Learning Objectives

After studying this unit, you will be able to

• Differentiate between amorphous and crystalline solids

• Explain the different types of crystalline solids based on the attractive forces between the particles

• Differentiate between unit cell, crystal lattice and Bravais Lattices

• Explain the close packing pattern in the solid state

• Describe different types of voids present in the close packed structures

• Calculate the packing efficiency of different types of cubic unit cells

• Find the relation between the density of the substance and characteristics of its unit cell

• Describe the imperfections present in the solids and explain their effect on some properties

• Explain the range of electrical and magnetic properties exhibited by the solids

• Explain the types of doping and their effect on the electrical properties

Solid State

• One of the state of matter characterized by the rigidity of the constituent particles

• Constituent particles have fixed positions and they can only oscillate about their mean position.

• Constituent particles arranged in a regular pattern, the spatial arrangement is completely ordered (crystalline solid) unlike gaseous or liquid state.

• Because of the highly ordered arrangement, solids have lower potential energy, so energy is required to convert solid to a liquid

• Properties of the solids can be changed by adding some impurities or due to some structural imperfections or defects.

Classification of Crystalline solids

On the basis of nature of intermolecular forces

a) Ionic solids : - constituent particles are ions

• coulombic or electrostatic forces exist between the ions

• Insulators in solid state but conductors in molten state and in aqueous solutions

• e.g. NaCl, $\mathrm{MgO}$

b) Metallic Solids : - Constituent particles are positive ions immersed in a sea of delocalized electrons.

• Operative bonding between the ions is metallic

• Conductors in solid state as well as in molten state

• e.g. $\mathrm{Cu}, \mathrm{Fe}, \mathrm{Mg}$.

c) Covalent Solids or Network Solids :

• Constituent particles are atoms.

• Covalent bonding between the atoms

• Insulators, graphite is an exception

• e.g. quartz, diamond

d) Molecular Solids: - Constituent particles are molecules

(3 types) $~~$ i). Non-polar: Dispersion or London forces e.g. $\mathrm{Ar} _{2}, \mathrm{CCl} _{4}$

$\hspace{1.5cm}$ii). Polar : Dipole-Dipole interactions between molecules e.g. HCl, $\mathrm{SO} _{2}$

$\hspace{1.5cm}$iii). Hydrogen bonded: Hydrogen bonding exist e.g. $\mathrm{H} _{2} \mathrm{O}$ (ice)

$\hspace{1.5cm}$- Insulators (all types)

$\rightarrow$ Bragg’s Law - to determine the crystal structure by X-ray diffraction

$$ \mathrm{n} \lambda=2 \mathrm{~d} \operatorname{Sin} \theta $$

$\mathrm{n}=$ order of diffraction, can be 1, 2, $3 \ldots$… (No Unit, it’s a number)

$\lambda=$ wavelength of $x$-ray

$d=$ distance between 2 planes ( $\lambda$ and $d$ both have same units)

$\theta=$ angle of diffraction

$\rightarrow$ Crystal Lattice:

Regular arrangement of constituent particles of a crystal in a three dimensional space.

$\rightarrow$ Unit cell:

smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.

Six Parameters of unit cell:

3 - Edge lengths $\quad\quad \quad \qquad$ - $\quad a, b, c$ (may or may not be mutually perpendicular)

3 - Angles between the edges - $\quad \alpha$ (between $b$ and $c$ ), $\beta$ (between $a$ and $c$ ), $\gamma$ (between $a$ and $b$ )

$\rightarrow$ Types of Unit Cells :

Primitive simple unit cells (constituent particles are present only at corners)

Non-Primitive/centred unit cells

i). Body-centred - particles are present at the corners and body centre

ii). Face - Centred - particles are present at the corners and centre of each face.

iii). End centred - particles are present at the corners and centre of any two opposite faces.

Note: constituent particle may be an atom, a molecule (group of atoms) or an ion.

$\rightarrow$ Crystal systems and Bravais Lattice

There are seven crystal systems and fourteen Bravais Lattice (Unit cells) corresponding to these seven crystal systems.

Number of atoms in different unit cells:

Contribution of particles at

Corner $=1 / 8$, Face centre $=1 / 2$, Body Centre $=1$,

Edge Centre $=1 / 4$

Coordination Number (C.No.): Number of nearest neighbours of a particle with which it is in contact with.

Packing of constituents in a crystal:

a). Close packing in 1-D - One type of arrangement is possible i.e. in a row C. No. $=2$.

b). Close packing in 2-D

i). Square close packing

• AAA type of arrangement

• $52.4 \%$ of available space occupied by spheres.

• C.No. $=4$

ii). Hexagonal close packing

• ABAB type of arrangement

• $60.4 \%$ of available space occupied by spheres

• C.No. $=6$

c). Close packing in 3-D

i) Hexagonal Close packed (hcp) - ABAB……. type of arrangement

ii) Cubic Close Packed (сcp) or face-centred cubic (fcc):

• ABCABC type of arrangement
• In both of these packings (hcp and ccp/fcc), $74 \%$ of the space is occupied.
• C.No. $=12$ (for hcp and ccp/fcc)

Note : In bcc, C. No. = 8

$\rightarrow$ Voids are Interstitial Sites :

Empty spaces left in the close packed structures.

a) In 2-D Packing : Trigonal voids are present

b) In 3-D packing : Two types of voids

i) Tetrahedral void - Simple triangular void surrounded by 4 spheres.

C.No. $=4$

ii) Octahedral void - Double triangular void surrounded by six spheres

C.No. $=6$

$\rightarrow$ Relation between the number of voids and the number of spheres in close packing

In close packing the number of spheres $=\mathrm{N}$ spheres

Number of octahedral voids $=\mathrm{N}$

Number of tetrahedral voids $=2 \mathrm{~N}$.

$\rightarrow$ Total voids in ccp per unit cell $=4$ (octahedral) +8 (tetrahedral)$=12$

$\rightarrow$ Total voids in hcp per unit cell $=6$ (octahedral) +12 (tetrahedral)$=18$

$\rightarrow$ Relation between radius ( $r$ ) of a void and the radius (R) of the sphere in close packing:

Tetrahedral void : $r=0.225 \mathrm{R}$

Octahedral void : $r=0.414 \mathrm{R}$

Packing Fraction

Fraction of total space filled by the constituent particles.

Packing fraction $=\dfrac{\text { Total Volume of spheres }}{\text { Volume of the unit cell }}$

$r=$ atomic radius; $a=$ edge length of a unit cell

$\rightarrow$ Density of the crystal:

$d=\dfrac{Z \times M}{a^{3} \times N _{A}} \mathrm{~g} / \mathrm{cm}^{3}$

$z=$ Number of particles present per unit cell

$M=$ Molar mass of element $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$

$a=$ Edge length of the unit cell $(\mathrm{cm})$

$N _{A}=$ Avogadro number

For lonic crystals, the formula used is the same.

$z=$ number of formula units in one unit cell

$M=$ Formula mass of the compound

$a=$ edge length which is $2 x$ distance between the ions in a crystal

Radius Ratio $=\dfrac{\text { radius of cation }(\mathrm{r}+)}{\text { radius of anion }(\mathrm{r}-)}$

$\rightarrow \quad$ Structure of some simple ionic solids:

Effect of temperature and pressure on crystal structure

• In $\mathrm{NaCl}$, there are 4 octahedral voids, all of which are occupied by $\mathrm{Na}^{+}$.

• In ZnS, there are 8 tetrahedral voids, half of which are occupied by $\mathrm{Zn}^{2+}$.

• In $\mathrm{CaF} _{2}$, there are 8 tetrahedral voids, all are occupied by F ions.

$\rightarrow$ Imperfections in solids

• Irregularities in the arrangement of constituent particles

$\rightarrow$ Types of Defects

(a) Line defects / crystal defects: deviations from the ideal arrangement in entire rows of lattice points.

(b) Point defects : deviation from ideal arrangement around a point or an atom in a crystalline substance.

$\rightarrow$ Types of point defects:

(a) Stoichiometric Defects

• These defects are also known as Intrinsic or thermodynamic defects.

• In this defect ratio between the cations and anions (stoichiometry) remain the same

i) Vacancy defect - When some of the lattice sites are vacant; density decreases in this defect.

ii) Interstitial defect - When some constituent particles occupy an interstitial site.

• Density increases in this defect

• Electrical Neutrality should be maintained

For Ionic Solids

$\rightarrow$ Stoichiometric defects in lonic Solids

iii) Frenkel defect / Dislocation defect:

• Cations are missing from lattice site and occupy interstitial sites.

• Occurs in compounds with low coordination number and large difference in size of cation and anions.

• Density remains the same

• e.g. Agl, $\mathrm{AgCl}, \mathrm{ZnS}$

• $\mathrm{AgBr}$ shows both Schottky as well as Frenkel defect.

iv) Schottky defect -

• Equal number of cations and anions are missing

• Occurs in compounds having high coordination number and small difference in size of cations and anions.

• Density decreases e.g. NaCl, $\mathrm{KCl}$
  
  

(b) Non stoichiometric defect: In this defect stoichiometriy does not remain the same

Two types

i) Metal Excess Defect

• Due to anionic vacancies - When crystals of alkali halides are heated in the vapour of alkali metal, some anions leave lattice sites in which electrons get trapped forming $\mathrm{F}$ - centre which impart colour to the crystals.

$\mathrm{NaCl}$ Crystals - Yellow colour

LiCI Crystals - Pink Colour

KCI Crystals-Violet/Lilac Colour

(Due to the excess of alkali (Na, Li & K) metal)

• Due to the presence of extra cations at interstitial Sites:

$\mathrm{ZnO} \xrightarrow{\Delta} \mathrm{Zn}^{2+} \quad+\dfrac{1}{2} \mathrm{O} _{2}+2 \mathrm{e}$

White Yellow

$\mathrm{Zn}^{2+}$ ions and electrons move to interstitial sites.

ii) Metal Deficiency Defect - occurs when metal shows variable valency.

Few metal ions with lower valency are missing from their site but metal ions with higher valency are present at some sites thus balancing the charge. However, this leaves some vacancies at the sites of metal ions.

e.g. $\mathrm{FeO}$ exist as $\mathrm{Fe} _{0.95}$

(c) Impurity defect:

When foreign atoms are present at the lattice site in place of host atoms or at the vacant interstitial sites e.g. $\mathrm{SrCl} _{2}$ (little amount) is present in $\mathrm{NaCl}$ or Solid solution of $\mathrm{CdCl} _{2}$ (impurity) and $\mathrm{AgCl}$

• In this case cationic vacancies are produced which are equal to the number of $\mathrm{Sr}^{2+}$ or $\mathrm{Cd}^{2+}$ ions.

Electrical Properties - Solids shows a large range of electrical conductivities, $10^{-20}$ to $10^{7} \mathrm{ohm}^{-1} \mathrm{~m}^{-1}$

• Conductors - Valence band is partially filled or it overlaps with the conduction band, movement of electrons is easy. Conductivity is between $10^{4}$ to $10^{7} \mathrm{ohm}^{-1} \mathrm{~m}^{-1}$.

Conductivity decreases with the rise in temperature e.g. metals.

• Semiconductors -

  • There is small enery gap between the valence and conduction band, some electrons may jump to the conduction band. Conductivity range $=10^{-6}$ to $10^{4}$ $\mathrm{ohm}^{-1} \mathrm{~m}^{-1}$

  • Conductivity increases with rise in temperature e.g. $\mathrm{Si}$ and $\mathrm{Ge}$. (Known as intrinsic semiconductors).

• Insulators

  • There is large energy gap between the valence and conduction band, electrons cannot jump to the conduction band.

  • Conductivity range $=10^{-20}$ to $10^{-10} \mathrm{ohm}^{-1} \mathrm{~m}^{-1}$ e.g. diamond, Quartz

Doping

• Process of adding impurities to a crystalline substance to change its properties.

(a) n-types semiconductor - doping of group 14 elements (Si, Ge) with group 15 elements like P or As. Free electrons are present, responsible for increased conductivity.

(b) p-type semiconductor - Doping of group 14 elements with group 13 elements like B, Al & Ga.

Electron holes are created which are responsible for increased conductivity.

$\rightarrow$ Compounds of Groups 13-15

• Act like semiconductors

• These are formed by combination of group 13 and 15 elements.

• e.g. InSb, AIP and GaAs

$\rightarrow$ Compounds of groups 12-16

• Act like semiconductors.

• These are formed by combination of group 12 and 16 elements.

• e.g. ZnS, CdS, CdSe and $\mathrm{HgTe}$

$\rightarrow$ Transition metal Oxides

(Show marked differences in electrical properties).

• $\mathrm{TiO}, \mathrm{CrO} _{2}$ and $\mathrm{ReO} _{3}$ - behave like metals
  

Magnetic Properties

1. Diamagnetic substances - Weakly repelled by magnetic field. They do not have unpaired electrons e.g. $\mathrm{TiO} _{2}, \mathrm{H} _{2} \mathrm{O}, \mathrm{NaCl}$.

2. Paramagnetic Substances - Attracted by magnetic field. They contain unpaired electrons e.g. $\mathrm{O} _{2}, \mathrm{Cu}^{2+}, \mathrm{Fe}^{3+}$. They lose their magnetism when magnetic field is removed.

3. Ferromagnetic substances - show permanent magnetism even in the absence of magnetic field. All domains are aligned in one direction eg. $\mathrm{Fe}, \mathrm{Co}, \mathrm{Ni}, \mathrm{CrO} _{2}$

4. Anti-ferromagnetic substances - Domains are aligned in parallel and anti parallel directions in equal numbers. Net magnetic moment is zero e.g. MnO.

5. Ferrimagnetic substances - Domains are aligned in parallel and antiparallel directions in unequal numbers. These are weakly attracted by magnetic field. So, small net magnetic moment. e.g. $\mathrm{Fe} _{3} \mathrm{O} _{4}, \mathrm{MgFe} _{2} \mathrm{O} _{4}$ and $\mathrm{ZnFe} _{2} \mathrm{O} _{4}$.

Solved Examples

Question 1. Experimentally it was found that a metal oxide has formula $\mathrm{M} _{0.98} \mathrm{O} _{1}$. $\mathrm{M}$ is present as $\mathrm{M}^{2+}$ and $\mathrm{M}^{3+}$ in its oxide. Fraction of metal which exists as $\mathrm{M}^{3+}$ would be

(a) $7.01 \%$

(b) $4.08 \%$

(c) $6.05 \%$

(d) $5.08 \%$

Question 2. If $\mathrm{NaCl}$ is doped with $10^{-4} \mathrm{~mol} \% \mathrm{SrCl} _{2}$ then concentration of cation vacancies will be

(a) $6.022 \times 10^{14} \mathrm{~mol}^{-1}$

(b) $6.022 \times 10^{15} \mathrm{~mol}^{-1}$

(c) $6.022 \times 10^{6} \mathrm{~mol}^{-1}$

(d) $6.022 \times 10^{17} \mathrm{~mol}^{-1}$

(a) $5.14 \times 10^{21}$

(b) $1.28 \times 10^{21}$

(c) $1.771 \times 10^{21}$

(d) $2.57 \times 10^{21}$

(a) $39.27 \%$

(b) $68.02 \%$

(c) $74.05 \%$

(d) $78.54 \%$

Show Answer

Solution (d).

Contact between the spheres is along the diagonal.

$4 R=\sqrt{ } 21$

$\mathrm{I}=2 \sqrt{ } 2 \mathrm{R}$

Area of Square $=I^{2}=(2 \sqrt{2} R)^{2}=8 R^{2}$

Area of atoms present in 1 unit cell $=\pi \mathrm{R}^{2}+4 \times\left(\dfrac{\pi \mathrm{R}^{2}}{4}\right)$

$$ \begin{aligned} & =2 \pi R^{2} \\ \% \text{of packing efficiency} & =\dfrac{2 \pi R^{2}}{8 R^{2}} \times 100 \\ & =78.54 \% \end{aligned} $$

Note : There are 2 atoms or effectively 2 circle in one unit cell.

PRACTICE QUESTIONS

Question 1. Structure of a mixed oxide is ccp. Cubic unit cell of mixed oxide is composed of oxide ions.

One Fourth of tetrahedral voids are occupied by divalent metal A and all the octahedral voids are occupied by monovalent metal $B$. Formula of oxide is

(a) $\mathrm{A} _{2} \mathrm{~B} _{3} \mathrm{O} _{4}$

(b) $\mathrm{AB} _{2} \mathrm{O} _{2}$

(c) $\mathrm{ABO} _{2}$

(d) $\mathrm{A} _{2} \mathrm{BO} _{2}$

Question 2. In face centred cubic lattice, A occupies corners and B occupies face centres. If one atoms of $B$ is missing from one of the face centre, formula of the compound is

(a) $\mathrm{AB} _{2}$

(b) $\mathrm{A} _{2} \mathrm{~B} _{2}$

(c) $\mathrm{A} _{2} \mathrm{~B} _{5}$

(d) $\mathrm{A} _{2} \mathrm{~B}$

Question 3. No. of tetrahedral voids in the unit cell of a face centred cubic lattice of similar atoms is

(a) 4

(b) 6

(c) 8

(d) 12

Question 4. In a solid ’ $A B$ ’ having the $\mathrm{NaCl}$ structure, ‘A’ atoms occupy the corners of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

(a) $\mathrm{AB} _{2}$

(b) $\mathrm{A} _{2} \mathrm{~B}$

(c) $A _{4} B _{3}$

(d) $\mathrm{A} _{3} \mathrm{~B} _{4}$

Question 5. A substance $A _{x} B _{y}$ crystallizes in a face centred cubic (fcc) lattice in which atoms ’ $A$ ’ occupy each corner of the cube and atoms ’ $B$ ’ occupy the centres of each face of the cube. Identify the correct composition of the substance $A _{x} B _{y}$

(a) $\mathrm{AB} _{3}$

(b) $\mathrm{A} _{4} \mathrm{~B} _{3}$

(c) $\mathrm{A} _{3} \mathrm{~B}$

(d) composition cannot be specified

Question 6. An ionic crystalline solid $\mathrm{MX} _{2}$, has a cubic unit cell. Which of the following arrangement of ions is consistent with the stoichiometry of the compound?

(a) $\mathrm{M}^{2+}$ ions at the corners and at each face centres $8 \mathrm{X}^{-}$ions in the tetrahedral holes.

(b) $\mathrm{M}^{2+}$ ions at the each face centres $8 \mathrm{X}$ ions at the corners

(c) $\mathrm{M}^{2+}$ ions at the corners and $\mathrm{X}$ at each face centres $2 \mathrm{X}$ ions completely inside the unit cell.

(d) $\mathrm{M}^{2+}$ ions at the corners and at each face centres, $4 \mathrm{X}$ ions completely inside the unit cell.

Question 7. Which of the following fcc structure contains cations in alternate tetrahedral voids?

(a) $\mathrm{NaCl}$

(b) ZnS

(c) $\mathrm{Na} _{2} \mathrm{O}$

(d) $\mathrm{CaF} _{2}$

Question 8. A compound $M _{n} X _{a}$ has cubic close packing ( $(c c p$ ) arrangement of $X$. Its unit cell structure is shown below. The empirical formula of the formula of the compound, is

(a) $M X$

(b) $\mathrm{MX} _{2}$

(c) $\mathrm{M} _{2} \mathrm{X}$

(d) $\mathrm{M} _{5} \mathrm{X} _{14}$

Question 9. If the positions of $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$are interchanged in $\mathrm{NaCl}$, having fcc arrangement of $\mathrm{Cl}^{-}$ions then in the unit cell of $\mathrm{NaCl}$

(a) $\mathrm{Na}^{+}$ions will decrease by 1 while $\mathrm{Cl}^{-}$ions will increase by 1

(b) $\mathrm{Na}^{+}$ions will increase by 1 while $\mathrm{Cl}^{-}$ions will decrease by 1

(c) Number of $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions will remain the same

(d) The crystal structure of $\mathrm{NaCl}$ will change.

Question 10. A compound consisting of the monovalent ions $\mathrm{A}^{+}$and $\mathrm{B}^{-}$crystallizes in the body -centre cubic lattice. If one of $\mathrm{A}^{+}$ions from the corner is replaced by a monovalent ion $\mathrm{C}^{+}$, what would be the simplest formula of the resulting compound?

(a) $\mathrm{A} _{7} \mathrm{~B} _{8} \mathrm{C}$

(b) $\mathrm{A} _{8} \mathrm{BC} _{8}$

(c) $\mathrm{A} _{6} \mathrm{~B} _{4} \mathrm{C} _{8}$

(d) $A _{6} B _{4} C _{6}$

Question 11.

Column I $\hspace{5cm}$ Column II
(Distribution of particles $\mathrm{X}$ and $\mathrm{Y}$ ) $\hspace{1.5cm}$ (Formula)

(a) $\mathrm{X}=$ At the corners
$\quad \mathrm{Y}=$ At face centres $\hspace{3cm}$ (p) $\mathrm{XY}$

(b) $\mathrm{X}=$ At the corners and face centres
$\quad \mathrm{Y}=\ln$ all tetrahedral voids $\hspace{2cm}$ (q) $\mathrm{Xy}_2$

(c) $\mathrm{X}=$ At the corners
$\quad \mathrm{Y}=$ One on each body diagonal $\hspace{1.4cm}$(r) $\mathrm{XY}_3$

(d) $X=$ At the corners and face centres
$\quad \mathrm{Y}=\mathrm{At}$ the edge centres and body centre $\hspace{0.2cm}$(s) $\mathrm{XY}_4$

(a) A-r B-q, C-p, D-s

(b) A-s, B-p, C-r, D-q

(c) A-q, B-r, C-p, D-s

(d) A-r, B-q, C-s, D-p

Question 12.

Column I $\hspace{5cm}$ Column II
(Position of particles $\mathrm{X}$ and $\mathrm{Y}$ ) $\hspace{2cm}$ (Formula)

(a) $Y=$ in cubic close packing
$\quad X=$ in tetrahedral voids $\hspace{2.5cm}$ (p) $X Y$

(b) $X=$ in cubic close packing $\hspace{2cm}$ (q) $\mathrm{X} _{2} \mathrm{Y} _{3}$
$\quad Y=$ equally distributed between octahedral and tetrahedral voids

(c) $Y=$ in cubic close packing
$\quad X=$ in octahedral voids $\hspace{2.5cm}$ (r) $X _{2} Y$

(d) $Y=$ in hexagonal close packing
$\quad X=$ in $2 / 3$ rd of octahedral voids $\hspace{1.2cm}$ (s) $XY_4$

(a) $A-p, B-q, C-r, D-s$

(b) $A-q, B-p, C-s, D-r$

(c) $A-r, B-s, C-p, D-q$

(d) $A-s, B-p, C-r, D-q$

Question 13. If an atom is placed on each lattice point, the number of atoms per unit cell for primitive, body -centered cubic and face-centred cubic lattice, respectively, are ?

(a) 1,1 and 4

(b) 1,2 and 4

(c) 1,4 and 2

(d) 1,1 and 1

Question 14. A solid has a structure in which platinum ions are located at the corners of unit cell; oxide ions are located at the cube edges and sodium ions at the cube centres. The chemical composition of this solid is

(a) $\mathrm{NaPtO} _{3}$

(b) $\mathrm{Na} _{3} \mathrm{PtO}$

(c) $\mathrm{NaPt} _{3} \mathrm{O}$

(d) $\mathrm{Na} _{2} \mathrm{PtO}$

Question 15. A compound of ’ $A$ ’ and ’ $B$ ’ crystallizes in a cubic lattice in which ’ $A$ ’ atoms occupy the lattice points at the corners of the cube. The atoms ‘B’ occupy the centre of each face of the cube. The probable empirical formula of the compound is

(a) $\mathrm{AB} _{2}$

(b) $\mathrm{A} _{3} \mathrm{~B}$

(c) $A B$

(d) $\mathrm{AB} _{3}$

Question 16. Arrangement of $X^{-}$ions around $A^{+}$ion in solid $A X$ is given. If the radius of $X^{-}$is $250 \mathrm{pm}$, radius of $A^{+}$is

(a) $104 \mathrm{pm}$

(b) $125 \mathrm{pm}$

(c) $183 \mathrm{pm}$

(d) $57 \mathrm{pm}$

Question 17. The arrangement of $X$ ions around $A^{+}$ions in solid $A X$ is given in the figure (not drawn to scale). If the radius of $X^{-}$is $250 \mathrm{pm}$, the radius of $A^{+}$is

(a) $104 \mathrm{pm}$

(b) $183 \mathrm{pm}$

(c) $125 \mathrm{pm}$

(d) $57 \mathrm{pm}$

Question 18. If spheres of radius ’ $r$ ’ are arranged in ccp fashion (ABC ABC.) The vertical distance between any two consecutive $A$ layers is

(a) $4 r \sqrt{\dfrac{2}{3}}$

(b) $4 r \sqrt{\dfrac{3}{2}}$

(c) $6 \mathrm{r}$

(d) $r \sqrt{6}$

Question 19. The number of atoms in $100 \mathrm{~g}$ of a face crystal with density $=10 \mathrm{~g} / \mathrm{cm}^{3}$ and cell edge equal to $200 \mathrm{pm}$ is equal to

(a) $5 \times 10^{24}$

(b) $5 \times 10^{25}$

(c) $6 \times 10^{23}$

(d) $2 \times 10^{25}$

Question 20. Ice crystallises in a hexagonal lattice having a volume of the unit cell as $132 \times 10^{-24} \mathrm{~cm}^{3}$. If density of ice at the given temperature is $0.92 \mathrm{~g} \mathrm{~cm}^{-3}$, then the number of $\mathrm{H} _{2} \mathrm{O}$ molecules per unit cell is

(a) 1

(b) 2

(c) 3

(d) 4

Question 21. The edge length of a face centered cubic cell of an ionic substance is $508 \mathrm{pm}$. If the radius of the cation is $110 \mathrm{pm}$, the radius of the anion is

(a) $144 \mathrm{pm}$

(b) $288 \mathrm{pm}$

(c) $398 \mathrm{pm}$

(d) $618 \mathrm{pm}$

Question 22. Percentages of free space in cubic close packed structure and in body centred packed structure are respectively

(a) $48 \%$ and $36 \%$

(b) $30 \%$ and $26 \%$

(c) $26 \%$ and $32 \%$

(d) $32 \%$ and $48 \%$

Question 23. Copper crystallizes in fcc with a unit cell length of $361 \mathrm{pm}$. What is the radius of copper atom?

(a) $108 \mathrm{pm}$

(b) $157 \mathrm{pm}$

(c) $127 \mathrm{pm}$

(d) $181 \mathrm{pm}$

Question 24. Number of unit cells in $4 \mathrm{~g}$ of ’ $\mathrm{x}$ ’ (atomic mass $=40)$ which crystallize in bcc pattern is : $\left(N _{\mathrm{A}}=\right.$ Avogadro number)

(a) $0.1 N _{\mathrm{A}}$

(b) $2 \times 0.1 N _{A}$

(c) $0.1 / 2 N _{A}$

(d) $2 \times N _{A}$

Question 25. In an ionic compound $A^{+} X^{-}$, the radii of $A^{+}$and $X^{-}$ions are $1.0 \mathrm{pm}$ and $2.0 \mathrm{pm}$ respectively. It shows $\mathrm{NaCl}$ type of crystal structure. The volume of the unit cell of the crystal of $A X$ will be

(a) $27 \mathrm{pm}^{3}$

(b) $64 \mathrm{pm}^{3}$

(c) $125 \mathrm{pm}^{3}$

(d) $216 \mathrm{pm}^{3}$

Question 26. Schottky defect in crystals is observed

(a) when unequal number of cations and anions are missing from the lattice

(b) when equal number of cations and anions are missing from the lattice

(c) when the density of the crystal is increased

(d) when an ion leaves its normal site and occupies an interstitial site.

Question 27. The colours of some otherwise colourless compounds are due to

(a) Frenkel defect

(b) Metal deficiency defect

(c) Schottky defect

(d) Metal excess defect (F-Centres)

Question 28. Crystal with metal deficiency defect is

(a) $\mathrm{NaCl}$

(b) $\mathrm{FeO}$

(c) $\mathrm{KCl}$

(d) $\mathrm{ZnO}$