-}}}=\frac{1.69}{1.89}=0.9337$
Since, radius ratio is greater than $0.732,$ the coordination number of cation $(Cs^+)$ is $8.$

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Chemistry Answer the following in brief.

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